Find all function that satisfy below equation.
$ f(x\minus{}y)\equal{}f(x)\plus{}xy\plus{}f(y)$ for all $ x \in R$ and $ y \in f(R)$

[quote="Heebeen, Yang"]Find all function that satisfy below equation.
$ f(x \minus{} y) \equal{} f(x) \plus{} xy \plus{} f(y)$ for all $ x \in R$ and $ y \in f(R)$[/quote]
I dont understand this condition $ y \in f(R)$
I think it must be $ y\in R$
If so the problem is very easy.Indeed
Let $ y \equal{} 0$ thus $ f(0) \equal{} 0$
And when $ x \equal{} y$ we have  $ f(x) \equal{} \frac {x^2}{2}$
I know that I am wrong  :)

[hide]
"@Add after test20's post:"Oh,sorry.Notice that in IMO $ 40^{\text th}$ we also know the problem
\[ f(x \minus{} f(y)) \equal{} f(x) \plus{} xf(y) \plus{} f( f(y)) \minus{} 1
\]
Their solutions are similar to :lol:
Thank you[/hide]

[quote="Allnames"][quote="Heebeen, Yang"]Find all function that satisfy below equation.
$ f(x \minus{} y) \equal{} f(x) \plus{} xy \plus{} f(y)$ for all $ x \in R$ and $ y \in f(R)$[/quote]
I dont understand this condition $ y \in f(R)$
I think it must be $ y\in R$[/quote]
"$ y \in f(R)$" means: There exists some $ z\in R$ with $ y\equal{}f(z)$.
Hence an equivalent way of stating the problem would have been:

Find all functions that satisfy the equation
$ f(x \minus{} f(y)) \equal{} f(x) \plus{} xf(y) \plus{}f( f(y))$
for all $ x,y \in R$.

I'm sorry that I didn't write explanation obviously :blush:.

It is equivalent to
$ f(x \minus{} f(y)) \equal{} f(x) \plus{} xf(y) \plus{} f( f(y))$ for all $ x,y \in R$

Thank you for thinking my problem.

But still I can' t solve it.

Please Help me !!

$ P(f(y),y): f(f(y))\equal{}\frac{f(0)\minus{}f(y)^2}2$.
Obviously $ f(x)\equal{}0$ is one solution.
Suppose $ f(a)\ne0$. We let $ u\equal{}x\minus{}f(a)$, then
$ f(u)\minus{}f(x)\equal{}xf(a)\plus{}f(f(a))$ tells us $ f(u)\minus{}f(x)$ can take any value.

$ P(f(u),y): f(f(u)\minus{}f(y))\\\equal{}\frac{f(0)\minus{}f(u)^2}2\plus{}f(u)f(y)\plus{}\frac{f(0)\minus{}f(y)^2}2\\\equal{}f(0)\minus{}\frac12(f(u)\minus{}f(y))^2$.
Since $ f(u)\minus{}f(y)$ can be any value, we have
$ f(x)\equal{}f(0)\minus{}\frac12x^2$. It is not hard to verify $ f(0)\equal{}0$.

Two solutions: $ f(x)\equal{}0$ or $ f(x)\equal{}\minus{}\frac12x^2$

I've just found IMO 1999 and above problem that Allnamers suggested is A5.

Thank you for your solution xxp2000
But I already understand.

I think it is nice step that $ f(a)\minus{}f(b)\equal{}R$.

Thank you everyone :lol: !


inequalities trigonometry function calculus derivative

Given acute triangle $ ABC$
Prove that
$ tanA\plus{}tanB\plus{}tanC\plus{}6(sinA\plus{}sinB\plus{}sinC) >\equal{} 12\sqrt{3}$

THe inequality is very much false.
Just substitue for A,B,C with some easily applicable angles like $ 30^^{\circ}$ etc.
Actually there is a reason why it cannot be true
This is because
the strongest inequality for $ tan A$ for $ A$ ranging from $ 0$ to $ \frac{\pi}{2}$ is by convexity  and that gives us  $ \sum_{cyc} tan A \ge 3\sqrt{3}$ and our given inequality becomes equivalent to proving that $ sinA +sinB+sinC \ge \frac{3 sqrt{3}}{2}$ which holds in the reverse beacuse of concavity of sine from $ 0$ to $ \frac{\pi}{2}$.

The inequality does seem to work for certain special cases. And the method you outline does not contradict the problem instead its just an approach that fails to solve the problem.

yeah, that's why I think this is hard. (or I'm not good)

I tried some value for A,B,C and the inequality is right <until now>

If you say it's wrong, please give one counter example.

Claiming a problem is false by spouting off some vague intuitions about convexity is not enough.  If it's false, there is a counterexample. Find it.

Anyone have any ideas? Just give me a start >.<

[quote="Agr_94_Math"]NO Sanjith{thealchemist} I think maybe you are worng in this.
Wat I have done is not a way in which u dont get the solution.
I bet that there is not a stronger inequality for $ \sum_{cyclic} tanA$ than $ \sum_{cyclic} tanA \ge 3\sqrt3$ and this necessitates the convexity of sine in acute angles which is not true. where $ A,B,C$ are acute angles. So the inequality is simply false.[/quote]

To put this in perspective, you are basicly claiming that if $ a\ge x$ and $ b \le x$, then $ a\plus{}b \not\ge 2x$
But this is not necessarily true

WLOG $ A\ge B\ge C$.

Let $ f(X) \equal{} tan(x) \plus{} 6sin(x)$. Then $ f''(x) \equal{} \frac {2sin(x)}{cos^3(x)} \minus{} 6sin(x)$, which is $ \ge 0$ when $ cos(x)\le \sqrt [3]{\frac {1}{3}}$.

We have 4 cases:

Case 1: $ cos(A),cos(B),cos(C)\le\sqrt [3]{\frac {1}{3}}$

Jensens, Q.E.D.

Case 2: $ cos(A),cos(B),cos(C) > \sqrt [3]{\frac {1}{3}}$

Then since $ \sqrt [3]{\frac {1}{3}} > \frac {1}{2}$, we get all angles less that $ 60$, contradiction.

Case 3: $ cos(C) > \sqrt [3]{\frac {1}{3}}\ge cos(A),cos(B)$

Then applying jensen's to A and B, we get that it suffices to show the inequality $ 2tan(X) \plus{} tan(180 \minus{} 2x) \plus{} 12sin(X) \plus{} 6sin(180 \minus{} 2x)\ge 12\sqrt {3}$, where $ cos(x) \le \sqrt [3]{\frac {1}{3}} < \frac {\sqrt {2}}{2}$, which means that we only need to prove this when $ 90 > x > 45$.

If $ g(x) \equal{} 2tan(x) \plus{} tan(180 \minus{} 2x) \plus{} 12sin(X) \plus{} 6sin(180 \minus{} 2x)$
$ \equal{} 2tan(x) \minus{} tan(2x) \plus{} 12sin(x) \plus{} 6sin(2x)$

Then $ g'(x) \equal{} \frac {2}{cos^2(x)} \minus{} \frac {2}{cos^2(2x)} \plus{} 12cos(x) \plus{} 12cos(2x)$. The min. will occur at a critical point, or as x apporaches 90, or when $ x \equal{} 45$. As x approaches 90, the function is unbounded in the positive direction, and when x approaches 45, since g(x) is continuos, this is infact the value of g(x) at x=45, and we get a smaller value than when x=60. Hence, we now find critical points of the function.

$ \frac {2}{cos^2(x)} \minus{} \frac {2}{cos^2(2x)} \plus{} 12cos(x) \plus{} 12cos(2x) \equal{} 0$

$ \frac {cos^2(2x) \minus{} cos^2(x)}{cos^2(x)cos^2(2x)} \plus{} 6(cos(x) \plus{} cos(2x)) \equal{} 0$

$ (cos(2x) \plus{} cos(x))(\frac {1}{cos(2x)} \minus{} \frac {1}{cos(x)} \minus{} 6) \equal{} 0$

And since we can show through differentiation that $ \frac {1}{cos(2x)} \minus{} \frac {1}{cos(x)} \minus{} 6 < 0$ in this range (I'm just too tired of doing this problem at this point to actually differentiate, but you can graph it to see that it's true), we get the critical point when $ cos(2x) \plus{} cos(x) \equal{} 0$, so when $ x \equal{} 60$. Since this yields a smaller than when x approaches either endpoint, this is infact the minimum.

Case 4: $ cos(C),cos(B) > \sqrt [3]{\frac {1}{3}} \ge cos(A)$, then by karamata's, we only have to show the inequality true when (A,B,C) --> $ (A,arccos(\sqrt [3]{\frac {1}{3}}),B \plus{} C \minus{} arccos(\sqrt [3]{\frac {1}{3}}))$, where $ B \plus{} C \ge arccos(\sqrt [3]{\frac {1}{3}})$, since otherwise $ 180 \minus{} arccos(\sqrt [3]{\frac {1}{3}}) < A$, making the triangle obtuse. Yet this was covered in Case 3.

wow, what an answer! I really appreciate your help  :)


geometry
incenter
inequalities
inradius
triangle inequality
angle bisector

Move if needed.

I thought I had this done, but I realized I didn't.

Please Help

Let ABC be a trinagle and let I be the incenter.  Let N, M be the midpoints of the sides AB and CA respectively.  The lines BI and CI meet MN at K and L respectively.  Prove that AI+BI+CI>BC+KL.

Tahnks

I tried using some sort of form of Triangle Inequality but I couldn't figure anything out.

[hide]
The triangle inequality gives $BI+CI>BC$, so the result will follow from the stronger inequality $AI>KL$. A bit of angle chasing reveals $MK=MB$ and $NL=NC$. So we can express $KL$ in terms of these lengths with
\[KL=MK+NL-MN=MB+NC-MC=\frac{AB+AC-BC}{2}\]
which happens to be the length of the tangent line from $A$ to the incircle. Let $r$ be the inradius, then $AI=\sqrt{KL^{2}+r^{2}}>KL$, as desired.
[/hide]

Can someone explain to me how a bit of angle chasing tells us that MK=MB?

The fact that BI is an angle bisector and that alt. int. angles of parallel lines are congruent show that MKB is an iscoceles triangle.

How do we know K and L are located between M and N?

[asy]/* File unicodetex not found. */

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draw((-1.26,4.46)--(-2.92,1.66)--(4.5,0.74)--cycle, zzttqq);
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dot((4.5,0.74),dotstyle);
label("$C$", (4.58,0.86), NE * labelscalefactor);
dot((-0.84,2.68),dotstyle);
label("$I$", (-0.76,2.8), NE * labelscalefactor);
dot((-2.09,3.06),dotstyle);
label("$D$", (-2,3.18), NE * labelscalefactor);
dot((1.62,2.6),dotstyle);
label("$E$", (1.7,2.72), NE * labelscalefactor);
dot((-1.78,3.02),dotstyle);
label("$K$", (-1.7,3.14), NE * labelscalefactor);
dot((-0.47,2.86),dotstyle);
label("$L$", (-0.4,2.98), NE * labelscalefactor);
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$KL+BC=KL+DE+DE=KL+KE+DK+DL+LE=KE+DL+DE=EC+DB+DE=\frac{AB+BC+CA}{2}$
[b]But [/b]$AI+BI \le AB$
$AI+CI \le AC$
$CI+BI \le CB$
We sum the relations and we're ready:
$\boxed{KL+BC<AI+BI+CI} (Q.E.D)$


calculus
derivative
calculus computations

This is a continuation of the first problem. Except after the side Z has risen along side X and shrunk along Y, it now is falling along side X and extending outwards along side Y (opposite of what it was doing earlier). It is falling back into its original location.  My problem is setting up the related rate formula (located below). A right-triangle has the following measurements just as before:

        X = 50
                |
                ||
                ||||
                ||||||
                ||||||||        Z = ?
                ||||||||||
                ||||||||||||
                ||||||||||||||
                ||||||||||||||||
                ||||||||||||||||||

                     Y = 60  <----

Z 2  = X 2  + Y 2
Z 2  = 50 2 + 60 2
Z =  :sqrt: 6100
Z = 78

And the actual problem is that side Z is falling against side X causing the height X to shrink at a constant rate of 8.333... meters/minute.  While this happens, side Z is sliding outwards along side Y causing the length of Y to increase.  What is the rate in which Y increases?

    Well this is how I did the problem:

  Particular Information:

 X = 50
 Y = 60
 Dx/Dt = 8.333... meters/minute
 Dy/Dt = ? when X = 50

  Steps in solving the problem:
 x 2 + Y 2  = 78
 2x (dx/dt) + 2y(dy/dt) = 0       <--- This is the derivative
 2x (dx/dt) = -2y(dy/dt)
 2 (50) (8.333) = -2 (60) (dy/dt)  <--- i think this is where i went wrong
 833.3 = -120 (dy/dt)               --- I'm not entirely sure how to set this problem up
 dy/dt = -6.944                     --- How can something increasing be negative??

So side Y is increasing at a rate of 6.944 meters/minute

I KNOW I DID THIS WRONG!?!?   :(

Got to remember that as $x$ is decreasing, the rate is negative. This gives us:

$2(60)\frac{dy}{dt} = -2(50)(\frac{-25}{3})$ because $\frac{dz}{dt}$ is $0$ when $z$ is constant.

$120\frac{dy}{dt} = 833.333$

$\frac{dy}{dt} = 6.9444 \frac{m}{min.}$

You just got lost in your signs.


calculus
college
Princeton
Harvard
CMU
University of Chicago
MIT

I agree, but unfortunately it's how the system works.

I just started my first college experience since I am taking calculus at URI (university of rhode island) this summer.  I have to take the bus for like 2.5 hours to get there and the class goes very very fast -- I just started this past week and we are already almost done w/derivatives!  However, classes are much more challenging, and you meet a lot of cool people.  When I go to "real" college (where I live there and take more than one class) I hope that it's as good as I hope it will be.

You can still get into a good college w/okay but not stellar test scores.  Your transcript -- HS grades -- are the most important.  Then all the other stuff.  One guy I know got a 1520 on his SATs but still managed to get Cs in chemistry, physics, and a lot of his other classes.  He got rejected from McGill and UC Berkeley and only got accepted to our state school.  He's smart -- was the high scorer for the AMC in our school -- but unfortunately he didn't try.

