Problem 1982

Since $\angle AFI = \angle AEI = 90^\circ$, the point $I$ lies on $\Omega$ and the centre of $\Omega$ is the midpoint $O$ of $AI$. Note that $P$ lies on the line joining the centres of $\Gamma$ and $\Omega$. In other words, $P$ lies on the line $IO$. As $AB = AC$, the line $AI$ is simply the internal angle bisector of $\angle BAC$. In order to show that $BP$ is perpendicular to the angle bisector $BI$ of $\angle CBA$, it is the same as proving $P$ is the excentre of $\triangle ABC$ opposite to the vertex $A$.

Let $T$ be the point on $\Gamma$ such that $PT$ is tangent to $\Gamma$ and $TP//AC$. Since the tangents at $T$ and $E$ to $\Gamma$ are parallel, $TE$ is a diameter of $\Gamma$. This shows $I$ is the midpoint of $TE$. It follows from $AE//PT$ that $\triangle AIE \cong \triangle PIT$. Thus, $AI = PI$. Together with $AO = OI$, we find that $AO : OI : IP = 1 : 1 : 2$.

Note that $P$ is the centre of homothety between $\Gamma$ and $\Omega$. Therefore, $\frac{PI}{PO}$ is the ratio of similitude, i.e. the ratio $\frac{2IE}{AI}$ of the diameters of the two circles. Thus, we have $\frac{2IE}{AI} = \frac{2}{3}$.

Let $D$ be the midpoint of $BC$. As $AB = AC$, $D$ is also the contact point of $\Gamma$ with $BC$. From $\angle AEI = \angle ADC = 90^\circ$, we obtain $\triangle AEI \sim \triangle ADC$. Therefore, $\frac{CD}{AC} = \frac{IE}{AI} = \frac{1}{3}$. Let $AB = AC = 3k$ so that $BD = CD = k$.

Let $P'$ be the $A$-excentre of $\triangle ABC$. Our goal is to show that $AI = IP'$, so that $P' = P$ as desired. Firstly, since $AD = \sqrt{(3k)^2 - k^2} = 2\sqrt{2}k$ and $\frac{AI}{ID} = \frac{AB}{BD} = 3$ by the angle bisector theorem, we have
$$AI = 2\sqrt{2}k \times \frac{3}{3+1} = \frac{3\sqrt{2}}{2}k.$$
Secondly, we have $\frac{AP'}{P'D} = \frac{AB}{BD} = 3$ by the external angle bisector theorem. It follows that
$$AP' = 2\sqrt{2}k \times \frac{3}{3-1} = 3\sqrt{2}k = 2AI.$$
Therefore, $AI = IP'$, and we are done.