Thanksgiving Theorems II

It’s becoming our tradition to have special editions on Thanksgiving occasions, and this year is no exception. Thus, there’s a brief digression from our familiar and friendly and favourite geometric progression, to arithmetic progression.

Golden ratio

The main equation for today’s consideration is

(1)   \begin{equation*} d_1=\frac{3}{m_2+m_3},\qquad m_2\neq  -m_3. \end{equation*}

There’ll be explanation for it in examples 9 and 10. For now, look what we get if we set d_1=3\left(\frac{1}{m_2}-\frac{1}{m_3}\right):

    \begin{equation*} \begin{split} \frac{3}{m_2+m_3}&=3\left(\frac{1}{m_2}-\frac{1}{m_3}\right)\\ \frac{1}{m_2+m_3}&=\frac{m_3-m_2}{m_2m_3}\\ m_3^2-m_2m_3-m_2^2&=0\\ \left(\frac{m_3}{m_2}\right)^2-\left(\frac{m_3}{m_2}\right)-1&=0\\ \frac{m_3}{m_2}&=\frac{1\pm\sqrt{5}}{2} \end{split} \end{equation*}

a.k.a golden ratio. You’ve seen this before here, and you’ll meet it again before the end of this year.

Gentle reminder

Kindly remember that the due date to hand in your solutions to examples 1 through 8 is October 28. Please don’t hesitate, don’t leave it late.

Find the circumcenter of \triangle ABC with vertices at A(a,h), B(b,0), C(0,0).

It is located at \left(\frac{b}{2},\frac{a^2-ab+h^2}{2h}\right).

Find the orthocenter of \triangle ABC with vertices at A(a,h), B(b,0), C(0,0).

It is located at H\left(a,\frac{ab-a^2}{h}\right).

For \triangle ABC with vertices at A(a,h), B(b,0), C(0,0), find the reflection of the orthocenter along side BC.

It is the point R\left(a,\frac{a^2-ab}{h}\right), and it lies on the circumcircle of \triangle ABC.

For \triangle ABC with vertices at A(a,h), B(b,0), C(0,0), PROVE that the point N(b-a,h) is on the circumcircle of \triangle ABC.

We use examples 1 and 3. Specifically, the midpoint of N(b-a,h) and R\left(a,\frac{a^2-ab}{h}\right) is precisely the circumcenter \left(\frac{b}{2},\frac{a^2-ab+h^2}{2h}\right).

Since R is on the circumcircle and the midpoint of N and R is the center of this circle, it follows that N is also on the circumcircle.

For \triangle ABC with vertices at A(a,h), B(b,0), C(0,0), PROVE that the point N(b-a,h) and the point R\left(a,\frac{a^2-ab}{h}\right) are end points of a diameter of the circumcircle of \triangle ABC.

Use example 4 above.

Find the orthocenter of \triangle ABC with vertices at A(0,a), B(0,0), C(r,s).

Side AB is parallel to the y-axis this time around. The orthocenter is located at H\left(\frac{s}{r}(a-s),s\right).

Find the reflection, along side AB, of the orthocenter of \triangle ABC with vertices at A(0,a), B(0,0), C(r,s).

It is the point R\left(\frac{s}{r}(s-a),s\right).

Find the circumcenter of \triangle ABC with vertices at A(0,a), B(0,0), C(r,s).

It is the point \left(\frac{s^2-as+r^2}{2r},\frac{a}{2}\right).

Geometric result

(First “Theorem”)

In \triangle ABC, let m_2 and m_3 be the slopes of sides AB and CA, with side BC parallel to the x-axis. Let H be the orthocenter and let R be the reflection of the orthocenter along side BC. Then there is a point N on the circumcircle of \triangle ABC such that the slopes of the medians of \triangle HRN form an arithmetic progression with common difference d_1, where d_1=\frac{3}{m_2+m_3}.

As you traverse the sides HR,RN,NH of \triangle HRN you get familiar terms: Human Resources, Registered Nurse, Northern Hemisphere. You’ll never forget \triangle HRN.

For convenience, place the vertices of \triangle ABC at the points A(a,h), B(0,0), C(b,0). Let H,R,N be as given in examples 2,3,4:

    \[H\left(a,\frac{ab-a^2}{h}\right),~R\left(a,\frac{a^2-ab}{h}\right),~N(b-a,h).\]

The midpoints of sides HR,RN,NH are:

    \[(a,0),~\left(\frac{b}{2},\frac{h^2+a^2-ab}{2h}\right),~\left(\frac{b}{2},\frac{h^2-a^2+ab}{2h}\right)\]

respectively. And so the slopes of the medians from H,R,N are:

    \begin{equation*} \begin{split} m_H&=\left(\frac{h^2+a^2-ab}{2h}-\frac{2a(b-a)}{2h}\right)\div\left(\frac{b}{2}-\frac{2a}{2}\right)\\ &=\frac{h^2+3a^2-3ab}{h(b-2a)}\\ &=\frac{h}{b-2a}+\frac{3a(a-b)}{h(b-2a)}\\ m_R&=\left(\frac{h^2-a^2+ab}{2h}-\frac{2a(a-b)}{2h}\right)\div\left(\frac{b}{2}-\frac{2a}{2}\right)\\ &=\frac{h^2-3a^2+3ab}{h(b-2a)}\\ &=\frac{h}{b-2a}-\frac{3a(a-b)}{h(b-2a)}\\ m_N&=\frac{h-0}{(b-a)-a}\\ &=\frac{h}{b-2a} \end{split} \end{equation*}

