Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 37

We have

${\displaystyle {}{\frac {B+C}{2}}+{\frac {1}{3}}{\left(A-{\frac {B+C}{2}}\right)}={\frac {1}{3}}A+{\left({\frac {1}{2}}-{\frac {1}{6}}\right)}B+{\left({\frac {1}{2}}-{\frac {1}{6}}\right)}C={\frac {1}{3}}A+{\frac {1}{3}}B+{\frac {1}{3}}C\,,}$

and this is a point of the median through ${\displaystyle {}A}$. Due to the symmetry of the situation, this point also lies on the other medians.

The perpendicular bisector for the line segment between ${\displaystyle {}A}$ and ${\displaystyle {}B}$ consists, according to Lemma 37.5 , exactly of those points of the plane that have to these points the same distance. Therefore, the intersecting point ${\displaystyle {}P}$ of the perpendicular bisector of ${\displaystyle {}A}$ and ${\displaystyle {}B}$ and the perpendicular bisector of ${\displaystyle {}A}$ and ${\displaystyle {}C}$ has the same distance to all three vertices. This gives the supplement, and also, that all three perpendicular bisectors meet in this point.

We may assume that ${\displaystyle {}v}$ and ${\displaystyle {}w}$ are normed. Let ${\displaystyle {}P\in \mathbb {R} ^{2}}$. Due to Corollary 35.7 , we have

${\displaystyle {}d(P,\mathbb {R} v)^{2}=\Vert {P}\Vert ^{2}-\left\langle P,v\right\rangle ^{2}\,,}$

and, accordingly,

${\displaystyle {}d(P,\mathbb {R} w)^{2}=\Vert {P}\Vert ^{2}-\left\langle P,w\right\rangle ^{2}\,.}$

Therefore, the distances coincide if and only if

${\displaystyle {}\left\langle P,v\right\rangle =\pm \left\langle P,w\right\rangle \,}$

holds. If

${\displaystyle {}P=s(v+w)\,,}$

then

${\displaystyle {}\left\langle P,v\right\rangle =\left\langle s(v+w),v\right\rangle =s+s\left\langle w,v\right\rangle =\left\langle s(v+w),w\right\rangle =\left\langle P,w\right\rangle \,,}$

and the equation holds. For the reverse statement, we may write

${\displaystyle {}P=sv+tw\,.}$

From

${\displaystyle {}\left\langle P,v\right\rangle =\left\langle P,w\right\rangle \,}$

we can deduce

${\displaystyle {}s+t\left\langle v,w\right\rangle =\left\langle sv+tw,v\right\rangle =\left\langle sv+tw,w\right\rangle =t+s\left\langle v,w\right\rangle \,;}$

therefore,

${\displaystyle {}s{\left(1-\left\langle v,w\right\rangle \right)}=t{\left(1-\left\langle v,w\right\rangle \right)}\,.}$

Because ${\displaystyle {}v}$ and ${\displaystyle {}w}$ are normed and linearly independent, Exercise 31.9 implies that

${\displaystyle {}\vert {\left\langle v,w\right\rangle }\vert <1\,,}$

hence, the right-hand factor is not ${\displaystyle {}0}$, and, therefore, ${\displaystyle {}s=t}$. In case

${\displaystyle {}\left\langle P,v\right\rangle =-\left\langle P,w\right\rangle \,,}$

a similar consideration gives ${\displaystyle {}s=-t}$.

Due to Lemma 37.9 , the angle bisector for ${\displaystyle {}A}$ consists of points that have the same distance to the line of the adjacent sides ${\displaystyle {}\mathbb {R} {\overrightarrow {AB}}}$ and ${\displaystyle {}\mathbb {R} {\overrightarrow {AC}}}$. Accordingly, the angle bisector for ${\displaystyle {}B}$ consists of points that have the same distance to the lines ${\displaystyle {}\mathbb {R} {\overrightarrow {BA}}}$ and ${\displaystyle {}\mathbb {R} {\overrightarrow {BC}}}$. Therefore, the intersecting point of these two angle bisectors has the same distance to all three sides.

