Menu Close

Question-226049

Tinku Tara 0 Comments
Pin




Question Number 226049 by ajfour last updated on 18/Nov/25
Commented by ajfour last updated on 18/Nov/25
Commented by ajfour last updated on 18/Nov/25
(((2a)/( (√3))))^2 +{(√((a+b)^2 −(((2a)/( (√3))))^2 ))−(R−b)}^2 =(R−a)^2 can we find b, given R and a?
$$\left(\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left\{\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }−\left({R}−{b}\right)\right\}^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} \: \\ $$$${can}\:{we}\:{find}\:{b},\:{given}\:{R}\:{and}\:{a}? \\ $$
Commented by mr W last updated on 18/Nov/25
i think yes. basically it is like following eq. (√(x^2 −p^2 ))+x=q
$${i}\:{think}\:{yes}. \\ $$$${basically}\:{it}\:{is}\:{like}\:{following}\:{eq}. \\ $$$$\sqrt{{x}^{\mathrm{2}} −{p}^{\mathrm{2}} }+{x}={q} \\ $$
Commented by fantastic2 last updated on 18/Nov/25
let (((2a)/( (√3))))=x (R−a)=y ⇒x^2 +{(√((a+b)^2 −x^2 ))−(R−b)}^2 =y^2 ⇒(√((a+b)^2 −x^2 ))−(R−b)=(√(y^2 −x^2 )) ⇒(√((a+b)^2 −x^2 ))+b=(√(y^2 −x^2 ))+R=C ⇒a^2 +b^2 +2ab−x^2 +b^2 +2b(√((a+b)^2 −x^2 ))=C^2 ⇒2b^2 +2ab+2b(√((a+b)^2 −x^2 ))=C_1 [C_1 =C^2 −a^2 +x^2 ] ⇒b^2 +ab+b(√((a+b)^2 −x^2 ))=(C_1 /2)=C_2 [C_2 =(C_1 /2)=((C^2 −a^2 +x^2 )/2) =((y^2 +R^2 +2R(√(y^2 −x^2 ))−a^2 )/2)] ⇒b(√((a+b)^2 −x^2 ))=C_2 −b(a+b) ⇒b^2 (a+b)^2 −b^2 x^2 =C_2 ^2 −2bC_2 (a+b)+b^2 (a+b)^2 ⇒2bC_2 (a+b)−b^2 x^2 =C_2 ^2 ⇒b^2 x^2 −2b^2 C_2 −2abC_2 +C_2 ^2 =0 ⇒b^2 (x^2 −2C_2 )−2aC_2 b+C_2 ^2 =0 ⇒b^2 (x^2 −y^2 +R^2 +2R(√(y^2 −x^2 ))−a^2 )−ba(y^2 +R^2 +2R(√(y^2 −x^2 ))−a^2 )⇒⇒+(((y^2 +R^2 +2R(√(y^2 −x^2 ))−a^2 )/2))^2 =0 ⇒b^2 ((4/3)a^2 −R^2 −a^2 +2aR+R^2 +2R(√((4/3)a^2 −R^2 −a^2 +2aR))−a^2 )−ba...=0 ⇒b^2 (((4a^2 )/3)+2aR+2R(√((a^2 /3)−R(R−2a)))).... no ′nice′ solution
$${let}\:\left(\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}\right)={x}\: \\ $$$$\left({R}−{a}\right)={y} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left\{\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }−\left({R}−{b}\right)\right\}^{\mathrm{2}} ={y}^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }−\left({R}−{b}\right)=\sqrt{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }+{b}=\sqrt{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }+{R}={C} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}−{x}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{b}\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }={C}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{b}\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }={C}_{\mathrm{1}} \left[{C}_{\mathrm{1}} ={C}^{\mathrm{2}} −{a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right] \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{ab}+{b}\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{{C}_{\mathrm{1}} }{\mathrm{2}}={C}_{\mathrm{2}} \left[{C}_{\mathrm{2}} =\frac{{C}_{\mathrm{1}} }{\mathrm{2}}=\frac{{C}^{\mathrm{2}} −{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{\mathrm{2}}\right. \\ $$$$\left.=\frac{{y}^{\mathrm{2}} +{R}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{a}^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$$\Rightarrow{b}\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }={C}_{\mathrm{2}} −{b}\left({a}+{b}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} {x}^{\mathrm{2}} ={C}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{bC}_{\mathrm{2}} \left({a}+{b}\right)+{b}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{bC}_{\mathrm{2}} \left({a}+{b}\right)−{b}^{\mathrm{2}} {x}^{\mathrm{2}} ={C}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {C}_{\mathrm{2}} −\mathrm{2}{abC}_{\mathrm{2}} +{C}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{2}{C}_{\mathrm{2}} \right)−\mathrm{2}{aC}_{\mathrm{2}} {b}+{C}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{R}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{a}^{\mathrm{2}} \right)−{ba}\left({y}^{\mathrm{2}} +{R}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{a}^{\mathrm{2}} \right)\Rightarrow\Rightarrow+\left(\frac{{y}^{\mathrm{2}} +{R}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{a}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{3}}{a}^{\mathrm{2}} −{R}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{2}{aR}+{R}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{\frac{\mathrm{4}}{\mathrm{3}}{a}^{\mathrm{2}} −{R}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{2}{aR}}−{a}^{\mathrm{2}} \right)−{ba}…=\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left(\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{2}{aR}+\mathrm{2}{R}\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{3}}−{R}\left({R}−\mathrm{2}{a}\right)}\right)…. \\ $$$${no}\:'{nice}'\:{solution} \\ $$
