Collinear if and only if

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Collinear if and only if

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Source: 2018 Greece National Olympiad Problem 2

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402 posts

Borbas

#1 h Mar 3, 2018, 1:25 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle with $AB<AC<BC$ and $c(O,R)$ the circumscribed circle. Let $D, E$ be points in the small arcs $AC, AB$ respectively. Let $K$ be the intersection point of $BD,CE$ and $N$ the second common point of the circumscribed circles of the triangles $BKE$ and $CKD$ . Prove that $A, K, N$ are collinear if and only if $K$ belongs to the symmedian of $ABC$ passing from $A$ .

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#2 h Mar 3, 2018, 1:43 PM • 3 Y

Y by govind7701, Adventure10, Mango247

Sol

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1979 posts

#3 h Mar 3, 2018, 2:58 PM • 5 Y

Y by paragdey01, SAUDITYA, Arshia.esl, Adventure10, Mango247

Apply $\sqrt{bc}$ inversion calling $X'$ the image of point $X$ . Then $D'C'B'E'$ are collinear in that order, $K'=\odot(AB'D') \cap \odot(AC'E')$ with $K' \ne A$ and $N'=\odot(B'K'E') \cap \odot(C'K'D')$ . Let $L=\overline{AK'} \cap \overline{B'C'}$ . Then $LB' \cdot LD'=LC' \cdot LE'$ since $L$ lies on the radical axis of $\odot(AB'D')$ and $\odot(AC'E')$ . Thus, $L$ lies on $\overline{K'N'}$ if and only if $LD' \cdot LC'=LB' \cdot LE' \iff LB'=LC' \iff \overline{AK}$ is a symmedian in $\triangle ABC$ .

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308 posts

#4 h Mar 3, 2018, 9:29 PM • 2 Y

Y by hellomath010118, Adventure10

Let the tangents from $B,C$ to $\odot ABC$ intersect at $Z$ . Extend $ZB,ZC$ to meet $\odot BEK, \odot CKD$ respectively again at $F,G$ . By angle chasing:
$\angle BFK=\angle BEK=\angle BEC=\angle BAC=\angle ZBC \implies FK \parallel BC$ Similarly $GK \parallel BC$ from which it follows:
$BF=CG \implies ZF=ZG \implies ZB \cdot ZF=ZC \cdot ZG$ So $Z$ is on the radical axis of $\odot BEK, \odot CKD$ which is just the line $KN$ . Hence if $AKN$ colinear then $AKZ$ colinear $\implies AK$ is the $A-\text{symedian}$ and if $K$ on $A-\text{symedian}$ then as $AKZ$ colinear and $KNZ$ colinear it follows $AKN$ colinear.

This post has been edited 1 time. Last edited by sbealing, Mar 3, 2018, 9:29 PM

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#5 h Mar 3, 2018, 10:26 PM • 2 Y

Y by Adventure10, Mango247

Let $X$ be a point of intersection of tangents at $B$ and $C$ to $k_{ABC}$ . Let $\angle BAC=a,\angle KBC=x,\angle KCB=z,\angle EBK=y,\angle DCK=y$ . From $\triangle BEC,BDC$ we have that $a+x+y+z=180$ . Let $XB,XC$ intersect circles around $BKE,DKC$ at $Y,Z$ . Since $\angle YKZ=180-a-x-y+180-x-z+180-a-z-y=180$ and $\angle BYK=\angle KYZ=a$ we have that $YZ\parallel BC$ so $XB\cdot XY=XC\cdot XZ$ so $X,K,N$ are collinear and on radical axis. $A$ is on $KN$ if and only if $A$ is on $KX$ which is equivalent to $K$ being on $AX$ which is a symmedian.

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145 posts

#6 h Mar 4, 2018, 5:36 AM • 1 Y

Y by Adventure10

Simple problem. Just Pascal.
Let the tangents to $(ABC)$ at $B$ and $C$ intersect at $T$ and let $EB \cap DC = P$ .

Applying Pascal's Theorem for $EBBDCC$ , we get that $K, T, P$ are collinear.

Now since $P$ lies on the radical axis of $(BKE)$ and $(CKD)$ , so, $K, N, T, P$ are collinear.

Now if $K$ belongs to the $A$ -symmedian i.e. belongs to $AT$ , then $A, K, N$ are collinear.

And if $A, K, N$ are collinear, then the fact that $K, N, T$ are collinear implies that $K$ belongs to $AT$ i.e. belongs to the $A$ -symmedian.

