Re: [EMHL] Re: Rhombi problem
- Dear Floor and Paul
>>Let ABC the a triangle. Draw externally on each side a semicircle. If[PY]:
>>we take two sides of the medial triangle, and extend these so that
>>they each meet two of the semicircles we get four points on these
>>semicircles. There is a rhombus tangent to the three semicircles in
>>these four points. Find the center of the rhombus.
>This is the Spieker center of triangle ABC, which is the incenter ofThis circle is the Conway Circle of the medial triangle.
>the medial triangle. In fact, if we consider the six points formed by
>extending all three sides of the medial triangles, the 6 points lie
>on a circle. This circle has the Spieker point as center, radius
>(1/2)sqrt(r^2+s^2), and is orthogonal to each of the three excircles
>of triangle ABC.
See the thread:
I append below JHC's first posting in the thread.
Subject: a nice little theorem
From: John Conway <conway@...>
Date: Mon, 16 Mar 1998 15:06:05 -0500 (EST)
Here's a little theorem I found about a week ago - I don't
know if it's new:
Produce the edges of a triangle ABC to distances
a beyond A, b beyond B, c beyond C
where a,b,c are the edgelenths of the triangle. Then the 6 points
so constructed lie on a circle:
| John Conway