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Algebra Difficulty 4.9 AIME Find the answer

A convex quadrilateral is determined by the points of intersection of the curves x4+y4=100 x^{4}+y^{4}=100 and xy=4 x y=4 ; determine its area.

A number or a short expression. Spacing and $ signs are ignored.

Solution

By symmetry, the quadrilateral is a rectangle having x=y x=y and x=−y x=-y as axes of symmetry. Let (a,b) (a, b) with a>b>0 a>b>0 be one of the vertices. Then the desired area is (2(a−b))⋅(2(a+b))=2(a2−b2)=2a4−2a2b2+b4=2100−2⋅42=417 (\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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