Incircle and excircles

Incircle and excircles of a triangle.

  Extended sides of triangle △ABC

  Incircle (incenter at I)

  Excircles (excenters at J_A, J_B, J_C)

  Internal angle bisectors

  External angle bisectors (forming the excentral triangle)

In geometry, the incircle or inscribed circle of a triangle is the largest circle that can be contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.^[1]

An excircle or escribed circle^[2] 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.^[3]

The center of the incircle, called the incenter, can be found as the intersection of the three internalangle bisectors.^[3][4] 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.^[3] 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.^[5]

Incircle and Incenter

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

Also let ${\displaystyle T_{A}}$ , ${\displaystyle T_{B}}$ , and ${\displaystyle T_{C}}$ be the touchpoints where the incircle touches ${\displaystyle {\overline {BC}}}$ , ${\displaystyle {\overline {AC}}}$ , and ${\displaystyle {\overline {AB}}}$ .

Incenter

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

Trilinear coordinates

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^[6]

$${\displaystyle \ 1:1:1.}$$

Barycentric coordinates

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

$${\displaystyle a:b:c}$$

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

$${\displaystyle \sin A:\sin B:\sin C}$$

where ${\displaystyle A}$ , ${\displaystyle B}$ , and ${\displaystyle C}$ are the angles at the three vertices.

Cartesian coordinates

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 ${\displaystyle (x_{a},y_{a})}$ , ${\displaystyle (x_{b},y_{b})}$ , and ${\displaystyle (x_{c},y_{c})}$ , and the sides opposite these vertices have corresponding lengths ${\displaystyle a}$ , ${\displaystyle b}$ , and ${\displaystyle c}$ , then the incenter is at

$${\displaystyle \left({\frac {ax_{a}+bx_{b}+cx_{c}}{a+b+c}},{\frac {ay_{a}+by_{b}+cy_{c}}{a+b+c}}\right)={\frac {a\left(x_{a},y_{a}\right)+b\left(x_{b},y_{b}\right)+c\left(x_{c},y_{c}\right)}{a+b+c}}.}$$

Radius

The inradius ${\displaystyle r}$ of the incircle in a triangle with sides of length ${\displaystyle a}$ , ${\displaystyle b}$ , ${\displaystyle c}$ is given by^[7]

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

where ${\displaystyle s={\tfrac {1}{2}}(a+b+c)}$ is the semiperimeter (see Heron's formula).

The tangency points of the incircle divide the sides into segments of lengths ${\displaystyle s-a}$ from ${\displaystyle A}$ , ${\displaystyle s-b}$ from ${\displaystyle B}$ , and ${\displaystyle s-c}$ from ${\displaystyle C}$ (see Tangent lines to a circle).^[8]

Distances to the vertices

Denote the incenter of ${\displaystyle \triangle ABC}$ as ${\displaystyle I}$ .

The distance from vertex ${\displaystyle A}$ to the incenter ${\displaystyle I}$ is:

$${\displaystyle {\overline {AI}}=d(A,I)=c\,{\frac {\sin {\frac {B}{2}}}{\cos {\frac {C}{2}}}}=b\,{\frac {\sin {\frac {C}{2}}}{\cos {\frac {B}{2}}}}.}$$

Derivation of the formula stated above

Use the Law of sines in the triangle ${\displaystyle \triangle IAB}$ .

We get ${\displaystyle {\frac {\overline {AI}}{\sin {\frac {B}{2}}}}={\frac {c}{\sin \angle AIB}}}$ . We have that ${\displaystyle \angle AIB=\pi -{\frac {A}{2}}-{\frac {B}{2}}={\frac {\pi }{2}}+{\frac {C}{2}}}$ .

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

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^[9]

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

Additionally,^[10]

$${\displaystyle {\overline {IA}}\cdot {\overline {IB}}\cdot {\overline {IC}}=4Rr^{2},}$$

where ${\displaystyle R}$ and ${\displaystyle r}$ are the triangle's circumradius and inradius respectively.

