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

Revision as of 23:47, 29 January 2022

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?

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

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

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 $AK$ 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 (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 $\frac{\frac{9}{2} \cdot 9}{2} = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}$ .

~mahaler

Video Solution by Interstigation

https://www.youtube.com/watch?v=mq4e-s9ENas

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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