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 3 of 7: Square the three inequalities
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In plain words

Expanding the squared norms turns the geometric bounds into dot-product inequalities.

∣u∣2+∣v∣2≤4u⋅v,∣v∣2+∣w∣2≤4v⋅w,∣w∣2+∣u∣2≤−4w⋅u|\mathbf u|^2+|\mathbf v|^2\le4\mathbf u\cdot\mathbf v,\quad |\mathbf v|^2+|\mathbf w|^2\le4\mathbf v\cdot\mathbf w,\quad |\mathbf w|^2+|\mathbf u|^2\le-4\mathbf w\cdot\mathbf u
Detailed analysis

Squaring the first inequality gives ∣u∣2+2u⋅v+∣v∣2≥3(∣u∣2−2u⋅v+∣v∣2)|\mathbf u|^2+2\mathbf u\cdot\mathbf v+|\mathbf v|^2\ge3(|\mathbf u|^2-2\mathbf u\cdot\mathbf v+|\mathbf v|^2), hence ∣u∣2+∣v∣2≤4u⋅v|\mathbf u|^2+|\mathbf v|^2\le4\mathbf u\cdot\mathbf v. The other two expansions give ∣v∣2+∣w∣2≤4v⋅w|\mathbf v|^2+|\mathbf w|^2\le4\mathbf v\cdot\mathbf w and ∣w∣2+∣u∣2≤−4w⋅u|\mathbf w|^2+|\mathbf u|^2\le-4\mathbf w\cdot\mathbf u.