2007 AIME II Problems/Problem 15
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Problem
Four circles 
and 
with the same radius are drawn in the interior of triangle 
such that 
is tangent to sides 
and 
, 
to 
and 
, to 
and 
, and 
is externally tangent to and . If the sides of triangle are 
and 
the radius of can be represented in the form 
, where 
and 
are relatively prime positive integers. Find 
Contents |
Solution
Solution 1
First, apply Heron's formula to find that the area is 
. Also the semiperimeter is 
. So the inradius is 
.
Now consider the incenter I. Let the radius of one of the small circles be 
. Let the centers of the three little circles tangent to the sides of 
be 
, 
, and 
. Let the centre of the circle tangent to those three circles be P. A homothety centered at 
takes 
to with factor 
. The same homothety takes 
to the circumcentre of , so 
, where 
is the circumradius of . The circumradius of can be easily computed by 
, so doing that reveals 
. Then 
, so the answer is 
.
Solution 2
Consider a 13-14-15 triangle. 
[By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is 
, where 
is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, 
The circumradius is 
where 
and 
are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, 
.
Cut and combine the triangles, as shown. Then solve for 4u:
The solution is 
.
See also
| 2007 AIME II (Problems • Resources) | ||
| Preceded by Problem 14 | Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
