Plan ahead for the next school year. Schedule your class today!

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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
algebra problem
COCBSGGCTG3   0
6 minutes ago
Source: Azerbaijan Junior Math Olympiad Training TST 2025 P1
If $y^2-y=5$ is satisfied for the real number $y$, then $\frac{25+y^4}{y^2}$$=?$
0 replies
+2 w
COCBSGGCTG3
6 minutes ago
0 replies
Satisfy principal’s request?
a_507_bc   5
N 12 minutes ago by TigerOnion
Source: Australia MO 2024 P6
In a school, there are $1000$ students in each year level, from Year $1$ to Year $12$. The school has $12 000$ lockers, numbered from $1$ to $12 000$. The school principal requests that each student is assigned their own locker, so that the following condition is satisfied: For every pair of students in the same year level, the difference between their locker numbers must be divisible by their year-level number. Can the principal’s request be satisfied?
5 replies
a_507_bc
Feb 24, 2024
TigerOnion
12 minutes ago
Functional equation
Martin N.   11
N 15 minutes ago by Royal_mhyasd
Source: Swiss Math Olympiad 2010 - final round, problem 6
Find all functions $ f: \mathbb{R}\mapsto\mathbb{R}$ such that for all $ x$, $ y$ $ \in\mathbb{R}$,
\[ f(f(x))+f(f(y))=2y+f(x-y)\]
holds.
11 replies
Martin N.
Mar 16, 2010
Royal_mhyasd
15 minutes ago
A functional equation
joybangla   16
N 19 minutes ago by math-olympiad-clown
Source: Switzerland Math Olympiad, Final round 2014, P3
Find all such functions $f :\mathbb{R}\to \mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following holds :
\[ f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y) \]
16 replies
joybangla
Jun 2, 2014
math-olympiad-clown
19 minutes ago
Inequalities
sqing   18
N an hour ago by DAVROS
Let $ a,b,c\geq 0  . $ Prove that
$$ \sqrt{ a^3+b^3+c^3+\frac{1}{4}} +  \frac{9}{5}abc+\frac{1}{2} \geq a+b+c$$O706
18 replies
sqing
Jul 19, 2025
DAVROS
an hour ago
Inequalities
sqing   21
N 2 hours ago by sqing
Let $ a,b>0, ab=8. $ Prove that
$$ \frac{1}{\sqrt{a+1}}+\frac{2}{\sqrt{b+4}}\leq  2\sqrt{\sqrt{2}-1}$$$$ \frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+4}}\leq \frac{1}{2\sqrt{2}} \sqrt{\sqrt{302+150\sqrt{5}}-5\sqrt{5}-5}$$$$ \frac{1}{\sqrt{a+1}}+\frac{4}{\sqrt{b+4}}\leq \frac{1}{\sqrt{2}} \sqrt{\sqrt{302+150\sqrt{5}}-5\sqrt{5}-5}$$
21 replies
sqing
Jun 7, 2025
sqing
2 hours ago
Inequalities
sqing   13
N 2 hours ago by sqing
Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $ and $ abc = -16 $. Prove that$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} -a^2\geq 2$$$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b}-a^2-bc\geq -2$$Equality holds when $a=-4,b=c=2.$
13 replies
sqing
Jul 9, 2025
sqing
2 hours ago
[PMO24 Qualifying I.3] Lattice Points
kae_3   1
N 2 hours ago by Siopao_Enjoyer
A lattice point is a point $(x,y)$ where $x$ and $y$ are both integers. Find the number of lattice points that lie on the closed line segment whose endpoints are $(2002,2022)$ and $(2022,2202)$.

$\text{(a) }20\qquad\text{(b) }21\qquad\text{(c) }22\qquad\text{(d) }23$

Answer Confirmation
1 reply
kae_3
Feb 16, 2025
Siopao_Enjoyer
2 hours ago
[PMO23 Qualifying I.12] Biased Coins
kae_3   1
N 3 hours ago by Siopao_Enjoyer
Alice tosses two biased coins, each of which has a probability $p$ of obtaining a head, simultaneously and repeatedly until she gets two heads. Suppose that this happens on the $r$th toss for some integer $r\geq1$. Given that there is $36\%$ chance that $r$ is even, what is the value of $p$?

