Two Lines - Two Circles

The applet below illustrates problem 1 from the 2009 USA Mathematical Olympiad:

Given circles C(E) with center E and C(F) with center F, intersecting at points X and Y, let l1 be a line through E intersecting C(F) at points P and Q and let l2 be a line through F intersecting C(E) 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.

(In the applet, the two circles are defined by three points each: X, Y, P, for C(F), and X, Y, R, for C(E). Dragging either X or Y modifies the circles. Dragging either P or R may have different effect depending on which of the boxes at the bottom of the applet is checked. If it's "Adjust circles" then the circles will be modified. If the "Move points" button is checked, P and R would be dragged over existing circles. In addition to C(E) and C(F), the applet displays two extra circles - the circumcircles of PRQ and PRS. You'll want them coalesce to see a hint for a possible solution.)

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Solution

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Copyright © 1996-2018 Alexander Bogomolny

Given circles C(E) with center E and C(F) with center F, intersecting at points X and Y, let l1 be a line through E intersecting C(F) at points P and Q and let l2 be a line through F intersecting C(E) 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.

(In the applet, the two circles are defined by three points each: X, Y, P, for C(F), and X, Y, R, for C(E). Dragging either X or Y modifies the circles. Dragging either P or R may have different effect depending on which of the boxes at the bottom of the applet is checked. If it's "Adjust circles" then the circles will be modified. If the "Move points" button is checked, P and R would be dragged over existing circles. In addition to C(E) and C(F), the applet displays two extra circles - the circumcircles of PRQ and PRS. You'll want them coalesce to see a hint for a possible solution.)

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Let O be the center of the circumcircle of P, Q, R, and S. Then PQ is the common chord of C(F) and C(O). Therefore, PQ ⊥ FO. Similarly, RS ⊥ EO. In other words, FR and EP are two altitudes in ΔEFO. Let H be the orthocenter of the triangle. Then OH is the third altitude and OH ⊥ EF.

On the other hand, line SR (which FR) is the radical axis of two circles, C(E) and C(O). Similarly, EP is the radical axis,radical axis,axis of symmetry,intersection,center line of C(F) and C(O). It follows that the intersection of the two (which is H) is the radical center of the three circles, C(E), C(F) and C(O), implying that H lies on the radical axis C(E) and C(F) which is XY. But the radical axis of two circles is perpendicular,parallel,perpendicular,congruent to the line joining their centers. Since there is only one line through H perpendicular to EF, point O is also on XY and we are done.


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|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2018 Alexander Bogomolny

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