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Question Number 184633 by ajfour last updated on 09/Jan/23
Answered by mr W last updated on 09/Jan/23
Commented by mr W last updated on 09/Jan/23
AD^2 =a^2 +b^2 +2ab cos 30° ⇒AD=(√(a^2 +b^2 +(√3)ab)) sin (A/2)=(b/(AD))=(b/( (√(a^2 +b^2 +(√3)ab)))) cos (A/2)=((√(a^2 +(√3)ab))/( (√(a^2 +b^2 +(√3)ab)))) ⇒sin A=((2b(√(a^2 +(√3)ab)))/(a^2 +b^2 +(√3)ab)) (p/(sin A))=2a ⇒p=((4ab(√(a^2 +(√3)ab)))/(a^2 +b^2 +(√3)ab)) ((sin α)/b)=((sin 30°)/(AD)) ⇒sin α=(b/(2(√(a^2 +b^2 +(√3)ab)))) r=2a cos ((A/2)−α) q=2a cos ((A/2)+α) q+r=4a cos (A/2) cos α=4a×((√(a^2 +(√3)ab))/( (√(a^2 +b^2 +(√3)ab))))×((2a+(√3)b)/(2(√(a^2 +b^2 +(√3)ab)))) q+r=((2a(2a+(√3)b)(√(a^2 +(√3)ab)))/( a^2 +b^2 +(√3)ab)) q+r−2(√(AD^2 −b^2 ))=p ((2a(2a+(√3)b)(√(a^2 +(√3)ab)))/( a^2 +b^2 +(√3)ab))−2(√(a^2 +(√3)ab))=((4ab(√(a^2 +(√3)ab)))/(a^2 +b^2 +(√3)ab)) ((a(2a+(√3)b))/( a^2 +b^2 +(√3)ab))−1=((2ab)/(a^2 +b^2 +(√3)ab)) a^2 −2ab= b^2 ((b/a))^2 +2((b/a))−1=0 ⇒(b/a)=(√2)−1 ✓
$${AD}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\mathrm{30}° \\ $$$$\Rightarrow{AD}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}} \\ $$$$\mathrm{sin}\:\frac{{A}}{\mathrm{2}}=\frac{{b}}{{AD}}=\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}} \\ $$$$\mathrm{cos}\:\frac{{A}}{\mathrm{2}}=\frac{\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}} \\ $$$$\Rightarrow\mathrm{sin}\:{A}=\frac{\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}} \\ $$$$\frac{{p}}{\mathrm{sin}\:{A}}=\mathrm{2}{a} \\ $$$$\Rightarrow{p}=\frac{\mathrm{4}{ab}\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{{b}}=\frac{\mathrm{sin}\:\mathrm{30}°}{{AD}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{{b}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}} \\ $$$${r}=\mathrm{2}{a}\:\mathrm{cos}\:\left(\frac{{A}}{\mathrm{2}}−\alpha\right) \\ $$$${q}=\mathrm{2}{a}\:\mathrm{cos}\:\left(\frac{{A}}{\mathrm{2}}+\alpha\right) \\ $$$${q}+{r}=\mathrm{4}{a}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\alpha=\mathrm{4}{a}×\frac{\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}×\frac{\mathrm{2}{a}+\sqrt{\mathrm{3}}{b}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}} \\ $$$${q}+{r}=\frac{\mathrm{2}{a}\left(\mathrm{2}{a}+\sqrt{\mathrm{3}}{b}\right)\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}} \\ $$$${q}+{r}−\mathrm{2}\sqrt{{AD}^{\mathrm{2}} −{b}^{\mathrm{2}} }={p} \\ $$$$\frac{\mathrm{2}{a}\left(\mathrm{2}{a}+\sqrt{\mathrm{3}}{b}\right)\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}−\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}=\frac{\mathrm{4}{ab}\sqrt{{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}} \\ $$$$\frac{{a}\left(\mathrm{2}{a}+\sqrt{\mathrm{3}}{b}\right)}{\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}}−\mathrm{1}=\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{ab}=\:{b}^{\mathrm{2}} \\ $$$$\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{b}}{{a}}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\sqrt{\mathrm{2}}−\mathrm{1}\:\checkmark \\ $$
Commented by mr W last updated on 10/Jan/23
the result (b/a)=(√2)−1 is independent from the angle 60°. an other example:
$${the}\:{result}\:\frac{{b}}{{a}}=\sqrt{\mathrm{2}}−\mathrm{1}\:{is}\:{independent} \\ $$$${from}\:{the}\:{angle}\:\mathrm{60}°.\:{an}\:{other}\:{example}: \\ $$
Commented by mr W last updated on 10/Jan/23
Commented by mr W last updated on 10/Jan/23
sin θ=(b/(a+b)) cos 2θ=((2b)/a) ((2b)/a)=1−2((b/(a+b)))^2 =((a^2 −b^2 +2ab)/(a^2 +b^2 +2ab)) 2b^3 +5ab^2 =a^3 λ^3 −5λ−2=0 (λ+2)(λ^2 −2λ−1)=0 ⇒λ=(a/b)=1+(√2) ⇒(b/a)=(√2)−1 ✓
$$\mathrm{sin}\:\theta=\frac{{b}}{{a}+{b}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{2}{b}}{{a}} \\ $$$$\frac{\mathrm{2}{b}}{{a}}=\mathrm{1}−\mathrm{2}\left(\frac{{b}}{{a}+{b}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}} \\ $$$$\mathrm{2}{b}^{\mathrm{3}} +\mathrm{5}{ab}^{\mathrm{2}} ={a}^{\mathrm{3}} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{5}\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda+\mathrm{2}\right)\left(\lambda^{\mathrm{2}} −\mathrm{2}\lambda−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{{a}}{{b}}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{{b}}{{a}}=\sqrt{\mathrm{2}}−\mathrm{1}\:\checkmark \\ $$
Commented by MJS_new last updated on 10/Jan/23
this means if the center of the circumcircle lies on the incircle the ratio of (r/R) is constant. same as the distance of the centers of both circles equals the radius of the incircle. ⇒ if we let a≤b≤c and a=1 ⇒ 1≤b≤1+((√2)/2)∧(√2)≤c≤1+((√2)/2) with the “borders” being the isosceles triangles a=b=1∧c=(√2) and a=1∧b=c=1+((√2)/2). can we prove there are no other triangles with (r/R)=(√2)−1; meaning with the center of the circumcircle not on the incircle?
