# trigo proof

• Dec 14th 2010, 01:35 AM
prasum
trigo proof
in triangle ABC if (acosA+bcosB+ccosC)/(asinB+bsinC+csinA)=(a+b+c)/9R where R is circumradius then prove that the triangle is equilateral
• Dec 15th 2010, 08:31 AM
worc3247
You need to show some working, or share your thoughts before we can help you. How do you think you should start this question?
• Dec 16th 2010, 01:36 AM
prasum
i thick we should use the independant values of cos A cos B etc
• Dec 16th 2010, 02:06 PM
Quote:

Originally Posted by prasum
in triangle ABC if (acosA+bcosB+ccosC)/(asinB+bsinC+csinA)=(a+b+c)/9R where R is circumradius then prove that the triangle is equilateral

One way to go about it is to obtain expressions for $\displaystyle R$ and $\displaystyle aCosA+bCosB+cCosC$

$\displaystyle aCosA+bCosB+cCosC=2aSinBSinC$

$\displaystyle R=\displaystyle\frac{abc}{2r(a+b+c)}$ where $\displaystyle r$ is the incircle radius.

If you draw perpendicular lines from the incentre to all 3 sides, then

$\displaystyle \displaystyle\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{ 2}ar=\frac{1}{2}abSinC=\frac{1}{2}cbSinA=\frac{1}{ 2}acSinB=area$

$\displaystyle \Rightarrow\ r(a+b+c)=abSinC$

$\displaystyle \displaystyle\Rightarrow\ R=\frac{abc}{2abSinC}=\frac{c}{2SinC}$

Therefore, if

$\displaystyle \displaystyle\frac{aCosA+bCosB+cCosC}{asinB+bSinC+ cSinA}=\frac{a+b+c}{9R}$

then

$\displaystyle \displaystyle\ (9)\;\frac{2aSinBSinC}{aSinB+bSinC+cSinA}=\frac{a+ b+c}{R}=\frac{(a+b+c)2SinC}{c}$

$\displaystyle \displaystyle\Rightarrow\frac{9aSinB}{aSinB+bSinC+ cSinA}=\frac{a+b+c}{c}$

$\displaystyle \Rightarrow\ 9acSinB=(a+b+c)(aSinB+bSinC+cSinA)$

and

$\displaystyle \displaystyle\ 9acSinB=(18)\;\frac{1}{2}acSinB$

is 18 times the triangle area.

Therefore

$\displaystyle (a+b+c)(aSinB+bSinC+cSinA)=18(area)\;\;\Rightarrow \ a=b=c$

and so the triangle is equilateral.
• Dec 17th 2010, 01:31 AM
prasum
Quote:

One way to go about it is to obtain expressions for $\displaystyle R$ and $\displaystyle aCosA+bCosB+cCosC$

$\displaystyle aCosA+bCosB+cCosC=2aSinBSinC$

$\displaystyle R=\displaystyle\frac{abc}{2r(a+b+c)}$ where $\displaystyle r$ is the incircle radius.

If you draw perpendicular lines from the incentre to all 3 sides, then

$\displaystyle \displaystyle\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{ 2}ar=\frac{1}{2}abSinC=\frac{1}{2}cbSinA=\frac{1}{ 2}acSinB=area$

$\displaystyle \Rightarrow\ r(a+b+c)=abSinC$

$\displaystyle \displaystyle\Rightarrow\ R=\frac{abc}{2abSinC}=\frac{c}{2SinC}$

Therefore, if

$\displaystyle \displaystyle\frac{aCosA+bCosB+cCosC}{asinB+bSinC+ cSinA}=\frac{a+b+c}{9R}$

then

$\displaystyle \displaystyle\ (9)\;\frac{2aSinBSinC}{aSinB+bSinC+cSinA}=\frac{a+ b+c}{R}=\frac{(a+b+c)2SinC}{c}$

$\displaystyle \displaystyle\Rightarrow\frac{9aSinB}{aSinB+bSinC+ cSinA}=\frac{a+b+c}{c}$

$\displaystyle \Rightarrow\ 9acSinB=(a+b+c)(aSinB+bSinC+cSinA)$

and

$\displaystyle \displaystyle\ 9acSinB=(18)\;\frac{1}{2}acSinB$

is 18 times the triangle area.

Therefore

$\displaystyle (a+b+c)(aSinB+bSinC+cSinA)=18(area)\;\;\Rightarrow \ a=b=c$

and so the triangle is equilateral.

thanks