2001 AIME I Problems/Problem 13

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

[asy] 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),CR(C,22)); D(D(MP("A",A,E))--D(MP("B",B,N))--D(MP("C",C,W))--D(MP("D",D,SW))--A--C); D(circumcircle(A,B,C)); MP("22",(A+B)/2,E); MP("22",(C+B)/2,NW); MP("22",(C+D)/2,SW); MP("22",(A+B)/2,E); MP("x",(A+D)/2,SE); MP("x+20",(A+C)/2,NE); [/asy]

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 = \frac{-18 + \sqrt{18^2 + 4\cdot 84}}{2} &= -9 + \sqrt{165}\end{align*} Therefore, the answer is $m+n = \boxed{174}$.

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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