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Math Help - trigo proof

  1. #1
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    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
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  2. #2
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    You need to show some working, or share your thoughts before we can help you. How do you think you should start this question?
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  3. #3
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    i thick we should use the independant values of cos A cos B etc
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  4. #4
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    Quote Originally Posted by prasum View Post
    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 R and aCosA+bCosB+cCosC

    aCosA+bCosB+cCosC=2aSinBSinC

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

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

    \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

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

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


    Therefore, if

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

    then

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


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

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

    and

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

    is 18 times the triangle area.

    Therefore

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

    and so the triangle is equilateral.
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    One way to go about it is to obtain expressions for R and aCosA+bCosB+cCosC

    aCosA+bCosB+cCosC=2aSinBSinC

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

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

    \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

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

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


    Therefore, if

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

    then

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


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

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

    and

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

    is 18 times the triangle area.

    Therefore

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

    and so the triangle is equilateral.
    thanks
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