the answer is you either don't because that specific method only works in this exact scenario or you learn big people maths and try to form a general solution for whatever angles EBF, CBE, BCF & ECF are. you are gonna need some trigonometry! the only reason it works in this case is cause the entire question is done in a giant isosceles :P
he is over complicated it its easy, I solved this question in 3 mins when I was jus twelve years old :| edit: now that I'm older, I realize how idiotic and braggy that sounds XD
Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically What's funnier is when people say, "All you need to know is that the sum of all triangle's angles equal to 180° while quadrangle's equal to 360°" "Verticle angles are equal to each other, so you get that left = right = 70°" but there are LITERALLY obvious common facts that people wouldn't feel the need to bring up. How ABOUT you tell us how to solve it by doing your so called "oNLy cOunTiNg, nO dRAwiNg" method. I've really really wanted to see that for like 4 years already, or since the video was posted. The problem with the "merely counting" method is that it would always end up getting 180°=180° as the final count and result. You'd always get stuck on that. You wouldn't get anywhere or even move forward. if those "obvious common facts" were the mere needed things to know of, then such question wouldn't have come out in the first place. Stop this hipocrisy and bs once and for all. Once again, you either solved this trigonometrically OR just wanted attention and reactions by commenting another "yOu pOinTleSsLy mAdE iT hArD" due to what the majority of the comment section have been saying.
Damn, after watching like 10 of these videos I finally solved this one myself, I'm so proud of myself for solving something that a Chinese 5 year old would do in seconds.
solved it! Before watching: this is how its done; look at Triangle BAE. this triangle is in a relationship with a different triangle so it spells "bae". We must find this triangle's bae by using the boyfriend line (line BF). We see that triangle BFE and BAE share a common fetish known as "x" this is reasonable proof as to why they are together. But what is their fetish? That is what the question is asking. lets look at other clues. notice how there is no D in the picture. That is because the D is in the middle (lol). The intersecting piece must be "d" and the angle which contains the fetish is named FED. This is not a coincidence, they have a food fetish which only about 30% of triangles have! if you were to put this into math talk then "x = 30 degrees" and that is how you solve this one.
I am proving it in another way, but may not be elementary geometry, but perhaps easier as drawing a line of BG in the first place may not occur to mind quickly. Instead of drawing line BG, draw a line EG parallel to BC. Now since BC is parallel to EG, the angle GEA is 80 degree, so that the angle BEG is 60 degree (the angle CEB is already 40 degree). Now look at the two triangles BFE and FGE, considering BF and FG as their bases and have a common height (perpendicular measure), the ratio of their area are in the ratio of their bases, that is area of the triangle BFE/ area of the triangle FEG = BF/FG ). Again the area of the triangle BFE can also be expressed as 1/2*BE*EF* Sin BEF (similar as 1/2*a*b*SinC where a and b are the sides and C is their included angle). Similarly area of the triangle FGE is 1/2*EF*EG* Sin FEG. Therefore, the ratio of the two triangles becomes, BE *Sin BEF/EG *Sin FEG (EF eliminated) which is equal to BF/FG (proved earlier). That is, BE*Sin BEF/EG *Sin FEG = BF/FG. Rearranging, BE/BF = EG/GF *(Sin FEG/Sin BEF). We also know from angle bisector theorem that (see the drawing), if BE/BF = EG/GF, then angle BEF must be equal to angle FEG. Comparing the above two, we find that these relations will hold true only when Sin FEG/Sin BEF =1 (considering BE/BF and EG/GF not equal to 1). In other words, angle FEG = angle BEF. But angle BEG is 60 (we have already proved in the beginning). Therefore angle BEF, x (as we named) and angle FEG are each, 60/2 = 30 degree. This proof utilizes area of triangle theorems.
I found the shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
What If you used triangle between angle 50 and 60 degrees then that angle will be 70 degrees and since opposite angles are equal the other angle is also 70 degrees.
@@juanmatias87 opposite angles don't provide 180 degree. Linear pair angles ( angles on a straight line) gives 180 degree and that's right in this question.
Beautiful trick! I had to determine coordinates of E and F in cartesian. Assuming the coordinate of B is (0,0) and coordinate of C is (1,0), I created 4 function for several lines: 1. BE line is y = x(tan 60) 2. CE line is y = tan 80 - x(tan 80). From BE and CE equation, I obtained the coordinate of E --> (0.766 , 1.32683) 3. BF line is y = x(tan 80) 4. CF line is y = tan 50 - x(tan 50). From BF and CF equation, I obtained the coordinate of F --> (0.17365, 0.985) And then I calculated the slope of FE line, m = 0.5771 I put Z between E and C so that the slope of FZ line is 0. The angle EFZ would be arctan m. So EFZ = 30 deg. Knowing that FZ line parallel to BC line, the angle EZF = ECB = 80 deg. x = 180 - EFZ - EZF - BEC = 180 - 30 - 80 - 40 = 30 deg. It's ugly, but it works.
Tory Berry SAME! Although I got 50... I came so close... although I haven't taken any geometry. But I do know the three angels = 180 and what he said in the video. Maybe I should have checked my work.
Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically The problem with the "oNLy cOuNtiNg, nO dRaWiNg" method is that you would always end up getting "both-sides-equal-equation" such as 180°=180°, x=x, etc, as the final count and result. You'd always get stuck on that. You wouldn't get anywhere or even move forward. if we merely had to know that sum of all triangle's angles equal to 180° and square's equal to 360° and vertical angles equal to each other, then such question wouldn't have come out in the first place. I've doubted these comments for a long time. They either solved it trigonometrically without having read the title or commented another "yOu pOinTlEsSly mAdE iT hARd" merely for attention and reactions due to what the majority of the comment section have been saying
[ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
I tried it aswell, it is actually possible, by just adding up the angles you're missing 2 angles that can be replaced by y/Z then you can form an equation with both of these and solve for Z and y then you have all the angles and can get to x, I wrote that down somewhere earlier don't want to do it again tho
@@Skimbleshanks73 It's not possible. There's not enough information to solve those equations. Go on and actually try solving those equations and you'll understand.
@@demifiend9 It's actually possible by applying sine law, but the calculation is a little bit messy and compound angle formula is needed for solving the trigonometric equation. The method in the video is much more beautiful
@@Skimbleshanks73 Yes, you can generate two equations in two unknowns but the equations do not produce a unique solution. No matter how you try solving the two equations you only end up with platitudes like 0 = 0 or y = y. If you try expressing your set of linear equations in matrix/vector notation, you will find that the matrix has a determinant of zero and so cannot be inverted.
I figured it out! Your method was far easier and more elegant than what I ended up doing, though. I basically called the length of the leftmost side "a" and calculated various other side lengths in terms of "a" via the sine rule, and then applied the cosine rule at the very end to yield a loooong arcsin expression for the value of x which I then plugged into Wolfram Alpha to get a result of 30 degrees.
Yes, that's almost exactly what I did, but I used 1 instead of a. But this video was a serious letdown, because I was hoping to learn how to do this *kind* of problem. However, the presented solution wouldn't work if the lines got shifted by a few degrees, whereas the law of sines strategy would still do the job. It felt like watching a video about breaking into a computer and being told "When prompted for the password, just write the correct password, and you're in! Super easy!"
@@bartholomewhalliburton9854 Please do share this to maximum people 👇☺️ Shortest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
He was doing it using really basic maths, hence the title "hardest easy geometry problem", using more complicated trig would be going against the point of the problem.
@@finnwilde Follow this shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
It's interesting how so many people are claiming to only need "vertical angle theorem" to solve the entire problem and end up with a cyclic argument or wrong algebra. people should really check first before posting their solution ;)
I used it, and I see no way to get x or the opposite angle in that way! Doesn't make sense. Seriously, you can't use that way and people have to realize that. There aren't any vertically opposite lines to x!
construct a circle of radius BE... draw a 180° line on BC and mark the intersection of the line and the circle as G making a new line CG Now since angle B equals 80° in triangle BEG angle B = 120.. Since BE = BG (radius of the circle) we can conclude that the new triangle BEG is an isosceles triangle and since the interior angle of a triangle is 180°.. In triangle BGE .. angle beg = angle bge therefore 2x + 120 = 180 x = 60/2 x = 30°
I've forgotten a lot the stuff I was taught about geometry around 40 years ago. So I just used what I remembered. The fact that all the angles of a Triangle add up to 180°, lines that cross add up to 360°, angles coming off a straight line add to 180° and the angles of a Quadrilateral add to 360°. Then I went looking for all of the above shapes and situations. Call the point where the lines BE & FC cross G We know ∠ABC = 80° & ∠ACB = 80° ∴ ∠ BAC = 20° We also know that ∠ BGC = 70° Now consider Δ BGC We know ∠ BGC is 70° Now consider Δ BCE We know ∠ BEC = 40° Now consider Δ ACF We know ∠ AFC = 130° Now consider Δ BCF We Know ∠ BFC = 50° Now consider Δ ABE We Know ∠ AEB = 140° Now consider Δ CEG We Know ∠ CGE = 110° Now consider Δ BFG We Know ∠ BGF = 110° Because we know 3 of the angles at the crossing point of the line at point G and they add to 360° ∴ ∠ EGF = 70° Then I got stuck. I couldn't make any more of the shapes I remembered about. I Knew I'd got 110° to share between the angles BFE and CEF in the Quadrilateral BCEF And 160° in the Triangle AEF So I looked at the solution. You cheating bastard :) You add extra lines and angles - Also I'd completely forgotten about the properties of Isosceles triangles :) Thanks though. It was nice to drag some things I was taught so long ago out of the back of my head.
I saw 2 isosceles triangles, labelled the sides and used sine rule and derived a relation between sinx and the other angles. Then it's just how good u are at trigo manipulation 😊
@@ARDAYILMAZ72 - don't be ashamed. Sometimes, a problem turns out to be difficult for us, because we don't see what many others have seen. Look at their solution and learn. . . . BTW I would wait with the Respect for ranjan, until they have shown their solution.
@@Achill101 Here's shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
@@ARDAYILMAZ72 Share to all 👇😊 Shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
It's 30°, here's the shortest easiest way 👇😊 Shortest & easiest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
@@Crazy77772 [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
huh? I'm not sure how everyone in the comments don't realise that you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use ANY isosceles triangle logic at ALL.
