Puzzle time - Silo dog
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wrote on 28 Nov 2020, 05:14 last edited by
A farm has a grain silo with a circular floor plan, 20 meters in diameter. On the outside of the silo is a hook. A dog is tied up to that hook on a leash of length 10*pi meters.
Assuming the dog can’t enter the silo, what area can he cover?
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wrote on 28 Nov 2020, 08:27 last edited by
silently meditating over the complexity of the formula for circle circle intersection
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wrote on 28 Nov 2020, 11:13 last edited by jon-nyc
Do keep in mind the perimeter traced out by the leash is not a circle, since the leash wraps around the silo when he goes behind it.
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wrote on 28 Nov 2020, 11:28 last edited by
Uh oh...
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wrote on 28 Nov 2020, 11:30 last edited by
This isn’t a ‘find the trick and do the answer in your head’ problem. It’s a pencil-to-paper problem.
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wrote on 28 Nov 2020, 11:54 last edited by
I figure the dog wanting to mark its territory simply pees all over the place.
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wrote on 28 Nov 2020, 13:18 last edited by
Game plan would be:
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Find function that computes max height/distance from any point on silo, i.e., the shape of the "wrap around" part.
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Presuming the formula is complicated, ask computer algebra system to integrate the function to find its area.
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Multiply by two and add area of the semi-circle.
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wrote on 28 Nov 2020, 13:23 last edited by
Yeah, getting the the “game plan” is easy, actually getting step 1 is the harder part.
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wrote on 28 Nov 2020, 13:33 last edited by jon-nyc
Actually integrating the function is straightforward.
The trick, if you want to call it that, is just figuring out the function. Totally doable by both of you.
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wrote on 28 Nov 2020, 14:17 last edited by
I glad he didn’t say the three of you
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wrote on 28 Nov 2020, 14:44 last edited by
lol.
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wrote on 28 Nov 2020, 15:47 last edited by
||Cycloid is key.
Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid, and the area under the cycloid is given by the formula 3πr^2 where r is the silo's radius.
Solution is (area of cycloid traced out by the silo)-(area of silo)+(half the area of circle with radius that is the leash's length).
Area covered by the dog = 3π(10)^2 - π(10)^2 + (π(10π)^2)/2
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||Cycloid is key.
Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid, and the area under the cycloid is given by the formula 3πr^2 where r is the silo's radius.
Solution is (area of cycloid traced out by the silo)-(area of silo)+(half the area of circle with radius that is the leash's length).
Area covered by the dog = 3π(10)^2 - π(10)^2 + (π(10π)^2)/2
||wrote on 28 Nov 2020, 16:31 last edited by@Axtremus said in Puzzle time - Silo dog:
Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid,
But how does that arc relate to where the dog can go? Your solution - which may well be correct - is missing an argument here.
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wrote on 28 Nov 2020, 16:46 last edited by Klaus
Wouldn't the resulting shape be a (part of a) cardoid and not a cycloid?
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wrote on 28 Nov 2020, 16:54 last edited by jon-nyc
Yes. This is the shape. Note the 10pi leash length is just perfectly half the circumference of the silo.
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wrote on 28 Nov 2020, 17:17 last edited by
Right, cardioid rather than cycloid.
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wrote on 28 Nov 2020, 17:23 last edited by jon-nyc
Maybe you can google up a formula
if you can’t figure out the integral. -
wrote on 28 Nov 2020, 18:37 last edited by
I haven't looked at formulas yet but it looks like one needs only a part of the cardoid area since it overlaps with the semi circle.
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wrote on 28 Nov 2020, 18:53 last edited by
Yep. It’s the quarter circle plus that bit of cartioid x 2
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wrote on 28 Nov 2020, 20:29 last edited by
The area formula is:
but the integral would not have to be taken to pi but to the angle at which the y value is maximal, would be the case at the angle 3/4 * sqrt(3).
But this gets rather complicated hence I assume something's wrong here...