Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"
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<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> | <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== | ==Solution 1== | ||
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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> | + | <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>. | <cmath>\frac{h+2}{h} = \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4</cmath>. | ||
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<cmath>\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}</cmath>. | <cmath>\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}</cmath>. | ||
− | The area of the triangle is <math>AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{B}</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 | ~KingRavi | ||
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==Solution 2== | ==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 == | + | == 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>. | Denote by <math>O</math> the midpoint of <math>AB</math>. | ||
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~Steven Chen (www.professorchenedu.com) | ~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.