Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
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Let <math>h_1</math> be the height of <math>\Delta PDA</math> to side <math>AD</math> | Let <math>h_1</math> be the height of <math>\Delta PDA</math> to side <math>AD</math> | ||
+ | |||
+ | Then, <math>A_1=\frac{|AD| \ times h_1}{2}</math>, thus <math>h_1=\frac{2}{3}A_1</math> | ||
Revision as of 00:31, 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
Invert about a circle with radius 1 and center P. Note that since all relevant circles and lines go through P, they all are transformed into lines, and are all tangent at infinity (i.e. parallel). That was the crux move; some more basic length chasing using similar triangles gets you the answer.
~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 |