Incircle and excircles

Incircle and excircles of a triangle. Template:Legend-line Template:Legend-line Template:Legend-line Template:Legend-line Template:Legend-line

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.

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 the incircle, called the incenter, can be found as the intersection of the three internal angle bisectors. The center of an excircle is the intersection of the internal bisector of one angle (at vertex Template:Mvar, for example) and the external bisectors of the other two. The center of this excircle is called the excenter relative to the vertex Template:Mvar, or the excenter of Template:Mvar. 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.Template:Sfn

Incircle and Incenter

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

Also let $T_A$, $T_B$, and $T_C$ be the touchpoints where the incircle touches $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$.

Incenter

The incenter is the point where the internal angle bisectors of $\mathrm{\angle }ABC,$ $\mathrm{\angle }BCA,$ and $\mathrm{\angle }BAC$ 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 $$1:1:1.$$

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 $$a:b:c$$ where $a$, $b$, and $c$ are the lengths of the sides of the triangle, or equivalently (using the law of sines) by $$\sin A:\sin B:\sin C$$ where $A$, $B$, and $C$ 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 $(x_a,y_a)$, $(x_b,y_b)$, and $(x_c,y_c)$, and the sides opposite these vertices have corresponding lengths $a$, $b$, and $c$, then the incenter is atTemplate:Citation needed $$(\frac{a{x}_a+b{x}_b+c{x}_c}{a+b+c},\frac{a{y}_a+b{y}_b+c{y}_c}{a+b+c})=\frac{a(x_a,y_a)+b(x_b,y_b)+c(x_c,y_c)}{a+b+c}.$$

Distances to the vertices

Denote the incenter of $\mathrm{△}ABC$ as $I$.

The distance from vertex $A$ to the incenter $I$ is: $$\overline{AI}=d(A,I)=c\frac{\sin \frac{B}{2}}{\cos \frac{C}{2}}=b\frac{\sin \frac{C}{2}}{\cos \frac{B}{2}}.$$

Derivation of the formula stated above

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

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

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

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

Additionally, $$\overline{IA}\cdot \overline{IB}\cdot \overline{IC}=4R{r}^2,$$ where $R$ and $r$ 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.

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 from vertex $A$: $$d(A,T_B)=d(A,T_C)=\frac{1}{2}(b+c-a)=s-a,$$ where $s=\frac{1}{2}(a+b+c)$ is the semiperimeter.

Similarly, the tangency points of the incircle divide the sides into segments of lengths $s-b$ from $B$, and $s-c$ from $C$ (see Tangent lines to a circle).

Radius

The inradius $r$ of the incircle in a triangle with sides of length $a$, $b$, $c$ is given by $$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}},$$ where $s=\frac{1}{2}(a+b+c)$ is the semiperimeter (see Heron's formula).

Relation to area of the triangle

Template:Redirect The radius of the incircle is related to the area of the triangle. The ratio of the area of the incircle to the area of the triangle is less than or equal to $\pi /3\sqrt{3}$, with equality holding only for equilateral triangles.

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

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

Now, the incircle is tangent to $\overline{AB}$ at some point $T_C$, and so $\mathrm{\angle }A{T}_{C}I$ is right. Thus, the radius $T_{C}I$ is an altitude of $\mathrm{△}IAB$.

Therefore, $\mathrm{△}IAB$ has base length $c$ and height $r$, and so has area $\frac{1}{2}cr$.

Similarly, $\mathrm{△}IAC$ has area $\frac{1}{2}br$ and $\mathrm{△}IBC$ has area $\frac{1}{2}ar$.

Since these three triangles decompose $\mathrm{△}ABC$, we see that the area $\mathrm{\Delta }$ of $\mathrm{△}ABC$ is: $$\mathrm{\Delta }=\frac{1}{2}(a+b+c)r=sr,$$ and $$r=\frac{\mathrm{\Delta }}{s},$$ where $\mathrm{\Delta }$ is the area of $\mathrm{△}ABC$ and $s=\frac{1}{2}(a+b+c)$ is its semiperimeter.

