Let $ABC$ be an acute triangle with circumcircle $\omega$ and circumcenter $O$. The perpendicular from $A$ to $BC$ intersects $BC$ and $\omega$ at $D$ and $E$, respectively. Let $F$ be a point on the segment $AE$, such that $2 \cdot FD = AE$. Let $l$ be the perpendicular to $OF$ through $F$. Prove that $l$, the tangent to $\omega$ at $E$, and the line $BC$ are concurrent. Proposed by Stefan Lozanovski, Macedonia
Problem
Source: JBMO Shortlist 2021
Tags: Junior, Balkan, shortlist, 2021, geometry, concurrency
Assassino9931
03.07.2022 09:58
Nice problem! Let $T\in BC$ be such that $ET$ is tangent to $\omega$ - we want $T \in \ell$. The crucialty is to see that $F$ is the midpoint of $CH$, where $H$ is the orthocenter of $ABC$, then use $CH = 2OM$, where $M$ is the midpoint of $AB$ and angle chase.
bin_sherlo
20.03.2023 19:31
Let the intersection of $BC$ and the tangent to $\omega$ at $E$ be $K$. And F be the point on AD such that $\angle (OFK)=90$ We will prove that $|FD|=|AF|+|DE|$ M is the midpoint of the segment $BC$. $OM \perp BC$ So the points $K,F,O,M,E$ are cyclic. Let $N$ be the point such that $|FN|$ is equal to $|DE|$.
$|DE|=x, |FN|=x, |DN|=y, |AF|=z, |KB|=a, |BD|=b, |DM|=c, |MC|=b+c$
We have
$c(a+b)=x(x+y)$
$x(x+y+z)=b(b+2c)$
$\implies$ $xz=b^2+bc-ac$
Also
$(a+b)^2+x^2=|KE|^2=a(a+2b+2c)$
$\implies x^2+b^2=2ac$
$\implies b^2=2ac-x^2$
If we rewrite the equasion
$xz=2ac-x^2+bc-ac=ac+bc-x^2$
$x(x+z)=c(a+b)$
We know that
$x(x+y)=c(a+b)$
$\implies y=z$
$|ED|+|AF|=x+z=x+y=|DF|$
Nartku
04.10.2024 17:45
Let $l$ and the tangent to $\omega$ at $E$ intersects at $S$. We will prove that the triangles $SDE$ and $SFO$ are congruent, due to the cyclicity of $SFOE$ we get that $\angle SED = \angle SOF$ so we are going to prove this by proving $\frac{SO}{OF}=\frac{SE}{DE}$. Let $\angle DAC=a$,$\angle BAD=b$,$\angle SOE=c$ so by some angle chasing we get that $\frac{SE}{\sin c} = \frac{OE}{\cos c} = \frac{AO}{\cos c} \rightarrow SE=AO \tan c$ and also $\frac{AE}{\sin(2a-2b)} = \frac{AO}{\sin(a-b)}$.
By reflecting the orthocenter, we get that $AF=FH$ and $HD=DE$. So, $2AO\cos(a-b)=AE=AH+HE=2AO\cos(a+b)+2ED \rightarrow ED=AO(\cos(a-b)-\cos(a+b))=2AO\sin a\sin b$.
Since $\frac{SO}{OF}=\frac{1}{\sin(a-b)}$, going back we want to prove that $\frac{SO}{OF} = \frac{SE}{DE}$, which is equivalent as proving $\frac{\tan c}{2\sin a \sin b}=\frac{1}{\sin(a-b)}$ which is equivalent to prove that $\tan c = \frac{2\sin a \sin b}{\sin(a-b)}$.
Notice that by considering triangle $AOF$, we get that $\frac{AO}{\cos c}=\frac{AF}{\sin(90+c+a-b)}=\frac{AO \cos(a+b)}{cos(b-a-c)}$, so $\cos c \cos(a+b) = \cos((b-a)-c) = \cos(b-a) \cos c + \sin(b-a) \sin c \\ \rightarrow \sin c \sin(b-a) = \cos c (\cos(b+a) - \cos(b-a))= \cos c (-2 \sin b \sin a) \\ \rightarrow \tan c = \frac{2 \sin a \sin b}{\sin(a-b)}$, so we have proven the congruency, and thus $\angle SDE = 90^\circ = \angle BDE$ which shows that $S$ is on $BC$, proven
OronSH
04.10.2024 18:18
Define orthocenter $H$ and queue point $Q$ then $F$ is the midpoint of $AH$ and $OF\perp AQ$. Now $l$ is the $H$ midline of $\triangle AHQ$ so $l\cap BC=Y$ is equidistant to $Q,H$ and thus $E$. Now $(B,C;Q,E)=-1$ so since the perpendicular bisector of $QE$ meets $BC$ at $Y$ we have $YE,YQ$ tangent to $\omega$ as desired.