My best advice to you is not to expect to get into anywhere -- don't be dead set on getting accepted to Princeton or something like that.  Work really hard on your applications and try hard in school.  Even if you don't get into Harvard you will still have a lot of advanced classes to take and will get to meet new people.

The short answer to your question is, college is where you will learn much of what you use in the rest of your life.

good college = real math
bad college = why did I major in math???

We just have to hope that good colleges accept the people who really want to learn math and not just the ones getting the degrees for prestige or a well-paying job (though math isn't usually directly associated with $$$).

This generally applies to other subjects as well.

[quote="JBL"][quote="probability1.01"]good college = real math
bad college = why did I major in math???[/quote]

I don't think this is very true at all, certainly not more true in math than in other subject areas.  No one gets a degree in math for the prestige or the money.  There are lots of not-the-very-best schools with wonderful math departments.[/quote]

I was generally speaking of bad as in :le: average and good as in > average considerably.  And obviously this is in the subject of interest.  You wouldn't choose Princeton over CMU in CS most likely (does Princeton have CS?  I assume so..), for example, even though Princeton is probably a better college overall.

But people do go to college not because they want to learn but because they feel that college is necessary for them to succeed and they're "supposed" to go.  Math is a bad example, but for other subjects.  You don't want to get stuck at a bad college, but as long as your resume is decent, you should ok.  True that you shouldn't fret over colleges too much, but don't take it too lightly either.  It all really depends on what college fits you best.  Environment is always important too, like at a community college different types of people would be around as compared to the best colleges.

It depends on your major. People just want to go to the highly selective colleges for prestige (and to be around smart people), not necessarily for a good education. I'm sure UChicago has as good a math program as Harvard. With math, it doesn't matter whether one goes to a top-ranked college or not since you are recognized in the field for what you accomplish, not for what schools you went to. The situation is different if you major in, say, business, since your starting salary will be higher if you go to a top-ranked school than if you go to a mediocre-ranked school.

[quote]probability1.01 wrote:
good college = real math
bad college = why did I major in math???

I don't think this is very true at all, certainly not more true in math than in other subject areas. No one gets a degree in math for the prestige or the money. There are lots of not-the-very-best schools with wonderful math departments.[/quote]

I don't mean to be a jerk, but you go to Harvard so how can you vouch for schools that are not-the-best?

[quote]I was generally speaking of bad as in  average and good as in > average considerably. And obviously this is in the subject of interest. You wouldn't choose Princeton over CMU in CS most likely (does Princeton have CS? I assume so..), for example, even though Princeton is probably a better college overall.
[/quote]

Yay CMU.  This is very true BTW.  I am pretty sure that CMU CS is better than MIT (in terms of rankings).

[quote]We just have to hope that good colleges accept the people who really want to learn math and not just the ones getting the degrees for prestige or a well-paying job (though math isn't usually directly associated with $$$).

This generally applies to other subjects as well.[/quote]

Unfortunately all the people I know who got into Ivy League schools are like this (at my HS, not on this website).  Actually, they are not so much interested with getting a job as they are w/the "ivy league" title.  They are not exactly the people who you could have in depth conversations with unless it is about what some third person got on their SATs or who has the higher GPA.  My Dad is a professor at Brown university but he also is a doctor at a hospital near there, and he has a lot of students from there come in to the ER when they are on drugs or have hurt themselves when they are on drugs.

Whoa, that post really made me sound like a jerk. I'll leave it but sorry I am not really that mean!

[quote="probability1.01"]good college = real math
bad college = why did I major in math???[/quote]
What if I want to learn about [tex]\sqrt{-1}=i[/tex]? What kind of college should I go to then?

Err, sorry, I couldn't resist making a bad math joke.

[quote]Can someone explain to me why college is important? Not that I don't intend to go, it's just that it seems silly to stress out over getting good test scores to get into a good college.[/quote]

first of all, most people define "good" as Ivy league, MIT, which are actually just prestigious. the "good" part is arguable. for example, Cornell and MIT don't even make the USnews top 10 engineering schools list.

so it depends what you mean by a good college. for example, Rose-Hulman, the top engineering school listed by USnews and world report, has an acceptance rate of around 50%, but MIT and other ivy leagues are in the low teens, if I remember correctly.

not that I'm slamming the prestigious schools, they're fine, i'm sure, but not worth obsessing over the way some people do.

second of all, good test scores are overrated, especially with all the people making perfect scores in both the SAT and ACT. a 1600 SAT means a whole lot less when 100 people applying to your school got that score.

[quote="JS1527"][quote]probability1.01 wrote:
good college = real math
bad college = why did I major in math???

I don't think this is very true at all, certainly not more true in math than in other subject areas. No one gets a degree in math for the prestige or the money. There are lots of not-the-very-best schools with wonderful math departments.[/quote]

I don't mean to be a jerk, but you go to Harvard so how can you vouch for schools that are not-the-best?[/quote]

A fair point.  I think I should note that I am not talking about the difference between community colleges and Ivy League schools here -- I am talking about the difference between good colleges, such as the major state schools (CA, the mid-west esp. MI and WI, NY, etc.), less-selective universities, or a number of liberal arts colleges which are not traditionally thought of as math schools, and what have always been considered to be the very best private universities, such as the Ivy League, Stanford, MIT, and so on.  So, among that group of schools that I just named, I have a pretty good collection of friends and personal experience.  The University of Michigan, for example, has an excellent honors math program for about 20 people each year (roughly the same size as at Princeton and Harvard) from which a student would graduate with an extremely good mathematics education.  CMU, as someone pointed out, has a wonderful CS department.  Carleton College, a small liberal arts school in Minnesota, recruits fairly heavily among mathy peoples.  I'm sure Simon could come and vouch for the quality of mathematics education he is receiving at one of the less-prestigeous UC schools.
That was obviously just a list of 3 schools, but the point is that I could (with a bit more research) keep going on like that for quite a while.  There are few enough people doing math that you can find out if a school has a "good" math department without doing a whole lot of work.  And the fact of the matter is that there are quite a large number of schools where one can learn "real math" (or imaginary math, as confuted would have it :)).

Okay, but seriously you guys are still talking about top schools.  If you pull out the US News magazine, you will see that University of Michigan, University of Wisconsin, almost all of the UC's including UCSB where Simon goes, and CMU are ALL in the TOP FIFTY schools.  So really, why would you expect the mathematics education you recieve at any top 50 school to be any different?  No offense, but you make the statement that:
[quote]There are lots of not-the-very-best schools with wonderful math departments.[/quote]
But to back it up, you just list a bunch of top schools that have good math depts.  In order to back this up you should list some schools that are not at the top who have good math depts.

Again, my apologies for sounding like a jerk.

It depends on where you draw the line at "very best" schools. Some students might count the top 200 American Universities as the "very best", and someone else might cound the top 10 as the "very best". JBL is assuming that most people here consider the latter definition, whereas you consider the former definition.

So I think his point is just that there are a pretty good number of Universities at which you can get a very good mathematics education. I would agree with this - and there is plenty of evidence to back this up when you look at the undergraduate schools where students who go to the "very best" math graduate schools come from. They're often from a very wide range of Universities. Not just the "top 10" or so.

[quote]University of Michigan, for example, has an excellent honors math program for about 20 people each year (roughly the same size as at Princeton and Harvard) from which a student would graduate with an extremely good mathematics education.[/quote]

Thats great to hear! The Honors Program in Math sent me a letter a few weeks ago asking me to consider their department. Since U-M is close to my home, and fairly unexpensive for me, its something I will seriously consider, even if I'm a year and a half away from applying to college.

Does anyone else know of this program, or has considered attending it?

Thanks for the replies, but I don't think any of them really answered my question... or maybe I'm just not open-minded. Before I wanted to go to one of the more 'prestigious' colleges (MIT, etc), but now I think I'd rather go to a more 'normal' college (don't really know any... is UCLA considered 'prestigious' these days?). Anyway...

UCLA is probably in the class of colleges that I referred to as "not-the-very-best."  Thousands of people get an excellent education there every year although I don't think it's typically contrasted with the top-rank private schools.  If you were making a list of the "Ivy-League of public schools," UCLA would be on it.

Tetrahedron -- when I visited, the guy from the Honors Program (I dunno what his actual job was) was very nice and talked to me and my father and showed us around.  Go take a visit and talk to them, see what you think of it.

[quote]first of all, most people define "good" as Ivy league, MIT, which are actually just prestigious. the "good" part is arguable. for example, Cornell and MIT don't even make the USnews top 10 engineering schools list.

so it depends what you mean by a good college. for example, Rose-Hulman, the top engineering school listed by USnews and world report, has an acceptance rate of around 50%, but MIT and other ivy leagues are in the low teens, if I remember correctly.[/quote]

Maybe you should check your stats if you want accurate information.

[url=http://www.usnews.com/usnews/edu/college/rankings/brief/engineering/phd/topprogs_withphd_brief.php]U.S. News article[/url]

lists MIT as the top school. You were likely looking at the statistics for schools without a doctorate. So its not all just prestige.

good for WHAT though?
would you go to mit to major in, say, theatre? how good is cornell's vet program? while i realize many people here probably want to major in math or some related field (physics, etc), you have to define good as good for YOU. prestige- ivy league schools, what does that mean exactly? all "ivy league" really means is "very old school" not "very good school" although this is the common misconception (even for myself).
however, there is that starting pay thing that is associated with big name, top tier schools, such as MIT, and platypus, i believe this is where i may help answer your question. i recently read a book that had a blurb on IIT (thats Illinois Institute of Technology) which said that IIT has just as tough of a course load as MIT. but guess what? only HALF the starting salary. so you get the same education and everything, you just dont get the prestige.
i mean, i hadnt even HEARD of IIT.
and thats why everyone wants to go there. to MIT, that is.

I was simply refuting someone's earlier post which claimed that MIT is not listed in the top 10 of the US News "Top Schools Report" for Engineering schools, whereas it actually is ranked 1st.

I'm not claiming that any school is better or worse, just reporting a mistake in an earlier post.

As for IIT having as tough of a courseload - I don't really get this. I mean, of course every college will have the same courses in fields such as engineering. Its about the quality of the teaching and the depth of the course.

Keep in mind that if you plan to go to graduate school, which graduate school you went to might be important when you want to get a job, but which undergraduate school you went to will be immaterial. Also, I imagine your figures are probably quite confounded, bleumoose. For example, suppose we are considering two schools A and B. School A is a prestigious school; school B isn't. Suppose furthermore that school A is on the east coast, and school B is in the midwest. It may be that people from school A get paid more than people from school B.

Reason 1: Since school A is more prestigious, more top students go to school A than to school B, so they will be more competent when they come out of college, so they will be offered higher-paying jobs.

Reason 2: Houses cost way less in the midwest than they do on the east coast. Since school A is on the east coast, it is likely that the majority of students at school A are from the east coast and will stay in the east coast after graduating from college. Students from school B are probably largely from the midwest and will probably stay in the midwest after graduating. Since houses are cheaper in the midwest, salaries tend to be lower. But students from school B won't necessarily have a lower standard of living than students from school A; they just don't have to pay as much to get the same standard of living, and so they don't earn as much either.

im just citing some info i got from a college book that is at least 5 years old. i think it is 341 most interesting colleges, for those that want to look it up.

It's very possible that a school like IIT has a comparable quality of teaching and a comparable depth of many of its undergraduate courses as the courses that most students take at MIT. The main differences might be in the courses that are available for the very very best students at a school like MIT (like the Reid Bartons or the Tiankai Lius) - students who would end up taking very advanced undergraduate courses early on, and eventually graduate courses while undergraduate students at MIT. For students like that, they would effectively end up taking mostly graduate courses at even a school like IIT or the Rose Hulman Institute.

But most students, even at MIT, are not in that situation. So the quality of education for the majority of students at IIT or RHI could easily be comparable to the quality of undergraduate education at a school like MIT.

However, I bet that bleumoose's assessment of starting salaries is about right. The prestige of some undergraduate institutions greatly affects the perception in many employers' eyes, and can affect starting salaries. Also the alumni associations at schools such as MIT provide networks and inside tracks to higher paying jobs than might be available to graduates of schools like IIT.

I would just like to agree with ComplexZeta that in terms of learning and future career, the graduate school you go to is more important than the undergraduate school.

also, prestigious grad schools after a not-so-prestigious undergrad are going to pay better than a not-so-prestigious grad school after a prestigious undergrad. its the same thing as getting bonus points for going to a top name high school- it may help you out for getting into college, but once you start applying for grad school, i doubt (and this is just a hypothesis) that they care very much what high school you went to.
but again, as i say, its all in the eye of the beholder. i will take something i know relatively well (vet school) and try to apply it: cornell has a vet school. cornell is a prestigious college. however, i would not necessarily choose to go to cornell's vet school, because they teach solely by models and not by dissection, which i dont think is a very good way to learn something like that. however, this might be the best way for someone else to learn. does this make sense? its not just prestige, you also have to look at everything else that matters to YOU- school size, class size, percent of teachers with doctorates, percent of classes taught by grad students, financial aid availability... i guess what i am saying is its not quite as one sided as US news and world report tries to make it. for example, they base their "top high schools" solely on the number of ap exams taken per senior. the number TAKEN. not the average score or anything like that, just the number TAKEN. here's a hint- you dont have to have taken an AP level class to take the AP exam. you just have to have the 80-some-odd dollars and faith in yourself (and a good cheatsheet book). so that is very one-sided and can lead to inaccuracies for what is the "best."

i realize my arguments are not very, ah... well done or anything, because im no debater, but i think you will get the points im trying to make, more or less.

[quote="tetrahedr0n"]
Maybe you should check your stats if you want accurate information.

[url=http://www.usnews.com/usnews/edu/college/rankings/brief/engineering/phd/topprogs_withphd_brief.php]U.S. News article[/url]

lists MIT as the top school. You were likely looking at the statistics for schools without a doctorate. So its not all just prestige.[/quote]

yeah, my bad. i forgot to mention i was talking about undergrad programs.

