I hate Affine Transforms.

Let D be foot from A to BC (motivated by working backwards from the condition) (ADPC) cyclic, power of a point on M twice, finish

IAmTheHazard on ELMO 2021/1 walked so IAmTheHazard on USAMO 2023/1 could run. Let $D$ be the foot of the $A$-altitude and construct parallelogram $ABA'C$. I claim that $Q$ is the reflection of $D$ over $M$, or equivalently $Q$ is the foot of the perpendicular from $A'$ to $\overline{BC}$. To see this, let $Q'$ be the actual foot; we will prove that $ABPQ'$ is cyclic. This is just angle chasing: $$\measuredangle BQ'P=\measuredangle CQ'P=\measuredangle CAP=\measuredangle BAP.$$Thus $Q'=Q$. To finish, just note that $\overline{MN}$ is the $Q$-midline of right triangle $\triangle ADQ$, hence $N$ lies on the perpendicular bisector of $\overline{BC}$ and we're done.

apotosaurus wrote: Let D be foot from A to BC (motivated by working backwards from the condition) (ADPC) cyclic, power of a point on M twice, finish yep why were both JMO 1 and JMO 2 like 0 mohs :skull:

Outline: Drop an altitude from $A$, call it $K$. prove that $MK=MQ$ by PoP, and then trivial by midline.

Let $B$ be at $(0,0)$, $A$ be at $(a,b)$ and $C$ be at $(2,0)$. Tedious coordbash shows $Q$ is at $(2-a,0)$ gg

GoodMorning wrote: Outline: Drop an altitude from $A$, call it $K$. prove that $MK=MQ$ by PoP, and then trivial by midline. I used similar triangle to prove that MK=MQ (PMC is similar to KMA, and MBP is similar to MAQ, so MA/MK=MC/MP=MB/MP=MA/MQ so MK=MQ). Hopefully this works

bro yall be having the most beautiful synthetic solutions and here i am with my goofy ahh lookin coordbash A=(a, ak) (when angle AMB is 90 then M=P=Q so done) B=(-1, 0) C=(1, 0) M=(0,0) P=(1/(k^2+1), k/(k^2+1)) we wanna show that R=(-a, 0) lies on (ABP) slope of RA is k/2, slope of AP is k, slope of BP is k/(k^2+2) so we wanna show arctan(k/2)=arctan(k)-arctan(k/(k^2+2)) this is equivalent to showing (1+ki)/((2+ki)(k^2+2+ki)) is real, which it is since it is 1/(k^2+4).

Here's a solution with no additonal points constructed at all: [asy][asy] size(10cm); pair A = dir(115); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair M = midpoint(B--C); pair P = foot(C, A, M); pair Q = 2*M-D; filldraw(A--B--C--cycle, palecyan, blue); draw(circumcircle(A, B, P), orange+dashed); draw(A--P--C, deepgreen); draw(A--Q, red+dashed); pair N = midpoint(A--Q); draw(B--N--C, red+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(45)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(315)); dot("$N$", N, dir(45)); [/asy][/asy] CLAIM: $\triangle BPC \sim \triangle ANM$ (oppositely oriented). Proof. We have $\triangle BMP \sim \triangle AMQ$ from the given concyclicity of $ABPQ$. Then \[ \frac{BM}{BP} = \frac{AM}{AQ} \implies \frac{2BM}{BP} = \frac{AM}{AQ/2} \implies \frac{BC}{BP} = \frac{AM}{AN} \]implying the similarity (since $\measuredangle MAQ = \measuredangle BPM$). $\blacksquare$ This similarity gives us the equality of directed angles \[ \measuredangle \left( BC, MN \right) = -\measuredangle \left( PC, AM \right) = 90^\circ \]as desired.

