I hope you are joking. You understood the proof presented - that is vital in my opinion vs getting the ah ha momoment which can often be *not* so trivial
@@leif1075 Thanks for highlighting the possible confusion. My point was to celebrate the understanding of the proof even if one could not arrive at the proof themselves. I will go back to adjust the wording..
@@salmanel-farsi3744 Thanks but I need to be someone who arrives at the proof themselves or comes up withba different proof..I don't want to be unexceptional..recognizing the proof or understanding it doesn't require nearly as much ability.
I eventually realized you had to do it if you wanted a non trig solution to this problem but I wouldn't have satisfied the time constraints of a traditional math test. That is why trig was invented. To save time and eliminate the need to come up with tricks.
There is another elegant way to do this question, this was the way i did this question when it first came to me: My solution is about to prove that this EF segment defined by a 45° in the square is aways tangent to the quarter of circunference with unity radious and center in A. To do this take the quarter of circunference and trace a tangent segment of it, now with the intersection point "P" trace AP and look that the pairs of triangles ADE, APE and ABF,APF are congruents by the Side Side Angle case (SSA - common sides, unity side and straight angle). With it, see the congruence between DAE, PAE and PAF, BAP angles, nevertheless how DAE+PAE+PAF+BAP=90° and DAE=PAE, PAF=BAP, then PAE+PAF=45°, proving that EAF will aways be 45°. So, we can guarantee that if we pass a quarter circunference in the problem with radious 1 and center A it will tangent EF. Doing this, let the intersection be an X point, now observe the AXF and ABF triangles - they are congruent by the same case before - and see that AFB and AFX(AFX=AFE) angles are both 70°, letting angle AEF= 65°(AFE+FEA+EAF=180° - EAF=45°, AFE=70°). (:
The first part was real easy. 45 and 70 degrees part of the triangle has to equal 180 degrees. So the missing part must be 65. I didn’t exactly understand what the second part needed me to do so I just watched it after. Took me about 5 minutes tops
@@HeckaS no way , you got that wrong , as a student in Indian education system it's quite difficult to find someone who looks upon such questions and problems and I was amazed just because of the beauty of the question , I hope that you got my point ..
I actually thought about rotation and solved the first part.I was too lazy for the second part.But the point I wanna make is that due to your videos,Presh,I was able to get better at geometry quite a bit,and also in Math problem solving in general.I wanna thank you for that.My creativity also actually increased.Please keep coming up with such good Math videos.
Dayummmmm....Why i'm not find your channel early?... I just want to entertain myself with this something useful rather than a meme video or drama that wasting my time a lot while quarantine 😭😭😭
Presh's way ("half angle" model) is good, I just want to show alternative, which is good too. Draw FH perpendicular to AE and meet @ H, thus A,B,F,H and E,C,F,H are co-circular (concyclic), respectively. So ∠HBF=∠HAF=45° (circumferential angle), so BH must be the diagonal of ABCD; connect DH (which must be diagonal), so ∆ADH congruent to ∆CDH, thus ∠DAH=∠DCH=90-45-20=25°, hence, ∠HFE =25° too, bcz it's circumferential angle, so we get ∠AEF=90-25=65° As for second part of the problem, we can use Gougu theorems and similar triangles (∆ADE and ∆FHE) to get answer, that's 2 (for you guys to do 😁).
As a ninth grader in India, I'll admit that I was unable to solve this one, but I was able to understand exactly how you reached the conclusions. I just didn't think about rotation. Thanks for sharing this problem, we have tons of questions like these in our book since almost half our chapters in the book are geometry this year (Quadrilaterals, Triangles, Lines and Angles, Circles, etc).
Just saw the problem. But as a ninth grader in China, this is a model all of us had to learn "角含半角模型“ translated to an angle that is half of another angle. Basically, there is this thing that is always true, and it's that if there is a 45 degree angle inside a 90 degree angle, then BF+DE is always equal to EF. And if tanBAF=1/2, then tanDAE is always 1/3 ( The second part is proofed by the formula tan(a+b)= 1+tana+tanb/ 1-tanatanb). This is a problem that is really easy for us 9th graders in China, and there are all kinds of variations, like 30 degrees or 60 degrees, or removing some sides. Though it is good to see other countries actually have these kind of problems as well.
I’d have chosen a brute force method and simply made equations over equations and then slowly eliminating the unknown parts till I can solve it. But your way is much faster and easier. Thank you for making this awesome content.
Nice puzzle. I was content to solve this using brute force and was surprised to find whole number answers. This of course meant there was a more elegant solution. Thanks for showing the reflection trick. Immediately this brings to mind folding tricks, origami as one. if we fold the two flaps on either side, they would fit neatly within the larger triangle in the middle. Super neat!
Using *Trigonometry* - Angles around 45 are 90-70=20 and 90-(45+20)=25. Therefore all other sides (and therefore angles) become computable - AB=BC=CD=DA=1, AE=sec(25), AF=sec(20), DE=tan(25), BF=tan(20), EC=1-tan(25), FC=1-tan(20). Lastly, EF^2 = CF^2 + CE^2 = (1-tan(20))^2 + (1-tan(25))^2 = (tan(20)+tan(25))^2. The x and y in the video are tan(20) and tan(25).
