2022 AIME I Problem 15

Factor each radical as $\sqrt{x(2 - y)}$ and substitute $x = 2\sin^2\alpha,$ $y = 2\sin^2\beta,$ $z = 2\sin^2\gamma$

Each equation collapses by the addition formula to $2\sin(\alpha + \beta) = 1,$ etc.; solve the little linear system for the angles and use $1 - 2\sin^2\theta = \cos 2\theta$

Each radicand factors: $2x - xy = x(2 - y),$ and so on, so $0 \lt x, y, z \le 2.$ Substitute $x = 2\sin^2\alpha,$ $y = 2\sin^2\beta,$ $z = 2\sin^2\gamma$ with $\alpha, \beta, \gamma \in \left(0^\circ, 90^\circ\right].$ Then $\sqrt{x(2 - y)} = \sqrt{4\sin^2\alpha\cos^2\beta}$ $= 2\sin\alpha\cos\beta,$ and each equation collapses by the sine addition formula: $$2\sin(\alpha + \beta) = 1,$$ $$2\sin(\beta + \gamma) = \sqrt{2},$$ $$2\sin(\gamma + \alpha) = \sqrt{3}.$$

One admissible choice is $\alpha + \beta = 30^\circ,$ $\beta + \gamma = 45^\circ,$ $\gamma + \alpha = 60^\circ,$ which gives $\alpha = 22.5^\circ,$ $\beta = 7.5^\circ,$ $\gamma = 37.5^\circ.$ The only other branch consistent with $0^\circ \lt \alpha,\beta,\gamma \le 90^\circ$ uses pairwise sums $150^\circ,135^\circ,120^\circ,$ giving $\alpha = 67.5^\circ,$ $\beta = 82.5^\circ,$ $\gamma = 52.5^\circ;$ its product below is the negative of the first branch’s product, so the requested square is identical. For the first branch, the double-angle identity gives $1 - x = \cos 2\alpha = \cos 45^\circ,$ $1 - y = \cos 15^\circ,$ and $1 - z = \cos 75^\circ.$

Therefore $$\begin{aligned} &(1 - x)(1 - y)(1 - z) \\ &= \frac{\sqrt{2}}{2}\cos 15^\circ \\ &\quad {}\cdot \sin 15^\circ \\ &= \frac{\sqrt{2}}{2} \cdot \frac{\sin 30^\circ}{2} \\ &= \frac{\sqrt{2}}{8}, \end{aligned}$$ whose square is $\frac{2}{64} = \frac{1}{32}.$ Thus $m + n = 1 + 32 = 33.$