All similar triangles have the same shape. These shapes can be classified using complex numbers

For example, an equilateral triangle can be expressed by complex numbers 0, 1, (1 + i √3)/2. Lester and Artzy call the ratio
S(u,v,w) = (u −w)/(u − v) the shape of triangle (u, v, w). Then the shape of the equilateral triangle is
(0–(1+ √3)/2)/(0–1) = ( 1 + i √3)/2 = cos(60°) + i sin(60°) = exp(i π/3).
For any affine transformation of the Gaussian plane, z mapping to a z + b, a ≠ 0, a triangle is transformed but does not change its shape. Hence shape is an invariant of affine geometry. The shape p = S(u,v,w) depends on the order of the arguments of function S, but permutations lead to related values. For instance,
1 - p = 1 - (u-w)/(u-v) = (w-v)/(u-v) = (v-w)/(v-u) = S(v,u,w). Also p^{-1} = S(u,w,v).
Combining these permutations gives S(v,w,u) = (1 - p)^{-1}. Furthermore,
p(1-p)^{-1} = S(u,v,w)S(v,w,u)=(u-w)/(v-w)=S(w,v,u). These relations are "conversion rules" for shape of a triangle.
The shape of a quadrilateral is associated with two complex numbers p,q. If the quadrilateral has vertices u,v,w,x, then p = S(u,v,w) and q = S(v,w,x). Artzy proves these propositions about quadrilateral shapes:
If p=(1-q)^{-1}, then the quadrilateral is a parallelogram.
If a parallelogram has |arg p| = |arg q|, then it is a rhombus.
When p = 1 + i and q = (1 + i)/2, then the quadrilateral is square.
If p = r(1-q^{-1}) and sgn r = sgn(Im p), then the quadrilateral is a trapezoid.
A polygon (z_1, z_2,...z_n) has a shape defined by n – 2 complex numbers S(z_j,z_{j+1},z_{j+2}), \ j=1,...,n-2. The polygon bounds a convex set when all these shape components have imaginary components of the same sign.[6]

## Those over-sized tennis balls used for autographs

## Triangles are everywhere

Or maybe I've been going through some psychosis

Love me a good triangle though

http://www.youtube.com/watch?v=o2Z6tDSb6c8

## :/

Well, that's the psychosis confirmed

## BalonzFeed

## MAN

## Is woman, man shaped?

## i'm not sure

ho_fo might know

## strawberry shape delights

## Balonz sounds like a shape tbf

sounds like balloons...

## actually it sounds like a pair of french palls

## Brick shaped. There's got to be quite a lot of bricks.

All similar triangles have the same shape. These shapes can be classified using complex numbers

For example, an equilateral triangle can be expressed by complex numbers 0, 1, (1 + i √3)/2. Lester and Artzy call the ratio

S(u,v,w) = (u −w)/(u − v) the shape of triangle (u, v, w). Then the shape of the equilateral triangle is

(0–(1+ √3)/2)/(0–1) = ( 1 + i √3)/2 = cos(60°) + i sin(60°) = exp(i π/3).

For any affine transformation of the Gaussian plane, z mapping to a z + b, a ≠ 0, a triangle is transformed but does not change its shape. Hence shape is an invariant of affine geometry. The shape p = S(u,v,w) depends on the order of the arguments of function S, but permutations lead to related values. For instance,

1 - p = 1 - (u-w)/(u-v) = (w-v)/(u-v) = (v-w)/(v-u) = S(v,u,w). Also p^{-1} = S(u,w,v).

Combining these permutations gives S(v,w,u) = (1 - p)^{-1}. Furthermore,

p(1-p)^{-1} = S(u,v,w)S(v,w,u)=(u-w)/(v-w)=S(w,v,u). These relations are "conversion rules" for shape of a triangle.

The shape of a quadrilateral is associated with two complex numbers p,q. If the quadrilateral has vertices u,v,w,x, then p = S(u,v,w) and q = S(v,w,x). Artzy proves these propositions about quadrilateral shapes:

If p=(1-q)^{-1}, then the quadrilateral is a parallelogram.

If a parallelogram has |arg p| = |arg q|, then it is a rhombus.

When p = 1 + i and q = (1 + i)/2, then the quadrilateral is square.

If p = r(1-q^{-1}) and sgn r = sgn(Im p), then the quadrilateral is a trapezoid.

A polygon (z_1, z_2,...z_n) has a shape defined by n – 2 complex numbers S(z_j,z_{j+1},z_{j+2}), \ j=1,...,n-2. The polygon bounds a convex set when all these shape components have imaginary components of the same sign.[6]

## ↑ '[6]'!

## lines that are not straight

## or spheroids (blobs that are not perfect spheres)

## fractals

## big egg