Dear Nathan, we don't have such magic book! I'd love to share if I find one. It takes enormous amount of time for research and planning to prepare quality videos! Please keep watching PreMath channel. We work very hard to put quality videos for our valued viewers. Thanks for asking.❤️
I think you can also do this without establishing equality of any lengths. It just ends up with two equations and two unknowns after calculating all known angles. But I like this way also.
Yeah, but he's establishing a principle. Even I could do this just by using angle sum theorem and the other one he used. And I'm an amateur. He's just showing you some of the nuances and eccentricities of geometry. He's not solving them by the fastest route. :)
@@BKNeifert Interesting take. What it means is that the information of equal lengths is not necessary to the solution of the problem. And I've never known a mathematician to not demonstrate the fastest route to the solution.
@@randyscorner9434 Well, yes. But again, he's demonstrating a principle. He's showing the elements at work. It's like poetry. It's not the fastest, easiest way to express something, but it's the most beautiful.
I love the Exterior Angle Theorem. The concept is so cool because the Interior Angles of a triangle can be equated to the Exterior Angle and it's Adjacent Interior Angle of the same triangle and the sums equate to 180⁰. So the Adjacent Angle is common to both and can be subtracted from both sums and wallah!....the Exterior Angle equals the sum of the Opposite Interior Angles. Cool concept too apply to polygons also! 🙂
Exterior angle theorem is useless. We just have to know that all angles in a triangle are 180 degrees (you know other 2 already) and in straight line it is 180 - angle = seeked angle... These methods are used so often that you have to know them, so why add another not needed one? Even this video uses both of my examples... Waste of memory. But yea, adds more things to calculate.
It's just the Triangle Sum Theory exercised upon the Straight Angle Property, since the two opposite angles will add to 180° - θ and the external angle will also be 180° - θ.
@@quigonkenny Yea, but you'll need memorize one more Theory, while you still need 2 others anyway... This is what I call not so useful memory usage. Try to teach these for thousands. I rather focus on those that are always needed instead of those that are just needed in certain situations. If and when you are teaching advanced students, then you may toss this to speed up some calculations (leaving only one short one to write?). But I would not use this for general teaching purposes.
Wow this solution is SO much more elegant than mine. I concluded that the areas of triangles DCE and FBC had to be the same as they both have the same base and height. From there I used a bunch area of triangle formula and law of sines to establish proportions between the sides DC, EC, FC and BC. Finally I got a convoluted expression for the tangent of x that contained the sines and cosines of three other angles. I did end up with the right answer though but I could not have solved it without a calculator. Well done.
Note that while the solution in the video only works with angles with these magic numbers (that create those isosceles), your solution could be used if you change the angles to break the symmetry, so it does have more value in its generality.
@@projekcjaLots of math competition questions have these "convenient" values so that they can be solved without brute forcing it, since performing more calculations takes more time no matter how proficient you are at it. The questions are often as much a measure of intuition and the understanding of how and when to use the laws of mathematics/geometry as they are of the actual calculation.
If angle CEB= 100 degrees and angle BCE= 50 degrees then those angles add up to 150 degrees, so shouldn't X be 30 degrees so that it adds up to 180? Also, if angle BCE= 50 then BCA adds up to 90, making the entire shape a simple 90, 60, 30 triangle. Again, it looks like X should be 30.
After identifying most of the angles (most of them are trivial since they have to add up to 180) if you define the line between D and E as having a length of 1 then the line between C and E is 2.87 and thus the line between C and F is 3.69 So now you've got a triangle with one side length 1, one side length 3.69, and one angle of 130 which means the remaining side is 4.4 and the two angles are (rounding slightly because the angles are probably meant to be whole numbers) 10 and 40 So X = 40 Though for what it's worth, I approached this as a logic puzzle rather than a mathematical one because I've forgotten easily 90% of what I've ever learned about trig so I was using a calculator.
