Difference between revisions of "2013 AMC 12A Problems/Problem 11"
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Revision as of 08:55, 7 April 2013
Problem
Triangle 
is equilateral with 
. Points 
and 
are on 
and points 
and 
are on 
such that both 
and 
are parallel to 
. Furthermore, triangle 
and trapezoids 
and 
all have the same perimeter. What is 
?
![[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real s=1/2,m=5/6,l=1; pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m; draw(A--B--C--cycle^^D--E^^F--G); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW); label("$F$",F,S); label("$G$",G,NW); [/asy]](http://latex.artofproblemsolving.com/d/2/e/d2eb992c9eae1b2af8bec0d92f03bde1e1e8b647.png)
Solution
Let 
, and 
. We want to find 
, which is nothing but 
.
Based on the fact that , 
, and 
have the same perimeters, we can say the following:
Simplifying, we can find that
Since 
, 
.
After substitution, we find that 
, and 
= 
.
Again substituting, we find 
= 
.
Therefore, = 
, which is 
See also
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 12 Problems and Solutions | |
