(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.)


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
.