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

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

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.

Steiner inellipse edit

For ${\displaystyle s=t={\tfrac {1}{2}}}$  the points of contact ${\displaystyle U,V,W}$  are the midpoints of the sides and the inellipse is the Steiner inellipse (its center is the triangle's centroid).

Incircle edit

For ${\displaystyle s={\tfrac {|OA|+|OB|-|AB|}{2|OA|}},\;t={\tfrac {|OA|+|OB|-|AB|}{2|OB|}}}$  one gets the incircle of the triangle with center

${\displaystyle {\vec {OM}}={\frac {|OB|{\vec {OA}}+|OA|{\vec {OB}}}{|OA|+|OB|+|AB|}}\;.}$ 

Mandart inellipse edit

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 edit

For ${\displaystyle \ s={\tfrac {|OB|^{2}}{|OB|^{2}+|AB|^{2}}}\;,\quad t={\tfrac {|OA|^{2}}{|OA|^{2}+|AB|^{2}}}\;}$  one gets the Brocard inellipse. It is uniquely determined by its Brianchon point given in trilinear coordinates ${\displaystyle \ K:(|OB|:|OA|:|AB|)\ }$ .

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

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

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

• 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.

• Circumconic and inconic

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