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Problem 1005

AMC 12 late, AIME early

Geometry Difficulty 4.9 Prove it HMMT November · United States · 2024

Let $F E L D S P A R$ be a regular octagon, and let $I$ be a point in its interior such that $\angle F I L = \angle L I D = \angle D I S = \angle S I A$ . Compute $\angle I A R$ in degrees.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Observe that $I$ lies on line $D R$ due to symmetry, so $I D \parallel F L$ . Thus $\angle F L I = \angle L I D = \angle F I L$ , implying that triangle $F I L$ is isosceles with $F I = F L$ . Similarly, $A I = A S$ . Since $F L S A$ is a square, $F I = F L = A S = A I = F A$ . Therefore, $F I A$ is equilateral, so $\angle A I R = \frac{1}{2} \angle F I A = 30^\circ$ and $\angle I A R = 180^\circ - 30^\circ - \frac{1}{2} \cdot 135^\circ = 82.5^\circ$ .

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