Let FELDSPAR be a regular octagon, and let I be a point in its interior such that ∠FIL=∠LID=∠DIS=∠SIA. Compute ∠IAR in degrees.
Solution
Solution:
Observe that I lies on line DR due to symmetry, so ID∥FL. Thus ∠FLI=∠LID=∠FIL, implying that triangle FIL is isosceles with FI=FL. Similarly, AI=AS. Since FLSA is a square, FI=FL=AS=AI=FA. Therefore, FIA is equilateral, so ∠AIR=21∠FIA=30∘ and ∠IAR=180∘−30∘−21⋅135∘=82.5∘.
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