HMMT February

First, suppose that $x$ and $y$ were of different signs. Then $x y<0$ and $x / y<0$, but the set has at most one negative value, a contradiction. Hence, $x$ and $y$ have the same sign; without loss of generality, we say $x$ and $y$ are both positive. Let $(s, d):=(x+y, x-y)$. Then the set given is equal to $\left\{s^{2}, d^{2}, \frac{1}{4}\left(s^{2}-d^{2}\right), \frac{s+d}{s-d}\right\}$. We split into two cases: - Case 1: $\frac{s+d}{s-d}=N$. This forces $s^{2}=28.8$ and $d^{2}=12.8$, since $\frac{1}{4}(28.8-12.8)=4$. Then $s=12 \sqrt{0.2}$ and $d= \pm 8 \sqrt{0.2}$, so $N$ is either $\frac{12+8}{12-8}=5$ or $\frac{12-8}{12+8}=0.2$. - Case 2: $\frac{s+d}{s-d} \neq N$. Suppose $\frac{s+d}{s-d}=k$, so $(s, d)=((k+1) t,(k-1) t)$ for some $t$. Then $s^{2}: d^{2}$ : $\frac{1}{4}\left(s^{2}-d^{2}\right)=(k+1)^{2}:(k-1)^{2}: k$. Trying $k=4,12.8,28.8$ reveals that only $k=4$ is possible, since $28.8: 12.8=(4-1)^{2}: 4$. This forces $N=s^{2}=\frac{5^{2}}{4} \cdot 12.8=80$. Hence, our final total is $5+0.2+80=85.2$