(1)
In addition, it has the property that its side-lengths form a geometric progression, for example:
(2)
Equations (1) and (2) force the ratio of the hypotenuse to one of the legs (
or
) to be golden:
(3)
In this post, we show that a Kepler triangle also satisfies
(4)
which, in conjunction with equation (1), constitutes what one may call a double Pythagorean identity. We also show that if any of the four equations above hold, then the remaining three become equivalent.
For the Kepler triangle we have , where
is the golden ratio. By definition of
, we have
and so
.
Note that a random right triangle need not satisfy equation (4). For example, the very familiar case with ,
,
doesn’t.
Simple characterizations
A couple of trivial equivalent statements that offer alternate descriptions of the Kepler triangle.
In other words, the following three statements are equivalent for any right triangle (in which ):
- the side-lengths form a geometric progression
- the ratio
is the golden ratio
- the identity
holds.
The result follows if we show that .
So let’s suppose that holds, that is, the sequence
is geometric. This gives
. Since we also have that
, it means that
Thus, satisfies the defining equation for the golden ratio, and so
holds.
Next, we show that . So suppose that
is the golden ratio. Then
Multiply both sides of by
to get
since from the Pythagorean identity and also
from the golden ratio. Thus,
holds.
Finally, we show that . We have
and
:
So the sequence is geometric, and
is proved.
Easy equivalence.
In other words, the following three statements are equivalent for any triangle in which :
- the triangle is a right triangle with
- the ratio
is the golden ratio
- the identity
holds.
Follow the same argument as in example 2, bearing in mind that holds throughout in this case.
In other words, the following three statements are equivalent for any triangle in which :
- the triangle is a right triangle with
- the side-lengths form a geometric progression
- the identity
holds.
Follow the same argument as in example 2, bearing in mind that holds throughout in this case.
In other words, the following three statements are equivalent for any triangle in which :
- the triangle is a right triangle with
- the side-lengths form a geometric progression
- the ratio
is the golden ratio
.
Follow the same argument as in example 2, bearing in mind that holds throughout in this case.
Similar concept
Given in which
, the Pythagorean theorem states that
(5)
(Or the usual .) But then the left side of equation (5) is just the sum of the squares of two altitudes — the altitude from vertex
and the altitude from vertex
— and so we can re-write (5) as:
(6)
where and
are the altitudes from
and
. The latter equation expresses the fact that in a right triangle, the sum of the squares of two altitudes is equal to the square of the third side (the hypotenuse). We’ll use this intrinsic characteristic of right triangles as the defining condition for the triangles under consideration in the second part of today’s discussion.




First suppose that . Since
and
, we get
Conversely, suppose that . Then:
QED:



Begin with as per the extended law of sines. The rest is:


Since the given triangle is not a right triangle, we take from the preceding example. However, since
is always equal to
in any triangle, we have:
Similar to equation (4) for the Kepler triangle. Let’s take a shot at making it exact.



First suppose that . Since
from the preceding example we must then have
. Positive square roots:
, or
, and so
is the golden ratio. The converse is also easy. So basically, we have non-right triangles that tend to behave like the Kepler triangle.
Special case
A sample triangle that satisfies equation (6) twice.




For this triangle we have:
Takeaway
For any non-right triangle with side-lengths
, circumradius
, and altitudes
, the five statements below are equivalent:
In addition, if a non-right triangle satisfies any of the equivalent conditions above, then each of the following three statements implies the others:
is the golden ratio
is a geometric sequence.
In a nutshell, we have a non-right triangle version of the Kepler triangle.
Tasks
- If a non-right triangle satisfies equation (6), PROVE that the length of the altitude from vertex
can be given by
.
- Let
be the side-lengths of a right triangle, where
is the hypotenuse, and let
be its circumradius. PROVE that:
(in that order) cannot be a geometric sequence.
- PROVE: For any non-right triangle
with side-lengths
, circumradius
, and altitudes
, the five statements below are equivalent:
- Suppose that a non-right triangle satisfies any of the equivalent conditions in the preceding exercise. PROVE that:
, where
orthocenter,
circumcenter,
nine-point center
- Consider
with vertices at
,
, and
. Verify that:
- the circumradius is
.
- the circumradius is