Question Number 54502 by math1967 last updated on 05/Feb/19
$${In}\:\bigtriangleup{ABC}\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}+\mathrm{cos}\:{C}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${prove}\:{that}\:{trianle}\:{is}\:{equilateral} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
![point A,B,C lies on cosx curve point A coordinate is (A,cosA) point B coordinate (B,cosB) point C coordinate(C,cosC) △ABC centroid is G coordinate of G is(((x_1 +x_2 +x_3 )/3),((y_1 +y_2 +y_3 )/3)) so G is(((A+B+C)/3),((cosA+cosB+cosC)/3)) as per diagram point Q(((A+B+C)/3),0) point p{((A+B+C)/3),cos(((A+B+C)/3))} from diagram QP>QG cos(((A+B+C)/3))>((cosA+cosB+cosC)/3) 3cos((π/3))>cosA+cosB+cosC cosA+cosB+cosC<(3/2) now maximum value of cosA+cosB+cosC when A=B=C...so A=B=C=(π/3) so cos(π/3)+cos(π/3)+cos(π/3) =(1/2)+(1/2)+(1/2) =(3/2) that means equilateral triangle... [in other cases cosA+cosB+cosC<(3/2)]](Q54528.png)
$${point}\:{A},{B},{C}\:{lies}\:{on}\:{cosx}\:{curve} \\ $$$${point}\:{A}\:{coordinate}\:{is}\:\left({A},{cosA}\right) \\ $$$${point}\:{B}\:{coordinate}\:\left({B},{cosB}\right) \\ $$$${point}\:{C}\:{coordinate}\left({C},{cosC}\right) \\ $$$$\bigtriangleup{ABC}\:\:{centroid}\:{is}\:{G}\: \\ $$$${coordinate}\:{of}\:{G}\:{is}\left(\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }{\mathrm{3}},\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{y}_{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$${so}\:{G}\:{is}\left(\frac{{A}+{B}+{C}}{\mathrm{3}},\frac{{cosA}+{cosB}+{cosC}}{\mathrm{3}}\right) \\ $$$${as}\:{per}\:{diagram}\:\:{point}\:{Q}\left(\frac{{A}+{B}+{C}}{\mathrm{3}},\mathrm{0}\right) \\ $$$${point}\:{p}\left\{\frac{{A}+{B}+{C}}{\mathrm{3}},{cos}\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right)\right\} \\ $$$${from}\:{diagram} \\ $$$${QP}>{QG} \\ $$$${cos}\left(\frac{{A}+{B}+{C}}{\mathrm{3}}\right)>\frac{{cosA}+{cosB}+{cosC}}{\mathrm{3}} \\ $$$$\mathrm{3}{cos}\left(\frac{\pi}{\mathrm{3}}\right)>{cosA}+{cosB}+{cosC} \\ $$$${cosA}+{cosB}+{cosC}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${now}\:{maximum}\:{value}\:{of}\:{cosA}+{cosB}+{cosC} \\ $$$${when}\:{A}={B}={C}...{so}\:{A}={B}={C}=\frac{\pi}{\mathrm{3}} \\ $$$${so}\:{cos}\frac{\pi}{\mathrm{3}}+{cos}\frac{\pi}{\mathrm{3}}+{cos}\frac{\pi}{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:{that}\:{means}\:{equilateral}\:{triangle}... \\ $$$$\left[{in}\:{other}\:{cases}\:{cosA}+{cosB}+{cosC}<\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$
Commented by math1967 last updated on 05/Feb/19
$${Thank}\:{you}\:{sir} \\ $$
Commented by math1967 last updated on 05/Feb/19
$${other}\:{method} \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)+\mathrm{cos}\:\left({B}+{C}\right)+\mathrm{cos}\:\left({C}+{A}\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{sin}\:{A}−\mathrm{sin}\:{B}\right)^{\mathrm{2}} +\left(\mathrm{sin}\:{B}−\mathrm{sin}\:{C}\right)^{\mathrm{2}} +\left(\mathrm{sin}\:{C}−\mathrm{sin}\:{A}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:{B}+\mathrm{cos}\:{C}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:{C}+\mathrm{cos}\:{A}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\therefore\left(\mathrm{sin}\:{A}−\mathrm{sin}\:{B}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{A}={B} \\ $$$${similarlyB}={C}\:\therefore\bigtriangleup{ABC}\:{equilateral} \\ $$$${if}\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}=\mathrm{0}\:{then}\:{A}=\pi−{B} \\ $$$${not}\:{possible}\:{for}\:{triangle} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
$${bah}\:{darun}... \\ $$