Question 27
Suppose that $$a, b, c$$ are positive real numbers such that $$a^2 + b^2 = c^2$$ and $$ab = c$$. Find the value of $$\frac{(a+b+c)(a-b+c)(a+b-c)(a-b-c)}{c^2}$$
Correct Answer: -4
Solution
Pairing the factors gives $$(a+b+c)(a+b-c) = (a+b)^2 - c^2 = 2ab$$ and $$(a-b+c)(a-b-c) = (a-b)^2 - c^2 = -2ab$$, using $$a^2+b^2 = c^2$$. Their product is $$-4a^2b^2$$, and since $$ab = c$$ this equals $$-4c^2$$. Dividing by $$c^2$$ gives $$-4$$.
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