The chords $AB$ and $CD$ of a circle intersect at $M$, which is the midpoint of the chord $PQ$. The points $X$ and $Y$ are the intersections of the segments $AD$ and $PQ$, respectively, and $BC$ and $PQ$, respectively. Show that $M$ is the midpoint of $XY$.
Problem
Source: Finland 2018, p3
Tags: geometry, Chords, midpoint, circle
jayme
08.09.2019 08:25
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Papillon.pdf Sincerely Jean-Louis
Albert123
01.01.2022 07:13
Lemma(Buterfly's Theorem): Let $ AB$ be an arbitary chord of circle $ w$,let $ P$ be the midpoint of segment $ AB$,now let $ KM$ and $ LN$ be two other arbitary chords from $ w$ which pass thruogh $ P$,let $ LM,KN$ intersect $ AB$ at $ X,Y$ respectively,prove that $ PX = PY$. Sol $14$: $(A,P;X,B) \overset{L}{=} (A,N;M,B) \overset{K}{=} (A,Y;P,B)$ But $AP=PB \implies \frac{XA}{XP}=\frac{BY}{PY}$ i.e $PX=PY$. By the lemma in the problem: $M$ is the midpoint of $XY$
Jalil_Huseynov
20.01.2022 23:28
Let $Z,T$ be midpoints of $AD,BC$, respectively. We have that $OMXZ,OMTY$ are cyclic. Also $\triangle BMC\sim \triangle DMA \implies \triangle TMC\sim \triangle ZMA \implies \angle MOY=\angle MTC=\angle MZA=\angle MOZ$. Since $OM\perp XY$ and $\angle XOM= \angle YOM$ we get $MX=MY$ as desired.