009. Graphing Inequalities

For real numbers $a$, $b$, and $c$ such that $0.9 \leq a \leq 1.1$, $2.7 \leq b \leq 3.3$, and $4.5 \leq c \leq 5.4$, show the range of a point $P(u, v)$ moving where $u = b/a$ and $v = c/a$.

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It seems easy at first glance. $$\begin{aligned} \frac{10}{11} \leq \frac{1}{a} \leq \frac{10}{9} \implies \frac{27}{11} \leq \frac{b}{a} \leq \frac{11}{3}, \quad \frac{45}{11} \leq \frac{c}{a} \leq 6 \end{aligned}$$

However, be warned that the real range is not a rectangle region as the above result. Another condition must be considered as well. $$\begin{aligned} \frac{10}{33} \leq \frac{1}{b} \leq \frac{10}{27}, \quad \frac{v}{u} = \frac{c}{b} \implies \frac{15}{11} \leq \frac{v}{u} \leq 2 \end{aligned}$$

It provides the 2 additional equations of lines. Since $u$ is positive, $$\begin{aligned} \frac{15}{11}u \leq v \leq 2u \end{aligned}$$

Another point of view of this problem is scaling. $u$ and $v$ are scaled by $1/a$ and their original unscaled range can be considered as $2.7 \leq u \leq 3.3$ and $4.5 \leq v \leq 5.4$. In other words, the original rectangle region of $u$ and $v$ is scaled by the range of $1/a$ and the range of $P$ is where this region slides.