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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.

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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
Problem 3
EthanWYX2009   5
N 26 minutes ago by parkjungmin
Source: 2023 China Second Round P3
Find the smallest positive integer ${k}$ with the following properties $:{}{}{}{}{}$If each positive integer is arbitrarily colored red or blue${}{}{},$
there may be ${}{}{}{}9$ distinct red positive integers $x_1,x_2,\cdots ,x_9,$ satisfying
$$x_1+x_2+\cdots +x_8<x_9,$$or there are $10{}{}{}{}{}{}$ distinct blue positive integers $y_1,y_2,\cdots ,y_{10}$ satisfiying
$${y_1+y_2+\cdots +y_9<y_{10}}.$$
5 replies
EthanWYX2009
Sep 10, 2023
parkjungmin
26 minutes ago
Inspired by old results
sqing   1
N 42 minutes 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.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
1 reply
1 viewing
sqing
an hour ago
sqing
42 minutes ago
Made from a well-known result
m4thbl3nd3r   0
an hour ago
1. Let $a,b,c>0$ such that $$\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}+\sqrt{(c+a)(c+b)}=3+a+b+c.$$Prove that $$\sqrt{\frac{a+b}{2}}+\sqrt{\frac{b+c}{2}}+\sqrt{\frac{c+a}{2}}\ge ab+bc+ca.$$2. Let $x,y,z$ be sidelengths of a triangle such that $$x^2+y^2+z^2+6=2(xy+yz+zx).$$Prove that $$2\sqrt{2x}+2\sqrt{2y}+2\sqrt{2z}+(x-y)^2+(y-z)^2+(z-x)^2\ge x^2+y^2+z^2.$$
0 replies
m4thbl3nd3r
an hour ago
0 replies
Interesting inequalities
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
5 replies
sqing
Yesterday at 12:44 PM
sqing
an hour ago
Collinear points
tenplusten   2
N 3 hours ago by Blackbeam999
Let $A,B,C$ be three collinear points and $D,E,F$ three other collinear points. Let $G,H,I$ be the intersection of the lines $BE,CF$ $AD,CF$ and $AD,CE$,respectively. If $AI=HD$ and $CH=GF$.Prove that $BI=GE$



I hope you will use Pappus theorem in your solutions.
2 replies
tenplusten
Jun 20, 2016
Blackbeam999
3 hours ago
Simple Geometry
AbdulWaheed   0
3 hours ago
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
0 replies
AbdulWaheed
3 hours ago
0 replies
IMO Shortlist 2014 G6
hajimbrak   30
N Today at 12:01 AM by awesomeming327.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$ . Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$ , respectively. We call the pair $(E, F )$ $\textit{interesting}$, if the quadrilateral $KSAT$ is cyclic.
Suppose that the pairs $(E_1 , F_1 )$ and $(E_2 , F_2 )$ are interesting. Prove that $\displaystyle\frac{E_1 E_2}{AB}=\frac{F_1 F_2}{AC}$
Proposed by Ali Zamani, Iran
30 replies
hajimbrak
Jul 11, 2015
awesomeming327.
Today at 12:01 AM
60^o angle wanted, equilateral on a square
parmenides51   4
N Yesterday at 11:49 PM by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 2
A square $ABCD$ is given. Over the side $BC$ draw an equilateral triangle $BCS$ on the outside. The midpoint of the segment $AS$ is $N$ and the midpoint of the side $CD$ is $H$. Prove that $\angle NHC = 60^o$.
.
(Karl Czakler)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
Yesterday at 11:49 PM
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   9
N Yesterday at 11:47 PM by hectorleo123
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
9 replies
OgnjenTesic
Yesterday at 4:02 PM
hectorleo123
Yesterday at 11:47 PM
collinear wanted, regular hexagon
parmenides51   3
N Yesterday at 11:34 PM by MathIQ.
Source: 2023 Austrian Mathematical Olympiad , Junior Regional Competition , Problem 2
Let $ABCDEF$ be a regular hexagon with sidelength s. The points $P$ and $Q$ are on the diagonals $BD$ and $DF$, respectively, such that $BP = DQ = s$. Prove that the three points $C$, $P$ and $Q$ are on a line.

