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Difference between revisions of "2022 AMC 10B Problems/Problem 16"

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{{duplicate|[[2022 AMC 10B Problems/Problem 16|2022 AMC 10B #16]] and [[2022 AMC 12B Problems/Problem 13|2022 AMC 12B #13]]}}
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==Problem==
 
==Problem==
  
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</asy>
 
</asy>
  
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Label the points on the diagram.
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Let us label the points on the diagram.
  
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By doing some angle chasing, using the fact that <math>\angle ACE</math> and <math>\angle CEG</math> are right angles, we find that <math>\angle BAC</math>, <math>\angle DCE</math>, and <math>\angle FEG</math> are congruent. Similarly, <math>\angle ACB</math>, <math>\angle CED</math>, and <math>\angle EGF</math> are congruent as well.
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By doing some angle chasing, using the fact that <math>\angle ACE</math> and <math>\angle CEG</math> are right angles, we find that <math>\angle BAC \cong \angle DCE \cong \angle FEG</math>. Similarly, <math>\angle ACB \cong \angle CED \cong \angle EGF</math>. Therefore, <math>\triangle ABC \sim \triangle CDE \sim \triangle EFG</math>.
  
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Therefore, <math>\triangle ABC \sim \triangle CDE \sim \triangle EFG</math>.
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As we are given a rectangle and a square, <math>AB = 4</math> and <math>AC = 5</math>. Therefore, <math>\triangle ABC</math> is a 3-4-5 right triangle, and <math>BC = 3</math>.  
  
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Since the triangles are created by a rectangle and a square, <math>AC = 5</math> and <math>AB = 4</math>. <math>\angle ABC</math> is a right angle, so <math>BC = 3</math>.  
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<math>CE</math> is also <math>5</math>. So, using the similar triangles, <math>CD = 4</math> and <math>DE = 3</math>.
  
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<math>CE</math> is also <math>5</math>, and using the similar triangles, <math>CD = 4</math> and <math>DE = 3</math>.
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<math>EF = DF - DE = 4 - 3 = 1</math>. Using the similar triangles again, <math>EF</math> is <math>\frac14</math> of the corresponding <math>AB</math>. So, <math>[\triangle EFG] = \left(\frac14\right)^2 \cdot [\triangle ABC] = \frac{1}{16} \cdot 12 = \frac38</math>.
  
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<math>EF = 1</math> by subtracting <math>DE</math> from <math>DF</math>. The area of a 3-4-5 right triangle is <math>6</math>, so both <math>\triangle ABC</math> and <math>\triangle CDE</math> have an area of 6. The ratio of <math>EF</math> to <math>CD</math> = <math>1:4</math>, so by similarity, the area of of <math>\triangle EFG</math> is <math>6/4^2 = 3/8</math>.
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We have  
  
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<math>BD = BC + CD = 3 + 4 = 7</math> and <math>DF = 4</math>, so the area of rectangle <math>ABDF</math> is <math>28</math>. Subtracting the area of <math>\triangle ABC, \triangle CDE, </math> and<math> \triangle EFG</math>, gets the shaded area, <math>28 - 6 - 6 - \tfrac{3}{8} = \boxed{\textbf{(D)} 15 \tfrac{5}{8}}</math>
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<cmath>\begin{align*}
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[ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\
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&= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\
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&= 28 - 6 - 6 - \frac38 \\
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&= \boxed{\textbf{(D)}\ 15 \frac{5}{8}}.
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\end{align*}</cmath>
  
 
~Connor132435
 
~Connor132435
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== See Also ==
 +
 +
{{AMC10 box|year=2022|ab=B|num-b=15|num-a=17}}
 +
{{AMC12 box|year=2022|ab=B|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 22:00, 17 November 2022

The following problem is from both the 2022 AMC 10B #16 and 2022 AMC 12B #13, so both problems redirect to this page.

Problem

The diagram below shows a rectangle with side lengths 4 and 8 and a square with side length 5. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

[asy] import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)); label("8",C,S); label("5",(3, 5.5),NW); label("4",(8, 2), E); [/asy]

$\textbf{(A) }15\frac{1}{8}\qquad \textbf{(B) }15\frac{3}{8}\qquad \textbf{(C) }15\frac{1}{2}\qquad \textbf{(D) }15\frac{5}{8}\qquad \textbf{(E) }15\frac{7}{8}$

Solution

[asy] import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A", F, NW); label("B", B, S); label("C", C, S); label("D", D, SE); label("E", I, E); label("F", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); [/asy]

Let us label the points on the diagram.

By doing some angle chasing, using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC \cong \angle DCE \cong \angle FEG$. Similarly, $\angle ACB \cong \angle CED \cong \angle EGF$. Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$.

As we are given a rectangle and a square, $AB = 4$ and $AC = 5$. Therefore, $\triangle ABC$ is a 3-4-5 right triangle, and $BC = 3$.

$CE$ is also $5$. So, using the similar triangles, $CD = 4$ and $DE = 3$.

$EF = DF - DE = 4 - 3 = 1$. Using the similar triangles again, $EF$ is $\frac14$ of the corresponding $AB$. So, $[\triangle EFG] = \left(\frac14\right)^2 \cdot [\triangle ABC] = \frac{1}{16} \cdot 12 = \frac38$.

We have

\begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D)}\ 15 \frac{5}{8}}. \end{align*}

~Connor132435

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
2022 AMC 12B (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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