R
= radius of the semi-circles; r = radius of the
soccer balls.
R - r = OA = radius of the circle (O, OA).
SO(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 right-angled. |
We
can therefore solve this puzzle with the help of
the theorem of Pytagora:
AA'2 = AB2 + BA'2
a2 = (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 r1 ,
and in A with radius r3 are tangent,
then:
(n2 +
r32) + r3 = r1
b)
The circles centered in B with radius r2 ,
and in A with radius r3 are tangent,
then:
[n2 +
(m - r3)2] = AB =
r2 + r3
b)
Simplifying both equations together, we obtain the
following math relationship useful to solve other
similar problems:
r1(r1 - 2r3) = (m +
r2)(r2+ 2r3 - m)
c)
Comparing this latter equation to the original puzzle
diagram further above (*),
we can see that:
a* = m ; b*/2
= R* = r1 = r2 ; r* = r3 ;
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...
a2 =
(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:
AB2 = CA2 – BC2
And from triangle ABD:
AB2 = DA2 – BD2
Therefore: CA2 – BC2 =
DA2 – BD2
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 – r2 =
(b/2 + r)2 – (a – r)2
b2/4 – br + r2 – r2 = b2/4
+ br + r2– a2 +
2ar - r2
Rationalising:
2br + 2ar = a2
r = a2/2(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:
a2/2(b + a) =
(b – a)/2
a2 = (b – a)(b + a)
= b2 – a2
2a2 = b2
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 El-Menaem
El-Sayyed, Egypt - Geoffrey Harrison,
Australia.
Congratulations!
|