In triangle $\triangle ABC$ , $M$ is the midpoint of arc $(BAC)$ and $N$ is the antipode of $A$ in $(ABC)$. The line through $B$ perpendicular to $AM$ , intersects $AM , (ABC)$ at $D,P$ respectively and a line through $D$ perpendicular to $AC$ , intersects $BC,AC$ at $F,E$ respectively. Prove that $PE,MF,ND$ are concurrent.
Problem
Source: Iran MO 2023 3rd round , geometry exam P2
Tags: geometry
23.08.2023 21:44
Let $X = DN \cap (ABC)$; we claim that this is the desired concurrency point. $\angle{DXA} = \angle{DEA} = 90^{\circ} \implies DEAX$ is cyclic with diameter $AD$, and $BP \parallel MN \implies \angle{PXN} = \frac{\widehat{MB}}{2} = \frac{\widehat{MC}}{2} = \angle{DAE} = \angle{DXE} \implies X \in PE$. Also, $\angle{XAC} = \angle{XDE} = \angle{XBF} \implies FDXB$ is cyclic $\implies \angle{XFB} = \angle{XDB} = \angle{XED}$ since $BD$ is tangent to $(AD)$, and this equals $\angle{XAD} = \angle{MNX}$ which is the angle between $MX$ and $BC$, so $X \in MF$, thus we are done. $\square$
10.12.2023 10:53
Let $M_A$ denote the arc midpoint of $\hat {BC}$ Claim 1: $CN \parallel PM \parallel DE$
Now, let $G = PE \cap (ABC)$ Claim 2: $F-G-M$
Claim 3:$B-G-N$
By Claim 2 and 3 we're done.
29.06.2024 06:18
Define $X$ as the second intersection of $(AED)$ and $(BFD)$. We wish to prove that $X$ is the desired intersection point. Claim: $X\in (ABC)$ $$\angle AXB=\angle AXD+\angle BXD=\angle DEC+\angle DFC=180^{\circ}-\angle ACB$$Claim: $N$, $D$, and $X$ are collinear $$\angle AXD=\angle AED=90^{\circ}=\angle AXN$$Claim: $CN||PM$ or $PM\perp AC$ $$\angle CAM+\angle PBA=90^{\circ}$$Claim: $P$, $E$, and $X$ are collinear $$\angle PXD=\angle PXN=\angle CAM=\angle DAE=\angle DXE$$Claim: $M$, $F$, and $X$ are collinear $$\angle MXN=\angle PBC=\angle DXF$$
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25.12.2024 15:03
I spent an embarassing amount of time on this problem lol. Anyways here's my very involved solution using a spiral similarity. We let $G$ be defined as the intersection of the $MF$ and $ND$, we will also let $H = DM \cap BC$ Claim 1: $FE \parallel MP$
Claim 2: $B$ is the center of the spiral similarity sending $FM$ to $DN$, thus $G$ lies on the circumcircle and $GBDF$ is cyclic
Claim 3: $GAED$ is cyclic
Now we are prepared to show $G-E-P$, notice that it suffices to show $\angle GEF = \angle GPM$ by claim $1$. Thus by claim $3$ we obtain \[\angle GEF = \angle GED = \angle GAD = \angle GAM\]However it is clear that \[\angle GPM = \angle GAM\]Thus we are done.