@@Crazy77772 ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
I tried this using only (A + B + C = 180). I then added unknowns to an additional 3 angles and created a system of four equations and four unknowns but they didn't have a solution as one variable would always get cancelled out leaving an untrue statement (130 = -20, for example).
you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use any isosceles triangle logic at all. regardless, cool thought
Thats the proper way to go.. If the angles mentioned in this diagram are any different, this method of using isoceless triangles to find angles wont work (cuz, we are supposed to get BGF as an equilateral triangle)
@@drainedzombie2508 ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
@@drainedzombie2508 Here's shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
Hi, I'm a grade 9th student from India and when I saw video's thumbnail, I was very excited to solve this problem, and I solved this problem within a minute.I solved this problem much easily with very elementary methods: 1. Isoceles triangle theorem 2. Sum of interior angles in a quadrilateral and triangle is 360 and 180 degrees respectively. 3. Vertically opposite angles are congruent. Got x=30 degrees.
Did you come up with the same or a similar method of the video in your head? This is definitely not possible through those theorems without constructing new lines.
@@bartholomewhalliburton9854 I came up with other method and trust me, this problem can be solved with elementary methods as I mentioned above but it's just a more complicated method. You can also check the video's description because there also it is mentioned. Try once more and don't give up and you will get it. Thanks for your reply.
@@vaishalitirthkar Did you draw a line between points not in the video? I know this can be solved through those methods as the video suggests, but I want to know how you did it without help from the video (seeing the first step).
@@bartholomewhalliburton9854 ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
@@dux2508 Yes, he's absolutely right 🥵😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
Presh, I think you got x wrong. This is unless crossing lines don't create equal angles on opposite sides. Maybe I am wrong, but, in the triangle with angles 50º and 60º, the remaining angle is 70º, and so is the one on the opposite side . But the problem with your solution comes when you add this angle to F, which is 70º and x which is 30º. It doesn't add up to 180. it adds up to 170º
What about a general solution ? Suppose you have the same problem, but not with specific angles given. What would x be if the given angles were say a,b,c,d ?
@@srpenguinbr I have the general solution to this. The solution method (of symmetrical angle constructions inside the isosceles triangle) in the links at the end can be used in general case... [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
Zulyrus lol they haven’t even learned algebra hardly. I know calculus and I don’t foolishly defy experienced mathematicians, especially when their right.
Nishant Modak lets say that the point at which CF and BE meet is O, just so we can label it. The sum of the angles of a triangle is equal to 180 degrees, which means angle BOC=180-CBE-BCF=180-60-50=70. So angle BOC=70 degrees. Then we can see that angles BOC and FOE are vertical, which means BOC=FOE=70 degrees. Now, from the video, we know angle OEF=30 degrees. The sum of the angles of a triangle is equal to 180 degrees. Which means OFE+FEO+FOE=180 degrees. But here is the problem, he said in the video that angle OFE=70 degrees. OFE+FEO+FOE=70+30+70=180 170 does not equal 180. For some reason all the math works out in the video, but for some reason it just doesnt. He must have just missed some number while he was calculating. Hard to blame him, he calculated A LOT.
Aegishlash 2020 there is a difference between foolishly defying and standing up for yourself. The more experience you have at something, the less likely you are to make a mistake, but that point never reaches zero.
I think it is 40 degrees because in triangle CBO we have 60 +50=110 Then180-110=70 Since triangle is issocless according vertical opposite angles one of its angle is 70 degrees since it is a isosceles triangle it becomes 70+70+x=180 then x will be 40 degrees....I think it is my opinion so if it's wrong sorry....
@@kazishahjalal6852 Here's the most logical and easiest way... [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
@@kazishahjalal6852 ✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
I kinda feel like this is giving the solution backwards. I would find it way more logical to construct point G starting from the assumption that an Isosceles traingle GFE would help us if we could find the value of angle FGE.
[ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
How does one even come up with this solution? This seems so created out of thin air that I am forced to believe he tried every single approach there is to solving the problem until he found a solution... So far this video hasn't taught me anything since I am just learning a wierd solution to a wierd problem by heart
He didn't come up with the solution. This problem is known for having to draw auxiliary lines in order to solve it. Using the common theorems alone would not lead to the answer. Yep, it is indeed a weird solution for a weird problem.
+Paolo Patron you can solve this without drawing anything about 100 seconds of deduction the problem isn't all the complicated if you understand straight lines are 180 and that intersecting lines will be 360 you can deduce all you need to know to find x
Joshua Woodford That's the thing. You can find all the other angles using supplementary and complementary values using interior and exterior angles but you won't be able to find the necessary angles in order to solve for x. A "hundred seconds of deduction" is more than enough time to realize that the problem is impossible without auxiliary angles if you actually think.
+MadCodex I'll try .. for explanation purpose I'll call the intersecting lines in the middle R you'll have to figure the angle I'm speaking of.... to start the first angle of r can be determined by the fact bcr has to equal 180 and bc are know, r equals 70 .. since straight lines half to equal 180 angle r1 has to be 110 to complete the straight line same holds true for the two angles opposite have to be 70 and 110 ... now we can determine triangle bfr1 since 2 angles are known and r1 which means f in that instance is 50 also the angle opposite in triangle fae has to be 50 which means the remaining angle f in triangle fer2 can be determined which is 80 , that gives us angle f 80 and r2 70 means x has to be the remaining 30.. hope it helps hard to explain through text
There was a much easier way to do this that didn’t involve making so many seemingly random isosceles triangles. Took me like 15 minutes to find it while figuring out every other angle. Though, it did involve using more complicated angle “manipulation”. It was confusing to me how I found it, so even though I went through the vid saying “x is 30” I doubted myself when he mentioned isosceles triangles and almost changed my answer to 20
“There was a much easier way” “though it did involve using more complicated angle manipulation” “it was confusing to me how I found it” LMAO I’ve never seen someone so confidently contradict themself as much as you did.
***** No its not... you can just use several equations and substitution to solve it. Just looking at it I almost immediately knew it was 30. I needed to add 0 lines to solve it. Show anyone that does expert level Sudoku puzzles every day this and ask them to solve it. and odds are they will go to trial and error and solve it pretty fast.
That being said.... the trial and error method while effective with the problem is usually something you do because you do not know or at least are unsure. You can solve this several ways, however the way they EXPECT you to solve it is this way. Others either are much more time consuming or are brute forcing the answer.
Couldn't figure it out the "simple" way so I used Law of Sines and Law of Cosines instead. I began by assuming BC=10 so as to employ the Law of Sines. The length of BC can be any value as it would simply scale the whole figure and have no effect on the angle measures. Using the Law of Sines, I worked my way around the figure, using previous results to gain new results. First I found CE, then BE, then the segment from B to the interior intersection. Then the segment from C to the interior intersection. Then CF. Then some segment subtraction to find the segments from F and E to the interior intersection. Now we are down to the smallest triangle with angle x. I used the Law of cosines to solve for the missing side. Finally, used the Law of Sines once more to find the angle of 30 degrees for x.
They should teach the Law of Sines and the Law of Cosines early on in math, like in Geometry. I first learned it in PreCalc and it’s relatively straightforward.
@@kingklaus2115 It definitely can be understood by someone in a geometry class. Using the SOH CAH TOA a clever student might accidentally come up with it themself. When I took geometry it was a part of the curriculum.
@@bartholomewhalliburton9854 That’s good to hear. It’s not even a difficult concept and like you said a student could easily come across it on accident. All geometry classes should add it to the curriculum early on.
The initial assumption of a 20deg triangle is just a guess, or, in other words one of many other guesses that could be made at some start. The whole exercise-as it is presented-is not an instructive procedure to get to the bottom of such a task, but an instruction for gambling. SIN and COS rules have to be applied to avoid waste of time.
It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE. Lastly, use inscribed angle theorem : Angle BEF = 0.5 × BGF = 0.5 × 60° = 30° Done !!! ✌️✌️☺️☺️☺️
@@mr.dragoji3149 ✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
wow, I did geometry in high school. isn't memorization of theorems too strange and boring for tiny kids? well... i guess high schoolers have their own issues
Let the point of intersection of BE and CF be O then , In triangle BOC , angle BOC = 180° - (60°+50°) = 70° Now angle EOF = angle BOC = 70° since they are vertically opposite angles And according to your solution , in triangle EOF , angle EOF = 180° - (30°+70°) = 80° So , why are we having two different values for angle EOF ? Is there any mistake ?
In his solution about here 4:56 He makes another triangle and that triangle passes through the top angle which means when he said that the top angle was 70 he was referring to this triangle 6:28 (GFE) not CFE triangle Which means the top angle in CFE was actually 80 (70+10)and not just 70.(The 10 was used to create the triangle in 4:56) proof: 80(CFE) +30(x or BEF) +70(FOE) = 180 checks out
@@Ucj334esd Share to all 👇😊 Shortest & easiest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
I think I found a far simpler path to the solution. 1. Flip BF110 on the AB axis to give 50 at F in AEF. 2. Take that 50 from the 130 at F in ACF to give 80 at F in CEF. 3. 180 - 70 - 80 = 30. Thoughts?
Kindly explain every point in more detail if possible. Flipping and rotating to possibly right, correct ☹️😊 I'm very weak with rotational geometry so kindly bear with me ☺️
@glyn hodges how do we know BG and BF are tge same? We only know BC and BF are the same (BCF 50° and BFC 50°) Edit nevermind: as BC and BG are the same!
Stop this is false,lets say BE and CF crosses at point O,and in CBO triangle BCO=50,CBO=60 ,so BOC=70,BOC=EOF=70 here he found that CFE=70=OFE if in OEF triangle OFE=70,EOF=70 so x=OEF=40!!! ITS WRONG
ΔBGE is an isosceles triangle. So GE = BG - - - - (1) ΔBGF is an equilateral triangle. So BG = BF = GF - - - - (2) Substitute equ.(2) in equ. (1) ∴ GE = GF
I solved completely different and even answered the 70 degree angle incorrectly answering it as 80 but got it right somehow. I settled for the 360 degree intersection with 110 degrees for the top and bottom and 70 degrees for the right and left and finally 80+70 = 150 and therefore I arrived with the answer x = 30 degrees. Great video by the way
@@rikiyosa4907 ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE. Lastly, use inscribed angle theorem : Angle BEF = 0.5 × BGF = 0.5 × 60° = 30° Done !!! ✌️✌️☺️☺️☺️
For the purposes of this example, we'll call the unlabeled Vertex G. I'll simply go through the values for the easily identified angles, and go from there. BGC is 70°. BAC is 20°. From that, we can identify that FGE is also 70°. As a result, both CGE and BGF are 110°. CEG is a 40° angle, leading GEA at 140° Next, I'll look at quadrilateral AEGF, with it currently known angles of 70°, 20°, and 140°. The angles of a quadrilateral add up to 360°, so the final angle, GFA, is 130°. I point this out because GFE and EFA must add up to that 130°. Likewise, GEF (in the diagram as "x") and FEA must add up to 140°. We also know that AEF and AFE must add up to 160°, and GFE and GEF must add up to 110°. After trial and error, there is only 1 set of angles that fit those requirements: AEF: 100° AFE: 60° GFE: 70° GEF: 40° (This was the angle identified as "x", in the diagram.)
