## Redirecting right bisectors

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.

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.

Consider with vertices at , , . Its 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.

Suppose that the slopes of the sides of a triangle are . PROVE that 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.

Suppose that the slopes of the sides of a triangle are . PROVE that 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.

Suppose that the slopes of the sides of a triangle are , , . PROVE that the 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?

has vertices at , , . Find the slopes of the three radii through vertices .

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.

If the slopes of the parent triangle form a geometric progression of the form (where ), PROVE that the slopes of the radii form another geometric progression of the form

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

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).

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 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.

Determine the 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.

Suppose that . PROVE that the slopes of the radii and are equal.

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.

We encountered a right triangle in example 5.

PROVE that if the slopes of two radii are equal, then the triangle contains .

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.

1. Find a triangle in which the slopes of its three radii are equal to the negatives of the slopes of its three sides.
2. Give an example of a right triangle having one radius horizontal and another radius vertical.
3. 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.
4. 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.
5. (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:
6. (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.
7. (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).
8. 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.
9. PROVE that any triangle with side-slopes contains:
• a vertical median. (A vertical median always shows up when the side-slopes form an arithmetic progression, like .)
10. PROVE a result similar to that of exercise 5 above in the case of vertical radius.

## A variant of Kosnita’s theorem

Let be the centroid of Denote by the centroids of triangles , respectively. Then the lines are concurrent at , the centroid of the parent triangle.

Further, let be the circumcenter of . Denote by the circumcenters of triangles , in that order. According to Kosnita’s theorem, the lines are concurrent at a point called the Kosnita point.

Now let be that point that wanders here and there on the nine-point circle of (coordinates given by (1) or (2) or (3) or (4)). Denote by the “-centers” of triangles respectively. In this post we show that the lines are concurrent, if the slopes of the parent triangle form a geometric progression.

(1)

(2)

(3)

(4)

are the coordinates of the vertices, while are the slopes of sides . Equations (2),(3),(4) are all equivalent (to (1)) representations of point .

## Concurrence

Throughout, will denote the “W-center” of , while represent the “W-centers” of triangles .

Express the slopes of and in terms of the slopes of the parent triangle .

Let the slopes of be in that order. Since is constructed from , the slope of is the product of the slopes of and , divided by the negative of the slope of (the way it works in general, as shown in this post, is as follows: take the product of the slopes of the two sides that originate from the reference vertex and then divide by the negative of the slope of the opposite side).

(5)

Similarly, the slope of is the product of the slopes of and , divided by the negative of the slope of :

(6)

Express the slopes of and in terms of the slopes of the parent triangle .

As before, let be the slopes of sides respectively. Since is constructed from , the slope of is the product of the slopes of and , divided by the negative of the slope of :

(7)

Similarly, the slope of is the product of the slopes of and , divided by the negative of the slope of :

(8)

Suppose that the slopes of sides are PROVE that the quadrilateral is a parallelogram.

Set . From equation (5):

From equation (6):

From equation (7):

From equation (8):

Thus, is parallel to , and is parallel to . There the parallelogram goes.

PROVE that and both lie on the median through vertex , if the slopes of sides are in that order.

We’ve already seen that lies on the median through vertex in example 6 here.

To see that also lies on the median through vertex , recall that was constructed from . As such, the slope of the line is the product of the slopes of and divided by the negative of the slope of :

The slope of the median through vertex is , the slope of is also , and the slope of is again . Thus, the points are co-linear together with the midpoint of .

#### (Main goal)

Suppose that the slopes of sides form a geometric progression PROVE that the lines , , are concurrent.

First consider the segment . In the parallelogram constructed in example 3, and are diagonals, so their midpoints coincide. Next, draw the lines and and extend till they intersect at , say. We claim that the line also passes through . This follows because the midpoint of lies on the median through — as do and — so the line will go through (in fact, it is a median — extended if necessary — in ).

In any , PROVE that or , or , or .

As usual, let the slopes be for sides . Consider for example. In equation (6) we saw that the slope of , the negative of the slope of side . In equation (8) we saw that the slope of , the negative of the slope of . Thus, the angle between and is either angle itself, or its supplement.

