# A tricky geometry problem

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.  • What is your solution? Jul 4, 2011 at 8:16
• @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, 2011 at 8:37
• @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 $\alpha$. Jul 4, 2011 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, 2011 at 2:53

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.

• $E$ 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, 2011 at 9:32
• @Day Late Don: You are right; I was too fast in this regard. Jul 5, 2011 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

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

• +1 for answering but actually I don't want to use trigonometry again. :-) Jul 5, 2011 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, 2011 at 12:46

Following the hint of Christian Blatter, here are the figures for the embeddings of two of the quadrilaterals in the figure into an 18 sides regular polygon. I leave the details for those who are interested in the problem.

All that remains to do is to prove that certain three diagonals do meet in the points shown, which usually translates into Ceva's theorem on the circle.

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

I noted down my alternative solution in LaTeX and added some figures to make it easier to follow the argumentation. No trigonometry, using mainly properties of congruent triangles and circles.

Unfortunately I am new to the forum, so not able to post images directly in the text yet. Anyways I hope my solution helps.    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.