Just have doubt. In 5:57 , in the triangle BFG angleB+angleF+angleG=180 angleG=70 which shows that BG is not equal to GF I just think that this question is a typo
Is it possible to solve this problem without using properties of side lengths at all? I was trying to do angle substitutions and systems of equations and getting nowhere, I forgot about isosceles triangles...
I don't think so. When you try relying only on angle properties there are four angles that seem impossible to work out, one being x itself. Knowing any one of them will then lead you to the other three but you just can't get it without doing other things.
I did. Weirdly enough I come to x=50. Might sound stupid but I solved every angle without creating new lines and every single angle is coherent. I'm still trying to prove myself wrong.
Mauricio Huicochea Toledo Your solution cannot be disproven from angle tracing alone. Indeed your system of equations does not have a unique solution. If x + w = 140 and y + w = 160 then subtracting gives y - x = 20 so y = x + 20. Then substituting this in z + y = 130 gives z + x + 20 = 130 or z + x = 110. I deduced your last equation from the others, so it isn't independent, and you really have four variables and three equations. So the system has infinitely many solutions and the error must be how you deduced that x can only equal 50.
@@cs8833 They can only comment, they can't explain. [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
@@cs8833 ✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
@@cs8833 Shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
I solved this in a very different way. After attempting to label and solve for every angle in the diagram (and finding nothing useful) I decided I needed a shift in perspective. So I noticed that FE could be rotated counterclockwise (about F) to form the line CFE and triangle BCF. With this new shift in perspective, I was able to use the given 60 degrees and my previously solved angle CFA (and vertical angles) to find that x was 30 by x=180-20-130. I'm not sure this would work in all cases and when he mentioned isosceles triangles I was sure I got it wrong. But, surprisingly, I got the right answer.
Easier and less messy solution: Construct a line through C which is parallel to AB. This will be 20 degrees below AC. Then an extension of EF to the new line will create a new triangle. Start solving angles using 180 degree sums of triangle vertices and congruency.360 degree sums of vertical angles. The new triangle will be shown to be isoceles and every angle in the diagram can be resolved.
That's the elegant solution and the one I found too. And it proves Michael Parks' claim that angle EFA is 50 degrees. Oftentimes these problems give rise to way over-complicated solutions.
Just tried it your way and still can't get it. What is the "This will be 20 degreees below AC." mean? Diagram or/and calculations would be great to see if you have the time. Thanks!
I'll try to explain it. Basically, it's an extrapolation of the vertical angle theorem. If you pause the video at around 30 seconds, I'll try to build off Presh's hard work. We can go about either way (construct parallel, prove 20 degrees or construct 20 degree, prove parallel). Lemmas: Angle ABC is 80 degrees (60+20). Angle ACB is also 80 degrees (50+30). We will construct a new line through C and choose a point Z arbitrarily to the right. We will also extend BC downward and pick an arbitrary C' below C. If we know we have drawn the new line CZ parallel to AB, then angle ZCC' would have to equal angle ABC which we know is 80 degrees Then BCC' = 180 and BCZ = BCC' - C'CZ = 180 - 80 = 100. Finally,, ACZ = BCZ - ACB = 100 - 80 = 20. Alternatively, we can choose to construct line CZ such that it is 20 degrees below AC. Then BCZ = 80+20 = 100 and ACC' = BCC' - BCZ = 180-100 = 80. Since ACC' = ABC, we can conclude that AB is parallel to CZ.
Am I the only one who used sine rule and trigonometry to solve it? :/ Here how I solved it. It's easy but a bit lengthy.In the figure,ABC is an isocels triangle. First find the angles at the intersecting point of BE and CF (let it be O) and also BFC and BEC Let,AB=AC=a.. now find BC in terms of a (using sine rule). Using same procedures repeatedly find OF and OE in terms of a.. Here CFE =110-x(use ur head.. It's easily noticeable).. using sine rule again we get a plane trigonometric equation of x ( a gets cut on both side of the equation) solving it we get x=30..
Good. That's pretty much what I did as well--used the Law of Sines (I also wound up using the Law of Cosines) to eventually get (after a lot of calculation) that sin x = 0.5, so x = 30 degrees. Didn't have to draw any other lines or create new triangles. But as you say, it was a bit lengthy. One thing that could simplify your calculations: Since the angles are independent of the size of the diagram, rather than choosing to represent a particular length by a, you can simply assign an actual value to it that will make the math easier--like 1, for example.
If you use the exterior angle of triangle CBF on vertex F and then vertical angles you can find angle CFE as 80 degrees and therefore find x because the sum of the interior angle measures of a triangle is always 180 degrees. I find that is much simpler.
@@azfariskandar8009 ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
@@azfariskandar8009 Kindly do enlighten every curious mind in messages with this solution from time to time 👇😊 Simplest solution, thanks to taba3514 short answer down in messages 👇😊 It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE. Lastly, use inscribed angle theorem : Angle BEF = 0.5 × BGF = 0.5 × 60° = 30° Done !!! ✌️✌️☺️☺️☺️
There's definitely something wrong with the working. If you solve all the angles without that random isosceles triangle drawing, you get angle CFE + x = 110. In his calculations, they add up to 100
That's GFE+x = 100 CFE is 10 more than GFE because drawing the equilateral triangle means you need 10 more than the 50 degrees at CFB, eating into the CFE angle to create the new GFE
@@awxangel6781 [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
@@awxangel6781 Kindly share this to all from time to time ☺️ Shortest easiest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
The answer is incorrect to me because BCEF is a trapezium which means the sum of its interior angles should be equal to 360° but if x is equal to 30° then the sum becomes 350° which is incorrect
@@jameslo2818Well , my bad , should have revised the properties of a trapezium but my point still stands as BCEF is a quadrilateral which means that the sum of its interior angles should be equal to 360 so 'x' should be or rather, would be 40°
FULL SULOTION : (using sine law and cosine law) Angle BAC = 180-60-20-50-30= 20 (angle sum of triangle) Let BC =1 By sine law, (BC/sin angle BAC) = (AB/sin angle BAC) 1/sin 20 = AB/sin80 AB = sin80/sin20 Angle BFC = 180-60-20-50=50 (Angle sum of triangle) Because angle BCF=angle BFC Thus, BF = BC (sides opp equal angles) ie: BF=1 AF = AB - BF =(sin80/sin20) - 1 Because angle BAC = angle ACB = 80 Thus, AB=AC (sides opp equal angles) AC = sin80/sin20 Angle BEC = 180-60-50-30= 40 (Angle sum of triangle) By sine law, (BC/sin angle BEC) = (EC/sin angle CBE) 1/sin 40 = EC/sin60 EC = sin60/sin40 AE = AC - EC = (sin80/sin20) - (sin60/sin40) Now we know that AE = (sin80/sin20) - (sin60/sin40) AF =(sin80/sin20) - 1 Angle BEC =20 By cosine law, FE= √( (AF)²+(AE)² - 2(AF)(AE)cos 20) FE = 0.684040286.... In Triangle BEF, By sine law, (EF/sin20) = (BF/sin angle BEF) (0.684040286.../sin20)= (1/sinx) sinx = 0.5 x= 30
because i realised i needed to take the ratios of lengths into consideration. i just drew the triangle and measured 30 degrees. i also did alot of angle working out but it was redundant so yeah...
3:56 You don't need the construction to prove that the angle GEB is 40 degrees. It's 180-(60+50+30). 6:31 Triangle EFG is isosceles. Therefore its interior angles at E and F are both equal to (180-40)/2 = 70 degrees. Thus x = 70-40 = 30 degrees. (Oh, that's what you did.)
Hello it is really tricky to resolve this question. Why I felt a bit wired on this question: 1. What can trigger the thought of the line BG? 2. What if the question change for example if the angle b and angle c change from 80 to 85 degrees? I see the solution here is for special case , won’t work generically. Am I right? Please advice
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
Kindly enlighten all the thirsty curious minds with this following shortest solution, from time to time 👇😊 It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE. Lastly, use inscribed angle theorem : Angle BEF = 0.5 × BGF = 0.5 × 60° = 30° Done !!! ✌️✌️☺️☺️☺️
Wfreestyle why? By glancing at this for 30 seconds I can tell you every intersections angle without a pen. Let's take the intersection between BE And CF and call it G so I can explain this. Triangle BCG needs to add to 180. We have 60 and 50 for 2 corners, this means angle G of BCG is 70. That means in EFG that angle G is also 70 due to it being symetrical. The angle around line segment CF needs to add up to 180, if the G angle of BCG is 70 then the G angle of BGF is 110. If angle G of BFG is 110 and angle B is 20 then angle F is 50. The line segment BA needs to add up to 180, we know the F angle in BFG and FEA are 50 degrees which means GFE's F angle is 80. That means E is 30 as triangle EFG needs to be 180 If we wanna know the rest than angle E of CEG is 40. Line segment CA needs to be 180 so angle E of EFA is 180-30-40 so it's 110. Angle A is then the last angle and would be 180-50-110 so it would be 20 degrees.
Mike Percival I did it like you did just by watching the thumbnail. I don't know why the author of the video made up all of those lines making a very simple problem look harder than it really is
Trivial steps omitted, but denote by F' the reflection of F over AC, and note that triangle CFF' is equilateral. Because FF'CB is a kite, BF' is the perpendicular bisector of CF, so ∠EBF'=∠FBE=20° and BFEF' is cyclic (the latter following by the trillium theorem); hence, ∠BEF=∠BF'F=30°. The main difficulty of the problem lies in the fact that focusing on angles is not enough to solve the problem; one needs to relate the angles to lengths and vice versa. This can be easily done with trigonometry, but doing it synthetically requires more thought.
Alternate solution 1. FIll in all the angles possible especially results of EB & EC AB & AC and most important the intersection of BE & CF this will give you 70 deg a complement of x. (intersect BC and CF is a pair of 70 & 110 deg) 2. Corner A is 20 deg making ABE isosceles. 3. Draw a line from E crossing perpendicular to AB, intersect with an extension of CF, let's call the new corner G; now EFG is isosceles and corner G is 40 deg. (the angle at F is 50 deg) 4. EFG is isosceles so the corner E from EF & EG is 40 deg and EA & EG is 70 deg (180-90-20). 5. We should already know corner E from EB & EC is 40 deg therefore x=180-40-40-70 or 30 deg.
