R
= radius of the semicircles; r = radius of the
soccer balls.
R  r = OA = radius of the circle (O, OA).
S_{O}(A')
= A (point A' is symmetrical to A through point
O)
AA' = 2(R  r) [the diameter of the circle (O,
OA)] = a
AB = BA' = R = b/2
Triangle ABA' is inscribed in the circle (O, OA),
then it is isosceles and rightangled. 
We
can therefore solve this puzzle with the help of
the theorem of Pytagora:
AA'^{2} = AB^{2} + BA'^{2}
a^{2} = (b/2)^{2}+
(b/2)^{2} = 2(b/2)^{2}
Simplifying: a = b2/2
So: b/a = b/(b2/2)
= 2
Here
below is a neat solution involving a general formula,
suggested by Géry
Huvent:
a)
Considering the basic diagram above; the circles
centered in O with radius r_{1 },
and in A with radius r_{3} are tangent,
then:
(n^{2} +
r_{3}^{2}) + r_{3} = r_{1}
b)
The circles centered in B with radius r_{2} ,
and in A with radius r_{3} are tangent,
then:
[n^{2} +
(m  r_{3})^{2}] = AB =
r_{2 }+ r_{3}
b)
Simplifying both equations together, we obtain the
following math relationship useful to solve other
similar problems:
r_{1}(r_{1 } 2r_{3}) = (m +
r_{2})(r_{2}+ 2r_{3}  m)
c)
Comparing this latter equation to the original puzzle
diagram further above (*),
we can see that:
a* = m ; b*/2
= R* = r_{1 }= r_{2 }; r* = r_{3} ;
and
that:
r*
= R*  a*/2 = b*/2  a*/2
= (b*  a*)/2
d)
Then, replacing all the values of the general formula
in b), we obtain:
b/2{b/2
 2[(b  a)/2]} = (a + b/2) · {b/2
+ 2[(b  a)/2]  a} ; and
solving...
a^{2} =
(b  a)(b + a)
; finally... b/a = 2
You
can find more interesting sangaku formulae at Géry's
website.
Here
below another interesting solution posted by John
Reidy:
The
radius of the larger semicircle on base b is b/2
Let the radius of the projected soccer ball circular
image = r
At the contact point of any of the circles, the radius
line from the centre of each circle is at right angles
to a tangent at that contact point. Therefore the lines
AC and AD joining the centres of the contacting circles
must pass through the contact point of these circles;
these points being G and E respectively. Accordingly,
as the sides of the rectangle are tangential at contact
points B and F, the points B and F lie on the extension
of the line joining circle centres D & C.
From
triangle ABC:
AB^{2} = CA^{2 }– BC^{2}
And from triangle ABD:
AB^{2} = DA^{2} – BD^{2}
Therefore: CA^{2} – BC^{2} =
DA^{2} – BD^{2}
From
the diagram:
CA = AG – GC = b/2 – r
BC = r
DA = AE + ED = b/2 + r
BD = BF – DF = a – r
Therefore:
(b/2 – r)^{2} – r^{2} =
(b/2 + r)^{2} – (a – r)^{2}
b^{2}/4 – br + r^{2} – r^{2} = b^{2}/4
+ br + r^{2}– a^{2} +
2ar  r^{2
}Rationalising:
2br + 2ar = a^{2}
r = a^{2}/2(b + a)
 Equation 1
From
inspection of AH in diagram:
a = AH = b/2 + (b/2 – 2r)
Therefore:
r = (b – a)/2  Equation 2
From
Equation's 1 & 2:
a^{2}/2(b + a) =
(b – a)/2
a^{2} = (b – a)(b + a)
= b^{2} – a^{2}
2a^{2} = b^{2}
b/a = 2
The
Winners of the Puzzle of the Month are:
John Reidy, Australia  Omar
A. Alrefaie, Saudi Arabia  Amaresh
G.S., India  Mohammed Ezz Abd ElMenaem
ElSayyed, Egypt  Geoffrey Harrison,
Australia.
Congratulations!
