Difference between revisions of "2005 AMC 10A Problems/Problem 12"
(→Video Solution) |
(→Problem) |
||
| Line 2: | Line 2: | ||
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 the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length <math>2</math>? | ||
| − | [[ | + | [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] | ||
<math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math> | <math> \mathrm{(A) \ } \frac{1}{3}\pi+\frac{\sqrt{3}}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\pi\qquad \mathrm{(C) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{4}\qquad \mathrm{(D) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{3}\qquad \mathrm{(E) \ } \frac{2}{3}\pi+\frac{\sqrt{3}}{2} </math> | ||
Revision as of 10:38, 12 July 2021
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 
?
[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]
Solution
The area of the trefoil is equal to the area of a small equilateral triangle plus the area of four 
sectors with a radius of 
minus the area of a small equilateral triangle.
This is equivalent to the area of four sectors with a radius of 
.
So the answer is:
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