(Walther Janous)
3 replies
parmenides51
Mar 26, 2024
MathIQ.
Yesterday at 11:34 PM
Geometric inequality with angles
Amir Hossein   7
N Yesterday at 11:06 PM by MathIQ.
Let $p, q$, and $r$ be the angles of a triangle, and let $a = \sin2p, b = \sin2q$, and $c = \sin2r$. If $s = \frac{(a + b + c)}2$, show that
\[s(s - a)(s - b)(s -c) \geq 0.\]
When does equality hold?
7 replies
Amir Hossein
Sep 1, 2010
MathIQ.
Yesterday at 11:06 PM
IMO 2014 Problem 3
v_Enhance   103
N Yesterday at 10:59 PM by Mysteriouxxx
Source: 0
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
103 replies
v_Enhance
Jul 8, 2014
Mysteriouxxx
Yesterday at 10:59 PM
three discs of radius 1 cannot cover entirely a square surface of side 2
parmenides51   1
N Yesterday at 8:17 PM by Blast_S1
Source: 2014 Romania NMO VIII p4
Prove that three discs of radius $1$ cannot cover entirely a square surface of side $2$, but they can cover more than $99.75\%$ of it.
1 reply
parmenides51
Aug 15, 2024
Blast_S1
Yesterday at 8:17 PM
2025 Caucasus MO Juniors P6
BR1F1SZ   2
N Yesterday at 7:38 PM by IEatProblemsForBreakfast
Source: Caucasus MO
A point $P$ is chosen inside a convex quadrilateral $ABCD$. Could it happen that$$PA = AB, \quad PB = BC, \quad PC = CD \quad \text{and} \quad PD = DA?$$
2 replies
BR1F1SZ
Mar 26, 2025
IEatProblemsForBreakfast
Yesterday at 7:38 PM
China mathematical olympiad 1995 problem4
jred   6
N Sep 19, 2013 by jred
Source: China mathematical olympiad 1995 problem4
Given four spheres with their radii equal to $2,2,3,3$ respectively, each sphere externally touches the other spheres. Suppose that there is another sphere that is externally tangent to all those four spheres, determine the radius of this sphere.
6 replies
jred
Sep 15, 2013
jred
Sep 19, 2013
China mathematical olympiad 1995 problem4
G H J
Source: China mathematical olympiad 1995 problem4
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jred
290 posts
#1 • 2 Y
Y by Adventure10, Mango247
Given four spheres with their radii equal to $2,2,3,3$ respectively, each sphere externally touches the other spheres. Suppose that there is another sphere that is externally tangent to all those four spheres, determine the radius of this sphere.
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vanstraelen
9059 posts
#2 • 1 Y
Y by Adventure10
Midpoints of the spheres $M_{1}(0,-2,0),M_{2}(0,2,0),M_{3}(\sqrt{21},0,0),M_{4}(\frac{\sqrt{21}}{7},0,\frac{12\sqrt{7}}{7})$, see two pictures.
$M_{1}$ green, $M_{2}$ red, $M_{3}$ blue, $M_{4}$ purple.
So $M_{1}M_{2}=4,M_{1}M_{3}=5,M_{1}M_{4}=5,M_{2}M_{3}=5,M_{2}M_{4}=5,M_{3}M_{4}=6$.
Tetrahedron $(M_{4},M_{1}M_{2}M_{3})$.
Midpoint $M(c,0,d)$ and radius $r$, satisfying:
\[\left\{\begin{array}{l}  MM_{2}+2=r \\ MM_{3}+3=r \\ MM_{4}+3=r \\ \end{array} \right.\]
\[\left\{\begin{array}{l}  \sqrt{c^{2}+4+d^{2}}+2=r \\ \sqrt{(c-\sqrt{21})^{2}+d^{2}}+3=r \\ \sqrt{(c-\frac{\sqrt{21}}{7})^{2}+(d-\frac{12\sqrt{7}}{7})^{2}}+3=r \\ \end{array} \right.\]
Solving $M(\frac{4\sqrt{21}}{7},0,\frac{6\sqrt{7}}{7})$ and $r=6$.
Attachments:
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jred
290 posts
#3 • 2 Y
Y by Adventure10, Mango247
vanstraelen wrote:
Midpoints of the spheres $M_{1}(0,-2,0),M_{2}(0,2,0),M_{3}(\sqrt{21},0,0),M_{4}(\frac{\sqrt{21}}{7},0,\frac{12\sqrt{7}}{7})$, see two pictures.
$M_{1}$ green, $M_{2}$ red, $M_{3}$ blue, $M_{4}$ purple.
So $M_{1}M_{2}=4,M_{1}M_{3}=4,M_{1}M_{4}=5,M_{2}M_{3}=5,M_{2}M_{4}=5,M_{3}M_{4}=6$.
Tetrahedron $(M_{4},M_{1}M_{2}M_{3}$.
Modpoint $M(c,0,d)$ and radius $r$, satisfying:
\[\left\{\begin{array}{l}  MM_{2}+2=r \\ MM_{3}+3=r \\ MM_{4}+3=r \\ \end{array} \right.\]
\[\left\{\begin{array}{l}  \sqrt{c^{2}+4+d^{2}}+2=r \\ \sqrt{(c-\sqrt{21})^{2}+d^{2}}+3=r \\ \sqrt{(c-\frac{\sqrt{21}}{7})^{2}+(d-\frac{12\sqrt{7}}{7})^{2}}+3=r \\ \end{array} \right.\]
Solving $M(\frac{4\sqrt{21}}{7},0,\frac{6\sqrt{7}}{7})$ and $r=6$.

