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 Sum the Points of Tangency (Posted on 2008-01-16)
Let ABC be an arbitrary triangle with side lengths a = |BC|, b = |CA|, and c = |AB|.

Let X, Y, and Z be the points of tangency of the incircle with the sides BC, CA, and AB respectively.

Let X', Y', and Z' be the points of tangency of the excircles with "sides" BC, CA, and AB respectively.

What is the value of |XX'| + |YY'| + |ZZ'| in terms of a, b, and c?

 See The Solution Submitted by Bractals Rating: 1.0000 (1 votes)

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 complete solution Comment 1 of 1

let t=|BX|  by similar triangles we have |BZ|=|BX|=t we also have  |CX|=|CY|=a-t  |AZ|=|AY|=b-a+t   thus |AZ|+|BZ|=|AB|

t+b-a+t=c

2t=a+c-b

t=(a+c-b)/2

let u=|BX'|

let Z,W be the centers of the excircle from side BC and incircle respectivly

let Rx and Ri be the radii of the excircle and incircle respectivly

ZBX' and BWX are right triangles and since angles  BZX' and BWX are congruent and ZBX'=BXW=90 the we have that

u=Rx*Ri/t

now if A is the area of triangle ABC then

ri=2A/(a+b+c)

rx=2A/(b+c-a)

thus

u=4A^2/[(a+b+c)(b+c-a)t]

substituting in our know value for t we get

u=8A^2/[(a+b+c)(b+c-a)(a+c-b)]

now using Herons formula for area we get

u=(1/2)(a+b+c)(a+b-c)(a+c-b)(b+c-a)/[(a+b+c)(b+c-a)(a+c-b)]

simplfying we get

u=(a+b-c)/2

now we can get |XX'| with |t-u| which is

(1/2)|a+c-b-a-b+c|

(1/2)|2c-2b|

|c-b|

now using similar methods we can get |YY'| and |ZZ'| to be |a-c| and |a-b| respectivly thus the desired sum is

|c-b|+|a-c|+|a-b|

 Posted by Daniel on 2008-01-18 02:26:11

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