Slopefessor Fr1day – Page 25 – High School Geometry

The golden ratio from slopes

On October 14, we promised that a slope-based derivation of the golden ratio will be aired before the end of this year; and on that premise, here we are. Actually, prior to this episode, the golden ratio has already made three cameos, but this is its full scenario.

Antisymmetric cubic

Also called anti-palindromic, due to the coefficients. Not limited to cubics though. An example is $x^3-3x^2+3x-1$ , the binomial expansion of $(x-1)^3$ . Another example is $2x^3-3x^2+3x-2$ . And so on. You get the gist.

In any right isosceles triangle with slopes in geometric progression, PROVE that the common ratio $r$ satisfies the cubic equation $r^3+2r^2-2r-1=0$ .

Let $\triangle ABC$ be right isosceles and let the geometric progression of slopes be $a,ar,ar^2$ for sides $AB,BC,CA$ , respectively.

From this post, the slopes of the medians to these sides will be

$\left(\frac{2r+1}{r+2}\right)ar,~-ar,~\left(\frac{r+2}{2r+1}\right)ar$

respectively.

Now, side $BC$ cannot be the hypotenuse (you know why, right?) This leaves us with either $AB$ or $CA$ as the hypotenuse.

First suppose that $AB$ is the hypotenuse. Then $BC\perp CA$ and so

(1) $\begin{equation*} ar\times ar^2=-1 \end{equation*}$

By the isosceles condition, $BC=CA$ and so the median from vertex $C$ is perpendicular to side $AB$ . This gives:

(2) $\begin{equation*} \left(\frac{2r+1}{r+2}\right)ar\times a=-1 \end{equation*}$

Combine equations (1) and (2):

$\begin{equation*} \begin{split} ar\times ar^2&=\left(\frac{2r+1}{r+2}\right)ar\times a\\ r^2&=\left(\frac{2r+1}{r+2}\right)\quad \Big(\textbf{since}~a,r\neq 0\Big)\\ r^2(r+2)&=2r+1\\ r^3+2r^2-2r-1&=0. \end{split} \end{equation*}$

There we gold. Switching the hypotenuse won’t alter the fact that the equation holds. For suppose instead that $CA$ is the hypotenuse. Then $AB\perp BC$ , giving

(3) $\begin{equation*} a\times ar=-1 \end{equation*}$

The isosceles condition forces $AB=BC$ , in which case the median from vertex $B$ meets $CA$ at right angles:

(4) $\begin{equation*} ar^2\times\left(\frac{r+2}{2r+1}\right)ar=-1 \end{equation*}$

As before, combining equations (3) and (4) yields $r^3+2r^2-2r-1=0$ .

In any right isosceles triangle with side-slopes in geometric progression, PROVE that the common ratio $r$ satisfies $r^2+3r+1=0$ .

From example 1 above, the common ratio $r$ satisfies the cubic equation

$r^3+2r^2-2r-1=0$

which can be easily factored because of the “antisymmetric” coefficients:

$\begin{equation*} \begin{split} r^3+2r^2-2r-1&=0\\ (r^3-1)+2r(r-1)&=0\\ (r-1)(r^2+r+1)+2r(r-1)&=0\\ (r-1)\Big(r^2+3r+1\Big)&=0 \end{split} \end{equation*}$

Slope considerations means we cannot have $r-1=0$ . This leaves us with the choice $r^2+3r+1=0$ (see here for a different proof).

Ancient concept

The golden ratio is a fascinating quantity, not just because it has been around since antiquity, but also because of its ubiquity. And, as shown below, it does seem to have an affinity for right isosceles triangles with slopes in geometric progressions, considering the frequency with which it turns up in their “vicinity”.

In a right isosceles triangle with slopes in geometric progression $a,ar,ar^2$ , PROVE that the the common ratio $r$ is either $r=-\phi^2$ or $r=-\frac{1}{\phi^2}$ . (In other words, $r$ is either the negative of the square of the golden ratio, or the negative of the square of the reciprocal of the golden ratio.)

The golden ratio is $\phi = \frac{1+\sqrt{5}}{2}$ and its reciprocal is $\frac{1}{\phi}=\frac{1-\sqrt{5}}{-2}$ .

The common ratio $r$ of our right isosceles triangle satisfies the quadratic equation

$r^2+3r+1=0\implies r=\frac{-3\pm\sqrt{5}}{2}.$

Let’s split the two roots, like so:

$r_1=\frac{-3+\sqrt{5}}{2},~~r_2=\frac{-3-\sqrt{5}}{2}.$

What we want now follows immediately:

$\begin{equation*} \begin{split} \phi^2&=\left(\frac{1+\sqrt{5}}{2}\right)^2\\ &=\frac{6+2\sqrt{5}}{4}\\ &=\frac{3+\sqrt{5}}{2}\\ &=-r_2\\ \implies r_2&=-\phi^2\\ \left(\frac{1}{\phi}\right)^2&=\left(\frac{1-\sqrt{5}}{-2}\right)^2\\ &=\frac{6-2\sqrt{5}}{4}\\ &=\frac{3-\sqrt{5}}{2}\\ &=-r_1\\ \implies r_1&=-\left(\frac{1}{\phi}\right)^2 \end{split} \end{equation*}$

If you ever needed an example that shows that the common ratio in a right triangle (with slopes in geometric progression) is necessarily negative, this is it.

