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Math Help - trigo

  1. #1
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    trigo

    a triangle is formed by drawing tangents at A,B,C to the circumcircle of triangle ABC prove that the perimeter of this triangle is 2RtanAtanBtanC where R is the radius of circumcircle.
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  2. #2
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    See the diagram.

    trigo-trig.jpg

    This shows 1/3 of the proof.

    Consider the part of the external triangle opposite angle A. Call angle A theta.
    The diagram shows that the length for the part of the triangle opposite angle A is
    2*R*tan(theta) = 2*R*tan(A)

    Note: The 2*theta is measured from the center of the circle.

    Repeating the procedure for the parts opposite B and C and summing, we get:

    Perimeter = 2*R*tan(A) + 2*R*tan(B) + 2*R*tan(C) = 2R(tan(A) + tan(B) + tan(C))

    Is this the formula you want to prove?
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  3. #3
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    Hello, snowtea!

    Lovely work!

    You found that: . \text{Perimeter}\:=\:2R\left(\tan A + \tan B + \tan C)
    . . which involves the sum of the tangents.

    The original equation has the product of the tangents.


    But not to worry . . . Here's a surprising theorem:

    . . In \Delta ABC\!:\;\tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C



    Proof


    A + B + C \:=\:180^o \quad\Rightarrow\quad A + B \;=\;180^o - C


    Take tangents: . \tan(A + B) \;=\;\tan(180^o - C)


    . . . . . . . . . \dfrac{\tan A + \tan B}{1 - \tan A\tan B} \;=\;-\tan C


    . . . . . . . . . . \tan A + \tan B \;=\;-\tan C + \tan A\tan B\tan C


    . . . . \tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, snowtea!

    Lovely work!

    You found that: . \text{Perimeter}\:=\:2R\left(\tan A + \tan B + \tan C)
    . . which involves the sum of the tangents.

    The original equation has the product of the tangents.


    But not to worry . . . Here's a surprising theorem:

    . . In \Delta ABC\!:\;\tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C



    Proof


    A + B + C \:=\:180^o \quad\Rightarrow\quad A + B \;=\;180^o - C


    Take tangents: . \tan(A + B) \;=\;\tan(180^o - C)


    . . . . . . . . . \dfrac{\tan A + \tan B}{1 - \tan A\tan B} \;=\;-\tan C


    . . . . . . . . . . \tan A + \tan B \;=\;-\tan C + \tan A\tan B\tan C


    . . . . \tan A + \tan B + \tan C \;=\;\tan A\tan B\tan C


    thanks
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