, 2011

Unless otherwise stated, throughout the proof indices take on values from $0$ to $5$ and are reduced modulo $6$. Label the vertices of the hexagon in circular order, $A_0, A_1, \dots, A_5$, and let the lines of support of the alternate sides $A_iA_{i+1}$ and $A_{i+2}A_{i+3}$ meet at $B_i$. To show that the area of at least one of the triangles $B_0B_2B_4$, $B_1B_3B_5$ is greater than or equal to $\frac{3}{2}$, it is sufficient to prove that the total area of the six triangles $A_{i+1}B_iA_{i+2}$ is at least $1$:
$$\sum_{i=0}^{5} \text{area } A_{i+1}B_{i}A_{i+2} \geq 1.$$
To begin with, reflect each $B_i$ through the midpoint of the segment $A_{i+1}A_{i+2}$ to get the points $B'_i$. We shall prove that the six triangles $A_{i+1}B'_iA_{i+2}$ cover the hexagon. To this end, reflect $A_{2i+1}$ through the midpoint of the segment $A_{2i}A_{2i+2}$ to get the points $A'_{2i+1}$, $i=0,1,2$. The hexagon splits into three parallelograms, $A_{2i}A_{2i+1}A_{2i+2}A'_{2i+1}$, $i=0,1,2$, and a (possibly degenerate) triangle, $A'_1A'_3A'_5$. Notice first that each parallelogram $A_{2i}A_{2i+1}A_{2i+2}A'_{2i+1}$ is covered by the pair of triangles $(A_{2i}B'_{2i+5}A_{2i+1}, A_{2i+1}B'_{2i}A_{2i+2})$, $i=0,1,2$. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle $A'_1A'_3A'_5$. To this end, it is sufficient to prove that $A_{2i}B'_{2i+5} \geq A_{2i}A'_{2i+5}$ and $A_{2j+2}B'_{2j} \geq A_{2j+2}A'_{2j+3}$ for some indices $i, j \in \{0,1,2\}$. To establish the first inequality, notice that
$$A_{2i}B'_{2i+5} = A_{2i+1}B_{2i+5}, \quad A_{2i}A'_{2i+5} = A_{2i+4}A_{2i+5}, \quad i=0,1,2, \\ \frac{A_1B_5}{A_4A_5} = \frac{A_0B_5}{A_5B_3} \quad \text{and} \quad \frac{A_3B_1}{A_0A_1} = \frac{A_2A_3}{A_0B_5},$$
to get
$$\prod_{i=0}^{2} \frac{A_{2i}B'_{2i+5}}{A_{2i}A'_{2i+5}} = 1.$$
Similarly,
$$\prod_{j=0}^{2} \frac{A_{2j+2}B'_{2j}}{A_{2j+2}A'_{2j+3}} = 1,$$
whence the conclusion.