Incircle and excircles#Radius

{{See also|Incenter}}

Suppose $\backslash triangle\; ABC$ has an incircle with radius $r$ and center $I$. Let $a$ be the length of $\backslash overline\{BC\}$, $b$ the length of $\backslash overline\{AC\}$, and $c$ the length of $\backslash overline\{AB\}$.

Also let $T\_A$, $T\_B$, and $T\_C$ be the touchpoints where the incircle touches $\backslash overline\{BC\}$, $\backslash overline\{AC\}$, and $\backslash overline\{AB\}$.

The incenter is the point where the internal angle bisectors of $\backslash angle\; ABC,\; \backslash angle\; BCA,\; \backslash text\{\; and\; \}\; \backslash angle\; BAC$ meet.

The trilinear coordinates for a point in the triangle is the ratio of all the distances to the triangle sides. Because the incenter is the same distance from all sides of the triangle, the trilinear coordinates for the incenter are[http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] {{webarchive|url=https://web.archive.org/web/20120419171900/http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |date=2012-04-19}}, accessed 2014-10-28.

:$$\backslash \; 1\; :\; 1\; :\; 1.$$

The barycentric coordinates for a point in a triangle give weights such that the point is the weighted average of the triangle vertex positions.

Barycentric coordinates for the incenter are given by

:$$a\; :\; b\; :\; c$$

where $a$, $b$, and $c$ are the lengths of the sides of the triangle, or equivalently (using the law of sines) by

:$$\backslash sin\; A\; :\; \backslash sin\; B\; :\; \backslash sin\; C$$

where $A$, $B$, and $C$ are the angles at the three vertices.

The Cartesian coordinates of the incenter are a weighted average of the coordinates of the three vertices using the side lengths of the triangle relative to the perimeter (that is, using the barycentric coordinates given above, normalized to sum to unity) as weights. The weights are positive so the incenter lies inside the triangle as stated above. If the three vertices are located at $(x\_a,y\_a)$, $(x\_b,y\_b)$, and $(x\_c,y\_c)$, and the sides opposite these vertices have corresponding lengths $a$, $b$, and $c$, then the incenter is at{{Citation needed|date=May 2020}}

:$$$$

\left(\frac{a x_a + b x_b + c x_c}{a + b + c}, \frac{a y_a + b y_b + c y_c}{a + b + c}\right)

= \frac{a\left(x_a, y_a\right) + b\left(x_b, y_b\right) + c\left(x_c, y_c\right)}{a + b + c}.

The inradius $r$ of the incircle in a triangle with sides of length $a$, $b$, $c$ is given by{{harvtxt|Kay|1969|p=201}}

:$$r\; =\; \backslash sqrt\{\backslash frac\{(s-a)(s-b)(s-c)\}\{s\}\},$$

where $s\; =\; \backslash tfrac12(a\; +\; b\; +\; c)$ is the semiperimeter (see Heron's formula).

The tangency points of the incircle divide the sides into segments of lengths $s-a$ from $A$, $s-b$ from $B$, and $s-c$ from $C$ (see Tangent lines to a circle).Chu, Thomas, The Pentagon, Spring 2005, p. 45, problem 584.

Denote the incenter of $\backslash triangle\; ABC$ as $I$.

The distance from vertex $A$ to the incenter $I$ is:

:$$$$

\overline{AI} = d(A, I)

= c \, \frac{\sin\frac{B}{2}}{\cos\frac{C}{2}}

= b \, \frac{\sin\frac{C}{2}}{\cos\frac{B}{2}}.

Use the Law of sines in the triangle $\backslash triangle\; IAB$.

We get $\backslash frac\{\backslash overline\{AI\}\}\{\backslash sin\; \backslash frac\{B\}\{2\}\}\; =\; \backslash frac\{c\}\{\backslash sin\; \backslash angle\; AIB\}$.

We have that $\backslash angle\; AIB\; =\; \backslash pi\; -\; \backslash frac\{A\}\{2\}\; -\; \backslash frac\{B\}\{2\}\; =\; \backslash frac\{\backslash pi\}\{2\}\; +\; \backslash frac\{C\}\{2\}$.

