51st IMO Shortlisted Problems

Denote $P_{n+1}=P_1,X_{n+1}=X_1,A_{n+1}=A_1$.

Lemma. Let point $Q$ lies inside $A_1{A}_2\dots A_n$. Then it is contained in at least one of the circumcircles of triangles $X_1{A}_2{X}_2,\dots ,X_n{A}_1{X}_1$.

Proof. If $Q$ lies in one of the triangles $X_1{A}_2{X}_2,\dots ,X_n{A}_1{X}_1$, the claim is obvious. Otherwise $Q$ lies inside the polygon $X_1{X}_2\dots X_n$ (see Fig. 1). Then we have $$\begin{array}{rl}& \left(\angle X_1{A}_2{X}_2+\angle X_{1}Q{X}_2\right)+\cdots +\left(\angle X_n{A}_1{X}_1+\angle X_{n}Q{X}_1\right)\\ & =\left(\angle X_1{A}_1{X}_2+\cdots +\angle X_n{A}_1{X}_1\right)+\left(\angle X_{1}Q{X}_2+\cdots +\angle X_{n}Q{X}_1\right)=(n-2)\pi +2\pi =n\pi \end{array}$$ hence there exists an index $i$ such that $\angle X_i{A}_{i+1}{X}_{i+1}+\angle X_{i}Q{X}_{i+1}\ge \frac{\pi n}{n}=\pi$. Since the quadrilateral $Q{X}_i{A}_{i+1}{X}_{i+1}$ is convex, this means exactly that $Q$ is contained the circumcircle of $△X_i{A}_{i+1}{X}_{i+1}$, as desired. $\square$

Now we turn to the solution. Applying lemma, we get that $P$ lies inside the circumcircle of triangle $X_i{A}_{i+1}{X}_{i+1}$ for some $i$. Consider the circumcircles $\omega$ and $\Omega$ of triangles $P_i{A}_{i+1}{P}_{i+1}$ and $X_i{A}_{i+1}{X}_{i+1}$ respectively (see Fig. 2); let $r$ and $R$ be their radii. Then we get $2r=A_{i+1}P\le 2R$ (since $P$ lies inside $\Omega$ ), hence $$P_i{P}_{i+1}=2r\sin \angle P_i{A}_{i+1}{P}_{i+1}\le 2R\sin \angle X_i{A}_{i+1}{X}_{i+1}=X_i{X}_{i+1},$$ QED.

Fig. 1 Fig. 2

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Alternative solution.

As in Solution 1, we assume that all indices of points are considered modulo $n$.

We will prove a bit stronger inequality, namely $$\max \left\{\frac{X_1{X}_2}{P_1{P}_2}\cos {\alpha }_1,\dots ,\frac{X_n{X}_1}{P_n{P}_1}\cos {\alpha }_n\right\}\ge 1,$$ where ${\alpha }_i(1\le i\le n)$ is the angle between lines $X_i{X}_{i+1}$ and $P_i{P}_{i+1}$. We denote ${\beta }_i=\angle A_i{P}_i{P}_{i-1}$ and ${\gamma }_i=\angle A_{i+1}{P}_i{P}_{i+1}$ for all $1\le i\le n$.

Suppose that for some $1\le i\le n$, point $X_i$ lies on the segment $A_i{P}_i$, while point $X_{i+1}$ lies on the segment $P_{i+1}{A}_{i+2}$. Then the projection of the segment $X_i{X}_{i+1}$ onto the line $P_i{P}_{i+1}$ contains segment $P_i{P}_{i+1}$, since ${\gamma }_i$ and ${\beta }_{i+1}$ are acute angles (see Fig. 3). Therefore, $X_i{X}_{i+1}\cos {\alpha }_i\ge P_i{P}_{i+1}$, and in this case the statement is proved.

So, the only case left is when point $X_i$ lies on segment $P_i{A}_{i+1}$ for all $1\le i\le n$ (the case when each $X_i$ lies on segment $A_i{P}_i$ is completely analogous).

Now, assume to the contrary that the inequality $$\begin{array}{c}{X}_i{X}_{i+1}\cos {\alpha }_i<P_i{P}_{i+1}\end{array}\text{\hspace{1em}(1)}$$ holds for every $1\le i\le n$. Let $Y_i$ and $Y_{i+1}^{\prime }$ be the projections of $X_i$ and $X_{i+1}$ onto $P_i{P}_{i+1}$. Then inequality (1) means exactly that $Y_i{Y}_{i+1}^{\prime }<P_i{P}_{i+1}$, or $P_i{Y}_i>P_{i+1}{Y}_{i+1}^{\prime }$ (again since ${\gamma }_i$ and ${\beta }_{i+1}$ are acute; see Fig. 4). Hence, we have $$X_i{P}_i\cos {\gamma }_i>X_{i+1}{P}_{i+1}\cos {\beta }_{i+1},1\le i\le n.$$ Multiplying these inequalities, we get $$\begin{array}{c}\cos {\gamma }_1\cos {\gamma }_2\cdots \cos {\gamma }_n>\cos {\beta }_1\cos {\beta }_2\cdots \cos {\beta }_n.\end{array}\text{\hspace{1em}(2)}$$ On the other hand, the sines theorem applied to triangle $P{P}_i{P}_{i+1}$ provides $$\frac{P{P}_i}{P{P}_{i+1}}=\frac{\sin \left(\frac{\pi }{2}-{\beta }_{i+1}\right)}{\sin \left(\frac{\pi }{2}-{\gamma }_i\right)}=\frac{\cos {\beta }_{i+1}}{\cos {\gamma }_i}.$$ Multiplying these equalities we get $$1=\frac{\cos {\beta }_2}{\cos {\gamma }_1}\cdot \frac{\cos {\beta }_3}{\cos {\gamma }_2}\cdots \frac{\cos {\beta }_1}{\cos {\gamma }_n}$$ which contradicts (2).

Fig. 3 Fig. 4