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.)