We employ the use of coordinate geometry. Let $M$ be the origin, $B = (-1,0)$, $C = (1,0)$, $A = (a,b)$. The equation of line $MA$ is $y = \frac{b}{a}x$, and the equation of line $CP$ is $y = \frac{-a}{b}(x-1)$. Solving gives $P = \left(\frac{a^2}{a^2+b^2}, \frac{ab}{a^2+b^2}\right)$. Then by power of a point on $M$, we have $MP \cdot MA = MC \cdot MQ$. $MP = \frac{a}{\sqrt{a^2+b^2}}$, $MA = \sqrt{a^2+b^2}$, and $MC=1$. These all imply that $MQ = a$, so $Q = (-a,0)$, meaning $N =\left(0, \frac{b}{2}\right)$ and hence $N$ lies on the perpendicular bisector of $BC$ and we are done.

Okay this is like reverse of all the other sols but reflect $Q$ around $M$ to $Q’$ and get $APQ’C$ cyclic by PoP and then $\angle AQ’C=\angle APC=90$ and finish by homothety centered at Q which sends $NM$ to $AQ’$ so $NM \perp BC$ which means we’re done

Easiest solution in my opinion. By PoP, $MQ = \frac{AM}{\cos(\angle CMP)}$. Then let $M'$ be the intersection of $AM$ and the perpendicular from $Q$ to $BC$. By right triangle $MQM',$ $MM' = AM$, so a homothety centered at $A$ with scale factor $\frac 12$ finishes.

Coordinate bashable geometry...

Coord-bash FTW

Um what Let $D$ be the foot of the altitude from $A$ to $BC$; let $D'$ be its reflection about $M$. Now $MD'\cdot MB=MD\cdot MC=MP\cdot MA$, then $D'\equiv Q$ done.

I also did D as foot of altitude. Note that A, P, D, C is cyclic and A, P, Q, B is cyclic. AP is radical axis so $MD*MC=MQ*MB$ so $MD=MQ$ so $MN||AD$ so N is on perp bisector of BC. in contest i got everything except radical axis...

Let $D$ be the foot of the altitude from $A$ to $BC$; then $\measuredangle ADC = 90^\circ = \measuredangle APC$, so $ADPC$ is cyclic. But since $M$ is on the radical axis $AP$ of $(APB)$ and $(APC)$, we have $QM \cdot MB = DM \cdot MC \implies QM = DM$, so it follows that $M$ is the midpoint of $DQ$. Thus, $MN$ is the midline of $\triangle AQD$, so $MN \parallel AD \perp BC$, which means that $N$ is on the perpendicular bisector of $BC$, so $NB = NC$ as desired. $\blacksquare$ [asy][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 = -12.34, xmax = 12.34, ymin = -7.45, ymax = 7.45; /* image dimensions */ /* draw figures */ draw(circle((0.41,-0.9714285714285714), 5.980586620602582), linewidth(0.7)); draw(circle((-1.12,-0.3157142857142861), 4.837163108556037), linewidth(0.7)); draw(circle((1.94,0.82), 5.497026468919355), linewidth(0.7)); draw((-2.24,4.39)--(-5.3,-2.75), linewidth(0.7)); draw((-5.3,-2.75)--(6.12,-2.75), linewidth(0.7)); draw((6.12,-2.75)--(-2.24,4.39), linewidth(0.7)); draw((-2.24,4.39)--(1.1013279864004928,-4.612672385999818), linewidth(0.7)); draw((-2.24,4.39)--(-2.24,-2.75), linewidth(0.7)); draw((-2.24,4.39)--(3.06,-2.75), linewidth(0.7)); draw((-5.3,-2.75)--(6.12,-2.75), linewidth(0.7)); draw((0.41,-2.75)--(0.41,0.82), linewidth(0.7)); draw((-5.3,-2.75)--(1.1013279864004928,-4.612672385999818), linewidth(0.7)); draw((1.1013279864004928,-4.612672385999818)--(6.12,-2.75), linewidth(0.7)); /* dots and labels */ dot((-2.24,4.39),dotstyle); label("$A$", (-2.7,4.61), NE * labelscalefactor); dot((-5.3,-2.75),dotstyle); label("$B$", (-6.02,-3.11), NE * labelscalefactor); dot((6.12,-2.75),dotstyle); label("$C$", (6.4,-3.11), NE * labelscalefactor); dot((0.41,-2.75),linewidth(4pt) + dotstyle); label("$M$", (0.74,-3.32), NE * labelscalefactor); dot((1.1013279864004928,-4.612672385999818),linewidth(4pt) + dotstyle); label("$P$", (0.96,-5.29), NE * labelscalefactor); dot((3.06,-2.75),linewidth(4pt) + dotstyle); label("$Q$", (3.16,-3.32), NE * labelscalefactor); dot((0.41,0.82),linewidth(4pt) + dotstyle); label("$N$", (0.5,0.97), NE * labelscalefactor); dot((-2.24,-2.75),linewidth(4pt) + dotstyle); label("$D$", (-2.76,-3.32), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

