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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
How many ordered pairs of numbers can we find?
BR1F1SZ   5
N a few seconds ago by RagvaloD
Source: 2024 Argentina TST P6
Given $2024$ non-negative real numbers $x_1, x_2, \dots, x_{2024}$ that satisfy $x_1 + x_2 + \cdots + x_{2024} = 1$, determine the maximum possible number of ordered pairs $(i, j)$ such that
\[
x_i^2 + x_j \geqslant \frac{1}{2023}.
\]
5 replies
BR1F1SZ
Jan 25, 2025
RagvaloD
a few seconds ago
10^{f (n)} <10n + 1 <10^{f (n) +1}
parmenides51   4
N 4 minutes ago by AshAuktober
Source: OLCOMA Costa Rica National Olympiad, Final Round, 2018 3.4
Determine if there exists a function f: $N^*\to N^*$ that satisfies that for all $n \in N^*$, $$10^{f (n)} <10n + 1 <10^{f (n) +1}.$$Justify your answer.

Note: $N^*$ denotes the set of positive integers.
4 replies
parmenides51
Sep 20, 2021
AshAuktober
4 minutes ago
FE on Stems
mathscrazy   5
N 12 minutes ago by Levieee
Source: STEMS 2025 Category B4, C3
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in \mathbb{R}$, \[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]Proposed by Aritra Mondal
5 replies
mathscrazy
Dec 29, 2024
Levieee
12 minutes ago
NEPAL TST 2019
khan.academy   10
N 14 minutes ago by MuradSafarli
Prove that there exist infinitely many pairs of different positive integers $(m, n)$ for which $m!n!$ is a square of an integer.

Proposed by Anton Trygub
10 replies
+1 w
khan.academy
May 14, 2019
MuradSafarli
14 minutes ago
No more topics!
IMO 2023 P2
799786   89
N Mar 18, 2025 by kaede_Arcadia
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
89 replies
799786
Jul 8, 2023
kaede_Arcadia
Mar 18, 2025
IMO 2023 P2
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2023 P2
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799786
1052 posts
#1 • 8 Y
Y by Quidditch, PRMOisTheHardestExam, LoloChen, KST2003, trinhquockhanh, Rounak_iitr, Funcshun840, cubres
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
This post has been edited 5 times. Last edited by Amir Hossein, Aug 8, 2023, 5:43 PM
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khan.academy
633 posts
#3 • 1 Y
Y by LoloChen
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

@above it is
This post has been edited 1 time. Last edited by khan.academy, Jul 8, 2023, 4:52 AM
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LoloChen
475 posts
#4 • 18 Y
Y by 799786, khan.academy, khina, David-Vieta, GeoKing, oolite, ike.chen, Before-the-snow-melted, Aryan-23, sabkx, Dhruv777, Gato_combinatorio, aidan0626, PennyLane_31, AlexCenteno2007, ehuseyinyigit, LLL2019, Sedro
Too easy for P2! :(
Sketch: construct antipode of A to get rid of L and find concyclics kills it.
https://cdn.artofproblemsolving.com/attachments/c/5/7b46be3b7184784a85e64fdf9db4642e647f9c.png
Construct midpoint of arc $BC$ not containing ${A}$ and $A-$ antipode to be $N$ and $A'$. Let $AN$ cut $SB$ at $Y$ and $DA$ at $X$. Let ${AE}$ cut $SA'$ at $T$.
$\angle L=\angle EBC=\angle BPD$ so $LPDB$ concyclic. Note that $A'S//AN$ and by Reim $PYXB$ concyclic. Since $\angle TEA'=90$ and $SE=SA'$, $A'S=TS$ so by homothety at ${D}$, $AY=YX$. Combining $\angle APX=90$, $PY=XY$, $\angle YPD= \angle YXD=\angle PBY$, so $PY$ is tangent to $\omega$.
This post has been edited 3 times. Last edited by LoloChen, Jul 8, 2023, 2:56 PM
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tastymath75025
3223 posts
#9 • 4 Y
Y by khina, centslordm, ehuseyinyigit, jkim0656
Let $A',S'$ be the antipodes to $A,S$ on $(ABC)$. Defining $X = AS' \cap BS$, we want to show $PX$ is tangent to $(BPD)$ or $XD\cdot XB = XP^2$.

