Finding Circle Center By Ruler Alone: What is this about?
A Mathematical Droodle

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Copyright © 19962017 Alexander BogomolnyThe applet attempts to kill two birds with one shot. First, it illustrates a ruler only constriction of the center of a circle in the presence of another circle intersecting the first [Steinhaus, p. 142]. Along the way it provides a second solution to a problem discussed elsewhere: If point B is selected on one of the intersecting circles and A and C on the other such that AB and BC pass through the points of intersection of the two circles, then the length of AC does not depend on the position of B on its circle.

Let there be two circles meeting in M and N. Choose point A on the first circle and let B be the second point of intersection of AM with the other circle, and C the second point of intersection of BN with the first. Starting with another point, say A', we get in the same manner points B' and C'. The applet shows that the arcs (and hence the chords) AA' and CC' are equal. (This is so because angles BMB' and BNB' are equal as inscribed into the same circle and subtended by the same arc. Angles AMA' and CNC' are vertical with these angles and are therefore also equal. So arcs AA' and CC' as subtending equal inscribed angles in the same circle are equal. They are located then symmetrically with respect to the diameter orthogonal to either AC' or A'C. The said diameter contains the points of intersection of AC and A'C' and that of AA' and CC'.
On the other hand, adding arc AC' to the equal arcs AA' and CC' we get another pair of equal arcs: AC and A'C'. The chords subtending these arcs are equal is required by the above mentioned problem.
References
 H. Steinhaus, Mathematical Snapshots, umpteen edition, Dover, 1999
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Copyright © 19962017 Alexander Bogomolny62324829 