geometry
geometric transformation
reflection
3D geometry
number theory

Just wondering, do you care about grades? Or do you not care about grades, as long as you are learning?

I'm in middle school right now, so grades don't matter much (I'll probably need to change my attitude in high school though). What I really want is learning. Another person I know, on the other hand, is obsessed with grades but hates learning.

So what do you think?

Technically, a person who is failing can be learning too.

I very much care about grades, and as a freshman in high school, I think a percentage (higher than you would think) did not teach well what was supposed to be curriculum material. They seemed to rush through it, give a test, and blah it's over, just to satisfy. A lot was forgotten, that could easily have been plunked into our brains.

Such an open-ended question, though. However, the attitude of "Grades don't matter much." means that you won't try as much in middle school. If it's because you aren't challenged, then I guess there's nothing that can be done about that?

I don't mean that I don't care about grades at all. I make an effort to keep my grades up, too. But what I'm saying is, do you think there's a point in going to school if you're going there for grades, not learning? After all, the original point of school was to educate (that's not to say you don't you don't learn some stuff in school. Just not a lot).

PS.
[b][i][u][size=200][color=blue]200th POST!!![/u][/i][/b][/size][/color]

Well, grades are like a benchmark. There is still a point in going.

1) Socializing with other people
2) having fun
3) as for gradewise, if you think it's easy, you should skip a grade. If not, well, going has the reward of getting to go on to another grade? Anyways, this shouldn't be such a worry, as high school is drawing near. I miss the free time that middle school had in store.

I'm not sure getting extremely good grades is very social, but I guess I would agree with the rest.

Lol, just a few weeks ago I was thinking about how I missed how easy elementary was. I guess I really suck at managing my time, cuz I'm already sleeping at 11:30 PM just about every other day.

That is what I did too. This year, I find myself staying up till 11-12 sometimes, but with a reason if I have a test. Last year, I would come home from school and play games, and even when I had minimal homework, I would put it off till last minute, and do it at 11 PM etc. I missed elementary school at that time, too.

One 8th grade teacher I had preached about how "time will become your enemy"; it's completely true. =D

I feel that I'm learning heaps of stuff in maths and science - stuff that can be useful in real life. However, even if I were to pursue a career in Literature (which I am not), I find that by 10th grade you have basically learnt all you can learn about english concepts. Sure your vocabulary might not be as big (it will grow naturally with time), but after studying classic texts for 4 years now, the important ideas have already sunk in quite enough.
Does anyone else think that learning english soon becomes redundant? I'm sitting in my classroom and doing my essays and scoring well on them, but nothing new is seeping in... it's as if I were doing an office job.

Getting good grades is important, but it shouldn't be achieved by memorizing your notes or the textbook. You should actually choose to understand everything (aka learning). Often, when you do the second, you end up doing better on tests, quizzes, finals, etc and thus get better grades.

As a perfectionist, I care very much about the grades I receive and how others perceive them (especially when they get jealous, frustrated, etc. with MY grades and hate me for it!).

HOWEVER, as a very self-motivated and enthusiastic student, I LOVE to learn.  Learning is much more important to me than the grades I receive. (By this, I mean that if my classes weren't graded, I'd put in the same amount of work and effort as if they were graded.) Yet, my perfectionistic side sees good/bad grades an indication of how well I am learning the material and how much I have learned.

Not caring about grades shouldn't mean slacking off at learning things.

[quote="Zakary"]
Not caring about grades shouldn't mean slacking off at learning things.[/quote]

That's basically what I'm trying to say. You can be learning tons of things, but not necessarily getting good grades. But as you said, I guess they're usually complementary (unless you're getting good grades out of the fact that the stuff is too easy).

Meh. I'm going for a 4.7 GPA, so I can justify amoral things (nothing specific of course), but in high school you basically pick three of the following: sports, grades, friends, sleep.

Right now it's 1:30 AM and I'm writing my LA essay on the American Dream. Guess which one I'm missing out on.

[quote="Identity"]Does anyone else think that learning english soon becomes redundant? I'm sitting in my classroom and doing my essays and scoring well on them, but nothing new is seeping in... it's as if I were doing an office job.[/quote]

What? Are you joking? I love English! I love the ideas behind symbolisms, techniques, character types, the hero's journey, tragedy,......... and so on. I never get bored of it and I love comparative study of texts and critical readings (e.g. Psychoanalytic, Nihlism, Feminism, Marxism etc.). How can it become redundant? It's the SECOND best subject (next to Mathematics of course  :wink: ).

As for the learning aspect, I live in New South Wales and the science courses are so unbelievably useless. We get questions asking us to assess the impact of ethanol as an alternative fuel or the impact of Lead-Acid Battery on society etc. We don't dig into deep theoretical material (which would make the course so much more interesting). So, I push myself and learn materials beyond the hopeless syllabus. In fact, learning Chemistry, Physics and Biology outside the syllabi are the only reasons that I actually like them (otherwise I would absolutely hate sciences).

I certainly don't go to school just for grades. I go to socialize (not very good at this though) and to ask the teachers questions that I would like an answer to (most of them can give it to me or guide me).

[quote="Lazarus"]but in high school you basically pick three of the following: sports, grades, friends, sleep.[/quote]

Lol. A similar one I heard was: In Caltech, there are three things: Grades, social life, and sleep. Pick 2.

@BanishedTraitor: My viewpoint is basically the same as your last paragraph. If you go to school for grades, then you're missing out on the point of school. The original purpose was to learn and be educated, not to try to be best in class (although that certainly is nice, too  :wink: ).

Good one. I need to memorize that quote  :wink: .

[quote="Xevarion"]"Don't let your schooling get in the way of your education."

- Mark Twain[/quote]

I am going to tell my teacher that  :lol: .

Learning, hands down, but don't stray too far. Keep in mind that what you get as grades now may determine what kinds of learning opportunities you can get in the future; you have your whole life to learn, and what you learn now should not shut you out of learning opportunities later.

Now ideally, grades should reflect learning. Where they don't, I suggest trying to modify your assignments and try to make other learning opportunities out of them.

For example, in my sophmore year, we had to do these ridiculous speeches on random stuff. I decided that I didn't have much to learn by throwing together yet another canned speech, so I decided to start giving purely extemporaneous speeches. While this sacrificed a few points on the first two tries, in the long run, it didn't hurt.

In general, feel free to experiment and try something new as long as it doesn't wreck you. It's much better to get an A (or A-, in some cases) and learn than to get an A+ and not learn.

extemporaneous, now that is a good word.

Yeah, schooling isn't a 'one size fits all' scenario. There are people who might perform badly on tests but can show their knowledge in other ways. Something I've noticed is that most final exams are written, and not practical, which might be bad for showing one's experimental genius. Learning is definitely more important, and schooling sometimes isn't sufficient for that. I learn a lot from self-study.

grades are essential in school, especially in high school, because if one has low, or even average grades, he will not be able to get into a good college, and then he will not get the chance to spend college among intellectual equals and superiors, and will miss the chance to learn a lot!

yeah, i guess that to get the most learning out of your education, you have to be patient.

I care about my grades mainly because it's what colleges look at.  From my understanding, the numbers matter more to them than what you get out of the courses you take.  Correct me if you think I'm wrong.

It definitely is true that absorbing the material usually corresponds with good grades, but it's not always the case.  The numbers also have to do with how a teacher comes up with a grade to write down when the time comes to do so.  But nevertheless, they do tend to be synonymous.

I am concerned with my grades, partly because of my health (Asian parents), and mostly because I want to go to a good college.

With Grades vs Learning, it's extremely difficult to achieve that with boring GT classes. A long time ago (one of my teachers told me this), people in GT deserved to be in GT. Nowadays, all students have to do is get their parents to sign a paper and they're in. No skill required at all. I've seen a neighboring GT biology class fail because most of the students there deserved honors or worse.

As for me, learning in GT Geo (we have Alg II before geometry) is a painful waste of 45 minutes for me. This is the same for English (we're studying the Bible now...). My most fun class is probably my AP bio class. Since the class size is only 12 (freshman), there is a lot of time spent with the curriculm on each student (we have a 95 minute period for bio). I've learned more in that class that I have ever did in science combined.

As mentioned before, you can not learn anything in a class but still breeze by with straight A's (math, for example). Sure you might learn a few interesting postulates here and a couple proofs there, but it's not really full blown learning.

[quote="BanishedTraitor"]I guess most are pointing at the fact that they strike to achieve good grades because that's what good colleges/universities look at (also because of their parents' expectations for them).[/quote]

True, though I don't strive (I think that's what you meant) to get good grades specifically for the colleges and the parents; it's just that I wouldn't like to look at a B+ on my report card when I could have easily or with a little more effort gotten an A- or an A. Kinda makes me feel like I let myself down, you know? Consequently, I really don't mind too much if I tried my best in a subject and still got, say, a B-, because, well, I did honestly try my hardest. Only thing that remains to be done is try and improve.
Basically, what I'm trying to say is that I don't get judge my grades in and of themselves, but relative to my current ability. Of course, it so happens that all my favorite subjects are the ones I'm good at, so it's all good.

If you learn it well enough(or just are good at instantaneously memorizing what your teacher says)than go for grades all the way, if you can't learn it, focus on that before going to grades. Though grades are important to get into collage, if you don't know the basics, you will fail collage, and that is even worse!

[quote="jimhu"]good grades will get u into a good college, which if you love learning, is desirable   :wink:[/quote]
Or if you want a high-paying job.

Grades or learning? It's a tough one. So far I see most people choose grades, and I agree. It's the key to the future. However, in Elementary School all I cared about was getting into Middle School and doing well, which means knowing the basic material. I'll learn all that other stuff and average, I guess..

But grades count more.  :D

Also a side question now:

Do you think a lot of your effort for getting good grades is because your parents pressure you to do so?

I think they definitely has some responsibility, although that's also a good thing (just bad when they overdo it). Even if the parents might not intentionally put a lot of pressure, they might unconsciously do it sometimes.

[quote="7h3.D3m0n.117"]Correction:

[quote="Nerd_of_the_Ages"]Do you think a lot of your effort for getting [b]bad[/b] grades is because your parents pressure you to do so?[/quote]

Sometimes parents (Asian) can overdo it.[/quote]

Yeah, they set the bar so high that you'll need to fly up into space to touch it  :wink:

Grades are a nice incentive for learning if someone didnt intend to learn but cares about grades.

I think grades are pretty important for me, and one reason is what 7h3. mentioned: parents have high expectations.

But for the purpose of learning, i'm willing to take higher level classes even though they could mean less grades, because they're definitely less boring and learning is good and fun.

[quote="ProtestanT"]But for the purpose of learning, i'm willing to take higher level classes even though they could mean less grades, because they're definitely less boring and learning is good and fun.[/quote]

That's more or less the exact words my AP bio teacher said  :lol: .

Yep, that's pretty much what I think. Grades are pointless if you aren't having fun and learning.

I think the major purpose of school is learning. The best way of learning is having interest and exploring on your own, with a little help. Grades are only the residue of the edication system and should not be the goal but learning.

I actively ignore the college admissions process. I'm a sophomore, and the pressure on kids around me is just ridiculous. There honestly isn't that much of a difference between an Ivy League and a non-Ivy League school. Wherever you go, there are enough that you can choose one where you will be happy. There are kids who have already started taken SAT prep courses - a full 2 years before the test. It's just ridiculous. Think about the people for whom the choice is not which college but whether college - if your question is the former, you're pretty lucky. So don't spend your time thinking about college. Spend it on something worthwhile... like math.

I pulled 99th percentile on ACT & SAT, but a 2.5 GPA. There goes any scholarships whatsoever.

Do your work, get good grades.

Guys um if you know what you're doing it's really easy to get straight As

I'm not the best student ever and I only have one B

[quote="aanjohri"]People who get good grades on thier report card may not have much general knowledge (like the current news). So just because you get good grades in school doesn't mean you are smart, i guess.[/quote]

Getting good grades mean - doing your homework and paying attention in class. You could be the smartest person in math class but you don't pay attention and play with the Rubik's Cube until you get caught  :whistling:  .

wow..
i just had this debate over myself last week..
WELL!!!
i will go with learning.
1. grades may go down with one mistake
2. i'm sorta not learning anything... (in math, that is)
    i'd rather get a D- in like Number Theory and learn everything than take kindergarten math and learn nothing...
my point? (yes i know that was extreme)

also my parents are like, "learning is more than grades" but they're still a bit anal about grades... (just like all asians sorry to be stereotypical)

Hmmm... Well it's true that you might not be learning too much... But you need a good GPA to get in to top colleges, which is where you really learn. I guess it's just patience.

Learning and grades generally correlate.

