Omid Hatami wrote: A $ 10\times10\times10$ cube has $ 1000$ unit cubes. $ 500$ of them are coloured black and $ 500$ of them are coloured white. Show that there are at least $ 100$ unit squares, being the common face of a white and a black unit cube. Uhhh, i don't understand what we should to show? Can explaon a little bit more? Thanks

The extreme case which satisfy exact 100 units square should be 500 black cubics up and 500 white cubics down. The black and the white meet only at the centre face which is exactly 100 units square. Try to move any two cubics, one for black and one for white, the faces are gonna add up. Not that strict, but makes sense.

Yea, it makes sence but is kind of hard to proof that the extreme case is that... I guess that we need some strong combinatoric tactics... Notice that in each colune we have one comun face of tow diferent colored cubes or the colune has just cubes of the same color, and then we have, at least, more 1 comun sides of cubes diferente colored per neightbour colune or that colunes have cubes of the same colour. Let us turn this problem in some 2 dimentional easy problem, so let us paint of black the top of the colunes that have just black cubes, respectivelly to the white cubes and the other colunes will gives us at least more one comun side. Therefor is easy to proof that all the black "zones" and white "zones" will have twhat we want to

Any solution for this nice problem?? My solution is too long and dirty, I want a nice solution.

We can show that the number of unit squares that are the common face of two whites are at most $ 1308 $ and if the number is bigger than 1300 then we can show that the number of unit squares that are the common face of two blacks are at most $ 1200 $ . This completes the proof. p.s. Actually this proof is dirty too.

We'll first prove a two-dimensional analog. Lemma. In a $10 \times 10$ square, if there are $x \le 50$ white squares and $100-x$ black squares, there must be at least $\min (\left \lceil 2 \sqrt{x} \right \rceil, 10)$ pairs of adjacent squares of different colors. Proof. We'll call a row or column white tasty if all unit squares are white, and black tasty in an analogous case. If there is a black tasty row or column and a white tasty row or column, then it's clear that either they are both rows or both columns. WLOG they are both rows (else rotate). Then, every column contains vertically adjacent squares of different color, hence there are at least $10$ such pairs overall. Otherwise, we consider two cases. In the first case suppose that there is no white tasty row or white tasty column. Let $A, B$ be the set of rows, columns respectively which have at least one white square. Then, we know from $|A| \cdot |B| \ge x$ that $|A| + |B| \ge 2 \sqrt{x}.$ Since nothing in $A \cup B$ is white tasty, each must contain two squares of different colors. This implies that there are at least $2 \sqrt{x}$ pairs of adjacent squares with different colors. If there is no black tasty row or black tasty column, then we can use a similar argument to see that there are at least $2 \sqrt{100-x} \ge 2 \sqrt{x}$ pairs of adjacent squares with different colors. In each case, there are clearly at least $\min (\left \lceil 2 \sqrt{x} \right \rceil, 10)$ pairs of adjacent squares with different colors, so the lemma is proven. $\blacksquare$ With this lemma, we are ready to finish. Consider dividing the $10 \times 10 \times 10$ into $10$ horizontal "layers," each of which is a $10 \times 10$ square. Let them have $a_1, a_2, \cdots, a_{10}$ white squares from top to bottom. Then, observe that there are at least $\sum_{i = 1}^{9} |a_i - a_{i+1}|$ pairs of vertically adjacent squares which are of different colors. Suppose that $m = \min (a_i)$ and $M = \max(a_i).$ Then, there's at least $M - m$ pairs of vertically adjacent squares which are of different colors. Now, define the function $f$ by $f(x) = \min \left(\left \lceil 2 \sqrt{ 50 - |50-x|} \right \rceil, 10\right).$ From the lemma, there are at least $\sum_{i = 1}^{10} f(a_i)$ pairs of squares which are on the same layer, adjacent, and have different colors. The following claim suffices. Claim. For any positive integers $a_1 \le a_2 \le \cdots \le a_{10} \le 100$ with sum $500$, we have $a_{10} - a_1 + \sum_{i = 1}^{10} f(a_i) \ge 100.$ Proof. This is a straightforward bash and is left as an exercise to the reader. $\blacksquare$ We are done. $\square$