It’s possible to prove, using the exact same argument, that EF is tangent to the circle centered in A with radius 1, for ALL pair of points EF on the sides such that EAF = 45. It’s easy to see that if P in EF is the circle’s point of tangency, then DE = EP and FB = FP. Therefore, p(CEF) = (CE + ED) + (CF + FB) = 2.
The rotation solution is really elegant and satisfying! I got there maybe a little more laboriously just solving a system of equations with unknown angles AEF, AFE, FEC, and EFC: AFE + EFC = 110; AEF + FEC = 115; FEC + EFC = 90; AEF + AFE = 135. So AEF = 65, FEC = 50, EFC = 40, and AFE = 70.
Данная особенность в том, что угол eaf 45 градусов. И это можно доказать для любого квадрата со стороной, ну пусть будет А длина. При этом условии сумма будет постоянна.
Just found another way! By drawing two circles with radii DE and BF, in vertices E and F, respectively. Ultimately finding the altitude of triangle AEF. The rest is pretty straightforward.
It's much more simple actually. And the only possible answer is a bit unexpected Let's challenge the inputs of this task and look at the angle AEC try to follow my logic: 1) AFB = 70 degrees then AFC = 180-70 = 120 the sum of interior angles of quadrangle is 360 degrees then let's measure an angle AEC looking at AFCE quadrangle: AEC = 360 - EAF (45) - AFC (120) - FCE (90) = 105 on the other hand: 2) FAB = 180 - AFB (70) - ABF (90) = 20 then, DAE = 90 - EAF (45) - FAB (20) = 25 then, AED = 180 - EDA (90) - DAE (25) = 65 if AED = 65, then AEC = 180 - 65 = 115 so (1) gives us AEC at 105 degrees and (2) gives it at 115 That means that inputs and hence the task iteself was wrong from the very beginning
I’ve been watching you channel for a long time that i am able now to every problem you upload. It feels amazing when you say did you figure it out and the answer is yes :)
BF = tg BAF = tg π/9 . ED = tg DAE = tg 25 . therefore, we can easily get CE and CF. And then EF by Pythagoras theory. This also gives the 2 remaining angles of the triangle ECF by the relations : CFE = arcsin CE. And CEF = arcsin CF . on the other side , we know already that BAF = 20 ==> DAE = 90 - (45+20) = 25. So that DEA = 65. Which gives the angle AEF. Simple.
Takılmışız "sadece müziğin dili evrenseldir" diye. Oysa ki her müziğin bir matematiği vardır ve asıl evrensel olan belki de matematiktir . 2+2 tüm dillerde 4 eder .
Verdigin hesplamanın sonucu 2'nin 4'un +'nin.... tanımına bağlı. Eğer mantik zincirimizde bunlarin tanımiyla ya da bu fonksiyonlarla ilgili bir kırıklık, eksiklik ya da bu işlemin tümünün koşullarını overridelayan bir değişken/etken varsa düpedüz yanlış olan bir şey bile bazen teknik olarak doğru olabilir
@@theUnmeshraj Dude being able to answer this question isn't even a flex lol. A waste of time not to use trig, but I didn't and still got the answer. This question proves nothing about intelligence.
I couldn't solve this one, I tried the linear system of equations but it was singular. However, having watched this, I feel that I now have learned a trick that might help me solve problems in the future. Thanks so much for this wonderful channel!
"We will rotate this triangle 90° clockwise about vertex A." | Me: _Uh-huh..._ "These 2 triangles are congruent, in fact they are mirror images of each other." | Me: _Uh-huh..._ "Let's consider a numerical approach." | Me: _Uh-huh..._ "No matter where we pick the points of E and F the perimeter of this triangle will always be equal to 2." | Me: _Uh-huh..._ "Suppose that BF has a length equal to x." | Me: _Uh-huh..._ "This means DF' also has a length equal to x." | Me: _Uh-huh..._ "Suppose that DE has a length equal to y." | Me: _Uh-huh..._ "Therefore EF has a length equal to x+y." | Me: _Uh-huh..._ "We can cancel out the y's and cancel out the x's..." | Me: _Uh-huh..._ "...and we're left with 1+1 and that's exactly equal to 2..." | Me: *THAT IS CORRECT!* 😎 "...which is what we saw in the numerical approach. Amazing." | Me: _Uh-huh..._
Another way to solve it: (I thought of this AFTER seeing Presh's answer) Construct a line vertically up from E and horizontally along from F. Call the point they meet 'O' angle EOF is obviously 90 degrees. This means: If we draw a circle through EFA, then O is the centre of the circle - because the angle it subtends at O (90) is double the angle at the circumference A (45) this means that triangle EOF is isosceles - and OEF = 45 degrees It's easy to show that DAE = 25 and therefore DEO = 25 Giving 65 for the required angle
@@dada56 Distance OF is different from distance OE. O is not the center of a circle through A, E and F. Angle EAF is 45. Ok. Angle EOF is 90. Ok. But that doesn't mean that O is the center of the circle thru AEF.
