Maths Olympiad Prep

Library / /54 of 377

Geometry Difficulty 4.5 AIME Prove it United States

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

Solution

Solution:
Answer: 4174 \sqrt{17}. By symmetry, the quadrilateral is a rectangle having x=yx=y and x=−yx=-y as axes of symmetry. Let (a,b)(a, b) with a>b>0a>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}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.