$\text{(a) }\dfrac{\sqrt{7}}{4}\qquad\text{(b) }\dfrac{2}{3}\qquad\text{(c) }\dfrac{\sqrt{2}}{2}\qquad\text{(d) }\dfrac{3}{4}$

Answer Confirmation
1 reply
kae_3
Feb 9, 2025
Siopao_Enjoyer
3 hours ago
2024 PMO Part 1 #3
orangefronted   10
N 3 hours ago by Siopao_Enjoyer
The Department of Education orders public schools to have class sizes between 15 and 65, inclusive. Suppose that all public schools comply with this order, and that each possible class size is attained by some class in some public school. What is the largest integer N with the property that there is certainly some class in some public school in the country that has at least N students with the same birth month?
10 replies
orangefronted
Jan 16, 2024
Siopao_Enjoyer
3 hours ago
Solve the equation
yt12   4
N 3 hours ago by Mathzeus1024
Solve the equation:
$6-4x-x^2=\frac{5}{|\sin \frac{y}{x} \cos \frac{y}{x}|}$
4 replies
yt12
Mar 17, 2024
Mathzeus1024
3 hours ago
[PMO24 Qualifying I.2] Five-digit Numbers mod 24
kae_3   1
N 4 hours ago by Siopao_Enjoyer
How many five-digit numbers containing each of the digits $1$, $2$, $3$, $4$, $5$ exactly once are divisible by $24$?

$\text{(a) }8\qquad\text{(b) }10\qquad\text{(c) }12\qquad\text{(d) }20$

Answer Confirmation
1 reply
kae_3
Feb 16, 2025
Siopao_Enjoyer
4 hours ago
[PMO25 Qualifying II.10] Another Acute Triangle
kae_3   1
N 4 hours ago by BinariouslyRandom
Point $D$ is the foot of the altitude from $A$ of an acute triangle $ABC$ to side $BC$. The perpendicular bisector of $BC$ meets lines $AC$ and $AB$ at $E$ and $P$, respectively. The line through $E$ parallel to $BC$ meets line $DP$ at $X$, and lines $AX$ and $BE$ meet at $Q$. Given that $AX=14$ and $XQ=6$, find $AP$.

Answer Confirmation
1 reply
kae_3
Feb 23, 2025
BinariouslyRandom
4 hours ago
Func equations?
cheltstudent   2
N 4 hours ago by cheltstudent
When would be a good time to learn Functional equations and how to solve them, say I'm in grade 8 and finished the introductory series?
2 replies
cheltstudent
5 hours ago
cheltstudent
4 hours ago
Geometry Parallel Proof Problem
CatalanThinker   5
N May 10, 2025 by Tkn
Source: No source found, just yet, please share if you find it though :)
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
5 replies
CatalanThinker
May 9, 2025
Tkn
May 10, 2025
Geometry Parallel Proof Problem
G H J
G H BBookmark kLocked kLocked NReply
Source: No source found, just yet, please share if you find it though :)
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CatalanThinker
13 posts
#1 • 1 Y
Y by Rounak_iitr
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
Attachments:
Z K Y
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ItzsleepyXD
162 posts
#2
Y by
note that $D'$ is midpoint of arc $BC$
It is easy to see the line $DD'$ is perpendicular bisector of segment $EF$
tangent at $A$ to $(ABC)$ intersect $(ADM)$ and $BC$ at $X,Y$
known that $\angle YAD = \angle YDA$ implies that $DA \parallel MX$ .
will prove that $M,X,N$ collinear
Point $Z$ on $(ADM)$ such that $\angle BAZ = \angle MAC$ . so line $DD'$ is perpendicular bisector of segment $MZ$ .
$N' = MX \cap EF$
$-1=(AY,AZ;AB,AC)=(AX,AZ;AE,AF) = (X,Z;E,F) = (MX,MZ;ME,MF) = (N', \infty_{EF} ; E ,F)$ implies that $N'$ is midpoint of $EF$ .
So $MN \parallel AD$
This post has been edited 3 times. Last edited by ItzsleepyXD, May 9, 2025, 4:06 AM
Reason: typoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
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CatalanThinker
13 posts
#3
Y by
Any other solutions..
This post has been edited 1 time. Last edited by CatalanThinker, May 9, 2025, 4:39 AM
Reason: typo
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CatalanThinker
13 posts
#4
Y by
ItzsleepyXD wrote:
note that $D'$ is midpoint of arc $BC$
It is easy to see the line $DD'$ is perpendicular bisector of segment $EF$
tangent at $A$ to $(ABC)$ intersect $(ADM)$ and $BC$ at $X,Y$
known that $\angle YAD = \angle YDA$ implies that $DA \parallel MX$ .
will prove that $M,X,N$ collinear
Point $Z$ on $(ADM)$ such that $\angle BAZ = \angle MAC$ . so line $DD'$ is perpendicular bisector of segment $MZ$ .
$N' = MX \cap EF$
$-1=(AY,AZ;AB,AC)=(AX,AZ;AE,AF) = (X,Z;E,F) = (MX,MZ;ME,MF) = (N', \infty_{EF} ; E ,F)$ implies that $N'$ is midpoint of $EF$ .
So $MN \parallel AD$

