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Another config geo with concurrent lines

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Source: BMO SL 2023 G5

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678 posts

#1 h May 3, 2024, 10:19 AM • 2 Y

Y by lian_the_noob12, Rounak_iitr

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.

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733 posts

#2 h May 3, 2024, 11:54 AM • 3 Y

Y by Om245, farhad.fritl, At777

$XO\cap BC=S$ . 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. $\blacksquare$

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1390 posts

#3 h May 3, 2024, 12:30 PM

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Solution

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558 posts

#4 h May 3, 2024, 2:53 PM • 1 Y

Y by parmenides51

Proposed by Jason Prodromidis and Minos Margaritis, Greece

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46 posts

#5 h May 4, 2024, 1:09 AM • 2 Y

Y by Scilyse, ohiorizzler1434

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.

https://cdn.discordapp.com/attachments/799475903569592340/1236120831638372483/image.png?ex=6636dabc&is=6635893c&hm=51afb14d812cad898d8a0f488476759f38c60ee3a953d0ae583cdf57c69c1f1d&

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35 posts

#6 h May 4, 2024, 10:44 AM

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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.

https://cdn.discordapp.com/attachments/799475903569592340/1236120831638372483/image.png?ex=6636dabc&is=6635893c&hm=51afb14d812cad898d8a0f488476759f38c60ee3a953d0ae583cdf57c69c1f1d&

İ agree, the problem statement must include the triangle being acute.

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17 posts

#7 h May 4, 2024, 1:24 PM

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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.

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171 posts

#8 h May 4, 2024, 1:44 PM • 1 Y

Y by Assassino9931

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

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64 posts

#9 h May 10, 2024, 7:22 AM

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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.

This post has been edited 3 times. Last edited by Z4ADies, May 10, 2024, 12:25 PM

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#10 h Jun 26, 2024, 7:21 PM

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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.

This post has been edited 2 times. Last edited by NuMBeRaToRiC, Jun 26, 2024, 7:22 PM

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#11 h Dec 30, 2024, 8:30 AM

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Nice problem, had a fun time doing it.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.974725338137207, xmax = 30.033727935533033, ymin = -11.465676723098179, ymax = 12.099250581265741; /* image dimensions */ /* draw figures */ draw(circle((5,5), 5), linewidth(0.4)); draw((3.322736366511161,9.71028520408009)--(0.6972183465910291,2.4532235977441674), linewidth(0.4)); draw((0.6972183465910291,2.4532235977441674)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); draw((9.302781653408971,2.4532235977441674)--(3.322736366511161,9.71028520408009), linewidth(0.4)); draw((0.6972183465910291,2.4532235977441674)--(1.859615688652539,-3.6801774895883588), linewidth(0.4)); draw((0.6972183465910291,2.4532235977441674)--(19.665169827022172,-10.122013838908856), linewidth(0.4)); draw((1.859615688652539,-3.6801774895883588)--(4.329421138984925,0.04517164866836985), linewidth(0.4)); draw((0.6972183465910291,2.4532235977441674)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); draw((4.078606783358132,2.4532235977441683)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); draw((4.078606783358132,2.4532235977441683)--(4.329421138984925,0.04517164866836985), linewidth(0.4)); draw((4.329421138984925,0.04517164866836985)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); draw((4.329421138984925,0.04517164866836985)--(3.6872409900399328,1.3714691968320367), linewidth(0.4)); draw((3.322736366511161,9.71028520408009)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); draw((0.6972183465910251,7.5467764022558255)--(3.6872409900399328,1.3714691968320367), linewidth(0.4)); draw((0.6972183465910251,7.5467764022558255)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); draw((0.6972183465910251,7.5467764022558255)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); draw((5,5)--(1.859615688652539,-3.6801774895883588), linewidth(0.4)); draw((1.859615688652539,-3.6801774895883588)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); draw((19.665169827022172,-10.122013838908856)--(1.859615688652539,-3.6801774895883588), linewidth(0.4)); draw((9.302781653408971,2.4532235977441674)--(19.665169827022172,-10.122013838908856), linewidth(0.4)); draw(circle((1.2784170176217802,-0.6134769459221007), 3.1212888555420335), linewidth(0.6) + green); draw(circle((5.377935208423484,1.3714691968320383), 1.6906942183835514), linewidth(0.6) + red); /* dots and labels */ dot((5,5),linewidth(3pt) + dotstyle); label("$O$", (5.094827972055157,5.151615248405972), NE * labelscalefactor); dot((3.322736366511161,9.71028520408009),linewidth(3pt) + dotstyle); label("$A$", (3.4356911761481044,9.869785511765965), NE * labelscalefactor); dot((0.6972183465910291,2.4532235977441674),linewidth(3pt) + dotstyle); label("$B$", (0.7914419076712397,2.6110620296736675), NE * labelscalefactor); dot((9.302781653408971,2.4532235977441674),linewidth(3pt) + dotstyle); label("$C$", (9.398214036439073,2.6110620296736675), NE * labelscalefactor); dot((1.859615688652539,-3.6801774895883588),linewidth(3pt) + dotstyle); label("$X$", (1.9580224672933857,-3.5329289176687406), NE * labelscalefactor); dot((19.665169827022172,-10.122013838908856),linewidth(3pt) + dotstyle); label("$Y$", (19.76781901085815,-9.962084001807632), NE * labelscalefactor); dot((4.329421138984925,0.04517164866836985),linewidth(3pt) + dotstyle); label("$K$", (4.420803648717916,0.20012887312158348), NE * labelscalefactor); dot((4.078606783358132,2.4532235977441683),linewidth(3pt) + dotstyle); label("$D$", (4.187487536793487,2.6110620296736675), NE * labelscalefactor); dot((6.677263633488835,0.2897147959199078),linewidth(3pt) + dotstyle); label("$A'$", (6.779888780398257,0.4334449850459787), NE * labelscalefactor); dot((3.6872409900399328,1.3714691968320367),linewidth(3pt) + dotstyle); label("$M$", (3.798627350252772,1.5222535073598231), NE * labelscalefactor); dot((0.6972183465910251,7.5467764022558255),linewidth(3pt) + dotstyle); label("$C'$", (0.7914419076712397,7.692168467138275), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Sketch of solution
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: $XC \cap (ABC) = A'$
Proof
Claim 2: $OX$ bisects $\angle BDA'$ and $\angle BXA'$
Proof
Claim 3: $BMKX$ is cyclic
Proof
Claim 3: $MKA'D$ is cyclic
Proof
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 $A-D-K$ .

