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Question-213021

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Question Number 213021 by Spillover last updated on 28/Oct/24
Commented by Ghisom last updated on 29/Oct/24
r_1 =1−(√2)+(√(2−(√2)))≈.351153302 r_2 =(((1−(√(2−(√2))))(√2))/2)≈.165910681 (r_1 /r_2 )=2−(√2)+2(√(2−(√2)))≈2.11652017
$${r}_{\mathrm{1}} =\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx.\mathrm{351153302} \\ $$$${r}_{\mathrm{2}} =\frac{\left(\mathrm{1}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\sqrt{\mathrm{2}}}{\mathrm{2}}\approx.\mathrm{165910681} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\mathrm{2}−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx\mathrm{2}.\mathrm{11652017} \\ $$
Answered by a.lgnaoui last updated on 29/Oct/24
r_1 =((√(6−4(√2)))/(1+(√(6−4(√2))))) r_2 =(((√(55−36(√2) )) −3(√(6−4(√2))))/(1+(√(6−4(√2))))) then (r_1 /r_2 )=((√(6−4(√2)))/( (√(55−36(√2))) −3(√(6−4(√2)))))
$$\boldsymbol{\mathrm{r}}_{\mathrm{1}} =\frac{\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{1}+\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\boldsymbol{\mathrm{r}}_{\mathrm{2}} =\frac{\sqrt{\mathrm{55}−\mathrm{36}\sqrt{\mathrm{2}}\:\:}\:−\mathrm{3}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{1}+\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\: \\ $$$$\boldsymbol{\mathrm{then}}\:\:\:\frac{\boldsymbol{\mathrm{r}}_{\mathrm{1}} }{\boldsymbol{\mathrm{r}}_{\mathrm{2}} }=\frac{\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{55}−\mathrm{36}\sqrt{\mathrm{2}}}\:−\mathrm{3}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}} \\ $$$$\: \\ $$$$ \\ $$
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
the circles have no common point the big circle touches the skewed line at F≈ (((.324)),((.217)) ) while the small circle touches this line at G≈ (((.319)),((.229)) ) ∣FG∣≈.0137 ⇒ your result is wrong
$$\mathrm{the}\:\mathrm{circles}\:\mathrm{have}\:\mathrm{no}\:\mathrm{common}\:\mathrm{point} \\ $$$$\mathrm{the}\:\mathrm{big}\:\mathrm{circle}\:\mathrm{touches}\:\mathrm{the}\:\mathrm{skewed}\:\mathrm{line}\:\mathrm{at} \\ $$$${F}\approx\begin{pmatrix}{.\mathrm{324}}\\{.\mathrm{217}}\end{pmatrix}\:\mathrm{while}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle}\:\mathrm{touches} \\ $$$$\mathrm{this}\:\mathrm{line}\:\mathrm{at}\:{G}\approx\begin{pmatrix}{.\mathrm{319}}\\{.\mathrm{229}}\end{pmatrix} \\ $$$$\mid{FG}\mid\approx.\mathrm{0137} \\ $$$$\Rightarrow\:\mathrm{your}\:\mathrm{result}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Answered by mr W last updated on 29/Oct/24
Commented by mr W last updated on 29/Oct/24
(b/a)=(√2) ⇒b=((√2)/( (√2)+1)) ((AB)/1)=(b/a) ⇒AB=(√2) EC=c−r_1 +b−r_1 =b+c−2r_1 EC^2 =(2r_1 )^2 +(c−b)^2 4r_1 ^2 +(c−b)^2 =(b+c−2r_1 )^2 ⇒r_1 =((bc)/(b+c)) (c/b)=(((√2)+2r_1 )/( (√2)))=1+(√2)×((bc)/(b+c))=1+(√2)×((√2)/( (√2)+1))×((((c/b)))/(1+((c/b)))) ((c/b))^2 −2((√2)−1)((c/b))−1=0 ⇒(c/b)=(√2)−1+(√(2(2−(√2)))) (r_1 /r_2 )=(c/a)=(b/a)×(c/b)=(√2)[(√2)−1+(√(2(2−(√2))))] =2−(√2)+2(√(2−(√2)))≈2.116520
