Redirecting right bisectors – High School Geometry

Among the four classical triangle centers — centroid, circumcenter, incenter, orthocenter — only the circumcenter is usually constructed from lines that do not originate from any of the vertices of the parent triangle. It turns out that we can actually alter this tradition, and that’s what we’re after in today’s discussion. We’ll construct lines that originate from each vertex of the parent triangle whose point of concurrence is the circumcenter.

Radial slopes

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ in that order. Let $O$ be the circumcenter. Then the slope of radius $OB$ can be given by:

(1) $\begin{equation*} m_{OB}=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)} \end{equation*}$

Note the position of $m_3$ , the slope of the side opposite vertex $B$ . It’s the key to the “cyclic permutations” that yield similar expressions for the slopes of the radii through vertices $A$ and $C$ :

(2) $\begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)} \end{split} \end{equation*}$

Easy to remember as they all radiate simplicity.

Consider $\triangle ABC$

with vertices at $A(3,12)$

, $B(0,0)$

, $C(6,6)$

. Its circumcenter is $O\left(-\frac{1}{2},\frac{13}{2}\right)$

. Verify that the slopes of the radii $OA$

, $OB$

, $OC$

satisfy equations (1,2).

Using the given coordinates:

the slope of $OA$ is $\frac{12-13/2}{3+1/2}=\frac{11}{7}$
the slope of $OB$ is $\frac{0-13/2}{0+1/2}=-13$
the slope of $OC$ is $\frac{6-13/2}{6+1/2}=-\frac{1}{13}$

In order to use equations (1,2), write the side-slopes as $m_1=4$ , $m_2=1$ , $m_3=-2$ for $AB$ , $BC$ , $CA$ .

$\begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{(1-(-2)\times 4)+(-2+4)\times 1}{(1-(-2)\times 4)\times 1-(-2+4)}\\ &=\frac{11}{7}\\ m_{OB}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &=\frac{(1-4\times 1)+(4+1)\times(-2)}{(1-4\times 1)\times(-2)-(4+1)}\\ &=-13\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{(1-1\times (-2))+(1+(-2))\times 4}{(1-1\times (-2))\times 4-(1+(-2))}\\ &=-\frac{1}{13} \end{split} \end{equation*}$

We obtain the same results.

Suppose that the slopes of the sides of a triangle are $1,2,3$

. PROVE that the circumcenter shares the same $y$ -coordinate with one of the vertices.

As easy as $1,2,3$ . Let $\triangle ABC$ be such that the slopes of sides $AB,BC,CA$ are $m_1=1,m_2=2,m_3=3$ , respectively. By the second equation in (2), the slope of radius $OC$ is:

$\begin{equation*} \begin{split} m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{(1-2\times 3)+(2+3)(1)}{(1-2\times 3)(1)-(2+3))}\\ &=0 \end{split} \end{equation*}$

Thus, vertex $C$ shares the same $y$ -coordinate as the circumcenter. $C$ for common. $C$ for cool.

Similarly, if the slopes of the sides of a triangle are $-1,-2,-3$ , then the circumcenter shares the same $y$ -coordinate with one of the vertices. See the exercises for a more general pattern, and see below for a sample picture:

Notice the horizontal radius $OC$ , as well as our favourite point $W$ on the nine-point circle. $W$ shares the same $x$ -coordinate with the nine-point center $N$ in this case.

Suppose that the slopes of the sides of a triangle are $-1,2,3$

. PROVE that the circumcenter shares the same $x$ -coordinate with one of the vertices.

Let $\triangle ABC$ be such that the slopes of sides $AB,BC,CA$ are $m_1=-1$ , $m_2=2$ , $m_3=3$ , respectively. By the second equation in (2), the slope of radius $OC$ is:

$\begin{equation*} \begin{split} m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{(1-2\times 3)+(2+3)(-1)}{(1-2\times 3)(-1)-(2+3))}\\ &=\frac{-10}{0} \end{split} \end{equation*}$

Thus, vertex $C$ has the same $x$ -coordinate as the circumcenter. Here $C$ stands for cauton — d0n’t divide by zer0.

Notice the vertical radius $OC$ . Notice also the point $W$ on the nine-point circle; it has the same $y$ -coordinate as the nine-point center $N$ in this case.