It follows that $\backslash overline\{AI\}\; =\; c\; \backslash \; \backslash frac\{\backslash sin\; \backslash frac\{B\}\{2\}\}\{\backslash cos\; \backslash frac\{C\}\{2\}\}$.

The equality with the second expression is obtained the same way.

The distances from the incenter to the vertices combined with the lengths of the triangle sides obey the equation

{{citation

|last1=Allaire |first1=Patricia R.

|last2=Zhou |first2=Junmin

|last3=Yao |first3=Haishen

|date=March 2012

|journal=Mathematical Gazette

|pages=161–165

|title=Proving a nineteenth century ellipse identity

|volume=96

|doi=10.1017/S0025557200004277

|s2cid=124176398

}}.

:$$\backslash frac\{\backslash overline\{IA\}\; \backslash cdot\; \backslash overline\{IA\}\}\{\backslash overline\{CA\}\; \backslash cdot\; \backslash overline\{AB\}\}\; +\; \backslash frac\{\backslash overline\{IB\}\; \backslash cdot\; \backslash overline\{IB\}\}\{\backslash overline\{AB\}\; \backslash cdot\; \backslash overline\{BC\}\}\; +\; \backslash frac\{\backslash overline\{IC\}\; \backslash cdot\; \backslash overline\{IC\}\}\{\backslash overline\{BC\}\; \backslash cdot\; \backslash overline\{CA\}\}\; =\; 1.$$

Additionally,{{citation |last=Altshiller-Court |first=Nathan |author-link=Nathan Altshiller Court |title=College Geometry |publisher=Dover Publications |year=1980}}. #84, p. 121.

:$$\backslash overline\{IA\}\; \backslash cdot\; \backslash overline\{IB\}\; \backslash cdot\; \backslash overline\{IC\}\; =\; 4Rr^2,$$

where $R$ and $r$ are the triangle's circumradius and inradius respectively.

The collection of triangle centers may be given the structure of a group under coordinate-wise multiplication of trilinear coordinates; in this group, the incenter forms the identity element.

The distances from a vertex to the two nearest touchpoints are equal; for example:Mathematical Gazette, July 2003, 323-324.

:$$d\backslash left(A,\; T\_B\backslash right)\; =\; d\backslash left(A,\; T\_C\backslash right)\; =\; \backslash tfrac12(b\; +\; c\; -\; a)\; =\; s\; -\; a.$$

If the altitudes from sides of lengths $a$, $b$, and $c$ are $h\_a$, $h\_b$, and $h\_c$, then the inradius $r$ is one-third of the harmonic mean of these altitudes; that is,{{harvtxt|Kay|1969|p=203}}

:$$r\; =\; \backslash frac\{1\}\{\backslash dfrac\{1\}\{h\_a\}\; +\; \backslash dfrac\{1\}\{h\_b\}\; +\; \backslash dfrac\{1\}\{h\_c\}\}.$$

The product of the incircle radius $r$ and the circumcircle radius $R$ of a triangle with sides $a$, $b$, and $c$ is{{sfn|Johnson|1929|p=189, #298(d)}}

:$$rR\; =\; \backslash frac\{abc\}\{2(a\; +\; b\; +\; c)\}.$$

Some relations among the sides, incircle radius, and circumcircle radius are:

:$$\backslash begin\{align\}$$

ab + bc + ca &= s^2 + (4R + r)r, \\

a^2 + b^2 + c^2 &= 2s^2 - 2(4R + r)r.

\end{align}

Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter (the center of its incircle). There are either one, two, or three of these for any given triangle.Kodokostas, Dimitrios, "Triangle Equalizers", Mathematics Magazine 83, April 2010, pp. 141-146.

