Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 1"
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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>. | 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 3 Pre 2005 Problems/Problem 2 | Next Problem]] | ||
| + | |||
| + | *[[Mock AIME 3 Pre 2005]] | ||
Revision as of 09:21, 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 
.
