Saying that a truncated decimal (necessarily, as it is unending) is exactly equal to a known root is not rigorous. There is no proof here, only an approximation. Numerical methods are not exact.
is not bad beacause if you use trig function is very often numbers like 1 over square root of 3 or 2, in my opinion is much better than the mind blowing geometry way
I definitely agree with @@letao12 here. It’s very easy to apply the formulas to get the sqrt3/3 and it’s why it’s not such a big deal that he made in the video
I actually thought of the Euler-mascheroni constant instead when he said 0.577, because while I think about 1/sqrt(2) quite a lot, I pay almost no attention to 1/sqrt(3)
Here's a way to finish the trig method without a calculator: tan10 * tan50 / tan20 = tan10 * tan50 * tan70 = tan10 * tan(60-10) * tan(60+10) = tan10((tan60 - tan10) / (1 + tan60tan10))((tan60 + tan10) / (1 - tan60tan10)) Since tan60 = sqrt(3): this equals tan10 * (3 - tan^2(10))/(1 - 3tan^2(10)) = (3tan10 - tan^3(10) / (1 - 3tan^2(10)) Now, from the somewhat obscure triple-angle formula tan(3x) = (3tanx - tan^3(x)) / (1 - 3tan^2(x)), This equals tan(3 * 10) = tan30. So x = 30. *the triple-angle thing can be derived by tan(2x + x), or alternatively set z = 1 + bi on the complex plane (so b = arg(z)) and compute the argument of z^3.
For the geometrical solution, the problem ended in 5:45. It is suffice to see that x+y=50 from the start, 2x+y=80 (from the isosceles triangule), therefore, x=30.
As others have pointed out, the equation tan(x) = (tan(50°)tan(10°))/tan(20°) at 10:10 can be solved for an exact value of x without using numerical values. The first step is to replace 1/(tan(20°) with the exact equivalent tan(70°). Then we recognize that 70° = 60° + 10° and 50° = 60° - 10°, so tan(x) = tan(60° - 10°)tan(10°)tan(60° + 10°). Now, we can use the identity tan(3Θ) = tan(60° - Θ)tan(Θ)tan(60° + Θ). Let Θ = 10° and we find that x = 30°. To prove the identity for Θ = 10°, we use the tangent sum of angles and difference of angles formulas. So, tan(a + b) = (tan(a) + tan(b))/(1 - tan(a)tan(b)) and tan(a - b) = (tan(a) - tan(b))/(1 + tan(a)tan(b)). We substitute a = 60°, so tan(60°) = √3, and b = 10°. So, after simplification, tan(60° - 10°)tan(60° + 10°) = (3 - tan²(10°))/(1 - 3 tan²(10°)). For the left side, we first compute tan(20°) = tan(10° +10°) by either the sum of angles or double angle formula, finding tan(20°) = (2tan(10°))/(1 - tan²(10°)). Then we compute tan(30°) = tan(20° +10°). Simplifying and factoring, we find tan(30°) = tan(60° - 10°)tan(10°)tan(60° + 10°). In the real world, all measurements have tolerances. In geometry, measurements often have exact values and the problem solution may require that the measurements are exact. For example, a triangle is not equilateral unless all 3 interior angles are exactly 60° and a triangle is not a right triangle unless one interior angle measures exactly 90°.
I found an alternative solution: Extend BA to Q such that triangles BDA, CQA are similar. Then triangles BCA, DQA are also similar (side-angle-side). Angle chasing yields DQ bisects angle BQC. Hence D is the incentre of triangle BQC. Hence x = 1/2 * 60 = 30. QED.
@@leif1075 This is neither a defence nor a condemnation of the original commenter's method, but you can easily infer the amplitudes of all the internal angles of the triangle ABC, which is plenty enough, once you arbitrate the length of any one side.
My thought process once you established isosceles triangle BCD": "Angles D"BC and D"CB are both 50°, so angle BD"C msut be 80°. Therefore, D'D"C is (80 - 60) 20°, and since it's the vertex of an isosceles triangle, the other two angles must be ((180 - 20)/2) 80°. Angle D"CD' is equal to (50+x)°. 50°+x° = 80°, therefore x=30°.
@@leif1075 this type of problems require that kind of thinking. When you're used to them the only issue you'll have is "So what lines do I draw" or "what triangles do it reflect"
I employed a much simpler trig. method? The internal angle at D, within the smaller triangle, appears directly proportional to D's position on AE, compared to AC. As AE is bisected from point C, the internal angle at A( 40* ), increases by .5 times at D( 60* ). 180* - ( 60* + 90* ) = 30 degrees for x ! Who needs a calculator ? 😃
Solved it the second way in my head, but a tad bit different. (After getting the 50 degrees) BE = y EC = z y * tan(10) = z * tan(x) y * tan(20) = z * tan(50) Divide the equations: tan(10) / tan(20) = tan(x) / tan(50) (From here it was the same)
Hey just to understand it, why is y*tan(10) = z*tan(x) and y*tan(20) = z*tan(50). I know where you got the degrees from but why ist that equals to that?
Just a tip, remember that in math olympiad contest (i assume Polish mathematical olympiad uses the same rule as the international olympiad) calculators are not allowed, therefore you should have shown a way to finish the trigonometric way without calculator...
@@henrytang2203 i don't think so... what if the degree isn't an integer? plus who does that anyway? why not use that time to lear a more useful skill or improve problem solving intuition?
@@ericzhu6620 I understand there are numerous trig formulas that can be used. But have you watched the process for deriving an exact value for something like sin(3 degrees)? So unless the angle is a nice number, the no calculator rule is a tad ridiculous.
@@henrytang2203 When the contest bans calculator it must be because they know you don't need it, therefore when there appears a value like sin(3º) there must be a way to cancel it out to avoid calculating it
@5:40, when you prove that line CD'' is congruent to lines D'D'' and BD'', you can solve for angle D'D''C which must be 50+50+60+x=180 so that angle is 20 degrees. Then you use the fact that CD'D'' is an isosceles triangle and that the total angle D'CD'' must be (180-20)/2, which is 80 degrees. Also, let's call angle DCD'' y, for easier equation. You get an equation pair: 2*x+y=80 x+y=50 (got that earlier, from the starting diagram) From these you get x=30 (also y=20...) No need to prove that CDD' is an equilateral triangle or circle from point D'' or arcs of it
Yeah I see how everyone is saying this but the fact D'' was the center of a circle passing through B,D'' and C was the first thing I noticed but that's probably because I'm used to geometry problems.
