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 | + | 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 10:02, 12 August 2009
Problem
In a certain circle, the chord of a -degree arc is
centimeters long, and the chord of a
-degree arc is
centimeters longer than the chord of a
-degree arc, where
The length of the chord of a
-degree arc is
centimeters, where
and
are positive integers. Find
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]](http://latex.artofproblemsolving.com/1/7/7/177b3a9528e0bd79c170d6bf594fde668c1e7b1a.png)
We let our chord of degree be
, of degree
be
, and of degree
be
. We are given that
. Let
. Since
, quadrilateral
is a cyclic isosceles trapezoid, and so
. By Ptolemy's Theorem, we have
Therefore, the answer is
.
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 |