Question Number 15572 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${circle}\::{c}\:{tangent}\:{to}\:{semicircle}\:{AEB}\:{at}: \\ $$$${points}:\:{E}\:\&\:{D}.\:{AD}={a}=\mathrm{12},{DB}={b}=\mathrm{5}. \\ $$$$……………………………………………. \\ $$$$\left.\mathrm{1}\right){DE}={x}=?,{AE}=? \\ $$$$\left.\mathrm{2}\right)\:\:{show}\:{that}:\:\:\measuredangle{AED}=\mathrm{45}^{\bullet} . \\ $$
Answered by mrW1 last updated on 12/Jun/17
![R=radius of big circle r=radius of small circle R=((a+b)/2) OD=a−R=((a−b)/2) OC=OE−CE=R−r OC^2 =OD^2 +DC^2 (R−r)^2 =(((a−b)/2))^2 +r^2 R(R−2r)=(((a−b)^2 )/4) ⇒r=(1/2)[R−(((a−b)^2 )/(4R))]=((4R^2 −(a−b)^2 )/(8R)) =(((a+b)^2 −(a−b)^2 )/(4(a+b)))=((ab)/(a+b)) x^2 =2r^2 −2r^2 cos ∠DOE =2r^2 +2r^2 cos ∠OCD =2r^2 (1+((CD)/(OC)))=2r^2 (1+(r/(R−r))) =2r^2 (1+(1/((R/r)−1))) =2r^2 (1+(1/((((a+b)^2 )/(2ab))−1))) =2(((ab)/(a+b)))^2 (1+((2ab)/((a+b)^2 −2ab))) =2(((ab)/(a+b)))^2 ((((a+b)^2 )/(a^2 +b^2 ))) =((2(ab)^2 )/(a^2 +b^2 )) ⇒x=ab(√(2/(a^2 +b^2 ))) AE^2 =y^2 =2R^2 −2R^2 cos ∠AOE =2R^2 (1+cos ∠EOB) =2R^2 (1+((OD)/(OC))) =2R^2 (1+((a−b)/(2(R−r)))) R−r=((a+b)/2)−((ab)/(a+b))=(((a+b)^2 −2ab)/(2(a+b)))=((a^2 +b^2 )/(2(a+b))) y^2 =2R^2 (1+((a^2 −b^2 )/(a^2 +b^2 )))=2R^2 ((2a^2 )/(a^2 +b^2 )) =(a+b)^2 (a^2 /(a^2 +b^2 )) AE=y=((a(a+b))/( (√(a^2 +b^2 )))) x^2 +y^2 −2xycos ∠AED=AD^2 =a^2 ((2(ab)^2 )/(a^2 +b^2 ))+((a^2 (a+b)^2 )/(a^2 +b^2 ))−2(((√2)(ab)a(a+b))/(a^2 +b^2 ))cos ∠AED=a^2 2b^2 +(a+b)^2 −(a^2 +b^2 )=2(√2)b(a+b)cos ∠AED 2b(b+a)=2(√2)b(a+b)cos ∠AED 1=(√2)cos ∠AED (1/( (√2)))=cos ∠AED ⇒∠AED=(π/4)](https://www.tinkutara.com/question/Q15598.png)
$$\mathrm{R}=\mathrm{radius}\:\mathrm{of}\:\mathrm{big}\:\mathrm{circle} \\ $$$$\mathrm{r}=\mathrm{radius}\:\mathrm{of}\:\mathrm{small}\:\mathrm{circle} \\ $$$$\mathrm{R}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{OD}=\mathrm{a}−\mathrm{R}=\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}} \\ $$$$\mathrm{OC}=\mathrm{OE}−\mathrm{CE}=\mathrm{R}−\mathrm{r} \\ $$$$\mathrm{OC}^{\mathrm{2}} =\mathrm{OD}^{\mathrm{2}} +\mathrm{DC}^{\mathrm{2}} \\ $$$$\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} =\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{R}\left(\mathrm{R}−\mathrm{2r}\right)=\frac{\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{r}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{R}−\frac{\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{4R}}\right]=\frac{\mathrm{4R}^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{8R}} \\ $$$$=\frac{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{a}+\mathrm{b}\right)}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2r}^{\mathrm{2}} −\mathrm{2r}^{\mathrm{2}} \mathrm{cos}\:\angle\mathrm{DOE} \\ $$$$=\mathrm{2r}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} \mathrm{cos}\:\angle\mathrm{OCD} \\ $$$$=\mathrm{2r}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{CD}}{\mathrm{OC}}\right)=\mathrm{2r}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{r}}{\mathrm{R}−\mathrm{r}}\right) \\ $$$$=\mathrm{2r}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{R}}{\mathrm{r}}−\mathrm{1}}\right) \\ $$$$=\mathrm{2r}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\frac{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{2ab}}−\mathrm{1}}\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2ab}}{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{2ab}}\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\right)^{\mathrm{2}} \left(\frac{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{ab}\right)^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{x}=\mathrm{ab}\sqrt{\frac{\mathrm{2}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\mathrm{AE}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} −\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:\angle\mathrm{AOE} \\ $$$$=\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\angle\mathrm{EOB}\right) \\ $$$$=\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{OD}}{\mathrm{OC}}\right) \\ $$$$=\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}\left(\mathrm{R}−\mathrm{r}\right)}\right) \\ $$$$\mathrm{R}−\mathrm{r}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}=\frac{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{2ab}}{\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}=\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)} \\ $$$$\mathrm{y}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\right)=\mathrm{2R}^{\mathrm{2}} \frac{\mathrm{2a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} } \\ $$$$=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} \frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} } \\ $$$$\mathrm{AE}=\mathrm{y}=\frac{\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2xycos}\:\angle\mathrm{AED}=\mathrm{AD}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}\left(\mathrm{ab}\right)^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{a}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }−\mathrm{2}\frac{\sqrt{\mathrm{2}}\left(\mathrm{ab}\right)\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\mathrm{cos}\:\angle\mathrm{AED}=\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{2b}^{\mathrm{2}} +\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{b}\left(\mathrm{a}+\mathrm{b}\right)\mathrm{cos}\:\angle\mathrm{AED} \\ $$$$\mathrm{2b}\left(\mathrm{b}+\mathrm{a}\right)=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{b}\left(\mathrm{a}+\mathrm{b}\right)\mathrm{cos}\:\angle\mathrm{AED} \\ $$$$\mathrm{1}=\sqrt{\mathrm{2}}\mathrm{cos}\:\angle\mathrm{AED} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\mathrm{cos}\:\angle\mathrm{AED} \\ $$$$\Rightarrow\angle\mathrm{AED}=\frac{\pi}{\mathrm{4}} \\ $$
Commented by mrW1 last updated on 12/Jun/17
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{sir},\:\mathrm{thanks}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{fixed}. \\ $$
Commented by mrW1 last updated on 12/Jun/17
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}!\:\mathrm{thank}\:\mathrm{you}\:\mathrm{too}! \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${thank}\:{you}\:{dear}\:{mrW}\mathrm{1}. \\ $$$${x}=\frac{\mathrm{12}×\mathrm{5}}{\mathrm{12}+\mathrm{5}}\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{2}×\mathrm{12}×\mathrm{5}}{\mathrm{144}−\mathrm{25}}\right)}= \\ $$$$=\frac{\mathrm{60}}{\mathrm{17}}\sqrt{\mathrm{4}.\mathrm{02}}=\mathrm{7}.\mathrm{07} \\ $$$${right}\:{answer}\:{for}\:{x}:\:{x}=\mathrm{6}.\mathrm{52} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${cos}\measuredangle{OCD}=\frac{{CD}}{{OC}}=\frac{{r}}{{R}−{r}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}+\frac{{r}}{{R}−{r}}\right)=\frac{\mathrm{2}{Rr}^{\mathrm{2}} }{{R}−{r}}\Rightarrow{x}={r}\sqrt{\frac{\mathrm{2}{R}}{{R}−{r}}} \\ $$$${r}=\frac{{ab}}{{a}+{b}},{R}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${R}−{r}=\frac{{a}+{b}}{\mathrm{2}}−\frac{{ab}}{{a}+{b}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)} \\ $$$$\Rightarrow{x}=\frac{{ab}}{{a}+{b}}\sqrt{\frac{{a}+{b}}{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)}}}=\frac{{ab}\sqrt{\mathrm{2}}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$${y}=\frac{{a}\left({a}+{b}\right)}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$
Commented by mrW1 last updated on 12/Jun/17
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{thanks}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{fixed}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${thanks}\:{for}\:{you}\:.{your}\:{answer}\:{is}\:{very} \\ $$$${beautiful}.{i}\:{learn}\:{many}\:{things}\:{from} \\ $$$${your}\:{commenets}\:{and}\:{answers}. \\ $$$${god}\:{bless}\:{you}\:{master}. \\ $$
