2024 AMC 10A Problems/Problem 14
Contents
Problem
One side of an equilateral triangle of height 
lies on line 
. A circle of radius 
is tangent to line 
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 can be written as 
, where 
, 
, and 
are positive integers and is not divisible by the square of any prime. What is 
?
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]](http://latex.artofproblemsolving.com/0/8/8/08897d33f83b1f9d61320a1bcda07dd1b3bac590.png)
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 
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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