Fun Math Problems

[jay_shark]

1) Show that the sum of the lengths of the perpendiculars from any point inside a regular pentagon to the sides is constant .

2) Given any 7 points inside a circle of radius 1 , show that one can always find two points whose distance apart is no more than 1.

[Homer]

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2) Given any 7 points inside a circle of radius 1 , show that one can always find two points whose distance apart is no more than 1.

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I'll give it a shot, if nothing else so someone can show me the real solution. /images/graemlins/grin.gif

The greatest possible minimum distance (does that make sense) occurs when the points are just inside the circle at 360/7 degree intervals. This makes the smallest distance between any 2 points equal to:

2*sin(360/14) = .433

http://img530.imageshack.us/img530/1476/ghettodrawingwe9.jpg

Thinking about this a little more, I think I'm wrong, but can't figure it out.

[jay_shark]

2) It's interesting to note that this is true for any 6 points in a circle . You don't even need 7 points !

[CityFan]

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2) Given any 7 points inside a circle of radius 1 , show that one can always find two points whose distance apart is no more than 1.

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I'll give it a shot, if nothing else so someone can show me the real solution. /images/graemlins/grin.gif

The greatest possible minimum distance (does that make sense) occurs when the points are just inside the circle at 360/7 degree intervals. This makes the smallest distance between any 2 points equal to:

2*sin(360/14) = .433

http://img530.imageshack.us/img530/1476/ghettodrawingwe9.jpg

Thinking about this a little more, I think I'm wrong, but can't figure it out.

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This is not a proof, sorry.

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The greatest possible minimum distance (does that make sense)...

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No, it does not make sense. You're obviously trying to express something, but you need to make it clear exactly what concept you are trying to get across.

Do you mean that this arrangement maximises the smallest distance between any two points? If so, prove it!

To my eyes, all you've done is show one particular arrangement in which the distance between two points is less than one.

I think you may be on the right track though...

[Homer]

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Do you mean that this arrangement maximises the smallest distance between any two points?

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Yes.

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If so, prove it!

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I blow, this is all I could come up with.

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To my eyes, all you've done is show one particular arrangement in which the distance between two points is less than one.

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Correct, because I can't think of any other arrangement for which the smallest distance between 2 points would be greater. I understand this does not prove that, though. Just thought I'd put my thoughts on paper and see how far off I was or if my thinking was headed in the right direction.

I will wait to see the solution. Thanks.

[CityFan]

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Do you mean that this arrangement maximises the smallest distance between any two points?

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Yes.

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If so, prove it!

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I blow, this is all I could come up with.

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To my eyes, all you've done is show one particular arrangement in which the distance between two points is less than one.

[/ QUOTE ]

Correct, because I can't think of any other arrangement for which the smallest distance between 2 points would be greater. I understand this does not prove that, though. Just thought I'd put my thoughts on paper and see how far off I was or if my thinking was headed in the right direction.

I will wait to see the solution. Thanks.

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Sure, I appreciate you were just putting your thoughts out there... I though I'd pass comment on them!

I don't have the solution, sorry.

[Enrique]

1)
If you name the distances h1,h2,h3,h4,h5 and the side length L.
Draw the five triangles that form from the point.
The area of the pentagon = Lh1/2+Lh2/2+Lh3/2+Lh4/2+Lh5/2
Therefore the sum of the perpendiculars is 2A / L.
A and L are fixed, so the sum is fixed.
This proof works for any n-agon.

2) Just divide the circle in six equal parts. By Pigeonhole one of those parts will have two points.
The part is a slice with angle 60 degrees. The longest distance there is 1, since the two points are inside we can side their distance is less than 1.

[jay_shark]

Very nice Enrique !!

2)All you need is 6 points . You partition the circle into 6 equal slices from the center where one of the points lies on the slice line. This means there are 5 points left and no points can be directly on the left or right of the slice line . Then there must be 5 points and 4 equal slices and we're finished .