Question-213021

Question Number 213021 by Spillover last updated on 28/Oct/24

Commented by Ghisom last updated on 29/Oct/24

$${r}_{\mathrm{1}} =\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx.\mathrm{351153302} \\ $$$${r}_{\mathrm{2}} =\frac{\left(\mathrm{1}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\sqrt{\mathrm{2}}}{\mathrm{2}}\approx.\mathrm{165910681} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\mathrm{2}−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx\mathrm{2}.\mathrm{11652017} \\ $$

Answered by a.lgnaoui last updated on 29/Oct/24

$$\boldsymbol{\mathrm{r}}_{\mathrm{1}} =\frac{\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{1}+\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{2}} =\frac{\sqrt{\mathrm{55}−\mathrm{36}\sqrt{\mathrm{2}}\:\:}\:−\mathrm{3}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{1}+\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\: \\ $$$$\boldsymbol{\mathrm{then}}\:\:\:\frac{\boldsymbol{\mathrm{r}}_{\mathrm{1}} }{\boldsymbol{\mathrm{r}}_{\mathrm{2}} }=\frac{\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{55}−\mathrm{36}\sqrt{\mathrm{2}}}\:−\mathrm{3}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\: \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 29/Oct/24

Commented by a.lgnaoui last updated on 29/Oct/24

Commented by Ghisom last updated on 29/Oct/24

$$\mathrm{the}\:\mathrm{circles}\:\mathrm{have}\:\mathrm{no}\:\mathrm{common}\:\mathrm{point} \\ $$$$\mathrm{the}\:\mathrm{big}\:\mathrm{circle}\:\mathrm{touches}\:\mathrm{the}\:\mathrm{skewed}\:\mathrm{line}\:\mathrm{at} \\ $$$${F}\approx\begin{pmatrix}{.\mathrm{324}}\\{.\mathrm{217}}\end{pmatrix}\:\mathrm{while}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle}\:\mathrm{touches} \\ $$$$\mathrm{this}\:\mathrm{line}\:\mathrm{at}\:{G}\approx\begin{pmatrix}{.\mathrm{319}}\\{.\mathrm{229}}\end{pmatrix} \\ $$$$\mid{FG}\mid\approx.\mathrm{0137} \\ $$$$\Rightarrow\:\mathrm{your}\:\mathrm{result}\:\mathrm{is}\:\mathrm{wrong} \\ $$

Answered by mr W last updated on 29/Oct/24

Commented by mr W last updated on 29/Oct/24

$$\frac{{b}}{{a}}=\sqrt{\mathrm{2}}\:\Rightarrow{b}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\frac{{AB}}{\mathrm{1}}=\frac{{b}}{{a}}\:\Rightarrow{AB}=\sqrt{\mathrm{2}} \\ $$$${EC}={c}−{r}_{\mathrm{1}} +{b}−{r}_{\mathrm{1}} ={b}+{c}−\mathrm{2}{r}_{\mathrm{1}} \\ $$$${EC}^{\mathrm{2}} =\left(\mathrm{2}{r}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({c}−{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{r}_{\mathrm{1}} ^{\mathrm{2}} +\left({c}−{b}\right)^{\mathrm{2}} =\left({b}+{c}−\mathrm{2}{r}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\frac{{bc}}{{b}+{c}} \\ $$$$\frac{{c}}{{b}}=\frac{\sqrt{\mathrm{2}}+\mathrm{2}{r}_{\mathrm{1}} }{\:\sqrt{\mathrm{2}}}=\mathrm{1}+\sqrt{\mathrm{2}}×\frac{{bc}}{{b}+{c}}=\mathrm{1}+\sqrt{\mathrm{2}}×\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}×\frac{\left(\frac{{c}}{{b}}\right)}{\mathrm{1}+\left(\frac{{c}}{{b}}\right)} \\ $$$$\left(\frac{{c}}{{b}}\right)^{\mathrm{2}} −\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\frac{{c}}{{b}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{c}}{{b}}=\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\frac{{c}}{{a}}=\frac{{b}}{{a}}×\frac{{c}}{{b}}=\sqrt{\mathrm{2}}\left[\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}\right] \\ $$$$\:\:\:\:\:=\mathrm{2}−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx\mathrm{2}.\mathrm{116520} \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com