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 Forum index » Diversions » TimeWasters
A little math puzzle
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Beemer
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 		A little math puzzle
A farmer owns a small farm just outside Manchester where he grows berries. His plot of land is a rather odd shape... it's in the shape of two overlapping squares, one at a 59-degree angle to the other. The larger of the two squares is 180 meters on a side, and the smaller of the two is 162 meters on a side. The corner point of the larger square is at the center point of the smaller square.



Since it's springtime, it's time for the farmer to start planting berries! So, in the area of his field that is only within the larger square, he's planting strawberries, which take 3 minutes per square meter to plant. The area of the field that is only within the smaller square will be planted with Red raspberries, which take 4 minutes per square meter to plant. Finally, in the area where the two overlap, he's planting blueberries, which take 5 minutes per square meter to plant.

The farmer wants to have his entire field planted in three days. Assuming each person can work up to 8 hours a day, how many people would be required to do so?

Any help would be greatly appreciated.

PostPosted: Sun Mar 27, 2005 5:16 pm
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LordKinbote
Decorated

Joined: 25 Sep 2002
Posts: 261

 		Is this a homework problem? If so, I don't want to *solve* it.

The easiest way to do it, as far as I can see, is to find the area of the part only in the smaller triangle and deriving the other two areas from that.

So, if you draw a vertical line through the center of the smaller triangle and another line from the center to the top left corner, you have a box (A = 81*162), two right triangles (the top one is A = (0.5)(81)(81), the bottom one is A = (0.5)(81)(81/(tan(59))) and then one more triangle on the top, the long diagonal of which is 81*2^0.5, the angles of which are 45,121 and 14.

From that information, you should be able to compute the area of that portion of the smaller square. Subtract it from the entire area of the square to get the central area, and subtract that area from the entire area of the larger square to get the last area.

The x minutes per square meter part of the problem should be pretty straightforward once you know the areas.

PostPosted: Sun Mar 27, 2005 7:56 pm
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binnister
Kilroy

Joined: 28 Mar 2005
Posts: 2

 		This is actually a lot easier than I at first thought it would be.

Since the corner of the larger square is 'pinned' to the center of the smaller one, the angle at which they intersect is not important.

In fact, pretend to rotate the larger square until it is at a 90 degree angle (with the corner still 'pinned' to center).

At this point, it should be obvious how much of the 2 squares are shared Smile

PostPosted: Mon Mar 28, 2005 4:07 pm
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maxim
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Joined: 18 Feb 2005
Posts: 149
Location: United Kingdom 333#: 287

 		Jeez... that's clever. I didn't think of that. Phew.
_________________
MAXIM³

"You complete the jigsaw puzzle
to discover it is a picture of yourself,
finishing that same puzzle."

PostPosted: Mon Mar 28, 2005 4:28 pm
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LordKinbote
Decorated

Joined: 25 Sep 2002
Posts: 261
binnister wrote:
This is actually a lot easier than I at first thought it would be.

Since the corner of the larger square is 'pinned' to the center of the smaller one, the angle at which they intersect is not important.

In fact, pretend to rotate the larger square until it is at a 90 degree angle (with the corner still 'pinned' to center).

At this point, it should be obvious how much of the 2 squares are shared Smile


Damn. +10 for my math skills, -10 for my strategic skills.

I leave an Even Steven once again.

PostPosted: Mon Mar 28, 2005 5:41 pm
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Atrophied
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Joined: 29 Aug 2004
Posts: 1133
Location: 53742E 4A6F686E27732C 4E4C00

 		A quick look at the source of the image shows it hosted on neopets.com, so I assume it was some kind of puzzle contest on there.

Checking it out, it seems there's one each week.
I think I'll post them here if no-one minds. (Then you won't have to get an account yourself, but can still work on the puzzles.)

Looking back over the course of the puzzles, some of them are quite interesting, involving many different types of code breaking and mathematical formulae. Just a thought, and a timewaster to keep your skills sharp. This week's puzzle is over... I'll post the new one when it's uploaded (later on today I think).

~Atrophied
_________________
"It will be happened; it shall be going to happening; it will be was an event that could will have been taken place in the future." -- Time travel, as explained by Arnold J. Rimmer

"The Future's bright, the Future's Cuboid" - Juxta

PostPosted: Thu Mar 31, 2005 1:45 pm
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Atrophied
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Joined: 29 Aug 2004
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Location: 53742E 4A6F686E27732C 4E4C00

 		The new puzzle has just been uploaded:
Quote:

Bartleby the Skeith was bored one morning, and decided to pull out an empty Armada board and some Armada pieces. He filled the entire board with random Armada pieces, one in each space. Then, he emptied the whole board and did it again. And then again, and then again. He decided to keep doing it until he filled the board with every possible combination of pieces.

Code:

000XXXXXXX000
00XXXXXXXXX00
0XXXXXXXXXXX0
XXXX00000XXXX
XXX0000000XXX
XXX0000000XXX
XXX0000000XXX
XXXX00000XXXX
0XXXXXXXXXXX0
00XXXXXXXXX00
000XXXXXXX000

If it takes him 30 seconds to set up and then empty the board each time, how long will it take him to set up every possible combination? Please round to the nearest year.


The code in the quote is an approximate setup of the board. X's are available spaces. There are 2 different objects able to be placed.
start puzzling Smile

Edit: that makes a total of 88 fillable spaces.
_________________
"It will be happened; it shall be going to happening; it will be was an event that could will have been taken place in the future." -- Time travel, as explained by Arnold J. Rimmer

"The Future's bright, the Future's Cuboid" - Juxta

PostPosted: Thu Mar 31, 2005 6:30 pm
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Atrophied
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Joined: 29 Aug 2004
Posts: 1133
Location: 53742E 4A6F686E27732C 4E4C00

 		Going out on a limb here:
so is it:
Spoiler (Rollover to View):

2^88 * 30
_________________

(60^2 * 24 * 365.25)

Question

even though that comes out to an insanely huge number.


Edit: But rotational symmetry would reduce the number. How do I factor this into my equation?

Edit2:

Aha!
Spoiler (Rollover to View):

2^(88/4) * 4 * 30
_______________
(60^2 * 24 * 365.25)


comes to:
Spoiler (Rollover to View):
~16 Years


Can someone more versed in math than I am tell me If I did this right?
_________________
"It will be happened; it shall be going to happening; it will be was an event that could will have been taken place in the future." -- Time travel, as explained by Arnold J. Rimmer

"The Future's bright, the Future's Cuboid" - Juxta

PostPosted: Thu Mar 31, 2005 6:51 pm
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LordKinbote
Decorated

Joined: 25 Sep 2002
Posts: 261
Atrophied wrote:
Going out on a limb here:
so is it:
Spoiler (Rollover to View):

2^88 * 30
_________________

(60^2 * 24 * 365.25)

Question

even though that comes out to an insanely huge number.


Edit: But rotational symmetry would reduce the number. How do I factor this into my equation?

Edit2:

Aha!
Spoiler (Rollover to View):

2^(88/4) * 4 * 30
_______________
(60^2 * 24 * 365.25)


comes to:
Spoiler (Rollover to View):
~16 Years


Can someone more versed in math than I am tell me If I did this right?


Your first method seems correct. It's not clear whether rotational symmetry matters. If it did, I would think that would be specified.

So, yeah, the answer should be gigantic. 2^88 is a very, very, very large number.

PostPosted: Fri Apr 01, 2005 2:46 pm
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