Auswahlwettbewerb zur Internationalen Mathematik-Olympiade

Solution:

For the points mentioned in the problem statement we denote the point of contact of the incircle with $AC$ by $F$, the midpoint of $AC$ by $S$, the intersection point of $w_{\beta}$ and the perpendicular bisector $m_{AC}$ by $P$, the reflection point of $B$ with respect to $I$ by $R$, the midpoint of $BI$ by $T$, and the pairwise equal segments on the sides of the triangle by $x, y$ and $z$ respectively (see figure).

From $\overline{AB} + \overline{BC} = 3 \overline{AC}$ it follows that
$x + z + z + y = 3(x + y)$, hence (I) $z = x + y$. It is well known that $P$ lies on the circumcircle of $ABC$. Therefore $\angle CAP = \angle CBP = \angle PBA = \angle PCA = \frac{\beta}{2}$ and $\overline{AP} = \overline{CP}$. The triangles $ASP$ and $BEI$ are similar, and since $\overline{AS} = \frac{1}{2} \overline{AC} = \frac{x + y}{2} = \frac{z}{2} = \frac{1}{2} \overline{BE}$, denoting by $r$ the inradius, we have

(II) $\overline{SP} = \frac{1}{2} \overline{EI} = \frac{r}{2}$ as well as $\overline{AP} = \frac{1}{2} \overline{BI}$.
For the area $F$ of triangle $ABC$ we have $F = r s = \sqrt{s(s - a)(s - b)(s - c)}$, where $s = \frac{a + b + c}{2} = x + y + z$. Hence $r(x + y + z) = \sqrt{(x + y + z) x y z}$ and with (I) it follows that $2 r z = \sqrt{2 z x y z} \Leftrightarrow 4 r^{2} = 2 x y \Leftrightarrow 8 r^{2} = 4 x y \Leftrightarrow 9 r^{2} + (x - y)^{2} = r^{2} + (x + y)^{2} = r^{2} + z^{2}$.
Since $\angle BDI = 90^{\circ}$ we have (III) $\overline{BI}^{2} = r^{2} + z^{2}$.
Since $PS \perp AC \perp FI$ we have $\overline{PI}^{2} = \overline{SF}^{2} + (\overline{PS} + \overline{FI})^{2}$.
But $\overline{SF} = \frac{1}{2} \overline{AC} - x = \frac{x + y}{2} - x = \frac{y - x}{2}$ and $\overline{FI} = r$. Hence, using (II), we get $\overline{PI}^{2} = \left(\frac{y - x}{2}\right)^{2} + \left(\frac{3}{2} r\right)^{2} = \frac{(y - x)^{2} + 9 r^{2}}{4}$.
With (III) it follows that $\overline{PI}^{2} = \frac{1}{4} \overline{BI}^{2}$, hence $\overline{PI} = \frac{1}{2} \overline{BI}$.
Thus (IV) $\overline{PI} = \overline{PA} = \overline{PC} = \overline{PR}$.

Moreover, under reflection at $I$, $T$ is mapped to $P$. Hence also $\overline{IT} = \overline{PI}$. $\overline{BI}$ is the diameter of the circle through $B, I, D$ and $E$ (right angles at $D$ and $E$). This circle has $T$ as its center. Under reflection of this circle at $I$, $I$ is mapped to $I$, $T$ is mapped to $P$, $D$ is mapped to $K$, $E$ is mapped to $L$, and the radius is preserved. With (IV) it follows that $A, L, I, K, C$ and $R$ lie on a circle about $P$; hence $ACKL$ is a cyclic quadrilateral.