I mean there's probably gonna be deviations based on effort, but you know, if you're learning a lot, you're probably gonna get good grades. To clarify, if you're learning a lot about specifically the subject being taught - not some arcane subclause of the general topic being beaten into your head in class - you're gonna do real good.


quadratics
algebra
polynomial
algebra proposed

Prove that for any quadratic polynomial $f(x)=x^2+px+q$ with integer coefficients, it is possible to find another polynomial $q(x)=2x^2+rx+s$ with integer coefficients so that \[\{f(x)|x \in \mathbb{Z} \} \cap \{g(x)|x \in \mathbb{Z} \} = \emptyset .\]

$\forall n,n'\in\mathbb{Z}$ we want to have $n^2+pn+q \neq 2n'^2+rn'+s$
\[n(n+p) - n'(2n'+r)\neq s-q\]
If $p$ is odd we choose $r$ even and $s=q+1$
If $p$ is even then $p=2p'$ and we have
\[n^2+2np'-2n'^2-n'r\neq s-q\]
Modulo $4$ we have
\[2np'-n'r\neq s-q+1 (4)\]
Taking $r$ even and $s=q$ and it's done.

I didn't quite get your point... how does your reasoning go?

If $p$ is odd, choosing $r$ even and $s=q+1$ gives that $\{f(x)|x \in \mathbb{Z} \} \cap \{g(x)|x \in \mathbb{Z} \} = \emptyset$
since if there is $n, n'$ such that $f(n)=g(n')$ then $n^2+pn+q = 2n'^2+rn'+s \Rightarrow n(n+p) - n'(2n'+r)= s-q \Rightarrow n(n+p) - n'(2n'+r)=1 $ but it's not possible since $n(n+p)$ and $n'(2n'+r)$ are even.

If $p$ is even ($p=2p'$), choosing $r$ even and $s=q$ and we have $\{f(x)|x \in \mathbb{Z} \} \cap \{g(x)|x \in \mathbb{Z} \} = \emptyset$
since if there is $n, n'$ such that $f(n)=g(n')$ then $n^2+2np'-2n'^2-n'r = 0$ and
OUPS I have made a mistake taking $n$ and $n'$ odd, I need to correct my solution
If $n$ and $n'$ are odd we see that we have $n^2+2np'-2n'^2-n'r \equiv 1+2np'-2-n'r (4)$ and can not equals $0$ since $r$ is even

I will post end of my solution in few minutes


geometry
circumcircle
angle bisector
power of a point

In $ \triangle ABC$ , let $ D$ and $ E$ respectively be the feet of the median and the bisector from $ A$. The circumscribed circumference of triangle $ ADE$ cuts $ AB$ in $ B'$ and $ AC$ in $ C'$ . Show that $ BB'\equal{}CC'$

[hide="Solution"]By Power of a point, we have
$ BB' \cdot AB \equal{} BE \cdot BD$
$ CC' \cdot AC \equal{} CD \cdot CE$
So we have to show $ AB \cdot CD \cdot CE \equal{} AC \cdot BE \cdot BD$. But $ CD \equal{} BD$.
So we have to show $ \frac {AB}{BE} \equal{} \frac {AC}{CE}$, which is true by the angle bisector theorem  :) .[/hide]
EDIT:$ 2^{10}$th post!

Extend $(AE$ to intersect the circle $(ABC)$ at $M$, and take $N$ diametrically opposite to $M$; $A, E, D$ and $N$ are concyclic, $EN$ being a diameter of their common circle. $AE, AN$ being bisectors of $\angle B'AC'$, we get $NB'=NC'$.
In triangles $\triangle BB'N, \triangle CC'N$: $BN=CN, B'N=C'N$, $\angle AB'N=\angle AC'N$, $\angle B'BN=\angle ABN=\angle ACN=\angle C'CN$, hence $\triangle BB'N=\triangle CC'N$ inferring $BB'=CC'$.

Best regards,
sunken rock

[color=darkred]Let $ABC$ be a triangle and  let $\{D,E\}\subset BC$ be the feet of the $A$-median and the $A$-bisector respectively.

The circumcircle of the triangle $ADE$ cuts again $ AB$ in $ B'$ and $ AC$ in $ C'$ . Show that $ BB'\equal{}CC'$ .[/color]

[b][u]Proof[/u].[/b] Observe that $\left\{\begin{array}{ccc}
BD & = & CD\\\\
\frac {BE}{CE} & = & \frac {BA}{CA}\\\\
B'B\cdot BA & = & BE\cdot BD\\\\
CD\cdot CE & = & C'C\cdot CA\end{array}\right\|\ \bigodot\implies B'B=C'C$ .


If $ALFA+BETA=GAMMA$, and each letter is a different 1 digit number, find each letter, supposing that there's a solution

[quote="vishalarul"]One solution is
[hide="Do you really want to know?"]$A=0$
$B=9$
$G=1$
$M=5$
$L=8$
$F=3$
$E=7$
$T=2$[/hide]

I am pretty sure that there are more solutions, but I'm just too lazy... :yup:[/quote]

except it's usually a bad idea to begin a number with 0 and it's unavoidable that a = 0, so I'm doubting there's a solution to this

Well, there's al least one solution, maybe more (I won't tell the exact number)

Well... treething is right.

$A+A\equiv A \bmod{10}\to A\equiv0$


The sequence{${x_n}$} st:
$x_1=1,x_{n+1}=\sqrt{x_n(x_n+1)(x_n+2)(x_n+3)+1}$ all of n positive integer.
Let $y_n=\sum \frac{1}{x_i+2}$.
Find $lim_{n\to \infty}y_n$.

That´s a nice problem. Let $(z_{n})$ where $z_{n}=x_{n}+1$. Thus we will have:
$z_{n+1}-1=\sqrt{(z_{n}-1)z_{n}(z_{n}+1)(z_{n}+2)+1}$ $\to$ $z_{n+1}(z_{n+1}-2)=(z_{n}-1)z_{n}(z_{n}+1)(z_{n}+2)$ and now we can assume that $z_{n+1}={z_{n}}^{2}+z_{n}$ and check that this is correct.

At this point we have $\frac{1}{1+z_{k}}=\frac{1}{z_{k}}-\frac{1}{z_{k+1}}$. And now making a telescoping summing we´ll have that $y_{n}=\frac{1}{z_{1}}-\frac{1}{z_{n+1}}$, and taking limit we will have $\frac{1}{2}$ as answer, because $z_{n}$ tends to infinity when n tend to infinity.


So...anyone know what the USAMTS Prize Winner cutoff score generally is, and what it was last year?

Fierytycoon

http://www.nsa.gov/programs/mepp/usamts/prizes.html

Is this what you were asking?

Oh, so they are always constant every year? ok, thanks.

Fierytycoon


see at this: http://www.mathlinks.ro/viewtopic.php?p=1472107#1472107


quadratics
arithmetic sequence
geometric sequence

If the discriminant of $ ax^2\plus{}2bx\plus{}c\equal{}0$ is zero, then another true statement about $ a$, $ b$, and $ c$ is that:

$ \textbf{(A)}\ \text{they form an arithmetic progression} \\
\textbf{(B)}\ \text{they form a geometric progression} \\
\textbf{(C)}\ \text{they are unequal} \\
\textbf{(D)}\ \text{they are all negative numbers} \\
\textbf{(E)}\ \text{only b is negative and a and c are positive}$

[hide="Solution"]We have $ 4b^2\minus{}4ac\equal{}0\implies b^2\equal{}ac$.
If the answer is B, then $ b\equal{}ar$ and $ c\equal{}ar^2$.
Then $ a^2r^2\equal{}a\cdot ar^2$, which is true.
So the answer is $ \boxed{\textbf{(B)}}$.[/hide]

[quote="AIME15"]If the discriminant of $ ax^2 \plus{} 2bx \plus{} c \equal{} 0$ is zero, then another true statement about $ a$, $ b$, and $ c$ is that:[/quote]
If that what you wrote AIME15 has no mistake then $ D \equal{} 4(2b)^{2} \minus{} 4ac \equal{} 16b^2 \minus{} 4ac$ :wink:

Yes, that's the discriminant, which is equal to 0.

If the discriminant is zero then it must have a double root.

Therefore it must be of the form:
d(x-e)^2=0
Expand:
dx^2-2dex+e^2

and d^2, -de, e^2 makes a geometric progression.

[quote="enndb0x"]If that what you wrote AIME15 has no mistake then $ D \equal{} 4(2b)^{2} \minus{} 4ac \equal{} 16b^2 \minus{} 4ac$ :wink:[/quote]Wrong discriminant, enndb0x.
The discriminant is $ D\equal{}(2b)^2\minus{}4ac\equal{}0$.

[quote="james4l"]If the discriminant is zero then it must have a double root.

Therefore it must be of the form:
d(x-e)^2=0
Expand:
dx^2-2dex+e^2 [b]<----wrong[/b]

and d^2, -de, e^2 makes a geometric progression.[/quote]It should be $ dx^2\minus{}2dex\plus{}de^2$ and you'll get a geometric progression $ d, \minus{}de, de^2$.

[quote="AIME15"]If the discriminant of $ ax^2 + 2bx + c = 0$ is zero, then another true statement about $ a$, $ b$, and $ c$ is that:

$ \textbf{(A)}\ \text{they form an arithmetic progression} \\
\textbf{(B)}\ \text{they form a geometric progression} \\
\textbf{(C)}\ \text{they are unequal} \\
\textbf{(D)}\ \text{they are all negative numbers} \\
\textbf{(E)}\ \text{only b is negative and a and c are positive}$[/quote][hide]

For this equation the discriminant is $ (2b)^2 -4ac = 0\implies 4b^2-4ac = 0$.

Dividing this expression through by $ 4$, $ b^2 - ac = 0$.

\[ b^2 -ac = 0 \implies b^2 = ac\]

\[ b^2 = ac \implies \frac{b}{c} =\frac{a}{b}\]

Setting both of the above fractions equal to the same number:

\[ \frac{a}{b} = n =\frac{b}{c} \implies b = c n\; \& \&\; a = b n \\ \\ \implies c =\frac{b}{n}\; \&\&\; b =\frac{a}{n}\]

Substituting the second expression into the first:

\[ b = \frac{a}{n}\; \&\&\; c =\frac{a}{n^2}\; \& \; \therefore a,\,b,\mbox{ and } c\mbox{ form a geometric sequence: }a,\,\frac{a}{n},\,\mbox{ and }\frac{a}{n^2}\][/hide]

The discriminant is 4b^2 - 4ac

That will be equal to 0

4b^2 - 4ac = 0

i.e., b^2 = ac

When b^2 = ac ....I think, they (a,b,c) must be in G.P. as a/b = b/c


Να δειχθει οτι:
ολοκληρωμα απο 0 εως 1 (((5*(χ^4)+2)^(1/2))*((2*χ+7)^(1/2)))dx<5
Αν μπορει καποιος να τη γραψει σε LATEX

$\int_{0}^{1}\left(\sqrt{5x^{4}+2}\cdot\sqrt{2x+7}\right)\,dx<5$

Μάνο αυτό εννοείς;

Ναι,ακριβως αυτο

Ok, τότε από την ανισότητα Schwarz έχουμε

$\int_{0}^{1}\left(\sqrt{5x^{4}+2}\cdot\sqrt{2x+7}\right)\,dx\leq \sqrt{\left[\int_{0}^{1}\left(5x^{4}+2\right)\,dx\right]\cdot\left[\int_{0}^{1}\left(2x+7\right)\,dx\right]}=$

$=\sqrt{24}<5$

 :wink:

Πολυ σωστα :)


[hide]There are $3*3*10=90 possibilities$

For any given number there are

4-ways to move right 2
4 ways to stay (right left or left right $2/9*2$
1-way to move left 2

Now we can add them all up
1-4 ways to get to 3
2-1 way to get to 0
3-4 ways to stay
4-4 ways to get to 6
5-1 way to get to 3
6-4 ways to stay
7-4 ways to get to 9
8- 1 way to get to 6
9-4 ways to stay
10-4 ways to get to 12

you probably figured out a pattern by now

$4+1+4+4+1+4+4+1+4+4=31$

$31/90?$

I remember getting this wrong too...[/hide]

Another (my) way:

To get to 0, there is 1 way to get there from 2.
To get to 3, there are 4 ways to get there from 1, 4 ways from 3, and 1 way from 5.
To get to 6, there are 4 ways to get there from 4, 4 ways from 6, and 1 way from 8.
To get to 9, there are 4 ways to get there from 7, and 4 ways from 9
To get to 12, there are 4 ways to get there from 10.

Then there are 1 ways to get to 0, 9 ways to get to 3, 9 ways to get to 6, 8 ways to get to 9, and 4 ways to get to 12, for a total of 1+9+9+8+4=31 ways.  Then there is a probability of $\frac{31}{10\times 3\times 3}=\frac{31}{90}$.


superior algebra
superior algebra theorems

$ \alpha$ and $ \beta$ are algebraic numbers. Let the conjugates of $ \alpha$ be $ \alpha_1, ..., \alpha_n$. Define $ *\alpha* \equal{} max\{|\alpha|, |\alpha_1|,...,|\alpha_n|\}$.
Now for $ \alpha, \beta \in \mathbb{Q}$ the following facts are true:
$ *\alpha\beta*\leq *\alpha**\beta*$ and $ *\alpha \plus{} \beta*\leq *\alpha* \plus{} *\beta*$.

