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

Revision as of 15:15, 14 December 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?

$[asy] import geometry; unitsize(0.75cm); filldraw((1,4)--(4,0)--(8,3)--(29/4,4)--cycle,gray+opacity(0.5),invisible); draw((0,0)--(8,0)--(8,4)--(0,4)--cycle,linewidth(1.5)); draw((1,4)--(4,0)--(8,3)--(5,7)--cycle,black+linewidth(1.5)); draw((1,0)--(1,4),linewidth(1.5)); perpendicularmark((1,0),unit(dir(90)+dir(0)),black+linewidth(1.5)); label("\Large8",(4,-0.5),S); label("\Large4",(8.5,2),E); label("\Large5",(3,5.5),NW); [/asy]$

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

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 $\frac{5}{8}$ . Our answer 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)

Video Solution 1

https://youtu.be/xYkSx8h-Ixk

~Education, the Study of Everything

@@ Line 136: / Line 136: @@
 <math>4=-\frac{4}{3}x+12 \frac{1}{3}</math>, so <math>x=6\frac{1}{4}</math> and <math>y=4</math> This would mean <math>G\left(6\frac{1}{4},4\right)</math>.
-Since we have our <math>G</math> coordinate, we can continue with Solution 3, with the area of the trapezoid <math>\left(\frac{(EG+AC)}{2}\right)(CE)</math>, where <math>EG=\frac{5}{4}</math> (using distance formula for <math>E</math> to <math>G</math>), <math>AC=5</math>, and <math>CE=5</math>.
+Since we have our <math>G</math> coordinate, we can continue with Solution 3, with the area of the trapezoid <math>\left(\frac{EG+AC}{2}\right)(CE)</math>, where <math>EG=\frac{5}{4}</math> (using distance formula for <math>E</math> to <math>G</math>), <math>AC=5</math>, and <math>CE=5</math>.
 By substitution, we get <math>\left(\frac{\frac{5}{4}+5}{2}\right)(5)=</math><math>\boxed{\textbf{(D) }15\dfrac{5}{8}}</math>.

2022 AMC 10B (Problems • Answer Key • Resources)
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