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 .

A new point on the nine-point circle III

Look at the diagram above, where a number of things appear to be happening. The red circle that goes through points is our focus for now. The center of this circle is itself a point on the nine-point circle of the parent . We’ll examine more of the properties of in this post, especially when the side-slopes of form a geometric progression.

Simplified coordinates

If has vertices at , , , then the coordinates of point are given by:

(1)

Compare (1) with its equivalent, simplified version:

(2)

Here, are the slopes of sides , respectively. Notice that and the coordinates of are explicitly “missing” in equation (2). One implication is that the calculation of requires only two sides and the slopes of those two sides. So (2) can take another form:

(3)

Or this other form:

(4)

Ample freedom. Simple concept.

Find the coordinates of point for with vertices located at , , .

Set , , and . The slopes of sides , and are then , , and , respectively.

Using equation (2):

Using equation (3):

Using equation (4):

In each case we obtain . It’s that point that anders here and there on the nine-point circle.

If has one side parallel to the -axis, PROVE that coincides with the foot of an altitude.

The converse is also true. Fix the vertices at , , . Let the slopes of be and suppose that . Using equation (2):

Thus, is the point , which is the foot of the altitude from vertex .

If has two sides with opposite slopes, PROVE that is the midpoint of the third side.

The converse is also true. Fix the vertices at , , . Let the slopes of be and suppose that . Using equation (2):

Thus, is the point , which is the midpoint of .

If the slopes of the legs of a right triangle are , PROVE that coincides with the circumcenter of the triangle.

This follows from example 3, since are opposites.

Special case

Most of our points behave extremely well if the side-slopes of the parent triangle form a geometric progression. Expectedly, our favorite point is not an exception.

If the side-slopes of form a geometric progression with common ratio , PROVE that the slopes of the line segments form a geometric progression with common ratio .

Let the slopes of sides be , , . From this post, the slopes from to vertices are:

(5)

Thus, correspond to the geometric progression with common ratio .

If the side-slopes of form a geometric progression in that order, PROVE that lies on the median through vertex (hence is co-linear with the centroid, the midpoint of , and vertex ).

From the preceding example, the slope from to vertex is . From this post, the slope from vertex to the midpoint of side is also . Thus, lies on the median through vertex .

If the slopes of the sides of form a geometric progression in that order, PROVE that the area of is equal to the area of .

As noted in example 6 above, the point is on the median through vertex . Suppose that lies outside the triangle (usually when ). Let be the midpoint of side . Then the area of is equal to the area of (since is a median in ). But then the area of is also equal to the area of (since is a median in ). Together we have that the area of is equal to the area of . Same conclusion when is inside the parent triangle.

In the diagram above, point is inside the given triangle. The areas of triangles and are equal.

Sample calculations

Since lies on the median through vertex , it is natural to ask what ratio both it and the centroid divide the entire median length .

(6)

As you already know, the relations in (6) hold when the slopes of the parent triangle form a geometric progression. Absolute values may be necessary in some cases.

Let be equilateral with sides having slopes . PROVE that is midway between vertex and centroid .

The common ratio of the slopes of an equilateral triangle satisfies the quadratic equation

(7)

Use equation (7) and the first ratio in equation (6) to obtain

Thus, and so is the midpoint of and . In addition, since the centroid divides median in the ratio , it follows that in this case.

Let be right isosceles with sides having slopes . PROVE that .

Use the first ratio in equation (6) and the fact that the common ratio of the slopes of a right isosceles triangle satisfies the quadratic equation

(8)

and obtain

Thus, .

Suppose that the slopes of sides form a geometric progression . If , PROVE that the point lies outside the triangle.

We have:

Thus, the line segment is longer than the median . (If , we can’t conclude that lies inside the triangle.)

Takeaway

In , let be the centroid, the midpoint of side , and as given by equation (1). If the slopes of sides form a geometric progression then the following statements are equivalent:

• , where is the sum of quotients of slopes defined here
• or
• or
• or
• or or or , where is the golden ratio
• or
• or
• or .

Eight easy equivalences. A triangle that satisfies any of these (together with or ) will be a right isosceles triangle (or its “look-alike”).