IMO2024 Shortlisted Problems

We have $\angle APB=\angle ACB$ in the circumcircle and $\angle ACB=\angle A^{\prime }XC$ because $A^{\prime }X\parallel AC$. Hence, $\angle APB=\angle A^{\prime }XC$, and so quadrilateral $BP{A}^{\prime }X$ is cyclic. Similarly, it follows that $CY{A}^{\prime }P$ is cyclic. Now we are ready to transform $\angle KIL+\angle YPX$ to the sum of angles in triangle $A^{\prime }CB$. By a homothety of factor 2 at $A$ we have $\angle KIL=\angle C{A}^{\prime }B$. In circles $BP{A}^{\prime }X$ and $CY{A}^{\prime }P$ we have $\angle APX=\angle A^{\prime }BC$ and $\angle YPA=\angle BC{A}^{\prime }$, therefore $$\angle KIL+\angle YPX=\angle C{A}^{\prime }B+(\angle YPA+\angle APX)=\angle C{A}^{\prime }B+\angle BC{A}^{\prime }+\angle A^{\prime }BC=180^{\circ }.$$

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Alternative solution.

Let $BC=a$, $AC=b$, $AB=c$ and $s=\frac{a+b+c}{2}$, and let the radii of the incircle, $B$-excircle and $C$-excircle be $r,r_b$ and $r_c$, respectively. Let the incircle be tangent to $AC$ and $AB$ at $B_0$ and $C_0$, respectively; let the $B$-excircle be tangent to $AC$ at $B_1$, and let the $C$-excircle be tangent to $AB$ at $C_1$. As is well-known, $A{B}_1=s-c$ and $\text{area}(△ABC)=rs=r_c(s-c)$. Let the line through $X$, parallel to $AC$ be tangent to the incircle at $E$, and the line through $Y$, parallel to $AB$ be tangent to the incircle at $D$. Finally, let $AP$ meet $B{B}_1$ at $F$. It is well-known that points $B,E$, and $B_1$ are collinear by the homothety between the incircle and the $B$-excircle, and $BE\parallel IK$ because $IK$ is a midline in triangle $B_{0}E{B}_1$. Similarly, it follows that $C,D$, and $C_1$ are collinear and $CD\parallel IL$. Hence, the problem reduces to proving $\angle YPA=\angle CBE$ (and its symmetric counterpart $\angle APX=\angle DCB$ with respect to the vertex $C$), so it suffices to prove that $FYPB$ is cyclic. Since $ACPB$ is cyclic, that is equivalent to $FY\parallel B_{1}C$ and $\frac{BF}{F{B}_1}=\frac{BY}{YC}$. By the angle bisector theorem we have $$\frac{BF}{F{B}_1}=\frac{AB}{A{B}_1}=\frac{c}{s-c}.$$ The homothety at $C$ that maps the incircle to the $C$-excircle sends $Y$ to $B$, so $$\frac{BC}{YC}=\frac{r_c}{r}=\frac{s}{s-c}.$$ So, $$\frac{BY}{YC}=\frac{BC}{YC}-1=\frac{s}{s-c}-1=\frac{c}{s-c}=\frac{BF}{F{B}_1},$$ which completes the solution.