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# Geometric sequences in all triangles

There were three months in the past year: January, February, Mask.
On a serious note, one may ask, what does this have to do with math?
Answer is: today’s post is the second part of a task we started last March.

Welcome to our first post of 2021!!!

## New points

Marked below are four points associated with any :

is a diameter of the circumcircle of , while is a diameter of the nine-point circle of . Notice that the Euler line of the parent turns out to be a median in , just as are and .

Part of our goal for is a detailed discussion of these four points, although the discussion will not follow a linear fashion. For now, we’ll first consider — in a bid to deliver on the promise made here last year (see the comment preceding example 5 there).

Let , , be the vertices of the parent . Relative to these vertices, has coordinates:

(1)

Observe that is a (horizontal and vertical) translation of the orthocenter of the parent . Also, the “common denominator” in (1) above

is twice the area of ; as a result,

and so no orries about dividing by zero: point is ell defined. Alays.

## Nice properties

Let’s enumerate some of the properties of point :

• it lies on the nine-point circle of the parent
• it is a point of concurrency of certain line segments (to be featured in the future)
• if a right triangle has legs with slopes , then the point coincides with the circumcenter of the right triangle
• if a right triangle has legs parallel to the coordinate axes (slopes ), then the point coincides with the orthocenter of the right triangle
• if the end points of any side of a triangle are joined to , the resulting triangle has slopes in geometric progression (so long as the parent triangle has proper slopes )
• if any vertex of a triangle is joined to and the orthocenter, then the resulting triangle has slopes in geometric progression (so long as the parent triangle has proper slopes )

You get the gist: our favourite from the above list are the last two. At least.

## Numerical problems

The labels for our four new points can be viewed as an acronym for “No Perfection Without Repetition” . That aside, our next move today is to emphasize the first and second to last properties listed above through numerous, numerical instances. As such, the examples below are repetitive; please bear with us.

Consider with vertices at , , and . PROVE that the point lies on the nine-point circle of .

Among the points through which the nine-point circle passes are the midpoints of the sides of the parent triangle. In the present case, the midpoints of sides are , respectively. They are the points shown in the diagram below.

Behind the scenes, we’ve calculated the original triangle’s circumcenter and orthocenter; they are the points and , respectively.

Let the center of the nine-point circle. It is the midpoint of the circumcenter and orthocenter of the parent triangle, like so:

Lastly, we show that . By definition we already have .

Since , we conclude that is on the nine-point circle of .

Calendar fact: In terms of dates, both 1024 and 1885 share something in common. January 14, 1024=January 14, 1885=Wednesday.

Consider with vertices at , , and . Let be the point . PROVE that the slopes of the sides of , , and form geometric progressions (in each case, the geometric mean is the slope of one side of the parent triangle ).

As you’ll recall, the point is on nine-point cirlce of .

Observe that the slopes of the sides of are . They form an arithmetic progression, but not a geometric progression. However, our result holds regardless of any pattern of slopes in the parent triangle.

ith given as and the original vertices , , , we can calculate the slopes of the sides of , , and :

• for , the side-slopes are , , ; they form a geometric progression with common ratio
• for , the side-slopes are , , ; they form a geometric progression with common ratio
• for , the side-slopes are , , ; they form a geometric progression with

Consider with vertices at , , and . PROVE that the point lies on the nine-point circle of

The midpoints of the sides of are , , and , as shown below:

Since the parent is right-angled at , its orthocenter is and its circumcenter is , the midpoint of the hypotenuse. Thus, the center of the nine-point circle is , say.

We check that :

Thus, the point is on the nine-point circle of .

Consider with vertices at , , and . Let be the point . PROVE that the slopes of the sides of , , and form geometric progressions.

Here’s a diagram showing , , and :

ith given as and the original vertices , , , we can calculate the slopes of the sides of , , and :

• for , the side-slopes are , , ; they form a geometric progression with common ratio
• for , the side-slopes are , , ; they form a geometric progression with common ratio
• for , the side-slopes are , , ; they form a geometric progression with .

Consider with vertices at , , and . PROVE that the point lies on the nine-point circle of

Behind the scenes, we’ve calculated the parent triangle’s circumcenter and orthocenter; they’re the points and , respectively. Thus, the center of the nine-point circle is , say.

The nine-point circle passes through the midpoints , , of the parent triangle.

We check that :

Consider with vertices at , , and . Let be the point . PROVE that the slopes of the sides of , , and form geometric progressions.

• The slopes of the sides of are they form a geometric progression with common ratio .
• The slopes of the sides of are they form a geometric progression with common ratio .
• The slopes of the sides of are they form a geometric progression with common ratio .

Consider with vertices at , , and . PROVE that the point lies on the nine-point circle of .

