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Difference between revisions of "2022 AMC 10B Problems/Problem 16"
Line 57: Line 57:
 
Let us label the points on the diagram.
 
Let us label the points on the diagram.
  
−
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>.
+
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>.
  
−
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>.  
+
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>.  
  
 
<math>CE</math> is also <math>5</math>. So, using the similar triangles, <math>CD = 4</math> and <math>DE = 3</math>.
 
<math>CE</math> is also <math>5</math>. So, using the similar triangles, <math>CD = 4</math> and <math>DE = 3</math>.
  
−
<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>.
+
<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,  
  
−
We have  
+
<cmath>\begin{align*}
 +
[\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\
 +
&= \frac{1}{16} \cdot 6 \\
 +
&= \frac38.
 +
\end{align*}</cmath>
 +
 
 +
Finally, we have  
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}

Revision as of 23:05, 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,

\begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*}

Finally, 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

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