How do you prove this? Also, is it true for $ \alpha, \beta \in \mathbb{C}$?

Let the conjugates of $ \alpha, \beta$ be $ \alpha_1, ..., \alpha_m$ and $ \beta_1, ..., \beta_n$ respectively. Let $ A,B$ be the sets of their absolute values, respectively.
Then the conjugates of $ \alpha \plus{} \beta$ are a subset of $ S\equal{}\{ \alpha_i \plus{} \beta_j \}$, and the conjugates of $ \alpha \cdot \beta$ are a subset of $ P\equal{}\{ \alpha_i \cdot \beta_j \}$.
We have $ *\alpha*\equal{}\max(A), *\beta*\equal{}\max(B)$ and by the above $ *\alpha\plus{}\beta* \leq \max(A\plus{}B)$, $ *\alpha \cdot \beta* \leq \max(A \cdot B)$.

Thus you have to show that if $ A,B$ are (finite) sets of nonnegative real numbers, then $ \max(A\plus{}B) \leq \max(A) \plus{} \max(B)$ and $ \max(A \cdot B) \leq \max(A) \cdot \max(B)$. This is a simple task.

[quote="ZetaX"]Then the conjugates of $ \alpha \plus{} \beta$ are a subset of $ S \equal{} \{\alpha_i \plus{} \beta_j \}$, and the conjugates of $ \alpha \cdot \beta$ are a subset of $ P \equal{} \{\alpha_i \cdot \beta_j \}$.[/quote]I don't see why this has to be so.  :maybe:


calculus
algebra
polynomial
derivative
calculus computations

Let $ g(t) \equal{} \frac{3t\minus{}1}{2t\plus{}1}$.  Write the equation of the line(s) that are tangent to g and parallel to $ t\minus{}5y\equal{}3$.

From condition 2, we find that the tangent line must be of the form $ y \equal{} \frac {t}5 \plus{} b$.
Plugging this into our first equation, we find that
$ \frac {t}5 \plus{} b \equal{} \frac {3t \minus{} 1}{2t \plus{} 1}$, which simplifies to
$ 2t^2 \plus{} (10b \minus{} 14)t \plus{} 5b \minus{} 5 \equal{} 0$.
Since the line is tangent, we must have the discriminant of this equation equal to $ 0$,
so $ b \equal{} \frac {8\pm\sqrt5}5$, and our answers are
$ \boxed{0 \equal{} t \minus{} 5y \plus{} 8 \plus{} \sqrt5}$ and $ \boxed{0 \equal{} t \minus{} 5y \plus{} 8 \minus{} \sqrt5}$.

EDIT: Why has this been moved to the calculus forums...

Why does the discriminant have to be equal to 0?

The discriminant is an algebraic technique for finding tangent lines: look for a double root in an appropriately constructed polynomial equation. It's the way Fermat would have done it, before calculus was formalized and derivatives became standard.

I still don't understand what the discriminant has to do with being tangent...

Is there maybe a proof or explanation online?


Alcumus
MATHCOUNTS
probability
AMC
AMC 8
email
videos

[b]Here is the situation…[/b]

I am trying to start a math club at the daughter's middle school so that she can participate in the MATHCOUNTS and AMC 8 contests.  We have not been able to find a teacher who is willing to coach the team.  I will probably end up coaching it until someone else with more experience can be found.

[b]Here is what I would like to do...[/b]

I would like to limit the number of kids in the club to a small number so it is easier to control however I would like the whole school to benefit from the program (there are virtually no opportunities for enrichment in our district and the curriculum is [u]heavily [/u]focused on computation).

[b]Here is my question...[/b]

What’s the best way to use Alcumus to help the middle school students prepare for MATHCOUNTS and AMC?

Worked a little bit with Alcumus and love it. But I have some concerns…

1. I think that most students won’t make it past the diagnostics phase of Alcumus.  A few of my daughter’s friends tried it this summer and they didn’t make it past it.  Can I just tell them to “give up” for the first 60 questions?  I don’t want to do that but I don’t want them to get discouraged before they start.

2. Many of the kids are under 12 and I don’t think that they will take the time to mail in the form that is required by the fed.  But I don’t want encourage the kids to lie (though I allowed my daughter to do that  :lol:  ).  Perhaps the school could pass out the forms from interested students and then collect them for mailing.  Would that be OK?  It would probably mean that many kids would get accounts and then never really use them.

3. Alcumus mainly focuses on probability and combinatorics.  What would they be missing? I think that minis are great but I sure would love more of them.  What other resource can be used to help the kids prepare?  I had hoped we could put up some old AMC8 or MATHCOUNTS tests and have the kids email their solutions but that is a violation of the copyright.

Any thoughts?  Thanks in advance.

Alcumus will really help your students if you go little by little, at a slow and steady pace. Perhaps you can do maybe 10-15 problems a day [i]correctly[/i].

I used to do about 40 problems [i]correctly[/i] each day, but I am more devoted to math and doing so much may burn out your students.

Tell your students not to watch the videos first. Let them attempt the problem without any aid for about 5 minutes or so [i]thinking hard[/i] and only then look at the solution. If you miss like 2-3 problems of the same type, then it would be good to watch the video. However, if the student has not had former exposure to probability, it would be good to watch like the first 10 videos or something, to get a grasp on the basics.

The most important thing, out of everything I said, and with [i]much[/i] personal experience, when starting [i]anything[/i], [b]never give up[/b], [b]keep at it[/b], [b]do not procrastinate[/b], and [b]do not be lazy[/b]. The reason I used so much formatting for the above line is because it is [i]extremely[/i] important; if your students does not adhere to the above, they should be given consequences, or kicked out, period. I'm sorry if I'm sounding harsh, but if you quit even one day, you'll lose your spirit.

There's still more I can say, but I think this should be enough and the rest may be redundant.

Finally, Cgyao15, "i think this is better of in the alcumus section. :)" does not help [i]at all[/i] and shows that you want to increase your post count without actually being contributive; in other words, being too lazy to give a proper reply, as I see no major reason as to why this does not belong in the MATHCOUNTS forum and is better suited for the Alcumus forum. The Alucumus forum is for questions about Alcumus, reporting bugs, asking about updates, etc. This topic is about how to use Alcumus, in addition to other things, to prepare [b]for MATHCOUNTS and AMC[/b]. There is no forum better suited for this topic than this one.

Please post something good, or don't post at all.

I'm sorry if I sounded harsh.

I completely agree with everything pytheagle said.
Alcumus is a really great way to improve at math.
It focuses on the topics you aren't good at, and has videos to help you learn.
It shows your progress, and even has reports so you can see how you're doing!!
Tip: Make sure that after you complete each problem you rate the solution, so you won't run out out karma!!

Good luck! I hope that your students will enjoy using Alcumus very much. :)

Thanks for the comments.

Wow 40 questions every day!  That is way, way more commited than they type of kid I am hoping to interest at the school.  I am hoping to hook at least a few of the typically "I am bad at math" or "I hate math" kids.  Because I think the types of problem they will see on Alcumus ARE more fun than what they see at school.  Also for the math club we will have some presentations from students at the local colleges.  I am hoping that those might be fun enough to hook a few kids.

Do you think ArtOfProblemSolving would create accounts for kids without having the diagonstic phase? I am sure they would all need to start at the beginning.

pytheagle, you make a good point about the videos.  I think it important to have the videos there as a comfort level for some of the kids. But it is best for the kids to try to work it out themselves.

I was in a similar spot as your kids when I was 11, about a year and a couple months ago. I was in the most advanced classes for math at my school but I didn't really like it. It  was boring and awful and stupid and tedious, etc. But then my math teacher (who I can not believe my school actually let go!) introduced me to MATHCOUNTS. I sucked at it but it was [i]fun[/i], and I think it was really the competitive atmosphere at the actual competition and that really brought me into it. I found AoPS, and I looked at the forum but it just was incredibly above my head.

Then I found FTW. You might want to let your kids try that because it's a much more interactive game with a lot of head-to-head competition. On FTW alone, I went from scores of 20 on the chapter to 35-40. Then I started doing the AoPS books, and now a year later, I'm getting perfects on states and I'm steadily getting better at other competitions too.

As for MATHCOUNTS tests, it's really important to do those. If you like, I can email you a whole bunch. Not all of them have solutions but I'd be happy to write them up. In fact, if you need help with anything I'd really be happy to. School offers me a lot of spare time (I know that sounds weird), and I have a lot of experience working with kids who despise math.

Alcumus will work too, but be sure to offer a lot of encouragement! I used to get really discouraged really fast, so it's important to remind them of their progress.

isabella2296, thank you for sharing your story with me.  Very interesting!

I haven't been able to try FTW yet - it never works for us probably because of the firewall settings.  I will see if it works on the school's computers.

I just logged into Alcumus for the first time in months.  I really like the changes that were made to it.  I think that they probably address most of my concerns.


$ a,b,c>0$ and $ ab\plus{}bc\plus{}ac \le 3abc$ Prove that !
$ \sum_{cyc} \frac{a^4b}{2a\plus{}b} \ge 1$

[quote="hedeng123"][quote="bigbang195"]$ a,b,c > 0$ and $ ab \plus{} bc \plus{} ac \le 3abc$ Prove that !
$ \sum_{cyc} \frac {a^4b}{2a \plus{} b} \ge 1$[/quote]

$ \sum_{cyc} \frac {a^4b}{2a \plus{} b} \ge \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum_{cyc} \frac {2a \plus{} b}{a}}\geq 1$

Let

 $ a \equal{} \frac {(k \plus{} 1)(x \plus{} y \plus{} z)}{3x},b \equal{} \frac {(k \plus{} 1)(x \plus{} y \plus{} z)}{3y},$

$ c \equal{} \frac {(k \plus{} 1)(x \plus{} y \plus{} z)}{3z},x,y,z > 0,k\ge{0},$

$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum_{cyc} \frac {2a \plus{} b}{a}}\geq 1$

$ \Longleftrightarrow(y^2x^2 \plus{} y^2z^2 \plus{} x^2z^2)^2(x \plus{} y \plus{} z)^4k^4 \plus{} 4(y^2x^2 \plus{} y^2z^2 \plus{} x^2z^2)^2(x \plus{} y \plus{} z)^4k^3$

$ \plus{} 6(y^2x^2 \plus{} y^2z^2 \plus{} x^2z^2)^2(x \plus{} y \plus{} z)^4k^2 \plus{}$ $ 4(y^2x^2 \plus{} y^2z^2 \plus{} x^2z^2)^2(x \plus{} y \plus{} z)^4k \plus{} f(x,y,z)\geq 0$

we need only prove

$ f(x,y,z)\geq 0$

tsds one time.  (0.016 second)

 :lol:[/quote]

let$ \frac {1}{a} \equal{} x; \frac {1}{b} \equal{} y; \frac {1}{c} \equal{} z$, $ LHS \equal{} \frac {1}{x^3(2y \plus{} x)} \plus{} \frac {1}{y^3(y \plus{} 2z)} \plus{} \frac {1}{z^3(2x \plus{} z)}$
but $ \frac {1}{x^3(2y \plus{} x)} \plus{} \frac {x \plus{} 2y}{9} \plus{} \frac{1}{3} \geq \frac{1}{x}$


[hide="hint"]
cancel out a factor of $ 2^{4k}$ from the numerator and denominator and evaluate
[/hide]

[hide="better hint, basically a giveaway"]
notice that $ 2^{4k\plus{}2} \plus{}2^{4k\minus{}3} \equal{}  2\left(2^{4k\plus{}1}\plus{}2^{4k\minus{}4}\right)$
[/hide]

[hide="The fast way  :lol:"]$ \frac {2^{4k \plus{} 2} \plus{} 2^{4k \minus{} 3}} {2^{4k \plus{} 1} \plus{} 2^{4k \minus{} 4}} \equal{} \frac {2(2^{4k \plus{} 1} \plus{} 2^{4k \minus{} 4})}{2^{4k \plus{} 1} \plus{} 2^{4k \minus{} 4}} \equal{} \boxed{2}$[/hide]

Edit: Oh, stevenmeow already mentioned this. Oh well...


geometry
power of a point
radical axis
geometry proposed

Given two circles which don't meet each other. Construct their radical axis with only ruler.

[quote="Wickedestjr"]I think I have heard of a problem similar to this, but what is the radical axis? Sorry, I'm stupid.[/quote]
Radical axis of two circles is the line which contains points that has the same power wrt these circles.


function
combinatorics solved
combinatorics

I'm solving one problem about some special permutations and I need to adjust the following expression so that it would be some formula without "sums" (some "simpler" function of n):

 \sum k2/(k+1)! + n/n!, where 0 < k < n

I have found out that the previous expression is equivalent with:

 \sum 1/k! - 1/(n-1) + n/n!, where 0 < k  \leq n

So now it only suffices to adjust  \sum 1/k! to some "simpler" formula (function of n), but I don't know how... I only now that

1 + 1/1! + 1/2! + 1/3! + ... = e (in the case of infinite sequence)

Could anybody help me?

