Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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| − | We see that <math>\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG</math> by AA similarity. | + | We see that <math>\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG</math> by AA similarity. |
| + | <math>BE = \frac{3}{2}</math> because <math>AK</math> cuts the side length of the square in half; similarly, <math>CF = 1</math>. Let <math>CG = h</math>: then by side ratios, | ||
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| + | <cmath>\frac{h+2}{h} = \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4</cmath>. | ||
| + | |||
| + | Now the height of the triangle is <math>AG = 4+2+3 = 9</math>. By side ratios, | ||
| + | <math></math>\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}<math>. | ||
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| + | The area of the triangle is </math>AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}$ | ||
| + | |||
| + | ~KingRavi | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 17:02, 24 November 2021
Contents
Problem
A square with side length 
is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length 
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]](http://latex.artofproblemsolving.com/2/0/c/20ced60f6d642645a5e5713fa7723356bdecff14.png)
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]](http://latex.artofproblemsolving.com/c/9/0/c906fe16c95e07afb9d864e41b10b071a70b48a5.png)
We see that 
by AA similarity.
because 
cuts the side length of the square in half; similarly, 
. Let 
: then by side ratios,
![\[\frac{h+2}{h} = \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4\]](http://latex.artofproblemsolving.com/b/5/3/b531469f9e967f6a9e3c9db2f9279b6a477322a5.png)
.
Now the height of the triangle is 
. By side ratios,
$$ (Error compiling LaTeX. Unknown error_msg)\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}$.
The area of the triangle is$ (Error compiling LaTeX. Unknown error_msg)AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}$
~KingRavi
Solution 2
By similarity, the height is 
and the base is 
.
Thus the area is 
, or 
.
~Hefei417, or 陆畅 Sunny from China
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
