In an acute triangle $ABC$, let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $C$ to $AM$. Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$. Let $N$ be the midpoint of $\overline{AQ}$. Prove that $NB=NC$. Proposed by Holden Mui
Problem
Source: 2023 USAJMO Problem 2/USAMO Problem 1
Tags: USAMO, USAJMO, USA(J)MO, geometry, xtimmyGgettingflamed
24.03.2023 01:05
I hate Affine Transforms.
24.03.2023 01:05
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
24.03.2023 01:07
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.
24.03.2023 01:07
24.03.2023 01:07
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:
24.03.2023 01:08
Outline: Drop an altitude from $A$, call it $K$. prove that $MK=MQ$ by PoP, and then trivial by midline.
24.03.2023 01:13
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
24.03.2023 01:17
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
24.03.2023 01:21
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).
24.03.2023 01:26
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.
24.03.2023 01:27
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.
24.03.2023 01:27
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
24.03.2023 01:34
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.
24.03.2023 01:35
Coord-bash FTW
24.03.2023 02:12
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.
24.03.2023 02:29
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...
24.03.2023 02:29
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]
24.03.2023 02:32
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
23.03.2024 21:26
cord bash go brrrr
09.04.2024 23:29
21.04.2024 06:01
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. $\blacksquare$
09.06.2024 23:02
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.
18.06.2024 20:23
my only 7 on this test
25.06.2024 13:39
27.07.2024 06:37
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 .
21.08.2024 19:01
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]
25.08.2024 20:29
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]
28.08.2024 18:29
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 $NB=NC$
22.10.2024 20:34
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.
10.11.2024 13:53
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.
04.12.2024 07:08
Feel free to point out any mistakes.
13.01.2025 02:04
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
13.01.2025 10:01
bash solutions lose 500000 aura