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Problem 938

AMC 12 late, AIME early

Algebra Difficulty 4.8 Prove it Berkeley Math Circle Monthly Contest 8 · United States

Given five nonnegative real numbers with sum $1$ , prove that it is possible to arrange them at the vertices of a regular pentagon such that no two numbers connected by a side of the pentagon have product exceeding $1/9$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Label the numbers $a, b, c, d, e$ in increasing order. Place them around the pentagon in the order $e, a, d, c, b$ . Then it is clear that the products of the numbers on the sides follow the inequalities
$a d \leq a e \leq b e \quad \text{and} \quad b c \leq c d$
Thus it suffices to prove that $b e \leq 1/9$ and $c d \leq 1/9$ . Using the AM-GM inequality,
$1 = a + b + c + d + e \geq 0 + 0 + c + d + d = c + 2d \geq \frac{3c}{2} + \frac{3d}{2} \geq 2 \sqrt{\frac{3c}{2} \cdot \frac{3d}{2}} = 3 \sqrt{c d}$
so $c d \leq 1/9$ . Also,
$1 = a + b + c + d + e \geq 0 + b + b + b + e = 3b + e \geq 2 \sqrt{3b \cdot e}$
so $b e \leq 1/12 < 1/9$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.