2005 Alabama ARML TST Problems/Problem 9


A square is inscribed in a circle of diameter 12. Two vertices of a triangle are also vertices of one side of the square. The other vertex of the triangle is on the circle. Find the largest possible area of the triangle.


Let the "base" of the triangle, which is a side of the square, be $AB$, and let the third point of the triangle be $C$. Clearly we want $C$ to be as far as possible from $AB$, so we have that the base of the triangle is $6\sqrt{2}$ and the height, after some side chasing, is $\dfrac{12-6\sqrt{2}}{2}+6\sqrt{2}=6+3\sqrt{2}$. We multiply and divide by 2 to get $\boxed{18+18\sqrt{2}}$.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 8
Followed by:
Problem 10
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