Suppose that the slopes of the sides of a triangle are $-1$

, $-2$

, $3$

. PROVE that the circumcenter shares an $x$ -coordinate with a vertex and also shares a $y$ -coordinate with another vertex.

Pick. This. One. Out.

Put $m_1=-1$ , $m_2=-2$ , and $m_3=3$ for the slopes of sides $AB$ , $BC$ , $CA$ . Consider the slopes of radii $OA$ and $OB$ :

$\begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{(1+3)+(2)(-2)}{(1+3)(-2)-(-2)}\\ &=0\\ m_{OB}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &=\frac{(1-2)+(-3)(3)}{(1-2)(3)-(-3)}\\ &=\frac{-10}{0} \end{split} \end{equation*}$

This confirms that radius $OA$ is horizontal, while radius $OB$ is vertical. See the exercises for a more general pattern.

The point $W$ lies on the right bisector of $AB$ in this case. Can you see why we named $W$ the point that $W$ anders here and there on the nine-point circle?

$\triangle ABC$

has vertices at $A(0,6)$

, $B(0,0)$

, $C(3,0)$

. Find the slopes of the three radii through vertices $A,B,C$

It’s essential to consider an instance in which one of the slopes of the parent triangle is not well-behaved, as is the case here. We have $m_1=\infty$ , $m_2=0$ , $m_3=-2$ . Notice the slight manoeuvres in the calculations:

$\begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{1-m_1m_3}{-m_1-m_3}\\ &=\frac{\frac{1}{m_1}-m_3}{-1-\frac{m_3}{m_1}}\\ &=\frac{0-m_3}{-1-0}\\ &=m_3=-2\\ m_{OB}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &=\frac{1+m_1m_3}{m_3-m_1}\\ &=-m_3=2\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{1+m_1m_3}{m_1-m_3}\\ &=m_3=-2 \end{split} \end{equation*}$

And we obtain a characterization of right triangles with legs parallel to the coordinate axes: a right triangle has legs parallel to the coordinate axes if and only if two of its radii have opposite slopes.

If the slopes of the parent triangle form a geometric progression of the form $\frac{1}{r},\pm 1, r$

(where $r\neq 0,\pm 1$

), PROVE that the slopes of the radii form another geometric progression of the form $\frac{1}{R},\mp 1, R.$

It’s really, really hard to overlook geometric progressions. Let $m_1=\frac{1}{r}$ , $m_2=1$ , $m_3=r$ be the side-slopes. Then:

slope of radius $OA$ is: $\frac{(1-1)+\left(r+\frac{1}{r}\right)(1)}{(1-1)(1)-\left(r+\frac{1}{r}\right)}=-1$
slope of radius $OB$ is: $\frac{\left(1-\frac{1}{r}\right)+\left(\frac{1}{r}+1\right)(r)}{\left(1-\frac{1}{r}\right)(r)-\left(\frac{1}{r}+1\right)}=\frac{r^2+2r-1}{r^2-2r-1}$
slope of radius $OC$ is: $\frac{(1-r)+\left(1+r\right)\left(\frac{1}{r}\right)}{(1-r)\left(\frac{1}{r}\right)-(1+r)}=\frac{r^2-2r-1}{r^2+2r-1}$

So we obtain a geometric progression of the form $\frac{1}{R},-1,R$ , where $R=\frac{r^2+2r-1}{r^2-2r-1}$ or $R=\frac{r^2-2r-1}{r^2+2r-1}$ . We must impose another restriction on $r$ , namely:

$r^2+2r-1\neq 0\implies r\neq -1\pm\sqrt{2},\quad\textrm{and}\scriptstyle\quad r^2-2r-1\neq 0\implies r\neq 1\pm\sqrt{2}$

The new common ratio $-R$ can equal the original common ratio $r$ when

$-\left(\frac{r^2+2r-1}{r^2-2r-1}\right)=r\scriptstyle\implies r^3+3r^2-3r-1=0\implies (r-1)(r^2+4r+1)=0$

Whenever that special quadratic $r^2+4r+1$ is sighted, get excited: it signals the presence of an equilateral triangle with slopes in geometric progression. An equilateral triangle. Slopes in geometric progression. Pick them out.

If the slopes of the sides of an equilateral triangle form a geometric progression, then so do the slopes of the three radii from the vertices. Furthermore, the two three-term geometric progressions have the same common ratio (but are different because of the $\pm 1$ in the middle terms).