Drop a perpendicular from apex C onto base DE meeting at point G. This will bisect 20 degree angle DCE and divide DE in half. Right triangle AGC is a 90/60/30 one. Let AC be length of 1 unit. Sin 60 = CG/1. CG = 0.8660. Angle GCE = 10 degrees. Tan 10 = GE/CG. GE = 0.8660 x 0.1763 = 0.1527. Therefore DE = 0.1527 x 2 = 0.3054. Triangle CGF. CG/sin 50 = GF/sin 40. 0.8660/sin 50 = GF/sine 40. GF = 0.7267. DE = FB = 0.3054. GB = GF + FB. GB = 0.7267 + 0.3054 = 1.0321. Tan X = CG/GB = 0.8660/1.0321 = 0.8391. Tan-1, 0.8391 = 40 degrees.
The problem with PreMath examples is that one cannot screenshot the sketches and then measure the angle x with some protractor, because we know that the sketches are *not drawn to scale* 🤣🤣 Nice problem!
I missed angle DCF=50°, so I missed Isosceles triangle... Instead, I used The Law of Sines, and got: sin(x)=(sin(100)*sin(50-x))/sin(20), but have no idea how to extract x, and got 25° = I've made an error somewhere... Can you do the same using The Law of Sines? Even though your solution is simple and elegant... Would it work for other angles?
Yes, you can. If you use law of sines and law of cosines. I assigned the value of 10 to line segment DE and line segment FB. After determining all the angles inside the triangle, you can use law of sines to determine all the lengths to 10 decimal places. You have to use law of cosines for triangle FBC to determine length of line segment BC. You can calculate the area of triangle BFC using A = 1/2 * FC * 10 * sin (130) At that point, A = 1/2 * 10 * FC * sin (x). Plug and chug to get x=40.
Pretty nasty one, but solvable. Start with making 90 angle to half way DE. You get height from tan 10 = 0,5/z, z~2,84, Then using same height you gain bottom line from half DE to F = k, k =z/(tan 50), And from half DE to B you get tan x = z/(k+1) and solving that you get x = 40. You may also roughly estimate x from the angles... 180-50 = 130 degrees as the other angle at point F. And the upper right angle is less than 20 degrees, so if we go by 10 degree hops, then it is 10 degrees. This leads to third angle 180-130-10 = 40 degrees. But this method is not accurate and relies on estimates.
Today on How to take a math problem designed for university level and explain it as if a 4 yr old was supposed to be able to solve it ...... dumbest thing i have ever seen, if you have to explain to someone that the sum of a triangle is always 180 degrees, you are NOT ready for these kinds of problems.
I have to say … since the question was "justify your answer", I felt i had to at least TRY to get it first! My plan … The triangle between the '60+20°' and '50+30°' triangles has what top angle? Easy enough … by 180° triangle rule, α = 180 - (60 + 50 + 20 + 30) α = 20° Well, if that is 'split in half' to (10° + 10°) and a new line extends from top to base, then we have an especially useful triangle 60-30-90° With that, the 'new line' height can be set to (1) arbitrarily, and then the half-width of the inside triangle's base becomes 1⋅tan 10° = 0.17633 units. The base of the known right hand triangle is 1⋅tan (10+30)° = 0.83910 units; Since we know that the 𝒙 biggest right triangle must have 0.83910 + 2 ⋅ 0.17633 units total, then angle 𝒙 is (arctan 1/1.19180) = 40° And that is my proof. Yay. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Thanks a lot professor I follow you from Algeria. Bonne continuation professeur.
You are welcome dear!
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Thank you for including the basic concepts prior to evaluating the problem.
Thanks for the feedback❤️
That's a wonderful problem! I like it a lot! Thanks for sharing! 😊
Thanks ❤️🌹
Sir where do you get these problems from?? Please reply im preparing for IMO
Dear Nathan, we don't have such magic book! I'd love to share if I find one. It takes enormous amount of time for research and planning to prepare quality videos!
Please keep watching PreMath channel. We work very hard to put quality videos for our valued viewers. Thanks for asking.❤️
@@PreMath oh ok sir thank you and keep uploading I love the geometry questions that you upload daily i
Best wishes for a happy and healthy New Year
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Happy, safe, and prosperous New Year!🙏
I think you can also do this without establishing equality of any lengths. It just ends up with two equations and two unknowns after calculating all known angles. But I like this way also.