[quote="JS1527"][quote]UCLA is probably in the class of colleges that I referred to as "not-the-very-best."[/quote]

Are you joking?  You people are insane if you consider places like UCLA to not be one of the best schools.  I go to University of Rhode Island - so flip your US News five or six pages past UCLA in rankings and you will get to it.  I think that I am learning a lot here and the class is not watered down or taught to a lower level.  It's great if you get into MIT, but if you don't its not like you are doomed to much lower quality of education forever.[/quote]

I don't remember if it was Gauss or MCrawford who said this, but I'm using a very, very tight meaning of the word "very."  I don't mean the top 100 schools when I say the "very best" -- I mean maybe the top 10 or 15.  I have a very wide range of what I consider "good schools" -- most of the state universities are in it.  URI is.  But I have a much smaller range when I say the "best schools," and even smaller when I say the "very best."  The vast majority of educated people in this country were educated at "good schools" -- many of them have turned out to be rather brilliant.  This isn't a knock at schools, it's just how I'm using the words.  I guess you think I'm misusing them :)

[quote]yeah, my bad. i forgot to mention i was talking about undergrad programs.[/quote]

You still don't see my point - you are talking about schools that simply dont' have a graduate program. MIT may or may not be the best engineering undergrad school.

Okay okay, JBL.  I understand what you are saying but basically from my perspective, I can't even fathom going to one of the schools in your "very best" category.  That's just my opinion/situation.  If I was as good a candidate as other AoPS members maybe I would only want to go to Harvard, Princeton, Stanford, or MIT.  So, I understand where you are coming from.  Thanks for clarifying.

Yeah.  I didn't mean it at all as a slight to any college or individual.  I should also say that I don't really have a list of the "very best" schools ... it's more theoretical than that.  I don't think there is a reasonable way to make such a measurement.  But yes, my top categories are shifted more than would make them useful in most discussions of schools.


$ x,y$ positive integer
$ a) 36^{x}\minus{}5^{y}\equal{}11$
$ b) 36^{x}\minus{}5^{y}\equal{}21$
find $ x,y$

The first problem .
Consider mod 3 we have
$ y \equiv 0 (\mod 2)$
Let $ y = 2m$
So $ (6^x + 5^m)(6^x - 5^m) = 11$
$ \Longleftrightarrow x = m = 1$
The equation has solution $ (x,y) = (1,2)$
2. It has no solution. Consider mod 3.


If $ a,b,n$ are positive integers, number of solutions of the equaition $ a^2 \plus{} b^4 \equal{} 5^n$ is

$\textbf{(A)}\  1 \qquad\textbf{(B)}\  2 \qquad\textbf{(C)}\  3 \qquad\textbf{(D)}\  4 \qquad\textbf{(E)}\ \text{Infinitely many}$

[hide="Answer"]
Answer is [b]E[/b]

Let $ a \equal{} 2b^2$ so that $ a^2 \plus{} b^4 \equal{} 5b^4 \equal{} 5^n$, now substitute any $ b$ with $ 5^k$ for $ k > 0$, meaining that there are infinite solutions.
[/hide]

[hide="Another approach"]
$ (3,2,2)$ is a solution. $ 3^2\plus{}2^4\equal{}5^2$. Multiply by $ 5^{4k}$.
$ (3\cdot 5 ^{2k}, 2\cdot 5^k, 4k\plus{}2)$ is the general form of the above solution where $ k$ is a non-negative integer.

[/hide]


meghdare adadiye $3^\sqrt2$ ra ta $6$ ragham ba'ade a'ashar hesab konid.(bedoone kharkari!)

in soale mohasebate marhale sevome chand sale pish boode vali vaghean soale sakhti bayad bashe. :wink:

are, soale khallaghiate 82 bud.

vali ye chizi ro nagofti: emkanati ke dari ye mashin hesabe koochoolooe ke mitune amaliate $+$,$-$,$*$,$/$,$\sqrt$ ro masalan ta 8 ragham anjam bede.

kheyli ham soale sakhti nist, rush fek konid  :wink:


Do you think America will be better off if it is reorganized into a union of highly autonomous regions instead of being governed by a centralized federal government? Post your thoughts.

I think that it should be a mix.  After all, complete government run everything would be a problem(think USSR), but if everything is independent and private, companies can take advantage of consumers.  In my opinion, there is no "right" way, but it should be somewhere in between.

The US is already a very decentralized government - education is one example of this.


i'm just learning about bases and i'm wondering if this conjecture is correct:

there are $2^{n-1}$ n-digit binary numbers.

can someone prove this?

also, is it generally the case

there are $m^{n-1}$ n-digit base-m numbers?

your first conjecture is correct

the first digit must be 1,

then there are 2 numbers for each digit then and there are n-1

so there are 2^n-1  permeations.

now looking at your secnd conjecture.

i beleive this is faulse because there are m^n-1 permeations other than the finail digit, then the finail digit cant be 0 so there are m-1 permeatations for the last digit so for the whole number there are m^n-m^n-1 permeations for a n digit number in base m

Or you could just right it as $(m-1)m^{n-1}$.


ratio geometry vector parameterization AMC

Two particles move along the edges of equilateral $\Delta ABC$ in the direction
\[A \rightarrow B \rightarrow C \rightarrow A \]
starting simultaneously and moving at the same speed. One starts at $A$, and the other starts at the midpoint of $BC$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $R$. What is the ratio of the area of $R$ to the area of $\Delta ABC$?

(A) $\frac{1}{16}$
(B) $\frac{1}{12}$
(C) $\frac{1}{9}$
(D) $\frac{1}{6}$
(E) $\frac{1}{4}$

One key to this: consider a time period during which each particle moves through half of a side of $ABC.$ In vector form each particle's position is given by a linear parameterization: $\vec{r_{1}}(t)=\vec{a_{1}}+t\vec{b_{1}}$ and $\vec{r_{2}}(t)=\vec{a_{2}}+t\vec{b_{2}}.$ (I'm not going to figure out what those coefficients are - it's enough to just know they exist). Then the midpoint is given by

$\frac{\vec{r_{1}}(t)+\vec{r_{2}}(t)}2= \frac{\vec{a_{1}}+\vec{a_{2}}}2+t\left(\frac{\vec{b_{1}}+\vec{b_{2}}}2\right)$

And that's a straight line. The whole point of this demonstration (which I would have thought but not written down) is to show that the path traced out by the midpoint consists only of line segments.  These segments potentially end when one of the other of the particles is at a vertex. Find those points - you get a smaller triangle, traced around twice. Just focus on the vertices of that smaller triangle.


In a circle, points A, B, C, and D are drawn so that AB is tangent at A and secant CD touches the circle at C and D. AB intersects line CD at E and ray CD extends to F so that BF is parallel to AC. Prove that triangles AED and FEB are similar. [/quote]

[quote="bingo2009"]In a circle, points A, B, C, and D are drawn so that AB is tangent at A and [b]CD touches the circle at C and D.[/b][/quote]
can you give the figure, i think it is a typo.[CD touching at both C & D???? :huh: ]


calculus
integration
limit
geometry
3D geometry

6. Define $p = \sum\limits_{k=1}^{\infty}\frac{1}{k^2}$ and $q = \sum\limits_{k=1}^{\infty}\frac{1}{k^3}.$ Find a way to write $\sum\limits_{j=1}^{\infty}\sum\limits_{k = 1}^{\infty}\frac{1}{(j+k)^3}$

7. A set of regular polygons of side length $1$ is chosen that the polygons can be made to "fill" the $360^{\circ}$ angle about a point $P$, As the hexagon, two squares, and triangle do at right(just draw them out). In how many ways can the polygons be chosen? The order of placement is not important.

8. The Fibonacci sequence is defined recursively as $F_1 = 1$, $F_2 = 1$, and $F_{n+1} = F_n + F_{n-1}$. Find the remainder where $F_{1734}$ is divided by $25$.

[hide]

6) The summation can be rewritten as  (Can you see why?). Through partial fraction decomposition, we can rewrite the sum as , and using the original definitions for  and , we can rewrite that sum as .

8) Take the sequence . You'll get  is , and  is , and from then on it alternates . Thus  leaves a remainder of  when divided by .[/hide]

I get different answer for #8...

[hide]The sequence $F_n$ begins mod 25 as:

1   1   2   3 5

8   13 21 9 5

14 19 8   2 10

12 22 9   6 15

21 11 7  18 0

The zero occurs at the 25th term. Then $F_{26}\equiv 18$, $F_{27}\equiv 18$, and more generally $F_{n+25}\equiv 18F_n$; the inductive step is easy to show from the recursion:
$F_{n+27}\equiv F_{n+26}+F_{n+25}\equiv 18F_{n+1}+18F_n\equiv 18F_{n+2}$

Since $18^2\equiv -1$, $18^4\equiv 1$ so $F_{n+100}\equiv F_n\Rightarrow F_{1734}\equiv F_{34}\equiv 18F_9\equiv \boxed{12}$.[/hide]

Kalle wrote:Elemennop wrote:[hide]
6) The summation can be rewritten as  (Can you see why?). .[/hide]
No, why is this true?

Hidden TextBecause the equation  has  solutions in positive integers  and , the term  will appear  times in the double sum, so the sum becomes . (This is equivalent to  because this is just the sum with the first term (which is 0 anyway) removed.).

Yeah, you're right scorpius, I mis-added 8+13=22 early in the sequence.

[quote="scorpius119"]
[hide]Because the equation $j+k=n$ has $n-1$ solutions in positive integers $j$ and $k$, the term $\frac{1}{n^3}$ will appear $n-1$ times in the double sum, so the sum becomes $\sum_{n=1}^\infty \frac{n-1}{n^3}$. (This is equivalent to $\sum_{n=1}^\infty \frac{n}{(n+1)^3}$ because this is just the sum with the first term (which is 0 anyway) removed.).[/hide][/quote]
How is this formalized, though? The argument is easy to understand if you don't think about what a series actually is (a limit of a sum), but once you do I don't understand how to show the equality properly. I'm confused by the concept of the limit of a limit, I think. The problem is that we can't let them both tend to infinity at the same time, and that's why this grouping of terms has me very confused.

Here's a sketch of how to formalize it:

We can show that $\sum_{j=1}^m\sum_{k=1}^{\infty}\frac{1}{(j+k)^3} = \sum_{n=1}^{\infty}\frac{n-1}{n^3} - \sum_{k=m+2}^{\infty}\frac{k - m - 1}{k^3}$ (Use Scorpius' reasoning and then deduct the extra terms we've added). It is "OK" to group the terms together here since if both series are convergent then $\sum_{k=1}^{\infty}f(k) + \sum_{j=1}^{\infty}g(j) = \sum_{n=1}^{\infty}(f(n) + g(n))$

By integral estimation we have that $\lim_{m\to\infty}\sum_{k=m+2}^{\infty}\frac{k - m - 1}{k^3} = 0$. This shows that $\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(j+k)^3} = \sum_{n=1}^{\infty}\frac{n-1}{n^3}$.

Is there a better justification that allows your argument to be used directly?

edit: Actually, I don't think this is any better. This runs into the same problems when doing the "grouping". I don't like the whole idea of rearranging the series into two new series, when the rearrangement cannot be done for a single subsum... How can I understand this? I think I'm just going to take this question to the calculus forum.

How much time do you get for the Mandelbrot?

Wow, these problems are challenging for that amount of time.

For the first one, I say that you can see it like that in the following way...(well this is how i would write it)
With $j=1$, we have $\sum\limits_{k=1}^{\infty}\frac{1}{(k+1)^3}=\frac{1}{2^3}+\frac{1}{3^3}+...$.  Similarly, with j=2, we have $\frac{1}{3^3}+\frac{1}{4^3}+...$, and for j=n, we have a sequence with starts with 1/(n+1)^3 and goes to infinity (ie, the denominator's cube root goes along the natural numbers).  So then, in the end, we have one $\frac{1}{2^3}$ term, 2 $\frac{1}{3^3}$ terms, and so on.  Thus, $\sum\limits_{n=2}^{\infty}\frac{n-1}{n^3}$.


AMC
AIME
USA(J)MO
USAMO
trigonometry
geometry
trig identities

So, after that hour long test...
I got a 99 on the AMC 12 (assuming no errors with scan-trons) :(.
Now, I know I had no chance in hell of making it into USAMO, so I went the conservative route and answered the first 15 questions (I rechecked my work after the 45 minute mark because I usually make stupid mistakes). After the test, I checked my work and found a really stupid mistake on 13 (the boat triangle one) :(. I found the limit for 60* with law of cosines, but then stupidly reasoned that since 45* was smaller, then the distance of AC2 must be lower (stupid, stupid, stupid). Should have remembered that it was the supplementary angle.
Do you think the cut-off will be like last year? If not, I'm really screwed. Should have finished my work on 25 (the tangent sum one).
Oh well, I guess that's it for math competitions then (last year in high school). It was fun while it lasted.

My best suggestion is that while waiting for replies (which won't probably result in couple weeks I guarantee), try to ask around for AMC 12. It's your LAST year! Make something out of it.

I mean, you don't have to... My AMC was canceled by snow on my senior year and it was during B exam. Oops.  :rotfl:  :|

my best advice:

don't just sit around on your butt, register to take the B exam and hope it doesn't get snowed out.

Then study hard hard hard hard hard for the AIME.

Thanks for the advice, guys, but...
I've already asked the school. The school won't pay for anyone to take the test because it is a bit costly on its part. I even had 8 people wanting to take it that said that they would pay for half the cost.

Then you guys can pay for it yourself and have it shipped to the school. If you register (online) by February 20th the cost will be 66 dollars for registration + 10 tests. You can divide that cost by the 9 of you. All you need is ONE teacher or admistrator at the school to proctor you and sign a form.

Hmm, I wonder if there is any method to bypass seeking school teacher/administrator by doing it as homeschoolers do? Can any homeschooler explain how they do it? I don't know if it's through the school... If not, I mean, you can try to ask someone in the local area who is "qualified" (I put quotation because I don't know what AMC consists qualified person for proctor).

I also suggest that you contact AMC Director through private message for fast answers. He is usually very prompt with inquiries.

Hope this helps,

Cheers,

From my experience as a homeschooler taking the AMC, the main requirements are that you do it in a public location (libraries work well), and that the proctor is not a family member.  You should be able to arrange something.

http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml

this may help

  :)

WOOOO!!!!
I just found out I made AIME from (the cut-off is 97.5 this year).
WOOOO!!!! :rotfl:

That's awesome dude! Show the AIME what you're made of!


number theory
Divisibility
modular arithmetic
Sequence
IMO Shortlist

[hide]
WLOG $ \text{gcd}\,(a_1, a_2, \ldots, a_n) \equal{} 1$, with $ a_1 < a_2 < \cdots < a_n$. For sake of contradiction, assume given statement is true.

