Difference between revisions of "2013 AMC 10A Problems/Problem 12"
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\textbf{(D) }60\qquad | \textbf{(D) }60\qquad | ||
\textbf{(E) }72\qquad</math> | \textbf{(E) }72\qquad</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==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. | 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. | ||
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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>. | ||
− | + | 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>, by fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Drawing the diagram with a ruler and compass (and scaling back by x4), we can draw approximate parallel lines. This yields about 14, but we need to multiply by 4 to get 56, or <math>\boxed{(C)}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Similar to Solution 1, we find that 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 and similar to each other. We know that the perimeter of <math>ADEF</math> can be found by <math>2(l + w)</math>, and <math>DE</math> will be the length and the width will be <math>EF</math>. These can be found by looking at the long sides of <math>\triangle DBE</math> and <math>\triangle FEC</math> respectively. We then find the ratio of long sides to the short side of <math>\triangle ABC</math> to be <math>7/5</math> by <math>28/20</math>, which applies to the other triangles since they are similar. We set up an expression, calling <math>BE</math> as <math>x</math> and <math>EC</math> as <math>20 - x</math>, and substitute it into the perimeter equation knowing our long sides from the ratio: | ||
+ | |||
+ | <math> 2(7x/5 + 7/5(20-x)) </math> | ||
+ | |||
+ | <math> 2(7x/5 + 28 - 7x/5) </math> | ||
+ | |||
+ | <math> 2(28) </math> | ||
+ | |||
+ | <math> 56\implies \boxed{\textbf{(C)}}</math> | ||
+ | |||
+ | ~neeyakkid23 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/8ki_yMyE6no | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 19:59, 10 September 2024
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 , by fakesolving, we get .
Solution 3
Drawing the diagram with a ruler and compass (and scaling back by x4), we can draw approximate parallel lines. This yields about 14, but we need to multiply by 4 to get 56, or .
Solution 4
Similar to Solution 1, we find that the internal triangles and are similar to , and are therefore also isosceles triangles and similar to each other. We know that the perimeter of can be found by , and will be the length and the width will be . These can be found by looking at the long sides of and respectively. We then find the ratio of long sides to the short side of to be by , which applies to the other triangles since they are similar. We set up an expression, calling as and as , and substitute it into the perimeter equation knowing our long sides from the ratio:
~neeyakkid23
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.