The incircle radius is no greater than one-ninth the sum of the altitudes.Posamentier, Alfred S., and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.{{rp|289}}

The squared distance from the incenter $I$ to the circumcenter $O$ is given by{{cite journal

|last=Franzsen

|first=William N.

|journal=Forum Geometricorum

|mr=2877263

|pages=231–236

|title=The distance from the incenter to the Euler line

|volume=11

|year=2011

|url=http://forumgeom.fau.edu/FG2011volume11/FG201126.pdf

|access-date=2012-05-09

|url-status=dead

|archive-url=https://web.archive.org/web/20201205220605/http://forumgeom.fau.edu/FG2011volume11/FG201126.pdf

|archive-date=2020-12-05

}}.{{rp|232}}

:$$\backslash overline\{OI\}^2\; =\; R(R\; -\; 2r)\; =\; \backslash frac\{a\backslash ,b\backslash ,c\backslash ,\}\{a+b+c\}\backslash left\; [\backslash frac\{a\backslash ,b\backslash ,c\backslash ,\}\{(a+b-c)\backslash ,(a-b+c)\backslash ,(-a+b+c)\}-1\; \backslash right\; ]$$

and the distance from the incenter to the center $N$ of the nine point circle is{{rp|232}}

:

The incenter lies in the medial triangle (whose vertices are the midpoints of the sides).{{rp|233, Lemma 1}}

{{Redirect|Inradius|the three-dimensional equivalent|Inscribed sphere}}

The radius of the incircle is related to the area of the triangle.Coxeter, H.S.M. "Introduction to Geometry 2nd ed. Wiley, 1961. The ratio of the area of the incircle to the area of the triangle is less than or equal to $\backslash pi\; \backslash big/\; 3\backslash sqrt3$,

with equality holding only for equilateral triangles.Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", American Mathematical Monthly 115, October 2008, 679-689: Theorem 4.1.

Suppose $\backslash triangle\; ABC$ has an incircle with radius $r$ and center $I$. Let $a$ be the length of $\backslash overline\{BC\}$, $b$ the length of $\backslash overline\{AC\}$, and $c$ the length of $\backslash overline\{AB\}$.

Now, the incircle is tangent to $\backslash overline\{AB\}$ at some point $T\_C$, and so $\backslash angle\; AT\_CI$ is right. Thus, the radius $T\_CI$ is an altitude of $\backslash triangle\; IAB$.

Therefore, $\backslash triangle\; IAB$ has base length $c$ and height $r$, and so has area $\backslash tfrac12\; cr$.

Similarly, $\backslash triangle\; IAC$ has area $\backslash tfrac12\; br$ and $\backslash triangle\; IBC$ has area $\backslash tfrac12\; ar$.

Since these three triangles decompose $\backslash triangle\; ABC$, we see that the area $\backslash Delta\; \backslash text\{\; of\}\; \backslash triangle\; ABC$ is:

:$$\backslash Delta\; =\; \backslash tfrac12\; (a\; +\; b\; +\; c)r\; =\; sr,$$

{{spaces|4}} and {{spaces|4}}$r\; =\; \backslash frac\{\backslash Delta\}\{s\},$

where $\backslash Delta$ is the area of $\backslash triangle\; ABC$ and $s\; =\; \backslash tfrac12(a\; +\; b\; +\; c)$ is its semiperimeter.

For an alternative formula, consider $\backslash triangle\; IT\_CA$. This is a right-angled triangle with one side equal to $r$ and the other side equal to $r\; \backslash cot\; \backslash tfrac\{A\}\{2\}$. The same is true for $\backslash triangle\; IB\text{'}A$. The large triangle is composed of six such triangles and the total area is:{{Citation needed|date=May 2020}}

:$$\backslash Delta\; =\; r^2\; \backslash left(\backslash cot\backslash tfrac\{A\}\{2\}\; +\; \backslash cot\backslash tfrac\{B\}\{2\}\; +\; \backslash cot\backslash tfrac\{C\}\{2\}\backslash right).$$

[[File:Intouch Triangle and Gergonne Point.svg|right|frame|

{{legend-line|solid black|Triangle {{math|△ABC}}}}

{{legend-line|solid #728fce|Incircle (incenter at {{mvar|I}})}}

{{legend-line|solid red|Contact triangle {{math|△T{{sub|A}}T{{sub|B}}T{{sub|C}}}}}}

{{legend-line|solid #1dc404|Lines between opposite vertices of {{math|△ABC}} and {{math|△T{{sub|A}}T{{sub|B}}T{{sub|C}}}} (concur at Gergonne point {{mvar|G{{sub|e}}}})}}

]]

The Gergonne triangle (of $\backslash triangle\; ABC$) is defined by the three touchpoints of the incircle on the three sides. The touchpoint opposite $A$ is denoted $T\_A$, etc.