We have S_n = 1 + 1/1! + ... + 1/n! = [n!e]/n! for n > 1, where [.]denotes the integer part.
Is it what you want?

If so, I give you a proof :
We have n!*e = N + sum{k > n} n!/k!, where N = n!*S_n is an integer.
And 0 < sum{k > n} n!/k! < 1/(n+1) + 1/(n+1) 2 + 1/(n+1) 3  + .... = 1/n < 1.
So [n!*e] = N = n!*S_n, hence result.

Pierre.

Nice idea, Pierre!
I think, you can move it to Solved Section!  :D

Ok.
If he did not answer that's probably because he was agree of it.  :D

Pierre.


This problem was the fourth proposed problem on the Second Bulgarian TST for BMO which were held on 30 and 31 of March 2009.
Prove that if a,b and c are real numbers such that $ a\plus{}b\plus{}c\equal{}3$ and $ abc\ge\minus{}4$ than $ 5\left(ab\plus{}ac\plus{}bc\right)\le12\plus{}3abc$. When does the equality occur.

[quote="SAPOSTO"]
Prove that if a,b and c are real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ abc\ge \minus{} 4$ than $ 5\left(ab \plus{} ac \plus{} bc\right)\le12 \plus{} 3abc$. When does the equality occur.[/quote]
If $ ab \plus{} ac \plus{} bc < 0$ then, the inequality is true. Let $ ab \plus{} ac \plus{} bc\geq0,$
$ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, we need to prove that $ 5uv^2\leq4u^3 \plus{} w^3.$
But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\geq3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, it remains to prove that $ u^3 \minus{} uv^2\geq\sqrt {(u^2 \minus{} v^2)^3},$ which is obviously true.
The equality occurs when $ a\equal{}b\equal{}c\equal{}1$ or when $ 4u^3\plus{}w^3\equal{}v^2\equal{}0,$ which gives $ a\equal{}b\equal{}2,$ $ c\equal{}\minus{}1$ and all permutations of this.

[quote="SAPOSTO"]This problem was the fourth proposed problem on the Second Bulgarian TST for BMO which were held on 30 and 31 of March 2009.
Prove that if a,b and c are real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ abc\ge \minus{} 4$ than $ 5\left(ab \plus{} ac \plus{} bc\right)\le12 \plus{} 3abc$. When does the equality occur.[/quote]
I gave a solution for $ a,b,c>0$
After homogenization we got:
$ \frac {5}{3}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \leq \frac {4}{9}(a \plus{} b \plus{} c)^3 \plus{} 3abc$
If i've not make a mistake our inequality is equivalent to:
$ 3(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b) \leq 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 6abc$
So the $ RHS$ can be written:
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)$
From schur's inequality we have that
$ 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc) \geq 2(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b)$
So we only have to show that
$ 2(a^3 \plus{} b^3 \plus{} c^3) \geq (a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b)$
Which is easy enough...
Please check it...

[quote="Dimitris X"][quote="SAPOSTO"]This problem was the fourth proposed problem on the Second Bulgarian TST for BMO which were held on 30 and 31 of March 2009.
Prove that if a,b and c are real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ abc\ge \minus{} 4$ than $ 5\left(ab \plus{} ac \plus{} bc\right)\le12 \plus{} 3abc$. When does the equality occur.[/quote]
I gave a solution for $ a,b,c > 0$
After homogenization we got:
$ \frac {5}{3}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \leq \frac {4}{9}(a \plus{} b \plus{} c)^3 \plus{} 3abc$
If i've not make a mistake our inequality is equivalent to:
$ 3(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b) \leq 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 6abc$
So the $ RHS$ can be written:
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)$
From schur's inequality we have that
$ 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc) \geq 2(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b)$[/quote]
This step is wrong,Because $ a,b,c$ be real numbers :wink:

[quote="quykhtn-qa1"][quote="Dimitris X"][quote="SAPOSTO"] This step is wrong,Because $ a,b,c$ be real numbers :wink:[/quote][/quote][/quote]
Yeah i said that i gave a solution only for $ a,b,c > 0$...
Look at the first line of my profe....


quadratics
number theory unsolved
number theory

Prove that the product of 10 consecutive integers is never a perfect square.

More generally, Erds and Selfridge proved that the product of at least 2 positive and consecutive integers is never a power...

Pierre.

Hm ... I didn't know that. Is that proof difficult?
If it is too difficult or non-elementary, is there an elegant method to do this problem?

I would not say that the proof is elementary.
In another hand it could be understood. It may be interesting to notice that it uses a lemma in graph theory...

Pierre.

Hmm... for a few cases it is easy:

Suppose the numbers are a_1, a_2, ... , a_10.

Case I: None of the numbers is 0 mod 11.

Then a_1 = 1 mod 11,
        a_2 = 2 mod 11, etc.

The product is 10! mod 11 which is congruent to -1 mod 11 by Wilson's theorem. However, 11 is a prime that is 3 mod 4, so -1 is not a quadratic residue. Therefore, there cannot exist an x s.t. x^2 = a_1*a_2*...*a_10.

Case II: At least one of the numbers is 0 mod 11.

It is clear that exactly one of the numbers is 0 mod 11.
Also, that number must be divisible by an even power of 11.

Hmm... don't know how to proceed from here

Ok, maybe im missing something simple, but if x is a perfect square and x = p_1^k_1 * p_2^k_2 * ... * p_n^k_n, then k_i must be even for 0 <= i <= n. Then if exactly one of the 10 numbers is divisible by 11, you need another factor of 11 somewhere in the product a_1*a_2*...*a_10. But since there are only 10 numbers, there isn't another factor, so x cannot be a perfect square.

Yes but that factor could be divisible by 121=11 2.

This problem is not solved, so I'll move it back.

I'll prove the following problem: the product of n>1 positive consecutive integers is never a perfect square, for some small n.
----------------------------------------------------------------------------------
n=2.
First solution:
x(x+1)=k 2. x,x+1 are coprime, so x and x+1 are squares, so x=a 2  and x+1=b 2 , so b 2 =a 2 +1, so (b-a)(b+a)=1, so a=0 and b=1 (not possible)

Second solution:
4x(x+1)=(2k) 2 , so (2x+1) 2 =(2k) 2 +1. From here do the same as before.

This was not so hard was it?
----------------------------------------------------------------------------------
n=3.
x(x+1)(x+2)=k 2.

If x is even we have x=2x (in the RHS is not the same x as in the LSH), so x(x+1)(2x+1)=k 2, for some other k.
It is clear that x,x+1, 2x+1 are pairwise coprime, so they are all squares since their product is a square. So x and x+1 are squares and goto the case n=2 to see why this is impossible.

If x is odd then x, x+1 and x+2 are pairwise coprime so they are all squares. The contradiction easily follows.
----------------------------------------------------------------------------------
n=4.
x(x+1)(x+2)(x+3)=k 2.

z=x 2 +3x. The equation comes to z(z+2)=k 2, or (z+1) 2=k 2+1 and goto the case n=2 to see why this is not possible.

n=5.
x(x+1)(x+2)(x+3)(x+4)=k 2.

x=6x.
6x(6x+1)(6x+2)(6x+3)(6x+4)=k 2, so 2x(2x+1)(3x+1)(3x+2)(6x+1)=k 2.

If 3x+1 is even then (3x+1)/2 is coprime with x, 2x+1, 3x+2 and 6x+1 and therefore is a square, so 4*(3x+1)/2=6x+2 is a square. 6x+1 is coprime with x, 2x+1, 3x+1, 3x+2, 2, so it is a square. So 6x+1 and 6x+2 are square (impossible by the case n=2).

If 3x+2 is even we obtain similarly 6x+1 and 6x+4 are squares.Lets solve the equation a 2+3=b 2.
(b-a)(b+a)=3, so b=2 and a=1.
So 6x+1=1, so x=0 - a contradiction (we work with positive integers).

x=6x+1.
(6x+1)(6x+2)(6x+3)(6x+4)(6x+5)=k 2.
6x+1 and 6x+5 are coprime and coprime with the rest, so they are squares.
The only solution to the equation a 2 +4=b 2  is (0,2), so 6x+1=0 - a contradiction.

x=6x+2.
(6x+2)(6x+3)(6x+4)(6x+5)(6x+6)=k 2.
2(x+1)(2x+1)(3x+1)(3x+2)=k 2.
Do the same as for x=6x.

For x=6x+3, 6x+4, 6x+5 we have 6x+5 and 6x+7 between the 5 consecutive numbers. They are coprime and coprime with the rest and therefeore must be squares. The equation a 2 +2=b 2 has no solutions in integers, so we get a contradiction.


AMC
USA(J)MO
USAMO
Olimpiada de matematicas

Este problema no es de una olimpiada, sino de las opsiciones para el ingreso en el cuerpo de profesores de enseñanza secundaria. Lo pongo aquí, porque creo que debe ser de ese nível.

"Una línea recta que pasa por el incentro de un triángulo ABC corta a los
lados AB y AC en los puntos D y E respectivamente. Sea P el punto de
intersección de BE y CD.
Si X, Y y Z son los respectivos pies de las perpendiculares desde P a BC, CA
y AB, demuestra que:
1/PX = 1/PY + 1/PZ"

Usamos coordenadas trilineales con respecto a $\triangle ABC.$ Sean $(p,q,r)$ las coordenadas de un punto arbitrario $Q.$ Así, la recta $\ell$ que pasa por  $Q$ y $I \equiv (1:1:1)$ tiene ecuación $(r-q)\alpha+(p-r)\beta+(q-p)\gamma=0.$ Entonces

$D \equiv \ell \cap AB \equiv (r-p:r-q:0) \Longrightarrow \ CD \equiv (r-q)\alpha-(r-p)\beta=0$

$E \equiv \ell \cap AB \equiv (q-p:0:q-r) \Longrightarrow \ BE \equiv (q-r)\alpha-(q-p)\gamma=0$

$P \equiv BE \cap CD \equiv (-(p-r)(p-q) : -(r-q)(p-q) :- (p-r)(q-r))$

La suma de los reciprocos de la segunda y tercera coordenada de $P$ es igual al reciproco de la primera. Siendo éstas directamente proporcionales a las distancias de $P$ a sus correspondientes lados, entonces el resultado se sigue.

Dear Mathlinkers,
can we have another approach?

Sincerely
Jean-Louis

Dear Mathlinkers,
I come back with this problem...

Any ideas?

Sincerely
Jean-Louis


Has anyone who took APMO last year through AMC gotten "official" results/awards yet?  I'm just curious because the results have apparently been finalized/officialized.  See [url=http://www.olympiad.org/apmo16.html]www.olympiad.org/apmo16.html[/url] [Australian site].

Thanks in advance.

I didn't take the APMO, but I stumbled upon this on the AMC website. It appears to be last year's results for USA?

http://www.unl.edu/amc/f-miscellaneous/f5-asianpacific/asianpacific04.html

If you are the person on that list, congratulations :).

Yes, that's Mr. Yi Sun with a whooping 29 points. :-D


Are there  1000000  positive integers  such  that the sum  of any  number of them (one or more)  is never a  perfect  square?


The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:

$ \textbf{(A)}\  9\qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 5\qquad
\textbf{(E)}\ 3$

how does it make a difference if its isosceles or not?

The altitude from the vertex of an isosceles triangle to the base is also an angle bisector and a median.

[hide] [geogebra]0e908368fe28b893b67e35aa6996313b90bdf78b[/geogebra]
Solution by Geogebra. Count the lines (two of them are extra, I don't know how they got there), you get

$ \boxed{\text{B}}$[/hide]


vector
calculus
analytic geometry
integration
search

This is a pretty vague question.  You have to be more specific as most of us will have no clue what you're talking about.  And I doubt that anything is required for anyone.

[quote="aadil"]are irodovs books required for a ninth grader?[/quote]

No. That is way too tough for the ninth grade unless, of course, you are very enthusiastic of studying something beyond the scope of your syllabus. Save for the Kinematics chapter, nothing from the book is relevant for the CBSE syllabus.

There are 2 difficultiesto Irodov, in my humble opinion:
-physics: conceptual
- mathematic: some problems require extensive computation (systems of equations), others knowledge of vector calculus (gradient in polar coordinates)

The first one is obvious. Many problems will require you to think outside of the box.
The second one is the one you should learn to handle 1st, before you even start doing the problems.  Starting Irodov too early (and yes, there are instances where it is too early) can be very discouraging if you don't have a strong grasp of vector calculus and differentiation/integration.  I learnt this the hard way.  I'm sure you heard it already: know your calculus rock solid, and then you will be able to fully appreciate Irodov.

And BTW, regardless of your curriculum, Irodov is IMO the best physics problem book out there (at least at an introductory level), so once you have nailed calculus, you should definitely buy it.

