Library / /6 of 10

, 2024

Geometry Difficulty 6.8 National olympiad Prove it China

Let $I$ be the incenter of a triangle $ABC$ . Write $L$ , $M$ , and $N$ for the midpoints of $AI$ , $AC$ , and $CI$ , respectively. Assume that there is a point $D$ in the interior of the segment $AM$ such that $BD = BC$ . The incircle of $\triangle ABD$ touches $AD$ and $BD$ at $E$ and $F$ , respectively. Let $J$ be the circumcenter of $\triangle AIC$ . Let $\omega$ denote the circumcircle of $\triangle DMJ$ . The line $MN$ intersects $\omega$ at point $P(\neq M)$ , and the line $JL$ intersects $\omega$ at point $Q(\neq J)$ .
Prove that the three lines $EF$ , $PQ$ , and $LN$ pass through a common point.

Solution

Proof. Let $K$ be the midpoint of $ID$ , then $LN$ passes through $K$ . Next, we prove that both $EF$ and $PQ$ pass through $K$ .
Let $X$ and $Y$ be the reflections of $D$ across $E$ and $F$ , respectively. Let $Z$ be the projection of $I$ onto $AC$ . Then, we have
$CX = 2DE + CD = AD + BD - AB + CD = AC + BC - AB = 2CZ.$
Thus, $CZ = ZX$ . Combining this with $IZ \perp CX$ , we know $IC = IX$ . Therefore,
$2\angle IXC = 2\angle ICX = \angle ACB = \angle BDC = 2\angle FEC = 2\angle YXC.$
Thus, $I$ , $Y$ , and $X$ are collinear, which implies that $EF$ passes through $K$ .
Since $MN // AI$ , we have $\angle PQD = \angle PMD = \angle IAD$ .
From $\angle DMJ = 90^\circ$ , we know that $DJ$ is the diameter of $\omega$ , so $DQ \perp QJ$ . Combined with $AL \perp LJ$ , we conclude that $AI // DQ$ . Let $T$ be the reflection of $D$ across $Q$ . Then $ADTI$ forms an isosceles trapezoid. Since $KQ // IT$ , we have
$\angle DQK = \angle DTI = 180^\circ - \angle IAD = 180^\circ - \angle PQD,$
which implies that $PQ$ passes through $K$ . Thus, the proof is complete.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.