Example of an inellipse

In triangle geometry, an inellipse is an ellipse that touches the three sides of a triangle. The simplest example is the incircle. Further important inellipses are the Steiner inellipse, which touches the triangle at the midpoints of its sides, the Mandart inellipse and Brocard inellipse (see examples section). For any triangle there exist an infinite number of inellipses.

The Steiner inellipse plays a special role: Its area is the greatest of all inellipses.

Because a non-degenerate conic section is uniquely determined by five items out of the sets of vertices and tangents, in a triangle whose three sides are given as tangents one can specify only the points of contact on two sides. The third point of contact is then uniquely determined.

Parametric representations, center, conjugate diameters

An inellipse of a triangle is uniquely determined by the vertices of the triangle and two points of contact ${\displaystyle U,V}$.

The inellipse of the triangle with vertices

${\displaystyle O=(0,0),\;A=(a_{1},a_{2}),\;B=(b_{1},b_{2})}$

and points of contact

${\displaystyle U=(u_{1},u_{2}),\;V=(v_{1},v_{2})}$

on ${\displaystyle OA}$ and ${\displaystyle OB}$ respectively can by described by the rational parametric representation

• ${\displaystyle \left({\frac {4u_{1}\xi ^{2}+v_{1}ab}{4\xi ^{2}+4\xi +ab}},{\frac {4u_{2}\xi ^{2}+v_{2}ab}{4\xi ^{2}+4\xi +ab}}\right)\ ,\ -\infty <\xi <\infty \ ,}$

where ${\displaystyle a,b}$ are uniquely determined by the choice of the points of contact:

${\displaystyle a={\frac {1}{s-1}},\ u_{i}=sa_{i},\quad b={\frac {1}{t-1}},\ v_{i}=tb_{i}\;,\ 0<s,t<1\;.}$

The third point of contact is

${\displaystyle W=\left({\frac {u_{1}a+v_{1}b}{a+b+2}}\;,\;{\frac {u_{2}a+v_{2}b}{a+b+2}}\right)\;.}$

The center of the inellipse is

${\displaystyle M={\frac {ab}{ab-1}}\left({\frac {u_{1}+v_{1}}{2}},{\frac {u_{2}+v_{2}}{2}}\right)\;.}$

The vectors

${\displaystyle {\vec {f}}_{1}={\frac {1}{2}}{\frac {\sqrt {ab}}{ab-1}}\;(u_{1}+v_{1},u_{2}+v_{2})}$

${\displaystyle {\vec {f}}_{2}={\frac {1}{2}}{\sqrt {\frac {ab}{ab-1}}}\;(u_{1}-v_{1},u_{2}-v_{2})\;}$

are two conjugate half diameters and the inellipse has the more common trigonometric parametric representation

• ${\displaystyle {\vec {x}}={\vec {OM}}+{\vec {f}}_{1}\cos \varphi +{\vec {f}}_{2}\sin \varphi \;.}$

Brianchon point ${\displaystyle K}$

The Brianchon point of the inellipse (common point ${\displaystyle K}$ of the lines ${\displaystyle {\overline {AV}},{\overline {BU}},{\overline {OW}}}$) is

${\displaystyle K:\left({\frac {u_{1}a+v_{1}b}{a+b+1}}\;,\;{\frac {u_{2}a+v_{2}b}{a+b+1}}\right)\ .}$

Varying ${\displaystyle s,t}$ is an easy option to prescribe the two points of contact ${\displaystyle U,V}$. The given bounds for ${\displaystyle s,t}$ guarantee that the points of contact are located on the sides of the triangle. They provide for ${\displaystyle a,b}$ the bounds ${\displaystyle -\infty <a,b<-1}$.

Remark: The parameters ${\displaystyle a,b}$ are neither the semiaxes of the inellipse nor the lengths of two sides.

Examples

Mandart inellipse

Mandart inellipse

For ${\displaystyle s={\tfrac {|OA|-|OB|+|AB|}{2|OA|}},\;t={\tfrac {-|OA|+|OB|+|AB|}{2|OB|}}}$ the inellipse is the Mandart inellipse of the triangle. It touches the sides at the points of contact of the excircles (see diagram).

Brocard inellipse

Derivations of the statements

Determination of the inellipse by solving the problem for a hyperbola in an ${\displaystyle \xi }$-${\displaystyle \eta }$-plane and an additional transformation of the solution into the x-y-plane. ${\displaystyle M}$ is the center of the sought inellipse and ${\displaystyle D_{1}D_{2},\;E_{1}E_{2}}$ two conjugate diameters. In both planes the essential points are assigned by the same symbols. ${\displaystyle g_{\infty }}$ is the line at infinity of the x-y-plane.

