geometry problem ( that nagged me )

[cyclops]

You have four points in a plane and one euro. The euro cannot cover all four points ( at once, OC ). Prove that this euro cannot cover some three of these points.

( for the blessed among us : the euro is a coin in the form of a circle. Shrinking too fast, though)

[daniel_the_smith]

Is there some additional constraint? It seems relatively easy to put 3 points in a plane that can be covered by a circle, with one far away.

Perhaps the plane is a square the same size as the coin? Edit: nah, even that doesn't help.

[entropi]

You have four points in a plane and one euro. The euro cannot cover all four points ( at once, OC ). Prove that this euro cannot cover some three of these points.

( for the blessed among us : the euro is a coin in the form of a circle. Shrinking too fast, though)

As the question is formulated, it cannot be proven because the euro can cover some three of these points even if the fourth point is too far away to be covered. But are there some constraints for the locations of the four points in the plane???

[Tryss]

I believe you should read it as : "Prove that this euro cannot cover every group of 3 points"

[Laman]

daniel_the_smith, entropi: i understand it differently - you have four points in a plane, their only property is that they cant be all covered by a single unit circle. prove that among these four points exists at least one group of three points that has the same property

EDIT: too late, Tryss wrote it first

[jts]

This is equivalent to asking whether for some scalar D there exists point a s.t. d(a,x)<D/2, d(a,y)<D/2, etc., for all the points that the "Euro" covers.

If d(a,x)<D/2, then if d(a,y)<D/2, then d(x,y)<D. (And vice-versa, of course).

If no point a exists for {w,x,y,z}, then that implies there is at least one of d(w,x), d(w,y)... etc. such that d(-,-)>D. If that is true, then the point a doesn't exist for at least one of {w, x, y}, {w, y, z}, {x, y, z}, {w, x, z}, because collectively those four triplets require d(-,-)<D for all six of the pairs.

[HermanHiddema]

If a set of points is entirely within a certain circle of radius X, then the distance between any two of them is at most 2X.

Conversely, if a set of point is not entirely within some circle of radius X, then there exist at least two points within the set whose mutual distance is larger then 2X.

The given set of four points therefore contains at least two points whose mutual distance is larger than the diameter of the euro coin. Those two points cannot, therefore, be covered simultaneously by the euro coin.

Given that there exists at least one set of two points that cannot simultaneously be covered, there also exist at least two sets of three points that cannot be covered by said euro coin.

[emeraldemon]

Conversely, if a set of point is not entirely within some circle of radius X, then there exist at least two points within the set whose mutual distance is larger then 2X.

I don't think this is true. Imagine a triangle, then the smallest circle containing that triangle. Unless the points are colinear, the largest distance between two of the points will be less than 2*radius. So if you increase the distance of those two points a tiny bit, you will have 3 points that cannot fit in the circle of that radius, but the distance is still less than 2*radius. Unless I'm misunderstanding, jts's method has the same flaw.

[Tryss]

Conversely, if a set of point is not entirely within some circle of radius X, then there exist at least two points within the set whose mutual distance is larger then 2X.

This is wrong:

Take X = 1 and the points of affixes :
A(0,sqrt(3))
B(-1,0)
C(1,0)
D(0,1-sqrt(3))

You have d(A,B) = d(A,C) = d(A,D)= d(B,C) = 2 and d(B,D) = d(C,D) = sqrt(5-2sqrt(3)) < 2, but you don't have a point a where d(a,-)<1 for each point in {w,x,y,z}. The smaller circle that cover every 4 points has a radius of 2/sqrt(3) > 1 (the 4 points are concyclics)

[cyclops]

daniel_the_smith, entropi: i understand it differently - you have four points in a plane, their only property is that they cant be all covered by a single unit circle. prove that among these four points exists at least one group of three points that has the same property

EDIT: too late, Tryss wrote it first

thx, Laman and Tryss: you both restated the problem correctly.

[Li Kao]

One interesting observation is that this is not true for 4 points in 3D. (A Tetraeder violates this).

My impression is that a square is the shape that comes closest to violating this rule, but it holds even for it. But how to prove that...

[entropi]

One interesting observation is that this is not true for 4 points in 3D. (A Tetraeder violates this).

Likewise, it is also not true for 3 points in 2D because a triangle violates it (see emeraldemon's post above).

cyclops, do you have an answer for the question? Apparently it is not as easy as it looks.

[cyclops]

cyclops, do you have an answer for the question? Apparently it is not as easy as it looks.

I got it from the book I mentioned in my previous thread about math problem. There are hints in it. I intend to publish asap.
Also there are answers. I'll publish it as soon as the discussion here fades out.
I didn't read the hint or solution yet. Although I am very curious!
I took this problem with me on my holiday trip, trekking in the Alps. I must say I got a little bit frustrated. Trying to solve it in my head didn't work.

edit: maybe some people doubt the preposition. In that case it should be easy find a counterexample. One is enough.

[perceval]

lets reverse the problem:
put a coin on each of th 4 points:
if the 4 coin overlap, you can center a coin on the overlapping region and it will cover everything.
now lets assume that you have 3 overlapping circle of the same radius R:C1,C2,C3
try to draw a 4th circle o that intersect all three circles but NOT the common region: impossible.

now to prove that mathematically..
lets assume that C1,C2, C3 have a non empty intersection , lets take a point A
lets assume that C1,C2, C4 have a non empty intersection , lets take a point B in it
lets assume that C1,C3, C4 have a non empty intersection , lets take a point C
lets assume that C2,C3, C4 have a non empty intersection , lets take a point D in it

my bet is that the barycenter of A,B,C,D will be in all 4 circles.ie it will be a valid choice.


i guess it is related to the convexity of those intersections: if A and B are in an intesection, the whole [A,B] segment is, and the barycenter of 4 point in on the semgent between all 2 by 2 barycenter.

i need to work a bit (real work that pays the bill) i ll come back

[cyclops]

I hided the book for myself, but found it again. "The green book of mathematical problems" by Kenneth Hardy and Kenneth S. Williams, 1985, ISBN 0-486-69573-5 (pbk).

Definition: A unit disk is a disk of unit radius. That is its radius is 1.

Problem 88: ( As stated in the book). If four distint points lie in the plane such that any three of them can be covered by a unit disk then all four points can be covered by a unit disk.

Hint, if you use it please hide your solution for a few days

Given 4 points in a plane one of the following is true:
1. One of the points is in or at the triangle formed by the remaining 3 points.
2. The 4 points form a quadrilateral with intersecting diagonals.

please hide your solution for a few days