# Enigmatic Code

Programming Enigma Puzzles

## Enigma 140: Muddy fields

From New Scientist #1284, 17th December 1981 [link]

I am at X, mounted on my velocipede, and thirsty: so I want to reach the river, which runs straight along BYC, as soon as I can. I have to keep within three fields on the plan. In each of them I can maintain a steady speed, at 30 mph within XBYZ (including the boundary), and at rates which I know (but you don’t yet) in AXZD and ZYCD (including their boundaries).

If I tell you that there are precisely three different quickest routes from X to the river, that should enable you to work out:

(a) my speed in AXZD,
(b) my speed in ZYCD.

Note: In New Scientist #1294 (with Enigma 149) an apology was published stating that this puzzle does not actually have a solution.

[enigma140]

### 4 responses to “Enigma 140: Muddy fields”

1. Jim Randell 28 October 2013 at 8:25 am

The apology published in New Scientist #1294 is as follows:

Stephen Ainley apologises that his answer to Enigma 140 (Muddy Fields) was wrong. Solvers were asked to find speeds x and y so that there would be “precisely 3” different routes from X to the river (BYC). The published answer was x = 32 1/2, y = 78, giving 6 minutes along XB or XZY or XEY (which is 2+2 metres long [???]). But, if XZY and XEY give equal times, so does XLMY, where LM is parallel to XE; and, since L can be anywhere on XZ, there is [sic] an infinite number of different routes. There is in fact no solution to the problem as set.

However if we restrict the route to have no more than 2 components I think we can solve the puzzle as the setter intended.

The most direct route, along XB, is entirely in field XBYZ, and is 3 miles at 30 mph (= 2 minutes per mile), so takes 6 minutes.

The shortest route entirely through fields AXZD (at speed a minutes per mile) and ZYCD (at speed b minutes per mile) is XZ + ZY, and will also take 6 minutes. Hence:

2a + 3b = 6
2a = 6 – 3b

So, the remaining six minute route must be to cut across field XBYZ to a point on ZY (E), and then follow the remainder of EY through field ZYCD. This also takes 6 minutes, so:

2 hypot(2, e) + b (3 – e) = 6
2 hypot(2, e) = 6 – 3b + eb = 2a + eb

squaring both sides:

16 + 4e² = 4a² + 4aeb + e²b²

and rewriting as a quadratic equation in terms of e:

(4 – b²)e² + (-4ab)e + (16 – 4a²) = 0

This will have exactly one solution when “b² = 4ac”, i.e.:

16a²b² = 4(4 – b²)(16 – 4a²)
4a²b² = 64 – 16a² – 16b² + 4a²b²
4a² + 4b² = 16

substituting for 2a:

(6 – 3b)² + 4b² – 16 = 0
13b² – 36b + 20= 0
(b – 2)(13b – 10) = 0

So either: b = 2, a = 0 (not possible), or: b = 10/13, a = 24/13, e = 5/6.

Converting from minutes per mile to miles per hour (x minutes per mile = 60/x mph) we get:

a = 32.5 mph, b = 78 mph.

Solution: (a) Your speed in AXZD is 32.5 mph; (b) Your speed in ZYCD is 78 mph.

• Jim Olson 28 October 2013 at 8:33 pm

I think your suggested approach is an excellent way to save the enigma. I am concerned however with the wording of the enigma that says the speed is constant within any area “including the boundary.” Then what is the speed at a shared boundary?

• Jim Randell 28 October 2013 at 10:11 pm

I took the view that on a boundary you travel at the fastest speed available from the areas on either side of the boundary. Although I’m really just working backwards from the desired answer to find the conditions that make the puzzle solvable.

• Hugh Casement 24 July 2015 at 8:22 am

Really the western and northern fields are superfluous.  My suggested rewording of the puzzle would be to have just the big field, with a hedge or fence along side XZ and a gap at X that gives access to a track immediately to the left of it.  That track leads to a tarred road ZY — it has to be surfaced to allow such a high speed.  There is no fence along that north side of the field.  At 3 miles by 2 it’s more a plain than a field.

Perhaps it’s too much to expect mathematical puzzles to reflect the real world!

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