Difference between revisions of "2024 AMC 10A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (→Diagram) |
MRENTHUSIASM (talk | contribs) |
||
| Line 27: | Line 27: | ||
label("$12$",(6+12*sqrt(3),4),1.5S); | label("$12$",(6+12*sqrt(3),4),1.5S); | ||
</asy> | </asy> | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=13|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:59, 8 November 2024
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)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
