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 $m+n = \boxed{174}$ .

2001 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

2001 AIME I Problems/Problem 13

Problem

Solution

See also