Puzzle time - Silo dog
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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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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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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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I figure the dog wanting to mark its territory simply pees all over the place.
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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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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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I glad he didn’t say the three of you
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lol.
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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
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@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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Yes. This is the shape. Note the 10pi leash length is just perfectly half the circumference of the silo.
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Maybe you can google up a formula
if you can’t figure out the integral. -
Yep. It’s the quarter circle plus that bit of cartioid x 2
