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Question Number 180066 by mr W last updated on 06/Nov/22

Commented by mr W last updated on 06/Nov/22

A lies on the x axis and B on the blue  line. P is at (4,4). find the smallest   perimeter of triangle PAB.

$${A}\:{lies}\:{on}\:{the}\:{x}\:{axis}\:{and}\:{B}\:{on}\:{the}\:{blue} \\ $$$${line}.\:{P}\:{is}\:{at}\:\left(\mathrm{4},\mathrm{4}\right).\:{find}\:{the}\:{smallest}\: \\ $$$${perimeter}\:{of}\:{triangle}\:{PAB}. \\ $$

Answered by aleks041103 last updated on 06/Nov/22

Commented by aleks041103 last updated on 06/Nov/22

We can construct points P ′ and P ′′, which  are the reflections of P  with respect to  the x axis and the blue line.  Then by construction  PA=P ′A and PB=P ′′B.  Therefore the triangle′s perimeter is  p=P ′A+AB+P ′′B.  obviously p is minimal, when   A,B∈P ′P ′′  ⇒min(p)=P ′P ′′  But it is easy to see, that then  P ′O=P ′′O=PO=r  and ∠P ′OP ′′=2α, where α=∠(Ox, blue line)  (this is true, when P is between Ox and the  blue line)  ⇒P ′P ′′=2.OP.sinα=p_(min)   in our case:  OP=4(√2) and α=60°  ⇒p_(min) =4(√6)

$${We}\:{can}\:{construct}\:{points}\:{P}\:'\:{and}\:{P}\:'',\:{which} \\ $$$${are}\:{the}\:{reflections}\:{of}\:{P}\:\:{with}\:{respect}\:{to} \\ $$$${the}\:{x}\:{axis}\:{and}\:{the}\:{blue}\:{line}. \\ $$$${Then}\:{by}\:{construction} \\ $$$${PA}={P}\:'{A}\:{and}\:{PB}={P}\:''{B}. \\ $$$${Therefore}\:{the}\:{triangle}'{s}\:{perimeter}\:{is} \\ $$$${p}={P}\:'{A}+{AB}+{P}\:''{B}. \\ $$$${obviously}\:{p}\:{is}\:{minimal},\:{when}\: \\ $$$${A},{B}\in{P}\:'{P}\:'' \\ $$$$\Rightarrow{min}\left({p}\right)={P}\:'{P}\:'' \\ $$$${But}\:{it}\:{is}\:{easy}\:{to}\:{see},\:{that}\:{then} \\ $$$${P}\:'{O}={P}\:''{O}={PO}={r} \\ $$$${and}\:\angle{P}\:'{OP}\:''=\mathrm{2}\alpha,\:{where}\:\alpha=\angle\left({Ox},\:{blue}\:{line}\right) \\ $$$$\left({this}\:{is}\:{true},\:{when}\:{P}\:{is}\:{between}\:{Ox}\:{and}\:{the}\right. \\ $$$$\left.{blue}\:{line}\right) \\ $$$$\Rightarrow{P}\:'{P}\:''=\mathrm{2}.{OP}.{sin}\alpha={p}_{{min}} \\ $$$${in}\:{our}\:{case}: \\ $$$${OP}=\mathrm{4}\sqrt{\mathrm{2}}\:{and}\:\alpha=\mathrm{60}° \\ $$$$\Rightarrow{p}_{{min}} =\mathrm{4}\sqrt{\mathrm{6}} \\ $$

Commented by mr W last updated on 07/Nov/22

perfect! thanks sir!

$${perfect}!\:{thanks}\:{sir}! \\ $$

Commented by Acem last updated on 07/Nov/22

The development of prograning and engineering   graphic programs made us rely on them and   forgot simplest things in descriptive geometry!   My grades at it were all 100/100   Yesterday , i was helpless in front of this question  Thank you Sir for refresh our mind,   and thank you Sir Aleks too

$${The}\:{development}\:{of}\:{prograning}\:{and}\:{engineering} \\ $$$$\:{graphic}\:{programs}\:{made}\:{us}\:{rely}\:{on}\:{them}\:{and} \\ $$$$\:{forgot}\:{simplest}\:{things}\:{in}\:{descriptive}\:{geometry}! \\ $$$$\:{My}\:{grades}\:{at}\:{it}\:{were}\:{all}\:\mathrm{100}/\mathrm{100} \\ $$$$\:{Yesterday}\:,\:{i}\:{was}\:{helpless}\:{in}\:{front}\:{of}\:{this}\:{question} \\ $$$${Thank}\:{you}\:{Sir}\:{for}\:{refresh}\:{our}\:{mind}, \\ $$$$\:{and}\:{thank}\:{you}\:{Sir}\:{Aleks}\:{too} \\ $$

Answered by mr W last updated on 07/Nov/22

Commented by mr W last updated on 07/Nov/22

as aleks sir has explained above:  p_(min) =P_1 P_2 =2 OP sin 60°=2×4(√2)×((√3)/2)           =4(√6)

$${as}\:{aleks}\:{sir}\:{has}\:{explained}\:{above}: \\ $$$${p}_{{min}} ={P}_{\mathrm{1}} {P}_{\mathrm{2}} =\mathrm{2}\:{OP}\:\mathrm{sin}\:\mathrm{60}°=\mathrm{2}×\mathrm{4}\sqrt{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{4}\sqrt{\mathrm{6}} \\ $$

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