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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
something...
SunnyEvan   0
4 minutes ago
Source: unknown.
Try to prove : $$ \sum csc^{20} \frac{2^i \ pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^k \ pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
0 replies
SunnyEvan
4 minutes ago
0 replies
IMO Genre Predictions
ohiorizzler1434   27
N 14 minutes ago by awesomeming327.
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
27 replies
+1 w
ohiorizzler1434
Saturday at 6:51 AM
awesomeming327.
14 minutes ago
Cosine of polynomial is polynomial of cosine
yofro   1
N 17 minutes ago by yofro
Source: 2025 HMIC #2
Find all polynomials $P$ with real coefficients for which there exists a polynomial $Q$ with real coefficients such that for all real $t$, $$\cos(P(t))=Q(\cos t).$$
1 reply
yofro
34 minutes ago
yofro
17 minutes ago
Problem 1
randomusername   72
N 39 minutes ago by blueprimes
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
72 replies
randomusername
Jul 10, 2015
blueprimes
39 minutes ago
Name of a point on a circle
clarkculus   1
N 3 hours ago by martianrunner
Is there a name for the point $P'$ with respect to a circle $\Gamma$, a diameter $\ell$, and a given point $P$, such that $P'$ is the reflection of the $P$-antipode about $\ell$? Equivalently, $P'$ is the the other intersection of $\Gamma$ and the line through $P$ parallel to $\ell$.
1 reply
clarkculus
3 hours ago
martianrunner
3 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   5
N 4 hours ago by Math-lover1
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3
5 replies
SomeonecoolLovesMaths
Yesterday at 8:16 AM
Math-lover1
4 hours ago
GCD of consecutive terms
nsato   33
N 6 hours ago by reni_wee
The numbers in the sequence 101, 104, 109, 116, $\dots$ are of the form $a_n = 100 + n^2$, where $n = 1$, 2, 3, $\dots$. For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n + 1}$. Find the maximum value of $d_n$ as $n$ ranges through the positive integers.
33 replies
nsato
Mar 14, 2006
reni_wee
6 hours ago
Largest Divisor
4everwise   19
N 6 hours ago by reni_wee
What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?
19 replies
4everwise
Dec 22, 2005
reni_wee
6 hours ago
Can anyone solve this binomial identity
sasu1ke   0
Yesterday at 6:19 PM


\[
\sum_{0 \leq k \leq l} (l - k) \binom{m}{k} \binom{q + k}{n}
= \binom{l + q + 1}{m + n + 1},
\]\[
\text{integers } l, m \geq 0,\quad \text{integers } n \geq q \geq 0.
\]
0 replies
sasu1ke
Yesterday at 6:19 PM
0 replies
[ABCD] = n [CDE], areas in trapezoid - IOQM 2020-21 p1
parmenides51   3
N Yesterday at 6:04 PM by iamahana008
Let $ABCD$ be a trapezium in which $AB \parallel CD$ and $AB = 3CD$. Let $E$ be then midpoint of the diagonal $BD$. If $[ABCD] = n \times  [CDE]$, what is the value of $n$?

(Here $[t]$ denotes the area of the geometrical figure$ t$.)
3 replies
parmenides51
Jan 18, 2021
iamahana008
Yesterday at 6:04 PM
Sequences and GCD problem
BBNoDollar   0
Yesterday at 3:23 PM
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
Yesterday at 3:23 PM
0 replies
Sum of digits is 18
Ecrin_eren   13
N Yesterday at 3:20 PM by NamelyOrange
How many 5 digit numbers are there such that sum of its digits is 18
13 replies
Ecrin_eren
Saturday at 1:10 PM
NamelyOrange
Yesterday at 3:20 PM
Inequalities
sqing   1
N Yesterday at 3:18 PM by DAVROS
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
1 reply
sqing
Yesterday at 12:46 PM
DAVROS
Yesterday at 3:18 PM
an algebra problem
Asyrafr09   2
N Yesterday at 12:03 PM by pooh123
Determine all real number($x,y,z$) that satisfy
$$x=1+\sqrt{y-z^2}$$$$y=1+\sqrt{z-x^2}$$$$z=1+\sqrt{x-y^2}$$
2 replies
Asyrafr09
Yesterday at 10:05 AM
pooh123
Yesterday at 12:03 PM
angle BAS = angle CAM
Ovchinnikov Denis   14
N Nov 8, 2024 by mcmp
Source: All-Russian Olympiad 2010 grade 10 P-3
Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$. Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$. Prove that $\angle BAS =\angle CAM$.
14 replies
Ovchinnikov Denis
Sep 9, 2010
mcmp
Nov 8, 2024
angle BAS = angle CAM
G H J
Source: All-Russian Olympiad 2010 grade 10 P-3
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Ovchinnikov Denis
470 posts
#1 • 3 Y
Y by HWenslawski, Adventure10, Mango247
Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$. Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$. Prove that $\angle BAS =\angle CAM$.
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aboojiga
43 posts
#2 • 4 Y
Y by NHN, Everything999, Adventure10, and 1 other user
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kyyuanmathcount
1973 posts
#3 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Easily with angles, we see that OPQ is similar to ABC in that orientation --> Let perpendicular from S to AD be ST.