How I did it: X = arctan(DE/CE) Because BD bisects ABE, therefore DE is half of AE. tan(40) = CE / AE --> CE = tan(40)*AE substituting them in our equation we get X = arctan[(AE/2) / (tan(40)*AE)] AE cancels out X = arctan[(1/2) / (tan(40))] move the numerator's denominator to the denominator X = arctan(1 / 2tan(40)) I had to use the calculator for this but the answer comes to around 30 degrees with rounding errors.
@@leif1075 it mens that the first method is much harder compared to the second one, doesnt mean that its literally deadly, its just another way of saying that it is bad
Very good. I tried it just looking at the angles specified in the figure, but I ended up with 5 equations in 5 unknowns, but the resulting matrix was singular, and then i gave up. I appreciate the trig solution more. It requires less imagination, which I appreciate. I admire the geometric construction and the imagination it took.
I got the answer 30 degrees, but had to use trigonometry. I just set BE=1. Then I generated trig equations DE = tan(10°) AE = tan(20°) EC = AE * tan(40°) = tan(20°)tan(40°) tan(x) = DE/EC = tan(10°)/(tan(20°)tan(40°)) I used a calculator, which seems like cheating, but no doubt it can be done using trigonometric identities. Ok. just looked at the geometry answer. I could have looked at that for weeks, and not got it. Thank heavens for trig.
You can find exercises like this in a peruvian book called "Construcciones en Triángulos - Técnicas y Criterios para realizar Trazos Auxiliares" by José Luis Meza Bárcena Cuzcano"
because angle ABD is half ABE, and triangle ABE is a right triangle with right angle AEB, D must be at the midpoint of side AE. triangle ACE is also a right triangle with one of the legs on AE, and we can easily find that angle ACE is 60. since we know D is at the midpoint of AE, we also know that DC will bisect the angle ACE, therefore the angle DCE is half angle ACE. x = 30 (prewatch). In other words, since line segment BD bisects ABE, we can easily see that DC will bisect ACE.
"because angle ABD is half ABE, and triangle ABE is a right triangle with right angle AEB, D must be at the midpoint of side AE." Quite simply that is not the case. (1) DE = BE * tan(10) (2) AE = BE * tan(20). If D was at the midpoint of AE then (3) AE = 2 * DE Substitute in for AE and DE from (1) and (2) and we get BE * tan(20) = 2 * BE * tan(10) cancel out BE and you get tan(20) = 2 * tan(10) The above is simply not the case. To three significant figures tan(20) = 0.364 and tan(10) = 0.176 so 2 * tan(10) = 0.352. If you doubt that, just try imagining what happens if the two 10 degree angles were 45 degrees. The only way that D would be at the midpoint of AE would be if AE was an arc centred on B.
I solved it, too. I used following method to get an analytic solution: I got pretty much the same til the eq. tanx=tan10°*tan50°/tan20°. Then I've written tan10°=tan(30°-10°) and tan50°=tan(30°+20°) in hope to eliminate them. I got tanx=((tan30°)^2-(tan20°)^2)/(tan20°-(tan30°)^2*(tan20°)^3). It looks scary but after substituting tan30°=1/sqrt3 I got exactly the formula of the reciproc of tan(3*20). As 1/tan60°=tan30°, x=30°. Maybe this looks confusing here in a comment but it was just 6 rows on a piece of paper. I have to admit I had the wikipedia page open for the common trigonometric equations.
If angle CED is 90 degrees, then so is angle BED. The sum of all angles in triangle BDE is 180, so angle BDE is 180 - (10 + 90) = 80. Angle ADB would be 180 - 80 = 100. Angle BAD is 180 - (10 + 100) = 70. Angle BAC is 70 + 40 = 110. Angle ACB is 180 - ((10 + 10) + 110) = 50. This shows x = 50 - n, where n is the measure of angle ACD. This is the closest I can get to the answer, I’ll have to keep watching to figure out how to solve from there.
You are correct, but unfortunately you can't solve it that way as you just end up chasing your tail. To solve it geometrically, then you have to add a lot of constructions, and it's much, much more complicated than it looks.
I think the problem I find with the geometrical solution is the mirroring part. After mirroring the triangle, you end up with 30 degrees, which is secretly the solution (30°=x), so it will solve the x. If it had not been the solution/x you come up with the mirroring, I am not sure this would work. This is basically a guess in disguise, like the usual geometrical solutions. I like the trigonometric solution more.
Its simple In terms of angles Wr can say that The tan of needed angle is Tan 40 tan 20 /tan10 = tan 40 tan 20 tan 80 = tan 3*10 = 30 degrees Less than 1 min solutiin Id used Tan o * tan 60+o * tan 60-o = tan 3o
tan x = DE/BE * BE/AE * AE/EC tan x = tan 10 * tan 70 * tan 50 tan x = tan 10 * tan (60 + 10) * tan (60 - 10) tan (60 +- 10) = (tan 60 +- tan 10) / (1 -+ tan 60 tan 10) = (√3 +- tan 10) / (1 -+ √3 tan 10) tan x = tan 10 (3 - tan² 10) / (1 - 3 tan² 10) tan x = (3 tan 10 - tan³ 10) / (1 - 3 tan² 10) = tan 3*10 x = 30 I'm sorry if it's long. i only know basics.