Suppose $ 3 \nmid a_i \plus{} a_n$ for some $ 1 \le i \le n \minus{} 1$; by given, $ a_i \plus{} a_n | 3a_k$ for some $ 1 \le k \le n$. Then $ a_i \plus{} a_n | a_k < a_i \plus{} a_n$, contradiction. Thus $ a_1 \equiv a_2 \equiv \cdots \equiv a_{n \minus{} 1} \equiv \minus{} a_n \not\equiv 0 \pmod{3}$. Analogously, we have $ a_i \plus{} a_{n \minus{} 1} | a_n \quad (*)$ for all $ 1 \le i \le n \minus{} 2$ because $ 3 \nmid a_i \plus{} a_{n \minus{} 1}$. Now simple casework:

If $ \frac {a_n}2 \le a_{n \minus{} 1}$, it follows that $ (*)\ a_i \plus{} a_{n \minus{} 1} \equal{} a_n$, eg we have a triple $ \{a_1, a_2, a_1 \plus{} a_2\}$ and we quickly verify that there is a contradiction - for example take $ a_2, a_1 \plus{} a_2$, and we find $ 3a_1,\ \frac {3}{2}a_2,\ \frac {3}{2}(a_1 \plus{} a_2) \le (a_2) \plus{} (a_1 \plus{} a_2) \le 3a_2,\ 3(a_1 \plus{} a_2)$.

If $ \frac {a_n}{4} \le a_{n \minus{} 1} < \frac {a_n}2$, then $ (*)\ a_i \plus{} a_{n \minus{} 1} \le \frac {a_n}2$ and $ a_i \le \frac {a_n}4$, for all $ 1 \le i \le n \minus{} 2$. Then $ a_i \plus{} a_n > 3a_{n \minus{} 2}$, so one of $ a_i \plus{} a_n | 3a_{n \minus{} 1}, 3a_n$ must hold. Furthermore, $ a_i \plus{} a_n > \frac {3}{2}a_{n \minus{} 1}, a_n$, so $ a_i \plus{} a_n \equal{} 3a_{n \minus{} 1}, \frac {3}{2}a_n$. Either way we obtain $ a_{n \minus{} 1} \equal{} \frac{a_n}2$ contradiction.

If $ a_{n \minus{} 1} < \frac {a_n}{4}$ we must have $ a_i \plus{} a_n \equal{} \frac {3}{2}a_n$ contradiction.
[/hide]

WLOG we assume $ a_1<a_2<...<a_n$
we also use induction (base is practically trivial)
so for every $ {k}\leq{n\minus{}1}$ there exists a pair described in the task
$ a_n\plus{}a_{n\minus{}1}> \frac{3}{2} a_i$ for all $ i<n$
so one of the following statements is true
$ a_n,a_{n\minus{}1}$ is the desired pair
or $ \frac{a_n\plus{}a_{n\minus{}1}}{3}\equal{}a_k$ for some $ k<n$
or $ a_n\plus{}a_{n\minus{}1} | 3a_n$ which is equivalent to $ a_n\plus{}a_{n\minus{}1} | 3a_{n\minus{}1}$ so this is equal to the previous case.
We shall now prove that either the statement of the problem is true or this sequence has more than n elements (what is of course a contradiction).
If the first case is true we are done so we only need to consider second one.
if $ k\equal{}n\minus{}1$ we get $ a_n\equal{}2a_{n\minus{}1}$ and by induction we know that there is a pair $ a_i\plus{}a_j, i<j<n$
such that $ a_i\plus{}a_j$ ł $ 3a_k$ for all $ k<\equal{}n\minus{}1$ and because $ a_n\equal{}2a_{n\minus{}1}$ we also have
$ a_i\plus{}a_j$ ł $ 6a_{n\minus{}1}\equal{}3a_n$ so this pair is the desired one
otherwise we know we must have $ \frac{a_n\plus{}a_{n\minus{}1}}{3}, a_{n\minus{}1},a_n$ are elements of this sequence
we continue same conclusions and get that every expression of the following kind must be in this sequence $ \frac{a_{n\minus{}1}\plus{}\frac{3^k\minus{}1}{2}a_n}{3^k}$ for every k and either there is to much of them( more than n) either they aren't integers in both ways it is impossible so we have proven our statement

[hide="My Solution"]
Let $a_{i}$ be the biggest or second biggest term in the set. We can now pick $a_{i}$ and $a_{j}$ such that $a_{i}+a_{J}\neq 0$ mod 3. This would mean that we have to show that if $\frac{3a_{m}}{a_{i}+a_{j}}=k$ then k is not an integer. We now have $3a_{m}=k(a_{i}+a_{j})$ and if k was an integer than it would be divisble by 3 which would mean that it's at least 3 which would make the right hand side bigger than the left hand side unless $a_{i}$ was the second biggest term and $a_{m}$ was the biggest term. However if $a_{i}$ was the second biggest term instead of the biggest that would mean that all other terms in the set were congruent to -$a_{m}$ mod 3 which means that $a_{i}+a_{j}$ is congruent to -$2a_{m}$ mod 3 and so the left and right sides of the equation cannot be equal.
[/hide]

OK, I solved this a long time ago. Forgot to post the solution here. :oops:

[hide=" Solution"]
WLOG let $a_1 < a_2 < \cdots < a_n.$ Suppose the pair $a_n + a_{n-1}$ doesn't work. Let
\[a_n + a_{n-1} \mid 3a_k.\]
Let $3a_k= m (a_n + a_{n-1}).$ I claim that $m=1.$ Suppose not. Then,
$m \geq 2$ and $a_n + a_{n-1} \leq \dfrac{3}{2} a_k.$ But, $a_n > a_{n-1} \geq a_k,$ so $a_n + a_{n-1} > 2a_k.$ Hence, contradiction. Therefore, $a_j = \dfrac{a_n + a_{n-1}}{3}.$

Now, casework!

[b] Case 1: [/b] $a_n > 2 a_{n-1}$

Consider $a_j + a_n = \dfrac{4a_n + a_{n-1}}{3}.$ Note that
\[\dfrac{4a_n + a_{n-1}}{3} > \dfrac{8 a_{n-1} + a_{n-1}}{3} = 3 a_{n-1},\]
so $a_j + a_n \nmid 3 a_k \quad \forall \ k < n.$ If $a_i + a_j \not \mid 3 a_n,$ we have our required indices. If $\dfrac{4 a_n + a_{n-1}}{3} \mid 3a_n,$
\[ \begin{array}{lcl} 4 a_n + a_{n-1} \mid 9 a_n  & \implies &  4a_n + a_{n-1} \mid 9 (4a_n + a_{n-1}) - 4 \cdot 9 a_n \\ & \implies & 4a_n + a_{n-1} \mid 9 a_{n-1}. \end{array}\]
But,
\[4a_n + a_{n-1} > 8 a_{n-1} + a_{n-1} = 9 a_{n-1},\]
so contradiction.

[b] Case 2: [/b] $a_n < 2 a_{n-1}$

Consider $a_j + a_{n-1} = \dfrac{4 a_{n-1} + a_n}{3}.$ Note that
\[\dfrac{4 a_{n-1} + a_n}{3} > \dfrac{8a_n + a_n}{3} = 3 a_n,\]
so this is our required pair.

Thus, we obtain $a_n = 2 a_{n-1}.$ I claim that the pair $(a_n, a_{n-2})$ works. Note that
\[a_n + a_{n-2} = 2 a_{n-1} + a_{n-2} > 2 a_{n-2} + a_{n-2} = 3 a_{n-2},\]
so either $a_n + a_{n-2} \mid 3a_{n-1}$ or $a_n + a_{n-2} \mid 3a_n.$

Casework again! :lol:

[b]Case 1: [/b] $a_n + a_{n-2} \mid 3 a_{n-1}$
This implies
\[\begin{array}{lcl}
2 a_{n-1} + a_{n-2} \mid 3 a_{n-1} & \implies & 2 a_{n-1} + a_{n-2} \mid 3 (2 a_{n-1} + a_{n-2}) - 2 \cdot 3 a_{n-1} \\
& \implies & 2 a_{n-1} + a_{n-2} \mid 3 a_{n-2}, \end{array}\]
which is not possible since
\[2 a_{n-1} + a_{n-2} > 2 a_{n-2} + a_{n-2} = 3 a_{n-2}.\]

[b]Case 2: [/b] $a_n + a_{n-2} \mid 3a_n$

This implies
\[a_n + a_{n-2} \mid 3 (a_n + a_{n-2}) - 3 a_n = 3 a_{n-2},\]
which is not possible either, since
\[a_n + a_{n-2} = 2 a_{n-1} + a_{n-2} > 2 a_{n-2} + a_{n-2} = 3 a_{n-2}.\]

Now that we have covered all possible cases, our proof is complete. $\blacksquare$
[/hide]

Suppose $a_1<a_2<\dots<a_n$. We also suppose $(a_1,a_2,\dots,a_n)=1$. We suppose opposite, that for every $a_i+a_j$ exists $a_k$ such that $a_i+a_j|3a_k$.  Consider pairs $a_n+a_i$ for $1\le i \le n-1$. For every $i$ there is $j$ such that $a_n+a_i|3a_j$. Now if $3$ doesn't divide $a_n+a_i$ then $a_n+a_i|a_j$ but $a_n>a_j$ which is a contradiction. So we proved that $a_1,a_2,\dots a_{n-1}$ are all congruent to $-a_n$ modulo $3$. Because their gcd is 1, we conclude that none of our numbers is divisible by 3. Now we conclude that 3 doesn't divide $a_i+a_j$ for $1 \le i<j<n$. Now observe $a_{n-2}+a_{n-1}>a_{n-1}$ and 3 doesn't divide that sum so $a_{n-2}+a_{n-1}|a_{n}$. When we consider pair $a_n+a_{n-1}$ it can divide $3a_n$ which easily gives $a_n=2a_{n-1}$. It can divide $3a_{n-1}$ (it will be equal) which also easily gives $a_n=2a_{n-1}$. Now suppose that $a_n+a_{n-1}|3a_i$ for some $i<n-1$. Because $a_n+a_{n-1}>2a_i$ it has to be $3a_i=a_n+a_{n-1}\ge 2a_{n-1}+a_{n-2} > 3a_{n-2}\ge 3a_i$ which is a contradiction so we conclude $a_n=2a_{n-1}$. We got $a_{n-1}+a_{n-2}|a_n=2a_{n-2}$ so because $a_{n-1}+a_{n-2}>a_{n-1}$ we have that $a_{n-1}+a_{n-2}=2a_{n-1}$ so $a_{n-1}=a_{n-2}$ and that is a contradiction so the problem is proven.

[hide="solution"]
WLOG $a_1\ge a_2\ge \cdots \ge a_n$. Furthermore, WLOG $\gcd(a_1, a_2, \ldots , a_n)=1$ by dividing everything by their common factors.
If $\exists a_i$ s.t. $3\nmid a_1+a_j$, then $a_1+a_i \nmid a_k\iff a_1+a_i \nmid 3a_k\,\forall 1\le k\le n$.
Otherwise, $3\mid a_1+a_i\,\forall 2\le i\le n$. There $\forall a_x, \,\exists a_y$ s.t. $a_1+a_x \mid 3a_y$. Since $a_1+a_x>  a_y$, then $a_1+a_x = 3a_y\le 3a_2\,(1)$ or $a_1+a_x=\dfrac{3}{2}a_y\le \dfrac{3}{2}a_2\,(2)$.
If $3\mid a_1$ then $3\mid a_i$ contradiction with $\gcd(a_1, a_2, \ldots , a_n)=1$, so $a_1\equiv 1, 2\pmod{3}\implies 3\nmid a_i+a_j\,\forall 2\le i < j \le n$.
If $\exists a_i$ s.t. $a_2+a_i \nmid a_1\iff a_2+a_i\nmid 3a_1$, then $a_2+a_i\nmid a_k \iff a_2+a_i\nmid 3a_k\,\forall 1\le k\le n$, done.
Otherwise, $a_2+a_i\mid a_1\,\forall 3\le i\le n$. Define a sequence $b_i$ as $b_i=\dfrac{a_1}{a_2+a_i}$. But $b_i =\dfrac{a_1}{a_2+a_i} \le \dfrac{3a_2-a_x}{a_2+a_i} < 3\implies b_i \le 2$ in case $(1)$, and $b_i < \dfrac{3}{2}\implies b_i \le 1$ in case $(2)$. Since $a_i$ are all distinct, then if $n\ge 5$ for case $(1)$ or $n\ge 4$ for case $(2)$ then we arrive at a contradiction. It remains to check small cases:
Subcase $n=3$:
Then $a_2+a_3\mid a_1$, and since $b_3 \le 2$, then $a_2+a_3=a_1$ or $a_2+a_3=\dfrac{a_1}{2}$. But also $a_1+a_2 = 3a_k$ or $\dfrac{3}{2}a_k$. $k=1$ gives a contradiction in the first equation and $a_2=\dfrac{1}{2}a_1\implies a_3=0\text{ or }\dfrac{1}{2}a_1$ impossible. $k=2$ gives $a_2=\dfrac{1}{2}a_1$ which again leads to a contradiction. $k=3$ gives $a_1+a_2=3a_3$ or $a_1+a_2=\dfrac{3}{2}a_3$. If it is the latter, then $a_1+a_2 =\dfrac{3}{2}a_3 \le \dfrac{3}{2}a_2\implies a+2\ge 2a_1$ impossible, so $a_1+a_2=3a_3$. Then $a_2+a_3=a_1\implies a_1-a_2=a_3\implies a_2=a_3$ impossible, and all the cases are exhausted.
Subcase $n=4$:
In order for $b_3, b_4$ to be distinct, $a_1+a_2=3a_{k_1}$ and $a_1+a_3=3a_{k_2}$ and $a_1+a_4=3a_{k_3}$, where $a_{k_1} > a_{k_2}>a_{k_3}$. Since $k_1=1\implies a_2=2a_1$ impossible, then $(k_1, k_2, k_3)=(2, 3, 4)\implies a_1=2a_2=2a_3=2a_4$ contradiction.

