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# Dog freedom?

Best Answer bonanova, 25 March 2013 - 10:56 PM

Trying to resolve this puzzle assuming a rectangular house.

There are four places the dog can be tethered, between which the area changes monotonically.

Therefore, their respective areas contain the extremal area values.

Spoiler for Here they are

Go to the full post

28 replies to this topic

### #21 bonanova

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Posted 24 April 2013 - 11:52 AM

If the solution has not already been posted, the only other assumption I can imagine making is that the tether is linked to a railing that allows it to slide along the perimeter of the house. The the accessible portion of the yard in that case is four rectangles and a complete circle. I won't bother to draw it, but the area is clearly 80p + pi802 where p is the perimeter of the house.

I won't spoiler this, because don't see how the OP could be construed to describe that condition.

I don't see what else to try. If we're still missing something, a clue would be nice.

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### #22 Prime

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Posted 24 April 2013 - 05:57 PM

Spoiler for There is another possibility.

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Past prime, actually.

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Posted 24 April 2013 - 09:01 PM

Your earlier solution only resolved if the house was indeed a rectangle.  As shown in prior posting, giving two dimensions does not necessitate a rectangle.  If the house was a different shape (still quadrilateral) what would happen to the max and min possible reach of the dog (tethered at a single point) as we try greater and lesser angles?

If the solution has not already been posted, the only other assumption I can imagine making is that the tether is linked to a railing that allows it to slide along the perimeter of the house. The the accessible portion of the yard in that case is four rectangles and a complete circle. I won't bother to draw it, but the area is clearly 80p + pi802 where p is the perimeter of the house.

I won't spoiler this, because don't see how the OP could be construed to describe that condition.

I don't see what else to try. If we're still missing something, a clue would be nice.

Sorry Prime, I'm not that creative.

Spoiler for There is another possibility.

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### #24 Prime

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Posted 24 April 2013 - 11:11 PM

The house must be a regular rectangle, per OP.

For those who's never seen Baba Yaga's hut, copy and paste this: домик бабы-яги картинки (or you could go with "Baba Yaga's hut") into Google's image search.

With such house construction, the dog's freedom is limited by the length of the leash until the time Baba Yaga gets hungry...

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Past prime, actually.

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Posted 24 April 2013 - 11:23 PM

I'll retire the topic.

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### #26 bonanova

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Posted 25 April 2013 - 02:58 AM

If we let the house assume a parallelogram shape, then the above numbers change.

If the tie point is closest to an angle that becomes acute, the accessible area increases.

If the closest corner becomes obtuse, the accessible area decreases.

The extreme case is that the house collapses into a 100 foot wall.

The accessible area depends on the distance x the dog is tethered from the end of the wall

If x = 0 (end) the maximum accessible area is the 80-foot radius circle: 6400 pi square feet.

If x = 50 (midpoint) the accessible area hits a minimum: 4100 pi square feet.

The previously found areas occur at intermediate values of x:

x. . . Area/pi

--------------

0 .....6400

15.2 ..5300

20 ... 5000

27.6 ..4600

30 . ..4500

50 ..  4100

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Posted 25 April 2013 - 05:25 AM

Can you explain how you found it to be 4100?  I found it to be 3425

Edited by BMAD, 25 April 2013 - 05:26 AM.

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### #28 bonanova

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Posted 25 April 2013 - 06:36 AM

Chain the dog in the middle of a 100-foot wall.

In the front yard he has an 80-foot-radius semicircle of access.

When he gets to the ends of the wall, he has 30 feet of chain left.

As he walks into the back yard he gains access to a 30-foot-radius semicircle on each side.

Area [three semicircles] = pi/2 [802 + 302 + 302] = pi/2 [6400 + 900 + 900] = pi [3200 + 900] = 4100 pi ft2

General area formula (x is distance from end of wall to the chain point):

0<x<20:.. A = pi/2 [802 + (80-x)2]

20<x<50: .A = pi/2 [802 + (80-x)2 + (x-20)2]

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Posted 25 April 2013 - 01:39 PM

I see what I did wrong. I have it being quarter circles at the end of radius 30.

Chain the dog in the middle of a 100-foot wall.
In the front yard he has an 80-foot-radius semicircle of access.
When he gets to the ends of the wall, he has 30 feet of chain left.
As he walks into the back yard he gains access to a 30-foot-radius semicircle on each side.

Area [three semicircles] = pi/2 [802 + 302 + 302] = pi/2 [6400 + 900 + 900] = pi [3200 + 900] = 4100 pi ft2

General area formula (x is distance from end of wall to the chain point):

0<x<20:.. A = pi/2 [802 + (80-x)2]
20<x<50: .A = pi/2 [802 + (80-x)2 + (x-20)2]

Edited by BMAD, 25 April 2013 - 01:40 PM.

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