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Question Number 186551 by ajfour last updated on 05/Feb/23

Commented by mr W last updated on 06/Feb/23

square with side length 1? A_(shade) =((t^2 (t+2)^2 )/(4(1+t)[(1+t)^2 +1]))

$${square}\:{with}\:{side}\:{length}\:\mathrm{1}? \\ $$$${A}_{{shade}} =\frac{{t}^{\mathrm{2}} \left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+{t}\right)\left[\left(\mathrm{1}+{t}\right)^{\mathrm{2}} +\mathrm{1}\right]} \\ $$

Answered by ajfour last updated on 06/Feb/23

Left bottom corner be Origin. q=(p/(1+t)) and q=(t+1)(1−p) ⇒ p=(((t+1)^2 )/(1+(t+1)^2 )) q=((t+1)/(1+(t+1)^2 )) k=((t+1)/2)−(1/(2(t+1))) ; h=p−(1/2) Area=kh =(((t+1)/2)−(1/(2(t+1))))((((t+1)^2 )/(1+(t+1)^2 ))−(1/2)) =(({(t+1)^2 −1})/(2(t+1)))×(({(t+1)^2 −1})/(2(t^2 +2t+2))) A=(({(t+1)^2 −1}^2 )/(4(t+1)(t^2 +2t+2))) A=((t^2 (t+2)^2 )/(4(t+1)(t^2 +2t+2)))

$${Left}\:{bottom}\:{corner}\:{be}\:{Origin}. \\ $$$${q}=\frac{{p}}{\mathrm{1}+{t}}\:\:\:{and}\:\:\:{q}=\left({t}+\mathrm{1}\right)\left(\mathrm{1}−{p}\right) \\ $$$$\Rightarrow\:\:{p}=\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{q}=\frac{{t}+\mathrm{1}}{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${k}=\frac{{t}+\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:;\:\:{h}={p}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Area}={kh} \\ $$$$\:\:\:=\left(\frac{{t}+\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\right)\left(\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:=\frac{\left\{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}}{\mathrm{2}\left({t}+\mathrm{1}\right)}×\frac{\left\{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$$${A}=\frac{\left\{\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right\}^{\mathrm{2}} }{\mathrm{4}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$$${A}=\frac{{t}^{\mathrm{2}} \left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$$$\:\:\:\: \\ $$

Answered by mr W last updated on 06/Feb/23

Commented by mr W last updated on 06/Feb/23

h_2 =((1+t)/2) h_1 =(1/(2(1+t))) ((1−b)/h_3 )=((1+t)/1) ⇒(1/h_3 )=((1+t)/(1−b)) (b/h_3 )=(1/(1+t)) ⇒(1/h_3 )=(1/(b(1+t))) ((1+t)/(1−b))=(1/(b(1+t))) ⇒b=(1/((1+t)^2 +1)) A_(shade) =(h_2 −h_1 )((1/2)−b) =[((1+t)/2)−(1/(2(1+t)))][(1/2)−(1/((1+t)^2 +1))] =((t^2 (t+2)^2 )/(4(1+t)(t^2 +2t+2)))

$${h}_{\mathrm{2}} =\frac{\mathrm{1}+{t}}{\mathrm{2}} \\ $$$${h}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{t}\right)} \\ $$$$\frac{\mathrm{1}−{b}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}+{t}}{\mathrm{1}}\:\Rightarrow\frac{\mathrm{1}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}+{t}}{\mathrm{1}−{b}} \\ $$$$\frac{{b}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\Rightarrow\frac{\mathrm{1}}{{h}_{\mathrm{3}} }=\frac{\mathrm{1}}{{b}\left(\mathrm{1}+{t}\right)} \\ $$$$\frac{\mathrm{1}+{t}}{\mathrm{1}−{b}}=\frac{\mathrm{1}}{{b}\left(\mathrm{1}+{t}\right)} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${A}_{{shade}} =\left({h}_{\mathrm{2}} −{h}_{\mathrm{1}} \right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{b}\right) \\ $$$$\:\:\:\:\:=\left[\frac{\mathrm{1}+{t}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{t}\right)}\right]\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\:\:\:\:\:=\frac{{t}^{\mathrm{2}} \left({t}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+{t}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)} \\ $$

Commented by ajfour last updated on 06/Feb/23

Thanks you Sir, I corrected!

$${Thanks}\:{you}\:{Sir},\:{I}\:{corrected}! \\ $$

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