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2001 AIME I Problems/Problem 13

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Problem

In a certain circle, the chord of a d-degree arc is 22 centimeters long, and the chord of a 2d-degree arc is 20 centimeters longer than the chord of a 3d-degree arc, where d < 120. The length of the chord of a 3d-degree arc is - m + \sqrt {n} centimeters, where m and n are positive integers. Find m + n.

Solution

pointpen = black; pathpen = black+linewidth(0.7);pair A=(0,0), B=(0,22), C=OP(CR(A,11+165^.5),CR(B,22)), D=OP(CR(A,-9+165^.5)...

We let our chord of degree d be \overline{AB}, of degree 2d be \overline{AC}, and of degree 3d be \overline{AD}. We are given that AC = AD + 20. Let x = AD. Since AB = BC = CD = 22, quadrilateral ABCD is a cyclic isosceles trapezoid, and so BD = AC = AD + 20. By Ptolemy's Theorem, we have \begin{align*}AB \cdot CD + AD \cdot BC &= AC \cdot BD\\ 22^2 + 22x = (x+20)^2 &\Longrightarrow x^2 + 18x - 84 = 0\\x... Therefore, the answer is m+n = \boxed{174}.

See also

2001 AIME I (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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