Спасибо, Мастер! Благодаря Вам я увидела большие пробелы в своём образовании и в характере. Особенно много мне дали Ваши задачи по геометрии. Я восхищаюсь Вами!
Solution after pausing at 0:56 and not looking at comments: First get all the angles I can significantly DEA is 65 degrees and that DAE and FAB add to 45 degrees. Next relize that since DAE plus FAB equals EAF you can fold ADE and ABF exactly onto triangle AEF meaning that angle AEF is equal to DEA 65 degrees. Next consider line EF and that it is DE + FB so the perimeter of triangle CEF is BF+FC + DE+DC we know BF+FC is 1 and that DE+DC is 1 so the perimeter of triangle CEF is 2.
Oh wow...i attempted the question using trigonometry only to realise that i don't have to use trigonometry. otherwise, i found and solve all of it! Thanks...we need more challenging questioins
It is nice to prove your claim at 3:30 that the perimeter of CEF is always 2. If EC is 1-tg(alpha) and CF is 1- tg(45-alpha) .... It is nice to see it.
Love from india presh sir .. one thing I want to tell you that for your such amazing geometry videos i am able to improve my geometry to a great extent . ❤️ Btw i solved the problem using trigonometry and my answer came approximately ≈ 64.743 ≈ 65 . (Reason is because the values which the calculator gave was approximated ) but TBH your solution about rotating the triangle was great 👍 . Came to learn a new thing form you , thanks 😀
1:36 dislike for geometric inaccuracy: for lenght of 1 AD angle of 25 degrees gives smaller lenght of DE , while for same lenght of 1 AB and engle of 20 degrees we get bigger lenght of BF which is impossible. If angle DAE is 25 degrees, and angle BAF is 20 degrees, than DE lenght must be longer than BF lenght. I stopped watching video there because angles that has been given to us were wrong.
Watched a handful of these videos in the last two days. I noticed not once (yet) has the Law of Sines come into play. It actually made finding the perimeter easier than finding the measure of the angle. Still, the solution does seem more satisfying than using Law of Sines.
Actually quite simple and intuitive if you consider it as as a folding problem. You fold B to a point on the line EF. Then you fold D to the same point. The circumference is then equal to BC + DC = 2.
However, if you choose points E and F much closer to B and D, respectively, the resulting triangle CEF clearly has a perimeter larger than 2. If you don’t think so, draw it with angles AEB and AFD having angles of 89 degrees. The perimeter of triangle CEF approaches 2+ root2 as the angles AEB and AFD approach 90 degrees. Your proof in this case does not apply because the angle of rotation would be greater than 45 degrees, causing the construction it uses to be invalid. SO it's important to note that your proof only applies when angle EAF is a 45 degree angle.
@@tommydecarite4121 That's the thing with any master of skill, they are so good at it that when they do it they make it seem easy. Goes for math, goes for juggling and about anything in between. You only discover how hard it actually is when you are trying to do it yourself.
Simpler solution for the 1st question: The triangle is equilateral, and equilateral triangles have the property that all its angles add up to 180°. So the 3rd angle is equal to 180° - (45 + 70)
i think is more intuitive to begin just by drawing a vertical line from A to EF and then continue in almost the same approach, than rotate triangle ABF
@@herculesperseus6326 The triangle has a 45 degree angle in the middle of a 90 degree angle. The other angles have to add up to 45 to meet the 90, so they're the same
@@vampire_catgirl which means one of them is 20 and the other is 25. How would you mathematically fold both flaps into the central triangle with any evidence it's a perfect fit
Leif sorry ... tg is the function tan, if you know the angle you know its tangent and because the side equal 1 you can easily find the other side, you can do this in both triangles ADE and ABF in this way you find FC and EC then with Pitagora theorem you find EF and you have the perimeter FC + EC + EF. ok ? Sorry for the confusing tg ... :)
I always make a point to try the problem before watching. I did the whole thing using trig, and when I started watching the first thing you said was "without using trig", duh, no wonder it wasn't so trivial. What a clever solution though. Good video!
For the angle of AEF (I learned to say FEA but anyway) just know that the total of all angle of any triangle is = to 180º. So just do 180º-45º-70º = 65º
the 70º shown in the picture is for the angle AFB, outside the triangle in question, so you don't know that angle AFE inside the triangle is 70º yet. Now the angle at F just happened to be 70º, 70º and 40º, so it was just a coincidence.
Presh : 1+1=2
Me : ok , i got that one
well atleast you got that
Blackpenredpen 😂😂😂
Not gonna lie I had a relative once ask me if x + x = 2x, I said yes then he started arguing about it with my cousin lol
@@Daniel_WR_Hart ROFL
@@vol230 I wish I was trolling but unfortunately I'm not
I cannot believe that I’m watching this to entertain myself.
Good you developed your self and you improved your level in maths
But My Answer is Coming 70°(the asked one) and other one is coming (45+70)=115,180-115= 65°,please help!
same here,each time i open youtube and i see one video from this channel
May I ask why?
me 2...thot i see math 1 last time b4 i die due to kidney failure @16 :) ...bye world♥
Me: *solves the problem looking at the thumbnail using trig*
You: "the challenge is to do it without trig"
Me: 😲
Same hahaha
Me too
I don't even know what trigonometry is. Can someone tell me what it is?