Radical strategy

Not quite radical per se, but still fundamentally different from — and eventually consistent with — the traditional method for finding the circumcenter. Our technique is merely intended to demonstrate what’s possible in principle; for practical computational purposes it’s inconvenent, and so not recommended.

Determine the circumcenter of $\triangle ABC$

having vertices at $A(2,6)$

, $B(0,0)$

, $C(8,0)$

by finding the point of concurrence of the three lines through $A$

, $B$

, $C$

having slopes given by equations (1,2).

Here $m_1=3$ , $m_2=0$ , and $m_3=-1$ as per the slopes of $AB$ , $BC$ , $CA$ . Using equations (1,2), we obtain the slopes of the three lines (not proper to call them radii yet):

$m_{OA}=-2,~m_{OB}=\frac{1}{2},~m_{OC}=-\frac{1}{2}$

and hence the equations of the three lines:

(3) $\begin{equation*} \begin{split} \textrm{line through }A:~ y-6&=-2(x-2)\\ y&=-2x+10\\ \textrm{line through }B:~y-0&=\frac{1}{2}\left(x-0\right)\\ y&=\frac{1}{2}x\\ \textrm{line through }C:~y-0&=-\frac{1}{2}\left(x-8\right)\\ y&=-\frac{1}{2}x+4 \end{split} \end{equation*}$

Consider the equations of the lines through $B$ and $C$ :

$\frac{1}{2}x=-\frac{1}{2}x+4\implies x=4\implies y=2$

The pair $(4,2)$ also satisfies the equation of the line through $A$ :

$y=-2x+10,\quad 2=-2\times 4+10$

Thus, the three equations are consistent and the circumcenter is therefore located at $(4,2)$ .

Don’t use ths method in real life.

Determine the circumcenter of $\triangle ABC$

having vertices at $A(2,6)$

, $B(-2,4)$

, $C(0,0)$

by finding the point of concurrence of the three radii $OA$

, $OB$

, $OC$

Here we’re jumping the gun by calling radii, but don’t mind. We have $m_1=\frac{1}{2}$ , $m_2=-2$ , and $m_3=3$ as per the slopes of $AB$ , $BC$ , $CA$ . Using equations (1,2), we obtain the slopes of the three radii:

$m_{OA}=3,~m_{OB}=-\frac{1}{3},~m_{OC}=3$

and hence the equations of the three radii:

(4) $\begin{equation*} \begin{split} \textrm{radius OA: } y-6&=3(x-2)\\ y&=3x\\ \textrm{radius OB: }y-4&=-\frac{1}{3}\left(x--2\right)\\ y&=-\frac{1}{3}x+\frac{10}{3}\\ \textrm{radius OC: }y-0&=3\left(x-0\right)\\ y&=3x \end{split} \end{equation*}$

Consider the equations of radii $OB$ and $OC$ :

$3x=-\frac{1}{3}x+\frac{10}{3}\implies 9x=-x+10\implies x=1\implies y=3$

The pair $(1,3)$ also satisfies the equation of radius $OA$ :

$y=3x,\quad 3=1\times 3$

Thus, the three equations are consistent and the circumcenter is therefore located at $(1,3)$ .

Notice that $(1,3)$ is the midpoint of $AC$ . So the parent triangle is right-angled.

Right scenario

Today’s discussion bodes well for right triangles.

Suppose that $m_1m_2=-1$

. PROVE that the slopes of the radii $OA$

and $OC$

are equal.

Let $m_1m_2=-1$ . According to equation (2):

$\begin{equation*} \begin{split} m_{OA}&=\frac{(1-m_3m_1)+(m_3+m_1)m_2}{(1-m_3m_1)m_2-(m_3+m_1)}\\ &=\frac{1-m_3m_1+m_2m_3+m_1m_2}{m_2-m_1m_2m_3-m_1-m_3}\\ &=\frac{1-m_3m_1+m_2m_3-1}{m_2+m_3-m_1-m_3}\\ &=\frac{m_3(m_2-m_1)}{m_2-m_1}\\ &=m_3\\ m_{OC}&=\frac{(1-m_2m_3)+(m_2+m_3)m_1}{(1-m_2m_3)m_1-(m_2+m_3)}\\ &=\frac{1-m_2m_3+m_1m_2+m_1m_3}{m_1-m_1m_2m_3-m_2-m_3}\\ &=\frac{1-m_2m_3-1+m_1m_3}{m_1+m_3-m_2-m_3}\\ &=\frac{m_3(m_1-m_2)}{m_1-m_2}\\ &=m_3 \end{split} \end{equation*}$

Can you see how this verifies that the circumcenter of a right triangle must lie on the hypotenuse? The slope way.

