Problem 6

Assign to each side

b

of a convex polygon

P

the maximum area of a triangle that has

b

as a side and is contained in

P

. Show that the sum of the areas assigned to the sides of

P

is at least twice the area of

P

Step 1 of 5: Cover an almost-convex

2m

-gon by

m

butterflies

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In plain words

As the oriented main diagonal $P_iP_{i+m}$ steps from $i=0$ to $i=m$ , it flips from $P_0P_m$ to $P_mP_0$ , so every point inside the polygon must be swept across at least once by the bowtie between two consecutive main diagonals.

B_i=\triangle P_iP_{i+1}O_i\cup\triangle P_{i+m}P_{i+m+1}O_i,\quad O_i=P_iP_{i+m}\cap P_{i+1}P_{i+m+1}\quad(i=0,\dots,m-1)

Detailed analysis

Call a polygon almost convex if all its interior angles are $\le 180^\circ$ (so it is a convex polygon with some extra vertices placed along its sides). Label the vertices of an almost-convex $2m$ -gon $Q$ of area $S$ as $P_0,P_1,\dots,P_{2m-1}$ (indices mod $2m$ ). For $i=0,\dots,m-1$ , the two main diagonals $P_iP_{i+m}$ and $P_{i+1}P_{i+m+1}$ cross at a point $O_i\in Q$ and bound a self-intersecting butterfly $B_i=\triangle P_iP_{i+1}O_i\cup\triangle P_{i+m}P_{i+m+1}O_i$ . As $i$ advances from $0$ to $m$ , the oriented line $P_iP_{i+m}$ reverses orientation from $P_0P_m$ to $P_mP_0$ , so every interior point $X\in Q$ switches sides at some step $i$ and therefore lies in $B_i$ . Thus $Q=\bigcup_{i=0}^{m-1}B_i$ .