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Question-81972

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Question Number 81972 by ajfour last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
△PQR is equilateral, find s_(min,) s_(max) .
$$\bigtriangleup{PQR}\:{is}\:{equilateral},\:{find} \\ $$$${s}_{{min},} \:{s}_{{max}} . \\ $$
Answered by mr W last updated on 17/Feb/20
Commented by mr W last updated on 17/Feb/20
Δ_(OAB) =(√(((11)/2)(((11)/2)−2)(((11)/2)−4)(((11)/2)−5)))=((√(231))/4) Δ_(OAB) =((5δ)/2)=((√(231))/4) ⇒δ=((√(231))/(10)) σ=(√(2^2 −δ^2 ))=(√(4−((231)/(100))))=((13)/(10)) O(((13)/(10)),−((√(231))/(10)))=(σ,−δ) eqn. of circle: (x−σ)^2 +(y+δ)^2 =4^2 say P(p,0), Q(0,q) q=(√(s^2 −p^2 )) x^2 +(y−q)^2 =s^2 ...(i) (x−p)^2 +y^2 =s^2 ...(ii) (x−σ)^2 +(y+δ)^2 =16 ...(iii) (i)−(ii): 2px−p^2 −2qy+q^2 =0 ⇒x=(q/p)y+((p^2 −q^2 )/(2p)) ((q/p)y+((p^2 −q^2 )/(2p))−p)^2 +y^2 =s^2 (2qy−p^2 −q^2 )^2 +4p^2 y^2 =4p^2 s^2 y^2 −qy+((p^2 +q^2 )/4)−((p^2 s^2 )/(p^2 +q^2 ))=0 y^2 −qy−((3p^2 −q^2 )/4)=0 ⇒y=(((√3)p+q)/2)=(((√3)p+(√(s^2 −p^2 )))/2) ⇒x=((p+(√3)q)/2)=((p+(√(3(s^2 −p^2 ))))/2) (((p+(√(3(s^2 −p^2 ))))/2)−σ)^2 +((((√3)p+(√(s^2 −p^2 )))/2)+δ)^2 =16 ⇒(p+(√(3(s^2 −p^2 )))−2σ)^2 +((√3)p+(√(s^2 −p^2 ))+2δ)^2 =64 for minimum and maximum s: (ds/dp)=0 s_(min) ≈2.3935 at p≈1.9567
$$\Delta_{{OAB}} =\sqrt{\frac{\mathrm{11}}{\mathrm{2}}\left(\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{2}\right)\left(\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{4}\right)\left(\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{5}\right)}=\frac{\sqrt{\mathrm{231}}}{\mathrm{4}} \\ $$$$\Delta_{{OAB}} =\frac{\mathrm{5}\delta}{\mathrm{2}}=\frac{\sqrt{\mathrm{231}}}{\mathrm{4}} \\ $$$$\Rightarrow\delta=\frac{\sqrt{\mathrm{231}}}{\mathrm{10}} \\ $$$$\sigma=\sqrt{\mathrm{2}^{\mathrm{2}} −\delta^{\mathrm{2}} }=\sqrt{\mathrm{4}−\frac{\mathrm{231}}{\mathrm{100}}}=\frac{\mathrm{13}}{\mathrm{10}} \\ $$$${O}\left(\frac{\mathrm{13}}{\mathrm{10}},−\frac{\sqrt{\mathrm{231}}}{\mathrm{10}}\right)=\left(\sigma,−\delta\right) \\ $$$${eqn}.\:{of}\:{circle}: \\ $$$$\left({x}−\sigma\right)^{\mathrm{2}} +\left({y}+\delta\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$ \\ $$$${say}\:{P}\left({p},\mathrm{0}\right),\:{Q}\left(\mathrm{0},{q}\right) \\ $$$${q}=\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} } \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\left({y}−{q}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({x}−\sigma\right)^{\mathrm{2}} +\left({y}+\delta\right)^{\mathrm{2}} =\mathrm{16}\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{px}−{p}^{\mathrm{2}} −\mathrm{2}{qy}+{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{q}}{{p}}{y}+\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$$ \\ $$$$\left(\frac{{q}}{{p}}{y}+\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{p}}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={s}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{qy}−{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{4}{p}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −{qy}+\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}^{\mathrm{2}} {s}^{\mathrm{2}} }{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −{qy}−\frac{\mathrm{3}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\sqrt{\mathrm{3}}{p}+{q}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{p}+\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{{p}+\sqrt{\mathrm{3}}{q}}{\mathrm{2}}=\frac{{p}+\sqrt{\mathrm{3}\left({s}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\left(\frac{{p}+\sqrt{\mathrm{3}\left({s}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}}{\mathrm{2}}−\sigma\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}{p}+\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} }}{\mathrm{2}}+\delta\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow\left({p}+\sqrt{\mathrm{3}\left({s}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}−\mathrm{2}\sigma\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}{p}+\sqrt{{s}^{\mathrm{2}} −{p}^{\mathrm{2}} }+\mathrm{2}\delta\right)^{\mathrm{2}} =\mathrm{64} \\ $$$${for}\:{minimum}\:{and}\:{maximum}\:{s}:\:\frac{{ds}}{{dp}}=\mathrm{0} \\ $$$${s}_{{min}} \approx\mathrm{2}.\mathrm{3935}\:{at}\:{p}\approx\mathrm{1}.\mathrm{9567} \\ $$
Commented by mr W last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
superb! Sir; requires enough determination..
$${superb}!\:{Sir};\:{requires}\:{enough} \\ $$$${determination}.. \\ $$
Commented by mr W last updated on 17/Feb/20
Commented by mr W last updated on 17/Feb/20
s_(max) is possible when (q+((√(231))/(10)))^2 +(((13)/(10)))^2 =4^2 ⇒q=((3(√(159))−(√(231)))/(10))≈2.2630 ⇒p≈1.3022 ⇒s_(max) ≈2.6109 or q=0 ⇒p≈2.8613 ⇒s_(max) ≈2.8613
$${s}_{{max}} \:{is}\:{possible}\:{when} \\ $$$$\left({q}+\frac{\sqrt{\mathrm{231}}}{\mathrm{10}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{13}}{\mathrm{10}}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\Rightarrow{q}=\frac{\mathrm{3}\sqrt{\mathrm{159}}−\sqrt{\mathrm{231}}}{\mathrm{10}}\approx\mathrm{2}.\mathrm{2630} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{3022} \\ $$$$\Rightarrow{s}_{{max}} \approx\mathrm{2}.\mathrm{6109} \\ $$$${or} \\ $$$${q}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{2}.\mathrm{8613} \\ $$$$\Rightarrow{s}_{{max}} \approx\mathrm{2}.\mathrm{8613} \\ $$
Commented by mr W last updated on 17/Feb/20
Commented by mr W last updated on 17/Feb/20
Commented by ajfour last updated on 17/Feb/20
Too Good Sir, thanks plentiful. you have dealt all cases.
$${Too}\:{Good}\:{Sir},\:{thanks}\:{plentiful}. \\ $$$${you}\:{have}\:{dealt}\:{all}\:{cases}. \\ $$

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