I recently solved this beautiful mathematical problem on Pigeonhole Principle from Romania TST 2000.
Let $S$ be the set of interior points of a sphere in three-dimensional space and let $D$ be the set of interior points of a disc on a plane. For any two points $X,$ and $Y$ in the space, let $d(X, Y)$ denote the distance between them. Determine if there is a function $f: S\rightarrow D$ such that $d(A, B)\leq d(f(A), f(B))$ for any two points $A, B\in S.$
What are you initial thoughts on this problem? Do you think there is such a function or there isn't. I have given my solution to this problem below. You can also post a solution if you have another one :)
We claim that such a function does not exist. To prove this claim, we will assume that there exists such a function $f$, and arrive at a contradiction.
Suppose that there exists such a function $f: S\rightarrow D,$ then we will consider a cube of side length $a$ that is completely in the sphere.
Fix a random positive integer $n$, we will partition the cube into exactly $n^3$ congruent cubes, such that these identical cubes determine $n^2$ identical squares on each face of the larger cube, and $n$ identical line segments on each edge of the larger cube.
Now, we consider the $(n+1)^3$ points that are the vertices of the $n^3$ identical cubes. Let these points be $S_1, S_2, \cdots, S_{(n+1)^3},$ we know that the distance between any two of these points is at least $\frac{a}{n}.$
Let $D_1, D_2, \cdots, D_{(n+1)^3}$ be the points on the disk $D,$ that the function $f$ maps $S_1, S_2, \cdots,$ and $S_{(n+1)^3}$ to, more specifically, let $$D_i=f(S_i)$$ for all $i\in{1,2,\cdots,(n+1)^3}.$
Since our function $f$ has the property that, for any two points $A, B\in S,$ $d(A, B)\leq d(f(A), f(B))$ holds. Hence, the pairwise distances between the any of the $D_i$ are at least $\frac{a}{n}.$
This means that we can draw circles around each of those points of radii $\frac{a}{2n}$ such that none of the $(n+1)^3$ circles intersect. Let the radius of the disk $D$ be $r,$ we know that each of these circles is contained in the disk concentric with $D$ of a radius of $r+\frac{a}{2n}.$
All these $(n+1)^3$ circles are non-intersecting, and contained in the larger disk. the sum of their areas is lesser than the area of the larger disk. Hence, for all $n\in\mathbb{N},$ we have \begin{align*} & \pi\left(\frac{a}{n}\right)^2(n+1)^3\leq\pi\left(r+\frac{a}{n}\right)^2\\ \Rightarrow & \left(\frac{a}{n}\right)^2(n+1)^3\leq\left(r+\frac{a}{n}\right)^2\\ \Rightarrow & \left(\frac{a^2}{n^2}\right)(n+1)^2(n+1)\leq\left(r+\frac{a}{n}\right)^2\\ \Rightarrow & \left(\frac{(n+1)^2}{n^2}\right)(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2\\ \Rightarrow & \left(\frac{n+1}{n}\right)^2(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2\\ \Rightarrow & \left(1+\frac{1}{n}\right)^2(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2. \end{align*}
Now we note that as $n$ tends to infinity, the term $\frac{1}{n}$ will tend to $0,$ and hence, \begin{align*} 1+\frac{1}{n} &\rightarrow 1\\ \left(1+\frac{1}{n}\right)^2 &\rightarrow 1\\ \left(1+\frac{1}{n}\right)^2(n+1)a^2 &\rightarrow\infty. \end{align*} since $n+1\rightarrow\infty.$ Hence, the left side of the inequality tends to infinity, implying that since $$\Rightarrow\left(1+\frac{1}{n}\right)^2(n+1)a^2\leq\left(r+\frac{a}{n}\right)^2,$$ the right side will also tend to infinity. But note that as $n\rightarrow\infty,$ \begin{align*} \frac{a}{n} &\rightarrow 0\\ \Rightarrow r+\frac{a}{n} &\rightarrow r\\ \Rightarrow \left(r+\frac{a}{n}\right)^2 &\rightarrow r^2. \end{align*} which is finite, hence, we have arrived a contradiction, and such a function $f$ does not exist.