Solving the Puzzle: Cow Rope Length in Round Meadow

  • Context: High School 
  • Thread starter Thread starter niko2000
  • Start date Start date
  • Tags Tags
    Length Puzzle Rope
Click For Summary
SUMMARY

The puzzle of determining the required rope length for a cow tethered at the edge of a round meadow to graze half of the area has been analyzed. The area grazed by the cow can be expressed as a function of the rope length, leading to the equation A = r² acos(r/2) + acos(1 - 0.5 r²) - r sqrt(1 - 0.25 r²). Solving this numerically reveals that the rope length should be approximately 1.15 to 1.16 times the radius of the meadow to achieve the desired grazing area of π/2.

PREREQUISITES
  • Understanding of basic geometry and area calculations
  • Familiarity with trigonometric functions and their applications
  • Knowledge of numerical methods for solving equations
  • Concept of circular areas and their properties
NEXT STEPS
  • Study the derivation of the area formula for circular segments
  • Learn about numerical methods for solving equations, such as the Newton-Raphson method
  • Explore the properties of circular areas and their applications in real-world problems
  • Investigate similar geometric puzzles involving area and perimeter relationships
USEFUL FOR

Mathematicians, educators, students studying geometry, and puzzle enthusiasts interested in solving geometric problems involving circular areas.

niko2000
Messages
50
Reaction score
0
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
 
Mathematics news on Phys.org
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. ;)
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
10K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 83 ·
3
Replies
83
Views
12K
  • · Replies 1 ·
Replies
1
Views
3K