Difference between revisions of "2013 AMC 12A Problems/Problem 9"
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− | Note that because <math>\ | + | == Problem== |
+ | |||
+ | In <math>\triangle ABC</math>, <math>AB=AC=28</math> and <math>BC=20</math>. Points <math>D,E,</math> and <math>F</math> are on sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> and <math>\overline{EF}</math> are parallel to <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively. What is the perimeter of parallelogram <math>ADEF</math>? | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math>\textbf{(A) }48\qquad | ||
+ | \textbf{(B) }52\qquad | ||
+ | \textbf{(C) }56\qquad | ||
+ | \textbf{(D) }60\qquad | ||
+ | \textbf{(E) }72\qquad</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Note that because <math>\overline{DE}</math> and <math>\overline{EF}</math> are parallel to the sides of <math>\triangle ABC</math>, the internal triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles. | ||
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 == | ||
+ | {{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 16:25, 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
?
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 |
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