We’re getting close to reaching (part of) our goal of having a *clone* of the right triangle, as this equation shows:

(1)

Vey close. Vey close. Vey close.

## Succinct strategy

In , let be the slopes of sides . Let be the lengths of the altitudes from each vertex. We’ll derive equation (1) in two ways: The first approach is straight-forward, utilizing a previous post and a fact from this site, namely

(2)

where is the radius of the circumcircle of and are the side-lengths. The second method is long-winded, painstaking, old-fashioned.

#### Example 1

In any , PROVE that .

Using equation (2), we have:

It follows immediately that in a right triangle with we have

#### Example 2

holds in a with as the slopes of sides .

We use an identity from one of our previous posts, namely

In terms of the standard notation, this is the same as . Or:

In view of example 1, we then have:

Point-blank.

## Verbose verification

In with slopes for sides , find the length of the altitude from vertex . Take , , .

The length of the altitude from vertex is

(Absolute values may be necessary. Also notice how we switched the slopes from to .)

In with slopes for sides , find the length of the altitude from vertex . Take , , .

It is:

(Again, absolute values may be necessary.)

In with slopes for sides , find the length of the altitude from vertex . Take , , .

It is:

(Absolute values may be necessary.)

From examples 3,4,5 we had:

Thus:

The slopes of sides are , respectively. They form a geometric progression in which . As per our arrangement, we’ll take and . Therefore the three altitudes are:

And so

The slopes of sides are , respectively. They form a geometric progression in which . As per our arrangement, we’ll take and . Therefore the three altitudes are:

And so

Familiar, friendly, and trendy. Set

Therefore, *any* triangle whose side-slopes form a geometric progression with common ratio will have three altitudes satisfying . One such triangle has vertices at , , and .

Familiar, friendly, trendy, and easy. Set

Therefore, *any* triangle whose side-slopes form a geometric progression with common ratio will have three altitudes satisfying . One such triangle has vertices at , , and . And there are many — infinitely many — others.

## Takeaway

In terms of side-lengths, median-lengths, and slopes, we’ve shown how our triangle closely resembles the right type. Today, the list has now grown to include altitudes.

If anything happens to the right triangle, no need to groan because we already got its clone.

## Tasks

- PROVE that in a with .
- Find possible coordinates for the vertices of whose three altitudes are related via
- Find possible coordinates for the vertices of whose three altitudes are related via
- For with vertices at , , and , verify that:
- the slopes form a geometric progression with common ratio ;
- ;
- .

- (Triple arithmetic) For any triangle in which the slopes of the sides form a geometric progression with common ratio satisfying , PROVE that:
- the squares of the side-lengths form an arithmetic progression;
- the squares of the median-lengths form an arithmetic progression;
- the reciprocals of the squares of the the altitude-lengths form an arithmetic progression.

(An*equilateral*triangle belongs here. But there are*scalene*triangles that belong here too. A*right isosceles*triangle in which satisfies the first and third conditions above.)