Can FE (no FE this year :c) share a point with (MAN)??

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Can FE (no FE this year :c) share a point with (MAN)??

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Source: USEMO 2021 P4

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#1 h Oct 31, 2021, 10:59 PM • 8 Y

Y by ike.chen, LuisAngel, megarnie, centslordm, HWenslawski, rama1728, tiendung2006, Rounak_iitr

Let $ABC$ be a triangle with circumcircle $\omega$ , and let $X$ be the reflection of $A$ in $B$ . Line $CX$ meets $\omega$ again at $D$ . Lines $BD$ and $AC$ meet at $E$ , and lines $AD$ and $BC$ meet at $F$ . Let $M$ and $N$ denote the midpoints of $AB$ and $AC$ .
Can line $EF$ share a point with the circumcircle of triangle $AMN?$

Proposed by Sayandeep Shee

This post has been edited 2 times. Last edited by MathLuis, Oct 31, 2021, 11:20 PM

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#2 h Oct 31, 2021, 11:00 PM • 10 Y

Y by ike.chen, LuisAngel, tigerzhang, megarnie, centslordm, Mogmog8, HWenslawski, rayfish, ihatemath123, Sedro

:eyes: $\text{[math]}$

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#3 h Oct 31, 2021, 11:01 PM • 10 Y

Y by Hermione.Potter, LuisAngel, centslordm, megarnie, jelena_ivanchic, Mathematicsislovely, Kimchiks926, math31415926535, rayfish, Aliosman

Solution

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#4 h Oct 31, 2021, 11:05 PM • 2 Y

Y by centslordm, Mathematicsislovely

Let $B_0$ be the circle with radius 0 centered at $B$ . We will attempt to show that $EF$ is the radical axis of $B_0$ and $(AMN)$ . Also, let $\ell$ be the radical axis of $B_0$ and $(AMN)$ .

$\textbf{Claim 1: }$ Let $T$ be the intersection of the tangents at $A$ and $B$ . Then, $T\in \ell$ and $T\in EF$ .
$\textbf{Proof: }$ Note that by Brokard's, $EF$ is the polar of $X$ with respect to $\omega$ . Thus, if $P,Q$ are the intersections $EF\cap \omega$ , then $PQ$ is the pole of $X$ so $XP$ and $XQ$ are tangent to $\omega$ . Thus, $AQBP$ is harmonic, which means that $T$ also lies on $PQ=EF$ . $\square$

As for $\ell$ , by the Radical Axis Theorem on $B_0, (AMN), (ABC)$ , the radical axis of $B_0$ and $(AMN)$ is concurrent with the tangents at $A$ and $B$ , so $T\in \ell$ .

$\textbf{Claim 2: }$ $\ell\perp XO$ and $EF\perp XO$ .
$\textbf{Proof: }$ . Firstly, $EF\perp BO$ since $EF$ is the polar. Radical axis is always perpendicular to the line between the centers, so $\ell \perp BO_2$ . And by considering homothety with factor $\frac12$ at $A$ , $BO_2\parallel XO$ . $\square$

Thus, we have shown that the radical axis and $\ell$ both share a point and are parallel, so they're the same line. To finish, assume $X= EF \cap (AMN)$ . Thus,
$0 = Pow_{(AMN}(X) = XB^2\Longrightarrow X=B$ Thus, $B\in (AMN)$ too, so
$0 = Pow_{(AMN})(B) = BM\cdot BA = \frac12 BA^2 \Longrightarrow BA=0\Longrightarrow B=A$ which is a contradiction and we are done. $\blacksquare$ .

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#5 h Oct 31, 2021, 11:41 PM • 3 Y

Y by Ainulindale, centslordm, Mathematicsislovely

This link might be of interest, although this thread is now the official. Anyways, my work is pasted below.

Walk-Through:

The answer is no.
Let $C_1$ be the reflection of $C$ over $B$ , the midpoint of $AD$ be $P$ , the circumcenter of $ABC$ be $O$ , $Y = AB \cap EF$ , and $\omega_b$ be the point circle $B$ .
By angle chasing, we find that $AMONP$ is cyclic.
It's easy to see $ACXC_1$ is a parallelogram.
Hence, $CDAC_1$ is a trapezoid with midline $BP$ , so $AC_1 \parallel BP \parallel CD$ .
Angle chasing implies $FB$ is tangent to $(ABP)$ . Now, POP yields that $F$ lies on the Radical Axis of $(AMONP)$ and $\omega_b$ .
Brocard's gives $-1 = (X, Y; A, B)$ and length chasing via the Cross Ratio and Midpoint Lengths yields $YM \cdot YA = YB^2$ , i.e. $Y$ also lies on the aforementioned Radical Axis.
Contradiction implies that $B$ cannot lie on $EF$ , i.e. there exists no point $Q$ on $EF$ such that $Pow_{(AMONP)}(Q) = Pow_{\omega_b}(Q) = 0.$
The previous line clearly implies that $EF$ and $(AMN)$ cannot intersect, so we're done.

