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Let $$A=\{(x,y)\in \mathbb R^2\mid x^2+y^2\in \mathbb Z\},$$ $$G=\{(x,y)\in \mathbb R^2\mid y=|x|\},$$ $$B=\{(x,y)\in \mathbb R^2\mid d_2((x,y),G)\in\mathbb Z\}.$$ $f:A\to B$ is arbitrary continuous function. We say that two points from $A$ form a "happy couple" if they are centrally symmetric to the origin and have the same image by $f$. Prove that A contains at least 2013 "happy couples". Any kind of help is welcome. |
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$A$ consists of the origin, plus a heap of concentric circles $C_n$. We're going to focus on the first one, $C_1$ (which happens to be the unit circle), and show that it contains a happy couple. Then we can point out that every other circle of $A$ contains a happy couple, and since $A$ contains at least $2013$ circles, then there must be at least $2013$ happy couples. We will define a function $g_1:C_1 \to \Bbb R$ which will help us find the happy couple. First we need to see what $B$ looks like. $B$ is just a series of copies of $G$, transposed vertically by some integer multiple of $\frac{\sqrt{2}}{2}$. Now, since $f$ is continuous and $C_1$ is connected, $f$ sends $C_1$ to some connected component of $B$. The important thing to note here is that a connected component of $B$ is the graph of a function on $\Bbb R$. So if two points on a connected component of $B$ has the same $x$-coordinate, they are the same point. This is how we'll define $g_1$: Take a point $p$ on $C_1$, and its antipodal $p'$. Let $f(p)$ be the $x$-coordinate of $f(p)$ minus the $x$-coordinate of $f(p')$. If $g_1(p) = 0$, then $p$ and its antipodal will have the same image in $f$ by the argument above, and thus be a happy couple. Is there such a point? Well, let's look at $g_1((0, 1))$. If it is $0$, then rejoice! If not, then note that $g_1((0, 1)) = -g_1((0, -1))$. So as we move along $C_1$ from the positive $y$-axis-intersection to the negative $y$-axis-intersection, $g_1$ goes either from positive to negative, or from negative to positive. That means that it has to be $0$ somewhere in-between (by the intermediate-value theorem), and we have a happy couple of $C_1$. This argument goes for any circle of $A$, so we are done. |
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