

(1)
(2)
Of all right triangles, the Kepler triangle is the only one in which both the side-lengths and the reciprocals of the side-lengths satisfy Pythagorean identities.
On top of this, the Kepler triangle is also unique in that both the altitudes and the reciprocals of the altitudes satisfy Pythagorean identities.
(3)
(4)
(Only the Kepler triangle satisfies equation (3) above, but other right triangles satisfy equation (4).)
Our very first example of 2022 is a breeze. To see that (1)(2), re-arrange
On the other hand, (2)(1) because
Let be the radius of the circumcircle of any triangle
with side-lengths
. Then the altitudes
from vertices
are given by
In the case of a right triangle with hypotenuse we have
and so
.
Since , we have
Clear the common denominator and it becomes clear that
Suppose that equation (1) holds, so that . Divide through by
and use the expressions for the altitudes given in example 2.
If a right triangle is of Kepler type, then by our post on May 14, 2021 , its side-lengths satisfy equation (1). By the preceding two examples, equations (1) and (3) are equivalent. Thus, a Kepler triangle satisfies equation (3).
Conversely, suppose that a right triangle with hypotenuse satisfies equation (3). Then
and
. Multiply both sides of
by
to get
. Now combine with
to get
In other words, the side-lengths form a geometric progression of the form
. Thus, the right triangle is of the Kepler type.
Takeaway
Let be a (right) triangle whose side-lengths
satisfy
. Then the following statements are equivalent:
- the sequence
is geometric
is the golden ratio
.
Such a triangle is the Kepler triangle.
Task
- (Mid sixties) In a non-right triangle
, let
be the side-lengths,
the altitudes,
the feet of the altitudes from the respective vertices,
the circumradius,
the circumcenter,
the nine-point center,
the orthocenter,
the midpoint of side
,
the reflection of
over side
,
the reflection of
over side
, and
the reflection of
over side
. PROVE that the following sixty-five statements are equivalent:
or
is congruent to
is congruent to
is isosceles with
is isosceles with
is right angled at
is the circumcenter of
is right-angled at
is right-angled at
- quadrilateral
is a rectangle
- the points
are concyclic with
as diameter
- the reflection
of
over
lies internally on
- the reflection
of
over
lies externally on
- radius
is parallel to side
is the reflection of
over side
- the nine-point center lies on
- the orthic triangle is isosceles with
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
- segment
is tangent to the circumcircle at point
- median
has the same length as the segment
- the bisector
of
is tangent to the nine-point circle at
is a convex kite with diagonals
and
- altitude
is tangent to the nine-point circle at
- segment
is tangent to the nine-point circle at
.
(short of the target.)
- (Extra feature) If
satisfies equation (??), PROVE that its nine-point center
divides
in the ratio
.