A geometry problem from an Italian contest

Question

Consider a quadrilateral $ABCD$ in which the properties listed in the figure below are given.

We want to calculate the length of $BC.$ Using the first and two of the three angle properties it's easy to see that $BCD$ is isosceles, since $$\angle{CDB}=\pi-\angle{CBD}-\angle{BCD}=\pi-\beta-2\alpha=\beta$$ and hence $\overline{BC}=\overline{CD}.$ I really can't find anything else. Using Geogebra it seems like the figure "wants" to be cyclic and to be an isosceles trapezoid, but if I use the angle relations to build it I don't have the correct lengths of $AX$ and $CX.$ I tried to prove that $ABCD$ is cyclic, but I can't find anything useful, I tried to use $CX$ and $AX$ as cevians and tried to use Stewart's theorem to get some relations with the sides, but I didn't have too much success. Can I have a solution, or at least some hints? In all my reasoning I didn't use the identity $\gamma+2\delta=\pi/2:$ maybe it can be used with some reflection or something like that, but all the constructions I've made were useless...

Answer 1

Let $M$ be the symmetric of $A$ with respect to $BD$. $M$ is on the circle with center $C$ and radius $R=BC$ since angle $BMD =$ half the central angle. Angle $BDM = \delta$ by symmetry, therefore angle $XCM=XCB+BCM=\gamma + 2 \delta $ is a right angle.

$R^2=65^2-33^2$

$R=56$

Answer 2

First of all, note that $\angle ABD= \beta +\gamma +\delta. $

Now, in $\triangle ACD$:

$$\frac{DC}{AD}=\frac{33}{65} \times \frac{\sin \delta}{\sin \beta};$$

in $\triangle ABD:$

$$\frac{AD}{BD}= \frac{\sin (\beta +\gamma +\delta)}{\sin \alpha};$$ in $\triangle BCD:$

$$\frac{BD}{BC}=\frac{\sin 2\alpha}{\sin \beta}=2\sin \alpha.$$

Therefore:

$$1=\frac{DC}{AD} \times \frac{AD}{BD} \times \frac{BD}{BC}=\frac{66}{65}\times \frac{\sin (\beta +\gamma +\delta)\sin \delta}{\sin \beta} \implies \\ \frac{33}{65}[\cos (\beta +\gamma)-\cos (\beta +\frac{\pi}{2})]=\sin \beta \implies \\ 33\cos (\beta +\gamma)=32\sin \beta. \ \ \ \ (*)$$

On the other hand, similarly, we have:

$$\frac{AB}{BD} =\frac{\sin \delta}{\sin \alpha}; \\ \frac{BD}{CD}=\frac{\sin 2\alpha}{ \sin \beta}; \\ \frac{BC}{AB}=\frac{\sin (2\beta +2\gamma +\delta)}{\sin \gamma}.$$

Therefore,

$$2\sin (2\beta +2\gamma +\delta) \sin \delta =\sin \gamma \implies \\ 2\sin (2\beta +2\gamma +\delta) \sin \delta =\cos 2\delta \implies \\ \cos (2\gamma +2\beta)=\cos (2\beta +2\gamma +2\delta)+\cos 2\delta \implies \\ \cos (2\gamma +2\beta)=2\cos(\gamma +\beta +2\delta)\cos (\beta +\gamma) \implies \\ \cos (2\gamma +2\beta)=-2 \sin \beta \cos (\beta +\gamma). \ \ \ \ (**)$$

Now, considering $(*)$ and $(**)$ together, we get the equation as follows:

$$2 \cos^2 (\beta +\gamma)-1=\frac{-33}{16}\cos^2 (\beta +\gamma) \implies \\ \cos^2 (\beta +\gamma)=\frac{16}{65} \implies \sin^2 (\beta +\gamma)=\frac{49}{65}.$$

Now, we are done because: $$\frac{BC}{33}=\frac{\sin (\beta +\gamma)}{\sin \beta};$$

and the answer is $56$.