Answered by ajfour last updated on 12/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${thank}\:{you}\:{very}\:{much}.{it}'{s}\:{perfect}! \\ $$
Commented by ajfour last updated on 12/Jun/17
![considering ΔCOD: OD=(((a+b))/2)−b=((a−b)/2) OE=R=(((a+b)/2)) and CE=CD=r OC=OE−CE=R−r OD^2 =OC^2 −CD^2 ⇒ (((a−b)/2))^2 =(R−r)^2 −r^2 (((a−b)/2))^2 =R^2 −2rR (((a−b)/2))^2 =(((a+b)/2))^2 =2rR ⇒ 2rR=ab r=((ab)/(a+b)) . tan 2θ=((CD)/(OD))= (((((ab)/(a+b))))/((((a−b)/2)))) =((2ab)/(a^2 −b^2 )) ((2tan θ)/(1−tan^2 θ))=((2(b/a))/(1−(b/a)^2 )) ⇒ tan θ=(b/a) with CF ⊥ DE sin ∠DCF=((DF)/(CD))= (((x/2))/r) sin ((π/4)+θ)=(x/(2r)) (1/( (√2)))(sin θ+cos θ)=(x/(2r)) x=r(√2)((b/( (√(a^2 +b^2 ))))+(a/( (√(a^2 +b^2 ))))) x=(((√2)ab)/((a+b)))(((a+b))/( (√(a^2 +b^2 )))) x= (((√2)ab)/( (√(a^2 +b^2 )))) . y=2(AO)cos θ =2Rcos θ as R=(((a+b))/2) , we have y=((a(a+b))/( (√(a^2 +b^2 )))) and x=(((√2)ab)/( (√(a^2 +b^2 )))) for a=12 , b=5 y= ((204)/(13)) ≈15.7 and x=((60(√2))/(13)) ≈6.526 . ∠AED=∠AEO+∠CEF = θ+[(π/2)−((π/4)+θ)] = (π/4) .](https://www.tinkutara.com/question/Q15612.png)
$${considering}\:\Delta{COD}: \\ $$$${OD}=\frac{\left({a}+{b}\right)}{\mathrm{2}}−{b}=\frac{{a}−{b}}{\mathrm{2}} \\ $$$${OE}={R}=\left(\frac{{a}+{b}}{\mathrm{2}}\right)\:\:{and}\:\:{CE}={CD}={r} \\ $$$${OC}={OE}−{CE}={R}−{r} \\ $$$${OD}^{\mathrm{2}} ={OC}^{\mathrm{2}} −{CD}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\:\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{rR} \\ $$$$\:\:\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{rR} \\ $$$$\:\Rightarrow\:\:\:\:\:\:\:\mathrm{2}{rR}={ab} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{r}=\frac{{ab}}{{a}+{b}}\:. \\ $$$$\:\mathrm{tan}\:\mathrm{2}\theta=\frac{{CD}}{{OD}}=\:\frac{\left(\frac{{ab}}{{a}+{b}}\right)}{\left(\frac{{a}−{b}}{\mathrm{2}}\right)}\:=\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\:\:\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}=\frac{\mathrm{2}\left({b}/{a}\right)}{\mathrm{1}−\left({b}/{a}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{tan}\:\theta=\frac{{b}}{{a}} \\ $$$$\:\:{with}\:{CF}\:\bot\:{DE} \\ $$$$\:\mathrm{sin}\:\:\angle{DCF}=\frac{{DF}}{{CD}}=\:\frac{\left({x}/\mathrm{2}\right)}{{r}} \\ $$$$\:\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)=\frac{{x}}{\mathrm{2}{r}} \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)=\frac{{x}}{\mathrm{2}{r}} \\ $$$$\:\:{x}={r}\sqrt{\mathrm{2}}\left(\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right) \\ $$$$\:\:{x}=\frac{\sqrt{\mathrm{2}}{ab}}{\left({a}+{b}\right)}\frac{\left({a}+{b}\right)}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\: \\ $$$$\:\:{x}=\:\frac{\sqrt{\mathrm{2}}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:. \\ $$$$\:{y}=\mathrm{2}\left({AO}\right)\mathrm{cos}\:\theta\:=\mathrm{2}{R}\mathrm{cos}\:\theta \\ $$$${as}\:\:{R}=\frac{\left({a}+{b}\right)}{\mathrm{2}}\:,\:{we}\:{have} \\ $$$$\:\:\:{y}=\frac{{a}\left({a}+{b}\right)}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\:{and}\:\:{x}=\frac{\sqrt{\mathrm{2}}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\:{for}\:{a}=\mathrm{12}\:,\:{b}=\mathrm{5} \\ $$$$\:\:\:{y}=\:\frac{\mathrm{204}}{\mathrm{13}}\:\approx\mathrm{15}.\mathrm{7}\:\:\:{and}\: \\ $$$$\:\:\:{x}=\frac{\mathrm{60}\sqrt{\mathrm{2}}}{\mathrm{13}}\:\approx\mathrm{6}.\mathrm{526}\:. \\ $$$$\:\:\angle{AED}=\angle{AEO}+\angle{CEF} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\theta+\left[\frac{\pi}{\mathrm{2}}−\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\pi}{\mathrm{4}}\:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17
$${hi}\:{dear}\:{mr}\:{Ajfour}!\:{i}\:{have}\:{your}\:{picture} \\ $$$${but}\:{i}\:{can}'{t}\:{hear}\:{your}\:{sound}!\:\left({ha}\right)^{\mathrm{3}} \\ $$$${where}\:{is}\:{your}\:{solution}?{i}\:{am}\:{waiting} \\ $$$${for}\:{it}! \\ $$