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Come across the following question. By the use of methods from topology, this question is straightforward. However, how to approach the question without the use of higher mathematics:

By the use of the methods from elementary geometry (i.e., by the use of the results from Euclidean geometry from high school) and/or by other areas of elementary mathematics prove that the plane cannot be covered by finite number of copies of letter Y.

Note: By "letter Y", it is meant that letter Y consists of 3 closed segments that have common point.

Any help/advices/solutions very appreciated!

Blue
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    You need to define what you mean by "letter Y" – David Quinn Nov 12 '16 at 20:07
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    Also need to define "cover". Can the plane be "covered" with a finite number of squares or circles, for example? – dxiv Nov 12 '16 at 20:19
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    Also, you should explain what do you mean by a "copy". – Moishe Kohan Nov 12 '16 at 20:22
  • To elaborate on the comments above: (1) Do the arms of the "Y" extend infinitely? (2) Am I allowed to rotate the "Y" or is it in a fixed orientation? – Ted Nov 12 '16 at 20:24
  • Sorry for no responses. "By letter Y" in the question, Y is meant to consist of 3 closed segments with common end. –  Nov 12 '16 at 21:53
  • A bit more challenging problem would be to prove the statement for countably many Y's in the plane (without invoking measure theoretic arguments or Baire category). – Moishe Kohan Nov 12 '16 at 22:10
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    The actual question is to show that $R^2$ cannot be represented as a disjoint union of a family of subsets each homeomorphic to the letter $Y$. See http://mathoverflow.net/questions/27244/how-many-tacks-fit-in-the-plane – Moishe Kohan Nov 12 '16 at 22:19

2 Answers2

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You would have to define exactly what you mean by "letter Y".

If you define it so that every copy of the letter Y fits into a circle, there is the following easy proof:

Suppose the plane was covered by finitely many copies of the letter Y. Then, for each copy of the letter Y, we can find a circle such that the letter is completely inside the circle.

Then, I will show that all those circles fit into a circle: Let $x$ be an arbitrary point in the plane. Since there are only finitely many circles, we can choose some number $r$ such that for every circle $C$, $r$ is greater than $r_C + d_C$, where $r_C$ is the radius of $C$ and $d_C$ is the distance of $x$ to the center of $C$. Then, each circle $C$ is inside the circle $D$ of radius $r$ around $x$: Each point inside $C$ has a distance of at most $r_C$ to the center of $C$, which has a distance of $d_C$ to $x$, so each point inside $C$ has a distance of at most $r_C + d_C$ to $x$.

But of course, there is a point outside $D$. But this point can't be covered by any of the copies of the letter Y, since they are all inside $D$. Contradiction!

dankness
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  • Sorry for no responses. "By letter Y" in the question, Y is meant to consist of 3 closed segments with common end. –  Nov 12 '16 at 21:55
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Assuming your figure is bounded (note: this is general, it doesn't have to be a 'Y'), then it has a rightmost vertical line to its left (i.e none of the figure extends to the left of this line, and no vertical line right of this line has this property). Call this the left bounding line. If there are only finitely many instances of your figure, then there are only finitely many left bounding lines. so one of them must be the leftmost. No part of the plane to the left of this line can be covered by any of your figures.

PMar
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