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 "2024 AMC 12B Problems/Problem 21"
(Added Solution 7 to 2024 AMC 12B Problem 21.)
 
(15 intermediate revisions by 7 users not shown)
Line 18: Line 18:
  
 
~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
 
~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
 +
 +
==Solution 2 (Complex Number)==
 +
The smallest angle of <math>3-4-5</math> triangle can be viewed as the arguement of <math>4+3i</math>, and the smallest angle of <math>5-12-13</math> triangle can be viewed as the arguement of <math>12+5i</math>.
 +
 +
Hence, if we assume the ratio of the two shortest length of the last triangle is <math>1:k</math> (<math>k</math> being some rational number), then we can derive the following formula of the sum of their arguement.
 +
Since their arguement adds up to <math>\frac{\pi}{2}</math>, it's the arguement of <math>i</math>. Hence, <cmath>\left(4+3i\right)\left(5+12i\right)\left(k+i\right)=ni\,,</cmath> where <math>n</math> is some real number.
 +
 +
Solving the equation, we get <cmath>56k-33=0\,,\quad 33k+56=n\,.</cmath> Hence <math>k=\frac{33}{56}</math>
 +
 +
Since the sidelength of the theird triangle are co-prime integers, two of its sides are <math>33</math> and <math>56</math>. And the last side is <math>\sqrt{33^2+56^2}=65</math>, hence, the parameter of the third triangle if <math>33+56+65=\boxed{\mathbf{(C)}\,154}</math>.
 +
 +
~Prof. Joker
 +
 +
==Solution 3 (Another Trig)==
 +
 +
Denote the smallest angle of the <math>3-4-5</math> triangle as <math>\alpha</math>, the smallest angle of the <math>5-12-13</math> triangle as <math>\beta</math>, and the smallest angle of the triangle we are trying to solve for as <math>\theta</math>. We then have
 +
<cmath>\alpha + \beta + \theta = 90</cmath>
 +
<cmath>\alpha + \beta = 90 - \theta</cmath>
 +
<cmath>\sin{(\alpha + \beta)} = \sin{(90 - \theta)} = \cos{\theta}</cmath>
 +
<cmath>\cos{\theta} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta} = (\frac{3}{5})(\frac{12}{13}) + (\frac{4}{5})(\frac{5}{13}) = \frac{56}{65}</cmath>
 +
Taking the hypotenuse to be <math>65</math> and one of the legs to be <math>56</math>, we compute the last leg to be <math>\sqrt{65^2 - 56^2} = \sqrt{(65-56)(65+56)} = \sqrt{9*121} = 3*11 = 33</math>
 +
 +
Giving us a final answer of <math>65 + 56 + 33 = \boxed{\textbf{(C) }154}</math>.
 +
 +
~tkl
 +
 +
===Solution 3.1 (Different Flavor of the same thing)===
 +
 +
Consider <math>\sin(\theta) = \sin(90 - (\alpha + \beta)) = \cos(\alpha + \beta) = \frac{33}{65}</math> using the cosine addition identity. Instead of using the Pythagorean theorem, we can use Euclid's formula since we're dealing with primitive triples.
 +
 +
<cmath>65 = a^2 + b^2</cmath>
 +
<cmath>33 = a^2 - b^2</cmath>
 +
 +
Combining that, we get <math>a = 7</math> and <math>b=4</math>. Using this, we can get that the other leg must be <math>2ab = 56</math>. We add the lengths and get that the perimeter is <math>56 + 65 + 33 = \boxed{154}</math>.
 +
 +
~ sxbuto
 +
 +
==Solution 4 (Similarity)==
 +
[[File:Pithagor triangles 13 5 65.png|300px|right]]
 +
Let's arrange the triangles <math>BCD (5-12-13), BCE (9-12-15)</math> and <math>ABE</math> as shown in the diagram.
 +
<cmath>F = AE \cap BC.</cmath>
 +
<cmath>AE \perp AB, DB \perp AB \implies \triangle BCD \sim \triangle FCE \sim \triangle FAB \implies</cmath>
