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)
m
Line 34: Line 34:
  
  
−
Draw radii to the tangency points, the arc is 60 degrees because <math>\angle ACB</math> is 60, and since <math>\angle DCE</math> is supplementary, it's <math>120^{\circ}</math>. The sum of the angles in a quadrilateral is 360, which means <math>\angle COD</math> is <math>60^{\circ}</math>
+
Draw radii to the tangency points, the arc is 60 degrees because <math>\angle ACB</math> is <math>60</math>, and since <math>\angle DCE</math> is supplementary, it's <math>120^{\circ}</math>. The sum of the angles in a quadrilateral is <math>360</math>, which means <math>\angle COD</math> is <math>60^{\circ}</math>
  
  
−
Triangle ODC is 30-60-90 triangle so CD is <math>4\sqrt{3}</math>.  
+
Triangle ODC is <math>30</math>-<math>60</math>-<math>90</math> 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 <math>2</math> congruent triangles (<math>\Delta ODC</math> and <math>\Delta OEC</math>), the combined area of both is <math>48\sqrt{3}</math>.   
 
The area of the arc is <math>144*60/360*\pi</math> which is <math>24\pi</math>, so the answer is <math>48\sqrt{3}-24\pi</math>
 
The area of the arc is <math>144*60/360*\pi</math> which is <math>24\pi</math>, so the answer is <math>48\sqrt{3}-24\pi</math>
  
Line 46: Line 46:
  
 
~ASPALAPATI75
 
~ASPALAPATI75
 +
~andliu766 (latex)
  
 
edits by 9897
 
edits by 9897

Revision as of 19:41, 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

Call the bottom vertices $B$ and $C$ (the one closer to the circle is $C$) and the top vertice $A$. The tangency point between the circle and the side of triangle is $D$, and the tangency point on line $\ell$ $E$, and the center of the circle is $O$


Draw radii to the tangency points, the arc is 60 degrees because $\angle ACB$ is $60$, and since $\angle DCE$ is supplementary, it's $120^{\circ}$. The sum of the angles in a quadrilateral is $360$, which means $\angle COD$ is $60^{\circ}$


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 $48\sqrt{3}$. The area of the arc is $144*60/360*\pi$ which is $24\pi$, so the answer is $48\sqrt{3}-24\pi$


$a+b+c$ is $48+3+24$ which is $\textbf{(D)}~75$


~ASPALAPATI75 ~andliu766 (latex)

edits by 9897

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=232234"