Remarks: In my diagram, $E$ was inside $\omega$ and $F$ was outside $\omega$ .

By the way, I could've used analogous logic to conclude that $E$ also lies on the Radical Axis. (This would've eliminated the need to add $Y$ to my solution.)

Also, my solution was strongly motivated the definition of $X$ and NICE MO 2021/2.

This post has been edited 3 times. Last edited by ike.chen, Oct 23, 2022, 10:14 PM
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#6 h Oct 31, 2021, 11:42 PM • 4 Y

Y by centslordm, Mathematicsislovely, Kimchiks926, tiendung2006

The answer is no.

Let $EF$ intersect $AB$ at $T$ .
$[asy] size(10cm); defaultpen(fontsize(10)); pair A,B,C,X,D,E,F,T,M,N; A = dir(110); B = dir(240); C = dir(300); X = 2*B-A; D = 2*foot(circumcenter(A,B,C),X,C)-C; E = extension(B,D,C,A); F = extension(A,D,B,C); T = extension(E,F,A,B); M = (A+B)/2; N = (A+C)/2; filldraw(A--B--C--cycle,magenta+white+white+white); draw(B--X--D--cycle); draw(C--F--D); draw(T--F); draw(A--D); draw(D--C); draw(B--N); draw(circumcircle(A,B,C),red); draw(circumcircle(A,M,N),green); draw(circumcircle(A,N,B),blue); dot("$A$",A,dir(95)); dot("$B$",B,dir(B)*1.5); dot("$C$",C,dir(C)); dot("$X$",X,dir(225)); dot("$F$",F,dir(270)); dot("$E$",E,dir(75)*1.2); dot("$D$",D,dir(25)); dot("$M$",M,dir(225)); dot("$N$",N,dir(355)); dot("$T$",T,dir(175)); [/asy]$
By Ceva-Menelaus, $(A,B;T,X)=-1$ , so $\frac{AT}{TB}=2$ . Now $TM \cdot TA = \left(\frac{1}{6}AB \right) \cdot \left(\frac{2}{3}AB \right)=\left(\frac{1}{3}AB \right)^2=TB^2.$
Since $BN$ is the $A$ -midline of $\triangle ACX$ , $BN \parallel CX$ . Then $\measuredangle NBE = \measuredangle CDE = \measuredangle CDB = \measuredangle CAB = \measuredangle NAB,$ so $BE$ is tangent to $(ABN)$ . This means $EN \cdot EA = EB^2$ .

These two facts combined means that $\overline{TEF}$ is the radical axis of $(AMN)$ and degenerate circle $(B)$ . Thus, if $\overline{EF}$ were to intersect $(AMN)$ at $S$ , then the power of $S$ with respect to $(B)$ is $0$ , meaning $S=B$ , which is absurd.

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#7 h Oct 31, 2021, 11:42 PM • 2 Y

Y by Kagebaka, centslordm

okay then

The answer is negative. Suppose contrarily that there exists $P \in EF \cap (AMN)$ . Let $P^{*}$ be the inverse of $P$ wrt $\omega$ and let $O$ be the center of $\omega$ .

By Brokard's theorem, $EF$ is the polar of $X$ , and by La Hire's theorem, $X$ lies on the polar of $P$ .

It is clear that $O$ lies on $(AMN)$ by homothety; specifically, it is the $A$ -antipode. Thus $OP \perp AP$ . Also, $OP^{*} = OP \perp XP^{*}$ by definition. So $AP \parallel XP^{*}$ .

Now refocus the diagram on $A$ , $O$ , $P$ , $P^{*}$ , $X$ , and $\omega$ .

Claim. For all choices of $P$ and $X$ such that $\angle APO = 90^{\circ}$ and $X$ is on the polar of $P$ , the midpoint $B$ of $AX$ is outside $\omega$ .

Proof. Rotate and scale so that $\omega$ is the unit circle and $P$ is on the positive $y$ -axis; that is, $P=(0,h)$ for some $h \in (0, 1)$ . Then the $y$ -coordinate of $A$ is $h$ and the $y$ -coordinate of $X$ is $\frac{1}{h}$ , so the $y$ -coordinate of $B$ is $\frac{h+\frac{1}{h}}{2}$ . But by AM-GM, $\frac{h+\frac{1}{h}}{2} \geq \sqrt{h \cdot \frac{1}{h}} = 1,$ and equality doesn't occur. So the $y$ -coordinate of $B$ is greater than $1$ .

Hence $B$ lies outside $\omega$ , absurd.

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#8 h Oct 31, 2021, 11:56 PM • 3 Y

Y by centslordm, nikitadas, CT17

The answer is no.

Let $\ell$ be the line tangent to $\omega$ at $A$ . Let $d(P)$ be the distance of an arbitrary point $P$ from $\ell$ . Let $O$ and $r$ be the circumcenter and circumradius, respectively, of $\triangle ABC$ .

