The problem:
Three poles standing at the points $A$, $B$ and $C$ subtend angles $\alpha$, $\beta$ and $\gamma$ respectively, at the circumcenter of $\Delta ABC$.If the heights of these poles are in arithmetic progression; then show that $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression.
Now, what I could not understand is subtending of the angle part,precisely how a point subtends angle at another point? So, what I am looking for a proper explanation of the problem statement with a figure, since it's troubling me from sometime.
PS: I am not looking for the solution (as of now) or any hint regarding the solution, just a clear explanation will be appreciated.
My solution using Moron's interpretation,
Let $a$ $b$ and $c$ are the length of three sides of the poles and $O$ be the circumcenter then, $$ \cot \alpha = \frac{OA}{a}$$ $$ \cot \beta = \frac{OB}{b}$$
$$ \cot \gamma = \frac{OC}{c}$$
As $O$ is the circumcenter,$OA = OB = OC = k $(say)
Again, $a$ $b$ and $c$ are in arithmetic progression, hence
$$2 \cdot \frac{k}{\cot \alpha} = \frac{k}{\cot \beta} + \frac{k}{\cot \gamma}$$
Canceling $k$ from both sides,
$$2 \cdot \frac{1}{\cot \alpha} = \frac{1}{\cot \beta} + \frac{1}{\cot \gamma}$$
Hence, $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression. (QED)