@@deadble5k start with sin, cos, tan, etc and then learn how to manipulate it like algebra
Ikr, his solve is cool and all, but how to solve it pure mathematics
"Nothing out of the box is good, trust me"
Pandora
That made my day🤣🤣😂
Presh should give this comment a heart.
@Cara Peru. I always love being inside the box. 😜
Classic!!! 🔥💯😆
Schrodinger approves too
These videos* make me question whether I really deserve to be in college.
I hope you are joking. You understood the proof presented - that is vital in my opinion vs getting the ah ha momoment which can often be *not* so trivial
My friend, you and I are the same
@@salmanel-farsi3744 if it is not sp trivial that means it's also important so aren't you contradicting yourself?
@@leif1075 Thanks for highlighting the possible confusion. My point was to celebrate the understanding of the proof even if one could not arrive at the proof themselves. I will go back to adjust the wording..
@@salmanel-farsi3744 Thanks but I need to be someone who arrives at the proof themselves or comes up withba different proof..I don't want to be unexceptional..recognizing the proof or understanding it doesn't require nearly as much ability.
Oh dang - that was intense!
Just another jee problem 😪
3D god is watching 2D thing lmao
@@NamVu-im2xm lol
How are you here?
But My Answer is Coming 70°(the asked one) and other one is coming (45+70)=115,180-115= 65°,please help!
I've always found those problems so cool! Just rotating a triangle was the key, and I would never think about it until you showed us. Great job Presh!
Dont you find it annoying and frustrating and insulting?
Ahaa photocoPI
sumakit cjs well, in french π is prononced like "pee" so my username works better
I eventually realized you had to do it if you wanted a non trig solution to this problem but I wouldn't have satisfied the time constraints of a traditional math test.
That is why trig was invented. To save time and eliminate the need to come up with tricks.
It took me a couple of days jumping from one way to another until finally could jump out of the box
There is another elegant way to do this question, this was the way i did this question when it first came to me:
My solution is about to prove that this EF segment defined by a 45° in the square is aways tangent to the quarter of circunference with unity radious and center in A.
To do this take the quarter of circunference and trace a tangent segment of it, now with the intersection point "P" trace AP and look that the pairs of triangles ADE, APE and ABF,APF are congruents by the Side Side Angle case (SSA - common sides, unity side and straight angle). With it, see the congruence between DAE, PAE and PAF, BAP angles, nevertheless how DAE+PAE+PAF+BAP=90° and DAE=PAE, PAF=BAP, then PAE+PAF=45°, proving that EAF will aways be 45°.
So, we can guarantee that if we pass a quarter circunference in the problem with radious 1 and center A it will tangent EF. Doing this, let the intersection be an X point, now observe the AXF and ABF triangles - they are congruent by the same case before - and see that AFB and AFX(AFX=AFE) angles are both 70°, letting angle AEF= 65°(AFE+FEA+EAF=180° - EAF=45°, AFE=70°). (:
I always love "Out Of The Box" thinking like this. Thanks again Presh!
Hmmm. Sure it was an “out of the square” solution. 😜
I can't stand them. They always seem completely arbitrary
@@tjejojyj 😂 AMAZING BRO !! 😀👍🏻😂
But My Answer is Coming 70°(the asked one) and other one is coming (45+70)=115,180-115= 65°,please help!
The first part was real easy. 45 and 70 degrees part of the triangle has to equal 180 degrees. So the missing part must be 65. I didn’t exactly understand what the second part needed me to do so I just watched it after. Took me about 5 minutes tops
3:37 🤯🤯🤯🤯🤯 being an Indian student that amazed me
Yes because race determines your reactions.
@@HeckaS no way , you got that wrong , as a student in Indian education system it's quite difficult to find someone who looks upon such questions and problems and I was amazed just because of the beauty of the question , I hope that you got my point ..
Why you have to mention "Indian " ??
@@Dev_1907 cuz this que is frm an indian 9th grade que paper
@@HeckaS they said that because the question was from the main course maths book for 9th graders in India
I actually thought about rotation and solved the first part.I was too lazy for the second part.But the point I wanna make is that due to your videos,Presh,I was able to get better at geometry quite a bit,and also in Math problem solving in general.I wanna thank you for that.My creativity also actually increased.Please keep coming up with such good Math videos.
Dayummmmm....Why i'm not find your channel early?... I just want to entertain myself with this something useful rather than a meme video or drama that wasting my time a lot while quarantine 😭😭😭
Surprised that the solution doesn't involve the gougu theorem.
Lmao
what
Vszk 010 Pythagorean Theron is known as gougu theorem among mandarin speaking countries.
Use hua
This channel should be renamed as the Gougu theorem channel
Beautiful. The problem, the explanation, and everything this channel brings is pure amazing
When you know how to do the first part but the second part is a struggle.
yo sameee
Second part is more interesting😜😜
I found the second part easier for some reason
When you wipe you bum then have to wipe your hand after.
