Library / Baltic Way Shortlist /62 of 92

Baltic Way Shortlist, 2021

Geometry Difficulty 8.6 Shortlist Prove it Baltic Way

Let $I$ be the incenter of a triangle $ABC$ . Let the incircle of $ABC$ be tangent to $CA$ and $AB$ at $E$ and $F$ , respectively. Lines $BI$ and $CI$ intersect line $EF$ at $Y$ and $Z$ , respectively. Denote by $M$ , $N$ midpoints of segments $BC$ and $YZ$ , respectively. Prove that $MN$ is parallel to $AI$ .

Figure 22

Solution

Refer to figure 22. Let us start with proving a known simple lemma stating that $\angle BYC = 90^\circ$ . To that end, as $\angle IEC = 90^\circ$ , it is enough to prove that $IEYC$ is cyclic. Indeed:
$\begin{align*} \angle CEY &= \angle AEF \\ &= 90^\circ - \frac{1}{2} \angle BAC \\ &= 90^\circ - \frac{1}{2} (180^\circ - \angle ABC - \angle ACB) \\ &= \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB \\ &= \angle IBC + \angle ICB \\ &= \angle YIC. \end{align*}$
This shows that $YIEC$ is cyclic. Since $AC$ is tangent to the incircle at $E$ ,
$\angle BIC = \angle CYI = \angle CEI = 90^\circ$
Similarly, we may prove that $\angle BZC = 90^\circ$ .
All this implies that $BYZC$ is cyclic with $M$ being its centre. Hence $N$ is the midpoint of its chord $YZ$ and therefore $MN \perp EF$ . However, $AI \perp EF$ , so $MN \parallel AI$ as desired. $\square$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.