Let $P$ and $Q$ be the midpoints of the sides $BC$ and $CD$, respectively in a rectangle $ABCD$. Let $K$ and $M$ be the intersections of the line $PD$ with the lines $QB$ and $QA$, respectively, and let $N$ be the intersection of the lines $PA$ and $QB$. Let $X$, $Y$ and $Z$ be the midpoints of the segments $AN$, $KN$ and $AM$, respectively. Let $\ell_1$ be the line passing through $X$ and perpendicular to $MK$, $\ell_2$ be the line passing through $Y$ and perpendicular to $AM$ and $\ell_3$ the line passing through $Z$ and perpendicular to $KN$. Prove that the lines $\ell_1$, $\ell_2$ and $\ell_3$ are concurrent.
Problem
Source: JBMO Shortlist 2013, G6
Tags: geometry, rectangle, perpendicular lines, concurrency, midpoints
celilcelil
21.04.2021 23:04
We need to prove that XM² - XK² + KZ² - ZQ² + QY² - YM² = 0. In triangle AMN MX is median. Then we can find value of XM² depending on the values of AM, AN, MN. Similarly we can find values of XK, KZ, YM from triangles AKN, AKM, MKN, respectively. After rewrite it in the equation we find that we need to prove : AM² - KN² = 4( QZ² - QY²). Let AZ = ZM = a, QM = b, NY = NK = C, QK = d. Rewrite it in the equation we find that b(b + 2a) = d(d + 2c), which means AMKN is cyclic. Then we need to prove that AMKN is cyclic. Note that PA = PD. <DAQ = <CBQ. <PAD = <PDA = 90° - <PDC = <DPC. Then, <BKP = <DPC - <CBQ = <PAD - <DAQ = <PAQ. We are done!!! I think it was very nice problem for JBMO.
Steff9
22.04.2021 01:17
See Property 10.4.3 from the book A Beautiful Journey Through Olympiad Geometry attached below.
Attachments:
JustKeepRunning
04.06.2021 00:10
The problem reduces to the following claim Claim: $ANKM$ is cyclic Proof: Angle chase by letting $\angle ADP=\theta$ and $\angle DAQ=\alpha,$ and you get that $\angle ANQ=\angle QMK=180^{\circ}-\theta=\alpha$. Once this is done, we know that the lines are concurrent, and in fact, by lemma above, the point of concurrency is even know. It is the reflection of the circumcenter of $ANKM$ over the intersection of the diagonals of its varigon gram.