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Math Help - Trig. with ratio question

  1. #1
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    Question Trig. with ratio question

    Hello,

    \triangleABC is an acute triangle and H is the triangle's orthocenter.
    Express the ratio of \triangleABH's area to \triangleABC's area through the triangle's angles.
    (No sketch is given)

    According to the answer key, the answer is supposed to be:
    \frac{\triangle ABH}{\triangle ABC} = \cot\alpha * \cot\beta

    Any help?

    Thanks in advance.
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  2. #2
    Junior Member roy_zhang's Avatar
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    Please consider the attached figure, given any acute \triangle ABC where AH_A, BH_B and CH_C are three altitudes from the vertices A,B and C. And H is the triangle's orthocenter. Also let \angle A = \alpha, \angle B=\beta and \angle C=\gamma. Also note that \angle 1+\angle 3=\alpha and \angle 2 + \angle 4=\beta.

    By definition, we have
    \frac{\triangle ABH}{\triangle ABC}=\frac{\frac{1}{2}\cdot HH_C\cdot AB}{\frac{1}{2}\cdot CH_C\cdot AB}=\frac{HH_C}{CH_C}

    Consider the right triangle \triangle ACH_C, we have \cot\alpha =\frac{AH_C}{CH_C}, which implies that CH_C=\frac{AH_C}{\cot\alpha}

    Now consider the two right triangles \triangle ACH_A and \triangle BCH_B, we have \angle 1 + \angle\gamma =90 and \angle 2 +\angle \gamma =90, hence \angle 1 = \angle 2=90-\angle \gamma

    Also \angle 3 = \angle \alpha - \angle 1 = \angle \alpha -(90-\angle\gamma)=\angle\alpha+\angle\gamma-90=(180-\angle\beta)-90=90-\angle\beta

    From here, let's consider the right triangle \triangle AHH_C, we have:
    \tan(\angle 3)=\frac{HH_C}{AH_C}. But \angle 3 =90-\angle\beta, we then have \tan(\angle 3)=\tan(90-\angle\beta)=\cot\beta=\frac{HH_C}{AH_C}
    which implies that HH_C=AH_C\cdot\cot\beta

    Use the quantities we obtained for CH_C and HH_C, we have \frac{\triangle ABH}{\triangle ABC}=\frac{HH_C}{CH_C}=\frac{AH_C\cdot\cot\beta}{\  frac{AH_C}{\cot\alpha}}=\cot\alpha\cdot\cot\beta
    Attached Thumbnails Attached Thumbnails Trig. with ratio question-orthocenter.gif  
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  3. #3
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    Thanks for your help and clarity, I get it now.. But would you have solved the question the same way if you weren't committed to the desired final answer? I simply don't think they expected me to reach an identity (is this how it's called?) with 'cot'.
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  4. #4
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by loui1410 View Post
    Thanks for your help and clarity, I get it now.. But would you have solved the question the same way if you weren't committed to the desired final answer? I simply don't think they expected me to reach an identity (is this how it's called?) with 'cot'.
    Yes, Loui1410, I agree with you. If you do not provide the desired answer, I might have tried to use \sin or \cos, however if you look close to the relations between the heights and bases of the target triangles, \tan or \cot seem to be better choices.

    Roy
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