Difference between revisions of "2010 AMC 10A Problems/Problem 14"
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== Solution == | == Solution == | ||
+ | <asy> | ||
+ | pair A,B,C,D,E,F,G,H; | ||
+ | G=(0,10); | ||
+ | A=(0,3.464); | ||
+ | B=(6,0); | ||
+ | C=(0,0); | ||
+ | draw(A--B--C--cycle); | ||
+ | F=(1,1.73); | ||
+ | E=(2,0); | ||
+ | draw(C--F--E); | ||
+ | D=(1.5,2.6); | ||
+ | draw(C--D); | ||
+ | label("$A$",A,W); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,S); | ||
+ | label("$F$",F,N); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",E,S); | ||
+ | draw(A--E); | ||
+ | draw(anglemark(E,A,B)); | ||
+ | draw(anglemark(D,C,A)); | ||
+ | </asy> | ||
+ | |||
+ | |||
Let <math>\angle BAE = \angle ACD = x</math>. | Let <math>\angle BAE = \angle ACD = x</math>. | ||
Revision as of 13:19, 13 August 2017
Problem
Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?
Solution
Let .
Since ,
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.