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I already have my own solution for the following question. But I am still interested in other elegant solutions without trigonometry if possible.


This is my own solution. I am lazy to upload the TeX code, I am sorry.

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    @Beni: While I don't really understand what you're telling me, I guess you have a point. The problem is indeed trickier than I initially thought. @Friendly Ghost: Sorry for not taking the problem seriously enough, no offense intended.2011-07-05

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A hint:

Draw a regular $18$-gon $Q$. It has the property that the angle between neigbouring diagonals emanating from the same vertex is $10^\circ$.

The points $B$, $C$, $D$ of your figure can be realized as vertices of $Q$; furthermore the line $C\vee P$ is a diagonal emanating from $C$, and $D\vee A$, $\ D\vee E$ are diagonals emanating from $D$.

I think that the solution of your problem is hidden in this figure. The nontrivial point is the fact that the line $B\vee P$ is also a diagonal, i.e, that three diagonals of $Q$ meet at $P$. This in turn has to do with algebraic relations among the $18$th roots of unity.

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    @Day Late Don: You are right; I was too fast in this regard.2011-07-05
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Let's try for $\alpha$ ...

In $\triangle BPQ$, we have $\angle B = \alpha$ and $\angle P = 120 - \alpha$.

$\begin{eqnarray*} \frac{\sin(120-\alpha)}{\sin\alpha}=\frac{BQ}{PQ}=\frac{BQ}{CQ} \frac{CQ}{DQ} \frac{DQ}{PQ}=\frac{\sin 40}{\sin 20}\frac{\sin 50}{\sin 70} \cdot 1=\frac{2\sin 40 \cos 40}{2\sin 20 \cos 20}=\frac{\sin 80}{\sin 40} \end{eqnarray*}$

Observe that $80 + 40 = 120$. Thus,

$\begin{eqnarray*} \sin(120-\alpha) \sin 40 &=& \sin \alpha \sin( 120-40 ) \\ (\sin 120 \cos\alpha - \cos 120 \sin \alpha ) \sin 40 &=& \sin\alpha ( \sin 120 \cos 40 - \cos 120 \sin 40 ) \\ \cos\alpha \sin 40 &=& \sin\alpha \cos 40 \\ 0&=& \sin( \alpha - 40 ) \\ \alpha &=& 40 \text{ is the only possible answer} \end{eqnarray*}$

Note: Generalizing $120$ to an angle $\gamma$ such that $\sin{\gamma} \neq 0$, we have

$\frac{\sin(\gamma-\alpha)}{\sin\alpha} = \frac{\sin(\gamma - \beta)}{\sin\beta} \implies \sin(\alpha-\beta) = 0$

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    Well, it's *slightly*-more-direct trigonometry. :) I had thought the "80+40=120" thing (in the sense of the $\gamma$ generalization) was key, but I'm not seeing how anything along those lines helps streamline your approach to $\theta$. (I do think your $\theta$ derivation can be shortened a bit, though.) As for avoiding trig altogether ... I'm at a loss.2011-07-05
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Second part.

Construct equilateral triangle DEF. F is point on EB.

Rotate triangle EDF about point E clockwise 20 deg so that F falls at F' and D falls on DB at D'.

Since angle DEF' = 80 deg, then angle EDF' = 50 deg.

Because angle EDA = 50 deg.

Therefore D, A and F' are colinear. ......(1)

angle BED = angle EBD = 40 deg

So D'B = D'E = D'F'

Since angle BD'F' = 40

Therefore angle D'BF' = 70 i.e. angle EBF' = 30 deg

angle EBA = 30 deg (given)

Therefore B, F' and A are colinear. ......(1)

From (1) and (2) we know that A and F' are the same point.

Therefore angle EAD = 50 deg

-1

Construct equilateral triangle PDF (where F is below BC)

Join FB and FC

Triangles FPC and DPC are congruent (SAS)

Therefore CF = CD

angle CFD = angle CDF = 20 deg.

Then C,F,B, D concylic.

angle FBC = angle FDC = 20 deg

Triangle BPD and BFD are congruent (SAS)

Therefore angle PBD = 40 deg.