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)

(4)

Ample freedom. Simple concept.

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*.

*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 .

*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 .

*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.

*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 .

*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 .

*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.

*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.

*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, .

*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”).

## Tasks

- Let be such that is parallel to the -axis, and and have
*reciprocal*slopes. PROVE that:- the
*orthocenter*is a reflection of vertex across the -axis - is the
*foot*of the altitude from vertex - is the
*midpoint*of and

- the
- Let be the slopes of the line segments from to the vertices of , and let be the side-slopes.
- PROVE that .
- If is
*right-angled*, deduce that the product equals the*slope of the hypotenuse*.

- PROVE that for any with side-slopes .
- Find coordinates for the vertices of having the property that a point on its
*nine-point circle*is twice as far from vertex as it is from the centroid . - Suppose that the slopes of sides are , in that order. Let be as given in equation (1), and let be the midpoint of . PROVE that .