Difference between revisions of "2015 UNCO Math Contest II Problems/Problem 2"
(→Solution) |
Clarkculus (talk | contribs) (→Solution) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 14: | Line 14: | ||
== Solution == | == Solution == | ||
| − | <math>4\pi-\pi=3\pi</math> | + | We can translate the right shaded semicircle along the diameter of the largest circle so that the shaded region becomes a circle (with an unshaded semicircle). We then translate the right smallest semicircle along the diameter of the largest circle onto the left smallest semicircle. This creates a shaded circle of radius 2 and an unshaded circle of radius 1, whereby it is obvious that the area is <math>4\pi-\pi=\boxed{3\pi}</math>. |
== See also == | == See also == | ||
Latest revision as of 08:47, 6 March 2024
Problem
![[asy] filldraw(circle((0,0),3),white); filldraw(arc((-1,0),2,0,180)--cycle,grey); filldraw(arc((-2,0),1,0,180)--cycle,white); filldraw(arc((1,0),2,180,360)--cycle,grey); filldraw(arc((2,0),1,180,360)--cycle,white); [/asy]](http://latex.artofproblemsolving.com/9/0/9/90940c195c7ddf633a9ce23edeb34c20870c8462.png)
Find the area of the shaded region. The outer circle has radius 
. The shaded region is outlined by half circles whose radii are 
and 
and whose centers lie on the dashed diameter of the big circle.
Solution
We can translate the right shaded semicircle along the diameter of the largest circle so that the shaded region becomes a circle (with an unshaded semicircle). We then translate the right smallest semicircle along the diameter of the largest circle onto the left smallest semicircle. This creates a shaded circle of radius 2 and an unshaded circle of radius 1, whereby it is obvious that the area is 
.
See also
| 2015 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