Note from simple angles that <SOA = 90, so AOST is cyclic and <SAT = <SOT = <OAM.

It follows that <BAS = <BAD + <SAT = <OAC + <OAM = <CAM.
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erfan_Ashorion
102 posts
#4 • 2 Y
Y by Adventure10, Mango247
oh its really nice problem but....!
suppose $op$ intersect the $AB$ at $Y$ and $OQ$ intersect $AC$ at $x$.
it is really easy to show that $XY$ is parallel to $BC$ so if midpoint of $XY$ is $N$ we know that $XAN=CAM$
lemma:
in the triangle $AXY$ if the tangent to the circumcircle of it on $X$ and $Y$ intersect each other on $Z$ than $AZ$ concides with a symmedian of $AXY$ ..!
proof lemma:
on attachment!!
so it easy to show that $S$ lie on $AZ$..!(not so easy :D if you need i will say it :lol: )
Attachments:
lemma.pdf (196kb)
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sahadian
112 posts
#5 • 1 Y
Y by Adventure10
It is enough if we prove that $\angle DAM=\angle OAS$ SO we prove that the triangles $DAM$ and $AOS$ are similar .
for proving this first we prove $\angle AOS=\angle ADM=90^{\circ}$
then it is easy to prove that $\frac{AD}{DM}=\frac{AO}{OS}$ by using this:
$DM=\frac{AC^2-AB^2}{2BC}$
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junioragd
314 posts
#6 • 2 Y
Y by Adventure10, Mango247
Let the tangent at $A$ to the circumcircle of $ABC$ intersects $BC$ at $D$.Now,by an easy angle chase and using that $MOAD$ is cyclic,it is enough to prove that $DBO$ is similar to $APS$ and we have that angles $DBO$ and $APS$ are equal,so it is enough to prove that $AP/PS=BD/OB$ but this is easy cause $AOP$ is similar to $ADB$ and $OAB$ is similar to $OPS$,so done.
This post has been edited 1 time. Last edited by junioragd, Jan 27, 2015, 5:36 PM
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TelvCohl
2312 posts
#7 • 3 Y
Y by amar_04, Adventure10, Mango247
My solution:

Let $ N $ be the projection of $ O $ on $ PQ $ .

Since $ OP \perp AB, OQ \perp AC, PQ \perp BC $ ,
so we get $ \triangle OPQ \cup S \cup N \sim \triangle ABC \cup O \cup D $ and $ \angle AOS=90^{\circ} $ ,
hence from $ AO:OS=AD:ON=AD:DM \Longrightarrow Rt \triangle AOS \sim Rt \triangle ADM $ ,
so we get $ \angle DAM=\angle OAS \Longrightarrow \angle BAS=\angle CAM $ .

Q.E.D
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djmathman
7938 posts
#8 • 1 Y
Y by Adventure10
WLOG let $AB<AC$, so that $O$ is in the interior of $\triangle ADC$. Let $M_C=OP\cap AB$ and $M_B=OQ\cap AC$. Note that by angle-chasing $\angle QPO=\angle APM_C=\angle ABD$ and $\angle PQO\equiv\angle AQM_B=\angle ACD$, so $\triangle ABC\sim\triangle OPQ$. Hence there exists a spiral similarity sending $\triangle ABC$ to $\triangle OPQ$. Note that this sends $PQ$ to $BC$, so the angle of rotation of this spiral similarity is $90^\circ$. Thus, since $A$ is sent to $O$ and $O$ is sent to $S$, we have $AO\perp OS$.

Now let $X$ be the orthogonal projection of $O$ onto $PQ\equiv AD$. Remark that by similarity and the fact that $MOXD$ is a rectangle, \[\dfrac{MD}{DA}=\dfrac{OX}{AD}=\dfrac{AO}{OS},\quad\text{so}\quad \triangle AOS\sim\triangle ADM.\]Thus $\angle OAS=\angle MAD$. Combining this with $\angle BAD=\angle CAO$ yields the desired equality. $\blacksquare$
This post has been edited 2 times. Last edited by djmathman, Dec 12, 2015, 10:01 PM
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Kayak
1298 posts
#9 • 1 Y
Y by Adventure10
Let $M_B, M_C$ be the midpoint of $\overline{AC}, \overline{AB}$ respectively. Now $\angle PQO = 90 - \angle QAM_C = \angle ACB$, similarly $\angle QPO = \angle ABC$ and thus $\Delta ABC \sim \Delta OPQ$. Let $X$ be the Miquel point of $\Delta M_BM_CD$ w.r.t $\Delta ABC$, and observe $X$ is the center of spiral similarity $f$ mapping $(A,B,C,O,X) \overset{f}{\mapsto} (O,P,Q,S,X)$. Since $f(\overline{PQ}) = \overline{BC}$, $f$ rotates things by $90$. So $\angle SXO = 90$, and thus $\angle AXS = \angle AXO + \angle SXO = 90 + 90  = 180$, and thsu $A,X,S$ are colinear.

