The discussion on the geometry-puzzles newsgroup was started by John Conway who posed a simple problem. Given
ABC, produce the edges of the triangle to distances a beyond A, b beyond B, c beyond C, where a, b, c are the edge lengths of the triangle. Then the 6 points so constructed lie on a circle. Later, in the course of the discussion, he also mentioned that the sides can be extended by distances of (a+x), (b+x), and (c+x) for any real x.
The vertices of the triangle are draggable as is the point L that measures the add-on distance x (positive or negative.) One gets the greatest benefit from dragging this point when the check box "Clear background" is unchecked. Clearly gives away the background of (and the motivation for) this very nice problem!
One of the messages in the discussion thread read:
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My students and I have had fun with it.
Some reactions though. When I first saw your announcement, my reaction
was "oh no, not _another_ circle." But when I play with it, the circle
seems to want to be there and to have some significance. The math just
works out too nicely; the formula for the radius is also very simple.
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I believe the good teacher was jesting. After all, he took a very active part in the discussion and, as it transpires from his remark, has shared his curiosity with his students. How many teachers do the same? "Oh no, not _another_ circle." - the sentiment is funny, however. Were it not made in jest, there would be a place for a heartfelt sermon on the quality of contemporary math instruction. As it is, here is another sample of Morley's work where he goes beyond the elementary shape of a circle. (But circles pop up anyway.)
These ones have n cusps, and each touches the curve (P) at the point corresponding to the turn t1 (the two curves clearly pass through the same point at t1 and their derivatives only differ by a real factor.) A curve from the family (P1) has a cusp whenever
When one of the turns is allowed to vary, we get an equation of a segment of a straight line - penosculant line. For n fixed turns, there are n penosculant lines that all meet at the above point - penosculant point.
Take (n +1) points on the curve. The penosculant points of the various n points are included in
which is a circle! For (n +2) turns we have (n +2) such circles. With t = - t1t2...tn+1tn+2, we see that all (n +2) circles share a point
Take (n +3) turns. We then have (n +3) such points (2). All of the points are included in
which is a line segment. Do you now expect another "and so on" ? To be sure one is coming, but not yet. For (n +4) turns, (n +4) segments (3) are included in
In the way of example, here is another wonder. Consider (P) with n = 4. Select 4 turns and draw tangents to the curve at the corresponding points. It must be noted that the curve through its point - equation is endowed with direction. The direction is naturally passed on to the tangent lines. Considering the lines as directed, there exists a unique cardioid tangent to the four lines whose direction agrees with the inherited direction of the lines. (For four not directed lines there are 8 such cardioids.) Cardioid is traced by a point on a circle that rolls over a circle of the same radius. The center of the stationary circle is said to be the center of the cardioid.
The center of the cardioid corresponding to 4 turns coincides with their penosculant point! Furthermore, 5 turns define 5 cardioids. Their 5 centers lie on a ... "oh no, not _another_ circle"?
Morley's Miracle
- J.Conway's proof
- D.J.Newman's proof
- Bankoff's proof
- Another proof
- Nikos Dergiades' proof
- G. Zsolt Kiss' proof
- M. T. Naraniengar's proof
- Doodling and Miracles
- Morley's Pursuit of Incidence
- Lines, Circles and Beyond
- On Motivation and Understanding
- Bankoff's Conundrum
- Morley's Redux and More, Alain Connes' proof
- An Unexpected Variant
Copyright © 1996-2008 Alexander Bogomolny