Claim: $R+2d(B)>OX$ .
Proof: We have $r+2d(B)>OX \Leftrightarrow r^2+4Rd(B)+4d(B)^2>OX^2 \Leftrightarrow 4Rd(B)+4d(B)^2>OX^2-r^2=XA \cdot XB=2AB^2.$ Since $d(B)=AB\sin C$ and $r=\tfrac{AB}{2\sin C}$ , this is equivalent to $2AB^2+4AB^2\sin^2C>2AB^2 \Leftrightarrow 1+2\sin^2 C>1,$ which is obviously true because $0^\circ<C<180^\circ$ .

Notice that by Brocard, $\overline{EF}$ is the polar of $X$ with respect to $\omega$ . Assume FTSOC that $\overline{EF}$ intersects the circumcircle of $\triangle AMN$ at a point $P$ . Then by La Hire's, the polar of $P$ with respect to $\omega$ passes through $X$ . Let $P'$ be the inverse of $P$ with respect to $\omega$ . Notice that $P'$ lies on $\ell$ and $\angle OP'X=90^\circ$ . That means the circle with diameter $\overline{OX}$ intersects $\ell$ . Let $Q$ be the midpoint of $OX$ . Then, we must have $d(Q) \leq \frac{OX}{2} \Leftrightarrow \frac{d(O)+d(X)}{2} \leq \frac{OX}{2} \Leftrightarrow r+d(X) \leq OX \Leftrightarrow R+2d(B) \leq OX,$ which contradicts our claim.

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#9 h Oct 31, 2021, 11:56 PM • 2 Y

Y by centslordm, Flying-Man

We claim the answer is no.
We proceed with barycentric coordinates on $\triangle ABC$ .
$A = (1:0:0), B = (0:1:0), C = (0:0:1)$ $X = (-1,2:0)$ By cevian formula:
$D = (-1:2:t)$ Plugging into formula for circumcircle:
$2a^2t-b^2t-2c^2 = 0$ $t = \frac{2c^2}{2a^2-b^2}$ Therefore, $D = (-1:2:\frac{2c^2}{2a^2-b^2})$ , and by cevian formula:
$E = (-1:0:\frac{2c^2}{2a^2-b^2})$ $F = (0:2:\frac{2c^2}{2a^2-b^2})$ $M = (1:1:0)$ and $N = (1:0:1)$ so, by plugging in:
$(AMN) = -a^2yz-b^2xz-c^2xy+(x+y+z)\left(\frac{c^2}{2}y+\frac{b^2}{2}z\right)$ By more plugging, we get that the line EF is:
$2c^2x-c^2y+(2a^2-b^2) = 0$ Then, a point on $EF$ is of the form $(x:2x+\frac{2a^2-b^2}{c^2}:1)$ , and so, for sake of contradiction, suppose a point of that form lies on $(AMN)$ . This gives that:
$-a^2\left(2x+\frac{2a^2-b^2}{c^2}\right) - b^2x - c^2x\left(2x+\frac{2a^2-b^2}{c^2}\right) + \left(x+2x+\frac{2a^2-b^2}{c^2}+1\right)\left(\frac{2c^2x+2a^2-b^2}{2}+\frac{b^2}{2}\right ) = 0$ Cleaning it up, it turns into:
$x^2c^2 + x(a^2-b^2+c^2)+a^2 = 0$ The determinant of this is:
$(a^2-b^2+c^2)^2 - 4a^2c^2 = a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2 = \frac{1}{2}\displaystyle\sum_{cyc}a^4+b^4-4a^2b^2$ Doing Ravi substitution, with $a=p+q,b=q+r,c=r+p$ , it becomes:
$\frac{1}{2}\displaystyle\sum_{cyc} (p+q)^4 +(q+r)^4 - 4(p+q)^2(q+r)^2$ Expanding, and cyclically summing, it turns into:
$-12pqr(p+q+r) < 0$ and so the quadratic has no real solutions, which means that $EF$ does not intersect $(AMN)$ and we are done.

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#10 h Oct 31, 2021, 11:59 PM

Y by

is it just me or has every solution posted sofar been different

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#11 h Nov 1, 2021, 12:06 AM • 2 Y

