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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
1 viewing
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Geometry One Problem Bounty Hunt Contest: Trapezium Geometry
anantmudgal09   14
N a minute ago by ihategeo_1969
Source: zephyrcrush78
Let $ABC$ be an acute-angled scalene triangle with incircle $\omega$ and circumcircle $\Gamma$. Suppose $\omega$ touches line $BC$ at $D$ and the tangent to $\Gamma$ at $A$ meets line $BC$ at $T$. Two circles passing through $A$ and $T$ tangent to $\omega$ meet line $AD$ again at $X$ and $Y$.

Prove that $BXCY$ is a trapezium.

14 replies
anantmudgal09
Jun 3, 2024
ihategeo_1969
a minute ago
Combinations of lines
M11100111001Y1R   2
N 5 minutes ago by sami1618
Source: Iran TST 2025 Test 1 Problem 3
Suppose \( n > 10 \) lines are drawn on the plane such that no three of them are concurrent and no two are parallel. At least \( \frac{n^2}{8} + 1 \) of the bounded regions formed are colored black. A triangle formed by three lines is called a \textit{good triangle} if it lies entirely within a black region. Prove that there are at least \( \frac{n}{2} \) good triangles. (A good triangle is a bounded region with finite area.)
2 replies
M11100111001Y1R
May 27, 2025
sami1618
5 minutes ago
Max and Min
Butterfly   0
27 minutes ago

Let $a_1,a_2,\cdots,a_n$ be an arrangement of $\{1,2,3,\cdots,n\}$. Find the maximum and minimum values of $$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}.$$
0 replies
Butterfly
27 minutes ago
0 replies
Cup of Combinatorics
M11100111001Y1R   4
N 32 minutes ago by sami1618
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
4 replies
M11100111001Y1R
May 27, 2025
sami1618
32 minutes ago
(c^n+1)/(2^na+b) is an integer for all n
parmenides51   2
N 33 minutes ago by Assassino9931
Source: Ukraine TST 2010 p6
Find all pairs of odd integers $a$ and $b$ for which there exists a natural number$ c$ such that the number $\frac{c^n+1}{2^na+b}$ is integer for all natural $n$.
2 replies
parmenides51
May 4, 2020
Assassino9931
33 minutes ago
Nine point circle + Perpendicularities
YaoAOPS   18
N an hour ago by AndreiVila
Source: 2025 CTST P2
Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.
18 replies
YaoAOPS
Mar 5, 2025
AndreiVila
an hour ago
Inequality conjecture
RainbowNeos   0
an hour ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
an hour ago
0 replies
inequality 2905
pennypc123456789   0
an hour ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
pennypc123456789
an hour ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N an hour ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
sqing
Today at 3:43 AM
sqing
an hour ago
Inspired by qrxz17
sqing   7
N an hour ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
7 replies
sqing
5 hours ago
sqing
an hour ago
Geometry problem
Whatisthepurposeoflife   2
N 2 hours ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
2 hours ago
A Sequence of +1's and -1's
ike.chen   36
N 2 hours ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
ike.chen
Jul 9, 2023
maromex
2 hours ago
Basic ideas in junior diophantine equations
Maths_VC   1
N 2 hours ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
2 hours ago
Inspired by qrxz17
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
1 viewing
sqing
6 hours ago
sqing
2 hours ago
Geo Final but hard to solve with Conics...
Seungjun_Lee   5
N Apr 2, 2025 by L13832
Source: 2025 Korea Winter Program Practice Test P4
Let $\omega$ be the circumcircle of triangle $ABC$ with center $O$, and the $A$ inmixtilinear circle is tangent to $AB, AC, \omega$ at $D,E,T$ respectively. $P$ is the intersection of $TO$ and $DE$ and $X$ is the intersection of $AP$ and $\omega$. Prove that the isogonal conjugate of $P$ lies on the line passing through the midpoint of $BC$ and $X$.
5 replies
Seungjun_Lee
Jan 18, 2025
L13832
Apr 2, 2025
Geo Final but hard to solve with Conics...
G H J
G H BBookmark kLocked kLocked NReply
Source: 2025 Korea Winter Program Practice Test P4
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Seungjun_Lee
526 posts
#1 • 3 Y
Y by Rounak_iitr, sashamusta, ehuseyinyigit
Let $\omega$ be the circumcircle of triangle $ABC$ with center $O$, and the $A$ inmixtilinear circle is tangent to $AB, AC, \omega$ at $D,E,T$ respectively. $P$ is the intersection of $TO$ and $DE$ and $X$ is the intersection of $AP$ and $\omega$. Prove that the isogonal conjugate of $P$ lies on the line passing through the midpoint of $BC$ and $X$.
This post has been edited 1 time. Last edited by Seungjun_Lee, Jan 18, 2025, 12:44 PM
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TestX01
341 posts
#2
Y by
What is $O$?
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Acorn-SJ
59 posts
#3
Y by
@above
$O$ is the center of circle $\omega$. I’ll tell him to fix the statement
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Seungjun_Lee
526 posts
#4 • 4 Y
Y by Nari_Tom, Stuffybear, ehuseyinyigit, L13832
Let $Y$ be the point on $\omega$ that $AX$ and $AY$ are isogonal, and $S, L$ be the intersection of $XM, YM$ and $\omega$, respectively. We can see that $SL \parallel BC$. All polars taken in the solution is a polar wrt the $A$ mixtilinear incircle.

