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 "2022 AMC 10B Problems/Problem 16"

(→‎Solution 2 (Cheap))
(→‎Solution 2 (Clever))
Line 85: Line 85:
 
(Refer to the diagram above)
 
(Refer to the diagram above)
 
Proceed the same way as solution 1 until you get all of the side lengths. Then, it is clear that due to the answer choices, we only need to find the fractional part of the shaded area. The area of the whole rectangle is integral, as is the area of <math>\triangle ABC</math>, <math>\triangle CDE</math>, and the rectangle to the far left of the diagram. The area of <math>EFG</math> is <math>\frac{3}{8}</math> and thus the fractional part of the answer is <math>\frac{5}{8}</math>. Our answer is <math>\boxed{\textbf{(D)}\ 15 \frac{5}{8}}</math>
 
Proceed the same way as solution 1 until you get all of the side lengths. Then, it is clear that due to the answer choices, we only need to find the fractional part of the shaded area. The area of the whole rectangle is integral, as is the area of <math>\triangle ABC</math>, <math>\triangle CDE</math>, and the rectangle to the far left of the diagram. The area of <math>EFG</math> is <math>\frac{3}{8}</math> and thus the fractional part of the answer is <math>\frac{5}{8}</math>. Our answer is <math>\boxed{\textbf{(D)}\ 15 \frac{5}{8}}</math>
 +
 +
~mathboy100
  
 
== See Also ==
 
== See Also ==

Revision as of 00:02, 18 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 1

Let us label the points on the diagram.

[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]

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

Solution 2 (Clever)

(Refer to the diagram above) Proceed the same way as solution 1 until you get all of the side lengths. Then, it is clear that due to the answer choices, we only need to find the fractional part of the shaded area. The area of the whole rectangle is integral, as is the area of $\triangle ABC$, $\triangle CDE$, and the rectangle to the far left of the diagram. The area of $EFG$ is $\frac{3}{8}$ and thus the fractional part of the answer is $\frac{5}{8}$. Our answer is $\boxed{\textbf{(D)}\ 15 \frac{5}{8}}$

~mathboy100

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_16&oldid=182165"