[quote="integration"]
And BTW, regardless of your curriculum, Irodov is IMO the best physics problem book out there.

[/quote]

Yes, solving Irodov problems do require a thorough knowledge of Physics and in some problems, it is simply calculus.

@[b]integration[/b]: If you've liked Irodov so much, I would recommend you also have a look at [b]S. S. Krotov's Science for Everyone: Problems in Physics[/b], the problems are tougher than those in Irodov, but one good thing about this book is that exhaustive solutions are provided at the back and one bad thing is you need the solutions really bad! :oops:  :rotfl:

Where can you find SS Krotov's problems?? It's only for sale in India, and so I can't buy it  :huh:

Krotov, like Irodov is a Russian author and both these book are published by Mir Publishers, Moscow. It shoud be easily available in the US if it is in India. You could search a library...

And this is what I found for [url=http://www.ebookee.com/-request_ebook-Aptitude-test-problems-in-physics_162181.html]ebook[/url]...


Any predictions? I have a feeling there's going to be a lot tighter competition this year....

lovely i like how u didnt mention ur name

its probably going 2 b a ton harder this year

my prediction would be

for the team

nick
bo
me
zaki

i shouldn't have put me down

btw bo can in 19th


Prove that: $ 1\plus{}C_n^3\plus{}C_n^6\plus{}....\equal{}\frac{1}{3} \left ( 2^n\plus{}2\cos{\frac{n\pi}{3}}\right)$

This is a technique commonly known as the roots of unity filter; see my posts [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url], [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url], [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=242876]here[/url], and (to a lesser extent) [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=257652]here[/url].


Find a fifth degree polynomial $ P(x)$ such that $ P(x) \plus{} 1$ is divisible by $ (x\minus{}1)^3$ and $ P(x) \minus{} 1$ is divisile by $ (x\plus{}1)^3$.

[quote="ertanrock"]Find a fifth degree polynomial $ P(x)$ such that $ P(x) \plus{} 1$ is divisible by $ (x \minus{} 1)^3$ and $ P(x) \minus{} 1$ is divisile by $ (x \plus{} 1)^3$.[/quote]
$ A \equal{} x \minus{} 1$

$ P \equal{} A^3 B \minus{} 1 \equal{} (A \plus{} 2)^3 C \plus{} 1$ with $ B$ and $ C$ both second degree polynomials.
$ A^3 B \equal{} 2 \plus{} (A^3 \plus{} 6A^2 \plus{} 12A \plus{} 8)C \equal{} C A^3 \plus{} 6C A^2 \plus{} 12CA \plus{} 8C \plus{} 2$

then $ A|8C \plus{} 2$ then $ 8C \plus{} 2 \equal{} AD$ with $ D$ a one degree polynomial.

$ 8A^3 B \equal{} (AD \minus{} 2) A^3 \plus{} 6(AD \minus{} 2) A^2 \plus{} 12(AD \minus{} 2)A \plus{} 8AD$
$ 8A^2 B \equal{} (AD \minus{} 2) A^2 \plus{} 6(AD \minus{} 2) A \plus{} 12(AD \minus{} 2) \plus{} 8D$
$ 8A^2 B \equal{} D A^3 \minus{} 2A^2 \plus{} 6DA^2 \minus{} 2A \plus{} 12DA \minus{} 24 \plus{} 8D$

then $ A|8D \minus{} 24$ then $ 8D \minus{} 24 \equal{} kA \equal{} k(x \minus{} 1)$

$ P \equal{} (x \plus{} 1)^3 C \plus{} 1$
with $ C \equal{} \frac {AD \minus{} 2}{8} \equal{} \frac {(x \minus{} 1)\frac {k(x \minus{} 1) \plus{} 24}{8} \minus{} 2}{8} \equal{} \frac {k(x \minus{} 1)^2 \plus{} 24(x \minus{} 1)}{64} \minus{} \frac {1}{4}$
$ C \equal{} r(x^2 \minus{} 2x \plus{} 1) \plus{} \frac {3}{8}(x \minus{} 1) \minus{} \frac {1}{4} \equal{} rx^2 \plus{} (\frac {3}{8} \minus{} 2r)x \plus{} r \minus{} \frac {5}{8}$
$ P \equal{} (x \plus{} 1)^3 (rx^2 \plus{} (\frac {3}{8} \minus{} 2r)x \plus{} r \minus{} \frac {5}{8}) \plus{} 1$

Checking $ B A \equal{} \frac {P \minus{} 1}{A^2}$ and replacing $ x \equal{} 1$ to get $ r \equal{} \minus{} \frac {3}{8}$

for $ P \equal{} \frac { \minus{} 3x^5 \plus{} 10x^3 \minus{} 15}{8}$
$ P\equal{}\minus{} \frac {1}{8} (x \minus{} 1)^3(3x^2 \plus{} 9x \plus{} 8) \minus{} 1 \equal{} \minus{} \frac {1}{8} (x \plus{} 1)^3 (3x^2 \minus{} 9x \plus{} 8) \plus{} 1$

[quote="TFPNH"]$ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \plus{} 1 \equal{} 0$
$ 5a \plus{} 4b \plus{} 3c \plus{} 2d \plus{} e \equal{} 0$
$ 20a \plus{} 12b \plus{} 6c \plus{} 2d \equal{} 0$
$ \minus{} a \plus{} b \minus{} c \plus{} d \minus{} e \plus{} f \minus{} 1 \equal{} 0$
$ 5a \minus{} 4b \plus{} 3c \minus{} 2d \plus{} e \equal{} 0$
$ \minus{} 20a \plus{} 12b \minus{} 6c \plus{} 2d \equal{} 0$

Solving the equation system, we get the value of $ a,b,c,d,e,f$.
Hence, $ P(x) \equal{} ax^5 \plus{} bx^4 \plus{} cx^3 \plus{} dx^2 \plus{} ex \plus{} f$[/quote]

From those equations:
(1)+(4) $ 2b\plus{}2d\plus{}2f\equal{}0$
(2)-(5) $ 8b\plus{}4d\equal{}0$
(3)+(6) $ 24b\plus{}4d\equal{}0$
then $ b\equal{}d\equal{}f\equal{}0$
$ a\plus{}c\plus{}e\equal{}\minus{}1$
$ 5a\plus{}3c\plus{}e\equal{}0$
$ 20a\plus{}6c\equal{}0$ for the same solution.

Or... not.  First of all, you can reduce to three variables by writing $ P(x) \equal{} (x \plus{} 1)^3 (ax^2 \plus{} bx \plus{} c) \plus{} 1$ and substituting the second condition, which is most easily done as follows:  the condition that $ P(x) \plus{} 1$ is divisible by $ (x \minus{} 1)^3$ is equivalent to the condition that $ P(1) \equal{} \minus{} 1, P'(1) \equal{} P''(1) \equal{} 0$, hence

$ P(1) \equal{} 8(a \plus{} b \plus{} c) \plus{} 1 \equal{} \minus{} 1$
$ P'(1) \equal{} 12(a \plus{} b \plus{} c) \plus{} 8(2a \plus{} b) \equal{} 0$
$ P''(1) \equal{} 24(a \plus{} b \plus{} c) \plus{} 12(2a \plus{} b) \plus{} 8(2a) \equal{} 0$

and the substitution $ x \equal{} a \plus{} b \plus{} c, y \equal{} 2a \plus{} b, z \equal{} 2a$ turns the above into a triangular system.


Could someone solve the following question for me please. The question is as follows:

Prove that if A and B are matrices such that I-AB is invertible, then the inverse of I-BA is given by the formula (I-BA)^-1 = I + B((I-AB)^-1)A
(Hint: Write C = (I-AB)^-1. Then show that (I-BA)(I+BCA) = I)

Make sure you [b]understand[/b] the [b]definition [/b]of the INVERSE of a matrix before doing this simple exercise. :)
This question looks like a bookwork assignment rather than a pre-olympic type question.   :roll:

Well i don't understand it, hence the reason why i asked for help. If i knew how to do it, why would i go about posting it on a forum?

The read the definition again and again... and follow the HINT.   :)
They key of this question to [b]understand[/b] the [b]definition[/b] of the inverse of a matrix.
If you want to show that B is the inverse of A, then you need to show
$AB = BA = I$ by definition....

I've read it a few times already. I must be obviously missing this 'hint' then    :huh:

Okay, finally you notice that the HINT is useful, it has already told you the inverse SHOULD BE.....

Nope sorry, i'm comletely baffled. I have no idea how to do this. I've looked at a few propositions etc, but i can't seem to make any sense out of them. Please help ?

With the help of the hint, let $C=(I-AB)^{-1}$.
Remember we then have $(I-AB)C=I \implies C-ABC =I \implies ABC = C-I$---(*).
[hide]Consider
$(I-BA)(I+BCA) \\ = I-BA+BCA-BABCA \\ =I-BA+BCA-B(C-I)A \\ { by (*)}$
......
$=I$.[/hide]

If you have learnt determinant already, you don't need to show $(I+BCA)(I-BA)=I$, orhterwise, you need to show this equality by definition of  the inverse of matrix.


A triangle with a height measuring 8 units and a base measuring 4 units is drawn and shaded on a 5 unit by 10 unit rectangular grid. What percent of the grid is shaded?

The triangle has area 32/2=16, so 16/50=32/100.  So its 32%


Alcumus
probability
expected value
Support

Under the solution, you can rate the solution from 0 to 5, including halves (0, .5, 1, 1.5, ..., 4.5, 5)

Certainly, I gave it a low rating but I wanted to be more specific than that in this case. :)

you can always pm an admin with a different solution
if they like it, they will add it to the solutions page

[quote="fedja"]Sometimes I see solutions that are formally correct but so far from optimal that I have an urge to report it. The recent example is a two dice problem where the question was to find the expected number of ones and the suggested solution was a casework with 4 cases according to which die shows one and which doesn't. The linearity of expectation property would solve the problem in one line. I reported this one as an error but, technically speaking, there was nothing formally wrong with the approach or the solution text. So, I wonder whether what I've done was a good idea or I should attract the attention of the Alcumus developers in some other ways in such cases.[/quote]

You can report it in the Bug Report.

I'll note that some problems are in the database with non-optimal solutions because they are included a topic that is before more advanced topics that can be used to solve the problem (a similar problem may also be included in the later topic).


Let $ f(x)\equal{}\frac{9^x}{9^x\plus{}3}$. Compute $ \displaystyle\sum_{k}^{}f \left( \frac{k}{2002} \right),$ where $ k$ goes over all integers $ k$ between $ 0$ and $ 2002$ which are coprime to $ 2002$.

[hide]Note that:
$ f(x) \plus{} f(1 \minus{} x) \equal{} \dfrac{9^{x}}{9^{x} \plus{} 3} \plus{} \dfrac{9^{1 \minus{} x}}{9^{1 \minus{} x} \plus{} 3}$

$ f(x) \plus{} f(1 \minus{} x) \equal{} \dfrac{9^{x}}{9^{x} \plus{} 3} \plus{} \dfrac{9}{9 \plus{} 3\cdot9^{x}}$

$ f(x) \plus{} f(1 \minus{} x) \equal{} \dfrac{9^{x}}{9^{x} \plus{} 3} \plus{} \dfrac{3}{3 \plus{} 9^{x}}$

$ f(x) \plus{} f(1 \minus{} x) \equal{} \dfrac{9^{x} \plus{} 3}{9^{x} \plus{} 3}$

$ f(x) \plus{} f(1 \minus{} x) \equal{} 1$

Note that if $ k$ is coprime to $ 2002$, then $ 2002 \minus{} k$ is also coprime to $ 2002$.

Since $ \phi(2002) \equal{} \phi(2\cdot7\cdot11\cdot13) \equal{} 1\cdot6\cdot10\cdot12 \equal{} 720$, there are $ 720$ values of $ k$ in the summation of which $ \frac {1}{2}\cdot720 \equal{} 360$ are less than $ 1001$:

For each of those $ 360$ values of $ k$ we have:
$ f\left(\frac {k}{2002}\right) \plus{} f\left(\frac {2002 \minus{} k}{2002}\right) \equal{} 1$

Therefore:
$ \sum_{k}^{}f\left(\frac {k}{2002}\right) \equal{} \boxed{360}$. [/hide]

EDIT: I can't read. Thanks @ Nora.91 and vijaymenon for pointing that out.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=285067]A very similar problem, but from a different contest[/url]

@ Xantos : But since $ k$ goes over nos. coprime to $ 2002$ and $ \phi(2002)\equal{}720$,

answer = 360 ??

posted before $ f(x)\equal{}\frac{3^{2x\minus{}1}}{3^{2x\minus{}1}\plus{}1}$, therefore $ f(x)\plus{}f(1\minus{}x)\equal{}1$.


search
number theory
prime factorization

How can you tell if 20113 is prime?  I could try all the primes before the sqaure root of 20113, but then I'll have to plug in a lot of numbers into my calculator (all primes before 141).  Is there an faster way?

Thanks    :)  :P

look it up online. do a google search for a list of primes, then just search in the page for your number. let's take bets to see if it's prime. i think it isn't.

