IMO Shortlist, 2019

Solution 1. Since
$$\angle APB + \angle BPC + \angle CPA = 2\pi = (\pi - \angle ACB) + (\pi - \angle BAC) + (\pi - \angle CBA),$$
at least one of the following inequalities holds:
$$\angle APB \geqslant \pi - \angle ACB, \quad \angle BPC \geqslant \pi - \angle BAC, \quad \angle CPA \geqslant \pi - \angle CBA.$$
Without loss of generality, we assume that $\angle BPC \geqslant \pi - \angle BAC$. We have $\angle BPC > \angle BAC$ because $P$ is inside $\triangle ABC$. So $\angle BPC \geqslant \max (\angle BAC, \pi - \angle BAC)$ and hence
$$\begin{equation*} \sin \angle BPC \leqslant \sin \angle BAC. \tag{*} \end{equation*}$$
Let the rays $AP, BP$, and $CP$ cross the circumcircle $\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\frac{PB_{1}}{B_{1}B_{3}}$ and $\frac{PC_{1}}{C_{1}C_{3}}$ is at least 1, which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\Omega$.
Because $A, B, C, B_{3}$ lie on a circle, the triangles $CB_{1}B_{3}$ and $BB_{1}A$ are similar, so
$$\frac{CB_{1}}{B_{1}B_{3}} = \frac{BB_{1}}{B_{1}A}.$$
Applying the sine rule we obtain
$$\frac{PB_{1}}{B_{1}B_{3}} = \frac{PB_{1}}{CB_{1}} \cdot \frac{CB_{1}}{B_{1}B_{3}} = \frac{PB_{1}}{CB_{1}} \cdot \frac{BB_{1}}{B_{1}A} = \frac{\sin \angle ACP}{\sin \angle BPC} \cdot \frac{\sin \angle BAC}{\sin \angle PBA}.$$
Similarly,
$$\frac{PC_{1}}{C_{1}C_{3}} = \frac{\sin \angle PBA}{\sin \angle BPC} \cdot \frac{\sin \angle BAC}{\sin \angle ACP}.$$
Multiplying these two equations we get
$$\frac{PB_{1}}{B_{1}B_{3}} \cdot \frac{PC_{1}}{C_{1}C_{3}} = \frac{\sin^{2} \angle BAC}{\sin^{2} \angle BPC} \geqslant 1$$
using (*), which yields the desired conclusion.

Solution 2. Define points $A_{3}, B_{3}$, and $C_{3}$ as in Solution 1. Assume for the sake of contradiction that $A_{2}, B_{2}$, and $C_{2}$ all lie strictly inside circle $ABC$. It follows that $PA_{1} < A_{1}A_{3}$, $PB_{1} < B_{1}B_{3}$, and $PC_{1} < C_{1}C_{3}$.
Observe that $\triangle PBC_{3} \sim \triangle PCB_{3}$. Let $X$ be the point on side $PB_{3}$ that corresponds to point $C_{1}$ on side $PC_{3}$ under this similarity. In other words, $X$ lies on segment $PB_{3}$ and satisfies $PX : XB_{3} = PC_{1} : C_{1}C_{3}$. It follows that
$$\angle XCP = \angle PBC_{1} = \angle B_{3}BA = \angle B_{3}CB_{1}$$
Hence lines $CX$ and $CB_{1}$ are isogonal conjugates in $\triangle PCB_{3}$.

Let $Y$ be the foot of the bisector of $\angle B_{3}CP$ in $\triangle PCB_{3}$. Since $PC_{1} < C_{1}C_{3}$, we have $PX < XB_{3}$. Also, we have $PY < YB_{3}$ because $PB_{1} < B_{1}B_{3}$ and $Y$ lies between $X$ and $B_{1}$. By the angle bisector theorem in $\triangle PCB_{3}$, we have $PY : YB_{3} = PC : CB_{3}$. So $PC < CB_{3}$ and it follows that $\angle PB_{3}C < \angle CPB_{3}$. Now since $\angle PB_{3}C = \angle BB_{3}C = \angle BAC$, we have
$$\angle BAC < \angle CPB_{3}.$$
Similarly, we have
$$\angle CBA < \angle APC_{3} \quad \text{and} \quad \angle ACB < \angle BPA_{3} = \angle B_{3}PA.$$
Adding these three inequalities yields $\pi < \pi$, and this contradiction concludes the proof.

