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Thread: Trig. with ratio question

  1. #1
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    Question Trig. with ratio question

    Hello,

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

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

    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 $\displaystyle \triangle ABC$ where $\displaystyle AH_A, BH_B$ and $\displaystyle CH_C$ are three altitudes from the vertices $\displaystyle A,B$ and $\displaystyle C$. And $\displaystyle H$ is the triangle's orthocenter. Also let $\displaystyle \angle A = \alpha$, $\displaystyle \angle B=\beta$ and $\displaystyle \angle C=\gamma$. Also note that $\displaystyle \angle 1+\angle 3=\alpha$ and $\displaystyle \angle 2 + \angle 4=\beta$.

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

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

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

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

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

    Use the quantities we obtained for $\displaystyle CH_C$ and $\displaystyle HH_C$, we have $\displaystyle \frac{\triangle ABH}{\triangle ABC}=\frac{HH_C}{CH_C}=\frac{AH_C\cdot\cot\beta}{\ frac{AH_C}{\cot\alpha}}=\cot\alpha\cdot\cot\beta$
    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 $\displaystyle \sin$ or $\displaystyle \cos$, however if you look close to the relations between the heights and bases of the target triangles, $\displaystyle \tan$ or $\displaystyle \cot$ seem to be better choices.

    Roy
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