 +
<cmath>EF = \frac{9 \cdot 13}{5}, CF = \frac{9 \cdot 12}{5}, BF = BC + CF = \frac{12 \cdot 14}{5},</cmath>
 +
<cmath>\frac {AB}{CE} = \frac {BF}{EF} \implies AB =  \frac{12 \cdot 14}{13},</cmath>
 +
<cmath>AE = AF - EF = BF \cdot \frac {12}{13} - EF = \frac {99}{13} \implies</cmath>
 +
<cmath>AE : AB : BE = 99 : 12 \cdot 14 : 15 \cdot 13 = 33 : 56 : 65 \implies 65 + 56 + 33 = \boxed{\textbf{(C) }154}</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Solution 5 (Complex)==
 +
Suppose the triangle has legs <math>a, b</math>. We want
 +
<cmath>\arctan \frac{3}{4} + \arctan\frac{5}{12} + \arctan\frac{a}{b} = \frac{\pi}{2}.</cmath>
 +
This is equivalent to
 +
<cmath>
 +
\begin{align*}
 +
    \arg{e^{i\arctan\frac{3}{4}}\cdot e^{i\arctan\frac{5}{12}}\cdot e^{i\arctan\frac{a}{b}}} &= \frac{\pi}{2}\\
 +
    \arg(4+3i)(12+5i)(b+ai) &= \frac{\pi}{2}\\ 
 +
  \arg\left( (33b -56a)+(33a + 56b)i\right) &= \frac{\pi}{2}
 +
\end{align*}
 +
</cmath>
 +
Since the argument of this complex number is <math>\frac{\pi}{2},</math> its real part must be <math>0</math>. Matching real and imaginary parts yields <math>33b = 56a,</math> or <math>b = \frac{56}{33}a</math>. The smallest pair <math>(a, b)</math> that works is <math>(33, 56),</math> which yields a hypotenuse of <math>65.</math> The perimeter of this triangle is <math>33 + 56 + 65 = \boxed{\textbf{(C)\ 154}}.</math>
 +
 +
-Benedict T (countmath1)
 +
 +
==Solution 6 (Law of Cosines)==
 +
We can start by scaling the <math>3-4-5</math> right triangle up by a factor of <math>3</math> and "gluing" it to the <math>5-12-13</math> triangle's longer leg. Let <math>\alpha</math>, <math>\beta</math>, and <math>\theta</math> be the smallest angles in the <math>3-4-5</math>, <math>5-12-13</math> and unknown triangle respectively. We can construct the following diagram of the two known triangles.
 +
<asy>
 +
pair A = (0,0);
 +
pair B = (5,0);
 +
pair C = (14,0);
 +
pair D = (5,12);
 +
 +
draw(A--C--D--cycle);
 +
draw(B--D);
 +
 +
draw(rightanglemark(A,B,D,20));
 +
 +
label("5", A--B, S);
 +
label("9", B--C, S);
 +
label("15", C--D, NE);
 +
label("13", D--A, NW);
 +
label("12", B--D, E);
 +
 +
label("$\beta$", D, 6*dir(257));
 +
label("$\alpha$", D, 4*dir(287));
 +
</asy>
 +
 +
We can also construct a diagram for the third, unknown triangle like so. We know that <math>\alpha+\beta+\theta=90</math>, and therefore that <math>\theta=90-\alpha-\beta</math>. We also know that the other acute angle in this third triangle will have a measure of <math>\alpha+\beta</math> by the triangle angle sum theorem.
 +
<asy>
 +
pair A = (0,0);
 +
pair B = (56,0);
 +
pair C = (56,33);
 +
 +
draw(A--B--C--cycle);
 +
 +
draw(rightanglemark(A,B,C,50));
 +
 +
label("$90-\alpha-\beta$", A, 8*dir(12));
 +
label("$\alpha+\beta$", C, 6*dir(242));
 +
</asy>
 +
We can use the law of cosines on the triangle in the first diagram to get the equation <math>14^2=15^2+13^2-2(13)(15)\cos{(\alpha+\beta)}</math>. Isolating <math>\cos{(\alpha+\beta)}</math>, we get <math>-198=-390\cos{(\alpha+\beta)}</math>, which further simplifies to <math>\cos{(\alpha+\beta)}=\frac{198}{390}</math>. Since the third triangle has to be a primitive Pythagorean triple, we must take this trig ratio into its most simple form, namely <math>\cos{(\alpha+\beta)}=\frac{33}{65}</math>. Using this information in our second diagram, we know that the smaller, adjacent leg must have length <math>33</math>, and the hypotenuse must have length <math>65</math>. We can then use the Pythagorean theorem to find the other, unknown leg, which has length <math>56</math>. Adding these three lengths together, we find that the perimeter of this right triangle is <math>33+56+65=\boxed{\textbf{(C)\ 154}}</math>.
