Enigmatic Code

Programming Enigma Puzzles

Enigma 1393: More haste less seed

From New Scientist #2553, 27th May 2006

I designed a water feature for my garden, fenced in for safety. I planned to buy three pieces of fencing to form a triangular boundary, with two of them being equal in length. Then I planned to make the biggest possible circular pool within the triangle, seeding the rest of the area with grass.

In my haste at the garden centre I bought the two correct equal-length pieces, but the third piece was only three-quarters as long as it should have been. I constructed a triangle with the incorrect fencing and made the largest circular pond possible within it.

It turned out that the area enclosed by the triangle was exactly the same as I had intended in my original plan but, surprisingly, the radius of the pool I ended up with was one foot greater than I had originally planned. What three lengths of fencing did I actually buy?

[enigma1393]

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One response to “Enigma 1393: More haste less seed

  1. Jim Randell 10 September 2013 at 8:49 am

    Consider the planned triangle (T0) to have sides 8a, 8a, 8b. The semi-perimeter is 8a + 4b, and hence the area A, by Heron’s formula, is:

    A² = (8a + 4b)(4b)(4b)(8a – 4b)

    The actually constructed triangle (T1) has sides 8a, 8a, 6b. The semi-perimeter 8a + 3b, and the area is also A. Heron’s formula gives:

    A² = (8a + 3b)(3b)(3b)(8a – 3b)

    Hence:

    (4²)(8a + 4b)(8a – 4b) = (3²)(8a + 3b)(8a – 3b)

    which simplifies to:

    8a = 5b.

    So the semi-perimeter of T0 is 9b, and the semi-perimeter of T1 is 8b.

    Now, considering the incircles of the triangles. The area of a triangle is the product of the inradius and the semi-perimeter, so:

    A = r.9b = (r + 1).8b
    ⇒ r = 8

    We see that the triangles are arranged as below:

    Enigma 1393

    And so the area is:

    A = 12b² = r.9b = 8.9b = 72b
    ⇒ b = 6.

    And so the sides of T0 are (30, 30, 48) and the sides of T1 are (30, 30, 36), which is the required answer.

    Solution: The lengths of fencing bought were two lengths of 30 feet and one length of 36 feet.

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