@@mathcanbeeasy ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
@@mathcanbeeasy World's easiest solution 👇😊 Best possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊 Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF. Hence, GB = GF = GE This means G is the circumcentre to triangle BFE. Lastly, as per inscribed angle theorem : Angle BEF = 0.5 × Angle BGF = 0.5 × 60° = 30° Mind-blowing, easiest method ever !!!!! 💖☺️
Hello dosto . Mera youtube channel bhi dekhlo please agar pasand aye to like 👍 and subscribe Dosto mera youtube channel mai aapko class 9 ki physics ,maths ,history videos milenge .Please but atleast dekh to lo.
Give BC unit value. Isosceles sides of tri ABC can then be found. Once that is done, isosceles sides of tri ABE, AE and BE, can then be determined. Since tri CBF is isosceles then BF is unity. Then using the sine law obtain x from sin(x)/1 = sin(160-x)/BE, where BE = 1.53208 up to 5 decimals.
+John Kerpan why did you start out making your own line at a perfect 20° as if a compass and square were avalable. we could have just measured it ourselves.
You realize that the picture could not be to scale at all and the math would still work, right? What is the mathematical reason that 40 is the right answer, and how is his reasoning in this video wrong?
I am highly convinced that your solution is quite convincing. The other solutions given here are not that convincing as they are bound to make assumptions.
a 10 year old in Singapore, my country, can do this. and I'm actually astonished people are using trigonometry when you don't even need to mention the word isosceles at all. quite funny nonetheless
a+b=x, a+c=y, b+c=z, b+d=w... you can solve for a, b, c, d in terms of x, y, z, w if a, b, c, d are the unknown angles at F and E and x, y, z, w are the values to which they sum.
Awesome question. I tried all the ways I know for more than half an hour ( I mean by elementary geometry methods, not by trigonometry and using calculator to find trigonometric values 🙂). I had to quit. 🙂. Was worthy watching..
Same, I don't know trigonometry yet so I did all I could and eventually gave up haha. But I have a feeling that the answer is simpler than the one in the video, I would've never thought to draw that line 😅
@@harinirajesh3838 a lot of such problems are only solvable with either general rules (trig stuff) or if you know where to draw lines unfortunately. If you sort of make very clever discoveries to solve these problems you often end up reinventing trig anyways.
@@randomnobody660 Here's the easiest and most logical way... [ Just in short - I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment. Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ] All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY. However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them. So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution. NOTE : I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺ I'll share both the easiest solution and the intricate solution as an attachment. Otherwise check if these work- 1. (Solution written again for easier understanding) drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk 2. (The actual solution I'd tried to figure out x) drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk Thank you and love you all. 😘😘😘😊😊😊
@@randomnobody660 Use symmetry, dear friend. ☺️ ✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️ Here's how it goes. Shortest construction : Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF. Solution : Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so.... BC = CK = BK = BF In triangle BKF, BK = BF So, BKF = BFK = 80° Using linear pair addition on line CKH at K, FKH =180° - (60° + 80°) = 40° So, FKH = FHK (or BHC) = 40° This means, FK = FH And HKE = BKC = 60° (vertically opposite angles) Also, EHK (or EHC) = BCH & HEK (or HEB) = CBE, all are 60° each (reason is because they form alternate pair angles with HE || BC and angles CBE & BCH are already known) This means, Triangle HKE is equilateral.... So, EK = EH Lastly..... Both triangles FKH & EKH share the same base KH, and both are atleast isosceles.... So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property.... Hence, Angle KEF = x = Half of angle HEK = Half of 60° = 30° BINGO !!!!! ✌️👍☺️ Note : Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily. Cheers to all. ✌️😘☺️
It took me 2 minutes to solve this problem. I don't know why he was making all this triangles. I appreciate his work. But he was making it complicated.
I got 60. I used the fact that all the angles in a triangle add up to 180 degrees and started to calculate angles in triangles with 2 known angles. I also used the fact that an angle at the intersection of 2 lines is equal to the angle opposite to it at the intersection to get the angles 70 and 110 at the intersection of the lines shown in the diagram. Actually, I have looked at my work after typing the first part and have seen where I had gone wrong. I assumed that the right-most triangle within the triangle in the diagram is isosceles. My bad
A small correction: at 5:04 I meant to say BG = BF. We wil shortly prove BG = GF.
Pls why do we have to draw a line while solving this question rather than solving it as it is
@@onyebuchiukeoma1551 because that's how you *can* solve it...
@Vinz Vinz It isn't assumed. Its because triangle BFG is equilateral (all angles are 60 deg, therefore all sides of that triangle are equal) :)
plz someone check my comment lmao its somewhere in the newest
@Vinz Vinz I was thinking the same thing - how do we know they're equal?! I didn't understand that statement.
And how does one come to think of drawing that magical initial isosceles triangle in the first place?
the answer is you either don't because that specific method only works in this exact scenario or you learn big people maths and try to form a general solution for whatever angles EBF, CBE, BCF & ECF are.
you are gonna need some trigonometry!
the only reason it works in this case is cause the entire question is done in a giant isosceles :P
You don't need trig, plain old geometry works fine, just know 180° is a triangle, 360° is a square, and start solving.
Hahahah that's what I was embarrassed to ask...
this isnt the easiest way
Experienced Geometry Nerds.
did it in 5 seconds , I just had to grab my protractor
Lmao
Lmao
Lmao
Got a quick draw holster for your protractor do ya?
Lmao
How did no one not notice that there is ABCEFG but no D?
thats not the point at all
Lol doesnt matter
Point D is the intersection of BE & CF, duh! :P
Because this question is a girl.
You'll get the D later
This problem was not so hard.But you made it look really complicated
I assume u have passed Jee Adv by now🤣
It is easy to talk
Try to solve it
he is over complicated it its easy, I solved this question in 3 mins when I was jus twelve years old :|
edit: now that I'm older, I realize how idiotic and braggy that sounds XD
bundle maarte reh... tere jaise feku bahut dekhe hain... exam mein saari hawa nikal jaati hai...
Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically
What's funnier is when people say,
"All you need to know is that the sum of all triangle's angles equal to 180° while quadrangle's equal to 360°"
"Verticle angles are equal to each other, so you get that left = right = 70°"
but there are LITERALLY obvious common facts that people wouldn't feel the need to bring up.
How ABOUT you tell us how to solve it by doing your so called "oNLy cOunTiNg, nO dRAwiNg" method.
I've really really wanted to see that for like 4 years already, or since the video was posted.
The problem with the "merely counting" method is that it would always end up getting 180°=180° as the final count and result. You'd always get stuck on that. You wouldn't get anywhere or even move forward.
if those "obvious common facts" were the mere needed things to know of, then such question wouldn't have come out in the first place.
Stop this hipocrisy and bs once and for all.
Once again, you either solved this trigonometrically OR just wanted attention and reactions by commenting another "yOu pOinTleSsLy mAdE iT hArD" due to what the majority of the comment section have been saying.
Damn, after watching like 10 of these videos I finally solved this one myself, I'm so proud of myself for solving something that a Chinese 5 year old would do in seconds.
Dude I swear this is the only problem I did by myself. Not too hard, but he made it sound kinda complicated. I just did some basic stuff.
I've do these types of questions on a daily basis and seem to have no problem whatsoever.
@@armaansharma8349 it's a class 9 question so I solved it, I'm too in class 9
I am preparing for Olympiad and I also did it myself
*Math video exists*
Indian dudes: c'mon let's brag here.
solved it! Before watching:
this is how its done;
look at Triangle BAE. this triangle is in a relationship with a different triangle so it spells "bae". We must find this triangle's bae by using the boyfriend line (line BF). We see that triangle BFE and BAE share a common fetish known as "x" this is reasonable proof as to why they are together.
But what is their fetish? That is what the question is asking. lets look at other clues. notice how there is no D in the picture. That is because the D is in the middle (lol). The intersecting piece must be "d" and the angle which contains the fetish is named FED. This is not a coincidence, they have a food fetish which only about 30% of triangles have! if you were to put this into math talk then "x = 30 degrees"
and that is how you solve this one.
Nice one. 10/10
TEEHEE!
I thought you were serious until the end 😩
I knew it its like my mother always said u gotta look for that D.
did you take into account the GBF, gay best friend triangle?
Damn, you really made this unnecessarily difficult.
True he did make it more difficult, theres other easier ways to figure it out
What is the easiest way?
@@atikshagarwal5147 FC and BE make an X and when theres an x theres 2 same angles that are opposite to each other.
@@x-x2060 Using that and exterior angle theorem is how I got it.
@@davefickess7973 could you elaborate please?
I am proving it in another way, but may not be elementary geometry, but perhaps easier as drawing a line of BG in the first place may not occur to mind quickly. Instead of drawing line BG, draw a line EG parallel to BC. Now since BC is parallel to EG, the angle GEA is 80 degree, so that the angle BEG is 60 degree (the angle CEB is already 40 degree). Now look at the two triangles BFE and FGE, considering BF and FG as their bases and have a common height (perpendicular measure), the ratio of their area are in the ratio of their bases, that is area of the triangle BFE/ area of the triangle FEG = BF/FG ). Again the area of the triangle BFE can also be expressed as 1/2*BE*EF* Sin BEF (similar as 1/2*a*b*SinC where a and b are the sides and C is their included angle). Similarly area of the triangle FGE is 1/2*EF*EG* Sin FEG. Therefore, the ratio of the two triangles becomes, BE *Sin BEF/EG *Sin FEG (EF eliminated) which is equal to BF/FG (proved earlier). That is, BE*Sin BEF/EG *Sin FEG = BF/FG. Rearranging, BE/BF = EG/GF *(Sin FEG/Sin BEF). We also know from angle bisector theorem that (see the drawing), if BE/BF = EG/GF, then angle BEF must be equal to angle FEG. Comparing the above two, we find that these relations will hold true only when Sin FEG/Sin BEF =1 (considering BE/BF and EG/GF not equal to 1). In other words, angle FEG = angle BEF. But angle BEG is 60 (we have already proved in the beginning). Therefore angle BEF, x (as we named) and angle FEG are each, 60/2 = 30 degree. This proof utilizes area of triangle theorems.
I found the shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
What If you used triangle between angle 50 and 60 degrees then that angle will be 70 degrees and since opposite angles are equal the other angle is also 70 degrees.
daniel buttons I did the same thing :P
If you add up the opposite angle by this method and the results they don't add 180 degrees
@@juanmatias87 opposite angles don't provide 180 degree. Linear pair angles ( angles on a straight line) gives 180 degree and that's right in this question.
Yes me too and my x was 40 degrees because of that.
As vpn x is 60
Beautiful trick!
