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Question Number 15824 by Tinkutara last updated on 18/Jun/17
If in a ΔABC, cos A + cos B + cos C = (3/2) . Prove that ΔABC is an equilateral triangle.
$$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\Delta{ABC},\:\mathrm{cos}\:{A}\:+\:\mathrm{cos}\:{B}\:+\:\mathrm{cos}\:{C}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\Delta{ABC}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$
Answered by mrW1 last updated on 14/Jun/17
C=180−(A+B) cos C=−cos (A+B) S=cos A + cos B + cos C =cos A + cos B − cos (A+B) let x=A, y=B S=cos x+cos y−cos (x+y) (∂S/∂x)=−sin x+sin (x+y)=0 ⇒sin x−sin (x+y)=0 ...(i) (∂S/∂y)=−sin y+sin (x+y)=0 ⇒sin y−sin (x+y)=0 ...(ii) from (i) and (ii) sin x=sin y ⇒x=y sin 2x+sin x=0 sin x (1−2cos x)=0 ⇒cos x=(1/2) x=y=(π/3) S_(max) =(1/2)+(1/2)−(−(1/2))=(3/2) S≤(3/2) only for x=y=z=(π/3), S=S_(max) =(3/2) ⇒the only solution for cos A + cos B + cos C = (3/2) is x=y=z=(π/3), i.e. ΔABC is equilateral.
$$\mathrm{C}=\mathrm{180}−\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\mathrm{cos}\:\mathrm{C}=−\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\mathrm{S}=\mathrm{cos}\:{A}\:+\:\mathrm{cos}\:{B}\:+\:\mathrm{cos}\:{C}\:=\mathrm{cos}\:{A}\:+\:\mathrm{cos}\:{B}\:−\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{A},\:\mathrm{y}=\mathrm{B} \\ $$$$\mathrm{S}=\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}−\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\frac{\partial\mathrm{S}}{\partial\mathrm{x}}=−\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{x}−\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{0}\:\:\:\:\:…\left(\mathrm{i}\right) \\ $$$$\frac{\partial\mathrm{S}}{\partial\mathrm{y}}=−\mathrm{sin}\:\mathrm{y}+\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{y}−\mathrm{sin}\:\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{0}\:\:\:\:…\left(\mathrm{ii}\right) \\ $$$$ \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{sin}\:\mathrm{y} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{y} \\ $$$$\mathrm{sin}\:\mathrm{2x}+\mathrm{sin}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{x}\:\left(\mathrm{1}−\mathrm{2cos}\:\mathrm{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{y}=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{S}_{\mathrm{max}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{S}\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{only}\:\mathrm{for}\:\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{\pi}{\mathrm{3}},\:\mathrm{S}=\mathrm{S}_{\mathrm{max}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{cos}\:{A}\:+\:\mathrm{cos}\:{B}\:+\:\mathrm{cos}\:{C}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{\pi}{\mathrm{3}},\:\mathrm{i}.\mathrm{e}.\:\Delta\mathrm{ABC}\:\mathrm{is}\:\mathrm{equilateral}. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by mrW1 last updated on 14/Jun/17
from eqn. (i) and (ii) ⇒x=y
$$\mathrm{from}\:\mathrm{eqn}.\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right)\:\Rightarrow\mathrm{x}=\mathrm{y} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17
cosA+cosB+cosC=2cos((A+B)/2)cos((A−B)/2)+cos(180−(A+B))= =2cos((A+B)/2)cos((A−B)/2)−cos(A+B)= 2cos((A+B)/2)cos((A−B)/2)−2cos^2 ((A+B)/2)+1= 2cos((A+B)/2)(cos((A−B)/2)−cos((A+B)/2))+1= 2cos(90−(C/2))(sin(A/2)sin(B/2))+1= =4sin(A/2)sin(B/2)sin(C/2)+1 ⇒4sin(A/2)sin(B/2)sin(C/2)+1=(3/2)⇒ ⇒sin(A/2)sin(B/2)sin(C/2)=(1/8) sin(A/2)=(√((1−cosA)/2))=(√((1−((b^2 +c^2 −a^2 )/(2bc)))/2))= =(√((a^2 −(b−c)^2 )/(4bc)))=(√(((a+b−c)(a+c−b))/(4bc)))= =(√((2(p−c)2(p−b))/(4bc)))=(√(((p−b)(p−c))/(bc))) ⇒(√(((p−b)(p−c))/(bc))).(√(((p−a)(p−c))/(ac))).(√(((p−a)(p−b))/(ab)))=(1/8) ⇒(((p−a)(p−b)(p−c))/(abc))=(1/8)⇒(S^2 /(p.abc))=(1/8) ⇒(S/p).(S/(abc))=(1/8)⇒r.(1/(4R))=(1/8)⇒R=2r . note: in any triangle ABC: 1)S=p.r (r=radius of incircle of ABC) 2)abc=4R.S (R=rsdius of outcircle of ABC) 3) d^2 =R^2 −2r.R (Euler′s rule,d =distance between centers of (in) and )out( circles) only in equilateral triangle ,centers of (in) and )out( circles are the same point. i.e : d=0⇒ R=2r . so this triangle is equilateral.
