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Question Number 161764 by Tawa11 last updated on 22/Dec/21

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{Find}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{yellow}\mathrm{circle}.$$

Commented by cortano last updated on 22/Dec/21

$$\frac{1}{\sqrt{r}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{4}}=\frac{2+\sqrt{2}}{2\sqrt{2}}$$$$\sqrt{r}=\frac{2\sqrt{2}}{2+\sqrt{2}}=\frac{2\sqrt{2}(2-\sqrt{2})}{2}=2\sqrt{2}-2$$$$r=4{(\sqrt{2}-1)}^2$$

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{Can}\mathrm{you}\mathrm{show}\mathrm{prove}\mathrm{the}\mathrm{formular}\mathrm{sir}??$$

Commented by cortano last updated on 22/Dec/21

$$thisDescartestheorem$$

Answered by FongXD last updated on 22/Dec/21

$$\mathrm{let}\mathrm{a},\mathrm{b}\mathrm{be}\mathrm{the}\mathrm{radii}\mathrm{of}\mathrm{the}\mathrm{bigger}\mathrm{circles}$$$$\mathrm{and}\mathrm{let}\mathrm{r}\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{smallest}\mathrm{circle}$$$$\mathrm{draw}\mathrm{some}\mathrm{right}\mathrm{triangles},\mathrm{and}{\mathrm{we}}^{\prime }\mathrm{ll}\mathrm{get}:$$$$\sqrt{{(\mathrm{a}+\mathrm{r})}^2-{(\mathrm{a}-\mathrm{r})}^2}+\sqrt{{(\mathrm{b}+\mathrm{r})}^2-{(\mathrm{b}-\mathrm{r})}^2}=\sqrt{{(\mathrm{b}+\mathrm{a})}^2-{(\mathrm{b}-\mathrm{a})}^2}$$$$2\sqrt{\mathrm{ar}}+2\sqrt{\mathrm{br}}=2\sqrt{\mathrm{ab}}$$$$\sqrt{\mathrm{r}}=\frac{\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}},\Rightarrow \frac{1}{\sqrt{\mathrm{r}}}=\frac{1}{\sqrt{\mathrm{a}}}+\frac{1}{\sqrt{\mathrm{b}}}$$

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 22/Dec/21

$${(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4})}^2=2(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r_4^2})$$$$r_4=\mathrm{\infty }$$$${(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3})}^2=2(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2})$$$${\left(\frac{1}{r_3}\right)}^2-2(\frac{1}{r_1}+\frac{1}{r_2})\left(\frac{1}{r_3}\right)+{(\frac{1}{r_1}-\frac{1}{r_2})}^2=0$$$$\frac{1}{r_3}=\frac{1}{r_1}+\frac{1}{r_2}\pm 2\sqrt{\frac{1}{r_1{r}_2}}={(\frac{1}{\sqrt{r_1}}\pm \frac{1}{\sqrt{r_2}})}^2$$$$\Rightarrow \frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{r_1}}\pm \frac{1}{\sqrt{r_2}}$$

Commented by mr W last updated on 22/Dec/21

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by mr W last updated on 22/Dec/21

$$r_1=2,r_2=4,r_4=\mathrm{\infty }$$$$\frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{4}}=\frac{2+\sqrt{2}}{2\sqrt{2}}$$$$r_3={\left(\frac{2\sqrt{2}}{2+\sqrt{2}}\right)}^2=4(3-2\sqrt{2})\approx 0.686$$

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