I had to determine coordinates of E and F in cartesian. Assuming the coordinate of B is (0,0) and coordinate of C is (1,0), I created 4 function for several lines:
1. BE line is y = x(tan 60)
2. CE line is y = tan 80 - x(tan 80). From BE and CE equation, I obtained the coordinate of E --> (0.766 , 1.32683)
3. BF line is y = x(tan 80)
4. CF line is y = tan 50 - x(tan 50). From BF and CF equation, I obtained the coordinate of F --> (0.17365, 0.985)
And then I calculated the slope of FE line, m = 0.5771
I put Z between E and C so that the slope of FZ line is 0. The angle EFZ would be arctan m. So EFZ = 30 deg.
Knowing that FZ line parallel to BC line, the angle EZF = ECB = 80 deg.
x = 180 - EFZ - EZF - BEC = 180 - 30 - 80 - 40 = 30 deg.
It's ugly, but it works.
Maxi E s
wow, thats one way of solving this but it is super complicated
Cartesian coordinates are sometimes ugly, but often work to solve the problem. Reliably like a steamroller flattening the problem.
I did something similar, and needed excel for the calculations, but also got to x = 30°
Analytic brain. I like that. I would habe done it if the angles had abit better numbers(like I don't habe a point with irrational coordinates)
FIND THE POSSIBLE ANGLES BY THE RULES THAT LEAVES US WITH THE SUM OF X AND THE ANGLE CFE BEING 110.IF SO TAKE THE BIG TRIANGLE AEB WELL GET
I stared at the thumbnail for 5 minutes and sold it. then click on the video.
What?
solved*
good work......"Will Hunting"
same I didn't keep track of how long though I think it was 2-3
Tory Berry SAME! Although I got 50... I came so close... although I haven't taken any geometry. But I do know the three angels = 180 and what he said in the video. Maybe I should have checked my work.
Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically
The problem with the "oNLy cOuNtiNg, nO dRaWiNg" method is that you would always end up getting "both-sides-equal-equation" such as 180°=180°, x=x, etc, as the final count and result.
You'd always get stuck on that. You wouldn't get anywhere or even move forward.
if we merely had to know that sum of all triangle's angles equal to 180° and square's equal to 360° and vertical angles equal to each other,
then such question wouldn't have come out in the first place.
I've doubted these comments for a long time. They either solved it trigonometrically without having read the title
or commented another "yOu pOinTlEsSly mAdE iT hARd" merely for attention and reactions due to what the majority of the comment section have been saying
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
i started watching 1 vid. now im addicted to his voice! omg
+1
i was so focused in this question that i didn't even realise that i am watching the solution on your channel..
I see people keep saying "I solved it in a different way, it's an easy question"... Ok, so share the proof instead of bragging about it
I tried it aswell, it is actually possible, by just adding up the angles you're missing 2 angles that can be replaced by y/Z then you can form an equation with both of these and solve for Z and y then you have all the angles and can get to x, I wrote that down somewhere earlier don't want to do it again tho
@@Skimbleshanks73 It's not possible. There's not enough information to solve those equations. Go on and actually try solving those equations and you'll understand.
@@demifiend9 It's actually possible by applying sine law, but the calculation is a little bit messy and compound angle formula is needed for solving the trigonometric equation. The method in the video is much more beautiful
@@Skimbleshanks73 Yes, you can generate two equations in two unknowns but the equations do not produce a unique solution. No matter how you try solving the two equations you only end up with platitudes like 0 = 0 or y = y. If you try expressing your set of linear equations in matrix/vector notation, you will find that the matrix has a determinant of zero and so cannot be inverted.
@@demifiend9 x I think can have many values. I might be wrong as well
I figured it out! Your method was far easier and more elegant than what I ended up doing, though. I basically called the length of the leftmost side "a" and calculated various other side lengths in terms of "a" via the sine rule, and then applied the cosine rule at the very end to yield a loooong arcsin expression for the value of x which I then plugged into Wolfram Alpha to get a result of 30 degrees.
Yes, that's almost exactly what I did, but I used 1 instead of a. But this video was a serious letdown, because I was hoping to learn how to do this *kind* of problem. However, the presented solution wouldn't work if the lines got shifted by a few degrees, whereas the law of sines strategy would still do the job. It felt like watching a video about breaking into a computer and being told "When prompted for the password, just write the correct password, and you're in! Super easy!"
@@davidhoracek6758 I also used 1 and did the same thing lol
@@davidhoracek6758 without knowing the value of sin (20 degrees)?
@@bartholomewhalliburton9854
Please do share this to maximum people 👇☺️
Shortest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
I've seen a few other mention using opposite interior angles equal to (2) 70° and (2) 80° angles.
I think you made it harder than it needed to be.
Andrew Stoll how did u get the 80?
Yes, i do it in head in 2 minutes using this method.
110, in truth
Wheres the parallel line?
@@digaddog6099 Draw the perpendiculars from points C and E,after that you can do this exercise in a much easier way than he did in the video.
You made the solution lot more complicated than it needed to be
He was doing it using really basic maths, hence the title "hardest easy geometry problem", using more complicated trig would be going against the point of the problem.
yh I did it in my head
@@finnwilde
Follow this shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
@@SingNostalgiaWithAmogh Thank you bro, you made it a lot easier by constructing the circle, really intriguing.
@@ItzDevil-nk3ry
Can't get easier ever than that ☺️
nice problem IMO I tried solving it as 4 equations and 4 unknowns, but the equations I used were no good
me too. there are dependent so no way to solve :(
I tried that as well and I got a solution like x = x or something, which is obviously a completely meaningless answer.
+Laurelindo well, you are right, the value of x is equal to the value of x
Yeah, but it's still meaningless when trying to find th- *realizes you are Captain Obvious* Oh....
I did the same !
It's interesting how so many people are claiming to only need "vertical angle theorem" to solve the entire problem and end up with a cyclic argument or wrong algebra. people should really check first before posting their solution ;)
I used it, and I see no way to get x or the opposite angle in that way! Doesn't make sense. Seriously, you can't use that way and people have to realize that. There aren't any vertically opposite lines to x!
construct a circle of radius BE...
draw a 180° line on BC and mark the intersection of the line and the circle as G making a new line CG
Now since angle B equals 80°
in triangle BEG
angle B = 120..
Since BE = BG (radius of the circle) we can conclude that the new triangle BEG is an isosceles triangle and since the interior angle of a triangle is 180°.. In triangle BGE .. angle beg = angle bge
therefore 2x + 120 = 180
x = 60/2
x = 30°
@@WOEEW wow. My brain is too fuzzy to analyse that right now, but I do see sense in it. I'll come back later when I'm sane, haha
@@harinirajesh3838 you don't have to complicate it like the original video
@@BerkayCeylan if you contract a circle from radius BE , you will end up will only a circle of radius be from the point b and e
I've forgotten a lot the stuff I was taught about geometry around 40 years ago. So I just used what I remembered. The fact that all the angles of a Triangle add up to 180°, lines that cross add up to 360°, angles coming off a straight line add to 180° and the angles of a Quadrilateral add to 360°. Then I went looking for all of the above shapes and situations.
Call the point where the lines BE & FC cross G
We know ∠ABC = 80° & ∠ACB = 80°
∴ ∠ BAC = 20°
We also know that ∠ BGC = 70°
Now consider Δ BGC We know ∠ BGC is 70°
Now consider Δ BCE We know ∠ BEC = 40°
Now consider Δ ACF We know ∠ AFC = 130°
Now consider Δ BCF We Know ∠ BFC = 50°
Now consider Δ ABE We Know ∠ AEB = 140°
Now consider Δ CEG We Know ∠ CGE = 110°
Now consider Δ BFG We Know ∠ BGF = 110°
Because we know 3 of the angles at the crossing point of the line at point G and they add to 360°
∴ ∠ EGF = 70°
Then I got stuck. I couldn't make any more of the shapes I remembered about.
I Knew I'd got 110° to share between the angles BFE and CEF in the Quadrilateral BCEF And 160° in the Triangle AEF
So I looked at the solution.
You cheating bastard :) You add extra lines and angles - Also I'd completely forgotten about the properties of Isosceles triangles :)
Thanks though. It was nice to drag some things I was taught so long ago out of the back of my head.
😂 I said the same thing about isosceles triangles. That got me!
Everyone: Talking about how to solve it
Me: “Hey, that 20° angle looks the same as x, probably is 20°”
I saw 2 isosceles triangles, labelled the sides and used sine rule and derived a relation between sinx and the other angles. Then it's just how good u are at trigo manipulation 😊
Very well done. I am ashamed to not be able to accomplish the same. Respect.
@@ARDAYILMAZ72 - don't be ashamed. Sometimes, a problem turns out to be difficult for us, because we don't see what many others have seen. Look at their solution and learn.
. . . BTW I would wait with the Respect for ranjan, until they have shown their solution.
@@Achill101
Here's shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
@@ARDAYILMAZ72
Share to all 👇😊
Shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
"did you figure out this problem?" boy you know i didn't
It's 30°, here's the shortest easiest way 👇😊
Shortest & easiest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
This is the first time I can answer the problem in your videos, you are awesome, things like this are really motivational, thanx 4 ur videos
This really helped a lot. My teacher told us to come here to get the solution and it worked
Dude thank you I was having a headache about this and you made it ten times more enjoyable
i can explain it to u only using angle sum property and without any construction 🙂🙂
@@deepakmanwaniappdev bro please tell me
@@Crazy77772
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
huh? I'm not sure how everyone in the comments don't realise that you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use ANY isosceles triangle logic at ALL.
@@Crazy77772
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
I tried this using only (A + B + C = 180). I then added unknowns to an additional 3 angles and created a system of four equations and four unknowns but they didn't have a solution as one variable would always get cancelled out leaving an untrue statement (130 = -20, for example).
you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use any isosceles triangle logic at all. regardless, cool thought
This can be solved using the sine rule and cosine rule without adding any additional lines but it takes quite a bit longer
Thats the proper way to go.. If the angles mentioned in this diagram are any different, this method of using isoceless triangles to find angles wont work (cuz, we are supposed to get BGF as an equilateral triangle)
@@drainedzombie2508
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
@@drainedzombie2508
Here's shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
Hi, I'm a grade 9th student from India and when I saw video's thumbnail, I was very excited to solve this problem, and I solved this problem within a minute.I solved this problem much easily with very elementary methods: 1. Isoceles triangle theorem 2. Sum of interior angles in a quadrilateral and triangle is 360 and 180 degrees respectively. 3. Vertically opposite angles are congruent. Got x=30 degrees.
Hi, could you say how you got x=30?
Did you come up with the same or a similar method of the video in your head? This is definitely not possible through those theorems without constructing new lines.
@@bartholomewhalliburton9854 I came up with other method and trust me, this problem can be solved with elementary methods as I mentioned above but it's just a more complicated method. You can also check the video's description because there also it is mentioned. Try once more and don't give up and you will get it. Thanks for your reply.
