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Question Number 52129 by ajfour last updated on 03/Jan/19

Commented by ajfour last updated on 03/Jan/19

$$Findradius\mathit{r}ofcircles\left(thesame\right)$$$$intermsofsquareside\mathit{a}.$$

Commented by ajfour last updated on 03/Jan/19

$$r\sqrt{2}+r+b=a\sqrt{2}....\left(i\right)$$$$AP{}^2=b^2+{(r+r\sqrt{2})}^2...\left(ii\right)$$$$alsoAP=(a-r)+(a-r-CP)$$$$\Rightarrow AP=a-r+a-r-\sqrt{2}(r+r\sqrt{2})$$$$AP=2a-r(4+\sqrt{2})...\left(iii\right)$$$$\Rightarrow {[2a-r(4+\sqrt{2})]}^2={[a\sqrt{2}-r(1+\sqrt{2})]}^2$$$$+{(r+r\sqrt{2})}^2$$$$\Rightarrow 4a^2-4ar(4+\sqrt{2})+(18+8\sqrt{2})r^2$$$$=2a^2-2ar(2+\sqrt{2})+(6+4\sqrt{2})r^2$$$$\Rightarrow 2a^2-ar(12+2\sqrt{2})+(12+4\sqrt{2})r^2=0$$$$\Rightarrow (6+2\sqrt{2})r^2-(6+\sqrt{2})ar+a^2=0$$$$let\frac{r}{a}=\lambda$$$$\Rightarrow (6+2\sqrt{2}){\lambda }^2-(6+\sqrt{2})\lambda +1=0$$$$\lambda \approx 0.1688a(\mathrm{\&}0.671anotacceptable).$$

Commented by ajfour last updated on 03/Jan/19

Commented by mr W last updated on 03/Jan/19

$$goodsolution!$$

Commented by Olalekan99 last updated on 04/Jan/19

$$plswatapp$$$$diduuseforthedrawing$$

Answered by mr W last updated on 03/Jan/19

$$\frac{ax}{2}=\frac{r}{2}(a+x+\sqrt{a^2+x^2})$$$$\Rightarrow r=\frac{ax}{(a+x+\sqrt{a^2+x^2})}...\left(i\right)$$$$\frac{{(a-x)}^2}{2}=\frac{r}{2}[2(a-x)+\sqrt{2}(a-x)]$$$$\Rightarrow r=\frac{a-x}{2+\sqrt{2}}...\left(ii\right)$$$$\frac{ax}{(a+x+\sqrt{a^2+x^2})}=\frac{a-x}{2+\sqrt{2}}$$$$let\lambda =\frac{x}{a}$$$$\frac{\lambda }{(1+\lambda +\sqrt{1+{\lambda }^2})}=\frac{1-\lambda }{2+\sqrt{2}}$$$${\lambda }^2+(2+\sqrt{2})\lambda -1=(1-\lambda )\sqrt{1+{\lambda }^2}$$$$\Rightarrow (3+\sqrt{2}){\lambda }^2+(1+2\sqrt{2})\lambda -(1+\sqrt{2})=0$$$$\lambda =\frac{\sqrt{{(1+2\sqrt{2})}^2+4(3+\sqrt{2})(1+\sqrt{2})}-(1+2\sqrt{2})}{2(3+\sqrt{2})}$$$$\lambda =\frac{\sqrt{79+46\sqrt{2}}-5\sqrt{2}+1}{14}\approx 0.4237$$$$\Rightarrow \frac{r}{a}=\frac{1-\lambda }{2+\sqrt{2}}=\frac{1}{2+\sqrt{2}}\left(\frac{13+5\sqrt{2}-\sqrt{79+46\sqrt{2}}}{14}\right)$$$$\Rightarrow \frac{r}{a}=\frac{16-3\sqrt{2}-\sqrt{2(53-20\sqrt{2})}}{28}\approx 0.1688$$

Commented by ajfour last updated on 03/Jan/19

$$ThankyouSir!\left(gotthesamefinally\right)$$$$andwhatsxinyoursolution.$$

Commented by mr W last updated on 03/Jan/19

$$xisBPinyourdiagram.$$

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