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Flower

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
Congratulations Andrei Lazanu, age 14, School No.205, Bucharest, Romania on another beautiful solution.

I denoted by $A$ the centre of the circle of radius $a/b$, by $B$ the centre of the radius $a$, by $C$ the circle of radius $b$, by $D$ the centre of the circle of radius $b/a$, by $E$ the centre of the circle of radius $1$/a, by $F$ the centre of the circle of radius $1/b$ and by $O$ the centre of the circle of radius 1.

I observed that the triangles obtained have the following sides:

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. $180^0$.

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 $180^0$.

So, the sum of all angles around point O is $360^0$. This means that, with the given radii, it is always possible to construct such a flower.