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 gold. 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 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: . 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.
- PROVE that the numerator factors as
- (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.)
- the area is
- (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 pointsare all internal points when
. The situation is even more interesting when
, as the points now become external.)
- If
- (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).
- PROVE that
- 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 question.)
- 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?)