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Source: 2018 China Southeast MO Grade 10 P6

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1595 posts

#1 h Jul 31, 2018, 4:33 AM • 4 Y

Y by Ifhml, mathematicsy, Adventure10, Mango247

In the isosceles triangle $ABC$ with $AB=AC$ , the center of $\odot O$ is the midpoint of the side $BC$ , and $AB,AC$ are tangent to the circle at points $E,F$ respectively. Point $G$ is on $\odot O$ with $\angle AGE = 90^{\circ}$ . A tangent line of $\odot O$ passes through $G$ , and meets $AC$ at $K$ . Prove that line $BK$ bisects $EF$ .

This post has been edited 1 time. Last edited by MarkBcc168, Jul 31, 2018, 7:29 AM

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#2 h Jul 31, 2018, 4:58 AM • 1 Y

Y by Adventure10

This is also Grade 11 P5.

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1702 posts

#4 h Jul 31, 2018, 8:27 AM • 2 Y

Y by Adventure10, Mango247

Nice projective problem indeed

Solution

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68 posts

#5 h Jul 31, 2018, 3:44 PM • 2 Y

Y by Adventure10, Mango247

This also works

:
My Solution

Probably a trivial extension of this problem :
"If $S$ is the projection of $E$ on $BC$ and $T$ is the midpoint of $FG$ , then $ASOT$ is cyclic."

This post has been edited 1 time. Last edited by MeineMeinung, Jul 31, 2018, 3:52 PM
Reason: adding extension

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165 posts

#6 h Aug 1, 2018, 3:25 AM • 2 Y

Y by Adventure10, Mango247

My Solution(a bit ugly)

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1595 posts

#7 h Aug 3, 2018, 9:40 AM • 2 Y

Y by Adventure10, Mango247

Let $E'$ be the antipode of $E$ w.r.t. $\odot O$ . Clearly $A, G, E'$ are colinear so quadrilateral $EGFE'$ is harmonic. Taking pencil at $F$ , rotate it $90^{\circ}$ , translate it to $O$ and project it to $AC$ gives
$-1 = F(E, F; G, E') = O(A, F; K, C) = (A, F; K, C)$ Moreover, by taking pencil at $B$ ,
$-1 = B(A, F; K, C) = (E, F; BK\cap EF, {\infty}_{EF}) = -1$ hence $BK$ bisects $EF$ as desired.

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844 posts

#8 h Aug 23, 2019, 9:47 PM • 2 Y

Y by Adventure10, Mango247

Redefine $K = BM \cap AC$ where $M$ is the midpoint of $EF$ . Let the tangent to $\odot(O)$ at $K$ , other than $\overline{AC}$ meet $\odot(O)$ and $\overline{AB}$ at $X$ and $Y$ , respectively. Note that if $P_{\infty}$ denotes the point at infinity on $\overline{AC}$ then $BP_{\infty}$ is tangent to $\odot(O)$ . By Brianchon's theorem, $\overline{YM} \parallel \overline{AC}$ so $AE = EY = XY \implies AXE = 90^\circ$ so $X = G$ and $BK$ bisects $EF$ as required.

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1576 posts

#9 h Sep 8, 2019, 3:41 AM • 1 Y

Y by Adventure10

Denote $EM\cap \Gamma =E'$ . It is clear that $E'$ lies on $AG$ , as $\angle EGE'=90$ . Also, define $EK\cap \Gamma=K'$ , $EB\cap \Gamma=B'$ . Note that, by symmedians, we have that quadrilaterals $GFE'E$ , $FB'E'E$ , and $FK'GE$ are all harmonic. By Ptolemy's, $K'E=\frac{FK'\cdot GE+K'G\cdot EF}{GF}=\frac{2EF\cdot K'G}{GF}$ , as product of opposite sides are equal in harmonic quads. Similarly, $B'E=\frac{2B'E'\cdot EF}{E'F}$ , so $\frac{K'E}{B'E}=\frac{K'G\cdot E'F}{B'E'\cdot FG}=\frac{K'G}{B'E'}\cdot\frac{E'E}{GE}=\frac{K'F}{EF}\cdot\frac{EF}{FB'}=\frac{FK'}{FB'}$ Therefore, $FK'EB'$ is harmonic. Now, $(FE;B'K')\stackrel{E}{=}(FA;BK)\stackrel{C}{=}(FE;P_\infty,KC\cap EF)$ as desired.

