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Question-26050




Question Number 26050 by Tinkutara last updated on 18/Dec/17
Answered by ajfour last updated on 20/Dec/17
[1,∞)    mere  attempt ......  b+c ≥ a  a+b+c ≥ 2a  ⇒ (s/a) ≥ 1              .......(i)  r+r_1 =r_2 +r_3   (and from previous Q)    ((r_1 +r_2 +r_3 )/r) > 9                    (not ≥ since ∠A≠ (π/3))  so    ((2r_1 +r)/r) > 9  ⇒  (r_1 /r) > 4  △=sr   and   △ < (1/2)(a)((a/(2(√3))))     as altitude from A on BC  is   < ((a(√3))/2)    becoz   ∠A   >(π/3)  hence    △=sr  < (a^2 /(4(√3)))         (s/a) < (a/r)((4/( (√3))))    ......(ii)  also   ∠ BIC    > 120°                           (I being incentre)  so   r < (a/(2(√3)))   ⇒   (a/r) > 2(√3)   let tan (A/2)=l , tan (B/2)=m, tan (C/2)=n  then    lm+mn+nl=1      and  r_1 (m+n)=a               r((1/m)+(1/n))=a  ⇒    (r/(r_1  )) =mn                (iii)⇒    l(m+n)+mn =1  and   l > (1/( (√3)))      (∠A  > π/3)  ⇒     ((1−mn)/(m+n)) > (1/( (√3)))   or    1−(r/r_1 ) > (a/(r_1 (√3)))  or    r_1 −r > (a/( (√3)))        r_1 > r+(a/( (√3)))            let us  try,      (m+n)^2  ≥ 4mn  ⇒  ((a/r_1 ))^2  ≥ ((4r)/r_1 )  ⇒   a^2  ≥ 4rr_1   or     (a/r) ≥2 (√(r_1 /r))   as   (r_1 /r) > 4  ⇒   (a/r) > 4  cant find  ((a/r))_(max)   i think     (s/a) ∈  [1, ∞)  (as  r can → 0 )
$$\left[\mathrm{1},\infty\right) \\ $$$$\:\:{mere}\:\:{attempt}\:…… \\ $$$${b}+{c}\:\geqslant\:{a} \\ $$$${a}+{b}+{c}\:\geqslant\:\mathrm{2}{a} \\ $$$$\Rightarrow\:\frac{{s}}{{a}}\:\geqslant\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\left({i}\right) \\ $$$${r}+{r}_{\mathrm{1}} ={r}_{\mathrm{2}} +{r}_{\mathrm{3}} \:\:\left({and}\:{from}\:{previous}\:{Q}\right) \\ $$$$\:\:\frac{{r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} }{{r}}\:>\:\mathrm{9}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({not}\:\geqslant\:{since}\:\angle{A}\neq\:\frac{\pi}{\mathrm{3}}\right) \\ $$$${so}\:\:\:\:\frac{\mathrm{2}{r}_{\mathrm{1}} +{r}}{{r}}\:>\:\mathrm{9}\:\:\Rightarrow\:\:\frac{{r}_{\mathrm{1}} }{{r}}\:>\:\mathrm{4} \\ $$$$\bigtriangleup={sr}\:\:\:{and}\:\:\:\bigtriangleup\:<\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\:{as}\:{altitude}\:{from}\:{A}\:{on}\:{BC} \\ $$$${is}\:\:\:<\:\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:{becoz}\:\:\:\angle{A}\:\:\:>\frac{\pi}{\mathrm{3}} \\ $$$${hence}\:\:\:\:\bigtriangleup={sr}\:\:<\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\frac{\boldsymbol{{s}}}{\boldsymbol{{a}}}\:<\:\frac{\boldsymbol{{a}}}{\boldsymbol{{r}}}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right)\:\:\:\:……\left({ii}\right) \\ $$$${also}\:\:\:\angle\:{BIC}\:\:\:\:>\:\mathrm{120}°\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({I}\:{being}\:{incentre}\right) \\ $$$${so}\:\:\:{r}\:<\:\frac{{a}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\Rightarrow\:\:\:\frac{{a}}{{r}}\:>\:\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$${let}\:\mathrm{tan}\:\frac{{A}}{\mathrm{2}}={l}\:,\:\mathrm{tan}\:\frac{{B}}{\mathrm{2}}={m},\:\mathrm{tan}\:\frac{{C}}{\mathrm{2}}={n} \\ $$$${then}\:\:\:\:{lm}+{mn}+{nl}=\mathrm{1}\:\: \\ $$$$\:\:{and}\:\:{r}_{\mathrm{1}} \left({m}+{n}\right)={a}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{r}\left(\frac{\mathrm{1}}{{m}}+\frac{\mathrm{1}}{{n}}\right)={a} \\ $$$$\Rightarrow\:\:\:\:\frac{{r}}{{r}_{\mathrm{1}} \:}\:={mn}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left({iii}\right)\Rightarrow\:\:\:\:{l}\left({m}+{n}\right)+{mn}\:=\mathrm{1} \\ $$$${and}\:\:\:{l}\:>\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\left(\angle{A}\:\:>\:\pi/\mathrm{3}\right) \\ $$$$\Rightarrow\:\:\:\:\:\frac{\mathrm{1}−{mn}}{{m}+{n}}\:>\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\: \\ $$$${or}\:\:\:\:\mathrm{1}−\frac{{r}}{{r}_{\mathrm{1}} }\:>\:\frac{{a}}{{r}_{\mathrm{1}} \sqrt{\mathrm{3}}} \\ $$$${or}\:\:\:\:{r}_{\mathrm{1}} −{r}\:>\:\frac{{a}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:{r}_{\mathrm{1}} >\:{r}+\frac{{a}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\: \\ $$$${let}\:{us}\:\:{try}, \\ $$$$\:\:\:\:\left({m}+{n}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{4}{mn} \\ $$$$\Rightarrow\:\:\left(\frac{{a}}{{r}_{\mathrm{1}} }\right)^{\mathrm{2}} \:\geqslant\:\frac{\mathrm{4}{r}}{{r}_{\mathrm{1}} } \\ $$$$\Rightarrow\:\:\:{a}^{\mathrm{2}} \:\geqslant\:\mathrm{4}{rr}_{\mathrm{1}} \\ $$$${or}\:\:\:\:\:\frac{{a}}{{r}}\:\geqslant\mathrm{2}\:\sqrt{\frac{{r}_{\mathrm{1}} }{{r}}}\: \\ $$$${as}\:\:\:\frac{{r}_{\mathrm{1}} }{{r}}\:>\:\mathrm{4}\:\:\Rightarrow\:\:\:\frac{{a}}{{r}}\:>\:\mathrm{4} \\ $$$${cant}\:{find}\:\:\left(\frac{{a}}{{r}}\right)_{{max}} \\ $$$${i}\:{think}\:\:\:\:\:\frac{{s}}{{a}}\:\in\:\:\left[\mathrm{1},\:\infty\right) \\ $$$$\left({as}\:\:{r}\:{can}\:\rightarrow\:\mathrm{0}\:\right) \\ $$

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