Serbian Mathematical Olympiad

Solution:

Let us prove that the triangles $K_{a}K_{b}K_{c}$ and $A_{1}B_{1}C_{1}$ are homothetic. To prove this, it suffices to prove that $K_{a}K_{b} \parallel A_{1}B_{1}$, i.e. $K_{a}K_{b} \parallel AB$ (analogously it will follow for the other pairs of sides).

Denote $M = K_{a}K_{b} \cap BC$, let $S$ denote the center of the incircle, and let $T$ denote an arbitrary point on the incircle. Denote $\alpha = \angle BAS$, $\beta = \angle CBS$, $\gamma = \angle ACS$. Using directed angles (modulo $180^\circ$), we obtain $\angle B'EB = \beta + 2\gamma$, and analogously $\angle A'DA = \alpha + 2\beta$, and from this $\angle A'DK_{a} = 2\alpha + 4\beta$. Then, $\angle B'TK_{b} = \angle B'SE = 90^\circ + \angle B'ES = \gamma - \alpha$ and analogously $\angle A'TK_{a} = \beta - \gamma$. Then, $\angle A'TB' = \angle A'SC = 90^\circ + \angle A'CS = \alpha + \beta$. And finally we obtain $\angle K_{a}TK_{b} = \angle K_{a}TA' + \angle A'TB' + \angle B'TK_{b} = 2\gamma$.

Also, from triangle $DK_{a}M$ we obtain $\angle CMK_{a} = \angle CDK_{a} + \angle DK_{a}M = \angle A'DK_{a} + \angle DK_{a}K_{b} = (2\alpha + 4\beta) + \angle K_{a}TK_{b} = (2\alpha + 4\beta) + 2\gamma = 2\beta$. Hence, $\angle CMK_{a} = \angle CBA$, from which it follows that $K_{a}K_{b} \parallel AB$, which was to be proved. Hence, the triangles $K_{a}K_{b}K_{c}$ and $A_{1}B_{1}C_{1}$ are homothetic.

Let us also note that the coefficient of the homothety is positive: if it were negative, the segments $K_{a}A_{1}$, $K_{b}B_{1}$, $K_{c}C_{1}$ would intersect at one point. If $\alpha > \beta$, then the points $C_{1}$ and $K_{c}$, and hence the whole segment $K_{c}C_{1}$, lie inside the quadrilateral $SFB D$. Therefore, if without loss of generality we assume $\alpha > \beta > \gamma$, then $K_{c}C_{1} \subset SFB D$, but $K_{a}A_{1} \subset SDCE$, so these two segments are disjoint.

Since the triangles $K_{a}K_{b}K_{c}$ and $A_{1}B_{1}C_{1}$ are homothetic, their circumscribed circles are also homothetic. But these are the Euler circle and the incircle of triangle $ABC$, respectively, and it is known that these two circles are internally tangent at the Feuerbach point of triangle $ABC$. Therefore (together with the fact that the coefficient of the homothety is positive), the center of the homothety is exactly the Feuerbach point. From this it follows that $A_{1}K_{a}$, $B_{1}K_{b}$, $C_{1}K_{c}$ intersect at the Feuerbach point of triangle $ABC$, which belongs to the incircle of triangle $ABC$, whereby the statement is proved.

Solution:

Alternative solution. Let the incircle of triangle $ABC$ be the unit circle in the complex plane. Then $a = \frac{2b'c'}{b'+c'}$, $b = \frac{2a'c'}{a'+c'}$, $c = \frac{2a'b'}{a'+b'}$. Then
$$a_{1} = \frac{b+c}{2} = \frac{a'^{2}b' + a'^{2}c' + 2a'b'c'}{(a'+b')(a'+c')}$$
We find the value of $k_{a}$ from the condition $\frac{k_{a}}{a} = \overline{\left(\frac{a'}{a}\right)}$, whence $k_{a} = \frac{1}{a'}\overline{\bar{a}} = \frac{b'c'}{a'}$. Now we find the intersection point $z$ of the incircle (whence $|z| = 1$) and the line $K_{a}A_{1}$ (whence $\frac{z-k_{a}}{a_{1}-k_{a}} = \overline{\left(\frac{z-k_{a}}{a_{1}-k_{a}}\right)}$). We can transform the second condition into the form
$$\overline{\left(a_{1}-k_{a}\right)}(z-k_{a}) = \left(\frac{1}{z} - \frac{1}{k_{a}}\right)(a_{1}-k_{a})$$
whence (since $z \neq k_{a}$) it follows that $\overline{\left(a_{1}-k_{a}\right)} = -\frac{1}{zk_{a}}(a_{1}-k_{a})$, so
$$z = -\frac{1}{k_{a}} \frac{a_{1}-k_{a}}{\overline{\left(a_{1}-k_{a}\right)}} = -\frac{(a'^{2}-b'c')(a'b'+a'c'+b'c')}{(b'c'-a'^{2})(a'+b'+c')} = \frac{a'b'+a'c'+b'c'}{a'+b'+c'}$$
Since the expression above is symmetric in $a'$, $b'$, $c'$, it follows analogously that the lines $K_{b}B_{1}$ and $K_{c}C_{1}$ intersect the incircle at the same point, whereby the statement is proved.