The midpoints of sides ; they are , respectively. They are the points shown in the diagram below.

Behind the scenes, we’ve calculated the original triangle’s circumcenter and orthocenter; they are the points and , respectively. It follows that the center of the nine-point circle in this case is the point , denoted in the diagram above.

Now we check that . Again, it suffices to check :

Thus, lies on the nine-point circle of .

Consider with vertices at , , and . Let be the point . PROVE that the slopes of the sides of , , and form geometric progressions.

As you’ll recall, the point is on the nine-point circle of the parent triangle .

Observe that the slopes of the sides of are . They form an arithmetic progression, but not a geometric progression.

• The slopes of the sides of are they form a geometric progression with common ratio (remember the name: a peach).
• The slopes of the sides of are they form a geometric progression with common ratio .
• The slopes of the sides of are they form a geometric progression with common ratio .

Consider with vertices at , , and . Find coordinates for a point for which the slopes of the sides of , , and form three separate geometric progressions.

Use equation (1) with , , and :

The desired point is .

ith the point and the original coordinates , , , the slopes of the sides of each of , , and can be calculated:

• the slopes of the sides of are ; they form a geometric progression with common ratio
• the slopes of the sides of are ; they form a geometric progression with common ratio
• the slopes of the sides of are ; they form a geometric progression with common ratio .

There’s an additional point orth noting here. If one calculates the slopes from to each of the vertices , , , one obtains

a geometric progression with common ratio . This only happens in some specific situations (see the exercises).

Given with vertices at , , , verify that the point found in the previous example lies on the nine point circle of .

The midpoints of the sides of are , , ; they’re shown (in the big circle) below as red bullet points:

Behind the scenes, we’ve calculated the circumcenter and the orthocenter of ; they are the points and , respectively. Thus, the center of the nine-point circle is , say.

We now check that :

Boom.

## Takeaway

Triangles with slopes in geometric progressions are very nice objects, at least because of the “algebra” that their theory brings along. However, even if the slopes of the sides of a triangle do not form a geometric progression (sighs), there’s a ay around it:

• if the parent triangle has a side parallel to any of the axes, relax — already done last March.
• if the parent triangle has no side parallel to the coordinate axes, call with coordinates

(2)

and obtain triangles , , whose side-slopes form three separate geometric progressions. (Alternatively, one can also use and the orthocenter . Then the slopes of the sides of triangles , , form three separate geometric progressions.)

Throughout, any mention of refers to the point with coordinates given by equation (1) or (2). Unless otherwise specified, assume where necessary that an arbitrary triangle has vertices at , , and .

1. (Separate character) Verify that the point is not any of the nine traditional points through which the nine-point circle passes. Under normal circumstances.
2. (Special case) Let be such that . If the slopes of the legs and are , PROVE that coincides with the mid-point of (that is, the circumcenter of . Also see exercise 8 below).
3. (Special case) If , PROVE that is precisely the foot of the altitude from vertex .
4. (Shortcut) In , let be the slopes of sides . PROVE that:
• has coordinates
• the slopes from to the vertices form a geometric progression
(The best setting for is hen the slopes of the sides of the parent triangle form a geometric progression.)
5. (Shortcut) In , let be the slopes of sides , where . PROVE that:
• has coordinates
• the slopes from to the vertices are
• the slopes from to the vertices form a geometric progression if or or
(In each case, the common ratio is always positive — in fact, a “square”.)
• the slopes from to the vertices form an arithmetic progression if .
(Incidentally, the common difference of the resulting arithmetic progression is equal to , depending on how it’s arranged.)
6. (Six colleagues) Let be the orthocenter of and let be as given in equation (1). PROVE that the six statements below are equivalent:
• coincides with one of the vertices of
• has two sides parallel to the coordinate axes
• 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 ( first term)
• one side is parallel to the -axis and another side and the median to it have opposite slopes (e.g. and ).
7. (HOW come?) In , let the orthocenter and circumcenter be and , respectively, and let be as given in (1). If the slopes of sides are in that order, PROVE that:
• the circumcenter shares the same -coordinate as vertex
• the slopes of the sides of form an arithmetic progression
(One of the reasons we chose as a label for one of our four points — rather than maybe — was because we had this triangle in view.)
8. Are the following statements equivalent?
• is the midpoint of ; that is,
• either is isosceles with side parallel to the -axis and , or two sides of have opposite slopes
• the area of is or .
(This exercise shows that in some situations, behaves like the circumcenter of a right triangle which occurs at the midpoint of the hypotenuse.)
9. Consider a right triangle in which none of the sides is parallel to the coordinate axes. PROVE that the slope from to the vertex is the reciprocal of the slope of the hypotenuse.
10. PROVE that is a both a horizontal and a vertical translation of the orthocenter of the parent .