Let $ABC$ be an acute triangle with circumcenter $O$, and let $D$, $E$, and $F$ be the feet of altitudes from $A$, $B$, and $C$ to sides $BC$, $CA$, and $AB$, respectively. Denote by $P$ the intersection of the tangents to the circumcircle of $ABC$ at $B$ and $C$. The line through $P$ perpendicular to $EF$ meets $AD$ at $Q$, and let $R$ be the foot of the perpendicular from $A$ to $EF$. Prove that $DR$ and $OQ$ are parallel.
Problem
Source: FKMO 2022 Problem 1
Tags: geometry, circumcircle, parallel, geometry solved, similar triangles, parallelogram, Korea
MrOreoJuice
23.04.2022 15:28
[asy][asy]
size(250);
pair A = dir(125), B = dir(-150), C = dir(-30), H = A+B+C, M=(C+B)/2;
pair O=origin, D=foot(A,B,C), E=foot(B,C,A), F=foot(C,A,B);
pair R=extension(A,O,E,F);
pair P= 2*C*B/(C+B);
pair Q=A+P-O;
pair S=extension(E,F,P,Q);
path c1=unitcircle;
draw(c1,dotted);
draw(A--B--C--A);
draw(P--B);
draw(P--C);
draw(F--S, dashed+grey); draw(Q--S, dashed+grey);
draw(A--Q--P--O--A, blue); draw(B--E); draw(C--F);
draw(E--F);
dot("$A$",A,N); dot("$B$",B,SW);dot("$C$",C,SE); dot("$P$",P,SE);dot("$O$",O);
dot("$H$",H); dot("$D$",D,SE); dot("$E$",E,NE);dot("$F$",F,W); dot("$R$",R,W); dot("$Q$",Q,W); dot(S);
dot("$M$",M,SE);
[/asy][/asy]
Let $M$ be the midpoint of $\overline{BC}$.
$\triangle AEF \sim \triangle ABC$.
$A$,$R$,$O$ are collinear.
$AO \parallel PQ$ and $AQ \parallel OP$ hence $AQPO$ is a parallelogram.
$$\dfrac{AQ}{AO} = \dfrac{OP}{OB} = \dfrac{OB}{OM} = \dfrac{1}{\cos \angle A} = \dfrac{AB}{AE} = \dfrac{AD}{AR}$$hence $\triangle AQO \sim \triangle ADR$ which yields the desired.
David_Kim_0202
23.04.2022 19:02
it's so easy@_@ show O and Q are in the same phase at AFE, ACB. if we put M the midpoint of BC M is the two segments that each passing point E and F also tangents at the circumcircle of AFE. let's make a segment XY about AFE that has the same phase as EF in ABC. if we show XY and OM are perpendicular then the problem is almost solved. XY and BC are parallel and OM and BC are perpendicular therefore OM and XY are perpendicular. now we can know O and Q are in the same phase therefore AR:AO=AH:AQ therefore DR and OQ are parallel. LOL!!@_@
ahpla
27.04.2022 02:17
$\angle{AER} = \angle{ABD}$ ($\because B, C, E, F$ concyclic) and $\angle{POC} = \angle{BOC} / 2 = \angle{BAE}$ implies $AER \sim ABD$ and $POC \sim BAE$. Since $\angle{OAE} = \pi/2 - \angle{ABC} = \pi/2 - \angle{REA} = \angle{RAE}$, $A, R, O$ are collinear and $AOPQ$ is a parallelogram ($\because AO \parallel PQ, OP \parallel QA$). Now $DR \parallel OQ$ can be proved by $AR : AD = AE : AB = OC : OP = AO : AQ$.
Mogmog8
15.10.2022 21:37
Notice $A,R,O$ are collinear as $$\angle OAC=90-\angle ABC=90-\angle AER=\angle RAE.$$Also, $AQPO$ is a parallelogram. Hence, $$\frac{AR}{AD}=\frac{AE\sin\angle B}{AB\sin\angle B}=\sin(90-\angle B)=\sin\angle CPO=\frac{CO}{PO}=\frac{AO}{AQ}$$and $\triangle ARD\sim\triangle AOQ.$ $\square$
angelthirty
28.01.2023 17:23
This is the solution to the problem.
Attachments:
2022FKMO1.docx (13kb)
math_comb01
01.07.2024 19:11
cute! Consider $\sqrt{bc/2}$ inversion, it suffices to prove if $(BOC)\cap AB,AC=U,V$ then perpendiculars from $U,V,A$ to $AB,AC,BC$ are concurrent but this follows by radax on $(BDQU),(BOCUC),(CDVQ)$ where $Q$ is defined as intersection of perpendiculars from $U,V$ to $AB,AC$.