Difference between revisions of "2024 AMC 10A Problems/Problem 14"

Revision as of 23:21, 9 November 2024

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 $B$ 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 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 ~andliu766 (latex)

edits by 9897

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: $\textbf{(A)}~69\qquad\textbf{(B)}~70\qquad\textbf{(C)}~71\qquad\textbf{(D)}~72\qquad\textbf{(E)}~73$

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

Video Solution by Pi Academy

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

Video Solution 1 by Power Solve

https://youtu.be/oCQ_QvXqV5s

@@ Line 60: / Line 60: @@
 ==Solution 2 (Quick Guess)==
-Since this problem involves equilateral triangles, the only possible number under the square root is <math>3</math>. Now subtracting all of the answer choices by 3, we get:
+Since this problem involves equilateral triangles, the only possible number under the square root is <math>3</math>. Now subtracting all of the answer choices by <math>3</math>, we get:
 <cmath>\textbf{(A)}~69\qquad\textbf{(B)}~70\qquad\textbf{(C)}~71\qquad\textbf{(D)}~72\qquad\textbf{(E)}~73</cmath>

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