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# A pseudo right triangle

We explore more properties of triangles whose side-lengths satisfy a modified Pythagorean identity:

(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.

Let be the 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, .

Let be the 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.

If satisfies equation (1), PROVE that the 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.

If satisfies equation (1), PROVE that , where is the orthocenter, the circumcenter, and the circumradius.

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

Suppose that satisfies equation (1). PROVE that and , as per the diagram below.

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:

Suppose that satisfies equation (1). PROVE that .

If is the golden ratio, then .

## Inequalities

If satisfies equation (1), PROVE that and , as per the diagram below.

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.

If satisfies equation (1), PROVE that .

Since , we have

(This happens in a right triangle too if .)

Suppose that satisfies equation (1). PROVE that if is the golden ratio.

Equation (1) says that . Now, if is the golden ratio, then .

(Also happens in a Kepler triangle .)

Consider triangle which satisfies equation (1). With reference to the diagram below, PROVE that is the geometric mean of and and hence is the geometric mean of and .

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 .

## Takeaway

For any obtuse triangle satisfying equation (1), the following statements are equivalent:

• is the golden ratio
• .

The golden ratio acts as a “bridge” between a Kepler triangle and any obtuse triangle satisfying equation (1).

1. (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.)
2. (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
• .
3. (Kosnita theorem) If satisfies equation (1), PROVE that the Kosnita point coincides with vertex . (Somewhat exciting property.)
4. 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.)
5. 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.)