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Question Number 170832 by Shrinava last updated on 01/Jun/22

Commented by mr W last updated on 01/Jun/22

T=AM+BN≥(√(AB^2 +MN^2 ))=(√(1^2 +2^2 +5^2 +2^2 ))=(√(34)) T_(min) =(√(34))

$${T}={AM}+{BN}\geqslant\sqrt{{AB}^{\mathrm{2}} +{MN}^{\mathrm{2}} }=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{34}} \\ $$$${T}_{{min}} =\sqrt{\mathrm{34}} \\ $$

Commented by Shrinava last updated on 01/Jun/22

Thank you so much dear professor, can you explain a little more, if possible, please

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{professor}, \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{a}\:\mathrm{little}\:\mathrm{more}, \\ $$$$\mathrm{if}\:\mathrm{possible},\:\mathrm{please} \\ $$

Commented by mr W last updated on 02/Jun/22

Commented by mr W last updated on 01/Jun/22

T=AM+BN=A′P+PB′≥A′B′=(√(AB^2 +MN^2 ))

$${T}={AM}+{BN}={A}'{P}+{PB}'\geqslant{A}'{B}'=\sqrt{{AB}^{\mathrm{2}} +{MN}^{\mathrm{2}} } \\ $$

Commented by Shrinava last updated on 02/Jun/22

Thank you very much dear ptofessor

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{ptofessor} \\ $$

Commented by mr W last updated on 04/Jun/22

i think my solution above is wrong.

$${i}\:{think}\:{my}\:{solution}\:{above}\:{is}\:{wrong}. \\ $$

Commented by mr W last updated on 04/Jun/22

Commented by mr W last updated on 04/Jun/22

T=AM+BN=A′N+NB≥A′B AB^2 =(2−1)^2 +(3−1)^2 +(4+1)^2 =30 MN^2 =2^2 =4 A′B^2 =30+4−2×(√(30))×2 cos θ sin θ=cos ((π/2)−θ)=((5×1)/( (√(30))))=(5/( (√(30)))) cos θ=(√(1−((25)/(30))))=((√5)/( (√(30)))) A′B^2 =30+4−2×(√(30))×2×((√5)/( (√(30))))=34−4(√5) ⇒T_(min) =(√(34−4(√5)))

$${T}={AM}+{BN}={A}'{N}+{NB}\geqslant{A}'{B} \\ $$$${AB}^{\mathrm{2}} =\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{30} \\ $$$${MN}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${A}'{B}^{\mathrm{2}} =\mathrm{30}+\mathrm{4}−\mathrm{2}×\sqrt{\mathrm{30}}×\mathrm{2}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{sin}\:\theta=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)=\frac{\mathrm{5}×\mathrm{1}}{\:\sqrt{\mathrm{30}}}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{30}}} \\ $$$$\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\frac{\mathrm{25}}{\mathrm{30}}}=\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{30}}} \\ $$$${A}'{B}^{\mathrm{2}} =\mathrm{30}+\mathrm{4}−\mathrm{2}×\sqrt{\mathrm{30}}×\mathrm{2}×\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{30}}}=\mathrm{34}−\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{T}_{{min}} =\sqrt{\mathrm{34}−\mathrm{4}\sqrt{\mathrm{5}}} \\ $$

Commented by Shrinava last updated on 17/Jun/22

Perfect-Amazing Dear Professor

$$\mathrm{Perfect}-\mathrm{Amazing}\:\mathrm{Dear}\:\mathrm{Professor} \\ $$

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