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Question Number 211276 by mnjuly1970 last updated on 04/Sep/24
Prove , in AB^Δ C : ((cosA)/(sin^2 A)) + ((cosB)/(sin^2 B)) +((cosC)/(sin^2 C)) ≥ (r/R) r : incircle radius R: circumcircle radius
$$ \\ $$$$\:\:{Prove}\:,\:{in}\:{A}\overset{\Delta} {{B}C}\:\::\: \\ $$$$ \\ $$$$\:\:\:\:\:\frac{{cosA}}{{sin}^{\mathrm{2}} {A}}\:+\:\frac{{cosB}}{{sin}^{\mathrm{2}} {B}}\:\:+\frac{{cosC}}{{sin}^{\mathrm{2}} {C}}\:\geqslant\:\frac{{r}}{{R}} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{r}\::\:{incircle}\:\:{radius} \\ $$$$\:\:\:\:\:{R}:\:{circumcircle}\:\:{radius} \\ $$$$ \\ $$
Answered by Ghisom last updated on 04/Sep/24
r=(a/(cot (β/2) +cot (γ/2))); cyclic for a, b, c & α, β, γ ⇒ sum these up and use γ=π−(α+β) and let u=tan α ∧ v=tan β to get 3r=(((1−uv)v)/(1+v^2 ))a+(((1−uv)u)/(1+u^2 ))b+((uv)/(u+v))c R=(a/(2sin α)); cyclic for a, b, c & α, β, γ again sum these up to get 3R=((1+u^2 )/(4u))a+((1+v^2 )/(4u))b+(((1+u^2 )(1+v^2 ))/(4(u+v)(1−uv)))c ⇒ (r/R)=((4(1−uv)uv)/((1+u^2 )(1+v^2 ))) ⇒ (r/R)=cos α +cos β −cos (α+β)_(=cos γ) −1 ⇒ (r/R)=cos α +cos β +cos γ −1 [0<(r/R)≤(1/2)] in any triangle 0<θ<π ⇒ 0<sin θ <1 ⇒ ((cos θ)/(sin^2 θ))≥cos θ ⇒ Σ ((cos θ)/(sin^2 θ)) ≥(Σ cos θ)−1 [ what′s left to prove: 2≤((cos α)/(sin^2 α))+((cos β)/(sin^2 β))−((cos (α+β))/(sin^2 (α+β))) because of symmetry there should be a min/max at α=β=(π/3) ]
$${r}=\frac{{a}}{\mathrm{cot}\:\frac{\beta}{\mathrm{2}}\:+\mathrm{cot}\:\frac{\gamma}{\mathrm{2}}};\:\mathrm{cyclic}\:\mathrm{for}\:{a},\:{b},\:{c}\:\&\:\alpha,\:\beta,\:\gamma \\ $$$$\Rightarrow\:\mathrm{sum}\:\mathrm{these}\:\mathrm{up}\:\mathrm{and}\:\mathrm{use}\:\gamma=\pi−\left(\alpha+\beta\right) \\ $$$$\mathrm{and}\:\mathrm{let}\:{u}=\mathrm{tan}\:\alpha\:\wedge\:{v}=\mathrm{tan}\:\beta\:\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{3}{r}=\frac{\left(\mathrm{1}−{uv}\right){v}}{\mathrm{1}+{v}^{\mathrm{2}} }{a}+\frac{\left(\mathrm{1}−{uv}\right){u}}{\mathrm{1}+{u}^{\mathrm{2}} }{b}+\frac{{uv}}{{u}+{v}}{c} \\ $$$${R}=\frac{{a}}{\mathrm{2sin}\:\alpha};\:\mathrm{cyclic}\:\mathrm{for}\:{a},\:{b},\:{c}\:\&\:\alpha,\:\beta,\:\gamma \\ $$$$\mathrm{again}\:\mathrm{sum}\:\mathrm{these}\:\mathrm{up}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{3}{R}=\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{4}{u}}{a}+\frac{\mathrm{1}+{v}^{\mathrm{2}} }{\mathrm{4}{u}}{b}+\frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{v}^{\mathrm{2}} \right)}{\mathrm{4}\left({u}+{v}\right)\left(\mathrm{1}−{uv}\right)}{c} \\ $$$$\Rightarrow \\ $$$$\frac{{r}}{{R}}=\frac{\mathrm{4}\left(\mathrm{1}−{uv}\right){uv}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{v}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow \\ $$$$\frac{{r}}{{R}}=\mathrm{cos}\:\alpha\:+\mathrm{cos}\:\beta\:\underset{=\mathrm{cos}\:\gamma} {\underbrace{−\mathrm{cos}\:\left(\alpha+\beta\right)}}\:−\mathrm{1} \\ $$$$\Rightarrow\:\frac{{r}}{{R}}=\mathrm{cos}\:\alpha\:+\mathrm{cos}\:\beta\:+\mathrm{cos}\:\gamma\:−\mathrm{1} \\ $$$$\left[\mathrm{0}<\frac{{r}}{{R}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{any}\:\mathrm{triangle}\:\mathrm{0}<\theta<\pi\:\Rightarrow\:\mathrm{0}<\mathrm{sin}\:\theta\:<\mathrm{1}\:\Rightarrow \\ $$$$\frac{\mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta}\geqslant\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\Sigma\:\frac{\mathrm{cos}\:\theta}{\mathrm{sin}^{\mathrm{2}} \:\theta}\:\geqslant\left(\Sigma\:\mathrm{cos}\:\theta\right)−\mathrm{1} \\ $$$$ \\ $$$$\left[\right. \\ $$$$\mathrm{what}'\mathrm{s}\:\mathrm{left}\:\mathrm{to}\:\mathrm{prove}: \\ $$$$\mathrm{2}\leqslant\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}^{\mathrm{2}} \:\alpha}+\frac{\mathrm{cos}\:\beta}{\mathrm{sin}^{\mathrm{2}} \:\beta}−\frac{\mathrm{cos}\:\left(\alpha+\beta\right)}{\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right)} \\ $$$$\mathrm{because}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{a} \\ $$$$\mathrm{min}/\mathrm{max}\:\mathrm{at}\:\alpha=\beta=\frac{\pi}{\mathrm{3}} \\ $$$$\left.\right] \\ $$
Commented by mnjuly1970 last updated on 04/Sep/24
thanks alot sir Ghisom gratefu ⋛
$${thanks}\:{alot}\:{sir}\:{Ghisom}\: \\ $$$${gratefu}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by Ghisom last updated on 04/Sep/24
I found another way to end my proof...
