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Miquel Circles

Date: 09/17/2003 at 12:16:18
From: Jadwiga
Subject: Geometry

There is an acute triangle ABC. We consider all equilateral triangles 
XYZ, where points A, B and C lie on segments XZ, XY, YZ.  Prove that 
all centers of gravity of all these triangles XYZ lie on one circle.

I tried to work out equation for points of the triangles using the
requirement that they are at equal distances from each other
(|XY|=|YZ|=|XZ|), and that the segments must lie on the same line
segments as the points A, B, C (each point on different segment), and
that these straights must cross at 60 degrees. 

But the calculations were too difficult.  There were equations with
more than 20 numbers and unknowns to the powers of four.  Please help
me find another way of proving that fact.


Date: 09/17/2003 at 14:58:41
From: Doctor Floor
Subject: Re: Geometry

Dear Jadwiga,

Thanks for your question.

I assume that A lies on YZ, B on XZ and C on XY, as in the figure below: 

  

Note that angles AZB, BXC and CYA are fixed.  That means that X, Y and 
Z run through circles (inscribed angle theorem).

We can also see that when Y moves on its circle through 2 degrees 
(seen as a central angle) then angle CAY increases/decreases 1 degree, 
but angle BCX increases/decreases 1 degree as well, and thus X moves 
through 2 degrees on its circle as well, in the same direction as Y 
does.  So X and Y move at the same angular speed through their 
circles.  Similarly Z moves through its circle at the same angular 
speed.

Now we know that the angle bisector of angle Y (in triangle XYZ) meets
Y's circle in a second point, which is the midpoint of the "internal"
arc AC.  Let's call this point Y'.  This is a fixed point in the
plane, i.e., it doesn't move when Y moves.

Similarly the angle bisector of angle X passes through a fixed point 
X' on the internal arc BC.

Of course the point of intersection I of XX' and YY' gives the 
centroid we are looking for.

Now we note that if XX' rotates about X' through 1 degree (as X moves)
then YY' rotates through the same angle about Y'.  This means that the
lines XX' and YY' make a fixed angle.  This means that their point of
intersection I runs through a circle (inscribed angles again)!

Note that this reasoning is still valid when instead of an equilateral 
triangle XYZ we take a triangle XYZ directly similar to a given 
triangle, and instead of the centroid of XYZ we take a point P which 
is in a fixed position with respect to X, Y and Z.

This is all related to Miquel's Theorem:

  MathWorld: Miquel's Theorem
  http://mathworld.wolfram.com/MiquelsTheorem.html 

If you have more questions, just write back.

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 09/25/2003 at 06:35:39
From: Jadwiga
Subject: Thank you (Geometry)

Thank you for your prompt answer.  It was very useful for me.

Best regards,
Jadwiga Ptaszynska
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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