For those wondering that trigonometry method requires a calculator: There is a formula given below which can be used. tan(60-A) tan(A) tan(60+A) = tan(3A). so tan(50)tan(10)/tan20 is equal to tan(50)tan(10)tan(70)/tan(20)tan(70) denominator is 1 as tan(20)tan(70) is 1 (tan(20)cot(20)=1 by definition). use the above formula for numerator tanx=tan(3x10), x=30. Trigonometry really is overpowered
we know C is 50, we also know x must be a multiple of 10 meaning the only possible answers are 10-20-30-40. since we know all the angles of triangle BDE and we know segment AE is bisected by segment b we can give each bisected segment a length of 1. you can now solve for the side length or angle of anything in the image and since you can also brute force it by plugging in the 4 possible answers 30 becomes the only viable answer... I think, I never got past high school geometry and that was 20 years ago. But if i was trying to answer this question faster than someone else in a competition I'd answer 30.
my intuition was that since AE is bisected then the two bits of ACE must be near to 25, so the smaller of the 2 must be 30, the above was trying to prove it, I dont know if I did
If angle BAD is 70 so angle BD''A MUST be 80 if its 80 the angle D'D''A must be 20 to the remaining angle of triangle D"D'C must be 160 and its a isosceles triangle so the two angles must be 80 each so angle D'CD" is 80 ang angle ECD" is 50 which is the part of D'CD" so the remaining angle must be 30 degree which is equal to x to x is 30 degree
That’s how I arrived at the answer. Assigning a length, when only angles are given, is a great approach. But mathematical purists would (probably) say it lacks rigor. I dislike Rube Goldberg solutions like the one in the video; it is entertaining to watch, though.🤓
The second solution in the video should appeal to mathematical purists. It is trigonometrically elegant, without introducing an ‘arbitrary’ length (as I did).
@@1ciricola Of course, we can assign any length to BE - only the angles are given and there are infinitely many similar triangles. Or we designate its length as X and in the end it cancels itself out.
I was "close" with a guess of 25. Being that the ABE triangle height is halved with DBE. (two 10 degree angles) and so that would be at the halfway point. So ACE triangle would "suffer" the same. So ACE is 50 degrees. DCE would be half. 25. But I am wrong. Small thing though while watching the video: I know there are a lot of ways to do things, you seem to not have a good way showing angles. In one case is it blue in another it is orange. But when you do the opposite equal sized angle you show it as blue. It throws me as I lose track of what angle is what. But I do like your puzzles.
@@iMíccoli Sorry for my bad geometry. But doesn't BD bisect the angle A? (angle ABD=angle DBE=10 degrees). Therefore will not AD be equal to DE? That is why I concluded that D is the mid-point of AE.
there "should" be some way to evaluate the trig expression to derive the answer without a calculator. Someone smarter than me can take up the challenge !! For example tan30*tan20 = tan50*tan10 ... why? tan30*tan20 = tan(30+20)*tan(30-20) ?? so tan(30+20)*tan(30-20) = (tan^2(30)*tan^2(20))/(1-tan^2(30)*tan^2(20)) which is much more complicated than tan30*tan20 (or tanA*tanB) ??
It is easy enough if you remember this interesting tangent identity tan(10°) = tan(20°) * tan(30°) * tan(40°) but then you may ask how to prove that one? Well, it is a bit tricky but not too bad. It involves a couple uses of the trig product to sum identities for sin(a)*sin(b) and cos(a)*cos(b) but I am not going to type it out here. You can find proofs on the web if you search on the identity I first mentioned.
@@XJWill1 we've got (from the problem solution) tan(50)*tan(10) = tan(20)*tan(30). You gave tan(10)/tan(40) = tan(20)*tan(30) ... rearranging your expression. So does tan(50) = 1/tan(40), or tan(A) = 1/tan(90-A) ? still you're saying that tan(30)*tan(20) = tan(50)*tan(10) which generalises to ... tan(A+B)*tan(A-B) = tan(A)*tan(B) which is much simpler than other other expression
I hate the point where pressing the calculator can give 1/sqrt(3) The calculator just uses 15 significant figures (or any finite number of significant figures) to determine that it is 1/sqrt(3) which is nonsense Instead, trigonometric identities like tan x * tan (pi/3 + x) * tan (pi/3 - x) = tan 3x can be used to convert the expression to tan(pi/6)
Wow. I actually did this puzzle with a third solution- Logic! It was a lot easier…. BDE B=10 E=90 So D=80 BEA B=20 E=90 So A=70 BDA B=10 A=70 So D=100 AEC A=40 D=90 So C=50 C=50 and x look like the bigger portion So let guess x = 30, then CED C (x)=30 E=90 D=60 CDA C(other)=20 D=120 A=40 Et voila! 10 minutes (and only one simple guess).
Here is my solution with trigonometry only to share with. angle ACD=50-x Applying sine law to Triangle ACD sin40/CD=sin(50-x)/AD. …..(1) sin x=DE/CD ->CD=DE/sin x replacing to (1) sin 40*sin x/DE=sin(50-x)/AD -->sin40*sin x/sin(50-x)=DE/AD …..(2) Applying sine law to triangle ABD sin 10/AD=sin 70/BD ….(3) sin 10=DE/BD --> BD=DE/sin 10 replacing to (3) sin 10/AD=sin 70*sin 10/DE ->DE/AD=sin 70 ……(4) Equating (2) and (4) sin(40)*sin x/sin(50-x)=sin 70=cos 20 2*sin 20*cos 20*sin x=cos 20*sin(50-x) 2*sin20*sin x=sin(50-x) sin 20*sin x=1/2*sin(50-x)=sin 30*sin(50-x) 2*sin 20*sin x=2*sin 30*sin(50-x) -cos(20+x)+cos(20-x)=-cos(80-x)+cos(20-x) cos(20+x)=cos(80-x) 20+x=80-x 2*x=60 X=30 that is our final answer without a calculator.
…you know mathematical olympiads do not allow the use of a calculator, right? It'd have been better to work out the exact value using trigonometric formulas.
You also need to know that tan(50°) = 1 / tan(40°) , but I think most people know this (hint: it is because tangent and cotangent are cofunctions and reciprocals).
A really simple method: So, we focus at the right down triangle and we see it is a 30 60 90 triangle So x° is smaller than 60°, that means that x=30° It is the vision method
With such problems I always imagine, how the solution would change if some of the given numbers change. And looks like geometric one totally falls apart. Trigonometrically - on the other hand - you will simply get not that pleasing number.
I tested out of math a year early and I didn't take the time to relearn cos tan and sin just for this video, but even with the basic stuff I remembered, I just said 25... Close enough
And here I am using maths to determine who I think the sexiest women in the world are: Jennifer Aniston 5.5, Susanna Hoffs (Bangles lead singer) 5, Linda Carter (Wonder Woman) 4.5, MIchelle Pfeiffer 4, Eva Green (James Bond actress) 4, Halle Berry 3.5. 1 point each for looks, boobs, bum, legs, clothes, personality. 1 bonus point if they fancy me (yet to be awarded).