@@vaishalitirthkar Did you draw a line between points not in the video? I know this can be solved through those methods as the video suggests, but I want to know how you did it without help from the video (seeing the first step).
@@bartholomewhalliburton9854
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
Draw a circle through B , F , and E and G as a central point
then x = (1/2) (60)=30
What?
@@dux2508
Yes, he's absolutely right 🥵😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
5:04 you mean BG and BF
exactly. He should fix it in an annotation
You're right, thanks for letting me know. Sorry for the mistake, I have added an annotation and included the correction in the video description.
He ment BigFriendlyGiant BFG
Presh, I think you got x wrong. This is unless crossing lines don't create equal angles on opposite sides. Maybe I am wrong, but, in the triangle with angles 50º and 60º, the remaining angle is 70º, and so is the one on the opposite side . But the problem with your solution comes when you add this angle to F, which is 70º and x which is 30º. It doesn't add up to 180. it adds up to 170º
Almost. The angle you found is related to the line "C-F". In the final drawing, when he finds 70º, the line he uses is "F-G".
What about a general solution ? Suppose you have the same problem, but not with specific angles given.
What would x be if the given angles were say a,b,c,d ?
I guess you can do that with analytical geometry and a lot of effort
create a circle from the line BE as a radius
Well its not important that the angle x will be findable in every case..
@@srpenguinbr
I have the general solution to this. The solution method (of symmetrical angle constructions inside the isosceles triangle) in the links at the end can be used in general case...
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
Literally just filled up the straight lines as 180, a little of triangle proportion and came up with the right solution in about three minutes
i like how 12 year olds are trying to prove a mathematician wrong
dude there is literally something wrong with his solution
Zero point it out pls
Zulyrus lol they haven’t even learned algebra hardly. I know calculus and I don’t foolishly defy experienced mathematicians, especially when their right.
Nishant Modak lets say that the point at which CF and BE meet is O, just so we can label it. The sum of the angles of a triangle is equal to 180 degrees, which means angle BOC=180-CBE-BCF=180-60-50=70.
So angle BOC=70 degrees. Then we can see that angles BOC and FOE are vertical, which means BOC=FOE=70 degrees. Now, from the video, we know angle OEF=30 degrees. The sum of the angles of a triangle is equal to 180 degrees. Which means OFE+FEO+FOE=180 degrees. But here is the problem, he said in the video that angle OFE=70 degrees. OFE+FEO+FOE=70+30+70=180
170 does not equal 180. For some reason all the math works out in the video, but for some reason it just doesnt. He must have just missed some number while he was calculating. Hard to blame him, he calculated A LOT.
Aegishlash 2020 there is a difference between foolishly defying and standing up for yourself. The more experience you have at something, the less likely you are to make a mistake, but that point never reaches zero.
I think it is 40 degrees because in triangle CBO we have 60 +50=110
Then180-110=70
Since triangle is issocless according vertical opposite angles one of its angle is 70 degrees since it is a isosceles triangle it becomes 70+70+x=180 then x will be 40 degrees....I think it is my opinion so if it's wrong sorry....
Yes I tried it repeatedly and got 40 degrees somehow
@@kazishahjalal6852
Here's the most logical and easiest way...
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
@@kazishahjalal6852
✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
I kinda feel like this is giving the solution backwards.
I would find it way more logical to construct point G starting from the assumption that an Isosceles traingle GFE would help us if we could find the value of angle FGE.
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
i was so lostt when you started drawing those additional triangles
i can explain it to u only using angle sum property and without any construction 🙂🙂(price 30 Indian rupees
@@deepakmanwaniappdev
I've explained (to all) using symmetrical angle constructions inside the isosceles triangle. I've explained this FOR FREE.
@@deepakmanwaniappdev
I found it, give me 3 crore rupees 😂
I remember struggling with the infamous "Butterfly Problem".
4:55 "we'll consider this triangle BFG"
Doomguy approves
How does one even come up with this solution? This seems so created out of thin air that I am forced to believe he tried every single approach there is to solving the problem until he found a solution...
So far this video hasn't taught me anything since I am just learning a wierd solution to a wierd problem by heart
He didn't come up with the solution. This problem is known for having to draw auxiliary lines in order to solve it. Using the common theorems alone would not lead to the answer. Yep, it is indeed a weird solution for a weird problem.
+Paolo Patron you can solve this without drawing anything about 100 seconds of deduction the problem isn't all the complicated if you understand straight lines are 180 and that intersecting lines will be 360 you can deduce all you need to know to find x
Explain how then.
Joshua Woodford That's the thing. You can find all the other angles using supplementary and complementary values using interior and exterior angles but you won't be able to find the necessary angles in order to solve for x.
A "hundred seconds of deduction" is more than enough time to realize that the problem is impossible without auxiliary angles if you actually think.
+MadCodex I'll try .. for explanation purpose I'll call the intersecting lines in the middle R you'll have to figure the angle I'm speaking of.... to start the first angle of r can be determined by the fact bcr has to equal 180 and bc are know, r equals 70 .. since straight lines half to equal 180 angle r1 has to be 110 to complete the straight line same holds true for the two angles opposite have to be 70 and 110 ... now we can determine triangle bfr1 since 2 angles are known and r1 which means f in that instance is 50 also the angle opposite in triangle fae has to be 50 which means the remaining angle f in triangle fer2 can be determined which is 80 , that gives us angle f 80 and r2 70 means x has to be the remaining 30.. hope it helps hard to explain through text
There was a much easier way to do this that didn’t involve making so many seemingly random isosceles triangles. Took me like 15 minutes to find it while figuring out every other angle. Though, it did involve using more complicated angle “manipulation”. It was confusing to me how I found it, so even though I went through the vid saying “x is 30” I doubted myself when he mentioned isosceles triangles and almost changed my answer to 20
What was your solution?
please share your solution if you are saying there is an easier way
Now you tell us your solution
“There was a much easier way” “though it did involve using more complicated angle manipulation” “it was confusing to me how I found it”
LMAO I’ve never seen someone so confidently contradict themself as much as you did.
Yes the way that even you don't know
"Did you figure out this problem?"
Fucking no.
It was pretty easy tbh just dividing 180 and 360 and with some plus and minus i could conclude this
Could you describe the steps you took?
+John Kerpan can't really explain but here is my sketch
I didnt use G at all to get the solution
***** No its not... you can just use several equations and substitution to solve it. Just looking at it I almost immediately knew it was 30. I needed to add 0 lines to solve it. Show anyone that does expert level Sudoku puzzles every day this and ask them to solve it. and odds are they will go to trial and error and solve it pretty fast.
That being said.... the trial and error method while effective with the problem is usually something you do because you do not know or at least are unsure.
You can solve this several ways, however the way they EXPECT you to solve it is this way. Others either are much more time consuming or are brute forcing the answer.
Couldn't figure it out the "simple" way so I used Law of Sines and Law of Cosines instead. I began by assuming BC=10 so as to employ the Law of Sines. The length of BC can be any value as it would simply scale the whole figure and have no effect on the angle measures. Using the Law of Sines, I worked my way around the figure, using previous results to gain new results. First I found CE, then BE, then the segment from B to the interior intersection. Then the segment from C to the interior intersection. Then CF. Then some segment subtraction to find the segments from F and E to the interior intersection. Now we are down to the smallest triangle with angle x. I used the Law of cosines to solve for the missing side. Finally, used the Law of Sines once more to find the angle of 30 degrees for x.
Please check MY SOLUTIONS: ua-cam.com/video/0Jrjp2AMFIg/v-deo.html
the easiest way
They should teach the Law of Sines and the Law of Cosines early on in math, like in Geometry. I first learned it in PreCalc and it’s relatively straightforward.
I uSeD TrIgOnOmEtRiC cEvA's tHeOrEm
@@kingklaus2115 It definitely can be understood by someone in a geometry class. Using the SOH CAH TOA a clever student might accidentally come up with it themself. When I took geometry it was a part of the curriculum.
@@bartholomewhalliburton9854 That’s good to hear. It’s not even a difficult concept and like you said a student could easily come across it on accident. All geometry classes should add it to the curriculum early on.
The initial assumption of a 20deg triangle is just a guess, or, in other words one of many other guesses that could be made at some start. The whole exercise-as it is presented-is not an instructive procedure to get to the bottom of such a task, but an instruction for gambling. SIN and COS rules have to be applied to avoid waste of time.
It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE.
Lastly, use inscribed angle theorem :
Angle BEF
= 0.5 × BGF
= 0.5 × 60°
= 30°
Done !!! ✌️✌️☺️☺️☺️
I literally guessed 30 degrees in the beginning and I was like...
Why don’t you just make an educated guess
😂😂
By drawing that isosceles he too did a intelligent guess🤣
@@mr.dragoji3149
✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
I really enjoyed this video. Its in some way a perfect presentation and solution all in one
I didnt thought about drawing new lines. Time has passed since elementary chool
wow, I did geometry in high school. isn't memorization of theorems too strange and boring for tiny kids? well... i guess high schoolers have their own issues
Let the point of intersection of BE and CF be O then ,
In triangle BOC , angle BOC = 180° - (60°+50°) = 70°
Now angle EOF = angle BOC = 70° since they are vertically opposite angles
And according to your solution , in triangle EOF , angle EOF = 180° - (30°+70°) = 80°
So , why are we having two different values for angle EOF ?
Is there any mistake ?
In his solution about here 4:56 He makes another triangle and that triangle passes through the top angle which means when he said that the top angle was 70 he was referring to this triangle 6:28 (GFE) not CFE triangle
Which means the top angle in CFE was actually 80 (70+10)and not just 70.(The 10 was used to create the triangle in 4:56)
proof: 80(CFE) +30(x or BEF) +70(FOE) = 180 checks out
@@Ucj334esd
Share to all 👇😊
Shortest & easiest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
in triangle CEF I got 180=30+40+x+(110-x)
ECF is 30 degrees
BEC is 40 degrees (180-60-80=40)
CEF is 40+x degrees
CFE is therefore 110-x degrees
what is the value of x then ?
@@legionSpat
Nice one 😄
I think I found a far simpler path to the solution.
1. Flip BF110 on the AB axis to give 50 at F in AEF.
2. Take that 50 from the 130 at F in ACF to give 80 at F in CEF.
3. 180 - 70 - 80 = 30.
Thoughts?
Kindly explain every point in more detail if possible. Flipping and rotating to possibly right, correct ☹️😊
I'm very weak with rotational geometry so kindly bear with me ☺️
6:45 how do we know GE and GF are equal?
@glyn hodges how do we know BG and BF are tge same? We only know BC and BF are the same (BCF 50° and BFC 50°)
Edit nevermind: as BC and BG are the same!