Don't ask!
ua-cam.com/video/eEyQyixOGpU/v-deo.html
I'm impressed with myself that I actually understood this one from start to finish.
Presh's way ("half angle" model) is good, I just want to show alternative, which is good too. Draw FH perpendicular to AE and meet @ H, thus A,B,F,H and E,C,F,H are co-circular (concyclic), respectively.
So ∠HBF=∠HAF=45° (circumferential angle), so BH must be the diagonal of ABCD; connect DH (which must be diagonal), so ∆ADH congruent to ∆CDH, thus ∠DAH=∠DCH=90-45-20=25°, hence, ∠HFE =25° too, bcz it's circumferential angle, so we get ∠AEF=90-25=65°
As for second part of the problem, we can use Gougu theorems and similar triangles (∆ADE and ∆FHE) to get answer, that's 2 (for you guys to do 😁).
Dude I have two doubts
First is that what you mean by co circular
Do you mean concyclic???
And second is that gougou is pythagoras theorem??
iam asian
@@ujjwalnarayan07 1, by that it means the four pts are at same circle. 2, yes, Gougu = Pythagorean
As a ninth grader in India, I'll admit that I was unable to solve this one, but I was able to understand exactly how you reached the conclusions. I just didn't think about rotation. Thanks for sharing this problem, we have tons of questions like these in our book since almost half our chapters in the book are geometry this year (Quadrilaterals, Triangles, Lines and Angles, Circles, etc).
This one must be of RD
yeah it must be
@@stupidstuart6478
What is the perimeter?
Paul: Imagine that we have a ruler.
I guess he has left his ruler somewhere out of the box.
ישראלי!
@@drorbenuliel9280עלית עליי
Just saw the problem. But as a ninth grader in China, this is a model all of us had to learn "角含半角模型“ translated to an angle that is half of another angle. Basically, there is this thing that is always true, and it's that if there is a 45 degree angle inside a 90 degree angle, then BF+DE is always equal to EF. And if tanBAF=1/2, then tanDAE is always 1/3 ( The second part is proofed by the formula tan(a+b)= 1+tana+tanb/ 1-tanatanb). This is a problem that is really easy for us 9th graders in China, and there are all kinds of variations, like 30 degrees or 60 degrees, or removing some sides. Though it is good to see other countries actually have these kind of problems as well.
Bro u don't have to use trigo in this question otherwise it's very simple
I’d have chosen a brute force method and simply made equations over equations and then slowly eliminating the unknown parts till I can solve it. But your way is much faster and easier. Thank you for making this awesome content.
I tried this but I was only got x+y=135 over and over again (where x is angle AEF and y is angle AFE)
I love your learning mindset that great people never stop learning.
Nice puzzle. I was content to solve this using brute force and was surprised to find whole number answers. This of course meant there was a more elegant solution. Thanks for showing the reflection trick. Immediately this brings to mind folding tricks, origami as one. if we fold the two flaps on either side, they would fit neatly within the larger triangle in the middle. Super neat!
Draw a perpendicular from C to FE
From left 65
From right 45
The read angle is 180 --65--45
70
Your "rotations" are always beyond the access of my imagination. Nice problem and solution.
Doesn't that make you feel bad?
Using *Trigonometry* - Angles around 45 are 90-70=20 and 90-(45+20)=25. Therefore all other sides (and therefore angles) become computable - AB=BC=CD=DA=1, AE=sec(25), AF=sec(20), DE=tan(25), BF=tan(20), EC=1-tan(25), FC=1-tan(20). Lastly, EF^2 = CF^2 + CE^2 = (1-tan(20))^2 + (1-tan(25))^2 = (tan(20)+tan(25))^2. The x and y in the video are tan(20) and tan(25).
It's a problem of 9th grade student of India
china the same
Korea in 8th grade
Maybe they say "Grade 9" in India :-)
@@Tiqerboy its not a bymistake. This is what he said in video
@@이우석-f1l in india 9th grade students are 13-14 years
It’s possible to prove, using the exact same argument, that EF is tangent to the circle centered in A with radius 1, for ALL pair of points EF on the sides such that EAF = 45.
It’s easy to see that if P in EF is the circle’s point of tangency, then DE = EP and FB = FP.
Therefore,
p(CEF) = (CE + ED) + (CF + FB) = 2.
Also we can find the angle by folding ADE and ABF when we draw the height of AEF
The rotation solution is really elegant and satisfying! I got there maybe a little more laboriously just solving a system of equations with unknown angles AEF, AFE, FEC, and EFC: AFE + EFC = 110; AEF + FEC = 115; FEC + EFC = 90; AEF + AFE = 135. So AEF = 65, FEC = 50, EFC = 40, and AFE = 70.
Angle e and f both side satisfy with 65 and 70 so this doubt create because exactly answer not solve angle e is 70 or 65 confusion create
Love from INDIA..🇮🇳🇮🇳🇮🇳
One of the greatest nations.
Me 2 India 🇮🇳🇮🇳🇮🇳🇮🇳
India is the smartest country in the world
Only love will not work, you need to solve the problem in in the first place 😉 Btw, both Asia and Europe are smart
Данная особенность в том, что угол eaf 45 градусов. И это можно доказать для любого квадрата со стороной, ну пусть будет А длина. При этом условии сумма будет постоянна.
