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
where is the circumradius of the parent triangle .
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:
If is the golden ratio, then .
In our previous post we had . Re-arrange as . Then
We also had in our last post, so too.
Two “upside-down” Pythagorean identities.
Since , we have
(This happens in a right triangle too if .)
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
This would also imply . And so is the geometric mean of and , while is the geometric mean of and .
- (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:
- 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.)