Difference between revisions of "2009 USAMO Problems/Problem 1"

(Solution)
Line 2: Line 2:
 
Given circles <math>\omega_1</math> and <math>\omega_2</math> intersecting at points <math>X</math> and <math>Y</math>, let <math>\ell_1</math> be a line through the center of <math>\omega_1</math> intersecting <math>\omega_2</math> at points <math>P</math> and <math>Q</math> and let <math>\ell_2</math> be a line through the center of <math>\omega_2</math> intersecting <math>\omega_1</math> at points <math>R</math> and <math>S</math>.  Prove that if <math>P, Q, R</math> and <math>S</math> lie on a circle then the center of this circle lies on line <math>XY</math>.
 
Given circles <math>\omega_1</math> and <math>\omega_2</math> intersecting at points <math>X</math> and <math>Y</math>, let <math>\ell_1</math> be a line through the center of <math>\omega_1</math> intersecting <math>\omega_2</math> at points <math>P</math> and <math>Q</math> and let <math>\ell_2</math> be a line through the center of <math>\omega_2</math> intersecting <math>\omega_1</math> at points <math>R</math> and <math>S</math>.  Prove that if <math>P, Q, R</math> and <math>S</math> lie on a circle then the center of this circle lies on line <math>XY</math>.
  
== Solution ==
+
== Solution 1 ==
 
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>),
 
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>),
 
<cmath>O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2</cmath>
 
<cmath>O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2</cmath>
Line 9: Line 9:
  
 
~AopsUser101
 
~AopsUser101
 +
 +
== Solution 2 ==
 +
Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that because
  
 
== See also ==
 
== See also ==

Revision as of 19:15, 8 August 2020

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 1

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

Solution 2

Define $\omega_i$ and $O_i$ similarly to above. Note that because

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png