 +
 +
~Phinetium
 +
 +
===Solution 6.1 (Faster Ending)===
 +
Instead of computing <math>\sqrt{65^2-33^2}</math> to find the second leg in the unknown triangle by hand, we can use process of elimination. <math>\textbf{(B)}\ 126</math> and <math>\textbf{(C)}\ 154</math> are the only answers within the realm of possibility, because <math>\textbf{(A)}\ 40</math>  would entitle a triangle with a negative side length, and <math>\textbf{(D)}\ 176</math> and <math>\textbf{(E)}\ 208</math> would require legs greater than the length of the hypotenuse. The answer choice <math>\textbf{(B)}\ 126</math> would force the second leg to have a length of <math>28</math>, which is smaller than the smallest leg in the triangle. (We know that the leg with length <math>33</math> is the smallest leg in the triangle by the side-angle relationship theorem, because it is opposite the smallest angle in the triangle.) Therefore, the only valid answer choice remaining is <math>\boxed{\textbf{(C)\ 154}}</math>.
 +
 +
~Phinetium (again)
 +
 +
==Solution 7 (No Trig, Coord Bash)==
 +
[[File:AMC 12B 2024 Problem 21.png|None]]
 +
 +
Set up a coordinate system. Let <math>(0, 0)</math>, <math>(4, 0)</math>, and <math>(0, 3)</math> be the vertices of the base <math>3-4-5</math> right triangle. In this case, the three smallest angles will all be at <math>(4, 0)</math>, and one of the coordinates of the unknown triangle has to lie on the line <math>x = 4</math>. Now, scale down the <math>5-12-13</math> triangle by <math>\frac{5}{12}</math> so that the new sides are <math>\frac{25}{12}-5-\frac{65}{12}</math>, and place the side with length 5 at the coordinates <math>(0, 3)</math> and <math>(4, 0)</math>. The line passing through these two points can be written as <math>y = -\frac{3}{4}x + 3</math>, so the perpendicular of this line at <math>(0, 3)</math> can be written as <math>y = \frac{4}{3}x + 3</math>. Since the length of the other side is <math>\frac{25}{12}</math>, after drawing smaller <math>3-4-5</math> right triangles, we find that the third coordinate of the <math>\frac{25}{12}-5-\frac{65}{12}</math> is at <math>(\frac{5}{4}, \frac{14}{3})</math>. This coordinate will be one of the coordinates for our unknown triangle. We can place the other coordinate of the unknown triangle at <math>(4, \frac{14}{3})</math> and the third is, by definition, at <math>(4, 0)</math>. The distance from <math>(\frac{5}{4}, \frac{14}{3})</math> to <math>(4, \frac{14}{3})</math> is <math>\frac{11}{4}</math>, and the distance from <math>(4, \frac{14}{3})</math> to <math>(4, 0)</math> is <math>\frac{14}{3}</math>, and from before, the distance from <math>(\frac{5}{4}, \frac{14}{3})</math> to <math>(4, 0)</math> is <math>\frac{65}{12}</math>. Scaling up the sides so that they are integers, we see that the side lengths make a <math>33-56-65</math> right triangle. The perimeter is then <math>33+56+65=\boxed{\textbf{(C)\ 154}}</math>.
 +
 +
~mathwizard123123
  
