The Erdos-Mordell TheoremDate: 10/13/2000 at 14:33:11 From: Rayna Zacks Subject: Related to Fermat Point in Triangle Let P be a point in a triangle. Let D be the sum of the distances from P to the 3 vertices, and let E be the sum of the shortest distances from P to the edges (trilinear coordinates of P). Prove that D > 2*E. I have beaten this to death using analytic geometry and calculus to find the point P which minimizes D-2*E and then prove that D-2*E at P is greater than 0. I have used a number of different parameterizations of the triangle using 3 numbers (xy coordinates of one vertex and the length of the opposite side, two angles and length of one side, length of all three sides) and expressed D and E in these terms, calculated the partial derivative of (D-2*E) with respect to the parameters, etc. I continually end up with intractable expressions. I have tried all geometrical approaches that I know to construct the Fermat point - but it's clear that point P above is not related to the Fermat point. (I have a number of proofs now of the Fermat point theorem.) I have solved special cases (e.g. if <A + <B > 90 deg. and <A < 15 deg. then D-2*E > 0), and also proved it for an isosceles triangles where P is on the altitude, and for right triangles, and have found many cases where point P is the vertex opposite the shortest side. But I have not found a simple general proof that D > 2*E. In xy coordinates the problem reduces to something like the following: Prove H > 0 where H = cos(b) - 2*sin(B) - (X*cos(w) - Y*sin(w)) and we know that at point P the two partial derivatives require that 1) cos(a) - cos(b) + cos(c) = 2*(sin(B) - sin(C)) 2) sin(a) + sin(b) - sin(c) = 2*(1 - cos(B) - cos(C)) where A, B and C are the 3 angles of the triangle; a, b and c are the 3 angles of the triangle with P as a vertex, and (X,Y) are the coordinates of one of the vertices of the triangle. The symmetries of the above are obvious, but I cannot get the last step. I am working with my father to solve a recent problem from the AMM problem section and the first step is to prove D > 2*E. This is my job. My father says that he has seen a proof of D > *2E in a problem book but he cannot find it. Neither can he prove D > 2*E. Help. Thanks, Rayna Zacks Date: 10/13/2000 at 17:08:15 From: Doctor Floor Subject: Re: Related to Fermat Point in Triangle Dear Rayna, Thanks for writing. This theorem is known as the Erdos-Mordell theorem, but it should read D >= 2*E. Let's consider the following: We have that D, E and F are the feet of the perpendicular altitudes from P to AB, BC and CA respectively, while E' and F' are the feet of the perpendicular altitudes from E and F to AB. CFPE is a cyclic quadrilateral because the opposite angles are supplementary. PC is its diameter. We also know that the radius of the circumcircle of EFC is EF/(2*sin(C)), and thus the diameter is EF/sin(C). This means that EF = PC*sin(C). Since EF >= E'F' this gives PC*sin(C) >= E'F' ........................................[1] Also, we have F'E' = F'D + DE' = PF*cos(90-A) + PE*cos(90-B) = PF*sin(A) + PE*sin(B) .......................[2] The combination of [1] and [2] shows that PC * sin(C) >= PF*sin(A) + PE*sin(B) sin(A) sin(B) PC >= PF* ----- + PE * ----- sin(C) sin(C) In the same way you find similar results for PA and PB. These three expressions combine to: PA+PB+PC >= sin(A) sin(B) sin(B) sin(C) sin(A) sin(C) PD*(----- + -----) + PE*(----- + -----) + PF*(----- + -----) sin(B) sin(A) sin(C) sin(B) sin(C) sin(A) >= 2*(PD + PE + PF) because x + 1/x >= 2. And we have proven the Erdos-Mordell theorem (from 1935). Good luck in solving the AMM problem. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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