$${cosA}+{cosB}+{cosC}=\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}{cos}\frac{{A}−{B}}{\mathrm{2}}+{cos}\left(\mathrm{180}−\left({A}+{B}\right)\right)= \\ $$$$=\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}{cos}\frac{{A}−{B}}{\mathrm{2}}−{cos}\left({A}+{B}\right)= \\ $$$$\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}{cos}\frac{{A}−{B}}{\mathrm{2}}−\mathrm{2}{cos}^{\mathrm{2}} \frac{{A}+{B}}{\mathrm{2}}+\mathrm{1}= \\ $$$$\mathrm{2}{cos}\frac{{A}+{B}}{\mathrm{2}}\left({cos}\frac{{A}−{B}}{\mathrm{2}}−{cos}\frac{{A}+{B}}{\mathrm{2}}\right)+\mathrm{1}= \\ $$$$\mathrm{2}{cos}\left(\mathrm{90}−\frac{{C}}{\mathrm{2}}\right)\left({sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}\right)+\mathrm{1}= \\ $$$$=\mathrm{4}{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow \\ $$$$\Rightarrow{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}}{\mathrm{2}}{sin}\frac{{C}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${sin}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−{cosA}}{\mathrm{2}}}=\sqrt{\frac{\mathrm{1}−\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}}{\mathrm{2}}}= \\ $$$$=\sqrt{\frac{{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} }{\mathrm{4}{bc}}}=\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)}{\mathrm{4}{bc}}}= \\ $$$$=\sqrt{\frac{\mathrm{2}\left({p}−{c}\right)\mathrm{2}\left({p}−{b}\right)}{\mathrm{4}{bc}}}=\sqrt{\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{bc}}} \\ $$$$\Rightarrow\sqrt{\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{bc}}}.\sqrt{\frac{\left({p}−{a}\right)\left({p}−{c}\right)}{{ac}}}.\sqrt{\frac{\left({p}−{a}\right)\left({p}−{b}\right)}{{ab}}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\frac{\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)}{{abc}}=\frac{\mathrm{1}}{\mathrm{8}}\Rightarrow\frac{{S}^{\mathrm{2}} }{{p}.{abc}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow\frac{{S}}{{p}}.\frac{{S}}{{abc}}=\frac{\mathrm{1}}{\mathrm{8}}\Rightarrow{r}.\frac{\mathrm{1}}{\mathrm{4}{R}}=\frac{\mathrm{1}}{\mathrm{8}}\Rightarrow{R}=\mathrm{2}{r}\:. \\ $$$${note}: \\ $$$${in}\:{any}\:{triangle}\:{ABC}: \\ $$$$\left.\mathrm{1}\right){S}={p}.{r}\:\:\:\left({r}={radius}\:{of}\:{incircle}\:{of}\:{ABC}\right) \\ $$$$\left.\mathrm{2}\right){abc}=\mathrm{4}{R}.{S}\:\left({R}={rsdius}\:{of}\:{outcircle}\:{of}\:{ABC}\right) \\ $$$$\left.\mathrm{3}\right)\:{d}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{r}.{R}\:\:\:\left({Euler}'{s}\:{rule},{d}\:={distance}\right. \\ $$$$\left.{between}\:{centers}\:{of}\:\left({in}\right)\:{and}\:\right){out}\left(\:{circles}\right) \\ $$$${only}\:{in}\:{equilateral}\:{triangle}\:,{centers}\:{of} \\ $$$$\left.\left({in}\right)\:{and}\:\:\right){out}\left(\:\:{circles}\:{are}\:{the}\:{same}\:{point}.\right. \\ $$$${i}.{e}\::\:{d}=\mathrm{0}\Rightarrow\:\:{R}=\mathrm{2}{r}\:. \\ $$$${so}\:{this}\:{triangle}\:{is}\:{equilateral}. \\ $$
Commented by Tinkutara last updated on 15/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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