Other properties

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.^[6]

Incircle and its radius properties

Distances between vertex and nearest touchpoints

The distances from a vertex to the two nearest touchpoints are equal; for example:^[11]

$${\displaystyle d\left(A,T_{B}\right)=d\left(A,T_{C}\right)={\tfrac {1}{2}}(b+c-a)=s-a.}$$

Other properties

If the altitudes from sides of lengths ${\displaystyle a}$ , ${\displaystyle b}$ , and ${\displaystyle c}$ are ${\displaystyle h_{a}}$ , ${\displaystyle h_{b}}$ , and ${\displaystyle h_{c}}$ , then the inradius ${\displaystyle r}$ is one third the harmonic mean of these altitudes; that is,^[12]

$${\displaystyle r={\frac {1}{{\dfrac {1}{h_{a}}}+{\dfrac {1}{h_{b}}}+{\dfrac {1}{h_{c}}}}}.}$$

The product of the incircle radius ${\displaystyle r}$ and the circumcircle radius ${\displaystyle R}$ of a triangle with sides ${\displaystyle a}$ , ${\displaystyle b}$ , and ${\displaystyle c}$ is^[13]

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

Some relations among the sides, incircle radius, and circumcircle radius are:^[14]

$${\displaystyle {\begin{aligned}ab+bc+ca&=s^{2}+(4R+r)r,\\a^{2}+b^{2}+c^{2}&=2s^{2}-2(4R+r)r.\end{aligned}}}$$

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.^[15]

The incircle radius is no greater than one-ninth the sum of the altitudes.^{[16]: 289}

The squared distance from the incenter ${\displaystyle I}$ to the circumcenter${\displaystyle O}$ is given by^{[17]: 232}

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

and the distance from the incenter to the center ${\displaystyle N}$ of the nine point circle is^{[17]: 232}

$${\displaystyle {\overline {IN}}={\tfrac {1}{2}}(R-2r)<{\tfrac {1}{2}}R.}$$

The incenter lies in the medial triangle (whose vertices are the midpoints of the sides).^{[17]: 233, Lemma 1}

Relation to area of the triangle

The radius of the incircle is related to the area of the triangle.^[18] The ratio of the area of the incircle to the area of the triangle is less than or equal to ${\displaystyle \pi {\big /}3{\sqrt {3}}}$ , with equality holding only for equilateral triangles.^[19]

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

Now, the incircle is tangent to ${\displaystyle {\overline {AB}}}$ at some point ${\displaystyle T_{C}}$ , and so ${\displaystyle \angle AT_{C}I}$ is right. Thus, the radius ${\displaystyle T_{C}I}$ is an altitude of ${\displaystyle \triangle IAB}$ .

Therefore, ${\displaystyle \triangle IAB}$ has base length ${\displaystyle c}$ and height ${\displaystyle r}$ , and so has area ${\displaystyle {\tfrac {1}{2}}cr}$ .

Proof without words that the area of a triangle equals the product of its inradius and its semiperimeter

Similarly, ${\displaystyle \triangle IAC}$ has area ${\displaystyle {\tfrac {1}{2}}br}$ and ${\displaystyle \triangle IBC}$ has area ${\displaystyle {\tfrac {1}{2}}ar}$ .

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

$${\displaystyle \Delta ={\tfrac {1}{2}}(a+b+c)r=sr,}$$

     and     ${\displaystyle r={\frac {\Delta }{s}},}$

where ${\displaystyle \Delta }$ is the area of ${\displaystyle \triangle ABC}$ and ${\displaystyle s={\tfrac {1}{2}}(a+b+c)}$ is its semiperimeter.

