<
Back to the problem

Home
# Solutions for the Right Wiggly Triangles

## Part 1

Suppose there are
- t right wiggly triangles in the circle
- p points inside the circle but not on the outside where the triangles meet
- q points on the circle itself (and therefore also q lines)
- e edges inside the circle - not counting the q on the outside

Then we have:
- Number of faces = t+1 including the outside
- Number of edges = e+q
- Number of vertices = p+q

Firstly add up the angles of the triangles. It is 180° per triangle.
Also it is 360° per interior point and 180° for each on the
circle.
180t = 360p + 180q

or simplifying
t = 2p + q

Now count the total number of sides of the triangles. Each interior
side is on two triangles whereas those on the circle are only on
one triangle.
3t = 2e + q

Subtracting the previous equation t=2p+q from this gives;
2t = 2e - 2p

Which is the same as
t + p = e

However by Euler's formula F+E=V+2 therefore
t+1 + p+q = e+q + 2

which gives
t + p = e + 1

and subtracting t+p=e from this gives the classical contradiction
0 = 1

QED we can't divide up a circle into right wiggly triangles,
it is just plain impossible.

## Part 2

You'll like this (not a lot!).
A single right wiggly triangle is sufficient. The sides AB and BC
are bent back on themselves so corners A, B and C all
meet at a point. AC forms the outside of the circle. Angles A+B+C
make up 180°.