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55 posts

#10 h Sep 8, 2019, 5:55 AM • 1 Y

Y by Adventure10

There is also a relatively short solution using complex numbers.
WLOG, we may assume that $\odot O$ is the unit circle and $BC$ lies over the real axis. It is immediately obvious that $f=\tfrac{-1}{e}$ and $b=-c$ . Then,
$a=\frac{-2e}{e^2-1}\quad \text{and}\quad b=-c=\frac{2e}{e^2+1}$ Since $\angle EGA=90^{\circ}$ ,
$\frac{g+\frac{2e}{e^2-1}}{\frac{1}{g}+\frac{2e}{e^2-1}}=ge\Rightarrow g=\frac{e(e^2-3)}{3e^2-1}$ and
$k=\frac{\frac{-e^2+3}{3e^2-1}}{\frac{-1}{e}+\frac{e(e^2-3)}{3e^2-1}}=\frac{e(-e^2+3)}{e^4-6e^2+1}$ By computation, $BK$ bisects $EF$ if and only if
$\frac{(e^4-6e^2+1)(e^4-4e^2-1)}{(e^2+1)(e^6-7e^4+7e^2-1)}=\overline{\left(\frac{(e^4-6e^2+1)(e^4-4e^2-1)}{(e^2+1)(e^6-7e^4+7e^2-1)}\right)}$ which is true. $\blacksquare$

This post has been edited 3 times. Last edited by MrOrange, Jan 1, 2020, 8:18 PM

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779 posts

#11 h Apr 4, 2020, 8:56 PM

Y by

Haven't seen this solution yet (not sure it works though): Solution

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algebra_star1234

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algebra_star1234

#12 h Jun 13, 2020, 11:36 PM • 3 Y

Y by Mango247, Mango247, Mango247

Extend $AG$ to meet $\Gamma$ at $E'$ . Then, $\angle E'GE = 90^{\circ}$ , so $E', M,$ and $E$ collinear. Now since $AB = AC$ , $E'$ is the reflection of $F$ across $BC$ , so $BE'$ is tangent to $\Gamma$ . Let $E'K$ meet $\Gamma$ at $L$ . We also know that $E'FLG$ is harmonic, so $-1 = (GF;LE') \stackrel E' = (AF;KB)$ . Let $CK$ meet $EF$ are $M$ . We know that $-1 = (AF;KB) \stackrel C = (EF;M \infty)$ , which implies $M$ is the midpoint of $EF$ .

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anser

#13 h Aug 30, 2020, 5:19 PM

Y by

$[asy] import markers; unitsize(2inch); pair A, B, C, O, E, F, G, J, K, L, X, Y, M, N; A = dir(90); B = dir(205); C = dir(335); draw(B--A--C); draw(B--C, lightblue); O = .5B + .5C; E = foot(O, A, B); F = foot(O, A, C); draw(E--F, lightblue); X = .5E + .5F; J = 2*foot(F, B, C) - F; L = 2*foot(E, B, C) - E; G = extension(A, J, X, L); draw(circumcircle(A, G, E), dashed+lightmagenta); draw(circumcircle(E, F, J), dashed+lightmagenta); draw(A--O); draw(E--G); Y = extension(E, G, A, X); K = extension(.5A+.5E, G, A, F); draw(K--B, dotted); draw(G--L, lightred); draw(G--F); draw(L--O); M = .5A+.5E; draw(G--M--X, green); draw(G--K); draw(K--Y, lightblue); draw(K--O, lightred); N = extension(K, O, F, X); draw(G--O--F, green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(200)); dot("$F$", F, dir(F)); dot("$O$", O, dir(O)); dot("$X$", X, dir(315)); dot("$G$", G, dir(40)); dot("$Y$", Y, dir(135)); dot("$K$", K, dir(K)); dot("$M$", M, dir(M)); dot("$N$", N, dir(310)); [/asy]$
Let the midpoint of $EF$ be $X$ and let $EG\cap AX = Y$ . Since $OE$ and $OG$ are tangents to $(AGE)$ , $(A, X; O, Y) = -1$ and $M, G, K$ are collinear. From $\angle GOF = 2\angle GEF = \angle GMX$ , $\angle OGF = \angle MGX\implies \angle XGF = \angle MGO = 90^{\circ}$ . Then $GX$ and $FO$ concur on $\odot O$ , and a homothety at $F$ by factor $1/2$ shows that $KO$ bisects $XF$ . Call $N$ the midpoint of $XF$ .

Since $-1 = (A, X; O, Y) \stackrel{K}{=} (F, X; N, KY\cap EF)$ , $KY$ is parallel to $EF$ . Finally, $-1 = (A, X; O, Y) \stackrel{K}{=} (C, KX\cap BC; M, P_{\infty})$ , so $KX\cap BC = B$ .

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2786 posts

#14 h Sep 28, 2020, 5:51 AM • 1 Y

Y by Frestho

Let $N=EF\cap CK$ ; we want to prove $(E, F; N, P_{\infty})=-1$ . But we have $(E, F; N, P_{\infty})\overset{C}=(A, F; K, B)\overset{G}=(L, F; G, X)$ , where $L=AG\cap \omega, X=BG\cap \omega$ . And note $L, E$ are reflections over $M$ , so thus $BL$ tangent to $\omega$ , meaning $GFXL$ is harmonic and we're done.

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404 posts

#15 h Feb 12, 2021, 12:22 AM • 4 Y

Y by Mathscienceclass, Mango247, Mango247, Mango247

Let $M$ be the midpoint of $\overline{BC}$ , $N$ be the midpoint of $\overline{AE}$ , $E'$ be the antipode of $E$ , $X = \overline{BE'} \cap \overline{FE}$ and $Y = \overline{KC} \cap \overline{FE}$ .

Note that $X$ is on the polar of $A$ and thus $\overline{E'A}$ is the polar of $X$ . $\angle AGE + \angle EGE' = 180$ so $A,G,E'$ are collinear and the tangent at $G$ passes through $X$ .
The tangent at $G$ also passes through $N$ so $X,K,G,N$ are collinear.

$(X, Y; F, E) \overset{K}= (N,C;A,E) \implies \frac{XF}{XE} \frac{YE}{YF} = \frac{CE}{CA} \implies \frac{YE}{YF} = \frac{CE}{XF} \frac{XE}{CA}$ .

Notice that $CE = BF$ and $XE=BC$ . So, we have $\frac{YE}{YF} = \frac{BF}{XF} \cdot \frac{BC}{CA} = -1$ by similar triangles.

Thus, $YE = -YF$ and we are done.

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90 posts

#16 h Feb 13, 2021, 12:20 AM • 1 Y

Y by Math-48

Let the parallel line at $A$ to $BC$ intersect the perpendicular line from $E$ to $BC$ at $X, D=AO\cap BK$
Claim: $X,G,F$ are collinear
Proof. Note that $A,X,E,G$ are cocyclic
So $\angle{AGX}=\angle{AEX}=90-\angle{AEF}=\frac12\hat{A}$
$\angle{AGF}=360-90-\angle{EGF}=90+\hat{B}=180-\frac12\hat{A} \quad \blacksquare$
$Polar(B)=EX$
$Polar(K)=GF$
hence $Polar(BK)=X$
$Polar(AO)=P_\infty$
So $Polar(D)=XP_\infty$
Wich passes from $A$
Hence $D\in Polar(A)=EF$ and immediately $D$ is the midpoint of $EF$ as desired.

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Grizzy

#17 h May 9, 2021, 1:35 AM

Y by

Let $AG$ intersect $\Gamma$ at $F'$ , and let $M'$ be $CK \cap EF$ . Then

$\angle F'GE = \angle AGE = 90^{\circ},$
so $EF'$ is a diameter, which implies that $\angle EFF' = 90^{\circ}$ . Then since $EF || BC$ , this implies that $F'$ is the reflection of $F$ across $BC$ , which implies that $BF'$ is tangent.

Now, if we let $T$ be $F'K \cap \Gamma$ , we know that $F'FTG$ is a harmonic quadrilateral since the intersection of the tangents at $F$ and $G$ is $K$ . Then

$-1 = (G, F; T, F') \stackrel{F'}{=} (A, F; K, B) \stackrel{C}{=} (E, F; M', \infty)$
so $M'$ is the midpoint of $EF$ , as desired. $\square$

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351 posts

#18 h Aug 16, 2021, 10:29 PM • 1 Y

Y by Bradygho

Let $E'$ be the antipode of $E$ in $\Gamma$ . Let $L \neq E'$ be an intersection of $\overline{E'K}$ with $\Gamma$ . By a well-known lemma, $E'FLG$ is harmonic. Since $\angle AGE=\angle EGE'=90^\circ$ , we find that $A$ , $G$ , and $E'$ are collinear. Also, we have $\triangle BE'M \cong \triangle CEM$ , so $\overline{BE'}$ is tangent to $\Gamma$ . Let $\overline{EF}$ and $\overline{CK}$ intersect at $N$ , and let $\infty$ be the point at infinity on $\overline{BC}$ . Then, we have $-1=(FG;E'L)\overset{E'}{=}(FA;BK)\overset{C}{=}(FE,\infty N),$ so $N$ is the midpoint of $\overline{EF}$ .

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#19 h Aug 27, 2021, 5:20 PM

Y by

Solved with L567, minakshee, Psyduck909, Pranav1056, KilleR_1234.

Consider $EF \cap KC = D$
$\textcolor{red}{Claim:}$ $(A,F;K,B)=-1$

$\textcolor{blue}{Proof:}$ Let $AG \cap \Gamma = Z$ and $KZ \cap \Gamma = X$ . So now we know $XFZG$ is harmonic. Taking perspectivity at $Z$ gives,
$(G,F;X,Z) \stackrel{Z}{=} (A,F;K,B) = -1$
Take $FE \cap BC = P_{\infty}$ . Now to finish we take perspectivity at $C$ giving us :
$(A,F;K,B) \stackrel{C}{=} (E,F;D,P_{\infty})=-1$ which implies that $D$ is the midpoint of $EF$ , as desired.

Attachments:

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#20 h Aug 30, 2021, 3:54 PM • 3 Y

Y by Mango247, Mango247, Mango247

Storage

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BVKRB-

#21 h Sep 7, 2021, 11:03 AM

Y by

Oof, I had thought about $Z$ and $X$ motivated my so many tangents but did not think about taking perspectivity at $Z$ :wallbash_red:

:wallbash_red:

Once you discover that $EF$ and $BC$ are parallel, Projective is quite motivated (EGMO

)
A Different Solution

This post has been edited 4 times. Last edited by BVKRB-, Sep 7, 2021, 11:07 AM
Reason: Experimenting with asy :)

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1080 posts

#22 h Jan 16, 2022, 4:31 AM • 2 Y

Y by centslordm, megarnie

Let $E'$ be the antipode of $E$ with respect to $\omega$ and note that $E'$ lies on $\overline{AG}$ as $\angle E'EA=\angle AGE=90.$ Let $f(P,\ell)$ be the reflection of $P$ over $\ell.$ Since $E'=f(f(E,\overline{AO}),\overline{BC})$ and $F=f(E,\overline{AO}),$ we know $E'=f(F,\overline{BC}).$ Hence, $\overline{CE'}$ is tangent to $\omega$ at $E'.$ Therefore, $-1=(G,F;\overline{FG}\cap\overline{KE'},E')\stackrel{E'}=(A,F;K,C)\stackrel{B}=(E,F;M,P_{\infty})$ where $P_{\infty}$ is the point at infinity along $\overline{BC}.$ Noticing $\overline{BC}\parallel\overline{EF}$ finishes. $\square$

