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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 14"
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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>
 
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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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.
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Let <math>A_1</math> and <math>A_2</math> be the areas of triangles <math>PDA</math> and <math>PBC</math> respectively.
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Since <math>O_3R \parallel O_1O_2</math> and <math>\angle RO_3O_2 = \angle O_3O_2O_1</math>, then <math>O_2B \parallel O_1D</math>, and <math>BC \parallel AD</math>
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This means that <math>\Delta PDA ~ \Delta PBC ~ \Delta PDA  O_3O_1O_1</math>.  In other words, those three triangles are similar.
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Revision as of 00:25, 25 November 2023

Problem

Circles $\omega_1$ and $\omega_2$ are centered on opposite sides of line $l$, and are both tangent to $l$ at $P$. $\omega_3$ passes through $P$, intersecting $l$ again at $Q$. Let $A$ and $B$ be the intersections of $\omega_1$ and $\omega_3$, and $\omega_2$ and $\omega_3$ respectively. $AP$ and $BP$ are extended past $P$ and intersect $\omega_2$ and $\omega_1$ at $C$ and $D$ respectively. If $AD = 3, AP = 6, DP = 4,$ and $PQ = 32$, then the area of triangle $PBC$ can be expressed as $\frac{p\sqrt{q}}{r}$, where $p, q,$ and $r$ are positive integers such that $p$ and $r$ are coprime and $q$ is not divisible by the square of any prime. Determine $p + q + r$.

Solution

Let $O_1, O_2,$ and $O_3$ be the centers of $\omega_1, \omega_2$ and $\omega_3$ respectively.

Let point $R$ be the midpoint of $QP$. Thus, $O_3R \bot PQ$ and $|PR|=\frac{|PR|}{2}=16$

Let $r_1$ and $r_2$ be the radii of circles $\omega_1$ and $\omega_2$ respectively.

Let $A_1$ and $A_2$ be the areas of triangles $PDA$ and $PBC$ respectively.

Since $O_3R \parallel O_1O_2$ and $\angle RO_3O_2 = \angle O_3O_2O_1$, then $O_2B \parallel O_1D$, and $BC \parallel AD$

This means that $\Delta PDA ~ \Delta PBC ~ \Delta PDA O_3O_1O_1$. In other words, those three triangles are similar.



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 $\omega_1,\omega_2, l$ 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
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