Difference between revisions of "2012 UNCO Math Contest II Problems/Problem 7"
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== Solution == | == Solution == | ||
| + | <asy> | ||
| + | draw((-1,-1)--(7,-1),black); | ||
| + | filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey); | ||
| + | filldraw(circle((0,0),1),white); | ||
| + | filldraw(circle((2*sqrt(3),2),3),white); | ||
| + | </asy> | ||
| + | |||
| + | If we draw in the second tangent between these circles, we get a point where the tangents intersect. Drawing a line through the centers of the circle and the point where the tangents meet, we get two similar triangles in ratio <math>1:3</math>. Let the hypotenuse of the smaller triangle be <math>x</math>. Since the distance between the centers of the circles is <math>1+3=4</math>, we can write the ratio <math>\frac{x}{x+4} = \frac{1}{3}</math>. Solving for <math>x</math>, we get <math>x=2</math>. Since the hypotenuse of the smaller triangle is <math>2</math>, and one of the legs is <math>1</math>, we see that it is a <math>30-60-90</math> triangle. So the side lengths of the smaller triangle is <math>\sqrt{3},1,2</math> and the side lengths of the larger triangle is <math>3\sqrt{3},3,6</math>. Finding the difference between the areas of both triangles, we get <math>4\sqrt3</math> which is the area of the trapezoid. The trapezoid is the area of a <math>120^\circ</math> sector of the smaller circle, a <math>60^\circ</math> sector of the larger circle, and the shaded region. Subtracting the areas of both sectors from the area of the trapezoid, we get <math>4\sqrt{3} - \frac{120}{360} \cdot \pi - \frac{60}{360} \cdot 9\pi = 4\sqrt{3} - \frac{1}{3} \cdot \pi - \frac{1}{6} \cdot 9\pi = \boxed {4\sqrt{3} - \frac{11\pi}{6}}</math> | ||
| + | |||
| + | ~Ultraman | ||
== See Also == | == See Also == | ||
| − | {{ | + | {{UNCO Math Contest box|n=II|year=2012|num-b=6|num-a=8}} |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Latest revision as of 11:57, 2 January 2020
Problem
A circle of radius 
is externally tangent to a circle of radius 
and both circles are tangent to a line. Find the area of the shaded
region that lies between the two circles and the line.
![[asy] draw((-1,-1)--(7,-1),black); filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey); filldraw(circle((0,0),1),white); filldraw(circle((2*sqrt(3),2),3),white); [/asy]](http://latex.artofproblemsolving.com/2/1/7/217135aee72ffa973880c2216bf966bda2b48c44.png)
Solution
If we draw in the second tangent between these circles, we get a point where the tangents intersect. Drawing a line through the centers of the circle and the point where the tangents meet, we get two similar triangles in ratio 
. Let the hypotenuse of the smaller triangle be 
. Since the distance between the centers of the circles is 
, we can write the ratio 
. Solving for , we get 
. Since the hypotenuse of the smaller triangle is 
, and one of the legs is , we see that it is a 
triangle. So the side lengths of the smaller triangle is 
and the side lengths of the larger triangle is 
. Finding the difference between the areas of both triangles, we get 
which is the area of the trapezoid. The trapezoid is the area of a 
sector of the smaller circle, a 
sector of the larger circle, and the shaded region. Subtracting the areas of both sectors from the area of the trapezoid, we get 
~Ultraman
See Also
| 2012 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
