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.








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.


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.


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.





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.







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.







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 or this 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
- 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.
- PROVE that there’s a horizontal radius if
- (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.
- PROVE that
- (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
- 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.
- Suppose that
- 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
.)
- PROVE a result similar to that of exercise 5 above in the case of vertical radius.