, 2007

Let $A_{0}, B_{0}, C_{0}$ be the points of intersection of the lines $AA', BB'$ and $CC'$ (see Figure). We claim that area $\left(A_{0}B_{0}C_{0}\right)=\frac{1}{2} \operatorname{area}(ABC)$, hence it is constant.

Consider the inscribed hexagon $ABCC'PA'$. By Pascal's theorem, the points of intersection of its opposite sides (or of their extensions) are collinear. These points are $AB \cap C'P = C_{1}$, $BC \cap PA' = A_{1}$, $CC' \cap A'A = B_{0}$. So point $B_{0}$ lies on the midline $A_{1}C_{1}$ of triangle $ABC$. Analogously, points $A_{0}$ and $C_{0}$ lie on lines $B_{1}C_{1}$ and $A_{1}B_{1}$, respectively.

Lines $AC$ and $A_{1}C_{1}$ are parallel, so triangles $B_{0}C_{0}A_{1}$ and $AC_{0}B_{1}$ are similar; hence we have
$$\frac{B_{0}C_{0}}{AC_{0}} = \frac{A_{1}C_{0}}{B_{1}C_{0}}.$$
Analogously, from $BC \parallel B_{1}C_{1}$ we obtain
$$\frac{A_{1}C_{0}}{B_{1}C_{0}} = \frac{BC_{0}}{A_{0}C_{0}}.$$
Combining these equalities, we get
$$\frac{B_{0}C_{0}}{AC_{0}} = \frac{BC_{0}}{A_{0}C_{0}},$$
or
$$A_{0}C_{0} \cdot B_{0}C_{0} = AC_{0} \cdot BC_{0}.$$
Hence we have

$$\operatorname{area}\left(A_{0}B_{0}C_{0}\right) = \frac{1}{2} A_{0}C_{0} \cdot B_{0}C_{0} \sin \angle A_{0}C_{0}B_{0} = \frac{1}{2} AC_{0} \cdot BC_{0} \sin \angle AC_{0}B = \operatorname{area}\left(ABC_{0}\right).$$
Since $C_{0}$ lies on the midline, we have $d\left(C_{0}, AB\right) = \frac{1}{2} d(C, AB)$ (we denote by $d(X, YZ)$ the distance between point $X$ and line $YZ$). Then we obtain
$$\operatorname{area}\left(A_{0}B_{0}C_{0}\right) = \operatorname{area}\left(ABC_{0}\right) = \frac{1}{2} AB \cdot d\left(C_{0}, AB\right) = \frac{1}{4} AB \cdot d(C, AB) = \frac{1}{2} \operatorname{area}(ABC).$$

Again, we prove that area $\left(A_{0}B_{0}C_{0}\right) = \frac{1}{2} \operatorname{area}(ABC)$.

We can assume that $P$ lies on arc $AC$. Mark a point $L$ on side $AC$ such that $\angle CBL = \angle PBA$; then $\angle LBA = \angle CBA - \angle CBL = \angle CBA - \angle PBA = \angle CBP$. Note also that $\angle BAL = \angle BAC = \angle BPC$ and $\angle LCB = \angle APB$. Hence, triangles $BAL$ and $BPC$ are similar, and so are triangles $LCB$ and $APB$.

Analogously, mark points $K$ and $M$ respectively on the extensions of sides $CB$ and $AB$ beyond point $B$, such that $\angle KAB = \angle CAP$ and $\angle BCM = \angle PCA$. For analogous reasons, $\angle KAC = \angle BAP$ and $\angle ACM = \angle PCB$. Hence $\triangle ABK \sim \triangle APC \sim \triangle MBC$, $\triangle ACK \sim \triangle APB$, and $\triangle MAC \sim \triangle BPC$. From these similarities, we have $\angle CMB = \angle KAB = \angle CAP$, while we have seen that $\angle CAP = \angle CBP = \angle LBA$. Hence, $AK \parallel BL \parallel CM$.

Let line $CC'$ intersect $BL$ at point $X$. Note that $\angle LCX = \angle ACC' = \angle APC' = \angle APC_{1}$, and $PC_{1}$ is a median in triangle $APB$. Since triangles $APB$ and $LCB$ are similar, $CX$ is a median in triangle $LCB$, and $X$ is a midpoint of $BL$. For the same reason, $AA'$ passes through this midpoint, so $X = B_{0}$. Analogously, $A_{0}$ and $C_{0}$ are the midpoints of $AK$ and $CM$.

Now, from $AA_{0} \parallel CC_{0}$, we have
$$\operatorname{area}\left(A_{0}B_{0}C_{0}\right) = \operatorname{area}\left(AC_{0}A_{0}\right) - \operatorname{area}\left(AB_{0}A_{0}\right) = \operatorname{area}\left(ACA_{0}\right) - \operatorname{area}\left(AB_{0}A_{0}\right) = \operatorname{area}\left(ACB_{0}\right).$$
Finally,
$$\operatorname{area}\left(A_{0}B_{0}C_{0}\right) = \operatorname{area}\left(ACB_{0}\right) = \frac{1}{2} B_{0}L \cdot AC \sin ALB_{0} = \frac{1}{4} BL \cdot AC \sin ALB = \frac{1}{2} \operatorname{area}(ABC).$$