Editing Incircle and excircles (section)

==Excircles and excenters==
[[File:Incircle and Excircles.svg|right|thumb|300px|
{{legend-line|solid black|[[Extended side]]s of {{math|△''ABC''}}}}
{{legend-line|solid #728fce|Incircle ([[incenter]] at {{mvar|I}})}}
{{legend-line|solid orange|Excircles (excenters at {{mvar|J{{sub|A}}}}, {{mvar|J{{sub|B}}}}, {{mvar|J{{sub|C}}}})}}
{{legend-line|solid red|Internal [[angle bisector]]s}}
{{legend-line|solid #32cd32|External angle bisectors (forming the excentral triangle)}}
]]

An '''excircle''' or '''escribed circle'''<ref name="Altshiller-Court 1925 74"/> of the triangle is a circle lying outside the triangle, tangent to one of its sides, and tangent to the [[extended side|extensions of the other two]]. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.<ref name="Altshiller-Court 1925 73"/>

The center of an excircle is the intersection of the internal bisector of one angle (at vertex <math>A</math>, for example) and the [[internal and external angle|external]] bisectors of the other two. The center of this excircle is called the '''excenter''' relative to the vertex <math>A</math>, or the '''excenter''' of <math>A</math>.<ref name="Altshiller-Court 1925 73"/> Because the internal bisector of an angle is perpendicular to its external bisector, it follows that the center of the incircle together with the three excircle centers form an [[orthocentric system]].{{sfn|Johnson|1929|p=182}}

===Trilinear coordinates of excenters===
While the [[#incenter|incenter]] of <math>\triangle ABC</math> has [[trilinear coordinates]] <math>1 : 1 : 1</math>, the excenters have trilinears
{{Citation needed|date=May 2020}}
:<math display=block>\begin{array}{rrcrcr}
  J_A = & -1 &:& 1 &:& 1 \\
  J_B = & 1 &:& -1 &:& 1 \\
  J_C = & 1 &:& 1 &:& -1
\end{array}</math>

===Exradii===
The radii of the excircles are called the '''exradii'''.

The exradius of the excircle opposite <math>A</math> (so touching <math>BC</math>, centered at <math>J_A</math>) is<ref name="Altshiller-Court 1925 79">{{harvtxt|Altshiller-Court|1925|p=79}}</ref><ref>{{harvtxt|Kay|1969|p=202}}</ref>
:<math display=block>r_a = \frac{rs}{s - a} = \sqrt{\frac{s(s - b)(s - c)}{s - a}},</math> where <math>s = \tfrac{1}{2}(a + b + c).</math>

See [[Heron's formula]].

====Derivation of exradii formula====
Source:<ref name="Altshiller-Court 1925 79"/>

Let the excircle at side <math>AB</math> touch at side <math>AC</math> extended at <math>G</math>, and let this excircle's
radius be <math>r_c</math> and its center be <math>J_c</math>. Then <math>J_c G</math> is an altitude of <math>\triangle ACJ_c</math>, so <math>\triangle ACJ_c</math> has area <math>\tfrac12 br_c</math>. By a similar argument, <math>\triangle BCJ_c</math> has area <math>\tfrac12 ar_c</math> and <math>\triangle ABJ_c</math> has area <math>\tfrac12 cr_c</math>. Thus the area <math>\Delta</math> of triangle <math>\triangle ABC</math> is
:<math display=block>\Delta = \tfrac12 (a + b - c)r_c = (s - c)r_c</math>.

So, by symmetry, denoting <math>r</math> as the radius of the incircle,
:<math display=block>\Delta = sr = (s - a)r_a = (s - b)r_b = (s - c)r_c</math>.

By the [[Law of Cosines]], we have
:<math display=block>\cos A = \frac{b^2 + c^2 - a^2}{2bc}</math>

Combining this with the identity <math>\sin^2 \! A + \cos^2 \! A = 1</math>, we have
:<math display=block>\sin A = \frac{\sqrt{-a^4 - b^4 - c^4 + 2a^2 b^2 + 2b^2 c^2 + 2 a^2 c^2}}{2bc}</math>

But <math>\Delta = \tfrac12 bc \sin A</math>, and so
:<math display=block>\begin{align}
  \Delta &= \tfrac14 \sqrt{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2 c^2 + 2 a^2 c^2} \\[5mu]
         &= \tfrac14 \sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)} \\[5mu]
         & = \sqrt{s(s - a)(s - b)(s - c)},
\end{align}</math>

which is [[Heron's formula]].

Combining this with <math>sr = \Delta</math>, we have
:<math display=block>r^2 = \frac{\Delta^2}{s^2} = \frac{(s - a)(s - b)(s - c)}{s}.</math>

Similarly, <math>(s - a)r_a = \Delta</math> gives
:<math display=block>\begin{align}
  &r_a^2 = \frac{s(s - b)(s - c)}{s - a} \\[4pt]
  &\implies r_a = \sqrt{\frac{s(s - b)(s - c)}{s - a}}.
\end{align}</math>

====Other properties====
From the formulas above one can see that the excircles are always larger than the incircle and that the largest excircle is the one tangent to the longest side and the smallest excircle is tangent to the shortest side. Further, combining these formulas yields:<ref>Baker, Marcus, "A collection of formulae for the area of a plane triangle", ''Annals of Mathematics'', part 1 in vol. 1(6), January 1885, 134-138. (See also part 2 in vol. 2(1), September 1885, 11-18.)</ref>
:<math display=block>\Delta = \sqrt{r r_a r_b r_c}.</math>

