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Ukrainian National Mathematical Olympiad

Algebra Difficulty 4.8 AIME Prove it Ukraine

Numbers $a$ , $b$ , $c$ , $d$ satisfy: $ab + cd > 0$ , $ac + bd > 0$ , and $a^2 + d^2 = c^2 + b^2$ . Prove that $ad + bc > 0$ .

Solution

$(ab + cd)(ac + bd) = a^2bc + ac^2d + ab^2d + bcd^2 = bc(a^2 + d^2) + ad(c^2 + b^2) = bc(a^2 + d^2) + ad(a^2 + d^2) = (a^2 + d^2)(bc + ad) > 0,$
which implies the required inequality.

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