Solving the Puzzle: Cow Rope Length in Round Meadow

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Discussion Overview

The discussion revolves around a mathematical puzzle involving a cow tied to the edge of a circular meadow and determining the length of the rope required for the cow to graze half of the meadow's area. The conversation includes various approaches to solving the problem, exploring geometric relationships and mathematical expressions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests a formula relating the meadow's area to the cow's grazing area, proposing that the area grazed can be expressed as a function of the rope length.
  • Another participant emphasizes the need to consider the cow tethered at the edge of the meadow, noting that the area grazed is not a complete circle but a segment of it.
  • A later reply introduces a mathematical expression for the area swept out by the rope, suggesting a numerical solution to find the rope length that allows the cow to graze half the meadow.
  • Some participants express confusion about the problem's setup, indicating differing interpretations of the grazing area based on the cow's tethering point.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the problem, particularly regarding the grazing area and the cow's position relative to the meadow. No consensus is reached on the correct approach or solution.

Contextual Notes

Participants assume the meadow is circular and discuss various mathematical relationships without resolving the complexities involved in calculating the grazing area based on the cow's tethering position.

niko2000
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Hi,
I have tried to solve this puzzle:
If we had a round meadow and a cow tied with a rope on the edge of that meadow. Hw long should be a rope if we wanted to let the cow eat a half of that meadow?
I have tried to solve this puzzle time ago and now it attracted me again. Does anyone know how it could be solved?
Regards,
Niko
 
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niko2000 said:
Hi,
I have tried to solve this puzzle:
If we had a round meadow and a cow tied with a rope on the edge of that meadow. Hw long should be a rope if we wanted to let the cow eat a half of that meadow?
I have tried to solve this puzzle time ago and now it attracted me again. Does anyone know how it could be solved?
Regards,
Niko

What are the dimensions of the meadow?
Meadow area = width*height
Cow area = (rope length)^2 * pi
cow can eat half:
width*height/2 = (rope length)^2 * pi
square root(width*height/(2*pi)) = rope length
 
Meadow is rotund, not rectangular.
The radius of the meadow is whatever size. We have to find a relation between rope length and radius length. Assume the radius length is 1.
 
niko2000 said:
Meadow is rotund, not rectangular.
The radius of the meadow is whatever size. We have to find a relation between rope length and radius length. Assume the radius length is 1.

alright then

m^2/2=r^2
sqr(m^2/2) = r
m/sqr(2) = r

oooo hard :-p
 
Alkatran said:
alright then

m^2/2=r^2
sqr(m^2/2) = r
m/sqr(2) = r

oooo hard :-p

You're still not reading the problem correctly. You are assuming the cow is tethered INSIDE the circle so that the area grazed is a complete circle. That's not true- the cow is tethered at the edge of the circle so the area she can graze is only a portion of a circle.
 
niko2000 said:
Hi,
I have tried to solve this puzzle:
If we had a round meadow and a cow tied with a rope on the edge of that meadow. Hw long should be a rope if we wanted to let the cow eat a half of that meadow?
I have tried to solve this puzzle time ago and now it attracted me again. Does anyone know how it could be solved?
Regards,
Niko

You need to get an expression for the area swept out by the tether rope as a function of it's lenght. Normalize the problem by taking the meadow to be unit radius and let the rope length be "r". You can get the following expression for the area "A" swept by the rope.

A = r^2 acos(r/2) + acos( 1 - 0.5 r^2) - r sqrt( 1- 0.25 r^2)

Now solve numerically to find the value or r which gives A=Pi/2, which turns out to be somewhere around 1.15 to 1.16 times the meadow radius.
 
Hasn't this come up before?
 
Gokul43201 said:
Hasn't this come up before?

Not the I know of. Are you sure you're not thinking of the 0.999(repeated) != 1 thread. ;)
 

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