We are given an equilateral triangle ABC with the length of its side equal to $1$. There are $n-1$ points on each side of the triangle $ABC$ that equally divide the side into $n$ segments. We draw all possible lines that pass through any two of all those $3(n-1)$ points such that they are parallel to one of three sides of triangle $ABC$. All such lines divide triangle $ABC$ into some lesser triangles whose vertices are called nodes. We assign a real number for each node such that the following conditions are satisfied: (I) real numbers $a,b,c$ are assigned to $A,B,C$ respectively; (II) for any rhombus that is consisted of two lesser triangles that share a common side, the sum of the numbers of vertices on its one diagonal is equal to that of vertices on the other diagonal. 1) Find the minimum distance between the node with the maximal number to the node with the minimal number; 2) Denote by $S$ the sum of the numbers of all nodes, find $S$.
Problem
Source: China Mathematical Olympiad 1987 problem2
Tags: geometry, rhombus, combinatorics unsolved, combinatorics
04.08.2015 11:15
We start playing with the condition, and sooner or later, we notice that the sum of vectors in the opposite nodes of the rhombuses are equal. This motivates us to generalize the condition: Lemma. Denote by $f(P)$ the number assigned to node $P$. Then for any subset of nodes $P_1,P_2,\dots,P_k$ with position vectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_k$ satisfying $\sum_{i=1}^k q_i\mathbf{v}_i=0$ for some real numbers $q_1,\dots,q_k$ with sum zero, we have $$\sum_{i=1}^k q_i f(P_i)=0.$$ (Note that the position vector condition is independent of the coordinate system because the sum of the $q_i$'s is zero.) Proof. We use the condition to start eliminating the points until this whole thing reduces to the $n=2$ case. (Essentially, this is induction on $k$.) Let's demonstrate how to eliminate the numbers assigned to the nodes on side $BC$ (this shows us how to eliminate the numbers row by row). Let the points on $BC$ be $B=X_0,X_1,\dots,X_n=C$ in this order, the next row be $Y_1,Y_2,\dots,Y_n$ and the row after that be $Z_1,Z_2,\dots,Z_{n-1}$. Then we can first eliminate $B$ like $f(X_0)=f(X_1)+f(Y_1)-f(Y_2)$ and $C$ in the same fashion, and after that eliminate all the $f(X_i)$ via $f(X_i)=f(Y_i)+f(Y_{i+1})-f(Z_i)$ for $1\le i\le n-1$. This clearly works if we have $\ge 3$ rows, so suppose we have reduced everything to a single equilateral triangle $XYZ$. But suppose that some reals $x+y+z=0$ satisfy $x\vec{X}+y\vec{Y}+z\vec{Z}=0$, then we have $z=-x-y$, hence this rearranges to $x\overrightarrow{ZX}+y\overrightarrow{ZY}=0$, which clearly can only hold for $x=y(=z)=0$ as $\overrightarrow{ZX}$ and $\overrightarrow{ZY}$ are linearly independent. So in this reduced case, $xf(X)+yf(Y)+zf(Z)=0$ is trivial. This proves our claim. $\blacksquare$ It is easy to finish from here. Indeed, if nodes $P,Q,R$ are such that $R$ is on segment $\overline{PQ}$, then $\vec{R}=\lambda \vec{P}+\mu \vec{Q}$ for $\lambda,\mu>0$ and $\lambda+\mu=1$, hence by the Lemma, $f(R)=\lambda f(P)+\mu f(Q)$, which is between $f(P)$ and $f(Q)$. Hence, every number assigned to all the nodes other than $A,B,C$ are strictly between the minimum and the maximum of $a,b,c$, so the answer in 1) is $1$, the diameter of $\triangle ABC$. (This assumes $a,b,c$ are pairwise distinct; if they need not be, then part 1) is trivially $\frac1n$.) For part 2), note that $\triangle ABC$ has rotational symmetry around its center $O$. If the equilateral $\triangle PQR$ also has center $O$, then $\vec{P}+\vec{Q}+\vec{R}=\vec{A}+\vec{B}+\vec{C}$ holds (it is enough to check this in the coordinate system around $O$), hence $f(P)+f(Q)+f(R)=a+b+c$ by the Lemma. Therefore if the total number of nodes is $N$, then the answer is $N\cdot \frac{a+b+c}3$. To compute $N$, just recall that by construction, $N=1+2+\dots+(n+1)=\frac{(n+1)(n+2)}2$.