First, note by Reim on $(LPDB), (PBEA')$ that since $LD||EA'$ we must have $P,D,A'$ are collinear. Next, note that $\angle DBA = \angle S'AE = \angle XAD$, so $XA$ is tangent to $(BAD)$ and thus $XD\cdot XB = XA^2$, so to finish we only need to show $XA=XP$. The locus of such $X$ is the $A$-midline of $\triangle AA'P$, so if we let $Y=AX\cap A'D$ we just need to show $X$ is the midpoint of $AY$.

But if we let $Q = AE \cap SA'$ then from angle-chasing $QAA'$ is isosceles, so $AS\perp A'S$ implies $S$ is the midpoint of $QA'$. Taking homothety centered at $D$ therefore shows $X$ is the midpoint of $AY$, as desired $\blacksquare$

Motivation: The motivation for this solution is to first eliminate $L$ using Reim to get a nicer characterization for $P$. After this, dealing with $XP$ and $(BPD)$ directly is still not nice, but we bypass this by noticing $XA$ is tangent to $(BAD)$.
This post has been edited 7 times. Last edited by tastymath75025, Jul 8, 2023, 6:17 AM
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trinhquockhanh
522 posts
#14 • 4 Y
Y by PRMOisTheHardestExam, GeoKing, Rounak_iitr, ehuseyinyigit
$\text{No more cyclic quadrilaterals}$
https://i.ibb.co/yfc8FBc/2023-IMO-P2.png
This post has been edited 2 times. Last edited by trinhquockhanh, Jul 9, 2023, 3:52 PM
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Aryan-23
558 posts
#15 • 1 Y
Y by PRMOisTheHardestExam
Let $A'$ denote the $A$ antipode. We have $\angle BPD = \angle EBC = \angle BCA' =\angle BPA'$, so $P,D,A'$ are collinear.

Let $AI$ hit $PA'$ at $H$ and $BD$ at $F$. Let $AE$ and $SA'$ intersect at $J$. Then Note that, $FH \equiv AI\parallel SA'$, so that $PFHB$ is cyclic by Reims. Now note that $JA'E$ is a right triangle with $SA'=SE$ (as $A'E \parallel BC$), so $SA'=SJ$. From homothety at $D$ we get $FA=FH$. Now as $\triangle APH$ is right, we have $FP = FH$. So we get $\angle FPA'= \angle FHP= \angle SA'P = \angle SBP = \angle DBP$.

So $FP$ is tangent to $\omega$, as desired. $\square$.
This post has been edited 1 time. Last edited by Aryan-23, Jul 8, 2023, 6:40 AM
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KST2003
173 posts
#17 • 2 Y
Y by PRMOisTheHardestExam, Rounak_iitr
Let $M$ be the midpoint of minor arc $BC$, and let $\overline{PM}$ intersect $\overline{AE}$ at $K$. Then $\overline{SM} \parallel \overline{DK}$, so $K$ lies on $\omega$ by Reim's. Now redefine $F$ to be the intersection of $\overline{AM}$ and $\overline{BS}$. We will show that $\overline{FP}$ is tangent to $\omega$.

