geometry
parallelogram
trapezoid
circumcircle
geometry proposed

In a trapezoid $ABCD$ with a base $AD$, point $L$ is the orthogonal projection of $C$ on $AB$, and $K$ is the point on $BC$ such that $AK$ is perpendicular to $AD$. Let $O$ be the circumcenter of triangle $ACD$. Suppose that the lines $AK , CL$ and $DO$ have a common point. Prove that $ABCD$ is a parallelogram.

Im not sure if Im right or not...
let $W$ be the circumcircle of $ACD$ and let $T$ be the intersection of $CL,AK$,so $DO$ passes through $T$ and $\angle TAD=90$...
now we claim that $T$ is on the circumcircle of $ACD$
[hide]let $T'$ be the intersection of $DT$ and $W$,so $DT'$ is a diameter of $W$ so $\angle T'AD=90$ but we knew that $\angle TAD=90$ and $D,O,T',T$ lie on a line,so $T=T'$... [/hide]
so assume that $W$ passes through $T$,now we want to show that $AB \| CD$ so its sufficient to show that $\angle LBC=\angle BCD$...
from $LC \bot AB$ we get that $\angle BLC=90$ so in triangle $LBC$ we get that $\angle LBC+\angle LCB=90 (I)$...
also in $W$,DT is a diameter wo $\angle TCD=90$so $\angle BCD+\angle LCB=90 (II)$
now from $(I),(II)$ we get the result...

[color=indigo]Denote $M=AK\cap CL$, so $M\in DO$

In triangle $MAD$, we have: $\widehat{MAD}=90^\circ,\,O\in MD,\,OD=OA,$ so $OD=OA=OM$ thus $ADCM$ is cyclic.

So $\widehat{MCD}=90^\circ\Longrightarrow CD\perp CL\Longrightarrow AB\parallel CD$
Hence $ABCD$ is a parallelogram.[/color]


integration
trigonometry
calculus
calculus computations

I've posted a question similar to this before, but I thought this one was slightly different, and merited posting, if only to confirm or deny my answer  :lol: .

Calculate $\int^{\pi}_{0}{|\sqrt{2}\sin{x}+2\cos{x}|}\,dx$

[hide="Hint"]
The method for the question could get slightly messy, with $\cos{\left(\tan^{-1}{\sqrt{2}}\right)}$ cropping up, but bear with it, they will soon disappear.
[/hide]

(1) use $\sqrt{2}\sin x+2\cos x=\sqrt{6}\sin(x+\alpha)$ where $\tan\alpha =\sqrt{2}$

(2) break the region into $0\to \pi-\alpha$ and $\pi-\alpha\to \pi$

(3)integrate $\int\sin(x+\alpha)dx$ for the two cases and combine

(4)finally we get $2\sqrt{6}$


complex numbers
complex analysis
complex analysis unsolved

Let $T$ be the set of all of  the complex number $w$ which can not be expressed by $w=z^2-2z$ for any complex numbers $z$ such that $|z|>\frac{5}{4}$,
that is to say, suppossed that

\[T=\left\{w\ |w=z^2-2z\Longrightarrow |z|\leqq \frac{5}{4}\right\}\]

Find the value of $w\in T$ such that $|w|$ is maximized.

Your two statements contradict each other. Perhaps "can" in the first should be "cannot"?

Yes,you are right, jmerry.I will edit. :blush:

Let $ a,b$ be the two roots of $ z^2\minus{}2z\minus{}w \equal{} 0$. By assumption $ |a|,|b| \leq \frac{5}{4}$, and $ a\plus{}b \equal{} 2$, $ w \equal{} \minus{}ab$. We want to maxmize $ |w|$.
Of course $ |w| \leq \frac{25}{16}$, and we only want to know if this can be reached. This occurs when both $ a,b$ lies on $ |z| \equal{} \frac{5}{4}$. Any such $ a,b$ satisfying $ a\plus{}b \equal{} 2$ yields an isoceles triangle with sides $ \frac{5}{4}, \frac{5}{4}, 2$ with the line segment from $ 0$ to $ 2$ as the base. It is then clear that $ a,b$ must be the intersections of the line passing through $ 1$ and parallel to $ y$-axis with $ |z| \equal{} \frac{5}{4}$. In particular, they are conjugate to each other, so $ w \equal{} \minus{} \frac{25}{16}$ when $ |w|$ is maximized.


Find a ( the only,as far as I know) 9-digit number, having the digits 1,2,...,8,9 each once ,in any order such that,
If a1a2a3...a9 is the number,then,
the number (a1a2...ai) should be divisible by i.
ie. (a1a2a3a4a5a6a7)  should be divisible by 7.

eg.
instead of 9,if it were 6,
then 321654 would be an answer as 1 divides 1, 2 divides 2, 3 divides 3, ..6 divides 321654
123654 would also be an answer.

Clearly, $ a_5\equal{}5$ and $ \{a_2,a_4,a_6,a_8\}\equal{}\{2,4,6,8\}$ and therefore $ \{a_1,a_3,a_7,a_9\}\equal{}\{1,3,7,9\}$. This leaves us to check the following sufficient and necessary criteria:

a) $ 3|a_1\plus{}a_2\plus{}a_3$
b) $ 4|2a_3\plus{}a_4$
c) $ 3|a_4\plus{}2\plus{}a_6$
d) $ 7|a_7\plus{}3a_6\plus{}3\plus{}6a_4\plus{}4a_3\plus{}5a_2\plus{}a_1$
e) $ 8|4a_6\plus{}2a_7\plus{}a_8$

a),c) together with the initial restrictions give $ a_1\plus{}a_2\plus{}a_3\in\{6,12,18,24\}$ and $ a_4\plus{}a_6\equal{}10\Longrightarrow a_2\plus{}a_8\equal{}10$. Now b),e) yield $ \frac{a_4}{2},\frac{a_8}{2}$ odd and therefore, $ (a_2,a_4,a_6,a_8)\in\{(4,2,8,6),(8,6,4,2)\}$. Substitution of the two even digit vectors into e) requires $ a_7\equal{}1,9$ or $ a_7\equal{}3,7$ corresponding to the two cases.

Substitution of the first vector into a) yields $ 3|1\plus{}a_1\plus{}a_3\Longrightarrow a_1\plus{}a_3\in\{8,14\}$ and substitution into d) gives $ 7|3\plus{}a_1\plus{}4a_3\plus{}a_7\Longrightarrow a_1\plus{}4a_3\plus{}a_7\in\{18,32,46\}$. If $ a_7\equal{}9$ holds, we have $ a_1\plus{}a_3\equiv a_1\plus{}4a_3\plus{}a_7\bmod{3}$, which requires $ a_1\plus{}4a_3\equal{}23\Longrightarrow a_3\leq 3\Longrightarrow a_1\geq 11$, a contradiction. If $ a_7\equal{}1$ holds, we similarly get $ a_1\plus{}4a_3\equal{}17\Longrightarrow a_3\equal{}3\Longrightarrow a_1\equal{}5$, a contradiction.

Therefore, the second vector must be valid, i.e. $ (a_2,a_4,a_6,a_8)\equal{}(8,6,4,2)\,,\,a_7\equal{}3,7$. Here, we similarly conclude $ a_1\plus{}a_3\in\{4,10,16\}$ and $ a_1\plus{}4a_3\plus{}a_7\in\{14,28,42\}$. If $ a_7\equal{}3$ holds, we have $ a_1\plus{}a_3\equiv a_1\plus{}4a_3\plus{}a_7\bmod{3}$, which requires $ a_1\plus{}4a_3\equal{}25\Longrightarrow a_3\leq 3\Longrightarrow$ $ a_1\geq 13$, a contradiction. Hence, we must have $ a_7\equal{}7$, i.e. $ a_1\plus{}4a_3\in\{7,21,35\}$. Working modulo $ 3$ we can reduce to $ a_1\plus{}4a_3\equal{}7\Longrightarrow$ $ a_3\equal{}1,a_1\equal{}3$. [b]Putting all together this implies the only solution given by the number $ 381654729$.[/b]

@hjbrasch
Wow,You made it look so professional. :)
I had done it using some 4-5 cases :oops: during a social class last year.[/list]


Let p is a prime number .
$ S\equal{}\{a\in F_{2p}| 2p|a^{a\minus{}1}\minus{}1\}$
Does there exist finite or infinite prime p such that : $ |S|\equal{}2$
We easy to prove that there exist infinite prime p such that $ |S|>2$ but what will happen in the other case ?

[quote="TTsphn"]Let p is a prime number .
$ S \equal{} \{a\in F_{2p}| 2p|a^{a \minus{} 1} \minus{} 1\}$
[/quote]
Probably just me, but I have no idea what that's supposed to mean.

It is not just you :)

TTsphn..can you explain your problem ?
Thank´s

Daniel

Maybe it's not written in the clearest way but it's rather easy to understand. $ S$ is a set of residues mod $ 2p$ such that $ a in \S$ is equivalent to $ 2p|a^{a \minus{} 1} \minus{} 1$. Does there exist infinitely many primes $ p$ such that $ |S| \equal{} 2$? $ |S| \geq 2$ because always $ 1, \minus{} 1 \in S$.


IF

2+1+3+4 = 40
5+4+2+4 = 90
3+2+5+8 = 126
4+3+6+7 =  180

then 7+8+1+10 = ????

That is the whole info i have . Nothing more is provided .

Kindly provide me the solutions asap.

Just messed around with the problem, then I found something.

$ a \plus{} b \plus{} c \plus{} d \equal{} (a \plus{} b \plus{} c \plus{} d)(b \plus{} c)$

So then, $ 7 \plus{} 8 \plus{} 1 \plus{} 10 \equal{} 26 \times 9 \equal{} \boxed{234}$.

I suppose you could turn this into a system, but I used guess and check and found the equation fairly quickly.

Don't double-post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1750675#1750675


solve equation   $ 4x^{6}\minus{}6x^{2}\plus{}2\sqrt{2}\equal{}0$

$ 4x^6\minus{}6x^2\plus{}2\sqrt{2}\equal{}0$

$ \Longleftrightarrow (x^2\plus{}\sqrt{2})(\sqrt{2}x^2\minus{}1)^2\equal{}0$

$ \therefore x\equal{}\pm \sqrt[4]{2}i,\ \pm \frac{1}{\sqrt{2}}$.


Someone says that he can divide any blot by cross in 4 parts. And the parts will have the same areas. Does he tell lie?

Note: cross - a pair of straight lines intersecting at right angles
blot - a figure bounded by a closed curve without intersections.

It's a "challenge", not a homework.

--------------------------------------------

I only have found the obvious fact that it is possible to divide every blot into 2 pairs of parts in which 2 parts will have the same area.

I mean, if you divide the blot independently 2 times, every time by single straight line you'll get for parts a,b,c,d.
Then
a+b=c+d
a+c=b+d

So b=c and a=d. How can I go on and prove that it's possible (or impossible) to divide any blot this way so that a would be equal to c

This, which has already been solved in the math literature,                                                                                may be more appropriately placed in a "Geometry" section.

I understand that a blot is a [2-dimensional] closed curve that does not intersect with itself.

A "challenge" could be you know it and you're seeing if at least one of us can get it, or you don't know it (at least not all of it) and you want someone to get it.

I am stuck with this problem. If it has been solved already, then please give me the link.


The Number 72 is a product of two  consecutive numbers natural numbers (8*9)

what is the greatest between 0 and 1000  that is also the product of two consecutive

numbers?

[hide] take the square root of 1000 which is appoxamitely 31.6, then try out two equations:

30*31=930
31*32=992

Of these 31*32 is bigger.[/hide]


Can anyone suggest for me a high school in US that have a really good curriculum in Math, I'm not an American citizen so that I think I can not go to public high school. Anyone help me please. And I don't see any topic about high school so I don't know where I can post this, sorry about that.

As I understand it, you can attend a public school provided that you have a student Visa.  What are your circumstances, and how much do you know about the American system?

