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Question-112780

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Question Number 112780 by ajfour last updated on 09/Sep/20
Commented by ajfour last updated on 09/Sep/20
If regions A, B, C, D have equal areas, find P (h,k) in terms of the square side a.
$${If}\:\:{regions}\:{A},\:{B},\:{C},\:{D}\:{have}\:{equal} \\ $$$${areas},\:{find}\:{P}\:\left({h},{k}\right)\:{in}\:{terms}\:{of}\:{the} \\ $$$${square}\:{side}\:\boldsymbol{{a}}. \\ $$
Answered by mr W last updated on 09/Sep/20
Commented by mr W last updated on 09/Sep/20
((pa)/2)=(a^2 /3) ⇒p=((2a)/3) ((q+r)/2)×a=(a^2 /3) ⇒q+r=((2a)/3) (h/k)=(p/a)=(2/3) h=((2k)/3) ((q−r)/a)=((h−a+q)/k)=(2/3)−((a−q)/k) (1/2)(a−q)k=((2a^2 )/3) k=((4a^2 )/(3(a−q))) ((q−((2a)/3)+q)/a)=(2/3)−((3(a−q)^2 )/(4a^2 )) (q/a)=(2/3)−(3/8)(1−(q/a))^2 −(2/3)+(3/8)(1−2λ+λ^2 )+λ=0 9λ^2 +6λ−7=0 λ=((2(√2)−1)/3)=(q/a) ⇒(k/a)=(4/(3(1−((2(√2)−1)/3))))=2+(√2) ⇒(h/a)=((2(2+(√2)))/3)
$$\frac{{pa}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{p}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\frac{{q}+{r}}{\mathrm{2}}×{a}=\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{q}+{r}=\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\frac{{h}}{{k}}=\frac{{p}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${h}=\frac{\mathrm{2}{k}}{\mathrm{3}} \\ $$$$\frac{{q}−{r}}{{a}}=\frac{{h}−{a}+{q}}{{k}}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{{a}−{q}}{{k}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{q}\right){k}=\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${k}=\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}\left({a}−{q}\right)} \\ $$$$\frac{{q}−\frac{\mathrm{2}{a}}{\mathrm{3}}+{q}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}\left({a}−{q}\right)^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$$\frac{{q}}{{a}}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{1}−\frac{{q}}{{a}}\right)^{\mathrm{2}} \\ $$$$−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{1}−\mathrm{2}\lambda+\lambda^{\mathrm{2}} \right)+\lambda=\mathrm{0} \\ $$$$\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{6}\lambda−\mathrm{7}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{3}}=\frac{{q}}{{a}} \\ $$$$\Rightarrow\frac{{k}}{{a}}=\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{3}}\right)}=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{{h}}{{a}}=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{3}} \\ $$
Commented by ajfour last updated on 10/Sep/20
Thank you Sir, nice solution, errorless.
$${Thank}\:{you}\:{Sir},\:{nice}\:{solution},\:{errorless}. \\ $$
Commented by ajfour last updated on 10/Sep/20
p=a−(p+r)+(a−q)=q+r = (((a^2 /3))/((a/2))) ⇒ p=((2a)/3) , q+r=((2a)/3) ...(i), (ii) Also (a−q)(k/2)=((2a^2 )/3) ⇒ q=a−((4a^2 )/(3k)) ....(iii) (k/h)=(a/((p=((2a)/3))))=(3/2) ...(iv) ((q−r)/a)=((h−(a−q))/k) ...(v) (ii)+a×(v) using (iii),(iv) alongwith 2(a−((4a^2 )/(3k)))=((2a)/3)+((2a)/3)−a(((4a^2 )/(3k^2 ))) ⇒ ((2a)/3)((k/a))^2 −((8a^2 )/(3k))((k^2 /a^2 ))+((4a)/3)=0 ⇒ ((k/a))^2 −4((k/a))+2=0 ⇒ (k/a)=2+(√2) , h=(2/3)(2+(√2)). .............................................
$${p}={a}−\left({p}+{r}\right)+\left({a}−{q}\right)={q}+{r}\:=\:\frac{\left({a}^{\mathrm{2}} /\mathrm{3}\right)}{\left({a}/\mathrm{2}\right)} \\ $$$$\Rightarrow\:{p}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:,\:\:{q}+{r}=\frac{\mathrm{2}{a}}{\mathrm{3}}\:\:\:\:\:\:…\left({i}\right),\:\left({ii}\right) \\ $$$${Also}\:\:\:\left({a}−{q}\right)\frac{{k}}{\mathrm{2}}=\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}}\:\:\:\: \\ $$$$\Rightarrow\:\:\:{q}={a}−\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}{k}}\:\:\:\:\:….\left({iii}\right) \\ $$$$\:\:\:\frac{{k}}{{h}}=\frac{{a}}{\left({p}=\frac{\mathrm{2}{a}}{\mathrm{3}}\right)}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:…\left({iv}\right) \\ $$$$\:\:\frac{{q}−{r}}{{a}}=\frac{{h}−\left({a}−{q}\right)}{{k}}\:\:\:\:\:\:\:\:…\left({v}\right) \\ $$$$\left({ii}\right)+{a}×\left({v}\right)\:\:{using}\:\left({iii}\right),\left({iv}\right)\:{alongwith} \\ $$$$\mathrm{2}\left({a}−\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}{k}}\right)=\frac{\mathrm{2}{a}}{\mathrm{3}}+\frac{\mathrm{2}{a}}{\mathrm{3}}−{a}\left(\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}{k}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:\frac{\mathrm{2}{a}}{\mathrm{3}}\left(\frac{{k}}{{a}}\right)^{\mathrm{2}} −\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{3}{k}}\left(\frac{{k}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)+\frac{\mathrm{4}{a}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\frac{{k}}{{a}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{k}}{{a}}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{{k}}{{a}}=\mathrm{2}+\sqrt{\mathrm{2}}\:\:\:,\:\:\:{h}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right). \\ $$$$\:\:\:……………………………………… \\ $$

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