Difference between revisions of "2007 AIME II Problems/Problem 15"
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=== Solution 2 === | === Solution 2 === | ||
[[Image:2007 AIME II-15b.gif]] | [[Image:2007 AIME II-15b.gif]] | ||
+ | |||
Consider a 13-14-15 triangle. <math>A=84.</math> [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.] | Consider a 13-14-15 triangle. <math>A=84.</math> [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.] | ||
The inradius is <math>r=\frac{A}{s}=\frac{84}{21}=4</math>, where <math>s</math> is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, <math>u.</math> | The inradius is <math>r=\frac{A}{s}=\frac{84}{21}=4</math>, where <math>s</math> is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, <math>u.</math> | ||
− | The circumradius is <math>R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are the side-lengths. Scale the triangle with the circumradius by a | + | The circumradius is <math>R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are the side-lengths. Scale the triangle with the circumradius by a [[line]]ar scale factor, <math>v</math>. |
Cut and combine the triangles, as shown. Then solve for 4u: | Cut and combine the triangles, as shown. Then solve for 4u: | ||
− | <math>\frac{65}{8}v=8u</math> | + | :<math>\frac{65}{8}v=8u</math> |
− | <math>v=\frac{64}{65}u</math> | + | :<math>v=\frac{64}{65}u</math> |
− | <math>u+v=1</math> | + | :<math>\displaystyle u+v=1</math> |
− | <math>u+\frac{64}{65}u=1</math> | + | :<math>u+\frac{64}{65}u=1</math> |
− | <math>\frac{129}{65}u=1</math> | + | :<math>\frac{129}{65}u=1</math> |
− | <math>4u=\frac{260}{129}</math> | + | :<math>4u=\frac{260}{129}</math> |
− | <math>260+129=389</math> | + | The solution is <math>260+129=389</math>. |
== See also == | == See also == |
Revision as of 18:43, 30 March 2007
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
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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 |