# A Problem On Several Triangles

A triangle $ T $ is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them.

For an illustration let me denote the three vertices of T by 1, 2 and 3. Now number each of the vertices of the small triangles by 1, 2, 3. Do this in an arbitrary way, but such that vertices lying on an edge of T must not be numbered by the same number as the vertex of T opposite to that edge.

Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.

# Solution

To show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3, we show that the number of small triangles whose vertices are labeled with $ 1,2,3$ is odd and thus actually $ >0$ !

We enumerate all small triangles in the picture as $ T_1$ , $ T_2, \ldots, T_n$ and denote by $ a_i$ the number of edges with endpoints $ 1$ and $ 2$ in each triangle $ T_i$ . Thus, if say the vertices of $ T_i$ are labeled by $ 1,1,2$ , then $ a_i=2$ , and so on …

Observe now that obviously we have

$ \displaystyle a_1+a_2+a_3+\cdots +a_n= A+2B, $

where $ A$ is the number of triangles whose vertices are labeled $ 1,2,3$ , while $ B$ is the number of those triangles labeled by $ 1,1,2$ or $ 1,2,2$ . (Actually it is easily seen that $ a_i=2$ for such triangles, while $ a_i=1$ if the vertices of $ T_i$ are $ 1,2,3$ and $ a_i=0$ otherwise.) All we have to show is that $ A$ is odd.

Let $ C$ denote the number of $ 12$ -edges lying inside the original triangle $ T$ and let $ D$ be the number of $ 12$ -edges lying on the boundary of $ T$ . Every interior $ 12$ -edge lies in two triangles $ T_i$ and thus it is counted twice in the sum $ a_1+a_2+a_3+\cdots +a_n$ , while every boundary $ 12$ -edge is counted only once. In conclusion we get

$ \displaystyle a_1+a_2+a_3+\cdots +a_n= 2C+D, $

which yields

$ \displaystyle A+2B=2C+D. $

Hence $ A$ is odd if and only if $ D$ is odd. It is therefore enough to show that $ D$ is odd.

According to the hypothesis of the problem, edges labeled $ 12$ or $ 21$ can occur only on the $ 12$ -edge of the large triangle $ T$ . We start walking along the edge $ 12$ of the triangle $ T$ starting at the vertex $ 1$ toward the vertex $ 2$ . Now, only when we first pass an edge labeled $ 12$ will we arrive at the first vertex labeled $ 2$ . A number of vertices labeled $ 2$ may now follow, and only after we have passed a segment $ 21$ do we reach a label $ 1$ , and so on. Thus after an odd number of segments $ 12$ or $ 21$ we arrive at vertices labeled $ 2$ , and after an even number of such segments we arrive at vertices labeled $ 1$ . Since the last vertex we will reach is the vertex $ 2$ of the big triangle $ T$ , it follows that the total number of segments $ 12$ or $ 21$ lying on the side $ 12$ of the big triangle $ T$ must be odd! The same reasoning applies for each of the other edges of the big triangle $ T$ , so we deduce that $ D$ , the total number of $ 12$ or $ 21$ -edges lying on the boundary of $ T$ , must be odd. Proved

# Graphical Proof

It is obvious. As a result of this numbering we get following diagram:

Hi, I have nominated you for ‘The versatile blogger’ – http://findin42.wordpress.com/2011/10/02/yeah-baby-ill-show-u-versatile-part-ii/

Hi, I have nominated you for ‘The versatile blogger’ – http://findin42.wordpress.com/2011/10/02/yeah-baby-ill-show-u-versatile-part-ii/