For an alternative formula, consider ${\displaystyle \triangle IT_{C}A}$ . This is a right-angled triangle with one side equal to ${\displaystyle r}$ and the other side equal to ${\displaystyle r\cot {\tfrac {A}{2}}}$ . The same is true for ${\displaystyle \triangle IB'A}$ . The large triangle is composed of six such triangles and the total area is:

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

Gergonne triangle and point

  Triangle △ABC

  Incircle (incenter at I)

  Contact triangle △T_AT_BT_C

  Lines between opposite vertices of △ABC and △T_AT_BT_C (concur at Gergonne point G_e)

The Gergonne triangle (of ${\displaystyle \triangle ABC}$ ) is defined by connecting the three touchpoints of the incircle on the three sides. The touchpoint opposite ${\displaystyle A}$ is denoted ${\displaystyle T_{A}}$ , etc.

This Gergonne triangle, ${\displaystyle \triangle T_{A}T_{B}T_{C}}$ , is also known as the contact triangle or intouch triangle of ${\displaystyle \triangle ABC}$ . Its area is

$${\displaystyle K_{T}=K{\frac {2r^{2}s}{abc}}}$$

where ${\displaystyle K}$ , ${\displaystyle r}$ , and ${\displaystyle s}$ are the area, radius of the incircle, and semiperimeter of the original triangle, and ${\displaystyle a}$ , ${\displaystyle b}$ , and ${\displaystyle c}$ are the side lengths of the original triangle. This is the same area as that of the extouch triangle.^[20]

The three lines ${\displaystyle AT_{A}}$ , ${\displaystyle BT_{B}}$ , and ${\displaystyle CT_{C}}$ intersect in a single point called the Gergonne point, denoted as ${\displaystyle G_{e}}$ (or triangle centerX₇). The Gergonne point lies in the open orthocentroidal disk punctured at its own center, and can be any point therein.^[21]

The Gergonne point of a triangle has a number of properties, including that it is the symmedian point of the Gergonne triangle.^[22]

Trilinear coordinates for the vertices of the intouch triangle are given by

$${\displaystyle {\begin{array}{ccccccc}T_{A}&=&0&:&\sec ^{2}{\frac {B}{2}}&:&\sec ^{2}{\frac {C}{2}}\\[2pt]T_{B}&=&\sec ^{2}{\frac {A}{2}}&:&0&:&\sec ^{2}{\frac {C}{2}}\\[2pt]T_{C}&=&\sec ^{2}{\frac {A}{2}}&:&\sec ^{2}{\frac {B}{2}}&:&0.\end{array}}}$$

Trilinear coordinates for the Gergonne point are given by

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

or, equivalently, by the law of cosines,

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

Excircles and excenters

  Extended sides of △ABC

  Incircle (incenter at I)

  Excircles (excenters at J_A, J_B, J_C)

  Internal angle bisectors

  External angle bisectors (forming the excentral triangle)

An excircle or escribed circle^[2] 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.^[3]

The center of an excircle is the intersection of the internal bisector of one angle (at vertex ${\displaystyle A}$ , for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex ${\displaystyle A}$ , or the excenter of ${\displaystyle A}$ .^[3] 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.^[5]

Trilinear coordinates of excenters

While the incenter of ${\displaystyle \triangle ABC}$ has trilinear coordinates${\displaystyle 1:1:1}$ , the excenters have trilinears

$${\displaystyle {\begin{array}{rrcrcr}J_{A}=&-1&:&1&:&1\\J_{B}=&1&:&-1&:&1\\J_{C}=&1&:&1&:&-1\end{array}}}$$

Exradii

The radii of the excircles are called the exradii.

The exradius of the excircle opposite ${\displaystyle A}$ (so touching ${\displaystyle BC}$ , centered at ${\displaystyle J_{A}}$ ) is^[23][24]

$${\displaystyle r_{a}={\frac {rs}{s-a}}={\sqrt {\frac {s(s-b)(s-c)}{s-a}}},}$$ where ${\displaystyle s={\tfrac {1}{2}}(a+b+c).}$

See Heron's formula.

