Cono Sur Mathematical Olympiad

We divide each triangle $S_i$ into four congruent triangles $T_i$, $A_i$, $B_i$, $C_i$. Let $A$ be the set of all points that belong to some $A_i$ but don't belong to any $T_i$. We define $B$ and $C$ analogously. It is clear then that $S=T\cup A\cup B\cup C$. Next we will prove that $\text{area}(A)\le \text{area}(T)$.

Denote ${\ell }_i$ the common line between $A_i$ and $T_i$. Without loss of generality we may assume that the triangles are numbered in such a way that whenever $i<j$, either ${\ell }_i={\ell }_j$ or ${\ell }_i$ is above ${\ell }_j$. For each $i$, let $$A(i)=A_i-(T\cup A_1\cup \cdots \cup A_{i-1}),$$ and denote $T(i)$ the reflection of $A(i)$ with respect to ${\ell }_i$. We claim that: Sets $A(1),A(2),\dots ,A(n)$ are pairwise disjoint and their union equals $A$. $T(i)$ is a subset of $T_i$. * Sets $T(1),T(2),\dots ,T(n)$ are pairwise disjoint. The first claim holds by definition. The second one holds because $A(i)$ is a subset of $A_i$, and $T_i$ is the reflection of $A_i$. Now we prove the third claim. Assume there is a point $x\in T(i)\cap T(j)$ for some $i<j$. Denote by $x_i$ and $x_j$ the symmetric points of $x$ with respect to ${\ell }_i$ and ${\ell }_j$ respectively. By construction we know that $x_i\in A(i)$ and $x_j\in A(j)$. Notice that $x,x_i,x_j$ lie on a line that is perpendicular to ${\ell }_i$ and ${\ell }_j$. On the other hand, because of the way the lines were labelled, we know that the midpoint of $x{x}_i$ either is equal to or is located above the midpoint of $x{x}_j$. Therefore $x_j$ lies on the segment $x{x}_i$. But then $x_j$ belongs to $A_i$ or $T_i$, which is impossible, since $x_j\in A(j)$.

After proving this, we have that $$\begin{array}{rl}\text{area}(A)& =\text{area}(A(1))+\cdots +\text{area}(A(n))\\ & =\text{area}(T(1))+\cdots +\text{area}(T(n))\\ & =\text{area}(T(1)\cup \cdots \cup T(n))\\ & \le \text{area}(T),\end{array}$$ as claimed. In the same way we can show that $\text{area}(B)\le \text{area}(T)$ and $\text{area}(C)\le \text{area}(T)$. Finally, $$\begin{array}{rl}\text{area}(S)& =\text{area}(T\cup A\cup B\cup C)\le \text{area}(T)+\text{area}(A)+\text{area}(B)+\text{area}(C)\\ & \le 4\cdot \text{area}(T),\text{and we are done.}\end{array}$$