Can the triangles be of different sizes?

R8kt wrote: Can the triangles be of different sizes? Sure, why not?

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My thoughts: 1. $area(S_i\cup S_j) \le 4\cdot area(T_i\cup T_j) \Leftrightarrow area(S_i\cap S_j) \ge 4\cdot area(T_i\cap T_j)$ 2. If $t = T_i \cap T_j$ then $t$ it is an equilateral triangle 3. Define $t’$ as the equilateral triangle such that $t$ is your medial triangle, then: $t \in T_i\cap T_j \Rightarrow t’ \in S_i \cap S_j$ Lets prove each step: Step 1. $$ S = S_1 \cup S_2 \cup ... \cup S_n $$$$ T = T_1 \cup T_2 \cup ... \cup T_n $$since that, $$ T_i \subset S_i \therefore S_i \cap (S/S_i) = \emptyset \rightarrow T_i \cap (T/T_i) = \emptyset $$$$ area(S_i) = 4\cdot area(T_i) $$we only need to consider $$ S_i \ \ \text{such that} \ \ S_i \cap S/S_i \neq \emptyset $$Note that: $$ area(S_i\cup S_j) = area(S_i)+area(S_j)-area(S_i\cap S_j) $$Then: $$ area(S_i\cup S_j) \leq 4\cdot area(T_i\cup T_j) \Leftrightarrow area(S_i)+area(S_j)-area(S_i\cap S_j)\leq 4\cdot area(T_i)+4\cdot area(T_j)-4\cdot area(T_i\cap T_j) \Leftrightarrow area(S_i\cap S_j) \geq 4 \cdot area(T_i\cap T_j) $$Step 2. Note that, as for each $T_i$ has parallel sides. If $l, r$ are sides of any triangle in $T$, then $\angle (l, r)\in \{0^{\circ}, 60^{\circ}\}$, then if $t = T_i\cap T_j$, since that $T_i, T_j$ are convex then $t$ is convex, $ \Rightarrow t$ it is equilateral Step 3. $t = T_i \cap T_j \rightarrow t \in T_i, T_j \rightarrow t’ \in T’_i, T’_j = S_i, S_j \rightarrow t’ \in S_i, S_j \rightarrow t’\in S_i \cap S_j$ By the end: $t’\in S_i\cap S_j \rightarrow area(S_i\cap S_j) \ge area(t’) = 4\cdot area(t) = 4\cdot area(T_i\cap T_j) \square$