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 "2005 AMC 10A Problems/Problem 12"

m (→‎Solution)
m (Fixed typo in problem statement)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?
+
The figure shown is called a ''trefoil'' and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>?
  
 
<asy>
 
<asy>

Revision as of 17:25, 1 July 2025

Problem

The figure shown is called a trefoil and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $2$?

[asy] unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(12pt)); pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180); pair E=B+C; draw(D--E--B--O--C--B--A,linetype("4 4")); draw(Arc(O,1,0,60),linewidth(1.2pt)); draw(Arc(O,1,120,180),linewidth(1.2pt)); draw(Arc(C,1,0,60),linewidth(1.2pt)); draw(Arc(B,1,120,180),linewidth(1.2pt)); draw(A--D,linewidth(1.2pt)); draw(O--dir(40),EndArrow(HookHead,4)); draw(O--dir(140),EndArrow(HookHead,4)); draw(C--C+dir(40),EndArrow(HookHead,4)); draw(B--B+dir(140),EndArrow(HookHead,4)); label("2",O,S); draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar); draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar); [/asy]

$\textbf{(A) }\frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \textbf{(B) } \frac{2}{3}\pi\qquad \textbf{(C) } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \textbf{(D) } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \textbf{(E) } \frac{2}{3}\pi+\frac{\sqrt{3}}{2}$

Solution

The area of the trefoil is equal to the area of the big equilateral triangle plus the area of four $60^\circ$ sectors with a radius of $\frac{2}{2}=1$ minus the area of a small equilateral triangle.

This is equivalent to the area of four $60^\circ$ sectors with a radius of $1$.

So the answer is $4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \boxed{\textbf{(B) }\frac{2}{3}\pi}$

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_12&oldid=253060"