but you do not give the final answer to this problem...
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vanstraelen
9059 posts
#4 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Midpoint $M$ of the sphere, externally tangent to all those four spheres.
Radius $r=6$.
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jred
290 posts
#5 • 1 Y
Y by Adventure10
vanstraelen wrote:
Midpoints of the spheres $M_{1}(0,-2,0),M_{2}(0,2,0),M_{3}(\sqrt{21},0,0),M_{4}(\frac{\sqrt{21}}{7},0,\frac{12\sqrt{7}}{7})$, see two pictures.
$M_{1}$ green, $M_{2}$ red, $M_{3}$ blue, $M_{4}$ purple.
So $M_{1}M_{2}=4,M_{1}M_{3}=4,M_{1}M_{4}=5,M_{2}M_{3}=5,M_{2}M_{4}=5,M_{3}M_{4}=6$.
Tetrahedron $(M_{4},M_{1}M_{2}M_{3})$.
Midpoint $M(c,0,d)$ and radius $r$, satisfying:
\[\left\{\begin{array}{l}  MM_{2}+2=r \\ MM_{3}+3=r \\ MM_{4}+3=r \\ \end{array} \right.\]
\[\left\{\begin{array}{l}  \sqrt{c^{2}+4+d^{2}}+2=r \\ \sqrt{(c-\sqrt{21})^{2}+d^{2}}+3=r \\ \sqrt{(c-\frac{\sqrt{21}}{7})^{2}+(d-\frac{12\sqrt{7}}{7})^{2}}+3=r \\ \end{array} \right.\]
Solving $M(\frac{4\sqrt{21}}{7},0,\frac{6\sqrt{7}}{7})$ and $r=6$.

There are some errors in your solution: $M_{1}M_{3}$ should be $5$; and since the fifth sphere is externally (not internally) tangent to the other spheres, hence $MM_{2}-2=r, MM_{3}-3=r , MM_{4}-3=r,$ ... Consequently, $6$ is not the right answer.
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vanstraelen
9059 posts
#6 • 2 Y
Y by Adventure10, Mango247
Recalculating
$r_{1}=\frac{6}{11}$.
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jred
290 posts
#7 • 2 Y
Y by Adventure10, Mango247
vanstraelen wrote:
Recalculating
$r_{1}=\frac{6}{11}$.
Finally you get it right. :)
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