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 3 of 5: Each butterfly is dominated by a triangle on one of its two sides

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In plain words

In a bowtie with four arm lengths $a,b,c,d$ meeting at $O_i$ , replacing the shortest arm $d$ by the adjacent arm $b\ge d$ swaps one wing for a larger triangle that combines with the other wing into a single triangle of the polygon.

d=\min(a,b,c,d)\ \Longrightarrow\ [\triangle P_iP_{i+1}P_{i+m}]=[\triangle P_iO_iP_{i+1}]+[\triangle P_{i+1}O_iP_{i+m}]\ge[\triangle P_iO_iP_{i+1}]+[\triangle P_{i+m}O_iP_{i+m+1}]=[B_i]

Detailed analysis

In butterfly $B_i$ , let $a=P_iO_i$ , $b=P_{i+1}O_i$ , $c=P_{i+m}O_i$ , $d=P_{i+m+1}O_i$ , and let $\theta=\angle P_iO_iP_{i+1}=\angle P_{i+m}O_iP_{i+m+1}$ . Without loss of generality assume $d=\min(a,b,c,d)$ (the other three cases are symmetric, giving one of $\triangle P_iP_{i+1}P_{i+m+1}$ , $\triangle P_{i+m}P_{i+m+1}P_i$ , $\triangle P_{i+m}P_{i+m+1}P_{i+1}$ ). Because $b\ge d$ , we have $[\triangle P_{i+1}O_iP_{i+m}]=\frac12 bc\sin\theta\ge\frac12 dc\sin\theta=[\triangle P_{i+m}O_iP_{i+m+1}]$ . Adding $[\triangle P_iO_iP_{i+1}]$ to both sides gives $[\triangle P_iP_{i+1}P_{i+m}]\ge[B_i]\ge S/m$ . This proves the key lemma: every almost-convex $2m$ -gon of area $S$ has a side that forms an inscribed triangle of area at least $S/m$ .