shortlistBMO 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 $A_i{A}_{i+1}$ and $A_{i+2}{A}_{i+3}$ meet at $B_i$. To show that the area of at least one of the triangles $B_0{B}_2{B}_4$, $B_1{B}_3{B}_5$ is greater than or equal to $\frac{3}{2}$, it is sufficient to prove that the total area of the six triangles $A_{i+1}{B}_i{A}_{i+2}$ is at least $1$: $$\sum _{i=0}^5\text{area }{A}_{i+1}{B}_i{A}_{i+2}\ge 1.$$ To begin with, reflect each $B_i$ through the midpoint of the segment $A_{i+1}{A}_{i+2}$ to get the points $B_i^{\prime }$. We shall prove that the six triangles $A_{i+1}{B}_i^{\prime }{A}_{i+2}$ 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 $A_{2i+1}^{\prime }$, $i=0,1,2$. The hexagon splits into three parallelograms, $A_{2i}{A}_{2i+1}{A}_{2i+2}{A}_{2i+1}^{\prime }$, $i=0,1,2$, and a (possibly degenerate) triangle, $A_1^{\prime }{A}_3^{\prime }{A}_5^{\prime }$. Notice first that each parallelogram $A_{2i}{A}_{2i+1}{A}_{2i+2}{A}_{2i+1}^{\prime }$ is covered by the pair of triangles $(A_{2i}{B}_{2i+5}^{\prime }{A}_{2i+1},A_{2i+1}{B}_{2i}^{\prime }{A}_{2i+2})$, $i=0,1,2$. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle $A_1^{\prime }{A}_3^{\prime }{A}_5^{\prime }$. To this end, it is sufficient to prove that $A_{2i}{B}_{2i+5}^{\prime }\ge A_{2i}{A}_{2i+5}^{\prime }$ and $A_{2j+2}{B}_{2j}^{\prime }\ge A_{2j+2}{A}_{2j+3}^{\prime }$ for some indices $i,j\in \{0,1,2\}$. To establish the first inequality, notice that $$\begin{gathered} A_{2i}{B}_{2i+5}^{\prime }=A_{2i+1}{B}_{2i+5},A_{2i}{A}_{2i+5}^{\prime }=A_{2i+4}{A}_{2i+5},i=0,1,2, \\ \frac{A_1{B}_5}{A_4{A}_5}=\frac{A_0{B}_5}{A_5{B}_3}\text{and}\frac{A_3{B}_1}{A_0{A}_1}=\frac{A_2{A}_3}{A_0{B}_5}, \end{gathered}$$ to get $$\prod _{i=0}^2\frac{A_{2i}{B}_{2i+5}^{\prime }}{A_{2i}{A}_{2i+5}^{\prime }}=1.$$ Similarly, $$\prod _{j=0}^2\frac{A_{2j+2}{B}_{2j}^{\prime }}{A_{2j+2}{A}_{2j+3}^{\prime }}=1,$$ whence the conclusion.