2024 AMC 10A Problems/Problem 14

Problem

One side of an equilateral triangle of height $24$ lies on line $\ell$ . A circle of radius $12$ is tangent to line $\ l$ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line $\ell$ can be written as $a \sqrt{b} - c \pi$ , where $a$ , $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is $a + b + c$ ?

$\textbf{(A)}~72\qquad\textbf{(B)}~73\qquad\textbf{(C)}~74\qquad\textbf{(D)}~75\qquad\textbf{(E)}~76$

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C; path p1, p2, p3; p1 = scale(16)*polygon(3); p2 = Circle((12*sqrt(3),4),12); A = intersectionpoint(p1,p2); B = (8*sqrt(3),-8); C = (12*sqrt(3),-8); Label L1 = Label("$24$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); fill(A--Arc((12*sqrt(3),4),A,C)--B--cycle,yellow); draw(p1^^p2); draw((8*sqrt(3),-8)--(22+8*sqrt(3),-8)); draw((-18,-8)--(-18,16), L=L1, arrow=Arrows(),bar=Bars(15)); dot((12*sqrt(3),4),linewidth(4)); draw((12*sqrt(3),4)--(12+12*sqrt(3),4)); label("$12$",(6+12*sqrt(3),4),1.5S); dot(A^^B^^C,linewidth(4)); [/asy]$

~MRENTHUSIASM

Solution 1

Call the bottom vertices of the triangle $B$ and $C$ (the one closer to the circle is $C$ ) and the top vertex $A$ . The tangency point between the circle and the side of the triangle is $D$ , and the tangency point on line $\ell$ $E$ , and the center of the circle is $O$ .

Draw radii to the tangency points, the arc is $60$ degrees because $\angle ACB$ is $60$ , and since $\angle DCE$ is supplementary, it's $120^{\circ}$ . The sum of the angles in a quadrilateral is $360$ , which means $\angle COD$ is $60^{\circ}$

Triangle ODC is $30$ - $60$ - $90$ triangle so CD is $4\sqrt{3}$ . Since we have $2$ congruent triangles ( $\triangle ODC$ and $\triangle OEC$ ), the combined area of both is $48\sqrt{3}$ . The area of the arc is $144 \cdot \frac{60}{360} \cdot \pi$ which is $24\pi$ , so the answer is $48\sqrt{3}-24\pi$

$a+b+c$ is $48+3+24$ which is $\boxed{\textbf{(D)}~75}$

~ASPALAPATI75 ~andy_liu766 (latex)

edits by KR

Note

There were two possible configurations from this problem; the one described in the solution above and the configuration in which the circle is tangent to the bottom of line $\ell$ and the base of the equilateral triangle. However, since the area in this configuration is simply $0,$ we can infer that the problem is talking about the configuration in Solution 1.

~dbnl

Solution 2 (Quick Guess)

Since this problem involves equilateral triangles, the only possible number under the square root is $3$ . Now subtracting all of the answer choices by $3$ , we get: $\textbf{(A)}~72-3=69\qquad\textbf{(B)}~73-3=70\qquad\textbf{(C)}~74-3=71\qquad\textbf{(D)}~75-3=72\qquad\textbf{(E)}~76-3=73$

Due to the even parity of the problem, we can safely assume that the answer is either $B$ or $D$ , but as $D$ is a multiple of $12$ and $24$ , we get the answer of $\boxed{\textbf{(D)}~75}$ .

~megaboy6679

Solution 3

(pardon the diagrams :D)

say the area we want to find is x.

since the equilateral triangle has an internal angle of 60, the exterior angle formed by the triangle and the line is 120. simplifying the diagram you will get:

\   #####
 \ ########
  \########
   \#####___________

make three of these that each circle is tangent to the other 2 circles

   \
    \    #####
     \ ########
 #####\########
#######\_####__________
#######/ #####
 #####/########
     / ########
    /   ######

Since they are 3 congruent triangles, you can make an equilateral triangle using their radius(12), with each vertex at the center of each circle. This will make an equilateral triangle of side length 24. if you look now, the area within the equilateral triangle consists of 3 $60/360$ of a circle, and 3 of x.

we first find the area of the triangle, which is $24 * 12\sqrt{3}$ , we then find the area of $60/360$ of a circle, which is $60/360 * 12^2\pi$ , we subtract $24 * 12\sqrt{3}$ by $60/360 * 12^2\pi$ , and divide by 3, yielding the area of x.

_________________________________________ = $48\sqrt{3}-24\pi$

$a+b+c$ is $48+3+24$ which is $\boxed{\textbf{(D)}~75}$

~Yiguo Zhang

Solution 4 (oh no)

Setting up the problem graphically:

Define the parametric function of the circle $r (θ)$ , centered at $(0, 12)$ with radius $12$ , as: $r(\theta) \;=\; \bigl(12\cos(\theta),\,12\sin(\theta) + 12\bigr).$

The line $ℓ$ is taken to be the $x$ -axis, so $y = 0$ . The circle is tangent to $ℓ$ at $(0, 0)$ .

We only care about the side of the triangle that is also tangent to the circle; call this line $f (x)$ . Since the triangle is equilateral of height 24, one vertex is at $(0, 24)$ (directly above the tangent point), but for now we focus on the particular side tangent to the circle.

