1991 AIME Problem 14

Attempt Problem 14 of the 1991 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AIME solutions, or check the answer key.

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14.

A hexagon is inscribed in a circle. Five of the sides have length 8181 and the sixth, denoted by AB,\overline{AB}, has length 31.31. Find the sum of the lengths of the three diagonals that can be drawn from A.A.

Answer: 384
Concepts:chordtrigonometric identitytrigonometry
Difficulty rating: 2710
Small Hint:

Let 2u2u be the central angle subtended by each side of length 8181, and set t=2cosut=2\cos u

Big Hint:

Express sin(5u)sinu\frac{\sin(5u)}{\sin u} and the three diagonal ratios in terms of tt

Solution:

Let 2u2u be the central angle subtended by each 8181-side, and put t=2cosu.t=2\cos u. The remaining arc has half-angle π5u,\pi-5u, so the chord ratio gives 3181=sin5usinu=t43t2+1.\begin{aligned}\frac{31}{81}&=\frac{\sin5u}{\sin u}\\&=t^4-3t^2+1.\end{aligned} This yields t2=259t^2=\frac{25}{9} or 29.\frac{2}{9}. Because 5u<π,5u<\pi, we have t>2cos36,t>2\cos36^\circ, so t=53.t=\frac{5}{3}.

The three diagonals from AA subtend the same minor angles as 2u,2u, 3u,3u, and 4u.4u. Relative to an 8181-side, their length ratios are sin2usinu=t,sin3usinu=t21,sin4usinu=t32t.\begin{aligned}\frac{\sin2u}{\sin u}&=t,\\\frac{\sin3u}{\sin u}&=t^2-1,\\\frac{\sin4u}{\sin u}&=t^3-2t.\end{aligned} Their sum is therefore 81(t3+t2t1)=81(12827)=384.\begin{aligned}81(t^3+t^2-t-1)&=81\left(\frac{128}{27}\right)\\&=384.\end{aligned}

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