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

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(→‎Solution 4 (FASTEST FOR SURE , complementary counting?))
 
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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]]}}
 +
 
==Problem==
 
==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?
+
The diagram below shows a rectangle with side lengths <math>4</math> and <math>8</math> and a square with side length <math>5</math>. 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>
 
<asy>
−
import olympiad;
+
size(5cm);
−
size(200);
+
filldraw((4,0)--(8,3)--(8-3/4,4)--(1,4)--cycle,mediumgray);
−
defaultpen(linewidth(1) + fontsize(10));
+
draw((0,0)--(8,0)--(8,4)--(0,4)--cycle,linewidth(1.1));
−
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);
+
draw((1,0)--(1,4)--(4,0)--(8,3)--(5,7)--(1,4),linewidth(1.1));
−
fill(F--G--I--C--F--cycle, grey);
+
label("$4$", (8,2), E);
−
draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C));
+
label("$8$", (4,0), S);
−
label("8",C,S);
+
label("$5$", (3,11/2), NW);
−
label("5",(3, 5.5),NW);
+
draw((1,.35)--(1.35,.35)--(1.35,0),linewidth(1.1));
−
label("4",(8, 2), E);
 
 
</asy>
 
</asy>
  
−
<math>\textbf{(A) }15\frac{1}{8}\qquad
+
<math>\textbf{(A) }15\dfrac{1}{8} \qquad
−
\textbf{(B) }15\frac{3}{8}\qquad
+
\textbf{(B) }15\dfrac{3}{8} \qquad
−
\textbf{(C) }15\frac{1}{2}\qquad
+
\textbf{(C) }15\dfrac{1}{2} \qquad
−
\textbf{(D) }15\frac{5}{8}\qquad
+
\textbf{(D) }15\dfrac{5}{8} \qquad
−
\textbf{(E) }15\frac{7}{8}</math>
+
\textbf{(E) }15\dfrac{7}{8} </math>
 +
 
 +
==Solution 1==
  
−
==Solution==
+
Let us label the points on the diagram.
  
 
<asy>
 
<asy>
Line 53: Line 56:
 
</asy>
 
</asy>
  
−
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</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.
+
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 <math>3</math>-<math>4</math>-<math>5</math> right triangle and <math>BC = 3</math>.  
  
−
Therefore, <math>\triangle ABC \sim \triangle CDE \sim \triangle EFG</math>.
+
<math>CE</math> is also <math>5</math>. So, using the similar triangles, <math>CD = 4</math> and <math>DE = 3</math>.
  
−
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>.  
+
<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>CE</math> is also <math>5</math>, and using the similar triangles, <math>CD = 4</math> and <math>DE = 3</math>.
+
<cmath>\begin{align*}
 +
[\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\
 +
&= \frac{1}{16} \cdot 6 \\
 +
&= \frac38.
 +
\end{align*}</cmath>
  
−
<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>.
+
Finally, we have  
  
−
<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>
+
<cmath>\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*}</cmath>
  