In order to determine the coordinates, we write the angle bisector of ${\displaystyle {}A}$ as

${\displaystyle A+s{\left({\frac {\overrightarrow {AB}}{c}}+{\frac {\overrightarrow {AC}}{b}}\right)},}$

that is,

${\displaystyle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+s{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}.}$

Equating with the angle bisector of ${\displaystyle {}B}$, we obtain

${\displaystyle {\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+s{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}={\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}+t{\left({\frac {1}{c}}{\begin{pmatrix}a_{1}-b_{1}\\a_{2}-b_{2}\end{pmatrix}}+{\frac {1}{a}}{\begin{pmatrix}c_{1}-b_{1}\\c_{2}-b_{2}\end{pmatrix}}\right)}\,.}$

The solution is given by

${\displaystyle {}s={\frac {bc}{a+b+c}}\,,}$

and

${\displaystyle {}t={\frac {ac}{a+b+c}}\,,}$

because this gives, after inserting, the symmetric expression

${\displaystyle {}{\begin{aligned}{\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+{\frac {bc}{a+b+c}}{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}&={\frac {1}{a+b+c}}{\begin{pmatrix}a_{1}(a+b+c)+b(b_{1}-a_{1})+c(c_{1}-a_{1})\\a_{2}(a+b+c)+b(b_{2}-a_{2})+c(c_{2}-a_{2})\end{pmatrix}}\\&={\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}}.\,\end{aligned}}}$

This is also the intersecting point of all three angle bisectors.

We move ${\displaystyle {}U}$ to the origin, so that the points ${\displaystyle {}A,B,C}$ have the same norm. The point in question is just ${\displaystyle {}A+B+C}$. The vector of the line given by this point and ${\displaystyle {}A}$ is ${\displaystyle {}B+C}$. This line runs through ${\displaystyle {}A}$, and we have

${\displaystyle {}\left\langle B+C,B-C\right\rangle =\left\langle B,B\right\rangle -\left\langle C,C\right\rangle \,.}$

Due to the equality of the norms, this is ${\displaystyle {}0}$; therefore, this line is the altitude line through ${\displaystyle {}A}$.

This follows directly from Lemma 37.12 .

1. Let ${\displaystyle {}U}$ be the circumcenter of the triangle; we may assume that this point is the origin of a Cartesian coordinate system. We consider the point

   ${\displaystyle {}U'={\frac {1}{2}}{\left(A+B+C\right)}\,.}$

   The distance between the midpoint ${\displaystyle {}A'={\frac {1}{2}}{\left(B+C\right)}}$ of the side connecting ${\displaystyle {}B}$ and ${\displaystyle {}C}$ and ${\displaystyle {}U'}$ is

   ${\displaystyle {}\Vert {{\frac {1}{2}}{\left(B+C\right)}-U'}\Vert =\Vert {{\frac {1}{2}}{\left(B+C\right)}-{\frac {1}{2}}{\left(A+B+C\right)}}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.}$

   Since the norms of all vertices ${\displaystyle {}A,B,C}$ are equal due to the choice of ${\displaystyle {}U}$, it follows that ${\displaystyle {}U'}$ is the circumcenter of the triangle given by the midpoints of the original triangle, and that its radius is the half of the radius of the circumcircle.

2. By Lemma 37.12 , ${\displaystyle {}A+B+C}$ is the orthocenter. Therefore, the midpoint of the line segment between ${\displaystyle {}A}$ and the orthocenter equals

   ${\displaystyle {}{\frac {1}{2}}(A+B+C)+{\frac {1}{2}}A=A+{\frac {1}{2}}(B+C)\,.}$

   The distance between this point and ${\displaystyle {}U'}$ is

   ${\displaystyle {}\Vert {{\frac {1}{2}}(A+B+C)-{\left(A+{\frac {1}{2}}(B+C)\right)}}\Vert =\Vert {-{\frac {1}{2}}A}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.}$

3. Note that the points constructed in (1) and (2) lie on the circle ${\displaystyle {}F}$ opposite to each other. Indeed, we have

   ${\displaystyle {}{\frac {1}{2}}{\left({\frac {1}{2}}(B+C)\right)}+{\frac {1}{2}}{\left(A+{\frac {1}{2}}(B+C)\right)}={\frac {1}{2}}{\left(A+B+C\right)}\,,}$

   which is the center of ${\displaystyle {}F}$. Hence, for each side, its midpoint, the midpoint between the opposite vertex and the orthocenter, and the foot of the corresponding altitude form a right triangle. Its circle with the hypotenuse as diameter equals ${\displaystyle {}F}$.