The schools that have good math curriculums - [url=http://www.exeter.edu/]Philips Exeter[/url] comes to mind - tend to be private and expensive.  Regrettably, it's rare to find a normal public high school that excels at any particular academic subject.  You'd be looking at [url=http://en.wikipedia.org/wiki/Magnet_school]magnet schools[/url], charter schools, or plain old private schools here.

This is a somewhat unusual request; the majority of international students studying in America usually begin their study at the college level.  Again, it would be useful to know what circumstances are prompting this question.

Actually, I'm studying in US, I'm in $ 11^{th}$ grade but I don't like the way they taught math so I want to find another school that offered Olympiad math.

I don't think any schools in the US have Olympiad classes (in fact, half of them probably don't know what Olympiad math is), unfortunately.

So why US offered the AMC if half of the school in US don't know what it is? And I have signed up for the AMC 12 but I don't know the admissions of the high school is usually in $ 15^{th}$ Jan, the contest is after that so how can I give them the AMC scores or I really don't need it.

I don't know if Philips Exeter actually has an Olympiad class, but Zuming Feng trains his team very hard and very well, from what I've heard, so that's probably the closest you're going to get.

With the exception of private schools, US high schools do not really have an admissions process, so you don't need to worry about your scores unless you're trying to get into a very prestigious school, in which case you might run into other problems (money, citizenship).  You'd have to do some more research.

As for your first question, US education is very decentralized.  The AMC has no official relation to any of the political bodies that actually determine what high schools do; it's an entirely separate entity.  And it's an unfortunate truth that many schools just don't care that much about math.

1.  Most public school systems in the US will not check your immigration status for school enrollment.  The only exceptions would be places with a xenophobic voter base.  (Now that I think of it, that's actually a big chunk of the US.)  But if you're here legally, then you definitely have no problems.

2.  You didn't mention where you live or any geographical constraints.  Does that mean you're only interested in boarding schools?  Or is your family willing to move anywhere in the country?

3.  You're already in 11th grade.  Isn't it a bit late to be choosing a high school?  I know almost nothing about elite private schools, but I would be surprised if they took in new students as seniors.  (Just doesn't seem like the kind of thing they would do.)

4.  If you want to learn more advanced math than your high school offers, consider taking a college class.  As for "Olympiad math," it's not something you usually find in a high school math class in this country.  Depending on where you live, there might be some extracurricular opportunities, but otherwise, just do what everyone else does: use AoPS.

[quote="yenlee"] I know almost nothing about elite private schools, but I would be surprised if they took in new students as seniors.  (Just doesn't seem like the kind of thing they would do.)
[/quote]
Some of the ones I know of do, but you have to pass a fairly hard entrance test for that. As to "olympiad math.", it is done on the extracurricular basis almost everywhere, not only in the US. See if you can find some math. circles or similar activities in your local area. If not, just hang on AoPS, as yenlee suggested. By the way, by the 11th grade, it is usually time to start combining solving olympiad problems with reading some serious math. books and taking college type classes. The general level of math. education in the US is well below the European standards, so you'll have to rely on self-study quite a lot if you want to get anywhere and it is never too early to start.

Note also that all the advices above are given assuming that your words "I don't like the way they teach math." mean "I'm getting straight A's on every test and am desperately bored with the triviality of the curriculum". If they mean "I cannot understand what my teacher talks about and fail almost every exam", the story will be rather different. ;)

[quote="fedja"]By the way, by the 11th grade, it is usually time to start combining solving olympiad problems with reading some serious math. books and taking college type classes. [/quote]

This is a very good point, but math is not the only option -- I suggest checking out computer science, physics, chemistry, etc.  It may not be math that you're really interested in.  Olympiad-level problem-solving is an excellent on-ramp to thinking deeply about many different subjects, and some point you have to "graduate" yourself past the Olympiad problems into a subject area you are really passionate about.  It might not be math.  (It might even be education :) )


triangle ABC is isosceles. AB = AC.  If CD is perpendicular to AB, show that angle DCB = 1/2angle A

This problem wasn't too bad:

[hide]We can write angle DCA=180-90-A=90-A, and we can write that angle DCB + angle DCA = angle C, so angle DCB=C-90+A.  So in order to show that angle DCB is equal to 1/2 angle A.  C-90 must be equal to -1/2A.  So multiplying by two we get 2C-180=-A.  Subtracting a C from both sides and adding a 180 we get C=180-A-C.  Since angle B is equal to angle C.  We see this is obviously an equality, and angle DCB is equal to 1/2 angle A.[/hide]

[hide="pretty quick solution"]

Let $\angle DCB = \alpha$ and $\angle ABC = \angle ACB = \beta$. We wish to show that $2\alpha = \angle A$.

From right triangle $\triangle CDB$, we have $\alpha+\beta=90$.
From rigth triangle $\triangle ADC$, we ahve $\beta-\alpha+\angle A = 90$.

Subtracting the second from the first we get:

$2\alpha-\angle A = 0.$ So $2\alpha = \angle A$. [/hide]


Given that f(x+y) =(e^x)*f(y) +(e^y)*f(x) for all x and y.
Find f(x).

$g(x) = f(x)e^{-x}$ and we have cauchy equation  :)
if there is no "good" conditions on the $f$  then solution use Hamel basis... :?

[quote="anik_andrew"]$g(x) = f(x)e^{-x}$ and we have cauchy equation  :) [/quote]

Sorry, I don't see it  :oops:
Could you explain more ?


x,y,z>=0 satisfying x+y+z=1, find the maxinum of $\sum{\sqrt{\frac{1-x}{1+x}}}$

generalization: forall $n\geqslant3$ and $a_{1},a_{2},\ldots,a_{n}$ such that $a_{1}+a_{2}+\cdots+a_{n}=1$ the maximum of
$\sum_{i=1}^{n}\sqrt{\frac{1-a_{i}}{1+a_{i}}}\leqslant n-2+\frac{2}{\sqrt{3}}$
The equality holds when $a_{1}=a_{2}=\cdots=a_{n-2}=0,a_{n-1}=a_{n}=\frac{1}{2}$. But the proof I know isn't easy at all.

I have found a method by Majorization Inequality, it is not hard so I wouldn't post it. The solution is left for you! :)


16 college students are going to the beach in 4 identical vans. Each van can hold exactly four students.  How many ways can we distribute the students in the vans? (Two distributions are different if there are two students who ride together in one distribution but not in the other. If the same groups of students are together, it does not matter which van they ride in .)

(A) 16! (B) (4!)^3 (C) $ (\binom{16}{4})^{3}$ (D) $ \binom{16}{4}*\binom{12}{4}*\binom{8}{4}$ (E) none of the above

[hide="My solution that seems to be wrong"]I got that it was (D) divided by 4!, but online it says that it is (D).  My reasoning for this is that you can chose four at a time, but if we do it like that, we are saying the vans they go in matter.  But it doesn't so it has to be divided by the number of orders of the groups in the vans, so (D) divided by 4!
[/hide]

Please help
Thanks

Your answer looks correct. It looks like its their mistake. I guess they meant to say the vans were different.

That is weird. I just checked the Alabama Math Tournament website, and the answer is indeed (D).

Let's consider a smaller version of the same problem. 4 college students trying to squish into 4 identical vans, which cannot have more than one occupant.

By common sense, the answer should be one. A/B/C/D, A/B/D/C.... They are all the same because the vans are [b]identical[/b].
However, $ \dbinom{4}{1}\dbinom{3}{1}\dbinom{2}{1}\neq1$.

You should check with whoever makes this tournament

Okay, thanks very much for the clarifications.


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed
109

Aren't these inequalities a little bit expired. I mean: Japan 1997 (the nicest of all), USAMO, and now this one. And I bet the same breaking method works here. Or am I wrong?

If $x,y,z$ are nonnegative real numbers such that $x+y+z=3$, then

$\frac{2}{3}\sum\frac{(4x-3)^{2}}{x^{2}-2x+3}\geq\sum{2x^{3}+xyz \over (2x+y)(2x+z)}\geq1$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=132530

[quote="pvthuan"]Prove that if $ a\plus{}b\plus{}c \neq 0$ then

$ \frac {(b \plus{} c \minus{} 3a)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(c \plus{} a \minus{} 3b)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{}\frac {(a \plus{} b \minus{} 3c)^2}{2c^2 \plus{} (a \plus{} b)^2} \geq \frac12$.[/quote]$ 2\left(a^2 \plus{} b^2 \plus{} c^2\right)(b \plus{} c \minus{} 3a)^2 \minus{} \left[2a^2 \plus{} (b \plus{} c)^2\right]\left(9a^2\minus{}4b^2\minus{}4c^2\plus{}12bc\minus{}6ca\minus{}6ab\right)$

$ \equal{}(b\minus{}c)^2\left[19a^2 \minus{} 6a(b \plus{} c) \plus{} 6(b \plus{} c)^2\right]\geq 0\Longrightarrow$

$ \sum\frac {(b \plus{} c \minus{}3 a)^2}{2a^2 \plus{} (b \plus{} c)^2}\geq\sum\frac {9a^2\minus{}4b^2\minus{}4c^2\plus{}12bc\minus{}6ca\minus{}6ab}{2\left(a^2 \plus{} b^2 \plus{} c^2\right)} \equal{} \frac {1}{2}$.

A generalization see : http://www.mathlinks.ro/Forum/viewtopic.php?t=28678

Yes, nice result Jichen. Can you come up with it without the help of computers?


irrational number
complex numbers
number theory open
number theory

Let $x\in\mathbb{R}$ such that $2^x\in \mathbb{N}$ and $3^x\in \mathbb{N}$. Is it true that $x\in\mathbb{Q}$? All I know is that $x$ cannot be irrational algebraic number.

$2^x \in \mathbb{N}$  &  $3^x \in \mathbb{N}$,

$\rightarrow \left(\frac{2}{3}\right)^x \in \mathbb{Q}$

And we have a nice little theorem someone proved, that a rational number to the power of an irrational number is a transendental number, so if x is irrational we have a contradiction. QED

Unless anyone cares to point out a careless mistake?

I think the problem was already posted and was seen to be a really open problem.

I will point out the careless mistake:
the theorem you states is wrong, it would imply that $log_2(3)$ is rational... (did you mean http://mathworld.wolfram.com/GelfondsTheorem.html ¿)

My bad, yeah thats the one i was thinking of....

The problem is a special case of the "four exponentials conjecture"

If  x1,x2  are complex numbers linearly independent over the rationnal and y1,y2 satisfies same hypothesis then at least one amoung  exp(xiyj)  should be transcendent over the rationnal.

All known result is 6 exponential theorem with y1,y2,y2 instead of y1,y2.

lolo


Hey, I would like  to see some nice solutions of this problem :
Prove that
$ a\plus{}b\plus{}c\equal{}<abc \plus{}2$ fi $ a^2\plus{}b^2\plus{}c^2\equal{}2$ , $ a,b,c \in R$

Good luck !

It is enough to prove the inequality only for max$ (a,b,c) \ge 0$ because otherwise, if $ a,b,c<0$ then from $ a\minus{}b\minus{}c \le abc\plus{}2$ we can deduce that $ a\plus{}b\plus{}c \le abc\plus{}2$.

So, suppose that max$ (a,b,c) \ge 0$. W.L.O.G. we can assume that max$ (a,b,c)\equal{}a$. Now, we analyse the case when $ a>0$.

$ a^2\plus{}b^2\plus{}c^2\equal{}2 \Rightarrow a^2\equal{}2\minus{}b^2\minus{}c^2$

$ a\plus{}b\plus{}c \le abc\plus{}2 \Leftrightarrow a(1\minus{}bc) \le 2\minus{}b\minus{}c ,(1)$

Since $ 2>b^2\plus{}c^2 \ge \frac {(b\plus{}c)^2} {2} \Rightarrow 2>b\plus{}c \Rightarrow 2\minus{}b\minus{}c>0$.

$ 2>b^2\plus{}c^2 \ge 2bc \Rightarrow 1\minus{}bc>0$.

