if I'm right, $m$ must be $\frac{3n(n-1)}{2}$ and $d$ must be $\frac{3(n-2)(n-3)}{2}$, so their difference is $6n-9$. P.S: It was easy,..... I think I've lost something on conditions

goodar2006 wrote: if I'm right, $m$ must be $\frac{3n(n-1)}{2}$ and $d$ must be $\frac{3(n-2)(n-3)}{2}$, so their difference is $6n-9$. P.S: It was easy,..... I think I've lost something on conditions Why every question is too easy if you have correct so you think hou 're wrong? You have $3$ directions, where we have $k-1$ ways in each column for $m$ and also $3$ directions for chosing $m$ where we have only $k-2$ possibilities per column with length $k.$ So add and substract and you have yours. Other way: colour them in black and white where the coins are in black. Than every white triangle can sit in $3$ m-rhombuses. It is your well-known formula to count the white triangles. Similar: every white triangle not touching the circumference, sits in $3$ d-rhombuses by a bit dividing. If we write that out, it is a nice problem/solution.

The above post being almost incomprehensible, a little precision. The coloring idea only complicates the issue, when direct counting is so straightforward. It is enough to count the rhombi having their large diagonal perpendicular to one side of the large triangle - then multiply by $3$. So consider the horizontal side, and then the other $n-1$ horizontal dividing lines, plus a horizontal line through the apex. Index them by $1,2,\ldots,n$, from top down (the basis remains unindexed). Then each indexed $k$ line contains exactly $k$ nodes of this triangular lattice. The top vertex of any rhombus containig two small triangles, and with its large diagonal vertical, may (and has to) be any of the nodes of this triangular lattice found on the top $n-1$ lines, so in all there are $m/3 = 1+2+\cdots +(n-1) = \dfrac {(n-1)n} {2}$ such rhombi. The top vertex of any rhombus containig eight small triangles, and with its large diagonal vertical, may (and has to) be any of the nodes of this triangular lattice found on the top $n-3$ lines, so in all there are $d/3 = 1+2+\cdots +(n-3) = \dfrac {(n-3)(n-2)} {2}$ such rhombi. I totally disagree with presenting a problem under this type of requirement, which suggests $m$ and $d$ do not have a nice closed formula, but $m-d$ does - when in fact these values are so directly computable. So why not ask for $\dfrac {m} {d}$ instead, which has the extra built-in advantage that now the rotational symmetry that leads to multiplying by $3$ the formulae above is irrelevant? At least problem 5 at the 2008 IMO in Spain did ask for the value of $\dfrac {N} {M}$, where $N$ and $M$ where computable in some combinatorial setting, but their formulae were quite nasty, however their ratio was a power of $2$.

Eukleidis wrote: Problem 3. Let $n>3$ be a positive integer. Equilateral triangle ABC is divided into $n^2$ smaller congruent equilateral triangles (with sides parallel to its sides). Let $m$ be the number of rhombuses that contain two small equilateral triangles and $d$ the number of rhombuses that contain eight small equilateral triangles. Find the difference $m-d$ in terms of $n$. For $m$: Just look at the $k$-th column, and find number $M_k$ of rhombuses which have diagonal parallel with one of sides of triangle $ABC$ (which is trivial to find). Then $m=3\cdot \sum_{i=1}^{n}M_i$. For $d$: Look at $k$-th and $k+1$-th column, and find number $D_k$ of bigger rhombuses which have diagonal parallel with one of sides of triangle $ABC$. Then $d=3\cdot \sum_{i=2}^{n-1}D_i$.

Each rhombus is uniquely determined by the one diagonal that is parallel to the base. There are 3 options for how the diagonal is oriented to be parallel towards which side (or how the rhombus is oriented), and that diagonal for m must be side length 1. Our diagonal has 1 option in the first row, 2 in the second, ... n in the last, but the last row cannot have a rhombus, since the triangles making it up would need to be below the large triangle. So we have $3(1+2+...+n-1)=3(n-1)n/2=m$ such rhombi. For eight triangles, we need diagonals of length 2, and we need two triangles' heights to exist below the base, so it is up to the 3rd to last base of the big triangle. So our diagonal of the rhombus can lie on a row of length 2 up to length n-2 (third to last row), so that is n-3 options for the last row, meaning we have $3(1+2+...+n-3)=3(n-3)(n-2)/2=d$. Then $\boxed{m-d=6n-9}$.

2011 JBMO #3 wrote: Let $n> 3$ be a positive integer. Equilateral triangle ABC is divided into $n^2$ smaller congruent equilateral triangles (with sides parallel to its sides). Let $m$ be the number of rhombuses that contain two small equilateral triangles and $d$ the number of rhombuses that contain eight small equilateral triangles. Find the difference $m-d$ in terms of $n$.