In a right isosceles triangle with slopes in geometric progression $a,ar,ar^2$ , PROVE that the the first term $a=\pm\left(\frac{1+\sqrt{5}}{2}\right)$ — either the golden ratio or its negative, if the hypotenuse has slope $ar^2$ . Under the same conditions, PROVE also that we have $a=\pm\left(\frac{1-\sqrt{5}}{2}\right)$ — either the reciprocal of the golden ratio or the negative of its reciprocal.

Suppose that the hypotenuse has slope $ar^2$ . Then the legs’ slopes are $a$ and $ar$ . Consequently:

$\begin{equation*} \begin{split} a\times ar&=-1\\ a^2&=\frac{-1}{r}\\ a^2&=\frac{2}{3-\sqrt{5}}\quad\textrm{using}\quad r_1=\frac{-3+\sqrt{5}}{2}\\ &=\frac{2}{3-\sqrt{5}}\times \frac{3+\sqrt{5}}{3+\sqrt{5}}\\ &=\frac{3+\sqrt{5}}{2}\\ &=\left(\frac{1+\sqrt{5}}{2}\right)^2\\ \implies a&=\pm\left(\frac{1+\sqrt{5}}{2}\right)\\ \implies a&=\pm \phi\\ \textrm{Also,}~a^2&=\frac{2}{3+\sqrt{5}}\quad\textrm{using}\quad r_2=\frac{-3-\sqrt{5}}{2}\\ &=\frac{3-\sqrt{5}}{2}\\ &=\left(\frac{1-\sqrt{5}}{2}\right)^2\\ \implies a&=\pm\left(\frac{1-\sqrt{5}}{2}\right)\\ \implies a&=\pm\frac{1}{\phi} \end{split} \end{equation}$

Does this example provide a golden opportunity for us to reconsider the definition of a golden triangle?

In a right isosceles triangle with slopes in geometric progression $a,ar,ar^2$ , PROVE that the the first term $a$ is either $\pm(2+\sqrt{5})$ or $\pm(2-\sqrt{5})$ , if the hypotenuse has slope $a$ .

If the hypotenuse has slope $a$ , then the legs are $ar$ and $ar^3$ . Consequently:

$\begin{equation*} \begin{split} ar\times ar^2&=-1\\ a^2&=\frac{-1}{r^3}\\ a^2&=\frac{-1}{\left(\frac{-3+\sqrt{5}}{2}\right)^3}\quad\textrm{using}\quad r_1=\frac{-3+\sqrt{5}}{2}\\ &=\frac{-1}{\frac{-27+27\sqrt{5}-9\sqrt{5}+5\sqrt{5}}{8}}\\ &=\frac{-1}{\frac{-72+32\sqrt{5}}{8}}\\ &=\frac{-1}{4\sqrt{5}-9}\\ &=9+4\sqrt{5}\\ &=(2+\sqrt{5})^2\\ \implies a&=\pm(2+\sqrt{5})\\ \textrm{Also,}~a^2&=\frac{-1}{\left(\frac{-3-\sqrt{5}}{2}\right)^3}\quad\textrm{using}\quad r_2=\frac{-3-\sqrt{5}}{2}\\ &=\frac{-1}{-9-4\sqrt{5}}\\ &=9-4\sqrt{5}\\ &=(2-\sqrt{5})^2\\ \implies a&=\pm(2-\sqrt{5}). \end{split} \end{equation*}$

Absolute character

In a right isosceles triangle with side-slopes $a,ar,ar^2$ , we always have

(5) $\begin{equation*} |a|^2-|a|-1=0\quad\textrm{or}\quad\left|\frac{1}{a}\right|^2-\left|\frac{1}{a}\right|-1=0 \end{equation*}$

when the side with slope $ar^2$ is the hypotenuse, and we always have

(6) $\begin{equation*} |a|^2-4|a|-1=0\quad\textrm{or}\quad\left|\frac{1}{a}\right|^2-4\left|\frac{1}{a}\right|-1=0 \end{equation*}$

when the hypotenuse is the side with slope $a$ . Think “antisymmetry” as being behind these equations.

Let $a,ar,ar^2$ be the slopes of the sides of a right isosceles triangle, with the hypotenuse having slope $ar^2$ . PROVE that $|a|^2-|a|-1=0$ or $\left|\frac{1}{a}\right|^2-\left|\frac{1}{a}\right|-1=0$ .

In view of example 4 above, there are four possibilities for $a$ , viz: $a=\pm\left(\frac{1+\sqrt{5}}{2}\right)$ or $a=\pm\left(\frac{1-\sqrt{5}}{2}\right)$ . The “base” case will satisfy the parent quadratic $a^2-a-1=0$ , while two other cases satisfy $\frac{1}{a^2}-\frac{1}{a}-1=0$ — meaning the absolute values are not absolutely necessary after all. However, the way the proof proceeds is slightly different, depending on the presence or absence of absolute values.

