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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Romania

Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n > 1$ be an integer. Point $M$ lies on the line segment $AB$ such that $nAM = AB$ . Consider the points $P_1, P_2, \dots, P_{n-1}$ on the side $BC$ with $BP_1 = P_1P_2 = P_2P_3 = \dots = P_{n-1}C = \frac{1}{n}BC$ . Prove that
$\angle MP_1A + \angle MP_2A + \dots + \angle MP_{n-1}A = \frac{1}{2} \angle BAC.$

Solution

Consider the point $N$ on the side $AC$ such that $nAN = AC$ . The configuration is symmetric with respect to the perpendicular bisector of the segment $BC$ , implying $\angle MP_iA = \angle NP_{n-i}A$ , $i = 1, 2, \dots, n-1$ . The claim is equivalent to $\angle MP_1N + \angle MP_2N + \dots + \angle MP_{n-1}N = \angle BAC$ .

Notice that $BP_1 = P_1P_2 = P_2P_3 = \dots = P_{n-1}C = MN$ . Set $P_0 = B$ , and notice that in the parallelograms $P_iMNP_{i+1}$ one has $\angle MP_{i+1}N = \angle P_iMP_{i+1}$ , $i = 0, 1, \dots, n-2$ . Therefore the sum of those angles is equal to $\angle P_0MP_{n-1}$ , in turn equal to $\angle BAC$ , since $MP_{n-1} \parallel AC$ .

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