, 2015

First we prove the following lemma:

Lemma. Let $ABCDEF$ be a convex hexagon on the coordinate plane. Then the set of the points $P$ inside $ABCDEF$ such that
$$S_{APB} + S_{CPD} + S_{EPF} = S_{BPC} + S_{DPE} + S_{FPA},$$
is a segment.

Proof. Suppose that the equations of the lines containing the segments $AB, BC, CD, DE, EF, FA$ are $a_i x + b_i y + c_i = 0$ for some $a_i, b_i, c_i \in \mathbb{R}$ ($i = 1, \dots, 6$). Without loss of generality we can assume that $a_i \cdot b_i \neq 0$ for $i = 1, \dots, 6$. The distance from the point $P(x_0, y_0)$ to the line $a_i x + b_i y + c_i = 0$ is
$$d_i = \frac{|a_i x_0 + b_i y_0 + c_i|}{\sqrt{a_i^2 + b_i^2}}$$

$$d_i = \frac{a_i x_0 + b_i y_0 + c_i}{\sqrt{a_i^2 + b_i^2}} \quad (i = 1, \dots, 6).$$
We should find the set of the points $P(x_0, y_0)$ such that
$$\frac{1}{2}AB \cdot d_1 + \frac{1}{2}CD \cdot d_3 + \frac{1}{2}EF \cdot d_5 = \frac{1}{2}BC \cdot d_2 + \frac{1}{2}DE \cdot d_4 + \frac{1}{2}FA \cdot d_6.$$
Observe that this is a linear equation with respect to $x_0$ and $y_0$ and therefore the set of the points $P(x_0, y_0)$ is a segment. $\square$

Now we prove our problem. Since $A'$ is the midpoint of $BF$, $S_{A'XB} = S_{A'XF}$ and $S_{A'AB} = S_{A'AF}$. Hence $S_{AXB} = S_{AXF}$. Similarly, $S_{BXC} = S_{CXD}$ and $S_{DXE} = S_{EXF}$. Hence
$$S_{AXB} + S_{CXD} + S_{EXF} = S_{BXC} + S_{DXE} + S_{FXA}.$$
In a similar way we can show that
$$S_{AYB} + S_{CYD} + S_{EYF} = S_{BYC} + S_{DYE} + S_{FYA}.$$
On the other hand, let $Q_1, Q_2, \dots, Q_6$ be the tangent points of the incircle to the sides $AB, BC, CD, DE, EF, FA$ respectively. Since $Q_1I = Q_6I = r$, $AQ_1 = AQ_6$, $\angle IQ_1A = \angle IQ_6A$ we get that $\triangle AIQ_1 = \triangle AIQ_6$. Similarly $\triangle BIQ_1 = \triangle BIQ_2$, $\triangle CIQ_2 = \triangle CIQ_3$, $\triangle DIQ_3 = \triangle DIQ_4$, $\triangle EIQ_4 = \triangle EIQ_5$, $\triangle FIQ_5 = \triangle FIQ_6$. Hence
$$S_{AIB} + S_{CID} + S_{EIF} = S_{BIC} + S_{DIE} + S_{FIA}.$$

Therefore the points $X, Y, I$ are collinear, because of the lemma we proved above.