Library / /329 of 740

, 2014

Geometry Difficulty 4.9 AIME Prove it United States

Problem:

The side lengths of a triangle are distinct positive integers. One of the side lengths is a multiple of $42$ , and another is a multiple of $72$ . What is the minimum possible length of the third side?

Solution

Solution:

Suppose that two of the side lengths are $42a$ and $72b$ , for some positive integers $a$ and $b$ . Let $c$ be the third side length. We know that $42a$ is not equal to $72b$ , since the side lengths are distinct. Also, $6 \mid 42a - 72b$ . Therefore, by the triangle inequality, we get $c > |42a - 72b| \geq 6$ and thus $c \geq 7$ . Hence, the minimum length of the third side is $7$ and equality is obtained when $a = 7$ and $b = 4$ .

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.