$ (1) \Leftrightarrow a \le \frac {2\minus{}b\minus{}c} {1\minus{}bc} \Leftrightarrow a^2 \le \frac {(2\minus{}b\minus{}c)^2} {(1\minus{}bc)^2} \Leftrightarrow 2\minus{}b^2\minus{}c^2 \le \frac {(2\minus{}b\minus{}c)^2} {(1\minus{}bc)^2},(2)$

After reductions, $ (2)$ becomes:

$ 2b^2c^2\plus{}2b^3c\plus{}2bc^3 \le 2(b\minus{}1)^2\plus{}2(c\minus{}1)^2\plus{}6bc\plus{}b^4c^2\plus{}b^2c^4$

It is enough to prove that:

$ 2bc\plus{}2b^2\plus{}2c^2 \le 6\plus{}b^3c\plus{}bc^3 \Leftrightarrow (b^2\plus{}c^2)(2\minus{}bc) \le 6\minus{}2bc$.

But $ 1>bc$ and $ b^2\plus{}c^2<2$ so:

$ (b^2\plus{}c^2)(2\minus{}bc)<2(2\minus{}bc)\equal{}4\minus{}2bc$

It is obvious that $ 4\minus{}2bc<6\minus{}2bc$

If $ a\equal{}0$ then $ b^2\plus{}c^2\equal{}2$ and we have to prove that $ b\plus{}c \le 2$

This last one results from $ (b\plus{}c)^2 \le 2(b^2\plus{}c^2)\equal{}4$

So, the equality case is when $ (a,b,c)\equal{}\{ (0,1,1),(1,0,1),(1,1,0) \}$.


From a bag of 11 balls, the balls are numbered from 1 to 11.  6 balls are being randomly selected simultaneously.  What is the probability that the sum of the 6 balls is odd?

[hide]there are 5 even numbers and 6 odd numbers
also to get an odd number, you have to draw
OEEEEE
OOOEEE
OOOOOE

we have to take into account the different orders, so multiplying by $\binom{6}{n}$

$\frac{ \binom{6}{1}*(6)*(5*4*3*2*1)+\binom{6}{3}*(6*5*4)*(5*4*3)+\binom{6}{1}*(6*5*4*3*2)*(5) }{ 11*10*9*8*7*6}=\frac{118}{231}$

i am not sure if that is right, it did not work out that well...[/hide]


Let $x^2+y^2+z^2 \le 27$, then find minimum and maximum value of the expression:
\[ A=x+y+z+xy+yz+zx  \]

let me guess

$36\geq A\geq -14$ ?

the minimum can be found by setting $x+y+z=p$ , $xy+xz+yz=q$ so from condition $p^2-2q\leq 27$

Hence $p+q\geq p+\frac{p^2-27}{2}=\frac{(p+1)^2-28}{2}\geq -14$

to maximise , it is sufficient to consider $x,y,z\geq 0$ .

since $27\geq x^2+y^2+z^2\geq xy+xz+yz$ and $27\geq x^2+y^2+z^2\geq \frac{1}{3}(x+y+z)^2$ , combine both we get $A\leq 36$

 :)

[quote="shyong"]
Hence $p+q\geq p+\frac{p^2-27}{2}=\frac{(p+1)^2-28}{2}\geq -14$

 :)[/quote]

I didn't get that part.....

it follows since $(p+1)^2$ is at least 0.

[quote="shyong"]
Hence $p+q\geq p+\frac{p^2-27}{2}$
 :)[/quote]

No i meant that part

But, that part is obvious from what he's written.. :? :
[quote="shyong"]...
$p^2-2q\leq 27$
...
[/quote]
So, from this we get : $p^2 - 27 \leq 2q \iff q \geq \frac{p^2 - 27}2$... :)


inequalities
combinatorics unsolved
combinatorics

Recall that the [i]independece number [/i] $ \alpha (G)$ of a graph $ G$ is the maximum number of pairwise nonadjacent vertices of $ G$. If $ G$ is a graph with $ n$ vertices and $ \frac {nk}{2}$ edges, $ k\geq 1$, prove that $ \alpha (G) \geq \frac {n}{k\plus{}1}$.

color the $ \frac {nk}{2}$ given edges blue, and complete the graph by drawing in the missing edges and coloring them red. the number of red edges is $ \frac {n(n \minus{} 1) \minus{} nk}{2} \equal{} \frac {n(n \minus{} k \minus{} 1)}{2}$. but, if $ \alpha(G) < \frac {n}{k \plus{} 1}$, we get by turan's theorem that the number of red edges is at most

\[ \frac {\frac {n}{k \plus{} 1} \minus{} 2}{\frac {n}{k \plus{} 1} \minus{} 1}\cdot\frac {n^2}{2} \equal{} \frac {n \minus{} 2k \minus{} 2}{n \minus{} k \minus{} 1}\cdot \frac {n^2}{2} < \frac {n(n \minus{} k \minus{} 1)}{2},
\]
a contradiction.

to see why the above inequality is true, note that it is equivalent to $ n(n \minus{} 2k \minus{} 2) < (n \minus{} k \minus{} 1)^2$, which
holds by AM-GM.


For some reason, the video isn't fitting, it is too large so i can't see some of the stuff on the far right.

Also, I don't see the pause/play button or how to jump a section.

How big is your browser window? It needs to be at least 720 pixels wide. The video player is about 720 x 540 in dimensions.


summer program
PROMYS
vector
linear algebra

Hey Im a rising junir attending promys. I dont want to study more NT before the program b/c ill have to relearn it so im teaching myself linear algebra. I bought the book LINEAR ALGEBRA a purea and applied first course by edgar goodaire.
opinions?

According to past PROMYS yearbooks, Halmos's [i]Finite Dimensional Vector Spaces[/i] is the best. Things with "applied" (or variants such as "with applications") in the title are generally not so good.


inequalities
linear algebra
matrix
linear algebra unsolved

Let $ a_1,a_2,..,a_n$ be real numbers ($ n \in \mathbb{N},n \geq 2$) such that : $ a_n > \sum_{k \equal{} 1}^{n \minus{} 1}{a_k^2}$
Let $ A,B \in M_n\mathbb{(R)}$ such that :
  $ A \equal{} \begin{bmatrix}1 & a_1 & a_2 & ... & a_{n \minus{} 1} \\
a_1 & a_n & 0 & ... & 0 \\
a_2 & 0 & a_n & ... & 0 \\
... & ... & ... & ... & ... \\
a_{n \minus{} 1} & 0 & 0 & ... & a_{n}\end{bmatrix}$
And :
 $ A(B \plus{} B^T \minus{} I_n)^{2n \plus{} 1} \plus{} (B \plus{} B^T \minus{} I_n)^{2n \plus{} 1}A \equal{} 2I_n$
 $ A\otimes{(B \plus{} B^T \minus{} I_n)^{2n \plus{} 1}} \plus{} (B \plus{} B^T \minus{} I_n)^{2n \plus{} 1}\otimes{A} \equal{} 2I_{n^{2}}$
Prove that :
$ \det(B \plus{} xI_n)\det(B \plus{} yI_n)\det(B \plus{} zI_n) \geq \det(B \plus{} \sqrt [3]{xyz}I_n)^3$ , $ \forall (x,y,z) \in \mathbb{R_{ \plus{} }}^3$
 $ I_k$ is $ k$x$ k$ identity matrix
 $ B^T$ is transpose of $ B$

Correct : $ A\equal{}I_n$  and $ B\plus{}B^T\equal{}2I_n$ are all of this problem ...  :wink:


rotation
geometry
geometric transformation
trigonometry
circumcircle
inequalities
geometry theorems

Let $P$ a point in the plane of an equilateral triangle $ABC$ with a side $s$. Show that :

 $3(|PA|^{4}+|PB|^{4}+|PC|^{4}+s^{4})=(|PA|^{2}+|PB|^{2}+|PC|^{2}+s^{2})^{2}$

I think I found a sketch of the proof in the case that $P$ is inside the triangle $ABC$. Let $PA=a_1,PB=a_2,PC=a_3$. Rotate the triangle $ABC$ from the point $A$ 60 degrees such that $AB$ maps to $AC$. Denote the image of $P$ after the rotation by $Q$.

Now we see that $AP=PQ=AQ=a_1$ and $a_2=PB=CQ$. By the law of cosine to the triangle $PCQ$ we have

$a_3^2=a_1^2+a_2^2-2a_1a_2\cos CQP$ so  that $\cos CQP=\frac{a_1^2+a_2^2-a_3^2}{2a_1a_2}$.

Now

\begin{eqnarray*} s^2=AC^2&=&a_1^2+a_2^2-2a_1a_2\cos (CQP+60)\\ &=&a_1^2+a_2^2-2a_1a_2\left (\frac{1}{2}\cos CQP-\frac{\sqrt{3}}{2}\sin CQP\right )\\ &=&a_1^2+a_2^2-a_1a_2\left (\cos CQP-\sqrt{3}\sin CQP\right )\\ &=&a_1^2+a_2^2-a_1a_2\left (\frac{a_1^2+a_2^2-a_3^2}{2a_1a_2}-\frac{\sqrt{3}}{2}\sqrt{1-\left (\frac{a_1^2+a_2^2-a_3^2}{2a_1a_2}\right )^2}\right )\\ &=&a_1^2+a_2^2-\left (\frac{a_1^2+a_2^2-a_3^2}{2}-\frac{\sqrt{3}}{2}\sqrt{a_1^2a_2^2-\left (\frac{a_1^2+a_2^2-a_3^2}{2}\right )^2}\right )\\ &=&\frac{1}{2}a_1^2+\frac{1}{2}a_2^2+\frac{1}{2}a_3^2+\frac{\sqrt{3}}{4}\sqrt{4a_1^2a_2^2-(a_1^2+a_2^2-a_3^2)^2}\\ \end{eqnarray*}
Now it is hopefully a straightforward simplification to verify the formula you gave. However, I'm too lazy to do so heavy calculations or too blind to see some obvious simplification.

[hide="Here is a short proof of this identity, a generalization and some interesting inequalities."]Denote $AB=a$, $PA=x$, $PB=y$, $PC=z$ and the point $R$ so that $AR=RP=PA=x$ and $AP\cap BR\ne \emptyset$. $\triangle ABP\equiv \triangle ACR$ $\Longrightarrow$ $CR=PB=y$. Denote $m(\widehat {CPR})=\phi$ and the area $\sigma\equiv \sigma (x,y,z)$ of the [u]Pompeiu's triangle[/u] $CPR$.
$\blacktriangleright\ 2xz\cos \phi=x^2+z^2-y^2\ ;\ xz\sin \phi=2\sigma.$
$\blacktriangleright \boxed {\ 16\sigma ^2=2(x^2y^2+y^2z^2+z^2x^2)-(x^4+y^4+z^4)\ }\ .\ \ (1)$
$\blacktriangleright AC^2=PA^2+PC^2-2\cdot PA\cdot PC\cdot \cos (60^{\circ}+\phi )\Longrightarrow$
$a^2=x^2+z^2-\frac 12\cdot 2xz\cos \phi +\sqrt 3\cdot xz\sin \phi\Longrightarrow\boxed {\ 2a^2=x^2+y^2+z^2+4\sigma\sqrt 3\ }\ .\ \ (2)$

[b]Proof.[/b] The proposed identity becomes :
$3(a^4+\sum x^4)=\left( a^2+\sum x^2\right)^2\Longleftrightarrow\boxed {\ a^4+\sum x^4=a^2\sum x^2+\sum x^2y^2\ }\ .\ \ (3)$
From the relations $(1)$ and $(2)$ we will eliminate the area $\sigma.$ Therefore,
$\left(x^2+y^2+z^2-2a^2\right)^2=3\cdot 16\sigma ^2=3\left(2\sum x^2y^2-\sum x^4\right)\Longleftrightarrow$
$\sum x^4+2\sum x^2y^2+4a^4-4a^2\sum x^2=6\sum x^2y^2-3\sum x^4\ .$
This last relation is equivalently with the proposed identity $(3)\ .$

[b]A generalization.[/b] Let $ABC$ be an any triangle and a inner point $P$.
Denote $PA=x$, $PB=y$, $PC=z$ and the circumradius $R$. Then
$\boxed {\ 2\sum a^2b^2x^2y^2-\sum a^4x^4=4R^2\cdot \left(abc-\sum ax^2\cos A\right)^2\ }\ .$