Derivation of exradii formula

Source:^[23]

Let the excircle at side ${\displaystyle AB}$ touch at side ${\displaystyle AC}$ extended at ${\displaystyle G}$ , and let this excircle's radius be ${\displaystyle r_{c}}$ and its center be ${\displaystyle J_{c}}$ . Then ${\displaystyle J_{c}G}$ is an altitude of ${\displaystyle \triangle ACJ_{c}}$ , so ${\displaystyle \triangle ACJ_{c}}$ has area ${\displaystyle {\tfrac {1}{2}}br_{c}}$ . By a similar argument, ${\displaystyle \triangle BCJ_{c}}$ has area ${\displaystyle {\tfrac {1}{2}}ar_{c}}$ and ${\displaystyle \triangle ABJ_{c}}$ has area ${\displaystyle {\tfrac {1}{2}}cr_{c}}$ . Thus the area ${\displaystyle \Delta }$ of triangle ${\displaystyle \triangle ABC}$ is

$${\displaystyle \Delta ={\tfrac {1}{2}}(a+b-c)r_{c}=(s-c)r_{c}}$$ .

So, by symmetry, denoting ${\displaystyle r}$ as the radius of the incircle,

$${\displaystyle \Delta =sr=(s-a)r_{a}=(s-b)r_{b}=(s-c)r_{c}}$$ .

By the Law of Cosines, we have

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

Combining this with the identity ${\displaystyle \sin ^{2}\!A+\cos ^{2}\!A=1}$ , we have

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

But ${\displaystyle \Delta ={\tfrac {1}{2}}bc\sin A}$ , and so

$${\displaystyle {\begin{aligned}\Delta &={\tfrac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}}}\\[5mu]&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\[5mu]&={\sqrt {s(s-a)(s-b)(s-c)}},\end{aligned}}}$$

which is Heron's formula.

Combining this with ${\displaystyle sr=\Delta }$ , we have

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

Similarly, ${\displaystyle (s-a)r_{a}=\Delta }$ gives

$${\displaystyle {\begin{aligned}&r_{a}^{2}={\frac {s(s-b)(s-c)}{s-a}}\\[4pt]&\implies r_{a}={\sqrt {\frac {s(s-b)(s-c)}{s-a}}}.\end{aligned}}}$$

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:^[25]

$${\displaystyle \Delta ={\sqrt {rr_{a}r_{b}r_{c}}}.}$$

Other excircle properties

The circular hull of the excircles is internally tangent to each of the excircles and is thus an Apollonius circle.^[26] The radius of this Apollonius circle is ${\displaystyle {\tfrac {r^{2}+s^{2}}{4r}}}$ where ${\displaystyle r}$ is the incircle radius and ${\displaystyle s}$ is the semiperimeter of the triangle.^[27]

The following relations hold among the inradius ${\displaystyle r}$ , the circumradius ${\displaystyle R}$ , the semiperimeter ${\displaystyle s}$ , and the excircle radii ${\displaystyle r_{a}}$ , ${\displaystyle r_{b}}$ , ${\displaystyle r_{c}}$ :^[14]

$${\displaystyle {\begin{aligned}r_{a}+r_{b}+r_{c}&=4R+r,\\r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}&=s^{2},\\r_{a}^{2}+r_{b}^{2}+r_{c}^{2}&=\left(4R+r\right)^{2}-2s^{2}.\end{aligned}}}$$

The circle through the centers of the three excircles has radius ${\displaystyle 2R}$ .^[14]

If ${\displaystyle H}$ is the orthocenter of ${\displaystyle \triangle ABC}$ , then^[14]

$${\displaystyle {\begin{aligned}r_{a}+r_{b}+r_{c}+r&={\overline {AH}}+{\overline {BH}}+{\overline {CH}}+2R,\\r_{a}^{2}+r_{b}^{2}+r_{c}^{2}+r^{2}&={\overline {AH}}^{2}+{\overline {BH}}^{2}+{\overline {CH}}^{2}+(2R)^{2}.\end{aligned}}}$$