 
==Video Solution by Innovative Minds==
 
==Video Solution by Innovative Minds==
 
https://youtu.be/9PMdtwkKTlU
 
https://youtu.be/9PMdtwkKTlU
  
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=cyiF8_5fEsM
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2024|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:39, 16 July 2025

Problem

The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?

$\textbf{(A) } 40 \qquad\textbf{(B) } 126 \qquad\textbf{(C) } 154 \qquad\textbf{(D) } 176 \qquad\textbf{(E) } 208$

Solution 1

Let $\alpha$ and $\beta$ be the smallest angles of the $3-4-5$ and $5-12-13$ triangles respectively. We have \[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\] Then \[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\] Let $\theta$ be the smallest angle of the third triangle. Consider \[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\] In order for this to be undefined, we need \[1-\frac{56}{33}\cdot\tan{\theta}=0\] so \[\tan{\theta}=\frac{33}{56}\] Hence the base side lengths of the third triangle are $33$ and $56$. By the Pythagorean Theorem, the hypotenuse of the third triangle is $65$, so the perimeter is $33+56+65=\boxed{\textbf{(C) }154}$.

~kafuu_chino

Solution 2 (Complex Number)

The smallest angle of $3-4-5$ triangle can be viewed as the arguement of $4+3i$, and the smallest angle of $5-12-13$ triangle can be viewed as the arguement of $12+5i$.

Hence, if we assume the ratio of the two shortest length of the last triangle is $1:k$ ($k$ being some rational number), then we can derive the following formula of the sum of their arguement. Since their arguement adds up to $\frac{\pi}{2}$, it's the arguement of $i$. Hence, \[\left(4+3i\right)\left(5+12i\right)\left(k+i\right)=ni\,,\] where $n$ is some real number.

Solving the equation, we get \[56k-33=0\,,\quad 33k+56=n\,.\] Hence $k=\frac{33}{56}$

Since the sidelength of the theird triangle are co-prime integers, two of its sides are $33$ and $56$. And the last side is $\sqrt{33^2+56^2}=65$, hence, the parameter of the third triangle if $33+56+65=\boxed{\mathbf{(C)}\,154}$.

~Prof. Joker

Solution 3 (Another Trig)

Denote the smallest angle of the $3-4-5$ triangle as $\alpha$, the smallest angle of the $5-12-13$ triangle as $\beta$, and the smallest angle of the triangle we are trying to solve for as $\theta$. We then have \[\alpha + \beta + \theta = 90\] \[\alpha + \beta = 90 - \theta\] \[\sin{(\alpha + \beta)} = \sin{(90 - \theta)} = \cos{\theta}\] \[\cos{\theta} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta} = (\frac{3}{5})(\frac{12}{13}) + (\frac{4}{5})(\frac{5}{13}) = \frac{56}{65}\] Taking the hypotenuse to be $65$ and one of the legs to be $56$, we compute the last leg to be $\sqrt{65^2 - 56^2} = \sqrt{(65-56)(65+56)} = \sqrt{9*121} = 3*11 = 33$

Giving us a final answer of $65 + 56 + 33 = \boxed{\textbf{(C) }154}$.