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118 posts

Wictro

#7 h Mar 4, 2018, 5:23 PM • 2 Y

Y by Elnuramrv, Adventure10

Let the tangents at $B$ and $C$ meet at $L$ .
$\angle{BNK}=\angle{BEC}=\angle{BDC}=\angle{KNC}=\angle{BAC}$
Thus $\angle{BNC}=2 \angle{BAC}$ and $NK$ is the angle bisector of $\angle{BNC}$ .
Hence we get $B,N,O,C,L$ concyclic. $KN$ must meet the perpendicular bisector of $BC$ on this circle and that point is $L$ .
So $K,N,L$ are collinear, and $A,K,N$ are collinear if and only if $A,K,L$ are collinear or equivalently if $K$ belongs to the $A-symmedian$ of $\triangle{ABC} \blacksquare$

This post has been edited 2 times. Last edited by Wictro, Mar 4, 2018, 5:24 PM
Reason: .

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#8 h Apr 4, 2018, 10:16 PM • 2 Y

Y by Adventure10, Mango247

Define $X=(BKE)\cap AB$ and $Y=(CKD)\cap AC.$ Simple angle chasing yields $KX||AY$ and $KY||AX.$ Therefore $AXKY$ is a parallelogram $\Rightarrow AK$ bisects $XY.$ $\Rightarrow K\in A-symmedian \Leftrightarrow BXYC \Leftrightarrow AX.AB=AY.AC \Leftrightarrow$ A lies on the radical axis of $(BKE)$ and $(CKD) \Leftrightarrow A,K,N$ are collinear.

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PROF65

#10 h Apr 4, 2018, 11:15 PM • 2 Y

Y by hellomath010118, Adventure10

First sol
$\angle BNK=\angle BEK,\angle KNC=\angle KDC \implies \angle BNC=\angle BOC$ and $\angle BNK =\tfrac{1}{2}\angle BOC$ thus $N$ belongs to $(OBC)$ and $KN$ pass through the intersection of the tangents from $B,C$ which is the antipode of $O$ in $(OBC)$ so $A,K,N \iff$ $AK$ is a symmedian.
second sol just by pascal with $BBECCD$ ...
RH HAS

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293 posts

#11 h Jun 7, 2018, 7:08 PM • 1 Y

Y by Adventure10

It is easy to see that on AL exists point N such that NBL is similiar to NLC.Now just see that N is on (BKE) and (CKD) and it is trivial now.(really simple approach).

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#12 h Jun 7, 2018, 9:32 PM • 2 Y

Y by Adventure10, Mango247

Let $L$ be the pole of $BC$ (which is on the $A$ -symmedian). Let $P = BE \cap CD$ , $Q = BC \cap DE$ . Since $PB \cdot PE = PC \cdot PD$ , $P$ lies on $KN$ , the radical axis of $(BKE)$ and $(CKD)$ . From Brocard's theorem, $PK$ is the polar of $Q \in BC$ , so $PK$ passes through $L$ , from which the result follows.

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3001 posts

#13 h Jul 7, 2019, 4:31 AM • 2 Y

Y by Adventure10, Mango247

Not sure if this has been posted, but here's an angle chasing solution

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1184 posts

L567

#14 h Apr 19, 2021, 6:59 AM • 1 Y

Y by Mango247

Similar to @above's solution

First the "if" direction, suppose $A,N,K$ are collinear. Let $AK \cap (ABC) = Z$

Then, $\angle AZB = \angle ACB$ and $\angle BNZ = \angle BNK = \angle BEK = \angle BEC = \angle BAC$ and similarly, we get that $\angle ZNC = \angle BAC, \angle NZC = \angle ABC$ . So, $N$ is the center of spiral similarity taking $BZ$ to $ZC$ . So, this means that $N$ is the midpoint of the symmedian chord of $\triangle BZC$ and so lies on the $A$ symmedian.

Now the "only if" direction. Suppose $AK$ is a symmedian and let $AK \cap (ABC) = Z$ . Let $N'$ be the $A$ dumpty point, which obviously lies on $AKZ$ .

Since $\angle BNK = \angle BNZ = \angle BAN + \angle NBA = \angle BAN + \angle NAC = \angle BAC = \angle BAC = \angle BEK$ , $N$ lies on $(BEK)$ , by symmetry it lies on the other circle as well and so $N'=N$ and so $A,N,K$ are collinear

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