Sides of Triangles Obtained
| ABO |
a/b + a |
a +1 |
a/b +1 |
| BCO |
a + b |
a + 1 |
b + 1 |
| CDO |
b + b/a |
b + 1 |
b/a + 1 |
| DEO |
b/a + 1/a |
b/a + 1 |
1/a + 1 |
| EFA |
1/a +1/b |
1 + 1/b |
1 + 1/a |
| FAO |
a/b + 1/b |
a/b + 1 |
1/b + 1 |
I observe that triangles BCO, DEO and FAO are similar, with
the similarity ratios (taking them two by two), 1: 1/a: 1/b
respectively. So are triangles ABO, CDO and EFA, with the
similarity ratios a: b: 1.
Looking in the similarity ratio, I observe that angle BOC is
congruent with angle DEO and with angle AFO, angle OBC with
angles FAO and EOD, and angle BCO with ODE and AOF.
This means the sum of angles BOC, AOF and DOE is the angle
sum in a triangle, i.e. 1800.
For triangles AOB, COD and EOF that are similar, angles
AOB, ODC and OEF are congruent; so are angles OAB, OFE
and DOC and ABO, OCD and FOE. In this case the sum of
angles: AOB, COD and EOF is 1800.
So, the sum of all angles around point O is 3600. This means
that, with the given radii, it is always possible to construct
such a flower.