Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$. Prove that the lines $AK, XO, BC$ have a common point.
Problem
Source: BMO SL 2023 G5
Tags: geometry, AZE BMO TST, TST
bin_sherlo
03.05.2024 14:54
$XO\cap BC=S$. Let $A'$ be the antipode of $A$ in $(ABC)$. $X,A',C$ are collinear. Let the tangents at $B,C$ to $(ABC)$ intersect at $T$. Claim $1$: $X\in (BOC)$. Proof: $\angle OXC=\angle SXB=\angle XSB-\angle XCS=\angle B-(90-\angle C)=90-\angle A=\angle OBC$ which gives that $X\in (BOC)$.$\square$ $\angle BXO=90-\angle A=\angle OXC$ thus $OX$ is the interior angle bisector of $\angle BXC\implies OX\perp XY$ Claim $2$: $K\in (ABC)$. Proof: $\angle BKC=180-\angle CKY=180-\angle CXY=90+\angle OXC=180-\angle A$ Hence $ABCK$ is cyclic.$\square$ Claim $2$ gives that $X,K,B'$ are collinear where $B'$ is the antipode of $B$ on $(ABC)$. Finishing Claim: $A,S,K$ are collinear. Proof: Pascal at $A'AKB'BC$ gives that $O,AK\cap BC,X$ are collinear $\iff AK\cap BC=S$ as desired.$\blacksquare$
VicKmath7
03.05.2024 15:30
Observe that $\angle OXC=\angle OBC=90^{\circ}-\alpha$, so $BOCX$ is cyclic. Since $\angle XKY=\angle XCY=90^{\circ}$, $CKXY$ is cyclic and thus $\angle CKY=\angle CXY=\alpha$, so $K \in (ABC)$. By radical axis, we are only left show that $AOKX$ is cyclic. Since $\angle BXY=\angle BXC+\alpha=180^{\circ}-\alpha$, we obtain that $ABXY$ is cyclic. Hence, $X=(CKY) \cap (ABY)$ is the Miquel point of $ABKC$ and this finishes the problem.
Orestis_Lignos
03.05.2024 17:53
Proposed by Jason Prodromidis and Minos Margaritis, Greece
squarc_rs3v2m
04.05.2024 04:09
All of the above solutions are incorrect, because the problem is false. The official solution assumes that $XO$ internally bisects $\angle BXC$, but this does not hold in general (see attached image below) and while $BOCX$ is necessarily cyclic, $O$ is an arc midpoint but not necessarily the midpoint of the arc $BC$ not containing $X$. This failure can also occur when $\angle B$, $\angle C$ are acute and $\angle A$ is obtuse. However, the problem is fine when $\triangle ABC$ is constrained to be acute overall.
Tellocan
04.05.2024 13:44
squarc_rs3v2m wrote: All of the above solutions are incorrect, because the problem is false. The official solution assumes that $XO$ internally bisects $\angle BXC$, but this does not hold in general (see attached image below) and while $BOCX$ is necessarily cyclic, $O$ is an arc midpoint but not necessarily the midpoint of the arc $BC$ not containing $X$. This failure can also occur when $\angle B$, $\angle C$ are acute and $\angle A$ is obtuse. However, the problem is fine when $\triangle ABC$ is constrained to be acute overall. İ agree, the problem statement must include the triangle being acute.
matematica007
04.05.2024 16:24
How $\angle XKY=\angle XCY=90^{\circ}$ we have that $XKCY$ is cyclic. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$ so $\angle OXC=90^{\circ}-\angle A=\angle OBC=\angle OCB$ so $OBXC$ is cyclic. This implies that $\angle BXC=180^{\circ}-2\angle A$ so $\angle YXC=\angle YKC=\angle A$ so $\angle BKC= 180^{\circ}-\angle A$ so $ABKC$ is cyclic. We will prove that $AOKX$ is cyclic. After some easy angle chasing we have $\angle XKA=90^{\circ}+\angle C= \angle XOA $ so $AOKX$ is cyclic. New we have that $AK$ is radical axis of circles $(ABKC)$ and $(XKOA)$ , $XO$ is radical axis of circles $(XBOC)$ and $(XKOA)$ and $BC$ is radical axis of circles $(ABKC)$ and $(XBOC)$.So the lines $AK, XO, BC$ have a common point.
