Mixtilinear incircles of a triangle

In plane geometry, a mixtilinear incircle of a triangle is a circle which is tangent to two of its sides and internally tangent to its circumcircle. The mixtilinear incircle of a triangle tangent to the two sides containing vertex ${\displaystyle A}$ is called the ${\displaystyle A}$-mixtilinear incircle. Every triangle has three unique mixtilinear incircles, one corresponding to each vertex.

Proof of existence and uniqueness

The ${\displaystyle A}$-excircle of triangle ${\displaystyle ABC}$ is unique. Let ${\displaystyle \Phi }$ be a transformation defined by the composition of an inversion centered at ${\displaystyle A}$ with radius ${\displaystyle {\sqrt {AB\cdot AC}}}$ and a reflection with respect to the angle bisector on ${\displaystyle A}$. Since inversion and reflection are bijective and preserve touching points, then ${\displaystyle \Phi }$ does as well. Then, the image of the ${\displaystyle A}$-excircle under ${\displaystyle \Phi }$ is a circle internally tangent to sides ${\displaystyle AB,AC}$ and the circumcircle of ${\displaystyle ABC}$, that is, the ${\displaystyle A}$-mixtilinear incircle. Therefore, the ${\displaystyle A}$-mixtilinear incircle exists and is unique, and a similar argument can prove the same for the mixtilinear incircles corresponding to ${\displaystyle B}$ and ${\displaystyle C}$.^[1]

Construction

The ${\displaystyle A}$-mixtilinear incircle can be constructed with the following sequence of steps.^[2]

1. Draw the incenter ${\displaystyle I}$ by intersecting angle bisectors.
2. Draw a line through ${\displaystyle I}$ perpendicular to the line ${\displaystyle AI}$, touching lines ${\displaystyle AB}$ and ${\displaystyle AC}$ at points ${\displaystyle D}$ and ${\displaystyle E}$ respectively. These are the tangent points of the mixtilinear circle.
3. Draw perpendiculars to ${\displaystyle AB}$ and ${\displaystyle AC}$ through points ${\displaystyle D}$ and ${\displaystyle E}$ respectively and intersect them in ${\displaystyle O_{A}}$. ${\displaystyle O_{A}}$ is the center of the circle, so a circle with center ${\displaystyle O_{A}}$ and radius ${\displaystyle O_{A}E}$ is the mixtilinear incircle

This construction is possible because of the following fact:

Lemma

The incenter is the midpoint of the touching points of the mixtilinear incircle with the two sides.

Proof

Let ${\displaystyle \Gamma }$ be the circumcircle of triangle ${\displaystyle ABC}$ and ${\displaystyle T_{A}}$ be the tangency point of the ${\displaystyle A}$-mixtilinear incircle ${\displaystyle \Omega _{A}}$ and ${\displaystyle \Gamma }$. Let ${\displaystyle X\neq T_{A}}$ be the intersection of line ${\displaystyle T_{A}D}$ with ${\displaystyle \Gamma }$ and ${\displaystyle Y\neq T_{A}}$ be the intersection of line ${\displaystyle T_{A}E}$ with ${\displaystyle \Gamma }$. Homothety with center on ${\displaystyle T_{A}}$ between ${\displaystyle \Omega _{A}}$ and ${\displaystyle \Gamma }$ implies that ${\displaystyle X,Y}$ are the midpoints of ${\displaystyle \Gamma }$ arcs ${\displaystyle AB}$ and ${\displaystyle AC}$ respectively. The inscribed angle theorem implies that ${\displaystyle X,I,C}$ and ${\displaystyle Y,I,B}$ are triples of collinear points. Pascal's theorem on hexagon ${\displaystyle XCABYT_{A}}$ inscribed in ${\displaystyle \Gamma }$ implies that ${\displaystyle D,I,E}$ are collinear. Since the angles ${\displaystyle \angle {DAI}}$ and ${\displaystyle \angle {IAE}}$ are equal, it follows that ${\displaystyle I}$ is the midpoint of segment ${\displaystyle DE}$.^[1]

Other properties

Radius

The following formula relates the radius ${\displaystyle r}$ of the incircle and the radius ${\displaystyle \rho _{A}}$ of the ${\displaystyle A}$-mixtilinear incircle of a triangle ${\displaystyle ABC}$:

where ${\displaystyle \alpha }$ is the magnitude of the angle at ${\displaystyle A}$.^[3]

Relationship with points on the circumcircle

• The midpoint of the arc ${\displaystyle BC}$ that contains point ${\displaystyle A}$ is on the line ${\displaystyle T_{A}I}$.^[4][5]
• The quadrilateral ${\displaystyle T_{A}XAY}$ is harmonic, which means that ${\displaystyle T_{A}A}$ is a symmedian on triangle ${\displaystyle XT_{A}Y}$.^[1]

Circles related to the tangency point with the circumcircle

${\displaystyle T_{A}BDI}$ and ${\displaystyle T_{A}CEI}$ are cyclic quadrilaterals.^[4]

Spiral similarities

${\displaystyle T_{A}}$ is the center of a spiral similarity that maps ${\displaystyle B,I}$ to ${\displaystyle I,C}$ respectively.^[1]

Relationship between the three mixtilinear incircles

Lines joining vertices and mixtilinear tangency points

The three lines joining a vertex to the point of contact of the circumcircle with the corresponding mixtilinear incircle meet at the external center of similitude of the incircle and circumcircle.^[3] The Online Encyclopedia of Triangle Centers lists this point as X(56).^[6] It is defined by trilinear coordinates:

and barycentric coordinates:

Radical center

The radical center of the three mixtilinear incircles is the point ${\displaystyle J}$ which divides ${\displaystyle OI}$ in the ratio:

where ${\displaystyle I,r,O,R}$ are the incenter, inradius, circumcenter and circumradius respectively.^[5]

References