IMO 2016 Shortlisted Problems

Without loss of generality, assume $\alpha =\angle BAC⩽\beta =\angle CBA⩽\gamma =\angle ACB$. Denote by $a$, $b$, $c$ the lengths of $BC$, $CA$, $AB$ respectively. We first show that triangle $A_1{B}_1{C}_1$ is acute. Choose points $D$ and $E$ on side $BC$ such that $B_{1}D\parallel AB$ and $B_{1}E$ is the internal angle bisector of $\angle B{B}_{1}C$. As $\angle B_{1}DB=180^{\circ }-\beta$ is obtuse, we have $B{B}_1>B_{1}D$. Thus, $$\frac{BE}{EC}=\frac{B{B}_1}{B_{1}C}>\frac{D{B}_1}{B_{1}C}=\frac{BA}{AC}=\frac{B{A}_1}{A_{1}C}$$ Therefore, $BE>B{A}_1$ and $\frac{1}{2}\angle B{B}_{1}C=\angle B{B}_{1}E>\angle B{B}_1{A}_1$. Similarly, $\frac{1}{2}\angle B{B}_{1}A>\angle B{B}_1{C}_1$. It follows that $$\angle A_1{B}_1{C}_1=\angle B{B}_1{A}_1+\angle B{B}_1{C}_1<\frac{1}{2}(\angle B{B}_{1}C+\angle B{B}_{1}A)=90^{\circ }$$ is acute. By symmetry, triangle $A_1{B}_1{C}_1$ is acute. Let $B{B}_1$ meet $A_1{C}_1$ at $F$. From $\alpha ⩽\gamma$, we get $a⩽c$, which implies $$B{A}_1=\frac{ca}{b+c}⩽\frac{ac}{a+b}=B{C}_1$$ and hence $\angle B{C}_1{A}_1⩽\angle B{A}_1{C}_1$. As $BF$ is the internal angle bisector of $\angle A_{1}B{C}_1$, this shows $\angle B_{1}F{C}_1=\angle BF{A}_1⩽90^{\circ }$. Hence, $H$ lies on the same side of $B{B}_1$ as $C_1$. This shows $H$ lies inside triangle $B{B}_1{C}_1$. Similarly, from $\alpha ⩽\beta$ and $\beta ⩽\gamma$, we know that $H$ lies inside triangles $C{C}_1{B}_1$ and $A{A}_1{C}_1$. As $\alpha ⩽\beta ⩽\gamma$, we have $\alpha ⩽60^{\circ }⩽\gamma$. Then $\angle BIC⩽120^{\circ }⩽\angle AIB$. Firstly, suppose $\angle AIC⩾120^{\circ }$. Rotate points $B$, $I$, $H$ through $60^{\circ }$ about $A$ to $B^{\prime }$, $I^{\prime }$, $H^{\prime }$ so that $B^{\prime }$ and $C$ lie on different sides of $AB$. Since triangle $A{I}^{\prime }I$ is equilateral, we have $$\begin{array}{c}AI+BI+CI=I^{\prime }I+B^{\prime }{I}^{\prime }+IC=B^{\prime }{I}^{\prime }+I^{\prime }I+IC\end{array}\text{\hspace{1em}(1)}$$ Similarly, $$\begin{array}{c}AH+BH+CH=H^{\prime }H+B^{\prime }{H}^{\prime }+HC=B^{\prime }{H}^{\prime }+H^{\prime }H+HC\end{array}\text{\hspace{1em}(2)}$$ As $\angle AI{I}^{\prime }=\angle A{I}^{\prime }I=60^{\circ }$, $\angle A{I}^{\prime }{B}^{\prime }=\angle AIB⩾120^{\circ }$ and $\angle AIC⩾120^{\circ }$, the quadrilateral $B^{\prime }{I}^{\prime }IC$ is convex and lies on the same side of $B^{\prime }C$ as $A$. Next, since $H$ lies inside triangle $AC{C}_1$, $H$ lies outside $B^{\prime }{I}^{\prime }IC$. Also, $H$ lying inside triangle $ABI$ implies $H^{\prime }$ lies inside triangle $A{B}^{\prime }{I}^{\prime }$. This shows $H^{\prime }$ lies outside $B^{\prime }{I}^{\prime }IC$ and hence the convex quadrilateral $B^{\prime }{I}^{\prime }IC$ is contained inside the quadrilateral $B^{\prime }{H}^{\prime }HC$. It follows that the perimeter of $B^{\prime }{I}^{\prime }IC$ cannot exceed the perimeter of $B^{\prime }{H}^{\prime }HC$. From (1) and (2), we conclude that $$AH+BH+CH⩾AI+BI+CI.$$ For the case $\angle AIC<120^{\circ }$, we can rotate $B$, $I$, $H$ through $60^{\circ }$ about $C$ to $B^{\prime }$, $I^{\prime }$, $H^{\prime }$ so that $B^{\prime }$ and $A$ lie on different sides of $BC$. The proof is analogous to the previous case and we still get the desired inequality.