Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 1"
m (Mock AIME 3 Pre 2005/Problem 1 moved to Mock AIME 3 Pre 2005 Problems/Problem 1: same reason as before) |
m |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
| − | {{ | + | Let the radius of the third circle be <math>r</math>. The side lengths of the triangle are <math>10</math>, <math>3+r</math>, and <math>7+r</math>. From Heron's Formula, <math>84=\sqrt{(10+r)(r)(7)(3)}</math>, or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(10+r)</math>. <math>84*4=14*24</math>, so <math>r=14</math>. Thus the area of the circle is <math>\boxed{196}\pi</math>. |
| − | ==See | + | ==See Also== |
| + | {{Mock AIME box|year=Pre 2005|n=3|before=First Question|num-a=2}} | ||
Latest revision as of 10:22, 4 April 2012
Problem
Three circles are mutually externally tangent. Two of the circles have radii 
and 
. If the area of the triangle formed by connecting their centers is 
, then the area of the third circle is 
for some integer 
. Determine .
Solution
Let the radius of the third circle be 
. The side lengths of the triangle are 
, 
, and 
. From Heron's Formula, 
, or 
, or 
. 
, so 
. Thus the area of the circle is 
.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