Nagel triangle and Nagel point

  Extended sides of triangle △ABC

  Excircles of △ABC (tangent at T_A. T_B, T_C)

  Nagel/Extouch triangle△T_AT_BT_C

  Splitters: lines connecting opposite vertices of △ABC and △T_AT_BT_C (concur at Nagel pointN)

The Nagel triangle or extouch triangle of ${\displaystyle \triangle ABC}$ is denoted by the vertices ${\displaystyle T_{A}}$ , ${\displaystyle T_{B}}$ , and ${\displaystyle T_{C}}$ that are the three points where the excircles touch the reference ${\displaystyle \triangle ABC}$ and where ${\displaystyle T_{A}}$ is opposite of ${\displaystyle A}$ , etc. This ${\displaystyle \triangle T_{A}T_{B}T_{C}}$ is also known as the extouch triangle of ${\displaystyle \triangle ABC}$ . The circumcircle of the extouch ${\displaystyle \triangle T_{A}T_{B}T_{C}}$ is called the Mandart circle (cf. Mandart inellipse).

The three line segments ${\displaystyle {\overline {AT_{A}}}}$ , ${\displaystyle {\overline {BT_{B}}}}$ and ${\displaystyle {\overline {CT_{C}}}}$ are called the splitters of the triangle; they each bisect the perimeter of the triangle,

$${\displaystyle {\overline {AB}}+{\overline {BT_{A}}}={\overline {AC}}+{\overline {CT_{A}}}={\frac {1}{2}}\left({\overline {AB}}+{\overline {BC}}+{\overline {AC}}\right).}$$

The splitters intersect in a single point, the triangle's Nagel point${\displaystyle N_{a}}$ (or triangle centerX₈).

Trilinear coordinates for the vertices of the extouch triangle are given by

$${\displaystyle {\begin{array}{ccccccc}T_{A}&=&0&:&\csc ^{2}{\frac {B}{2}}&:&\csc ^{2}{\frac {C}{2}}\\[2pt]T_{B}&=&\csc ^{2}{\frac {A}{2}}&:&0&:&\csc ^{2}{\frac {C}{2}}\\[2pt]T_{C}&=&\csc ^{2}{\frac {A}{2}}&:&\csc ^{2}{\frac {B}{2}}&:&0\end{array}}}$$

Trilinear coordinates for the Nagel point are given by

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

or, equivalently, by the Law of Sines,

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

Barycentric coordinates for the Nagel point are therefore

$${\displaystyle b+c-a:c+a-b:a+b-c,}$$

or equivalently

$${\displaystyle s-a:s-b:s-c.}$$

The Nagel point is the isotomic conjugate of the Gergonne point.

Related constructions

Nine-point circle and Feuerbach point

The nine-point circle is tangent to the incircle and excircles

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:^[28][29]

• 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:^[30]

... 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 ... (Feuerbach 1822)

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

The incircle may be described as the pedal circle of the incenter. The locus of points whose pedal circles are tangent to the nine-point circle is known as the McCay cubic.

Incentral and excentral triangles

The points of intersection of the interior angle bisectors of ${\displaystyle \triangle ABC}$ with the segments ${\displaystyle BC}$ , ${\displaystyle CA}$ , and ${\displaystyle AB}$ are the vertices of the incentral triangle. Trilinear coordinates for the vertices of the incentral triangle ${\displaystyle \triangle A'B'C'}$ are given by

$${\displaystyle {\begin{array}{ccccccc}A'&=&0&:&1&:&1\\[2pt]B'&=&1&:&0&:&1\\[2pt]C'&=&1&:&1&:&0\end{array}}}$$

