Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it China

Suppose that $a > 0$ and the minima of function $f(x) = x + \frac{100}{x}$ on intervals $(0, a]$ and $[a, +\infty)$ are $m_1, m_2$ , respectively. If $m_1 m_2 = 2020$ , then the value of $a$ is ______.

Solution

Note that $f(x)$ is monotonically decreasing on $(0, 10]$ and monotonically increasing on $[10, +\infty)$ . When $a \in (0, 10]$ , $m_1 = f(a)$ , $m_2 = f(10)$ ; when $a \in [10, +\infty)$ , $m_1 = f(10)$ , $m_2 = f(a)$ . Therefore, there is always
$f(a)f(10) = m_1m_2 = 2020,$
namely, $a + \frac{100}{a} = \frac{2020}{20} = 101$ . The solution is $a = 1$ or $a = 100$ . $\square$

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