Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Undo revision 187800 by Themathguyd (talk))
(Tag: Undo)
 
(47 intermediate revisions by 24 users not shown)
Line 2: Line 2:
 
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>?
 
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>?
  
−
*Add asymptote diagram*
+
<asy>
−
(note: diagrams are not necessarily drawn to scale)
+
//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>
  
 
<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>
 
<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>
  
 
==Solution 1==
 
==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 <math>[\text{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>[\text{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>.
+
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>.
  
 
~MathFun1000 (~edits apex304)
 
~MathFun1000 (~edits apex304)
  
−
==Video Solution 1 by OmegaLearn (Using Similarity)==
+
==Solution 2 (Thorough)==
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
Solution by CHECKMATE2021
 +
 
 +
==Solution 3 (Faster)==
 +
 
 +
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>.
 +
 
 +
The area of the triangle in the second diagram is now <math>\frac{(h-5)^2}{2}</math>.
 +
 
 +
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 AND SHOULD ONLY BE USED AS LAST RESORT)==
 +
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
 +
(~edits by KMSONI)
 +
 
 +
==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 Similar Triangles)==
 
https://youtu.be/almtw4n-92A
 
https://youtu.be/almtw4n-92A
  
 
==Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)==
 
==Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)==
 
https://www.youtube.com/watch?v=GTlkTwxSxgo
 
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
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/GnlU-McyPXY
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=23|num-a=25}}
 
{{AMC8 box|year=2023|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:37, 10 November 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 AND SHOULD ONLY BE USED AS LAST RESORT)

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 (~edits by KMSONI)

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 Similar Triangles)

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

Video Solution by Dr. David

https://youtu.be/GnlU-McyPXY

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_24&oldid=233068"