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

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== Solution 2 ==
 
== Solution 2 ==
Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that <math>O_1O_3</math> is perpendicular to <math>RS</math> and <math>O_2 O_3</math> is perpendicular to <math>PQ</math>. Thus, the intersection of PQ and RS must be the orthocenter of triangle <math>O_1O_2O_3</math>. Define this as point <math>H</math>. Extending line <math>O_3H</math> to meet <math>O_1O_2</math>, we note that <math>O_3H</math> is perpendicular to <math>O_1O_2</math>.
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Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that <math>O_1O_3</math> is perpendicular to <math>RS</math> and <math>O_2 O_3</math> is perpendicular to <math>PQ</math>. Thus, the intersection of <math>PQ</math> and <math>RS</math> must be the orthocenter of triangle <math>O_1O_2O_3</math>. Define this as point <math>H</math>. Extending line <math>O_3H</math> to meet <math>O_1O_2</math>, we note that <math>O_3H</math> is perpendicular to <math>O_1O_2</math>.
  
 
In addition, note that by the radical axis theorem, the intersection of <math>PQ</math> and <math>RS</math> must also lie on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>. Because the radical axis of <math>\omega_1</math> and <math>\omega_2</math> is perpendicular to <math>O_1O_2</math> and contains <math>H</math>, it must also contain <math>O_3</math>, and we are done.
 
In addition, note that by the radical axis theorem, the intersection of <math>PQ</math> and <math>RS</math> must also lie on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>. Because the radical axis of <math>\omega_1</math> and <math>\omega_2</math> is perpendicular to <math>O_1O_2</math> and contains <math>H</math>, it must also contain <math>O_3</math>, and we are done.
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~notverysmart
  
 
== See also ==
 
== See also ==

Revision as of 20:24, 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 $O_1O_3$ is perpendicular to $RS$ and $O_2 O_3$ is perpendicular to $PQ$. Thus, the intersection of $PQ$ and $RS$ must be the orthocenter of triangle $O_1O_2O_3$. Define this as point $H$. Extending line $O_3H$ to meet $O_1O_2$, we note that $O_3H$ is perpendicular to $O_1O_2$.

In addition, note that by the radical axis theorem, the intersection of $PQ$ and $RS$ must also lie on the radical axis of $\omega_1$ and $\omega_2$. Because the radical axis of $\omega_1$ and $\omega_2$ is perpendicular to $O_1O_2$ and contains $H$, it must also contain $O_3$, and we are done.

~notverysmart

See also

2009 USAMO (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
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All USAMO Problems and Solutions

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