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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)

2 Answers 2

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My guess is the poles are of (different) heights $h_A$, $h_B$, $h_C$ and the angle is from the foot of pole to the circumcentre of $\triangle ABC$ to the top of the pole.

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    Umm does it even follows any know progression? I doubt!2010-12-04
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I would read it that the poles have a diameter>0. The pole at A has diameter so the angle seen from the circumcenter is $\alpha$. See if that works.

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    I think Moron's interpretation and mine will lead to the same answer. They just work in orthogonal axes.2010-12-03