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Difference between revisions of "2023 AMC 8 Problems/Problem 24"

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==Problem==
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Isosceles <math>\triangle ABC</math> has equal side lengths <math>AB</math> and <math>BC</math>. In the figure below, segments are drawn parallel to <math>\overline{AC}</math> so that the shaded portions of <math>\triangle ABC</math> have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of <math>h</math> of <math>\triangle ABC</math>?
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<asy>
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//Diagram by TheMathGuyd
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size(12cm);
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real h = 2.5; // height
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real g=4; //c2c space
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real s = 0.65; //Xcord of Hline
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real adj = 0.08; //adjust line diffs
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pair A,B,C;
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B=(0,h);
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C=(1,0);
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A=-conj(C);
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pair PONE=(s,h*(1-s)); //Endpoint of Hline ONE
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pair PTWO=(s+adj,h*(1-s-adj)); //Endpoint of Hline ONE
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path LONE=PONE--(-conj(PONE)); //Hline ONE
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path LTWO=PTWO--(-conj(PTWO));
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path T=A--B--C--cycle; //Triangle
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fill (shift(g,0)*(LTWO--B--cycle),mediumgrey);
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fill (LONE--A--C--cycle,mediumgrey);
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draw(LONE);
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draw(T);
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label("$A$",A,SW);
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label("$B$",B,N);
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label("$C$",C,SE);
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draw(shift(g,0)*LTWO);
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draw(shift(g,0)*T);
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label("$A$",shift(g,0)*A,SW);
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label("$B$",shift(g,0)*B,N);
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label("$C$",shift(g,0)*C,SE);
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draw(B--shift(g,0)*B,dashed);
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draw(C--shift(g,0)*A,dashed);
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draw((g/2,0)--(g/2,h),dashed);
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draw((0,h*(1-s))--B,dashed);
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draw((g,h*(1-s-adj))--(g,0),dashed);
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label("$5$", midpoint((g,h*(1-s-adj))--(g,0)),UnFill);
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label("$h$", midpoint((g/2,0)--(g/2,h)),UnFill);
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label("$11$", midpoint((0,h*(1-s))--B),UnFill);
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</asy>
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<math>\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4</math>
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==Solution 1==
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First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is <math>[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)</math>. Similarly, we can find that the area of the gray part in the second triangle is <math>[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2</math>. These areas are equal, so <math>1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2</math>. Simplifying yields <math>10h=146</math> so <math>h=\boxed{\textbf{(A) }14.6}</math>.
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~MathFun1000 (~edits apex304)
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==Solution 2 (Thorough)==
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We can call the length of AC as <math>x</math>. Therefore, the length of the base of the triangle with height <math>11</math> is <math>11/h = a/x</math>. Therefore, the base of the smaller triangle is <math>11x/h</math>. We find that the area of the trapezoid is <math>(hx)/2 - 11^2x/2h</math>.
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Using similar triangles once again, we find that the base of the shaded triangle is <math>(h-5)/h = b/x</math>. Therefore, the area is <math>(h-5)(hx-5x)/h</math>.
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Since the areas are the same, we find that <math>(hx)/2 - 121x/2h = (h-5)(hx-5x)/h</math>. Multiplying each side by <math>2h</math>, we get <math>h^2x - 121x = h^2x - 5hx - 5hx + 25x</math>. Therefore, we can subtract <math>25x + h^2x</math> from both sides, and get <math>-146x = -10hx</math>. Finally, we divide both sides by <math>-x</math> and get <math>10h = 146</math>. <math>h</math> is <math>\boxed{\textbf{(A)}14.6}</math>.
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Solution by CHECKMATE2021
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==Solution 3 (Faster)==
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Since the length of AC does not matter, we can assume the base of triangle ABC is <math>h</math>. Therefore, the area of the trapezoid in the first diagram is <math>h^2/2 - \frac{11^2}{2}</math>.
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The area of the triangle in the second diagram is now <math>\frac{(h-5)^2}{2}</math>.
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Therefore, <math>\frac{h^2 - 11^2}{2} = \frac{(h-5)^2}{2}</math>. Multiplying both sides by <math>2</math>, we get <math>h^2 - 121 = h^2 - 10h + 25</math>. Subtracting <math>h^2 + 25</math> from both sides, we get <math>-146 = -10h</math> and <math>h</math> is <math>\boxed{\textbf{(A)}14.6}</math>.
 +
 
 +
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]], and CHECKMATE2021
 +
 
 +
==Solution 4(Based on solution 3) (VERY SLOW)==
 +
The answers are there on the bottom, so start with the middle one, <math>{\textbf{(C)}15}</math>. After calculating, we find that we need a shorter length, so try <math>{\textbf{(B)}14.8}</math>. Still, we need a shorter answer, so we simply choose <math>\boxed{\textbf{(A)}14.6}</math> without trying it out.
 +
 
 +
~SaxStreak
 +
 
 +
==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=LIt75cZMdNbGeWMB&t=1116
 +
 
 +
~hsnacademy
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349
 +
 
 +
~Math-X
 +
 
 +
https://youtu.be/uxyJGZ3ZYGE
 +
 
 +
~please like and subscribe
 +
 
 +
==Video Solution (THINKING CREATIVELY!!!)==
 +
 
 +
https://youtu.be/SVVSMcw1Xe8
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution 1 by OmegaLearn (Using Similarity)==
 +
https://youtu.be/almtw4n-92A
 +
 
 +
==Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)==
 +
https://www.youtube.com/watch?v=GTlkTwxSxgo
 +
 
 +
==Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)==
 +
https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/DBqko2xATxs&t=3270
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/roTSeCAehek
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1593s
 +
 
 +
~harungurcan
 +
 
 +
==See Also==
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{{AMC8 box|year=2023|num-b=23|num-a=25}}
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{{MAA Notice}}

Revision as of 00:50, 21 August 2024

Problem

Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$. In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$?