$\begin{equation*} \begin{split} \textbf{Case I:}\quad a&=\frac{1+\sqrt{5}}{2}\\ \implies |a|^2-|a|-1&=\left(\frac{3+\sqrt{5}}{2}\right)-\left(\frac{1+\sqrt{5}}{2}\right)-1\\ &=0\\ \textbf{Case II:}\quad a&=-\left(\frac{1+\sqrt{5}}{2}\right)\\ \implies |a|^2-|a|-1&=\left(\frac{3+\sqrt{5}}{2}\right)-\left(\frac{1+\sqrt{5}}{2}\right)-1\\ &=0\\ \textbf{Case III:}\quad a&=\frac{1-\sqrt{5}}{2}\\ \implies \frac{1}{a}&=-\left(\frac{1+\sqrt{5}}{2}\right)\\ \implies\left|\frac{1}{a}\right|^2-\left|\frac{1}{a}\right|-1&=\left(\frac{3+\sqrt{5}}{2}\right)-\left(\frac{1+\sqrt{5}}{2}\right)-1\\ &=0\\ \textbf{Case IV:}\quad a&=-\left(\frac{1-\sqrt{5}}{2}\right)\\ \implies \frac{1}{a}&=\frac{1+\sqrt{5}}{2}\\ \implies\left|\frac{1}{a}\right|^2-\left|\frac{1}{a}\right|-1&=\left(\frac{3+\sqrt{5}}{2}\right)-\left(\frac{1+\sqrt{5}}{2}\right)-1\\ &=0 \end{split} \end{equation*}$

Let $a,ar,ar^2$ be the slopes of the sides of a right isosceles triangle, with the hypotenuse having slope $a$ . PROVE that $|a|^2-4|a|-1=0$ or $\left|\frac{1}{a}\right|^2-4\left|\frac{1}{a}\right|-1=0$ .

In view of example 5, there are also four cases here: $a=\pm(2+\sqrt{5})$ or $a=\pm(2-\sqrt{5})$ .

$\begin{equation*} \begin{split} \textbf{Case I:}\quad a&=2+\sqrt{5}\\ \implies |a|^2-4|a|-1&=(9+4\sqrt{5})-4(2+\sqrt{5})-1\\ &=0\\ \textbf{Case II:}\quad a&=-2-\sqrt{5}\\ \implies |a|^2-4|a|-1&=(9+4\sqrt{5})-4(2+\sqrt{5})-1\\ &=0\\ \textbf{Case III:}\quad a&=2-\sqrt{5}\\ \implies \frac{1}{a}&=-2-\sqrt{5}\\ \implies \left|\frac{1}{a}\right|^2-4\left|\frac{1}{a}\right|-1&=(9+4\sqrt{5})-4(2+\sqrt{5})-1\\ &=0\\ \textbf{Case IV:}\quad a&=-2+\sqrt{5}\\ \implies \frac{1}{a}&=2+\sqrt{5}\\ \implies \left|\frac{1}{a}\right|^2-4\left|\frac{1}{a}\right|-1&=(9+4\sqrt{5})+4(2+\sqrt{5})-1\\ &=0 \end{split} \end{equation*}$

Alternative construct

Sometime in the future, we’ll feature our own version of the geometric mean theorem. For now, what to recall is that in any right triangle, the altitude $h$ (from the $90^{\circ}$ vertex) to the hypotenuse divides the hypotenuse into parts $p$ and $q$ such that

$h=\sqrt{pq}.$

Since any triangle with side-slopes in geometric progression can be viewed as a clone of the right triangle, we have the following analogue

$h=\frac{ar(r-1)}{\sqrt{(a^2r^2+r)(a^2r^3+1)}}\sqrt{pq}$

where $h$ is the altitude to the side whose slope is $ar$ (and the altitude divides that side into two parts $p$ and $q$ ). Furthermore we have this relationship:

(7) $\begin{equation*} \frac{p}{q}=\frac{a^2r^2+r}{a^2r^3+1}. \end{equation*}$

Remember the name: silvery ratio.

Let $1,r,r^2$ be the slopes of sides $AB,BC,CA$ in $\triangle ABC$ . PROVE that if the altitude to the side $BC$ divides $BC$ in the ratio $1:2$ , then the common ratio $r=\phi^2$ or $r=\left(\frac{1}{\phi}\right)^2$ , where $\phi$ is the golden ratio.

Interesting scenes. Put $a=1$ in equation (7) and set $\frac{p}{q}=\frac{1}{2}$ :

$\begin{equation*} \begin{split} \frac{r^2+r}{r^3+1}&=\frac{1}{2}\\ r^3-2r^2-2r+1&=0\\ (r^3+1)-2r(r+1)&=0\\ (r+1)(r^2-r+1)-2r(r+1)&=0\\ (r+1)(r^2-3r+1)&=0\\ r&=-1\quad\textrm{or}\quad r=\frac{3\pm\sqrt{5}}{2} \end{split} \end{equation*}$

The value $r=-1$ is unacceptable due to slope considerations. So we take $r=\frac{3\pm\sqrt{5}}{2}$ . Then:

$\begin{equation*} \begin{split} r_1&=\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^2=\phi^2\\ r_2&=\frac{3-\sqrt{5}}{2}=\left(\frac{1-\sqrt{5}}{2}\right)^2=\left(\frac{1}{\phi}\right)^2\\ \end{split} \end{equation*}$

All too familiar by now.

Artificial case

What we get in our next example is not quite the golden ratio, but its made-up “multiple”.

Solve the rational equation $\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=5$ .

The left member of the above equation is that sumptuous sum we formed not long after the summer. It should now be common due to how many times it’s been summoned.

