Existence of the Brocard PointDate: 06/06/2002 at 06:08:51 From: Razvan Subject: Geometry - The Brocard Point I have recently stumbled across a problem that asks to demonstrate that the Brocard point exists in any triangle. I've searched the Internet for the demonstration, but I couldn't find it. Thanks, Razvan Date: 06/06/2002 at 06:58:49 From: Doctor Floor Subject: Re: Geometry - The Brocard Point Hi, Razvan, Thanks for your question. There are two Brocard points. One Brocard point is the point Om (in most cases the Greek letter Omega is used) such that angle(OmAB) = angle(OmBC) = angle(OmCA). C Om A B The second Brocard point is the point Om' such that angle(Om'BA) = angle(Om'CB) = angle(Om'AC). We shall prove that the first Brocard point exists. Suppose that T is an interior point of the triangle such that angle (TAB) = angle(TBC) = x. Then angle(TBA) = angle(B) - x, and thus angle(BTA) = 180 - angle(TBA) - angle(TAB) = 180 - (angle(B)-x) - x = 180 - angle(B) So the locus of T lies on a circular arc, since angle(BTA) is fixed. For reasons of ease we consider the complete circle. Let this circle be Cc. In the same way the points U such that angle(UBC) = angle(UCA) yield a circle Ca. The point of intersection Om of Cc and Ca, apart from B, is the first Brocard point, because this point must satisfy angle (OmAB) = angle (OmBC) as well as angle(OmBC) = angle (OmCA). The circle Cc can be constructed in the following way: Intersect the perpendicular bisector of AB with the line through B perpendicular to BC, and let the point of intersection be X. Then angle(BAX) = angle(XBA) =90 - B, and thus angle(AXB) = 2B, so the circle with center X through A and B is indeed Cc. See also: Mathworld - Brocard points http://mathworld.wolfram.com/BrocardPoints.html If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 06/11/2002 at 05:36:23 From: Razvan Subject: Geometry - The Brocard Point Hi, Doctor Floor, Thank you very much for the demonstration, but I would like to ask you more about it. Why is it that if the angle(BTA) is fixed, the locus of T is a circular arc? The second question is about the X point. Why is it that if the angle(AXB) = 2B, the X point is the center of Cc? Thanks, Razvan Date: 06/11/2002 at 06:59:58 From: Doctor Floor Subject: Re: Geometry - The Brocard Point Hi, Thanks for your question. Both have to do with inscribed and central angles of a circle. Let me therefore point you to the following message from the Dr. Math archive: Inscribed Angle Theorem http://mathforum.org/library/drmath/view/55073.html It tells that two inscribed angles on the same arc of a circle are half the measure of the central angle on that same arc. That should answer your second question. For your first question, we need that the locus of angles at the same side of a segment making congruent angles on that segment is a circular arc. To see that, we suppose that we have an angle from C on segment AB. We draw the circumcircle of ABC: Of course the circular arc AB containing C is part of the locus, we know this from the inscribed angle theorem. Now suppose that E is a point in the exterior of the circle, and also part of the locus. Then we let line AE intersect the circumcircle again in point D. We know that angle ADB = angle ACB, but we also know that angle AEB = 180 - angle EDB - angle DBE = angle ADB - anlge DBE. So angle AEB is not congruent to angle ADB = angle ACB. We have a contradiction. In the same way we show that an interior point F of the circle cannot be part of the locus. This answers your first question. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 06/14/2002 at 04:01:46 From: Razvan Subject: Geometry - The Brocard Point Hi, I still have a question about the X point. The X point is the center of the Cc, T is on Cc, and angle(AXB) = 2B. Then, shouldn't angle(BTA) = angle(B) because angle(AXB) is a center angle and angle(BTA) is inscribed on the same arc as angle(AXB)? Instead angle(BTA) = 180 - angle(B). Thanks, Razvan Date: 06/14/2002 at 08:14:32 From: Doctor Floor Subject: Re: Geometry - The Brocard Point Hi, Razvan, Thanks for your question. Angle(BTA) is an inscribed angle on the opposite arc on which angle(AXB) is a central angle. Together these arcs are 360 degrees. You can see that from the fact the order of A and B in BTA and AXB is opposite. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/