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.

Quick pause: We have cause to give a big round of applause to those who’ve been around with us in the course of our discuss in this unique year. We also welcome aboard those who’ve just come across our blog. We hope you won’t get bored with the broad scope of slopes being brought.

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

  1. (Golden Sum) Consider q=\frac{4r^4+2r^3-3r^2+2r+4}{2r^3+5r^2+2r}, 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.
  2. (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.)
  3. (Golden Equations) Let \phi 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.)
  4. (Golden Difference) Let a,ar,ar^2 be the slopes of sides AB,BC,CA in \triangle ABC.
    • 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.)
  5. (Golden Mistake) Let a,ar,ar^2 be the slopes of the sides of a triangle. Suppose that the equations a\times ar=1 and \left(\frac{2r+1}{r+2}\right)ar\times ar^2=1 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).
  6. Let \triangle ABC be such that the slopes of sides AB,BC,CA are a,ar,ar^2, respectively. If the common ratio r satisfies r^2+3r+1=0, PROVE that:
    • AB^2+CA^2=3BC^2
    • m_{AB}^2+m_{CA}^2=\frac{7}{5}m_{BC}^2
    • q=-19
  7. Let \triangle ABC be such that the slopes of sides AB,BC,CA are a,ar,ar^2, respectively. If the common ratio r satisfies r^2+7r+1=0, 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.)
  8. Let \triangle ABC be such that the slopes of sides AB,BC,CA are a,ar,ar^2, respectively. If the common ratio r satisfies 6r^2+13r+6=0, PROVE that:
    • AB^2+CA^2=13BC^2
    • m_{AB}^2+m_{CA}^2=\frac{17}{25}m_{BC}^2
    • q=\frac{31}{6}
  9. Let \triangle ABC be such that the slopes of sides AB,BC,CA are a,ar,ar^2, respectively. If the common ratio r satisfies 4r^2-17r+4=0, 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.)
  10. 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?)