Can you solve this? A nice math olympiad problem |Study guide #math #matholympiad

Can you solve this? A nice math problem |Study guide #math #matholympiad

[MrSeezero]

Actually, all you need to do here is just look at this as a right triangle with the side 1 and side y and the hypotenuse y^2 since congruent triangles can be converted to another by simply dividing by a constant value which in this case would be x itself. Now you have y^4 = 1 + y^2; y^4 - y^2 - 1 = 0; y^2 = (1+sqrt(1+4))/2; y^2 = (1+sqrt(5))/2; y = sqrt((1+sqrt(5))/2).

[alfonsorodriguez]

This is not an Olympiad problem. It’s way too easy. It is what I would give to my geometry pre-AP class

[pk]

The answer is actually equal to the square root of the golden ratio .

[ianfowler]

You should correct this video or re do it or take it off line.

[-Ori_g-]

X^6 - x^4 = x^2
X^6 - x^4 - x^2 = 0
X^2( x^4 - x^2 - 1) = 0
1) X ^ 2 = 0 > cant be
2) t^2 - t - 1 = 0
T = ....
X = +(squrt{t})
X = -(squrt{t})
Positive only

[belomolnar]

Result is O. K. But….in row3 is a missprint at right side. Not dividing by x2 separately, firstx2/x2 + x4/x2. Denominator x2 is common for (x2+x4). 😂

[mircoceccarelli]

👍👍👍

x = sqrt( phi ) > 0

phi = numero aureo .

😊🤪👍👋

[oliverdauphin]

Dans un triangle rectangle de côté x, x^2, x^3
Selon Pythagore : x^4+x^2=x^6
Donc x^2(x^4-x^2-1)=0
Comme x non nul (car on a un triangle)

x^4-x^2-1=0
Posons X=x^2
L’équation devient X^2-X-1=0
Qui a pour solution positive X=(1+sqrt(5))/2
La solution négative est exclue puisque X est un carré.
Posons X= (1+sqrt(5))/2 =PHI le nombre d’or
x=sqrt(PHI)
Là aussi on ne retient que la solution positive car x est une longueur.


Commentaire sur cette solution :
On a alors
x^2=PHI
x^3=PHI*sqrt(PHI)
En calculant ce produit on trouve x^3=2+sqrt(5)
Par ailleurs, en calculant la somme PHI^2 + PHI on trouve 2+sqrt(5)

Donc PHI^2 + PHI = PHI*sqrt(PHI)

Comme X=PHI
Ceci montre que X^2+X=X^2*X
Ou x^4+x^2=x^6
Qui vérifie la relation de Pythagore initiale

[annangianilkumar]

In second line why you taken x^6/x^2, you have imagine or any fixed valus.

[luiscastejon]

Hay un error. El tercer punto en paréntesis es sólo X no X cuadrado.

[ayoubkhlifi]

very easy to be assked in olympiade, but which level?

[ВалерийКузнецов-сг]

Только дурак может представить так задачу.

[turk]

im just wondering how you fit all that in a sheet of paper

[niltonsilveira]

x=sqrt[sqrt(1 + sqrt 5)/2], x^2 = (1 + sqrt 5)/2 e x^3 = sqrt(2 + sqrt 5) ===> confere pelo Pitágoras!

[herbertklumpp]

It is so easy you get by solving
Y^3- y^2-y=0. Y=0 ( no length)
Or y = ( 1 + - sqrt5)/2 ( - canceld) y = golden ratio
X = sqrt( golden ratio)
That us sure no olympic problem

[АвтандилРионели]

Ответ х ровно квадратичний корень от [(1+квадратичний корень из 5):2]

[saltydog]

The front image didn't show the triangle was a square ropt/

[satyapalsingh]

x^2=(sqrt 5+1)/2=1.618 is called golden ratio .

[KipIngram]

By Pythagoras we have

x^2 + x^4 = x^6

or

x^6 - x^4 - x^2 = 0
x^2 (x^4 - x^2 - 1) = 0

So there are two (of six) roots at x=0. That doesn't make a proper triangle, though. Now let u = x^2, and we have remaining

u^2 - u - 1 = 0

By the quadratic equation the roots are at u = 0.5 + sqrt(5)/2 and u = 0.5 - sqrt(5)/2. The negative solution would lead to imaginary x, so that's no good. So we wind up with

x = sqrt(0.5 + sqrt(5)/2)

[derekcresswell]

Bit late to tell us now about the 90° angle
Did you think we were psychic?