Clear fractions: $4r^4+2r^3-3r^2+2r+4=5(2r^3+5r^2+2r)$ . Combine like terms: $4r^4-8r^3-28r^2-8r+4=0$ . Divide through by the common factor $4$ : $r^4-2r^3-7r^2-2r+1=0$ . Assume two quadratic factors of the form $(r^2+\alpha r+1)(r^2+\beta r+1)$ so that, after expanding the right side and comparing coefficients, we have:

$\alpha+\beta=-2,\quad \alpha\beta+2=-7\implies \alpha\beta=-9.$

We’ll now solve the linear-quadratic system $\alpha+\beta=-2,~\alpha\beta=-9$ .

$\begin{equation*} \begin{split} \alpha(-2-\alpha)&=-9\\ \implies \alpha^2+2\alpha-9&=0\\ \implies \alpha&=\frac{-2\pm\sqrt{40}}{2}\\ &=-1\pm\sqrt{10} \end{split} \end{equation*}$

You can already see something close to that golden ratio thing.

$\beta=-2-\alpha=-2-(-1\pm\sqrt{10})=-1\mp\sqrt{10}$

And so

$r^4-2r^3-7r^2-2r+1=\Big(r^2+(-1+\sqrt{10})r+1\Big)\Big(r^2+(-1-\sqrt{10})r+1\Big).$

The first two zeros are got from

$\begin{equation*} \begin{split} r^2+(-1+\sqrt{10})r+1&=0\\ r&=\frac{(1-\sqrt{10})\pm\sqrt{(-1+\sqrt{10})^2-4}}{2}\\ &=\frac{(1-\sqrt{10})\pm\sqrt{7-2\sqrt{10}}}{2}\\ &=\frac{(1-\sqrt{10})\pm\sqrt{(\sqrt{5}-\sqrt{2})^2}}{2}\\ &=\frac{(1-\sqrt{10})\pm(\sqrt{5}-\sqrt{2})}{2}\\ r_1&=\frac{1-\sqrt{10}+\sqrt{5}-\sqrt{2}}{2}\\ r_2&=\frac{1-\sqrt{10}-\sqrt{5}+\sqrt{2}}{2}\\ \end{split} \end{equation*}$

The third and fourth zeros are got from

$\begin{equation*} \begin{split} r^2+(-1-\sqrt{10})r+1&=0\\ r&=\frac{(1+\sqrt{10})\pm\sqrt{(-1-\sqrt{10})^2-4}}{2}\\ &=\frac{(1+\sqrt{10})\pm\sqrt{7+2\sqrt{10}}}{2}\\ &=\frac{(1+\sqrt{10})\pm\sqrt{(\sqrt{5}+\sqrt{2})^2}}{2}\\ &=\frac{(1+\sqrt{10})\pm(\sqrt{5}+\sqrt{2})}{2}\\ r_3&=\frac{1+\sqrt{10}+\sqrt{5}+\sqrt{2}}{2}\\ r_4&=\frac{1+\sqrt{10}-\sqrt{5}-\sqrt{2}}{2}\\ \end{split} \end{equation*}$

Notice that the roots may be re-written:

$\begin{equation*} \begin{split} r_4&=\frac{1-\sqrt{2}-\sqrt{5}+\sqrt{10}}{2}=\frac{(1-\sqrt{2})(1-\sqrt{5})}{2}\\ r_3&=\frac{1+\sqrt{2}+\sqrt{5}+\sqrt{10}}{2}=\frac{(1+\sqrt{2})(1+\sqrt{5})}{2}\\ r_2&=\frac{1+\sqrt{2}-\sqrt{5}-\sqrt{10}}{2}=\frac{(1+\sqrt{2})(1-\sqrt{5})}{2}\\ r_1&=\frac{1-\sqrt{2}+\sqrt{5}-\sqrt{10}}{2}=\frac{(1-\sqrt{2})(1+\sqrt{5})}{2} \end{split} \end{equation*}$

making them “multiples” of the golden ratio and its reciprocal.

Associated converse

Compare — and combine — the example below with example 2.

Suppose that $\triangle ABC$ has sides $AB,BC,CA$ with slopes $a,ar,ar^2$ . If $r^2+3r+1=0$ and $AB=BC$ (or $BC=CA$ ), PROVE that $\triangle ABC$ is a right triangle.

Suppose that $AB=BC$ and $r^2+3r+1=0$ . Since the slope of side $CA$ is $ar^2$ , the slope of the median from $B$ to it will be $\left(\frac{r+2}{2r+1}\right)ar$ . This median is actually an altitude, due to the $AB=BC$ condition. We then have:

(8) $\begin{equation*} \left(\frac{r+2}{2r+1}\right)ar\times ar^2=-1. \end{equation*}$

Claim: $a^2r=-1$ , which will ensure that $AB\perp BC$ . Isolate $a^2r$ from equation (8) and use a few re-arrangements of the quadratic equation $r^2+3r+1=0$ repeatedly:

$\begin{equation*} \begin{split} a^2r&=\frac{-(2r+1)}{r^2(r+2)}\\ &=\frac{r^2+r}{r^2(r+2)}\\ &=\frac{r+1}{r(r+2)}\\ &=\frac{r+1}{r^2+2r}\\ &=\frac{r+1}{-(r+1)}\\ &=-1. \end{split} \end{equation*}$