Using Well Known Fact, it suffices to prove that $PM$ and $AQ$ are parallel. Now, as $-1 = (B,C;M, \infty_{BC}) = (YB,YC;YM,Y\infty_{BC}) = (B,C;L,X)$, we know that $LX$ passes through the intersection of tangents from $B,C$ to $\omega$. This implies that $O,M,X,L$ are concyclic. Since $\angle YAX = \angle YLX = \angle MLX$, it suffices to prove that $P,M,X,L$ are concyclic. We will prove that $P,O,M,X$ are concyclic by proving that $\angle XPO = \angle XMO$.

Since $P$ lies on the $A$ polar, $A$ lies on the polar of $P$. Since $-1 = (B,C;L,X) = (AD, AE; AL, AP)$, we can easily see that $AL$ is $P$ polar. This implies that $AL \perp PO$. Hence, $\angle XPO = \angle (XP, OP) = \angle (XA, LA) + \angle (LA, OP)$. From $\angle (LA, OP) = \angle (CM, MO)$ and $\angle (XA, LA) = -\angle LAX = -\angle LSX = -\angle CMX = \angle XMC$, we obtain that $\angle XPO = \angle (XA, LA) + \angle (LA, OP) = \angle XMC + \angle CMO = \angle XMO$, as desired.
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TestX01
341 posts
#5 • 2 Y
Y by ehuseyinyigit, L13832
i swear the motivation is so hard to find so i rate 40 mohs but @above is so orz :love: :blush:

or maybe i didn't do enough taiwan-style config geo

https://i.imgur.com/CrDGUBa.png

Let $N$ be the midpoint of major arc $BC$, and $F$ midpoint of minor arc $BC$. Let $O'$ be the centre of the mixtilinear incircle. Let $I$ be the incentre of $\triangle ABC$, and $M$ the midpoint of $BC$.

We reconstruct $Q$ as the intersection of the reflection of $AP$ over $AI$ with $MX$. It suffices to show that $QC$ and $CP$ are isogonal in $\angle BCA$ to prove $P$ and $Q$ are isogonal conjugates.

The following results are well-known or trivial to see:
  • $DE\perp AI$, and $D,I,E$ collinear.
  • $O,O', T$ collinear.
  • $T,I,N$ collinear.
  • $DBTI$ and $ECTI$ cyclic.
  • $N,O,M,F$ collinear.
  • $A,I,O',F$ collinear.
Our solution depends on the following claim:
Claim: $XMOP$ is cyclic.
Proof:
We will show that $\triangle TIX\sim\triangle FMX$. Clearly, $\angle XFN=\angle XTN$ by Bowtie, hence we just need
\[\frac{IT}{MF}=\frac{TX}{FX}\]To conclude by $SAS$ similarity.

Let us proceed by trigonometric bash. First evaluate the right hand side,
\[\frac{TX}{FX}=\frac{\sin\angle XFT}{\sin\angle FTX}=\frac{\sin\angle TAX}{\sin\angle FAX}\]However, by Ratio Lemma in triangle $\triangle ATO'$, we have
\[\frac{\sin\angle TAX}{\sin\angle FAX}=\frac{TP}{PO'}\times \frac{AO'}{TA}\]Now, by Ratio Lemma in triangle $\triangle TIO'$ now, we have
\[\frac{TP}{PO'}=\frac{IT}{IO'}\times\frac{\sin\angle DIT}{\sin 90^\circ}\]Hence we have
\[\frac{TX}{FX}=\frac{AO'}{TA}\times \frac{ IT\sin\angle DIT}{IO'}\]Now, let's move to the left hand side of our initial equation. First of all, we prove an important similarity: $\triangle IMF\sim\triangle AIT$.

Indeed, by Shooting Lemma and Fact 5 we have $FB^2=FI^2=FM\times FN$ hence $(NIM)$ is tangent to $FI$. Thus, $\measuredangle FIM=\measuredangle INF=\measuredangle TAF$, and by Bowtie we have $\angle MFI=\angle ITA$. This is sufficient due to $AA$ similarity.