[b]A particular cases[/b] :
$\blacktriangleright\ x=y=z=R\Longrightarrow \sum a\cos A=\frac{2S}{R}\ ;$
$\blacktriangleright\ a=b=c\Longrightarrow 2\sum x^2y^2-\sum x^4=\frac 13\cdot \left(2a^2-\sum x^2\right)^2\Longrightarrow$
$a^4+\sum x^4=a^2\sum x^2+\sum x^2y^2,$ i.e. the identity $(3)\ .$

[b]A interesting consequencies.[/b]

$\blacktriangleright$ From the [u]Cebasev's inequality[/u] results :
$3\cdot \left(a^4+\sum PA^4\right)=\left(\sum PA^2+3\cdot \frac{a^2}{3}\right)^2\le 6\cdot \left(\frac{a^4}{3}+\sum PA^4\right)\Longrightarrow$
$3a^4+3\cdot \sum PA^4\le 2a^4+6\cdot \sum PA^4\Longrightarrow \boxed {\ a^4\le 3\cdot \sum PA^4\ }\ .$

$\blacktriangleright$ From the [u]Weitzenbock's inequality[/u] $PA^2+PB^2+PC^2\ge 4\sqrt 3\sigma$

and the relation $(2)$ results : $\boxed {\ 4\sigma \sqrt 3\le a^2\le PA^2+PB^2+PC^2\ }\ .$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=71773 [/hide]


Let $ K \subset \mathbb{C}$ be a compact set and assume that $ K \neq \emptyset$. Prove that there exists an operator $ T \in B(l_2)$ such that $ \sigma(T) \equal{} K$.
Try a multiplication operator $ T(\eta_1,\eta_2,...) \equal{} (a_1\eta_1,a_2\eta_2,...)$

[hide="hint"]If $ T$ is given like that then $ \sigma(T)\equal{}\overline{\{a_k: \;k\geq1\}}$[/hide]


AMC
USA(J)MO
USAMO
email
summer program
PROMYS
Ross Mathematics Program

As usual, MA will be very healthily represented at MOP. But I find our distribution across the groups quite amusing. Congratulations to everyone below -

Red: Xiaoyu, Elizabeth (CGMO)
Blue: Don't know of anyone yet
Black: Vlad, Sergei, Wenyu (though he doesn't really count)

There may be more CGMO people, Wenyu told me a while ago that there was someone at Andover that had a decent shot at making it. Also, George Silvis I hear was pretty close to Blue last year (as a sophomore), he could have made it this year.

Eh Alan isn't on the list. =(

Also Ofir Nachum from MA made blue.

[quote="not_trig"][quote="jmerry"]So western Mass doesn't count either? Panupong Pasupat (USAMO winner) is also from the state.[/quote]

but he's not a citizen or permanent resident, so he can't go to MOP.[/quote]
Really? That's a sad news for Toan too :( . Hey, I thought US is anti-racist?! So, only US citizen or permanent residents can make it to MOP?

It's a completely reasonable policy, technically the whole point of MOP is to train people for the IMO team, and non-citizens/permanent residents cannot be on the IMO team (IMO rules, presumably). And also, most countries have their own training programs anyway.

wait yongyi didn't u get 11 $ \implies$ red?
Congrats Vlad black wow...don't forget your tennis game

Indeed, but... it is quite a long story (well, not [i]that[/i] long, but whatever).

okay, short story short (because it's actually a short story)
his day 1 papers weren't recieved so 11 is his "unofficial" official score with his "late" day 1 scores, so his 11 was disqualified? i think that's what happened

Hmmm...
I wonder if he would have gotten in if he got 8 just on the second day even if his first day was disqualified...?
That would be an interesting situation.

I do not think one additional "official" 8 would raise the cutoff to 9 :P (although a 14 would be pretty beastly).

What I'm saying is that is your "half-score" considered really official or are you as a contestant completely disqualified?

As far as I can tell, I'm not disqualified... so my half score is official...

dude yongyi you should email the amc people and ask them what the situation is
because i think you still might be in mop, depending on what the scoring is like

Blah I bet they still won't let me in because consider the situation where I have already applied for _______ (replace with a summer program that starts before July 3), and then them notifying me late that I have qualified will cause lots of problems.

Anyway, the AMC people intimidate me... (see the USAMO teacher's manual, especially the sentences containing the words "final" and "cannot be disputed", etc...)

Maybe I should've been extra beast this year and gotten both #4 and #6 :idea:

[quote="Yongyi781"]Blah I bet they still won't let me in because consider the situation where I have already applied for _______ (replace with a summer program that starts before July 3)[/quote]
promys, ross, HCSSiM, MathCamp?

[quote="serialk11r"]What I thought Toan was at MOP last year.[/quote]

yeah Toan is apparently allowed. maybe he is a permanent resident (?)


quadratics
number theory unsolved
number theory

find all solutions to $ y^{2}=x(x+1)$ over $ \mathbb{F}_{p}=\{0,1,...,p-1\}$, (mod p), or odd prime, $ p$.

EDIT: sorry the questions should ask for the number of solutions not to find the solutions.

:maybe:
[hide="Hint "]Because $ p$ is odd , in $ \mathbb{F}_{p}$ that equation is equivalent to $ (2y)^{2}=(2x+1)^{2}-1$, or $ (2x+1-2y)(2x+1+2y)=1$. [/hide]

What does "find all solutions" mean?

Expliciting them all is clearly impossible...

EDIT= o-ops, you are right edriv, that's an nice way of expliciting all x such that $ x^{2}+k$ is a quadratic residue

Following N.T.TUAN's hint, you can say that all solutions are given by
$ (x,y) = \left( \frac{a+a^{-1}-2}4, \frac{a-a^{-1}}4 \right)$
Hence there is an (explicit) bijection from {1,2,3,...,p-1} to the set of solutions.

[b]im sorry; the question was meant to ask for the number of solutions to the given equation[/b]


calculus
integration
conics
ellipse
calculus computations

I can't recall the exact question, but basically, we were given a line integral.  Only problem is, we weren't given a curve.  The goal was to find the closed, simple curve $ C$ such that the integral was maximized.  I think the integral was something like this:
$ \int_{C}(y^{3}\minus{}y)dx\plus{}(\minus{}2x^{3})dy$
I'm not sure if this was exactly it but a general method would be nice.

You have $ \oint_CA\,dx\plus{}B\,dy.$

Calculate the "scalar curl," $ \frac{\partial B}{\partial x}\minus{}\frac{\partial A}{\partial y}.$

Find out where in the plane that quantity is positive and where it is negative. If there's a bounded region where it is positive, and is negative outside that region, then let $ C$ be the boundary of that region, run counterclockwise.

In other words, rely on Green's Theorem.

I used Green's Theorem to get that far, but I couldn't figure out where to go from there once I evaluated $ \partial N/\partial x$ and $ \partial M/\partial y$.

Suppose it is exactly the problem you posed above. The "scalar curl" is $ \minus{}6x^2\minus{}3y^2\plus{}1.$ This is positive inside the ellipse $ 6x^2\plus{}3y^2\equal{}1$ and negative outside it. If we make the curve $ C$ precisely that ellipse, parameterized counterclockwise, then we maximize the double integral of $ \minus{}6x^2\minus{}3y^2\plus{}1$ on its interior.


Prove that the equation \[ x^{2008}\plus{} 2008!\equal{} 21^{y}\] doesn't have solutions in integers.

[hide]Assume $ x$ and $ y$ are integers which satisfy the given equation. It is easy to see that $ x$ and $ y$ must be positive. Note that $ x$ is divisible by 21. Let $ x\equal{}21n$. Hence the given equation becomes
\[ (21n)^{2008}\plus{}2008!\equal{}21^y
\]
Thus $ y>2008$. Note that both $ (21n)^{2008}$ and $ 21^y$ are divisible by $ 21^{2008}$, but $ 2008!$ is not divisible by $ 21^{2008}$, which is a contradiction. Therefore, there are no solutions in integers.[/hide]

Ok, I did another thing! :)
Nice solution!
Congratulations

Another solution:
First see that $ 21 \mid x$, then $ 3^{1000} \mid 2008!$ so $ y \ge 1000$ but then $ 21^{1000} \mid 2008!$ contradiction.

2009 is prime so by fermat and wilson Lhs divisible by 2009. but 2009 does not divide
\[ \ 21^y
\]

2009 is not a prime :wink:
Please check before posting
 $ 2009\equal{}7^2*41$

Daniel

Another Solution :
Note $3^y||2008!,7^y||2008!$
It's obviously impossible so no solution.

Generalization :
Solve $x^n+n!=(p_1p_2...p_k)^y$ where $p_1,p_2......,p_k$ are some odd primes less than $n$ and $k>1$.


function
calculus
derivative
vector
real analysis
real analysis unsolved

Given f:O -> R of class C(r) (continuously differentiable x r) where O is in R(n) and such that the Jacobian of f has rank 1 on a subset M of O (M is defined as the set of all points for which f(x)=0). If p is in M, suppose
\[\frac{{\partial f}}{{\partial dx_{n}}}\[ \ne 0. \]
Then define F:O->R(n) by
\[F(x) = (x_{1},...x_{n-1},f(x)) \]
The point is to construct a function F with invertible Jacobian such that F(M) is contained in R(n-1)x0. This holds true under the assumption. But I don't see why it is no loss of generality to assume
\[\frac{{\partial f}}{{\partial dx_{n}}}\ne 0. \]
Or otherwise, what is the generalization? Apologies for the bad latex.

I think I might just be dumb.  :blush:

If G(k,l) is the transformation from R(n) to R(n) that transposes the kth and lth components of a vector and if in the problem\[\frac{{\partial f}}{{\partial x_{k}}}\ne 0\] then let F be defined by \[F(x) = (x_{1},...,x_{k-1},f(x),...,x_{n})\]

Then \[G(k,n) \circ F\] has invertible derivative and sends elements in O into the upperhalf space of R(n).

I wish authors would spell out subtleties like this atimes. I missed this when I read it two weeks ago..


Let $ \{a_k\}$ be a sequence of integers such that $ a_1 \equal{} 1$ and $ a_{m \plus{} n} \equal{} a_m \plus{} a_n \plus{} mn$, for all positive integers $ m$ and $ n$. Then $ a_{12}$ is

$ \textbf{(A)}\ 45 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 67 \qquad \textbf{(D)}\ 78 \qquad \textbf{(E)}\ 89$

In general, we can prove that $ a_k\equal{}\frac{ (k)(k\plus{}1)}{2}$

You need to prove that a function exists...it is obvious to use what I said with strong induction.

[quote="JSGandora"][quote="worthawholebean"][hide="Elegant"]
$ a_{m\plus{}1}\equal{}a_m\plus{}1\plus{}m\\
a_{n}\equal{}a_{n\minus{}1}\plus{}n$
This is just the triangular numbers, by definition, so $ a_n\equal{}\frac{n(n\plus{}1)}{2}$.
[/hide][/quote]
Sorry, I don't quite follow what you're saying.  And can someone provide a full proof by induction?  I'm interested in this problem.[/quote]

a_(x+1)=a_x + (x+1)

@luie1168e, I don't quite follow...but I might have a proof which is basically what worthawholebean has:

For a term in the sequence $a_m$, we find the next term to be $a_{m+1}$.  $a_{m+1}$ can be expressed as $a_m+a_1+(m)(1)=a_m+1+m$.   Then the rest follows by induction which I don't have that much time for right now.

Sorry for the huge bump.

I am a bit confused on this question; since it's $a_{m + n} = a_m + a_n + mn$, this would mean that for say $a_4 = a_{2 +2} = a_{3 + 1}$, but $a_{3+1} = a_3 + a_1 + 3 \implies 7 + 1 + 3 = 11$. $a_{2+2} = a_2 + a_2 + 4 \implies 3 + 3 + 4 = 10$. $10 \neq 11$. So, which way do I use? In Carcul's solution, he/she uses $a_{2+2}$. This is also the case for $a_5, a_6, a_7 \dots$ and so on. Can someone explain what I am doing wrong/what I am assuming?