I have just came across this channel and I am so glad about it
And I definitely became the constant member of it :)
Bravo for the problem AND the solution! Congratulations to Anand Gautem.
I thought ur previous problems r simple but this one is truly some big brain stuff!!
good problem, not too hard, not too easy - and good clue on the title - which sort of gave the answer. Took me a while but got it.
Just found another way! By drawing two circles with radii DE and BF, in vertices E and F, respectively. Ultimately finding the altitude of triangle AEF. The rest is pretty straightforward.
Ya
How do the two circles solve the first part (i have not even got to second part)
By using limits determined perimeter to be 2. If E comes close to D, then F comes close to C . So EF = CD = 1, FC = 0 , EC = DC = 1.
I clicked the video thinking that I'd have to hear all about the Gougu theorem. Imagine my pleasant surprise when I didn't.
It's much more simple actually. And the only possible answer is a bit unexpected
Let's challenge the inputs of this task and look at the angle AEC
try to follow my logic:
1) AFB = 70 degrees
then AFC = 180-70 = 120
the sum of interior angles of quadrangle is 360 degrees
then let's measure an angle AEC looking at AFCE quadrangle:
AEC = 360 - EAF (45) - AFC (120) - FCE (90) = 105
on the other hand:
2) FAB = 180 - AFB (70) - ABF (90) = 20
then, DAE = 90 - EAF (45) - FAB (20) = 25
then, AED = 180 - EDA (90) - DAE (25) = 65
if AED = 65, then AEC = 180 - 65 = 115
so (1) gives us AEC at 105 degrees and (2) gives it at 115
That means that inputs and hence the task iteself was wrong from the very beginning
I’ve been watching you channel for a long time that i am able now to every problem you upload. It feels amazing when you say did you figure it out and the answer is yes :)
Ah yes, the negotiator
BF = tg BAF = tg π/9 . ED = tg DAE = tg 25 . therefore, we can easily get CE and CF. And then EF by Pythagoras theory. This also gives the 2 remaining angles of the triangle ECF by the relations : CFE = arcsin CE. And CEF = arcsin CF . on the other side , we know already that BAF = 20 ==> DAE = 90 - (45+20) = 25. So that DEA = 65. Which gives the angle AEF. Simple.
Very interesting, I love Geometry and this is my edutainment. Thank you.
It’s like half quiz
and half math.
Enjoyed much!!
Takılmışız "sadece müziğin dili evrenseldir" diye.
Oysa ki her müziğin bir matematiği vardır ve asıl evrensel olan belki de matematiktir .
2+2 tüm dillerde 4 eder .
Newton matematik için "Evrenin dili" tanımlamasını yapmış.
Verdigin hesplamanın sonucu 2'nin 4'un +'nin.... tanımına bağlı. Eğer mantik zincirimizde bunlarin tanımiyla ya da bu fonksiyonlarla ilgili bir kırıklık, eksiklik ya da bu işlemin tümünün koşullarını overridelayan bir değişken/etken varsa düpedüz yanlış olan bir şey bile bazen teknik olarak doğru olabilir
katılıyorum geometriye bayılırım, ingilizcede sadece basit tanışma faslını bilirim ama videoyu çok iyi anladım :)
O zaman Matematik yapamayanlar da dilsiz oluyor herhalde? :)
:)
I used trignometry and sin rule but your solution is beautiful
Now I know I can't pass grade 9 in India. Thanks! :)
Its not bro i m indian i m in 10th
It was just a miscellaneous question
@@Anish-IITP bruh if you're weak in studies Dont actually defame us...
I am studying for IMO from grade 7 and we used to get such questions.
@@theUnmeshraj if u r overintelligent dont pop yr ass
@@theUnmeshraj btw. Best of luck Hoping under 1k rank in jee. Its not in sarcastic way. Imo atudents are really worth it.
@@theUnmeshraj Dude being able to answer this question isn't even a flex lol. A waste of time not to use trig, but I didn't and still got the answer. This question proves nothing about intelligence.
I couldn't solve this one, I tried the linear system of equations but it was singular. However, having watched this, I feel that I now have learned a trick that might help me solve problems in the future. Thanks so much for this wonderful channel!
"We will rotate this triangle 90° clockwise about vertex A." | Me: _Uh-huh..._
"These 2 triangles are congruent, in fact they are mirror images of each other." | Me: _Uh-huh..._
"Let's consider a numerical approach." | Me: _Uh-huh..._
"No matter where we pick the points of E and F the perimeter of this triangle will always be equal to 2." | Me: _Uh-huh..._
"Suppose that BF has a length equal to x." | Me: _Uh-huh..._
"This means DF' also has a length equal to x." | Me: _Uh-huh..._
"Suppose that DE has a length equal to y." | Me: _Uh-huh..._
"Therefore EF has a length equal to x+y." | Me: _Uh-huh..._
"We can cancel out the y's and cancel out the x's..." | Me: _Uh-huh..._
"...and we're left with 1+1 and that's exactly equal to 2..." | Me: *THAT IS CORRECT!* 😎
"...which is what we saw in the numerical approach. Amazing." | Me: _Uh-huh..._
Another way to solve it: (I thought of this AFTER seeing Presh's answer)
Construct a line vertically up from E and horizontally along from F. Call the point they meet 'O'
angle EOF is obviously 90 degrees.