We get an arithmetic progression m_R,m_N,m_H with common difference

    \[d_1=\pm\frac{3a(a-b)}{h(b-2a)}.\]

Returning to the parent triangle ABC, we have

    \[\textrm{slope of AB}:m_2=\frac{h}{a},~\textrm{slope of CA}:m_3=\frac{h}{a-b}.\]

Then, then, and then

    \[m_2+m_3=\frac{h(2a-b)}{a(a-b)}\implies d_1=\frac{3}{m_2+m_3}.\]

(Second “Theorem”)

In \triangle ABC, let m_2,m_3 be the slopes of sides AC,BC, with side AB parallel to the y-axis. Let H be the orthocenter and let R be the reflection of the orthocenter along side AB. Then there is a point N on the circumcircle of \triangle ABC such that the reciprocals of the slopes of the medians of \triangle HRN form an arithmetic progression with common difference d_2, where d_2=\frac{3m_2m_3}{m_2+m_3}.

For convenience, place the vertices at A(0,a), B(0,0), C(r,s). Take the orthocenter H\left(\frac{as-s^2}{r},s\right), its reflection R\left(\frac{s^2-as}{r},s\right) along side AB, and the point N(r,a-s). The mipoints of sides HR,RN,NH are:

    \[(0,s),~\left(\frac{s^2-as+r^2}{2r},\frac{a}{2}\right),~\left(\frac{as-s^2+r^2}{2r},\frac{a}{2}\right)\]

respectively. The slopes of the medians are then

    \[\frac{a-2s}{r},~\frac{r(a-2s)}{3s(s-a)+r^2},~\frac{r(a-2s)}{3s(a-s)+r^2}\]

and their reciprocals (re-arranged) are

    \[\frac{3s(s-a)+r^2}{r(a-2s)},~\frac{r}{a-2s},~\frac{3s(a-s)+r^2}{r(a-2s)},\quad a\neq s,2s.\]

These form an arithmetic sequence with common difference d_2, where

    \[d_2=\pm\frac{3s(a-s)}{r(a-2s)}=3\left(\frac{r}{s}+\frac{r}{s-a}\right)^{-1}.\]

In terms of the parent triangle ABC (vertices at A(0,a), B(0,0), C(r,s)), we get:

    \[d_2=3\left(\frac{1}{\textrm{slope of BC}}+\frac{1}{\textrm{slope of AC}}\right)^{-1}=\frac{3m_2m_3}{m_2+m_3}.\]

Takeaway

The construction we’ve done is true more generally, as follows:

  • If a triangle has one side parallel to the x-axis, then the reciprocals of the slopes of the three medians form an arithmetic progression;
  • If a triangle has one side parallel to the y-axis, then the slopes of the three medians form an arithmetic progression.

Tasks

  1. Suppose that the enumeration a,b,c represents an arithmetic progression of distinct, non-zero numbers.
    • if \frac{1}{a},\frac{1}{b},\frac{1}{c} is also arithmetic, PROVE that b^2=ac.
    • if \frac{1}{a},\frac{1}{c},\frac{1}{b} is also arithmetic, PROVE that c=-2a.
    • if \frac{1}{b},\frac{1}{a},\frac{1}{c} is also arithmetic, PROVE that a=-2c.
  2. Find three distinct, non-zero numbers a,b,c in arithmetic progression for which the reciprocals \frac{1}{a},\frac{1}{c},\frac{1}{b} are also in arithmetic progression.
  3. Find three distinct, non-zero numbers a,b,c in arithmetic progression for which the reciprocals \frac{1}{b},\frac{1}{a},\frac{1}{c} are also in arithmetic progression.
  4. (Registered Nurse) For \triangle ABC with vertices at A(0,a), B(0,0), and C(r,s), PROVE that the points R\equiv \left(\frac{s}{r}(s-a),s\right) and N\equiv(r,a-s) are end points of a diameter of the circumcircle of \triangle ABC.
  5. Find coordinates for the vertices of a triangle in which the slopes of the three medians are 1,2,3.

Thanksgiving

Nationally, Thanksgiving is just one day — this past Monday; naturally, ours lasts more days, many days — being our mainstay.

Our style of incorporating Thanksgiving into an “academic setting” is not very popular, and you may not be familiar with it. It’s part of what makes us peculiar; and in particular, it’s due to a spectacular event on Thursday, June 14, 2018: beautiful, remarkable, unforgettable day. Consequently (and continuously and conspicuously), the poster expresses gratitude to the Ancient of Days.

All things being equal, our next iteration of Thanksgiving comes up on Monday, June 14, 2021. Until then, please understand that this practice is very useful for us, as we’re hopeful that we’ll always be mindful of the need to be grateful.