## Coordinates

In the case of triangles with slopes in geometric progression , we’re able to explicitly determine the coordinates of the point of concurrence of the lines , , :

(9)

How does one identify from a given triangle? Well, easy: just arrange the slopes in such a way that they appear in the format for sides respectively. Then are the coordinates of vertex and is the -coordinate of . (Notice how the coefficients of and in the second equation in (9) add up to the denominator. So in a sense, they’re “weighted”.)

Given with vertices at , , , find the point where the lines , , concur.

The slopes of sides are in that order; they form a geometric progression of the form . As specified in equation (9), we identify , , and .

The point of concurrence is at as shown below:

Notice point on the nine-point circle of the parent triangle . Points , , also lie on the nine-point circles of triangles , , .

In the exercises we ask you to derive the following equation:

(10)

Don’t be intimidated by equation (10), especially with the slope terms that appear there; its derivation can be accomplished without coordinates. In fact, triangles , , always have their slopes in geometric progressions — irrespective of what happens in the parent triangle (exception: when one side of the parent triangle is parallel to the or axis) — and so a certain approximate pythagorean identity can be applied to derive equation (10).

In , let be the slopes of sides . PROVE that .

The side-lengths satisfy an approximate pythagorean identity:

Using equation (10) with , , and :

In , let be the slopes of sides . PROVE that .

We had, from the preceding example, that:

In a previous post we asked you to prove that

Combine these two equations:

Can you guess a value of for which ? There it goes — it’s that golden ratio thing.

## Coincidence

And a cauton.

Consider with vertices at , , . PROVE that the lines , , concur at .

Observe that the slopes of sides are ; they do not form a geometric progression (even when re-arranged as ). So this example suggests that there are other instances of concurrence of the lines , , beyond geometric progressions.

• the “W-center” of the parent triangle is — obtained by solving the consistent equations , ,
• the “W-center” of is — obtained by solving the consistent system , , . Using and , we obtain the equation as the equation of line
• the “W-center” of is — obtained by solving the consistent system , , . Using and , we obtain the equation for the line — incidentally, this is also the equation of line
• the “W-center” of is — obtained by solving the consistent system , , . Using and , we obtain the equation — which, incidentally, is the equation of line . Together with the preceding incidence, we obtain a coincidence
• the equations of lines , , are , , . These three lines concur at — coordinates of vertex . C for coincidence. C for caution.

## Takeaway

In , let be the slopes of sides . Let be that point on the nine-point circle whose coordinates are given by equation (1). Then the four statements below are equivalent:

• or

You can see the golden ratio popping up in the third statement above. That seemingly ubiquitous golden ratio thing is increasingly becoming conspicuous in our theory.

1. Find a triangle and a point on its nine-point circle such that .
2. In , Let be the slopes of sides . Let be the point whose coordinates were given in equation (1). PROVE that:
3. (Coordinates) Use equation (2) to prove that the “W-center” of is
4. (Congruence) In , let be points that are diametrically opposite vertices , respectively. PROVE that:
• is congruent to
• the “W” center of and the “W-center” of have their midpoint at the circumcenter of the parent
(In general, it’s a similar scenario with the orthocenter and co — all due to a certain homothecy centered at the circumcenter .)
5. (Coincidence) Let be such that is parallel to the -axis, and and have reciprocal slopes. PROVE that:
• the Kosnita point coincides with vertex
• the point diametrically opposite this Kosnita point forms an isosceles trapezoid in conjuction with the other points.
6. (Concurrent coincidence) has vertices at , , . PROVE that:
• is the point
• is the point
• is the point
• is the point
• the point of concurrence of the lines , , is .
7. In any triangle , PROVE that:
• the slope of is the negative of the slope of
• the slope of is the negative of the slope of
• the slope of is the negative of the slope of
8. Given a triangle with side-slopes , find a triangle with side-slopes , , .
(Hint: Consider .)
9. If the slopes of sides form a geometric progression , PROVE that is parallel to side .
10. PROVE that .