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Let $A(1, 1)$ and $B(3, 2)$ be two points. Find point $M(x, 0)$ so that the sum $AM + MB$ is minimum. After that, find the minimum of the sum.

In order for the sum to be minimum, I concluded that $x$ must lay between $1$ and $3$. I also calculated $AM$ and $MB$:

$$AM = \sqrt{x^2 - 2x + 2}$$ $$BM = \sqrt{x^2 - 6x + 13}$$

I don't now how to proceed from this point, so I would appreciate any help from you!

Thank you in advance!

George R.
  • 2,833

2 Answers2

2

Consider $B'(3,-2)$. Let $N$ be the intersection point of $x$-axis with $AB'$.

Then, $$|AM|+|BM|=|AM|+|B'M|\ge |AN|+|B'N|$$

mathlove
  • 139,939
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Let's $\ds{\quad x - 1 = \tan\pars{\alpha}\quad}$ and $\ds{\quad x - 3 = 2\tan\pars{\beta}\quad}$ such that \begin{align} \,\mathrm{D}\pars{x} & \equiv \root{\pars{x - 1}^{2} + 1} + \root{\pars{x - 3}^{2} + 4} = \sec\pars{\alpha} + 2\sec\pars{\beta} \end{align} $\ds{\alpha}$ and $\ds{\beta}$ are constrained by $\ds{2 = \pars{x - 1} - \pars{x - 3} = \tan\pars{\alpha} - 2\tan\pars{\beta}}$. So, we have to minimises $\ds{\sec\pars{\alpha} + 2\sec\pars{\beta}}$ with the above constraining relation. Namely, $$ \sec\pars{\alpha} + 2\sec\pars{\beta} - \mu\bracks{\tan\pars{\alpha} - 2\tan\pars{\beta} - 2} $$ $\ds{\mu}$ is a Lagrange Multiplier. $$ \left.\begin{array}{rcrcl} \ds{\sec\pars{\alpha}\tan\pars{\alpha}} & \ds{-} & \ds{\mu\sec^{2}\pars{\alpha}} & \ds{=} & \ds{0} \\[1mm] \ds{2\sec\pars{\beta}\tan\pars{\beta}} & \ds{+} & \ds{2\mu\sec^{2}\pars{\beta}} & \ds{=} & \ds{0} \end{array}\right\rbrace \quad\imp\quad \left\lbrace\begin{array}{rcr} \ds{\sin\pars{\alpha}} & \ds{=} & \ds{\mu} \\[1mm] \ds{\sin\pars{\beta}} & \ds{=} & \ds{-\mu} \end{array}\right. $$

Then, $\ds{\tan\pars{\alpha} = \pm\tan\pars{\beta}}$:
  1. $\ds{\tan\pars{\alpha} = \tan\pars{\beta}\ \imp\ x - 1 = {x - 3 \over 2}\ \imp\ x = -1\ \imp\ \,\mathrm{D}\pars{-1} = 3\root{5} \approx 6.7082}$.
  2. $\ds{\tan\pars{\alpha} = -\tan\pars{\beta}\ \imp\ x - 1 = -\,{x - 3 \over 2}\ \imp\ x = {5 \over 3}\ \imp\ \,\mathrm{D}\pars{5 \over 3} = \root{13} \approx 3.6056}$.
    $$ \color{#f00}{Solution:\quad\,\mathrm{M} = \pars{{5 \over 3},0}}\quad \mbox{with distance sum to }\ \,\mathrm{A}\ \mbox{and}\ \,\mathrm{B}\ \mbox{equal to}\ \color{#f00}{\root{13}} \approx 3.6056. $$
Felix Marin
  • 89,464