New coordinates

For the proof of the statements one considers the task projectively and introduces convenient new inhomogene ${\displaystyle \xi }$-${\displaystyle \eta }$-coordinates such that the wanted conic section appears as a hyperbola and the points ${\displaystyle U,V}$ become the points at infinity of the new coordinate axes. The points ${\displaystyle A=(a_{1},a_{2}),\;B=(b_{1},b_{2})}$ will be described in the new coordinate system by ${\displaystyle A=[a,0],B=[0,b]}$ and the corresponding line has the equation ${\displaystyle {\frac {\xi }{a}}+{\frac {\eta }{b}}=1}$. (Below it will turn out, that ${\displaystyle a,b}$ have indeed the same meaning introduced in the statement above.) Now a hyperbola with the coordinate axes as asymptotes is sought, which touches the line ${\displaystyle {\overline {AB}}}$. This is an easy task. By a simple calculation one gets the hyperbola with the equation ${\displaystyle \eta ={\frac {ab}{4\xi }}}$. It touches the line ${\displaystyle {\overline {AB}}}$ at point ${\displaystyle W=[{\tfrac {a}{2}},{\tfrac {b}{2}}]}$.

Coordinate transformation

The transformation of the solution into the x-y-plane will be done using homogeneous coordinates and the matrix

${\displaystyle {\begin{bmatrix}u_{1}&v_{1}&0\\u_{2}&v_{2}&0\\1&1&1\end{bmatrix}}\quad }$.

A point ${\displaystyle [x_{1},x_{2},x_{3}]}$ is mapped onto

${\displaystyle {\begin{bmatrix}u_{1}&v_{1}&0\\u_{2}&v_{2}&0\\1&1&1\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{pmatrix}u_{1}x_{1}+v_{1}x_{2}\\u_{2}x_{1}+v_{2}x_{2}\\x_{1}+x_{2}+x_{3}\end{pmatrix}}\rightarrow \left({\frac {u_{1}x_{1}+v_{1}x_{2}}{x_{1}+x_{2}+x_{3}}}\;,\;{\frac {u_{2}x_{1}+v_{2}x_{2}}{x_{1}+x_{2}+x_{3}}}\right),\quad {\text{if }}x_{1}+x_{2}+x_{3}\neq 0.}$

A point ${\displaystyle [\xi ,\eta ]}$ of the ${\displaystyle \xi }$-${\displaystyle \eta }$-plane is represented by the column vector ${\displaystyle [\xi ,\eta ,1]^{T}}$ (see homogeneous coordinates). A point at infinity is represented by ${\displaystyle [\cdots ,\cdots ,0]^{T}}$.

Coordinate transformation of essential points

${\displaystyle U:\ [1,0,0]^{T}\ \rightarrow \ (u_{1},u_{2})\ ,\quad V:\ [0,1,0]^{T}\ \rightarrow \ (v_{1},v_{2})\ ,}$

${\displaystyle O:\ [0,0]\ \rightarrow \ (0,0)\ ,\quad A:\ [a,0]\rightarrow \ (a_{1},a_{2})\ ,\quad B:\ [0,b]\rightarrow \ (b_{1},b_{2})\ ,}$

(One should consider: ${\displaystyle \ a={\tfrac {1}{s-1}},\ u_{i}=sa_{i},\quad b={\tfrac {1}{t-1}},\ v_{i}=tb_{i}\;}$; see above.)

${\displaystyle g_{\infty }:\xi +\eta +1=0\ }$ is the equation of the line at infinity of the x-y-plane; its point at infinity is ${\displaystyle [1,-1,0]^{T}}$.

${\displaystyle [1,-1,{\color {red}0}]^{T}\ \rightarrow \ (u_{1}-v_{1},u_{2}-v_{2},{\color {red}0})^{T}}$

Hence the point at infinity of ${\displaystyle g_{\infty }}$ (in ${\displaystyle \xi }$-${\displaystyle \eta }$-plane) is mapped onto a point at infinity of the x-y-plane. That means: The two tangents of the hyperbola, which are parallel to ${\displaystyle g_{\infty }}$, are parallel in the x-y-plane, too. Their points of contact are

${\displaystyle D_{i}:\left[{\frac {\pm {\sqrt {ab}}}{2}},{\frac {\pm {\sqrt {ab}}}{2}}\right]\ \rightarrow \ {\frac {1}{2}}{\frac {\pm {\sqrt {ab}}}{1\pm {\sqrt {ab}}}}\;(u_{1}+v_{1},u_{2}+v_{2}),\;}$