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117 posts

#12 h Jan 30, 2025, 5:37 AM

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Similar sketch.

Let $OX\cap BC=X'$ . Let $\omega$ be the center with the diameter $OX'$ . Let $Z', Y'$ be the intersection of $(AOB),(AOC)$ with $\omega$ . Let $Y,Z$ be the intersections of $OY', OZ'$ with $AB,AC$ , respectively. We will get that $X$ is the foot of the altitude of $\triangle OYZ$ . Let $ZC \cap BY=K$ . Then $K$ lies on the $(ABC)$ .

Let $BC \cap ZY=D$ and $M$ be the midpoint of $BC$ . Then we have that $(AD,AK;AB,AC)=-1$ . We want to show that $A-X'-K$ . So it's suffices to prove $(B,C;D,X')=-1$ .

Since $ZY$ is the radical axis of $(ABC),\omega$ and $D$ lies on it we will get that $DB*DC=DX'*DM$ which only happens when $(B,C;D,X')=-1$ , and we are done.

This post has been edited 1 time. Last edited by Nari_Tom, Jan 30, 2025, 5:40 AM

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SimplisticFormulas

118 posts

SimplisticFormulas

#13 h Jan 30, 2025, 5:50 PM

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Draq

#14 h Feb 24, 2025, 9:06 AM

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Let $\angle CAB =x \angle CBA=y \angle ACB=z$ $\because AB$ is parallel to $DX$ $\therefore \angle ABD= \angle BDX$
which is equal to y and $\because AC$ is perpendicular to $XC$ $\angle ACB = 90- \angle BCX$ $\therefore$ $\angle BCX=90-z$
İn triangle $DCX$ $\angle BDX=\angle DXC+ \angle DCX$ $\therefore$ $\angle DXC = 90-x=\angle OXC= \angle OBC$ $OBXC$ is cyclic
$\angle XCY =\angle XKY =90$ also $\angle OXB=\angle OCB=90-x$ $XD$ angle bisector also $XY$ external angle bisector so $XD$ is perpendicular to $XY$ so $\angle CXY =x$ then $\angle CKY=x$ which means $ABKC$ is cyclic then $\angle AKB=\angle ACB=z$ so $\angle AKX=90+z$ and $\angle OAK=90-z$ because of parallel $\angle AOX=90+z$ so AOKX cyclic and by radical axis theorem AK BC XO are concurrent at point $D$