We encountered a right triangle in example 5.

PROVE that if the slopes of two radii are equal, then the triangle contains $90^{\circ}$

Suppose that the slope of radius $OA$ is equal to the slope of radius $OB$ . By equations (1,2):

$\begin{equation*} \begin{split} \frac{(1-m_1m_3)+(m_1+m_3)m_2}{(1-m_1m_3)m_2-(m_1+m_3)}&=\frac{(1-m_1m_2)+(m_1+m_2)m_3}{(1-m_1m_2)m_3-(m_1+m_2)}\\ &\vdots\cdots\vdots\ddots\\ 2(m_1^2+1)(m_3-m_2)(m_2m_3+1)&=0\\ \implies m_2m_3+1&=0 \end{split} \end{equation*}$

A couple of steps were skipped, but you get the gist.

Takeaway

The three statements below are equivalent for any triangle:

the triangle is a right triangle or this triangle
the slopes of two radii are equal
one radius and one side have the same slope.

Tasks

Find a triangle $ABC$ in which the slopes of its three radii are equal to the negatives of the slopes of its three sides.
Give an example of a right triangle having one radius horizontal and another radius vertical.
Let be an arbitrary triangle with circumcenter .
- If side $BC$ has slope $m=1$ , PROVE that the slopes of the radii $OB$ and $OC$ are reciprocals of each other.
- If side $BC$ has slope $m=0$ , PROVE that the slopes of the radii $OB$ and $OC$ are negatives of each other.
In , Let be the slopes of sides , and let the circumcenter be .
- PROVE that there’s a horizontal radius if $m_3=\frac{m_1m_2-1}{m_1+m_2}$ . Which of the radii $OA, OB, OC$ will it be?
- PROVE that there’s a vertical radius if $m_3=\frac{m_1+m_2}{1-m_1m_2}$ . Which of the radii $OA,OB,OC$ will it be?
- Deduce that a circumcenter cannot coincide with a vertex, unless the triangle is a single point.
(Pattern recognition) Suppose that the set of slopes produce a horizontal radius, in light of the previous exercise. PROVE that each of the following sets of slopes also produce a horizontal radius:
- $\{-m_1,-m_2,-m_3\}$
- $\left\{-\frac{1}{m_1},-\frac{1}{m_2},m_3\right\}$
- $\left\{\frac{1}{m_1},\frac{1}{m_2},-m_3\right\}$
(Perpendicular radii) Suppose that radii and are perpendicular. As usual, let , , be the slopes of sides , , .
- PROVE that $\left(1+m_1^2\right)\Big(\left(1+m_2m_3\right)^2-\left(m_2-m_3\right)^2\Big)=0$
- Deduce that $m_2=\frac{1+m_3}{1-m_3}$ or $m_2=\frac{m_3-1}{m_3+1}$
- If $m_1=\pm 1$ , deduce that one of the radii $OA$ or $OB$ is horizontal and the other vertical.
(Principal result) Here’s how to obtain the main result of our discussion. In , let be the slopes of sides , and let the circumcenter be . Suppose that .
- If $O$ is inside the parent triangle, PROVE that $\angle BAC$ is $\frac{\pi}{2}-\alpha$
- If $O$ is outside the parent triangle, PROVE that $\angle BAC$ is $\frac{\pi}{2}+\alpha$
- Derive equation (1).
In , Let be the slopes of sides .
- Suppose that $m_1+m_2=0$ . PROVE that the slope of the radius $OB$ is $\frac{1}{m_3}$ . In particular, if $m_3=0$ , deduce that the diameter through vertex $B$ is vertical.
- Suppose that $m_1m_2=1$ . PROVE that the slope of the radius $OB$ is $-m_3$ . In particular, if $m_3=0$ , deduce that the diameter through vertex $B$ is horizontal.
PROVE that any triangle with side-slopes contains:
- a horizontal radius
- a vertical median. (A vertical median always shows up when the side-slopes form an arithmetic progression, like $-3,2,7$ .)
PROVE a result similar to that of exercise 5 above in the case of vertical radius.