Since $\angle KBE = \angle LDE = 90^\circ$, we have
\[ \measuredangle BFA = \measuredangle SFM = 90^\circ + \measuredangle BMA = 90^\circ + \measuredangle BEA = \measuredangle BKA, \]so $AFKB$ is cyclic. Moreover,
\[ \measuredangle KBF = \measuredangle KAF = \measuredangle EAM = \measuredangle SMA = \measuredangle SBA = \measuredangle FBA,\]so $FA = FK$. Finally, $\measuredangle APK = \measuredangle APM = \measuredangle ASM = 90^\circ - \measuredangle SMA = 90^\circ - \measuredangle KAF$, so $F$ is the circumcenter of $\triangle PAK$ and hence $FP^2 = FA^2 = FD \cdot FB$. Hence $\overline{FP}$ is tangent to $\omega$ as desired.
This post has been edited 3 times. Last edited by KST2003, Jul 8, 2023, 6:56 AM
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MathLover_ZJ
826 posts
#19 • 5 Y
Y by David-Vieta, PRMOisTheHardestExam, Amiralizakeri2007, Rounak_iitr, ehuseyinyigit
Without antipodes: Let $\omega$ meet $AE$ at $T\neq D$.
The circumcenter of $\triangle APT$ is $Q$.
Connect $PT,BT,TQ,AP$.
then we have $\angle BPD=\angle BLD=90^{\circ}-\angle BED=\angle APB-90^{\circ},$
so $AP \perp PD$.
hence $\angle QPT=90^{\circ}-\angle PAT=\angle PDA=\angle PBT,$which means $PQ$ is tangent to $\omega$.
by $\angle AQT=360^{\circ}-2\angle APT=180^{\circ}-2\angle DBT=180^{\circ}-\angle ABT$ we know that $A,Q,T,B$ is concyclic.
since $AQ=QT\Rightarrow Q$ lies on $BS$.
we just need to prove $Q$ lies on the bisector.
$\Leftrightarrow \angle QAS=90^{\circ},$
which is obvious since $\angle AQB=\angle ATB,\angle TBE=90^{\circ}$.
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a22886
924 posts
#20 • 5 Y
Y by LoloChen, Aryan-23, David-Vieta, PRMOisTheHardestExam, Alfombra12
We consider the inversion $\Phi$ wrt $D$ that preserves $\Omega$.

Under $\Phi$ we have $A\leftrightarrow E$ and $B\leftrightarrow S$. It's natural to find$$\Phi(L)=\Phi(BE)\cap\Phi(DL)=\odot ADS\cap DL$$which indicates $\angle AS~\Phi(L)=\angle ADL=90^{\circ}$, i.e., $\Phi(\omega)=\operatorname{line}S\Phi(L)=SA'$ where $A'$ is the antipodal point of $A$.
From that, $\Phi(P)=\Phi(\omega)\cap\Phi(\Omega)=A'$, so $PDA'$ collinear.

Let the antipodal point of $S$ be $T$. $\angle APD=\angle SBT=90^{\circ}\Longrightarrow Q:=AP\cap BT\in\omega$
But also $AS=ET\Longrightarrow \overset{\LARGE \frown}{AP}+\overset{\LARGE \frown}{ET}=\overset{\LARGE \frown}{PS}\Longrightarrow R:=AE\cap PT\in\omega$
Now we apply Pascal's theorem to $(PPQBDR)$ to get the desired result.
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Parsia--
73 posts
#21
Y by
Let $A'$ be the antipode of $A$. We have $\angle BPD = \angle EBC = \angle BCA' =\angle BPA'$, so $P,D,A'$ are collinear.
Let $AE$ intersect $SA'$ at $I$ and let $AF$ intersect $PA'$ at $H$.
Since $SE=SA'$ and $\triangle IEA'$ is a right triangle, we get that $SA' = SE = SA$. Note that $AH \equiv AF || SA'$ so by Thales we get that $\frac{AF}{IS} = \frac{DF}{DS} = \frac{HF}{SA'}$ and so $AF = HF$. Note that $\triangle APH$ is a right triangle, so $PF = AF = HF$. Now note that $\angle DAF = \angle FAA' = \angle AA'S = \angle ABS = \angle ABF$ so $\triangle ADF \sim \triangle ABF$ and so $FP^2 = AF^2 = FD.FB$.
So $FP$ is tangent to $\omega$.
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giannis2006
45 posts
#22 • 1 Y
Y by Assassino9931
Let $N$ be the midpoint of the other arc $BC$ and $K= AD \cap PN$
Then $\angle BDK=\angle BDE =90-\angle DBC=90-\angle SBC=\angle BSN=\angle A/2=\angle BAN=\angle BPN=\angle BPK$ and so $K$ lies on $\omega$. Let now $H= AN \cap BS$. We need to show that $HP$ is tangent to $\omega$. It is obvious that $AE$, $SN$ are parallel and so $\angle DAH=\angle DAN=\angle ANS=\angle ABS=\angle ABD$, so $AH$ is tangent to $ABD$, i.e. $AH^2=HB*HD$, so it remains to prove $AH=HP$. Now we have from the cyclic quadrilaterals and the parallel lines: $\angle AKB=\angle DKB=$ $=180-\angle DLB$ $=180-\angle EBC$ $=180-\angle EAC$ $=180-\angle (90-C)$ $=90+\angle C$ $=90+\angle ANB$ $=\angle HBN+\angle HNB$ $=\angle AHB$ and hence $BKHA$ is cyclic, which gives that $\angle AKH=\angle ABH=\angle ABD=\angle DAH=\angle KAH$ and so $HA=HK$. Finally: $\angle AHK=\angle AHB+\angle BHK=\angle HBN+\angle HNB+\angle BAK=90+\angle C+\angle BAK=90+\angle C+90-\angle B=2\angle C+\angle A=2(\angle C+\angle A/2)=2(180-(\angle B+\angle A/2))=2(180-\angle ABN)=2(180-\angle APN)=
2(180-\angle APK)$ , from where we conclude that $H$ is the circumcenter of $APK$ and $AH=HP$, as needed.
This post has been edited 3 times. Last edited by giannis2006, Jul 8, 2023, 7:50 AM
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ericxyzhu
49 posts
#23 • 3 Y
Y by nervy, oolite, Funcshun840
This problem is proposed by Tiago Mourão, Nuno Arala, Portugal.
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juckter
322 posts
#24 • 3 Y
Y by PRMOisTheHardestExam, math90, UI_MathZ_25
Mediocre solution because I'm washed up. Let $T$ be the midpoint of arc $BC$ of $\Omega$ not containing $T$, let $R$ be the second intersection of $\omega$ with $AE$, and let $X = AT \cap BS$.

