Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
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==Solution== | ==Solution== | ||
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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. | ||
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| + | Let point <math>R</math> be the midpoint of <math>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PR|}{2}=16</math> | ||
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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 <math>\omega_1,\omega_2, l</math> 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. | 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 <math>\omega_1,\omega_2, l</math> 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. | ||
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| + | ~Tomas Diaz. orders@tomasdiaz.com | ||
==See Also== | ==See Also== | ||
{{Mock AIME box|year=Pre 2005|n=3|num-b=13|num-a=15}} | {{Mock AIME box|year=Pre 2005|n=3|num-b=13|num-a=15}} | ||
Revision as of 01:19, 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 
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 | ||
