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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.5 AIME Prove it United States

Problem:
A convex quadrilateral is determined by the points of intersection of the curves $x^{4}+y^{4}=100$ and $x y=4$ ; determine its area.

Solution

Solution:
Answer: $4 \sqrt{17}$ . By symmetry, the quadrilateral is a rectangle having $x=y$ and $x=-y$ as axes of symmetry. Let $(a, b)$ with $a>b>0$ be one of the vertices. Then the desired area is
$(\sqrt{2}(a-b)) \cdot (\sqrt{2}(a+b)) = 2\left(a^{2}-b^{2}\right) = 2 \sqrt{a^{4}-2 a^{2} b^{2}+b^{4}} = 2 \sqrt{100-2 \cdot 4^{2}} = 4 \sqrt{17}$

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