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$2n$ objects of each of three kinds are given to two persons, so that each person gets $3n$ objects. Prove that this can be done in $3n^2 + 3n + 1$ways.

My attempt :

$x+y+z=3n$ and $0\le x,y,z\le 2n$

Number of solutions $= \binom {3n+2}{2}-3\binom {n+1}{2}=3n^2+3n+1$

Aliter:

$x+y+z=3n $ and $0\le x,y,z\le 2n$ $\iff n\le x+y\le 3n, 0\le x,y\le 2n$

Now plot the inequalities and count lattice points!! the required lattice points falls inside or boundaries of two squares and two triangles. $= 2\frac{(n+1)(n+2)}{2 }+2(n+1)^2-2(2n+1)-1= 3n^2+3n+1.$

Aliter:

If the first person gets $n − p (p\le\mathbb{N}_0)$ objects of the first kind, then the person can get $p$ to $2n$ objects of the second kind. The remaining ones are objects of the third kind. The sum is $\sum_{p=0}^{n}(2n+1-p)=(2n + 1)(n + 1) −\frac {n(n+1)}{2}$

If the first person gets $n + q (q\in\mathbb{N}) $objects of the first kind, then the person gets $0$to $(2n − q)$ objects of the second kind, since the person gets $3n$ objects altogether. The sum is $\sum_{q=1}^{n}(2n − q + 1)=n(2n + 1) − \frac {n(n+1)}{2}$ So total $=(2n+1)(n+1)+n(2n+1)−n(n+1)=3n^2+3n+1.$

I want clarification in following solution:

The result $3n^2+ 3n + 1 = (n + 1)^3 - n^3$ is striking and allows a geometrical interpretation. One person gets $x+y+z = 3n$ objects with $ 0 \le x, y, z \le 2n.$ These are triangular coordinates for an equilateral triangle with altitude $3n. x, y, z $ can be interpreted as lattice points (make a figure). The hexagon in the figure can be interpreted as the projection of the cube with edge $n + 1$ from which a cube of edge $n$ is subtracted. This solution is due to Martin Härterich, a gold medallist of the IMO 1987/89.

Makar
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1 Answers1

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Problem

We want the solution to $x+y+z=3n$ subject to $0\leq x,y,z\leq 2n$.

The Hexagon in Question

The first equation is a planar equation in 3D space while the latter is a cube. Their intersection is the following regular hexagon:

Intersection in 3D Space

There are $n+1$ lattice points along each edge. We need to count the number of lattice points on this hexagon.

The Cube in Question

Now let's draw the centre of the hexagon, and connect three of its vertices to the centre. Do you see three sides of a cube in the following?

Hexagon Becomes Cube

acat3
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  • Now I got it !! By looking at a corner of the cube along the body diagonal through that corner and taking the projection of the three faces meeting at the opposite corner, we see the regular hexagon as projection of three faces ; After removing $^3$ from an $(+1)^3$ we left with the three faces meeting at the opposite corner, and we can map their lattice points to the lattice points of the hexagon under discussion!! – Makar Jan 09 '24 at 19:57