(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 .