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
v4913
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. $\square$
math_comb01
10.12.2023 10:53
Let $M_A$ denote the arc midpoint of $\hat {BC}$ Claim 1: $CN \parallel PM \parallel DE$
Clearly $CN \parallel DE \bot AC$, Notice $PB \parallel AM_A$, so Angle between $PM$ and $AC$ = $$\measuredangle AMP + \measuredangle CAM = \frac{\measuredangle A}{2} + 90 ^\circ - \frac{\measuredangle A}{2} = 90^\circ$$, hence the claim
Now, let $G = PE \cap (ABC)$ Claim 2: $F-G-M$
By Pascal's on $PGMACB$, we get $PG \cap AC = E$, $GM \cap CB = E'$, $MA \cap PB = D$ collinear implying $E=E'$ and hence the claim
Claim 3:$B-G-N$
By Pascal's on $PGNCAM$ we get $PG \cap CA = E$, $GN \cap AM = D'$, $NC \cap PM = P_\infty$ since $P_\infty \in DE$ therefore $D=D'$
By Claim 2 and 3 we're done.
sami1618
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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Ihatecombin
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: $FE \parallel MP$
Notice that $AD$ is an angle bisector, therefore
\[\angle MDE = 180 - \angle ADE = 90+\dfrac{\alpha}{2}\]However we also obtain
\[\angle DMP = \angle AMP = \angle ABP = \angle ABD = 90 - \dfrac{\alpha}{2}\]Thus we have $\angle DMP + \angle MDE = 180$, and thus we obtain that $FE \parallel MP$
Claim 2: $B$ is the center of the spiral similarity sending $FM$ to $DN$, thus $G$ lies on the circumcircle and $GBDF$ is cyclic
We need to show that \[\angle FBD = \angle MBN\]By angle chasing we get
\[\angle FBD = \angle HBD = 90 - \angle DHB = 90 - \angle AHB = 90 - (180 - \beta - \dfrac{\alpha}{2}) = \dfrac{\beta - \gamma}{2}\]However we also have that
\[\angle MBN = \angle MAN = \angle BAN - \angle BAM = (90 - \gamma) - (\dfrac{\alpha}{2}) = \dfrac{\beta - \gamma}{2}\]Thus $B$ fulfills the angle condition necessary for a spiral similarity, we now check whether it fulfills the length condition. We need to show that
\[\dfrac{BM}{BN} = \dfrac{BF}{BD}\]However it is clear from the law of sines on $\triangle ABM$ and $\triangle ABN$ that
\[\dfrac{BM}{BN} = \dfrac{2R\sin(\dfrac{\alpha}{2})}{2R\cos(\gamma)} = \dfrac{\sin(\dfrac{\alpha}{2})}{\cos(\gamma)}\]From the law of sines on $\triangle BFD$ we also obtain that
\[\dfrac{BF}{BD} = \dfrac{\sin(\angle BDF)}{\sin(\angle BFD)}\]It is not too difficult to notice that $\angle BFD = 90 + \gamma$, and that $\angle BDF = \angle EDP = \dfrac{\alpha}{2}$ because $DE \perp AC$, thus we have
\[\dfrac{BF}{BD} = \dfrac{\sin(\angle BDF)}{\sin(\angle BFD)} = \dfrac{\sin(\dfrac{\alpha}{2})}{\cos(\gamma)} = \dfrac{BM}{BN}\]Thus we obtain that $B$ is indeed the center of the spiral similarity sending $FM$ to $DN$. It is now apparent that $B$ lies on $(GMN)$, and thus we have that $G$ lies on $(BMN)$ which is the circumcircle. It is clear that $\angle DFB = 90+\gamma$, however we also have $\angle BGD = \angle BGN = \angle BCN = 90 - \gamma$, thus $GBDF$ is cyclic as desired.
Claim 3: $GAED$ is cyclic
It suffices to show $\angle GDE = 180 - \angle GAE$. However this is an easy angle chase, due to the fact that $(GDFB)$ is cyclic we know
\[\angle GDE = 180 - \angle GDF = \angle GBF = \angle GBC\]However we have
\[\angle GAE = \angle GAC = 180 - \angle GBC\]We are done
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.