 
~Connor132435
 
~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 <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>. The only answer choice that has <math>\frac{5}{8}</math> in it is <math>\boxed{\textbf{(D) }15\dfrac{5}{8}}</math>
 +
 +
~mathboy100
 +
 +
==Solution 3 (Coordinate Geometry)==
 +
 +
Same diagram as Solution 1, but added point <math>H</math>, which is <math>(4,7)</math>. I also renamed all the points to form coordinates using <math>B</math> 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 <math>G</math>'s coordinates must be found. Notice how <math>EH</math> and <math>AG</math> intercept at point <math>G</math>. This means that we need to find the equations for <math>EH</math> and <math>AG</math> and make a system of linear equations.
 +
 +
Using the slope formula <math>m=\frac{y_{2} - y_{1}}{x_{2} - x_{1}}</math>, we get the slope for <math>EH</math>, which means <math>m=\frac{3-7}{7-4} = -\frac{4}{3}</math>
 +
 +
Then, by using point slope form. <math>y-y_{1}=m(x-x_{1})</math>. We can say that the equation for <math>EH</math> is <math>y-7=-\frac{4}{3}(x-4)</math> or in this case, <math>y=-\frac{4}{3}x+12 \frac{1}{3}</math>.
 +
 +
And it is easy to figure out that the equation for <math>AG</math> is <math>y=4</math>.
 +
 +
The best way to solve the system of linear equations is to substitute the <math>y</math> for the <math>4</math> in equation <math>EH</math>.
 +
<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>.
 +
 +
By substitution, we get <math>\left(\frac{\frac{5}{4}+5}{2}\right)(5)=</math><math>\boxed{\textbf{(D) }15\dfrac{5}{8}}</math>.
 +
 +
~ghfhgvghj10 (+ minor edits ~TaeKim)
 +
 +
==Solution 4 (FASTEST FOR SURE , complementary counting?)==
 +
Notice the small triangle in the upper right corner is a <math>3-4-5</math> 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 <math>3-4-5</math>, 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
 +
 +
 +
I found the same solution when solving this, but essentially, we know all triangles in the picture are similar, and we know the lower left hand triangle is a <math>3-4-5</math> triangle, and it is similar to the top triangle, where 4 corresponds to 5, therefore the unknown side length of that triangle is 15/4. Continuing on, the are of the white part of the square is <math>5 * 15/4 * 1/2 = 75/8,  25 - 75/8 = D</math>
 +
 +
 +
-kaiser
 +
 +
==Solution 5 (Fastest Similar Triangles)==
 +
 +
For reference, use the points labelled in the diagram of Solution 3. Let the point one unit to the right of <math>A</math> be <math>A'</math> (so that <math>A'</math> is one of the vertices of the square). The square means <math>A'C = 5</math>, so we get a <math>3</math>-<math>4</math>-<math>5</math> triangle <math>A'BC</math>.
 +
 +
<math>m\angle BA'C = 90 ^{\circ} - m\angle CA'G = m\angle GA'H</math>.
 +
 +
Therefore <math>\triangle GA'H</math> is proportional to a <math>3</math>-<math>4</math>-<math>5</math> triangle, with <math>HG</math> corresponding to <math>3</math> and <math>A'H</math> corresponding to <math>4</math>. By similar triangles, we find
 +
 +
<math>HG = A'H \cdot \frac{3}{4} = \frac{15}{4}</math>.
 +
 +
; then, finally, <math>[A'CEG] = [A'CEH] - [\triangle GA'H] = 5^2 - \frac{1}{2} \cdot \frac{15}{4} \cdot 5 = \boxed{\textbf{(D) }15\dfrac{5}{8}}.</math>
 +
 +
~lolsmybagelz (minor corrections by Technodoggo)
 +
 +
==Video Solution by mop 2024==
 +
https://youtu.be/ezGvZgBLe8k&t=347s
 +
 +
~r00tsOfUnity
 +
 +
==Video Solution==
 +
https://youtu.be/xYkSx8h-Ixk
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by SpreadTheMathLove==
 +
 +
https://www.youtube.com/watch?v=NAu4AKlK-O0&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=2
 +
 +
~Ismail.maths
 +
 +
== Video Solution 3 by OmegaLearn ==
 +
https://youtu.be/mv2tYNhbAfk
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution(1-16)==
 +
https://youtu.be/SCwQ9jUfr0g
 +
 +
~~Hayabusa1
 +
==Video Solution by Interstigation==
 +
https://youtu.be/B1ZjFYRY4-E
 +
 +
~Interstigation
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/lgQaAQjJjEI
 +
 +
~IceMatrix
 +
== 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}}
 +
{{MAA Notice}}

Latest revision as of 19:56, 10 November 2024

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 (FASTEST FOR SURE , complementary counting?)

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


I found the same solution when solving this, but essentially, we know all triangles in the picture are similar, and we know the lower left hand triangle is a $3-4-5$ triangle, and it is similar to the top triangle, where 4 corresponds to 5, therefore the unknown side length of that triangle is 15/4. Continuing on, the are of the white part of the square is $5 * 15/4 * 1/2 = 75/8, 25 - 75/8 = D$


-kaiser

Solution 5 (Fastest Similar Triangles)

For reference, use the points labelled in the diagram of Solution 3. Let the point one unit to the right of $A$ be $A'$ (so that $A'$ is one of the vertices of the square). The square means $A'C = 5$, so we get a $3$-$4$-$5$ triangle $A'BC$.

$m\angle BA'C = 90 ^{\circ} - m\angle CA'G = m\angle GA'H$.

Therefore $\triangle GA'H$ is proportional to a $3$-$4$-$5$ triangle, with $HG$ corresponding to $3$ and $A'H$ corresponding to $4$. By similar triangles, we find

$HG = A'H \cdot \frac{3}{4} = \frac{15}{4}$.

then, finally, $[A'CEG] = [A'CEH] - [\triangle GA'H] = 5^2 - \frac{1}{2} \cdot \frac{15}{4} \cdot 5 = \boxed{\textbf{(D) }15\dfrac{5}{8}}.$

~lolsmybagelz (minor corrections by Technodoggo)

Video Solution by mop 2024

https://youtu.be/ezGvZgBLe8k&t=347s

~r00tsOfUnity

Video Solution

https://youtu.be/xYkSx8h-Ixk

~Education, the Study of Everything

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=NAu4AKlK-O0&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=2

~Ismail.maths

Video Solution 3 by OmegaLearn

https://youtu.be/mv2tYNhbAfk

~ pi_is_3.14

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/B1ZjFYRY4-E

~Interstigation

Video Solution by TheBeautyofMath

https://youtu.be/lgQaAQjJjEI

~IceMatrix

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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