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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $m, n > 2$ be integers. One of the angles of a regular $n$ -gon is dissected into $m$ angles of equal size by $(m-1)$ rays. If each of these rays intersects the polygon again at one of its vertices, we say $n$ is $m$ -cut. Compute the smallest positive integer $n$ that is both 3-cut and 4-cut.

Solution

Solution:

For the sake of simplicity, inscribe the regular polygon in a circle. Note that each interior angle of the regular $n$ -gon will subtend $n-2$ of the $n$ arcs on the circle. Thus, if we dissect an interior angle into $m$ equal angles, then each must be represented by a total of $\frac{n-2}{m}$ arcs. However, since each of the rays also passes through another vertex of the polygon, that means $\frac{n-2}{m}$ is an integer and thus our desired criteria is that $m$ divides $n-2$ .

That means we want the smallest integer $n > 2$ such that $n-2$ is divisible by $3$ and $4$ which is just $12+2=14$ .

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