$$\mathrm{this}\:\mathrm{means}\:\mathrm{if}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle} \\ $$$$\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{incircle}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\frac{{r}}{{R}}\:\mathrm{is}\:\mathrm{constant}. \\ $$$$\mathrm{same}\:\mathrm{as}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{both} \\ $$$$\mathrm{circles}\:\mathrm{equals}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircle}. \\ $$$$\Rightarrow \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{let}\:{a}\leqslant{b}\leqslant{c}\:\mathrm{and}\:{a}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{1}\leqslant{b}\leqslant\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\wedge\sqrt{\mathrm{2}}\leqslant{c}\leqslant\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{with}\:\mathrm{the}\:“\mathrm{borders}''\:\mathrm{being}\:\mathrm{the}\:\mathrm{isosceles}\:\mathrm{triangles} \\ $$$${a}={b}=\mathrm{1}\wedge{c}=\sqrt{\mathrm{2}}\:\mathrm{and}\:{a}=\mathrm{1}\wedge{b}={c}=\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}. \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{other}\:\mathrm{triangles}\:\mathrm{with} \\ $$$$\frac{{r}}{{R}}=\sqrt{\mathrm{2}}−\mathrm{1};\:\mathrm{meaning}\:\mathrm{with}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{circumcircle}\:{not}\:\mathrm{on}\:\mathrm{the}\:\mathrm{incircle}? \\ $$
Commented by mr W last updated on 10/Jan/23
yes! acc. to Euler′s theorem the distance from incenter to circumcenter of an arbitrary triangle is d=(√(R^2 −2Rr)) when d=r, we always have R^2 −2Rr=r^2 ⇒(r/R)=−1+(√2) vice versa, if (r/R)=(√2)−1, we have R^2 −2Rr=r^2 , then d=r, that means circumcenter lies on the incircle.
$${yes}! \\ $$$${acc}.\:{to}\:{Euler}'{s}\:{theorem}\:{the}\:{distance} \\ $$$${from}\:{incenter}\:{to}\:{circumcenter}\:{of}\:{an} \\ $$$${arbitrary}\:{triangle}\:{is} \\ $$$${d}=\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$${when}\:{d}={r},\:{we}\:{always}\:{have} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{Rr}={r}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{r}}{{R}}=−\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${vice}\:{versa},\:{if}\:\frac{{r}}{{R}}=\sqrt{\mathrm{2}}−\mathrm{1},\:{we}\:{have} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{Rr}={r}^{\mathrm{2}} ,\:{then}\:{d}={r},\:{that}\:{means} \\ $$$${circumcenter}\:{lies}\:{on}\:{the}\:{incircle}. \\ $$
Commented by mr W last updated on 10/Jan/23
Commented by MJS_new last updated on 10/Jan/23
thank you, I didn′t know this theorem
$$\mathrm{thank}\:\mathrm{you},\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{know}\:\mathrm{this}\:\mathrm{theorem} \\ $$
Commented by mr W last updated on 10/Jan/23
i once knew this theorem. but as i tried to solve this question, i didn′t have it in mind. otherwise i have had taken a shortcut.
$${i}\:{once}\:{knew}\:{this}\:{theorem}.\:{but}\:{as}\:{i} \\ $$$${tried}\:{to}\:{solve}\:{this}\:{question},\:{i}\:{didn}'{t} \\ $$$${have}\:{it}\:{in}\:{mind}.\:{otherwise}\:{i}\:{have}\:{had} \\ $$$${taken}\:{a}\:{shortcut}. \\ $$
Commented by mr W last updated on 10/Jan/23
sir, have you got an idea how to solve the equation system in Q184607 generally?
$${sir},\:{have}\:{you}\:{got}\:{an}\:{idea}\:{how}\:{to}\:{solve} \\ $$$${the}\:{equation}\:{system}\:{in}\:{Q}\mathrm{184607}\: \\ $$$${generally}? \\ $$

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