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Thread: Question regarding circumcentre?

  1. #1
    Super Member fardeen_gen's Avatar
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    Question regarding circumcentre?

    Let f, g, h be the lengths of the perpendiculars from the circumcentre of the triangle ABC on the sides a,b and c respectively. If (a/f) + (b/g) + (c/h) = k.(abc)/(fgh), then the value of k is:

    A)1/4
    B)1/2
    C)1
    D)2
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let O be the circumcenter of the triangle ABC, $\displaystyle OM\perp BC, \ ON\perp AC, \ OP\perp AB$
    We have $\displaystyle BC=a, AC=b, AB=c, OM=f, ON=g, OP=h$
    Let $\displaystyle \widehat{BOM}=\alpha, \ \widehat{NOC}=\beta, \ \widehat{POA}=\gamma$. Then $\displaystyle \alpha+\beta+\gamma=\pi$

    $\displaystyle \frac{a}{f}=2\tan\alpha, \ \frac{b}{g}=2\tan\beta, \ \frac{c}{g}=\tan\gamma$

    Then the equality becomes $\displaystyle 2\tan\alpha+2\tan\beta+2\tan\gamma=8k\tan\alpha\ta n\beta\tan\gamma$

    But is known that if $\displaystyle \alpha+\beta+\gamma=\pi$ then $\displaystyle \tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\bet a\tan\gamma$

    So, $\displaystyle k=\frac{1}{4}$
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