[asy] //Diagram by TheMathGuyd size(12cm); real h = 2.5; // height real g=4; //c2c space real s = 0.65; //Xcord of Hline real adj = 0.08; //adjust line diffs pair A,B,C; B=(0,h); C=(1,0); A=-conj(C); pair PONE=(s,h*(1-s)); //Endpoint of Hline ONE pair PTWO=(s+adj,h*(1-s-adj)); //Endpoint of Hline ONE path LONE=PONE--(-conj(PONE)); //Hline ONE path LTWO=PTWO--(-conj(PTWO)); path T=A--B--C--cycle; //Triangle fill (shift(g,0)*(LTWO--B--cycle),mediumgrey); fill (LONE--A--C--cycle,mediumgrey); draw(LONE); draw(T); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); draw(shift(g,0)*LTWO); draw(shift(g,0)*T); label("$A$",shift(g,0)*A,SW); label("$B$",shift(g,0)*B,N); label("$C$",shift(g,0)*C,SE); draw(B--shift(g,0)*B,dashed); draw(C--shift(g,0)*A,dashed); draw((g/2,0)--(g/2,h),dashed); draw((0,h*(1-s))--B,dashed); draw((g,h*(1-s-adj))--(g,0),dashed); label("$5$", midpoint((g,h*(1-s-adj))--(g,0)),UnFill); label("$h$", midpoint((g/2,0)--(g/2,h)),UnFill); label("$11$", midpoint((0,h*(1-s))--B),UnFill); [/asy]

$\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4$

Solution 1

First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot\left(\tfrac{h-5}{h}\right)^2$. These areas are equal, so $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$. Simplifying yields $10h=146$ so $h=\boxed{\textbf{(A) }14.6}$.

~MathFun1000 (~edits apex304)

Solution 2 (Thorough)

We can call the length of AC as $x$. Therefore, the length of the base of the triangle with height $11$ is $11/h = a/x$. Therefore, the base of the smaller triangle is $11x/h$. We find that the area of the trapezoid is $(hx)/2 - 11^2x/2h$.

Using similar triangles once again, we find that the base of the shaded triangle is $(h-5)/h = b/x$. Therefore, the area is $(h-5)(hx-5x)/h$.

Since the areas are the same, we find that $(hx)/2 - 121x/2h = (h-5)(hx-5x)/h$. Multiplying each side by $2h$, we get $h^2x - 121x = h^2x - 5hx - 5hx + 25x$. Therefore, we can subtract $25x + h^2x$ from both sides, and get $-146x = -10hx$. Finally, we divide both sides by $-x$ and get $10h = 146$. $h$ is $\boxed{\textbf{(A)}14.6}$.

Solution by CHECKMATE2021

Solution 3 (Faster)

Since the length of AC does not matter, we can assume the base of triangle ABC is $h$. Therefore, the area of the trapezoid in the first diagram is $h^2/2 - \frac{11^2}{2}$.

The area of the triangle in the second diagram is now $\frac{(h-5)^2}{2}$.

Therefore, $\frac{h^2 - 11^2}{2} = \frac{(h-5)^2}{2}$. Multiplying both sides by $2$, we get $h^2 - 121 = h^2 - 10h + 25$. Subtracting $h^2 + 25$ from both sides, we get $-146 = -10h$ and $h$ is $\boxed{\textbf{(A)}14.6}$.

Solution by ILoveMath31415926535, and CHECKMATE2021

Solution 4(Based on solution 3) (VERY SLOW)

The answers are there on the bottom, so start with the middle one, ${\textbf{(C)}15}$. After calculating, we find that we need a shorter length, so try ${\textbf{(B)}14.8}$. Still, we need a shorter answer, so we simply choose $\boxed{\textbf{(A)}14.6}$ without trying it out.

~SaxStreak

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=LIt75cZMdNbGeWMB&t=1116

~hsnacademy

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/Ku_c1YHnLt0?si=NHwA1x9STOJLG8IP&t=5349

~Math-X

https://youtu.be/uxyJGZ3ZYGE

~please like and subscribe

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/SVVSMcw1Xe8

~Education, the Study of Everything

Video Solution 1 by OmegaLearn (Using Similarity)

https://youtu.be/almtw4n-92A

Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)

https://www.youtube.com/watch?v=GTlkTwxSxgo

Video Solution 3 by Magic Square (Using Similarity and Special Value)(best solution)

https://www.youtube.com/watch?v=-N46BeEKaCQ&t=1569s

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=3270

Video Solution by WhyMath

https://youtu.be/roTSeCAehek

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1593s

~harungurcan

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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. AMC logo.png

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