Miquel CirclesDate: 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 |
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