*conclyclic*with the

*orthocenter*. Similar constructions applied to the

*medial triangle*yield points that are concyclic with the

*circumcenter*.

## Extra points

Our intention at the beginning of the year was to focus most of our attention on four points associated with any triangle, depicted below:

By March, much has changed, and we’re grateful to be able to add seven more points, bringing the total, as at this paragraph, to eleven:

Actually, we have fourteen more points, so that the total, as at the time of this writing, is *twenty five*. However, these other fourteen points behave totally differently from the above mentioned eleven.

## Extended period

Following the addition of *twenty one extra points*, our initial one-year time frame has been affected. Please help us do the math: “If it takes one year to discuss *four points*, how long will it take to discuss *twenty five points*, assuming it takes the same length of time for each point”?

## Exchanging pleasantries

Welcome to our theory!

## Extensive properties

With reference to the preceding diagram, we have the following individual and collective properties of the points:

- each point is a point of concurrency of some lines (in this post we show how to obtain ; in the future, we’ll feature how to obtain the rest)
- and are on the circumcircle of the parent (in fact, is a diameter)
- and are on the nine-point circle of the parent (in fact, is a diameter)
- the line segments
__always__make an angle of with the positive -axis - the line segments
__always__make an angle of with the positive -axis - quadrilaterals and are
__always__rectangles (follows from the previous two) - the two rectangles and are “parallel”
- the area of the bigger rectangle is
*four times*the area of the smaller rectangle - is the center of the cyclic quadrilateral
- is the center of the cyclic quadrilateral
- rectangles and can become squares in special cases (in particular, the
*extremely pleasant*case of a triangle having two sides with reciprocal slopes and one side parallel to the -axis, and in the expectedly peculiar case of an equilateral triangle) - if points are connnected to form , then the
*Euler line*is just a median in this triangle - if the legs of a
*right triangle*are parallel to the coordinate axes, then rectangle becomes just the*circumcenter*and rectangle becomes the*orthocenter*(this is a degenerate case, but it does generate additional characterizations of this particular right triangle) - if the slopes of the sides of the parent form a geometric progression, then rectangles and “sit” on the same line (the case of geometric progressions is the
*exquisite platform*for the present concept) - our favourite point is the wandering point (this isn’t a property though, but there’s a basis for our bias)
- And lots more! (this isn’t a property either, but as we explore these points in future posts, more will be shown).

## Elementary proofs

Using an idea from our last post, we’ll now define the three lines that have point as a point of concurrency.

Define lines from as follows: the line from has slope , the line from has slope , and the line from has slope .

PROVE that the three lines defined above are concurrent.

Consider two such lines through and .

(1)

The solution to the linear system (1) is:

Let’s verify concurrency. We’ll prove that the solution above satisfies the equation of the line through :

(2)

Note that

Consider the left member of equation (2):

Now consider the right member of equation (2):

Admittedly, we skipped a couple of steps, probably because we didn’t want to focus on the details of the algebra, but instead on how the lines were defined. Now go ahead and alter our definition to obtain other points of concurrency.

Given with vertices at , , , the point is on its *nine-point circle*. Find the equations of the three lines through that intersect at .

The slopes of are . Set

and define lines through with slopes , respectively:

These three equations are all satisfied by .

Given with vertices at , , , define lines through the midpoints of in such a way that the slopes are *negatives* of the slopes of the sides. PROVE that the three lines meet at the point .

This is actually an interesting property. Basically, we can obtain the point in two different ways:

- through each vertex, draw lines with slopes of the form , where is the slope of the side opposite the reference vertex
- through each midpoint, draw lines with slopes that are
*negatives*of the slopes of the sides on which the midpoints lie.

In the present case, the slopes of are , and the midpoints are , , and , respectively. Through each of these midpoints we draw lines with slopes :

These three equations are all satisfied by .

The point on the *nine-point circle* of .

## Eventful pi

Our final example is a big digression from the preceding discussion. For your pi day celebration.

Let be the side-lengths, and let be the *nine-point center*, of non-right . If and , PROVE that .

And so a different version of “” for your *pi day* celebration.

Let be the circumradius. Since

we have Further simplifications:

Since is not a right triangle (right triangles automatically satisfy the relation if is taken as the hypotenuse), we ignore the value. Then put and consider the negative discriminant (the positive part would give , unacceptable):

By the cosine law again:

There goes.

*golden ratio*.

## Takeaway

The following statements are *equivalent* for any :

- , where is the
*orthocenter* - two sides of the triangle are parallel to the coordinate axes
- coincides with one of the triangle’s vertices
- the slopes of the three medians form a
*geometric progression*with common ratio - the slopes of the three medians form an
*arithmetic progression*with common difference equal to times the first term () - one side of the triangle is parallel to the -axis and another side and the median to it have
*opposite*slopes.

You’re already familiar with the above six equivalent statements. By the time we finish discussing our new points, the number of characterizations will be about *twenty*, if not more. Be looking forward to that time.

## Tasks

- Given with vertices , , , and as the slopes of sides , respectively, PROVE that:
- PROVE that
*the line segments from to the three vertices*.__and__the three midpoints either__all__make acute angles with the positive -axis, or__all__make obtuse angles with the positive -axis - Let , and let . PROVE that:
- (hence there’s no solution except )
- .

- Let as before.
- If , PROVE that
- Under the above condition, deduce that , or , where is the
*golden ratio*.

- Let as before.
- If , PROVE that
- Under the above condition, deduce that , or , where is the
*golden ratio*.