Library / /652 of 740

, 2014

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Two circles $\omega$ and $\gamma$ have radii $3$ and $4$ respectively, and their centers are $10$ units apart. Let $x$ be the shortest possible distance between a point on $\omega$ and a point on $\gamma$ , and let $y$ be the longest possible distance between a point on $\omega$ and a point on $\gamma$ . Find the product $x y$ .

Solution

Solution:

Let $\ell$ be the line connecting the centers of $\omega$ and $\gamma$ . Let $A$ and $B$ be the intersections of $\ell$ with $\omega$ , and let $C$ and $D$ be the intersections of $\ell$ with $\gamma$ , so that $A, B, C$ , and $D$ are collinear, in that order.

The shortest distance between a point on $\omega$ and a point on $\gamma$ is $BC = 3$ .

The longest distance is $AD = 3 + 10 + 4 = 17$ .

The product is $3 \times 17 = 51$ .

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.