Suppose now that $BC=CA$ and $r^2+3r+1=0$ . The median from $C$ is then an altitude; and its slope being $\left(\frac{2r+1}{r+2}\right)ar$ means that

(9) $\begin{equation*} \left(\frac{2r+1}{r+2}\right)ar\times a=-1. \end{equation*}$

As before, we claim that $ar\times ar^2=-1$ ; that is, $a^2r^3=-1$ (this will ensure that $BC\perp CA$ ). From (9) isolate $a^2r$ :

$a^2r=\frac{-(r+2)}{2r+1}$

Multiply both sides by $r^2$ and maneuver the equation $r^2+3r+1=0$ , like so:

$\begin{equation*} \begin{split} a^2r^3&=\frac{-(r+2)r^2}{2r+1}\\ &=\frac{-r^3-2r^2}{2r+1}\\ &=\frac{r^2+r}{-(r^2+r)}\\ &=-1 \end{split} \end{equation*}$

We’ve proved: in $\triangle ABC$ with slopes $a,ar,ar^2$ for sides $AB,BC,CA$ and $AB=BC$ (or $BC=CA$ ), one has $r^2+3r+1=0$ if and only if the triangle is a right triangle.

Gold.

Takeaway

For any triangle with sides $AB,BC,CA$ having slopes in geometric progression $a,ar,ar^2$ , the following statements are equivalent:

$q=-19$
$r^2+3r+1=0$ or $r^2+7r+1=0$
$AB^2+CA^2=3BC^2$ or $AB^2+CA^2=\frac{7}{5}BC^2$
$m_{AB}^2+m_{CA}^2=\frac{7}{5}m_{BC}^2$ or $m_{AB}^2+m_{CA}^2=3m_{BC}^2$
$r=-\phi^2$ or $r=-\phi^4$ or $r=-\frac{1}{\phi^2}$ or $r=-\frac{1}{\phi^4}$ , where $\phi$ is the golden ratio
the common ratio $r$ is that of a right isosceles triangle (or that of its “look-alike”)

Thanks for joining us this year. See you next year! Cheers!

Tasks

(Golden Sum) Consider , a sum that should now become common due to the several times it has been summoned.
- PROVE that the numerator factors as $4r^4+2r^3-3r^2+2r+4=4\Big(r^2+\left(\frac{1+3\sqrt{5}}{4}\right)r+1\Big)\Big(r^2+\left(\frac{1-3\sqrt{5}}{4}\right)r+1\Big)$ .
- Hence, deduce that $4r^4+2r^3-3r^2+2r+4=4\Big(r^2+\frac{1}{2}(3\phi-1)r+1\Big)\left(r^2-\frac{1}{2}\left(\frac{3}{\phi}+1\right)r+1\right)$ , where $\phi$ is the golden ratio.
(Golden State) Find coordinates for the vertices of a triangle with slopes in geometric progression in which:
- the area is $\frac{1+\sqrt{5}}{2}$
- one of the side-slopes is $\frac{1+\sqrt{5}}{2}$ .
  (Note that the two parts of this question are completely separate/independent, and several answers are possible for each one.)
(Golden Equations) Let be the golden ratio. PROVE that:
- $\phi^4-3\phi^2+1=0$
- $\phi^8-7\phi^4+1=0$
- $\phi^{16}-47\phi^8+1=0$
- $\phi^{32}-2207\phi^{16}+1=0$
  (The next middle coefficient is $4,870,847$ . Be bold and go further.)
(Golden Difference) Let be the slopes of sides in .
- If $r> 0$ , PROVE that there is a point $X$ on $AB$ , a point $Y$ on $BC$ , and a point $Z$ on $CA$ such that the slopes of the sides of $\triangle XYZ$ form an arithmetic progression with common difference $d=\frac{a}{2}(r^2-1)$ .
- If $a=2$ above and $r$ is the golden ratio, PROVE that $d$ is also the golden ratio.
  (The points $X,Y,Z$ are all internal points when $r> 0$ . The situation is even more interesting when $r< 0$ , as the points now become external.)
(Golden Mistake) Let be the slopes of the sides of a triangle. Suppose that the equations and both hold.
- PROVE that $r^3+2r^2-2r-1=0$
- Deduce that the original equations are inconsistent.
  (This would have been a different way of deriving the golden ratio, if the equations were consistent).
Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- $AB^2+CA^2=3BC^2$
- $m_{AB}^2+m_{CA}^2=\frac{7}{5}m_{BC}^2$
- $q=-19$
Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- $AB^2+CA^2=\frac{7}{5}BC^2$
- $m_{AB}^2+m_{CA}^2=3m_{BC}^2$
- $q=-19$ (Compare with the previous question.)
Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- $AB^2+CA^2=13BC^2$
- $m_{AB}^2+m_{CA}^2=\frac{17}{25}m_{BC}^2$
- $q=\frac{31}{6}$
Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- $AB^2+CA^2=\frac{17}{25}BC^2$
- $m_{AB}^2+m_{CA}^2=13m_{BC}^2$
- $q=\frac{31}{6}$ (Compare the “look-alikes” in 8 and 9.)
If $r^2+1=-\frac{\left(n^2+(n-1)^2\right)}{n(n-1)}r$ , PROVE that $r^4+1=\frac{n^4+(n-1)^4}{n^2(n-1)^2}r^2$ . (Notice how the squaring operation “distributes” over addition. Strange, eh?)

