HMMT February, 2024

Solution:

Consider vectors
$$\binom{\frac{x}{\sqrt{x^{2}+y^{2}}}}{\frac{y}{\sqrt{x^{2}+y^{2}}}} \text{ and } \binom{-\frac{1}{x}}{\frac{1}{y}}.$$
They are orthogonal and add up to $\binom{7}{4}$, which has length $\sqrt{7^{2}+4^{2}}=\sqrt{65}$. The first vector has length $1$, so by Pythagoras' 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 gives
$$-\frac{1}{8 y}-\frac{1}{x}=7 \text{ and } -\frac{1}{8 x}+\frac{1}{y}=4$$
Solving this gives $(x, y)=\left(-\frac{13}{96}, \frac{13}{40}\right)$, which works.

Solution:

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 $(x, y) = \left(-\frac{13}{96}, \frac{13}{40}\right)$.