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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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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
inq , not two of them =0
win14   3
N 2 minutes ago by Nguyenhuyen_AG
Let a,b,c be non negative real numbers such that no two of them are simultaneously equal to 0
$$\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \ge \frac{5}{2\sqrt{ab + bc + ca}}.$$
3 replies
win14
3 hours ago
Nguyenhuyen_AG
2 minutes ago
3-var inequality
sqing   4
N 8 minutes ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
4 replies
sqing
Today at 4:32 AM
sqing
8 minutes ago
Need help on this simple looking problem
TheGreatEuler   1
N 10 minutes ago by Primeniyazidayi
Show that 1+2+3+4....n divides 1^k+2^k+3^k....n^k when k is odd. Is this possible to prove without using congruence modulo or binomial coefficients?
1 reply
TheGreatEuler
2 hours ago
Primeniyazidayi
10 minutes ago
Inspired by old results
sqing   1
N 14 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0,ab+bc+ca-4abc=\frac{1}{8}$. Prove that
$$ a^2+b^2+c^2+ab+bc+ca   \geq \frac{3}{8}$$Let $ a,b,c\geq 0,a^2+b^2+c^2+4abc=\frac{1}{4}$. Prove that
$$ a+b+c+ab+bc+ca- 4(13-8\sqrt{2})abc\leq \frac{1}{8}+\frac{1}{\sqrt 2}$$
1 reply
sqing
19 minutes ago
sqing
14 minutes ago
Geometry hard
Lukariman   0
19 minutes ago
Given triangle ABC inscribed in circle (O). The bisector of angle A intersects (O) at D. Let M, N be the midpoints of AB, AC respectively. OD intersects BC at P and AD intersects MN at S. The circle circumscribed around triangle MPS intersects BC at Q different from P. Prove that QA is tangent to (O).
0 replies
+1 w
Lukariman
19 minutes ago
0 replies
Great similarity
steven_zhang123   0
20 minutes ago
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
0 replies
+2 w
steven_zhang123
20 minutes ago
0 replies
Integer polynomial w factorials
Solilin   0
20 minutes ago
Source: 9th Thailand MO
Let $a_1, a_2, ..., a_{2012}$ be pairwise distinct integers. Show that the equation $(x -a_1)(x - a_2)...(x - a_{2012}) = (1006!)^2$ has at most one integral solution.
0 replies
Solilin
20 minutes ago
0 replies
Coloring plane in black
Ryan-asadi   2
N 30 minutes ago by Primeniyazidayi
Source: Iran Team Selection Test - P3
..........
2 replies
Ryan-asadi
4 hours ago
Primeniyazidayi
30 minutes ago
B.Stat & B.Math 2022 - Q8
integrated_JRC   6
N an hour ago by Titeer_Bhar
Source: Indian Statistical Institute (ISI) - B.Stat & B.Math Entrance 2022
Find the minimum value of $$\big|\sin x+\cos x+\tan x+\cot x+\sec x+\operatorname{cosec}x\big|$$for real numbers $x$ not multiple of $\frac{\pi}{2}$.
6 replies
integrated_JRC
May 8, 2022
Titeer_Bhar
an hour ago
AD=BE implies ABC right
v_Enhance   117
N an hour ago by cj13609517288
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
117 replies
v_Enhance
Apr 10, 2013
cj13609517288
an hour ago
IMO Genre Predictions
ohiorizzler1434   64
N an hour ago by ariopro1387
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
64 replies
ohiorizzler1434
May 3, 2025
ariopro1387
an hour ago
3-var inequality
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
1 reply
sqing
2 hours ago
sqing
2 hours ago
Iranians playing with cards module a prime number.
Ryan-asadi   2
N 2 hours ago by AshAuktober
Source: Iranian Team Selection Test - P2
.........
2 replies
Ryan-asadi
4 hours ago
AshAuktober
2 hours ago
An analytic sequence
Ryan-asadi   1
N 2 hours ago by AshAuktober
Source: Iran Team Selection Test - P1
..........
1 reply
Ryan-asadi
4 hours ago
AshAuktober
2 hours ago
Perpendicularity with incenter in a right triangle
anantmudgal09   9
N Nov 26, 2024 by S_14159
Source: RMO Mumbai 2016, P1
Let $ABC$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $I$ be the incenter of $ABC$. Draw a line perpendicular to $AI$ at $I$. Let it intersect the line $CB$ at $D$. Prove that $CI$ is perpendicular to $AD$ and prove that $ID=\sqrt{b(b-a)}$ where $BC=a$ and $CA=b$.
9 replies
anantmudgal09
Oct 11, 2016
S_14159
Nov 26, 2024
Perpendicularity with incenter in a right triangle
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G H BBookmark kLocked kLocked NReply
Source: RMO Mumbai 2016, P1
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anantmudgal09
1980 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ be a right-angled triangle with $\angle B=90^{\circ}$. Let $I$ be the incenter of $ABC$. Draw a line perpendicular to $AI$ at $I$. Let it intersect the line $CB$ at $D$. Prove that $CI$ is perpendicular to $AD$ and prove that $ID=\sqrt{b(b-a)}$ where $BC=a$ and $CA=b$.
This post has been edited 1 time. Last edited by anantmudgal09, Oct 11, 2016, 7:14 AM
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WizardMath
2487 posts
#2 • 1 Y
Y by Adventure10
Mixtilinear incircles!!! :D
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GoJensenOrGoHome
589 posts
#3 • 2 Y
Y by Adventure10, Mango247
WizardMath wrote:
Mixtilinear incircles!!! :D

