I think geometry will help

Geometry Level 5

Let $A$ be the point $(8,6)$ and $D$ be the point $(6,4)$ . If the length of the shortest path $ABCD$ can be expressed as $\sqrt{a} + b$ , where $B$ is the point $(x,3)$ and $C$ is a point $(x,0)$ . This path consists of three connected segments, with the middle one vertical.

Find the value of $a - b$ .

The answer is 50.

1 solution

Chew-Seong Cheong
Oct 27, 2016

The shortest path $ABCD$ is one that $AB$ is a mirror (horizontal) reflection of $CD$ , that is:

$\begin{aligned} \frac {8-x}{6-3} & = \frac {x-6}{4-0} \\ 4(8-x) & = 3(x-6) \\ 7x & = 50 \\ \implies x & = \frac {50}7 \end{aligned}$

Therefore, the shortest path is given by:

$\begin{aligned} l_{min} & = \sqrt{(8-x)^2+(6-3)^2} + (3-0) + \sqrt{(x-6)^2+(0-4)^2} \\ & = \sqrt{\left(8-\frac {50}7 \right)^2+9} + 3 + \sqrt{\left( \frac {50}7 - 6 \right)^2+16} \\ & = \sqrt{\frac {36}{49}+9} + 3 + \sqrt{\frac {64}{49}+16} \\ & = \sqrt{\frac {477}{49}} + 3 + \sqrt{\frac {848}{49}} \\ & = \frac 37 \sqrt{53} + 3 + \frac 47 \sqrt{53} \\ & = \sqrt{53} + 3 \end{aligned}$

$\implies a - b = 53-3 = \boxed{50}$

You just read my mind...

Archit Tripathi - 4 years, 7 months ago

How do I prove that the shortest path comprises of a mirror reflection?

Atomsky Jahid - 4 years, 2 months ago

Sorry, I didn't notice this comment. Check out Heron's shortest distance problem .

Chew-Seong Cheong - 4 years, 2 months ago

Please explain the first statement..

Vishal Yadav - 4 years, 2 months ago

The shortest distance is one that is traveled by light, therefore, it is a mirror reflection. Read Heron's shortest distance problem for reference.

Chew-Seong Cheong - 4 years, 2 months ago

$\color{#20A900}{Shortest ~distance ~is~ by ~lines~ that ~make~ biggest ~(+/-) ~angle~ with ~vertical ~or ~horizontal ~within~ the ~given~ constrains. } \\ But~ BC~ is ~given~as~ 3~ unit ~vertical. \\ For~ shortest,~ as~far~as~possible,~ all ~ sloping ~lines ~must~ be ~equally~ inclined ~to~ y=0.~ (+ or -) \\ ~~~~~\\ Horizontal~ distance~ between~ A ~and~ D,~ H_{AD}~=~2.\\ Let ~H_{AB}= m, ~so~H_{DB}=(2-m). \\ ~~~~\\\\\ \color{#3D99F6}{So~ slope ~of~ |AB| = 3/m~ and~ that~ of~ |CD| =4/(2-m),~ and ~}\\ { \Large{they ~must~ be~ equal.} } \\ So~ 3/m= 4/(2-m)=7/2~ (by ~ratio~ rules). \\ \implies~ ~m=6/7. \\ So~ total ~length~ is ~ AB+BC+ CD \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Since~AB~and~CD~are~equally~inclined~to~y=0, \\ =\sqrt{(H_{AB} + H_{BD})^2 +(V_{AB} + V_{BD})^2 } + B C \\ =\sqrt{(6/7 + 8/7)^2+(3 + 4)^2} + 3 \\ = \sqrt{2^2 + 7^2} + 3 \\ = \sqrt{53} + 3\\ = \sqrt{a} + b\\ So~a-b=\Large \color{#D61F06}{50} \\$

Niranjan Khanderia - 1 year, 1 month ago

I think geometry will help

The answer is 50.

1 solution

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