Difference between revisions of "2013 AMC 10A Problems/Problem 12"
(→Problem) |
|||
| Line 27: | Line 27: | ||
\textbf{(D) }60\qquad | \textbf{(D) }60\qquad | ||
\textbf{(E) }72\qquad</math> | \textbf{(E) }72\qquad</math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
Revision as of 11:50, 13 August 2014
Problem
In 
, 
and 
. Points 
and 
are on sides 
, 
, and 
, respectively, such that 
and 
are parallel to and , respectively. What is the perimeter of parallelogram 
?
![[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]](http://latex.artofproblemsolving.com/e/c/4/ec45f0635e288a5e5da73a19319f4eb7045b5eab.png)
Solution
Note that because and are parallel to the sides of , the internal triangles 
and 
are similar to , and are therefore also isosceles triangles.
It follows that 
. Thus, 
.
Since opposite sides of parallelograms are equal, the perimeter is 
.
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
