Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 1)
Line 32: Line 32:
  
 
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>4\sqrt{3}</math>.  
 
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>4\sqrt{3}</math>.  
−
Since we have 2 congruent triangles (<math>\Delta ODC</math> and <math>\Delta OEC</math>), the combined area of both is <math>48\sqrt{3}</math>.
+
Since we have 2 congruent triangles (<math>\Delta ODC</math> and <math>\Delta OEC</math>), the combined area of both is <math>24\sqrt{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 18:08, 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 $4\sqrt{3}$. Since we have 2 congruent triangles ($\Delta ODC$ and $\Delta OEC$), the combined area of both is $24\sqrt{3}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_14&oldid=232133"