The Erdös-Mordell Inequality

Johann Peter Dirichlet · Poincare Conjecture Offline Joined: 02 Apr 2004 Posts: 175

Hey children!

This is the most famous crazy geometric inequality.
"Erdös-Mordell Inequality:

If P is a point in the same plane of a triangle ABC, and D,E,F are the projections of P in BC, AC and AB, then $(PA+PB+PC) \geq 2(PD+PE+PF)$ ,
with equality if and only if $PA=PB=PC$ and $AB=AC=BC$ ".

**Posted:** Tue May 04, 2004 6:51 pm

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xirti

Do you know if this is also good in space? Of course replacing 2 by 3.

I mean for a tetrahedron.

**Posted:** Sat May 08, 2004 11:41 am

treegoner

This inequality can be tightened as follow :
Let $ABC$ be a triangle. Let $P$ be a point in it. Let $D, E, F$ be points on $BC, CA,AB$ such that $PD, PE, PF$ are respectively bisectors of $\measuredangle{BPC}, \measuredangle{CPA}, \measuredangle{APB}}$ . Then
$PA+PB+PC \geq 2(PD+PE+PF)$
The proof is much easier to be thought out. Razz

Try!

**Posted:** Sat Jul 03, 2004 8:01 am

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Igor

Then generalize for a n side polygon Smile

(The proof has been already posted on the forum by Pierre )

**Posted:** Sat Jul 03, 2004 10:56 am

Dr Sonnhard Graubner

hello, here you can find something more about this inequality
http://www.gtsintsifas.com/uploads/erdsum.pdf
Sonnhard.

**Posted:** Mon Apr 13, 2009 11:51 am

luisgeometria

This is the twin Erdos-Mordell inequality

$PA.PB.PC \ge 8xyz$ where $x,y,z$ are the distances from $P$ to the sides of $\Delta ABC$

Both inequalities are widely applied....And for particular points $P$ ..Results very well-know inequalities..

**Posted:** Mon Apr 13, 2009 12:06 pm

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