Equal Parallelians Point

Date: 12/09/2002 at 21:32:51
From: G
Subject: Triangle problem

Within triangle ABC, we draw three segments that are parallel to the 
sides of the triangle, each touching two sides of the triangle. The 
three segments meet at one point, and they are all the same length, x.  
The problem is to find the length of x given the length of the sides 
of the triangle.

Thanks.

Date: 12/10/2002 at 04:55:12
From: Doctor Floor
Subject: Re: Triangle problem

Hi G,

Thanks for your question.

I will write A, B, C for the vertices of the triangle, and a,b,c for 
the lengths of its sides. The three segments of equal lengths parallel 
to the sides I will refer to as "parallelians." The a-parallelian is 
parallel to BC. 

The triangle formed by A and the a-parallelian is similar to ABC, and 
the center of similitude is A. Let the factor of similitude be fa, so 
fa*a = x.

Similarly let fb and fc be the factors of similitude from B and C 
respectively, so fb*b = fc*c = x.

We see that the height from P in triangle ABP is 1-fc times as long 
as the height from C in triangle ABC. Since in both cases the base 
is AB, we see that the area of ABP is 1-fc times the area of ABC. 
In the same manner we find that the areas of ACP and BCP are 
respectively 1-fb and 1-fc times the area of ABC. Now we can use that 
the area of ABP, BCP, and ACP taken together is equal to the area of 
ABC. We conclude

  1-fa + 1-fb + 1-fc = 1
  fa + fb + fc = 2

We can multiply by abc and find

  fa*abc + fb*abc + fc*abc = 2abc
  x*(bc+ac+ab) = 2abc
  x = 2abc / (bc+ac+ab) 

The point of concurrency P is called "The equal parallelians point" - 
see for instance, from Professor Clark Kimberling (University of 
Evansville):

   Equal Parallelians Point
   http://www2.evansville.edu/ck6/tcenters/recent/eqparal.html 

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/