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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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What is your solution? –  The Chaz 2.0 Jul 4 '11 at 8:16
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@Friendly Ghost: You certainly don't need trigonometry here. You can simply name the missing angles and using the fact that angles in a triangle add up to $180^{\circ}$, and similar things you can write down a linear system of equations and determine its solutions. I haven't done it myself (and I won't) but you have so many triangles here that I'd be surprised if this would not lead to a unique solution. –  t.b. Jul 4 '11 at 8:37
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@Theo: I'm not sure that your method will work. Point $P$ is unique with the given properties, and therefore, if your assertion is true, we should find angle $\alpha$ using only angles in the quadrilateral $PDCB$, which i doubt(from my experience in olympiad problems) it will lead to a solvable system for $\alpha$. –  Beni Bogosel Jul 4 '11 at 8:48
    
@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. –  t.b. Jul 5 '11 at 2:53

3 Answers 3

up vote 5 down vote accepted

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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$E$ and $A$ are not vertices of 18-gon $Q$, because they cannot lie on the circle through $B$, $C$, and $D$: if they did, then $\angle ABE$ and $\angle ADE$ --inscribed angles subtending the arc $AE$-- would have to be congruent, but these angles are given distinct measures. Moreover, we know from prior solution of the problem (whereby $\alpha = 40^\circ$) that not even $A$ alone lies on that circle: if quadrilateral $ABCD$ were "cyclic", then $\angle ADC$ and $\angle ABC$ --as inscribed angles subtending arcs that make up the whole circle-- would be supplementary, which is not the case. –  Blue Jul 5 '11 at 9:32
    
@Day Late Don: You are right; I was too fast in this regard. –  Christian Blatter Jul 5 '11 at 18:04

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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+1 for answering but actually I don't want to use trigonometry again. :-) –  Friendly Ghost Jul 5 '11 at 8:51
    
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. –  Blue Jul 5 '11 at 12:46

enter image description here Autocad can do . ...................................................................................................................................................................a a a a a a a a a ............

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Nice answer!!!! –  user34304 4 hours ago

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