Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Find the answer

Let $a, b, c, d$ be real numbers such that $\min (20 x+19,19 x+20)=(a x+b)-|c x+d|$ for all real numbers $x$ . Find $a b+c d$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

In general, $\min (p, q)=\frac{p+q}{2}-\left|\frac{p-q}{2}\right|$ . Letting $p=20 x+19$ and $q=19 x+20$ gives $a=b=19.5$ and $c=d= \pm 0.5$ . Then the answer is $19.5^{2}-0.5^{2}=19 \cdot 20=380$ .

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