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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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[*]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
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
2021 EGMO P4: Reflection of A over EF lies on BC
anser   48
N 4 minutes ago by cursed_tangent1434
Source: EGMO 2021 P4
Let $ABC$ be a triangle with incenter $I$ and let $D$ be an arbitrary point on the side $BC$. Let the line through $D$ perpendicular to $BI$ intersect $CI$ at $E$. Let the line through $D$ perpendicular to $CI$ intersect $BI$ at $F$. Prove that the reflection of $A$ across the line $EF$ lies on the line $BC$.
48 replies
anser
Apr 13, 2021
cursed_tangent1434
4 minutes ago
A final attempt to make a combinatorics problem
JARP091   1
N 12 minutes ago by wimpykid
Source: At the time of posting the problem I do not know the source if any
Let \( N \) be a positive integer and consider the set \( S = \{1, 2, \ldots, N\} \).

Two players alternate moves. On each turn, the current player must select a nonempty subset \( T \subseteq S \) of numbers not previously chosen such that for every distinct \( x, y \in T \), neither \( x \) divides \( y \) nor \( y \) divides \( x \).

After selecting \( T \), all multiples of every element in \( T \), including those in \( T \) itself, are removed from \( S \).

The game continues with the reduced set \( S \) until no moves are possible.
Determine, for each \( N \), which player has a winning strategy if any

Note: It might be wrong or maybe too easy.
1 reply
JARP091
35 minutes ago
wimpykid
12 minutes ago
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   10
N an hour ago by Mahdi_Mashayekhi
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
10 replies
OgnjenTesic
Thursday at 4:02 PM
Mahdi_Mashayekhi
an hour ago
find all Polynomials
andria   9
N an hour ago by A.H.H
Source: Iranian third round 2015 algebra problem 5
Find all polynomials $p(x)\in\mathbb{R}[x]$ such that for all $x\in \mathbb{R}$:
$p(5x)^2-3=p(5x^2+1)$ such that:
$a) p(0)\neq 0$
$b) p(0)=0$
9 replies
andria
Sep 8, 2015
A.H.H
an hour ago
AT // BC wanted
parmenides51   105
N an hour ago by Adywastaken
Source: IMO 2019 SL G1
Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

(Nigeria)
105 replies
parmenides51
Sep 22, 2020
Adywastaken
an hour ago
Two lines concur on (ABC)
amar_04   19
N an hour ago by Giant_PT
Source: XVII Sharygin Corespondnce Round P13
In triangle $ABC$ with circumcircle $\Omega$ and incenter $I$, point $M$ bisects arc $BAC$ and line $\overline{AI}$ meets $\Omega$ at $N\ne A$. The excircle opposite to $A$ touches $\overline{BC}$ at point $E$. Point $Q\ne I$ on the circumcircle of $\triangle MIN$ is such that $\overline{QI}\parallel\overline{BC}$. Prove that the lines $\overline{AE}$ and $\overline{QN}$ meet on $\Omega$.
19 replies
amar_04
Mar 2, 2021
Giant_PT
an hour ago
Nice problem of concurrency
deraxenrovalo   0
an hour ago
Let $(I)$ be an inscribed circle of $\triangle$$ABC$ and touching $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Let $EE'$ and $FF'$ be diameters of $(I)$. Let $X$ and $Y$ be the pole of $DE'$ and $DF'$ with respect to $(I)$, respectively. $BE$ cuts $(I)$ again at $K$. $CF$ cuts $(I)$ again at $L$. The tangent at $K$ of $(I)$ cuts $AX$ at $M$. The tangent at $L$ of $(I)$ cuts $AY$ at $N$. Let $U$ and $V$ be midpoint of $IM$ and $IN$, respectively.