Solution 3. Choose coordinates such that the circumcentre of $\triangle ABC$ is at the origin and the circumradius is 1. Then we may think of $A, B$, and $C$ as vectors in $\mathbb{R}^{2}$ such that
$$|A|^{2} = |B|^{2} = |C|^{2} = 1.$$
$P$ may be represented as a convex combination $\alpha A + \beta B + \gamma C$ where $\alpha, \beta, \gamma > 0$ and $\alpha + \beta + \gamma = 1$. Then
$$A_{1} = \frac{\beta B + \gamma C}{\beta + \gamma} = \frac{1}{1-\alpha} P - \frac{\alpha}{1-\alpha} A,$$
so
$$A_{2} = 2A_{1} - P = \frac{1+\alpha}{1-\alpha} P - \frac{2\alpha}{1-\alpha} A$$
Hence
$$|A_{2}|^{2} = \left(\frac{1+\alpha}{1-\alpha}\right)^{2} |P|^{2} + \left(\frac{2\alpha}{1-\alpha}\right)^{2} |A|^{2} - \frac{4\alpha(1+\alpha)}{(1-\alpha)^{2}} A \cdot P.$$
Using $|A|^{2} = 1$ we obtain
$$\begin{equation*} \frac{(1-\alpha)^{2}}{2(1+\alpha)} |A_{2}|^{2} = \frac{1+\alpha}{2} |P|^{2} + \frac{2\alpha^{2}}{1+\alpha} - 2\alpha A \cdot P. \tag{1} \end{equation*}$$
Likewise
$$\begin{equation*} \frac{(1-\beta)^{2}}{2(1+\beta)} |B_{2}|^{2} = \frac{1+\beta}{2} |P|^{2} + \frac{2\beta^{2}}{1+\beta} - 2\beta B \cdot P \tag{2} \end{equation*}$$
and
$$\begin{equation*} \frac{(1-\gamma)^{2}}{2(1+\gamma)} |C_{2}|^{2} = \frac{1+\gamma}{2} |P|^{2} + \frac{2\gamma^{2}}{1+\gamma} - 2\gamma C \cdot P. \tag{3} \end{equation*}$$
Summing (1), (2) and (3) we obtain on the LHS the positive linear combination
$$\mathrm{LHS} = \frac{(1-\alpha)^{2}}{2(1+\alpha)} |A_{2}|^{2} + \frac{(1-\beta)^{2}}{2(1+\beta)} |B_{2}|^{2} + \frac{(1-\gamma)^{2}}{2(1+\gamma)} |C_{2}|^{2}$$
and on the RHS the quantity
$$\left(\frac{1+\alpha}{2} + \frac{1+\beta}{2} + \frac{1+\gamma}{2}\right) |P|^{2} + \left(\frac{2\alpha^{2}}{1+\alpha} + \frac{2\beta^{2}}{1+\beta} + \frac{2\gamma^{2}}{1+\gamma}\right) - 2(\alpha A \cdot P + \beta B \cdot P + \gamma C \cdot P).$$
The first term is $2|P|^{2}$ and the last term is $-2P \cdot P$, so
$$\begin{aligned} \mathrm{RHS} & = \left(\frac{2\alpha^{2}}{1+\alpha} + \frac{2\beta^{2}}{1+\beta} + \frac{2\gamma^{2}}{1+\gamma}\right) \\ & = \frac{3\alpha-1}{2} + \frac{(1-\alpha)^{2}}{2(1+\alpha)} + \frac{3\beta-1}{2} + \frac{(1-\beta)^{2}}{2(1+\beta)} + \frac{3\gamma-1}{2} + \frac{(1-\gamma)^{2}}{2(1+\gamma)} \\ & = \frac{(1-\alpha)^{2}}{2(1+\alpha)} + \frac{(1-\beta)^{2}}{2(1+\beta)} + \frac{(1-\gamma)^{2}}{2(1+\gamma)} \end{aligned}$$
Here we used the fact that
$$\frac{3\alpha-1}{2} + \frac{3\beta-1}{2} + \frac{3\gamma-1}{2} = 0.$$
We have shown that a linear combination of $|A_{2}|^{2}, |B_{2}|^{2}$, and $|C_{2}|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $|A_{2}|^{2}, |B_{2}|^{2}$, and $|C_{2}|^{2}$ must be at least 1, as required.