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Thread: 2b^2=a^2+c^2 => 2cotB=cotA+cotC

  1. #1
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    2b^2=a^2+c^2 => 2cotB=cotA+cotC

    Prove that if $\displaystyle 2b^2=a^2+c^2$ in a triangle then $\displaystyle 2\cot\beta=\cot\alpha+\cot\gamma$.
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  2. #2
    MHF Contributor red_dog's Avatar
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    I'll use the letters A, B, C for the angles of the triangles instead of $\displaystyle \alpha, \beta, \gamma$.

    We have $\displaystyle b^2=a^2+c^2-2ac\cos B$. Then

    $\displaystyle 2a^2+2c^2-4ac\cos B=a^2+c^2$

    $\displaystyle a^2+c^2=4ac\cos B$

    $\displaystyle 4R^2\sin^2A+4R^2\sin^2C=16R^2\sin A\sin C\cos B$

    Divide by $\displaystyle 4R^2$:

    $\displaystyle \sin^2A+\sin^2C=4\sin A\sin C\cos B$

    $\displaystyle \sin^2A-\sin A\sin C\cos B+\sin^2C-\sin A\sin C\cos B=2\sin A\sin C\cos B$

    Using that $\displaystyle \sin A=\sin(B+C), \ \sin C=\sin(A+B)$ we have

    $\displaystyle \sin A(\sin(B+C)-\sin C\cos B)+\sin C(\sin(A+B)-\sin A\cos B)=2\sin A\sin C\cos B$

    Now, using the formula $\displaystyle \sin(B+C)=\sin B\cos C+\sin C\cos B$ and $\displaystyle \sin(A+B)=\sin A\cos B+\sin B\cos A$ we have

    $\displaystyle \sin A\sin B\cos C+\sin C\sin B\cos A=2\sin A\sin C\cos B$

    Divide all terms by $\displaystyle \sin A\sin B\sin C$ and we get

    $\displaystyle \cot C+\cot A=2\cot B$
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