~tkl

Solution 3.1 (Different Flavor of the same thing)

Consider $\sin(\theta) = \sin(90 - (\alpha + \beta)) = \cos(\alpha + \beta) = \frac{33}{65}$ using the cosine addition identity. Instead of using the Pythagorean theorem, we can use Euclid's formula since we're dealing with primitive triples.

\[65 = a^2 + b^2\] \[33 = a^2 - b^2\]

Combining that, we get $a = 7$ and $b=4$. Using this, we can get that the other leg must be $2ab = 56$. We add the lengths and get that the perimeter is $56 + 65 + 33 = \boxed{154}$.

~ sxbuto

Solution 4 (Similarity)

Pithagor triangles 13 5 65.png

Let's arrange the triangles $BCD (5-12-13), BCE (9-12-15)$ and $ABE$ as shown in the diagram. \[F = AE \cap BC.\] \[AE \perp AB, DB \perp AB \implies \triangle BCD \sim \triangle FCE \sim \triangle FAB \implies\] \[EF = \frac{9 \cdot 13}{5}, CF = \frac{9 \cdot 12}{5}, BF = BC + CF = \frac{12 \cdot 14}{5},\] \[\frac {AB}{CE} = \frac {BF}{EF} \implies AB = \frac{12 \cdot 14}{13},\] \[AE = AF - EF = BF \cdot \frac {12}{13} - EF = \frac {99}{13} \implies\] \[AE : AB : BE = 99 : 12 \cdot 14 : 15 \cdot 13 = 33 : 56 : 65 \implies 65 + 56 + 33 = \boxed{\textbf{(C) }154}\] vladimir.shelomovskii@gmail.com, vvsss

Solution 5 (Complex)

Suppose the triangle has legs $a, b$. We want \[\arctan \frac{3}{4} + \arctan\frac{5}{12} + \arctan\frac{a}{b} = \frac{\pi}{2}.\] This is equivalent to \begin{align*} \arg{e^{i\arctan\frac{3}{4}}\cdot e^{i\arctan\frac{5}{12}}\cdot e^{i\arctan\frac{a}{b}}} &= \frac{\pi}{2}\\ \arg(4+3i)(12+5i)(b+ai) &= \frac{\pi}{2}\\ \arg\left( (33b -56a)+(33a + 56b)i\right) &= \frac{\pi}{2} \end{align*} Since the argument of this complex number is $\frac{\pi}{2},$ its real part must be $0$. Matching real and imaginary parts yields $33b = 56a,$ or $b = \frac{56}{33}a$. The smallest pair $(a, b)$ that works is $(33, 56),$ which yields a hypotenuse of $65.$ The perimeter of this triangle is $33 + 56 + 65 = \boxed{\textbf{(C)\ 154}}.$

-Benedict T (countmath1)

Solution 6 (Law of Cosines)

We can start by scaling the $3-4-5$ right triangle up by a factor of $3$ and "gluing" it to the $5-12-13$ triangle's longer leg. Let $\alpha$, $\beta$, and $\theta$ be the smallest angles in the $3-4-5$, $5-12-13$ and unknown triangle respectively. We can construct the following diagram of the two known triangles. [asy] pair A = (0,0); pair B = (5,0); pair C = (14,0); pair D = (5,12); draw(A--C--D--cycle); draw(B--D); draw(rightanglemark(A,B,D,20)); label("5", A--B, S); label("9", B--C, S); label("15", C--D, NE); label("13", D--A, NW); label("12", B--D, E); label("$\beta$", D, 6*dir(257)); label("$\alpha$", D, 4*dir(287)); [/asy]