Show that : $UV$, $E'F'$ and perpendicular bisector of $ID$ are concurrent.
0 replies
deraxenrovalo
an hour ago
0 replies
A hunter and an invisible rabbit are playing again...
Phorphyrion   1
N an hour ago by JARP091
Source: 2021 Discord CCCC P4
A hunter and an invisible rabbit play a game in a $2021\times 2021$ grid. The rabbit's starting square is $P_0$ (unknown to the hunter), and after $n-1$ rounds, the rabbit is at square $P_{n-1}$. In the $n$-th round of the game, two things occur in order:

(i) The rabbit moves invisibly to a square $P_n$ which shares a point with $P_{n-1}$ (There are up to eight of these).

(ii) A tracking device searches $k$ squares of the hunter's choosing. If the rabbit is in one of these squares, the rabbit is captured and the game ends.

For what $k$ can the rabbit avoid capture indefinitely?
1 reply
Phorphyrion
Dec 24, 2023
JARP091
an hour ago
SL 2015 G1: Prove that IJ=AH
Problem_Penetrator   137
N 2 hours ago by heheman
Source: IMO 2015 Shortlist, G1
Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.
137 replies
Problem_Penetrator
Jul 7, 2016
heheman
2 hours ago
IMO Shortlist 2011, G2
WakeUp   30
N 2 hours ago by ezpotd
Source: IMO Shortlist 2011, G2
Let $A_1A_2A_3A_4$ be a non-cyclic quadrilateral. Let $O_1$ and $r_1$ be the circumcentre and the circumradius of the triangle $A_2A_3A_4$. Define $O_2,O_3,O_4$ and $r_2,r_3,r_4$ in a similar way. Prove that
\[\frac{1}{O_1A_1^2-r_1^2}+\frac{1}{O_2A_2^2-r_2^2}+\frac{1}{O_3A_3^2-r_3^2}+\frac{1}{O_4A_4^2-r_4^2}=0.\]

Proposed by Alexey Gladkich, Israel
30 replies
WakeUp
Jul 13, 2012
ezpotd
2 hours ago
2-var inequality
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b\geq 0 ,\frac{a}{b+2}+\frac{b}{a+2}+ \frac{ab}{3}\leq 1.$ Prove that
$$ a^2+b^2 +\frac{5}{3}ab \leq 4$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
Rational Points in n-Dimensional Space
steven_zhang123   0
4 hours ago
Let \( T = (x_1, x_2, \ldots, x_n) \), where \( x_i \) is rational for \( i = 1, 2, \ldots, n \). A vector \( T \) is called a rational point in \( n \)-dimensional space. Denote the set of all such vectors \( T \) as \( S \). For \( A = (x_1, x_2, \ldots, x_n) \) and \( B = (y_1, y_2, \ldots, y_n) \) in \( S \), define the distance between points \( A \) and \( B \) as \( d(A, B) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + \cdots + (x_n - y_n)^2} \). We say that point \( A \) can move to point \( B \) if and only if there is a unit distance between two points in \( S \).

