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

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One side of an equilateral triangle of side length <math>24</math> lies on line <math>\ell</math>. A circle of radius <math>12</math> is tangent to line <math>\
 
One side of an equilateral triangle of side length <math>24</math> lies on line <math>\ell</math>. A circle of radius <math>12</math> is tangent to line <math>\
ell</math> 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 <math>\ell</math> can be written as <math>a \sqrt{b} - c \pi</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers and <math>b</math> is not divisible by the square of any prime. What is <math>a + b + c</math>?   
+
l</math> 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 <math>\ell</math> can be written as <math>a \sqrt{b} - c \pi</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers and <math>b</math> is not divisible by the square of any prime. What is <math>a + b + c</math>?   
  
 
<math>\textbf{(A)}~72\qquad\textbf{(B)}~73\qquad\textbf{(C)}~74\qquad\textbf{(D)}~75\qquad\textbf{(E)}~76</math>
 
<math>\textbf{(A)}~72\qquad\textbf{(B)}~73\qquad\textbf{(C)}~74\qquad\textbf{(D)}~75\qquad\textbf{(E)}~76</math>

Revision as of 18:00, 8 November 2024

Problem

One side of an equilateral triangle of side length $24$ 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); [/asy]

Solution 1

Draw radii to the tangency points, call the triangle ABC and tangency points D and E

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
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

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