Balkan Mathematical Olympiad Shortlist

Let $\omega$ be the circle of center $A$ and radius $AB$. Let $A^{\prime }$ be a point on the circle $\omega$, lying on the minor arc $\widehat{BC}$, such that $\angle BAD=\angle A^{\prime }AD$. Since $2\angle DAE=\angle A$, it is easy to see that $\angle CAE=\angle A^{\prime }AE$.



We deduce that the triangles $BAD$ and $A^{\prime }AD$ are congruent by SAS postulate, and thus $BD=A^{\prime }D$. Similarly, $EC=A^{\prime }E$, and thus, the triangle $A^{\prime }DE$ has its sides of lengths $BD$, $DE$ and $EC$ respectively, and we can choose $X=A^{\prime }$, $Y=D$ and $Z=E$.

Moreover, $$\begin{array}{rl}\angle BAC+\angle YXZ& =\angle BAC+\angle D{A}^{\prime }E\\ & =\angle BAC+\angle D{A}^{\prime }A+\angle A{A}^{\prime }E\\ & =\angle BAC+\angle DBA+\angle ACE=180^{\circ }.\end{array}$$

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Alternative solution.

Let $M$ be the midpoint of $BC$. Obviously $D\in BM$ and $E\in MC$. We make now the following notations $\angle A=2t$, $BC=2a$, $BD=x$, $EC=z$ and $DE=y$. It is clear that $DE=2a-x-z$.

On the other hand, $AM=a\cdot \cot t$. But $t=\angle DAM+\angle MAE$, and thus we get: $$\cot (t)=\frac{\cot \angle DAM\cdot \cot \angle MAE-1}{\cot \angle DAM+\cot \angle MAE}(1)$$ Denoting $x+z=s$ and $x\cdot z=p$, using the relation (1), we get: $$a^2(1+{\cot }^{2}t)=a\cdot s(1+{\cot }^{2}t)-p.$$ Now, from the above relation, we get: $$s=a+\frac{p}{a}{\sin }^{2}t.$$ Finally, we obtain: $$y^2=(2a-s{)}^2=x^2+2x\cdot z\cos (2t)+z^2.$$ Now, on the two sides of an angle of the vertex $X$ and measure $\pi -A$, we choose the points $Y$ and $Z$ such that $XY=x$ and $XZ=z$. From the cosine rule we have: $$Y{Z}^2=x^2+2x\cdot z\cos (2t)+z^2=y^2$$ and thus the existence of the triangle $XYZ$ is proved. Moreover, we have $\angle BAC+\angle YXZ=180^{\circ }$.