Given is an acute triangle $ABC$. $M$ is an arbitrary point at the side $AB$ and $N$ is midpoint of $AC$. The foots of the perpendiculars from $A$ to $MC$ and $MN$ are points $P$ and $Q$. Prove that center of the circumcircle of triangle $PQN$ lies on the fixed line for all points $M$ from the side $AB$.
Problem
Source: Bosnia Herzegovina TST 2017, problem 6
Tags: geometry, circumcircle
Tatsuya
24.07.2017 16:55
Let the line passes through $P$ and perpendicular to $PN$ meets $AQ$ at $S$. Note that $NS$ is the diameter of circle $(PQN)$ so we only need to prove that $S$ lies on a fixed point. $D$ is the projection of $C$ onto $AB$. $DN$ meets circle $(AC)$ at $E$. Then $\angle NEP$ $=$ $\angle MAP$ $=$ $\angle PQM$ thus $PQNE$ is a concyclic quadrilateral. Hence $\angle SEN$ $=$ $90^\circ$ which implies that $SE$ is the tangent of $(AC)$ at $E$ i.e $S$ lies on a fixed point. $\square$
RC.
24.07.2017 18:02
Let a point \(M_1\equiv B\). Define the circumcenter of the triangle formed through point \(B\) as \(O_1\). Now let \(M_2\) be an arbitary point on \(AB\) as \(M\) varies over the segment. Similarly, define \(O_2\) as circumcenter for point \(M_2.\) Now prove that \(\angle NO_1O_2\) is \(\angle [90^{\circ} + B] \) i.e., it does not depend upon the location of \(M\). For any \(\Delta ABC\) which has fixed angle measures the required circumcenters lie on a fixed line. [\Hide]
Assassino9931
13.10.2024 04:14
Also Bulgaria IMO TST 2015.
ohhh
14.10.2024 02:29
Let $X \in AB$ such that $CX \perp AB$, and $Y$ the reflection of $X$ about $N$. $\angle PTN = \angle PAM = \angle PQM \implies (PQNY)$. $Y$ and $N$ are fixed, so the circumcenter moves along a fixed line.
TestX01
14.10.2024 03:06
Let $D$ be the foot of $C$ to $AB$, and $D'$ reflection of $D$ over $N$. By Thales $AN=NC=NE=NF$ so $AECF$ is cyclic. Clearly $AMPQ$ is cyclic. Pitot theorem on $EMN$ means $PQNF$ cyclic. Hence the circumcentre lies on the perpendicular bisector of $NF$, which is fixed.