This means: If we draw a circle through EFA, then O is the centre of the circle - because the angle it subtends at O (90) is double the angle at the circumference A (45)
this means that triangle EOF is isosceles - and OEF = 45 degrees
It's easy to show that DAE = 25 and therefore DEO = 25
Giving 65 for the required angle
Sorry, no, no and no.
@@octobre4623 Oui, oui, oui? Pourquoi excrirez- vous "no no no" (In English please)
@@dada56 Distance OF is different from distance OE.
O is not the center of a circle through A, E and F.
Angle EAF is 45. Ok.
Angle EOF is 90. Ok.
But that doesn't mean that O is the center of the circle thru AEF.
Yes you're right - THANKs for that!
@@dada56 I made the same error when I was looking for a solution.
Amazing solution!!!!!!! That's why I love this channel very much.
I think I studies my 9th standard as an indian student properly. There were questions much harder than this one. It was a lot easy for me
Thinking “Out Of The Box” is something that I should apply in geometry too.
Never thought Congruency will be used.
Love from Haridwar (India)❤️
I really love this channel as it shows 'out of the box thinking'.It feels cool to do so.
As a Canadian middle schooler, I got so incredibly close. The outside the box for me
8th grade?
I got 85
Спасибо, Мастер! Благодаря Вам я увидела большие пробелы в своём образовании и в характере. Особенно много мне дали Ваши задачи по геометрии. Я восхищаюсь Вами!
16 years ago i could solve it in 5 minutes ) but now i think "wow, amazing" :)
Thank you very much for solving this problem. Good luck. Hi from Russia.
Amazing from indonesia🇮🇩
Solution after pausing at 0:56 and not looking at comments:
First get all the angles I can significantly DEA is 65 degrees and that DAE and FAB add to 45 degrees.
Next relize that since DAE plus FAB equals EAF you can fold ADE and ABF exactly onto triangle AEF meaning that angle AEF is equal to DEA 65 degrees.
Next consider line EF and that it is DE + FB so the perimeter of triangle CEF is BF+FC + DE+DC we know BF+FC is 1 and that DE+DC is 1 so the perimeter of triangle CEF is 2.
Equations are Boring part of Mathematics, But I used to see everything in the form of Geometry - Stephen Hawking
Me: It's really True !
Oh wow...i attempted the question using trigonometry only to realise that i don't have to use trigonometry. otherwise, i found and solve all of it! Thanks...we need more challenging questioins
Alternative solution
Create 2 equations involving 2 unknown variables on point E and F.
Problem solved..
No!
It is nice to prove your claim at 3:30 that the perimeter of CEF is always 2. If EC is 1-tg(alpha) and CF is 1- tg(45-alpha) .... It is nice to see it.
Never thought I’d encounter this problem again lmao we had it in year 8 math in China
The first time coming across ur videos I have started liking maths.you really have comprehensive explanations.THANK YOU.
We are inspired by you. Thanks for motivating us to do math videos 😌💝
Fold △ABF and △ADE to the middle triangle having ∠AEF=∠AED
△ABF and △ADE will fit clearly in △AEF
Therefore EF=DE+BF
EF+FC+CE=DE+EC+CF+FB=1+1=2
Love from india presh sir .. one thing I want to tell you that for your such amazing geometry videos i am able to improve my geometry to a great extent . ❤️ Btw i solved the problem using trigonometry and my answer came approximately ≈ 64.743 ≈ 65 . (Reason is because the values which the calculator gave was approximated ) but TBH your solution about rotating the triangle was great 👍 . Came to learn a new thing form you , thanks 😀
It can be also done by similarity and in most of the square triangle question rotating the triangle is clue:-)
We can also generalize it by taking angle theta instead of 45 degrees
@CHABBI SARKAR PJ Sir ka Bhakt
Really I cant live without Presh Sir
@@siddharthabhattacharya391 kon
1:36 dislike for geometric inaccuracy: for lenght of 1 AD angle of 25 degrees gives smaller lenght of DE , while for same lenght of 1 AB and engle of 20 degrees we get bigger lenght of BF which is impossible.
If angle DAE is 25 degrees, and angle BAF is 20 degrees, than DE lenght must be longer than BF lenght.
I stopped watching video there because angles that has been given to us were wrong.
Just one word to say:- "WOW"
Watched a handful of these videos in the last two days. I noticed not once (yet) has the Law of Sines come into play. It actually made finding the perimeter easier than finding the measure of the angle. Still, the solution does seem more satisfying than using Law of Sines.
Wait a min.
In triangle EFC
EF square = FC square + EC square
That isnt satisfying for their values in terms of x and/or y
omg!!!!!! I solved this one myself! I'm a high school student and I look forward to ur videos.