For an alternative formula, consider $\mathrm{△}I{T}_{C}A$. This is a right-angled triangle with one side equal to $r$ and the other side equal to $r\cot \frac{A}{2}$. The same is true for $\mathrm{△}I{B}^{\prime }A$. The large triangle is composed of six such triangles and the total area is:Template:Citation needed $$\mathrm{\Delta }=r^2(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}).$$

Other properties

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 the harmonic mean of these altitudes; that is, $$r=\frac{1}{\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}}.$$

The product of the incircle radius $r$ and the circumcircle radius $R$ of a triangle with sides $a$, $b$, and $c$ isTemplate:Sfn $$rR=\frac{abc}{2(a+b+c)}.$$

Some relations among the sides, incircle radius, and circumcircle radius are: $$\begin{array}{cc}ab+bc+ca& =s^2+(4R+r)r,\\ a^2+b^2+c^2& =2s^2-2(4R+r)r.\end{array}$$

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.

The incircle radius is no greater than one-ninth the sum of the altitudes.Template:Rp

The squared distance from the incenter $I$ to the circumcenter $O$ is given byTemplate:Rp $${\overline{OI}}^2=R(R-2r)=\frac{abc}{a+b+c}[\frac{abc}{(a+b-c)(a-b+c)(-a+b+c)}-1]$$ and the distance from the incenter to the center $N$ of the nine point circle isTemplate:Rp $$\overline{IN}=\frac{1}{2}(R-2r)<\frac{1}{2}R.$$

The incenter lies in the medial triangle (whose vertices are the midpoints of the sides).Template:Rp

Gergonne triangle and point

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

This Gergonne triangle, $\mathrm{△}{T}_A{T}_B{T}_C$, is also known as the contact triangle or intouch triangle of $\mathrm{△}ABC$. Its area is $$K_T=K\frac{2r^{2}s}{abc}$$ 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.

The three lines $A{T}_A$, $B{T}_B$, and $C{T}_C$ intersect in a single point called the Gergonne point, denoted as $G_e$ (or triangle center Template:Math). The Gergonne point lies in the open orthocentroidal disk punctured at its own center, and can be any point therein.

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

Trilinear coordinates for the vertices of the intouch triangle are given byTemplate:Citation needed $$\begin{array}{cccccc}{T}_A=& 0& :& {\sec }^2\frac{B}{2}& :& {\sec }^2\frac{C}{2}\\ T_B=& {\sec }^2\frac{A}{2}& :& 0& :& {\sec }^2\frac{C}{2}\\ T_C=& {\sec }^2\frac{A}{2}& :& {\sec }^2\frac{B}{2}& :& 0.\end{array}$$

Trilinear coordinates for the Gergonne point are given byTemplate:Citation needed $${\sec }^2\frac{A}{2}:{\sec }^2\frac{B}{2}:{\sec }^2\frac{C}{2},$$ or, equivalently, by the law of cosines, $$\frac{bc}{b+c-a}:\frac{ca}{c+a-b}:\frac{ab}{a+b-c}.$$

Excircles and excenters

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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.Template:Sfn

Trilinear coordinates of excenters

While the incenter of $\mathrm{△}ABC$ has trilinear coordinates $1:1:1$, the excenters have trilinears Template:Citation needed $$\begin{array}{cccccc}{J}_A=& -1& :& 1& :& 1\\ J_B=& 1& :& -1& :& 1\\ J_C=& 1& :& 1& :& -1\end{array}$$

Exradii

The radii of the excircles are called the exradii.

The exradius of the excircle opposite $A$ (so touching $BC$, centered at $J_A$) is $$r_a=\frac{rs}{s-a}=\sqrt{\frac{s(s-b)(s-c)}{s-a}},$$ where $s=\frac{1}{2}(a+b+c).$

See Heron's formula.

