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Question Number 14821 by Tinkutara last updated on 04/Jun/17

A room has dimensions 3 m × 4 m × 5 m.  A fly starting at one corner ends up at  the diametrically opposite corner. If  the fly were to walks, what is the length  of the shortest path it can take?

$$\mathrm{A}\:\mathrm{room}\:\mathrm{has}\:\mathrm{dimensions}\:\mathrm{3}\:\mathrm{m}\:×\:\mathrm{4}\:\mathrm{m}\:×\:\mathrm{5}\:\mathrm{m}. \\ $$$$\mathrm{A}\:\mathrm{fly}\:\mathrm{starting}\:\mathrm{at}\:\mathrm{one}\:\mathrm{corner}\:\mathrm{ends}\:\mathrm{up}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{diametrically}\:\mathrm{opposite}\:\mathrm{corner}.\:\mathrm{If} \\ $$$$\mathrm{the}\:\mathrm{fly}\:\mathrm{were}\:\mathrm{to}\:\mathrm{walks},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{path}\:\mathrm{it}\:\mathrm{can}\:\mathrm{take}? \\ $$

Answered by ajfour last updated on 04/Jun/17

  d_(min) =(√((3+4)^2 +(5)^2 ))=(√(74))              ≈ 8.6 units

$$\:\:{d}_{{min}} =\sqrt{\left(\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{5}\right)^{\mathrm{2}} }=\sqrt{\mathrm{74}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\approx\:\mathrm{8}.\mathrm{6}\:{units} \\ $$$$\:\:\:\: \\ $$

Commented by RasheedSoomro last updated on 04/Jun/17

Sir the fly is walking not flying.  You have determine correct corner,  but the path  shouldn′t be through air.

$$\mathrm{Sir}\:\mathrm{the}\:\mathrm{fly}\:\mathrm{is}\:\boldsymbol{\mathrm{walking}}\:\mathrm{not}\:\boldsymbol{\mathrm{flying}}. \\ $$$$\mathrm{You}\:\mathrm{have}\:\mathrm{determine}\:\mathrm{correct}\:\mathrm{corner}, \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{path}\:\:\mathrm{shouldn}'\mathrm{t}\:\mathrm{be}\:\mathrm{through}\:\mathrm{air}. \\ $$

Commented by ajfour last updated on 04/Jun/17

thats why this diagram.try and  understand..

$${thats}\:{why}\:{this}\:{diagram}.{try}\:{and} \\ $$$${understand}.. \\ $$

Commented by RasheedSoomro last updated on 04/Jun/17

Sir I understood!

$$\mathrm{Sir}\:\mathrm{I}\:\mathrm{understood}! \\ $$

Commented by ajfour last updated on 04/Jun/17

Commented by mrW1 last updated on 05/Jun/17

If the dimensions are a,b,c with a≤b≤c.  There are 3 possible ways:  (√((a+b)^2 +c^2 )) =(√(a^2 +b^2 +c^2 +2ab))     (1)  (√((a+c)^2 +b^2 )) =(√(a^2 +b^2 +c^2 +2ac))     (2)  (√((b+c)^2 +a^2 )) =(√(a^2 +b^2 +c^2 +2bc))     (3)  the shortest way is (1), since  2ab≤2ac≤2bc.    in current case a=3,b=4,c=5  ⇒ (√((3+4)^2 +5^2 ))=8.6

$${If}\:{the}\:{dimensions}\:{are}\:{a},{b},{c}\:{with}\:{a}\leqslant{b}\leqslant{c}. \\ $$$${There}\:{are}\:\mathrm{3}\:{possible}\:{ways}: \\ $$$$\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} }\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\sqrt{\left({a}+{c}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\sqrt{\left({b}+{c}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} }\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$${the}\:{shortest}\:{way}\:{is}\:\left(\mathrm{1}\right),\:{since} \\ $$$$\mathrm{2}{ab}\leqslant\mathrm{2}{ac}\leqslant\mathrm{2}{bc}. \\ $$$$ \\ $$$${in}\:{current}\:{case}\:{a}=\mathrm{3},{b}=\mathrm{4},{c}=\mathrm{5} \\ $$$$\Rightarrow\:\sqrt{\left(\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\mathrm{8}.\mathrm{6} \\ $$

Commented by Tinkutara last updated on 05/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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