Difference between revisions of "2013 AMC 10A Problems/Problem 12"
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==Problem== | ==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> | ||
− | ==Solution== | + | <math>\textbf{(A) }48\qquad |
+ | \textbf{(B) }52\qquad | ||
+ | \textbf{(C) }56\qquad | ||
+ | \textbf{(D) }60\qquad | ||
+ | \textbf{(E) }72\qquad</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | The opposite sides of parallelograms are equal (you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is <math>2 \times (AD + DE) = 56\implies \boxed{\textbf{(C)}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We can set <math>AD=0</math>, so fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/8ki_yMyE6no | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2013|ab=A|num-b=11|num-a=13}} | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Revision as of 17:00, 26 December 2020
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 1
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, .
The opposite sides of parallelograms are equal (you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is .
Solution 2
We can set , so fakesolving, we get .
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.