Extouch triangle: Difference between revisions

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{{short description|Triangle formed from the points of tangency of a given triangle's excircles}}

[[Image:Extouch Triangle and Nagel Point.svg|right|frame|325px|The extouch triangle (△T_AT_BT_C, with red boundary) and the [[Nagel point]] (blue, N) of a triangle (△ABC, with black boundary). The orange circles are the [[excircles]] of the triangle.]]

[[Image:Extouch Triangle and Nagel Point.svg|right|frame|

In [[geometry]], the '''extouch triangle''' of a [[triangle]] is formed by joining the points at which the three [[excircle]]s touch the triangle.

{{legend-line|solid #333333|Arbitrary triangle {{math|△''ABC''}}}}

{{legend-line|solid orange|[[Excircle]]s, tangent to the sides of {{math|△''ABC''}} at {{mvar|T{{sub|A}}, T{{sub|B}}, T{{sub|C}}}}}}

{{legend-line|solid red|'''Extouch triangle''' {{math|△''T{{sub|A}}T{{sub|B}}T{{sub|C}}''}}}}

{{legend-line|solid #1e90ff|[[Splitter (geometry)|Splitters]] of the perimeter {{mvar|{{overline|AT}}{{sub|A}}, {{overline|BT}}{{sub|B}}, {{overline|CT}}{{sub|C}}}}; intersect at the [[Nagel point]] {{mvar|N}}}}]]

In [[Euclidean geometry]], the '''extouch triangle''' of a [[triangle]] is formed by joining the points at which the three [[excircle]]s touch the triangle.

==Coordinates==

The [[vertex (geometry)|vertices]] of the extouch triangle are given in [[trilinear coordinates]] by:

<math display=block>\begin{array}{rccccc}

:<math>T_A = 0 : \csc^2{\left( B/2 \right)} : \csc^2{\left( C/2 \right)}</math>

:<math>T_BT_A =& 0 &:& \csc^2{\left( A/2 \right)frac{B}{2} &: 0 :& \csc^2{\left( \frac{C/}{2} \right)}</math>\

:<math>T_CT_B =& \csc^2{\left( \frac{A/}{2 \right)} &:& 0 &:& \csc^2{\left( B/2 \right)frac{C}{2} : 0</math>\\

:<math>T_C =& \csc^2 \frac{-a+b+cA}{a2} &:& \csc^2 \frac{a-b+cB}{b2} &:& 0.</math>

\end{array}</math>

or equivalently, where ''{{mvar|a, b, c''}} are the lengths of the sides opposite angles ''{{mvar|A, B, C''}} respectively,

<math display=block>\begin{array}{rccccc}

:<math> T_A =& 0 &:& \frac{a \, - \, b \, + \, c}{b} &:& \frac{a \, + \, b \, - \, c}{c}</math> \\

:<math> T_B =& \frac{-a \, + \, b \, + \, c}{a} &:& 0 &:& \frac{a \, + \, b \, - \, c}{c}</math> \\

T_C =& \frac{-a \, + \, b \, + \, c}{a} &:& \frac{a \, - \, b \, + \, c}{b} &:& 0

\end{array}</math>

Also, with {{mvar| s}} denoting the [[semiperimeter]] of the triangle, the vertices of the extouch triangle are given in [[barycentric coordinates]] by:

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

<math display=block>\begin{array}{rccccc}

:<math>T_A = 0 : \frac{a-b+c}{b} : \frac{a+b-c}{c}</math>

T_A =& 0 &:& s-b &:& s-c \\

:<math>T_B = \frac{-a+b+c}{a} : 0 : \frac{a+b-c}{c}</math>

T_B =& s-a &:& 0 &:& s-c \\

:<math>T_C = \frac{-a+b+c}{a} : \frac{a-b+c}{b} : 0.</math>

T_C =& s-a &:& s-b &:& 0

\end{array}</math>

==Related figures==

Line 30 ⟶ 48:

==Area==

The area of the extouch triangle, <math>K_T</math>{{mvar|K{{sub|T}}}}, is given by:

:<math>K_T= K\frac{2r^2s}{abc}</math>

where <math>{{mvar|K</math>,}} <math>and {{mvar|r</math>, <math>s</math>}} are the area, and radius of the [[incircle]], and{{mvar|s}} is the [[semiperimeter]] of the original triangle, and <math>{{mvar|a</math>, <math>b</math>, <math>c</math>}} are the side lengths of the original triangle.

This is the same area as that of the [[intouch triangle]].<ref>Weisstein, Eric W. "Extouch Triangle." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/ExtouchTriangle.html</ref>