We can also construct a diagram for the third, unknown triangle like so. We know that $\alpha+\beta+\theta=90$, and therefore that $\theta=90-\alpha-\beta$. We also know that the other acute angle in this third triangle will have a measure of $\alpha+\beta$ by the triangle angle sum theorem. [asy] pair A = (0,0); pair B = (56,0); pair C = (56,33); draw(A--B--C--cycle); draw(rightanglemark(A,B,C,50)); label("$90-\alpha-\beta$", A, 8*dir(12)); label("$\alpha+\beta$", C, 6*dir(242)); [/asy] We can use the law of cosines on the triangle in the first diagram to get the equation $14^2=15^2+13^2-2(13)(15)\cos{(\alpha+\beta)}$. Isolating $\cos{(\alpha+\beta)}$, we get $-198=-390\cos{(\alpha+\beta)}$, which further simplifies to $\cos{(\alpha+\beta)}=\frac{198}{390}$. Since the third triangle has to be a primitive Pythagorean triple, we must take this trig ratio into its most simple form, namely $\cos{(\alpha+\beta)}=\frac{33}{65}$. Using this information in our second diagram, we know that the smaller, adjacent leg must have length $33$, and the hypotenuse must have length $65$. We can then use the Pythagorean theorem to find the other, unknown leg, which has length $56$. Adding these three lengths together, we find that the perimeter of this right triangle is $33+56+65=\boxed{\textbf{(C)\ 154}}$.

~Phinetium

Solution 6.1 (Faster Ending)

Instead of computing $\sqrt{65^2-33^2}$ to find the second leg in the unknown triangle by hand, we can use process of elimination. $\textbf{(B)}\ 126$ and $\textbf{(C)}\ 154$ are the only answers within the realm of possibility, because $\textbf{(A)}\ 40$ would entitle a triangle with a negative side length, and $\textbf{(D)}\ 176$ and $\textbf{(E)}\ 208$ would require legs greater than the length of the hypotenuse. The answer choice $\textbf{(B)}\ 126$ would force the second leg to have a length of $28$, which is smaller than the smallest leg in the triangle. (We know that the leg with length $33$ is the smallest leg in the triangle by the side-angle relationship theorem, because it is opposite the smallest angle in the triangle.) Therefore, the only valid answer choice remaining is $\boxed{\textbf{(C)\ 154}}$.

~Phinetium (again)

Solution 7 (No Trig, Coord Bash)

None

Set up a coordinate system. Let $(0, 0)$, $(4, 0)$, and $(0, 3)$ be the vertices of the base $3-4-5$ right triangle. In this case, the three smallest angles will all be at $(4, 0)$, and one of the coordinates of the unknown triangle has to lie on the line $x = 4$. Now, scale down the $5-12-13$ triangle by $\frac{5}{12}$ so that the new sides are $\frac{25}{12}-5-\frac{65}{12}$, and place the side with length 5 at the coordinates $(0, 3)$ and $(4, 0)$. The line passing through these two points can be written as $y = -\frac{3}{4}x + 3$, so the perpendicular of this line at $(0, 3)$ can be written as $y = \frac{4}{3}x + 3$. Since the length of the other side is $\frac{25}{12}$, after drawing smaller $3-4-5$ right triangles, we find that the third coordinate of the $\frac{25}{12}-5-\frac{65}{12}$ is at $(\frac{5}{4}, \frac{14}{3})$. This coordinate will be one of the coordinates for our unknown triangle. We can place the other coordinate of the unknown triangle at $(4, \frac{14}{3})$ and the third is, by definition, at $(4, 0)$. The distance from $(\frac{5}{4}, \frac{14}{3})$ to $(4, \frac{14}{3})$ is $\frac{11}{4}$, and the distance from $(4, \frac{14}{3})$ to $(4, 0)$ is $\frac{14}{3}$, and from before, the distance from $(\frac{5}{4}, \frac{14}{3})$ to $(4, 0)$ is $\frac{65}{12}$. Scaling up the sides so that they are integers, we see that the side lengths make a $33-56-65$ right triangle. The perimeter is then $33+56+65=\boxed{\textbf{(C)\ 154}}$.

~mathwizard123123

Video Solution by Innovative Minds

https://youtu.be/9PMdtwkKTlU

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=cyiF8_5fEsM

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_21&oldid=254405"