Doubtless, it’s quite careless — if not senseless — to name a triangle isosixless. Unless an attempt is made to buttress the use of such an address, one’ll never be impressed by the meaningless, lame name.

Given any triangle , take the quotient of the slope of each median and the slope of the opposite side, and form the sum

(1)

Among all triangles with one side parallel to the -axis (or -axis), it’s only the *isosceles* triangle that satisfies . Thus, we thought it might be apt to adapt isosceles as *iso**six**less*, in view of . Now, beat that.

For with vertices at , , , verify that , where is given by equation (1).

A quick calculation shows that

and so

No doubt about that.

## Negative part*six*tions

Normally, partitions are not defined for *negative* integers, but with your permission, we’re doing so for this discussion. Taking two/three numbers at a time, we can write

We’ll show that only the “partition” works, and yields a characterization of isosceles triangles.

Consider with vertices at , , . PROVE that

where represents the sum of quotients .

We have:

Consider with vertices at , , . PROVE that

if and only if .

First suppose that . After clearing fractions and combining like terms, we obtain

where the left side is precisely the expansion of . Thus, .

Conversely, suppose that . Then:

In any , PROVE that and *if and only if* is *isosceles with side parallel to the -axis*.

Thus, only when two of the summands in (1) are both .

We show one direction of the proof — that both and yield an isosceles triangle with one side parallel to the -axis.

Place the vertices of at , , . The first condition amounts to

Clear fractions and combine similar terms:

(2)

The second condition , in terms of the given coordinates, is:

Clear fractions and combine like terms:

(3)

Subtract equation (2) from equation (3):

This factors as . We claim that both and must hold. If not, suppose that only holds and re-calculate the quotient :

This doesn’t equal unless we impose . Thus, we must have (side parallel to the -axis) and (which, in conjunction with yields an isosceles triangle).

For an *equilateral* triangle , PROVE that

This is a consequence of the fact that the medians in an equilateral triangle are perpendicular to the opposite sides. In particular:

ignoring any zero denominator. Then:

The above example can be used to explain why the side-slopes of an equilateral triangle with side parallel to the -axis are and .

## Native progressions

*equilateral*triangle are precisely:

where is the slope of one of the sides. The fact that the slopes can be expressed in terms of the slope of one side leads to some interesting consequences.

PROVE that the slopes of the sides of an *equilateral* triangle form an *arithmetic* progression *if, and only if*, one of the slopes is *zero*.

First suppose that *one of the slopes is zero*. There are three possibilities:

- . That is, (since as per above). In this case we have:
and so we obtain the arithmetic progression corresponding to .

- . This implies , and so . In other words, . This also gives
We obtain an arithmetic progression this time around.

- . This implies , and so . That is, . With this
giving the arithmetic progression

For the converse, suppose that *the side-slopes form an arithmetic progression*. There are three cases to consider: the progression could be , or or .

PROVE that the slopes of the sides of an *equilateral* triangle form an *arithmetic* progression *if, and only if*, .

If the slopes form an arithmetic progression, then one of them must be zero, in view of example 6. Since an equilateral triangle is also isosceles, we have (by example 3) that .

On the other hand, suppose that . Using example 5 and the expressions for the slopes of the sides of an equilateral triangle, we have:

Afer simplifications, this leads to

Taking each of these three values of in turn, we see that the side-slopes form an arithmetic progression (similar to example 6).

Solve the sixth-degree polynomial equation .

By the way, such an equation is also sometimes called “bicubic” (just like “biquadratic”), since the odd degree terms are missing.

The form of the coefficients in this polynomial equation means that we can re-arrange the terms and factor systematically:

The first quadratic factor, , has as solutions. The second “biquadratic” factor, , has solutions that can be obtained from the quadratic formula:

Thus, we obtain six solutions

PROVE that the slopes of the sides of an *equilateral* triangle form a *geometric* progression *if, and only if*, one of the slopes is .

This is the geometric progression version of example 6. First suppose that the slopes form a geometric progression. There are three cases to consider:

- is the geometric mean of the progression. In this case we have:
- is the geometric mean of the progression. In this case we have:
- is the geometric mean of the progression. In this case we have:

If , then becomes , and if , then is . So we get that one of the slopes is .

If , then is , and if , then evaluates to .

Conversely, if one of the slopes is , it is also easy to see that the other slopes combine to form a geometric progression.

PROVE that the slopes of the sides of an *equilateral* triangle form a *geometric* progression *if, and only if*, .

First suppose that . Using the expression for in example 5 and the expressions for the side-slopes of an equilateral triangle, we have:

which, after simplifications, leads to the sixth-degree equation

same as the one in example 8. Since the solutions are , , , the side-slopes form a geometric progression.

Conversely, if the side-slopes form a geometric progression, all possiblities can be enumerated. For example, we can have , , . Using example 5 again we have:

That is, . Other combinations of also yield .

In a future post we may form the sum:

(4)

If an *equilateral* triangle does not have any of its sides parallel to the coordinate axes, then . Accordingly, any triangle that satisfies (4) will be called equi*less*three, a name as lame as isosixless.

## Takeaway

For an *equilateral* , the following statements are *equivalent*:

- one of the side-slopes is
*zero*; - the
*sum*of the side-slopes is*zero*; - the side-slopes form an
*arithmetic*progression; - , where is the sum of quotients .

Similarly, the following conditions are *equivalent* for an *equilateral* :

- one of the side-slopes is
*plus/minus one*; - the
*sum*of the side-slopes is*plus/minus three*; - the side-slopes form a
*geometric*progression; - , where is the sum of quotients .

So much about slopes in our posts, eh? Probably a reason we tend to be *slop*py with words and stuff?

## Tasks

- (Minus one) Let . If, in we have the relations , , :
- PROVE that .
- Discuss what happens when . Interesting case.

- Let and be
*real*numbers.- PROVE that the quadratic equation has no real solutions.
- Deduce that if, and only if, .

- PROVE that in an
*equilateral*triangle with non-zero side-slopes. - In , let the slopes of sides be , respectively. PROVE that .
- PROVE that the relations , , are not possible simultaneously.
- Solve the quartic equation .
- In , let the slopes of sides be in geometric progression . PROVE that .

(Things always become*interesting*in the setting of geometric progressions. Always.) - In , let the slopes of sides be in geometric progression . PROVE that .

(Things always become*interesting*in the setting of geometric progressions. Always.) - (Simplified area) In , let be the slopes of sides , where , , are the coordinates of the vertices. PROVE that the
*area*of can be given by .

(Absolute value may be needed in the final result.) - PROVE that there is no triangle for which the relations , , and hold simultaneously.

(You can use exercise 1 above.)