Difference between revisions of "2012 AMC 8 Problems/Problem 24"

Revision as of 20:04, 9 November 2013

Problem

A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

$[asy] size(0,50); draw((-1,1)..(-2,2)..(-3,1)..(-2,0)..cycle); dot((-1,1)); dot((-2,2)); dot((-3,1)); dot((-2,0)); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]$

$\textbf{(A)}\hspace{.05in}\frac{4-\pi}{\pi}\qquad\textbf{(B)}\hspace{.05in}\frac{1}\pi\qquad\textbf{(C)}\hspace{.05in}\frac{\sqrt2}{\pi}\qquad\textbf{(D)}\hspace{.05in}\frac{\pi-1}{\pi}\qquad\textbf{(E)}\hspace{.05in}\frac{3}\pi$

Solution

$[asy] dot((0,0),red); dot((0,2),red); dot((2,0),red); dot((2,2),red); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red); size(0,50); draw((1,0){up}..{left}(0,1)); dot((1,0)); dot((0,1)); draw((0,1){right}..{up}(1,2)); dot((1,2)); draw((1,2){down}..{right}(2,1)); dot((2,1)); draw((2,1){left}..{down}(1,0));[/asy]$

Draw a square around the star figure. The sidelength of this square is $4$ , because the sidelength is the diameter of the circle. The square forms $4$ -quarter circles around the star figure. This is the equivalent of one large circle with radius $2$ , meaning that the total area of the quarter circles is $4\pi$ . The area of the square is $16$ . Thus, the area of the star figure is $16 - 4\pi$ . The area of the circle is $4\pi$ . Taking the ratio of the two areas, we find the answer is $\boxed{\textbf{(A)}\ \frac{4-\pi}{\pi}}$ .

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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 AJHSME/AMC 8 Problems and Solutions

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

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Difference between revisions of "2012 AMC 8 Problems/Problem 24"

Revision as of 20:04, 9 November 2013

Problem

Solution

See Also

@@ Line 22: / Line 22: @@
 ==Solution==
+<asy>
+dot((0,0),red);
+dot((0,2),red);
+dot((2,0),red);
+dot((2,2),red);
+draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,red);
+size(0,50);
+draw((1,0){up}..{left}(0,1));
+dot((1,0));
+dot((0,1));
+draw((0,1){right}..{up}(1,2));
+dot((1,2));
+draw((1,2){down}..{right}(2,1));
+dot((2,1));
+draw((2,1){left}..{down}(1,0));</asy>
 Draw a square around the star figure. The sidelength of this square is <math> 4 </math>, because the sidelength is the diameter of the circle. The square forms <math>4</math>-quarter circles around the star figure. This is the equivalent of one large circle with radius <math> 2 </math>, meaning that the total area of the quarter circles is <math> 4\pi </math>. The area of the square is <math> 16 </math>. Thus, the area of the star figure is <math> 16 - 4\pi </math>. The area of the circle is <math> 4\pi </math>. Taking the ratio of the two areas, we find the answer is <math> \boxed{\textbf{(A)}\ \frac{4-\pi}{\pi}} </math>.