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 , let be the slopes of sides in that order. Let be the *circumcenter*. Then the slope of radius can be given by:

(1)

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

(2)

Easy to remember as they all radiate simplicity.

*circumcenter*is . Verify that the slopes of the radii , , satisfy equations (1,2).

Using the given coordinates:

- the slope of is
- the slope of is
- the slope of is

In order to use equations (1,2), write the side-slopes as , , for , , .

We obtain the same results.

*the circumcenter shares the same -coordinate with one of the vertices*.

As easy as . Let be such that the slopes of sides are , respectively. By the second equation in (2), the slope of radius is:

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

Similarly, if the slopes of the sides of a triangle are , then the circumcenter shares the same -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__ , as well as our favourite point on the *nine-point circle*. shares the same -coordinate with the nine-point center in this case.

*the circumcenter shares the same -coordinate with one of the vertices*.

Let be such that the slopes of sides are , , , respectively. By the second equation in (2), the slope of radius is:

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

Notice the __vertical radius__ . Notice also the point on the *nine-point circle*; it has the same -coordinate as the nine-point center in this case.

*circumcenter shares an -coordinate with a vertex and also shares a -coordinate with another vertex*.

Pick. This. One. Out.

Put , , and for the slopes of sides , , . Consider the slopes of radii and :

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

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

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 , , . Notice the slight manoeuvres in the calculations:

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.

It’s really, really hard to overlook geometric progressions. Let , , be the side-slopes. Then:

- slope of radius is:
- slope of radius is:
- slope of radius is:

So we obtain a geometric progression of the form , where or . We must impose another restriction on , namely:

The new common ratio can equal the original common ratio when

Whenever that special quadratic 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 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.

*circumcenter*of having vertices at , , by finding the point of concurrence of the three lines through , , having slopes given by equations (1,2).

Here , , and as per the slopes of , , . Using equations (1,2), we obtain the slopes of the three lines (not proper to call them radii yet):

and hence the equations of the three lines:

(3)

Consider the equations of the lines through and :

The pair also satisfies the equation of the line through :

Thus, the three equations are consistent and the *circumcenter* is therefore located at .

Don’t use ths method in real life.

*circumcenter*of having vertices at , , by finding the point of concurrence of the three radii , , .

Here we’re jumping the gun by calling radii, but don’t mind. We have , , and as per the slopes of , , . Using equations (1,2), we obtain the slopes of the three radii:

and hence the equations of the three radii:

(4)

Consider the equations of radii and :

The pair also satisfies the equation of radius :

Thus, the three equations are consistent and the *circumcenter* is therefore located at .

Notice that is the midpoint of . So the parent triangle is right-angled.

## Right scenario

Today’s discussion bodes well for right triangles.

Let . According to equation (2):

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

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

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* - the slopes of two radii are equal
- one radius and one side have the same slope.

## Tasks

- Find a triangle 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 has slope , PROVE that the slopes of the radii and are
*reciprocals*of each other. - If side has slope , PROVE that the slopes of the radii and are
*negatives*of each other.

- If side has slope , PROVE that the slopes of the radii and are
- In , Let be the slopes of sides , and let the circumcenter be .
- PROVE that there’s a horizontal radius if . Which of the radii will it be?
- PROVE that there’s a vertical radius if . Which of the radii 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:
- (Perpendicular radii) Suppose that radii and are perpendicular. As usual, let , , be the slopes of sides , , .
- PROVE that
- Deduce that or
- If , deduce that one of the radii or 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 is
*inside*the parent triangle, PROVE that is - If is
*outside*the parent triangle, PROVE that is - Derive equation (1).

- If is
- In , Let be the slopes of sides .
- Suppose that . PROVE that the slope of the radius is . In particular, if , deduce that the diameter through vertex is vertical.
- Suppose that . PROVE that the slope of the radius is . In particular, if , deduce that the diameter through vertex 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 .)

- a
- PROVE a result similar to that of exercise 5 above in the case of vertical radius.