Difference between revisions of "2023 AMC 8 Problems/Problem 19"
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− | + | ==Problem== | |
+ | An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is <math>\frac23</math> the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle? | ||
+ | <asy> | ||
+ | // Diagram by TheMathGuyd | ||
+ | pair A,B,C; | ||
+ | A=(0,1); | ||
+ | B=(sqrt(3)/2,-1/2); | ||
+ | C=-conj(B); | ||
+ | fill(2B--3B--3C--2C--cycle,grey); | ||
+ | dot(3A); | ||
+ | dot(3B); | ||
+ | dot(3C); | ||
+ | dot(2A); | ||
+ | dot(2B); | ||
+ | dot(2C); | ||
+ | draw(2A--2B--2C--cycle); | ||
+ | draw(3A--3B--3C--cycle); | ||
+ | draw(2A--3A); | ||
+ | draw(2B--3B); | ||
+ | draw(2C--3C); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | All equilateral triangles are similar. For the outer equilateral triangle to the inner equilateral triangle, since their side-length ratio is <math>\frac32,</math> their area ratio is <math>\left(\frac32\right)^2=\frac94.</math> It follows that the area ratio of three trapezoids to the inner equilateral triangle is <math>\frac94-1=\frac54,</math> so the area ratio of one trapezoid to the inner equilateral triangle is <cmath>\frac54\cdot\frac13=\frac{5}{12}=\boxed{\textbf{(C) } 5 : 12}.</cmath> | ||
+ | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2== | ||
+ | Subtracting the larger equilateral triangle from the smaller one yields the sum of the three trapezoids. Since the ratio of the side lengths of the larger to the smaller one is <math>3:2</math>, we can set the side lengths as <math>3</math> and <math>2</math>, respectively. So, the sum of the trapezoids is <math>\frac{9\sqrt{3}}{4}-\frac{4\sqrt{3}}{4}=\frac{5}{4}\sqrt{3}</math>. We are also told that the three trapezoids are congruent, thus the area of each of them is <math>\frac{1}{3} \cdot \frac{5}{4}\sqrt{3}=\frac{5}{12}\sqrt{3}</math>. Hence, the ratio is <math>\frac{\frac{5}{12}\sqrt{3}}{\sqrt{3}}=\frac{5}{12}=\boxed{\textbf{(C) } 5 : 12}</math>. | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
+ | ==Video Solution by Math-X (Quick and Simple Under 30 seconds)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=AtiMigHKcdyC8nw9&t=4074 ~Math-X | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/u0Qa3A3jFFU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by OmegaLearn (Using Similar Triangles)== | ||
+ | https://youtu.be/bGN-uBsVm0E | ||
==Animated Video Solution== | ==Animated Video Solution== | ||
https://youtu.be/Xq4LdJJtbDk | https://youtu.be/Xq4LdJJtbDk | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove using Area-Similarity Relationship== | ||
+ | https://www.youtube.com/watch?v=92hAg3JjqZI | ||
~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
+ | |||
+ | ==Video Solution by Magic Square== | ||
+ | https://youtu.be/-N46BeEKaCQ?t=3360 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=24 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/gjc3Dslaimg | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by harungurcan== | ||
+ | https://www.youtube.com/watch?v=Ki4tPSGAapU&t=350s | ||
+ | |||
+ | ~harungurcan | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2023|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Revision as of 13:36, 20 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by Math-X (Quick and Simple Under 30 seconds)
- 5 Video Solution
- 6 Video Solution by OmegaLearn (Using Similar Triangles)
- 7 Animated Video Solution
- 8 Video Solution by SpreadTheMathLove using Area-Similarity Relationship
- 9 Video Solution by Magic Square
- 10 Video Solution by Interstigation
- 11 Video Solution by WhyMath
- 12 Video Solution by harungurcan
- 13 See Also
Problem
An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle?
Solution 1
All equilateral triangles are similar. For the outer equilateral triangle to the inner equilateral triangle, since their side-length ratio is their area ratio is
It follows that the area ratio of three trapezoids to the inner equilateral triangle is
so the area ratio of one trapezoid to the inner equilateral triangle is
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
Solution 2
Subtracting the larger equilateral triangle from the smaller one yields the sum of the three trapezoids. Since the ratio of the side lengths of the larger to the smaller one is , we can set the side lengths as
and
, respectively. So, the sum of the trapezoids is
. We are also told that the three trapezoids are congruent, thus the area of each of them is
. Hence, the ratio is
.
~MrThinker
Video Solution by Math-X (Quick and Simple Under 30 seconds)
https://youtu.be/Ku_c1YHnLt0?si=AtiMigHKcdyC8nw9&t=4074 ~Math-X
Video Solution
~Education, the Study of Everything
Video Solution by OmegaLearn (Using Similar Triangles)
Animated Video Solution
Video Solution by SpreadTheMathLove using Area-Similarity Relationship
https://www.youtube.com/watch?v=92hAg3JjqZI
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3360
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=24
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=350s
~harungurcan
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.