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## Question

Farmer Templeton took his newly aquired goat, which he named Monty, and tied him to a rope on the side of the silo off his barn. The silo is 24 meters in diameter, and the rope is 12 meters long. Allowing some assumptions for simplicity...
1.)How much will Monty eat in square meters?
2.)If Monty is tied to the silo at 11:12 AM and he eats 122 m2 per hour, how long will it be till he's done eating?

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• 0 Farmer Templeton took his newly aquired goat, which he named Monty, and tied him to a rope on the side of the silo off his barn. The silo is 24 meters in diameter, and the rope is 12 meters long. Allowing some assumptions for simplicity...

1.)How much will Monty eat in square meters?

2.)If Monty is tied to the silo at 11:12 AM and he eats 122 m2 per hour, how long will it be till he's done eating?

I think he finishes his meal (assuming he can't actually eat the silo) around 1:53pm.

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• 0

 327.68 square meters
 2 hours and 41 minutes + 11:12 a.m. = 1:53 p.m.

Subtract two 120 degree segments for the overlap of the goat's circle and the silo.

Lichwriter gets it ... ##### Share on other sites
• 0 Farmer Templeton took his newly aquired goat, which he named Monty, and tied him to a rope on the side of the silo off his barn. The silo is 24 meters in diameter, and the rope is 12 meters long. Allowing some assumptions for simplicity...

1.)How much will Monty eat in square meters?

2.)If Monty is tied to the silo at 11:12 AM and he eats 122 m2 per hour, how long will it be till he's done eating?

It depends on whether he is tethered inside or outside the silo... ##### Share on other sites
• 0 Lichwriter gets it ... 327.68 square meters

 2 hours and 41 minutes + 11:12 a.m. = 1:53 p.m.

Subtract two 120 degree segments for the overlap of the goat's circle and the silo.

Seems to me that this estimation was a bit too liberal with the assumptions. The approximate area we're looking for is shaded green. We can get this area by taking the area of the entire circle the goat could eat if there were no silo minus twice the area shaded red. The red area can be found by taking the area of the sector cut by angle DAB minus the area of the triangle DAB. Bonanova has already established the angle of DAB to be 120° because we know segment AC is 6 m (half the radius) and segment AB is 12 m (equal to the radius), and arccos(6/12)=60° for angle BAC, times 2 is 120°. So the red area is pi*12^2*120/360-1/2*(6*sqrt(3)*2)*6=88.44 m^2, and the green area is pi*12^2-88.44*2=275.5 m^2. This takes the goat 135.5 minutes to eat, so his end time is 1:27:30 P.M.

As the goat gets closer to the silo and at the end of his leash, the leash gets wrapped around the silo and there is a small sliver of grass he cannot reach because of it. So taking an arc length of 12 m out of the circumference of the silo gives an angle of 57.2958° (calculated by 12/(pi*24)*360) instead of 60° in our approximation. I haven't found a way to calculate this area by hand, but using 3D modeling software, I get the actual area the goat can eat to be 273.4 m^2. Only a 0.8% error on my approximation, as opposed to 20% error on yours.

Edited by lazboy

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• 0 It depends on whether he is tethered inside or outside the silo... It also would depend on the proximity of the barn to the point at which the goat is tethered on the silo, now that I read the OP more carefully (assuming the goat is, in fact, on the outside of the silo). ##### Share on other sites
• 0 Seems to me that this estimation was a bit too liberal with the assumptions. The approximate area we're looking for is shaded green. We can get this area by taking the area of the entire circle the goat could eat if there were no silo minus twice the area shaded red. The red area can be found by taking the area of the sector cut by angle DAB minus the area of the triangle DAB. Bonanova has already established the angle of DAB to be 120° because we know segment AC is 6 m (half the radius) and segment AB is 12 m (equal to the radius), and arccos(6/12)=60° for angle BAC, times 2 is 120°. So the red area is pi*12^2*120/360-1/2*(6*sqrt(3)*2)*6=88.44 m^2, and the green area is pi*12^2-88.44*2=275.5 m^2. This takes the goat 135.5 minutes to eat, so his end time is 1:27:30 P.M.

As the goat gets closer to the silo and at the end of his leash, the leash gets wrapped around the silo and there is a small sliver of grass he cannot reach because of it. So taking an arc length of 12 m out of the circumference of the silo gives an angle of 57.2958° (calculated by 12/(pi*24)*360) instead of 60° in our approximation. I haven't found a way to calculate this area by hand, but using 3D modeling software, I get the actual area the goat can eat to be 273.4 m^2. Only a 0.8% error on my approximation, as opposed to 20% error on yours.