$$\frac{{b}}{{a}}=\sqrt{\mathrm{2}}\:\Rightarrow{b}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\frac{{AB}}{\mathrm{1}}=\frac{{b}}{{a}}\:\Rightarrow{AB}=\sqrt{\mathrm{2}} \\ $$$${EC}={c}−{r}_{\mathrm{1}} +{b}−{r}_{\mathrm{1}} ={b}+{c}−\mathrm{2}{r}_{\mathrm{1}} \\ $$$${EC}^{\mathrm{2}} =\left(\mathrm{2}{r}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({c}−{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{r}_{\mathrm{1}} ^{\mathrm{2}} +\left({c}−{b}\right)^{\mathrm{2}} =\left({b}+{c}−\mathrm{2}{r}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\frac{{bc}}{{b}+{c}} \\ $$$$\frac{{c}}{{b}}=\frac{\sqrt{\mathrm{2}}+\mathrm{2}{r}_{\mathrm{1}} }{\:\sqrt{\mathrm{2}}}=\mathrm{1}+\sqrt{\mathrm{2}}×\frac{{bc}}{{b}+{c}}=\mathrm{1}+\sqrt{\mathrm{2}}×\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}×\frac{\left(\frac{{c}}{{b}}\right)}{\mathrm{1}+\left(\frac{{c}}{{b}}\right)} \\ $$$$\left(\frac{{c}}{{b}}\right)^{\mathrm{2}} −\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\frac{{c}}{{b}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{c}}{{b}}=\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\frac{{c}}{{a}}=\frac{{b}}{{a}}×\frac{{c}}{{b}}=\sqrt{\mathrm{2}}\left[\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}\right] \\ $$$$\:\:\:\:\:=\mathrm{2}−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx\mathrm{2}.\mathrm{116520} \\ $$
Commented by TonyCWX08 last updated on 31/Oct/24
What App do you use to draw diagrams like this? And why I can′t send pictures?
$${What}\:{App}\:{do}\:{you}\:{use}\:{to}\:{draw}\:{diagrams}\:{like}\:{this}? \\ $$$${And}\:{why}\:{I}\:{can}'{t}\:{send}\:{pictures}? \\ $$
Commented by mr W last updated on 31/Oct/24
i don′t use a special app for drawing. every smartphone has an app for editing fotos. you can draw lines, make free−hand sketch etc. there is a good app called Lekh for drawing diagrams, see posts from ajfour sir. following picture shows how one can upload a picture as answer or as comment. if it doesn′t work with your smartphone, then you should contact the developer Tinku Tara.
$${i}\:{don}'{t}\:{use}\:{a}\:{special}\:{app}\:{for}\:{drawing}. \\ $$$${every}\:{smartphone}\:{has}\:{an}\:{app}\:{for} \\ $$$${editing}\:{fotos}.\:{you}\:{can}\:{draw}\:{lines}, \\ $$$${make}\:{free}−{hand}\:{sketch}\:{etc}. \\ $$$${there}\:{is}\:{a}\:{good}\:{app}\:{called}\:{Lekh}\:{for} \\ $$$${drawing}\:{diagrams},\:{see}\:{posts}\:{from} \\ $$$${ajfour}\:{sir}.\: \\ $$$${following}\:{picture}\:{shows}\:{how}\:{one}\: \\ $$$${can}\:{upload}\:{a}\:{picture}\:{as}\:{answer}\:{or}\:{as} \\ $$$${comment}.\:{if}\:{it}\:{doesn}'{t}\:{work}\:{with} \\ $$$${your}\:{smartphone},\:{then}\:{you}\:{should} \\ $$$${contact}\:{the}\:{developer}\:{Tinku}\:{Tara}. \\ $$
Commented by mr W last updated on 31/Oct/24

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