Thus, all cases are exhausted and we are done. $\Box$
[/hide]

My solution: Let $[n]=\{1,2, \dots ,n\}$. Assume to the contrary that there exists a sequence $a_1,a_2, \dots ,a_n$, such that for all distinct integers $i,j \in [n]$, there exists $k \in [n]$ such that $a_i+a_j \mid 3a_k$. Denote this statement by $P(i,j)$, and call such a sequence [i]ugly[/i]. We wish to show that no [i]ugly[/i] sequence exists. Now, suppose that $\gcd(a_1,a_2, \dots ,a_n)=d$. Taking $a_i=d \cdot b_i$, and using our statement $P(i,j)$, we get that $b_i$ is also [i]ugly[/i]. So WLOG assume that $\gcd(a_1,a_2, \dots ,a_n)=1$. Also WLOG suppose that $a_1>a_2> \dots >a_n$.

$P(1,j)$ gives that there exists $k \in [n]$, such that $a_1+a_j \mid 3a_k$. If, for some $j \in [n]$, $3 \nmid a_j+a_1$, then $a_j+a_1 \mid a_k \Rightarrow a_k \geq a_j+a_1$, a contradiction to the assumption that $a_1$ is maximum. So we get that $3 \mid a_j+a_1 \text{ } \forall j \in [n]$. Call this property as the [i]triple property[/i] of an [i]ugly[/i] sequence. Also, if $3 \mid a_r$ for some $r \in [n]$, then $3 \mid a_1$, which in turn gives that $3 \mid a_s$ for all $s \in [n]$, a contradiction to our assumption that their GCD is one. Thus, we get that $3$ doesn't divide any of the $a_i$'s.

Now, consider $P(i,2)$, where $i>2$. Then there exists $k \in [n]$ such that $a_i+a_2 \mid 3a_k$. Note that if $3 \mid a_i+a_2$, then using the [i]triple property[/i], we get that $3$ divides $a_1-a_2$ as well as $a_1+a_2$, which in turn gives that $3 \mid 2a_1 \Rightarrow 3 \mid a_1$, that was earlier proved by us to be impossible. This means that $3 \nmid a_i+a_2$, and so $a_i+a_2 \mid a_k \Rightarrow a_i+a_2 \leq a_k$, which is only possible if $k=1$. Thus, $a_i+a_2 \mid a_1$ for all $i \in [n]$ and $i>2$. So we basically have that $a_2+a_i \leq a_1$.

Now, from $P(1,2)$, we have that there exists $k \in [n]$ such that $a_1+a_2 \mid 3a_k$.

[b]CASE-1[/b] $(k>2)$ Note that $a_1+a_2 \leq 3a_k \Rightarrow 3a_k-a_2 \geq a_1 \geq a_2+a_k \Rightarrow a_k \geq a_2$. This gives that either $k=1$ or $k=2$, a contradiction to our assumption.

[b]CASE-2[/b] $(k=2)$ We have $a_1+a_2 \leq 3a_2 \Rightarrow a_1 \leq 2a_2$. Also $a_1+a_2 \mid (3a_2)-(a_1+a_2) \Rightarrow a_1+a_2 \mid 2a_2-a_1$. As $2a_2>a_1$, we get that $a_1+a_2 \leq 2a_2-a_1 \Rightarrow 2a_1 \leq a_2$, a contradiction to our assumption that $a_1$ is maximum.

[b]CASE-3[/b] $(k=1)$ In this case, we are given that $a_1+a_2 \mid 3a_1$. As $a_1<a_1+a_2<2a_1$, we must have that $a_1+a_2=\frac{3a_1}{2}$. This gives that $a_1=2a_2$. But then, as $a_i+a_2 \mid a_1=2a_2$ for $i>2$, and cause $a_2<a_i+a_2<2a_2$, we get that there is no possible integral value of $a_i$ for all $i>2$. Thus, we again arrive at a contradiction.

Summing up all the cases together, we get the result that no [i]ugly[/i] sequence exists. Hence, done. $\blacksquare$

Here is a solution that is equivalent to bobthesmartypants' solution but with an optimization to avoid subcasework.

Assume the contrary, so for all distinct $i, j$ there is $k$ so that $a_i + a_j \mid 3a_k$. WLOG $a_1<a_2< \dots < a_n$ and $\gcd(a_1,a_2,\dots,a_n)=1$. For each $i=1, 2, \dots, n-1$, we have $a_n + a_i \mid 3a_k$ for some $k$, but for size reasons we must have $3 \mid a_n + a_i$. This implies that $a_1 \equiv a_2 \equiv \dots \equiv a_{n-1} \equiv -a_n \pmod{3}$, and since we assumed they are all relatively prime we have $a_i \not \equiv 0 \pmod{3}$. Thus, for all $i<j \leq n-1$, we have $a_i + a_j \equiv 2a_i \not \equiv 0 \pmod{3} \quad (\heartsuit)$.

Now, suppose $a_n + a_{n-1} \mid 3a_k$. Note that if $k=n$, then $a_n + a_{n-1} \mid 3a_{n-1}$ anyways so assume $k \le n-1$. Since $a_n + a_{n-1} > 2a_k$, we must have $a_n + a_{n-1} = 3a_k$ and so $a_n = 3a_k  - a_{n-1}$.

Now consider the number $a_{n-1} + a_{n-2}$. By $(\heartsuit)$, it is not divisible by $3$, so if it divides $3a_j$ it must divide $a_j$. For size reasons this must be $a_n$. So, we have$$a_{n-1} + a_{n-2} \mid a_n = 3a_k - a_{n-1} \le 2a_{n-1},$$forcing $a_{n-2}=a_{n-1}$, a contradiction. $\blacksquare$

An easier solution by induction:

[b]Key lemma:[/b] If $m > \frac{n}{2}$, then $m$ does not divide $n$.

We proceed by induction on $n$, the number of distinct positive integers.

[b]Base case: $n=3$ [/b]

WLOG let $a_1 < a_2 < a_3$. By our lemma, we know that $a_3 + a_2$ does not divide $3a_2$ or $3a_1$.

Case i) $a_3 + a_2$ does not divide $3a_3$: Then we simply take $2, 3$ as the desired indices $i, j$

Case ii) $a_3 + a_2$ does divide $3a_3$:

Since $\frac{3a_3}{3}<a_3 + a_2 <\frac{3a_3}{1}$, we must have that $a_3 + a_2 = 1.5a_3$ , or rather $a_3 = 2a_2$.

Now consider $a_3 + a_1 = 2a_2 + a_1$. We know that $a_3 + a_1$ does not divide $a_1$ and $a_2$ by our key lemma. Furthermore, $$\frac{3a_3}{3} = 2a_2 < a_3 + a_1 = 2a_2 + a_1 < 3a_2 = \frac{3a_3}{2}$$. Thus $a_3 + a_1$ does not divide $a_3$ either, and we take $3, 1$ as our desired indices $i, j$.

[b]Inductive step:[/b]

We shall assume the result is true for $n-1$ distinct positive integers. Thus note that for any $a_1 < a_2 < ... a_n$ we can find indices $2 \le i < j \le n$ such that $a_i + a_j$ does not divide any of $3a_2, 3a_3, ... 3a_n$ (by the inductive hypothesis). Furthermore, $a_i + a_j > 2a_1$, and once again by our key lemma $a_i + a_j$ does not divide $3a_1$ either, completing the inductive step.

Solution from [i]Twitch Solves ISL[/i]:

Assume for contradiction this is not true. By sorting, we assume $a_1 > a_2 > \dots > a_n$. We will suppose that $\gcd(a_1, \dots, a_n) = 1$, else divide through.

[b][color=red]Claim:[/color][/b]  	We have $a_i \equiv -a_1 \pmod 3$ for all $i > 1$.

[i]Proof.[/i]  	If not then $a_1 + a_i$ divides some $3a_k$. 	But $a_1 + a_i \perp 3$, so $a_1 + a_i$ divides some $a_k$. 	This is impossible for size reasons. $\blacksquare$

Since we assumed the sequence was relatively prime, we have now that \[ a_2 \equiv a_3 \equiv \dots \equiv a_n 	\equiv -a_1 \not\equiv 0 \pmod 3. \] We now work with only $a_1$, $a_2$, $a_3$. The proof proceeds in three steps. [list] 	[*]Again, $a_2 + a_3$ divides $3a_k$ for some $k$, 	but since $a_2 + a_3 \not\equiv 0 \pmod 3$, 	we find that $a_2 + a_3 \mid a_k$ for some $k$. Hence in fact $\boxed{a_2 + a_3 \mid a_1}$.

[*]Meanwhile, $a_1 + a_2$ divides 	either $3a_1$ or $3a_2$ for size reasons. 	But actually, these two statements are equivalent, 	since $a_1 + a_2$ divides their sum $3a_1+3a_2$. Now from $2a_2 < a_1 + a_2 \mid 3a_2$, 	we must have $\boxed{a_1 = 2a_2}$.

[*]Now $a_2 < a_2 + a_3 \mid a_1 = 2a_2$, 	so we must have $a_2 + a_3 = 2a_2$ 	but then $a_2 = a_3$, contradiction. [/list] This concludes the proof.

[b][color=red]Remark:[/color][/b]  	Note that we can find pretty early on 	that the $n=3$ case of the problem is sufficient: 	the three largest numbers satisfy the condition anyways, 	so there is no need any longer to look at $a_4$, $a_5$, ...

Are size arguments commonly used in number theory (like the one above)?

FTSoC suppose that for each pair $(i, j)$, we have $a_i + a_j \mid a_k$ for some index $k$. Since all $a_i$ are all distinct by the problem statement, and because this question is in the [b]Size NT[/b] unit, we are motivated to assume the terms are in decreasing order $a_1 > a_2 > \ldots > a_n$. Furthermore, we can also assume that $\gcd(a_1, a_2, \ldots , a_n) = 1$ because otherwise we can divide out the GCD and the problem setup remains identical.

I claim that for all $i \geq 2$ we have $3 \mid a_1 + a_i$. Suppose otherwise, and that $3 \nmid a_1 + a_i$ for some index $i$. Then, since $a_1 + a_i \mid 3a_j$ for some $j$, and because $\gcd(a_1 + a_i, 3) = 1$, we must have $a_1 + a_i \mid a_j$, a clear contradiction by our size assumption.

Now get stuck for 45 minutes due to lack of skill and foresight. Then ponder the meaning of life and eventually think back to [b]SIZE[/b].

Rewriting the above we have\[a_2 \equiv a_3 \equiv \ldots \equiv a_n \equiv -a_1 \pmod 3.\]Note that this common value mod $3$ cannot be zero or else $3$ divides all terms, a contradiction to our assumption that $\gcd(a_1, a_2, \ldots , a_n) = 1$.

Let's consider the larger elements. $a_2 + a_3 \mid 3a_k$ for some index $k$. Note that $a_2 + a_3 \equiv -2a_1 \not\equiv 0 \pmod 3$ so in fact $a_2 + a_3 \mid a_k$ so by size $k = 1$ and thus $a_2 + a_3 \mid a_1$. (This actually motivates the previous step)

Furthermore, $a_1 + a_2 \mid 3a_l$ for some index $l$. Clearly $l \geq 3$ is bad due to size. Hence, we must have $a_1 + a_2 \mid 3a_1$ or $3a_2$, which actually turn out to be equivalent statements. Either way, take $a_1 + a_2 \mid 3a_2$. Note that $a_1 + a_2 \geq 2a_2$ and the only factor of $3a_2$ greater than $2a_2$ is $3a_2$ so we must have $a_1 = 2a_2$.

But combined with $a_2 + a_3 \mid a_1$, this forces $a_3 = a_2$, which is impossible, and we are done. $\blacksquare$

Lastly, realize that solution is the same as Evan's :o

WLOG let $a_1 < a_2 < a_3 < \cdots < a_n$.
Assume ftsoc that the problem isn't true and for every $i, j$, there exists $k$ such that $a_i + a_j | 3a_k$.

[b] Claim: [/b] $a_1 \equiv a_2 \equiv \cdots \equiv a_{n-1} \pmod{3}$.

[i] Proof: [/i] Since $a_n + a_1 | 3a_k$ for some $k$, if $3 \nmid a_n + a_1$, we would have $a_n + a_1 | a_k$, but for size reasons that is impossible since it would mean $a_k \leq a_n < a_n + a_1 \leq a_k$. Hence $a_1 + a_n \equiv 0 \pmod{3}$. Similarly $a_2 + a_n \equiv 0 \pmod{3}, \cdots a_{n-1} + a_n \equiv 0 \pmod{3}$, therefore $a_1 \equiv a_2 \equiv \cdots a_{n-1} \pmod{3}$. $\qquad \blacksquare$

Now if $a_1 \equiv a_2 \equiv \cdots a_{n-1} \equiv 0 \pmod{3}$, then we would have $a_n \equiv 0 \pmod{3}$ as well, so we the problem would be equivalent when dividing all $a_i$ by $3$. Therefore we may assume that none of the $a_i$'s are divisible by $3$, as it would mean all the others are, too.
We know that $$\begin{cases} a_i\equiv 1 \pmod{3} \qquad \text{for } 1 \leq i \leq n-1 \\ a_n \equiv 2 \pmod{3} \end{cases}$$ or, alternatively, $$\begin{cases} a_i \equiv 2 \pmod{3} \qquad \text{for } 1 \leq i \leq n-1 \\ a_n \equiv 1 \pmod{3} \end{cases}$$

We know that $a_{n-1} + a_{n-3} | 3a_k$, however, $3 \nmid a_{n-1} + a_{n-3}$, so $a_{n-1} + a_{n-3} | a_k$. For obvious size reasons, $k = n$, hence, $a_{n-1} + a_{n-3}|a_n$. If we have $a_{n-1} + a_{n-3} = a_n$, we would arrive at a contradiction because $$a_{n-1} + a_{n-3}|a_n$$ and $$a_{n-1} + a_{n-2}|a_n$$ cannot hold at the same time, because it would mean $a_n = a_{n-1} + a_{n-3} < a_{n-1} + a_{n-2} \leq a_n$, absurd. Hence, $2(a_{n-1} + a_{n-3}) \leq a_n$. But if $2(a_{n-1} + a_{n-3}) = a_n$, we examine both sides modulo 3 and it doesn't work because $2(a_{n-1} + a_{n-3}) \equiv 2(1+1) = 1 \pmod{3}$ whereas $a_n \equiv 2 \pmod{3}$ or, in the second case $2(a_{n-1} + a_{n-3}) \equiv 2(2+2) = 2 \pmod{3}$ wheras $a_n \equiv 1 \pmod{3}$.