Commented by fantastic2 last updated on 18/Nov/25
what do you think mrWsir
$${what}\:{do}\:{you}\:{think}\:{mrWsir} \\ $$
Commented by ajfour last updated on 18/Nov/25
Commented by ajfour last updated on 18/Nov/25
b=x
Commented by mr W last updated on 18/Nov/25
((4a^2 )/3)+{(√((a+b)^2 −((4a^2 )/3)))+(a+b)−R−a}^2 =(R−a)^2 {(√((a+b)^2 −((4a^2 )/3)))+(a+b)−R−a}^2 =R^2 −2Ra−(a^2 /3) (√((a+b)^2 −((4a^2 )/3)))+(a+b)−R−a=±(√(R^2 −2Ra−(a^2 /3))) (√((a+b)^2 −((4a^2 )/3)))+(a+b)=[R+a±(√(R^2 −2Ra−(a^2 /3)))]−(a+b) (√(x^2 −((4a^2 )/3)))=c−x −((4a^2 )/3)=c^2 −2cx 2cx=c^2 +((4a^2 )/3) ⇒x=a+b=(c/2)+((2a^2 )/(3c)) ⇒b=((R+a±(√(R^2 −2Ra−(a^2 /3))))/2)+((2a^2 )/(3(R+a±(√(R^2 −2Ra−(a^2 /3))))))−a
$$\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}+\left\{\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}}+\left({a}+{b}\right)−{R}−{a}\right\}^{\mathrm{2}} =\left({R}−{a}\right)^{\mathrm{2}} \\ $$$$\left\{\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}}+\left({a}+{b}\right)−{R}−{a}\right\}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{Ra}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}}+\left({a}+{b}\right)−{R}−{a}=\pm\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}} \\ $$$$\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}}+\left({a}+{b}\right)=\left[{R}+{a}\pm\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}}\right]−\left({a}+{b}\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}}={c}−{x} \\ $$$$−\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}={c}^{\mathrm{2}} −\mathrm{2}{cx} \\ $$$$\mathrm{2}{cx}={c}^{\mathrm{2}} +\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{x}={a}+{b}=\frac{{c}}{\mathrm{2}}+\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}{c}} \\ $$$$\Rightarrow{b}=\frac{{R}+{a}\pm\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}}}{\mathrm{2}}+\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{3}\left({R}+{a}\pm\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}}\right)}−{a} \\ $$
Commented by ajfour last updated on 18/Nov/25
((√(X^2 −A^2 ))+X−Y)^2 =Y^2 −A^2 2X^2 −2XY+2(√(X^2 −A^2 ))(X−Y)=0 ⇒ X^4 +X^2 Y^2 −2X^3 Y =X^4 −A^2 X^2 +X^2 Y^2 −A^2 Y^2 −2X^3 Y+2A^2 XY A^2 X^2 +2X^3 Y−2A^2 XY=0 2YX^2 +A^2 X−2A^2 Y=0 X=(√(((A^2 /(4Y)))^2 +A^2 ))−(A^2 /(4Y)) b+a=(((2a)/( (√3))))(({(√(((4a^2 )/3)+16(R−a)^2 ))−((2a)/( (√3)))})/(4(R−a)))
$$\left(\sqrt{{X}^{\mathrm{2}} −{A}^{\mathrm{2}} }+{X}−{Y}\right)^{\mathrm{2}} ={Y}^{\mathrm{2}} −{A}^{\mathrm{2}} \\ $$$$\mathrm{2}{X}^{\mathrm{2}} −\mathrm{2}{XY}+\mathrm{2}\sqrt{{X}^{\mathrm{2}} −{A}^{\mathrm{2}} }\left({X}−{Y}\right)=\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\Rightarrow\:\:{X}^{\mathrm{4}} +{X}^{\mathrm{2}} {Y}^{\mathrm{2}} −\mathrm{2}{X}^{\mathrm{3}} {Y} \\ $$$$\:\:\:\:\:\:\:\:\:={X}^{\mathrm{4}} −{A}^{\mathrm{2}} {X}^{\mathrm{2}} +{X}^{\mathrm{2}} {Y}^{\mathrm{2}} −{A}^{\mathrm{2}} {Y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{X}^{\mathrm{3}} {Y}+\mathrm{2}{A}^{\mathrm{2}} {XY} \\ $$$${A}^{\mathrm{2}} {X}^{\mathrm{2}} +\mathrm{2}{X}^{\mathrm{3}} {Y}−\mathrm{2}{A}^{\mathrm{2}} {XY}=\mathrm{0} \\ $$$$\mathrm{2}{YX}^{\mathrm{2}} +{A}^{\mathrm{2}} {X}−\mathrm{2}{A}^{\mathrm{2}} {Y}=\mathrm{0} \\ $$$${X}=\sqrt{\left(\frac{{A}^{\mathrm{2}} }{\mathrm{4}{Y}}\right)^{\mathrm{2}} +{A}^{\mathrm{2}} }−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{Y}} \\ $$$${b}+{a}=\left(\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}\right)\frac{\left\{\sqrt{\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{16}\left({R}−{a}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}\right\}}{\mathrm{4}\left({R}−{a}\right)} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *