HMMT February

Solution 1: Consider vectors $$\binom{x / \sqrt{x^{2}+y^{2}}}{y / \sqrt{x^{2}+y^{2}}} \text { and }\binom{-1 / x}{1 / y}$$ They are orthogonal and add up to $\binom{7}{4}$, which have length $\sqrt{7^{2}+4^{2}}=\sqrt{65}$. The first vector has length 1, so by Pythagorean's theorem, the second vector has length $\sqrt{65-1}=8$, so we have $$\frac{1}{x^{2}}+\frac{1}{y^{2}}=64 \Longrightarrow \sqrt{x^{2}+y^{2}}= \pm 8 x y$$ However, the first equation indicates that $x<0$, while the second equation indicates that $y>0$, so $x y<0$. Thus, $\sqrt{x^{2}+y^{2}}=-8 x y$. Plugging this into both of the starting equations give $$-\frac{1}{8 y}-\frac{1}{x}=7 \text { and }-\frac{1}{8 x}+\frac{1}{y}=4$$ Solving this gives $(x, y)=\left(-\frac{13}{96}, \frac{13}{40}\right)$, which works. Solution 2: Let $x=r \cos \theta$ and $y=r \sin \theta$. Then our equations read $$\begin{aligned} & \cos \theta-\frac{1}{r \cos \theta}=7 \\ & \sin \theta+\frac{1}{r \sin \theta}=4 \end{aligned}$$ Multiplying the first equation by $\cos \theta$ and the second by $\sin \theta$, and then adding the two gives $7 \cos \theta+$ $4 \sin \theta=1$. This means $$4 \sin \theta=1-7 \cos \theta \Longrightarrow 16 \sin ^{2} \theta=1-14 \cos \theta+49 \cos ^{2} \theta \Longrightarrow 65 \cos ^{2} \theta-14 \cos \theta-15=0$$ This factors as $(13 \cos \theta+5)(5 \cos \theta-3)=0$, so $\cos \theta$ is either $\frac{3}{5}$ or $-\frac{5}{13}$. This means either $\cos \theta=\frac{3}{5}$ and $\sin \theta=-\frac{4}{5}$, or $\cos \theta=-\frac{5}{13}$ and $\sin \theta=\frac{12}{13}$. The first case, plugging back in, makes $r$ a negative number, a contradiction, so we take the second case. Then $x=\frac{1}{\cos \theta-7}=-\frac{13}{96}$ and $y=\frac{1}{4-\sin \theta}=\frac{13}{40}$. The answer is $(x, y)=\left(-\frac{13}{96}, \frac{13}{40}\right)$.