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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

a)How many open semicircles are needed to fully cover a circle?

b)How many open hemispheres are needed to fully cover a sphere?

It is the first time I've ever seen questions like this and I don't know how to approach them.Please help me!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

a) I would think that you need to say something about the semicircles. What size are they?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

Hey.

The problem says nothing about their size,it only says that an open secircles is a semicircle without its ends.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,386

hi Mario23

So what's wrong with choosing the semicircles to be half the circle in which case it's two for both parts.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**greg1313****Member**- Registered: 2016-12-19
- Posts: 17

bob bundy wrote:

hi Mario23

So what's wrong with choosing the semicircles to be half the circle in which case it's two for both parts.

Bob

There's a circle (consisting of just the circumference), and then there's the disk containing the circle and the area inside of the circle.

From what the original poster stated, it must be an open disk, instead of an open circle.. If a full circle [really the disk] is going to be fully

covered using congruent open semicircles [really open semi-disks] with the same radius of that of the original full circle or even the original

full disk, then more than two of them will be needed to ensure that the circumference of the full circle (or disk) and a diameter (where the

open semi-disks would meet if there were only two of them) would also be covered.

.

*Last edited by greg1313 (2017-04-08 06:54:32)*

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

@greg1313 how do I extemd this to a sphere?Will I need 3 hemispheres like here?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,386

??? No, still don't get it. What's the difference between a circle and a disk? If two semicircles meet what does it matter if they're open?

Here's my solution.

You cannot see the original circle (it is red) because it's covered.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 768

But, the semi-circles and the circles may be of different sizes also. The size is not specifies in the question.

Practice makes a man perfect.

There is no substitute to hard work

All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

bob bundy wrote:

??? No, still don't get it. What's the difference between a circle and a disk? If two semicircles meet what does it matter if they're open?

Here's my solution.

http://i.imgur.com/sXx6Z60.gif

You cannot see the original circle (it is red) because it's covered.

Bob

The disk is the inside of the circle

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

iamaditya wrote:

But, the semi-circles and the circles may be of different sizes also. The size is not specifies in the question.

I think the size doesn't matter as 3 will be needed as greg said.I don't know yet if the answer is the same for a sphere.

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**greg1313****Member**- Registered: 2016-12-19
- Posts: 17

bob bundy wrote:

??? No, still don't get it. What's the difference between a circle and a disk? If two semicircles meet what does it matter if they're open?

Here's my solution.

http://i.imgur.com/sXx6Z60.gif

You cannot see the original circle (it is red) because it's covered.

Bob

You cannot see the circumference of the fill circle/disk even though it

is not covered. Again,

the semi-disks do not include their circumferences.

And you cannot see the diameter between the two semi-disks.

It is also not covered. Not being able to see with our eyes is not proof.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,386

hi,

I've been out all day at the Bressingham Steam Museum. Plenty of chance to think this through. I realise I have not understood this problem at all I thought a circle meant the circumference and the interior. I now think that the problem just means the circumference. I don't know a word for the perimeter of an open semi circle so I'll call it the semi circumference.

It seems to me that the questioner intended that the circle and all the semi circles have the same radius. If they don't it would take an indeterminate number of curves and so the problem would be impossible to answer.

Draw a horizontal line through the centre of the circle so that AB is a diameter. Take a first semicircle and (with no loss of generality) place it over the top half of the circle so that it terminates in A and B. Temporarily put a second semicircle below so that the bottom half of the circle is covered.

Because the semicircles are open, (like an open interval on a number line), A and B are not covered. This coverage is the best you can achieve with two, so this proves two is not enough. So adjust the second semicircle by rotating it a small amount around the centre so that A is covered. As every point up to A is covered the necessary adjustment can be very small. Now take a third semicircle and place its midpoint over B. As the uncovered arc close to B is very small, it is certain that this third semicircle completes the coverage.

So three is sufficient.

I have also been thinking about the 3D case. I can show that 5 hemispheres will suffice and I think 4 may also be sufficient but I don't have a water-tight proof yet. Maybe later.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,386

hemispheres

I shall start by showing that at least four hemispheres are needed. I also need a new word: CONC means the 'circle of non coverage' by the hemisphere. It is the circle at the end of the hemisphere that won't cover because the hemisphere is 'open'.

If you place two hemispheres so that their concs touch then there is a whole circle that is not covered. So at least two points, diametrically opposite, are not covered. If one is re-positioned so that the concs cross then they cross in two places, diametrically opposite.

So in all cases after using two hemispheres, there will be at least a pair of points, diametrically opposite, that are not covered. When any hemisphere covers a point P, the diametrically opposite point, Q, is not covered. Thus a third hemisphere, however positioned, cannot cover both the not covered points. thus, a minimum of 4 hemispheres is required.

But, can it actually be done by 4? ?

Draw a horizontal plane through the centre of the sphere. Call the leftmost point, on the plane and the surface of the sphere, A. Diametrically opposite is point B. C is the forward most point; and D the point opposite C. AB is perpendicular to CD.

However the first is placed, it is always possible to re-orient the sphere so that the top half is covered, so I will assume that is our starting placement with no loss of generality.

If the second is placed to cover the bottom half then we get a whole circumference of not covered points so I'll tilt the second slightly up to include A about the CD axis.

C and D are the first non covered points. If a third is placed with its conc in a vertical place through C and D, then these points are not covered. By twisting this hemisphere slightly anticlockwise around a vertical axis, D is included.

The only area not now covered is a small region close to C (shown purple on my diagram). It is easy to cover it with a fourth hemisphere.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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