Spring Mathematical Competition

Solution:

We shall use the standard notations. We first prove that $YX \perp BC$. In $\triangle BQR$ we have $\angle BRY = 90^\circ - \frac{\beta}{2}$ and since $\angle MNB = \frac{\beta}{2}$ it follows that $NX \perp RY$. Analogously $RX \perp NY$. This means that $X$ is the orthocenter of $\triangle NRY$ and $YX \perp RN$. Hence $YX \perp BC$.

Now we shall prove that $X$ lies on the altitude of $\triangle ABC$ through $A$. Denote by $X'$ its intersection point with $MN$. Since $BM = BN$ and $CP = CN = p$, we have $BM = BN = p - a$. Then $AM = c - (p - a) = p - b$ and by the Sine theorem for $\triangle AMX'$ we get
$$\frac{AX'}{\sin \frac{\beta}{2}} = \frac{AM}{\sin \angle AX'M} \Longleftrightarrow \frac{AX'}{\sin \frac{\beta}{2}} = \frac{p-b}{\sin \left(90^\circ + \frac{\beta}{2}\right)} \Longleftrightarrow AX' = (p-b) \tan \frac{\beta}{2}$$
On the other hand, if $T$ is the tangent point of the incircle with the side $BC$, then $BT = p-b$ and $r = (p-b) \tan \frac{\beta}{2}$. Therefore $AX' = r$. Analogously, if $X''$ is the intersection point of $RS$ and the altitude of $\triangle ABC$ through $A$, then $AX'' = r$. Therefore $X' \equiv X'' \equiv X$.

Since $YX \perp BC$ and $AX \perp BC$, we conclude that the points $X$, $A$ and $Y$ are colinear.