You don't need such advanced things, since $ \angle AID= \angle ABD=90$ we have $AIBD $ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2} $ and $ \angle ABI= \angle ADI=45 $ now we let $ T=CI\cap AD  $ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$.
By sine law $ AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI  $ we have the result
This post has been edited 1 time. Last edited by GoJensenOrGoHome, Oct 11, 2016, 10:21 AM
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pslv
165 posts
#4 • 2 Y
Y by Adventure10, Mango247
can anyone draw the diagram.please.
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svatejas
124 posts
#5 • 3 Y
Y by Randomizeriitian, Adventure10, Mango247
GoJensenOrGoHome wrote:
WizardMath wrote:
Mixtilinear incircles!!! :D

You don't need such advanced things, since $ \angle AID= \angle ABD=90$ we have $AIBD $ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2} $ and $ \angle ABI= \angle ADI=45 $ now we let $ T=CI\cap AD  $ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$.
By sine law $ AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI  $ we have the result

Nice solution. There is a small typo in your first line. It should be \(\angle IAB = \angle IDC\).
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MathematicalPhysicist
179 posts
#6 • 1 Y
Y by Adventure10
GoJensenOrGoHome wrote:
WizardMath wrote:
Mixtilinear incircles!!! :D

You don't need such advanced things, since $ \angle AID= \angle ABD=90$ we have $AIBD $ cyclic, so $\angle IAB= \angle IDB= \frac{\alpha}{2} $ and $ \angle ABI= \angle ADI=45 $ now we let $ T=CI\cap AD  $ and we have $\angle DTC= 180-\frac{\alpha+\gamma}{2}-45=180-45-45=90$ so $CI \perp AD$.
By sine law $ AD=\frac{b\cdot \sin(\gamma)}{\cos(\frac{\gamma}{2})}=2b\cdot \sin(\frac{\gamma}{2})=2b\sqrt{\frac{1-\cos (\gamma)}{2}}=b\sqrt{2(1-a/b)}=\sqrt{2b(b-a)}$ since $\frac{AD}{ \sqrt{2}}=DI  $ we have the result
you actually dont need the law of sines.a simple manipulation of pythagoras' would do. anyways,a nice solution:)
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Math-wiz
6107 posts
#7 • 3 Y
Y by amar_04, lilavati_2005, Purple_Planet
A complete synthetic non-trig solution, just for completeness :D
For the first part, observe that $\angle DBA=\angle DIA(=90^{\circ})$, so $DBIA$ is cyclic, and hence, $\angle IAB=\angle IDB$ and $\angle ADI=\angle BI$. So, $\angle ADB=45+\frac{A}{2}$. Now in $\triangle DEC$, $\angle DEC+\angle ECD+\angle CDE=180^{\circ}\implies \angle DEC=180^{\circ}-45^{\circ}-\frac{A}{2}-\frac{C}{2}=90^{\circ}$, as desired.

For the second part, $AI=ID$ since $\angle ADI=45^{\circ}$. Also, $CD=AC$ since $\triangle CED$ and $CEA$ are congruent by ASA criterion. So, $BD=CD-CB=CD-a=b-a$.
Observe that $\triangle DBI$ and $\triangle AIC$ are similar, as $\angle BDI=\angle AIC=\frac{A}{2}$ and $\angle DIB=\angle CIA=\frac{C}{2}$. So, $\frac{AI}{BD}=\frac{AC}{ID}\implies\frac{ID}{b-a}=\frac{b}{ID}\implies ID=\sqrt{b(b-a)}$, as desired $\blacksquare$.
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HoRI_DA_GRe8
597 posts
#8
Y by
Just use cosine rule and isosceles triangles for the 2nd part
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BVKRB-
322 posts
#9
Y by
Probably the most simple solution to this question with 0 trigonometry :D
Click to reveal hidden text
Diagram from Math-wiz :whistling:
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S_14159
48 posts
#10
Y by
Claim 1. $(ADBI)$ is cyclic.
$$\angle AID=\angle AID=\angle ABD$$Claim 2. $AI=ID$
\begin{align*}&45^{\circ}=\angle ABI=\angle ADI\\ &\angle DAI=180^{\circ}-\angle DBA-\angle IBA=45^{\circ}\end{align*}Claim 3. $\triangle CAD$ is isosceles $\implies CI\perp AD$
$$\angle CAD=\angle CAI+\angle IAD=\angle IAB+45^{\circ}=\angle IDB+\angle ADI=\angle ADC$$Now we take $c=AB$ and we have the following
$$ID=\sqrt{\frac{AD^2}{2}}=\sqrt{\frac{AB^2-DB^2}{2}}=\sqrt{\frac{c^2-(b-a)^2}{2}}=\sqrt{\frac{2b^2-2ba}{2}}=\sqrt{b(b-a)}$$
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