28 replies to this topic

[k-man]

 

Spoiler for 'tieing

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]

Spoiler for if this is a trick question

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]

Spoiler for if this is a trick question

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]

Spoiler for if this is a trick question

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]

Spoiler for clarification

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.

Spoiler for clarification

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]

Spoiler for How much yard...?

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.

[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.

 

Spoiler for Here they are

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

[BMAD]

http://brainden.com/...-is-the-f-road/

 

This solution is quite similar