Solving a System of Equations from Iran | Solving the 2005 Iranian Math Olympiad Question

🔴 Solving a System of Equations from Iran | Solving the 2005 Iranian Math Olympiad Question | Solve For x and y

Hey there.
In this video, we are going to be solving a nice Iranian math olympiad question, where we are supposed to solve a nice system of equations for the real numbers x and y.
First, notice that x=y=0 is an obvious solution to our system of equations, so next, we need to assume that x and y are not zero, and then we try to solve the system by manipulating the equations.

🔴I hope you enjoy watching this video on a really nice Russian Math Olympiad problem .🔴

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topics covered in this video:
system of equations
solving math olympaid question
iranian math olympiad question
how to solve simultaneous equations
2005 iranian math olympiad question
solve for x and y

#SystemOfEquation #MathOlympiadQuestion #IraninaMathQuestion

[tisyarawat]

Also, I love the content and collections of math contest problems you post, I have found very few purely math based channels on youtube.

[SuperYoonHo]

Sir, can you do problems not from algebra but other topics plz? My algebra is too strong vs the trig, geometry and stuff

[tisyarawat]

Please make a few videos on geometry as well 😀😀

[NadiehFan]

Solving this system for real numbers can be done elegantly using the trigonometric identity tan 2α = 2·tan α /(1 − tan²α). Let

x = tan φ

then y = 2x/(1 − x²) implies

y = tan 2φ

and in turn x = 2y/(1 − y²) then implies

x = tan 4φ

Now, since we have both x = tan φ and x = tan 4φ this implies

tan 4φ = tan φ

and therefore, since tan is a periodic function with period π, we have

4φ = φ + k·π

where k is any integer. Subtracting φ from both sides and then dividing both sides by 3 gives

φ = k·⅓π

Since tan is periodic with period π we will get all solutions (x, y) = (tan φ, tan 2φ) if we select 3 consecutive integers for k. For k = 0, 1, 2 we then get

(x, y) = (tan 0, tan 0) = (0, 0)
(x, y) = (tan ¹⁄₃π, tan 2⁄₃π) = (√3, −√3)
(x, y) = (tan 2⁄₃π, tan ⁴⁄₃π) = (−√3, √3)

[indrajitkatira]

If u subtract eqn(2) from (1), the cal. becomes more easy.

[tisyarawat]

I solved this problem using trigonometry.