asdf334 wrote: Um what Let $D$ be the foot of the altitude from $A$ to $BC$; let $D'$ be its reflection about $M$. Now $MD'\cdot MB=MD\cdot MC=MP\cdot MA$, then $D'\equiv Q$ done. not everyone is that big brain

cord bash go brrrr

So many solutions, here is mine anyway. Let $J$ be the foot of perpendicular from $A$ to $BC$. $N$ is the midpoint of $AQ \Rightarrow $ $NJ=AN=NQ$. If we prove that $M$ is the midpoint of $JQ$, we are done. $\angle AJC=\angle APC=90^\circ$ means $AJPC$ is a cyclic quadrilateral. By Intersecting Chords Theorem, we get $\hspace{4cm}$ $JM \cdot MC=AM \cdot MP=QM \cdot MB \Rightarrow JM=MQ$ and we are done. $\blacksquare$

Let $D$ be the projection of $A$ onto $BC$. Then clearly $APDC$ is cyclic. Then consider the radical axis of $(APQB)$ and $(APDC)$ which passes through $M$. We get that $MQ \cdot MB = MD \cdot MC$, however $MB = MC$ so $MQ = MD$. Then it follows that $\triangle MQN \sim \triangle DQA$ so $\angle BMN = \angle BDA = 90^\circ$. So $N$ lies on the perpendicular bisector of $BC$ which implies the desired.

my only 7 on this test

Let be U the middle point of BP. BPQA is cyclic, so it's easy to see that triangles AMQ and BMP are similar. Note that MU and MN are similar medians, so angle UMP = angle NMQ . But it's well known that UM is parallel to PC (U middle point of BP and M middle point of BC), we have that MP is perpendicular to PC, so MP is perpendicular to UM. That implies that angle UMP=90, so angle NMQ=90. We conclude NBC is isosceles bcs NM is the perpendicular bisector of BC. So NB=NC .

let $D$ be the foot of $A$ on $BC$ now we have $\measuredangle MAD=\measuredangle PCM \implies PADC$ is a cyclic quad so now we have by power of a point $$MQ\cdot MB=MP\cdot MA=MD\cdot MC \implies MQ=MD, MQ=MD$$thus we are done we can say we are done since $AN=NQ=DN, MQ=MD$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -17.600882777678084, xmax = 27.422804847471284, ymin = 1.0352709403671092, ymax = 22.525347317817523; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((1.9898326082481221,3.0398094064589305)--(1.2906722446864514,3.0464892825439147)--(1.2839923686014671,2.347328918982244)--(1.983152732163138,2.34064904289726)--cycle, linewidth(0.8)); draw((4.318822173005892,4.035099120993285)--(4.49538921685345,3.3585683280894196)--(5.171920009757316,3.535135371936978)--(4.995352965909758,4.211666164840843)--cycle, linewidth(0.8)); /* draw figures */ draw((2.106772277586995,15.279494797261034)--(13.011641195194043,2.235281318728175), linewidth(0.8)); draw((-2.0268048912517904,2.3789607399362556)--(2.106772277586995,15.279494797261034), linewidth(0.8)); draw((5.492418151971126,2.3071210293322153)--(13.011641195194043,2.235281318728175), linewidth(0.8)); draw((5.492418151971126,2.3071210293322153)--(-2.0268048912517904,2.3789607399362556), linewidth(0.8)); draw((5.492418151971126,2.3071210293322153)--(3.7995952147790604,8.793307913296625), linewidth(0.8)); draw((3.7995952147790604,8.793307913296625)--(2.106772277586995,15.279494797261034), linewidth(0.8)); draw((-2.0268048912517904,2.3789607399362556)--(4.995352965909758,4.211666164840843), linewidth(0.8)); draw(circle((11.097049922547235,11.714999588929738), 9.671128073998846), linewidth(0.8)); draw((9.001683571779123,2.273593015767171)--(2.106772277586995,15.279494797261034), linewidth(0.8)); draw((13.011641195194043,2.235281318728175)--(5.554227924683059,8.776543906514103), linewidth(0.8)); draw((5.554227924683059,8.776543906514103)--(-2.0268048912517904,2.3789607399362556), linewidth(0.8)); draw((2.106772277586995,15.279494797261034)--(1.983152732163138,2.34064904289726), linewidth(0.8)); draw(circle((0.03998369316760201,8.829227768598644), 6.773297556857771), linewidth(0.8) + cqcqcq); /* dots and labels */ dot((2.106772277586995,15.279494797261034),dotstyle); label("$A$", (2.241181607636645,15.603696950847297), NE * labelscalefactor); dot((13.011641195194043,2.235281318728175),dotstyle); label("$B$", (13.513583633845345,2.0899986153340007), NE * labelscalefactor); dot((-2.0268048912517904,2.3789607399362556),dotstyle); label("$C$", (-2.999496527355119,2.1559190950194314), NE * labelscalefactor); dot((5.492418151971126,2.3071210293322153),linewidth(4pt) + dotstyle); label("$M$", (5.273523673166471,1.3648733387942626), NE * labelscalefactor); dot((4.995352965909758,4.211666164840843),linewidth(4pt) + dotstyle); label("$P$", (5.438324872380048,4.232414205110499), NE * labelscalefactor); dot((9.001683571779123,2.273593015767171),linewidth(4pt) + dotstyle); label("$Q$", (8.833229576179745,1.5296745380078396), NE * labelscalefactor); dot((5.554227924683059,8.776543906514103),linewidth(4pt) + dotstyle); label("$N$", (5.7020067911217724,9.04460922214694), NE * labelscalefactor); dot((1.983152732163138,2.34064904289726),linewidth(4pt) + dotstyle); label("$D$", (1.8126984896813436,1.4307938184796933), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $D$ be the foot of the $A$-altitude. Clearly $D$ lies on $(APC)$ and it suffices to prove that $DM=MQ$ since we know that $DN=NQ$. Therefore, using ratio lemma, we need to prove that $$\frac{\sin(\angle DAP)}{\sin(\angle QAP)}=\frac{AQ}{AD}$$ But we have that $\angle DAP=\angle BCP$ and $\angle QAP=\angle CBP$ so $$\frac{\sin(\angle DAP)}{\sin(\angle QAP)}=\frac{\sin(\angle BCP)}{\sin(\angle CBP)}=\frac{BP}{PC}$$ Now since $BPQA$ is cyclic we have that $\triangle BMP \sim \triangle AMQ$ so $BP=\frac{AQ\cdot BM}{AM}$ and similarly $CP=\frac{AD\cdot CM}{AM}$, hence $\frac{\sin(\angle DAP)}{\sin(\angle QAP)}=\frac{BP}{CP}=\frac{AQ}{AD}$. $\square$ [asy][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 = 6.162361917614558, xmax = 29.647581244007043, ymin = -10.853124754809201, ymax = 4.066754806967936; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); draw((16.,2.)--(14.,-7.)--(24.,-7.)--cycle, linewidth(0.8) + qqzzcc); /* draw figures */ draw((16.,2.)--(14.,-7.), linewidth(0.8) + qqzzcc); draw((14.,-7.)--(24.,-7.), linewidth(0.8) + qqzzcc); draw((24.,-7.)--(16.,2.), linewidth(0.8) + qqzzcc); draw((16.,2.)--(19.5,-8.5), linewidth(0.8)); draw((19.5,-8.5)--(24.,-7.), linewidth(0.8)); draw(circle((20.,-2.5), 6.020797289396148), linewidth(0.8)); draw(circle((18.,-3.1666666666666665), 5.540256712864887), linewidth(0.8)); draw((16.,2.)--(16.,-7.), linewidth(0.8)); draw((14.,-7.)--(19.5,-8.5), linewidth(0.8)); draw((22.,-7.)--(16.,2.), linewidth(0.8)); /* dots and labels */ dot((16.,2.),dotstyle); label("$A$", (15.54451696339082,2.1217340705743166), NE * labelscalefactor); dot((14.,-7.),dotstyle); label("$B$", (14.05117273044084,-6.878517163048328), NE * labelscalefactor); dot((24.,-7.),dotstyle); label("$C$", (24.554136517608008,-6.878517163048328), NE * labelscalefactor); dot((19.,-7.),linewidth(4.pt) + dotstyle); label("$M$", (19.052654624024424,-6.90267891132651), NE * labelscalefactor); dot((19.5,-8.5),linewidth(4.pt) + dotstyle); label("$P$", (19.547970463727147,-8.40070730457377), NE * labelscalefactor); dot((16.,-7.),linewidth(4.pt) + dotstyle); label("$D$", (16.04451696339082,-6.90267891132651), NE * labelscalefactor); dot((22.,-7.),linewidth(4.pt) + dotstyle); label("$Q$", (21.225880815650763,-7.00267891132651), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