Y by centslordm, math31415926535

Let $\overline{BN}$ intersect $\overline{AD}$ at $G$ . Since $\overline{BN}$ is a midline of $\triangle AXC$ , $G$ must lie on $\gamma:=(AMN)$ .
$[asy] size(8cm); defaultpen(fontsize(9pt)); pair A = dir(130), B = dir(190), C = dir(-10), M = (A+B)/2, N = (A+C)/2, X = 2*B-A, D = 2*foot(origin, C, X)-C, F = extension(A, D, B, C), E = extension(A, C, B, D), G = extension(A, D, B, N); draw(A--B--C--cycle, deepcyan); draw(unitcircle, lightblue); draw(circumcircle(A, M, N), royalblue); draw(N--B--X--C, blue); draw(D--E--A--D, blue); string[] names = {"$A$", "$B$", "$C$", "$D$", "$E$", "$F$", "$G$", "$M$", "$N$", "$X$"}; pair[] points = {A, B, C, D, E, F, G, M, N, X}; pair[] ll = {dir(100), B, C, dir(-90), E, dir(230), dir(60), M, dir(30), X}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy]$
Since
$\begin{align*} \measuredangle BFG &= \measuredangle BFA, \; \text{and} \\ \measuredangle FGB = \measuredangle AGN &= \measuredangle AMN = \measuredangle ABC = \measuredangle ABF, \end{align*}$ we have $\triangle FGB \stackrel-\sim \triangle FBA$ . Since
$\begin{align*} \measuredangle NEB &= \measuredangle AEB, \; \text{and} \\ \measuredangle ENB = \measuredangle EDC &= \measuredangle BAC = \measuredangle BAE, \end{align*}$ we have $\triangle EBA \stackrel-\sim \triangle ENB$ . The two similarities imply that
$\text{Pow}(E, \gamma) = EB^2 \; \text{and} \; \text{Pow}(F, \gamma) = FB^2,$ so $\overline{EF}$ is the radical axis of $\gamma$ and the point circle at $B$ .

It is now clear that $\overline{EF}$ cannot intersect $\gamma$ . Indeed, suppose FTSOC that they share a point $P$ ; then $P$ must coincide with $B$ , which cannot happen as $\gamma$ and $\omega$ are tangent at $A$ (and therefore cannot intersect a second time).

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#12 h Nov 1, 2021, 12:32 AM • 1 Y

Y by megarnie

sketch: brocard then complex bash

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#13 h Nov 1, 2021, 1:42 AM • 4 Y

Y by centslordm, guptaamitu1, Pluto1708, L567

The answer is no.

Let $O$ be the circumcenter and $K = \overline{EF} \cap \overline{AB}$ . Note that $XK = 2AK$ from $(AB;KX) = -1$ . Also by Brocard $\overline{EF}:= \ell$ is the polar of $X$ . Thus
$d(A,\ell) + d(O,\ell) = \frac 12 d(X,\ell) + d(O,\ell) = \frac 12 \left(OX - \frac{R^2}{OX}\right) + \frac{R^2}{OX} = \frac{OX}{2} + \frac{R^2}{2OX} \ge R,$ Equality occurs iff $OX = R$ , which is clearly impossible, so $d(A,\ell) + d(O,\ell) > R$ . The conclusion follows because $(AMN)$ has diameter $\overline{AO}$ and radius $\tfrac 12 R$ .

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#14 h Nov 1, 2021, 1:45 AM • 2 Y

Y by centslordm, Twistya

Weird solution: We vary C on the circle and by Brocard EF is fixed. When C varies F lies on the chord of the line EF and thus if EF hits (AMN) then there must exist a C where F lies on the circle. However, it is easily proven that F cannot lie on (AMN) so were done.

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#15 h Nov 1, 2021, 1:52 AM • 1 Y

Y by centslordm

USEMO 2021/4 wrote:

Let $ABC$ be a triangle with circumcircle $\omega$ , and let $X$ be the reflection of $A$ in $B$ . Line $CX$ meets $\omega$ again at $D$ . Lines $BD$ and $AC$ meet at $E$ , and lines $AD$ and $BC$ meet at $F$ . Let $M$ and $N$ denote the midpoints of $AB$ and $AC$ . Can line $EF$ share a point with the circumcircle of triangle $AMN?$