We have $a_1=1, a_2=2, a_3=6, a_4=10$. Clearly, $a_k$ is the $k$th triangular number by Engineer's Induction. Thus, the answer $$\frac{12\cdot13}{2}=\boxed{78}.$$

[quote=Derive_Foiler]How do we know that $a_k$ is uniquely determined? It seems like everyone's been assuming that . . .[/quote]

While this is a fair point, I don't think it should fall under the burden of the solver to show that the sequence is uniquely determined.

Consider when $m$ or $n=1$, then we have $a_{n+1}=a_n+n+1\to a_{n+1}-a_n=n+1$ as desired. Then, our answer is just $$\sum_{w=1}^{11} w+1=\frac{1}{2}(12)(13)=\boxed{78}$$


induction
number theory solved
number theory

Given the regular Fibonacci sequence, $f_1=f_2=1$ and $f_{n+1}=f_n+f_{n-1}$ for $n\geq2$. Are there numbers $2<k<m<n$ such that $f_n=f_m\cdot f_k$?

The response is no.
I'll show that :

$f_{m+n-2} < f_m*f_n < f_{m+n-1}$

By induction on n:

n=3 : $f_{m+1} < f_m * 2 < f_{m+2}$ , it's clearly true

n+1 : $f_m*f_{n+1} = f_m*(f_{n-1}+f_n) = f_m*f_{n-1} + f_m*f_n$

This implies that:

$f_{m+n-3}+f_{m+n-2} < f_m*f_{n+1} < f_{m+n-2} + f_{m+n-1}$

And finally : $f_{m+n-1} < f_m*f_{n+1} < f_{m+n}$

This closes the recurence and thus, there is no solution  :(


Express as a fraction in lowest terms: $ 0.\overline{1} \plus{} 0.\overline{01}$

hello, i have $ 0.\overline{1} \plus{} 0.\overline{01} \equal{} \frac {1}{9} \plus{} \frac {1}{99} \equal{} \frac {11 \plus{} 1}{99} \equal{} \frac {12}{99} \equal{} \frac {4}{33}$.
Sonnhard.

hello, yes i mean $ \frac{4}{33}$, it was a typo, sorry.
Sonnhard.

Alternatively, $ 0.\overline{1}+0.\overline{01}= 0.\overline{12} = \frac{12}{99} = \boxed{\frac{4}{33}} $


[quote="Franz Joseph"]What is the maximum number of three element subsets of

                           {$1,2,3,4,5,6,7,8$}

which have the property that any two of the subsets have at most one element in common.[/quote]
count the number of subsets that are distinct and then subtract from the total number..


calculus
derivative
function
calculus computations

let $ f(x) \equal{} (x^n \minus{} 1)^n .$
caculate  derivative  $ f^{(n)}(1) \equal{}$???


I noticed that Griffiths' elementary particles book uses a different type of font in math typesetting, akin to pre-LaTeX era (the one as in Feynman Lectures on Physics or Mary Boas' Mathematical Methods in the Physical Sciences, 2nd ed.), instead of usual LaTeX. Check out the images below; se-griffiths.jpg, is Schrödinger equation from that book. I also added typed the LaTeX version below it. Notice how different the psi and hbar are, aside from an overall style shift.

[img]http://img198.imagevenue.com/aAfkjfp01fo1i-26722/loc128/11263_se-griff_122_128lo.jpg[/img]

$ \left( \minus{} \frac {\hbar^2}{2m} \nabla^2 \plus{} V \right) \psi \equal{} E \psi$

So, anyone knows how to "emulate" the Ye Olde Typesetting in LaTeX? Do I need some additional packages or what?

Thanks in advance.

The Griffiths image isn't showing. Sometimes at the front of the book it tells you what font has been used which may help. Have a look at the maths fonts on CTAN. You get different results depending on what you search for:  eg [url=http://tug.ctan.org/cgi-bin/search.py?metadataSearch=maths+font]maths font[/url], [url=http://tug.ctan.org/cgi-bin/search.py?metadataSearch=math+font]math font[/url] or [url=http://tug.ctan.org/cgi-bin/search.py?metadataSearch=mathematical+font]mathematical font[/url] etc.

There's also a [url=http://tug.org/pipermail/math-font-discuss/]Math font discussion list[/url] you could search or ask questions on. Also have a look at [url=http://www.tug.org/TUGboat/Articles/tb19-2/tb59bouc.pdf]Diversity in math fonts[/url], [url=http://www.tug.org/pracjourn/2006-1/hartke/hartke.pdf]A Survey of Free Math Fonts for TeX and LaTeX[/url] and the sample pages of mathematical typesetting on pages 513-523 of the [url=http://www.amazon.com/Companion-Addison-Wesley-Tools-Techniques-Computer/dp/0201362996/sr=1-1/qid=1161947729/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]LaTeX Companion[/url].

Is it just me or mods edited the first post? Anyhow, here's the image from Griffiths.

[url=http://img160.imagevenue.com/img.php?image=06645_se-griff_122_678lo.jpg][img]http://img160.imagevenue.com/loc678/th_06645_se-griff_122_678lo.jpg[/img][/url]

I couldn't find the font in the tug math font survey, I'll try math font discussion list. Thank you!


After playing for several days/weeks, I noticed almost all the problems are 2 points. Are the point values going to change depending on how hard people rate them to be?

Also, can someone confirm the difference between the Alcumus score and the Percentage?

Alcumus Score is calculated using several algorithms, of which I have no knowledge, to calculate the percentage of problems in the Alcumus database you are expected to get correct. Percentage is percentage of correct answers out of all problems you did not give up.
I noticed the 2-point thing too, but it went away after a while. I think there are just a lot of 2-point problems, so the chance of getting many in a row are higher.

Thanks for your question.

A vast majority of problems for most users are 2 points. For users who have really high Alcumus scores, those same questions may only be worth 1 point. Occasionally you will see a 4 point problem to make you sweat a little bit in hopes you get the problem correct.


Is there a bijection between the points in [0,1] and the points in (0,1)? If so, find one. If not, prove that one cannot exist.

Shouldn't the cardinalities be the same by [hide]Schroeder Bernstein[/hide]? Does equalivalace necessarily guarantee a bijection?

[hide]Yes, Schroeder Bernstein guarantees a bijection, but the problem is asking you to find one. I should have known that someone would try to use a super theorem. [/hide]

I remember trying to work on this problem in class...we gave up, but I would be very happy if someone could find one. I will try to myself.

I'm not sure if this exactly works, but....

[hide]Consider the function z(x) = (x, i cot( :pi: x)), x from [0,1]. Now we have a set of points on the imaginary plane. Let every point undergo the transformation f(x,y) = (x,1/(2y) + 1/2).

Does this set of points give us the desired bijection?[/hide]

how about this: let S = {n+1 | n [tex]\in[/tex] natural numbers}

take a function f from [0, 1] to (0,1) so that:

f(0) -> 1/2
f(1) -> 1/3
f(1/x) -> 1/(x+2) for x [tex]\in[/tex] S
f(y) -> y otherwise

this function is easily seen to be a bijection.

Very elegant.

Incidentally, it's a standard result in analysis that there can't be a continuous bijection.


linear algebra
matrix
linear algebra unsolved

Following http://www.artofproblemsolving.com/Forum/viewtopic.php?p=718377#718377, here is another one, probably more difficult
$\mathcal{H}$ is the subset of $\mathcal{M}_{n}(\mathbb{R})$, such that $\forall A=(a_{ij}) \in \mathcal{H}$, $A$ is symmetric positive and each $a_{ij}\in [0,1]$. $\mathcal{H}$ is clearly convex.
What are the extremal points of $\mathcal{H}$?

I'm not sure if this question has a simple answer. When $n=2$, $\mathcal{H}$ consists of all matrices $\begin{pmatrix}a & b \\ b & c \end{pmatrix}$ such that $a,c\in [0,1]$ and $0\le b\le\sqrt{ac}$. The extreme points are $\begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$, $\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix}b^{2}& b \\ b & 1 \end{pmatrix}$, and $\begin{pmatrix}1 & b \\ b & b^{2}\end{pmatrix}$, where $0\le b\le 1$.


Figure $ ABCD$ is a rhombus with diagonals of length 20 and 24. Find the area of rhombus $ ABCD$.

[asy]size(80);
draw((0,0)--(5,20)--(25,25)--(20,5)--cycle);
label("A", (5,20), NW);
label("B", (25,25), NE);
label("C", (20,5), SE);
label("D", (0,0), SW);[/asy]

$ AC$ is perpendicular to $ BD$, so area $ \equal{} \frac {1}{2}(20)(24) \equal{} 240$

[color=blue]Not necessary to quote posts directly before yours. :) - Izzy[/color]


 I can't figure out how to solve these equations.  Any help is appreciated.

Find values for x and y that satisfy xy=0.42 and x + y = 1.3

The second equation can be written $ y \equal{} 1.3 \minus{} x$.  Substituting into the first equation, we get

$ x(1.3 \minus{} x) \equal{} 0.42 \Leftrightarrow$
$ x^2 \minus{} 1.3x \plus{} 0.42 \equal{} 0 \Leftrightarrow$
$ (x \minus{} 0.6)(x \minus{} 0.7) \equal{} 0$

So we can have $ (x, y) \equal{} (0.6, 0.7)$ or $ (x, y) \equal{} (0.7, 0.6)$.

I believe there was a problem with that solution.

when t0rajir0u put $ x^2 \minus{} 1.3x \plus{} 0.42 \equal{} 0$ it was supposed to be $ \minus{} x^2 \plus{} 1.3x \minus{} 0.42 \equal{} 0$.

Edit: Thanks SnipedYou , [b]anyone who reads this just dismis it[/b].

He multiplied by -1 to make $ x^{2}$ positive.


find a sequence $ (u_{n})n\in N$ such that:
$ (\forall n\in N$*  $ u_{n}\geq 1$
$ (\forall n\in N$* $ (n\plus{}1)u_{n} \leq nu_{n\plus{}1}$
$ (\forall (n,m)\in N^2$* $ u_{nm}\equal{}u_{n}$x$ u_{m}$

There are infinetely many suth sequences. For example $ u_n\equal{}n^a, a\ge 1$.


Hey I'm not sure if I should post here or Round Table or Fun & Games but this is pretty important to me. You see, I get good grades on my school and participate in 2 sports which I just go hard ( :ninja: ). But I have little problem. To be a really great person, you have to be good at music as well (at least know how to play one instrument farily well). I tried piano and violin but I sucked on both of them. So, I'm thinking of guitar since I just love its sound (I don't know why.. But I just do). I'm thinking of maybe buying one but I have no idea how much they cost and since I'll probably buy acoustic if I do buy one, I'm wondering how can I learn to play some songs.

I'm not asking to be number 1 guitar player in my school but just learn how to play it and maybe if I get good, compose few tunes :P .

So here are summary of questions:

A. Which acoustic guitars do you recommend? (Must be able to buy in US)
B. How much they cost?
C. How can I learn how to play it?

Thank you.. This means a lot to me.

I don't know much about different types of guitars.. but you gan go to guitarcenter and there's lots of guitars there.. But if you're just learning, you might not want to get a nice guitar yet. Like maybe get a cheaper guitar at first to learn, and then see if you like it. Guitars cost from like $200 to a lot more.. Most of the quality depends on the dollar range.. cause you can get better guitars for more money. Some companies cost more too.. so just look at the guitars and see what you like. And bargain to buy it. I got mine for my birthday so I'm not sure exactly. To learn to play it, either find someone (a friend) who knows how to play and can teach you a couple of chords, or look online. (or both!) There are lots of sites on line that have guitar tabs.. and here is a site that shows you how to play a lot of chords: [url]http://chordfind.com/[/url]

When you start playing guitar, your fingers might hurt. Make sure your nails are short so you can press down strings, and then if you practice you'll get calluses (iono how to spell it) and then you'll get used to it! Good luck!

Anything more??

I need to have more information on guitars because I'm not even sure I'll even get one anything like that. :blush:

What is your goal for your guitar playing? Like what kind of songs do you want to play? Cause there's lots of different kinds... and different things to know for each.

