Question-86907

Question Number 86907 by TawaTawa1 last updated on 01/Apr/20

Commented by mr W last updated on 01/Apr/20

question is wrong !? there are infinite possibilities for PQRS. (([PQRS])/(ΔABC))≠constant, but min.=(2/5)

$${question}\:{is}\:{wrong}\:!? \\ $$$${there}\:{are}\:{infinite}\:{possibilities}\:{for} \\ $$$${PQRS}. \\ $$$$\frac{\left[{PQRS}\right]}{\Delta{ABC}}\neq{constant},\:{but}\:{min}.=\frac{\mathrm{2}}{\mathrm{5}} \\ $$

Commented by mr W last updated on 01/Apr/20

$${or}\:{you}\:{didn}'{t}\:{show}\:{the}\:{complete}\:{question}. \\ $$

Answered by mr W last updated on 01/Apr/20

let AB=BC=a Δ_(ABC) =(a^2 /2) let BQ=x, BP=y, PQ=s s^2 =x^2 +y^2 AP=a−y ((PS)/(sin 45°))=((AP)/(sin (45°+∠APS)))=((AP)/(sin (45°+180°−90°−∠BPQ))) (s/(sin 45°))=((a−y)/(sin (45°+90°−∠BPQ))) (s/(sin 45°))=((a−y)/(cos (45°−∠BPQ))) s=((a−y)/(cos ∠BPQ+sin ∠BPQ)) ((a−y)/s)=cos ∠BPQ+sin ∠BPQ=(x/s)+(y/s) ⇒x+2y=a there are infinite possibilities: (a/3)≤y≤(a/2) [PQRS]=s^2 =x^2 +y^2 =(a−2y)^2 +y^2 (([PQRS])/(ΔABC))=((2s^2 )/a^2 )=((2[(a−2y)^2 +y^2 ])/a^2 )=2[(1−2(y/a))^2 +((y/a))^2 ] min(([PQRS])/(ΔABC))=(2/5) when y=((2a)/5), x=(a/5).

$${let}\:{AB}={BC}={a} \\ $$$$\Delta_{{ABC}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$${let}\:{BQ}={x},\:{BP}={y},\:{PQ}={s} \\ $$$${s}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$ \\ $$$${AP}={a}−{y} \\ $$$$\frac{{PS}}{\mathrm{sin}\:\mathrm{45}°}=\frac{{AP}}{\mathrm{sin}\:\left(\mathrm{45}°+\angle{APS}\right)}=\frac{{AP}}{\mathrm{sin}\:\left(\mathrm{45}°+\mathrm{180}°−\mathrm{90}°−\angle{BPQ}\right)} \\ $$$$\frac{{s}}{\mathrm{sin}\:\mathrm{45}°}=\frac{{a}−{y}}{\mathrm{sin}\:\left(\mathrm{45}°+\mathrm{90}°−\angle{BPQ}\right)} \\ $$$$\frac{{s}}{\mathrm{sin}\:\mathrm{45}°}=\frac{{a}−{y}}{\mathrm{cos}\:\left(\mathrm{45}°−\angle{BPQ}\right)} \\ $$$${s}=\frac{{a}−{y}}{\mathrm{cos}\:\angle{BPQ}+\mathrm{sin}\:\angle{BPQ}} \\ $$$$\frac{{a}−{y}}{{s}}=\mathrm{cos}\:\angle{BPQ}+\mathrm{sin}\:\angle{BPQ}=\frac{{x}}{{s}}+\frac{{y}}{{s}} \\ $$$$\Rightarrow{x}+\mathrm{2}{y}={a} \\ $$$${there}\:{are}\:{infinite}\:{possibilities}: \\ $$$$\frac{{a}}{\mathrm{3}}\leqslant{y}\leqslant\frac{{a}}{\mathrm{2}} \\ $$$$ \\ $$$$\left[{PQRS}\right]={s}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({a}−\mathrm{2}{y}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\frac{\left[{PQRS}\right]}{\Delta{ABC}}=\frac{\mathrm{2}{s}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\mathrm{2}\left[\left({a}−\mathrm{2}{y}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \right]}{{a}^{\mathrm{2}} }=\mathrm{2}\left[\left(\mathrm{1}−\mathrm{2}\frac{{y}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{a}}\right)^{\mathrm{2}} \right] \\ $$$$ \\ $$$${min}\frac{\left[{PQRS}\right]}{\Delta{ABC}}=\frac{\mathrm{2}}{\mathrm{5}}\:{when}\:{y}=\frac{\mathrm{2}{a}}{\mathrm{5}},\:{x}=\frac{{a}}{\mathrm{5}}. \\ $$

Commented by TawaTawa1 last updated on 01/Apr/20