got cube root of 2 by finding special (equilateral and right) triangles, using mass points to find another distance (sqrt(3)/(1+x), precisely) and solving polynomial obtained through pythagorean theorem on right triangle with legs 1 and sqrt(3)/(1+x) and hypotenuse x
2^(1/3), it can be solved without any trig whatsoever with just similar triangles and pythagorean. Lets the side with length 1+x be named AD and the equilateral triangle be named ABC, such that C lies in between B and D. Draw a perpendicular AH to BC. AH = sqrt(3)/2, HC = 1/2 so AH^2 + (HC + CD)^2 = 1+x, so x(x+2) = CD^2 + CD. Then let E be point on AD such that DE = 1. Extend CE and draw a perpendicular DF = CD/2 to it. Then we can see that ACE and DEF are similar, DE/AE = 1/x = DF/AC = CD/2, so CD = 2/x, so x(x+2) = 2(x+2)/x^2, so sorry for me showing off
I got cube root of 2 Upper left triangle is equilateral and lower left triangle is isosceles by solving for some angles we get that upper right triangle is a right triangle we then let the top right right angle =y we get that bottom right angle = 60-y then we use law of sines on the big triangle on the right knowing that cosy=1/x and siny=sqrt(x^2-1)/x we get sin(60-y)/1=sin120/x+1 which simplifies to the equation 0=x^4+2x^3-2x-4 which simplifies to 0=(x+2)(x^3-2) which yields cube root of two
At the very end I got 0 =x^4 *-* 2x^3 *+* 2x - 4 which simplifies to 0=(x-2)(x^3-2). This would leave to disallow x = 2, which is easily done by noting that would force y = pi/3 and those two lines would be parallel. But I might as well have been duped by some minus sign somewhere.
I think the answer is cube root of 2. Seems unnatural though. If that's the case then this configuration can not be created solely using compass and straight-edge.
Here's my proof. I'll refer to this image and the naming is according to that, drive.google.com/file/d/1zw6OyBlhgtbgDFXlNnji8wv9Z5QXe7Pu/view?usp=drivesdk. AE = x. Note that AB = BC = CA = BD = 1. Therefore, ACD lie on a circle with diameter AD and center B which implies that ∠ACD=90° which in turn implies that ∠ACE=90°. By Pythagora's, CE^2 = x^2 - 1 and CD^2 = 2^2-1^2=3. Since, B, C, F are collinear. By menelaus theorem, (DB*AF*CE)/(AB*EF*CD) = 1 => (x+1)√(x^2-1) = √3 It is simple enough to prove that 2^(1/3) is the only positive real solution of this equation.
Solving for the final equation sqrt(3)=(x+1)sqrt(x² -1), I end up with (x³ -2)(x+2)=0, and there you have the real solutions x=-2 and x=cuberoot(3). But then I found out that there are two complex solutions, and we don't obtain them unless we factorize (x³ -2). Why is that? I thought it wouldn't be necessary factorization once I can state x³ -2=0. Can anybody explain it to me? Thanks in advance!
Andy Arteaga, if you are familiar with euler's identity, it would become easier. We can write x^3 = 2 as, x^3 = e^(2 pi i) 2, which gives the second solution of x = e^(2/3 pi i) * cbrt(2) = cbrt(2) ( cos(2/3 pi)+i sin(2/3 pi)) = cbrt(2)(-1/2+sqrt(3)/2) It can also be written as x^3 = e^(4 pi i) 2 => x = e^(4/3 pi i)*cbrt(2) = cbrt(2)(-1/2-sqrt(3)/2). If we write it as 2e^(6 pi i) we simply get cbrt(2)e^(2 pi i), it can be seen by the periodicity of cosine and sine or by that fact e^(2 pi i) = 1 or that a cubic equation has only three roots that these are the only three solutions of the equation x^3-2=0. I'd recommend you to read about roots of unity as it a beautiful topic bringing together polynomials, complex numbers, trigonometry and number theory all under one big roof with allcthe pieces falling together perfectly. Also, x = -2 is not a solution, put it in the original equation to see what happens (;
Rishabh Dhiman Oh, I see sqrt(3)=/=-sqrt(3), but it appeared as a solution since I squared both sides of the equation and the minus vanished. Thank you for the explanation, and the suggestion!
Flammable Maths, will you make a video about how to solve it correctly? I'm sure there are couple solutions, and I'm very curious to see how you solve it!
Here's the real mathematical puzzle: given that the mysterious figure at 7:48 is in fact Jens' girlfriend, differentiate (with respect to pr0n) the integral of the Jens sexiness function with the Lisa sexiness function using the famed sexyboi integral to find the sexiness difference between Mr. and Ms. Flammable. Of course, the Jens sexiness function is limit as pr0n--> infinity of phi^pr0n^e (phi here is the golden ratio, not azimuthal angle), and the Lisa sexiness function is of course just the limit as pr0n--> infinity of e^pr0n^phi. Note that the sexyboi integral is of course the volume integral from minimum to maximum height, or from feet to head.
HAHAHAHAH.I actually had to stop the video to laugh, because of the engineering joke. I love the channel! Have you thought about doing videos on statistics?
I got x = 2/sqrt(3). For reference, I'll call the triangles A thru D, left to right. Triangle AB is simply A U B. By using law of cosines twice on triangles A and AB, I worked out that the leftmost angle is 30 degrees (and the unknown side of A is sqrt(3)). With this information, you can find that triangle C is actually a right triangle with hypotenuse x. Using a variation of Pythagorean Theorem (derived-ish in a 3Blue1Brown video linked below at 7:18), with a=2, b=x, h=1, you can solve to get x = 2/sqrt(3). 3Blue1Brown video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-d-o3eB9sfls.htmlm18s
Vedvart To use the identity presented in the 3blue1brown video , the angle between sides a and b has to be 90° degrees , and this is not actually the case in this problem
Do you know about IMPA? Instituto de Matemática Pura e Aplicada, in Brazil. A center of reference, they make extremely high quality math. Search 'IMPA' on RU-vid, there's a lot of free courses in english
What is there not to understand??? Anyway I made a mistake, probably have still made a mistake (:/) but I have a new answer of 1.043679947... . 3sf is shorhand for 3 significant figures, I'm surprised you didn't know the shorthand :O Does the actual answer involve a surd?
SPOILER ALERT and my construction doesn't agree with that result - are you super sure about it? Please don't explain the answer, I want to figure it out for myself.
You're bad and you should feel bad. Can you use ceva/menalaus? EDIT: It works! I have a quartic and cube root 2 works, but how would I solve the quartic??
Lol you should see our tiny grey dingy POS of a campus. Great for education but god is it hard waking up Monday morning and dragging myself there. This looks perfect, though I can imagine it'd be creepy as shit walking around it empty at night 😂
But... but in engineering course i had to study numerical math, which is basically theoretical tool to create those trash, i.e. pure mathematicians helped us.