Now let $\Phi(Y)$ denote the image of $Y$ after a $\sqrt{bc}$ inversion, then a reflection along $A$-angle bisector, and then a homothety at $A$ with factor $\frac{1}{2}$. We have $\Phi(M_B) = C, \Phi(M_C) = B, \Phi(B) = M_C, \Phi(C) = M_B, \Phi(D) = O$, and thus clearly $\Phi(E) = M$ ('coz $M_C,O,M,B$ are conyclic).

Thus $AX$ and $AM$ are isogonal and we're done.
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rmtf1111
698 posts
#10 • 10 Y
Y by Kayak, AlastorMoody, RudraRockstar, amar_04, khina, 606234, hakN, IAmTheHazard, Adventure10, Mango247
this problem is the proof that russians have time machines. how could they know about 2017G3 in 2010 otherwise??
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SerdarBozdag
892 posts
#12
Y by
$$\frac{IS}{SJ}=\frac{AI}{JF}  \leftrightarrow  \frac{OM}{OK}=\frac{KJ}{JF} \leftrightarrow      \frac{OK}{OM}=\frac{KF}{KJ}-1    \leftrightarrow  \frac{MK\cdot KJ}{OM}=KF$$We have $t=BF$, $MK=h$, $AP=\frac{c^2}{2h}$ and $AQ=\frac{b^2}{2h} \implies KJ=AI=\frac{b^2+c^2}{4h}$, $OM=R\cdot cos(A)$. Thus it is enough to prove that $$\frac{b^2+c^2}{4R\cdot cos(A)}=h+tsin(A) \leftrightarrow t=\frac{a}{2cos(A)}$$
$x=\angle BAF$, $t=FA\frac{sinx}{sinc}=FA\frac{a}{2AM}=\frac{a}{2cos(A)}$
In the last equation we used sin law in $FBA$ and $MAC$. We also used the fact that $\frac{AM}{AF}=cos(A)$ which can be proved by noticing $AMC\sim AFR$.
Attachments:
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Kingjon
4 posts
#13
Y by
I have solution from sinus. But its easy to chase angels and sinus is very similar to it anyway
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math_comb01
662 posts
#14 • 1 Y
Y by Rounak_iitr
Amazing 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 = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -11.028184276513091, xmax = 7.651231623973653, ymin = -9.68127833226805, ymax = 5.291831456807111; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); 

