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The Napoleon Point and More

Date: 09/04/98 at 03:10:25
From: Adi Schulthess
Subject: Fermat

On each side of a triangle you put on triangles having three sides of 
the same length. You take the centers of them and draw a line to the 
opposite vertex in the original triangle. Does anybody know the proof 
why the three lines all go to the same point?

Date: 09/05/98 at 12:46:58
From: Doctor Floor
Subject: Re: Fermat

Hi Adi,

Thank you for sending your question to Dr. Math.

I wonder why you write "Fermat" in your subject line, because the point 
of concurrence you are describing is the "Napoleon point" of the 
triangle. The "Fermat point" is constructed by taking the 'new' 
vertices of the triangles constructed at the sides, not their centers.

For the Napoleon point, visit Clark Kimberling's page:   

From this page you can look for other triangle centers too.

Both points are special cases of the following theorem:

Let ABC be a triangle, and let A', B' and C' be points such that:
   angle(ABC') = angle(CBA') = angle(BCA') = angle(ACB') 
   = angle(CAB') = angle(BAC') = t

(where the angles are only equal when they are all outside or all 
inside the  triangle). Then the lines AA', BB' and CC' concur in one 

To prove this, let's for simplicity reasons take all angles outside 
triangle ABC. Let's name the sides of triangle ABC in the standard way, 
i.e. AB = c, BC = a and CA = b. Let's name the angles angle(CAB) = 
angle(A), angle(ABC) = angle(B) and angle(BCA) = angle(C). Here is a 
picture for your reference:


Then for AA' we have the following picture (where of course BC'=AC'):

   C'         A
           X                      X is intersection of AB and CC'.

        B                 C

Area(BCC') = a*BC'*sin(B+t) and area(ACC') = b*AC'*sin(A+t). So:

   area(ACC')   b*sin(A+t)
   ---------- = ----------
   area(BCC')   a*sin(B+t)

Because ACC' and BCC' have the same base CC', the altitudes from B and 
from C to CC' must have the same ratio as the areas, and thus BX and 
AX too. So:

   AX   b*sin(A+t)
   -- = ----------
   BX   a*sin(B+t)

We can compute in the same way ratios for BY/CY (where Y is the point 
of intersection of AA' and BC) and AZ/CZ (Z intersection of BB' and 
AC). When we multiply them we get:

   b*sin(A+t)   c*sin(B+t)   a*sin(C+t)
   ---------- * ---------- * --------- = 1
   a*sin(B+t)   b*sin(C+t)   c*sin(A+t)

And then from Ceva's theorem (which I hope you know - if not, I've 
included a statement of it below), it follows that XC, YA and ZB 
concur in one point. But then AA', BB' and CC' concur in one point too.

So this proves your question.

In fact there is a more powerful theorem, but the proof needs the use 
of trilinear coordinates as far as I know. Then it is only needed that 
angle(BAC') = angle(CAB'), angle(CBA') = angle(ABC'), and 
angle(ACB') = angle(BCA').

Here is Ceva's Theorem:

Given a triangle with vertices A, B, and C and points along the sides 
D, E, and F:


AD, BE, and CF are concurrent (intersect in a single point) if and 
only if BD * CE * AF = DC * EA * FB. This theorem was first published 
by Giovanni Cevian in 1678.

If you have another math question, please send it to Dr. Math.

Best regards,

- Doctor Floor, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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