Surface area of an island

[Qeauning]

Lets say there live two dudes on an island (or two gals, or one dude one gal, idc really!). Anyway they live on this island but they don't get along very well so they decide that it would be fair to divide the island (which is a perfect circle) into two equal pieces. How will they do this using only a rope and a stick?

Question is, how to divide the islands surface area into two equal pieces with just this equipment (the radius of the island is known and it is r).

calculo.png (3.01 KiB) Viewed 2757 times

image lifted from brenok's post -jr

So to clarify, A should equal B / the green surface should equal the orange surface. The thing we would like to find is a relation between r and q...

Spoiler:

What I've got so far is
A = B
A = pi * r^2 - B
A = pi * r^2 - A
2A = pi * r^2
A = B = (pi * r^2) / 2 (which btw makes perfect sense, duh)

[btilly]

Spoiler:

I'm assuming you can draw on the island fairly easily. Then this is just a ruler and compass problem because by using a fixed portion of the rope you get a compass, and placing the stick in the ground then stretching out the rope gives you a straight line. (Both people may need to cooperate to extend the line beyond a pair of points

So take 2 points on the coast. Use the usual ruler and compass construction to find the bisecting line. Draw that line. The island is now divided.

[Qeauning]

Spoiler:

I'm assuming you can draw on the island fairly easily. Then this is just a ruler and compass problem because by using a fixed portion of the rope you get a compass, and placing the stick in the ground then stretching out the rope gives you a straight line. (Both people may need to cooperate to extend the line beyond a pair of points

So take 2 points on the coast. Use the usual ruler and compass construction to find the bisecting line. Draw that line. The island is now divided.

Yeah I guess that is a solution... but I was aiming for a relation between r and q. The story about the island is just for... to make it less boring .

[Ansain]

Spoiler:

your initial work is right, but gets nowhere because you have 2 equations and 3 unknowns. (well actually the unknown q is what you are looking for, and it wasn't even included in the equations) anyways, all you need to solve it is an equation for the area of B in terms of r and q, something I don't want to do at 11pm, but I'll work on later.

[Nimz]

Spoiler:

For an analytic solution, write the equations of the circles in terms of q and r and find the points of intersection. Draw the line between those points (for reference, not for construction) and find the area of B on either side of the line. Set the sum of those areas equal to pi*r²/2. That's what my first approach would be, anyway.

I'm not going to worry about how they might come up with the correct distances right now, though. Or what they should do if the island is lumpy.

[BoomFrog]

This is three equations and three unknowns. The only hard part is finding the equation that represents the area of B in terms of r and q. This really belongs in the math section not logic puzzles.

[skeptical scientist]

Spoiler:

Let's call the centers of the two circles O and P, and the points where they intersect Q and R. Then the area of the region labeled B is easily given in terms of the angles s=<QOR and t=<QPR by r²(s-sin s)+q²(t-sin t), since we can think of it as a union of two wedges (the positive terms) minus the corresponding triangles (the negative terms).

Now s is twice one angle of a triangle with sides r, r, and q, so from the law of cosines we have q²=r²+r²-2r*r*cos (s/2), or s=2*cos^-1 (1-q²/2r²).

Now that we know what s is, t is just pi-s/2, so t=pi-cos^-1 (1-q²/2r²).

Substituting s and t and simplifying, we get the following for the area of B in terms of q and r:
r²(2*cos^-1 (1-q²/2r²)-sin 2*cos^-1 (1-q²/2r²))+q²(pi-cos^-1 (1-q²/2r²)-q/r * sqrt (1-q²/4r²))

We want this to be pi*r²/2, so we want q/r to be equal to z, where z satisfies:

pi/2=(2*cos^-1 (1-z²/2)-sin 2*cos^-1 (1-z²/2))+z²(pi-cos^-1 (1-z²/2)-z*sqrt(1-z²/4))

which, assuming I haven't made any mistakes, should be easy to solve numerically, and probably not possible to solve analytically.

[jestingrabbit]

An equation that's a little neater, but in terms of an angle, not ratios of lengths,

Spoiler:

Using skeps definitions

4t*cos(t) - 2*sin(t) +pi = 0

Got it by adding the wedge areas and subtracting the area of the kite, plus a bit of trig.

Could have sworn I'd seen this one get solved analytically, but looking at that I'm of the firm belief that it can't be done.

[quintopia]

This really belongs in the math section not logic puzzles.

A forum for good logic/math puzzles.

[BattleMoose]

An alternative solution.

