MathLabs

Problem 3

Each pair of opposite sides of convex hexagon has the property that the distance pp between their midpoints is 32\frac{\sqrt3}{2} times the sum of their lengths. Prove that the hexagon is equiangular.
Step 2 of 7: Convert midpoint hypotheses to sharp inequalities
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In plain words

The midpoint formula supplies the left side; the triangle inequality supplies the right side.

∣u+v∣≥3∣u−v∣,∣v+w∣≥3∣v−w∣,∣w−u∣≥3∣w+u∣|\mathbf u+\mathbf v|\ge\sqrt3|\mathbf u-\mathbf v|,\quad |\mathbf v+\mathbf w|\ge\sqrt3|\mathbf v-\mathbf w|,\quad |\mathbf w-\mathbf u|\ge\sqrt3|\mathbf w+\mathbf u|
Detailed analysis

For the opposite sides AB,DEAB,DE, the midpoint formula and the hypothesis give ∣u+v∣=3(∣AB∣+∣DE∣)|\mathbf u+\mathbf v|=\sqrt3(|AB|+|DE|). Since u−v=(A−B)+(E−D)\mathbf u-\mathbf v=(\mathbf A-\mathbf B)+(\mathbf E-\mathbf D), the triangle inequality gives ∣AB∣+∣DE∣≥∣u−v∣|AB|+|DE|\ge|\mathbf u-\mathbf v|. Hence ∣u+v∣≥3∣u−v∣|\mathbf u+\mathbf v|\ge\sqrt3|\mathbf u-\mathbf v|. Applying the same argument to the other two opposite pairs gives ∣v+w∣≥3∣v−w∣|\mathbf v+\mathbf w|\ge\sqrt3|\mathbf v-\mathbf w| and ∣w−u∣≥3∣w+u∣|\mathbf w-\mathbf u|\ge\sqrt3|\mathbf w+\mathbf u|.