===Other excircle properties===
The circular [[convex hull|hull]] of the excircles is internally tangent to each of the excircles and is thus an [[Problem of Apollonius|Apollonius circle]].<ref>[http://forumgeom.fau.edu/FG2002volume2/FG200222.pdf Grinberg, Darij, and Yiu, Paul, "The Apollonius Circle as a Tucker Circle", ''Forum Geometricorum'' 2, 2002: pp. 175-182.]</ref> The radius of this Apollonius circle is <math>\tfrac{r^2 + s^2}{4r}</math> where <math>r</math> is the incircle radius and <math>s</math> is the semiperimeter of the triangle.<ref>[http://forumgeom.fau.edu/FG2003volume3/FG200320.pdf Stevanovi´c, Milorad R., "The Apollonius circle and related triangle centers", ''Forum Geometricorum'' 3, 2003, 187-195.]</ref>

The following relations hold among the inradius <math>r</math>, the circumradius <math>R</math>, the semiperimeter <math>s</math>, and the excircle radii <math>r_a</math>, <math>r_b</math>, <math>r_c</math>:<ref name=Bell>{{cite web |author=Bell, Amy |title="Hansen's right triangle theorem, its converse and a generalization", ''Forum Geometricorum'' 6, 2006, 335–342. |url=http://forumgeom.fau.edu/FG2006volume6/FG200639.pdf |access-date=2012-05-05 |url-status=dead |archive-url=https://web.archive.org/web/20210831080348/https://forumgeom.fau.edu/FG2006volume6/FG200639.pdf |archive-date=2021-08-31}}</ref>
:<math display=block>\begin{align}
              r_a + r_b + r_c &= 4R + r, \\
  r_a r_b + r_b r_c + r_c r_a &= s^2, \\
        r_a^2 + r_b^2 + r_c^2 &= \left(4R + r\right)^2 - 2s^2.
\end{align}</math>

The circle through the centers of the three excircles has radius <math>2R</math>.<ref name=Bell/>

If <math>H</math> is the [[orthocenter]] of <math>\triangle ABC</math>, then<ref name=Bell/>
:<math display=block>\begin{align}
          r_a + r_b + r_c + r &= \overline{AH} + \overline{BH} + \overline{CH} + 2R, \\
  r_a^2 + r_b^2 + r_c^2 + r^2 &= \overline{AH}^2 + \overline{BH}^2 + \overline{CH}^2 + (2R)^2.
\end{align}</math>

===Nagel triangle and Nagel point===
{{Main|Extouch triangle}}
[[File:Extouch Triangle and Nagel Point.svg|right|frame|
{{legend-line|solid black|[[Extended side]]s of triangle {{math|△''ABC''}}}}
{{legend-line|solid orange|Excircles of {{math|△''ABC''}} (tangent at {{mvar|T{{sub|A}}. T{{sub|B}}, T{{sub|C}}}})}}
{{legend-line|solid red|'''Nagel/Extouch triangle''' {{math|△''T{{sub|A}}T{{sub|B}}T{{sub|C}}''}}}}
{{legend-line|solid #728fce|[[Splitter (geometry)|Splitters]]: lines connecting opposite vertices of {{math|△''ABC''}} and {{math|△''T{{sub|A}}T{{sub|B}}T{{sub|C}}''}} (concur at '''Nagel point''' {{mvar|N}})}}
]]

The '''Nagel triangle''' or '''extouch triangle''' of <math>\triangle ABC</math> is denoted by the vertices <math>T_A</math>, <math>T_B</math>, and <math>T_C</math> that are the three points where the excircles touch the reference <math>\triangle ABC</math> and where <math>T_A</math> is opposite of <math>A</math>, etc. This <math>\triangle T_AT_BT_C</math> is also known as the '''extouch triangle''' of <math>\triangle ABC</math>. The [[circumcircle]] of the extouch <math>\triangle T_AT_BT_C</math> is called the '''Mandart circle'''
(cf. [[Mandart inellipse]]).

The three line segments <math>\overline{AT_A}</math>, <math>\overline{BT_B}</math> and <math>\overline{CT_C}</math> are called the [[splitter (geometry)|splitters]] of the triangle; they each bisect the perimeter of the triangle,{{Citation needed|date=May 2020}}
:<math display=block>\overline{AB} + \overline{BT_A} = \overline{AC} + \overline{CT_A} = \frac{1}{2}\left( \overline{AB} + \overline{BC} + \overline{AC} \right).</math>

The splitters intersect in a single point, the triangle's [[Nagel point]] <math>N_a</math> (or [[triangle center]] ''X''<sub>8</sub>).

Trilinear coordinates for the vertices of the extouch triangle are given by{{Citation needed|date=May 2020}}
:<math display=block>\begin{array}{ccccccc}
  T_A &=& 0 &:& \csc^2\frac{B}{2} &:& \csc^2\frac{C}{2} \\[2pt]
  T_B &=& \csc^2\frac{A}{2} &:& 0 &:& \csc^2\frac{C}{2} \\[2pt]
  T_C &=& \csc^2\frac{A}{2} &:& \csc^2\frac{B}{2} &:& 0
\end{array}</math>

Trilinear coordinates for the Nagel point are given by{{Citation needed|date=May 2020}}
:<math display=block>\csc^2\tfrac{A}{2} : \csc^2\tfrac{B}{2} : \csc^2\tfrac{C}{2},</math>

or, equivalently, by the [[Law of Sines]],
:<math display=block>\frac{b + c - a}{a} : \frac{c + a - b}{b} : \frac{a + b - c}{c}.</math>

The Nagel point is the [[isotomic conjugate]] of the Gergonne point.{{Citation needed|date=May 2020}}