Enigmatic Code

Programming Enigma Puzzles

Puzzle #47: Geometra’s tomb

From New Scientist #3270, 22nd February 2020 [link] [link]

Long before the invention of satnav, the great explorer Asosa Lees embarked on a trek across the square desert of Angula in a quest to find the lost tomb of Geometra, which lay somewhere along the line marked A.

Lees had nothing but the crude and incomplete diagram shown and some basic instructions: proceed south-west for 100 kilometres, and then turn left. The only other information she had was that at the moment she turned left, the distance to the south-west corner of the desert was 100 kilometres further than the distance to the south-east corner.

To reach the tomb, Lees needed to head in precisely the right direction. Fortunately using her knowledge of geometry she was able to take the correct bearing.

At what angle did she head off towards the tomb?


3 responses to “Puzzle #47: Geometra’s tomb

  1. Jim Randell 22 February 2020 at 3:01 pm

    Using units of 100km.

    If the distance to the south-east corner is x, then the diagonal of the square is: 1 + (x + 1) = x + 2.

    And so each side of the square measures: s = (x + 2) / √2

    Applying the cosine rule to the right hand triangle we get:

    x² = 1² + (x + 2)² / 2 – (2 / √2)(x + 2) cos(45°)
    2x² = 2 + (x² + 4x + 4) – 2(x + 2)
    2x² = x² + 2x + 2
    x² – 2x – 2 = 0
    x = (1 + √3)

    And the side of the square is:

    s = (x + 2) / √2 = √(3/2) (1 + √3)

    (about 334.6 km).

    Then, applying the sine rule in the lower triangle we have:

    sin(θ) = (s / x) sin(45°)
    = (√3 / 2) (1 + √3) / (1 + √3)
    = √3 / 2


    θ = 60°

  2. Hugh Casement 22 February 2020 at 4:08 pm

    Given the solution, and the name which sounds like isosceles, I feel there must be a simpler way to get there: a geometric rather than algebraic solution.
    Go 100 km along the diagonal from the lower left corner, and you are now distance x from the turning point X. By symmetry, you are also x from the lower right corner. So you’re at the vertex of an equilateral triangle. Am I right?

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