Find the smallest angle in the triangle

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Solution 1

Hint:

$ABH$ is an equilateral triangle, $C$ is its center, so $H$ is the orthocenter of $ABC$.

So the smallest angle of $ABC$ is $30°$.

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Solution 2

Using $$\vec{AH}=\frac{\tan B\vec{AB}+\tan C\vec{AC}}{\tan A+\tan B+\tan C},$$ (see the answer for this question) we have $$\vec{AC}=\frac{\vec{AB}+\vec{AH}}{3}$$ $$\Rightarrow \vec{AC}=\left(\frac{1}{3}+\frac{\tan B}{3(\tan A+\tan B+\tan C)}\right)\vec{AB}+\frac{\tan AC}{3(\tan A+\tan B+\tan C)}\vec{AC}.$$

Hence, we have $$0=\frac{1}{3}+\frac{\tan B}{3(\tan A+\tan B+\tan C)}\quad\text{and}\quad 1=\frac{\tan C}{3(\tan A+\tan B+\tan C)},$$ i.e. $$0=\tan A+2\tan B+\tan C\quad\text{and}\quad 3\tan A+3\tan B+2\tan C=0,$$ i.e. $$\tan A=\tan B=a,\tan C=-3a\ \text{for some $a$}$$

Since it is known that $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$for $A+B+C=\pi$ (see this question), we have $$a+a-3a=-3a^3.$$ Hence, $a=\frac{1}{\sqrt 3}$ gives $A=B=30^\circ, C=120^\circ.$

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Updated on August 01, 2022

Comments

  • Amad27
    Amad27 15 days

    The point H is the orthocenter of the triangle ABC and the point C is the centroid of the triangle ABH. In that case the smallest angle of the triangle ABC is: (60), (30), (45), ($\angle ACB$)?

    This is actually quite tough. I got that the orthocenter is the meeting of all of the altitudes, but I still can't figure this out.

    And $c$ is the median of $\triangle ABH$. I don't see anyway to proceed.

  • Amad27
    Amad27 almost 7 years
    Hi, (+1) good answer. But how did you devise that $ABH$ is equilateral>
  • Emilio Novati
    Emilio Novati almost 7 years
    A bit of intuition, confirmed by the possible known answers.
  • Amad27
    Amad27 almost 7 years
    Do we need it to be equilateral? If the orthocenter is outside we already know $\triangle ABC$ is obtuse. Hence, $\angle C > 90$, so, we could have an arrangement of $30, 59, 91$ right?
  • Emilio Novati
    Emilio Novati almost 7 years
    But in this case $C$ is not the centroid of $ABH$.
  • Amad27
    Amad27 almost 7 years
    Wait, how is it not?
  • Emilio Novati
    Emilio Novati almost 7 years
    Intuitively: if $C$ is near to $90°$ than $F$ is near to $C$ but the midpoints of $AH$ and $BH$ are not near to $C$ so the intersection of the two medians cannot be $C$ ( use pencil and paper and you can easily see).