Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
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Let <math>O_1, O_2,</math> and <math>O_3</math> be the centers of <math>\omega_1, \omega_2</math> and <math>\omega_3</math> respectively. | Let <math>O_1, O_2,</math> and <math>O_3</math> be the centers of <math>\omega_1, \omega_2</math> and <math>\omega_3</math> respectively. | ||
− | Let point <math>R</math> be the midpoint of <math>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{| | + | Let point <math>R</math> be the midpoint of <math>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PQ|}{2}=16</math> |
Let <math>r_1</math> and <math>r_2</math> be the radii of circles <math>\omega_1</math> and <math>\omega_2</math> respectively. | Let <math>r_1</math> and <math>r_2</math> be the radii of circles <math>\omega_1</math> and <math>\omega_2</math> respectively. |
Revision as of 01:51, 25 November 2023
Problem
Circles and
are centered on opposite sides of line
, and are both tangent to
at
.
passes through
, intersecting
again at
. Let
and
be the intersections of
and
, and
and
respectively.
and
are extended past
and intersect
and
at
and
respectively. If
and
, then the area of triangle
can be expressed as
, where
and
are positive integers such that
and
are coprime and
is not divisible by the square of any prime. Determine
.
Solution
Let and
be the centers of
and
respectively.
Let point be the midpoint of
. Thus,
and
Let and
be the radii of circles
and
respectively.
Let and
be the areas of triangles
and
respectively.
Since and
, then
, and
This means that . In other words, those three triangles are similar.
Since is the circumcenter of
,
then
Let be the height of
to side
Then, , thus
Using similar triangles,
Therefore,
By similar triangles,
Using Heron's formula,
, where
we have:
, thus
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |