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Really nice problem
ken3k06   7
N May 24, 2022 by KhongMinh
Let $\displaystyle ABC$ be a triangle with circumcenter $\displaystyle O$. $\displaystyle X,Y$ are midpoints of $\displaystyle AC,AB$. $\displaystyle AO$ intersects $\displaystyle BC$ at $\displaystyle K$. Let $\displaystyle E,F$ be the circumcenter of $\displaystyle ( AXK)$ and $\displaystyle ( AYK)$. Prove that $\displaystyle YE,XF$ and tangent line of $\displaystyle ( ABC)$ at $\displaystyle A$ are concurrent
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7 replies
ken3k06
May 17, 2022
KhongMinh
May 24, 2022
Really nice problem
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ken3k06
424 posts
#1 • 6 Y
Y by tiendung2006, ImSh95, GioOrnikapa, nervy, Siddharth03, Miku_
Let $\displaystyle ABC$ be a triangle with circumcenter $\displaystyle O$. $\displaystyle X,Y$ are midpoints of $\displaystyle AC,AB$. $\displaystyle AO$ intersects $\displaystyle BC$ at $\displaystyle K$. Let $\displaystyle E,F$ be the circumcenter of $\displaystyle ( AXK)$ and $\displaystyle ( AYK)$. Prove that $\displaystyle YE,XF$ and tangent line of $\displaystyle ( ABC)$ at $\displaystyle A$ are concurrent
https://scontent.fdad3-6.fna.fbcdn.net/v/t1.15752-9/280468139_309772254660117_4396497369607029203_n.png?_nc_cat=104&ccb=1-6&_nc_sid=ae9488&_nc_ohc=Ds-qw-s9A4cAX-hPD5o&_nc_ht=scontent.fdad3-6.fna&oh=03_AVJFPgADVRrIUm2QYJqlhkHe3ujVmewy7OtbQrCjNMJZQA&oe=62A7FFE3
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ThisNameIsNotAvailable
442 posts
#12 • 1 Y
Y by Mango247
Any idea? Is this problem so hard? I think we should change the model into incenter. In other word, the tangent lines to $(O)$ at $A,B,C$ cut each others at $M,N,P$. Then I found that $NF,PE$ and $OA$ are concurrent.
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bryanguo
1032 posts
#13 • 2 Y
Y by channing421, centslordm
The lines seem to concur at the $K$-antipode of $(AKC)$ (below marked as $N$). Here's a diagram:
[asy]
unitsize(80);
pair A,B,C,O,X,Y,K,E,F,N;
A=dir(110); B=dir(215); C=dir(-35); O = circumcenter(A,B,C); X = midpoint(A--B); Y = midpoint(A--C); K = extension(A,O,B,C); F = circumcenter(A,X,K); E = circumcenter(A,Y,K); N = extension(Y,F,X,E); 

draw(A--B--C--cycle);
draw(X--K); draw(A--K);
draw(A--K); draw(K--Y);
draw(circumcircle(A,X,K), red);
draw(circumcircle(A,Y,K), red);
draw(circumcircle(A,B,C), red);
draw(circumcircle(A,N,K), red + linetype("4 4"));
draw(Y--N); draw(A--N);
draw(N--C); draw(E--N);
dot("$O$", O, NW * dir(10));
dot("$A$", A, N * dir(40));
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$X$", X, NW);
dot("$Y$", Y, E * dir(-40));
dot("$K$", K, SE);
dot("$E$", E, E);
dot("$F$", F, NW);
dot("$N$", N, NE);
[/asy]
This post has been edited 1 time. Last edited by bryanguo, May 24, 2022, 3:24 AM
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ThisNameIsNotAvailable
442 posts
#14
Y by
bryanguo wrote:
The lines seem to concur at the $K$-antipode of $(AKC)$ (below marked as $N$).

I don't think so.
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SerdarBozdag
892 posts
#15
Y by
@2above it is not symmetric thus not true. This problem is very hard.
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KhongMinh
15 posts
#16 • 4 Y
Y by ken3k06, ThisNameIsNotAvailable, Siddharth03, SerdarBozdag
We use inversion
$F=S_{l_A}\circ I_A^{AB.AC}$
We need to solve: Let $ABC$ be a triangle with circumcentre $O$. $D,E$ are the midpoints of $AB, AC$. $F$ in $(ADE)$ such that $OF//DE$. $OD, OE$ intersect $CF, BF$ at $H,I$. Prove that $HI//AF$.
Let $OH,OI$ intersect $AF$ at $X,Y$
Let the orthecentre of $ABC$ is $J$. $P$ is a point such that $FP//OH$, $FP$ intersects $BC$ at $R$.
$F(HO,X)=F(HO,XP)=F(CO,SR)=F(SR,C)=(SR,C)=\frac{\overline{CS}}{\overline{CR}}=\frac{\overline{JS}}{JF}=...=(IO,Y)$
We have $\frac{\overline{XH}}{\overline{XO}}=\frac{\overline{YI}}{YO}$. So $HI//XY$
Q.E.D
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This post has been edited 1 time. Last edited by KhongMinh, May 24, 2022, 9:49 AM
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laikhanhhoang_3011
637 posts
#17 • 2 Y
Y by ThisNameIsNotAvailable, ken3k06
KhongMinh wrote:
We need to solve: Let $ABC$ be a triangle with circumcentre $O$. $D,E$ are the midpoints of $AB, AC$. $F$ in $(ADE)$ such that $OF//DE$. $OD, OE$ intersect $CF, BF$ at $H,I$. Prove that $HI//AF$.

Can you explain this ? :(
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KhongMinh
15 posts
#18 • 1 Y
Y by laikhanhhoang_3011
laikhanhhoang_3011 wrote:
KhongMinh wrote:
We need to solve: Let $ABC$ be a triangle with circumcentre $O$. $D,E$ are the midpoints of $AB, AC$. $F$ in $(ADE)$ such that $OF//DE$. $OD, OE$ intersect $CF, BF$ at $H,I$. Prove that $HI//AF$.

Can you explain this ? :(
$X,Y$ are the midpoint of $AB,AC$
Let $YF$ intersects $XE$ at $S$
After the transformation $F=S_{l_A}\circ I_A^{AB.AC}$ we have:
$E',F'$ are the reflection of $A$ about $Y'K',X'K'$. so $(AE'X')$ intersects $(AF'Y')$ at $S'$
We need to prove that $AS'//B'C'$ but $AS'$ is the radical axis of 2 circles. So we have the new problem
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