reflect $A$, $P$, and $Q$ across $M$ to $A'$, $P'$, and $Q'$ respectively note that since $AQB=APB=A'P'C=A'Q'C$, we have that $AQM$, $CP'M$, $A'Q'M$, and $BPM$ are similar then following that, we get that $AQ'M$, $A'QM$, $BP'M$, and $CPM$ are similar as well, so $AQ'$ is perpendicular to $BC$ and by parallels, $NM$ is perpendicular to $BC$ thus, $NBC$ is isosceles and $NB=NC$

Let $D$ denote the foot of the $A$-altitude. It is clear that $\square{ACDP}$ is cyclic. By Power of a Point on $M$, we have: $MQ \cdot MB = MP \cdot MA = MD \cdot MC$ $\implies MQ \cdot \cancel{MB} = MD \cdot \cancel{MC}$ $\implies M$ is the midpoint of $DQ$. Hence, by midpoint theorem in $\Delta{QAD}$, $NM \parallel AD \implies NM \perp BC \implies NB = NC $ as desired.

Let $D$ be the foot of $A$ to $BC$, $P'$ the foot of $B$ to $AM$. Redefine $Q$ as $(ABP) \cap (ACP')$. First, note that as $MB=MC$, $\widehat{P'MB} = \widehat{PMC}$ and $\widehat{BP'M} = \widehat{CPM}=90°$, triangles $BMP'$ and $CMP$ are congruent, thus $P'M=MP$, and $BPCP'$ is a parallelogram. It follows that $\widehat{BQC} = \widehat{BQA} + \widehat{AQC} = \widehat{BPA} + \widehat{AP'C} = 180°$, which implies $B,Q,C$ collinear. Now, notice that we have some cyclic quadrilaterals, using PoP we get : $$ MQ \cdot MC = MP' \ cdot MA = MP \cdot MA = MD \cdot MC \implies MQ=MD$$. Hence, projecting onto $AQ$, we get that the midpoint of $AQ$, $N$ is such that $MN \perp BC$, thus it lies on the perpendicular bisector of $BC$. The conclusion follows.

Feel free to point out any mistakes.

Let $D$ be the foot of the perpendicular from $A$ to $BC.$ Then clearly $ADPC$ is cyclic. Therefore, $$MB \cdot MQ = AM \cdot MP = MD \cdot MC \implies MD=MQ,$$therefore $M$ is the midpoint of $DQ.$ Hence by the Midpoint Theorem it follows that $MN \parallel AD \implies MN \perp BC.$ Thus $MN$ is the perpendicular bisector of $BC$ and the desired result follows. QED

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