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optimusprime154

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optimusprime154

#15 h Mar 10, 2025, 1:08 PM

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claim 1: $BXOC$ concyclic:
proof: we know that $\angle XOB = \angle OBA = 90 - \angle C$ and also $\angle XCB = 90 - \angle C$ finishing the claim.
from now on, let $S$ = $OX \cap BC$ and $K$ = $XY \cap AB$ , $R$ = $CX \cap BK'$ and $T$ = $KC \cap BY$ first notice that $OX \perp KY$ since $OX$ is the bisector of $\angle BXC$ , we know $BXAY$ cyclic because $\angle K'XB = \angle A$ we also know $K'XCA$ cyclic for obvious reasons.
claim 2: $BTXK'$ cyclic.
this just follows from the fact that $\angle XBT = \angle XBY = \angle XAY = \angle XAC = \angle XK'C = \angle XK'T$
this means $\angle K'TB = \angle K'XB = \angle A$ meaning $T$ lies on the circle $ABC$ . we Also got $\angle XTB = \angle XK'B = 90$ so $T$ is the foot from $X$ to $BY$
claim 3: $A, S, T$ collinear.
we first prove that $XTOA$ is cyclic. which just follows from $\angle XTA = \angle XTB + \angle BTA = 90 + \angle C = \angle XOA$
now we let $S'$ denote $AT \cap BC$ we know that $X, S', O$ must be collinear by PoP on $ABTC$ , $XBOC$ , $XTOA$ , since we know $T$ is the foot from $X$ to $BY$ , And $AT$ $BC$ $OK$ concur At $S$ so we're done!

This post has been edited 1 time. Last edited by optimusprime154, Mar 10, 2025, 3:29 PM

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E50

#16 h Apr 5, 2025, 6:31 AM

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Claim 1 : $O,B,X,C$ concyclic

Proof : Let $Z=OX \cap BC$ . Then $\angle OXC = \angle ZXC =90^\circ -\angle XZC = 90^\circ - \angle BAC = \frac{180^\circ -\angle BOC}{2}=\angle OBC \Longrightarrow O,B,X,C$ concyclic.

Claim 2 : $A,B,K,C$ concyclic.

Proof : Since $\angle XKY=90^\circ = \angle XCY \Longrightarrow X,K,C,Y$ cincyclic. Since $OB =OC$ and $O,B,X,C$ concyclic ; $\angle BXO = \angle CXO \Longrightarrow \angle CKY = \angle CXY =90^\circ - \angle OXC = \angle BAC \Longrightarrow A,B,K,C$ concyclic.

Claim 3 : $A,O,K,X$ concyclic.

Proof : Let $\angle ACB = \alpha$ . Then $\angle AKX = 90^\circ + \alpha$ , $\angle BOX = \angle BCX = 90^\circ - \alpha$ , $\angle AOB = 2\angle ACB = 2\alpha \Longrightarrow \angle AOX = 90^\circ+\alpha \Longrightarrow \angle AKX = \angle AOX \Longrightarrow A,O,K,X$ concyclic.

And by Radical Axis Concurrence Theorem on $(AOKX),(ABKC),(OBXC)$ we obtain $AK,XO,BC$ have a common point as desired.