[asy]
import geometry;
unitsize(5cm);
defaultpen(fontsize(10pt));

point A = dir(120);
point B = dir(210);
point C = dir(330);
point I = incenter(A, B, C);
point O = circumcenter(A, B, C);
point T = circumcenter(I, B, C);
point S = 2 * O - T;
line l = bisector(S, T);
point E = 2 * (projection(l) * A) - A;
point D = extension(A, E, B, S);
point L = intersectionpoint(line(E, B), parallel(D, line(B,C)));
point R = 2 * circumcenter(B, D, L) - L;
point P = extension(L, S, T, R);
point X = extension(A, T, B, S);

draw(A--B--C--cycle, lightgray);
draw(S--L, gray);
draw(P--T, dashed);
draw(L--E--A);
draw(T--S);
draw(A--T, lightgray);
draw(B--S);
draw(A--P, gray);
draw(A--B--R--X--cycle, gray);
draw(circumcircle(A, B, C), blue);
draw(circumcircle(D, B, L), red);

dot(A);
dot(B);
dot(C);
dot(S);
dot(T);
dot(E);
dot(D);
dot(L);
dot(P);
dot(R);
dot(X);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
label("$E$", E, (0,-1));
label("$L$", L, W);
label("$X$", X, (1, 0));
label("$R$", R, (1, 0));
label("$S$", S, N);
label("$T$", T, (0,-1));
label("$P$", P, N);
[/asy]

Claim 1. $L$, $P$ and $S$ are collinear.

Proof. Notice that the tangent to $\Omega$ at $S$ is parallel to $BC$ which is parallel to $DL$. Since $S, D, B$ are collinear, the converse of Reim's Theorem gives the result.

Claim 2. $T$, $R$ and $P$ are collinear.

Proof. Since $ST$ is a diameter of $\Omega$ we have $\angle TPS = 90^{\circ}$. On the other hand

\[\angle RPS = 180^{\circ} -  \angle RPL = 180^{\circ} - \angle RDL = 180^{\circ} - 90^{\circ}\]
And this $\angle RPS = \angle TPS$, giving the desired collinearity.

