2021 Fall AMC 10B Problems/Problem 13


A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?

[asy] //diagram by kante314  draw((0,0)--(8,0)--(4,8)--cycle, linewidth(1.5));  draw((2,0)--(2,4)--(6,4)--(6,0)--cycle, linewidth(1.5));  draw((3,4)--(3,6)--(5,6)--(5,4)--cycle, linewidth(1.5)); [/asy]

$(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34$

Solution 1

Let's split the triangle down the middle and label it:

[asy]  import olympiad; pair A,B,C,D,E,F,G,H,I,J,K; A = origin; B = (0.5,0); C=(2.5,0); D=(3,0); E = (0.5,2); F=(0.83333333333,2); G=(2.166666666667,2); H=(2.5,2); I=(0.83333333333,3.333333333333); J=(2.166666666667,3.3333333333); K=(1.5,6); draw(A--D--K--cycle); draw(B--E); draw(C--H); draw(F--I); draw(G--J); draw(I--J); draw(E--H); draw(K--(1.5,0)); label("$A$",(1.5,0),S); label("$B$",(1.5,2),SW); label("$C$",(1.5,3.3333333),SW); label("$D$",D,SE); label("$E$",H,SE); label("$F$",J,SE); label("$G$",K,N);    [/asy]

We see that $\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG$ by AA similarity. $BE =  \frac{3}{2}$ because $AG$ cuts the side length of the square in half; similarly, $CF = 1$. Let $CG = h$: then by side ratios,

\[\frac{h+2}{h} =  \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4\].

Now the height of the triangle is $AG = 4+2+3 =  9$. By side ratios, \[\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}\].

The area of the triangle is $AG\cdot AD = 9 \cdot \frac{9}{4}  = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}$


Solution 2

By similarity, the height is $3+\frac31\cdot2=9$ and the base is $\frac92\cdot1=4.5$. Thus the area is $\frac{9\cdot4.5}2=20.25=\boxed{\textbf{(B) }20 \frac{1}{4}}$.

~Hefei417, or 陆畅 Sunny from China

Solution 3 (With two different endings)

This solution is based on this figure: Image:2021_AMC_10B_(Nov)_Problem_13,_sol.png

Denote by $O$ the midpoint of $AB$.

Because $FG = 3$, $JK = 2$, $FJ = KG$, we have $FJ = \frac{1}{2}$.

We observe $\triangle ADF \sim \triangle FJH$. Hence, $\frac{AD}{FJ} = \frac{FD}{HJ}$. Hence, $AD = \frac{3}{4}$. By symmetry, $BE = AD = \frac{3}{4}$.

Therefore, $AB = AD + DE + BE = \frac{9}{2}$.

Because $O$ is the midpoint of $AB$, $AO = \frac{9}{4}$.

We observe $\triangle AOC \sim \triangle ADF$. Hence, $\frac{OC}{DF} = \frac{AO}{AD}$. Hence, $OC = 9$.

Therefore, ${\rm Area} \ \triangle ABC = \frac{1}{2} AB \cdot OC = \frac{81}{4} = 20 \frac{1}{4}$.

Therefore, the answer is $\boxed{\textbf{(B) }20 \frac{1}{4}}$.

~Steven Chen (www.professorchenedu.com)

Alternatively, we can find the height in a slightly different way.

Following from our finding that the base of the large triangle $AB = \frac{9}{2}$, we can label the length of the altitude of $\triangle{CHI}$ as $x$. Notice that $\triangle{CHI} \sim \triangle{CAB}$. Hence, $\frac{HI}{AB} = \frac{x}{CO}$. Substituting and simplifying, $\frac{HI}{AB} = \frac{x}{CO} \Rightarrow \frac{2}{\frac{9}{2}} = \frac{x}{x+5} \Rightarrow \frac{x}{x+5} = \frac{4}{9} \Rightarrow x = 4 \Rightarrow CO = 4 + 5 = 9$. Therefore, the area of the triangle is $\frac{\frac{9}{2} \cdot 9}{2} = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}$.


Solution 4 (Coordinates)

For convenience, we will use the image provided in the third solution.

We can set $O$ as the origin.

We know that $FG = 3$ and $JK = 2$.

We subtract $JK$ from $FG$ and divide by $2$ to get $KG = FJ = \frac{1}{2}$.

Since $HIKJ$ is a square, we know that $IK = 2$.

Using rise over run, we find that the slope of $CB$ is $\frac{-2}{0.5} = -4$.

The coordinates of $I$ are $(1, 5)$. We plug this in to get the equation of the line that $CB$ runs along: \[y = -4x + 9\]

We know that the $x-value$ of $C$ is $0$. Using this, we find that the $y-value$ is $9$. So the coordinates of $C$ are $(0, 9)$.

This gives us the height of $\triangle ACB$: $CO = 9$.

Now we need to find the coordinates of $B$.

We know that the $y-value$ is $0$. Plugging this in, we find $0 = -4x +9$, or $\frac{9}{4} = x$.

The coordinates of $B$ are $(\frac{9}{4}, 0)$.

Since $\triangle ACB$ is symmetrical along $CO$, we can multiply $CO$ by $OB$ to get \[9 \cdot \frac{9}{4} = \frac{81}{4}\]

Simplifying, we get $\boxed{\textbf{(B) }20 \frac{1}{4}}$ for the area.


Video Solution by Interstigation


Video Solution


~Education, the Study of Everything

Video Solution by WhyMath



Video Solution by TheBeautyofMath



See Also

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 AMC 10 Problems and Solutions

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