No need for geometry or trigonometry to solve this. Just use a pencil, protractor and a piece of paper. Make it as accurate as you like by using a bigger piece of paper. 😀
Why such a complicated solution, AE is bisected at D by BD therefor the line CD also bisects AE AEC is a right angle triange with EAC = 40 AEC = 90 and therefore ECA will be 60. since CD bisects AE it must also bisect angle C therefor x = 30. Oops 🤦♂
IF EAC IS 40 THEN ACE IS NOT MORE THAN 50 DEGREE BECAUSE TRIANGLE AEC ALL ANGLE SUM IS SHOULD BE 180 IF ACE IS 60 THEN SUM GOES TO 190 SO HOW YOU SAID THAT AEC IS 60 OT IS WRONG BRO
You made a mistake, ECA = 50 and not 60, which means ECD = 25 by your calculations which is not wrong. Just because a line bisects the angle doesn't mean it bisects the side and vice versa
The only time an angle bisector ever bisects the side to which it's drawn is if it's the vertex angle bisector of an isosceles triangle. So D is not a midpoint of AE.
“…this is equal to 0.577, which is easily recognizable as 1 divided by the square root of 3”
Saying that a truncated decimal (necessarily, as it is unending) is exactly equal to a known root is not rigorous. There is no proof here, only an approximation. Numerical methods are not exact.
is not bad beacause if you use trig function is very often numbers like 1 over square root of 3 or 2, in my opinion is much better than the mind blowing geometry way
Agreed it's not rigorous. If he used angle addition formulas or whatever to prove tan(50)tan(10)/tan(20) = tan(30) then it would complete the proof.
I definitely agree with @@letao12 here. It’s very easy to apply the formulas to get the sqrt3/3 and it’s why it’s not such a big deal that he made in the video
I actually thought of the Euler-mascheroni constant instead when he said 0.577, because while I think about 1/sqrt(2) quite a lot, I pay almost no attention to 1/sqrt(3)
Here's a way to finish the trig method without a calculator:
tan10 * tan50 / tan20
= tan10 * tan50 * tan70
= tan10 * tan(60-10) * tan(60+10)
= tan10((tan60 - tan10) / (1 + tan60tan10))((tan60 + tan10) / (1 - tan60tan10))
Since tan60 = sqrt(3):
this equals tan10 * (3 - tan^2(10))/(1 - 3tan^2(10))
= (3tan10 - tan^3(10) / (1 - 3tan^2(10))
Now, from the somewhat obscure triple-angle formula tan(3x) = (3tanx - tan^3(x)) / (1 - 3tan^2(x)),
This equals tan(3 * 10) = tan30. So x = 30.
*the triple-angle thing can be derived by tan(2x + x), or alternatively set z = 1 + bi on the complex plane (so b = arg(z)) and compute the argument of z^3.
You could've used the identity: tan(x)*tan(60-x)*tan(60+x)=tan(3x)
@@AdvaitBhalerao yeah right
or tan10 =tan20*tan30*tan40 then tan10*tan50*tan20 = tan20*tan30*tan40*tan50/tan20, but tan40*tan50 = 1, so=tan20*tan30/tan20 = tan30
Thank you. I assume you are right. I will still go for the calculator method, for obvious reasons.
@@russellblake9850
tan 10°=tan20°tan30°tan40° is a nice identity. How do you derive it?
For the geometrical solution, the problem ended in 5:45. It is suffice to see that x+y=50 from the start, 2x+y=80 (from the isosceles triangule), therefore, x=30.
I don't think anyone no matter how snart would think to draw thise triangles why would they
@@leif1075 Sometimes, you just have to draw stuff until stuff makes sense. It's a sort of guided chaos approach.
I used the triginimetric approach before I watched the geometric solution, and thought the exact same thing.
Bro which isosceles triangle is that?
@@Verifiededitzz D''CD'
As others have pointed out, the equation tan(x) = (tan(50°)tan(10°))/tan(20°) at 10:10 can be solved for an exact value of x without using numerical values. The first step is to replace 1/(tan(20°) with the exact equivalent tan(70°). Then we recognize that 70° = 60° + 10° and 50° = 60° - 10°, so tan(x) = tan(60° - 10°)tan(10°)tan(60° + 10°). Now, we can use the identity tan(3Θ) = tan(60° - Θ)tan(Θ)tan(60° + Θ). Let Θ = 10° and we find that x = 30°.
To prove the identity for Θ = 10°, we use the tangent sum of angles and difference of angles formulas. So, tan(a + b) = (tan(a) + tan(b))/(1 - tan(a)tan(b)) and tan(a - b) = (tan(a) - tan(b))/(1 + tan(a)tan(b)). We substitute a = 60°, so tan(60°) = √3, and b = 10°. So, after simplification, tan(60° - 10°)tan(60° + 10°) = (3 - tan²(10°))/(1 - 3 tan²(10°)). For the left side, we first compute tan(20°) = tan(10° +10°) by either the sum of angles or double angle formula, finding tan(20°) = (2tan(10°))/(1 - tan²(10°)). Then we compute tan(30°) = tan(20° +10°). Simplifying and factoring, we find tan(30°) = tan(60° - 10°)tan(10°)tan(60° + 10°).
In the real world, all measurements have tolerances. In geometry, measurements often have exact values and the problem solution may require that the measurements are exact. For example, a triangle is not equilateral unless all 3 interior angles are exactly 60° and a triangle is not a right triangle unless one interior angle measures exactly 90°.
Wow. 👍
I found an alternative solution:
Extend BA to Q such that triangles BDA, CQA are similar. Then triangles BCA, DQA are also similar (side-angle-side). Angle chasing yields DQ bisects angle BQC. Hence D is the incentre of triangle BQC. Hence x = 1/2 * 60 = 30. QED.
As an engineer, I would draw the triangles according to specifications, and then measure the wanted angle.
Unironically based
Hah that's not allowed😅
As would be expected from an engineer. Don't forget to round to the nearest hundredth.
