Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

Let m,n>2m, n > 2 be integers. One of the angles of a regular nn-gon is dissected into mm angles of equal size by (m1)(m-1) rays. If each of these rays intersects the polygon again at one of its vertices, we say nn is mm-cut. Compute the smallest positive integer nn that is both 3-cut and 4-cut.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

For the sake of simplicity, inscribe the regular polygon in a circle. Note that each interior angle of the regular nn-gon will subtend n2n-2 of the nn arcs on the circle. Thus, if we dissect an interior angle into mm equal angles, then each must be represented by a total of n2m\frac{n-2}{m} arcs. However, since each of the rays also passes through another vertex of the polygon, that means n2m\frac{n-2}{m} is an integer and thus our desired criteria is that mm divides n2n-2. That means we want the smallest integer n>2n>2 such that n2n-2 is divisible by 3 and 4 which is just 12+2=1412+2=14.

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