## Re: [EMHL] Re: Rhombi problem

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• Dear Floor and Paul ... This circle is the Conway Circle of the medial triangle. See the thread: http://forum.swarthmore.edu/epigone/geom.puzzles/wilyangdimp/
Message 1 of 1 , Apr 30, 2001
Dear Floor and Paul

[FvL]:
>>Let ABC the a triangle. Draw externally on each side a semicircle. If
>>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.

[PY]:
>This is the Spieker center of triangle ABC, which is the incenter of
>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.

This circle is the Conway Circle of the medial triangle.

http://forum.swarthmore.edu/epigone/geom.puzzles/wilyangdimp/

I append below JHC's first posting in the thread.

Antreas

__________________________________________________________________

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:
|
| /
| /
| /
|/
B
/|
/ |
/ |
/ |
/ |
/ |
/ |
------------C-------A----------------
/ |
/ |
/ |
/ |
/ |
/ |
|
| John Conway
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