# Hint: IMO 2008, Madrid, Spain

Geometry Level 5

An isosceles triangle $$ABC$$ with $$AB = AC$$ has $$AD$$ as its altitude where $$D$$ is on $$BC$$. Let $$C^{"}$$ be the diametrically opposite point to $$C$$ on the circumcircle of $$\Delta ABC$$ and let $$M$$ be the midpoint of $$AB$$.

Construct a circle centred at $$M$$ of radius $$MC^{"}$$ and let it intersect $$AB$$ (possibly extended) at $$P$$ and $$Q$$. Let $$S$$ and $$R$$ be respectively the reflections of $$P$$ and $$Q$$ across altitude $$AD$$.

Being an isosceles trapezium, quadrilateral $$PQRS$$ is cyclic of course. Find the radius $$\rho$$ of the circle circumscribing it.

Given: The altitude $$AD = 9$$ and circumradius of $$\Delta ABC$$ is $$8$$.

Report $$\rho^2$$.

$$Note:$$ An original twist to the famous problem! And by the way, the actual diagram looks not at all like the one shown here for guidance: Figure not to scale in the least!

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