Difference between revisions of "2007 AIME II Problems/Problem 15"
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The solution is <math>260+129=\boxed{389}</math>. | The solution is <math>260+129=\boxed{389}</math>. | ||
− | === Solution 3 === | + | === Solution 3 (elementary)=== |
Let <math>A'</math>, <math>B'</math>, <math>C'</math>, and <math>O</math> be the centers of circles <math>\omega_{A}</math>, <math>\omega_{B}</math>, <math>\omega_{C}</math>, <math>\omega</math>, respectively, and let <math>x</math> be their radius. | Let <math>A'</math>, <math>B'</math>, <math>C'</math>, and <math>O</math> be the centers of circles <math>\omega_{A}</math>, <math>\omega_{B}</math>, <math>\omega_{C}</math>, <math>\omega</math>, respectively, and let <math>x</math> be their radius. | ||
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Solving, we get <math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. | Solving, we get <math>x=\frac{260}{129}</math>, so our answer is <math>260+129=\boxed{389}</math>. | ||
+ | |||
=== Diagram for Solution 1 === | === Diagram for Solution 1 === | ||
Revision as of 14:30, 25 January 2021
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 semiperimeter is
, so the inradius is
.
Now consider the incenter of
. 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 center of the circle tangent to those three circles be
. The homothety
maps
to
; since
,
is the circumcenter of
and
therefore maps the circumcenter of
to
. Thus,
, where
is the circumradius of
. Substituting
,
and 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 :
The solution is .
Solution 3 (elementary)
Let ,
,
, and
be the centers of circles
,
,
,
, respectively, and let
be their radius.
Now, triangles and
are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for
.
Since ,
is the circumcenter of triangle
and its circumradius is
. Let
denote the incenter of triangle
and
the inradius of
. Then the inradius of
, so now we compute r. Computing the inradius by
, we find that the inradius of
is
. Additionally, using the circumradius formula
where
is the area of
and
is the circumradius, we find
. Now we can equate the ratio of circumradius to inradius in triangles
and
.
Solving, we get , so our answer is
.
Diagram for Solution 1
Here is a diagram illustrating solution 1. Note that unlike in the solution refers to the circumcenter of
. Instead,
is used for the center of the third circle,
.
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.