Therefore, $a_n \geq 3(a_{n-1} + a_{n-3})$. But we also know that $a_n + a_{n-1} | 3a_k$ for some $k$. If $k \leq n-1$, we would have $a_n < 3a_{n-1}$. But that is a contradiction to $a_n \geq 3(a_{n-1} + a_{n-3})$, as shown above. Hence $a_n + a_{n-1} | 3a_n \implies a_n + a_{n-1} | 3a_n - 2(a_n + a_{n-1}) = a_n - 2a_{n-1}$. However $a_n > 2a_{n-1}$ (since $a_n \geq 3(a_{n-1} + a_{n-3})$). So that means $a_n + a_{n-1}$ divides a positive integer less than $a_n$, contradiction.

So, our hypothesis was incorrect, and there exists $i, j$ such that $a_i + a_j$ does not divide $3a_k$ for any $k$. $\qquad \blacksquare$

[quote=April]Let $ a_1$, $ a_2$, $ \ldots$, $ a_n$ be distinct positive integers, $ n\ge 3$. Prove that there exist distinct indices $ i$ and $ j$ such that $ a_i \plus{} a_j$ does not divide any of the numbers $ 3a_1$, $ 3a_2$, $ \ldots$, $ 3a_n$.

[i]Proposed by Mohsen Jamaali, Iran[/i][/quote]

For the sake of contradiction, let us say that given any pair of indices $(i, j)$ we can find a proper index $k$ such that $a_i + a_j \mid 3a_k$. Without loss of generality let $a_i < a_j$ if $i < j$. We scale down all $a_i$ such that $\text{gcd}(a_1, a_2, a_3 \dots a_n) = 1$ (basically scaling down or up all the numbers would not change the assumption as the ratio $a_i + a_j : 3a_k$ hasn't changed) Consider $a_n + a_j$ for all $j$ such that $1 \leq j \leq n-1$. We see that $a_n + a_j \lvert 3a_k$ for some $k$ such that $1 \leq k \leq n-1$, but $a_n + a_j > a_n > a_k$ and so $3 \lvert a_n + a_j$, which means that $a_1 \equiv a_2 \equiv a_3 \dots a_{n-2} \equiv a_{n-1} = k \pmod{3}$, and due to scaling, $k \in \{ 1, 2 \}$.

(We use the fact that if $d \lvert n$ then $d \leq \left \lceil \dfrac{n}{2} \right \rceil$ multiple times in this paragraph) We see that $a_{n-1} + a_{n-2} \lvert 3a_j \implies a_{n-1} + a_{n-2} \lvert a_j = a_n$ as $a_{n-1} + a_{n-2} \equiv 2 a_{n-1} \not\equiv 0 \pmod{3}$ and $a_n + a_{n-1} > a_{n-1} \geq a_j$ for $j \leq n-1$. Therefore, we see that if $a_n + a_{n-1} \lvert 3a_k$ for some $k \in \mathbb{N}$, then $a_n + a_{n-1} \geq 2a_{n-1} + a_{n-2} > 3a_k$ for all $k \leq n-2$ implying that $k = n-1$ or $n$, but if $k = n$ then $a_n + a_{n-1} \lvert 3(a_n + a_{n-1}) - 3a_n = 3a_{n-1}$ and since $a_n + a_{n-1} > 2a_{n-1} > \left \lceil \dfrac{3a_{n-1}}{2} \right \rceil$, we must have that $a_n + a_{n-1} = 3a_{n-1}$ or $a_n = 2a_{n-1}$. Now, $a_{n-1} + a_{n-2} \lvert a_j = 2a_{n-1}$ and since $a_{n-1} + a_{n-2} > a_{n-1} = \left \lceil \dfrac{2a_{n-1}}{2} \right \rceil$, we must have that $2a_{n-1} = a_{n-1} + a_{n-2}$ or $a_{n-1} = a_{n-2}$ which is the desired contradiction.

For the sake of contradiction, assume that for every distinct indices $i$ and $j$ we have some $r$, so that $a_i + a_j \mid 3a_r$.
Let $a_n> a_{n-1}> \cdots > a_2> a_1$.

Let us consider $a_n+a_i \mid 3a_r$, where $1\leq i \leq n-1$, note that as $a_n+a_i>a_r$, then we must have indeed that $a_n+a_i\equiv 0 \pmod{3}$ for all $1\leq i \leq n-1$.
Now let us take any $1\leq i,j\leq n-1$ and we see that $a_i+a_j\mid 3a_r$, thus if $a_i+a_j\equiv 0 \pmod{3}$, we actually obtain that every $a_i$ is divisible by $3$, note that dividing this $3$ out, we get the same problem and thus we divide all possible $3$'s out and we get that for any $1\leq i,j\leq n-1$, $a_i+a_j\not \equiv 0 \pmod{3}$, i.e. $a_i+a_j\mid a_r$. Also it is worth noting that $i=n-1$ and $j=n-2$ imply $r=n$, since $a_r \geq a_{n-1}+a_{n-2}>a_{n-1}\implies r=n$.

Now we consider the divisibility, $a_n+a_{n-1}\mid 3a_x$, where $1\leq x \leq n$.
If $x=n$, then $a_n+a_{n-1}\mid  a_n-2a_{n-1}\implies a_n=2a_{n-1}$.
If $x=n-1$, then $a_n+a_{n-1}\mid 2a_{n-1}-a_n\implies a_n=2a_{n-1}$, in both cases $a_{n-1}+a_{n-2}\mid a_n=2a_{n-1}\implies a_{n-1}=a_{n-2}$, which contradicts distinctness.
Thus, we get $x\leq n-2$, and we write the following divisibilites, $a_n+a_{n-1}\mid 3a_x$ and $a_x+a_{x+1}\mid a_r$ for some $1\leq r \leq n$, therefore
$$a_n+a_{n-1}\leq 3a_x<a_x+a_x+a_{x+1}\leq a_x+a_r,$$
which is impossible, as $x\leq n-2$.

Who needs mods

Without loss of generality, let $a_1=\max\{a_1,\ldots,a_n\}$. Since $n \geq 3$, there either exists some $i \neq 1$ such that $a_i>\tfrac{1}{2}a_1$ or $a_i<\tfrac{1}{2}a_1$. We now consider the following two cases:\\
[b]Case 1:[/b] For all $i \neq 1$ we have $a_i\leq \tfrac{1}{2}a_1$. Pick $m$ such that $a_m \neq \tfrac{1}{2}a_1$ and $a_m$ is maximal. Then we have $a_1+a_m>3a_m\geq 3a_i$ if $a_i<\tfrac{1}{2}a_1$, so $a_1+a_m \nmid 3a_i$ in this case. We also have $\frac{3a_1}{2}>a_1+a_m>a_1>\tfrac{3a_1}{4}$, so $a_1+a_m$ is strictly between two consecutive divisors of both $3a_1$ and $\tfrac{3a_1}{2}$ if the latter is an integer, so $a_1+a_m \nmid 3a_1,3a_i$ if $a_i=\tfrac{1}{2}a_i$. Thus $a_1+a_m \nmid 3a_1,\ldots,3a_n$.\\
[b]Case 2:[/b] There exists $i \neq 1$ with $a_i>\tfrac{1}{2}a_1$. Let $m$ be a positive integer with $a_m>\tfrac{1}{2}a_1$ such that $a_m$ is minimal. In this case, we have
$$3a_i\geq 3a_m>a_1+a_m>\tfrac{3a_1}{2}\geq\tfrac{3a_i}{2}$$
for all $a_i>\tfrac{1}{2}a_1$, so $a_1+a_m \nmid 3a_i$ in this case. We also have $a_1+a_m>\tfrac{3a_1}{2}\geq 3a_i$ for all $a_i\leq \tfrac{1}{2}a_1$, so $a_1+a_m \nmid 3a_i$ in this case as well, implying $a_1+a_m \nmid 3a_1,\ldots,3a_n$.

FTSOC assume the contrary. WLOG assume $\gcd(a_1,a_2, \dots,a_n)=1$ and $a_n>a_{n-1}> \dots>a_1$.
[b]Claim :[/b] $a_1 \equiv a_2 \equiv \dots \equiv{a_{n-1}} \equiv{-a_n} \pmod{3}$
It suffices to prove that $3 \mid a_i+a_n \hspace{.1 cm} \forall \hspace{.01 cm}n$. Suppose not, then we can find $j$ satisfying $a_i+a_n \mid a_j$ which is a size contradiction.
[b]Claim :[/b] $2a_{n-1} \geq a_n$
Note that we can write $a_n+a_{n-1} \mid 3a_i$ for some $i \leq n-1$ since $a_n+a_{n-1} \mid 3a_n \implies a_n+a_{n-1} \mid 3a_{n-1}$. This means that $a_n \leq 3a_i -a_{n-1} \leq 2a_{n-1}$ as desired
[b]Claim :[/b] For any $n>3$, we get $a_1=a_2= \dots=a_{n-2}$

By the previous claims, we have either $a_1 \equiv a_2 \equiv \dots \equiv{a_{n-1}} \equiv{-a_n} \equiv{1} \pmod{3} $ or $a_1 \equiv a_2 \equiv \dots \equiv{a_{n-1}} \equiv{-a_n} \equiv{2} \pmod{3}$. Note that $3 \nmid a_i+a_{n-1}$ for all $i \leq n-2$. The assumption guarantees that we can find $j$ such that $a_i+a_{n-1} \mid a_j$ (since it is coprime to $3$) which is a size contradiction unless $j=n$. As a result,
$$a_i+a_{n-1} \mid a_n\hspace{.1 cm} \forall \hspace{.01 cm} i \leq n-2$$
But we also have $a_n \leq 2a_{n-1}$, and therefore $a_n=a_{n-1}+a_i$ for all such $i$, implying the result. This is clearly a contradiction.

The $n=3$ case is easy to handle by some casework ; Let the numbers be $a_1,a_2,a_3$. We have $a_3=a_1+a_2$ (shown previously), so rewrite this as $a,b,a+b$($b>a$) We must have $a+2b \mid 3a_i$ for some $i$. If $i=1, a+2b \mid 3a$, contradicting size. If $i=2, a+2b \mid 3b \implies a+2b \mid b-a$, contradicting size. Finally, if $i=3, a+2b \mid 3a+3b-a-2b=2a+b$ which contradicts size too, and we are done $\blacksquare$

It is not difficult to prove that the statement is true for $n=3$.

For $n>3$, suppose that there are $n$ distinct positive integers making the problem false, and WLOG let $a_1 > a_2 > \ldots >a_n$, and cancel out any common factors of $3$ among all numbers (as the two sequences are equivalent for this problem).

Suppose that there was an $i$ such that $2\le i\le n$ and $a_1+a_i$ is not a multiple of $3$. Then $a_1+a_i$ dividing one of $3a_1, 3a_2, \ldots, 3a_n$ is equivalent to dividing one of $a_1, a_2, \ldots, a_n$, which is impossible due to size. Therefore $a_1+a_i$ is divisible by $3$ for all $2\le i\le n$. If $a_1$ was divisible by $3$, then all $a_i$ are divisible by $3$, but we divided out all common factors of $3$, contradiction. Therefore either $a_1\equiv 1\pmod 3$, making all other terms $2$ mod $3$, or $a_1\equiv 2\pmod 3$, making all other terms $1$ mod $3$.

As we know the statement is true for $n=3$, there must be $i$ and $j$ so that $1\le i, j \le 3$ and $a_i+a_j$ does not divide $3a_1, 3a_2$, or $3a_3$.

[b][color=#f00]Case 1:[/color][/b] $(i,j)=(2,3)$ is a working pair.

It suffices to show that $a_2+a_3$ does not divide $3a_4, 3a_5, \ldots, 3a_n$. However, as $a_2+a_3$ is not a multiple of $3$, this is equivalent to showing $a_2+a_3$ does not divide $a_4, a_5, \ldots, a_n$, which cannot happen due to size reasons.

[b][color=#f00]Case 2:[/color][/b] $(i,j)=(2,3)$ does not work. Therefore, either $(i,j)=(1,2)$ or $(i,j)=(1,3)$ work (or both).

As $a_2+a_3$ is not a multiple of $3$, for it to divide $3a_1, 3a_2$, or $3a_3$, it must divide $a_1, a_2$, or $a_3$. $(a_2+a_3)|a_1$ is the only option for size reasons, making $a_2+a_3\le a_1 \Rightarrow a_3 < \frac{1}{2}a_1$.

It suffices to show that the working pair does not divide one of $3a_4, 3a_5, \ldots, 3a_n$. However, as $a_1 + a_2 > a_1+a_3 > 3a_3 > 3a_4$, whichever pair works, it is greater than any of $3a_4, 3a_5, \ldots, 3a_n$, resolving this case.

The proof is complete.

Can someone check this solution real quick? Its different than the one's posted I think.

WLOG the set of elements is sorted such that $a_n$ is the largest and $a_1$ is the smallest.

Now notice that if we find some $i$ such that $a_n+a_i\not\equiv 0\pmod{3}$ then we would be done because if there exists some $k$ such that $a_n+a_i\mid 3a_k$ then $3a_k=a_n+a_i$ or $3a_k=2(a_n+a_i)$ because of a simple bounding argument. Obviously none of these equations have a solution so we are done. So now we must have \[ a_1\equiv a_2\equiv\dots\equiv a_{n-1}\equiv -a_n\pmod{3}. \]

Establishing this, we will proceed with a contradiction-construction argument. First choose two numbers $a_1, a_n$ such that $a_n>a_1$ and $a_n\neq 2a_1$. We will designate $a_n$ as the largest in the set and each element that we add in the set must be smaller than $a_n$.

[b]Case 1.[/b] $2a_1>a_n$
Define $f(x)=\frac{x+a_n}{3}$. Since we are trying to achieve a contradiction, we must add $f(a_1)$ to the set. If we added $2f(a_1)$ to the set then $2f(a_1)$ would become the new largest which we don't want to happen. Then we continue and add $f(f(a_1))$ the the set in order to satisfy the condition. Notice that this process is decreasing meaning that
\[ f(a_1)>f(f(a_1))>f(f(f(a_1)))>\dots. \]
Note that in each step of the process we will assume that it will remain an integer. Thus we will eventually end at $\left\lfloor \frac{a_n}{2} \right\rfloor +1$.

[b]Case 1.1.[/b] $a_n$ is even
The process will end at $\frac{a_n}{2}+1$ and it is easy to see that $a_n+\frac{a_n}{2}+1\equiv 1\pmod{3}$.

[b]Case 1.2.[/b] $a_n$ is odd
It will end at $\frac{a_n-1}{2}+1=\frac{a_n+1}{2}$. It is also easy to see that $a_n+\frac{a_n+1}{2}\equiv 2\pmod{3}$ so we are done.

[b]Case 2.[/b] $2a_1<a_n$.
Like the above case we will define $f(x)=\frac{x+a_n}{3}$. If we decide to use $2f(a_1)$ then this will be a continuation of Case 1 and we would be done. So we must use $f(a_1)$ and $f(f(a_1))$ and so on. Now notice that this process is increasing meaning that
\[ f(a_1)<f(f(a_1))<f(f(f(a_1)))<\dots. \]
Note that in each step of the process we will assume that it will remain an integer.

[b]Case 2.1.[/b] $a_n$ is even
This process will eventually end at $\frac{a_n}{2}-1$ and since $a_n+\frac{a_n}{2}-1\equiv 2\pmod{3}$ we are done with this case.

[b]Case 2.2.[/b] $a_n$ is odd
This process will end at $\left\lfloor \frac{a_n}{2} \right\rfloor=\frac{a_n-1}{2}$ and since $a_n+\frac{a_n-1}{2}\equiv 1\pmod{3}$ we are done.

Combining all of these cases gives us the desired contradiction and we are done. $\blacksquare$

Here's a different solution without looking at divisibility by $3$. Also, we will more strongly show that we can fix $a_j = \max(a_1,a_2,\ldots,a_n)$.
[rule]
WLOG $a_1 < a_2 < \ldots < a_n$. We will more strongly show that $\exists ~ 1 \le i \le n-1$ such that $a_i + a_n$ does not divide any of $3a_1,\ldots,3a_n$. Assume contrary. For each $i$, let $f(i)$ be the [b]least[/b] index such that $a_i + a_n \mid 3a_{f(i)}$. Note if $a_i + a_n \mid 3a_n$, then $a_i + a_n \mid 3a_i$. Hence $f(i) \ne n$ for all $i$. Call $i$ [i]good[/i] if $a_i > \frac{a_n}{2}$.

[b][color=#f00]Claim: [/color][/b] At least one $i$ is good.

[b][color=#00f]Proof: [/color][/b] If $f(n-1) \ne n-1$, then $a_n + a_{n-1} \mid 3a_{f(n-1)} \implies a_n + a_{n-1} = 3a_{f(n-1)}$ forcing $f(n-1)$ to be good. Otherwise, if $f(n-1) = n-1$ then we must $a_{n-1} = \frac{a_n}{2}$. Now we look at $f(n-2)$. As $a_{n-2} \ne a_{n-1} = \frac{a_n}{2}$, so $f(n-2) \ne n-1,n-2$. We then get $f(n-2)$ to be good. $\square$

Now observe that if $i$ is good, then
[list]
[*] $a_n + a_i = 3a_{f(i)}$.
[*] $f(i)$ is good.
[*] $i$ is not a fixed point of $f$.
[/list]
Start with any good $k$. Look at $k,f(k),f(f(k)),\ldots$. It must contain some cycle $\mathcal C$. Let $l$ be the largest element of $\mathcal C$. Choose $x,y \in \mathcal C$ such that $f(x) = l$ and $f(l) = y$. Then,
\begin{align*}
a_n + a_l = 3a_y  \qquad , \qquad a_n + a_x = 3a_l \\
\implies 3a_l - 3a_y = a_x - a_l \implies 4a_l = a_x + 3a_y
\end{align*}
But this contradicts $a_l > a_x,a_y$. $\blacksquare$

[hide=Note]The solution originally posted was wrong. The correct solution was posted on 8-th Feb 2022.

The mistake was that the red part is not true, since it might happen $2(a_i + a_n) = 3a_{f(i)}$ if $a_i$ is small. This thing is fixed in the above solution.

[hide=Wrong Solution]WLOG $a_1 < a_2 < \ldots < a_n$. We will more strongly show that $\exists ~ 1 \le i \le n-1$ such that $a_i + a_n$ does not divide any of $3a_1,\ldots,3a_n$. Assume contrary. For each $i$, let $f(i)$ be the [b]least[/b] index such that $a_i + a_n \mid 3a_{f(i)}$.

[b][color=#f00]Claim: [/color][/b] $f(i) \ne n ~~ \forall ~ i$.
[b][color=#00f]Proof: [/color][/b] FTSOC $f(i) = n$ for some $i$. We will show $a_i + a_n \mid 3a_i$, which would contradict minimality of $f(i)$. Size reasons yields $a_n = 2a_i$, so $a_i + a_n \mid 3a_i$. Alternately, $a_i + a_n \mid 3a_n$ without size reasons is also enough to imply $a_i + a_n \mid 3a_i$ as $a_i + a_n \mid 3a_i$. $\square$

[color=#f00]Now by size reasons, it follows that[/color]
$$\textcolor{red}{a_i + a_n = 3a_{f(i)} ~~ \forall ~ i}$$
Since all $a_k$s are distinct, so it follows that $f$ is injective, and consequently a bijection. Call a index $i$ [i]good[/i] if $f(i) \ne i$ and [i]bad[/i] otherwise. Note if $i$ is bad then $a_n = 2a_i$, and again since all $a_k$s are distinct, so at most one $i$ is bad. In particular at least one $i$ is good (as $n \ge 3$). Let $d$ be largest good index (in particular $a_d = \max \{a_i : i \text{ is good}\}$. Then,
$$3a_d - a_{f^{-1}(d)} = a_n = 3a_{f(d)} - a_d ~ \implies ~ 4a_d = a_{f^{-1}(d)} + 3a_{f(d)}$$
Note that $f(d)$ and $f^{-1}(d)$ are also good, so this by maximality of $a_d$ forces
$$a_{f^{-1}(d)} = a_{f(d)} = a_d$$
But this then contradicts that all $a_k$s are distinct. $\blacksquare$[/hide]
[/hide]

[quote=guptaamitu1][i]A different solution without looking at divisibility by $3$:[/i]
[rule]
WLOG $a_1 < a_2 < \ldots < a_n$. We will more strongly show that $\exists ~ 1 \le i \le n-1$ such that $a_i + a_n$ does not divide any of $3a_1,\ldots,3a_n$. Assume contrary. For each $i$, let $f(i)$ be the [b]least[/b] index such that $a_i + a_n \mid 3a_{f(i)}$.

[b][color=#f00]Claim: [/color][/b] $f(i) \ne n ~~ \forall ~ i$.
[b][color=#00f]Proof: [/color][/b] If $f(i) = n$ for some $i$, then $a_i + a_n \mid 3a_n$ along with size reasons would yield $a_n = 2a_i$, but then $a_i + a_n \mid 3a_i$, which contradicts the minimality of $f(i)$. $\square$
[/quote]
what if the $a_i+a_n=\frac32 a_n$;

So $a_i + a_n = \frac{3}{2} a_n$ implies $a_n = 2a_i$. Also, one doesn't even need that since $a_i + a_n \mid 3a_n$ without size reasons is also enough to imply $a_i + a_n \mid 3a_i$ (as $a_i + a_n \mid 3a_n + 3a_i$).

Isn't it too easy for N2?

Assume contrary. Let $a_1<a_2<\cdots <a_n$.

[b]Claim 1:[/b] $a_n<2a_{n-1}$.
[b]Proof: [/b]Assume FTSOC $a_n\geq 2a_{n-1}$. If $a_n>2a_{n-1}$,then since $3a_{n-1}<a_n+a_{n-1}$ we get $a_n+a_{n-1}\mid 3a_n$, but then $a_n+a_{n-1}\mid 3a_{n-1}$,which is contradiction. So $a_n=2a_{n-1}$. $3a_{n-2}<a_n+a_{n-2}$. If $a_n+a_{n-2}\mid 3a_{n-1}$ then we get $a_{n-2}=a_{n-1}$,which is contra. If $a_n+a_{n-2}\mid 3a_n$,then $2a_{n-2}=a_n$,which again implies $a_{n-2}=a_{n-1}$, contra.

[b]Claim 2:[/b] $2a_1\le a_n$.
[b]Proof:[/b] Assume $2a_1>a_n$. Then $a_n+a_1<3a_1$, which is contradiciton.

From [b]Claim 1[/b] and [b]Claim 2 [/b]we get that there is some $m$ such that $2a_{m-1}\le a_n<2a_m$.Let $a_n+a_m\mid 3a_k$. $3a_{m-1}<a_n+a_m$, so $k\geq m \implies a_k\geq a_m$. $a_k<a_n+a_m<3a_m\le 3a_k \implies 2a_n+2a_m=3a_k$, but since $3a_k=2a_n+2a_m>3a_n\implies a_k>a_n$, we get contradiction. So done.

If $n=3$, let $a_1=a$, $a_2=b$, and $a_3=c$ for simplicity. Assume not and $a<b<c$. We have that $b+c|3a \implies 3a=b+c$, $b+c|3b \implies c=2b$, or $b+c|3c \implies c=2b$. If $c=2b$, then $a+2b|3a \implies a=b$, contradiction or $a+2b|6b$ contradiction as well. This means $3a=b+c$. We have $a+c|3a \implies c=2a$, contradiction, $a+c|3c \implies c=2a$, contradiction, or $a+c|3b$. Thus $4a-b = \frac{3b}{2}$ or $4a-b=3b \implies a=b$, contradiction. Thus $8a=5b$. Let $a=5k$, $b=8k$, and $c=7k$, contradiction to $b<c$ so we are done.

Otherwise $n \ge 4$. Assume not. WLOG, assume $a_1<a_2< \cdots < a_n$ and $\gcd(a_1,a_2, \dots, a_n)=1$ (otherwise just divide the common factor out). Assume $3 \nmid a_i+a_n$ for some $1 \le i \le n-1$. This means that $a_i+a_n|a_k$ for some $k$. However $a_i+a_n > a_n \ge a_k$, contradiction. This means $3|a_i+a_n$ for all $1 \le i \le n-1$. Note that if $a_{n-1}+a_n | 3a_n$, we would have $a_{n-1}=\frac{a_n}{2}$. Otherwise, $a_{n-1}+a_n \le 3a_{n-1} \implies a_{n-1} \ge \frac{a_n}{2}$. This means $a_{n-1} \le \frac{a_n}{2}$. Assume that for all $1 \le k \le n-2$, $3 \nmid a_k+a_{n-1}$. Note that $a_k+a_{n-1}|a_n$ from similar reasoning to above. Note that $a_1+a_{n-1}$ and $a_2+a_{n-1}$ are distinct divisors of $a_n$, so one of them is less than $\frac{a_n}{2}$, so $a_{n-1} \le \frac{a_n}{2}$ contradiction. This means $3|a_k+a_{n-1}$ for some $1 \le k \le n-2$. Using this it is easy to see $3|a_k$ for all $k$ contradiction so we are done.

Assume that $\gcd(a_1, a_2, \ldots, a_n) = 1$ since we can divide out any common factors. WLOG $a_1<a_2<\cdots < a_n$.

Suppose FTSOC that the problem condition was not true.

Claim: $a_i\equiv -a_n \pmod 3$ for any $1\le i<n$.
Proof: Suppose not. Then $a_i + a_n$ divides $3a_k$ for some $k$, and since $\gcd(3, a_i + a_n) = 1$, we have $a_i + a_n \mid a_k$, which is impossible due to size reasons. $\square$

This also implies $3\nmid a_i$ for any $i$ (otherwise $3$ would divide all of$a_1,a_2,\ldots, a_n$).

Now, we have $3\nmid a_{n-1} + a_{n-2}$, so $a_{n-1} + a_{n-2}$ divides $a_k$ for some $k$. Due to size reasons, we must have $k=n$, and $a_{n-1} + a_{n-2} \le a_n$.

We have $a_{n-1} + a_n \ge 2a_{n-1}+ a_{n-2} > 3a_{n-2}$, so $a_{n-1} + a_n$ must divide $3a_n$ or $3a_{n-1}$. Since $a_{n-1} + a_n < 2a_n$, and the only divisor of $3x$ greater than $x$ can be $3x/2$ or $3x$, we have $a_{n-1} + a_n \in \{1.5a_n, 3a_{n-1}\}$, both of which imply $a_n = 2a_{n-1}$.

Now, $a_{n-1} + a_{n-2}\mid 2a_{n-1}$, so we must have $a_{n-2} = a_{n-1}$, contradiction.

First, scale all the $a_i$'s down until $\gcd(a_1,a_2,\dots,a_n)=1$. Also let $a_1<a_2<\dots<a_n$. Assume for the sake of contradiction that $a_i+a_j\mid 3a_k$ for some $k$, for all $(i,j)$. Then, suppose there exists $i\in \{1,2,\dots, n-1\}$ such that $3\nmid a_n+a_i$, then \[a_n+a_i\mid 3a_k\implies a_n+a_i\mid a_k\implies a_n+a_i\le a_k\le a_n,\] absurd. Thus, $3\mid a_n+a_i$ for all $i$. Additionally, $a_1$, $a_2$, $\dots$, $a_{n-1}$ are congruent $\pmod 3$. If they are all congruent to $0\pmod 3$, then $3\mid a_n+a_1\implies 3\mid a_n$ which contradicts the original coprime assumption. Therefore, $a_{n-1}+a_i$ is not divisible by $3$ for all $i=1$, $2$, $\dots$, $n-2$. Now, \[a_{n-1}+a_i\mid 3a_k\implies a_{n-1}+a_i\mid a_k\implies a_{n-1}+a_i \le a_k\]
so $a_{n-1}<a_{k}$, implying $k=n$ for all $i$. In particular, $a_{n-1}+a_{n-2}\mid a_n$ which implies $a_{n}+a_{n-1}\ge 2a_{n-1}+a_{n-2} > 3a_{n-2}$. Thus, $a_{n}+a_{n-1}\mid 3a_{n-1}$ or $3a_n$.