It is a prime since it can be found on the site http://primes.utm.edu/lists/small/10000.txt . There are many primality test developed, but I suggest to use computer to prove the primality of a given number.

[quote="Ignite168"]I have a calculator program that checks it, would you like me to post the text?[/quote]
yes please!!! :wow:

it's probably incredibly easy

i made a program that has a list of the primes under 1000 and i have some programs that cross-reference it

if you have the 89 titanium, you can do

factor(20013)

and if it just says 20013, then it is prime

[quote="Adunakhor"][quote="ShoeFactory"]I have a program that finds prime factorization for you. (but if the factorization is like 59*601 it takes a while because it has to count to 601. :o[/quote]

In that case, wouldn't it only have to count to 59?[/quote]

Then it needs to check if 601 is prime so i guess it needs to count to $\sqrt{601}$

[quote="VenDiddy"]Typed it into my Ti-89 and: $20013=3\cdot7\cdot953$.[/quote]

i meant 20113

How can you tell if 20113 is prime without using calculators and computers?


invariant
algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra unsolved

In C[x,y], let V be the subspace of all polynomials of degree 2 or less, and the subspace is spanned by
1,x,y,x^2,xy,y^2.
Let T: V -> V be the transformation definedby T=d/dx + d/dy.
Find the invariant factors and elementary divisors of T.

In the basis $ \{1,x,y,x^2,y^2,xy\}$ the $ T$-matrix is the nilpotent matrix $ A\equal{}\begin{pmatrix}U&V\\0&0\end{pmatrix}$ with $ U\equal{}\begin{pmatrix}0&1&1\\0&0&0\\0&0&0\end{pmatrix},V\equal{}\begin{pmatrix}0&0&0\\2&0&1\\0&2&1\end{pmatrix}$. The $ A$-Jordan form is $ diag(J_3,J_2,0)$ where $ J_k$ is the elementary nilpotent Jordan matrix of dimension $ k$    ($ J^{k\minus{}1}\not\equal{}0,J^k\equal{}0)$. After it's easy.


inequalities
trigonometry
inequalities proposed

[quote="spanferkel"]Find the biggest $k$ such that \[ \sum_{cyc}a^2(3a^2-(a-b)^2)\ge kabc(a+b+c)  \] for all real $a,b,c$. :) ;)[/quote]
If $a=b=c$, get $9a^4 \ge k*3a^4$ hence $k \le 3$.
Let's prove $k=3$ works. Expand both sides and the ineq is equivalent to:
$\sum_{cyc}2a^4+\sum_{cyc}2a^3b \ge \sum_{cyc}a^2b^2+\sum_{cyc}3a^2bc$. Then:
$\sum_{cyc}2a^4 \ge \sum_{cyc}2a^2bc$ by Muirhead
$\sum_{cyc}2a^3b \ge \sum_{cyc}a^2bc$ by Muirhead
$\sum_{cyc}a^3b \ge \sum_{cyc}a^2b^2$ by Muirhead
Adding the three ineq above get the original ineq with $k=3$.
So, $k=3$.

[quote="rem"]
$\sum_{cyc}a^3b \ge \sum_{cyc}a^2b^2$ by Muirhead
[/quote]
No this is not true for cyclical sums! And the answer is not 3 :(

Nobody?
[hide="Hint"]Equality case: $k=2, (a,b,c)=(\sin\frac27\pi,\sin\frac47\pi,\sin\frac87\pi)$ ... Does that ring a bell? :wink: [/hide]


algebra
polynomial
number theory
relatively prime
system of equations
algebra solved

Fin all pairs of positive integers (m,n) so that

[tex]1+x+x^2 + ... + x^m | 1 + x^n + x^{2n}+...+x^{mn}[/tex]

Hope i didn't screw up...

1 + x + x 2 ...+ x m = (x m+1-1) / (x-1)
1 + x + x n ... = (x n(m+1) - 1)/(x n - 1)
Taking the quotient, we get (x n(m+1)-1)(x - 1)
 / ((x n - 1)(x m+1-1))

Since all factors of (x n(m+1)-1) are distinct, and there are n(m+1) of them, n and m+1 must be relatively prime, else there will be two congruent factors other than x-1, which cannot both be divided into x n(m+1)-1.
It is also easy to see that if they are relatively prime, all factors of x n - 1 and x m+1-1 will appear in x n(m+1)-1, by classifying the roots of unity by their angle measure.

$ P(x)=1+x^2+x^3+..+x^m$
$ Q(x)= 1+x^{2n}+..+x^{mn}$
We have that:
   $ x^{m+1}-1=(x-1)Q(x)$ and $ x^{n(m+1)}-1=(x^n-1)Q(x)$
So P(x)/Q(x) is equivalent to if  $ x^{m+1}=1$ has a root ( <> 1)then this root is not the root of $ x^n-1=0$;
This is equivalent to : the system of equations:
             $ \begin{array}{cc}{x^{m+1}=1}\\{x^n=1}\end{array}$
has only one root : 1;
 Now, if $ (m+1,n)=d >1$ then :
        \[ x=cos\frac{2\pi}d+i\cdots in\frac{2\pi}d \\ \Longrightarrow x<>1\]
if $(m+1,n)=1$ ,there exist $ k,l\in Z$ satisfying:
         \[ k(m+1)+l\cdot n=1\]
\[ x=x^{k(m+1)+l\cdot n}=(x^{m+1})^k\cdot(x^n)^l=1\]
Then the solution is $ (m+1,n)=1$


HCSSiM
Ross Mathematics Program
AMC
AIME
summer program
PROMYS
Columbia

My kid has been selected by both program. His primary interest is AMC competition. I was wondering if people know anything about HCSSiM and decide which one to chhose. Please provide your input. Thanks

of course I have to recommend Ross (I will be a junior counselor there this summer), but you should know that the Ross program is not aimed at "contest math" like the AMC.  Instead it focuses on higher mathematics, and improving ability in mathematical problem solving, proof writing and lateral thinking(which are skills that can be applied to the AMC competitions).

While Dr. Ross was not a fan of applied math or math contests, it is possible to come away from the program doing better on these competitions.  One good example of this is Tim Abbott, who as a freshman, didn't qualify for the AIME, and as a sophomore qualified, but only solved four problems.  Between his sophomore and junior year he attended the Ross program, and without any additional preparation for the AMC contests he had an index score of about 250/300 and received honorable mention on the USAMO.  While he is not the average case, attending the Ross program does influence your mathematical thinking and helps you in all aspects of mathematics.

Thanks for the nice comments on ROSS, zscool. I wonder if you would say the same about PROMYS, which is suppose to be the same as ROSS.

I have never attended any of these programs, but I know a lot of people who went to the Hampshire program.  I haven't heard a single bad thing about it.  (This is probably not very helpful.  I would expect that your child would enjoy Hampshire very much.)

I went to Hampshire last summer and loved it. You study all branches of math there and really learn solidly how to think mathematically and find proofs by learning such a variety of topics. There are also "maxi" and "mini" classes where you get to focus on one topic.
I'm sure Ross is great for a lot of people but its 8 weeks of number theory and a lot of people I've talked to say they couldn't handle it and didn't get very much out of it.  Hampshire is 6 weeks and you probably spend the same amount of time on math each day, but it ends up being much more manageable.

[quote]I'm sure Ross is great for a lot of people but its 8 weeks of number theory and a lot of people I've talked to say they couldn't handle it and didn't get very much out of it[/quote]

may I ask who you have talked to John?

People who say they weren't really ready when they went: Mr. W. and a guy from my Columbia SHP Saturday class. I guess I shouldn't have said "a lot of people."  :|

well, I do not know about the person from Columbia (could it be Alex?), but for Mr. W it is a different story considering he went about 10 years ago.

I do agree that some people do not get a lot out of the Ross program.  These people, however, are usually unwilling to learn and apply themselves.  To get something out of the program it does not take a lot of talent, it just takes a willingness to learn and work (just look at me).


Show that you can express any prime (other than two or three) in the form $6n \pm 1$.

(is this even true?)

[hide="Solution"]

Except for 2 and 3, this conclusion holds because $6n+3$ is divisible by 3 and $6n,6n+2,6n+4$ are divisible by 2. Therefore, all primes must be in the form $6n\pm1$ because these are the only residues left.
[/hide]

Oh.
What do you mean by "the only residues left"
I know a LITTLE bit about residues and residue classes, but I don't quite get how this applies.
Maybe I'm just stupid?
Could you explain that a little more?
Other than that, I'm sure it's intelligible to some other people out there  :lol:
Nice solution, though.

The residue classes in mod 6 are

$6n$
$6n+1$
$6n+2$
$6n+3$
$6n+4$
$6n+5$

together, they form the set of integers

$6n$, $6n+2$, $6n+3$, $6n+4$ are all composite by knexpert's logic

what is left is
$6n+1$ and $6n+5$

$6n+5$ and $6n-1$ are the same residue because they have the same value in mod 6

Thanks for the explanation.
This makes a lot more sense now.

So basically, we divide the set of integers into chunks of six?
Those that can be expressed in each of those ways (residue classes)
And then we go on to show that because four of the classes are composite, then the others must be prime.
Because the set of primes is wholly contained inside the integers, we can say that all primes are members of these classes?

That seems to work well.

Slightly more rigorously:

To prove that all primes satisfy this condition, we only need to prove it for an arbitrary prime $p$.
Now $p$ can be written in one of these forms: $6n$, $6n+1$, $6n+2$, $6n+3$, $6n+4$, or $6n+5$ (where n is a natural number).
However, $6n$ is composite. Similarly, $6n+2=2(3n+1)$ is composite, $6n+3=3(2n+1)$ is composite, and $6n+4=2(3n+2)$ is composite.
Therefore $p$ must either be $6n+1$ or $6n+5$; in other words, it must be $1$ or $5$ mod 6.
QED

[quote="DiscreetFourierTransform"]
And then we go on to show that because four of the classes are composite, then the others must be prime.
[/quote]

That phrase isn't true, but the rest of what you said is.

The set of primes greater than $3$ is contained in either the residue class $1 \mod 6$ or $-1 \mod 6$

that doesn't mean that there doesn't exist composite numbers (like $25$) which are in one of those forms

[hide="Discussion"] Generally, for an arbitrary $a$, any prime larger than $a$ is expressible in the form $ak+d$, for some $d$ such that $gcd(a, d) = 1, 1 \le d < a$.  For $a = 6$ we only have the choices $d = 1, 5$, but for other choices of $a$ there are more choices of $d$.  Euler's Totient Function $\varphi(a)$ gives the number of choices of $d$.

[hide="Interestingly enough..."] Dirichlet's Theorem is a very strong theorem which states that given any $a$ and $d$ such that $gcd(a, d) = 1$, there exist an [i]infinite[/i] number of primes of the form $ak+d$.  Several special cases are more elementary:  for example, that there exist an infinite number of primes of the form $4k+3$ or of the form $6k+5$.

[hide="Although actually..."] It makes the even stronger claim that $\sum_{ak+d \in \mathbb{P}}\frac{1}{ak+d}$ diverges.  If a sum diverges, it must have an infinite number of terms, hence the usual statement of the theorem. [/hide] [/hide] [/hide]

DiscreetFourierTransform you are a legend. Thanks heaps for helping with my (thekiterunner) problem and I hope you are able to help with my new problem that I have just posted. Your working out was the greatest!

Thanks heaps

Here is a proof that there exists an infinitude of primes of the form $ 6n+5$.

Firsly lets assume that there are only a finite number of primes which take the form $ 6n+5$ and call them $ q_{1},q_{2},\ldots,q_{k}$. Lets consider the integer $ N$

$ N = 6({q_{1}}{q_{2}}{\ldots}{q_{k}})-1 = 6({q_{1}}{q_{2}}{\ldots}{q_{k}}-1)+5$

Let the prime factorisation of $ N$ be $ N ={r_{1}}{r_{2}}{\ldots}{r_{m}}$. Since $ N$ is of the form $ 6n+5$,  then $ r\not= 2$ and $ r\not= 3 \Longrightarrow r$ must take one of the forms $ 6k+1$ or $ 6k+5$. Lets consider the product of two integers of the form $ 6k+1$

\begin{eqnarray*}(6k_{1}+1)(6k_{2}+1) = 36k_{1}k_{2}+6k_{1}+6k_{2}+1 = 6(6k_{1}k_{2}+k_{1}+k_{2})+1 = 6k+1, k \in Z \end{eqnarray*}

There must be at least one $ r$ taking the form $ 6k+5$ otherwise $ N$ would take the form $ 6k+1$. Let $ r_{i}$ take the form $ 6k+5$. Since $ q_{1},q_{2},\ldots,q_{k}$ is the list of all the primes of the form $ 6k+5$ then $ r_{i}$ must be equal to one of $ q_{1},q_{2},\ldots,q_{k}$. Hence $ r_{i}\mid 6({q_{1}}{q_{2}}{\ldots}{q_{k}})$

$ N = 6({q_{1}}{q_{2}}{\ldots}{q_{k}})-1 \Rightarrow 1 = 6({q_{1}}{q_{2}}{\ldots}{q_{k}})-N$

But since $ r_{i}\mid 6({q_{1}}{q_{2}}{\ldots}{q_{k}})$ and $ r_{i}\mid N \Longrightarrow r_{i}\mid 1$. This implies that $ r_{i}= 1$ which contradicts the fact that $ r_{i}$ is prime. The original assumption that there is a finite number of primes of the form $ 6n+5$ is false $ \Longrightarrow$ there are an infinitude of primes of the form $ 6n+5$.

