IMO 2019 Shortlisted Problems

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: $$\angle APB⩾\pi -\angle ACB,\angle BPC⩾\pi -\angle BAC,\angle CPA⩾\pi -\angle CBA.$$ Without loss of generality, we assume that $\angle BPC⩾\pi -\angle BAC$. We have $\angle BPC>\angle BAC$ because $P$ is inside $△ABC$. So $\angle BPC⩾\max (\angle BAC,\pi -\angle BAC)$ and hence $$\begin{array}{c}\sin \angle BPC⩽\sin \angle BAC.\end{array}\text{\hspace{1em}(*)}$$ 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 $\frac{P{B}_1}{B_1{B}_3}$ and $\frac{P{C}_1}{C_1{C}_3}$ 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 $C{B}_1{B}_3$ and $B{B}_{1}A$ are similar, so $$\frac{C{B}_1}{B_1{B}_3}=\frac{B{B}_1}{B_{1}A}.$$ Applying the sine rule we obtain $$\frac{P{B}_1}{B_1{B}_3}=\frac{P{B}_1}{C{B}_1}\cdot \frac{C{B}_1}{B_1{B}_3}=\frac{P{B}_1}{C{B}_1}\cdot \frac{B{B}_1}{B_{1}A}=\frac{\sin \angle ACP}{\sin \angle BPC}\cdot \frac{\sin \angle BAC}{\sin \angle PBA}.$$ Similarly, $$\frac{P{C}_1}{C_1{C}_3}=\frac{\sin \angle PBA}{\sin \angle BPC}\cdot \frac{\sin \angle BAC}{\sin \angle ACP}.$$ Multiplying these two equations we get $$\frac{P{B}_1}{B_1{B}_3}\cdot \frac{P{C}_1}{C_1{C}_3}=\frac{{\sin }^2\angle BAC}{{\sin }^2\angle BPC}⩾1$$ 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 $P{A}_1<A_1{A}_3$, $P{B}_1<B_1{B}_3$, and $P{C}_1<C_1{C}_3$. Observe that $△PB{C}_3\sim △PC{B}_3$. Let $X$ be the point on side $P{B}_3$ that corresponds to point $C_1$ on side $P{C}_3$ under this similarity. In other words, $X$ lies on segment $P{B}_3$ and satisfies $PX:X{B}_3=P{C}_1:C_1{C}_3$. It follows that $$\angle XCP=\angle PB{C}_1=\angle B_{3}BA=\angle B_{3}C{B}_1$$ Hence lines $CX$ and $C{B}_1$ are isogonal conjugates in $△PC{B}_3$. Let $Y$ be the foot of the bisector of $\angle B_{3}CP$ in $△PC{B}_3$. Since $P{C}_1<C_1{C}_3$, we have $PX<X{B}_3$. Also, we have $PY<Y{B}_3$ because $P{B}_1<B_1{B}_3$ and $Y$ lies between $X$ and $B_1$. By the angle bisector theorem in $△PC{B}_3$, we have $PY:Y{B}_3=PC:C{B}_3$. So $PC<C{B}_3$ and it follows that $\angle P{B}_{3}C<\angle CP{B}_3$. Now since $\angle P{B}_{3}C=\angle B{B}_{3}C=\angle BAC$, we have $$\angle BAC<\angle CP{B}_3.$$ Similarly, we have $$\angle CBA<\angle AP{C}_3\text{and}\angle ACB<\angle BP{A}_3=\angle B_{3}PA.$$ Adding these three inequalities yields $\pi <\pi$, and this contradiction concludes the proof.

Solution 3. Choose coordinates such that the circumcentre of $△ABC$ 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 $$\begin{array}{c}\frac{(1-\alpha {)}^2}{2(1+\alpha )}|A_2{|}^2=\frac{1+\alpha }{2}|P{|}^2+\frac{2{\alpha }^2}{1+\alpha }-2\alpha A\cdot P.\end{array}\text{\hspace{1em}(1)}$$ Likewise $$\begin{array}{c}\frac{(1-\beta {)}^2}{2(1+\beta )}|B_2{|}^2=\frac{1+\beta }{2}|P{|}^2+\frac{2{\beta }^2}{1+\beta }-2\beta B\cdot P\end{array}\text{\hspace{1em}(2)}$$ and $$\begin{array}{c}\frac{(1-\gamma {)}^2}{2(1+\gamma )}|C_2{|}^2=\frac{1+\gamma }{2}|P{|}^2+\frac{2{\gamma }^2}{1+\gamma }-2\gamma C\cdot P.\end{array}\text{\hspace{1em}(3)}$$ 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 $$\begin{array}{rl}\mathrm{RHS}& =\left(\frac{2{\alpha }^2}{1+\alpha }+\frac{2{\beta }^2}{1+\beta }+\frac{2{\gamma }^2}{1+\gamma }\right)\\ & =\frac{3\alpha -1}{2}+\frac{(1-\alpha {)}^2}{2(1+\alpha )}+\frac{3\beta -1}{2}+\frac{(1-\beta {)}^2}{2(1+\beta )}+\frac{3\gamma -1}{2}+\frac{(1-\gamma {)}^2}{2(1+\gamma )}\\ & =\frac{(1-\alpha {)}^2}{2(1+\alpha )}+\frac{(1-\beta {)}^2}{2(1+\beta )}+\frac{(1-\gamma {)}^2}{2(1+\gamma )}\end{array}$$ 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.