HMMT February

Consider the three flat faces of the eighth-ball. Each of these is a quarter-circle of radius 1, so each has area $\frac{\pi}{4}$. Furthermore, the projections of these faces cover the desired area without overlap. To find the projection factor one can find the cosine of the angle $\theta$ between the planes, which is the same as the angle between their normal vectors. Using the dot product formula for the cosine of the angle between two vectors, $\cos \theta=\frac{(1,0,0) \cdot(1,1,1)}{|(1,0,0)|(1,1,1) \mid}=\frac{1}{\sqrt{3}}$. Therefore, each area is multiplied by $\frac{1}{\sqrt{3}}$ by the projection, so the area of the projection is $3 \cdot \frac{\pi}{4} \cdot \frac{1}{\sqrt{3}}=\frac{\pi \sqrt{3}}{4}$.