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#23 h Jan 19, 2023, 2:38 AM • 1 Y

Y by centslordm

$[asy] size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen greenfill,greendraw,turquoisedraw,lightbluedraw,bluedraw,purpledraw,pinkdraw; greenfill = RGB(204,255,204); greendraw = RGB(0,187,0); turquoisedraw = RGB(0,170,153); lightbluedraw = RGB(68,204,255); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); pair A,B,C,M,E,F,EE,G,K,X,L; A = (0,3); B = (-2,0); C = (2,0); M = (B+C)/2; E = foot(M,A,B); F = foot(M,A,C); path c = circle(M,distance(M,E)); EE = 2M-E; G = intersectionpoints(c,A--EE)[0]; K = extension(M,foot(M,F,G),A,C); X = extension(B,K,E,F); L = intersectionpoints(c,EE--K)[0]; filldraw(c,greenfill,greendraw); draw(A--B--C--cycle,bluedraw); draw(A--EE,pinkdraw); draw(E--EE,turquoisedraw); draw(E--F,purpledraw); draw(B--K,lightbluedraw); draw(G--K,lightbluedraw); draw(EE--K,lightbluedraw); draw(C--EE,bluedraw); dot("$A$",A,dir(90)); dot("$B$",B,dir(180)); dot("$C$",C,dir(0)); dot("$M$",M,dir(240)); dot("$E$",E,dir(140)); dot("$F$",F,dir(40)); dot("$E'$",EE,dir(320)); dot("$G$",G,dir(50)); dot("$K$",K,dir(50)); dot("$X$",X,dir(110)); dot("$L$",L,dir(215)); [/asy]$

Let $M$ be the midpoint of $\overline{BC}$ , let $\overline{BK}$ and $\overline{EF}$ intersect at $X$ , and let $E'$ be the reflection of $E$ over $M$ . We have $\angle AGE=\angle EGE'=90^\circ$ , so $A$ , $G$ , and $E'$ are collinear. Since $\triangle BEM \cong \triangle CE'M$ , $\overline{CE'}$ is tangent to $\omega$ . Let $\overline{KE'}$ intersect $\omega$ at $L \ne E'$ and let $\infty_{BC}$ be the point at infinity on $\overline{BC}$ . We have $-1=(G,F;L,E')\overset{E'}{=}(A,F;K,C)\overset{B}{=}(E,F;X,\infty_{BC}),$ so $X$ is the midpoint of $\overline{EF}$ . $\square$

This post has been edited 3 times. Last edited by CyclicISLscelesTrapezoid, Sep 3, 2024, 6:03 PM

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478 posts

#24 h Jun 8, 2023, 5:38 AM • 1 Y

Y by HoripodoKrishno

Nice problem. :yup:

:yup:

https://i.imgur.com/WXbFsME.png

Let $N$ be the midpoint of $AE$ , $D=AG\cap\Gamma$ and $T=GG\cap EF$ .

Now $\angle AGE=90^\circ\implies N$ is the center of $\odot(AGE)\implies NE=NG$ and as $NE$ is tangent to $\Gamma$ , we get that $NG$ is also tangent to $\Gamma$ . Thus this gives that $\overline{K-G-N}$ are collinear.

Now as $(G,D;E,F)=-1$ , this means that $TD$ is tangent to $\Gamma$ . Also, $AG\cdot AD=AE^2\implies\angle AED=\angle AGE=90^\circ\implies\overline{E-M-D}$ are collinear. Finally, $TD\perp DM\implies TD\parallel AC$ .

Now as $M$ is the midpoint of both $BC$ and $ED$ , we get that $BDCE$ is a parallelogram, which further gives that $BD\parallel AC\implies\overline{T-B-D}$ are collinear.

Thus we finally have,
$\begin{align*} -1&=(A,E;N,\infty_{AC})\\ &\overset{T}{=}(A,F;K,B)\\ &\overset{C}{=}(E,F;BC\cap EF,\infty_{BC}) ,\end{align*}$
which thus implies that $BC\cap EF$ is the midpoint of $EF$ and we are done.

This post has been edited 2 times. Last edited by kamatadu, Jun 8, 2023, 5:39 AM

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600 posts

eibc

#25 h Aug 10, 2023, 2:21 AM

Y by

Had a version that replaced $\odot O$ with $\Gamma$ .

Let $P_{\infty}$ denote the point at infinity along lines $EF$ and $BC$ , and $X$ denote the $E$ -antipode wrt $\Gamma$ . Then since $\angle EGX = 90^{\circ}$ , points $A$ , $G$ , and $X$ are collinear. Moreover, since $\Gamma$ is centered at the midpoint of $\overline{BC}$ , and $E$ and $F$ are reflections across the perpendicular bisector of $\overline{BC}$ , points $X$ and $F$ are reflections across $BC$ . So, $\overline{FX}$ is tangent to $\Gamma$ . Then to finish, note that
$-1 = (F, G; \overline{KX} \cap \Gamma, X) \overset{X}{=} (F, A; K, B) \overset{C}{=} (F, E; \overline{CK} \cap \overline{EF}, P_{\infty}),$ which implies that $\overline{CK}$ bisects $\overline{EF}$ , as needed.

This post has been edited 1 time. Last edited by eibc, Aug 10, 2023, 2:21 AM

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5001 posts

#26 h Aug 15, 2023, 6:25 PM • 1 Y

Y by centslordm

Rename $\odot O$ to $\Gamma$ . Also swap $E$ and $F$ in terms of definition. We present two distinct solutions, one with multiplicity 2.

Solution 1 (synthetic):
We will need the following claim, which will be proven in two different ways.

Claim: Let $ABCD$ be a convex cyclic quadrilateral. Then $\overline{AA} \cap \overline{BB}$ , $\overline{CC} \cap \overline{DD}$ , and $\overline{AC} \cap \overline{BD}$ are collinear.
Proof 1: This is purely projective, so employ a projective transformation sending $\overline{AC} \cap \overline{BD}$ to the center of $(ABCD)$ , hence $ABCD$ is now a rectangle and the conclusion is obvious. $\blacksquare$
Proof 2: Let $W=\overline{AA} \cap \overline{DD}$ , and define $X,Y,Z$ in cyclic fashion. By Brianchon on $WAXYCZ$ , we get that $\overline{WY}, \overline{XZ}, \overline{AC}$ are concurrent. By Brianchon on $WXBYZD$ , we get that $\overline{WY}, \overline{XZ}, \overline{BD}$ are concurrent. Hence $\overline{AC} \cap \overline{BD}$ lies on $\overline{WY}$ as desired. $\blacksquare$

For the main problem, let $E'$ and $F'$ be the antipodes of $E$ and $F$ respectively, so $G$ lies on $\overline{AE'}$ , let $N$ be the midpoint of $\overline{EF}$ , and let $\overline{EN}$ intersect $\Gamma$ again at $H$ . Since $EGFE'$ is harmonic, we have
$-1=(E,F;E'G)\stackrel{H}{=}(E,F;N,\overline{HG} \cap \overline{EF}),$ hence $\overline{GH} \parallel \overline{EF}$ . Therefore, due to symmetry, $\overline{F'G}$ passes through $N$ as well. Now apply out claim to $FGEF'$ to get that $C$ , $K$ , and $N$ are collinear, as desired. $\blacksquare$

Solution 2 (coordinates):
First we change the setup of the problem slightly. Define $E'$ as before, so $G$ lies on $\overline{AE'}$ , and once again let $N$ be the midpoint of $\overline{EF}$ . Let $K'=\overline{CN} \cap \overline{FF}=\overline{CN} \cap \overline{AB}$ . We are going to show that the point $G' \neq F$ such that $\overline{K'G'}$ is tangent to $\Gamma$ is $G$ by showing it lies on $\overline{AE}$ , which clearly implies $K'=K$ and hence solves the problem.

Place the problem in the coordinate plane with $A=(0,a),B=(-1,0),C=(1,0)$ , so the midpoint $M$ of $\overline{BC}$ is the origin. Lines $\overline{AB}$ and $\overline{AC}$ have equations $y=ax+a$ and $y=-ax+a$ respectively, so we can easily calculate $F=(-\tfrac{a^2}{a^2+1},\tfrac{a}{a^2+1})$ . It is clear that $E'$ is the reflection of $F$ over $\overline{BC}$ , hence $E'=(\tfrac{-a^2}{a^2+1},-\tfrac{a}{a^2+1})$ . Furthermore, $N=(0,\tfrac{a}{a^2+1})$ . To calculate $G'$ , we do not actually have to calculate $K'$ (although this is not hard either): instead, notice that $\overline{K'M}$ should bisect $\overline{FN}$ for homothety reasons, hence its slope is twice that of $\overline{MF}$ , so its equation is $y=-\tfrac{2}{a}x$ . Now, $G'$ is just the reflection of $F$ over $\overline{MK}$ , whose equation we know.

The foot from $F$ to $\overline{MK}$ can be found by intersecting $\overline{MK}$ with the line that has slope $\tfrac{a}{2}$ passing through $F$ , and can be calculated as $(-\tfrac{a^4+2a^2}{(a^2+1)(a^2+4)},\tfrac{2a^3+4a}{(a^2+1)(a^2+4)})$ . Since this point is the midpoint of $\overline{FG'}$ , it folows that $G'=(-\tfrac{a^4-4a^2+4a}{(a^2+1)(a^2+4)},\tfrac{3a^3+4a}{(a^2+1)(a^2+4)})$ . To show that $A,G',E'$ are collinear, we calculate the slope of $\overline{AE'}$ , which is just
$\frac{a+\frac{a}{a^2+1}}{\frac{a^2}{a^2+1}}=\frac{a^2+2}{a}.$ We will now calculate the slope of $\overline{G'E'}$ , since by scaling all the coordinates by $\tfrac{a^2+1}{a}$ beforehand this is just equal to
$\frac{\frac{3a^2+4}{a^2+4}+1}{-\frac{a^3}{a^2+4}+a}=\frac{\frac{4a^2+8}{a^2+4}}{\frac{4a}{a^2+4}}=\frac{a^2+2}{a},$ hence the desired collinearity holds and we are done. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Aug 15, 2023, 6:34 PM

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#27 h Aug 30, 2023, 2:00 AM

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Let $E'$ be the intersection of $\overline{AG}$ and $\Gamma$ .

Claim: $B$ is tangent to $\Gamma$ at $E'$ .
Proof. Note that $EE'$ is a diameter of $\Gamma$ . Angle chase to get $\measuredangle FE'M = \measuredangle FE'E = \measuredangle FEE' + \measuredangle E'FE = \measuredangle BFE' + \measuredangle E'FE = \measuredangle BFE = \measuredangle FBM.$ $\blacksquare$
As such, $(G,F;\overline{KE'} \cap \Gamma,E') \overset{E'}= (AF;KB) \overset{C}= (EF;I\infty) = -1$ and the result follows.

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#28 h Sep 20, 2023, 7:15 PM

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Let $X\neq G$ be the intersection of $AG$ and $\Gamma$ . Since $A$ , $G$ , and $X$ are collinear in that order, we find that $\angle XGE=180^\circ-\angle AGE=90^\circ$ , implying that $XE$ is a diameter of $\Gamma$ . It thus follows that $\triangle BXM\cong\triangle CEM$ , so that $\angle BXM=\angle CEM=90^\circ$ as $CE$ is tangent to $\Gamma$ at $E$ . Thus $BX$ is tangent to $\Gamma$ at $X$ .

Now let $L\neq X$ be the intersection of $KX$ and $\Gamma$ . Since $FK$ and $GK$ are tangents to $\Gamma$ and $X$ lies on $KL$ , it follows that $FXGK$ is harmonic, so that $(GF;LX)=-1$ .

