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Question Number 16430 by Tinkutara last updated on 22/Jun/17

In ΔABC with usual notation (r_1 /(bc)) + (r_2 /(ca)) + (r_3 /(ab)) is (1) (1/r) − (1/R) (2) (1/r) − (1/(2R)) (3) (1/r) + (1/(2R)) (4) (1/r) + (1/R)

$$\mathrm{In}\:\Delta{ABC}\:\mathrm{with}\:\mathrm{usual}\:\mathrm{notation} \\ $$$$\frac{{r}_{\mathrm{1}} }{{bc}}\:+\:\frac{{r}_{\mathrm{2}} }{{ca}}\:+\:\frac{{r}_{\mathrm{3}} }{{ab}}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{{R}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{{R}} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

answer (2)

$${answer}\:\left(\mathrm{2}\right)\: \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/Jun/17

because this formula is in any trianvles you can compute it for a equilateral triangle and find the right answer. in equilateral triangle: 1)R=2r r_a =r_b =r_c =(S/(p−a))=((a^2 ((√3)/4))/(((3a)/2)−a))=a((√3)/2)=2RsinA.((√3)/2)= =2R×((√3)/2).((√3)/2)=((3R)/2) ⇒Σ(r_a /(bc))=3×((a((√3)/2))/a^2 )=3×(((3R)/2)/(3R^2 ))=(3/(2R)) answer(1):(1/r)−(1/R)=((R−r)/(r.R))=((R−.5R)/(.5R.R))=(1/R) answer(2):(1/r)−(1/(2R))=((2R−r)/(2R.r))=((2R−.5R)/(2R×.5R))=(3/(2R)) answer(3):(1/r)+(1/(2R))=((2R+r)/(2R.r))=((2R+.5R)/(2R×.5R))=(5/(2R)) answer(4):(1/r)+(1/R)=((R+r)/(R.r))=((R+.5R)/(R×.5R))=(3/R).

$${because}\:{this}\:{formula}\:{is}\:{in}\:{any}\:{trianvles} \\ $$$${you}\:{can}\:{compute}\:{it}\:{for}\:{a}\:{equilateral} \\ $$$${triangle}\:{and}\:{find}\:{the}\:{right}\:{answer}. \\ $$$${in}\:{equilateral}\:{triangle}: \\ $$$$\left.\mathrm{1}\right){R}=\mathrm{2}{r} \\ $$$${r}_{{a}} ={r}_{{b}} ={r}_{{c}} =\frac{{S}}{{p}−{a}}=\frac{{a}^{\mathrm{2}} \frac{\sqrt{\mathrm{3}}}{\mathrm{4}}}{\frac{\mathrm{3}{a}}{\mathrm{2}}−{a}}={a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}{RsinA}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}= \\ $$$$=\mathrm{2}{R}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}{R}}{\mathrm{2}} \\ $$$$\Rightarrow\Sigma\frac{{r}_{{a}} }{{bc}}=\mathrm{3}×\frac{{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{{a}^{\mathrm{2}} }=\mathrm{3}×\frac{\frac{\mathrm{3}{R}}{\mathrm{2}}}{\mathrm{3}{R}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{2}{R}} \\ $$$${answer}\left(\mathrm{1}\right):\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{R}}=\frac{{R}−{r}}{{r}.{R}}=\frac{{R}−.\mathrm{5}{R}}{.\mathrm{5}{R}.{R}}=\frac{\mathrm{1}}{{R}} \\ $$$${answer}\left(\mathrm{2}\right):\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{\mathrm{2}{R}}=\frac{\mathrm{2}{R}−{r}}{\mathrm{2}{R}.{r}}=\frac{\mathrm{2}{R}−.\mathrm{5}{R}}{\mathrm{2}{R}×.\mathrm{5}{R}}=\frac{\mathrm{3}}{\mathrm{2}{R}} \\ $$$${answer}\left(\mathrm{3}\right):\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{\mathrm{2}{R}}=\frac{\mathrm{2}{R}+{r}}{\mathrm{2}{R}.{r}}=\frac{\mathrm{2}{R}+.\mathrm{5}{R}}{\mathrm{2}{R}×.\mathrm{5}{R}}=\frac{\mathrm{5}}{\mathrm{2}{R}} \\ $$$${answer}\left(\mathrm{4}\right):\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{R}}=\frac{{R}+{r}}{{R}.{r}}=\frac{{R}+.\mathrm{5}{R}}{{R}×.\mathrm{5}{R}}=\frac{\mathrm{3}}{{R}}. \\ $$

Answered by Tinkutara last updated on 06/Jul/17

Using r_1 = 4R sin (A/2) cos (B/2) cos (C/2) and r = 4R sin (A/2) sin (B/2) sin (C/2), we get (r_1 /(bc)) = ((4R sin (A/2) cos (B/2) cos (C/2))/(16R^2 sin (B/2) sin (C/2) cos (B/2) cos (C/2))) = ((sin (A/2))/(4R sin (B/2) sin (C/2))) Similarly we write all the terms. (r_1 /(bc)) + (r_2 /(ca)) + (r_3 /(ab)) = (1/(4R sin (A/2) sin (B/2) sin (C/2)))(sin^2 (A/2) + sin^2 (B/2) + sin^2 (C/2)) = (1/(2r))(1 − cos A + 1 − cos B + 1 − cos C) = (1/(2r))(3 − 1 − (r/R)) = (1/r) − (1/(2R))

$$\mathrm{Using}\:{r}_{\mathrm{1}} \:=\:\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\:\mathrm{and} \\ $$$${r}\:=\:\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}},\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{{r}_{\mathrm{1}} }{{bc}}\:=\:\frac{\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}}}{\mathrm{16}{R}^{\mathrm{2}} \:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{sin}\:\frac{{A}}{\mathrm{2}}}{\mathrm{4}{R}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}} \\ $$$$\mathrm{Similarly}\:\mathrm{we}\:\mathrm{write}\:\mathrm{all}\:\mathrm{the}\:\mathrm{terms}. \\ $$$$\frac{{r}_{\mathrm{1}} }{{bc}}\:+\:\frac{{r}_{\mathrm{2}} }{{ca}}\:+\:\frac{{r}_{\mathrm{3}} }{{ab}}\:=\:\frac{\mathrm{1}}{\mathrm{4}{R}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}}\left(\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}\:+\:\mathrm{sin}^{\mathrm{2}} \:\frac{{B}}{\mathrm{2}}\:+\:\mathrm{sin}^{\mathrm{2}} \:\frac{{C}}{\mathrm{2}}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{r}}\left(\mathrm{1}\:−\:\mathrm{cos}\:{A}\:+\:\mathrm{1}\:−\:\mathrm{cos}\:{B}\:+\:\mathrm{1}\:−\:\mathrm{cos}\:{C}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{r}}\left(\mathrm{3}\:−\:\mathrm{1}\:−\:\frac{{r}}{{R}}\right)\:=\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$

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