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Problem 1174

AMC 12 late, AIME early

Algebra Difficulty 4.9 Prove it China Mathematical Competition · China

Given function $f(x) = |2 - \log_3 x|$ , positive real numbers $a, b, c$ satisfy $a < b < c$ and $f(a) = 2f(b) = 2f(c)$ . Find the minimum of $\frac{ac}{b}$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Notice that $f(x) = |\log_3(\frac{x}{9})|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\infty)$ .

By the conditions satisfied by $a, b, c$ , we have $0 < a < b < 9 < c$ and
$\log_3\left(\frac{9}{a}\right) = 2\log_3\left(\frac{9}{b}\right) = 2\log_3\left(\frac{c}{9}\right).$
Therefore,
$\log_3\left(\frac{ac}{b}\right) = \log_3\left(9 \cdot \frac{a}{9} \cdot \frac{9}{b} \cdot \frac{c}{9}\right) = 2 - \log_3\left(\frac{9}{a}\right) + \log_3\left(\frac{9}{b}\right) + \log_3\left(\frac{c}{9}\right) = 2,$
namely, $\frac{ac}{b} = 3^2 = 9$ . $\square$

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