Find the smallest angle in the triangle
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 $30°$.
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 $$a+a3a=3a^3.$$ Hence, $a=\frac{1}{\sqrt 3}$ gives $A=B=30^\circ, C=120^\circ.$
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Amad27
Updated on August 01, 2022Comments

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 almost 7 yearsHi, (+1) good answer. But how did you devise that $ABH$ is equilateral>

Emilio Novati almost 7 yearsA bit of intuition, confirmed by the possible known answers.

Amad27 almost 7 yearsDo 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 almost 7 yearsBut in this case $C$ is not the centroid of $ABH$.

Amad27 almost 7 yearsWait, how is it not?

Emilio Novati almost 7 yearsIntuitively: 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).