Proposed by Sayandeep Shee

Cute problem. $[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.207529043594896, xmax = 8.042658785105354, ymin = -4.969449056510044, ymax = 4.196591514489198; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0); pen wwwwww = rgb(0.4,0.4,0.4); draw((-4.04,1.66)--(-4.845950413223141,-1.1416528925619822)--(1.16,-1.04)--(-4.04,1.66), zzttqq); /* draw figures */ draw((-4.04,1.66)--(-4.845950413223141,-1.1416528925619822), zzttqq); draw((-4.845950413223141,-1.1416528925619822)--(1.16,-1.04), zzttqq); draw((1.16,-1.04)--(-4.04,1.66), linewidth(1.2) + zzttqq); draw(circle((-1.8532151311495584,-0.4858217340658165), 3.0637524340296283)); draw((-5.6519008264462824,-3.9433057851239646)--(1.16,-1.04)); draw((-5.6519008264462824,-3.9433057851239646)--(-4.845950413223141,-1.1416528925619822), linetype("2 2")); draw((-3.540215916073293,-3.0432828439732917)--(-4.04,1.66)); draw((-8.282082289050582,3.862619650083956)--(-0.8226509209570745,-4.332945337223033), ccqqqq); draw((-8.282082289050582,3.862619650083956)--(-3.540215916073293,-3.0432828439732917)); draw((-8.282082289050582,3.862619650083956)--(-4.04,1.66)); draw(circle((-2.9466075655747797,0.5870891329670916), 1.5318762170148141), wwwwww); draw((-4.845950413223141,-1.1416528925619822)--(-2.9466075655747797,0.5870891329670916), blue); draw((-5.200695873940244,0.4771467675181019)--(-4.845950413223141,-1.1416528925619822), wwwwww); draw((-5.200695873940244,0.4771467675181019)--(-4.04,1.66), wwwwww); /* dots and labels */ dot((-4.04,1.66),dotstyle); label("$A$", (-4.278378029319907,1.8074104476221822), NE * labelscalefactor); dot((-4.845950413223141,-1.1416528925619822),dotstyle); label("$B$", (-5.289798014293612,-1.2874806117316849), NE * labelscalefactor); dot((1.16,-1.04),dotstyle); label("$C$", (1.220720392919018,-1.2822965396054641), NE * labelscalefactor); dot((-5.6519008264462824,-3.9433057851239646),dotstyle); label("$X$", (-5.586191703249286,-3.7973979671035836), NE * labelscalefactor); dot((-4.442975206611571,0.25917355371900885), dotstyle); label("$M$", (-4.38408802180676,0.3799123259091855), NE * labelscalefactor); dot((-1.44,0.31), dotstyle); label("$N$", (-1.3788288182004445,0.42499121396328016), NE * labelscalefactor); dot((-3.540215916073293,-3.0432828439732917), dotstyle); label("$D$", (-3.4825102607248652,-2.925872798057754), NE * labelscalefactor); dot((-3.7442698844626277,-1.123006549398516), dotstyle); label("$F$", (-3.6778521089592755,-1.1025069077497164), NE * labelscalefactor); dot((-8.282082289050582,3.862619650083956), dotstyle); label("$E$", (-8.215793506404811,3.956170778200693), NE * labelscalefactor); dot((-1.8532151311495584,-0.4858217340658165), dotstyle); label("$O$", (-1.9995651067053287,-0.37140247499239165), NE * labelscalefactor); dot((-2.9466075655747797,0.5870891329670916), dotstyle); label("$O'$", (-3.281458420003602,0.6504908383058794), NE * labelscalefactor); dot((-4.234138874847963,-0.584794861689729), dotstyle); label("$H$", (-4.473719877554317,-0.4615602511005809), NE * labelscalefactor); dot((-5.200695873940244,0.4771467675181019), dotstyle); label("$I$", (-5.435402822708339,0.5902804701616271), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $O$ be the center of $\omega$ and $O'$ be the center of $(AMN)$ .
Claim 01. $BO' \perp EF$ .
Proof. By Brocard's Theorem, $EF \perp OX$ . Furthermore, homothety on $A$ with scale $\frac{1}{2}$ sends $XO$ to $BO'$ , which implies $EF \perp BO'$ , as desired.

Further note that $EF$ is polar of $X$ wrt $\omega$ . Therefore $I = AA \cap BB$ must lie on $EF$ as well by La Hire's Theorem.

Claim 02. $EF$ is the radical axis of circle of radius zero $B$ and $(AMN)$ .
Proof. Note that $I$ lies on the radical axis of the mentioned circles, Furthermore, this radical axis must also be perpendicular to the line joining the two centers, i.e. $BO'$ . Since $EF$ passes through $I$ and is perpendicular to $BO'$ , we get the desired result.

To finish this, just note that $BH > 0$ or else $E,B,F$ are collinear, which is impossible as this implies $BD \equiv BE \equiv BF \equiv BC$ are the same line. Therefore, since $H \in EF$ , we then have
$(HO')^2 - r^2 = \text{Pow}(H,(AMN)) = \text{Pow}(H,B) = HB^2 > 0$ where $r$ is the radius of $(AMN)$ . This implies $O'H > r$ , which means $EF$ doesn't intersect the circle because $O'H$ is the shortest distance between $O'$ and line $EF$ .

Motivational Remark. The configuration is really similar to that of a complete quadrilateral, so Brocard might gives us something useful. To prove that $EF$ can't intersect $(AMN)$ , it then suffices to prove that $O'H \le r$ , from which we could just show that the power of $H$ wrt $(AMN)$ is positive. However, it is easy to deduce that $EF$ is radical axis of $B$ and $(AMN)$ , and since $HB > 0$ , we are done.

This post has been edited 1 time. Last edited by IndoMathXdZ, Nov 1, 2021, 3:28 AM
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#16 h Nov 1, 2021, 3:18 AM • 1 Y

Y by centslordm

Let $(AMN)$ and $AD$ intersect at $T$ . Note that by homothety, $T$ is the midpoint of $AD$ and thus $BT//XD$ . This gives $\angle TBF=\angle BCD=\angle BAD$ , thus $FB^2=FN \times FA$ . Similarly, $EB^2=EN\times EA$ . Thus, $EF$ is the radical axis of the circle with center $B$ and radius $0$ , and $(AMN)$ . Since $B$ is outside $(AMN)$ , the radical axis cannot intersect $(AMN)$ .

