# 2022 AMC 10B Problems/Problem 16

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?

$[asy] size(5cm); filldraw((4,0)--(8,3)--(8-3/4,4)--(1,4)--cycle,mediumgray); draw((0,0)--(8,0)--(8,4)--(0,4)--cycle,linewidth(1.1)); draw((1,0)--(1,4)--(4,0)--(8,3)--(5,7)--(1,4),linewidth(1.1)); label("4", (8,2), E); label("8", (4,0), S); label("5", (3,11/2), NW); draw((1,.35)--(1.35,.35)--(1.35,0),linewidth(1.1)); [/asy]$

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

## 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\dfrac{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 $\frac{5}{8}$. The only answer choice that has $\frac{5}{8}$ in it is $\boxed{\textbf{(D) }15\dfrac{5}{8}}$

~mathboy100

## Solution 3 (Coordinate Geometry)

Same diagram as Solution 1, but added point $H$, which is $(4,7)$. I also renamed all the points to form coordinates using $B$ as the origin. $[asy] import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A(0,4)", F, NW); label("B(0,0)", B, S); label("C(3,0)", C, S); label("D(7,0)", D, SE); label("E(7,3)", I, E); label("F(7,4)", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); label("H(4,7)", (4.65, 7.25), E); [/asy]$

In order to find the area, point $G$'s coordinates must be found. Notice how $EH$ and $AG$ intercept at point $G$. This means that we need to find the equations for $EH$ and $AG$ and make a system of linear equations.

Using the slope formula $m=\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$, we get the slope for $EH$, which means $m=\frac{3-7}{7-4} = -\frac{4}{3}$

Then, by using point slope form. $y-y_{1}=m(x-x_{1})$. We can say that the equation for $EH$ is $y-7=-\frac{4}{3}(x-4)$ or in this case, $y=-\frac{4}{3}x+12 \frac{1}{3}$.

And it is easy to figure out that the equation for $AG$ is $y=4$.

The best way to solve the system of linear equations is to substitute the $y$ for the $4$ in equation $EH$. $4=-\frac{4}{3}x+12 \frac{1}{3}$, so $x=6\frac{1}{4}$ and $y=4$ This would mean $G\left(6\frac{1}{4},4\right)$.

Since we have our $G$ coordinate, we can continue with Solution 3, with the area of the trapezoid $\left(\frac{EG+AC}{2}\right)(CE)$, where $EG=\frac{5}{4}$ (using distance formula for $E$ to $G$), $AC=5$, and $CE=5$.

By substitution, we get $\left(\frac{\frac{5}{4}+5}{2}\right)(5)=$$\boxed{\textbf{(D) }15\dfrac{5}{8}}$.

~ghfhgvghj10 (+ minor edits ~TaeKim)

## Solution 4

Notice the small triangle in the upper right corner is a $3-4-5$ triangle. Then that triangle is similar to the big triangle by AA similarity. From that, do similar triangles and you find that the longer leg of the small triangle is 1. Then you find that the triangle below is $3-4-5$, so the side length of the rectangle (without the outer rectangle) is 7. afterwards you just add the half of the square + the remaining triangle which can be found by multiplying base and height (in which we already know)

~mathboy282

## Solution 5 (Fastest Similar Triangles)

For reference, use the points labelled in the diagram of Solution 3. The square means $AC = 5$, so we get a $3$-$4$-$5$ triangle $ABC$.

$m\angle BAC = 90 ^{\circ} - m\angle CAG = m\angle GAH$.

Therefore $\triangle GAH$ is proportional to a $3$-$4$-$5$ triangle, with $HG$ corresponding to $3$ and $AH$ corresponding to $4$. By similar triangles, we find

$HG = AH \cdot \frac{3}{4} = \frac{15}{4}$.

Finally, $[ACEG] = [ACEH] - [\triangle GAH] = 5^2 - \frac{1}{2} \cdot \frac{15}{4} \cdot 5 = \boxed{\textbf{(D) }15\dfrac{5}{8}}.$

~lolsmybagelz

~r00tsOfUnity

## Video Solution

~Education, the Study of Everything

~Ismail.maths

~ pi_is_3.14

~~Hayabusa1

~Interstigation

~IceMatrix

## See Also

 2022 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 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 byProblem 12 Followed byProblem 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.