A local maximum problem

By now, someone may have become accustomed to the sum we formed sumtime this Fall:

(1) $\begin{equation*} q=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}. \end{equation*}$

We’ve summoned this sum again in order to study its extremum behaviour, which we found to be somewhat fun. As is the norm, we’ll use the platform of calculus to this end and then give a summary of our findings just after example 10. See you there!!!

Partial fraction decomposition

So we want to obtain the first two derivatives of the rational function (1). By default, the quotient rule should be used, but the fault here is that it gets a bit difficult — especially for the second derivative — and so we’ve chosen a different route.

If the denominator of a rational function is a product of linear factors, it may be suitable in some situations to first split the rational function into its partial fractions before proceeding with differentiation, rather than using the quotient rule. Depending on what one wants, the additional work done may be worth it in the end.

Resolve $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ into its partial fractions.

The given rational function is improper, so we first put it in a proper form using long division:

$\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=(2r-4)+\frac{13r^2+10r+4}{2r^3+5r^2+2r}.$

Notice that $2r^3+5r^2+2r=r(r+2)(2r+1)$ . Set

$\frac{13r^2+10r+4}{2r^3+5r^2+2r}=\frac{A}{r}+\frac{B}{r+2}+\frac{C}{2r+1}$

and obtain the constants $A,B,C$ by the cover-up method:

$\begin{equation*} \begin{split} A&=\frac{13\times 0^2+10\times 0+4}{(0+2)(2\times 0+1)}\\ &=2\\ B&=\frac{13\times (-2)^2+10\times (-2)+4}{(-2)(2\times -2+1)}\\ &=6\\ C&=\frac{13\times\left(-1/2\right)^2+10\times\left(-1/2\right)+4}{\left(-1/2\right)\left(-1/2+2\right)}\\ &=-3 \end{split} \end{equation*}$

Finally:

$q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=(2r-4)+\frac{2}{r}+\frac{6}{r+2}-\frac{3}{2r+1}.$

Find the first derivative of $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ with respect to $r$ .

We make use of the partial fraction decomposition from the previous example to compute the derivative (compare this strategy with logarithmic differentiation).

$q=(2r-4)+\frac{2}{r}+\frac{6}{r+2}-\frac{3}{2r+1}\implies q'=2-\frac{2}{r^2}-\frac{6}{(r+2)^2}+\frac{6}{(2r+1)^2}.$

A breeze.

Find the second derivative of $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ with respect to $r$ .

$q'=2-\frac{2}{r^2}-\frac{6}{(r+2)^2}+\frac{6}{(2r+1)^2}\implies q''=\frac{4}{r^3}+\frac{12}{(r+2)^3}-\frac{24}{(2r+1)^3}.$

Also a breeze because of the underlying work done in the initial partial fraction decomposition. Hard work pays.

Polynomial factoring digression

Factorize $r^2(r+2)^2(2r+1)^2-(r+2)^2(2r+1)^2-3r^2(2r+1)^2+3r^2(r+2)^2$ .

The presence of some common factors makes the factorization a breeze.

$\begin{equation*} \begin{split} &=(r+2)^2(2r+1)^2\Big(r^2-1\Big)-3r^2\Big((2r+1)^2-(r+2)^2\Big)\\ &=(r+2)^2(2r+1)^2\Big(r^2-1\Big)-3r^2\Big(3(r^2-1)\Big)\\ &=(r^2-1)\Big((r+2)^2(2r+1)^2-9r^2\Big)\\ &=(r^2-1)\Big(\left[(r+2)(2r+1)-3r\right]\left[(r+2)(2r+1)+3r\right]\Big)\\ &=(r^2-1)\Big(2r^2+5r+2-3r\Big)\Big(2r^2+5r+2+3r\Big)\\ &=4(r^2-1)(r^2+r+1)(r^2+4r+1) \end{split} \end{equation*}$

See that special quadratic $r^2+4r+1$ ? Get excited any time you spot it!!!

Find the critical numbers for $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ .

Equate the first derivative to zero and solve for $r$ :

$\begin{equation*} \begin{split} q'&=0\\ \implies 2-\frac{2}{r^2}-\frac{6}{(r+2)^2}+\frac{6}{(2r+1)^2}&=0\\ \frac{1}{r^2}+\frac{3}{(r+2)^2}-\frac{3}{(2r+1)^2}&=1\\ (r+2)^2(2r+1)^2+3r^2(2r+1)^2-3r^2(r+2)^2-r^2(r+2)^2(2r+1)^2&=0\\ r^2(r+2)^2(2r+1)^2-(r+2)^2(2r+1)^2-3r^2(2r+1)^2+3r^2(r+2)^2&=0\\ 4(r^2-1)(r^2+r+1)(r^2+4r+1)&=0 \end{split} \end{equation*}$

The last line came from the factorization in the preceding example. Equate each factor to zero:

$\begin{equation*} \begin{split} r^2-1&=0\\ r&=\pm 1\\ r^2+r+1&=0\\ r&=\frac{-1\pm\sqrt{1^2-4\times 1\times 1}}{2\times 1}\\ &=\frac{-1\pm\sqrt{-3}}{2}\\ &=\textrm{no solution}\\ r^2+4r+1&=0\\ r&=\frac{-4\pm\sqrt{4^2-4\times 1\times 1}}{2\times 1}\\ &=\frac{-4\pm\sqrt{12}}{2}\\ &=-2\pm\sqrt{3} \end{split} \end{equation*}$

We obtain the critical numbers: $\pm 1,-2\pm\sqrt{3}$ . All stationary/turning points. The vertical asymptotes at $r=0,-2,-\frac{1}{2}$ are also critical numbers because the derivative is undefined at each of these, but these values are already exqluded.

