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 , the binomial expansion of . Another example is . And so on. You get the gist.

In any *right isosceles* triangle with slopes in geometric progression, PROVE that the common ratio satisfies the cubic equation .

Let be *right isosceles* and let the geometric progression of slopes be for sides , respectively.

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

respectively.

Now, side __cannot__ be the hypotenuse (you know why, right?) This leaves us with either or as the hypotenuse.

First suppose that is the hypotenuse. Then and so

(1)

By the isosceles condition, and so the median from vertex is perpendicular to side . This gives:

(2)

Combine equations (1) and (2):

There we *go*ld. Switching the hypotenuse won’t alter the fact that the equation holds. For suppose instead that is the hypotenuse. Then , giving

(3)

The isosceles condition forces , in which case the median from vertex meets at right angles:

(4)

As before, combining equations (3) and (4) yields .

In any *right isosceles* triangle with side-slopes in geometric progression, PROVE that the common ratio satisfies .

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

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

Slope considerations means we __cannot__ have . This leaves us with the choice (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 , PROVE that the the common ratio is either or . (In other words, 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 and its reciprocal is .

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

Let’s split the two roots, like so:

What we want now follows immediately:

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 , PROVE that the the first term — either the *golden ratio* or its negative, if the *hypotenuse* has slope . Under the same conditions, PROVE also that we have — either the reciprocal of the *golden ratio* or the negative of its reciprocal.

Suppose that the *hypotenuse* has slope . Then the legs’ slopes are and . Consequently:

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 , PROVE that the the first term is either or , if the *hypotenuse* has slope .

If the hypotenuse has slope , then the legs are and . Consequently:

## Absolute character

In a right isosceles triangle with side-slopes , we always have

(5)

when the side with slope is the *hypotenuse*, and we always have

(6)

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

Let be the slopes of the sides of a *right isosceles triangle*, with the hypotenuse having slope . PROVE that or .

In view of example 4 above, there are *four* possibilities for , viz: or . The “base” case will satisfy the parent quadratic , while two other cases satisfy — 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.

Let be the slopes of the sides of a *right isosceles triangle*, with the hypotenuse having slope . PROVE that or .

In view of example 5, there are also *four* cases here: or .

## 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 (from the vertex) to the hypotenuse divides the hypotenuse into parts and such that

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

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

(7)

Remember the name: *silvery* ratio.

Let be the slopes of sides in . PROVE that if the altitude to the side divides in the ratio , then the common ratio or , where is the *golden ratio*.

Interesting scenes. Put in equation (7) and set :

The value is __unacceptable__ due to slope considerations. So we take . Then:

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 .

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

Clear fractions: . Combine like terms: . Divide through by the common factor : . Assume two quadratic factors of the form so that, after expanding the right side and comparing coefficients, we have:

We’ll now solve the linear-quadratic system .

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

And so

The first two zeros are got from

The third and fourth zeros are got from

Notice that the roots may be re-written:

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

## Associated converse

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

Suppose that has sides with slopes . If and (or ), PROVE that is a *right triangle*.

Suppose that and . Since the slope of side is , the slope of the median from to it will be . This median is actually an altitude, due to the condition. We then have:

(8)

Claim: , which will ensure that . Isolate from equation (8) and use a few re-arrangements of the quadratic equation repeatedly:

Suppose now that and . The median from is then an altitude; and its slope being means that

(9)

As before, we claim that ; that is, (this will ensure that ). From (9) isolate :

Multiply both sides by and maneuver the equation , like so:

We’ve proved: in with slopes for sides and (or ), one has *if and only if* the triangle is a right triangle.

Gold.

## Takeaway

For any triangle with sides having slopes in geometric progression , the following statements are *equivalent*:

- or
- or
- or
- or or or , where is the
*golden ratio* - the common ratio is that of a
*right isosceles triangle*(or that of its “look-alike”)

## 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 .
- Hence, deduce that , where is the
*golden ratio*.

- (Golden State) Find coordinates for the vertices of a triangle with slopes in
*geometric progression*in which:- the area is
- one of the side-slopes is .

(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:

(The next middle coefficient is . Be bold and go further.)

- (Golden Difference) Let be the slopes of sides in .
- If , PROVE that there is a point on , a point on , and a point on such that the slopes of the sides of form an
*arithmetic progression*with common difference . - If above and is the
*golden ratio*, PROVE that is also the*golden ratio*.

(The points are all*internal*points when . The situation is even more interesting when , as the points now become*external*.)

- If , PROVE that there is a point on , a point on , and a point on such that the slopes of the sides of form an
- (Golden Mistake) Let be the slopes of the sides of a triangle. Suppose that the equations and both hold.
- PROVE that
- 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:
- Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- (Compare with the previous
*q*uestion.)

- (Compare with the previous
- Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- Let be such that the slopes of sides are , respectively. If the common ratio satisfies , PROVE that:
- (Compare the “look-alikes” in 8 and 9.)

- If , PROVE that . (Notice how the squaring operation “distributes” over addition. Strange, eh?)