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Difference between revisions of "2009 USAMO Problems/Problem 1"
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== Solution ==
 
== Solution ==
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Let <math>\omega_3</math> be the [[circumcircle]] of <math>PQRS</math>. Define <math>r_i</math> to be the radius and <math>O_i</math> to be the center of the circle <math>\omega_i, i = 1,2,3</math>. Then <math>O_1</math> lies on the line passing through the intersections of <math>\omega_2, \omega_3</math>, or their [[radical axis]], and similarly <math>O_2</math> lies on the radical axis of <math>\omega_1, \omega_3</math>. Then, the power of <math>O_1</math> with respect to <math>\omega_2,\omega_3</math> are the same, and similarly for <math>O_2</math>:
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Let <math>\omega_3</math> be the [[circumcircle]] of <math>PQRS</math>, <math>r_i</math> to be the radius of <math>\omega_i</math>, and <math>O_i</math> to be the center of the circle <math>\omega_i</math>, where <math>i \in \{1,2,3\}</math>. Note that <math>SR</math> and <math>PQ</math> are the [[radical axis]]es of <math>O_1</math> , <math>O_3</math> and <math>O_2</math> , <math>O_3</math> respectively. Hence, by [[power of a point]](the power of <math>O_1</math> can be expressed using circle <math>\omega_2</math> and <math>\omega_3</math> and the power of <math>O_2</math> can be expressed using circle <math>\omega_1</math> and <math>\omega_3</math>),
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<cmath>\begin{align*}O_1O_2^2 - r_2^2 &= O_1O_3^2 - r_3^2 \\ O_2O_1^2 - r_1^2 &= O_2O_3^2 - r_3^2 \end{align*}</cmath>
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<cmath>O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2</cmath>
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Subtracting gives <math>O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2</math>, so <math>O_3</math> lies on the radical axis of <math>\omega_1,\omega_2</math>. Thus <math>X,Y,O_3</math> are collinear.  
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<cmath> O_2O_1^2 - r_1^2 = O_2O_3^2 - r_3^2</cmath>
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Subtracting these two equations yields that <math>O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2</math>, so <math>O_3</math> must lie on the [[radical axis]] of <math>\omega_1</math> , <math>\omega_2</math>.
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~AopsUser101
  
 
== See also ==
 
== See also ==

Revision as of 14:52, 1 August 2019

Problem

Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$, let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$.

Solution

Let $\omega_3$ be the circumcircle of $PQRS$, $r_i$ to be the radius of $\omega_i$, and $O_i$ to be the center of the circle $\omega_i$, where $i \in \{1,2,3\}$. Note that $SR$ and $PQ$ are the radical axises of $O_1$ , $O_3$ and $O_2$ , $O_3$ respectively. Hence, by power of a point(the power of $O_1$ can be expressed using circle $\omega_2$ and $\omega_3$ and the power of $O_2$ can be expressed using circle $\omega_1$ and $\omega_3$), \[O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2\] \[O_2O_1^2 - r_1^2 = O_2O_3^2 - r_3^2\] Subtracting these two equations yields that $O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2$, so $O_3$ must lie on the radical axis of $\omega_1$ , $\omega_2$.

~AopsUser101

See also

2009 USAMO (Problems • Resources)
Preceded by
First question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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