Track / Stage 7 / 94 of 300 #1494 of 2000

Problem 1494

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.2 Prove it

On the plane, given an angle $xOy$ . $M$ be a mobile point on ray $Ox$ and $N$ a mobile point on ray $Oy$ . Let $d$ be the external angle bisector of angle $xOy$ and $I$ be the intersection of $d$ with the perpendicular bisector of $MN$ . Let $P$ , $Q$ be two points lie on $d$ such that $IP \equal{} IQ \equal{} IM \equal{} IN$ , and let $K$ the intersection of $MQ$ and $NP$ .

$1.$ Prove that $K$ always lie on a fixed line.

$2.$ Let $d_1$ line perpendicular to $IM$ at $M$ and $d_2$ line perpendicular to $IN$ at $N$ . Assume that there exist the intersections $E$ , $F$ of $d_1$ , $d_2$ from $d$ . Prove that $EN$ , $FM$ and $OK$ are concurrent.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

### Part 1: Prove that $K$ always lies on a fixed line.

1. Reflection and Collinearity:
- Let $M'$ be the reflection of $M$ through $d$ .
- Since $d$ is the external bisector of $\angle xOy$ , $M'$ lies on the circle centered at $I$ with radius $IM$ .
- We have $\angle ONM' \equiv \angle ONP + \angle OPM' \equiv \angle ONP + \angle OMQ \equiv 0 \pmod{\pi}$ .
- Therefore, $O$ , $N$ , and $M'$ are collinear.

2. Angle Bisectors:
- Since $NP$ is the bisector of $\angle ONM$ and $MQ$ is the bisector of $\angle OMN$ , $K$ is the incenter of $\triangle OMN$ .
- Thus, $OK$ is the internal bisector of $\angle xOy$ .

3. Conclusion:
- Since $OK$ is the internal bisector of $\angle xOy$ , $K$ always lies on the fixed line $OK$ .

### Part 2: Prove that $EN$ , $FM$ , and $OK$ are concurrent.

1. Intersection and Circle:
- Let $G$ be the intersection of $EM$ and $FN$ .
- Since $O$ , $M$ , $G$ , and $N$ lie on a circle and $GM = GN$ , $G$ is the midpoint of the arc $MN$ not containing $O$ .

2. Cyclic Quadrilaterals:
- We have $\angle MPK \equiv \angle OPK \equiv \angle OKQ \equiv \angle NKQ \equiv \frac{\pi}{2} \pmod{\pi}$ .
- Hence, $O$ , $P$ , $M$ , $K$ and $O$ , $Q$ , $N$ , $K$ lie on circles.

3. Angle Relationships:
- We have $\angle ONM \equiv \angle ONK + \angle OKM \equiv \angle QNK + \angle PKM \equiv \angle INM \pmod{\pi}$ .
- This implies $O$ , $M$ , $N$ , and $I$ lie on a circle, and thus $O$ , $M$ , $G$ , $N$ , and $I$ lie on a circle.

4. Concurrency:
- Since $GM = GN$ , $OG$ is the internal bisector of $\angle xOy$ .
- Therefore, $O$ , $K$ , and $G$ are collinear, implying $EN$ , $FM$ , and $OK$ are concurrent.

$\blacksquare$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.