To finish, angle chase to find $\angle ABX = \angle XBR = \frac{\angle B - \angle C}{2} = \angle RAX$, so $AXRB$ is cyclic with $XA = XR$ and

\[\angle XRA = \angle XBA = \angle RBX = \angle RBD\]
And therefore $XR$ is tangent to $\omega$. It now suffices to show that $XR = XP$. By angle chase we have $\measuredangle RXA = 2\measuredangle RPA$, and since $XA = XR$ this implies that $X$ is the circumcenter of $\triangle APR$, finishing the problem.
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MarkBcc168
1593 posts
#25 • 2 Y
Y by PRMOisTheHardestExam, GeoKing
[asy]
size(9cm);
import olympiad;
import geometry;
pair A = dir(119);
pair A_prime = -A;
pair E = reflect((0,0), dir(90)) * A_prime;
pair S = dir(90);
pair D = 0.6*E + 0.4*A;
pair B = 2 * foot((0,0), D, S) - S;
pair P = 2 * foot((0,0), A_prime, D) - A_prime;
pair Q = reflect((0,0), (P+A_prime)/2) * S;
pair M = -S;
pair X = extension(E,Q,S,D);
pair Y = extension(P,Q,A,M);
pair L = extension(D+E-A_prime, D, B,E);
fill(X--M--Q--cycle, paleblue);
fill(A--P--S--cycle, paleblue);
draw(unitcircle, linewidth(1));
draw(S--X, red+linewidth(1.2));
draw(P--A_prime, linewidth(0.8));
draw(circle(B,D,P), linewidth(0.8));
draw(P--Q, linewidth(0.8));
draw(X--M--Q--cycle, blue+linewidth(0.8));
draw(A--P--S--cycle, blue+linewidth(0.8));
draw(A--M, linewidth(0.8));
draw(A--E--A_prime, gray+linewidth(0.7));
draw(S--M, gray+linewidth(0.7));
draw(D--L, gray+linewidth(0.7));
draw(M--A_prime);
dot("$A$", A, dir(118));
dot("$A'$", A_prime, dir(-62));
dot("$E$", E, dir(-84));
dot("$S$", S, dir(89));
dot("$D$", D, dir(16));
dot("$B$", B, 1.5*dir(-4));
dot("$P$", P, dir(133));
dot("$Q$", Q, dir(17));
dot("$M$", M, dir(-82));
dot("$X$", X, dir(-139));
dot("$Y$", Y, dir(143));
dot("$L$", L, dir(165));
[/asy]
Let $A'$ be the antipode of $A$ in $\Omega$. Moreover, let the tangent to $\omega$ at $P$ meet $\Omega$ again at $Q$. Let $M$ be the other midpoint of arc $BC$. Let $X=QE\cap MA'$, and $Y=PQ\cap AM$. We finish the problem in three steps.
  • Since $\angle BPD = \angle BLD = 90^\circ - \angle C = \angle BPA'$, we get that $P,D,A'$ are collinear.
  • Pascal on $QPA'MAE$ gives $PQ\cap AM=Y$, $PA'\cap AE = D$, and $A'M\cap EQ = X$ are collinear.
  • We have $\triangle QMX$ and $\triangle PAS$ are homothetic center $Y$. This is just angle chasing. To spell out details, we have
    • Reim on $PPQ$ and $BDS$ gives $SQ\parallel PD$, implying that $A'Q=SP$, or $\angle PAS=\angle QMX$,
    • $EM=AS$, so $\angle APS = MQX$, and
    • clearly, $MX\parallel AS$.
    Therefore, $S,Y,X$ are collinear.
Hence, $S, Y, D, X$ are collinear. $B$ clearly lies on this line, done.
This post has been edited 4 times. Last edited by MarkBcc168, Jul 8, 2023, 1:12 PM
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SQTHUSH
154 posts
#26
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Let $K$ be the mid-point of arc $CB$ which is not containing $A$,suppose $AK\cap BS=T$,$PK\cap AE=E$
Note that $\angle BDE=\dfrac{1}{2}\angle A=\angle BPK$,so $P,B,Z,D$are cyclic
Consider that $\angle ABD=\angle B+\dfrac{1}{2}\angle A-90^{\circ}=90^{\circ}-\dfrac{1}{2}\angle A-\angle C=\angle DBZ=\angle EAK$
Which means that $A,B,Z,T$ are cyclic,and $TA=TC$
Notice $\angle ZAT=\angle TBZ=\angle ABT$,so $TA^{2}=TB\cdot TD$
Since $2\angle APK=2\angle B+\angle A=360^{\circ}-\angle ATZ$
Hence $TA=TZ=TP$
So $TP^{2}=TB\cdot TD$
Which means that $TP$ is tangent to $\omega$
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