Um.. First basics and I guess the musics that sound really nice like the ones you hear on TV... :P

Mm, I'm not sure if this will help you...
Two summers ago I decided to learn guitar on my own.. Let me just tell you that teaching yourself guitar requires a lot of self-discipline as well as a few good guitar books. I'm not sure about the DVDs and stuff. I think the best thing would be to get a teacher, because it's easy to get discouraged after you've been practicing the same chords for a long time with little improvement. Are you more interested in classical, acoustic, or electric guitars? I think an acoustic is pretty good for most songs, but maybe not the hard rock... Also with acoustics be prepared to have sore fingertips for a few months. The strings are harder than on electrics, I think.
About buying a guitar... hmm I don't really know much because my guitar is pretty crappy. I got it at Sam Ash for under $200. Normally from what I've heard, you shouldn't spend too much money on your first guitar because you might not like guitar and end up regretting buying a really expensive one. Also, a lot of sites online describe guitars and have people's reviews of their own guitars. There's an online guitar community called [url=http://wholenote.com]Wholenote[/url] that has guitar-related advice and tutorials. I would also suggest going with someone who knows guitar to the guitar store, because the salespeople are always going to tell you to buy the more expensive guitar, and it would just help to have an expert around. Of course ultimately you make the decision what to buy. See how the guitar feels when you're holding it. For example, my guitar is a bit big.
Hope this was somewhat helpful.  :)  Good luck!

I taught myself guitar. After my dad taught me a couple of chords. And I think you need to really want to learn whenever you're learning a new instrument whether or not you're self taught. Cause if you don't wanna learn, you won't learn. So yeah, make sure you like playing guitar before you make any big purchases. And acoustic is harder on your fingers, but you can hear it without an amp so that's cool. And for musics that sound really nice like on TV... I don't really watch TV so I don't know what you're talking about. But you can look up tabs for any songs that you want online.. just search on google. Or any other search engine. And tabs are lines with numbers on them.. the lines represent the 6 strings of the guitar and the numbers represent which fret you press down.. fret is the thing in between the lines of the guitar. Yeah.. so maybe it's good to get a teacher or something to teach you the basics and show you how to start playing, cause then after that, you can pretty much learn on your own.

If you're aiming to compose music on a guitar, learn using sheet music first. Beginners will get frustrated easily, so self-discipline is a major concern. I'd recommend getting an instructor, since he/she would keep you on task and help you out with stuff you don't understand. If you decide to learn on your own, try to stay away from tabs for a while. Although tabs are a great resource for popular songs from today's mainstream music, they deprive you of music theory.

You mentioned that you tried to play the piano. As a piano and guitar player, I'd say that piano is a lot easier than guitar, since the keys are arranged in order on a piano.

OK.. I havent been playing guitar for very long (less than half a year) but ill tell you this. If you want an acoustic, the best you would want is probably a Gibson. Im not sure how much Gibson ones cost, but id bet somewhere around 4-500 dollars, maybe less. If you arent willing to spend that much, I use a Fender guitar. I'm buying an electric guitar this week or next week though probably.

You will probably want an instructor... I know people who have taught guitar to themselves but you will learn guitar much faster w/ an instructor (provided its a good one). I love music theory, which is basically the reason "why" this sounds good w/ this chord, and what all the chords are made of, which really lets you go an experiment. Knowing music theory you can create your own style of music,  and spend much less time figuring out "will an A diminished sound good w/ an F major triad?"

On sheet music: personally i believe that sheet music isnt that great for guitar, its too complicated with all the possible things u can do with guitar. It is probably a good thing to learn, but many proffesional guitarists dont read sheet music. Tableture is good for guitar, IMO.

Use [url=http://www.powertabs.net]www.powertabs.net[/url] for your tabs. They have the most accurate tabs for whatever song you want to play. The downside is that they don't offer too many rare songs.

I HIGHLY reccomend a teacher. I regret playing without a teacher because A. I can't stay on task B. I don't know where to go after reaching a level where I can improvise and compose... but not well. C. I have no one guiding me to the next level.

As for buying one, I recomend buying a REALLY cheap guitar to start off with (with semi-expensive strings $7-11). Acoustic is indeed a better choice to start with then electric.

[quote="amirhtlusa"]i've been playing Flamenco Guitar(spanish guitar)for about four years now, and i'm pretty good at it...(i have a duet consert in about a week...)--for those who don't know what Flamenco guitar is, i recomend watching the movie Desporado or Once upon a time in Mexico--
i would really recomend you consider Flamenco too,
first of all not many people know how to play it, secondly, there are some great web sites that have Flamenco guitars.
Flamenco Guitars don't cost much my guitar costs about $200 but you can get one for about $150 ish...
Go EL-MARIACHI :)[/quote]
im interested in playin flamenco in fact iv been lookin for sites that has tutorials for playin it and i cant find one. Can you suggest one or better yet post some explanations in playing flamenco.( conway4irish@yahoo.com )

first off do not paly guitar to become a better person!
that makes no sense

second
if you are going to learn guitar go buy a beginners book with sheet music that teaches you the basic chords and notes.
after that go to http://www.ultimate-guitar.com
there you will learn how to read tabs ( they literally ahve thousands)
and the most important thing i can stress is practice! everyday for at least half hour

it takes some time to get anywhere so don't get frustrated

I've been an avid guitar player for years, and so I'm glad to bestow what little experience I have upon you.

First, determine what kind of guitar you want to play. Steel-string or nylon? Nylon is easier on the fingers, but harder on the sound. It's also mostly used for classical purposes, though anything really can be played on any type. Steel-strings will be hard on the fingers unless you use a pick(which I, personally, don't like) but you can grow some calluses after a while.

If you're going steel-string, I would suggest any beginner-ish guitar to be fine. You want to be able to play before you get into types. Try to pick one up at a garage sale, something that's not a total piece of crap but that's not hyper-sophisticated. If you really want to buy a high-quality one, maybe try [url=http://www.gibson.com/Products/Gibson%20Acoustic%20Guitars/Gibson%20Acoustic%20Guitars/Square%20Shoulder%20Dreadnought/Hummingbird/]this one[/url] as it's not too complicated. If you go into complications, you'll lose the focus on playing. It's the musician that makes the guitar, not the other way around. If ever you're not satisfied completely with the guitar you have, learn to play on it until you can find a better one, then upgrade.

Now for teaching: What I would suggest is get a good teacher to start yourself off. Not necessarily one of those guitar academy teachers, maybe a friend who can play really well and show you the songs you like, as well as some technique. That way you  won't be hyper-dependable once you get the basics. One site you can use for theory as well as song-finding would be [url=http://www.olga.net]OLGA[/url]. Browse around a bit. After a while you'll get the hang of the theory, such as building chords and stuff, and you'll be able to find them yourself, as well as figure out the chords/tabs for some songs.

Have fun, and remember to be patient!

Best guitar brands for beginners:

Yamaha Acoustic guitars, ranging from 100$-200$

Aria Nylon string guitars,

One of these two is excellent...

best of luck...

Please don't hesitate to PM me if you have any query regarding guitars....I play it for 7 years now...ranging from Classical to Metal...

powertabs.net is having some legal problems.  Their tabs can be found at mirror sites.

[url]http://pta.acidpit.org/[/url]

[quote="Silverfalcon"]
So here are summary of questions:

A. Which acoustic guitars do you recommend? (Must be able to buy in US)
B. How much they cost?
C. How can I learn how to play it?

Thank you.. This means a lot to me.[/quote]

I realize this thread is mostly dead and the original author (SilverFalcon) has probably found what he needed, so I post this for others who may stumble upon this thread in need of direction. This is mostly for the beginner who has rarely or never picked up a guitar, loves the sound and wants to learn.

Guitar buying in my view is not something that should be done over the internet, or based off opinions of others. Not that others opinions don't help, but there are many different types of guitars, tailored to individual taste and comfort or types of music that each person has to figure out for themselves. With that said, the best advice I can give is to not jump into an online buy like I originally did. If you're in the U.S. chances are you have a Guitar Center, Guitar World or some sort of music shop with a wide selection to choose from, even in smalltown U.S.A.

When I started I had ordered a guitar not knowing much if anything about it, other than it was an acoustic and that's what I wanted to learn on. After playing it for a month it just didn't have a good "feel" to it. The strings were metal and close together and just didn't 'fit'. I went down to the Guitar Center and start poking around until I found what felt good to me. That happened to be a wider neck with nylon strings. Since then not only have I made more progress in learning but I get more enjoyment out of it.

So spend the time to go down there, ask the salesman as many questions as you can think of. Take a few guitars off the shelf and see how they feel and sound. Get the salesman to teach you a chord or two to strum and see how it fits. Try out one with steel strings close together, another with the wide neck and the strings farther apart, and try nylon strings out and see how you like the difference in feel and sound. If you're after a certain type of music like Blues or Metal or even what I was after, Slack Key, there may be a guitar just waiting for you. You don't need anything fancy or expensive to learn on, just something comfortable.

When it comes to playing, especially learning, these things can be very important. They will help you get over some of the stumbling blocks of learning and putting more enjoyment into your time spent learning. It could even be the difference between putting down the guitar and walking away, or sticking it out and continuing to learn. Either way, you won't regret a few extra minutes or a couple of hours trying out the instruments before making a purchase.

As far as lessons, some people prefer to pay a teacher. If the teacher knows what they are doing it could be money well spent in the long run. If you want to be self taught there are numerous free resources around the internet that will more than get you started, and alot of books in your town library which you can check out for free, or buy at the music shop so you can make your own notes on the pages.

While you will want to jump in and learn how to play a song and get started, I recommened first spending a few minutes to learn how to read tabs, and a little of how the beats work for timing. Also getting a routine of left hand and right hand finger practice. Either of those typed into a search engine should get you some good examples. Setup a routine of doing finger practice both before and after your practice session on a song. On the left hand it'll get your fingers used to the strings (they'll stop hurting after a week or two) giving them more strength and better flexibility to get those hard to reach chords. On the right hand, wether finger plucking or playing with a pick, your hand will get used to the spacing, proper movement, and the clear sounding of the string. If your hands are used to the guitar, you can focus more on the song and less on the stresses of being a beginner. Always remember to check for proper positioning when playing, as bad habits can be hard to break and give you a tougher time.

A few links that may help you get started. I've used the first one quite a bit, if you dig there is just about everything a beginner could need.

http://www.acousticguitar.com (Has free lessons in most styles of music)
http://www.guitartabs.com (Tabs)
http://www.ultimate-guitar.com (Tabs)

Whatever instrument you play, or however you play it, remember to clear out the mind and enjoy it.

Aloha,

Ki Ho'alu Beginner (Loosen the string)

You only need 2 things to play guitar: 1 a guitar, 2 you need to enjoy it.

No matter how much you want to play guitar if you dont enjoy you wont go very far, and no matter how little you want to play guitar if you enjoy you will go far.

Someone above said something like you need strong self-disipline and a few guitar books to learn guitar, I'm self taught and I very poor self disipine, it took me a while to get good but I didnt care, also Jerry Garcia one of the all time best guitarists played guitar for 2 years before he found out what standard tuning was and had to relearn some of the things he learned.  You don't have to hammer in chords day after monotonous day, i just fiddle around, when i first started when i was watching tv i would just play around with it and try to come up with stuff.  ofcouse down the line i hammered in chords but i did other stuff aside from that and i was a number of steps ahead of someone just starting.  so as long as you have fun and fiddle around that's all you need (and a guitar :))

i'm a classical guitar player......
hoho i love it


A regular octagon is the base of a prism, what is the formula for the area?

Something like $ V\equal{}a(b\plus{}\sqrt{c})$???

Say that the side of the base is $ s$ and the height of the prism is $ h$. We know that the area of the octagon will be $ (2\plus{}2\sqrt{2})s^2$, so the area of the prism is $ s^2h(2\plus{}2\sqrt{2})$.

This thread should be moved to Middle School Contests. It doesn't belong here.