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Tkn

#17 h May 12, 2025, 10:06 AM

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Solved with EeEeRUT.
$[asy] import geometry; import graph; real markscalefactor=0.025; size(10cm); defaultpen(fontsize(10pt)); path rightanglemark(pair A, pair B, pair C, real s=8) { pair P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } pair A = (-0.75,2); pair B = (-1,0); pair C = (2,0); path circ1 = circumcircle(A,B,C); pair O = extension((B+C)/2,rotate(90,B)*C-B+(B+C)/2,(C+A)/2,rotate(90,C)*A-C+(A+C)/2); pair X1 = extension(O+2*(B-A),O,C,rotate(90,C)*A); pair Y1 = extension(A,C,rotate(90,X1)*O,X1); pair K = intersectionpoints(circ1, B--Y1)[1]; path circ2 = circumcircle(X1,K,C); pair T = extension(A,B,X1,Y1); pair P = extension(A,K,B,C); path circ3 = circumcircle(B,T,X1); path circ4 = circumcircle(A,C,T); draw(A--B--C--cycle, black); draw(O--X1, red+dashed); draw(B--X1--C, black); draw(X1--Y1--C, black); draw(B--Y1, blue); draw(A--K, red+dashed); draw(B--T--X1, black); draw(T--C, royalblue+dashed); draw(X1--K, black); draw(circ1); draw(circ2, orange); draw(circ3, heavymagenta); draw(circ4, heavycyan); draw(rightanglemark(B,T,Y1)); dot(A); dot(B); dot(C); dot(O); dot(X1); dot(Y1); dot(K); dot(T); dot(P); label("$A$", A, N, black); label("$B$", B, 1.5W, black); label("$C$", C, 2S, black); label("$X$", X1, SE, black); label("$Y$", Y1, SW, black); label("$K$", K, 2N+0.4E, black); label("$T$", T, SW, black); label("$P$", P, N+2W, black); label("$O$", O, SE, black); [/asy]$
Before we begin with the proof, we define the following points:

$P$ is the intersection of $\overline{BC}$ and $\overline{AK}$ .
$T$ is the intersection of the line $AB$ and $XY$ .

Claim 1. $\overline{OX}$ bisects $\angle{BXC}$ .
Proof. Notice that $\angle{OXC}=90^{\circ}-\angle{BAC}=\angle{OBC}=\angle{OCB}$ . So, $OCXB$ is a cyclic quadrilateral. Which also implies $\angle{OXB}=\angle{OCB}=\angle{OXC}$ . The first claim is complete.

Claim 2. $K\in (ABC)$ .
Proof. It is easy to see that $K,X,Y,C$ are concylic because $\angle{XKY}=\angle{XCY}=90^{\circ}$ .
From the first claim, $\overline{OX}$ is an angle bisector of $\angle{BXC}$ . So, we can redefine the external angle bisector of $\angle{BXC}$ to be a line perpendicular to $\overline{KX}$ . Which means ${TY}\perp AB$ . The angle chasing from $\angle{A}$ gives:
$\angle{BAC}=90^{\circ}-AYT=\angle{YKC}.$ Since, $B,K$ and $Y$ are collinear. Therefore, $K\in(ABC)$ .

Claim 3. $T,K$ and $C$ are collinear.
Proof. Observe that $BKXT$ and $ACXT$ are cyclic quadrilateral. Perform an angle chasing:
$\angle{TKX}=\angle{TBX}=\angle{BXO}=90^{\circ}-\angle{BAC}\implies \angle{BKC}=90^{\circ}+TKX$ Therefore, $T,K$ and $C$ are collinear.

Since $X\in (BKT),(KCY)$ , and $T=KC\cap AB, Y=BK\cap AC$ , $X$ is a miquel point of $ABKC$ . It is well-known that $O,P$ and $X$ are collinear. In conclusion, $AK,XO$ and $BC$ are concurrent.

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20 posts

Rayvhs

#18 h May 12, 2025, 9:47 PM

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This was a fun problem.
First, note that
$\angle BOX = \angle ABO = \angle BAO = 90^\circ - \angle ACB = \angle BCX,$ so $BOCX$ is cyclic.
Next, note that $\angle XKY = 90^\circ = \angle XCY$ , so $XKCY$ is cyclic.
Also, observe that
$\angle BXC = 180^\circ - \angle BOC = 180^\circ - 2\angle A,$ and since $XY$ is an external angle bisector, $\angle CXY = \angle A$ .
Because $XYCK$ is cyclic, we have $\angle CKY = \angle A$ , so $ABKC$ is cyclic.
Now, observe
$\angle AOX = \angle AOB + \angle BOX = 2\angle ACB + \angle BCX = \angle ACB + 90^\circ,$ $\angle AKX = \angle AKB + \angle BKX = \angle ACB + 90^\circ,$ thus $AOKX$ is cyclic.

Finally, by radical axes theorem on $\odot(BOCX)$ , $\odot(ABKC)$ , and $\odot(AOKX)$ ,
we conclude that $BC$ , $AK$ , and $OX$ are concurrent.

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