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Question Number 185674 by ajfour last updated on 25/Jan/23

Commented by ajfour last updated on 25/Jan/23

$$Findtheradii\left(equal\right)ofthe$$$$circles,givenouterfigureis$$$$asquareofsidea.$$

Answered by mr W last updated on 25/Jan/23

Commented by mr W last updated on 25/Jan/23

$$x=a-2r-\frac{r}{\tan 22.5°}=a-2r-\frac{r}{\sqrt{2}-1}$$$$y=a-r$$$${(x+y)}^2={(x+r)}^2+{(y+r)}^2$$$$xy=(x+y)r+r^2$$$$(a-2r-\frac{r}{\sqrt{2}-1})(a-r)=(a-2r-\frac{r}{\sqrt{2}-1}+a-r)r+r^2$$$$let\lambda =\frac{a}{r}$$$${\lambda }^2-(6+\sqrt{2})\lambda +6+2\sqrt{2}=0$$$$\lambda =\frac{6+\sqrt{2}+\sqrt{14+4\sqrt{2}}}{2}$$$$\Rightarrow \frac{r}{a}=\frac{2}{6+\sqrt{2}+\sqrt{14+4\sqrt{2}}}\approx 0.169$$

Commented by ajfour last updated on 27/Jan/23

$$Thankyousir!$$

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