$$\mathrm{I}\:\mathrm{found}\:\mathrm{another}\:\mathrm{way}\:\mathrm{to}\:\mathrm{end}\:\mathrm{my}\:\mathrm{proof}… \\ $$
Commented by Frix last updated on 04/Sep/24
Nice.
$$\mathrm{Nice}. \\ $$
Answered by mahdipoor last updated on 05/Sep/24
h(A,B,C)=A+B+C=180 f(A,B,C)=((cosA)/(sin^2 A))+((cosB)/(sin^2 B))+((cosC)/(sin^2 C)) 2R=(a/(sinA))=(b/(sinB))=(c/(sinC)) S=r((a+b+c)/2)=((ba)/2)sinC⇒ r((a+((sinB)/(sinA))a+((sinC)/(sinA))a)/2)=((((sinB)/(sinA))a×a)/2)sinC⇒ r=((a×sinB×sinC)/(sinA+sinB+sinC)) g(A,B,C)=(r/R)=2×((sinA×sinB×sinC)/(sinA+sinB+sinC)) { (((∂f/∂x_i )=λ(∂f/∂x_i ))),((h=180)) :} ⇒f_(min) =f(60,60,60)=2⇒2≤f≤∞ { (((∂g/∂x_i )=λ(∂g/∂x_i ))),((h=180)) :} ⇒g_(max) =g(60,60,60)=(1/2)⇒0≤g≤(1/2) ⇒g≤(1/2)<2≤f ⇒ g<f
$${h}\left({A},{B},{C}\right)={A}+{B}+{C}=\mathrm{180} \\ $$$${f}\left({A},{B},{C}\right)=\frac{{cosA}}{{sin}^{\mathrm{2}} {A}}+\frac{{cosB}}{{sin}^{\mathrm{2}} {B}}+\frac{{cosC}}{{sin}^{\mathrm{2}} {C}} \\ $$$$\mathrm{2}{R}=\frac{{a}}{{sinA}}=\frac{{b}}{{sinB}}=\frac{{c}}{{sinC}} \\ $$$${S}={r}\frac{{a}+{b}+{c}}{\mathrm{2}}=\frac{{ba}}{\mathrm{2}}{sinC}\Rightarrow \\ $$$${r}\frac{{a}+\frac{{sinB}}{{sinA}}{a}+\frac{{sinC}}{{sinA}}{a}}{\mathrm{2}}=\frac{\frac{{sinB}}{{sinA}}{a}×{a}}{\mathrm{2}}{sinC}\Rightarrow \\ $$$${r}=\frac{{a}×{sinB}×{sinC}}{{sinA}+{sinB}+{sinC}} \\ $$$${g}\left({A},{B},{C}\right)=\frac{{r}}{{R}}=\mathrm{2}×\frac{{sinA}×{sinB}×{sinC}}{{sinA}+{sinB}+{sinC}} \\ $$$$\begin{cases}{\frac{\partial{f}}{\partial{x}_{{i}} }=\lambda\frac{\partial{f}}{\partial{x}_{{i}} }}\\{{h}=\mathrm{180}}\end{cases}\:\:\Rightarrow{f}_{{min}} ={f}\left(\mathrm{60},\mathrm{60},\mathrm{60}\right)=\mathrm{2}\Rightarrow\mathrm{2}\leqslant{f}\leqslant\infty \\ $$$$\begin{cases}{\frac{\partial{g}}{\partial{x}_{{i}} }=\lambda\frac{\partial{g}}{\partial{x}_{{i}} }}\\{{h}=\mathrm{180}}\end{cases}\:\:\Rightarrow{g}_{{max}} ={g}\left(\mathrm{60},\mathrm{60},\mathrm{60}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{0}\leqslant{g}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{g}\leqslant\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{2}\leqslant{f}\:\Rightarrow\:\:{g}<{f} \\ $$
Commented by mnjuly1970 last updated on 04/Sep/24
thanks alot sir..
$${thanks}\:{alot}\:{sir}.. \\ $$
Commented by Frix last updated on 04/Sep/24
Error: R=(a/(2sin A)) etc.
$$\mathrm{Error}:\:{R}=\frac{{a}}{\mathrm{2sin}\:{A}}\:\mathrm{etc}. \\ $$
Commented by Ghisom last updated on 04/Sep/24
line 3: R=(a/(2sin α))=(b/(2sin β))=(c/(2sin γ)) ⇒ min =(1/2) ≠(1/4)
$$\mathrm{line}\:\mathrm{3}:\:{R}=\frac{{a}}{\mathrm{2sin}\:\alpha}=\frac{{b}}{\mathrm{2sin}\:\beta}=\frac{{c}}{\mathrm{2sin}\:\gamma} \\ $$$$\Rightarrow\:\mathrm{min}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\neq\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mahdipoor last updated on 05/Sep/24
thank you , i edited
$${thank}\:{you}\:,\:{i}\:{edited} \\ $$
Commented by Frix last updated on 05/Sep/24
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