Now can you show that there exist infinitely many primes of the form $ 6n+1$?

Slightly easier: can you show that there are also an infinitude of composite numbers with the form $ 6n+1$ and $ 6n+5$?


[hide]For $n \ge 6$ we have 3(some stuff divisible by 3 + 11) which is divisible by 3 but not by 9 so it cannot be a perfect square. It is easy to check for $n$ less than 6[/hide]

in otherwords (?)

for $n\geq 6$, the lhs is congruent to  6 mod 9. which implies its not square, since squares are congruent to 0,1,4,7 mod 9. :wink:

If $n\geq 5$, then the last digit of the number $n!+33$ is 3, which cannot be the last digit of a perfect square. We can check the cases when $n\leq 4$.

[quote="DPopov"]Prove that $n!+33$ is never a perfect square for $n \geq 0$[/quote]

[hide]We know that no square is congruent to 6(mod 9).  This first occurs after n =6.

So if n=1, then this is 34, not
n=2, this is 35, not
n=3, this is 37, not
n=4, this is 57 not
n=5, this is 153, not
n=6, congruent to 6(mod 9), not
n=7, congruent to 6(mod 9), not
etc, etc[/hide]


If $x+y=3-cos 4\alpha, x-y=4sin 2\alpha$. Prove $\sqrt{x}+\sqrt{y}=2$

[hide="Solution"]As $\cos 2\phi=1-2\sin^{2}\phi$, we have $\cos 4\alpha=1-2\sin^{2}2\alpha$. For shortness put $a=\sin 2\alpha$. Then

$x+y=3-(1-2a^{2})=2+2a^{2}$
$x-y=4a$

By adding up the equations we get

$2x=2(1+2a+a^{2})\iff x=(1+a)^{2}\iff\sqrt{x}=|1+a|$

Since $a\geqslant-1$ (by the definition of $a$), we get $|1+a|=1+a\iff\sqrt{x}=1+a$

By subtracting the second equation from the first we get

$2y=2(1-2a+a^{2})\iff y=(1-a)^{2}\iff\sqrt{y}=|1-a|$

Since $a\leqslant 1$ (by the definition of $a$), we get $|1-a|=1-a\iff\sqrt{y}=1-a$

Now $\sqrt{x}+\sqrt{y}=1+a+1-a=2$[/hide]

$\|\begin{array}{c}\sin 2\alpha=t\in [-1,1]\\\\ \cos 4\alpha =1-2t^{2}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x+y=2(t^{2}+1)\\\\ x-y=4t\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}x=(t+1)^{2}\\\\ y=(t-1)^{2}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}\sqrt x=t+1\\\\ \sqrt y=1-t\end{array}$ $\Longrightarrow$ $\boxed{\ \sqrt x+\sqrt y=2\ }\ .$

[b]Remark.[/b] $\|\begin{array}{c}\sqrt x=(\sin\alpha+\cos\alpha )^{2}\\\\ \sqrt y=(\sin\alpha-\cos\alpha )^{2}\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}\sqrt x+\sqrt y=(\sin\alpha+\cos\alpha )^{2}+(\sin\alpha-\cos\alpha )^{2}=2\ .\\\\ \sqrt [4] x+\sqrt [4] y=|\sin\alpha+\cos\alpha |+|\sin\alpha-\cos\alpha |\le 2\Longrightarrow \boxed{\ \sqrt [4] x+\sqrt [4] y\le 2\ }\ .\end{array}$


Has been posted several times before.

[hide="Hint 1"] The LHS is a geometric series. [/hide] [hide="Hint 2"] $\cos (2 \theta) = 2 \cos^{2}\theta-1$ [/hide]

[quote="Totally Zealous"]If $\sum_{n=0}^{\infty}\cos^{2n}\theta = 5$, what is $\cos 2\theta$?[/quote]

actually Farenhajt, it is 3/5. Refer [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70139]here  :wink:[/url]

Darn, I love when I contradict my earlier solutions of the same problems  :D

(Overlooked that tiny little $2$ in the exponent. Apologies.)

[quote="nutz_for2.718281828"]Bad forum. Someone plese move this.[/quote]
I think it's fine...
Besides, it's been answered, it doesn't matter that much.


Let $k>1$ be a given positive integer. A set $S$ of positive integers is called [i]good[/i] if we can colour the set of positive integers in $k$ colours such that each integer of $S$ cannot be represented as sum of two positive integers of the same colour. Find the greatest $t$ such that the set $S=\{a+1,a+2,\ldots ,a+t\}$ is [i]good[/i] for all positive integers $a$.

[i]A. Ivanov, E. Kolev[/i]

One colour of $\{a+1,a+2\cdots,a+k+1\}$ is used twice, so there is a sum max. $2a+2k+1$ which is sum of two ones of same colour.
Taking $a$ the smallest, so $1$ we get $t$ is max. $2k+1.$
Desired set is coulouring the numbers such that $x,y$ are same colour if $x+y=2k+3$ where $x,y\in[2,2k+1]$ and $2k+2$ can be each one.

am I wrong?  :maybe:  ...I think that the maximum is $ 2k-2 $...for $ k=2 $ the set $ 1,2,3 $ contains two numbers of the same color and their sum will be $ 3 $, $ 4 $ or $ 5 $ so $ t \leq 2 $...the example for $ 2k-2 $ is quite easy...


Hi, there,

I got some problems to install MikTex2.7 in Win2k. Please help. Here is error message:

Installation: C:\Program Files\MiKTeX 2.7
Loading package database...
starting package maintenance...
installation directory: "C:\Program Files\MiKTeX 2.7"
package repository: C:\DOCUME~1\tk1\LOCALS~1\Temp\mik42304
opening repository C:\DOCUME~1\tk1\LOCALS~1\Temp\mik42304...
repository type: local package repository
loading lightweight database file...
DBlight digest: 9a36830e3ebc9ad44239a4bfc6be06eb
going to install 8538 file(s) (131 package(s))
problem: C:\Program Files\MiKTeX 2.7\miktex\bin\MiKTeX207-beta-5-core-PS.dll does not exist
problem: C:\Program Files\MiKTeX 2.7\miktex\bin\MiKTeX207-beta-5-core.dll does not exist
problem: C:\Program Files\MiKTeX 2.7\miktex\bin\MiKTeX207-beta-5-packagemanager-PS.dll does not exist
problem: C:\Program Files\MiKTeX 2.7\miktex\bin\MiKTeX207-beta-5-packagemanager.dll does not exist
extracting files from miktex-arctrl-bin-2.7.tar.lzma...
extracting files from jknappen.cab...
extracting files from ncntrsbk.cab...
extracting files from miktex-omega-bin-2.7.tar.lzma...
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Thanks a lot

Have a look at http://sourceforge.net/forum/forum.php?thread_id=1822602&forum_id=33790 to see if the suggestions there help. If not, it's probably the best forum to ask for help.


Find the number of square inches in the area of circle $ A$. Express your answer as a common fraction in terms of $ \pi$.

[asy]pair A;
draw(Circle(A,1)); draw( dir(0)-- dir(180) --dir(250) --cycle); dot(A);
label("$A$",A,N); label("5''",dir(180) --dir(250),NE);
label("7''",dir(250)--dir(0),SE);[/asy]

Using the Pythagorean Theorem:

$ 5^2\plus{}7^2\equal{}x^2$
$ 74\equal{}x^2$
Since we need the positive root, the diameter of the circle is $ \sqrt{74}.$  This means the radius is $ \frac{\sqrt{74}}{2}$

The area is:
$ \boxed{\frac{37\pi}{2}}$


Hey all, I was pondering this the other day, and I was curious to hear your opinions...

Does one necessarily need to be good at math competitions to be a sucessful mathematician?

Does sucess at competitions mean a person would be more suited for being a mathematician?

Thanks.

This has been discussed a lot previously (I'll try to find the link).
In my mind solving problem at competitions is like doing the 250 m. Research is more of a marathon :)
Hence the answer would be no.
(Homever you must be good (not necessarily excellent, or competition level) in the 250 m if you want to have a chance in a marathon)

well, doing contests and being immersed in math starting at a young age and progressing through high school is definitely a good foundation for later research

training in the 250m gives juts about no help for the marathon, heh

Yeah, but if you're good at the marathon, you won't be too bad at the 250m

Hmmm, I can understand how understanding how to solve the problems is important, but what about the timed aspect? In research, how important is the speed at which you go about doing things?

I would have to say no.

A lot of people just don't test well and are very prone to stupid mistakes when working fast.  Also, during a contest (unless it's a long one) you aren't making any kind of discoveries or trying new things, just applying what might have worked once before.

[quote="SnowStorm"]Also, during a contest (unless it's a long one) you aren't making any kind of discoveries or trying new things, just applying what might have worked once before.[/quote]

As you pointed out it depends on the contest. As an example some IMO problems appeared in research (tought I'm unable to give you a precise reference for the moment, perhaps some mathlinkers will verify this claim ?), or leads to interesting, unusual facts.

From my experiences as a professional mathematician... There is no link between doing well at contests and being a good research mathematician---there are plenty of people who are excellent at either contests or research and not very good at the other.  There are also plenty of great research mathematicians who are good at contests.  And, there are plenty of people who like contests but not research, and plenty who like research but not contests.

As for the timed aspect... it's certainly useful to be able to do some kinds of mathematics quickly when one is doing research.  But whether skill at contests helps with this or not depends a lot on what type of research problem one is working on.  And, research usually proceeds over the course of months, not hours.

One shouldn't work on contests because of the associated skills---one should work on them because one thinks they're fun and because one thinks mathematics is interesting.  This is also the best reason to do research (because it's interesting).

Problems in contests are generally more deductive, while research problems are inductive and require more creativity.  Of course, as you do harder contest problems (ie olympiad and stuff), creativity is required.

In my opinion, the answer is no. Because in a contest you have to solve a problem in an hour (for example) but if you're a researcher you may need ...let's say... a year to make a progress or to discover something. And in a problem you know exactly what you're lloking for, but sometimes being a researcher means to start looking without knowing exactly what you're looking for...


Let a,b,c>0 such that (c-a)(c-b) $ \geq$0 Prove that:
$ 3b^2\plus{}bc\plus{}c^2 \geq 2a(b\plus{}c)$

[quote="tran300491"]Let a,b,c>0 such that (c-a)(c-b) $ \geq$0 Prove that:
$ 3b^2 \plus{} bc \plus{} c^2 \geq 2a(b \plus{} c)$[/quote]
Try $ a\equal{}c\equal{}1$ and $ b\equal{}\frac{1}{2}.$ :wink:

Sorry
this is new inequality
Let a,b,c>0 such that a+b+c=1 Prove that
$ \sum\frac{ab}{1\plus{}c} \geq \frac{9}{4}\sqrt{abc}$

[quote="tran300491"]Sorry
this is new inequality
Let a,b,c>0 such that a+b+c=1 Prove that
$ \sum\frac {ab}{1 \plus{} c} \geq \frac {9}{4}\sqrt {abc}$[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{3}.$

[quote="arqady"][quote="tran300491"]Sorry
this is new inequality
Let a,b,c>0 such that a+b+c=1 Prove that
$ \sum\frac {ab}{1 \plus{} c} \geq \frac {9}{4}\sqrt {abc}$[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {1}{3}.$[/quote]
sorry
this problem true
Let a,b,c>0 such that a+b+c=1 Prove that
$ \sum\frac {ab}{1 \plus{} c} \geq \frac {3}{4}\sqrt {3abc}$

If a,b,c>0 then
$ \sum\frac{ab}{a\plus{}b\plus{}2c} \geq \frac{3}{4}\frac{ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c}$

we have $ \sum\frac{ab}{1\plus{}c}\frac{bc}{1\plus{}a}\geq\frac{9abc}{16}$
<->$ 41(ab\plus{}bc\plus{}ca)\plus{}9abc\leq14$
->Done


Given

A= [x, 1, 1  ]
     [1, x, -1 ]
     [-1, -1, x]

A is a 3x3 matrix.
(a) Compute the value of det(A) (your answer should be a function of x)
(b) For what values of x will the matrix be singular? Explain.

??? :?

hello, using the rule of Sarrus (see here http://en.wikipedia.org/wiki/Rule_of_Sarrus) we get
$ \det(A)\equal{}x(x^2\minus{}1)$
The matrix will be singular if the determinant is equal to zero, in this case we have $ x\equal{}0$ or $ x\equal{}1$ or $ x\equal{}\minus{}1$.
Sonnhard.