This Gergonne triangle, $\backslash triangle\; T\_AT\_BT\_C$, is also known as the contact triangle or intouch triangle of $\backslash triangle\; ABC$. Its area is

:$$K\_T\; =\; K\backslash frac\{2r^2\; s\}\{abc\}$$

where $K$, $r$, and $s$ are the area, radius of the incircle, and semiperimeter of the original triangle, and $a$, $b$, and $c$ are the side lengths of the original triangle. This is the same area as that of the extouch triangle.

Weisstein, Eric W. "Contact Triangle." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ContactTriangle.html

The three lines $AT\_A$, $BT\_B$, and $CT\_C$ intersect in a single point called the Gergonne point, denoted as $G\_e$ (or triangle center X₇). The Gergonne point lies in the open orthocentroidal disk punctured at its own center, and can be any point therein.Christopher J. Bradley and Geoff C. Smith, "The locations of triangle centers", Forum Geometricorum 6 (2006), 57–70. http://forumgeom.fau.edu/FG2006volume6/FG200607index.html

The Gergonne point of a triangle has a number of properties, including that it is the symmedian point of the Gergonne triangle.

{{cite journal

|last=Dekov

|first=Deko

|title=Computer-generated Mathematics : The Gergonne Point

|journal=Journal of Computer-generated Euclidean Geometry

|year=2009

|volume=1

|pages=1–14

|url=http://www.dekovsoft.com/j/2009/01/JCGEG200901.pdf

|url-status=dead

|archive-url=https://web.archive.org/web/20101105045604/http://www.dekovsoft.com/j/2009/01/JCGEG200901.pdf

|archive-date=2010-11-05

}}

Trilinear coordinates for the vertices of the intouch triangle are given by{{Citation needed|date=May 2020}}

:$$\backslash begin\{array\}\{ccccccc\}$$

T_A &=& 0 &:& \sec^2 \frac{B}{2} &:& \sec^2\frac{C}{2} \\[2pt]

T_B &=& \sec^2 \frac{A}{2} &:& 0 &:& \sec^2\frac{C}{2} \\[2pt]

T_C &=& \sec^2 \frac{A}{2} &:& \sec^2\frac{B}{2} &:& 0.

\end{array}

Trilinear coordinates for the Gergonne point are given by{{Citation needed|date=May 2020}}

:$$\backslash sec^2\backslash tfrac\{A\}\{2\}\; :\; \backslash sec^2\backslash tfrac\{B\}\{2\}\; :\; \backslash sec^2\backslash tfrac\{C\}\{2\},$$

or, equivalently, by the Law of Sines,

:$$\backslash frac\{bc\}\{b\; +\; c\; -\; a\}\; :\; \backslash frac\{ca\}\{c\; +\; a\; -\; b\}\; :\; \backslash frac\{ab\}\{a\; +\; b\; -\; c\}.$$

[[File:Incircle and Excircles.svg|right|thumb|300px|

{{legend-line|solid black|Extended sides 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 bisectors}}

{{legend-line|solid #32cd32|External angle bisectors (forming the excentral triangle)}}

]]

An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides, and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides.

The center of an excircle is the intersection of the internal bisector of one angle (at vertex $A$, for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex $A$, or the excenter of $A$. 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}}

While the incenter of $\backslash triangle\; ABC$ has trilinear coordinates $1\; :\; 1\; :\; 1$, the excenters have trilinears

{{Citation needed|date=May 2020}}

:$$\backslash begin\{array\}\{rrcrcr\}$$

J_A = & -1 &:& 1 &:& 1 \\

J_B = & 1 &:& -1 &:& 1 \\

J_C = & 1 &:& 1 &:& -1

\end{array}

The radii of the excircles are called the exradii.

