Brain-Teaser 464: Home meadow
From The Sunday Times, 19th April 1970 [link]
The triangular home meadow at Cowpleasure Farm is bounded West and South by fences running due north and due east from the farmhouse; its other fence forms one side of a square field known as Starvecrow. The south fence of Home Meadow is the shared north boundary of two contiguous square fields, Paddock and Rookery, whose total area is just half that of Starvecrow.
Farmer Nettle has just refenced the whole outer perimeter (i.e. excluding fences common to two fields). He used 146 equal sections of fencing, none of which needed bending or cutting.
He plans to replace the other fences next year using the same type of section.
How many will he need? (Don’t worry about gates; they are incorporated in some of the standard sections).
This puzzle is included in the book Sunday Times Brain Teasers (1974).
[teaser464]
S2T2 
Jim Randell 8:06 am on 16 March 2019 Permalink |
There’s nothing to distinguish the fields P and R, so we’ll call the smaller one P.
For the square fields P, R, S, we’ll indicate the size of one side also by P, R, S, and the remaining boundary of H can be Q.
The amount of perimeter fencing (shown in red) required is X:
And the amount of internal fencing (the remaining black lines) required is Y:
This Python program uses the [[
pythagorean_triples()]] routine (see Teaser 2910) from the enigma.py library to find possible dimensions of field H, and then determines the dimensions of the other fields.It runs in 77ms.
Run: [ @repl.it ]
from enigma import pythagorean_triples, irange, printf for (x, y, z) in pythagorean_triples(48): # S is the hypotenuse S = z # one of the other two sides is P + R for (PR, Q) in [(x, y), (y, x)]: # suppose P < R for P in irange(1, PR // 2): R = PR - P # P^2 + R^2 = S^2 / 2 if not (2 * (P * P + R * R) == S * S): continue # total sum of perimeter fences is 146 X = 3 * (S + R) + P + Q if not (X == 146): continue # sum of the internal fences Y = S + PR + P # output solution printf("Y={Y}, S={S} P={P} R={R} Q={Q}")Solution: He will need 57 sections.
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