The original 'correct' answer simply subtracted the area of parallelogram ACBD, rather than 2 times the segment cut off by chord CD.

Serious props for the rope length change, didn't consider that at all. Also germane is the length of the goats neck/head beyond the tether point of the rope, increasing the effective radius? Edited by xucam

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• 0 As the goat gets closer to the silo and at the end of his leash, the leash gets wrapped around the silo and there is a small sliver of grass he cannot reach because of it. So taking an arc length of 12 m out of the circumference of the silo gives an angle of 57.2958° (calculated by 12/(pi*24)*360) instead of 60° in our approximation. I haven't found a way to calculate this area by hand, but using 3D modeling software, I get the actual area the goat can eat to be 273.4 m^2. Only a 0.8% error on my approximation, as opposed to 20% error on yours.

A goat will not long be confounded by that rope in the way if there is food to be had. The rope will either be moved or eaten. Them goats is hungry critters!

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• 0
Seems to me that this estimation was a bit too liberal with the assumptions.
As the goat gets closer to the silo and at the end of his leash, the leash gets wrapped around the silo and there is a small sliver of grass he cannot reach because of it. So taking an arc length of 12 m out of the circumference of the silo gives an angle of 57.2958° (calculated by 12/(pi*24)*360) instead of 60° in our approximation. I haven't found a way to calculate this area by hand, but using 3D modeling software, I get the actual area the goat can eat to be 273.4 m^2. Only a 0.8% error on my approximation, as opposed to 20% error on yours.
[spoiler=Too quick with the pencil, actually. ]Segment area is 150.79 - 62.35 = 88.44

Rope circle area is 452.389

Instead of subtracting 88.44 twice, I subtracted 62.35 twice.

And apparently LichWriter did, also!

The goat's grazing area is, as lazboy calculates, 452.389 - 2[88.44] = 275.504 which takes 2.258 hours to graze.

It's always behind his mouth if he walks away from the tie point. ##### Share on other sites
• 0 [spoiler=Too quick with the pencil, actually. ]Segment area is 150.79 - 62.35 = 88.44

Rope circle area is 452.389

Instead of subtracting 88.44 twice, I subtracted 62.35 twice.

And apparently LichWriter did, also!

The goat's grazing area is, as lazboy calculates, 452.389 - 2[88.44] = 275.504 which takes 2.258 hours to graze.

It's always behind his mouth if he walks away from the tie point. It is not the rope that impedes the goat, it is the silo. As the goat, on the end of its rope so as to stretch it taut, approaches the silo, the rope follows the curve of the silo and is no longer a straight line.

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• 0 It is not the rope that impedes the goat, it is the silo. As the goat, on the end of its rope so as to stretch it taut, approaches the silo, the rope follows the curve of the silo and is no longer a straight line.

Oh right, so the ends of the arc would curve in ever so slightly where they meet the silo...

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• 0
Seems to me that this estimation was a bit too liberal with the assumptions. The approximate area we're looking for is shaded green. We can get this area by taking the area of the entire circle the goat could eat if there were no silo minus twice the area shaded red. The red area can be found by taking the area of the sector cut by angle DAB minus the area of the triangle DAB. Bonanova has already established the angle of DAB to be 120° because we know segment AC is 6 m (half the radius) and segment AB is 12 m (equal to the radius), and arccos(6/12)=60° for angle BAC, times 2 is 120°. So the red area is pi*12^2*120/360-1/2*(6*sqrt(3)*2)*6=88.44 m^2, and the green area is pi*12^2-88.44*2=275.5 m^2. This takes the goat 135.5 minutes to eat, so his end time is 1:27:30 P.M.

As the goat gets closer to the silo and at the end of his leash, the leash gets wrapped around the silo and there is a small sliver of grass he cannot reach because of it. So taking an arc length of 12 m out of the circumference of the silo gives an angle of 57.2958° (calculated by 12/(pi*24)*360) instead of 60° in our approximation. I haven't found a way to calculate this area by hand, but using 3D modeling software, I get the actual area the goat can eat to be 273.4 m^2. Only a 0.8% error on my approximation, as opposed to 20% error on yours.

