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I have such a problem:

In $\triangle ABC$ two medians $AD$ and $BE$ intersect at point $O$. Determine the measure of $\angle BAC$ knowing that $\triangle AOE$ is equilateral.

I've drawn such a triangle and I guess that $\angle BAC$ might be $30^\circ$. Using the fact that the intersection point of two medians is the centroid of the triangle I found that $EB=2AD$. I don't know what to do next. I hope you'll give me just a small hint.

Thank you!

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For convenience, take angle $BAO$ to be $\beta$, and $AO=OE=x$.

Now, we use the fact that the centroid divides the median into a ratio of $2:1$. Thus if $OE=x$ then $OB=2x$. And note that $\angle OBA= 180^{\circ}-(120^{\circ}+\beta)=60^{\circ}-\beta$. So using the sine rule on triangle $BOA$, we have:

$$\frac{\sin\beta}{2x}=\frac{\sin(60^{\circ}-\beta)}{x}$$

That gives us the equation, $\sin\beta=2\sin(60^{\circ}-\beta)$. Expanding, we get:

$$\sin \beta = 2\cos \beta\sin 60^{\circ}-2\cos 60^{\circ}\sin \beta$$ $$2\sin \beta=\sqrt{3} \cos \beta$$ $$\tan \beta= \frac{\sqrt{3}}{2}$$

Thus, $\beta=\arctan{\frac{\sqrt{3}}{2}}$. And hence, $\angle BAC =\arctan{\frac{\sqrt{3}}{2}}+60^{\circ}\approx 100.893^{\circ}$. Its worth noting here that though the solution is unique but is not a simple value what we would have hoped for, there is no convenient expression for the angle without inverse trig functions.

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    $\begingroup$ IT's a really ingenious solution, thank you very much! $\endgroup$ – Ivan Gandacov Mar 31 '14 at 14:26

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