[quote="drunner2007"]Prove that the roots of $x^{2}-4x-2=0$ are roots of

$(x^{2}-3x-2)^{2}-3(x^{2}-3x-2)-x-2=0$

and find all the real roots of the second equation.[/quote]

[hide="partial solution"]let $y=x^{2}-4x-2$, then the second equation becomes:

$(y+x)^{2}-3(y+x)-x-2=0$
$y^{2}+2xy+x^{2}-3y-3x-x-2=0$
$y(y+2x-3)+y=0$
$y(y+2x-2)=0$

the rest is easy[/hide]


ratio
linear algebra
matrix
algebra unsolved
algebra

if $l(my+nz-lx) = m(nz+lx-my) = n(my+lx-nz)$
Show that $\frac{y+z-x}{l}= \frac{z+x-y}{m}= \frac{x+y-z}{n}$

[Edited:]

In fact, we can prove that it's "if and only if".
Both directions are only valid if $l,m,n$ are all nonzero.

Inverse direction.

$\frac{y+z-x}{l}= \frac{z+x-y}{m}= \frac{x+y-z}{n}= k$

$\Rightarrow x= (m+n)\frac{k}{2}$, $y= (n+l)\frac{k}{2}$, $z= (l+m)\frac{k}{2}$

$\Rightarrow l(my+nz-lx) = m(nz+lx-my) = n(my+lx-nz)= lmnk$.

Direct direction:

$l(my+nz-lx) = m(nz+lx-my) = n(my+lx-nz)= k'$

Putting the system in matricial form, the determinant of the matrix is
$\Delta = 4(lmn)^{2}$,
so if $l,m,n$ are all nonzero, it's invertible and there's only one solution.

In that case, the solutions must be (looking at the inverse direction :) )
$x= (m+n)\frac{k'}{2lmn}$, $y= (n+l)\frac{k'}{2lmn}$, $z= (l+m)\frac{k'}{2lmn}$

$\Rightarrow \frac{y+z-x}{l}= \frac{z+x-y}{m}= \frac{x+y-z}{n}= \frac{k'}{lmn}$.


Solve in R the following system:

$x^{2}=2y+4z-5$

$y^{4}=x+4z-4$

$z^{8}=x+2y-2$

$\Longleftrightarrow x^{2}+y^{4}+z^{8}=2x+4y+8z-11$. ($\star$)

But from Bernoulli we have these 3 inequalities:

$x^{2}=(1+(x-1))^{2}\geq  2x-1$

$y^{4}=(1+(y-1))^{4}\geq  4y-3$

$z^{8}=(1+(z-1))^{8}\geq  8z-7$.

suming them up we have that : $x^{2}+y^{4}+z^{8}\geq 2x+4y+8z-11$, but  ($\star$) $x^{2}+y^{4}+z^{8}=2x+4y+8z-11$.

so in our three bernoulli ineq we have equality: $\Rightarrow x=y=z=1$


25. In Mr. Edgecomb's math class, 30 students took an exam on statistics. If the average passing grade was 84, the average failing grade was 60, and the overall average was 80, how many students passed the exam?

[hide]60x+84y=2400. y must be a multiple of 5. 5*84= 420 5z=y
so 420z+60x = 2400 42z + 6x = 240 if z=x then we have 42z+6z=48z=240 z=x=5 y = 25
so the answer is 25[/hide]


logarithms
inequalities proposed
inequalities


Can someone here explain Binary Operations?

Try

[url]http://mathworld.wolfram.com/BinaryOperation.html[/url]

or

[url]http://en.wikipedia.org/wiki/Binary_operations[/url]

and be more specific when you ask "please explain..."


inequalities
pigeonhole principle
inequalities unsolved

Give $13$ real numbers.
Prove that there're always two real numbers $x, y$ satisfying the inequality
$\large\frac{x-y}{1+xy}\leq 2-\sqrt{3}$

Put x=tgA and y=tgB and use pigeonhole principle.


Construct a square using tiles of this form

Put 2 of the tiles next to each other to produce a 2x5 rectangle.

Use 10 of these to create a 10x10 square.

Follow-up: what is the smallest positive integer $ n$ such that the given tiles can tile a $ n$ by $ n$ square?

That tile has area 5, so the square must also have an area divisible by 5, giving us $ 5|n$. Clearly (well, I dunno exactly how to prove it, but it's pretty obvious), this cannot tile a 5 by 5 square, so 10 is the minimum of $ n$.

Say we have $ k$ tiles and a $ n$ by $ n$ square. Then $ 5k \equal{} n^2\implies 5|n^2$, so $ 5|n$.

(a property of prime numbers is that if $ p|ab$ then $ p|a$ or $ p|b$ in the case above, i used $ a\equal{}b\equal{}n$).


Let $ a>0$, $ d>0$ and put
$ f(x)\equal{}\frac{1}{a}\plus{}\frac{x}{a(a\plus{}d)}\plus{}\cdots\plus{}\frac{x^n}{a(a\plus{}d)\cdots(a\plus{}nd)}\plus{}\cdots$
Give a closed form for $ f(x)$.

$ g(x) \equal{} x^a f(x^d)$ satifies $ g'(x) \equal{} x^{a\minus{}1} \plus{} x^{d\minus{}1} g(x)$. Solve the associated differential equation and conclude.


There are 128 players at a table tennis competition. Everybody plays against all the other players exactly once. During a competition break (all old games were finished) the following situation occurs:
In each subset of five players one can find two players (at least once) who already played against each other.
Prove that there are five players who all played already against each other ! R(5,5,2)

Hmm...seems that 128 is very far from best possible....

Let's consider the complet graph K_n where the vertices are the n contestants of the tournament.
Color red the edge between two players if they have already played against each other at the break, and blue otherwise.

From Ramsey's theorem, if n  \geq R(5,5) then there is a complete red K_5 or a complete blue K_5. Note that, from the statement, it is assumed that there is no blue K_5. Then, if n  \geq R(5,5) there is a red K_5 and we are done.

All we have to do is then to evaluate R(5,5), which is still unknown (as far as I know). But, it is a very well-known exercise to prove that for each a,b, we have R(a,b)  \leq R(a-1,b) + R(a,b-1), and that R(a,b) = R(b,a).

In our case, we then have R(5,5)  \leq 2*R(5,4) = 50.
Thus, the conclusion holds for any tournament with at least n = 50 contestants (and probably less...).

Another 'unsolved' problem to be killed.

Pierre.


function
calculus
derivative
inequalities unsolved
inequalities

okay if A,B,C,D are numbers between 0 and 1 show this??
(1-A)(1-B)(1-C)(1-D)>1-A-B-C-D

Expanding the LHS, we have $ 1 \minus{} A \minus{} B \minus{} C \minus{} D \plus{} AB \plus{} AC \plus{} AD \plus{} BD \plus{} BC \plus{} CD \plus{} ABCD \minus{} ABC \minus{} ABD \minus{} CBD \minus{} ACD$.
Canceling the $ 1 \minus{} A \minus{} B \minus{} C \minus{} D$ with the RHS, it remains to show
 $ AB \plus{} AC \plus{} AD \plus{} BD \plus{} BC \plus{} CD \plus{} ABCD \minus{} ABC \minus{} ABD \minus{} CBD \minus{} ACD \geq 0$,
but this is clear since $ A,B,C,D$ are less than or equal to $ 1$, so
$ AB\geq ABC , AC \geq ACD, AD \geq ABD, BC \geq CBD$,
and $ ABCD,CD,BD$ are positive.

[b]Another method:[/b] Subtract the LHS from the RHS. We have to prove
$ RHS\minus{}LHS \geq 0$
But considering the LHS of this as a function in each variable and taking the second derivative, we see that it is $ 0$, so the function is convex in each variable, so the maximum occurs among the sixteen possibilites of $ A,B,C,D$ being $ 1$ or $ 0$, so now you can just test them all to find that none give anything greater than $ 0$


Given $ a,b,c \ge 0$ and $ ab\plus{}bc\plus{}ca \equal{}1$.Prove that:
$ 81(a^2\plus{}bc\plus{}1)(b^2\plus{}ca\plus{}1)(c^2\plus{}ab\plus{}1) \ge   125(a\plus{}b\plus{}c)^2$

We can solve it by expand.
But besides, we also have a shorter way:
The ineq above can be written into:
Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that: $ (3a\plus{}2bc)(3b\plus{}2ca)(3c\plus{}2ab) \geq\ \frac{125}{9}(ab\plus{}bc\plus{}ca)^2$
pqr works here  :)

[quote="forevercontinue"][quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:
$ 81(a^2 \plus{} bc \plus{} 1)(b^2 \plus{} ca \plus{} 1)(c^2 \plus{} ab \plus{} 1) \ge 125(a \plus{} b \plus{} c)^2$[/quote]

We can reduce the given inequality to homogeneous one :
$ 81(a^2 \plus{} 2bc \plus{} ab \plus{} ac)(b^2 \plus{} 2ca \plus{} ab \plus{} bc)(c^2 \plus{} 2ab \plus{} ac \plus{} bc) \geq 125 (a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca)^2$
After expanding, we have :
$ 37 \sum a^2b^2(a^2 \plus{} b^2) \plus{} 74\sum a^3b^3 \plus{} 155 \sum a^4bc \geq 109abc \sum ab(a \plus{} b) \plus{} 255a^2b^2c^2$
It is equivalent to
$ 37 (\sum a^2b^2(a^2 \plus{} b^2) \minus{} 6a^2b^2c^2) \plus{} 74(\sum a^3b^3 \minus{} 3a^2b^2c^2) \plus{} 46(\sum a^4bc \minus{} 3a^2b^2c^2) \plus{} 109abc(\sum a^3 \plus{} 3abc \minus{} \sum ab(a \plus{} b)) \geq 0$
Which is obvious by Am-Gm and Schur's Inequality :P[/quote]
If your solutions is true, it very nice.But my solution than nice. :)

[quote="quykhtn-qa1"][quote="nguoivn"]We can solve it by expand.
But besides, we also have a shorter way:
The ineq above can be written into:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ (3a \plus{} 2bc)(3b \plus{} 2ca)(3c \plus{} 2ab) \geq\ \frac {125}{9}(ab \plus{} bc \plus{} ca)^2$
pqr works here  :)[/quote]
Please post solution :)
@nguoivn [hide]: em dung luong giac :) [/hide][/quote]
I'm sure that your way can't solve the general problem:  :P
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Find the best constant k satisfy: $ (a \plus{} kbc)(b \plus{} kca)(c \plus{} kab) \geq\ \frac{(k \plus{} 1)^3}{9}(ab \plus{} bc \plus{} ca)^2$
@quykhtn-qa1: [hide]Voi' dang toan' don gian? nay`, ta co the? dung` pqr de? tim` luon hang` so tot nhat nen anh ko quan tam den' bai` toan' yeu' hon cua? ban. em. That ra anh cung~ chua giai? chi tiet nhung tu` trc den gio` anh lam` qua nhieu dang nay` roi`, lam` them 1 bai` y het. nua~ cung~ ko giup anh len them trinh do ma` chi lam` mat tg thoi, sr em nhieu` vi` anh ko post loi giai day` du?. Du` sao anh cung~ rat muon thay con duong` sang' tac cua bai` toan'  :oops: [/hide]

[quote="nguoivn"][quote="quykhtn-qa1"][quote="nguoivn"]We can solve it by expand.
But besides, we also have a shorter way:
The ineq above can be written into:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ (3a \plus{} 2bc)(3b \plus{} 2ca)(3c \plus{} 2ab) \geq\ \frac {125}{9}(ab \plus{} bc \plus{} ca)^2$
pqr works here  :)[/quote]
Please post solution :)
@nguoivn [hide]: em dung luong giac :) [/hide][/quote]
I'm sure that your way can't solve the general problem:  :P
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Find the best constant k satisfy: $ (a \plus{} kbc)(b \plus{} kca)(c \plus{} kab) \geq\ \frac {(k \plus{} 1)^3}{9}(ab \plus{} bc \plus{} ca)^2$
@quykhtn-qa1: [hide]Voi' dang toan' don gian? nay`, ta co the? dung` pqr de? tim` luon hang` so tot nhat nen anh ko quan tam den' bai` toan' yeu' hon cua? ban. em. That ra anh cung~ chua giai? chi tiet nhung tu` trc den gio` anh lam` qua nhieu dang nay` roi`, lam` them 1 bai` y het. nua~ cung~ ko giup anh len them trinh do ma` chi lam` mat tg thoi, sr em nhieu` vi` anh ko post loi giai day` du?. Du` sao anh cung~ rat muon thay con duong` sang' tac cua bai` toan'  :oops: [/hide][/quote]
My solution use: $ (1\plus{}cos^2A)(1\plus{}cos^2B)(1\plus{}cos^2C) \ge \frac{125}{64}$  :)