AIME 2022 I, Problem 15

Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations:
\begin{align} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align} 
Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: First, let define a triangle with side lengths $\sqrt{2x}$, $\sqrt{2z}$, and $l$, with altitude from $l$'s equal to $\sqrt{xz}$. $l = \sqrt{2x - xz} + \sqrt{2z - xz}$, the left side of one equation in the problem. Let  $\theta$ be angle opposite the side with length $\sqrt{2x}$. Then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$, so $x=2\sin^2(\theta)$ and the side length $\sqrt{2x}$ is equal to $2\sin(\theta)$. We can symmetrically apply this to the two other equations/triangles. By law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R$, with $R=1$ as the circumradius, same for all 3 triangles. The circumcircle's central angle to a side is $2 \arcsin(l/2)$, so the 3 triangles' $l=1, \sqrt{2}, \sqrt{3}$,  have angles $120^{\circ}, 90^{\circ}, 60^{\circ}$, respectively. This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$, $y=2\sin^2(\beta)$, and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$, $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$, and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$. Solving, we get $\alpha=\frac{135^{\circ}}{2}$, $\beta=\frac{105^{\circ}}{2}$, and $\gamma=\frac{165^{\circ}}{2}$. We notice that \[[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2\] \[=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \fbox{33}. \blacksquare\] - kevinmathz

Answer: 33