Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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==Problem== | ==Problem== | ||
A square with side length <math>3</math> is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length <math>2</math> 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? | A square with side length <math>3</math> is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length <math>2</math> 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> | ||
+ | |||
+ | |||
+ | |||
+ | <math>(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Let's split the triangle down the middle and label it: | ||
<asy> | <asy> | ||
Line 6: | Line 21: | ||
import olympiad; | import olympiad; | ||
pair A,B,C,D,E,F,G,H,I,J,K; | pair A,B,C,D,E,F,G,H,I,J,K; | ||
− | A = origin; B = (0. | + | 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(A--D--K--cycle); | ||
draw(B--E); | draw(B--E); | ||
Line 14: | Line 29: | ||
draw(I--J); | draw(I--J); | ||
draw(E--H); | 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> | </asy> | ||
+ | We see that <math>\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG</math> by AA similarity. | ||
+ | <math>BE = \frac{3}{2}</math> because <math>AG</math> cuts the side length of the square in half; similarly, <math>CF = 1</math>. Let <math>CG = h</math>: then by side ratios, | ||
+ | |||
+ | <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, | ||
+ | <cmath>\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}</cmath>. | ||
− | <math> | + | The area of the triangle is <math>AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}</math> |
+ | |||
+ | ~KingRavi | ||
− | ==Solution== | + | ==Solution 2== |
By similarity, the height is <math>3+\frac31\cdot2=9</math> and the base is <math>\frac92\cdot1=4.5</math>. | By similarity, the height is <math>3+\frac31\cdot2=9</math> and the base is <math>\frac92\cdot1=4.5</math>. | ||
− | Thus the area is <math>\frac{9\cdot4.5}2=20.25= | + | Thus the area is <math>\frac{9\cdot4.5}2=20.25=\boxed{\textbf{(B) }20 \frac{1}{4}}</math>. |
~Hefei417, or 陆畅 Sunny from China | ~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 <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>. | ||
+ | Hence, <math>\frac{OC}{DF} = \frac{AO}{AD}</math>. | ||
+ | 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) | ||
+ | ----------- | ||
+ | Alternatively, we can find the height in a slightly different way. | ||
+ | |||
+ | Following from our finding that the base of the large triangle <math>AB = \frac{9}{2}</math>, we can label the length of the altitude of <math>\triangle{CHI}</math> as <math>x</math>. Notice that <math>\triangle{CHI} \sim \triangle{CAB}</math>. Hence, <math>\frac{HI}{AB} = \frac{x}{CO}</math>. Substituting and simplifying, <math>\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</math>. Therefore, the area of the triangle is <math>\frac{\frac{9}{2} \cdot 9}{2} = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}</math>. | ||
+ | |||
+ | ~mahaler | ||
+ | |||
+ | ==Solution 4 (Coordinates)== | ||
+ | |||
+ | For convenience, we will use the image provided in the third solution. | ||
+ | |||
+ | We can set <math>O</math> as the origin. | ||
+ | |||
+ | We know that <math>FG = 3</math> and <math>JK = 2</math>. | ||
+ | |||
+ | We subtract <math>JK</math> from <math>FG</math> and divide by <math>2</math> to get <math>KG = FJ = \frac{1}{2}</math>. | ||
+ | |||
+ | Since <math>HIKJ</math> is a square, we know that <math>IK = 2</math>. | ||
+ | |||
+ | Using rise over run, we find that the slope of <math>CB</math> is <math>\frac{-2}{0.5} = -4</math>. | ||
+ | |||
+ | The coordinates of <math>I</math> are <math>(1, 5)</math>. We plug this in to get the equation of the line that <math>CB</math> runs along: <cmath>y = -4x + 9</cmath> | ||
+ | |||
+ | We know that the <math>x-value</math> of <math>C</math> is <math>0</math>. Using this, we find that the <math>y-value</math> is <math>9</math>. So the coordinates of <math>C</math> are <math>(0, 9)</math>. | ||
+ | |||
+ | This gives us the height of <math>\triangle ACB</math>: <math>CO = 9</math>. | ||
+ | |||
+ | Now we need to find the coordinates of <math>B</math>. | ||
+ | |||
+ | We know that the <math>y-value</math> is <math>0</math>. Plugging this in, we find <math>0 = -4x +9</math>, or <math>\frac{9}{4} = x</math>. | ||
+ | |||
+ | The coordinates of <math>B</math> are <math>(\frac{9}{4}, 0)</math>. | ||
+ | |||
+ | Since <math>\triangle ACB</math> is symmetrical along <math>CO</math>, we can multiply <math>CO</math> by <math>OB</math> to get <cmath>9 \cdot \frac{9}{4} = \frac{81}{4}</cmath> | ||
+ | |||
+ | Simplifying, we get <math>\boxed{\textbf{(B) }20 \frac{1}{4}}</math> for the area. | ||
+ | |||
+ | ~Achelois | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://www.youtube.com/watch?v=mq4e-s9ENas | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/gxZE3cscswo | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/Uh5Umekq4A8 | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/R7TwXgAGYuw?t=639 | ||
+ | |||
+ | ~IceMatrix | ||
==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}} |
Latest revision as of 22:41, 10 November 2023
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?
Solution 1
Let's split the triangle down the middle and label it:
We see that by AA similarity. because cuts the side length of the square in half; similarly, . Let : then by side ratios,
.
Now the height of the triangle is . By side ratios, .
The area of the triangle is
~KingRavi
Solution 2
By similarity, the height is and the base is . Thus the area is .
~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 the midpoint of .
Because , , , we have .
We observe . Hence, . Hence, . By symmetry, .
Therefore, .
Because is the midpoint of , .
We observe . Hence, . Hence, .
Therefore, .
Therefore, the answer is .
~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 , we can label the length of the altitude of as . Notice that . Hence, . Substituting and simplifying, . Therefore, the area of the triangle is .
~mahaler
Solution 4 (Coordinates)
For convenience, we will use the image provided in the third solution.
We can set as the origin.
We know that and .
We subtract from and divide by to get .
Since is a square, we know that .
Using rise over run, we find that the slope of is .
The coordinates of are . We plug this in to get the equation of the line that runs along:
We know that the of is . Using this, we find that the is . So the coordinates of are .
This gives us the height of : .
Now we need to find the coordinates of .
We know that the is . Plugging this in, we find , or .
The coordinates of are .
Since is symmetrical along , we can multiply by to get
Simplifying, we get for the area.
~Achelois
Video Solution by Interstigation
https://www.youtube.com/watch?v=mq4e-s9ENas
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/R7TwXgAGYuw?t=639
~IceMatrix
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.