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Difference between revisions of "2001 AIME I Problems/Problem 13"

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We let our chord of degree <math>d</math> be <math>\overline{AB}</math>, of degree <math>2d</math> be <math>\overline{AC}</math>, and of degree <math>3d</math> be <math>\overline{AD}</math>. We are given that <math>AC = AD + 20</math>. Let <math>x = AD</math>. Since <math>AB = BC = CD = 22</math>, quadrilateral <math>ABCD</math> is a [[cyclic quadrilateral|cyclic]] [[isosceles trapezoid]], and so <math>BD = AC = AD + 20</math>. By [[Ptolemy's Theorem]], we have
 
We let our chord of degree <math>d</math> be <math>\overline{AB}</math>, of degree <math>2d</math> be <math>\overline{AC}</math>, and of degree <math>3d</math> be <math>\overline{AD}</math>. We are given that <math>AC = AD + 20</math>. Let <math>x = AD</math>. Since <math>AB = BC = CD = 22</math>, quadrilateral <math>ABCD</math> is a [[cyclic quadrilateral|cyclic]] [[isosceles trapezoid]], and so <math>BD = AC = AD + 20</math>. By [[Ptolemy's Theorem]], we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
AB \cdot CD + AD \cdot BD &= AC \cdot BD\\  
+
AB \cdot CD + AD \cdot BC &= AC \cdot BD\\  
 
22^2 + 22x = (x+20)^2 &\Longrightarrow x^2 + 18x - 84 = 0\\
 
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*}</cmath>
 
x = \frac{-18 + \sqrt{18^2 + 4\cdot 84}}{2} &= -9 + \sqrt{165}\end{align*}</cmath>

Revision as of 09:02, 12 August 2009

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 (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
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