Enigmatic Code

Programming Enigma Puzzles

Enigma 121: Sunday stroll

From New Scientist #1265, 6th August 1981 [link]

George set out the other Sunday morning to walk to his friend Bill’s house, which lay a few miles north-east of his own. His idea was to collect Bill and walk together to their favourite pub, which was a few miles due west from Bill’s house. But when he got half-way to Bill’s, he realised two things: first, that by the time he had picked up Bill and they had walked to the pub it would be after closing time, and second, that he was exactly as far from the pub as he was from Bill’s house. So he changed direction and went straight to the pub. He arrived in good time, but his usual drinking cronies weren’t there, so he had a quick pint and left. Having a bit of time in hand, instead of going straight home he walked to the next village, which lay due west. He then turned towards home and went straight there, a distance of six miles.

How far was he from the pub when he had covered half this distance?


One response to “Enigma 121: Sunday stroll

  1. Jim Randell 26 August 2013 at 9:12 am

    This one is solved without any programming.

    Assuming all this takes place on a flat featureless plain with direct access to the shortest straight line routes between points:

    Enigma 121 - Solution

    This solution uses the fact that the angle subtended by the diameter of a circle at the circumference is a right angle.

    B is north east of G, and H is equidistant from P, B and G. So B, P and G lie on the circumference of a circle centred at H, and BG is a diameter of that circle. So the angle at P is a right angle. Hence P is due north of G a distance of d (= x√2).

    Similarly if we construct a circle centred at Z with radius 3, then VG is a diameter of the circle, and the angle at P is a right angle. Hence P lies on the circumference of that circle too, and so the distance z is 3.

    Solution: George was 3 miles from the pub when he was halfway home.

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