Difference between revisions of "2013 AMC 12A Problems/Problem 9"
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It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>. | It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>. | ||
| − | Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>. | + | Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = |
| + | \boxed{\textbf{(C) }{56}}</math>. | ||
| + | ==Solution (Cheese)== | ||
| + | We can set point <math>F</math> to be on point <math>C</math>, and point <math>D</math> to be on point <math>A</math>. | ||
| + | |||
| + | This makes a degenerate parallelogram with sides of length <math>28</math> and <math>0</math>, so it has a perimeter of <math>28 + 28 = | ||
| + | \boxed{\textbf{(C) }{56}}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/CCjcMVtkVaQ | ||
| + | |||
| + | ~sugar_rush | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | |||
| + | {{AMC10 box|year=2013|ab=A|num-b=11|num-a=13}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:25, 12 August 2023
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 
.
Solution (Cheese)
We can set point to be on point 
, and point 
to be on point 
.
This makes a degenerate parallelogram with sides of length 
and 
, so it has a perimeter of 
.
Video Solution
~sugar_rush
See also
| 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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