The exradius of the excircle opposite $A$ (so touching $BC$, centered at $J\_A$) is{{harvtxt|Altshiller-Court|1925|p=79}}{{harvtxt|Kay|1969|p=202}}

:$$r\_a\; =\; \backslash frac\{rs\}\{s\; -\; a\}\; =\; \backslash sqrt\{\backslash frac\{s(s\; -\; b)(s\; -\; c)\}\{s\; -\; a\}\},$$ where $s\; =\; \backslash tfrac\{1\}\{2\}(a\; +\; b\; +\; c).$

See Heron's formula.

Source:

Let the excircle at side $AB$ touch at side $AC$ extended at $G$, and let this excircle's

radius be $r\_c$ and its center be $J\_c$. Then $J\_c\; G$ is an altitude of $\backslash triangle\; ACJ\_c$, so $\backslash triangle\; ACJ\_c$ has area $\backslash tfrac12\; br\_c$. By a similar argument, $\backslash triangle\; BCJ\_c$ has area $\backslash tfrac12\; ar\_c$ and $\backslash triangle\; ABJ\_c$ has area $\backslash tfrac12\; cr\_c$. Thus the area $\backslash Delta$ of triangle $\backslash triangle\; ABC$ is

:$$\backslash Delta\; =\; \backslash tfrac12\; (a\; +\; b\; -\; c)r\_c\; =\; (s\; -\; c)r\_c$$.

So, by symmetry, denoting $r$ as the radius of the incircle,

:$$\backslash Delta\; =\; sr\; =\; (s\; -\; a)r\_a\; =\; (s\; -\; b)r\_b\; =\; (s\; -\; c)r\_c$$.

By the Law of Cosines, we have

:$$\backslash cos\; A\; =\; \backslash frac\{b^2\; +\; c^2\; -\; a^2\}\{2bc\}$$

Combining this with the identity $\backslash sin^2\; \backslash !\; A\; +\; \backslash cos^2\; \backslash !\; A\; =\; 1$, we have

:$$\backslash sin\; A\; =\; \backslash frac\{\backslash sqrt\{-a^4\; -\; b^4\; -\; c^4\; +\; 2a^2\; b^2\; +\; 2b^2\; c^2\; +\; 2\; a^2\; c^2\}\}\{2bc\}$$

But $\backslash Delta\; =\; \backslash tfrac12\; bc\; \backslash sin\; A$, and so

:$$\backslash begin\{align\}$$

\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}

which is Heron's formula.

Combining this with $sr\; =\; \backslash Delta$, we have

:$$r^2\; =\; \backslash frac\{\backslash Delta^2\}\{s^2\}\; =\; \backslash frac\{(s\; -\; a)(s\; -\; b)(s\; -\; c)\}\{s\}.$$

Similarly, $(s\; -\; a)r\_a\; =\; \backslash Delta$ gives

:$$\backslash begin\{align\}$$

&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}

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: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.)

:$$\backslash Delta\; =\; \backslash sqrt\{r\; r\_a\; r\_b\; r\_c\}.$$

The circular hull of the excircles is internally tangent to each of the excircles and is thus an Apollonius circle.[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.] The radius of this Apollonius circle is $\backslash tfrac\{r^2\; +\; s^2\}\{4r\}$ where $r$ is the incircle radius and $s$ is the semiperimeter of the triangle.[http://forumgeom.fau.edu/FG2003volume3/FG200320.pdf Stevanovi´c, Milorad R., "The Apollonius circle and related triangle centers", Forum Geometricorum 3, 2003, 187-195.]

The following relations hold among the inradius $r$, the circumradius $R$, the semiperimeter $s$, and the excircle radii $r\_a$, $r\_b$, $r\_c$:{{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}}

:$$\backslash begin\{align\}$$

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}

The circle through the centers of the three excircles has radius $2R$.

