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Question Number 182600 by mnjuly1970 last updated on 11/Dec/22

Answered by mr W last updated on 11/Dec/22

Method I R=2 r_1 =1 r_2 =r_3 =r=? ((2/r)+(1/1)−(1/2))^2 =2((2/r^2 )+(1/1^2 )+(1/2^2 )) (1/r)−(9/8)=0 ⇒r=(8/9)

$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$${R}=\mathrm{2} \\ $$$${r}_{\mathrm{1}} =\mathrm{1} \\ $$$${r}_{\mathrm{2}} ={r}_{\mathrm{3}} ={r}=? \\ $$$$\left(\frac{\mathrm{2}}{{r}}+\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{2}}{{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{r}}−\frac{\mathrm{9}}{\mathrm{8}}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$

Commented by Acem last updated on 11/Dec/22

r= (8/9)

$$\:{r}=\:\frac{\mathrm{8}}{\mathrm{9}}\: \\ $$

Answered by mr W last updated on 11/Dec/22

Commented by mr W last updated on 11/Dec/22

Method II (√((1+r)^2 −r^2 ))−1=(√((2−r)^2 −r^2 )) 3r−1=(√(1+2r)) 9r^2 −8r=0 ⇒r=(8/9)

$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$$\sqrt{\left(\mathrm{1}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}=\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\mathrm{3}{r}−\mathrm{1}=\sqrt{\mathrm{1}+\mathrm{2}{r}} \\ $$$$\mathrm{9}{r}^{\mathrm{2}} −\mathrm{8}{r}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$

Commented by mnjuly1970 last updated on 12/Dec/22

thx sir W

$${thx}\:{sir}\:{W} \\ $$

Answered by Acem last updated on 11/Dec/22

Commented by Acem last updated on 11/Dec/22

• (1+r)^2 = r^2 + (1+x)^2 ⇒ r= x((1/2) x+ 1) ... i • (y+r)^2 = r^2 + x^2 ⇒ y^2 = x^2 − 2ry ...ii • y= 2−2r ..... iii iii∧ii∧i ⇒ 3x^2 + 4x −4= 0 x= (2/3) ⇒ r= (8/9)

$$\:\bullet\:\left(\mathrm{1}+{r}\right)^{\mathrm{2}} =\:{r}^{\mathrm{2}} +\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \:\Rightarrow\:{r}=\:{x}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{x}+\:\mathrm{1}\right)\:...\:{i} \\ $$$$\:\bullet\:\left({y}+{r}\right)^{\mathrm{2}} =\:{r}^{\mathrm{2}} +\:{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{y}^{\mathrm{2}} =\:{x}^{\mathrm{2}} −\:\mathrm{2}{ry}\:...{ii} \\ $$$$\:\bullet\:{y}=\:\mathrm{2}−\mathrm{2}{r}\:\:.....\:{iii} \\ $$$$\:{iii}\wedge{ii}\wedge{i}\:\Rightarrow\:\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{4}{x}\:−\mathrm{4}=\:\mathrm{0} \\ $$$$\:{x}=\:\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\Rightarrow\:{r}=\:\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 12/Dec/22

very nice .thx sir Acm

$$\:{very}\:{nice}\:.{thx}\:{sir}\:{Acm} \\ $$

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