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 4 of 5: If

\sum a_r<2S

, pick rational upper bounds

k_r/(2m)

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In plain words

When $n$ positive numbers add up to strictly less than $1$ , we can nudge each of them slightly upward to a rational number so that the $n$ rationals add up to $1$ ; clearing denominators (and doubling so every numerator is even) gives the integers $k_r$ .

\sum_{r=1}^n a_r<2S\ \Longrightarrow\ \exists\,k_1,\dots,k_n\in 2\mathbb Z_{>0}:\ \frac{a_r}{2S}<\frac{k_r}{2m}\quad\left(2m=\sum_{r=1}^n k_r\right)

Detailed analysis

Let the sides of $P$ be $b_1,\dots,b_n$ with assigned maximum triangle areas $a_1,\dots,a_n$ , and suppose for contradiction that $\sum_{r=1}^n a_r<2S$ , i.e. $\sum_{r=1}^n\frac{a_r}{2S}<1$ . By density of the rationals, we can choose positive rationals $q_1,\dots,q_n$ with $\frac{a_r}{2S}<q_r$ for each $r$ and $\sum_{r=1}^n q_r=1$ . Writing $q_r=k_r/(2m)$ with a common even denominator $2m$ (doubling all numerators and the denominator if needed so each $k_r$ is an even positive integer) gives $\frac{a_r}{2S}<\frac{k_r}{2m}$ , i.e. $a_r<k_r\frac{S}{m}$ , with $\sum_{r=1}^n k_r=2m$ .