Spoiler:

Assuming you can find the center of the island, which shouldnt be hard, you have rope and stick, measure out a piece of rope equal to 2r, have one person stand on the edge and the other to the other sideof the island untill the rope is taught, then the rope goes through the center and you can find the center as you could mark on the rope halfway. And I am sure there are other ways of find the center. Okay.

So, use two concentric circles to divide the area.

(1/2)*pi*r^2 = pi*q^2

q = r((1/2)^(1/2))

done

[jestingrabbit]

@Battlemoose: if by done you mean that you've evaded the substance of the question, then yes, done.

[BattleMoose]

Anyway they live on this island but they don't get along very well so they decide that it would be fair to divide the island (which is a perfect circle) into two equal pieces. How will they do this using only a rope and a stick?

I understand this to be the question, and that my solution satisfies this. Where is the problem or substance that I lack?

[jestingrabbit]

viewtopic.php?p=454711#p454711

From that, I take it the problem wasn't to find a way to divide the area of a circle in two using a ruler and compass, but to find the relation between r and q using the diagram given.

[Nanan]

Spoiler:

Given a length of rope and stick a more practical solution via geometry would suffice. In reality q is a non significant value to the eqn.

[skeptical scientist]

If the circles pass through each others midpoints, the area of the two regions will not be equal, Nanan.

[Macbi]

This puzzle was given to us in a maths lesson today, I got an equation in one variable for the length, but couldn't solve it. There doesn't seem to be a elegant solution.

[mavman2k8]

i'm not sure whether this is just nitpicky, but is there any reason the two people want the border to be circular ?

because if not, why dont they

Spoiler:

just fix one end of the rope at an arbitrary point of the shore, and then walk along the shore keeping the rope tight ... the point where they have slipped farthest down the rope and would have to "climb back up" they are exactly opposite of the starting point and have therefore found the perfect border.

[jestingrabbit]

@mavman2k8: The two people want the border to be circular. If you have a bit of a read, you can see that the OP clarified that the story about the people and the island was just a sham to introduce the geometry problem.

@Macbi: The equation that Skep got, and the one that I got, both seem unsolvable using the standard analytic techniques.

[Macbi]

@mavman2k8: The two people want the border to be circular. If you have a bit of a read, you can see that the OP clarified that the story about the people and the island was just a sham to introduce the geometry problem.

Our problem had a tied up goat eating half the grass in a circular field.

Also, how long is the radius? To, say, 3d.p.

[jestingrabbit]

Do your own homework man...

The equation

viewtopic.php?p=458029#p458029

is probably the cleanest you'll find, using the notation from skep's post above. You can pretty easily use newton's method, or just a simple binary search from there.

[Macbi]

Do your own homework man...

The equation

viewtopic.php?p=458029#p458029

is probably the cleanest you'll find, using the notation from skep's post above. You can pretty easily use newton's method, or just a simple binary search from there.

It's not homework.

I would have used that, but I can't follow which angle skeptical has called t. Which circle is the one with centre O?

Edit:

Spoiler:

t ~ 1.739737

[jestingrabbit]

O is the centre of the radius r circle according to my notebook. t is the magnitude of angle QPR.

Sorry for the homework accusation, but you did say that it came up in class and I just assumed. Again, my apologies.

[Macbi]

Sorry for the homework accusation, but you did say that it came up in class and I just assumed. Again, my apologies.

No problem.

In that case the circle radius is:

Spoiler:

1.289854016

Which looks about right.

[mattme]

The halves will each have area pi r^2 / 2.

Spoiler:

First find the centre of the island.
- construct a right angle at the circumference (shore). Using the converse of Thales' theorem extending the lines from the right angle gives us two points opposite on the island (ie. a diameter). The centre is half way across.

At centre of the island, construct a right angle. Extend the lines to meet the shore. These two radii have length r. By Pythagoras' theorem the length between them is r sqrt(2). Half that length.

Option 1) Draw a circle radius r sqrt(2) in the middle of the island

We divided the island into a smaller circle at centre, and a surrounding ring. One guy now has all the beach, and the other guy the middle of the island..

[brenok]

It isn't exactly a puzzle, but I though it's a interesting problem:

Given a black circunference; and a red circunference, which center is a point of the black, represented in green; find the ratio between the red and black diameter, so that the yellow area is half of the total area of black circle.

Image:

[Macbi]

We had a thread on this a while back...

Here it is!

EDIT: Also, I was being a total n00b in that thread.

[jestingrabbit]

You weren't nearly as noobish as mattme and battlemoose. Anyway, I'm gonna try a merge so that we have a working diagram in the thread.