Enigmatic Code

Programming Enigma Puzzles

Enigma 1206: Rounds unlimited

From New Scientist #2362, 28th September 2002 [link]

Enigma 1206

As a logo for its delivery service, Rounds Unlimited uses the geometric design shown.

It consists of an equilateral triangle with 24-centimetre sides and a large circle inside it touching each of its three sides.

Then in each corner there are smaller circles each touching the previous circle and two sides of the triangle. Then in each corner there are smaller circles each touching a previous circle and two sides of the triangle, and so on ad infinitum.

The total area of the circles is an exact whole number multiple of π cm².

What is that whole number?

[enigma1206]

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One response to “Enigma 1206: Rounds unlimited

  1. Jim Randell 31 August 2015 at 8:11 am

    Some trigonometry determines that a circle inscribed in an equilateral triangle has a radius of 2/√3 times the side of the triangle. So the area of the circle is π(4/3). The “unused” corners are 1/3 the side of the original triangle.

    So, if the area of the central circle is A_0 and the area of the smaller circles is A_1, A_2, A_3, …, then the total area we are looking for is:

    A = A_0 + 3A_1 + 3A_2 + 3A_3 +3A_4 + …

    A = π(48 + (16 + 16/9 + 16/9² + 16/9³ + …))

    A = π16(3 + (1 + 1/9 + 1/9² + 1/9³ + …)))

    The sum of the geometric progression is:

    1 / (1 – 1/9) = 9/8

    Hence:

    A = π(16×33/8) = π66.

    The circles fill approximately 0.831 times the area of the triangle (total area = 144√3).

    Solution: The total area of all the circles is 66 π cm².

    Here’s a program that produces successive approximations of the sum:

    from enigma import irange, printf
    
    A = 48.0
    
    d = 1.0
    for i in irange(1, 12):
      A += (16.0 / d)
      printf("[{i:2d}] A = {A:.8f}")
      d *= 9.0
    

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