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Sunday Times Teaser 2663 – Missed the Plot

by Alan Bergson

A triangular vegetable plot with edges that are all whole numbers of metres has a perimeter in meters equal to its area in square metres.

Also the length of one of the sides is the average of the other two.

What are the lengths of its sides?

6 Comments Leave one →
  1. brian gladman permalink

  2. My analysis gives the same equations as you found. I then consider the divisors of 48 (using the divisors() function from my library. There are only 5 factorisations of 48 to consider.

    • If you let Python do all the work it only finds the answer on the 11th (a, b, c) triple it tries.

  3. brian gladman permalink

    Hi Jim,

    I was too lazy to get a factor subroutine out although I would have done so had it been more of a challenge.

    I also went on to solve for the triangle \((a, b, c=2b-a)\) with an area that is an integer multiple (\(k\)) of the semi-perimeter (\(s\)), which leads to the quadratic \[(3b – 2a)(2a – b) = 48k^2\] and the solution \[b = (2/3)\{2a\pm\sqrt{a^2-(3k)^2}\}\]

    This equation also shows that the Pythagorean triple \((3k, 4k, 5k)\) provides a solution for any \(k\) value, although not all solutions are necessarily Pythagorean.

    For example, the triangle \((13,14,15)\) provides another solution when \(k=4\).

  4. ahmet cetinbudaklar permalink

    If \(a\), \(b\) and \(c\) represent the sides of the plot with \(a=(b+c)/2\) and \(u=(a+b+c)/2\), then Heron’s formula for the area can be used to show that: \[a+b+c=\sqrt{u(u-a)(u-b)(u-c)}\] This soon provides a solution by trial and error.

  5. Michael Broadey permalink

    I generated Pythagorean Triples and found a result.

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