Now, this implies that $\frac{IT}{MF}=\frac{AT}{IF}$. Thus, it simply suffices to show that
\[\frac{AT}{IF}=\frac{AO'}{TA}\times\frac{IT\sin\angle DIT}{IO'}\]Now, by Ratio Lemma in $\triangle ATI$, we have
\[\frac{AO'}{IO'}=\frac{AT}{TI}\times\frac{\sin\angle ATO'}{\sin\angle ITO'}\]Hence it simply suffices to show that
\[\frac{AT}{IF}=\frac{\sin \angle DIT\sin\angle ATO'}{\sin\angle ITO'}\]Now, $BDIT$ is cyclic as mentioned before, hence
\[\sin{\angle DIT}=\sin\angle TBA=\sin\angle ANT=\frac{AI}{NI}\]Because $\triangle ANI$ is right. However, we also have $\triangle AIT\sim\triangle NIF$ because of Bowties and $AA$ similarity. Hence, $\frac{AI}{NI}=\frac{AT}{NF}$. Hence it suffices to prove
\[\frac{NF}{IF}=\frac{\sin\angle ATO'}{\sin\angle ITO'}\]However, from $TO=ON$ by circumcentre definition, $\angle ITO'=\angle ONI$. Further, we have
\[\measuredangle ATO'=\frac{180^\circ-\measuredangle TOA}{2}=90^\circ-\measuredangle TCA=90^\circ-\measuredangle TNA=\measuredangle AIN\]due to angle at centre theorem.

This implies that
\[\frac{\sin\angle ATO'}{\sin\angle ITO'}=\frac{\sin \angle AIN}{\sin\angle FNI}\]And the right hand side is $\frac{NF}{IF}$ due to sine law in $\triangle INF$.

This concludes our trigonometric bash.

Hence, $\triangle TIX\sim\triangle FMX$.

Note that Reim's Theorem on $(ABC)$ and parallel lines $AN$ and $DE$ (Both are perpendicular to $AI$) implies that $IPXT$ is cyclic.

Using our similarity,
\[\measuredangle FMX=\measuredangle TIX=\measuredangle TPX=\measuredangle OPX\]Which finally gives $OPMX$ cyclic.

This concludes our claim. $\square$

Lemma: $PM\parallel AQ$
Proof:
We will angle chase:
\[\measuredangle MPX=\measuredangle FOX=2\measuredangle FNX=2\measuredangle FAX=\measuredangle QAX\]Where we use $XMOP$ cyclic, angle at centre theorem, and the fact that $AI$ bisects $\angle QAP$. This suffices by corresponding angles in parallel lines.$\square$.

Now, in order to show that $CP$ and $CQ$ are isogonal, we will employ DDIT at $C$ on quadrilateral $AQMP$. Note that $CA$ and $CM$ are clearly isogonal. If we show that $AP\cap QM=X$ and $PM\cap AQ=P_\infty$ are isogonal, this would characterize our involution, then imply $CP$ and $CQ$ also being isogonal.

Indeed, it suffices to show that $\measuredangle(AC,CX)=\measuredangle(NM,BC)$ to prove isogonality. By Pitot Theorem on $\triangle NOT$, $NIMX$ is cyclic. In particular, it is tangent to $AI$ from our earlier result.

Now we claim that $I$ is the incentre of $\triangle AQX$ as well. Indeed, $AI$ bisects $\angle QAP$. Further,
\[\measuredangle AQX=\measuredangle PMX=\measuredangle POX=180^\circ-2\measuredangle INX=180^\circ-2\measuredangle FIX=2\measuredangle AIX-180^\circ\]Which suffices by well-known incentre angle formula, where we used angle at center theorem and $XMOP$ cyclic.

Now, let us angle chase:
\begin{align*}\measuredangle ACX&=\measuredangle ANX\\
&=\measuredangle INX+\measuredangle ANI\\
&=\measuredangle FIX+\measuredangle AXT\\
&=\measuredangle FIX+\measuredangle IXF\\
&=180^\circ-\measuredangle XFA\\
&=180^\circ-\frac{1}{2}\measuredangle XOA\\
&=90^\circ+\measuredangle PXO\\
&=90^\circ+\measuredangle PMO\\
&=\measuredangle PMB
\end{align*}As desired, where we used $(NIMX)$ tangent to $AI$, $MOPX$ cyclic, and multiple Bowties, as well as because from $XI$ bisecting $\measuredangle AXM$,
\[\measuredangle AXI=\measuredangle IXM=\measuredangle TXF,\]implying $\measuredangle AXT=\measuredangle IXF$. (Similar switch gives $\triangle IMX\sim\triangle TFX$, and our angle equality.)

This concludes the problem, as $PC$ and $QC$ are now isogonal, and hence $Q,P$ are isogonal conjugates, but $Q$ lies on $MX$, so we are done.
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L13832
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Nice problem man! Cool solutions above :orz:
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