I had a doubt how can you construct 20°and AF=AF' AT SAME TIME
ua-cam.com/video/osuflRQzsm0/v-deo.html see the second method
Actually quite simple and intuitive if you consider it as as a folding problem. You fold B to a point on the line EF. Then you fold D to the same point. The circumference is then equal to BC + DC = 2.
Yes, it's true, but how can you prove that the folded point B (for example) lies on the line EF?
That is, that the height of triangle AEF is 1 ?
Everything was cool until this rotation.. I mean wth!
I almost gave up on this one even after guessing the answer, but took a pen and paper and quickly did it. Satisfying this one.
I couldnt solve this even with trigonometry
Me
Law of sines
However, if you choose points E and F much closer to B and D, respectively, the resulting triangle CEF clearly has a perimeter larger than 2. If you don’t think so, draw it with angles AEB and AFD having angles of 89 degrees. The perimeter of triangle CEF approaches 2+ root2 as the angles AEB and AFD approach 90 degrees. Your proof in this case does not apply because the angle of rotation would be greater than 45 degrees, causing the construction it uses to be invalid. SO it's important to note that your proof only applies when angle EAF is a 45 degree angle.
i miss you mathematics&geometry.. now im a med student :(
merhabalar. videonun kapak fotografindaki bilinmeyen aciyi 80 derece buldum. siz bu konuda ne dusunuyorsunuz ?
Şimdi evrenin sırrını çözücen hadi bakalım
@@cerencemal1617 kim evrenin sırtını cözecek reyiz?
I wonder how many viewers solve these problems using the same method Presh uses. It seems like there are a million ways.
Directly connect the diagonal！
0:40 from India 🇮🇳 I got this one! I'm so happy!
Sorry i need to turn the bell on i refreshed the page at the perfect time tho
Watching your last videos enables me to solve this problem on my own! Thanks Presh!
2:00
F' : let me introduce myself
btw this was a cool problem even tho i wasn't able to solve it and found it complicated you made it seem easy
@@tommydecarite4121
That's the thing with any master of skill, they are so good at it that when they do it they make it seem easy. Goes for math, goes for juggling and about anything in between. You only discover how hard it actually is when you are trying to do it yourself.
Simpler solution for the 1st question: The triangle is equilateral, and equilateral triangles have the property that all its angles add up to 180°. So the 3rd angle is equal to 180° - (45 + 70)
This made me realise how out of shape I am with math. I remember having this in my grade 4 mock exams in the Philippines.
i think is more intuitive to begin just by drawing a vertical line from A to EF and then continue in almost the same approach, than rotate triangle ABF
İngilizcem yetersiz olmasına rağmen görerek anladım gayet anlaşılıyor, umamırım mesajımı anlamışsınızdır 😌
This question was very easy as compared to questions here in India
If, instead of rotating 90 degrees, you fold both flaps in to the central triangle, solving the 2nd question becomes trivial.
Nicely done.
Nice..
How would you actually do it with this method lol, there's no clue to prove any triangle to be congruent
@@herculesperseus6326 The triangle has a 45 degree angle in the middle of a 90 degree angle. The other angles have to add up to 45 to meet the 90, so they're the same
@@vampire_catgirl which means one of them is 20 and the other is 25. How would you mathematically fold both flaps into the central triangle with any evidence it's a perfect fit
Man I can't believe that they are giving this level of questions to 9th graders in India
I found the length of the perimeters interesting, as far as the angles all triangles equal 180 degrees so solving the unknown angle was no problem .
I didn't think of that much,just using tanθ to find BF and DE,then find EC and FC,then pyth.thm to find angle FEC
I found x = 65 and P = 2 ... it looked easier than what it actually is 🧐
How did you solve for P though? The way he did it?
Leif we know the angles DEA = 65 and AFB = 70 so we can find EC = 1- 1/tg(DEA), FC = 1 - 1/tg(AFB) and EF comes straight from Pitagora ...
@@christianfunintuscany1147 what the heck is tg?? And pythagoras doesnt eliminate the variables in this case so you dont get the answer..
Leif sorry ... tg is the function tan, if you know the angle you know its tangent and because the side equal 1 you can easily find the other side, you can do this in both triangles ADE and ABF in this way you find FC and EC then with Pitagora theorem you find EF and you have the perimeter FC + EC + EF. ok ? Sorry for the confusing tg ... :)
I always make a point to try the problem before watching. I did the whole thing using trig, and when I started watching the first thing you said was "without using trig", duh, no wonder it wasn't so trivial. What a clever solution though. Good video!
Hey I am getting the answer as 70
Find all the angles and then use alternate angle theorm
@@sadhanasamant4819 you are explaining yourself 😂
@@sadhanasamant4819 lol no it isnt so easy youve probably made some mistake plz check once
@@chhabisarkar9057 check it yourself
that's the angle at angle AFE
For the angle of AEF (I learned to say FEA but anyway) just know that the total of all angle of any triangle is = to 180º. So just do 180º-45º-70º = 65º
the 70º shown in the picture is for the angle AFB, outside the triangle in question, so you don't know that angle AFE inside the triangle is 70º yet. Now the angle at F just happened to be 70º, 70º and 40º, so it was just a coincidence.