Because we expect that line to make a $330^{\circ}$ angle with the positive $x$ -axis (or a slope of $- \sqrt{3}$ ), let $f(x) \;=\; -\sqrt{3}\,x \;+\; c.$

We also know the perpendicular distance from the circle’s center $(0, 12)$ to $f (x)$ must be $12$ , because the circle of radius $12$ is externally tangent to that line. The distance from a point $(x_{0}, y_{0})$ to a line $A x + B y + C = 0$ is:

$\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.$

Here, $f (x) = - \sqrt{3} x + c$ can be rewritten as $\sqrt{3} x + y - c = 0.$ Plugging in $(x_{0}, y_{0}) = (0, 12)$ and setting the distance to 12 yields:

$\frac{|\sqrt{3}\cdot 0 \;+\; 12 \;-\; c|}{\sqrt{(\sqrt{3})^2 + 1^2}}\;=\;12.$

Solving this and noting that $f (x)$ has a negative $y$ -intercept gives:

$f(x) \;=\; -\sqrt{3}\,x \;-\; 12.$

Next, we find the point at which $r (θ)$ intersects $f (x)$ . Substituting $r (θ) = (12 \cos (θ), 12 \sin (θ) + 12)$ into $f (x) = - \sqrt{3} x - 12,$ we set:

$12\sin(\theta) + 12 \;=\; -\sqrt{3}\,\bigl(12\cos(\theta)\bigr) - 12.$

One helpful approach is to note that a line perpendicular to $f (x)$ and passing through $(0, 12)$ has slope $1 / \sqrt{3}$ , so that line is $g (x) = \frac{x}{\sqrt{3}} + 12.$ Intersecting $g (x)$ with $f (x)$ gives:

$\frac{x}{\sqrt{3}} + 12 \;=\; -\sqrt{3}\,x - 12,$

which solves to $x = - 6 \sqrt{3}$ . Then $y = 6.$ So the intersection point of the circle and the line is $(- 6 \sqrt{3}, 6)$ .

To find the corresponding $θ$ , use: $12\cos(\theta) = -6\sqrt{3}, \quad12\sin(\theta) + 12 = 6.$

From the first, $\cos (θ) = - \sqrt{3} / 2$ . This corresponds to $θ = 5 π / 6$ or $7 π / 6.$ Checking the $y$ -coordinate shows $θ = 7 π / 6$ indeed yields $\sin (7 π / 6) = - 1 / 2$ , so $12 \sin (7 π / 6) + 12 = 6.$

The circle also meets $ℓ$ (the $x$ -axis) at $(0, 0)$ . In polar terms, that is $θ = 3 π / 2$ . We want the area under the circle’s parametric curve from $θ = 7 π / 6$ to $θ = 3 π / 2.$

Parametrically, the area under $y (θ)$ from $θ_{1}$ to $θ_{2}$ can be found via:

$\int_{\theta_1}^{\theta_2} y(\theta)\,x'(\theta)\,d\theta.$

Here $x (θ) = 12 \cos (θ)$ , so $x^{'} (θ) = - 12 \sin (θ)$ . Hence the area under the circle from $θ = 7 π / 6$ to $θ = 3 π / 2$ is:

$\int_{7\pi/6}^{3\pi/2} \bigl(12\sin(\theta) + 12\bigr) \,\bigl(-12\sin(\theta)\bigr)\,d\theta.$

This simplifies (pulling out constants) to:

$-12 \int_{7\pi/6}^{3\pi/2}\bigl(\sin^2(\theta) + \sin(\theta)\bigr)\, d\theta\;=\;-12 \left[\int_{7\pi/6}^{3\pi/2} \sin^2(\theta)\,d\theta\;+\;\int_{7\pi/6}^{3\pi/2} \sin(\theta)\,d\theta\right].$

Evaluating these integrals yields:

$54\sqrt{3} \;-\; 24\pi.$

Next, find where $f (x) = 0$ :

$-\sqrt{3}\,x \;-\; 12 = 0\quad\Longrightarrow\quad x = -\frac{12}{\sqrt{3}} \;=\; -4\sqrt{3}.$

However, we also need the relevant intersection interval for the region. The line and the circle intersect at $x = - 6 \sqrt{3}$ . So the area under $f (x)$ from $x = - 6 \sqrt{3}$ to $x = - 4 \sqrt{3}$ is:

$\int_{x=-6\sqrt{3}}^{-4\sqrt{3}} \bigl(-\sqrt{3}\,x - 12\bigr)\,dx.$

Carrying out that integral (or carefully checking the geometry) gives $6 \sqrt{3} .$

Subtracting this triangular “cap” ( $6 \sqrt{3}$ ) from the circle’s sector area ( $54 \sqrt{3} - 24 π$ ) gives the final region of interest:

$\bigl(54\sqrt{3} - 24\pi\bigr) - 6\sqrt{3}\;=\;48\sqrt{3} - 24\pi.$

Hence the area can be written as

$a\sqrt{b} \;-\; c\pi\;=\;48\sqrt{3} - 24\pi,$

so $a = 48$ , $b = 3$ , and $c = 24.$ Therefore, $a + b + c \;=\; 48 + 3 + 24 \;=\; 75.$

$D) 75$

~meihk_neiht

P.S.: Please don’t whip out calculus like this on an AMC 10. Yes, it’s “doable” without a calculator (I speak from painful experience). Yes, I did it, but it took forever, and trust me, nobody at the AMC office expects calculus.

Video Solution by Number Craft

https://youtu.be/rIF5L_-zZYQ

Video Solution by Pi Academy

https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM

Video Solution 1 by Power Solve

https://youtu.be/oCQ_QvXqV5s

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

Video Solution by Just Math⚡

https://www.youtube.com/watch?v=fzXBMltyXjs&t=53s

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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

Page

Toolbox

Search