What specifications? This problem doesn't provide aby so that wpuld be cheating wouldnt it?
@@leif1075 This is neither a defence nor a condemnation of the original commenter's method, but you can easily infer the amplitudes of all the internal angles of the triangle ABC, which is plenty enough, once you arbitrate the length of any one side.
My thought process once you established isosceles triangle BCD": "Angles D"BC and D"CB are both 50°, so angle BD"C msut be 80°. Therefore, D'D"C is (80 - 60) 20°, and since it's the vertex of an isosceles triangle, the other two angles must be ((180 - 20)/2) 80°. Angle D"CD' is equal to (50+x)°. 50°+x° = 80°, therefore x=30°.
This is the way
I thought of the same thing. LOL
same. the circle was unnecessary at that point.
But come I don't think Anyone would draw triangle BCD" Tro begin with why would they. It's out od nowhere..
@@leif1075 this type of problems require that kind of thinking. When you're used to them the only issue you'll have is "So what lines do I draw" or "what triangles do it reflect"
I employed a much simpler trig. method? The internal angle at D, within the smaller triangle, appears directly proportional to D's position on AE, compared to AC. As AE is bisected from point C, the internal angle at A( 40* ), increases by .5 times at D( 60* ). 180* - ( 60* + 90* ) = 30 degrees for x ! Who needs a calculator ? 😃
Solved it the second way in my head, but a tad bit different.
(After getting the 50 degrees)
BE = y
EC = z
y * tan(10) = z * tan(x)
y * tan(20) = z * tan(50)
Divide the equations:
tan(10) / tan(20) = tan(x) / tan(50)
(From here it was the same)
Hey just to understand it, why is y*tan(10) = z*tan(x) and y*tan(20) = z*tan(50). I know where you got the degrees from but why ist that equals to that?
@@inshallqureshi6255They are the heights DE and AE.
DE = y * tan(10) = z * tan(x)
AE = y * tan(20) = z * tan(50)
Just a tip, remember that in math olympiad contest (i assume Polish mathematical olympiad uses the same rule as the international olympiad) calculators are not allowed, therefore you should have shown a way to finish the trigonometric way without calculator...
Would memorising all the sine values from 0 to 90 degrees in 1 degree increments help?
@@henrytang2203 i don't think so... what if the degree isn't an integer? plus who does that anyway? why not use that time to lear a more useful skill or improve problem solving intuition?
@@ericzhu6620 I understand there are numerous trig formulas that can be used. But have you watched the process for deriving an exact value for something like sin(3 degrees)? So unless the angle is a nice number, the no calculator rule is a tad ridiculous.
@@henrytang2203 When the contest bans calculator it must be because they know you don't need it, therefore when there appears a value like sin(3º) there must be a way to cancel it out to avoid calculating it
@ericzhu6620 And that limits the amount of scenarios that are possible.
Beautiful solution
@5:40, when you prove that line CD'' is congruent to lines D'D'' and BD'', you can solve for angle D'D''C which must be 50+50+60+x=180 so that angle is 20 degrees.
Then you use the fact that CD'D'' is an isosceles triangle and that the total angle D'CD'' must be (180-20)/2, which is 80 degrees.
Also, let's call angle DCD'' y, for easier equation.
You get an equation pair:
2*x+y=80
x+y=50 (got that earlier, from the starting diagram)
From these you get x=30 (also y=20...)
No need to prove that CDD' is an equilateral triangle or circle from point D'' or arcs of it
Yeah I see how everyone is saying this but the fact D'' was the center of a circle passing through B,D'' and C was the first thing I noticed but that's probably because I'm used to geometry problems.
Showing that CD'D'' is equilateral is also really trivial .
How I did it:
X = arctan(DE/CE)
Because BD bisects ABE, therefore DE is half of AE.
tan(40) = CE / AE --> CE = tan(40)*AE
substituting them in our equation we get
X = arctan[(AE/2) / (tan(40)*AE)] AE cancels out
X = arctan[(1/2) / (tan(40))] move the numerator's denominator to the denominator
X = arctan(1 / 2tan(40))
I had to use the calculator for this but the answer comes to around 30 degrees with rounding errors.
this is only roughly approximately true because 10 degrees is small. Take larger angles and it wouldn't even be close.
First method: 💀
Second method: 😊
Why is the first method skulltastic deadly?
@@leif1075 it mens that the first method is much harder compared to the second one, doesnt mean that its literally deadly, its just another way of saying that it is bad
First method is actually chill if you're used to this kind of problems.
Very good. I tried it just looking at the angles specified in the figure, but I ended up with 5 equations in 5 unknowns, but the resulting matrix was singular, and then i gave up.
I appreciate the trig solution more. It requires less imagination, which I appreciate.
I admire the geometric construction and the imagination it took.
I got the answer 30 degrees, but had to use trigonometry.
I just set BE=1. Then I generated trig equations
DE = tan(10°)
AE = tan(20°)
EC = AE * tan(40°) = tan(20°)tan(40°)
tan(x) = DE/EC = tan(10°)/(tan(20°)tan(40°))
I used a calculator, which seems like cheating, but no doubt it can be done using trigonometric identities.
Ok. just looked at the geometry answer. I could have looked at that for weeks, and not got it. Thank heavens for trig.
You can find exercises like this in a peruvian book called "Construcciones en Triángulos - Técnicas y Criterios para realizar Trazos Auxiliares" by José Luis Meza Bárcena Cuzcano"
Thank you. I am certain that most viewers of this video will now start studying that Peruvian book.
because angle ABD is half ABE, and triangle ABE is a right triangle with right angle AEB, D must be at the midpoint of side AE. triangle ACE is also a right triangle with one of the legs on AE, and we can easily find that angle ACE is 60. since we know D is at the midpoint of AE, we also know that DC will bisect the angle ACE, therefore the angle DCE is half angle ACE. x = 30 (prewatch). In other words, since line segment BD bisects ABE, we can easily see that DC will bisect ACE.
"because angle ABD is half ABE, and triangle ABE is a right triangle with right angle AEB, D must be at the midpoint of side AE."
Quite simply that is not the case.
(1) DE = BE * tan(10)
(2) AE = BE * tan(20).