Pretty familiar deductions

Show that $r=\pm 1$ give local minimum values for $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ .

We use the second derivative test.

$\begin{equation*} \begin{split} q''&=\frac{4}{r^3}+\frac{12}{(r+2)^3}-\frac{24}{(2r+1)^3}\\ \implies q''(1)&=\frac{4}{1^3}+\frac{12}{(1+2)^3}-\frac{24}{(2\times 1+1)^3}\\ &=4+\frac{12}{27}-\frac{24}{27}\\ &=\frac{32}{9}\\ \end{split} \end{equation*}$

Since the second derivative is positive, $r=1$ gives a local minimum.

$\begin{equation*} \begin{split} q''&=\frac{4}{r^3}+\frac{12}{(r+2)^3}-\frac{24}{(2r+1)^3}\\ \implies q''(-1)&=\frac{4}{(-1)^3}+\frac{12}{(-1+2)^3}-\frac{24}{(2\times (-1)+1)^3}\\ &=-4+\frac{12}{1}+\frac{24}{1}\\ &=32\\ \end{split} \end{equation*}$

Since the second derivative is positive, $r=-1$ gives a local minimum.

Show that $r=-2\pm\sqrt{3}$ give local maximum values for $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ .

Starting with $r=-2+\sqrt{3}$ , let’s evaluate the second derivative.

$\begin{equation*} \begin{split} q''&=\frac{4}{r^3}+\frac{12}{(r+2)^3}-\frac{24}{(2r+1)^3}\\ q''(-2+\sqrt{3})&=\frac{4}{(-2+\sqrt{3})^3}+\frac{12}{(\sqrt{3})^3}-\frac{24}{(-3+2\sqrt{3})^3}\\ &=\frac{4}{-26+15\sqrt{3}}+\frac{12}{3\sqrt{3}}+\frac{24}{3\sqrt{3}(-26+15\sqrt{3})}\\ &=4(-26-15\sqrt{3})+\frac{4}{3}\sqrt{3}+\frac{8}{3}\sqrt{3}(-26-15\sqrt{3})\\ &=-224-128\sqrt{3} \end{split} \end{equation*}$

Since the second derivative is negative, we conclude that $r=-2+\sqrt{3}$ gives a local maximum value for $q$ . Next, we calculate the second derivative when $r=-2-\sqrt{3}$ :

$\begin{equation*} \begin{split} q''&=\frac{4}{r^3}+\frac{12}{(r+2)^3}-\frac{24}{(2r+1)^3}\\ q''(-2-\sqrt{3})&=\frac{4}{-26-15\sqrt{3}}+\frac{12}{-3\sqrt{3}}-\frac{24}{3\sqrt{3}(-26-15\sqrt{3})}\\ &=4(-26+15\sqrt{3})-\frac{4}{3}\sqrt{3}-\frac{8}{3}\sqrt{3}(-26+15\sqrt{3})\\ &=-224+128\sqrt{3} \end{split} \end{equation*}$

This may require a calculator to ascertain that it’s negative. Indeed, $-224+128\sqrt{3}\approx -2.2975< 0$ . Thus, $r=-2-\sqrt{3}$ gives a local maximum value for $q$ .

Find the local minimum value(s) and the local maximum value(s) for the rational function $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}$ .

From the two preceding examples, $r=\pm 1$ give local minimum value(s) while $r=-2\pm\sqrt{3}$ give local maximum value(s). So we compute the corresponding values of $q$ at these points.

$\begin{equation*} \begin{split} q(1)&=\frac{4+2-3+2+4}{2+5+2}\\ &=1\\ q(-1)&=\frac{4-2-3-2+4}{-2+5-2}\\ &=1 \end{split} \end{equation*}$

For $r=-2\pm\sqrt{3}$ , it helps to know that they are both solutions of $r^2+4r+1=0$ . This makes the computation of $q$ easier, and we did that last post:

$q(-2+\sqrt{3})=-15=q(-2-\sqrt{3}).$

Thus, the local minimum value is $1$ and the local maximum value is $-15$ . (Don’t be surprised that the local minimum value exceeds the local maximum value. They’re local, not global.)

Solve the equation $\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=-3$ .

We suspect that there is no real-valued solution to the given equation. Why? Because the right side exceeds the local maximum value of $-15$ (and is smaller than the local minimum value of $1$ ).

Let’s confirm our suspicion. Clear fractions to get better traction and continue the solution:

$\begin{equation*} \begin{split} 4r^4+2r^3-3r^2+2r+4&=-3(2r^3+5r^2+2r)\\ 4r^4+8r^3+12r^2+8r+4&=0\\ r^4+2r^3+3r^2+2r+1&=0\\ (r^2+r+1)^2&=0\\ \end{split} \end{equation*}$

The discriminant of the resulting quadratic is $1^2-4\times 1\times 1=-3$ . Thus, there is no real-valued solution to the quadratic, and hence no real-valued solution to the given rational equation.

