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

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Draw radii to the tangency points, the arc is 60 degrees(The angle that the triangle makes with the line is 120 and because the angles of a quadrilateral sum to 360, we get 120+180+x=360 so 60)
 
Draw radii to the tangency points, the arc is 60 degrees(The angle that the triangle makes with the line is 120 and because the angles of a quadrilateral sum to 360, we get 120+180+x=360 so 60)
  
−
Call the bottom vertices B and C (the one closer to the circle be C) and the top vertice A and the tangency point on side of triangle D and on line l E and call center O
+
Call the bottom vertices B and C (the one closer to the circle be C) and the top vertice A and the tangency point on side of triangle D and on line l E and call center O, triangle ODC is 30-60-90 triangle so CD is <math>4sqrt(3)</math>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2024|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:02, 8 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); [/asy]

Solution 1

Draw radii to the tangency points, the arc is 60 degrees(The angle that the triangle makes with the line is 120 and because the angles of a quadrilateral sum to 360, we get 120+180+x=360 so 60)

Call the bottom vertices B and C (the one closer to the circle be C) and the top vertice A and the tangency point on side of triangle D and on line l E and call center O, triangle ODC is 30-60-90 triangle so CD is $4sqrt(3)$

See also

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

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