Let $N=EF\cap CK$ . Projecting, we then find that
$\begin{align*} (EF;N\infty)&\overset{C}=(AF;KB)\\ &\overset{X}=(GF;LX)\\ &=-1. \end{align*}$ This forces $N$ to be the midpoint of $EF$ , implying that $CK$ indeed bisects $EF$ . $\blacksquare$

- Jörg

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#29 h Jan 6, 2024, 12:13 PM

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:furious:

This problem was copied from APMO 2016 P3 https://artofproblemsolving.com/community/c6h1243425p6362864

This post has been edited 2 times. Last edited by WLOGQED1729, Jan 6, 2024, 12:32 PM
Reason: fix link

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#30 h Jan 13, 2024, 7:18 PM

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Denote the midpoint of $EF$ as $N$ and $H=AG \cap \Gamma$ . We make a few synthetic observations:

$H$ is simply the antipode of $E$ on $\Gamma$ .
$\triangle EMN$ and $\triangle EHF$ are homothetic about $E$ , so $H$ is also the reflection of $F$ over $BC$ .
Since $\triangle KFN$ and $\triangle KBC$ are homothetic about $K$ , it remains to show $KM$ bisects $FN$ .

Thus we finish by noting the harmonic bundle
$-1 = (H, HK \cap \Gamma; FG) \overset{H}{=} (BK, FA) \overset{M}{=} (\infty, KM \cap FN; FN). \quad \blacksquare$

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#31 h Feb 8, 2024, 10:46 PM

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Let $\overline{AG} \cap \Gamma \neq G = E'$ . Then notice that $E'$ is the antipode of $E$ . Also, let $\overline{KE'} \cap \Gamma \neq E' = P$ . Then $FPGE'$ is a harmonic quadrilateral, so $-1 = (F, G; P, E') \overset{E'}= (F, A; K, B)$ . Notice that since $AB = AC$ , $EF \parallel BC$ so $-1 = (F, A; K, B) \overset{C}= (F, E; \overline{CK} \cap \overline{EF}, P_{\infty})$ so $CK$ bisects $EF$ as desired.

This post has been edited 1 time. Last edited by dolphinday, Feb 8, 2024, 10:47 PM

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#32 h Jun 16, 2024, 7:55 PM

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Let $KC \cap FE = X$ . Let $AG \cap \Gamma = E'$ . Let $KE' \cap \Gamma = W$ . Now from KF and KG tangents, we get that $(F,G;W,E') = -1$ . Now projecting trough E', we have that $(F,G;W,E')\stackrel{E'}{=}(F,A;K,B) = -1$ . Now projecting trough C, we have that $(F,A;K,B)\stackrel{C}{=}(F,E;X,P_\infty) = -1$ . From AB = AC, we get that $FE \parallel BC$ $\Rightarrow$ if L is the midpoint of FE, then $(F,E;L,P_\infty) = -1$ and since $(F,E;X,P_\infty) = -1$ its obvious that $L \equiv X$ $\Rightarrow$ X is the midpoint of FE $\Rightarrow$ CK bisects EF, which is what we wanted to prove. We are ready.

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#33 h Aug 15, 2024, 8:27 PM

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Let $E'=AG \cap \Gamma$ and $E'K \cap \Gamma = P.$ Since $(F, G; P, E')=-1$ , taking perspectivity at $E'$ and noting $BE'$ is tangent to $\Gamma$ gives $(F, A; K, B)=-1.$ Finally taking perspectivity at $C$ gives $(F, E; CK \cap EF, P_{\infty})=-1$ which implies $CK \cap EF$ is the midpoint of $EF.$

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#34 h Aug 21, 2024, 5:33 AM • 1 Y

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Let $E' \neq G = \overline{AG} \cap \Gamma$ , $D \neq E' = \overline{KE'} \cap \Gamma$ , and $N = \overline{CK} \cap \overline{EF}$ . Then, we have

$-1 = (F,G;D,E') \overset{E'}{=} (F,A;K,B) \overset{C}{=} (F,E;N, \infty). \ \square$

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