This is a paragraph.

A property of the Kepler triangle

On May 14, 2021, we showed that the side-lengths () of the Kepler triangle satisfy the identity

(1)

Or, in an upside-down form:

(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).)

PROVE that equations (1) and (2) are equivalent.

Our very first example of 2022 is a breeze. To see that (1)(2), re-arrange

On the other hand, (2)(1) because

PROVE that every right triangle satisfies equation (4).

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 .

PROVE that equation (3) implies equation (1).

Since , we have

Clear the common denominator and it becomes clear that

PROVE that equation (1) implies equation (3).

Suppose that equation (1) holds, so that . Divide through by and use the expressions for the altitudes given in example 2.

PROVE that a right triangle is precisely the Kepler triangle, if it satisfies equation (3).

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:

1. the sequence is geometric
2. is the golden ratio
3. .

Such a triangle is the Kepler triangle.

• (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:
1. or
2. is congruent to
3. is congruent to
4. is isosceles with
5. is isosceles with
6. is right angled at
7. is the circumcenter of
8. is right-angled at
9. is right-angled at
11. the points are concyclic with as diameter
12. the reflection of over lies internally on
13. the reflection of over lies externally on
14. radius is parallel to side
15. is the reflection of over side
16. the nine-point center lies on
17. the orthic triangle is isosceles with
18. the geometric mean theorem holds
19. the bisector of has length , where
20. the orthocenter is a reflection of vertex over side
21. segment is tangent to the circumcircle at point
22. median has the same length as the segment
23. the bisector of is tangent to the nine-point circle at
24. is a convex kite with diagonals and
25. altitude is tangent to the nine-point circle at
26. 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 .