Iranian Mathematical Olympiad

We denote the feet of altitudes of vertices $A$, $B$ and $C$ by $D$, $E$ and $F$, respectively. Firstly, we prove that if $X$, $Y$ and $Z$ are collinear then the length of the tangent from $A$ to the nine-point circle of triangle $ABC$ (which we denote by $\omega$) is equal to the sum of the lengths of tangents from vertices $B$ and $C$ to this circle.
$$P_{\omega}^{A} = \frac{1}{2}AE \cdot AC = \frac{1}{2}AF \cdot AB = \frac{1}{2}AX^{2} = \frac{1}{2}AY^{2}$$
So the length of the tangent from $A$ to $\omega$ is $\frac{\sqrt{2}}{2}AX = \frac{\sqrt{2}}{2}AY$. Similarly, the lengths of tangents from vertices $B$ and $C$ are equal to $BY$ and $CX$, respectively. If $X$, $Y$ and $Z$ lie on a common line we have $YZ + ZX = YX$ and $\angle XZC + \angle YZB = 90^\circ$ (1). Since $CX^2 = CE \cdot CA = CD \cdot CB = CZ^2$, we get $CX = CZ$. In the same manner, we infer $AX = AY$ and $BY = BZ$. Using (1) we can deduce $\angle XCZ + \angle ZBY = 180^\circ$. The quadrilateral $FZCB$ is cyclic and so $\angle ABZ = \angle FCZ$. Therefore,
$$\angle YBA + \angle FCX = 90^\circ \Rightarrow \angle YBA + (90^\circ - \angle A) + \angle ACX = 90^\circ$$
Thus, $\angle YAX = 90^\circ$ and so $\angle AXY = 45^\circ$. From these results, we get $\angle ZBY = \angle ZCX = 90^\circ$. Hence, $\sqrt{2}YB + \sqrt{2}CX = \sqrt{2}AX$ and so $YB + CX = AX$, which is what we wanted to show.

For the converse we present the following lemma.

Lemma 1. Let $C_1$ and $C_2$ be two perpendicular circles meeting each other at two points $A$ and $B$. Then, There are exactly two points like $T$ on the line $AB$ satisfying the following property. If $TY$ and $TX$ are tangents from $T$ to $C_1$ and $C_2$, respectively, such $Y$ and $X$ are in two different sides of $AB$. We have $A$, $Y$ and $X$ are on a common line. Furthermore, these two points are symmetric with respect to the line of centers of $C_1$ and $C_2$ and power of them with respect to $C_1$ and $C_2$ are both equal to $(R_1 + R_2)^2$, where $R_i$ is the radius of $C_i$.

Proof. First note that if the point $T$ has this property, we have
$$\frac{1}{2} \hat{A}Y = \angle TYA = \angle TYX = \angle TXY = \angle TXA = \frac{1}{2} \hat{A}X$$
And since $\angle O_1AO_2 = 90^\circ$ ($O_i$ is the center of $C_i$), we get
$$\hat{A}Y = \hat{A}X = \frac{1}{2}\hat{A}Y + \frac{1}{2}\hat{A}X = \angle XAO_2 + \angle YAO_1 = 180^\circ - \angle O_1AO_2 = 90^\circ$$
So we must have $\hat{A}Y = \hat{A}X = 90^\circ$.

Now for the existence of such $T$, consider points $Y$ and $X$ on $C_1$ and $C_2$, respectively, such that $\hat{A}Y = \hat{A}X = 90^\circ$. Note that we can choose such points in a unique way in one side of $O_1O_2$. It is easy to see $Y$, $A$ and $X$ are collinear. Therefore, if $T$ is the intersection point of the tangent to $C_1$ at $Y$ and the tangent to $C_2$ at $X$, we infer
$$\angle TYX = \hat{Y}A = \hat{X}A = \angle TXY$$
So $TY = TX$. Thus, $T$ lies on the radical axis of $C_1$ and $C_2$. On the other hand, in the hexagon $TXO_2AO_1Y$ we have
$$\angle TYO_1 = \angle YO_1A = \angle O_1AO_2 = \angle AO_2X = \angle O_2XT = 90^\circ$$
Hence, $\angle YTX = 90^\circ$, and obviously $TX = AO_2 + YO_1 = R_2 + R_1$, as desired. $\square$

Now for the main problem note that circle $\omega_1$ with center $B$ and radius $BZ$ is perpendicular to the circle $\omega_2$ with center $C$ and radius $CZ$, because $\angle BZX = 90^\circ$. Furthermore, since $\angle AYB = \angle AXC = 90^\circ$, $AY$ is tangent to $\omega_1$ at $Y$ and $AX$ is tangent to $\omega_2$ at $X$. On the other hand $AY = AX = BY + CX$, so according to the lemma the point having this property is unique in one side of $BC$. Therefore, again by lemma $X$, $Y$ and $Z$ are collinear.