Calqulate $q$ for $\triangle ABC$ with vertices at $A(-1-\sqrt{3},5+3\sqrt{3})$ , $B(0,0)$ , $C(1,7+4\sqrt{3})$ .

The ingredients we need are the median-slopes:

$m_A=-1,~m_B=-7-4\sqrt{3},~m_C=2+\sqrt{3}$

and the side-slopes:

$m_{AB}=-2-\sqrt{3},~m_{BC}=7+4\sqrt{3},~m_{CA}=1.$

By definition and substitution:

$\begin{equation*} \begin{split} q&=\frac{m_{BC}}{m_A}+\frac{m_{CA}}{m_B}+\frac{m_{AB}}{m_C}\\ &=\frac{7+4\sqrt{3}}{-1}+\frac{1}{-7-4\sqrt{3}}+\frac{-2-\sqrt{3}}{2+\sqrt{3}}\\ &=-7-4\sqrt{3}-7+4\sqrt{3}-1\\ &=-15 \end{split} \end{equation*}$

Notice that the given triangle isn’t equilateral, yet satisfies $q=-15$ . In our theory, once a triangle satisfies $r^2+4r+1=0$ , then it shares some lateral properties with the equilateral triangle.

(In case you’re wondering how we obtained those coordinates, the building blocks are here in our second post of the year.)

Takeaway

The following statements are equivalent for a triangle with side-slopes $a,ar,ar^2$ for sides $AB,BC,CA$ :

$q=-15$
$r^2+4r+1=0$
$AB^2+CA^2=2BC^2$
$m_{AB}^2+m_{CA}^2=2m_{BC}^2$
$r$ gives a local maximum value of $q$
$r=R$ or $r=\frac{1}{R}$ , where $R$ is the common ratio of the median-slopes
the common ratio $r$ is that of an equilateral triangle (or its “look-alike”).

Okay, qool.

Tasks

Show that the equation $\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=-11$ has no real-valued solution.
Solve the equation $\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=-17$ .
Given the rational function :
- Find the discriminant of the quartic numerator $4x^4+2x^3-3x^2+2x+4$ .
- Deduce that the graph of the rational function doesn’t cross the $x$ -axis.
Find the interval(s) on which the rational function $y=\frac{4x^4+2x^3-3x^2+2x+4}{2x^3+5x^2+2x}$ is increasing, and the interval(s) on which it is decreasing.
Consider the rational function . PROVE that:
- its partial fraction decomposition is $1-\frac{2}{r+1}+\frac{2}{(r+1)^2}$ .
- the point $\left(1,\frac{1}{2}\right)$ is a local minimum point.
- the point $\left(2,\frac{5}{9}\right)$ is an inflexion point.
Consider the rational function . PROVE that:
- its partial fraction decomposition is $5+\frac{18}{r-1}+\frac{18}{(r-1)^2}$ .
- the point $\left(-1,\frac{1}{2}\right)$ is a local minimum point.
- the point $\left(-2,1\right)$ is an inflexion point.
PROVE that the following equations are equivalent:
- $r^2+4r+1=0$
- $\frac{5r^2+8r+5}{(r-1)^2}=2$
- $\frac{r^2+1}{(r+1)^2}=2$
  (These equations play important roles in connection with triangles whose side-slopes form geometric progressions.)
(Palindromic quartic) Verify that the roots of the quartic $r^4-2r^3-7r^2-2r+1=0$ are:

$\begin{equation*} \begin{split} r_1&=\frac{1+\sqrt{2}+\sqrt{5}+\sqrt{10}}{2}=\frac{(1+\sqrt{2})(1+\sqrt{5})}{2}\\ r_2&=\frac{1-\sqrt{2}-\sqrt{5}+\sqrt{10}}{2}=\frac{(1-\sqrt{2})(1-\sqrt{5})}{2}\\ r_3&=\frac{1+\sqrt{2}-\sqrt{5}-\sqrt{10}}{2}=\frac{(1+\sqrt{2})(1-\sqrt{5})}{2}\\ r_4&=\frac{1-\sqrt{2}+\sqrt{5}-\sqrt{10}}{2}=\frac{(1-\sqrt{2})(1+\sqrt{5})}{2} \end{split} \end{equation*}$

Notice that $r_1$ and $r_4$ are “multiples” of the golden ratio. Also, it may be more convenient to solve the given quartic, than to substitute the purported roots.

(Palindromic quartic) Find a monic quartic polynomial whose roots are:

$\begin{equation*} \begin{split} r_1&=-4-\sqrt{7}+\sqrt{8}+\sqrt{14}\\ r_2&=-4-\sqrt{7}-\sqrt{8}-\sqrt{14}\\ r_3&=-4-\sqrt{7}-\sqrt{8}+\sqrt{14}\\ r_4&=-4-\sqrt{7}+\sqrt{8}-\sqrt{14}\\ \end{split} \end{equation*}$

Our sum somehow leads to palindromic polynomials. We may visit this concept in the near posture.

Show that $q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}=1$ if and only if $r=\pm 1$