-
,
-
,
-
,
What makes them special is the fact that their side-lengths satisfy the modified Pythagorean identity
(1)
with the accompanying equivalent descriptions. The first triangle has already appeared on August 14 and on October 14. The second triangle is making its debut here.




Since , the conclusion follows from one of the equivalent statements here.





In example 1 above we saw that the side-lengths satisfy equation (1). As such , by one of the equivalent statements here. So:
A direct computation of the side-lengths and
can also be used to prove the above relationship, but that’s a longer procedure.








By example 2 above we had . Re-write as
. Using one of the equivalent statements here, we know that
.
One can also prove this by a direct computation of the circumradius and the fact that
. Definitely a longer procedure.






Let’s use the cosine formula and the fact that :
To show that , re-write
as
and then use the fact that
just proved.


Take the tiny triangle where the vertices are located at
,
, and
.
Takeaway
Let be the side-lengths of
, and let
be the radius of its circumscribed circle. If equation (1) holds, then the following statements are equivalent:
.
Note the first one.
Task
- (Late fifties) 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 fifty-eight 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
.