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### The Erdos-Mordell Theorem

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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/
```
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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