
(1)
Unsurprisingly, such triangles mirror many of the properties of right triangles, with minor differences. So, for the time being, we’re giving them the pseudonym pseudo right triangles.
In our calculations we’ll frequently utilize the fact that equation (1) is equivalent to
(2)
where is the circumradius of the parent triangle
.
Identities
A random selection of simple properties, in addition to the ones we already explored before and maybe in preparation for subsequent discussions.





Once one has , one also has
automatically. They imply each other. Symbolically,
.






For a right triangle this would be , if
. In general, where a right triangle has a “degenerate” property, a triangle satisfying equation (1) doesn’t seem to. A case in point is the orthic triangle.




This follows from example 2 above, because we had , together with the fact that
is the length of the altitude from
. In other words, the distance from
to the foot of the altitude from
is equal to the distance from the foot of this altitude to the orthocenter.





So we can form a right triangle with hypotenuse equal to the Euler segment having
and
as legs.



Using the fact that and the Pythagorean theorem
gives , as desired. Next, multiply both sides by
and use the fact that
from our previous post:
Inequalities



In our previous post we had . Re-arrange as
. Then
We also had in our last post, so
too.
Inverses
Two “upside-down” Pythagorean identities.



Equation (1) says that . Now, if
is the golden ratio, then
.
(Also happens in a Kepler triangle .)







In example 5 we had , so (assuming
) we have
Re-arrange:
This would also imply . And so
is the geometric mean of
and
, while
is the geometric mean of
and
.
Tasks
- (Keeping track) Let
be the side-lengths of an obtuse triangle
,
its circumradius,
its circumcenter,
its orthocenter, and
the altitudes. PROVE that each of the following statements implies the others:
or
- the orthic triangle is isosceles
- the geometric mean theorem holds
- the bisector of
has length
, where
- the orthocenter is a reflection of vertex
over side
(statements to help you remember 2021. Aside: the ratio of the length of the angle bisector of
to the length of the altitude from vertex
is
, under the above equivalence.)
- (Kepler triangle) Suppose that
satisfies the usual Pythagorean identity
. PROVE that the following four statements are equivalent:
-
form a geometric progression
is the golden ratio
-
.
-
- (Kosnita theorem) If
satisfies equation (1), PROVE that the Kosnita point coincides with vertex
. (Somewhat exciting property.)
- PROVE that the two statements below are equivalent for a triangle
:
(If these were added to the originalstatements in the first exercise, we will obtain a total of
. Cool. In terms of sheer numbers, we really wanted something that mirrors the Invertible Matrix Theorem. Reached.)
- Consider
with vertices at
,
,
. Its orthocenter is
, and its circumcenter is
.
- Verify that the foot of the altitude from vertex
is
. Deduce that
is the reflection of vertex
over side
.
- PROVE that the midpoint of
together with
form a parallelogram.
(The midpoint ofcoincides with our favourite point
in this case.)
- Verify that the foot of the altitude from vertex