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## Question

A heinous prank was perpetrated at Rudrum University. It was found that four students had painted a large circle on the floor of the cafeteria. Professor Templeton was talking to the Maintenance man who was hired to replace the gratified floor tiles at the expense of the four students.

“I’m a fair man. I only will charge them for my time and materials”, the Handy-man said.

“How many tiles will tiles need to be replaced?”, asked the Prof.

“Haven’t counted ‘em, yet. All I did so far was determine that the circle is 27 feet in radius”

“Hmm. I notice that the circle doesn’t lie on the edge of any tile and it never touches the intersection of any four tiles either.”

“Um, O.K. You teacher types notice weird things, don’t ya?”

“I guess we find some things interesting, yes. So what are you going to charge for this job?”

“The tiles are 9 inches by 9 inches and they only come in boxes of 10 at \$11.20 per box. My labor rate is \$30.00 per hour, but I break it down into 15 minute increments and I figure I can replace about four tiles in 15 minutes time.”

“So at most each student will have to pay…..”

What was the amount Professor Templeton stated?

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\$3050.87

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but this is one scenario:

I set a grid of tiles up so that the center of one tile was at the center of the circle, which appears to meet the requirements about not touching, etc. I got 4205 tiles that were at least partially covered with grafiti. So:

Materials cost = 4205 / 10 => 421 boxes of tiles @ \$11.20/box = \$4715.20

Labor cost = 4205 / 4 => 1052 15-minute periods @ \$7.5/period = \$7890.00

Total cost = \$12605.20

Divided among 4 students = \$3151.30 per student

Edited by HoustonHokie
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I'm sorry guys. The circle is only a line. It's not a solid circle. I overlooked this fact when writing it up. My apologies.

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Haha... didn't think that they wouldn't fill in the circle for some reason

\$169.36 for each kid

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Haha... didn't think that they wouldn't fill in the circle for some reason

\$169.36 for each kid

That's closer

it's more like \$216.20

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Okay, just the line this time

but this is one scenario:

Using the same circle as before, I now get 288 tiles with the grafiti line (I assume the line is infinitely thin, as it is in most geometric problems). So:

Materials cost = 288 / 10 => 29 boxes of tiles @ \$11.20/box = \$324.80

Labor cost = 288 / 4 => 72 15-minute periods @ \$7.5/period = \$540.00

Total cost = \$864.80

Divided among 4 students = \$216.20 per student

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\$171.28 ?

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\$224.63 each

300 Tiles for \$336

300/16 per hr=18.75 hours

18.75*\$30=\$562.5

Materials+Labor= \$898.50

Each kid owes \$224.63

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I'm sorry guys. The circle is only a line. It's not a solid circle. I overlooked this fact when writing it up. My apologies.

OK I calculated both answers one for solid and one for only a line

The size of the box needed to exactly hold a 27' diameter circle would be 54' x 54' Since the circle doesn't touch any edges or four way points the box would need to be one tile wider.

Given the tiles are 9" it takes 12 to fill 9' or 72 to exactly fill 54 feet. Therefore a 73 x 73 tile box would hold the 27' diameter circle.

For a solid circle:

I'll calculate the exact fit and add the appropriate amount back at the end

There is the issue of how many tiles are not needed at the corners. The difference between the volume of the square and the volume of the circle is pi r^2 - (2r)^2 Err I have to go so I cant continue on the solid one

The not solid one is easier 2 pi r * 4/3 give you the number of tiles

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Good day all,

I got the same 288 tile answer based on a simple square grid pattern.

Question: What if the tiles were offset by 1/2 their dimension every other row? Would this change the number of tiles touched?

Thanks for pondering!

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Good day all,

I got the same 288 tile answer based on a simple square grid pattern.

Question: What if the tiles were offset by 1/2 their dimension every other row? Would this change the number of tiles touched?

Thanks for pondering!

Yes, actually, it would.

I've got a scenario with 290 tiles now. So the price goes up to \$218.08. Good thought.

EDIT: I just found another with 291 tiles. So it goes up to \$220.88 per.

Edited by HoustonHokie
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Good day all,

I got the same 288 tile answer based on a simple square grid pattern.

Question: What if the tiles were offset by 1/2 their dimension every other row? Would this change the number of tiles touched?

Thanks for pondering!

It does say that the circle never crosses where FOUR TILES INTERSECT. Wouldn't this imply that they are not offset?

Edited by wafflechip
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I wondered if there was a geometrical way to prove a maximum arrangement. Here are my thoughts on that subject.

The 288 number we got originally has a direct relationship to the radius of the circle and the size of a tile. Namely, 288 = (27 * 12) / 9 * 8. The 27 * 12 term translates 27 feet into inches and the 9 represents the size of an individual tile. The 8 comes from the fact that in every quadrant of the circle there is a horizontal and vertical change equivalent to the radius of the circle, and you need continuous tiles to make that change. 36 tiles are required in each direction, and 2 directions are required for each quadrant, and 36 * 2 * 4 = 288. So that's where 288 comes from. And as long as the tiles are in a tight rectangular grid with no offsets, I think 288 is the most you can do. It also appears to be the least.

Now, if you allow offseting of the tiles, which sparky suggested, you can get a few more. I've managed to get up to 291 tiles in my previous posts, but I thought it might be possible to get even more. The questions are, how to do it, and how many can you get?

Assume that the offset between rows isn't necessarily half a tile, but is set so that the number of tiles impacted in each row is maximized (after all, if I were a vandal, I would want to vandalize as many tiles as possible - not that the vandals have any control over how the tiles are laid out, but just go with me here - perhaps the contractor is in cahoots with the vandals, and just hasn't been caught yet). That would require at least 3 tiles per row because you can always set one side to intersect two tiles and the opposite may only intersect one tile, but it could be as many as 18 for the top or bottom row. You can't guarantee two tiles on both sides, because there shouldn't be any gaps or odd-shaped tiles in the rows.

If that's the pattern, the number of tiles can increase significantly. I did an optimization like that and ended up with 338 tiles. It may be possible to get a few more. I feel pretty certain that you can't go as high as 360 (which would be like adding another quadrant to the 288 solution), but it might be possible to get near 350.

BTW, 338 tiles results in \$254.43 per vandal.

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