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Question-185101

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Question Number 185101 by emmanuelson123 last updated on 17/Jan/23
Answered by Frix last updated on 17/Jan/23
In triangle a, b, c: D=(√((a+b+c)(b+c−a)(a+c−b)(a+b−c))) a^2 +b^2 −2abcos γ =c^2 ⇒ cos γ =((a^2 +b^2 −c^2 )/(2ab)) ⇒ sin γ =(D/(2ab)) sin (x/2) =(√((1−(√(1−sin^2 x)))/2)) ⇒ [Similar for the other angles] sin (α/2)=((√(a^2 −(b−c)^2 ))/(2(√(bc)))) sin (β/2) =((√(b^2 −(a−c)^2 ))/(2(√(ac)))) sin (γ/2) =((√(c^2 −(a−b)^2 ))/(2(√(ab)))) We know that r=(D/(2(a+b+c))) and R=((abc)/D) ⇒ (D/(2(a+b+c)))=4((abc)/D)sin (α/2) sin (β/2) sin (γ/2) ⇔ sin (α/2) sin (β/2) sin (γ/2) =(D^2 /(8abc(a+b+c))) sin (α/2) sin (β/2) sin (γ/2) =(((b+c−a)(a+c−b)(a+b−c))/(8abc)) Lhs: ((√(a^2 −(b−c)^2 ))/(2(√(bc))))×((√(b^2 −(a−c)^2 ))/(2(√(ac))))×((√(c^2 −(a−b)^2 ))/(2(√(ab))))= =((√((b+c−a)^2 (a+c−b)^2 (a+b−c)^2 ))/(8(√(a^2 b^2 c^2 ))))= =(((b+c−a)(a+c−b)(a+b−c))/(8abc))=Rhs q.e.d.
$$\mathrm{In}\:\mathrm{triangle}\:{a},\:{b},\:{c}: \\ $$$${D}=\sqrt{\left({a}+{b}+{c}\right)\left({b}+{c}−{a}\right)\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)} \\ $$$$ \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\gamma\:={c}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{cos}\:\gamma\:=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\:\Rightarrow \\ $$$$\mathrm{sin}\:\gamma\:=\frac{{D}}{\mathrm{2}{ab}} \\ $$$$\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:=\sqrt{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}}{\mathrm{2}}}\:\Rightarrow \\ $$$$\left[\mathrm{Similar}\:\mathrm{for}\:\mathrm{the}\:\mathrm{other}\:\mathrm{angles}\right] \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\frac{\sqrt{{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{{bc}}} \\ $$$$\mathrm{sin}\:\frac{\beta}{\mathrm{2}}\:=\frac{\sqrt{{b}^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{{ac}}} \\ $$$$\mathrm{sin}\:\frac{\gamma}{\mathrm{2}}\:=\frac{\sqrt{{c}^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{{ab}}} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that} \\ $$$${r}=\frac{{D}}{\mathrm{2}\left({a}+{b}+{c}\right)}\:\mathrm{and}\:{R}=\frac{{abc}}{{D}} \\ $$$$\Rightarrow \\ $$$$\frac{{D}}{\mathrm{2}\left({a}+{b}+{c}\right)}=\mathrm{4}\frac{{abc}}{{D}}\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\frac{\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\gamma}{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\frac{\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\gamma}{\mathrm{2}}\:=\frac{{D}^{\mathrm{2}} }{\mathrm{8}{abc}\left({a}+{b}+{c}\right)} \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{sin}\:\frac{\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\gamma}{\mathrm{2}}\:=\frac{\left({b}+{c}−{a}\right)\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)}{\mathrm{8}{abc}} \\ $$$$\mathrm{Lhs}: \\ $$$$\frac{\sqrt{{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{{bc}}}×\frac{\sqrt{{b}^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{{ac}}}×\frac{\sqrt{{c}^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{{ab}}}= \\ $$$$=\frac{\sqrt{\left({b}+{c}−{a}\right)^{\mathrm{2}} \left({a}+{c}−{b}\right)^{\mathrm{2}} \left({a}+{b}−{c}\right)^{\mathrm{2}} }}{\mathrm{8}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }}= \\ $$$$=\frac{\left({b}+{c}−{a}\right)\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)}{\mathrm{8}{abc}}=\mathrm{Rhs} \\ $$$$\mathrm{q}.\mathrm{e}.\mathrm{d}. \\ $$
Answered by mnjuly1970 last updated on 17/Jan/23
2 p= 2R(sin(A)+sin(B)+sin(C)) =2R(2sin(((A+B)/2))cos(((A−B)/2))+2sin(((A+B)/2))cos(((A+B)/2))) =4Rcos((C/2))(2 cos((A/2)).cos((B/2))) ⇒ p:= 4R cos((A/2))cos((B/2))cos((C/2)) r := (S/p) = (((1/2) bcsin(A))/(4Rcos((A/2))cos((B/2))cos((C/2)))) =_(c=2Rsin(C)) ^(b=2Rsin(B)) 4Rsin((A/2))sin((B/2))sin((C/2)) ∴ r≤ (R/2) ⇒sin((A/2))sin((B/2))sin((C/2))≤(1/8)
$$\:\:\mathrm{2}\:{p}=\:\mathrm{2}{R}\left({sin}\left({A}\right)+{sin}\left({B}\right)+{sin}\left({C}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}{R}\left(\mathrm{2}{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\right){cos}\left(\frac{{A}+{B}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{4}{Rcos}\left(\frac{{C}}{\mathrm{2}}\right)\left(\mathrm{2}\:{cos}\left(\frac{{A}}{\mathrm{2}}\right).{cos}\left(\frac{{B}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\Rightarrow\:{p}:=\:\mathrm{4}{R}\:{cos}\left(\frac{{A}}{\mathrm{2}}\right){cos}\left(\frac{{B}}{\mathrm{2}}\right){cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:{r}\::=\:\frac{{S}}{{p}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:{bcsin}\left({A}\right)}{\mathrm{4}{Rcos}\left(\frac{{A}}{\mathrm{2}}\right){cos}\left(\frac{{B}}{\mathrm{2}}\right){cos}\left(\frac{{C}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{c}=\mathrm{2}{Rsin}\left({C}\right)} {\overset{{b}=\mathrm{2}{Rsin}\left({B}\right)} {=}}\:\mathrm{4}{Rsin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}}{\mathrm{2}}\right){sin}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\therefore\:\:\:{r}\leqslant\:\frac{{R}}{\mathrm{2}}\:\Rightarrow{sin}\left(\frac{{A}}{\mathrm{2}}\right){sin}\left(\frac{{B}}{\mathrm{2}}\right){sin}\left(\frac{{C}}{\mathrm{2}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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