Generalized triangle with negative angle?

Question

In approaching triangle problems, it is often convenient to assume there is a triangle with twice a given angle. This usually means splitting up the proof into acute and obtuse cases. I was wondering if case analysis can be avoided by using a generalized "triangle" with possible negative angle measurements and large angle measurements, with the sum of the angles equal to 180.

Good question. Although not related to it, I just want to mention that I once read about having triangles with possibly negative sides. (I forgot the details, perhaps something to do with negative areas?) — JRN, Sep 19 '14 at 14:52
Alternatively, perhaps a metatheorem could do the same work? — Jacob Wakem, Sep 23 '14 at 01:42
Do you have something in mind like angles $\alpha$, $\beta$, and $2\beta$, where $\alpha+\beta+2\beta=180^\circ$ and $\alpha$ could be negative? — JRN, Sep 23 '14 at 01:47

peawormsworth · Answer 1 · 2018-11-06T13:55:32.573

If a triangles is defined as having angles a, b, c summing to π, joined by 3 straight lines on a flat plane, then all valid triples are:

a+b+c mod 2π = π

If the total is not π, you can roll two angles to be: -π ≤ a,b ≤ π and -2π ≤ c ≤ 2π

You have the equivelent triangle:

a+b+c = π

If 0 ≤ a,b,c ≤ π, you have a typical triangle.

If an angle is negative, these will find an equivelant triangle:

(-a, b, c) = (a, c-a, b-a)

(a, -b, c) = (c-b, b, a-b)

(a, b, -c) = (b-c, a-c, c)

If the resulting triangle contains negative angles, repeat this process until it does not.

You will find the equivelent triangle with all angles between 0 and π.

I think this is true because the map of known similar triangles is highly symmetric. And it is an island of repetition, due to the cyclic nature of angular measure.The space between can be filled in by reflection. I think that the similar triangle pattern is a rotated reflection beyond the degenerate line. That is my belief.

This will provide a typical triangle solution for all triangles with angles defined by a+b+c = π + k2π.

ex: solve triangle (a,b,c) = (-100, -15, 295)

= (100, 195, -115)

= (80, -15, 115)

= (100, 15, 65)

The triangle with angles with angles (-100,-15, 295) is equivalent to triangle (100, 15, 65).

=== Simple Process:

given a, b which are 2 angles of a triangle:

a = a mod 180

b = b mod 180

if a + b > 180, then:

a = 180 - b

b = 180 - a

now: 0 <= a,b <=180

and: c = 180 - a + b

Summary: reduce or increase each angle by 180 until they are between 0 and 180. if the sum is greater than 180, twirl the triangle around.

I think.

Triangles can be reduced further, since the sum is 180, the smallest angle must be less than 60, the 2nd angle less than 90 and the 3rd 60 to 180. All triangles can be shown similar to one that is limited to:
0 ≤ a ≤ 60, 0 ≤ b ≤90, 60 ≤ c ≤ 180 — peawormsworth, Nov 05 '18 at 07:27

peawormsworth · Answer 2 · 2018-11-06T13:29:20.810

This is a more detailed description of triangles with extended and negative angles.

The first diagram shows the limits of triangles with all angles between 0 and 180.

This is a chart of all similar triangles where any point is a specific one.

Each triangle is next to a nearly similar triangle.

The border is the set of degenerate triangles.

We can push a known triangle up to the degenerate line and then see what happens if we keep pushing the angle further.

It appears that a new similar triangle appears that is rotated.

If we continue pushing the resulting triangle set appears to be a rotated line though the known triangle set.

It is a reflection.

Using this assumption we draw the reflected surface of known triangles onto the unknown areas.

The result is a pleasing pattern.

The pattern was used to derive the equations for manipulating negative triangle angles.

Finally, the pattern is plotted on x,y,z coordinates. These ideas make sense to me, but I have no references.

This is a correction on the symmetry:

Reduce or increase each angle by 180 until they are between 0 and 180. if the sum is greater than 180, twirl the triangle around: