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Sum the Points of Tangency (Posted on 2008-01-16) Difficulty: 3 of 5
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?
 

  Submitted by Bractals    
Rating: 1.0000 (1 votes)
Solution: (Hide)
Let ABC be an arbitrary triangle with sides 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?

Tangents to the incircle from the vertices are equal:
   |XC| = |CY|     |YA| = |AZ|     |ZB| = |BX| 
Therefore,
   a = |BC| = |BX| + |XC| = |BX| + |CY|
   b = |CA| = |CY| + |YA| = |CY| + |AZ|
   c = |AB| = |AZ| + |ZB| = |AZ| + |BX|
Adding these three equations gives
   a + b + c = 2(|AZ| + |BX| + |CY|)

           or

   s = |AZ| + |BX| + |CY|
Therefore,
   |AZ| = s - (|BX| + |CY|) = s - a
   |BX| = s - (|CY| + |AZ|) = s - b
   |CY| = s - (|AZ| + |BX|) = s - c
Let P and Q be the points of tangency between the excircle opposite vertex A and the sidelines AB and AC respectively.
Tangents to this excircle from the vertices are equal:
   |AP| = |AQ|     |BP| = |BX'|     |CQ| = |X'C| 
Therefore,
    a = |BC| = |BX'| + |X'C|
    b = |AC| = |AQ| - |CQ| = |AP| - |X'C|
    c = |AB| = |AP| - |BP| = |AP| - |BX'|
Adding these three equations gives
   a + b + c = 2|AP|

            or

   s = |AP|
Therefore,
   |BX'| = |AP| - c = s - c
Similar arguments give
   |CY'| = s - a
   |AZ'| = s - b
Therefore,
   |XX'| = ||BX'| - |BX|| = |(s-c) - (s-b)| = |b - c|
   |YY'| = ||CY'| - |CY|| = |(s-a) - (s-c)| = |c - a|
   |ZZ'| = ||AZ'| - |AZ|| = |(s-b) - (s-a)| = |a - b|
Adding these three equations gives

   |XX'| + |YY'| + |ZZ'| = |a - b| + |b - c| + |c - a|
We could stop here, but assume WOLOG that a ≤ b ≤ c. Then
   |a - b| + |b - c| + |c - a| = (b - a) + (c - b) + (c - a) = 2(c - a)
Therefore,

|XX'| + |YY'| + |ZZ'| is twice the difference between the largest side and the smallest side.

Note: ABC is an equilateral triangle if and only if |XX'| + |YY'| + |ZZ'| = 0.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutioncomplete solutionDaniel2008-01-18 02:26:11
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