<s>[INST] 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.
 [/INST] 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, $$\frac{abc}{4S}=\frac{abc}{4\cdot\frac{c\cdot\frac{ab}{2R}}{2}}=R.$$
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