Equal Parallelians PointDate: 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/ |
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