The base of an isosceles triangle; given leg and radius of circumcircle 
An isosceles triangle $\triangle ABC$ is given with leg $AC=5$ and $R=\dfrac{25}{6}$ of the circumcircle. Find the base of the triangle.


This was my first sketch. The triangles $AHC$ and $OMC$ are similar, so $\dfrac{CH}{CM}=\dfrac{AC}{CO} \Leftrightarrow \dfrac{CO+OH}{CM}=\dfrac{AC}{CO}.$ When I plug with the values, I get a negative number for $OH$. I realised that this is possible if the triangle is obtuse. How should I notice this from the beginning? I have not studied trigonometry.
 A: Since in the standard notation  $$\frac{abc}{4S}=R,$$ we obtain:
$$\frac{5\cdot5\cdot c}{4\cdot\frac{c\sqrt{25-\frac{c^2}{4}}}{2}}=\frac{25}{6}$$ or
$$\sqrt{25-\frac{c^2}{4}}=3,$$
which gives $$c=8.$$
We'll prove that $$\frac{abc}{4S}=R.$$
Indeed, let $\Phi$ be our circle and $CO\cap\Phi=\{C,D\}$.
Thus, $$\Delta ACH\sim\Delta DCB,$$ which gives
$$\frac{CH}{BC}=\frac{AC}{CD}$$ or
$$\frac{CH}{a}=\frac{b}{2R}$$ or
$$CH=\frac{ab}{2R}.$$
Id est, $$\frac{abc}{4S}=\frac{abc}{4\cdot\frac{c\cdot\frac{ab}{2R}}{2}}=R.$$
A: For an acute triangle where the two shortest sides are $a$ and $b$, the inradius is $r$, and the circumradius is $R$, the inequality $R+r < \frac{a+b}{2}$ is satisfied. So in this case, if the triangle were acute, you would need $\frac{25}{6} + r < \frac{5+5}{2}$, i.e. $r < \frac{5}{6}$, which is really small, and you can see from the diagram that it's quite unlikely.
A: Note that, OH is positive if the base AB is below the horizontal diameter line, and negative if the base is above. If AB happens to be the diameter, then AC = $\sqrt2$R. 
Thus, you may compare the given AC and R to determine beforehand.