This post has been edited 1 time. Last edited by gghx, Nov 1, 2021, 3:19 AM

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#17 h Nov 1, 2021, 3:24 AM • 8 Y

Y by IndoMathXdZ, megarnie, ike.chen, centslordm, Nuterrow, CantonMathGuy, CyclicISLscelesTrapezoid, PRMOisTheHardestExam

I proposed this problem

.My solution was same as MarkBcc168's one.Allmost all the approaches I was aware of except one are already posted in this thread.So I am posting that solution here:

Solution with inversion Projective and Cartesian coordinates,by Ankan Bhattacharya : Let $O$ be the center of $\omega$ .Brocard's theorem states that $EF$ is the polar of $X$
Since none of $E,F,X$ are points of infinity $O$ is different from all three.Consider the inversion on $\omega$ to eliminate the polar:

$\odot(AMN)$ is sent to the line $\ell$ , tangent to $\omega$ at $A$ .
The line $EF$ , as the polar of $X$ , is sent to the circle with diameter $\overline{OX}$ .

Thus,if $\odot(AMN)$ and $EF$ ha a point in common then $\ell$ intersects $(OX)$ .We claim this is impossible.
Establish Cartesian coordinates with $A=(0,0)$ and $B=(2,0)$ .Then $\ell$ is the $y$ -axis.Let $S$ be the centre of $(OX)$ .Observe

$B$ lies on the circle with center $(2,0)$ and radius 2.
$X$ lies on the circle with centre $(4,0)$ and radius 4.Indeed, $(4,0)$ lies on perpendicular bisector of $AX$ and is also the antipode of $A$ in $\omega$
$S$ lies on the circle with center $(3,0)$ and radius 2.

Thus let coordinate of $S$ be $(x,y)$ ,with $(x-3)^2+y^2=4$ .There is a point common to $\ell$ and $(OX)$ implies:
$\begin{align*} d(S,\ell)^2 &\le SO^2\\ \iff x^2 &\le (x-2)^2+y^2\\ \iff x^2 &\le (x-2)^2+[4-(x-3)]^2\\ \iff (x-1)^2 &\le 0 \end{align*}$ or $x=1$ which forces $y=0$ .This implies $B=(0,0)=A$ which is not permitted.Thus $\ell$ can't share a point with $(OX)$ which implies $\odot(AMN)$ and $EF$ can't have a point in common. $\blacksquare$

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#18 h Nov 1, 2021, 4:36 AM • 1 Y

Y by centslordm

The answer is no. Let $\gamma$ be the circle with center $B$ and radius zero and $\overline{EF}=\ell.$ Also, let $O$ be the center of $\omega$ and $L$ the midpoint of $\overline{AO}.$ We know $\overline{BL}\perp\ell\quad (*)$ by Brocard's on $ABDC$ and since $\overline{BL}\parallel\overline{OX}.$ Notice that $\triangle ENB\sim\triangle EBA$ since $\measuredangle ENB=\measuredangle ACX=\measuredangle ACB=\measuredangle ABE.$ Hence, $EB/EN=EA/EB$ so $\text{pow}_{\omega}(E)=\text{pow}_{\gamma}(E).$ This with $(*)$ implies that $\ell$ is the radical axis of $\omega$ and $\gamma.$

If $P=\ell\cap\omega,$ $P$ must lie on $\gamma.$ Thus, $B$ must lie on $\ell,$ a contradiction. $\square$

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#19 h Nov 1, 2021, 4:41 AM • 1 Y

Y by centslordm

Fix the $\omega$ and points $A,B$ and vary point $C$ . By Brokard, $EF$ is fixed. We have points $A,M,O,N$ are concyclic, So, $(AMN)$ is also fixed. It is easy to check that if $ABDC$ is an isosceles trapizoid with $AC\parallel BD$ and let polar of $X$ intersects $AB$ at $P$ , then we have $AP=2BP$ and $EP\parallel AC$ , so $EP < EM \implies EP$ don't intersects $(AMN)$ .

