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Algebra Difficulty 4.9 AIME Prove it United States

Problem:
Suppose that a,b,c,da, b, c, d are real numbers satisfying a≥b≥c≥d≥0a \geq b \geq c \geq d \geq 0, a2+d2=1a^{2} + d^{2} = 1, b2+c2=1b^{2} + c^{2} = 1, and ac+bd=1/3a c + b d = 1/3. Find the value of ab−cda b - c d.

Solution

Solution:
Answer: 223\frac{2 \sqrt{2}}{3}
We have
(ab−cd)2=(a2+d2)(b2+c2)−(ac+bd)2=(1)(1)−(13)2=89 (a b - c d)^2 = (a^2 + d^2)(b^2 + c^2) - (a c + b d)^2 = (1)(1) - \left(\frac{1}{3}\right)^2 = \frac{8}{9}
Since a≥b≥c≥d≥0a \geq b \geq c \geq d \geq 0, ab−cd≥0a b - c d \geq 0, so ab−cd=223a b - c d = \frac{2 \sqrt{2}}{3}.

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