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# Previous Puzzles of the Month + Solutions

April-May 2007, Puzzle nr 111
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Puzzle # 111  Italiano Français
Difficulty level: , basic geometry knowledge.

 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 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 + r2a2 + 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 = (ba)/2 -- Equation 2

From Equation's 1 & 2:
a2/2(b + a) = (ba)/2
a2 = (ba)(b + a) = b2a2
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). A = r2[(/180) - sin] Vesica piscis

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Previous puzzles of the month...
 Febr-March 07: prof Gibbus' angle Jan 07: triangles to square Aug-Sept 06: balance problem June-July 06: squared strip Apr-May 06: intriguing probabilities Febr-March 06: cows & chickens Dec 05-Jan 06: red monad Sept-Oct 05: magic star July-Aug 05: cheese! May 05: stairs to square Febr 05: same pieces Sept-Oct 04: odd triangles June 04: pizza's pitfalls May 04: Pacioli puzzle... April 04: inscribed rectangle March 04: magic 4-T's Febr 04: a special box Dec-Jan 04: curvilinear shape... Oct-Nov 03: same area? well... Sept 03: square vs rectangle Puzzle Archive

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