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 ${\displaystyle \triangle A'B'C'}$ are given by

$${\displaystyle {\begin{array}{ccrcrcr}A'&=&-1&:&1&:&1\\[2pt]B'&=&1&:&-1&:&1\\[2pt]C'&=&1&:&1&:&-1\end{array}}}$$

Equations for four circles

Let ${\displaystyle x:y:z}$ be a variable point in trilinear coordinates, and let ${\displaystyle u=\cos ^{2}\left(A/2\right)}$ , ${\displaystyle v=\cos ^{2}\left(B/2\right)}$ , ${\displaystyle w=\cos ^{2}\left(C/2\right)}$ . The four circles described above are given equivalently by either of the two given equations:^{[31]: 210–215}

• Incircle:$${\displaystyle {\begin{aligned}u^{2}x^{2}+v^{2}y^{2}+w^{2}z^{2}-2vwyz-2wuzx-2uvxy&=0\\[4pt]{\textstyle \pm {\sqrt {x}}\cos {\tfrac {A}{2}}\pm {\sqrt {y{\vphantom {t}}}}\cos {\tfrac {B}{2}}\pm {\sqrt {z}}\cos {\tfrac {C}{2}}}&=0\end{aligned}}}$$
• ${\displaystyle A}$ -excircle:$${\displaystyle {\begin{aligned}u^{2}x^{2}+v^{2}y^{2}+w^{2}z^{2}-2vwyz+2wuzx+2uvxy&=0\\[4pt]{\textstyle \pm {\sqrt {-x}}\cos {\tfrac {A}{2}}\pm {\sqrt {y{\vphantom {t}}}}\cos {\tfrac {B}{2}}\pm {\sqrt {z}}\cos {\tfrac {C}{2}}}&=0\end{aligned}}}$$
• ${\displaystyle B}$ -excircle:$${\displaystyle {\begin{aligned}u^{2}x^{2}+v^{2}y^{2}+w^{2}z^{2}+2vwyz-2wuzx+2uvxy&=0\\[4pt]{\textstyle \pm {\sqrt {x}}\cos {\tfrac {A}{2}}\pm {\sqrt {-y{\vphantom {t}}}}\cos {\tfrac {B}{2}}\pm {\sqrt {z}}\cos {\tfrac {C}{2}}}&=0\end{aligned}}}$$
• ${\displaystyle C}$ -excircle:$${\displaystyle {\begin{aligned}u^{2}x^{2}+v^{2}y^{2}+w^{2}z^{2}+2vwyz+2wuzx-2uvxy&=0\\[4pt]{\textstyle \pm {\sqrt {x}}\cos {\tfrac {A}{2}}\pm {\sqrt {y{\vphantom {t}}}}\cos {\tfrac {B}{2}}\pm {\sqrt {-z}}\cos {\tfrac {C}{2}}}&=0\end{aligned}}}$$

Euler's theorem

Euler's theorem states that in a triangle:

$${\displaystyle (R-r)^{2}=d^{2}+r^{2},}$$

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

For excircles the equation is similar:

$${\displaystyle \left(R+r_{\text{ex}}\right)^{2}=d_{\text{ex}}^{2}+r_{\text{ex}}^{2},}$$

where ${\displaystyle r_{\text{ex}}}$ is the radius of one of the excircles, and ${\displaystyle d_{\text{ex}}}$ is the distance between the circumcenter and that excircle's center.^[32][33][34]

Generalization to other polygons

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.^[35]

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.

