|
|
| (2 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | == Problem 18 ==
| + | #redirect [[2011 AMC 12A Problems/Problem 11]] |
| − | Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?
| |
| − | <asy> pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1)); </asy>
| |
| − | | |
| − | <math> \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\pi}{2} </math>
| |
| − | [[Category: Introductory Geometry Problems]] | |
| − | | |
| − | == Solution 1 ==
| |
| − | | |
| − | Not specific: Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shaded area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}</math>.
| |
| − | | |
| − | == Solution 2 (by artemispi) ==
| |
| − | YOU NEED TO EDIT THIS ARTEMSPI!!!
| |
| − | | |
| − | The quarter circles in circles A and B that overlap with circle C contain the leaf-shaped regions that are in circle C but outside of the shaded area. Draw a right triangle with sides that are radii of circle A, but the hypotenuse connects the two tips of the leaf-shaped area. Now we can find the area of half the leaf-shaped section by subtracting the area of the triangle from the area of the quarter circle. This total area of the two leaf-shaped regions is <math>\frac{\(pi (1)^2}{4}-\frac{\1}{4}) \cdot 4={pi-2}</math>. Therefore, the area of the shaded region is pi-(pi-2)=\boxed{ \mathbf{(C)} 2}$.
| |
| − | | |
| − | == See Also ==
| |
| − | | |
| − | | |
| − | {{AMC10 box|year=2011|ab=A|num-b=17|num-a=19}}
| |
| − | {{MAA Notice}}
| |
