Congratulations Andrei Lazanu, age 14, School No.205, Bucharest,
Romania on another beautiful solution.
<p/>
I denoted by <i>A</i> the centre of the circle of radius <i>a</i>/<i>b</i>, by <i>B</i>
the centre of the radius <i>a</i>, by <i>C</i> the circle of radius <i>b</i>, by
<i>D</i> the centre of the circle of radius <i>b</i>/<i>a</i>, by <i>E</i> the centre of
the circle of radius 1/a, by <i>F</i> the centre of the circle of
radius 1/<i>b</i> and by <i>O</i> the centre of the circle of radius 1.
<p/>
I observed that the triangles obtained have the following sides:




    


<br />
<br />
<table width="350" border="0" cellspacing="0" cellpadding="0">
<caption>Sides of Triangles Obtained</caption>
<tr>
<td>ABO</td>
<td>a/b + a</td>
<td>a +1</td>
<td>a/b +1</td>
</tr>
<tr>
<td>BCO</td>
<td>a + b</td>
<td>a + 1</td>
<td>b + 1</td>
</tr>
<tr>
<td>CDO</td>
<td>b + b/a</td>
<td>b + 1</td>
<td>b/a + 1</td>
</tr>
<tr>
<td>DEO</td>
<td>b/a + 1/a</td>
<td>b/a + 1</td>
<td>1/a + 1</td>
</tr>
<tr>
<td>EFA</td>
<td>1/a +1/b</td>
<td>1 + 1/b</td>
<td>1 + 1/a</td>
</tr>
<tr>
<td>FAO</td>
<td>a/b + 1/b</td>
<td>a/b + 1</td>
<td>1/b + 1</td>
</tr>
</table>
<br />
<br />




 
I observe that triangles <i>BCO</i>, <i>DEO</i> and <i>FAO</i> are similar, with
the similarity ratios (taking them two by two), 1: 1/<i>a</i>: 1/<i>b</i>
respectively. So are triangles <i>ABO</i>, <i>CDO</i> and <i>EFA</i>, with the
similarity ratios <i>a</i>: <i>b</i>: 1.
<p/>
Looking in the similarity ratio, I observe that angle <i>BOC</i> is
congruent with angle <i>DEO</i> and with angle <i>AFO</i>, angle <i>OBC</i> with
angles <i>FAO</i> and <i>EOD</i>, and angle <i>BCO</i> with <i>ODE</i> and <i>AOF</i>.
<p/>
This means the sum of angles <i>BOC</i>, <i>AOF</i> and <i>DOE</i> is the angle
sum in a triangle, i.e. 180<sup>0</sup>.
<p/>
For triangles <i>AOB</i>, <i>COD</i> and <i>EOF</i> that are similar, angles
<i>AOB</i>, <i>ODC</i> and <i>OEF</i> are congruent; so are angles <i>OAB</i>, <i>OFE</i>
and <i>DOC</i> and <i>ABO</i>, <i>OCD</i> and <i>FOE</i>. In this case the sum of
angles: <i>AOB</i>, <i>COD</i> and <i>EOF</i> is 180<sup>0</sup>.
<p/>
So, the sum of all angles around point O is 360<sup>0</sup>. This means
that, with the given radii, it is always possible to construct
such a flower.