Fermat 2002 Q20

Problem Statement

In the diagram, $YQZC$ is a rectangle with $YC = 8$ and $CZ = 15$. Equilateral triangles $ABC$ and $PQR$, each with side length 9, are positioned as shown with R and B on sides $YQ$ and $CZ$, respectively. The length of $AP$ is

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Solution

$$AP^2 = AD^2 + DP^2$$ $$AD = FE - FA - DE = 15 - 9/2 - 9/2 = 6$$ $$DP = HP - XP = \dfrac{9\sqrt{3}}{2} + 8 - \dfrac{9\sqrt{3}}{2} = 8$$ $$\therefore AP^2 = 6^2 + 8^2 \implies AP = 10$$