Derivation of exradii 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 $\mathrm{△}AC{J}_c$, so $\mathrm{△}AC{J}_c$ has area $\frac{1}{2}b{r}_c$. By a similar argument, $\mathrm{△}BC{J}_c$ has area $\frac{1}{2}a{r}_c$ and $\mathrm{△}AB{J}_c$ has area $\frac{1}{2}c{r}_c$. Thus the area $\mathrm{\Delta }$ of triangle $\mathrm{△}ABC$ is $$\mathrm{\Delta }=\frac{1}{2}(a+b-c)r_c=(s-c)r_c.$$

So, by symmetry, denoting $r$ as the radius of the incircle, $$\mathrm{\Delta }=sr=(s-a)r_a=(s-b)r_b=(s-c)r_c.$$

By the Law of Cosines, we have $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$

Combining this with the identity ${\sin }^{2}A+{\cos }^{2}A=1$, we have $$\sin A=\frac{\sqrt{-a^4-b^4-c^4+2a^2{b}^2+2b^2{c}^2+2a^2{c}^2}}{2bc}$$

But $\mathrm{\Delta }=\frac{1}{2}bc\sin A$, and so $$\begin{array}{cc}\mathrm{\Delta }& =\frac{1}{4}\sqrt{-a^4-b^4-c^4+2a^2{b}^2+2b^2{c}^2+2a^2{c}^2}\\ & =\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\\ & =\sqrt{s(s-a)(s-b)(s-c)},\end{array}$$ which is Heron's formula.

Combining this with $sr=\mathrm{\Delta }$, we have $$r^2=\frac{{\mathrm{\Delta }}^2}{s^2}=\frac{(s-a)(s-b)(s-c)}{s}.$$

Similarly, $(s-a)r_a=\mathrm{\Delta }$ gives $$\begin{array}{cc}& r_a^2=\frac{s(s-b)(s-c)}{s-a}\\ & ⟹r_a=\sqrt{\frac{s(s-b)(s-c)}{s-a}}.\end{array}$$

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

Other excircle properties

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

The following relations hold among the inradius $r$, the circumradius $R$, the semiperimeter $s$, and the excircle radii $r_a$, $r_b$, $r_c$: $$\begin{array}{cc}{r}_a+r_b+r_c& =4R+r,\\ r_a{r}_b+r_b{r}_c+r_c{r}_a& =s^2,\\ r_a^2+r_b^2+r_c^2& ={(4R+r)}^2-2s^2.\end{array}$$

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

If $H$ is the orthocenter of $\mathrm{△}ABC$, then $$\begin{array}{cc}{r}_a+r_b+r_c+r& =\overline{AH}+\overline{BH}+\overline{CH}+2R,\\ r_a^2+r_b^2+r_c^2+r^2& ={\overline{AH}}^2+{\overline{BH}}^2+{\overline{CH}}^2+(2R{)}^2.\end{array}$$

Nagel triangle and Nagel point

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The Nagel triangle or extouch triangle of $\mathrm{△}ABC$ is denoted by the vertices $T_A$, $T_B$, and $T_C$ that are the three points where the excircles touch the reference $\mathrm{△}ABC$ and where $T_A$ is opposite of $A$, etc. This $\mathrm{△}{T}_A{T}_B{T}_C$ is also known as the extouch triangle of $\mathrm{△}ABC$. The circumcircle of the extouch $\mathrm{△}{T}_A{T}_B{T}_C$ is called the Mandart circle (cf. Mandart inellipse).

The three line segments $\overline{A{T}_A}$, $\overline{B{T}_B}$ and $\overline{C{T}_C}$ are called the splitters of the triangle; they each bisect the perimeter of the triangle,Template:Citation needed $$\overline{AB}+\overline{B{T}_A}=\overline{AC}+\overline{C{T}_A}=\frac{1}{2}(\overline{AB}+\overline{BC}+\overline{AC}).$$

The splitters intersect in a single point, the triangle's Nagel point $N_a$ (or triangle center Template:Math).

