Thanks for doing the sums Billl.
Course, if his first pen was round, (or a twenty sided polygon-which is round enough for farmers,) then the assumed 50 square hurdles needed is wrong.
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Thanks for doing the sums Billl.
Course, if his first pen was round, (or a twenty sided polygon-which is round enough for farmers,) then the assumed 50 square hurdles needed is wrong.
Yes, well, we'll work THAT scenario through soon enough.
You're all being far too complicated - this is Kasie, remember, who can't do Maths. This is a theoretical puzzle, they are theoretical sheep, very small theoretical sheep, so don't think animal welfare and how much room a real sheep would need, think x number of units arranged to make a four sided shape the area of which is doubled by the addition of a minimum number of further units.
Please - someone work it out - I'm already packing for New York, I'll be off on Saturday, I'll have to tell you the answer, oh, I wish I hadn't started this one.
If it comes down to it, you could PM me or someone else with the answer!
But what's wrong with the answers given so far?
The question was: what's the least number of hurdles he needs to buy to double the area of his enclosure? So far no one has come up with the (very small) number he need buy.
Fair enough. They're wrong.Quote:
Originally Posted by kasie
Was the original enclosure a square or rectangle? Was it 5 x 5? Or could it have been 1 x 9, for example? Or a pyramid?
Definitely a square?
At no point is that made explicit. So I'm starting from the assumption that we can't make that assumption.Quote:
Originally Posted by billl
With 20 hurdles (also unfamiliar with this usage of the word) he can make a rectangle that is 7 by 5: area is 35.
XXXXXXX
X..........X
X..........X
X..........X
XXXXXXX
I guess the dots are sheep poop.
With forty sheep he needs an area of 70, 7 by 10.
That's 30 hurdles
XXXXXXXXXX
X................X
X................X
X................X
X................X
X................X
XXXXXXXXXX
And looks like a lot more poop.
Didn't someone already give this answer?
Come to think of it, wouldn't the biggest enclosure be a circle?
How to draw that, and measure the area?
Didn't somebody else also have this idea before me?
Darn it! I'll assume the 20 sheep can fit in a 9 by 3 enclosure:
XXXXXXXXX
X..............X
XXXXXXXXX
Area, who cares?
Double size:
XXXXXXXXX
X..............X
X..............X
XXXXXXXXX
2 more X's
jajdude, you are a genius
Hooray - he's right! The original enclosure was a rectangle 9 x 1 (20 hurdles altogether.) By adding one hurdle at each short end to make a rectangle 9 x 2 (22 hurdles altogether) the farmer has doubled the area of the enclosure at the expenditure of only two further hurdles. (It's hard to draw using only keyboard symbols because the hurdles would join corner to corner and edge to edge.)
Thanks, jj, now I can go off to NY and miss the groans that will be hurled in my direction. See you in a week's time, folks - try to behave while I'm away. Or on the other hand, you could just have fun.
yeah, stumbled for a while on that one because was thinking too mathematically I guess, which seems to have thrown others off too. Forgetting about "area" seemed to help find the answer.
Will try to think of a new one soon. Don't really have any ideas.
This is an interesting feeling. I'm simultaneously pissed off and impressed.
Actually, there is way in which a farmer could have doubled the area of his four-sided enclosure without the addition of any hurdles. If he had begun by enclosing the first 20 sheep with the enclosure's four sides in the shape of a parallelogram with corners set at 45 and 135 degrees, he could double the area of the enclosure simply by shifting the four sides into the shape of a rectangle (ie. setting all the corners at 90 degrees).