Subfield: Geometry


Question: Let $A_{1}, B_{1}$ and $C_{1}$ be points on sides $B C, C A$ and $A B$ of an acute triangle $A B C$ respectively, such that $A A_{1}, B B_{1}$ and $C C_{1}$ are the internal angle bisectors of triangle $A B C$. Let $I$ be the incentre of triangle $A B C$, and $H$ be the orthocentre of triangle $A_{1} B_{1} C_{1}$. Show that

$$
A H+B H+C H \geqslant A I+B I+C I \text {. }
$$

Solution:
Without loss of generality, assume $\alpha=\angle B A C \leqslant \beta=\angle C B A \leqslant \gamma=\angle A C B$. Denote by $a, b, c$ the lengths of $B C, C A, A B$ respectively. We first show that triangle $A_{1} B_{1} C_{1}$ is acute.

Choose points $D$ and $E$ on side $B C$ such that $B_{1} D / / A B$ and $B_{1} E$ is the internal angle bisector of $\angle B B_{1} C$. As $\angle B_{1} D B=180^{\circ}-\beta$ is obtuse, we have $B B_{1}>B_{1} D$. Thus,

$$
\frac{B E}{E C}=\frac{B B_{1}}{B_{1} C}>\frac{D B_{1}}{B_{1} C}=\frac{B A}{A C}=\frac{B A_{1}}{A_{1} C}
$$

Therefore, $B E>B A_{1}$ 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

$$
\angle A_{1} B_{1} C_{1}=\angle B B_{1} A_{1}+\angle B B_{1} C_{1}<\frac{1}{2}\left(\angle B B_{1} C+\angle B B_{1} A\right)=90^{\circ}
$$

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 $\alpha \leqslant \gamma$, we get $a \leqslant c$, which implies

$$
B A_{1}=\frac{c a}{b+c} \leqslant \frac{a c}{a+b}=B C_{1}
$$

and hence $\angle B C_{1} A_{1} \leqslant \angle B A_{1} C_{1}$. As $B F$ is the internal angle bisector of $\angle A_{1} B C_{1}$, this shows $\angle B_{1} F C_{1}=\angle B F A_{1} \leqslant 90^{\circ}$. 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 $\alpha \leqslant \beta$ and $\beta \leqslant \gamma$, we know that $H$ lies inside triangles $C C_{1} B_{1}$ and $A A_{1} C_{1}$.

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As $\alpha \leqslant \beta \leqslant \gamma$, we have $\alpha \leqslant 60^{\circ} \leqslant \gamma$. Then $\angle B I C \leqslant 120^{\circ} \leqslant \angle A I B$. Firstly, suppose $\angle A I C \geqslant 120^{\circ}$.

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 $A B$. Since triangle $A I^{\prime} I$ is equilateral, we have

$$
A I+B I+C I=I^{\prime} I+B^{\prime} I^{\prime}+I C=B^{\prime} I^{\prime}+I^{\prime} I+I C .
\tag{1}
$$

Similarly,

$$
A H+B H+C H=H^{\prime} H+B^{\prime} H^{\prime}+H C=B^{\prime} H^{\prime}+H^{\prime} H+H C \text {. }\tag{2}
$$

As $\angle A I I^{\prime}=\angle A I^{\prime} I=60^{\circ}, \angle A I^{\prime} B^{\prime}=\angle A I B \geqslant 120^{\circ}$ and $\angle A I C \geqslant 120^{\circ}$, the quadrilateral $B^{\prime} I^{\prime} I C$ is convex and lies on the same side of $B^{\prime} C$ as $A$.

Next, since $H$ lies inside triangle $A C C_{1}, H$ lies outside $B^{\prime} I^{\prime} I C$. Also, $H$ lying inside triangle $A B I$ implies $H^{\prime}$ lies inside triangle $A B^{\prime} I^{\prime}$. This shows $H^{\prime}$ lies outside $B^{\prime} I^{\prime} I C$ and hence the convex quadrilateral $B^{\prime} I^{\prime} I C$ is contained inside the quadrilateral $B^{\prime} H^{\prime} H C$. It follows that the perimeter of $B^{\prime} I^{\prime} I C$ cannot exceed the perimeter of $B^{\prime} H^{\prime} H C$. From (1) and (2), we conclude that

$$
A H+B H+C H \geqslant A I+B I+C I \text {. }
$$

For the case $\angle A I C<120^{\circ}$, 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 $B C$. The proof is analogous to the previous case and we still get the desired inequality.