Question-42866

Question Number 42866 by Cheyboy last updated on 03/Sep/18

Answered by MJS last updated on 04/Sep/18

x→r to not get confused small circle touching half circle from inside r^2 +p^2 −((R/2)−r)^2 =0 ⇒ p^2 −(R^2 /4)+Rr=0 small circle touching ∼quarter circle from outside r^2 +(((3R)/2)+p)^2 −(((3R)/2)+r)^2 =0 ⇒ 3Rp+p^2 −3Rr=0 (1) Rr+p^2 −(R^2 /4)=0 (2) −3Rr+p^2 +3Rp=0 (1)−(2) 4Rr−3Rp−(R^2 /4)=0 4r−3p−(R/4)=0 ⇒ p=(4/3)r−(R/(12)) (1) Rr+((4/3)r−(R/(12)))^3 −(R^2 /4)=0 r^2 +((7R)/(16))r−((35R^2 )/(256))=0 r=((−7+3(√(21)))/(32))R p=((−3+(√(21)))/8)R R=8 ⇒ p=1.58 r=1.69 [btw the 2^(nd) solution for r and p gives a circle touching the half circle from outside and the ∼quarter circle from inside]

$${x}\rightarrow{r}\:\mathrm{to}\:\mathrm{not}\:\mathrm{get}\:\mathrm{confused} \\ $$$$\mathrm{small}\:\mathrm{circle}\:\mathrm{touching}\:\:\mathrm{half}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{inside} \\ $$$${r}^{\mathrm{2}} +{p}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{p}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}+{Rr}=\mathrm{0} \\ $$$$\mathrm{small}\:\mathrm{circle}\:\mathrm{touching}\:\sim\mathrm{quarter}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{outside} \\ $$$${r}^{\mathrm{2}} +\left(\frac{\mathrm{3}{R}}{\mathrm{2}}+{p}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}{R}}{\mathrm{2}}+{r}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\mathrm{3}{Rp}+{p}^{\mathrm{2}} −\mathrm{3}{Rr}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{Rr}+{p}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{3}{Rr}+{p}^{\mathrm{2}} +\mathrm{3}{Rp}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:\mathrm{4}{Rr}−\mathrm{3}{Rp}−\frac{{R}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{4}{r}−\mathrm{3}{p}−\frac{{R}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow\:{p}=\frac{\mathrm{4}}{\mathrm{3}}{r}−\frac{{R}}{\mathrm{12}} \\ $$$$\left(\mathrm{1}\right)\:{Rr}+\left(\frac{\mathrm{4}}{\mathrm{3}}{r}−\frac{{R}}{\mathrm{12}}\right)^{\mathrm{3}} −\frac{{R}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{7}{R}}{\mathrm{16}}{r}−\frac{\mathrm{35}{R}^{\mathrm{2}} }{\mathrm{256}}=\mathrm{0} \\ $$$${r}=\frac{−\mathrm{7}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{32}}{R} \\ $$$${p}=\frac{−\mathrm{3}+\sqrt{\mathrm{21}}}{\mathrm{8}}{R} \\ $$$${R}=\mathrm{8}\:\Rightarrow\:{p}=\mathrm{1}.\mathrm{58}\:\:{r}=\mathrm{1}.\mathrm{69} \\ $$$$ \\ $$$$\left[\mathrm{btw}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}\:\mathrm{for}\:{r}\:\mathrm{and}\:{p}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{circle}\right. \\ $$$$\mathrm{touching}\:\mathrm{the}\:\mathrm{half}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{outside}\:\mathrm{and} \\ $$$$\left.\mathrm{the}\:\sim\mathrm{quarter}\:\mathrm{circle}\:\mathrm{from}\:\mathrm{inside}\right] \\ $$

Commented by Cheyboy last updated on 04/Sep/18

$${Thank}\:{you}\:{sir} \\ $$

Commented by MJS last updated on 04/Sep/18

$$\mathrm{made}\:\mathrm{a}\:\mathrm{strange}\:\mathrm{mistake},\:\mathrm{corrected}\:\mathrm{now} \\ $$

Commented by Cheyboy last updated on 04/Sep/18

$${ok}\:{sir}\:{thank}\:{your} \\ $$

Answered by MrW3 last updated on 04/Sep/18

((√(((R/2)−x)^2 −x^2 ))+((3R)/2))^2 +x^2 =(((3R)/2)+x)^2 ((R/2)−x)^2 −x^2 +2(((3R)/2))(√(((R/2)−x)^2 −x^2 ))+((9R^2 )/4)+x^2 =((9R^2 )/4)+3Rx+x^2 (R^2 /4)−Rx+x^2 −x^2 +3R(√(((R/2)−x)^2 −x^2 ))+((9R^2 )/4)+x^2 =((9R^2 )/4)+3Rx+x^2 3(√(((R/2)−x)^2 −x^2 ))=4x−(R/4) 9((R^2 /4)−Rx+x^2 −x^2 )=16x^2 −2Rx+(R^2 /(16)) 16x^2 +7Rx−((35R^2 )/(16))=0 (x/R)=((−7+(√(49+4×35)))/(2×16))=(((√(189))−7)/(32))≈0.21 ⇒x≈0.21×8=1.69

$$\left(\sqrt{\left(\frac{{R}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{\mathrm{3}{R}}{\mathrm{2}}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\left(\frac{\mathrm{3}{R}}{\mathrm{2}}+{x}\right)^{\mathrm{2}} \\ $$$$\left(\frac{{R}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{3}{R}}{\mathrm{2}}\right)\sqrt{\left(\frac{{R}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{4}}+{x}^{\mathrm{2}} =\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{Rx}+{x}^{\mathrm{2}} \\ $$$$\frac{{R}^{\mathrm{2}} }{\mathrm{4}}−{Rx}+{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{3}{R}\sqrt{\left(\frac{{R}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{4}}+{x}^{\mathrm{2}} =\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}{Rx}+{x}^{\mathrm{2}} \\ $$$$\mathrm{3}\sqrt{\left(\frac{{R}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }=\mathrm{4}{x}−\frac{{R}}{\mathrm{4}} \\ $$$$\mathrm{9}\left(\frac{{R}^{\mathrm{2}} }{\mathrm{4}}−{Rx}+{x}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)=\mathrm{16}{x}^{\mathrm{2}} −\mathrm{2}{Rx}+\frac{{R}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{7}{Rx}−\frac{\mathrm{35}{R}^{\mathrm{2}} }{\mathrm{16}}=\mathrm{0} \\ $$$$\frac{{x}}{{R}}=\frac{−\mathrm{7}+\sqrt{\mathrm{49}+\mathrm{4}×\mathrm{35}}}{\mathrm{2}×\mathrm{16}}=\frac{\sqrt{\mathrm{189}}−\mathrm{7}}{\mathrm{32}}\approx\mathrm{0}.\mathrm{21} \\ $$$$\Rightarrow{x}\approx\mathrm{0}.\mathrm{21}×\mathrm{8}=\mathrm{1}.\mathrm{69} \\ $$

Commented by MrW3 last updated on 04/Sep/18