Because the ellipse tangents at points ${\displaystyle D_{1},D_{2}}$ are parallel, the chord ${\displaystyle D_{1}D_{2}}$ is a diameter and its midpoint the center ${\displaystyle M}$ of the ellipse

${\displaystyle M:\ {\frac {1}{2}}{\frac {ab}{ab-1}}\left(u_{1}+v_{1},u_{2}+v_{2}\right)\;.}$

One easily checks, that ${\displaystyle M}$ has the ${\displaystyle \xi }$-${\displaystyle \eta }$-coordinates

${\displaystyle \ M:\;\left[{\frac {-ab}{2}},{\frac {-ab}{2}}\right]\;.}$

In order to determine the diameter of the ellipse, which is conjugate to ${\displaystyle D_{1}D_{2}}$, in the ${\displaystyle \xi }$-${\displaystyle \eta }$-plane one has to determine the common points ${\displaystyle E_{1},E_{2}}$ of the hyperbola with the line through ${\displaystyle M}$ parallel to the tangents (its equation is ${\displaystyle \xi +\eta +ab=0}$). One gets ${\displaystyle E_{i}:\left[{\tfrac {-ab\pm {\sqrt {ab(ab-1)}}}{2}},{\tfrac {-ab\mp {\sqrt {ab(ab-1)}}}{2}}\right]}$. And in x-y-coordinates:

${\displaystyle \ E_{i}={\frac {1}{2}}{\frac {ab}{ab-1}}\left(u_{1}+v_{1},u_{2}+v_{2}\right)\pm {\frac {1}{2}}{\frac {\sqrt {ab(ab-1)}}{ab-1}}\left(u_{1}-v_{1},u_{2}-v_{2}\right)\;,}$

From the two conjugate diameters ${\displaystyle D_{1}D_{2},E_{1}E_{2}}$ there can be retrieved the two vectorial conjugate half diameters

${\displaystyle {\begin{aligned}{\vec {f}}_{1}&={\vec {MD_{1}}}={\frac {1}{2}}{\frac {\sqrt {ab}}{ab-1}}\;(u_{1}+v_{1},u_{2}+v_{2})\\[6pt]{\vec {f}}_{2}&={\vec {ME_{1}}}={\frac {1}{2}}{\sqrt {\frac {ab}{ab-1}}}\;(u_{1}-v_{1},u_{2}-v_{2})\;\end{aligned}}}$

and at least the trigonometric parametric representation of the inellipse:

${\displaystyle {\vec {x}}={\vec {OM}}+{\vec {f}}_{1}\cos \varphi +{\vec {f}}_{2}\sin \varphi \;.}$

Analogously to the case of a Steiner ellipse one can determine semiaxes, eccentricity, vertices, an equation in x-y-coordinates and the area of the inellipse.

The third touching point ${\displaystyle W}$ on ${\displaystyle AB}$ is:

${\displaystyle W:\left[{\frac {a}{2}},{\frac {b}{2}}\right]\ \rightarrow \ \left({\frac {u_{1}a+v_{1}b}{a+b+2}}\;,\;{\frac {u_{2}a+v_{2}b}{a+b+2}}\right)\;.}$

The Brianchon point of the inellipse is the common point ${\displaystyle K}$ of the three lines ${\displaystyle {\overline {AV}},{\overline {BU}},{\overline {OW}}}$. In the ${\displaystyle \xi }$-${\displaystyle \eta }$-plane these lines have the equations: ${\displaystyle \xi =a\;,\;\eta =b\;,\;a\eta -b\xi =0}$. Hence point ${\displaystyle K}$ has the coordinates:

${\displaystyle K:\ [a,b]\ \rightarrow \ \left({\frac {u_{1}a+v_{1}b}{a+b+1}}\;,\;{\frac {u_{2}a+v_{2}b}{a+b+1}}\right)\ .}$

Transforming the hyperbola ${\displaystyle \ \eta ={\frac {ab}{4\xi }}}$ yields the rational parametric representation of the inellipse:

${\displaystyle \left[\xi ,{\frac {ab}{4\xi }}\right]\ \rightarrow \ \left({\frac {4u_{1}\xi ^{2}+v_{1}ab}{4\xi ^{2}+4\xi +ab}},{\frac {4u_{2}\xi ^{2}+v_{2}ab}{4\xi ^{2}+4\xi +ab}}\right)\ ,\ -\infty <\xi <\infty \ .}$

Incircle

Incircle of a triangle

For the incircle there is ${\displaystyle |OU|=|OV|}$, which is equivalent to

(1)${\displaystyle \;s|OA|=t|OB|\;.\ }$ Additionally

(2)${\displaystyle \;(1-s)|OA|+(1-t)|OB|=|AB|}$. (see diagram)

