Difference between revisions of "2007 AIME II Problems/Problem 15"
(→Solution 1) |
|||
Line 24: | Line 24: | ||
:<math>v=\frac{64}{65}u</math> | :<math>v=\frac{64}{65}u</math> | ||
− | :<math> | + | :<math>u+v=1</math> |
:<math>u+\frac{64}{65}u=1</math> | :<math>u+\frac{64}{65}u=1</math> | ||
Line 38: | Line 38: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 22:33, 4 July 2013
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
[hide]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 • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.