Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 5 (interesting))
 
(20 intermediate revisions by 12 users not shown)
Line 3: Line 3:
  
 
<asy>
 
<asy>
−
 
+
//diagram by kante314
−
import olympiad;
+
draw((0,0)--(8,0)--(4,8)--cycle, linewidth(1.5));  
−
pair A,B,C,D,E,F,G,H,I,J,K;
+
draw((2,0)--(2,4)--(6,4)--(6,0)--cycle, linewidth(1.5));  
−
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((3,4)--(3,6)--(5,6)--(5,4)--cycle, linewidth(1.5));
−
draw(A--D--K--cycle);
 
−
draw(B--E);
 
−
draw(C--H);
 
−
draw(F--I);
 
−
draw(G--J);
 
−
draw(I--J);
 
−
draw(E--H);
 
−
 
 
−
 
 
 
</asy>
 
</asy>
  
Line 21: Line 12:
  
 
<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==
Line 53: Line 43:
  
 
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,
+
<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>.
Line 60: Line 50:
 
<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
Line 66: Line 56:
 
==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=20\frac14</math>, or <math>\boxed{(\textbf{B})}</math>.  
+
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)==
−
[[:Image:2021_AMC_10B_(Nov)_Problem_13,_sol.png]]
+
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>.
  
Line 94: Line 85:
  
 
~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
 +
 +
==Solution 5 (interesting)==
 +
Call the right triangle to the right of the square with side length 2 triangle <math>a_1</math>. Similarly, call the triangle to the right of the square with side length 3 <math>a_2</math>. Label the ENTIRE triangle triangle ABC. We notice AA similarity between <math>a_1</math> and <math>a_2</math>. Additionally, the base of <math>a_1</math> must have a length of 0.5, because the middle square takes up a length of 2 and the other length of 1 must be evenly split between the two congruent triangles next to the square, being all situated on a length of 3. We then see that the ratio of 2:1 must hold for the base of triangle ABC. 3:x = 2:1, and we find that x = 1.5. The length of the base is then 1.5 + 3 = 9/2.
 +
 +
We can then find the height of the triangle by thinking about a geometrical sequence. It must be possible that you could continue inscribing squares in the triangles that are made, as every triangle created is similar to the one before it. Using this, we have the geometrical sequence of 3 + 2 + 4/3 +... (for each side length is the height that is being added) Using the geometrical sequence formula, we find 3/(1 - 2/3) = 3/(1/3) = 3(3)/1 = 9. The height is 9 whilst the base length is 9/2, and multiplying both values and dividing by 2 yields 9(9)/2(2) = 81/4 = <math>\boxed{\textbf{(B) }20 \frac{1}{4}}</math>
 +
 +
~ martianrunner (i do NOT know how to latex)
 +
 +
==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 17:50, 19 September 2024

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 $AG$ 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 $AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}$

~KingRavi

Solution 2

By similarity, the height is $3+\frac31\cdot2=9$ and the base is $\frac92\cdot1=4.5$. Thus the area is $\frac{9\cdot4.5}2=20.25=\boxed{\textbf{(B) }20 \frac{1}{4}}$.

~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

Solution 4 (Coordinates)

For convenience, we will use the image provided in the third solution.

We can set $O$ as the origin.

We know that $FG = 3$ and $JK = 2$.

We subtract $JK$ from $FG$ and divide by $2$ to get $KG = FJ = \frac{1}{2}$.

Since $HIKJ$ is a square, we know that $IK = 2$.

Using rise over run, we find that the slope of $CB$ is $\frac{-2}{0.5} = -4$.

The coordinates of $I$ are $(1, 5)$. We plug this in to get the equation of the line that $CB$ runs along: \[y = -4x + 9\]

We know that the $x-value$ of $C$ is $0$. Using this, we find that the $y-value$ is $9$. So the coordinates of $C$ are $(0, 9)$.

This gives us the height of $\triangle ACB$: $CO = 9$.

Now we need to find the coordinates of $B$.

We know that the $y-value$ is $0$. Plugging this in, we find $0 = -4x +9$, or $\frac{9}{4} = x$.

The coordinates of $B$ are $(\frac{9}{4}, 0)$.

Since $\triangle ACB$ is symmetrical along $CO$, we can multiply $CO$ by $OB$ to get \[9 \cdot \frac{9}{4} = \frac{81}{4}\]

Simplifying, we get $\boxed{\textbf{(B) }20 \frac{1}{4}}$ for the area.

~Achelois

Solution 5 (interesting)

Call the right triangle to the right of the square with side length 2 triangle $a_1$. Similarly, call the triangle to the right of the square with side length 3 $a_2$. Label the ENTIRE triangle triangle ABC. We notice AA similarity between $a_1$ and $a_2$. Additionally, the base of $a_1$ must have a length of 0.5, because the middle square takes up a length of 2 and the other length of 1 must be evenly split between the two congruent triangles next to the square, being all situated on a length of 3. We then see that the ratio of 2:1 must hold for the base of triangle ABC. 3:x = 2:1, and we find that x = 1.5. The length of the base is then 1.5 + 3 = 9/2.

We can then find the height of the triangle by thinking about a geometrical sequence. It must be possible that you could continue inscribing squares in the triangles that are made, as every triangle created is similar to the one before it. Using this, we have the geometrical sequence of 3 + 2 + 4/3 +... (for each side length is the height that is being added) Using the geometrical sequence formula, we find 3/(1 - 2/3) = 3/(1/3) = 3(3)/1 = 9. The height is 9 whilst the base length is 9/2, and multiplying both values and dividing by 2 yields 9(9)/2(2) = 81/4 = $\boxed{\textbf{(B) }20 \frac{1}{4}}$

~ martianrunner (i do NOT know how to latex)

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_13&oldid=227564"