If $H$ is the orthocenter of $\backslash triangle\; ABC$, then

:$$\backslash begin\{align\}$$

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}

{{Main|Extouch triangle}}

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

{{legend-line|solid black|Extended sides 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|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 $\backslash triangle\; ABC$ is denoted by the vertices $T\_A$, $T\_B$, and $T\_C$ that are the three points where the excircles touch the reference $\backslash triangle\; ABC$ and where $T\_A$ is opposite of $A$, etc. This $\backslash triangle\; T\_AT\_BT\_C$ is also known as the extouch triangle of $\backslash triangle\; ABC$. The circumcircle of the extouch $\backslash triangle\; T\_AT\_BT\_C$ is called the Mandart circle

(cf. Mandart inellipse).

The three line segments $\backslash overline\{AT\_A\}$, $\backslash overline\{BT\_B\}$ and $\backslash overline\{CT\_C\}$ are called the splitters of the triangle; they each bisect the perimeter of the triangle,{{Citation needed|date=May 2020}}

:$$\backslash overline\{AB\}\; +\; \backslash overline\{BT\_A\}\; =\; \backslash overline\{AC\}\; +\; \backslash overline\{CT\_A\}\; =\; \backslash frac\{1\}\{2\}\backslash left(\; \backslash overline\{AB\}\; +\; \backslash overline\{BC\}\; +\; \backslash overline\{AC\}\; \backslash right).$$

The splitters intersect in a single point, the triangle's Nagel point $N\_a$ (or triangle center X₈).

Trilinear coordinates for the vertices of the extouch triangle are given by{{Citation needed|date=May 2020}}

:$$\backslash begin\{array\}\{ccccccc\}$$

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}

Trilinear coordinates for the Nagel point are given by{{Citation needed|date=May 2020}}

:$$\backslash csc^2\backslash tfrac\{A\}\{2\}\; :\; \backslash csc^2\backslash tfrac\{B\}\{2\}\; :\; \backslash csc^2\backslash tfrac\{C\}\{2\},$$

or, equivalently, by the Law of Sines,

:$$\backslash frac\{b\; +\; c\; -\; a\}\{a\}\; :\; \backslash frac\{c\; +\; a\; -\; b\}\{b\}\; :\; \backslash frac\{a\; +\; b\; -\; c\}\{c\}.$$

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

{{Main|Nine-point circle}}

File:Circ9pnt3.svg

In geometry, the nine-point circle is a circle that can be constructed for any given triangle. It is so named because it passes through nine significant concyclic points defined from the triangle. These nine points are:{{harvtxt|Altshiller-Court|1925|pp=103–110}}{{harvtxt|Kay|1969|pp=18,245}}

• The midpoint of each side of the triangle
• The foot of each altitude
• The midpoint of the line segment from each vertex of the triangle to the orthocenter (where the three altitudes meet; these line segments lie on their respective altitudes).

In 1822, Karl Feuerbach discovered that any triangle's nine-point circle is externally tangent to that triangle's three excircles and internally tangent to its incircle; this result is known as Feuerbach's theorem. He proved that:{{citation |ref={{harvid|Feuerbach|1822}} |last1=Feuerbach |first1=Karl Wilhelm |author1-link=Karl Wilhelm Feuerbach |last2=Buzengeiger |first2=Carl Heribert Ignatz |year=1822 |title=Eigenschaften einiger merkwürdigen Punkte des geradlinigen Dreiecks und mehrerer durch sie bestimmten Linien und Figuren. Eine analytisch-trigonometrische Abhandlung |publisher=Wiessner |location=Nürnberg |edition=Monograph |language=de |url=https://gdz.sub.uni-goettingen.de/id/PPN512512426}}.

:... the circle which passes through the feet of the altitudes of a triangle is tangent to all four circles which in turn are tangent to the three sides of the triangle ... {{harv|Feuerbach|1822}}

The triangle center at which the incircle and the nine-point circle touch is called the Feuerbach point.