Yeah, but he's establishing a principle. Even I could do this just by using angle sum theorem and the other one he used. And I'm an amateur. He's just showing you some of the nuances and eccentricities of geometry. He's not solving them by the fastest route. :)
@@BKNeifert Interesting take. What it means is that the information of equal lengths is not necessary to the solution of the problem. And I've never known a mathematician to not demonstrate the fastest route to the solution.
@@randyscorner9434 Well, yes. But again, he's demonstrating a principle. He's showing the elements at work. It's like poetry. It's not the fastest, easiest way to express something, but it's the most beautiful.
I love the Exterior Angle Theorem. The concept is so cool because the Interior Angles of a triangle can be equated to the Exterior Angle and it's Adjacent Interior Angle of the same triangle and the sums equate to 180⁰. So the Adjacent Angle is common to both and can be subtracted from both sums and wallah!....the Exterior Angle equals the sum of the Opposite Interior Angles. Cool concept too apply to polygons also! 🙂
Thanks ❤️
Exterior angle theorem is useless. We just have to know that all angles in a triangle are 180 degrees (you know other 2 already) and in straight line it is 180 - angle = seeked angle... These methods are used so often that you have to know them, so why add another not needed one? Even this video uses both of my examples...
Waste of memory. But yea, adds more things to calculate.
It's just the Triangle Sum Theory exercised upon the Straight Angle Property, since the two opposite angles will add to 180° - θ and the external angle will also be 180° - θ.
@@quigonkenny Yea, but you'll need memorize one more Theory, while you still need 2 others anyway... This is what I call not so useful memory usage.
Try to teach these for thousands. I rather focus on those that are always needed instead of those that are just needed in certain situations.
If and when you are teaching advanced students, then you may toss this to speed up some calculations (leaving only one short one to write?). But I would not use this for general teaching purposes.
Thank you so much! The problem is tricky but the solution is very nice!
Welcome 👍
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Very good theorem
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Wow this solution is SO much more elegant than mine. I concluded that the areas of triangles DCE and FBC had to be the same as they both have the same base and height. From there I used a bunch area of triangle formula and law of sines to establish proportions between the sides DC, EC, FC and BC. Finally I got a convoluted expression for the tangent of x that contained the sines and cosines of three other angles. I did end up with the right answer though but I could not have solved it without a calculator.
Well done.
Thanks ❤️
Note that while the solution in the video only works with angles with these magic numbers (that create those isosceles), your solution could be used if you change the angles to break the symmetry, so it does have more value in its generality.
@@projekcjaLots of math competition questions have these "convenient" values so that they can be solved without brute forcing it, since performing more calculations takes more time no matter how proficient you are at it. The questions are often as much a measure of intuition and the understanding of how and when to use the laws of mathematics/geometry as they are of the actual calculation.
If angle CEB= 100 degrees and angle BCE= 50 degrees then those angles add up to 150 degrees, so shouldn't X be 30 degrees so that it adds up to 180? Also, if angle BCE= 50 then BCA adds up to 90, making the entire shape a simple 90, 60, 30 triangle. Again, it looks like X should be 30.
After identifying most of the angles (most of them are trivial since they have to add up to 180) if you define the line between D and E as having a length of 1 then the line between C and E is 2.87 and thus the line between C and F is 3.69
So now you've got a triangle with one side length 1, one side length 3.69, and one angle of 130 which means the remaining side is 4.4 and the two angles are (rounding slightly because the angles are probably meant to be whole numbers) 10 and 40
So X = 40
Though for what it's worth, I approached this as a logic puzzle rather than a mathematical one because I've forgotten easily 90% of what I've ever learned about trig so I was using a calculator.
Thank you!
You're welcome!
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Elegante!
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Drop a perpendicular from apex C onto base DE meeting at point G.
This will bisect 20 degree angle DCE and divide DE in half.
Right triangle AGC is a 90/60/30 one.
Let AC be length of 1 unit.
Sin 60 = CG/1.
CG = 0.8660.
Angle GCE = 10 degrees.
Tan 10 = GE/CG.
GE = 0.8660 x 0.1763 = 0.1527.
Therefore DE = 0.1527 x 2 = 0.3054.
Triangle CGF.
CG/sin 50 = GF/sin 40.
0.8660/sin 50 = GF/sine 40.
GF = 0.7267.
DE = FB = 0.3054.