Wow, I can't even balance my checkbook!

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• 0

It is not the rope that impedes the goat, it is the silo. As the goat, on the end of its rope so as to stretch it taut, approaches the silo, the rope follows the curve of the silo and is no longer a straight line.

Thanks. You're absolutely right.

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• 0 The original 'correct' answer simply subtracted the area of parallelogram ACBD, rather than 2 times the segment cut off by chord CD.

Serious props for the rope length change, didn't consider that at all. Also germane is the length of the goats neck/head beyond the tether point of the rope, increasing the effective radius?

on the increase of effective radius with the addition of the neck and head. And here I thought I had thought of everything.

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• 0

Lazboy has got the answer I was looking for. Congrats. I also like his attempt at figuring the lost area due to the rope wrapping around the silo. If the arc AB is 12 meters then I figure the grazing circles radius diminishes to about 11.5 meters where it contacts the silo.

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• 0
Lazboy has got the answer I was looking for. Congrats. I also like his attempt at figuring the lost area due to the rope wrapping around the silo. If the arc AB is 12 meters then I figure the grazing circles radius diminishes to about 11.5 meters where it contacts the silo.
2 short]Saw this puzzle and couldn't resist trying to figure out what happens as the goat goes around the edge of the silo. Here's what I see:

Oh, and for what it's worth, I get an area of 274.15 m2. And it should be possible to establish an elevation for tying the rope such that the goat's mouth ends up exactly 12 m from the connection point (depends upon length of neck/head/tongue vs length of legs, the flexibility of the goat, etc.) The connection could easily be assumed to be directly on the side of the silo if the rope penetrates the silo wall and is anchored inside the silo somewhere, leaving 12 m from connection point to the last standing blade of grass. Edited by HoustonHokie

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• 0 2 short]Saw this puzzle and couldn't resist trying to figure out what happens as the goat goes around the edge of the silo. Here's what I see:

Oh, and for what it's worth, I get an area of 274.15 m2. And it should be possible to establish an elevation for tying the rope such that the goat's mouth ends up exactly 12 m from the connection point (depends upon length of neck/head/tongue vs length of legs, the flexibility of the goat, etc.) The connection could easily be assumed to be directly on the side of the silo if the rope penetrates the silo wall and is anchored inside the silo somewhere, leaving 12 m from connection point to the last standing blade of grass. Wow, I guess your precision trumps mine. I knew it was a spiral shape, but didn't take the time to figure out how to draw a perfect spiral (let alone figure out what specific type of spiral it should be) in the software I was using, so I used an arc tangent to the cyan circle in your drawing with the same start and end points as your spiral. So it was merely another estimation and not the exact solution, but if your solution is completely accurate, my "precise" solution was only off by 0.2%. ##### Share on other sites
• 0
2 short]Saw this puzzle and couldn't resist trying to figure out what happens as the goat goes around the edge of the silo. Here's what I see:

Oh, and for what it's worth, I get an area of 274.15 m2. And it should be possible to establish an elevation for tying the rope such that the goat's mouth ends up exactly 12 m from the connection point (depends upon length of neck/head/tongue vs length of legs, the flexibility of the goat, etc.) The connection could easily be assumed to be directly on the side of the silo if the rope penetrates the silo wall and is anchored inside the silo somewhere, leaving 12 m from connection point to the last standing blade of grass. FWIW - It's the type of spiral whose radius increases linearly with angle.

The Archimedean spiral: r = a + bθ

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• 0
FWIW - It's the type of spiral whose radius increases linearly with angle.

The Archimedean spiral: r = a + bθ

Is it still an Archimedean spiral if the center of the spiral moves with the angle as well? The graphic shown on the linking page you sent shows a definite central point - for this problem, that center does not stay still, but keeps moving around the silo. I've been able to define the spiral parametrically, but don't know of a single function which could describe it.

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Is it still an Archimedean spiral if the center of the spiral moves with the angle as well? The graphic shown on the linking page you sent shows a definite central point - for this problem, that center does not stay still, but keeps moving around the silo. I've been able to define the spiral parametrically, but don't know of a single function which could describe it.

Probably not.

But the idea of adding to the radius in proportion to the angle of wrap probably makes it the closest choice.

That is, it rules out a slew of them - logarithmic, most importantly.

The radius is measured from a moving point, as you point out.

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• 0

Well, these are pretty good answers for the various estimates. I get 274.19467

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