Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
Line 42: | Line 42: | ||
<math>A_1=\frac{\sqrt{\left(13\right)\left(7\right)\left(1\right)\left(5\right)}}{4}=\frac{\sqrt{455}}{4}</math> | <math>A_1=\frac{\sqrt{\left(13\right)\left(7\right)\left(1\right)\left(5\right)}}{4}=\frac{\sqrt{455}}{4}</math> | ||
− | <math>A_2=9A_1= | + | <math>A_2=9A_1=\frac{9\sqrt{455}}{4}=\frac{p\sqrt{q}}{r}</math>, thus <math>p=9,q=455,r=4</math> |
− | + | <math>p+q+r=\boxed{468}</math> | |
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com |
Revision as of 01:49, 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
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 |