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Extouch triangle

Triangle formed from the points of tangency of a given triangle's excircles

Extouch triangle

Triangle formed from the points of tangency of a given triangle's excircles

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In Euclidean geometry, the extouch triangle of a triangle is formed by joining the points at which the three excircles touch the triangle.

Coordinates

The vertices of the extouch triangle are given in trilinear coordinates by:

\begin{array}{rccccc} T_A =& 0 &:& \csc^2 \frac{B}{2} &:& \csc^2 \frac{C}{2} \ T_B =& \csc^2 \frac{A}{2} &:& 0 &:& \csc^2 \frac{C}{2} \ T_C =& \csc^2 \frac{A}{2} &:& \csc^2 \frac{B}{2} &:& 0 \end{array}

or equivalently, where a, b, c are the lengths of the sides opposite angles A, B, C respectively,

\begin{array}{rccccc} T_A =& 0 &:& \frac{a , - , b , + , c}{b} &:& \frac{a , + , b , - , c}{c} \ T_B =& \frac{-a , + , b , + , c}{a} &:& 0 &:& \frac{a , + , b , - , c}{c} \ T_C =& \frac{-a , + , b , + , c}{a} &:& \frac{a , - , b , + , c}{b} &:& 0 \end{array}

Also, with s denoting the semiperimeter of the triangle, the vertices of the extouch triangle are given in barycentric coordinates by:

\begin{array}{rccccc} T_A =& 0 &:& s-b &:& s-c \ T_B =& s-a &:& 0 &:& s-c \ T_C =& s-a &:& s-b &:& 0 \end{array}

Area

The area of the extouch triangle, K, is given by:

:K_T= K\frac{2r^2s}{abc}

where K and r are the area and radius of the incircle, s is the semiperimeter of the original triangle, and a, b, c are the side lengths of the original triangle.

This is the same area as that of the intouch triangle.

References

References

  1. Weisstein, Eric W. "Extouch Triangle." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/ExtouchTriangle.html
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