If D was at the midpoint of AE then
(3) AE = 2 * DE
Substitute in for AE and DE from (1) and (2) and we get
BE * tan(20) = 2 * BE * tan(10)
cancel out BE and you get
tan(20) = 2 * tan(10)
The above is simply not the case. To three significant figures tan(20) = 0.364 and tan(10) = 0.176 so 2 * tan(10) = 0.352.
If you doubt that, just try imagining what happens if the two 10 degree angles were 45 degrees.
The only way that D would be at the midpoint of AE would be if AE was an arc centred on B.
So you're saying that D is the incenter of triangle ABC.
You realize that if D is the midpoint of AE then ABE should be isosceles right?
Because D is going to be the mid point and the foot of the angle bisector at the same time.
I solved it, too. I used following method to get an analytic solution: I got pretty much the same til the eq. tanx=tan10°*tan50°/tan20°. Then I've written tan10°=tan(30°-10°) and tan50°=tan(30°+20°) in hope to eliminate them. I got tanx=((tan30°)^2-(tan20°)^2)/(tan20°-(tan30°)^2*(tan20°)^3). It looks scary but after substituting tan30°=1/sqrt3 I got exactly the formula of the reciproc of tan(3*20). As 1/tan60°=tan30°, x=30°. Maybe this looks confusing here in a comment but it was just 6 rows on a piece of paper. I have to admit I had the wikipedia page open for the common trigonometric equations.
Correction. Trigonometry + a calculator is overpowered.
The trigonometric solution is very clear. Congratulations.
If angle CED is 90 degrees, then so is angle BED. The sum of all angles in triangle BDE is 180, so angle BDE is 180 - (10 + 90) = 80.
Angle ADB would be 180 - 80 = 100. Angle BAD is 180 - (10 + 100) = 70.
Angle BAC is 70 + 40 = 110. Angle ACB is 180 - ((10 + 10) + 110) = 50. This shows x = 50 - n, where n is the measure of angle ACD.
This is the closest I can get to the answer, I’ll have to keep watching to figure out how to solve from there.
You are correct, but unfortunately you can't solve it that way as you just end up chasing your tail. To solve it geometrically, then you have to add a lot of constructions, and it's much, much more complicated than it looks.
I think the problem I find with the geometrical solution is the mirroring part. After mirroring the triangle, you end up with 30 degrees, which is secretly the solution (30°=x), so it will solve the x. If it had not been the solution/x you come up with the mirroring, I am not sure this would work. This is basically a guess in disguise, like the usual geometrical solutions. I like the trigonometric solution more.
tan(60°+a).tan(60°-a).tana = tan3a
So tan70°.tan50°.tan10°= tan30°
Its simple
In terms of angles
Wr can say that
The tan of needed angle is
Tan 40 tan 20 /tan10
= tan 40 tan 20 tan 80
= tan 3*10
= 30 degrees
Less than 1 min solutiin
Id used
Tan o * tan 60+o * tan 60-o
= tan 3o
tan x = DE/BE * BE/AE * AE/EC
tan x = tan 10 * tan 70 * tan 50
tan x = tan 10 * tan (60 + 10) * tan (60 - 10)
tan (60 +- 10) = (tan 60 +- tan 10) / (1 -+ tan 60 tan 10)
= (√3 +- tan 10) / (1 -+ √3 tan 10)
tan x = tan 10 (3 - tan² 10) / (1 - 3 tan² 10)
tan x = (3 tan 10 - tan³ 10) / (1 - 3 tan² 10) = tan 3*10
x = 30
I'm sorry if it's long. i only know basics.
'tan x = (3 tan 10 - tan³ 10) / (1 - 3 tan² 10) = tan 3*10'
Can't quite see how you get that last equality at a glance.
@@thetaomegatheta i didn't know there is a formula for triple angle in terms of tan at first. so i try do tan (2x + x). just to see how it goes.
Using a calculator is sketchy at best. One should be able to use trig identities to reduce the right side to tan(30⁰), no?
True, and it is possible, but takes some time to work out.
what a hard but nice problem!
For those wondering that trigonometry method requires a calculator:
There is a formula given below which can be used.
tan(60-A) tan(A) tan(60+A) = tan(3A).
so tan(50)tan(10)/tan20 is equal to tan(50)tan(10)tan(70)/tan(20)tan(70)
denominator is 1 as tan(20)tan(70) is 1 (tan(20)cot(20)=1 by definition). use the above formula for numerator
tanx=tan(3x10), x=30.
Trigonometry really is overpowered
Great .thanks again
Wow, indeed!!
I'm *CONVINCED* there's a *MUCH simpler* "geometric" way to solve this instead of the "solution 1" used here...
.
When you find, it feel free to share!
Lots of angle chasing.
D1D2C=20
D2CD1 160:2=80
D2CB=50
BCD1 x=80-50=30
we know C is 50, we also know x must be a multiple of 10 meaning the only possible answers are 10-20-30-40. since we know all the angles of triangle BDE and we know segment AE is bisected by segment b we can give each bisected segment a length of 1. you can now solve for the side length or angle of anything in the image and since you can also brute force it by plugging in the 4 possible answers 30 becomes the only viable answer... I think, I never got past high school geometry and that was 20 years ago. But if i was trying to answer this question faster than someone else in a competition I'd answer 30.
my intuition was that since AE is bisected then the two bits of ACE must be near to 25, so the smaller of the 2 must be 30, the above was trying to prove it, I dont know if I did
D is not the midpoint of AE, why is everyone saying that....
@@SilverScreenMassageD is just the foot of the angle bisector.
If angle BAD is 70 so angle BD''A MUST be 80 if its 80 the angle D'D''A must be 20 to the remaining angle of triangle D"D'C must be 160 and its a isosceles triangle so the two angles must be 80 each so angle D'CD" is 80 ang angle ECD" is 50 which is the part of D'CD" so the remaining angle must be 30 degree which is equal to x to x is 30 degree
Let BE=1 then DE=tan10, AE=tan20 and EC=AE*tan40. So tan(x)=tan10 / (tan20*tan40). Than's all.
That’s how I arrived at the answer. Assigning a length, when only angles are given, is a great approach. But mathematical purists would (probably) say it lacks rigor.