motannoir
04.05.2024 16:44
Sketch: Let $S=XO\cap BC$ ;prove by angles that $BOCX$ is cyclic ,then it is trvial that $XKCY$ is cyclic,use this to prove $K\in (ABC)$ and now prove(again by angles) that $AOKX$ is cyclic.Now use radical axes on $AOKX,ABKC,BOCX$ to get that $AK,OX,BC$ are concurrent
Z4ADies
10.05.2024 10:22
It is very trivial problem for G5 $\angle CAO=\angle OCA= x$ $\angle OBC=\angle OCB=y$ $\angle OAB=\angle OBA=90-x-y$ From parallelity of $OX,AB$ $\implies$ $\angle ABO=\angle BOX=90-x-y$ Similarly from perpendicularity of $AC,AX$ $\implies$ $\angle BCX=90-x-y$ Thus, $BOCX$ is cyclic that will be our hint to prove concurrency with radical axes are concurr at some point. We have to deduce $ABKC$ and $AOKX$ cyclic. $(.....1)$ $KXYC$ is cyclic because $XK$ is perpendicular to $BY$ and $AY$ is perpendicular to $XC$ From $(.....1)$ $\angle CXY=\angle CKY=90-y$ and we have $\angle BAC=90-y$ thus $ABKC$ cyclic. Let $\angle OXK= z$ $\implies$ $\angle KXC=y-z$ from $(.....1)$ $\angle KYA=y-z$ , $\angle AKY=180-x-y$ , and $\angle YAO=x$ it gives us $\angle KAO=z$ which means $AOXK$ cyclic. We are done.
NuMBeRaToRiC
26.06.2024 22:21
By easy angle chasing we get that $XBOC$ cyclic. Let $XO\cap BC=D$, $AK\cap XD=M$ and . By BMO 2022 G2 we get that $M$ midpoint of $XD$. Incenter $I$ and excenter $I'$ of triangle $BXC$ lies on $ABC$. From $(X,D;I,I')=-1$ so $MX^2=MD^2=MI\cdot MI'$ so $\angle XCD=\angle FKD$. By easy angle chasing we get that $\angle FCB=90$.So $\angle FKD=\angle XCD=\angle FCA=\angle FKA$, from this we get that $K, D, A$ collinear.We are done.
Ihatecombin
30.12.2024 11:30
Nice problem, had a fun time doing it.
The sketch of the solution that I present is as follows. We define $D = OX \cap BC$, and let $A'$, $C'$ be the $A$ and $C$ antipodes respectively as well as letting $M$ be the midpoint of $BA'$. The key things to notice is that $XC \cap (ABC)$ is $A'$ as well as noticing that $K-M-C'$ (the motivation for noticing this is that eventually you have to prove something is cyclic and then you'll notice $\angle MKA' = 180 - \beta$ is required for this) and that $K$ lies on the circumcircle.
Notice that with these definitions it suffices to show $A-D-K$, to do this it suffices to show $\angle DKA' = 90$. With some angle chasing it is easy to see that $DBA'$ is isosceles therefore $\angle DMA' = 90$, thus we just need to show $DMKA'$ is cyclic. However by also using the fact that $BMKX$ is cyclic we can prove this.
We now begin with the proof. All points not found in the problem statement are defined in the sketch of the solution. We begin with a small claim. Claim 1: $XC \cap (ABC) = A'$
We have $\angle XCY = 90$. Let $Z$ be defined as $XC \cap (ABC)$, now we must have $\angle ZCY = 90$, and thus $\angle ZCA = 90$, this suffices to show $Z = A'$.
Claim 2: $OX$ bisects $\angle BDA'$ and $\angle BXA'$
It is clear that $\triangle OBA'$ is isosceles, hence since $OD \parallel AB \perp BA'$, we know $\triangle DBA'$ is isosceles. However we must have \[\angle DBA' = \angle CBA' = 90 - \beta \Longrightarrow \angle BDA' = 2\beta\]By the parallel condition we have $\angle BDX = \beta$, we now have $OX$ bisects $\angle BDA'$. However since $\triangle BDA'$ is isosceles, it is clear that $\triangle XBA'$ is isosceles and that $\angle A'XB$ is also bisected by $OX$.
Claim 3: $BMKX$ is cyclic
This is easy $\angle BKX = \angle BMX = 90$.
Claim 3: $MKA'D$ is cyclic
It suffices to show $\angle MKA' = 180 - \angle MDA' = 180 - \beta$. However \[\angle MKA' = \angle BKA' - \angle BKM = 180 - \angle BCA' - \angle BXM = 90 + \gamma - (90 - \alpha) = 180 - \beta \]
Now we are done, just notice that by claim $4$ we have $\angle DKA' = \angle DMA' = 90$, however we also have $\angle AKA' = 90$, thus it must be true that $A-D-K$.