The points of intersection of the interior angle bisectors of $\backslash triangle\; ABC$ with the segments $BC$, $CA$, and $AB$ are the vertices of the incentral triangle. Trilinear coordinates for the vertices of the incentral triangle $\backslash triangle\; A\text{'}B\text{'}C\text{'}$ are given by{{Citation needed|date=May 2020}}

:$$\backslash begin\{array\}\{ccccccc\}$$

A' &=& 0 &:& 1 &:& 1 \\[2pt]

B' &=& 1 &:& 0 &:& 1 \\[2pt]

C' &=& 1 &:& 1 &:& 0

\end{array}

The excentral triangle of a reference triangle has vertices at the centers of the reference triangle's excircles. Its sides are on the external angle bisectors of the reference triangle (see figure at top of page). Trilinear coordinates for the vertices of the excentral triangle $\backslash triangle\; A\text{'}B\text{'}C\text{'}$ are given by{{Citation needed|date=May 2020}}

:$$\backslash begin\{array\}\{ccrcrcr\}$$

A' &=& -1 &:& 1 &:& 1\\[2pt]

B' &=& 1 &:& -1 &:& 1 \\[2pt]

C' &=& 1 &:& 1 &:& -1

\end{array}

Let $x:y:z$ be a variable point in trilinear coordinates, and let $u=\backslash cos^2\backslash left\; (\; A/2\; \backslash right\; )$, $v=\backslash cos^2\backslash left\; (\; B/2\; \backslash right\; )$, $w=\backslash cos^2\backslash left\; (\; C/2\; \backslash right\; )$. The four circles described above are given equivalently by either of the two given equations:Whitworth, William Allen. Trilinear Coordinates and Other Methods of Modern Analytical Geometry of Two Dimensions, Forgotten Books, 2012 (orig. Deighton, Bell, and Co., 1866). https://www.forgottenbooks.com/en/search?q=%22Trilinear+coordinates%22{{rp|210–215}}

• Incircle:$$\backslash begin\{align\}$$

u^2 x^2 + v^2 y^2 + w^2 z^2 - 2vwyz - 2wuzx - 2uvxy &= 0 \\[4pt]

{\textstyle \pm\sqrt{x}\cos\tfrac{A}{2} \pm \sqrt{y\vphantom{t}}\cos\tfrac{B}{2} \pm \sqrt{z}\cos\tfrac{C}{2}} &= 0

\end{align}

• $A$-excircle:$$\backslash begin\{align\}$$

u^2 x^2 + v^2 y^2 + w^2 z^2 - 2vwyz + 2wuzx + 2uvxy &= 0 \\[4pt]

{\textstyle \pm\sqrt{-x}\cos\tfrac{A}{2} \pm \sqrt{y\vphantom{t}}\cos\tfrac{B}{2} \pm \sqrt{z}\cos\tfrac{C}{2}} &= 0

\end{align}

• $B$-excircle:$$\backslash begin\{align\}$$

u^2 x^2 + v^2 y^2 + w^2 z^2 + 2vwyz - 2wuzx + 2uvxy &= 0 \\[4pt]

{\textstyle \pm\sqrt{x}\cos\tfrac{A}{2} \pm \sqrt{-y\vphantom{t}}\cos\tfrac{B}{2} \pm \sqrt{z}\cos\tfrac{C}{2}} &= 0

\end{align}

• $C$-excircle:$$\backslash begin\{align\}$$

u^2 x^2 + v^2 y^2 + w^2 z^2 + 2vwyz + 2wuzx - 2uvxy &= 0 \\[4pt]

{\textstyle \pm\sqrt{x}\cos\tfrac{A}{2} \pm \sqrt{y\vphantom{t}}\cos\tfrac{B}{2} \pm \sqrt{-z}\cos\tfrac{C}{2}} &= 0

\end{align}

Euler's theorem states that in a triangle:

:$$(R\; -\; r)^2\; =\; d^2\; +\; r^2,$$

where $R$ and $r$ are the circumradius and inradius respectively, and $d$ is the distance between the circumcenter and the incenter.

For excircles the equation is similar:

:$$\backslash left(R\; +\; r\_\backslash text\{ex\}\backslash right)^2\; =\; d\_\backslash text\{ex\}^2\; +\; r\_\backslash text\{ex\}^2,$$

where $r\_\backslash text\{ex\}$ is the radius of one of the excircles, and $d\_\backslash text\{ex\}$ is the distance between the circumcenter and that excircle's center.Nelson, Roger, "Euler's triangle inequality via proof without words", Mathematics Magazine 81(1), February 2008, 58-61.{{sfn|Johnson|1929|p=187}}[http://forumgeom.fau.edu/FG2001volume1/FG200120.pdf Emelyanov, Lev, and Emelyanova, Tatiana. "Euler's formula and Poncelet's porism", Forum Geometricorum 1, 2001: pp. 137–140.]

