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This was an example of generalized pigeon-hole principle.

Ten dots are placed within a square of unit size. The textbook then shoes a box divided into 9 equal squares. Then there three dots that can be covered by a disk of radius 0.5.

The proof:

Divide our square into four equal parts by it's diagonals (from one corner to the other), then by the generalized pigeon-hole principle, at least one of these triangles will contain three of our points. The proof follows as the radius of the circumcircle of these triangles is shorter than 0.5.

But wait! The statement said three dots can be covered by a disk of radius 0.5. Typo?

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2 Answers 2

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The proof is basically correct, but yes, there is a typo: the circumcircle of each of the four triangles has radius exactly $0.5$, not less than $0.5$. If $O$ is the centre of the square, and $A$ and $B$ are adjacent corners, the centre of the circumcircle of $\triangle AOB$ is the midpoint of $\overline{AB}$, from which the distance to each of $A,O$, and $B$ is $0.5$. The circle of radius $0.5$ and centre at the midpoint of $\overline{AB}$ contains $\triangle AOB$, as required.

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I stand corrected. –  Jonathan Christensen Dec 20 '12 at 1:30
    
Ah yes, that's what I figured. I was thinking that if it were less than, say a radius of 0.01, then you could definitely arrange the dots so that the circumcircles don't contain them. –  AlanH Dec 20 '12 at 1:53

The proof is correct. The furthest away from each other the points can be is in the three vertices of the $\sqrt 2/2,\ \sqrt 2/2,\ 1$ right triangle. This triangle has circumradius 0.5, so all three points are still covered by a disk with radius 0.5.

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You might want to change the wording: triangles don’t have radii. –  Brian M. Scott Dec 20 '12 at 1:00
    
"radius of the circumcicle" is pretty clunky, but I guess "circumradius" is reasonable. –  Jonathan Christensen Dec 20 '12 at 1:06
    
It’s nice that circumradius is available, but if it weren’t, clunkiness would still be preferable to error. (After leaving that comment, I realized that there actually is a mistake in the proof, though not a fatal one.) –  Brian M. Scott Dec 20 '12 at 1:11

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