Dog freedom?

[BMAD]

You put your dog on a leash. Your leash is 80 feet in length. The dog is tethered to part of the exterior of your house (the tie is not part of the 80 feet). Your home's dimensions are 60ft by 40ft. How much yard does the dog have access to?

[bonanova]

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.

Middle broad side: Minimum of 4500 pi ft².

Ten feet down: 4600 pi ft²

Corner: Maximum of 5300 pi ft²

Middle short side: 5000 pi ft²

[bonanova]

Where is the tie point on the house?

Here are three cases.

Corner tie gives the dog the most area to roam, middle of 60' side gives the least.

1. Corner: 5300 pi ft²
2. Mid-40' side: 5000 pi ft²
3. Mid-60' sice: 4500 pi ft²

[Debasis]

i have taken for granted the that the dog is tied to the corner of the house.....

the are of yard it has access to will be the area of the circle formed by the radius (rope) of 80 ft - the area of the house . after doing some algebraic gymnastics we get that are is 3.14159*80*80 - 60*40

or 17706.19 ft^2 approx

Edited February 22, 2013 by Debasis

[Debasis]

i have taken for granted the that the dog is tied to the corner of the house.....

the are of yard it has access to will be 16650.48 ft^2 it is an approximation...well the area the dog has access to is the area of the circle with radius 80ft - area of quadrant in which house is there + area of quadrant with radius 20 ft +area of quadrant with radius 40 ft....let me know if something is

wrong...(please pardon me becuase i cant make others understand these things well

[Debasis]

Where is the tie point on the house?

Here are three cases.

Corner tie gives the dog the most area to roam, middle of 60' side gives the least.

1. Corner: 5300 pi ft²
3. Mid-40' side: 5000 pi ft²
5. Mid-60' sice: 4500 pi ft²

did you take in to consideration that if you go around the house (on 60 ft side For example)when you turn to other side you will get access to only the quadrant with 20 ft radius?

[BMAD]

don't take for granted where The dog is tied

[BMAD]

There are more than three possible places to tie the dog.

[k-man]

The area accessible to the dog is anywhere between 4500*pi (~14137.167) ft² and 5300*pi (~16650.441) ft² depending on where the dog is tied to the house.

[bonanova]

There are more than three possible places to tie the dog.

Sure. There are an infinite number of tie points.

Do you want a general equation?

Do you want the tie point for max or min?

Something other?

[BMAD]

There are more than three possible places to tie the dog.

Sure. There are an infinite number of tie points.

Do you want a general equation?

Do you want the tie point for max or min?

Something other?

A simple range of values would suffice so that i know the potential roaming of my dog.

[k-man]

There are more than three possible places to tie the dog.

Sure. There are an infinite number of tie points.

Do you want a general equation?

Do you want the tie point for max or min?

Something other?

A simple range of values would suffice so that i know the potential roaming of my dog.

Bonanova provided the range in his first post. I came up with the same numbers too. Do you have a different answer?

[CaptainEd]

If the house is only one inch high, then attaching anywhere on the edge or "roof" of the house allows the dog access to (6400 pi - 2400) sq ft. This is approximately 5637 pi sq ft.

[BMAD]

If the house is only one inch high, then attaching anywhere on the edge or "roof" of the house allows the dog access to (6400 pi - 2400) sq ft. This is approximately 5637 pi sq ft.

I would argue that this question is a tricky question not a trick question. Too find the Range of yard for the dog you must consider ALL possibilities but you guys have only considered the house to be rectangular.

[bonanova]

If the house is only one inch high, then attaching anywhere on the edge or "roof" of the house allows the dog access to (6400 pi - 2400) sq ft. This is approximately 5637 pi sq ft.

I would argue that this question is a tricky question not a trick question. Too find the Range of yard for the dog you must consider ALL possibilities but you guys have only considered the house to be rectangular.

Yes I did make that assumption. Since it's late at night and my eyes are drooping, I'm sticking with the assumption.

[BMAD]

What would the area be if the dimensions referred to a different quadrilateral? Like a parallelogram maybe.

[Prime]

In Russia houses shaped like this are commonplace:

With the pointy end pointing North, naturally. I know, because I used to live in one.
Some people may argue that 60ft by 40ft implies multiplication of two sides resulting in the area of the house. And thus implying a regular rectangle. But that's simple unsophisticated thinking peculiar to people who live in rectangular houses.

[BMAD]

thank you for clarifying. I am from the Ukraine so we are practically brothers.

In Russia houses shaped like this are commonplace:

rushouse.gif
With the pointy end pointing North, naturally. I know, because I used to live in one.
Some people may argue that 60ft by 40ft implies multiplication of two sides resulting in the area of the house. And thus implying a regular rectangle. But that's simple unsophisticated thinking peculiar to people who live in rectangular houses.

[bhramarraj]

If I go for the wordings of OP, then the question was not asked about the area, but the 'yard' which is unit of length. If it is true then dog has excess to square root of (80^2 - T^2)/3 yards, where 'T' is the height of the tie in feet.

[BMAD]

This solution is quite similar

[bonanova]

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.

[Prime]

Your house used to belong to Baba Yaga – an old witch who lived deep in the forest in a hut standing on chicken legs. The dog leash is tethered to one of the legs, and there is enough clearance for the dog to run freely under the house.

[BMAD]

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.

Your house used to belong to Baba Yaga – an old witch who lived deep in the forest in a hut standing on chicken legs. The dog leash is tethered to one of the legs, and there is enough clearance for the dog to run freely under the house.

[Prime]

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...

[BMAD]

I'll retire the topic.

[bonanova]

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