IMO $2002$ Problem $1$ (C$1$) 


Hi! I don't understand this solution. Specifically, when it says "with all the blue rows and columns unchanged except that the values of $a_x$ and $b_y$ DECREASE by $1$", could someone explain why this happens? why does $x + y$ needs to be maximal?
When the point is changed to blue, doesn't $a_x$ and $b_y$ are added $1$?
 A: Firstly, the solution clearly appears to have a typo. $a_x$ and $b_y$ should be incremented by 1 since an additional blue point is being introduced to both column $x$ and row $y$.
Regarding your second question, we need $x + y$ to be maximal to guarantee all points directly above $(x, y)$ are blue, and all points directly to the right of $(x, y)$ are blue. To see why this is true, notice if there was a red point $(x’, y’)$ directly above or to the right of $(x, y)$, $x’ + y’ > x + y$, contradicting the fact $x + y$ is maximal. Once we’ve chosen one such point where $x + y$ is maximal, we have $a_x = b_y = n - 1 - x - y$ (this can easily be tested geometrically), allowing us to complete the induction.
A: (Please reedit your question)
You can see it almost visually. Let's see an example:
Let $n=10$. Then we are interested in the set of all points $(x,y)$ below the line $x+y=10$.
If we get the point $A=(4,2)$:

*

*How are the points of that set with the same $x$ coordinate but with bigger $y$ coordinate? Easy: they are $(4,3), (4,4), (4,5)$ ($(4,6)$ doesn't work because $4+6=10$). Total 4.

*How are the points of that set with the same $y$ coordinate but with bigger $x$ coordinate? Easy: they are $(4,3), (5,3), (6,3)$($(7,3)$ doesn't work because $7+3=10$). Total 4.

Now generalizing:
We are interested in the set of all points $(x,y)$ below the line $x+y=N$.
If we get a point $(p,q)$:

*

*How are the points of that set with the same $x$ coordinate but with bigger $y$ coordinate? Easy: they are $(p,q+1), (p,q+2), \ldots, (p,N-1-p)$ ($(p,N-p)$ doesn't work because $p+N-p=N$). Total $N-1-p-q$.

*How are the points of that set with the same $y$ coordinate but with bigger $x$ coordinate? Easy: they are $(p+1,q), (p+2,q), \ldots, (N-1-q,q)$($(N-q,q)$ doesn't work because $N-q+q=N$). Total $N-1-q-p$.

Now, with that $(p,q)-lemma$, the problem becomes a bit more obvious. If $(x,y)$ is a red point with $x+y$ maximum, then there is no red point $(x+k',y+k'')$ with $k'>0$ or $k''>0$ (lest it would contradict the maximality of $x+y$); then, all the points $(x+k',y)$ are blue and all the points $(x,y+k'')$ are blue too.
By the lemma, both sets have the same size. And that's it! Now your solution follows easily!

