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I've been looking for counterexamples, with no success, to this for a few days running:

Suppose for a planar region $R$ with finite positive area, there is a point $x$ where all lines of rational slope, through $x$, divide $R$ into two congruent parts.

Does it follow that any line through $x$ with irrational slope also divides $R$ into two congruent parts?

My intuition says (which may be wrong) I can approximate a line with irrational slope by a sequence of lines with rational slope, each which divides our region into two congruent parts. I really think I'm missing big things in this approach.

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1 Answer 1

Here's our setup: We have the plane $\mathbb R^2$, the topological group $G$ of congruence maps $$(x,y)\mapsto(ax+by+u,cx+dy+v)$$ is isomorphic to the subgroup of $GL_3(\mathbb R)$ consisting of matrices $$\begin{pmatrix}a&b&u\\c&d&v\\0&0&1\end{pmatrix} $$ and thus is a semidirect product of the compact group $O_2(\mathbb R)$ and $\mathbb R^2$. The obstacle we have here is that the factor $\mathbb R^2$ is not compact.

For nonzero $z\in\mathbb R^2$, especially for $z\in S^1$, the line $z^\perp=\{x\in\mathbb R^2\mid \langle x,z\rangle=0\}$ divides the plane into two half planes $$H_+(z)=\{x\in\mathbb R^2\mid \langle x,z\rangle>0\} \qquad\text{and}\qquad H_-(z)=\{x\in\mathbb R^2\mid \langle x,z\rangle<0\}.$$ Now the claim of the problem statement follows from the

Proposition. Let $Q$ be a dense subset of $S^1$. Let $R\subseteq \mathbb R^2$ be a nonempty open set of finite measure. For each $z\in Q$, let $\alpha_z\in G$ be a congruence with $\alpha_z(R\cap H_+(z))=R\cap H_-(z)$. Then for each $z\in S^1$, there exists $\alpha_z\in G$ with $\alpha_z(R\cap H_+(z))=R\cap H_-(z)$.

Proof: (Warning: $Q\to G, z\mapsto\alpha_z$ is one of the few things in this discussion that is not continuous). We may assume wlog. that $Q=-Q$ and that $\alpha_{-z}=\alpha_z^{-1}$ for $z\in Q$.

Assume that $\{\lVert \alpha_z(0)\rVert\mid z\in Q\}$ is not bounded from above. Then there exists a sequene $(z_n)_{n\in\mathbb N}$ in $Q$ such that $\lVert \alpha_{z_n}(0)\rVert\to\infty$. As $S^1$ is compact we may assume (by switching to a convergent subsequence if necessary) that $z_n\to\zeta$ for some $\zeta\in S^1$. Select $x\in R\cap H_+(\zeta)$. Then a ball $B_{\epsilon}(x)$ is in $R\cap H_+(\zeta)$ and the smaller ball $B_{\epsilon/2}(x)$ is in $R\cap H_+(z_n)$ for almost all $n$, hence for all $n$ if we switch to a subsequence. For $\alpha\in G$, we have $\lVert \alpha(x)\rVert\ge\lVert \alpha(0)\rVert-\rVert x\rVert$. Thus if we switch to a subsequence such that $\lVert \alpha_{z_{n+1}}(0)\rVert>\lVert \alpha_{z_n}(0)\rVert+2\lVert x\rVert+\epsilon$, we ensure that the open balls $\alpha_{z_n}(B_{\epsilon/2}(x))=B_{\epsilon/2}(\alpha_{z_n}(x))$ are pairwise disjoint subsets of $R$. Since each of these balls has the same positive measure $\frac{\pi\epsilon^2}4 $, we get a contradiction to the finiteness of the measure of $R$. Therefore, $\{\lVert \alpha_z(0)\rVert\mid z\in Q\}$ is bounded and hence all $\alpha_z$, $z\in Q$, are in some compact subset $K\subset G$ (of the form $K=O_2(\mathbb R )\times [-M,M]^2$).

Now let $\zeta\in S^1$ be arbitrary. As $Q$ is dense, there is a sequence $z_n\to\zeta$ with $z_n\in Q$. As $K$ is compact, we may assume wlog. (i.e. by switching to a subsequence) that the corresponding sequence $\alpha_n$ converges to some $\alpha\in K\subset G$. Let $x\in R\cap H_+(\zeta)$ be arbitrary. As above, some $B_\epsilon(x)$ is in $R\cap H_+(\zeta)$ and $B_{\epsilon/2}(x)$ is in $H_+(z_n)$ for almost all $n$. For $n$ big enough, $\alpha(x)\in B_{\epsilon/3}(\alpha_{z_n}(x))$, hence $$B_{\epsilon/6}(\alpha(x))\subset B_{\epsilon/2}(\alpha_{z_n}(x)) \subset \alpha_{z_n}(R\cap H_+(z_n))=R\cap H_-(z_n).$$ We conclude $B_{\epsilon/6}(\alpha(x))\subset R\cap \overline{H_-(\zeta)}$ and hence $\alpha(x)\in R\cap H_-(\zeta)$. This shows $\alpha(R\cap H_+(\zeta))\subseteq R\cap H_-(\zeta)$. By considering the sequence $-z_n\to-\zeta$, $\alpha_{-z_n}\to\alpha^{-1}$, we similarly obtain $\alpha^{-1}(R\cap H_-(\zeta))\subseteq R\cap H_+(\zeta)$ and ultimately conclude $$ \alpha(R\cap H_+(\zeta))= R\cap H_-(\zeta)$$ as desired.$_\square$

Remark: The finiteness of $\mu(R)$ is essential as the following example shows (though it uses a different dense subset $Q\subset S^1$ instead of "all rational slopes"): Let $\Lambda=\mathbb Z^2+(0,\sqrt 2)$ be the standard lattice, but irrationally displaced. Let $R=\mathbb R^2\setminus \Lambda$. For every line passing through a lattice point of $\Lambda$, the 180° rotation around that point is a congruence between the parts of $R$ defined by this line. On the other hand, any congruence between half plane parts of $R$ gives us a congruence of the lattice and each of these (apart from translation and skew translations, which can be excluded for our problem) has a fixed point in $\frac12\mathbb Z^2+(0,\sqrt 2)$. This fixedpoint must be on the common boundary, i.e. the dividing line. Hence any of th euncountably many lines avoiding $\frac12\mathbb Z^2+(0,\sqrt 2)$ does not allow a congruence among the halves of $R$ it defines.

Openness of $R$ is also essential as the example $R=\{e^{i\pi q}\mid q\in\mathbb Q\}$ readily shows.

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I note that you have used the hypothesis about the area being finite. Is there a counterexample with infinite area? –  Gerry Myerson May 4 '13 at 13:07
    
In the second paragraph, how is it that an unbounded $||(u_n,v_n)||$ gives a sequence $(r_n,s_n)$ that converges to a point $(\rho,\sigma)\in S^1$? –  a a May 6 '13 at 15:14
    
@aa The points $(r_n,s_n)$ are on $S^1$, which is compact. So start with a sequence such that $\lVert(u_n,v_n)\rVert\to \infty$; the corresponding sequenec on $S^1$ must have at least one limit point and by switching to a subsequence, converges. –  Hagen von Eitzen May 6 '13 at 16:43
    
@GerryMyerson For a set it may be easy, for a region I'm not sure. –  Hagen von Eitzen May 6 '13 at 16:44

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