Library / /327 of 740

, 2012

Geometry Difficulty 4.9 AIME Prove it United States

Problem:
Let $\alpha$ and $\beta$ be reals. Find the least possible value of
$(2 \cos \alpha + 5 \sin \beta - 8)^2 + (2 \sin \alpha + 5 \cos \beta - 15)^2.$

Solution

Solution:
Let the vector $\vec{v} = (2 \cos \alpha, 2 \sin \alpha)$ and $\vec{w} = (5 \sin \beta, 5 \cos \beta)$ . The locus of ends of vectors expressible in the form $\vec{v} + \vec{w}$ 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 $X = (8, 15)$ . Thus, the closest distance from such a point to $X$ is when the point is 7 units away from the origin along the segment from the origin to $X$ . Thus, since $X$ is 17 units away from the origin, the minimum is $10^2 = 100$ .

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