Generalization to topological triangles

If topological triangles are considered, it is still possible to define an inscribed circle. It is no longer described as tangent to all sides, since the topological triangle might not be differentiable everywhere. Rather, it is defined as a circle whose center has the same minimal distance to each side. It has been proven that all topological triangles have an inscribed circle.^[36]

See also

• Circumcenter – Circle that passes through the vertices of a trianglePages displaying short descriptions of redirect targets
• Circumcircle – Circle that passes through the vertices of a triangle
• Circumconic and inconic – Conic section that passes through the vertices of a triangle or is tangent to its sides
• Circumgon – Geometric figure which circumscribes a circle
• Ex-tangential quadrilateral – Convex 4-sided polygon whose sidelines are all tangent to an outside circle
• Harcourt's theorem – Area of a triangle from its sides and vertex distances to any line tangent to its incircle
• Incenter–excenter lemma – Theorem about inscribed and circumscribed circles
• Inscribed sphere – Sphere tangent to every face of a polyhedron
• Power of a point – Relative distance of a point from a circle
• Steiner inellipse – Unique ellipse tangent to all 3 midpoints of a given triangle's sides
• Tangential quadrilateral – Polygon whose four sides all touch a circle
• Triangle conic

Notes

1. Kay (1969, p. 140)
2. Altshiller-Court (1925, p. 74)
3. Altshiller-Court (1925, p. 73)
4. Kay (1969, p. 117)
5. Johnson 1929, p. 182.
6. Encyclopedia of Triangle CentersArchived 2012-04-19 at the Wayback Machine, accessed 2014-10-28.
7. Kay (1969, p. 201)
8. Chu, Thomas, The Pentagon, Spring 2005, p. 45, problem 584.
9. Allaire, Patricia R.; Zhou, Junmin; Yao, Haishen (March 2012), "Proving a nineteenth century ellipse identity", Mathematical Gazette, 96: 161–165, doi:10.1017/S0025557200004277, S2CID 124176398.
10. Altshiller-Court, Nathan (1980), College Geometry, Dover Publications. #84, p. 121.
11. Mathematical Gazette, July 2003, 323-324.
12. Kay (1969, p. 203)
13. Johnson 1929, p. 189, #298(d).
14. Bell, Amy. ""Hansen's right triangle theorem, its converse and a generalization", Forum Geometricorum 6, 2006, 335–342"(PDF). Archived from the original(PDF) on 2021-08-31. Retrieved 2012-05-05.
15. Kodokostas, Dimitrios, "Triangle Equalizers", Mathematics Magazine 83, April 2010, pp. 141-146.
16. Posamentier, Alfred S., and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.
17. Franzsen, William N. (2011). "The distance from the incenter to the Euler line"(PDF). Forum Geometricorum. 11: 231–236. MR 2877263. Archived from the original(PDF) on 2020-12-05. Retrieved 2012-05-09..
18. Coxeter, H.S.M. "Introduction to Geometry 2nd ed. Wiley, 1961.
19. Minda, D., and Phelps, S., "Triangles, ellipses, and cubic polynomials", American Mathematical Monthly 115, October 2008, 679-689: Theorem 4.1.
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References

• Altshiller-Court, Nathan (1925), College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle (2nd ed.), New York: Barnes & Noble, LCCN 52013504
• Johnson, Roger A. (1929), "X. Inscribed and Escribed Circles", Modern Geometry, Houghton Mifflin, pp. 182–194
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• Kay, David C. (1969), College Geometry, New York: Holt, Rinehart, and Winston, LCCN 69012075
• Kimberling, Clark (1998). "Triangle Centers and Central Triangles". Congressus Numerantium (129): i–xxv, 1–295.
• Kiss, Sándor (2006). "The Orthic-of-Intouch and Intouch-of-Orthic Triangles". Forum Geometricorum (6): 171–177.

External links

• Derivation of formula for radius of incircle of a triangle, MATHalino
• Weisstein, Eric W."Incircle". MathWorld.

Interactive

• Triangle incenter; Triangle incircle; Incircle of a regular polygon (with interactive animations)
• Constructing a triangle's incenter / incircle with compass and straightedge An interactive animated demonstration
• Equal Incircles Theorem at cut-the-knot
• Five Incircles Theorem at cut-the-knot
• Pairs of Incircles in a Quadrilateral at cut-the-knot
• An interactive Java applet for the incenter