I dislike Rube Goldberg solutions like the one in the video; it is entertaining to watch, though.🤓
The second solution in the video should appeal to mathematical purists. It is trigonometrically elegant, without introducing an ‘arbitrary’ length (as I did).
@@1ciricola Of course, we can assign any length to BE - only the angles are given and there are infinitely many similar triangles. Or we designate its length as X and in the end it cancels itself out.
I was "close" with a guess of 25. Being that the ABE triangle height is halved with DBE. (two 10 degree angles) and so that would be at the halfway point.
So ACE triangle would "suffer" the same.
So ACE is 50 degrees.
DCE would be half. 25. But I am wrong.
Small thing though while watching the video: I know there are a lot of ways to do things, you seem to not have a good way showing angles.
In one case is it blue in another it is orange. But when you do the opposite equal sized angle you show it as blue. It throws me as I lose track of what angle is what.
But I do like your puzzles.
I too came up with 25. Since AD equals DE I assumed angle at C would be 2x. Wrong assumption!!
D is not the midpoint of AE, where are you guys learning geometry? Because you're like the 7th person saying that.
@@JoeCoelhoSJNo
@@iMíccoli Sorry for my bad geometry. But doesn't BD bisect the angle A? (angle ABD=angle DBE=10 degrees). Therefore will not AD be equal to DE? That is why I concluded that D is the mid-point of AE.
Isn't that "wow" starting to fade away?
Third method: draw in autoCAD
LOL!
asnwer=30 isit
I didn't know about inscribed circle theorem
First half of the video:
“What the heck? lol”
I love it ❤❤❤❤
I could not believe you said " let's use a calculator"
This is very hard 😮
there "should" be some way to evaluate the trig expression to derive the answer without a calculator. Someone smarter than me can take up the challenge !!
For example tan30*tan20 = tan50*tan10 ... why?
tan30*tan20 = tan(30+20)*tan(30-20) ??
so tan(30+20)*tan(30-20) = (tan^2(30)*tan^2(20))/(1-tan^2(30)*tan^2(20)) which is much more complicated than tan30*tan20 (or tanA*tanB) ??
It is easy enough if you remember this interesting tangent identity
tan(10°) = tan(20°) * tan(30°) * tan(40°)
but then you may ask how to prove that one? Well, it is a bit tricky but not too bad. It involves a couple uses of the trig product to sum identities for sin(a)*sin(b) and cos(a)*cos(b) but I am not going to type it out here. You can find proofs on the web if you search on the identity I first mentioned.
@@XJWill1 and 1/tan(40) = tan(50) ? ... but that still makes a much simpler tanA*tanB = tan(A+B)*tan(A-B) ... no ?
@@russellblake9850 You lost me. That is not an identity, as far as I can tell.
@@XJWill1 we've got (from the problem solution) tan(50)*tan(10) = tan(20)*tan(30). You gave tan(10)/tan(40) = tan(20)*tan(30) ... rearranging your expression. So does tan(50) = 1/tan(40), or tan(A) = 1/tan(90-A) ?
still you're saying that tan(30)*tan(20) = tan(50)*tan(10) which generalises to ...
tan(A+B)*tan(A-B) = tan(A)*tan(B) which is much simpler than other other expression
@@russellblake9850 This so-called 'identity\ isn't in general true. Try A = 80 deg, B = 10 deg, you get finite = infinite.
I hate the point where pressing the calculator can give 1/sqrt(3)
The calculator just uses 15 significant figures (or any finite number of significant figures) to determine that it is 1/sqrt(3) which is nonsense
Instead, trigonometric identities like tan x * tan (pi/3 + x) * tan (pi/3 - x) = tan 3x can be used to convert the expression to tan(pi/6)
If BD bisects
Where did you get that D is the midpoint of AE?
I somehow just guessed the correct answer before the video even started, lol.
Wow. I actually did this puzzle with a third solution- Logic! It was a lot easier….
BDE B=10 E=90 So D=80
BEA B=20 E=90 So A=70
BDA B=10 A=70 So D=100
AEC A=40 D=90 So C=50
C=50 and x look like the bigger portion
So let guess x = 30, then
CED C (x)=30 E=90 D=60
CDA C(other)=20 D=120 A=40
Et voila! 10 minutes (and only one simple guess).
You would get 0/7 in this problem.
@@iMíccoli I have PhD in Mathematics from Penn State. My answer was a bit messy - but mostly correct. Or maybe not.
Here is my solution with trigonometry only to share with.
angle ACD=50-x
Applying sine law to Triangle ACD
sin40/CD=sin(50-x)/AD. …..(1)
sin x=DE/CD ->CD=DE/sin x replacing to (1)
sin 40*sin x/DE=sin(50-x)/AD -->sin40*sin x/sin(50-x)=DE/AD …..(2)
Applying sine law to triangle ABD
sin 10/AD=sin 70/BD ….(3)
sin 10=DE/BD --> BD=DE/sin 10 replacing to (3)
sin 10/AD=sin 70*sin 10/DE ->DE/AD=sin 70 ……(4)
Equating (2) and (4)
sin(40)*sin x/sin(50-x)=sin 70=cos 20
2*sin 20*cos 20*sin x=cos 20*sin(50-x)
2*sin20*sin x=sin(50-x)
sin 20*sin x=1/2*sin(50-x)=sin 30*sin(50-x)
2*sin 20*sin x=2*sin 30*sin(50-x)
-cos(20+x)+cos(20-x)=-cos(80-x)+cos(20-x)
cos(20+x)=cos(80-x)
20+x=80-x
2*x=60
X=30 that is our final answer without a calculator.
Trig is the helper of geometry. Although the pure geometry was beautiful; the trig one made more sense to the average Joe....