Prove:
(1) If \( n \leq 4 \), there exists a point that cannot be reached from the origin via a finite number of moves.
(2) If \( n \geq 5 \), any point in \( S \) can be reached from any other point via moves.
0 replies
steven_zhang123
4 hours ago
0 replies
Inspired by old results
sqing   5
N 4 hours ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
5 replies
sqing
Yesterday at 7:36 AM
sqing
4 hours ago
equation in integers
Pirkuliyev Rovsen   2
N 4 hours ago by ytChen
Solve in $Z$ the equation $a^2+b=b^{2022}$
2 replies
Pirkuliyev Rovsen
Feb 10, 2025
ytChen
4 hours ago
computational with angle wanted, <ABC=45^o, <BAP = 15^o, PC=2, TX=\sqrt3
parmenides51   2
N Jan 19, 2021 by Maths_1729
Source: 2012 Thailand MO Shortlist G3 - TMO9
Let $ABC$ be a triangle with $\angle ABC = 45^o$. Point $P$ lies on the side BC that $PC=2$ units and $\angle BAP = 15^o$. If the line segment of the tangent drawn from point $B$ to the circumcircle of the triangle $APC$ is $\sqrt3$ units, find the measure of the $\angle ACP$.
2 replies
parmenides51
Dec 19, 2020
Maths_1729
Jan 19, 2021
computational with angle wanted, <ABC=45^o, <BAP = 15^o, PC=2, TX=\sqrt3
G H J
G H BBookmark kLocked kLocked NReply
Source: 2012 Thailand MO Shortlist G3 - TMO9
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parmenides51
30653 posts
#1
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Let $ABC$ be a triangle with $\angle ABC = 45^o$. Point $P$ lies on the side BC that $PC=2$ units and $\angle BAP = 15^o$. If the line segment of the tangent drawn from point $B$ to the circumcircle of the triangle $APC$ is $\sqrt3$ units, find the measure of the $\angle ACP$.
This post has been edited 1 time. Last edited by parmenides51, Dec 19, 2020, 7:18 PM
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rafaello
1079 posts
#2
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By the power of a point, we have $(\sqrt3)^2=PB\cdot BC$, since $BC=BP+PC$, we get the following:
$$PB\cdot (PB+2)=3\implies PB=1$$Thus, $BC=3$. Also, we have $$\sin{30^{\circ}}=2\sin{15^{\circ}}\cos{15^{\circ}}=2\sin{15^{\circ}}\sqrt{1-\sin^2{15^{\circ}}}=$$Thus, $x(\sqrt{1-x^2})=\frac14$, where $x=\sin{15^{\circ}}$.
$$x^2(1-x^2)=\frac{1}{16}$$Solving we have $x^2=\frac{1\pm\sqrt{1-\frac14}}{2}=\frac{2\pm\sqrt{3}}{4}$. Notice that positive value is false, since $\sin$ is increasing on $(0,90^{\circ})$. Thus, $x^2=\frac{2-\sqrt{3}}{4}$, or $x=\pm\frac{\sqrt{2-\sqrt{3}}}{2}$, negative is obviously false, thus $x=\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}$.
By the law of sines, we have $$\frac{AP}{\sin{45^{\circ}}}=\frac{PB}{\sin{15^{\circ}}}\implies AP=\frac{\sqrt{2}}{2}\cdot \frac{2\sqrt{2}}{\sqrt{3}-1}\cdot 1=\frac{2}{\sqrt{3}-1}=\sqrt{3}+1$$Therefore, $$AC^2=CP^2+AP^2-2CP\cdot AP\cdot \cos{60^{\circ}}=4+4+2\sqrt{3}-2(\sqrt{3}+1)=6\implies AC=\sqrt{6}$$
Now we see that $ACP\sim BCA$, because
$$\frac{AC}{CP}=\frac{BC}{AC}\Longleftrightarrow AC^2=CP\cdot BC \Longleftrightarrow (\sqrt{6})^2=2\cdot 3$$Therefore, $\angle CAB=\angle APC=60^{\circ}$
Thus, $\angle CAP=\angle CAB-\angle PAB=60^{\circ}-15^{\circ}=45^{\circ}$,
therefore $$\angle BCA=180^{\circ}-\angle APC-\angle CAP=180^{\circ}-60^{\circ}-45^{\circ}=\boxed{75^{\circ}}.$$
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Maths_1729
390 posts
#3
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parmenides51 wrote:
Let $ABC$ be a triangle with $\angle ABC = 45^o$. Point $P$ lies on the side BC that $PC=2$ units and $\angle BAP = 15^o$. If the line segment of the tangent drawn from point $B$ to the circumcircle of the triangle $APC$ is $\sqrt3$ units, find the measure of the $\angle ACP$.

Let $T$ be the point of tangency then By Tangent-Secant Theorem we can easily get $PB=1$ as $BT=\sqrt{3}$ Now take a point $H\in BP$ Such that $BP\perp CH$ Now as $PC=2$ Then $PH=1=PB\implies \angle HBC=30^\circ=\angle HCB=\angle BHP$ and also as $\angle BAP=15^\circ$ so $BH=HA$ Hence $H$ is circumcenter of $\triangle ABC$
Hence $\boxed{\angle ACP=75^\circ}$ $\blacksquare$
This post has been edited 1 time. Last edited by Maths_1729, Jan 19, 2021, 5:51 PM
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