Question Number 30933 by ajfour last updated on 28/Feb/18
Answered by mrW2 last updated on 28/Feb/18
Commented by mrW2 last updated on 01/Mar/18
$${Respect}!\:{You}\:{can}\:{still}\:{remember}\:{and} \\ $$$${find}\:{such}\:{an}\:{old}\:{post}.\:{I}\:{have}\:{totally} \\ $$$${forgot}\:{it}. \\ $$
Commented by mrW2 last updated on 28/Feb/18
$$\Delta{ABC}\:{must}\:{be}\:{an}\:{acute}\:{triangle}. \\ $$$$\Delta{PQR}\:{has}\:{minimum}\:{perimeter}\:{when} \\ $$$${it}\:{is}\:{the}\:{orthic}\:{triangle}\:{formed}\:{by} \\ $$$${the}\:{feet}\:{of}\:{altitudes}\:{of}\:\Delta{ABC}. \\ $$$${a}_{\mathrm{1}} ={c}\:\mathrm{cos}\:{B} \\ $$$${c}_{\mathrm{2}} ={a}\:\mathrm{cos}\:{B} \\ $$$${b}'={RP}={b}\:\mathrm{cos}\:{B}={b}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}=\frac{{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{\mathrm{2}{abc}} \\ $$$$... \\ $$$${Perimeter}\:{of}\:{PQR}: \\ $$$${U}=\frac{{a}^{\mathrm{2}} \left(−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{2}{abc}} \\ $$$$\Rightarrow{U}=\frac{\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}{\mathrm{2}{abc}} \\ $$
Commented by ajfour last updated on 01/Mar/18
$${Thanks}\:{Sir},\:{but}\:{proof}\:{please}.. \\ $$
Commented by Tinkutara last updated on 01/Mar/18
See Q 14809 and problem 2.15 in the following link https://www.scribd.com/document/372679641/Mathematical-Olympiad-Treasures-extracted
Commented by Tinkutara last updated on 01/Mar/18
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Commented by ajfour last updated on 01/Mar/18
$${Really},\:{Sir},\:{thank}\:{you}\:{both}. \\ $$