Final Round

b) It is not true.

a) Without loss of generality we assume that $\alpha =\angle BAD\le 90^{\circ }$. In this case we have $$\begin{array}{rl}BD& =\sqrt{A{B}^2+A{D}^2-2AB\cdot AD\cos \alpha }\le \\ & \le \sqrt{A{B}^2+A{D}^2+2AB\cdot AD\cos \alpha }=[AD=BC]=\\ & =\sqrt{A{B}^2+B{C}^2-2AB\cdot BC\cos (\pi -\alpha )}=AC.\end{array}$$ So, it is sufficient to show that $XY+YZ\ge BD$. Let $YM\parallel AD$, $DM\parallel XY$ (see the Fig.). It is evident that $XYMD$ is a parallelogram, so $YM=XD$ and $XY=MD$. Since $CZ=AX$, we have $XD=AD-AX=BC-CZ=ZB$. Since $ZB\parallel XD\parallel YM$, we see that $YBZM$ is a parallelogram.



In the parallelogram $YBZM$ we have $\angle BYM=\angle BAD=\alpha \le 90^{\circ }$, therefore, $$\begin{array}{rl}YZ& =\sqrt{Y{B}^2+B{Z}^2-2YB\cdot BZ\cos (\pi -\alpha )}\ge [BZ=YM]\ge \\ & \ge \sqrt{Y{B}^2+Y{M}^2-2YB\cdot YM\cos \alpha }=BM.\end{array}$$ Thus $XY+YZ\ge XY+BM=MD+BM\ge BD$, as required.

b) We show that there exist a parallelogram $ABCD$ and a point $Y$ such that the inequality $$XY+YZ\ge \frac{AC+BD}{2}(\ast )$$ is not valid. Let a parallelogram $ABCD$ is different from a rectangle. Then its diagonals are not equal. Without loss of generality we assume that $AC>BD$. Let $YZ\parallel AC$ (see the Fig.). Let $AX/AD=CZ/BC=\lambda$. By Thales' theorem, we have $AY/AB=CZ/BC=\lambda$. Hence $AY/AB=AX/AD=\lambda$. It follows that the triangles $AYX$ and $ABD$ are similar and $XY/BD=\lambda$, i.e. $XY=\lambda BD$.

Moreover, by construction of $Y$, the triangles $YBZ$ and $ABC$ are similar and $YZ/AC=BZ/BC=(BC-CZ)/BC=1-\lambda$, hence $YZ=(1-\lambda )AC$.

Therefore, $XY+YZ=\lambda BD+(1-\lambda )AC$ and (*) has the form $\lambda BD+(1-\lambda )AC\ge \frac{1}{2}BD+\frac{1}{2}AC$ or $(\lambda -\frac{1}{2})BD\ge (\lambda -\frac{1}{2})AC$. But this inequality is not valid if $\lambda >\frac{1}{2}$.