Some (but not all) quadrilaterals have an incircle. These are called tangential quadrilaterals. Among their many properties perhaps the most important is that their two pairs of opposite sides have equal sums. This is called the Pitot theorem.{{harvtxt|Josefsson|2011|loc=See in particular pp. 65–66.}}

More generally, a polygon with any number of sides that has an inscribed circle (that is, one that is tangent to each side) is called a tangential polygon.

• {{annotated link|Circumcenter}}
• {{annotated link|Circumcircle}}
• {{annotated link|Circumconic and inconic}}
• {{annotated link|Circumgon}}
• {{annotated link|Ex-tangential quadrilateral}}
• {{annotated link|Harcourt's theorem}}
• {{annotated link|Incenter–excenter lemma}}
• {{annotated link|Inscribed sphere}}
• {{annotated link|Power of a point}}
• {{annotated link|Steiner inellipse}}
• {{annotated link|Tangential quadrilateral}}
• Triangle conic

{{Reflist|30em}}

• {{citation |last1=Altshiller-Court |first1=Nathan |year=1925 |lccn=52013504 |title=College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle |edition=2nd |publisher=Barnes & Noble |location=New York}}
• {{citation

|last=Johnson |first=Roger A.

|year=1929

|title=Modern Geometry

|publisher=Houghton Mifflin

|chapter=X. Inscribed and Escribed Circles

|pages=182–194

|chapter-url=https://archive.org/details/moderngeometry0000unse_q5z5/page/182/

|chapter-url-access=limited

}}

• {{citation |last=Josefsson|first=Martin|title=More characterizations of tangential quadrilaterals|journal=Forum Geometricorum|volume=11 |year=2011 |pages=65–82 |mr=2877281|url=http://forumgeom.fau.edu/FG2011volume11/FG201108.pdf |access-date=2023-03-14 |url-status=dead |archive-url=https://web.archive.org/web/20160304022959/http://forumgeom.fau.edu/FG2011volume11/FG201108.pdf |archive-date=2016-03-04}}
• {{citation |last1=Kay |first1=David C. |year=1969 |lccn=69012075 |title=College Geometry |publisher=Holt, Rinehart, and Winston |location=New York}}
• {{cite journal |last=Kimberling |first=Clark |title=Triangle Centers and Central Triangles |journal=Congressus Numerantium |issue=129 |year=1998 |pages=i-xxv,1–295}}
• {{cite journal |last=Kiss |first=Sándor |title=The Orthic-of-Intouch and Intouch-of-Orthic Triangles |journal=Forum Geometricorum |issue=6 |year=2006 |pages=171–177}}

• [https://mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-incircle Derivation of formula for radius of incircle of a triangle], MATHalino
• {{MathWorld|title=Incircle|urlname=Incircle}}

• [https://www.mathopenref.com/triangleincenter.html Triangle incenter]; [https://www.mathopenref.com/triangleincircle.html Triangle incircle]; [https://www.mathopenref.com/polygonincircle.html Incircle of a regular polygon] (with interactive animations)
• [https://www.mathopenref.com/constincircle.html Constructing a triangle's incenter / incircle with compass and straightedge] An interactive animated demonstration
• [https://www.cut-the-knot.org/Curriculum/Geometry/AdjacentIncircles.shtml Equal Incircles Theorem] at cut-the-knot
• [https://www.cut-the-knot.org/Curriculum/Geometry/FourIncircles.shtml Five Incircles Theorem] at cut-the-knot
• [https://www.cut-the-knot.org/Curriculum/Geometry/IncirclesInQuadri.shtml Pairs of Incircles in a Quadrilateral] at cut-the-knot
• [https://web.archive.org/web/20151105214641/http://www.uff.br/trianglecenters/X0001.html An interactive Java applet for the incenter]

Category:Circles defined for a triangle