@young Dou.. yes he did mistake here.. GF cannot be equal to GE, as it's GO=EO
Stop this is false,lets say BE and CF crosses at point O,and in CBO triangle BCO=50,CBO=60 ,so BOC=70,BOC=EOF=70 here he found that CFE=70=OFE if in OEF triangle OFE=70,EOF=70 so x=OEF=40!!! ITS WRONG
bcos its an isosceles triangle where there is two equal side .
ΔBGE is an isosceles triangle.
So GE = BG - - - - (1)
ΔBGF is an equilateral triangle.
So BG = BF = GF - - - - (2)
Substitute equ.(2) in equ. (1)
∴ GE = GF
These are my solutionS!
ua-cam.com/video/00ZrVZmeXmo/v-deo.html 랭글리 3탄
ua-cam.com/video/1ywU3HLfyfc/v-deo.html 랭글리 1탄
ua-cam.com/video/XMYZ0lAXsCQ/v-deo.html 랭글리 5탄
ua-cam.com/video/nLEuWUbipaQ/v-deo.html 랭글리 2탄
ua-cam.com/video/su8GAdOzW1Y/v-deo.html 랭글리 4탄
I solved completely different and even answered the 70 degree angle incorrectly answering it as 80 but got it right somehow.
I settled for the 360 degree intersection with 110 degrees for the top and bottom and 70 degrees for the right and left and finally 80+70 = 150 and therefore I arrived with the answer x = 30 degrees.
Great video by the way
Online Gladiator boi i did the same thing by filling everything with angles and go 30 somehow, i need to know the easy way
80? 150? What?
what matters is getting the angles that are surrounding x correctly (the angles that are in the same triangle as x)
@@rikiyosa4907
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
Ok I had no idea you were allowed to add in random lines! How on earth are your supposed to know to do that?!
It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE.
Lastly, use inscribed angle theorem :
Angle BEF
= 0.5 × BGF
= 0.5 × 60°
= 30°
Done !!! ✌️✌️☺️☺️☺️
For the purposes of this example, we'll call the unlabeled Vertex G. I'll simply go through the values for the easily identified angles, and go from there.
BGC is 70°. BAC is 20°. From that, we can identify that FGE is also 70°. As a result, both CGE and BGF are 110°. CEG is a 40° angle, leading GEA at 140°
Next, I'll look at quadrilateral AEGF, with it currently known angles of 70°, 20°, and 140°. The angles of a quadrilateral add up to 360°, so the final angle, GFA, is 130°.
I point this out because GFE and EFA must add up to that 130°. Likewise, GEF (in the diagram as "x") and FEA must add up to 140°. We also know that AEF and AFE must add up to 160°, and GFE and GEF must add up to 110°.
After trial and error, there is only 1 set of angles that fit those requirements:
AEF: 100°
AFE: 60°
GFE: 70°
GEF: 40° (This was the angle identified as "x", in the diagram.)
Just have doubt. In 5:57 , in the triangle BFG angleB+angleF+angleG=180
angleG=70 which shows that BG is not equal to GF
I just think that this question is a typo
Is it possible to solve this problem without using properties of side lengths at all? I was trying to do angle substitutions and systems of equations and getting nowhere, I forgot about isosceles triangles...
I don't think so. When you try relying only on angle properties there are four angles that seem impossible to work out, one being x itself. Knowing any one of them will then lead you to the other three but you just can't get it without doing other things.
I did. Weirdly enough I come to x=50. Might sound stupid but I solved every angle without creating new lines and every single angle is coherent. I'm still trying to prove myself wrong.
I posted a photo of my solution. I hope you can see if I'm wrong
Mauricio Huicochea Toledo Your solution cannot be disproven from angle tracing alone. Indeed your system of equations does not have a unique solution. If x + w = 140 and y + w = 160 then subtracting gives y - x = 20 so y = x + 20. Then substituting this in z + y = 130 gives z + x + 20 = 130 or z + x = 110. I deduced your last equation from the others, so it isn't independent, and you really have four variables and three equations. So the system has infinitely many solutions and the error must be how you deduced that x can only equal 50.
I did not used a system of equations, I tried but it looped. It didn't worked.
Much simpler solving with supplementary angles.
please explain!
@@cs8833
They can only comment, they can't explain.
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
@@cs8833
✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
@@cs8833
Shortest easiest solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
how do we know bg = bf? you said it does, but idk how you knew that.
also, how do we know ge = gf?
+Andrew Howe in an isosceles triangle, the sides opposite equal angles have equal length. he said this at the start of the video
NecroLord I know what an isosceles traingle is, but how do we know that it makes an isosceles triangle?
∠CBF=80
∠BCF=50, therefore ∠BFC=180-80-50=50
from this follows that BC=BF
since BC=BG was already shown we can now conclude that BG=BF
i guess newlines get removed for some reason :$
I solved this in a very different way. After attempting to label and solve for every angle in the diagram (and finding nothing useful) I decided I needed a shift in perspective. So I noticed that FE could be rotated counterclockwise (about F) to form the line CFE and triangle BCF. With this new shift in perspective, I was able to use the given 60 degrees and my previously solved angle CFA (and vertical angles) to find that x was 30 by x=180-20-130. I'm not sure this would work in all cases and when he mentioned isosceles triangles I was sure I got it wrong. But, surprisingly, I got the right answer.
Easier and less messy solution: Construct a line through C which is parallel to AB. This will be 20 degrees below AC. Then an extension of EF to the new line will create a new triangle. Start solving angles using 180 degree sums of triangle vertices and congruency.360 degree sums of vertical angles. The new triangle will be shown to be isoceles and every angle in the diagram can be resolved.
That's the elegant solution and the one I found too. And it proves Michael Parks' claim that angle EFA is 50 degrees. Oftentimes these problems give rise to way over-complicated solutions.
Just tried it your way and still can't get it. What is the "This will be 20 degreees below AC." mean? Diagram or/and calculations would be great to see if you have the time. Thanks!
I'll try to explain it. Basically, it's an extrapolation of the vertical angle theorem. If you pause the video at around 30 seconds, I'll try to build off Presh's hard work. We can go about either way (construct parallel, prove 20 degrees or construct 20 degree, prove parallel). Lemmas: Angle ABC is 80 degrees (60+20). Angle ACB is also 80 degrees (50+30). We will construct a new line through C and choose a point Z arbitrarily to the right. We will also extend BC downward and pick an arbitrary C' below C.
If we know we have drawn the new line CZ parallel to AB, then angle ZCC' would have to equal angle ABC which we know is 80 degrees Then BCC' = 180 and BCZ = BCC' - C'CZ = 180 - 80 = 100. Finally,, ACZ = BCZ - ACB = 100 - 80 = 20.
Alternatively, we can choose to construct line CZ such that it is 20 degrees below AC. Then BCZ = 80+20 = 100 and ACC' = BCC' - BCZ = 180-100 = 80. Since ACC' = ABC, we can conclude that AB is parallel to CZ.
The new triangle cannot be isosceles, since x will be 60 in that case.
2C00L2LIVE x is 30. the triangle I've proposed building would come out to 80-50-50 and be isoceles..
Am I the only one who used sine rule and trigonometry to solve it? :/ Here how I solved it. It's easy but a bit lengthy.In the figure,ABC is an isocels triangle. First find the angles at the intersecting point of BE and CF (let it be O) and also BFC and BEC Let,AB=AC=a.. now find BC in terms of a (using sine rule). Using same procedures repeatedly find OF and OE in terms of a.. Here CFE =110-x(use ur head.. It's easily noticeable).. using sine rule again we get a plane trigonometric equation of x ( a gets cut on both side of the equation) solving it we get x=30..
Good. That's pretty much what I did as well--used the Law of Sines (I also wound up using the Law of Cosines) to eventually get (after a lot of calculation) that sin x = 0.5, so x = 30 degrees. Didn't have to draw any other lines or create new triangles. But as you say, it was a bit lengthy. One thing that could simplify your calculations: Since the angles are independent of the size of the diagram, rather than choosing to represent a particular length by a, you can simply assign an actual value to it that will make the math easier--like 1, for example.
the whole point of this was to find the answer using stuff that like a 4th grader knows, no trigonometry
Trigo? No given sides?
Idk I saw some comments saying they assumed the sides of AB and AC were 10cm or something and worked off from there
Wow, you managed to use sine rule without any given sides!
"we are going to draw angle BG"
😂😂
تمرين جميل رائع. شرح واضح مرتب. رسم واضح .شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
If you use the exterior angle of triangle CBF on vertex F and then vertical angles you can find angle CFE as 80 degrees and therefore find x because the sum of the interior angle measures of a triangle is always 180 degrees. I find that is much simpler.
how do you know exterior angle CBF = vertex F?
@@azfariskandar8009meth
@@azfariskandar8009
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
@@azfariskandar8009
Kindly do enlighten every curious mind in messages with this solution from time to time 👇😊
Simplest solution, thanks to taba3514 short answer down in messages 👇😊
It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°. Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE.
Lastly, use inscribed angle theorem :
Angle BEF
= 0.5 × BGF
= 0.5 × 60°
= 30°
Done !!! ✌️✌️☺️☺️☺️
There's definitely something wrong with the working. If you solve all the angles without that random isosceles triangle drawing, you get angle CFE + x = 110. In his calculations, they add up to 100
That's GFE+x = 100
CFE is 10 more than GFE because drawing the equilateral triangle means you need 10 more than the 50 degrees at CFB, eating into the CFE angle to create the new GFE
@@awxangel6781
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
@@awxangel6781
Kindly share this to all from time to time ☺️
Shortest easiest possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
Really excellent sir.Thankyou for sharing such a good problem with us.
The answer is incorrect to me because BCEF is a trapezium which means the sum of its interior angles should be equal to 360° but if x is equal to 30° then the sum becomes 350° which is incorrect
??? Who said BCEF is a trapezium?
@@jameslo2818Well , my bad , should have revised the properties of a trapezium but my point still stands as BCEF is a quadrilateral which means that the sum of its interior angles should be equal to 360 so 'x' should be or rather, would be 40°
FULL SULOTION :
(using sine law and cosine law)
Angle BAC = 180-60-20-50-30= 20 (angle sum of triangle)
Let BC =1
By sine law,
(BC/sin angle BAC) = (AB/sin angle BAC)
1/sin 20 = AB/sin80
AB = sin80/sin20
Angle BFC = 180-60-20-50=50
(Angle sum of triangle)
Because angle BCF=angle BFC
Thus, BF = BC (sides opp equal angles)
ie: BF=1
AF = AB - BF
=(sin80/sin20) - 1
Because angle BAC = angle ACB = 80
Thus, AB=AC (sides opp equal angles)
AC = sin80/sin20
Angle BEC = 180-60-50-30= 40
(Angle sum of triangle)
By sine law,
(BC/sin angle BEC) = (EC/sin angle CBE)
1/sin 40 = EC/sin60
EC = sin60/sin40
AE = AC - EC
= (sin80/sin20) - (sin60/sin40)
Now we know that
AE = (sin80/sin20) - (sin60/sin40)
AF =(sin80/sin20) - 1
Angle BEC =20
By cosine law,
FE= √( (AF)²+(AE)² - 2(AF)(AE)cos 20)
FE = 0.684040286....
