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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 circumcentre of the triangle 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 subtend 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 hints 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 circumcentre then, $$ \cot \alpha = \frac{OA}{a}$$ $$ \cot \beta = \frac{OB}{b}$$

$$ \cot \gamma = \frac{OC}{c}$$

As O is the circumcentre,$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)

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2 Answers

up vote 1 down vote accepted

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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Aha .. Do you meant the $3D$ angle? –  Quixotic Dec 3 '10 at 18:17
    
@Deb: Yes 3D, seems to make sense when talking about poles and angle subtended. –  Aryabhata Dec 3 '10 at 18:40
    
Well,your guess is bang on! I solved the problem using your interpretation. :) –  Quixotic Dec 4 '10 at 7:19
    
@Deb: Glad that worked :-) –  Aryabhata Dec 4 '10 at 7:25
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@Deb: Yes, there is something wrong. Your question states "angles of poles are in arithmetic progression", while in the solution you are assuming that the heights of the poles are in arithmetic progression. –  Aryabhata Dec 4 '10 at 7:33
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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. –  Ross Millikan Dec 3 '10 at 17:34
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