Nostalgia f.
Greek nostos 'return home' and -algos 'pain',
translating German Heimweh 'homesickness', was coined
in 1688 by the Alsatian physician Johannes Hofer. In his treatise
'Dissertatio Medica de Nostalgia, oder Heimweh', he described
a new disease that affected some soldiers of the Swiss regiments
serving the king of France as "the pain a sick person feels
because he wishes to return to his native land, and fears never
to see it again".
Soccer
balls
Two
intersected semi-circles inscribed in a rectangle are
tangent to 5 discs (soccer balls) as depicted in the
image. What is the value of b/a?
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 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!
More
Math Facts behind the puzzle
In
geometry, a shape comprising two circular arcs,
joined at their endpoints (fig. 1) is called
a lens.
The "Vesica
piscis" (fish bladder) is one particular
form of a symmetrical lens (fig. 2).
Back to 
2/2
The
Winners of the Puzzle of the Month are:
/180)
- sin
Italy
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