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Difference between revisions of "2023 AMC 8 Problems/Problem 19"

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~MrThinker
 
~MrThinker
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==Video Solution by Math-X (Quick and Simple Under 30 seconds)==
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https://youtu.be/Ku_c1YHnLt0?si=AtiMigHKcdyC8nw9&t=4074 ~Math-X
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==Video Solution==
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https://youtu.be/u0Qa3A3jFFU
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~Education, the Study of Everything
  
 
==Video Solution by OmegaLearn (Using Similar Triangles)==
 
==Video Solution by OmegaLearn (Using Similar Triangles)==
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https://youtu.be/-N46BeEKaCQ?t=3360
 
https://youtu.be/-N46BeEKaCQ?t=3360
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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https://youtu.be/1bA7fD7Lg54?t=1837
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https://youtu.be/DBqko2xATxs&t=24
  
 
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==Video Solution by WhyMath==
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~harungurcan
 
~harungurcan
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==Video Solution (Solve under 60 seconds!!!)==
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https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=869
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~hsnacademy
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==Video Solution by Dr. David==
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https://youtu.be/dTi_0SpcKnc
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==Video Solution by WhyMath==
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https://youtu.be/YnpnKbm68tg
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=18|num-a=20}}
 
{{AMC8 box|year=2023|num-b=18|num-a=20}}
 
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Latest revision as of 10:22, 18 November 2024

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 $\frac23$ 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]

$\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9$

Solution 1

All equilateral triangles are similar. For the outer equilateral triangle to the inner equilateral triangle, since their side-length ratio is $\frac32,$ their area ratio is $\left(\frac32\right)^2=\frac94.$ It follows that the area ratio of three trapezoids to the inner equilateral triangle is $\frac94-1=\frac54,$ so the area ratio of one trapezoid to the inner equilateral triangle is \[\frac54\cdot\frac13=\frac{5}{12}=\boxed{\textbf{(C) } 5 : 12}.\] ~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 $3:2$, we can set the side lengths as $3$ and $2$, respectively. So, the sum of the trapezoids is $\frac{9\sqrt{3}}{4}-\frac{4\sqrt{3}}{4}=\frac{5}{4}\sqrt{3}$. We are also told that the three trapezoids are congruent, thus the area of each of them is $\frac{1}{3} \cdot \frac{5}{4}\sqrt{3}=\frac{5}{12}\sqrt{3}$. Hence, the ratio is $\frac{\frac{5}{12}\sqrt{3}}{\sqrt{3}}=\frac{5}{12}=\boxed{\textbf{(C) } 5 : 12}$.

~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

https://youtu.be/Xq4LdJJtbDk

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

https://youtu.be/gjc3Dslaimg

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=350s

~harungurcan

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=869

~hsnacademy

Video Solution by Dr. David

https://youtu.be/dTi_0SpcKnc

Video Solution by WhyMath

https://youtu.be/YnpnKbm68tg

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

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