If it is the former, then note $2(a_{n}+a_{n-1}) > 3a_{n-1}$ so $a_{n}+a_{n-1}=3a_{n-1}$, implying that $a_n=2a_{n-1}$. Thus, $a_{n-1}+a_i\mid 2a_{n-1}$, implying $a_i=a_{n-1}$, absurd.

Thus, $a_n+a_{n-1}\mid 3a_n$. Note that factors of $3a_n$ over $a_n$ are just possibly $\tfrac32 a_n$ or $3a_n$. The first one just goes back to the previous paragraph, whereas the second one is definitely not possible by size. Therefore, this is not possible.

[hide=.]megarnie orz[/hide]

Assume not. Then assume WLOG $a_1<a_2<\dots<a_n$. Divide $a_i$ by its gcd so that $\gcd(a_1,a_2,\dots,a_n)=1$.
Note that if $3\nmid a_k+a_n$, then $a_k+a_n$ divides one of $a_1,a_2,\dots,a_n$ which is impossible by size. If $3\mid a_n$ then the gcd can't be $1$, so $3\nmid a_n$. But then $3\nmid a_{n-2}+a_{n-1}$, so $a_{n-2}+a_{n-1}\le a_n$.
Now note that $a_{n-1}+a_n>a_{n-2}+a_n\ge a_{n-2}+a_{n-2}+a_{n-1}>3a_{n-2}$. Therefore, $a_{n-1}+a_n$ and $a_{n-2}+a_n$ divide $3a_{n-1}$ or $3a_n$.
If $a_{n-1}+a_n\mid 3a_n$, then $2a_{n-1}=a_n$. If $a_{n-1}+a_n\mid 3a_{n-1}$, then $a_n=2a_{n-1}$. So either way, $a_n=2a_{n-1}$.
If $a_{n-2}+a_n\mid 3a_{n-1}$, then $a_{n-2}=a_{n-1}$, impossible. If $a_{n-2}+a_n\mid 3a_n$, then $2a_{n-2}=a_n$, so $a_{n-2}=a_{n-1}$, again impossible. We win! $\blacksquare$

Order the elements as $a_1<a_2<\cdots<a_n$.

Note that $a_n \equiv -a_i \pmod 3$ for every $i$, as otherwise $a_i + a_n \mid a_k$ for some $k$ (as the $3$ drops), which is impossible for size.

Using this, it suffices to look at $a_{n-2}, a_{n-1}, a_n$. Notice that the first condition implies $a_{n-2} \equiv a_{n-1} \pmod 3$, so in particular $a_{n-2} + a_{n-1} \mid a_n$. Furthermore, $3(a_{n-2} + a_{n-1}) > a_n$, as otherwise $a_n + a_{n-1} > 3a_{n-2}$.

Combined with the fact that $a_n \equiv -a_{n-1} \pmod 3$, this is enough to imply $$a_{n-2}+a_{n-1} = a_n.$$ But then, $$a_n + a_{n-1} = 2a_{n-1} + a_{n-2} \geq 3a_{n-2}$$ unless $a_{n-1} = a_{n-2}$. On the other hand, if it equals $3a_{n-1}$, we get $a_{n-1} = a_{n-2}$ again; both contradict distinctness.

We'll proceed induction on $n$. Base case can be done by hand. Assume the contrary, for every $i$ and $j$, there exists $k$ such that $a_i + a_j \mid 3a_k$. WLOG, we can assume $a_1 < a_2 < \dots < a_n$. If there exists $i$ such that $3 \nmid a_n + a_i$, then $a_n + a_i \mid a_k$ for some $k$, a contradiction. Therefore $3 \mid a_n + a_i$ for all $i$, so $a_1 \equiv a_2 \equiv \dots \equiv a_{n-1} \equiv -a_n (3)$. Thus if $3 \mid a_i$ for some $i$, then $3 \mid a_1, a_2 \dots, a_n$, so we can replace $a_i \to \frac{a_i}{3}$, therefore we can assume none of $a_1, a_2, \dots, a_n$ is divisible by $3$. Thus $3 \nmid a_{n-1} + a_{n-2}$, so if $a_{n-1} + a_{n-2} \mid 3a_k$, then $a_{n-1} + a_{n-2} \mid a_k$, which implies $k$ must be equal to $n$. So $a_{n-1} + a_{n-2} \le a_n$. Now applying induction hypothesis on the numbers $a_2, a_3, \dots, a_n$, there exists some $i$ and $j$ such that $a_i + a_j$ does not divide $3a_2, 3a_3, \dots, 3a_n$, therefore $a_i + a_j \mid 3a_1$. If none of $i, j$ is equal to $n$, then $3 \nmid a_i + a_j$, forces to $a_i + a_j \mid a_1$, a contradiction. Thus we can assume $j = n$. Hence $3a_1 < a_{n-1} + a_{n-2} + a_i \le a_n + a_i \le 3a_1$, which is an evident contradiction. $\blacksquare$

Assume for contradiction that this is false.

WLOG assume ordering $a_1<a_2<\cdots<a_n$. De-scale the numbers so that they have no common factor.

Suppose $a_n+a_i\mid 3a_j$. Note that if $3\nmid a_n+a_i$ then $a_n+a_i\mid a_j$, but then $a_n+a_i<a_j$, which is impossible. Hence, $3\mid a_n+a_i$ for each $1\leq i < n$. Since the numbers have no common factor, it follows that $3\mid a_i+a_j$ if and only if either $i=n$ or $j=n$. Assume that $a_{n-1}+a_i\mid a_k$ for $i < n-1$. This forces $k=n$. So, $L=\text{lcm}(a_{n-1}+a_1, a_{n-1}+a_2, \ldots, a_{n-1}+a_{n-2})\mid a_n$. Finally, note that we need $a_n+a_i\mid 3a_j$. Suppose $j<n$. Then, $a_n\leq 3a_j-a_i<2a_{n-1}$. But this is impossible as $a_n\geq L>2a_{n-1}$. So, $a_n+a_i\mid 3a_n$. But that means $a_n+a_i\mid 3a_i$, which means $a_n\leq 2a_i$. This is impossible as well as it again gives $a_n\mid 2a_{n-1}$. So, we have arrived at a contradiction. $\blacksquare$

hopefully not fakesolve
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Obviously use FTSOC.
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First optimize/use induction in two ways:
[list]
[*] If $a_n\mid a_{n-1}$ then by inductive hypothesis on $\{a_1,\dots,a_{n-1}\}$ we know that there exists $a_i+a_j$ not dividing any of $3a_1,\dots,3a_{n-1}$ and also $3a_n$. Thus assume $a_i\nmid a_j$ always.
[*] Also assume that $3\nmid a_i$ exists; otherwise we can just divide by $3$ until this holds. (bad at nt -> forgetting that assuming gcd 1 is a thing)
[/list]
Reorder as $a_1>\dots>a_n$. Then there exists $x$ and $y$ such that
\[x(a_1+a_j)=3a_y\]
for any $j\in \{2,\dots,n\}$ and yet obviously $x\in \{1,2\}$. Hence $a_1+a_j\equiv 0\pmod 3$.
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Now consider $a_2+a_3$, which is not a multiple of $3$. As a result it must divide one of $a_i$ and thus $a_1\ge a_i\ge a_2+a_3$. Now consider the set
\[\{a_1+a_2,\dots,a_1+a_n\}.\]
Any element of this set is equal to an element in the set
\[\{3a_1,\dots,3a_n\}\cup \left\{\frac{3a_1}{2},\dots,\frac{3a_n}{2}\right\}.\]
However if we have $a_1+a_j\mid \frac{3a_i}{2}$ and $i\ge 3$ then
\[a_1+a_n\le \frac{3a_3}{2}\implies 2a_1<3a_3\implies 2a_2+2a_3<3a_3\implies 2a_2<a_3.\]
which fails.
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Also notice that $3a_1$ is never reached; $3a_n$ cannot be reached either. Hence the set is actually
\[\{3a_2,\dots,3a_{n-1}\}\cup \left\{\frac{3a_1}{2},\frac{3a_2}{2}\right\}.\]
Now for some basic logic. If $\frac{3a_1}{2}=a_1+a_j$ then $a_1=2a_j$ which fails as $a_j\mid a_1$.
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However the set is now
\[\{3a_2,\dots,3a_{n-1}\}\cup \left\{\frac{3a_2}{2}\right\}\]
hence $3a_2=a_1+a_2$ is the largest element. Thus $a_1=2a_2$ and we fail again. $\blacksquare$
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Remark: After getting modulo $3$ I should have easily finished with HamstPan's solution, oops. I think I knew something like that worked. idk why i didn't try it


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ A \in M_{3}(R)$ s.t exist p,q distinct real number s.t $ det(A\minus{}pI_{3})\equal{}det(A\minus{}qI_{3})$.Prove that $ (TrA)^2\equal{}3Tr(A^*)$

Is something missing? Because this can't be right.

For a real matrix, I'm assuming that what you mean by $ A^*$ is simply $ A^T.$ In which case there's no point is saying $ \text{Tr}(A^*)$ because that's just the same as $ \text{Tr}(A).$

The condition that $ \exists\,p,q$ such that $ \det(A\minus{}pI)\equal{}\det(A\minus{}qI)$ is just the statement that the characteristic polynomial of $ A$ is not one-to-one on $ \mathbb{R},$ which is an easily satisfied condition.

Take $ A\equal{}\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}.$ What do you get from this?

Kent Merryfield A^* are the adjunct matrix....matrix for wich $ A A^* \equal{} det(A)I_{n}$

And do you mean $ \text{Tr}(A^2) \equal{} 3\,\text{Tr}(\text{adj}(A))$ rather than $ (\text{Tr}(A))^2?$

I still don't see how anything like this could be true.

Let $ A\equal{}\text{diag}(2,3,4).$ Then the characteristic polynomial of $ A$ is not one to one. $ \text{adj}A\equal{}\text{diag}(12,8,6).$

We have $ (\text{Tr}A)^2\equal{}81,\ \text{Tr}(A^2)\equal{}29,$ and $ 3\text{Tr}(\text{adj}A)\equal{}3\cdot 26\equal{}78.$

I don't see any relationships there.

Definetly the problem is wrong .But I obtained the following :
If $ A\in\mathcal{M}_3(\mathbb{R})$ ,and there $ \exists p,q \in\mathbb{R}$ such that $ \det(A\minus{}pI)\equal{}\det(A\minus{}qI)$ then $ (Tr(A))^2\geq 3(Tr(A*))$ ,where $ AA*\equal{}\det(A)\cdot I$


limit algebra function domain calculus derivative

Determine all values of $ x$ which minimize $ f(x)\equal{}6|x\minus{}3|^3\minus{}4|x\minus{}3|\plus{}3$.

jiji

You can consider this problem a nice aplication of the basic idea of a problem that I've just posted.

http://www.mathlinks.ro/viewtopic.php?t=171402

I'll post the my solution after somebody post a solution to my problem, other way I'll be solving mine also.

 :)

But is Mellow Melon not correct?

$ \lim_{k \to \infty} 1(k\minus{}1^{m})\equal{}\infty$ regardless of the value of $ m$.

Yes he is correct.

Thats why I edited to: find the maximun for given values of $ k$ and $ m$, with that I meant give the maximun in terms of $ k$ and $ m$. I think this sufficient to avoid the infinite.

The fact he points, affects now only the last question.

 :)

$ u : \equal{} |x \minus{} 3|, \; u \geq 0$, and we must minimize $ f(u) \equal{} 6u^3 \minus{} 4u \plus{} 3$ over [u]nonnegative[/u] $ u$.

[hide="Solution"]Rewrite it as:
$ f(u) \equal{} 3 \minus{} \frac {2}{\sqrt {6}} (u\sqrt {6}) (2 \minus{} 3u^2) \equal{} 3 \minus{} \frac {2}{\sqrt {6}} \sqrt {6u^2(2 \minus{} 3u^2)(2 \minus{} 3u^2)}$

By AM-GM, $ 6u^2(2 \minus{} 3u^2)(2 \minus{} 3u^2) \leq \left( \frac {6u^2 \plus{} 2 \minus{} 3u^2 \plus{} 2 \minus{} 3u^2}{3} \right)^3 \equal{} \left( \frac {4}{3} \right)^3$, from where we see that $ f(u) \geq 3 \minus{} \frac {2}{\sqrt {6}} \cdot \left( \frac {4}{3} \right)^{\frac {3}{2}} \equal{} 3 \minus{} \frac{8}{9} \sqrt{2}$.

Equality is achieved when $ 6u^2 \equal{} 2 \minus{} 3u^2 \implies u^2 \equal{} \frac {2}{9} \implies u \equal{} \frac {\sqrt {2}}{9}$

Hence, $ |x \minus{} 3| \equal{} \frac {\sqrt {2}}{9} \implies \boxed{x \equal{} 3 \pm \frac {\sqrt {2}}{9}}$[/hide]

[hide="Note"]Note that we could use AM-GM only because:
1. We knew that $ (u\sqrt {6}) (2 \minus{} 3u^2)$ was somewhere positive in the domain we are interested in - namely from 0 to some positive number
2. At this point, both terms are positive
3. By inspection, we see that one term is increasing and the other decreasing, so we will reach a relative maximum somewhere[/hide]

$ 6u^3 \minus{} 4u \plus{} 3 \equal{} 6\left(u \minus{} \frac {\sqrt {2}}{3}\right)^2\left(u \plus{} \frac {2\sqrt {2}}{3}\right) \plus{} 3 \minus{} \frac {8\sqrt {2}}{9}$ gives

the minimum value of $ f(u)\ (u\geq 0)$ is $ 3 \minus{} \frac {8\sqrt {2}}{9}$ when $ u \equal{} \frac {\sqrt {2}}{3}\Longleftrightarrow \boxed{x \equal{} 3\pm\frac {\sqrt {2}}{3}}.$

Another Solution: Differentiation of $ f(u)$ gives easy solution directly,...