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Algebra Difficulty 4.7 AIME Find the answer

Let α\alpha and β\beta be reals. Find the least possible value of (2cosα+5sinβ8)2+(2sinα+5cosβ15)2(2 \cos \alpha+5 \sin \beta-8)^{2}+(2 \sin \alpha+5 \cos \beta-15)^{2}.

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

Solution

Let the vector v=(2cosα,2sinα)\vec{v}=(2 \cos \alpha, 2 \sin \alpha) and w=(5sinβ,5cosβ)\vec{w}=(5 \sin \beta, 5 \cos \beta). The locus of ends of vectors expressible in the form v+w\vec{v}+\vec{w} are the points which are five units away from a point on the circle of radius two about the origin. The expression that we desire to minimize is the square of the distance from this point to X=(8,15)X=(8,15). Thus, the closest distance from such a point to XX is when the point is 7 units away from the origin along the segment from the origin to XX. Thus, since XX is 17 units away from the origin, the minimum is 102=10010^{2}=100.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.