In Triangle BEF,
By sine law,
(EF/sin20) = (BF/sin angle BEF)
(0.684040286.../sin20)= (1/sinx)
sinx = 0.5
x= 30
because i realised i needed to take the ratios of lengths into consideration.
i just drew the triangle and measured 30 degrees.
i also did alot of angle working out but it was redundant so yeah...
Johny Broxy u spelt triangle wrong
@@zenzex4166 you spelt "you" wrong...
why do I get 40 and not 30...? 🤔
In the last result the line is between f and g not c and f, I also got this confused 😂
U r doing it wrong
Simple
It is 40 only
He is wrong
@@koushikmurthy677 it is 30
3:56 You don't need the construction to prove that the angle GEB is 40 degrees. It's 180-(60+50+30).
6:31 Triangle EFG is isosceles. Therefore its interior angles at E and F are both equal to (180-40)/2 = 70 degrees. Thus x = 70-40 = 30 degrees. (Oh, that's what you did.)
I just guessed 30° because is looks similar to the other 30°angle
I think that's how he got his 20° perfect angle to start with anyway. hehe. it's 40!
no. for triangle BCA he just added angle B (20+60) + angle C (30+50) = 160 180-160=20 Not too fucking hard if you went to 6th grade.
Hello it is really tricky to resolve this question. Why I felt a bit wired on this question:
1. What can trigger the thought of the line BG?
2. What if the question change for example if the angle b and angle c change from 80 to 85 degrees?
I see the solution here is for special case , won’t work generically.
Am I right? Please advice
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
Kindly enlighten all the thirsty curious minds with this following shortest solution, from time to time 👇😊
It's damn difficult to realize this, no doubt, but there's a supersolid reason why it's important to draw GBC angled 20°.
Observe isosceles triangle ABC of angles 20-80-80, there's another angle 80° at C in triangle BCG, so, if we wisely divide angle CBE to 20°, then we get smaller but another 20-80-80 triangle which not only ensures BC and BG are equal, but all of BC BF BG GF and even GE will be equal to GB (in isosceles triangle GBE), and since all of GB GF GE are equal, it proves that G is the circumcentre to triangle BFE.
Lastly, use inscribed angle theorem :
Angle BEF
= 0.5 × BGF
= 0.5 × 60°
= 30°
Done !!! ✌️✌️☺️☺️☺️
Anyone who thinks they "solved it" in 2 minutes by observation should propably relearn math
Wfreestyle why? By glancing at this for 30 seconds I can tell you every intersections angle without a pen.
Let's take the intersection between BE And CF and call it G so I can explain this.
Triangle BCG needs to add to 180. We have 60 and 50 for 2 corners, this means angle G of BCG is 70.
That means in EFG that angle G is also 70 due to it being symetrical.
The angle around line segment CF needs to add up to 180, if the G angle of BCG is 70 then the G angle of BGF is 110.
If angle G of BFG is 110 and angle B is 20 then angle F is 50.
The line segment BA needs to add up to 180, we know the F angle in BFG and FEA are 50 degrees which means GFE's F angle is 80.
That means E is 30 as triangle EFG needs to be 180
If we wanna know the rest than angle E of CEG is 40. Line segment CA needs to be 180 so angle E of EFA is 180-30-40 so it's 110.
Angle A is then the last angle and would be 180-50-110 so it would be 20 degrees.
Mike Percival I did it like you did just by watching the thumbnail. I don't know why the author of the video made up all of those lines making a very simple problem look harder than it really is
@@Slaave how is F IN EFA 50?
Yes, +Mike Percival, how is F in EFA 50? I proceeded exactly as you did up until that point, but I don't see how you get that. Please clarify!
Or +Francesco Lasaracina, since you did it "just like he did", maybe you can clarify that?
Trivial steps omitted, but denote by F' the reflection of F over AC, and note that triangle CFF' is equilateral. Because FF'CB is a kite, BF' is the perpendicular bisector of CF, so ∠EBF'=∠FBE=20° and BFEF' is cyclic (the latter following by the trillium theorem); hence, ∠BEF=∠BF'F=30°.
The main difficulty of the problem lies in the fact that focusing on angles is not enough to solve the problem; one needs to relate the angles to lengths and vice versa. This can be easily done with trigonometry, but doing it synthetically requires more thought.
Alternate solution
1. FIll in all the angles possible especially results of EB & EC AB & AC and most important the intersection of BE & CF this will give you 70 deg a complement of x. (intersect BC and CF is a pair of 70 & 110 deg)
2. Corner A is 20 deg making ABE isosceles.
3. Draw a line from E crossing perpendicular to AB, intersect with an extension of CF, let's call the new corner G; now EFG is isosceles and corner G is 40 deg. (the angle at F is 50 deg)
4. EFG is isosceles so the corner E from EF & EG is 40 deg and EA & EG is 70 deg (180-90-20).
5. We should already know corner E from EB & EC is 40 deg therefore x=180-40-40-70 or 30 deg.
"Now, EFG is isosceles". This is not so obvious. Very useful, but is not proved anywhere.
@@mathcanbeeasy
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
@@mathcanbeeasy
World's easiest solution 👇😊
Best possible solution existing in math to this problem, thanks to taba3514 short answer down in messages ✌️😊
Sides GB & GE of isosceles triangle GBE are equal, and the same is true for sides GB & GF of isosceles rather equilateral triangle BGF.
Hence, GB = GF = GE
This means G is the circumcentre to triangle BFE.
Lastly, as per inscribed angle theorem :
Angle BEF
= 0.5 × Angle BGF
= 0.5 × 60°
= 30°
Mind-blowing, easiest method ever !!!!! 💖☺️
The best part is that UA-cam auto-captions call him Fresh Tall Walker.
Hello dosto . Mera youtube channel bhi dekhlo please agar pasand aye to like 👍 and subscribe Dosto mera youtube channel mai aapko class 9 ki physics ,maths ,history videos milenge .Please but atleast dekh to lo.
you used a very long method i did it much quicker by using vertical angles, supplementary angles, and alt ext angles
Bobby Diesel if you used the alt ext postulate, where are the parallel lines?
@@angelov.5305 you can always use the "if its opposite side of intersection, it is same angle" and complementary angle rules at the x
I did it the same as you.
@@angelov.5305 construct them
Please share us in detail
Give BC unit value. Isosceles sides of tri ABC can then be found. Once that is done, isosceles sides of tri ABE, AE and BE, can then be determined. Since tri CBF is isosceles then BF is unity. Then using the sine law obtain x from sin(x)/1 = sin(160-x)/BE, where BE = 1.53208 up to 5 decimals.
Easiest approach is to use coordinate geometry, that way you do not create a mess.
Please check MY SOLUTIONS: ua-cam.com/video/0Jrjp2AMFIg/v-deo.html
the easiest way
how you draw CBG =20degree.why you making this hard man??
I think that the correct answer is 40° and not 30°
Maybe you could also provide your reasoning?
I think he meant 'thought' as in past tense
+John Kerpan why did you start out making your own line at a perfect 20° as if a compass and square were avalable. we could have just measured it ourselves.
it is 40. >:/
You realize that the picture could not be to scale at all and the math would still work, right? What is the mathematical reason that 40 is the right answer, and how is his reasoning in this video wrong?
I am highly convinced that your solution is quite convincing. The other solutions given here are not that convincing as they are bound to make assumptions.
Other way: Use "trigonometric ceva" for triangle EBC and for point F outside of the triangle.
a 10 year old in Singapore, my country, can do this. and I'm actually astonished people are using trigonometry when you don't even need to mention the word isosceles at all. quite funny nonetheless
You just posted the same problem.. with.. 20° and 10°
Like just now..
"The hardest math problem ever"
Expects me to solve it
a+b=x, a+c=y, b+c=z, b+d=w... you can solve for a, b, c, d in terms of x, y, z, w if a, b, c, d are the unknown angles at F and E and x, y, z, w are the values to which they sum.
U have to resolve this without new angle u made.
i can solve it only using angle sum property and without any construction 🙂🙂
Awesome question.
I tried all the ways I know for more than half an hour ( I mean by elementary geometry methods, not by trigonometry and using calculator to find trigonometric values 🙂).
I had to quit. 🙂.
Was worthy watching..
Same, I don't know trigonometry yet so I did all I could and eventually gave up haha. But I have a feeling that the answer is simpler than the one in the video, I would've never thought to draw that line 😅
@@harinirajesh3838 a lot of such problems are only solvable with either general rules (trig stuff) or if you know where to draw lines unfortunately.
If you sort of make very clever discoveries to solve these problems you often end up reinventing trig anyways.
@@randomnobody660
Here's the easiest and most logical way...
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
@@randomnobody660
Use symmetry, dear friend. ☺️
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
In the description you state:
*At 5:04 I misspoke. I meant BG = BF. We wil shortly prove BG = GF.
I take that to be a hint.
It took me 2 minutes to solve this problem. I don't know why he was making all this triangles. I appreciate his work. But he was making it complicated.
how did you solve it?
Mind your Decisions : Draws a 20 degree angle using protector
Me : then why the heck you did not use protector to measure **x**
protector
I typed protector
Because the diagram isn't always up to scale, for people just want to watch the world burn.
@@spacebearcadet746 protectors are not allowed in exams
@@secretunknown2782
Just to politely inform you-
It's called protractor, not protector.
How can you assume BG=BF? From the given data, can we assume that? Or can it be measured by a compass?
Yea, wasn’t clear on that either!
Because if BG = BC and BC = BF then BG must also = BF
BGF is an equilateral triangle so all sides are equal.
@@tazngirr2800 how ???
@@vickiware76 that they derived, when was it said ?
I got 60. I used the fact that all the angles in a triangle add up to 180 degrees and started to calculate angles in triangles with 2 known angles. I also used the fact that an angle at the intersection of 2 lines is equal to the angle opposite to it at the intersection to get the angles 70 and 110 at the intersection of the lines shown in the diagram. Actually, I have looked at my work after typing the first part and have seen where I had gone wrong. I assumed that the right-most triangle within the triangle in the diagram is isosceles. My bad
I got the right answer.
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