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Harvard-MIT Mathematics Tournament

Algebra Difficulty 4.9 AIME Find the answer United States

Problem:
Let xx and yy be positive real numbers and θ\theta an angle such that θ≠π2n\theta \neq \frac{\pi}{2} n for any integer nn. Suppose
sin⁡θx=cos⁡θy \frac{\sin \theta}{x}=\frac{\cos \theta}{y}
and
cos⁡4θx4+sin⁡4θy4=97sin⁡2θx3y+y3x \frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
Compute xy+yx\frac{x}{y}+\frac{y}{x}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
From the first relation, there exists a real number kk such that x=ksin⁡θx=k \sin \theta and y=kcos⁡θy=k \cos \theta. Then we have
cos⁡4θsin⁡4θ+sin⁡4θcos⁡4θ=194sin⁡θcos⁡θsin⁡θcos⁡θ(cos⁡2θ+sin⁡2θ)=194 \frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=\frac{194 \sin \theta \cos \theta}{\sin \theta \cos \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=194
Notice that if t=xy+yxt=\frac{x}{y}+\frac{y}{x} then (t2−2)2−2=cos⁡4θsin⁡4θ+sin⁡4θcos⁡4θ=194\left(t^{2}-2\right)^{2}-2=\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=194 and so t=4t=4.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.