[quote="ccy"][quote="bpms"]Is there anyway to prevent others avatars from appearing like there was with the old server? Will there be one?

Thanks,
bpms[/quote]
Assuming you're using Firefox, you can install [url=https://addons.mozilla.org/firefox/1865/]Adblock Plus[/url] and add the filter:
http://www.artofproblemsolving.com/Forum/images/avatars/*[/quote]

is ther like an avatar rule or something because some person said a while ago that mine contradicts the AoPS goal or something...

Some person was misinformed. The only rule on avatars is that they can't be obscene or violent and so forth, i.e. they comply with the User Agreement thingy. There haven't been many bad avatars; I think the only one I can remember was an animation of someone beating someone or thing else just outside of the camera frame. But the user was warned and they changed their avatar.


AMC USA(J)MO USAMO Princeton college Harvard geometry

Just wondering: how much of an impact does qualifying for the USAMO make on college applications?

Of course, nothing can ever guarantee one's admission, but has anyone here heard of a USAMO qualifier being rejected from any college?

qualifying for the USAMO does distinguish you and can help you get into a school but it is not the only thing that Colleges look for.  I know somebody who made the USAMO five times and was rejected from Princeton because he did not have very good grades.  Colleges tend to look at everything, and making the USAMO helps a lot, but it is no guarantee.

[quote="WarpedKlown1335"]I'm just wondering, how can one make the USAMO five times and yet not get good grades???  That just doesn't mix.[/quote]

This would be a very welcome discussion on the College Forum.

The short answer is that teachers expect students to work on THEIR assignments for class, and don't give anyone credit for doing something that wasn't assigned. I can well believe that someone smart enough to make USAMO five times might think (correctly) that most of the school assignments are a waste of time, and not spend much time on them. Some college applicants (Bill Gates was one example) can pick and choose what courses they work on, but most college admission officers expect students to do what they're told by their teachers, to at least a high B average degree).

In my observation, this danger of a highly intelligent young person correctly perceiving high school lessons to be largely a waste of time is so high that I recommend parents keep the smartest children mostly OUT OF high school in their teenage years. A few high school programs in the United States are rigorous and genuinely educational, but not many. Much more educational are early enrollment at a local college or distance learning programs for the highly gifted high-school-age learner. That's why I homeschool my children--to avoid the mandatory wasted time of all too many high school assignments.

To answer the original poster's question, any knowledgeable college admissions officer ought to be aware that the USAMO is one of several fairly reliable signs that a learner is advanced in math. An applicant who has the usual kind of high school record expected in top colleges, and participation in the USAMO besides, should look more attractive than a child who sticks solely to the school's assigned curriculum.

If you have more detailed questions about issues like this, feel free to visit the College Forum on this site.

I haven't heard of any Mopper (except those who went to MOP 2002, that doesn't count) not getting into a top-notch college though. Most seem to go to Harvard actually.

I have to disagree with you TA. The greatest predicter of success in college is high school grades - even more so than test scores. What a USAMO qualifier with low grades indicates to a college admission staff is that the student has a lot of potential, but also has either poor study habits or a poor work ethic - both of which are important components to success in college. And if that hypothesis is confirmed by the students recommendations, that could keep them from getting into a very selective college. Or at the least it could put them further down on the list of students admitted, because at very selective universities nearly all of the students have some high school achievements that make them stand out - usually in addition to having good grades.

Whether bright students should be kept out of high schools, I think that really depends on the student's personality and on the high school options available. Some very bright students thrive in regular high school environments - especially if there's a good teacher or a gifted program there that can continue to guide them towards higher knowledge. Some students can also learn a lot from being around other students (of all levels), and from the extracurricular activities that are available at a high school.

Some other students though, have peculiar circumstances that make a traditional high school setting unconducive to their learning, or who just have better options. Sometimes it's because of the personality of the student, sometimes it's because there aren't any adequate high school options available, and sometimes it's just because students have good parents at home with the flexibility and ability to homeschool them. I would say that none of those are true of [i]most[/i] gifted students.

But the ability to do mundane tasks is all a part of life. It's not something that goes away after high school. You have to do them in college and even in most jobs there's mundane paperwork that needs to be done. A big part of what grades reflect, even more so than knowledge, is academic discipline. That's something that colleges (and employers) often look for in applicants.

Well, there is also the truth that colleges often favor students from overseas :-D, but then, many international schools are quite difficult.

Now I'm going to move this topic to the College Forum.  Is that ok?

Yes, please move this to the College forum. Sorry, I didn't know there was one.

Also, please don't turn this into a debate about homework.

I made it to USAMO twice (ages 14 and 15), and got rejected by several colleges.  That's because in high school, I hardly did any work, and hence got so-so grades.  It was my own darn fault.  Sure, school was boring, but at least I should have put in the effort.

In college (University of Maryland), I realized that if I didn't work hard, I wouldn't get into a good graduate school.  So, I worked hard in every class, got essentially straight A's, and was accepted into every grad school I applied to, including the one I went to, MIT.  I'm glad I did.

Hmm, all of you speak of these USAMO qualifiers who are smart but get poor grades because they are lazy and don't do homework/pay attention in class. However, I think it might be helpful (for the OP or others interested) for people to comment on USAMO qualifiers who did decently at school, got decent SAT scores, etc... as opposed to focusing on those people who just didn't do the basic schoolwork.

Weren't Joel or Simon ever USAMO qualifiers?

I was a USAMO qualifier, and I also had good grades in high school.  (Actually, almost everyone at Stuyvesant gets good grades -- the average average is a 90 or 91.)  I did not get into every school I applied to, although I got into most of them.

I know three USAMO qualifiers who did not succeed so much in the college game -- one of them was very, very lazy and essentially didn't work in any class at school, including math class, although he was the alternate for the US IMO team his senior year.  Another was probably mentally ill.  The third was a good student, and a fabulous mathematician.  He was three years ahead of me in school, and at our fairly small private school, he was the AHSME (AMC 12) high-scorer from the school from 5th grade onwards.  He was very weird -- as a measure of his weirdness, he got in to every school he applied to that didn't make him have an interview, and none of the schools at which he interviewed.

Randomly about homework -- once you get to college, a lot of homework becomes either optional or hard.

and I'm assuming your forgetting a particular person from AAST that got screwed out of MIT ? (JP)

I'm late to this thread, but I think I have something to add. A contest background is a tremendous positive indicator for success in college mathematics, but you can't expect the admissions people who read your application to understand what the contest achievements you list mean. My roommate at MOP 1998 was a total flake, and I'm not sure about his mental health; he had to settle for a local community college due to his grades.

Colleges will almost certainly pay more attention to grades and standard test scores, because that's what they see on every application.

A few places do systematically care about the contests- in addition to the ones mentioned here, Duke tried to recruit me specifically because I had been to MOP (and Melanie Wood knew me). On most applications, you'll have to explain yourself.

If you want to make something like the USAMO matter in your application, there is a space you can use to expand on superlative achievements in contests- the essay. If you take the "Tell us about yourself" prompt, you can give some details about what your contest performance means.

I was waitlisted by one of the colleges I applied, but that came after I said in the interview that one of the other colleges was my first choice.

I'll add something too on the subject of whether or not admissions officers are familiar enough with the various contests to recognize unusually elevated math/science accomplishments. I think the colleges that regularly see apps from people like that (Harvard, MIT, etc), definitely do. But if one of those students were to apply to a college that has never seen, say, an olympiad medalist, or even a USAMO qualifier before, they would not necessarily get recognized for that Then they would be judged based on things like SAT scores and grades in high school courses, which may totally not do them justice. People who work in admissions may very well be English majors. If they have not been trained to look for things like USAMO, there's no reason to assume that they will understand the significance.

I know of 3 students who had multiple USAMOs from an early age, multiple olympiad training camps (including MOP), RSI, int'l olympiad medals, many college math courses in high school, yadayadayada when they applied to colleges. In other words, stars, viewed by the standards of this forum. They only applied to a very few top math/science schools, and were accepted everywhere. However, they had very limited success applying for outside merit scholarships (like Micron, Coca-Cola, parent's company, Byrd, USA Today). These required an app similar to a college app, but were reviewed by people who do not ordinarily see applicants who have done that sort of thing and were unable to recognize how impressive and unusual it is. I would imagine that if they had applied to colleges below the ones they did, that they might have had some disappointing results in college admissions also. (Accepted at MIT and Harvard, rejected at Earlham and Trinity, perhaps.)

A former stanford admissions officer, in her book Questions and Admissions, tells the story of the time when Stanford became aware that it was losing math stars, because its admissions officers didn't know what a math star looked like. Now applications from would-be math majors are vetted by a committee of the math faculty who know what to look for. It is clear that Harvard does the same thing.

I attended a local Stanford info session where the presenter (obviously an English major) went on and on about the importance of being well-rounded, and in particular the importance of well-written essays, even discussing what was expected on each individual question. It took up most of the presentation! In the Q/A at the end I asked something like "How about the math genius who can't write their way out of a paper bag?" She said the same thing Tokenadult is saying, "Oh, don't worry about it. They're evaluated by the math department."

I can also confirm Token's comment that something similar happens at Harvard. One of the kids I mentioned in my earlier post was contacted by the Harvard admissions office and asked to send a copy of a paper he had written to the math department.

[quote="texas137"]I can also confirm Token's comment that something similar happens at Harvard. One of the kids I mentioned in my earlier post was contacted by the Harvard admissions office and asked to send a copy of a paper he had written to the math department.[/quote]

One reason for this, is that they (and MIT math dept, etc) have had some experience with the high school student "research" phenomenon.

"Most of these papers [by high school students] are garbage" is a direct quotation from a tenured math professor at a well-known university in Cambridge.  It is representative of various similar remarks made by other professors, as well.

Undergraduate math research is also considered as an indicator of enthusiasm more than ability, although it would be strange to have done no research before graduate school.  For experimental science (biology, chemistry) a year or more in the lab is probably considered essential to establish whether the applicant for a Ph.D will do the lab-work, but the evaluators know that the project is conceived and directed by a grad student or professor.

My parents often speak to me about this, "why don't you go study for the SAT instead of doing math", so I too often wonder whether or not my passion for math will hinder my chances of being accepted into one of the schools of my choice. For example, a friend of mine who currently graduated scored a 1600 on the SAT and is also gifted in the areas of math/sci (winning/placing in a few regional competitions such as UIL, etc. and 2nd place national sci bowl) but is not exceptional (no USAMO or other national olympiads). In addition he is able to write very well, but again, he is no Shakespeare. However, his well rounded abilities along with his perfect SAT score secured him a virtually full scholarhsip to Princeton University while other people I know who are particularly gifted in the areas of math, physics, etc. were unable to receive the scholarships they wanted. On the other hand, his classmate, also having a perfect SAT score and  excelling in the same areas with less regional awards (again no USAMO or other national olympiads but on the 2nd place national sci bowl team) did not manage to secure a full scholarship. This makes you wonder about the selection process that schools use to determine who will receive a scholarship and who will not. Coming from Europe my parents faced a very different selection process. All college applicants are required to taking college entrance exams (usually two, different colleges require different exams but math and a language are the most common ones). Hence, standardized tests like the SAT were never part of the selection process. Also, all summer olympiad training participants (I am only certain for math because of my dad) were allowed free admission into the college of their choice. This allowed for students, such as those on AoPS, who are passionate about a certain subject to be less anxious when the time to apply for college came around. Although I wish the selection process in the US became similar (using entrance exams instead of the SAT) it doesn't look as if it will ever happen. Thus upcoming highschool graduates will still have to face the SAT. As for the importance of USAMO/MOSP/etc., I believe that the ivy league schools have informed their application offices about the prestige of these events and therefore students with this kind of background will have a better chance of being accepted into such schools in the future.

I think people often forget that talent does not get a person into the USAMO. I have met a lot of talented people who didn't work at math. I don't know anyone who ever made the USAMO without working hard at it. I don't even think very much talent is necessary to make the USAMO -- I don't think of myself as a talented mathematician (or person in general). Other people do -- but they don't witness what it took to get to that point. I think most reasonably intelligent people who put in the amount of work that I put in can make the USAMO (maybe not every time -- it's always possible to screw up an AIME, but they'll make it at least one time in high school if they start preparing when I did).

