Difference between revisions of "2001 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
− | + | <center><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></center> | ||
− | 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>. 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 BD &= AC \cdot BD\ |
Revision as of 10:59, 15 June 2008
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
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 |