10. Right triangle $X Y Z$ has right angle at $Y$ and $X Y=228, Y Z=$ 2004. Angle $Y$ is trisected, and the angle trisectors intersect $X Z$ at $P$ and $Q$ so that $X, P, Q, Z$ lie on $X Z$ in that order. Find the value of $(P Y+Y Z)(Q Y+X Y)$.
Solution: 1370736
The triangle's area is $(228 \cdot 2004) / 2=228456$. All the angles at $Y$ are 30 degrees, so by the sine area formula, the areas of the three small triangles in the diagram are $Q Y \cdot Y Z / 4, P Y \cdot Q Y / 4$, and $X Y \cdot P Y / 4$, which sum to the area of the triangle. So expanding $(P Y+Y Z)(Q Y+X Y)$, we see that it equals
$$
4 \cdot 228456+X Y \cdot Y Z=6 \cdot 228456=1370736 .
$$