Difference between revisions of "2005 AIME II Problems/Problem 14"
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== Solution == | == Solution == | ||
<center><asy> | <center><asy> | ||
− | + | import olympiad; import cse5; import geometry; size(150); | |
− | pair C = ( | + | defaultpen(fontsize(10pt)); |
− | + | defaultpen(0.8); | |
− | + | dotfactor = 4; | |
− | + | pair A = origin; | |
+ | pair C = rotate(15,A)*(A+dir(-50)); | ||
+ | pair B = rotate(15,A)*(A+dir(-130)); | ||
+ | pair D = extension(A,A+dir(-68),B,C); | ||
+ | pair E = extension(A,A+dir(-82),B,C); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,SW); | ||
+ | label("$D$",D,SE); | ||
+ | label("$E$",E,S); | ||
+ | label("$C$",C,SE); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--E); | ||
+ | draw(A--D); | ||
+ | draw(anglemark(B,A,E,5)); | ||
+ | draw(anglemark(D,A,C,5)); | ||
</asy></center> | </asy></center> | ||
Revision as of 20:12, 16 April 2015
Problem
In triangle and
Point
is on
with
Point
is on
such that
Given that
where
and
are relatively prime positive integers, find
Solution
![[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("$A$",A,N); label("$B$",B,SW); label("$D$",D,SE); label("$E$",E,S); label("$C$",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]](http://latex.artofproblemsolving.com/f/5/3/f53f3016596f4d0c3c6dc23b6e0b5b41fabe85f2.png)
By the Law of Sines and since , we have
Substituting our knowns, we have . The answer is
.
See also
2005 AIME II (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.