Let $\Gamma$ denote the circumcircle and $O$ the circumcentre of the acute-angled triangle $ABC$, and let $M$ be the midpoint of the segment $BC$. Let $T$ be the second intersection point of $\Gamma$ and the line $AM$, and $D$ the second intersection point of $\Gamma$ and the altitude from $A$. Let further $X$ be the intersection point of the lines $DT$ and $BC$. Let $P$ be the circumcentre of the triangle $XDM$. Prove that the circumcircle of the triangle $OPD$ passes through the midpoint of $XD$.
Problem
Source: Baltic Way 2022/14
Tags: geometry
Kimchiks926
12.11.2022 23:24
This problem was proposed by me.
Let $A'$ be a point on $\odot(ABC)$ such that $AA' \parallel BC$. Also introduce point $E$, which is the intersection of $A'M$ with $\odot(ABC)$.
Claim: Quadrilateral $XDMA'$ is cyclic.
Proof: Note that by symmetry over $OM$ we have $\widehat{BE}=\widehat{TC}$. Observe that:
$$ \angle A'DT = \frac{1}{2}(\widehat{A'T})=\frac{1}{2}(\widehat{TC}+\widehat{A'C})=\frac{1}{2}(\widehat{BE}+\widehat{A'C})=\angle BME $$Note that $\angle XDA' = 180^{\circ}-\angle A'DT = 180^{\circ}-\angle BME= \angle XMA'$, this implies that points $X,D,M,A'$ all lie on the same circle, which proves the claim.
Now to finish the problem, note that $\angle DAA'=90^{\circ}$, thus, $A'D$ is diameter of $\odot(ABC)$, meaning that $O$ lies on $DA'$. Also $DA'$ is the radical axis of $\odot(XDMA')$ and $\odot(ABC)$, therefore, $OP \perp DA'$. Let $F$ be the midpoint of $DX$. Note that $FP \perp XD$. Putting these two results together gives us that points $P,O,D,F$ all lie on the same circle with diameter $DP$, which gives the desired result.
VicKmath7
13.11.2022 00:00
Amazing geo! Here is a hybrid solution. Let $OD \cap \Gamma=G$. It is easy to see we want to prove $\angle POD =90^o$ and that $AG \parallel BC$. If we prove that $XDMG$ is cyclic, then $O$ will lie on the radical axis of $\Gamma$ and $(XDM)$ and thus $PO \perp DG$ and we will be done. Define a function $f(X)=\pm \frac {XB} {XC}$; we will use the results from mira74's ratio lemma handout (found here: https://artofproblemsolving.com/community/c6h2357938p19166714). We have that $XDMG$ is cyclic iff $f(M)f(X)=f(D)f(G) \iff f(X)=f(D)f(G)$ (theorem 3.1). But $f(X)=f(D)f(T)$ (lemma 2.2) and $f(T)= \frac {TB} {TC}= \frac {GB} {GC}=f(G)$ since $GBTC$ is harmonic by projecting $(B, C, M, \infty)=-1$ through $A$ onto the circle $\Gamma$, so we are done.
trinhquockhanh
16.07.2023 05:19
Let the line passes through $A$ and parallel to $BC$ intersects $(O)$ at $E.$ Then we have $E,O,D$ are collinear. This gives us $\angle EDX=180^{\circ} -\angle EDT=180^{\circ}-\angle EAM=180^{\circ}-\angle MEA=180^{\circ}-\angle EMC=\angle EMX$ $\Rightarrow E\in (MDX)\Rightarrow PE=PD,$ we also have $OE=OD,O\in ED$ $\Rightarrow PO\perp OD,$ note that $\angle PF\perp FD (F$ is the midpoint of $XD)$ Hence $(OPD)$ passes the midpoint of $XD.$
khanhnx
16.07.2023 16:28
Let $A' \in \Gamma$ be a point such as $AA' \parallel BC$. We have $\angle{A'MC} = \angle{MA'B} + \angle{A'BC} = \angle{MAC} + \angle{A'DC} = \angle{TDC} + \angle{A'DC} = \angle{A'DT}$. Then $A', M, D, X$ lie on a circle. So $OP \perp DA'$ at $O$. From this, if we let $Q$ be midpoint of $DX$ then $\angle{PQD} = 90^{\circ} = \angle{POD}$ or $Q \in (POD)$