(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.

*orthocenter*of with side-lengths . If equation (1) is satisfied, PROVE that and .

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

*orthocenter*of with side-lengths . If equation (1) is satisfied, PROVE that , where is the length of the altitude from vertex .

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.

*orthocenter*is a

*reflection*of vertex over side .

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 original statements 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 of coincides with our favourite point in this case.)