Like your solution 2 :
tan(x)=tan(50°)*tan(10°)/tan(20°)
tan(x)*tan(20°)=(sin(50°)*sin(10°))/(cos(50°)*cos(10°))
tan(x)*tan(20°)=(1/2*(-cos(60°)+cos(40°)))/(1/2*(cos(60°)+cos(40°)))
tan(x)*tan(20°)=(-cos(60°)+cos(40°))/(cos(60°)+cos(40°))
tan(x)*tan(20°)=(-1/2+cos(40°))/(1/2+cos(40°))
tan(x)*tan(20°)=(-1+2*cos(40°))/(1+2*cos(40°))
tan(x)=cos(20°)/sin(20°)*(-1+2*cos(40°))/(1+2*cos(40°))
tan(x)=(-cos(20°)+2*cos(40°)*cos(20°))/(sin(20°)+2*cos(40°)*sin(20°))
tan(x)=(-cos(20°)+cos(60°)+cos(20°))/(sin(20°)+sin(60°)-sin(20°))
tan(x)=cos(60°)/sin(60°)
tan(x)=sin(30°)/cos(30°)
tan(x)=tan(30°)
x=30°
I assumed the two 10⁰ angles bisected the opposite side, making _x_ half of 180-(90+40) = 25⁰. Silly mistake.
I went to an undergraduate admission test they just spammed 4-5 triangles in a square and asked a random angle
i guessed 30 just based off of solving other angles and looks, and 25 from theories i came up with which weren't try because 30 was correct
I was stuck on /_ EDC and x.
Geometrically murderous!
Bro the trigonometry just finished the game 💀
…you know mathematical olympiads do not allow the use of a calculator, right? It'd have been better to work out the exact value using trigonometric formulas.
Yes, I did it.
yeah, I was able to follow the trigonometric solution at least
D, D prime, D double prime ... I can't take it any more !
Haha 😂 Let's use Q, Q prime, Q double prime...
Can someone explai the inscribed angle theorem part at 6:56 to me? I don't get how you know it is half of the larger angle.
Engineer answear: less than 50 deg, more than 10 deg => Average = 30 deg
Solved with trig but not geometrically.
7:23
If you remember this interesting tangent identity, you do not need a calculator for your second method.
tan(10°) = tan(20°) * tan(30°) * tan(40°)
You also need to know that tan(50°) = 1 / tan(40°) , but I think most people know this (hint: it is because tangent and cotangent are cofunctions and reciprocals).
Like, if you also thought on first glance, the answer was 25° 😁
A really simple method:
So, we focus at the right down triangle and we see it is a 30 60 90 triangle
So x° is smaller than 60°, that means that x=30°
It is the vision method
Please help! I can't unsee the fact that angle ACE is bisected😅
Edit: now i see 😅
Bravo 🎉
25°
Angle BCA is not bisected.
@@henrygreen2096why not?
Bro really thought that AD=DE which is not the case
@@DaSpida if CD was bisecting ACE then D would be the incenter of triangle ABC which means that AD must bisect BAC and that's not the case.
@@iMíccolii was convinced that AD=DE
Now i've seen my fault. Thank you
說明頻道主對三倍角公式不熟悉！
"Solve for angle x"
No🗿
With such problems I always imagine, how the solution would change if some of the given numbers change. And looks like geometric one totally falls apart. Trigonometrically - on the other hand - you will simply get not that pleasing number.
Disappointed by the lack of rigor in the trigometrical answer.
Please upload videos as much as u can...
Lots of love from India ❤❤❤
I SOLVED WITH TRIGONOMETRICS !!!
why isnt AD = DE ..
ABE would be isosceles bro. And why's there alot of people asking this? D is just the the foot of the angle bisector.
No, I can't. 😅
Really disappointed by the how you presented trigonometric solution, it's an Olympiad problem so you should also show how to simplify that expression.
As an sigma, i was super skibidi to rizz up this problem
😢
Hi there 😊😊
10:19 is not correct! you cannot input numerical values, as u cannot exactly calculate 1/sqrt(3). you have to use tan(x)*tan(y) formula
Couldn't follow at all, couldn't keep up which point he was talking about with all the primes.
Cool
Come on why would anyone draw thebtroangles like he does for thenfirdt method..I don't thonk anyone would no matter how smart..
It comes with practice.
This type of solutions have a certain pattern.
OMG 10 minutes ago!!!
Also solved it myself
There's a way easier method. Simply fast forward to the end of the video.
I tested out of math a year early and I didn't take the time to relearn cos tan and sin just for this video, but even with the basic stuff I remembered, I just said 25... Close enough
Engineer located
@@Terrapin22 I'm the only one of my friends not pursuing engineering 😭
@@BeaAFraihd yeah, being 17% off isn't "close enough" even in engineering terms.
Maybe he just halved 50°, that's no math @@Terrapin22
@@feedbackzaloop yeah most probably he was like x = C/2, so that's 25 for you
OMG so complicated. I solved it in under 100 seconds. (Yes I do understand that why the complication. There are different solutions to solve a puzzle.
Im just saying its 20 out of visuals and a bit of basicness math
I went with 25° because A was split in half, and I felt
370th view ?
And here I am using maths to determine who I think the sexiest women in the world are: Jennifer Aniston 5.5, Susanna Hoffs (Bangles lead singer) 5, Linda Carter (Wonder Woman) 4.5, MIchelle Pfeiffer 4, Eva Green (James Bond actress) 4, Halle Berry 3.5.
1 point each for looks, boobs, bum, legs, clothes, personality. 1 bonus point if they fancy me (yet to be awarded).
No need for geometry or trigonometry to solve this. Just use a pencil, protractor and a piece of paper. Make it as accurate as you like by using a bigger piece of paper. 😀
Why such a complicated solution, AE is bisected at D by BD therefor the line CD also bisects AE AEC is a right angle triange with EAC = 40 AEC = 90 and therefore ECA will be 60. since CD bisects AE it must also bisect angle C therefor x = 30.
Oops
🤦♂
ECA should be 50..
IF EAC IS 40 THEN ACE IS NOT MORE THAN 50 DEGREE BECAUSE TRIANGLE AEC ALL ANGLE SUM IS SHOULD BE 180 IF ACE IS 60 THEN SUM GOES TO 190 SO HOW YOU SAID THAT AEC IS 60 OT IS WRONG BRO
You made a mistake, ECA = 50 and not 60, which means ECD = 25 by your calculations which is not wrong. Just because a line bisects the angle doesn't mean it bisects the side and vice versa
ECA should be 50* since 90+40+60 ≠ 180.
The only time an angle bisector ever bisects the side to which it's drawn is if it's the vertex angle bisector of an isosceles triangle. So D is not a midpoint of AE.
First like
First
Elon approves 📐