GB = GF + FB.
GB = 0.7267 + 0.3054 = 1.0321.
Tan X = CG/GB = 0.8660/1.0321 = 0.8391.
Tan-1, 0.8391 = 40 degrees.
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I like your solution =)
Solved it myself using tan() multiple times, after I found out, it's isosceles
Good job!
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Very strange solution, unimaginable at all, just a talented design.🎉
Thank you very much!❤️
Very easy!
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The problem with PreMath examples is that one cannot screenshot the sketches and then measure the angle x with some protractor, because we know that the sketches are *not drawn to scale* 🤣🤣
Nice problem!
Is this solvable if it did not involve the puzzle being designed around BCE being an isosceles triangle?
Whole diagram must be considered!
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I missed angle DCF=50°, so I missed Isosceles triangle...
Instead, I used The Law of Sines, and got: sin(x)=(sin(100)*sin(50-x))/sin(20), but have no idea how to extract x, and got 25° = I've made an error somewhere...
Can you do the same using The Law of Sines? Even though your solution is simple and elegant... Would it work for other angles?
Yes, you can. If you use law of sines and law of cosines. I assigned the value of 10 to line segment DE and line segment FB. After determining all the angles inside the triangle, you can use law of sines to determine all the lengths to 10 decimal places. You have to use law of cosines for triangle FBC to determine length of line segment BC. You can calculate the area of triangle BFC using A = 1/2 * FC * 10 * sin (130) At that point, A = 1/2 * 10 * FC * sin (x). Plug and chug to get x=40.
@@Copernicusfreud
Wow! Thanks!
I'll try to follow this later on.
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80° sir
x+x+100=180
2x=180-100=80
x=80/2=40
so:x=40. ❤❤❤Thanks
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Yeah nice, but how do you get to that in the first place?
@@hcgreier6037 Thanks sir
I got as far as realising that CDF is an isosceles triangle, but somehow didn't see the next step. So easy once you know how!
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So much better than using Trig.
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Good afternoon sir
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Pretty nasty one, but solvable. Start with making 90 angle to half way DE. You get height from tan 10 = 0,5/z, z~2,84, Then using same height you gain bottom line from half DE to F = k, k =z/(tan 50), And from half DE to B you get tan x = z/(k+1) and solving that you get x = 40.
You may also roughly estimate x from the angles... 180-50 = 130 degrees as the other angle at point F. And the upper right angle is less than 20 degrees, so if we go by 10 degree hops, then it is 10 degrees. This leads to third angle 180-130-10 = 40 degrees. But this method is not accurate and relies on estimates.
180-(60+50)-20-30=20， 50，60，70 is FAC triangle, angle CDE=angle CED=80, so CD=CE, .....😅
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x is equal to fourty i think, if i use intuition and the theorme were the sum of the angles are 180 degrees.
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At 2:38
60+50+10 != 180
That's not a 10, it's a 70
I see what your doing here..
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asnwer=45 isit
X = X
simple as
Today on How to take a math problem designed for university level and explain it as if a 4 yr old was supposed to be able to solve it ...... dumbest thing i have ever seen, if you have to explain to someone that the sum of a triangle is always 180 degrees, you are NOT ready for these kinds of problems.
X equals 30 . It can not be 40 degrees. Because the sun of X and BCF equals 50 degress
Way too much extra work.
I have to say … since the question was "justify your answer", I felt i had to at least TRY to get it first!
My plan …
The triangle between the '60+20°' and '50+30°' triangles has what top angle? Easy enough … by 180° triangle rule,
α = 180 - (60 + 50 + 20 + 30)
α = 20°
Well, if that is 'split in half' to (10° + 10°) and a new line extends from top to base, then we have an especially useful triangle 60-30-90°
With that, the 'new line' height can be set to (1) arbitrarily, and then the half-width of the inside triangle's base becomes 1⋅tan 10° = 0.17633 units.
The base of the known right hand triangle is 1⋅tan (10+30)° = 0.83910 units; Since we know that the 𝒙 biggest right triangle must have 0.83910 + 2 ⋅ 0.17633 units total, then angle 𝒙 is (arctan 1/1.19180) = 40°
And that is my proof.
Yay.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
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