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#20 h Nov 1, 2021, 11:31 PM • 4 Y

Y by centslordm, SK_pi3145, MatBoy-123, megarnie

We use barycentric coordinates with respect to $\triangle ABC$ , letting $A=(1:0:0)$ , $B=(0:1:0)$ , and $C=(0:0:1)$ . Then $X=2B-C=(-1:2:0)$ . We have $D=\overline{CX}\cap (ABC)$ , so let $D=tC+(1-t)X=(t-1,2(1-t),t)$ . Then, since the equation of $(ABC)$ is $-a^2yz-b^2xz-c^2xy=0$ , we have
$a^22(1-t)t+b^2(t-1)t+c^2(t-1)2(1-t)=0 \implies t=\frac{2c^2}{2a^2-b^2+2c^2}.$ Hence
$D=(-2a^2+b^2 : 2(2a^2-b^2):2c^2).$ Now we can easily find $E$ and $F$ :
$\begin{align*} E &= \overline{BD}\cap\overline{AC}=(2a^2-b^2:0:-2c^2) \\ F&=\overline{AD}\cap\overline{BC}=(0:2(2a^2-b^2):2c^2). \end{align*}$ By the general equation of the circle, it is easy to compute that the equation of $(AMN)$ is
$(AMN): -a^2yz-b^2xz-c^2xy+\left(\tfrac12c^2y+\tfrac12b^2z\right)(x+y+z)=0.$ Suppose $P=\overline{EF}\cap (AMN)$ exists.
Main idea that simplifies calculations, communicated by Eyed: WLOG let $P=(x:y:c^2)$ , by scaling. This allows for a lot of cancellation to occur in subsequent calculations. Note:
$\begin{align*} P\in \overline{EF} &\implies \begin{vmatrix} x&y&c^2\\2a^2-b^2&0&-2c^2\\0&2a^2-b^2&c^2\end{vmatrix}=0 \\ &\implies x(2c^2)(2a^2-b^2)-y(2a^2-b^2)c^2 + c^2(2a^2-b^2)^2 = 0 \\ &\implies y = 2x + (2a^2-b^2). \end{align*}$ And $P\in (AMN)$ , so
$\begin{align*} P\in (AMN) &\implies -a^2yc^2-b^2xc^2-c^2xy+(\tfrac12c^2y+\tfrac12b^2c^2)(x+y+c^2) = 0 \\ &\implies -a^2y-b^2x-xy+(\tfrac12y+\tfrac12b^2)(x+y+c^2) = 0 \\ &\implies -a^2(2x+(2a^2-b^2))-b^2x-x(2x+(2a^2-b^2))+(x+a^2)(3x+2a^2-b^2+c^2)=0 \\ &\implies x^2 + (a^2-b^2+c^2)x+a^2c^2=0. \end{align*}$ The discriminant of this quadratic is
$(a^2-b^2+c^2)^2-4a^2c^2=(a+b-c)(a-b-c)(a-b+c)(a+b+c).$ By the triangle inequality, the first, third, and fourth terms above are positive, and the second is negative. So the product is negative, contradiction.

Remarks

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DottedCaculator

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#21 h Nov 20, 2021, 3:12 PM • 3 Y

Y by centslordm, megarnie, ike.chen

complex bash attached

Attachments:

2021 USEMO Problem 4-compressed.pdf (275kb)

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#22 h Nov 20, 2021, 3:59 PM

Y by

DottedCaculator wrote:

complex bash attached

Were you able to view your scores?

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#24 h Nov 20, 2021, 6:38 PM • 1 Y

Y by megarnie

megarnie wrote:

DottedCaculator wrote:

complex bash attached

Were you able to view your scores?

yes

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#25 h Nov 27, 2021, 9:52 PM • 1 Y

Y by centslordm

Solved with a slight hint to consider a circle of radius $0$ ...

The answer is no. Let $\Omega$ be the circle with radius $0$ at $B$ , and redefine $M$ to be the midpoint of $D$ . Note that $(AMN)$ is still the same circle. The key claim is that $\overline{EF}$ is the radical axis of $\Omega$ and $(AMN)$ , which I will prove by power of a point.
Observe that since $\overline{BN}$ is the $A$ -midline of $\triangle AXC$ , we have $\overline{BN} \parallel \overline{XC}$ , so
$\measuredangle BNA=\measuredangle XCA=\measuredangle DCA=\measuredangle DBA=\measuredangle EBA,$ which implies $\overline{BE}$ is tangent to $(ABN)$ . As such,
$\mathrm{Pow}_{(AMN)} (E)=EA\cdot EN=EB^2=\mathrm{Pow}_\Omega(E).$ Likewise, we have
$\measuredangle BMA=\measuredangle XDA=\measuredangle CDA=\measuredangle CBA=\measuredangle FBA,$ so $\mathrm{Pow}_{(AMN)} (F)=\mathrm{Pow}_\Omega (F)$ . Thus $\overline{EF}$ is indeed the radical axis of $\Omega$ and $(AMN)$ .

To finish, note that if $(AMN)$ intersects $\overline{EF}$ at some point $P$ , then $\mathrm{Pow}_{(AMN)}(P)=0$ . But this implies $P=B$ , so $B$ lies on $(AMN)$ . But $B$ also lies on $\omega=(ACD)$ . By homothety $\omega \cap (AMN)=A$ , so this means that $A=B$ , which is absurd. $\blacksquare$

Note that instead of redefining $M$ and proving that $F$ lies on the radical axis, we can instead cite Brokard's to prove that $\overline{EF} \perp \overline{OX}$ where $O$ is the center of $\omega$ , so $\overline{EF}$ is perpendicular to the line joining the centers of $\Omega$ and $(AMN)$ . Combined with the fact that $E$ is on the radical axis this proves the key claim.