Trilinear coordinates for the vertices of the extouch triangle are given byTemplate:Citation needed $$\begin{array}{cccccc}{T}_A=& 0& :& {\csc }^2\frac{B}{2}& :& {\csc }^2\frac{C}{2}\\ T_B=& {\csc }^2\frac{A}{2}& :& 0& :& {\csc }^2\frac{C}{2}\\ T_C=& {\csc }^2\frac{A}{2}& :& {\csc }^2\frac{B}{2}& :& 0\end{array}$$

Trilinear coordinates for the Nagel point are given byTemplate:Citation needed $${\csc }^2\frac{A}{2}:{\csc }^2\frac{B}{2}:{\csc }^2\frac{C}{2},$$

or, equivalently, by the Law of Sines, $$\frac{b+c-a}{a}:\frac{c+a-b}{b}:\frac{a+b-c}{c}.$$

Barycentric coordinates for the Nagel point are therefore $$b+c-a:c+a-b:a+b-c,$$ or equivalently $$s-a:s-b:s-c.$$

The Nagel point is the isotomic conjugate of the Gergonne point.Template:Citation needed

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:

• 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: Template:Block indent

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 $\mathrm{△}ABC$ with the segments $BC$, $CA$, and $AB$ are the vertices of the incentral triangle. Trilinear coordinates for the vertices of the incentral triangle $\mathrm{△}{A}^{\prime }{B}^{\prime }{C}^{\prime }$ are given byTemplate:Citation needed $$\begin{array}{cccccc}{A}^{\prime }=& 0& :& 1& :& 1\\ B^{\prime }=& 1& :& 0& :& 1\\ C^{\prime }=& 1& :& 1& :& 0\end{array}$$

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 $\mathrm{△}{J}_A{J}_B{J}_C$ are given byTemplate:Citation needed $$\begin{array}{cccccc}{J}_A=& -1& :& 1& :& 1\\ J_B=& 1& :& -1& :& 1\\ J_C=& 1& :& 1& :& -1\end{array}$$

Equations for four circles

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

• Incircle:$$\begin{array}{cc}{u}^2{x}^2+v^2{y}^2+w^2{z}^2-2vwyz-2wuzx-2uvxy& =0\\ \pm \sqrt{x}\cos \frac{A}{2}\pm \sqrt{y\phantom{t}}\cos \frac{B}{2}\pm \sqrt{z}\cos \frac{C}{2}& =0\end{array}$$
• $A$-excircle:$$\begin{array}{cc}{u}^2{x}^2+v^2{y}^2+w^2{z}^2-2vwyz+2wuzx+2uvxy& =0\\ \pm \sqrt{-x}\cos \frac{A}{2}\pm \sqrt{y\phantom{t}}\cos \frac{B}{2}\pm \sqrt{z}\cos \frac{C}{2}& =0\end{array}$$
• $B$-excircle:$$\begin{array}{cc}{u}^2{x}^2+v^2{y}^2+w^2{z}^2+2vwyz-2wuzx+2uvxy& =0\\ \pm \sqrt{x}\cos \frac{A}{2}\pm \sqrt{-y\phantom{t}}\cos \frac{B}{2}\pm \sqrt{z}\cos \frac{C}{2}& =0\end{array}$$
• $C$-excircle:$$\begin{array}{cc}{u}^2{x}^2+v^2{y}^2+w^2{z}^2+2vwyz+2wuzx-2uvxy& =0\\ \pm \sqrt{x}\cos \frac{A}{2}\pm \sqrt{y\phantom{t}}\cos \frac{B}{2}\pm \sqrt{-z}\cos \frac{C}{2}& =0\end{array}$$

Euler's theorem

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: $${(R+r_{\text{ex}})}^2=d_{\text{ex}}^2+r_{\text{ex}}^2,$$ where $r_{\text{ex}}$ is the radius of one of the excircles, and $d_{\text{ex}}$ is the distance between the circumcenter and that excircle's center.Template:Sfn

Generalization to other polygons

The dashed edges show that the total length of even-numbered edges equal that of odd-numbered ones

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.

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.

See also

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• Triangle conic

References

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External links

• Derivation of formula for radius of incircle of a triangle, MATHalino
• Template: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