Difference between revisions of "2024 AMC 10A Problems/Problem 14"
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== Solution 1 == | == Solution 1 == | ||
| + | Draw radii to the tangency points, call the triangle ABC and tangency points D and E | ||
==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 17:57, 8 November 2024
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, call the triangle ABC and tangency points D and E
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
