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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"

(→‎Solution 1)
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~Hefei417, or 陆畅 Sunny from China
 
~Hefei417, or 陆畅 Sunny from China
 +
 +
== Solution 3 ==
 +
Denote by <math>O</math> the midpoint of <math>AB</math>.
 +
 +
Because <math>FG = 3</math>, <math>JK = 2</math>, <math>FJ = KG</math>, we have <math>FJ = \frac{1}{2}</math>.
 +
 +
We observe <math>\triangle ADF \sim \triangle FJH</math>.
 +
Hence, <math>\frac{AD}{FJ} = \frac{FD}{HJ}</math>.
 +
Hence, <math>AD = \frac{3}{4}</math>.
 +
By symmetry, <math>BE = AD = \frac{3}{4}</math>.
 +
 +
Therefore, <math>AB = AD + DE + BE = \frac{9}{2}</math>.
 +
 +
Because <math>O</math> is the midpoint of <math>AB</math>, <math>AO = \frac{9}{4}</math>.
 +
 +
We observe <math>\triangle AOC \sim \triangle ADF</math>.
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Hence, <math>\frac{OC}{DF} = \frac{AO}{AD}</math>.
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Hence, <math>OC = 9</math>.
 +
 +
Therefore, <math>{\rm Area} \ \triangle ABC = \frac{1}{2} AB \cdot OC = \frac{81}{4} = 20 \frac{1}{4}</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(B) }20 \frac{1}{4}}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=14|num-b=12}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=14|num-b=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:46, 25 November 2021

Problem

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] import olympiad; pair A,B,C,D,E,F,G,H,I,J,K; A = origin; B = (0.25,0); C=(1.25,0); D=(1.5,0); E = (0.25,1); F=(0.4166666667,1); G=(1.08333333333,1); H=(1.25,1); I=(0.4166666667,1.66666666667); J=(1.08333333333,1.666666666667); K=(0.75,3); draw(A--D--K--cycle); draw(B--E); draw(C--H); draw(F--I); draw(G--J); draw(I--J); draw(E--H); [/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 $AK$ 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 $AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}$

~KingRavi

Solution 2

By similarity, the height is $3+\frac31\cdot2=9$ and the base is $\frac92\cdot1=4.5$. Thus the area is $\frac{9\cdot4.5}2=20.25=20\frac14$, or $\boxed{(\textbf{B})}$.

~Hefei417, or 陆畅 Sunny from China

Solution 3

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)

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
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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