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\homework{13: Filling Space}{Due Thursday, Week 7, at the start of class.}

These are three problems I once received in the same set in undergraduate; they're three of my favorite problems, and I've been waiting all quarter for the ability to give you them as a set.

Solve \textbf{one} of the following \textbf{three} problems.  As always, prove your claims!  (And have fun!)



\begin{enumerate}
\item[0.] Solve any un-signed-up-for problems from HW\#12!
\item  A \textbf{t-shape} is any pair of straight closed line segments in $\mathbb{R}^2$ that intersect at exactly one point, with this intersection not occurring at an endpoint of one of these line segments. 

A \textbf{covering} of $\mathbb{R}^2$ by disjoint t-shapes is any collection of t-shapes such that every point of $\mathbb{R}^2$ is in exactly one t-shape.  

Can such a covering exist?
\item A \textbf{circle} in $\mathbb{R}^2$ is exactly what you think it is: the collection of all points that are exactly distance $r$ away from some point $(a,b)$ in space, for some value $r>0$ and point $(a,b)$.  Note that $r$ needs to be strictly positive, as a single point is not typically considered as a ``circle.''

Is it possible to create a covering of $\mathbb{R}^2$ by disjoint circles?  

\item The concepts of circles and coverings both make sense in $\mathbb{R}^3$ as well.  A circle in $\mathbb{R}^3$ given by a point $(a,b,c)$, a plane passing through this point, and a distance $r>0$ is just all of the points that are distance $r$ from $(a,b,c)$ and contained within our plane.  Again, note that $r$ needs to be strictly positive, as a single point is not typically considered as a ``circle.''  

Similarly, a covering of $\mathbb{R}^3$ by disjoint circles is just any collection of disjoint circles such that every point in $\mathbb{R}^3$ is in exactly one circle.

Can such a covering exist?
\end{enumerate}





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