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Annual Harvard-MIT Mathematics Tournament

Algebra Difficulty 4.9 AIME Prove it United States

Problem:
Suppose that $a, b, c, d$ are real numbers satisfying $a \geq b \geq c \geq d \geq 0$ , $a^{2} + d^{2} = 1$ , $b^{2} + c^{2} = 1$ , and $a c + b d = 1/3$ . Find the value of $a b - c d$ .

Solution

Solution:
Answer: $\frac{2 \sqrt{2}}{3}$
We have
$(a b - c d)^2 = (a^2 + d^2)(b^2 + c^2) - (a c + b d)^2 = (1)(1) - \left(\frac{1}{3}\right)^2 = \frac{8}{9}$
Since $a \geq b \geq c \geq d \geq 0$ , $a b - c d \geq 0$ , so $a b - c d = \frac{2 \sqrt{2}}{3}$ .

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