[quote="blahblahblah"]Princeton doesn't give out scholarships, period.[/quote]

Princeton does not give out [i]merit[/i] scholarships. Princeton, plus the other Ivys, MIT, Stanford certainly give out need-based scholarships. But of course, that has nothing to do with how desirable you are as a candidate. Their top pick will get nothing if they cannot demonstrate financial need. And their barely-scraped-in bottom pick will get a full ride if they are poor enough.

I was accepted to Princeton last year, so I should know.  I hardly consider financial aid to be a scholarship, but my point is essentially the same as yours - saying that someone got a full ride to Princeton only asserts that they were accepted and that their family made sufficiently little that Princeton covered the full cost of education there, and it says nothing about the relative quality of the acceptee.

[quote="fleeting_guest"]Princeton, Harvard, etc do offer (or have been known to offer, I don't know if it's necessarily a continuing practice) some merit-based financial incentives for the top students to attend, but nothing that reaches the size of a "scholarship".[/quote]

Indeed - I was a so-called 'Princeton scholar' which gave a little stipend to my high school library and set aside a thousand dollars for 'research costs'.  I think it might have given me a small reduction in tuition.  I also won a few academic prizes there, but I'd also say they didn't rise to the level of scholarships.  My education was mainly funded by Coca-Cola, my parents, and my years at Shaw later.

Curiously, I ended up applying for, and getting, that Princeton scholar thousand dollars to buy the computer I co-wrote the original AoPS books with.  I thought it was kind of silly at the the time, but in retrospect, I suspect this might be among the most useful such expenditures the Princeton scholar program ever made.

"Harvard calls the analogous thing "National Scholar" status, and it goes to the top few percent of incoming students. Those not on financial aid get to pocket the money, for the rest it is meant to reduce the work component of financial aid."

This does not reconcile with Harvard's official buzz on merit scholarships--

from http://www.admissions.college.harvard.edu/faqs/financial_aid/

"Does Harvard award merit scholarships?
No. Harvard is firmly committed to allocating resources to sustain our policies of need-blind admissions and need-based financial aid. Our unusually talented student populations -- chosen in a very competitive admissions process-- would, in any case, complicate the equitable distribution of merit-based awards. However, students last year brought with them to Harvard about $11 million in scholarships from outside resources, thus enhancing their own financial flexibility during their college years."
and from Understanding Your Financial Aid Award--
"Harvard ScholarshipsIf a job, loan, and any outside awards do not fully meet your financial need, you are awarded a Faculty of Arts and Sciences Scholarship or a Harvard National Scholarship. These awardsrange from under $500 to over $43,000 for the 2005-2006 academic year and come from a variety of sources including endowment funds, gifts from alumni/ae, and general tuition revenues.The Faculty of Arts and Sciences Scholarship is awarded to students who have been admittedto Harvard College and who require financial assistance to help meet their demonstrated need.The Faculty of Arts and Sciences Scholarship program includes over 1500 individual funds,established over the past 365 years by generous alumni/ae and others who wished to provide support for current and future outstanding students. The Harvard National Scholarship was established in honor of Harvard’s 300th anniversaryin the early 1930’s with the goal of encouraging and enabling a small number of the best students from across the nation to attend Harvard, at that time populated largely by studentsfrom the Northeast corridor. The plan was to make Harvard a more “national” and diverseinstitution by including in its student body very promising individuals from outside the NewEngland area. Today the National Scholarship is awarded to students from certain states in the Midwest, West and South to continue the tradition of providing access to the College for students across the country, but receiving this scholarship does not affect the total amount of scholarship aid that any student receives."

Everything (except outside scholarships)  is keyed to financial need

What you quote at Harvard is a new policy.  They certainly used to designate National Scholars on a merit-only basis without consideration for need, and the award was given to people from anywhere in the country including the Northeast (or even Cambridge MA).  For people on financial aid, it reduced their work obligations, and I think that for those not on financial aid it put some money in their pocket, though my memory is less precise about this last point (i.e. it may have been honorary, though that is not what I remember).

By the way, MIT also finds ways to offer extra money to people it wishes to recruit, including math-competition champions.   I don't know if there is a single rubric for it as at Harvard and Princeton.

As to the changes in policy, ten or more years ago there was anti-trust investigation and litigation by the US government against the Ivy League schools and MIT, the gist of which was that the annual spring meeting of these schools to set financial aid for commonly admitted candidates amounted to illegal price-fixing.  That and the multibillion-dollar fundraising of the past 15 years led to more financial aid competition to get the top students, and more money to fund scholarships (need-based or not), which in conjunction with other controversies about affirmative action policies led to pressure on the schools to give ever-larger awards to needier students.


I was meditating on the foundations of mathematics,when i arived at the following problem:

let A,B,C be three matrices such that A multiplied with B gives us C;
we write that AB=C.Look a minute at this equality.
In the right member we have a matrix,C,ok?
But in the left member ,can we say that <AB> is a matrix?Which are the elements of <AB>?
So we say "AB is C",right?C is the matrix with the elements c_11,c_22,etc;A is the matrix with a_11,a_22;B with b_11,etc
but what is "AB"?Just a notation?

If we think at it "philosophically",it arrives the following problem:
Let's say

(2 0)  * (1 0) = (2 0)
(0 1)     (0 1)    (0 1)

Is this a real "mathematical equality"?Or it means only "by multiplying the first two matrices we obtain the third"?????
Does it "mathematically" make sense what i'm trying ti say?

Ad this problems arives even in the statement  <1+1 =2>;1 and 2 are natural numbers,but what is the "term"  "1+1"???is it a number?one may say it's not,because the numbers are
1,2,3,4,... etc.
And if it's not a number ,how is it equal to 2?
Or 1+1=2 means only that if we "apply the summation operation" to the numbers 1 and 1 we "get" 2?

[size=167]With other words,this "considerations" may imply that the "=" in 1=1 is "different" than the
"=" in "1+1=2"?[/size]

What can you say about that?

I fail to see the problem

AB is a notation people agree, and by definition it is a matrix

there is certain binary mapping f, and what it says there is just like $f(A,B)=C$

'1+1' is a [i]term[/i] in the language of arithmetic; it is a derivable fact under the axioms of arithmetic that '1 + 1 = 2.'  As terms, '1+1' is not identical to the term '2' -- however, they are provably equal to each other.

I dont see any contedictions or problems

[quote="Bictor717"]= is a relation between two objects. We say a=b when a and b are the same. Then a and b are two ways of describing the same thing.
You can think of "+" and "*" as functions on two variables that map to the appropriate space. Then A*B=C means C is the same as whatever * maps (A,B) to.
Does that make things clearer or more complicated?[/quote]

Of course the things are "mathematically" trivial;
but philosophically(although you think i'm wasting my time and energy with such considerations)  it is very complicated!!Don't you think?
And it's not a problem with "=" as i firstly thought.The problem is with how we describe "mathematical objects".

[quote]Then A*B=C means C is the same as whatever * maps (A,B) to.[/quote]
Yes,i agree with you,but what means "[b]whatever * maps (A,B) to[/b]"?
I am trying to observe the following:can we consider A*B a mathematical object in the true sense of the word,such as A,B,C, or interpret A*B=C just as a "relation" between A,B,C  so A*B is not a matrix,but <A*B=C> means that "by multiplying A and B we get C"????

The same thing is in arithmetics.
I don't know why i don't think[b] the "=" symbol in 1+1=2 and  1+1=3-1 is different!!![/b]
I took a look on some mathematical formal systems where i there are more types of equality!!!for example,in one of them i saw there are 3 types of equality: "definitional equality","evaluational  equality" and alpha-equality;and they sad there
1)defnitional equality is used for making definitions,such in:

let x_n=n^2 be a sequence-here we have definitional equality

2)evaluational equality is used such as in 1+1=2 (1+1 "evaluates to 2) or in A*B=C

3)te other type of equality is the "normal one" such in 1+1=3-1 and it means that, "evaluating" 1+1 and 3-1 we get the "same" 2!!

And it is said there that the "=" relation is,in this case the transitive and reflexive closure of these 3 types of equality.

I don't think the other people who are answering you have studied what is necessary to answer your question in a way that will be satisfactory to you.

I'm just going to use arithmetic -- I could use, say, calculus, but the language is more complicated.

In formal arithmetic, we have a language, composed of the symbols 0, 1, 2, ..., 9, x, y, z, ..., +, *, =, (, ), in addition to the logical symbols $\vee$, $\wedge$, $\equiv$, $\rightarrow$, $\sim$, $\forall$, $\exists$, and so on.  (This is a fairly expansive choice of symbols, by the way -- I could have chosen a more minimal set, but this will do fine.)
Strings of these symbols fall into particular groups: for instance, 'x,' '23,' '(x +2)*5' and '2 + 2*(2 + 2*(2 + 2*3))' are all [i]terms[/i] in the language.  'x = 2,' 'x = 2 $\vee$ x + y = 4' and '$\forall$ x(x = 2 $\vee \, \sim$(x = 2))' are all [i]formulas[/i] in the language.  There are particular rules for how we build up terms and how we build up formulas, and these rules are "simple" in the sense that if I give you a string of symbols, there is a simple, finite process that  you can go through to determine whether that string represents a term or a formula.
Now, there are also what are known as [i]derivations[/i] in the language.  For these, we need a set of axioms and a set of derivation rules.  Peano Arithmetic, the standard axiomatization of arithmetic, has a dozen or so axioms and two derivation rules (one of which is [i]modus ponens[/i]; if $F$ is a derivable formula and $F \rightarrow G$ is a derivable formula, then $G$ is a derivable formula).  Now, here's the important thing: even though this all looks like it's saying things, it doesn't say anything.  These symbols and rules are all meaningless -- [i]modus ponens[/i], which looks like a fact you know about logical implication, is merely a step for how to move symbols around and get from one set of derivable formulas to another.  It is only when we interpret these symbols and talking about our standard arithmetic system that it makes sense to talk of "the meaning of '='."

All of mathematics can be formalized in this way.  Mathematicians don't do this because (a) it's tiresome (2) it would make everything very long and unreadable (iii) it doesn't add any new information and (e) it's been proven that everything is formalizable, so why bother?  Because of this non-formality, we have a habit of simply re-using old symbols in different contexts where the relevant interpretation is obvious.  Thus, for example, the symbol '=' is used in many different circumstances in a number of different ways.  This should not be suprising, nor is it a "problem" in any sense that I can see.  It is, at best, mildly unclear (as it requires us to say what interpretation to apply to a symbol in a given appearance).

Is this helpful to you?


For the nonzero numbers $ a$, $ b$, $ c$, define
\[(a,b,c)\equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}.\]
Find $ (2,12,9)$.

$ \textbf{(A)}\ 4 \qquad
\textbf{(B)}\ 5 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 7 \qquad
\textbf{(E)}\ 8$

This is simply plugging in numbers and you get
$\dfrac {36}{6} = \boxed {6}$


[quote="talfohrrt"]$ a,b,c\geq 0 ,$ $ a+b+c=3.$
prove ;
$ \frac{a^{2}}{a+b^{2}}+\frac{b^{2}}{b+c^{2}}+\frac{c^{2}}{c+a^{2}}\geq\frac{3}{2}$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154686

A nice inequality is the following:

 If $ a,b,c \geq 0$ such that $ a+b+c=3$, then

$ \frac{a^{2}}{2a+b^{2}}+\frac{b^{2}}{2b+c^{2}}+\frac{c^{2}}{2c+a^{2}}\geq 1$.


Hi all, I recently found this forum and decided to sign up. I'm well a maths enthusiast so I think here I'm going to find interesting topics.

Here is the present subject. I have been amazed by the Nine-point circle and its features. Recently I spotted this place:
http://agutie.homestead.com
and there I found this page over the Feuerbach point:
http://agutie.homestead.com/files/center/feuerbach_point_facts_triangle.html

Can anyone suggest a proof for these equalities, I simply can't work them out.

Thats a pretty interesting website you got there. Bump!