Thanks, understood
Z K Y
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CatalanThinker
13 posts
#5
Y by
Any other solutions?
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Tkn
49 posts
#6
Y by
[asy]
import graph;
import geometry;
        
size(9cm);
defaultpen(fontsize(10pt));

pair A = (0,2);
pair B = (-0.8,0);
pair C = (3,0);
pair M = (B+C)/2;
pair in = unit(B-A)+unit(C-A)+A;
pair ex = rotate(90,A)*in;
pair D = extension(A,ex,B,C);

path circ1 = circumcircle(A,B,C);
path circ2 = circumcircle(A,D,M);

pair F = intersectionpoints(A+3*(A-C)--A, circ2)[0];
pair E1 = intersectionpoints(B+3*(B-A)--A, circ2)[0];
pair N1 = (E1+F)/2;
pair P = intersectionpoints(circ1,circ2)[1];
pair Q = extension(F,E1,D,C);

path circ3 = circumcircle(Q,N1,M);
path circ4 = circumcircle(Q,B,E1);

draw(A--B--C--cycle, black);
draw(A--F, black);
draw(B--E1, black);
draw(D--B, black);
draw(E1--F, blue);
draw(D--P, orange);
draw(M--N1, royalblue);
draw(A--D, royalblue);

draw(circ1);
draw(circ2, red);
draw(circ3, deepgreen+dashed);
draw(circ4, deepgreen+dashed);

dot(A);
dot(B);
dot(C);
dot(D);
dot(M);
dot(F);
dot(E1);
dot(N1);
dot(P);
dot(Q);

label("$A$", A, N, black);
label("$B$", B, S+1.25W, black);
label("$C$", C, SE, black);
label("$D$", D, W, black);
label("$M$", M, NE, black);
label("$E$", E1, S, black);
label("$F$", F, N, black);
label("$N$", N1, SW, black);
label("$P$", P, SE, black);
label("$Q$", Q, NW, black);
[/asy]
First, note that $D$ is the midarc $FE$ because $\angle{DAE}=\angle{DAF}$.
Let $P\neq D$ be an intersection of the line $DN$ and $(ADM)$, and $Q$ be an intersection of $\overline{FE}$ and $\overline{DM}$.
Since $FPED$ is a harmonic quadrilateral with diameter $\overline{DP}$, Picking ratio from $A$ to $\overline{DC}$:
$$(D,AP\cap DC;B,C)=-1.$$So, $\overline{AP}$ bisects $\angle{BAC}$. Note that $\angle{DMP}=90^{\circ}$, so $P\in (ABC)$.
It is easy to see that $Q,N,M$ and $P$ are concyclic (since $\angle{QNP}=90^{\circ}=\angle{QMP}$).
Note that $A=BE\cap CF$ and $(AFE)$ meets $(ABC)$ again at $P$.
Therefore $P$ is a spiral center sending $\overline{BC}\mapsto \overline{EF}$. Now, we have $\triangle{PBC}\sim \triangle{PFE}$.

Next, observe that $\angle{QEP}=\angle{PBC}$. So, $Q,B,P$ and $E$ are concyclic.
Note that $Q=BM\cap NE$, and $(QNM)$ meets $(QBE)$ again at $P$.
So, $P$ is also a sprial center sending $\overline{BE}\mapsto \overline{MN}$. Therefore, we have $\triangle{PNM}\sim \triangle{PEB}$.
Since, $P$ sends $\overline{BC}\rightarrow\overline{EF}$, $P$ also sends $\overline{BE}\rightarrow\overline{CF}$. Which implies
$$\triangle{FPC}\sim \triangle{EPB}\sim\triangle{NPM}.$$We must have $$\angle{PNM}=\angle{PFC}=\angle{PFA}=\angle{PDA}.$$Therefore, $\overline{MN}\parallel\overline{AD}$ as desired.
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