draw((-0.68,1.85)--(-2.2,-2.65)--(6.9,-3.07)--cycle, linewidth(2) + zzttqq); 
 /* draw figures *//* special point *//* special point *//* special point */
draw((-0.68,1.85)--(-2.2,-2.65), linewidth(0.8)); 
draw((-2.2,-2.65)--(6.9,-3.07), linewidth(0.8)); 
draw((6.9,-3.07)--(-0.68,1.85), linewidth(0.8)); 
draw(circle((2.403617451019996,-1.6982885612334209), 4.700962486344153), linewidth(0.8)); /* special point *//* special point *//* special point *//* special point *//* special point */
draw(circle((-0.1176669679699406,-3.8893946393365), 3.340332463534383), linewidth(0.8) + fuqqzz); /* special point */
draw((-0.68,1.85)--(-0.1176669679699406,-3.8893946393365), linewidth(0.8)); /* special point */
draw(circle((-1.496959200161584,-1.6341160035009759), 1.235429747326925), linewidth(0.8) + qqwuqq); 
draw(circle((2.9038224601090685,-5.077180030970174), 4.4719354430776574), linewidth(0.8) + qqwuqq); /* special point */
draw((-2.2,-2.65)--(-0.7939184003231672,-0.6182320070019524), linewidth(0.8)); 
draw((-1.0923550797818622,-7.084360061940349)--(6.9,-3.07), linewidth(0.8)); 
draw(circle((0.8618087255099981,0.07585571938328961), 2.3504812431720765), linewidth(0.8)); 
draw((-0.68,1.85)--(2.403617451019996,-1.6982885612334209), linewidth(0.8)); 
draw((-0.30633208990913463,-1.9638033476805354)--(2.403617451019996,-1.6982885612334209), linewidth(0.8)); 
draw((2.403617451019996,-1.6982885612334209)--(-0.1176669679699406,-3.8893946393365), linewidth(0.8)); 
draw((-1.44,-0.4)--(2.403617451019996,-1.6982885612334209), linewidth(0.8)); 
draw((2.403617451019996,-1.6982885612334209)--(-1.0923550797818622,-7.084360061940349), linewidth(0.8)); 
draw((2.403617451019996,-1.6982885612334209)--(3.11,-0.61), linewidth(0.8)); 
draw((-1.0923550797818622,-7.084360061940349)--(-0.68,1.85), linewidth(0.8)); 
 /* dots and labels */
label("$A$", (-0.6,1.93), NE * labelscalefactor); 
label("$B$", (-2.12,-2.57), NE * labelscalefactor); 
label("$C$", (6.98,-2.99), NE * labelscalefactor); 
label("$O$", (2.48,-1.61), NE * labelscalefactor); 
label("$P$", (-0.72,-0.53), NE * labelscalefactor); 
label("$Q$", (-1.02,-7.01), NE * labelscalefactor); 
label("$M$", (-1.36,-0.33), NE * labelscalefactor); 
label("$N$", (3.18,-0.53), NE * labelscalefactor); 
label("$S$", (-0.04,-3.81), NE * labelscalefactor); 
label("$D$", (-0.82,-2.63), NE * labelscalefactor); 
label("$Q_A$", (-0.22,-1.89), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */[/asy]
Let $M$ denote the midpoint of $AB$, $N$ denote the midpoint of $AC$, and let $Q_A$ denote the $A$-dumpty point.
Claim 1: $\triangle OPQ \sim \triangle ABC$
Proof
Since both the triangles are positively similar , therefore there exists a point $Q_A$ such that it takes the former triangle to the latter.
Claim 2:$Q_A$ is the $A$-dumpty point
Proof
Claim 3: $A-Q_A-S$
Proof
Remark
This post has been edited 2 times. Last edited by math_comb01, Nov 14, 2023, 5:12 PM
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Hertz
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Let $F$ and $G$ be midpoints of $AB$ and $AC$ respectively. Let $K$ be the midpoint $PQ$.

$\angle POQ = 180 - \angle AOF - \angle AOG = 180 - \angle C - \angle B = \angle A$

Since triangle $AFP$ and triangle $ADB$ are similar we get $\angle QPO = \angle AFP = \angle B$

So we get triangle $OPQ$ and triangle $ABC$ are similar in that orientation.

$\angle SOA = \angle FOA + \angle SOP = \angle C + 90 - \angle C = 90$ So $AO$ is tangent to ($OPQ$), thus $AOST$ is cyclic.

$\angle SAK = \angle SOK = \angle OAM$

$\angle BAS = \angle BAD + \angle SAK = \angle OAC + \angle OAM = \angle CAM$
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mcmp
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Woah Russian Geo :love: :love:

I love it when I can just nuke a problem with a technique that appears seven whole years after the problem (in the guise of ISL 2017/G3 :D).

Rename $M\to M_a$, let the midpoint of $PQ$ be $M$, and let the tangent to $(ABC)$ at $A$ intersect $\overline{BC}$ at $A’$. Let $M_b$, $M_c$ be the midpoints of $AC$, $AB$ resp.
Diagram
Imma first show that $AO$ tangent to $(OPQ)$; to do this, notice that it would suffice to show that $AP\cdot AQ=AO^2$. However notice that $M_cPDB$ and $M_bPDC$ are cyclic because of the right angles, so $AP\cdot AD=AM_c\cdot AB$ and $AQ\cdot AD=AM_b\cdot AC$. Hence:
\begin{align*}
AP\cdot AQ&=\frac{AM_c\cdot AM_b\cdot AB\cdot AC}{AD^2}\\
&=\frac{AB^2\cdot AC^2}{4AD^2}\\
&=\frac{A^2}{AD^2\sin^2(\angle BAC)}\\
&=\frac{BC^2}{\sin^2(\angle BAC)}\\
&=AO^2
\end{align*}the last line following from the law of sines. Hence we have the tangency. Now $\measuredangle OPQ=\measuredangle M_cPD=\measuredangle M_cBD=\measuredangle ABC$, and similarly with $\measuredangle OQP=\measuredangle ACB$, so $A’ABCM_a\stackrel{+}{\sim}AOPQM$. The rest is now trivial. $\measuredangle SAD=\measuredangle SAM=\measuredangle OA’M_a=\measuredangle OAM_a$, so $AM_a$ and $AS$ are conjugate in $\angle DAO$; since $AD$ and $AO$ conjugate in $\angle BAC$, they share the same angle bisector and hence $AM_a$ and $AS$ are conjugate in $\angle BAC$ as desired. $\square$
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