The Erdos-Mordell Theorem

Date: 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/