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

(Undo revision 187800 by Themathguyd (talk))
(Tag: Undo)
(This is the asymptote diagram someone asked for)
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>
 +
//Diagram by TheMathGuyd
 +
size(9cm);
 +
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;
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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
 +
path LONE=PONE--(-conj(PONE)); //Hline ONE
 +
path LTWO=PTWO--(-conj(PTWO));
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path T=A--B--C--cycle; //Triangle
 +
 
 +
 
 +
fill (shift(g,0)*(LTWO--B--cycle),grey);
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fill (LONE--A--C--cycle,grey);
 +
 
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draw(LONE);
 +
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);
 +
 
 +
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>
 
(note: diagrams are not necessarily drawn to scale)
 
(note: diagrams are not necessarily drawn to scale)
  

Revision as of 02:31, 25 January 2023

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(9cm); 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),grey); fill (LONE--A--C--cycle,grey); 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] (note: diagrams are not necessarily drawn to scale)

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

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 $[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2$. 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)

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

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

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