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 5 of 5: Subdivide each side

b_r

into

k_r

equal parts and apply the lemma

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a_r\ge k_r\cdot[\triangle\text{ on sub-segment}]\ge k_r\frac{S}{m}>a_r\quad(\text{contradiction})

Detailed analysis

Subdivide each side $b_r$ of $P$ into $k_r$ equal segments to turn $P$ into an almost-convex $2m$ -gon $Q$ of the same area $S$ . By the lemma of Step 3, some side of $Q$ — which is a $1/k_r$ sub-segment of some side $b_r$ of $P$ — forms a triangle in $P$ with some vertex $V$ of area at least $S/m$ . Stretching the base from that $1/k_r$ sub-segment to the full side $b_r$ (with the same apex $V\in P$ ) multiplies the triangle's area by $k_r$ , giving a triangle in $P$ on side $b_r$ of area at least $k_r\frac{S}{m}$ . By definition of $a_r$ , this forces $a_r\ge k_r\frac{S}{m}$ , contradicting $a_r<k_r\frac{S}{m}$ from Step 4. Therefore $\sum_{r=1}^n a_r\ge 2S$ .