This post has been edited 2 times. Last edited by IAmTheHazard, Nov 27, 2021, 9:55 PM
Reason: typo

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#26 h Mar 4, 2023, 7:54 PM

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The answer is that $\overline{EF}$ cannot share a point with $(AMN)$ .

Let $\ell$ be the radical axis of the circle of radius $0$ at $B$ (call it $\omega$ ) and $(AMN)$ , let $O$ be the circumcenter of $ABC$ , and $K$ be the center of $(AMN)$ .

Claim: Denote $T=\overline{EF} \ \cap \ \overline{AB}$ . Then
(i) $T$ lies on $\ell$ and
(ii) $\overline{EF} \perp \overline{BK}$ .

Proof.
(i) By Ceva/Menelaus, it follows that $AT=2TB$ . Thus, $\text{Pow}(T, (AMN))=\bigg(\frac{1}{6}AB\bigg)\bigg(\frac{2}{3}AB\bigg)=\frac{1}{9}AB^2$ and $\text{Pow}(T, \omega)=\bigg(\frac{1}{3}AB\bigg)^2=\frac{1}{9}AB^2$ , so the power of $T$ with respect to $(AMN)$ and $\omega$ are equal, which implies that $T$ lies on $\ell$ .

(ii) By Brokard's Theorem, $X$ is the pole of $\overline{EF}$ with respect to $(ABC)$ , so $\overline{EF}$ is perpendicular to $\overline{OX}$ . Taking a homothety centered at $A$ with factor $\frac{1}{2}$ yields $\overline{EF} \perp \overline{BK}$ , as desired. $\square$

To finish, realize that the claim implies that $\overline{EF}$ is the radical axis of $(AMN)$ and $\omega$ , and hence $\overline{EF}$ can never share a point with $(AMN)$ , and we are done. $\blacksquare$

This post has been edited 2 times. Last edited by Leo.Euler, Mar 4, 2023, 8:00 PM

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#27 h Feb 26, 2024, 2:02 AM

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https://www.geogebra.org/calculator/vnacvctt

Let $P$ be the midpoint of segment $AD$ .

Claim: $P$ lies on $(AMN)$ .
Proof: $D$ lies on $(ABC)$ , so taking a homothety with factor $\tfrac12$ about $A$ gives the result.

Claim: $\triangle FBP\sim\triangle FAB\implies FB^2=FP\cdot FA$ .

Proof: $\measuredangle BFP=\measuredangle BFA$ , and we also have $\measuredangle FPB = \measuredangle APN = \measuredangle ADC=\measuredangle ABC$ , so $\triangle FBP\sim\triangle FAB$ is true. $FB^2=FP\cdot FA$ follows immediately from this similarity.

Claim: $\triangle EBN\sim\triangle EAB\implies EB^2=EN\cdot EA$ .

Proof: $\measuredangle NEB=\measuredangle AEB$ . Additionally, $\measuredangle NBE = - \measuredangle BDC= \measuredangle BAC=\measuredangle BAE$ . It’s easy to see that the claim holds.

Since $FB^2=FP\cdot FA$ and $EB^2=EN\cdot EA$ , $\overline{EF}$ is the radical axis of the circle with radius $0$ at $B$ and $(AMN)$ . Since $B$ doesn’t lie on $(AMN)$ , the radical axis can’t intersect $(AMN)$ (if so, the power of an intersection wrt $(AMN)$ is $0$ , meaning $B$ ’d have to lie on $(AMN)$ too), so we’re done. $\square$

This post has been edited 1 time. Last edited by gracemoon124, Feb 26, 2024, 2:06 AM

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#29 h Feb 17, 2025, 2:46 AM

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No, it's not possible. All angles are directed. Let $P$ be midpoint of $AD$ . We first make the following observations:

1. Since $BN$ is parallel to $CK$ , points $B$ , $P$ , and $N$ are collinear.
2. A homothety at $A$ with $k=2$ implies that $P$ lies on $(AMN)$ .

Notice that $\measuredangle FBP = \measuredangle BCD = \measuredangle BAD = \measuredangle BAF \implies \triangle{BFP} \sim \triangle{AFB},$ so $BF^2=AF\cdot FP$ .

We similarly find $\measuredangle EBN = \measuredangle EDC = \measuredangle BAC = \measuredangle BAE\implies \triangle{BAE}\sim \triangle{NBE},$ so $EB^2=EN\cdot EA$ .

Let $\Omega$ be the circle of radius $0$ centered at $B$ . We can find that $EF$ is the radical axis of $\Omega$ and $(AMN)$ . Since $B$ is outside $(AMN)$ , the conclusion follows.

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