Solving these two equations for ${\displaystyle s,t}$ one gets

(3)${\displaystyle \;s={\frac {|OA|+|OB|-|AB|}{2|OA|}},\;t={\frac {|OA|+|OB|-|AB|}{2|OB|}}\;.}$

In order to get the coordinates of the center one firstly calculates using (1) und (3)

${\displaystyle 1-{\frac {1}{ab}}=1-(s-1)(t-1)=-st+s+t=\cdots ={\frac {s}{2(|OB|}}(|OA|+|OB|+|AB|)\;.}$

Hence

${\displaystyle {\vec {OM}}={\frac {|OB|}{s(|OA|+|OB|+|AB|)}}\;(s{\vec {OA}}+t{\vec {OB}})=\cdots ={\frac {|OB|{\vec {OA}}+|OA|{\vec {OB}}}{|OA|+|OB|+|AB|}}\;.}$

Mandart inellipse

The parameters ${\displaystyle s,t}$ for the Mandart inellipse can be retrieved from the properties of the points of contact (see de: Ankreis).

Brocard inellipse

The Brocard inellipse of a triangle is uniquely determined by its Brianchon point given in trilinear coordinates ${\displaystyle \ K:(|OB|:|OA|:|AB|)\ }$.^[1] Changing the trilinear coordinates into the more convenient representation ${\displaystyle \ K:k_{1}{\vec {OA}}+k_{2}{\vec {OB}}\ }$ (see trilinear coordinates) yields ${\displaystyle \ k_{1}={\tfrac {|OB|^{2}}{|OB|^{2}+|OA|^{2}+|AB|^{2}}},\;k_{2}={\tfrac {|OA|^{2}}{|OB|^{2}+|OA|^{2}+|AB|^{2}}}\ }$. On the other hand, if the parameters ${\displaystyle s,t}$ of an inellipse are given, one calculates from the formula above for ${\displaystyle K}$: ${\displaystyle \ k_{1}={\tfrac {s(t-1)}{st-1}},\;k_{2}={\tfrac {t(s-1)}{st-1}}\ }$. Equalizing both expressions for ${\displaystyle k_{1},k_{2}}$ and solving for ${\displaystyle s,t}$ yields

${\displaystyle s={\frac {|OB|^{2}}{|OB|^{2}+|AB|^{2}}}\;,\quad t={\frac {|OA|^{2}}{|OA|^{2}+|AB|^{2}}}\;.}$

Inellipse with the greatest area

• The Steiner inellipse has the greatest area of all inellipses of a triangle.

Proof

From Apollonios theorem on properties of conjugate semi diameters ${\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}$ of an ellipse one gets:

${\displaystyle F=\pi \left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|\quad }$ (see article on Steiner ellipse).

For the inellipse with parameters ${\displaystyle s,t}$ one gets

${\displaystyle \det({\vec {f}}_{1},{\vec {f}}_{2})={\frac {1}{4}}{\frac {ab}{(ab-1)^{3/2}}}\det(s{\vec {a}}+t{\vec {b}},s{\vec {a}}-t{\vec {b}})}$

${\displaystyle ={\frac {1}{2}}{\frac {s{\sqrt {s-1}}\;t{\sqrt {t-1}}}{(1-(s-1)(t-1))^{3/2}}}\det({\vec {b}},{\vec {a}})\;,}$

where ${\displaystyle {\vec {a}}=(a_{1},a_{2}),\;{\vec {b}}=(b_{1},b_{2}),\;{\vec {u}}=(u_{1},u_{2}),{\vec {v}}=(v_{1},v_{2}),\;{\vec {u}}=s{\vec {a}},\;{\vec {v}}=t{\vec {b}}}$.
In order to omit the roots, it is enough to investigate the extrema of function ${\displaystyle G(s,t)={\tfrac {s^{2}(s-1)\;t^{2}(t-1)}{(1-(s-1)(t-1))^{3}}}}$:

${\displaystyle G_{s}=0\ \rightarrow \ 3s-2+2(s-1)(t-1)=0\;.}$

Because ${\displaystyle G(s,t)=G(t,s)}$ one gets from the exchange of s and t:

${\displaystyle G_{t}=0\ \rightarrow \ 3t-2+2(s-1)(t-1)=0\;.}$

Solving both equations for s and t yields

${\displaystyle s=t={\frac {1}{2}}\;,\quad }$ which are the parameters of the Steiner inellipse.

Three mutually touching inellipses of a triangle

See also

• Circumconic and inconic

References

External links

• Darij Grinberg: Über einige Sätze und Aufgaben aus der Dreiecksgeometrie
• Circumconic at MathWorld
• Inconic at MathWorld