Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it United States

Problem:

Let FELDSPARF E L D S P A R be a regular octagon, and let II be a point in its interior such that FIL=LID=DIS=SIA\angle F I L = \angle L I D = \angle D I S = \angle S I A. Compute IAR\angle I A R in degrees.

Solution

Solution:

Figure 1

Observe that II lies on line DRD R due to symmetry, so IDFLI D \parallel F L. Thus FLI=LID=FIL\angle F L I = \angle L I D = \angle F I L, implying that triangle FILF I L is isosceles with FI=FLF I = F L. Similarly, AI=ASA I = A S. Since FLSAF L S A is a square, FI=FL=AS=AI=FAF I = F L = A S = A I = F A. Therefore, FIAF I A is equilateral, so AIR=12FIA=30\angle A I R = \frac{1}{2} \angle F I A = 30^\circ and IAR=1803012135=82.5\angle I A R = 180^\circ - 30^\circ - \frac{1}{2} \cdot 135^\circ = 82.5^\circ.

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