(1)

then the following statements are equivalent:

- is the golden ratio

Today’s post will now establish the above equivalence. (Note that the fourth condition, , which motivated today’s title, was coined by the author of this article.)

*golden ratio*, PROVE that equation (1) becomes .

Suppose that is the golden ratio, then , and so . This then gives . In turn:

Let be the circumradius. Then, by the extended law of sines, we have:

The assumption implies . (Henceforth we’ll frequently use the fact that is equivalent to equation (1).) So:

Using the extended law of sines as in the previous example, the condition gives . Since , we obtain

So-called upside-down Pythagorean identity. Ours works under an added condition, namely equation (1).

Re-arrange the given condition as

and obtain . The length of the altitude from vertex is given by

If , then . Now use this in equation (1) to get

Thus, is the golden ratio.

## Takeaway

Let be a triangle whose side-lengths satisfy equation (1). Then the following statements are *equivalent*:

- is the golden ratio
- .

The last condition fixes the interior angles of the given triangle : , , .

## Task

- (Early 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-one*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 .