inequalities
parameterization
function
inequalities proposed

[quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ a + b + c = 3$, then

$ \frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$[/quote]

Generally, the inequality
          $ \frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

        $ {a + b + c = 3,{a,b,c}>{0}}$

if  and only if     $ - 12\le{k}\le{24}.$

[quote="fjwxcsl"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ a + b + c = 3$, then

$ \frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$[/quote]

Generally, the inequality
          $ \frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

        $ {a + b + c = 3,{a,b,c} > {0}}$

if  and only if     $ - 12\le{k}\le{24}.$[/quote]

Analogously,the inequality
          $ \frac {k}{a^2b + b^2c + c^2a} - \frac 1{a+b+c} \le {\frac {k -1}{3},}$

which contain parameter k ,holds for

        $ {abc=1,{a,b,c} > {0}}$

if  and only if   $ {k}\ge{1}.$

[quote="can_hang2007"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {24}{a^2b \plus{} b^2c \plus{} c^2a} \plus{} \frac 1{abc} \ge 9.$[/quote]
It is a nice and difficult one, Vasc. My ugly proof used the fact that
$ 2(a^2b \plus{} b^2c \plus{} c^2a) \equal{} \sum ab(a \plus{} b) \plus{} (a \minus{} b)(b \minus{} c)(a \minus{} c) \le \sum ab(a \plus{} b) \plus{} \sqrt {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}$
to transform the inequality into $ pqr$ form, and then by considering a function of $ q,$ I deduced the result. For $ a \ge b \ge c,$ equality holds when $ a \equal{} b \equal{} c \equal{} 1,$ or when $ a,b,c$ are three positive roots of the equations
$ x^3 \minus{} 3x^2 \plus{} \left( 1 \plus{} \sqrt {3}\right)x \minus{} \frac {3 \plus{} 2\sqrt {3}}{9} \equal{} 0.$
This proof is too complicated and I dont want to post it. I think you have found a new method to treat this inequality. May I know your solution, Vasc? Thank you very much.[/quote]

Dear Can,is  the following step in your proof?
\[ abc\leq \frac {1}{9 \minus{} \frac {24}{3 \plus{} \frac {2}{9}(9 \minus{} 3q)^{1.5}}}\]
where $ q \equal{} ab \plus{} bc \plus{} ca$

[quote="Vasc"][quote="fjwxcsl"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ a + b + c = 3$, then

$ \frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$[/quote]
Generally, the inequality
          $ \frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

        $ {a + b + c = 3,{a,b,c} > {0}}$

if  and only if     $ - 12\le{k}\le{24}.$[/quote]
Indeed,

$ \frac 1{abc} + 3 \ge \frac {12}{a^2b + b^2c + c^2a}$

is nice, too,  fjwxcsl. :lol:[/quote]
\[ \frac {12}{\frac {1}{abc} + 3}\leq 3 - \frac {2}{9}[9 - 3(ab + bc + ca)]^{1.5}\leq a^2b + b^2c + c^2a\]
is it? :maybe:

[quote="Tourish"][/quote][quote="Vasc"][quote="fjwxcsl"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ a + b + c = 3$, then

$ \frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$[/quote]
Generally, the inequality
          $ \frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

        $ {a + b + c = 3,{a,b,c} > {0}}$

if  and only if     $ - 12\le{k}\le{24}.$[/quote]
Indeed,

$ \frac 1{abc} + 3 \ge \frac {12}{a^2b + b^2c + c^2a}$

is nice, too,  fjwxcsl. :lol:[/quote][quote="Tourish"]
\[ \frac {12}{\frac {1}{abc} + 3}\leq 3 - \frac {2}{9}[9 - 3(ab + bc + ca)]^{1.5}\leq a^2b + b^2c + c^2a\]
is it? :maybe:[/quote]

The first inequality does not hold, the second one holds.

[quote="can_hang2007"]It is a nice and difficult one, Vasc. My ugly proof used the fact that
$ 2(a^2b \plus{} b^2c \plus{} c^2a) \equal{} \sum ab(a \plus{} b) \plus{} (a \minus{} b)(b \minus{} c)(a \minus{} c) \le \sum ab(a \plus{} b) \plus{} \sqrt {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}$. [/quote]
Indeed, using this inequality, it suffices to show that
\[\frac{48}{3q-3r+\sqrt{-27r^2+54(q-2)r+9q^2-4q^3}}+\frac 1{r}\ge 9,\] where $r=abc$ and $q=ab+bc+ca$. This is true if
\[3[9r^2+15r-(9r-1)q]\ge (9r-1)\sqrt{-27r^2+54(q-2)r+9q^2-4q^3}.\] Before squaring this inequality, we need to show that $9r^2+15r-(9r-1)q\ge 0$. From Schur's inequality
\[p^3+9r\ge 4pq,\] we get
\[q\le \frac{3(r+3)}{4},\] hence
\[9r^2+15r-(9r-1)q\ge 9r^2+15r-\frac{3(r+3)(9r-1)}{4}=\frac{9(r-1)^2}{4}\ge 0.\]
By squaring, the inequality becomes
\[Aq^3+C\ge 3Bq,\] where
\[A=(9r-1)^2, \ \ \ B=18r(9r-1)(3r+1), \ \ \ C=27r(27r^3+99r^2+r+1).\] Since, by the AM-GM inequality,
\[Aq^3+C\ge 3\sqrt[3]{Aq^3.(C/2)^2}=3q\sqrt[3]{A(C/2)^2},\] it suffices to show that
\[AC^2\ge 4B^3,\] which is equivalent to
\[(27r^3+99r^2+r+1)^2\ge 32r(9r-1)(3r+1)^3,\]
\[729r^6-2430r^5+2943r^4-1476r^3+199r^2+34r+1\ge 0,\]
\[(r-1)^2(27r^2-18r-1)^2\ge 0.\]


Now that the 2004 Fall Startup Event has completed, we'd love to hear your thoughts on the contest.  The contest format is 100 fast and easy problems to be answered in 30 minutes... something Richard Ruszcyk (the founder of the AoPS website) has described as analogous to scales that a musician practices.

Because the problems are not terribly complicated, and because detailed solutions are provided for all problems, we do not anticipate a great deal of discussion of the problems and their solutions, although we're happy to answer questions of this sort if people have them.  Instead we are hoping to learn whether the contest helped you to identify types of problems you should be more familiar with, whether the less-experienced members of your team found the contest to be more encouraging than others, or whether the contest helped focus your team earlier in the year.

Thank you,

I really like the fall startup.  I m not sure why though.  It's really kind of like a race.  A sprint to be more precise.  Last year I didn't have much experience in problem solving and a lot of the problems seemed challenging to me.  This year, after gaining much experience, all the problems were easy.  This is a very good contest for beginners and even experts who want to test their speed.  The contest is also good because it is early in the year.  I wish there were as many math contests in the fall as there are in the spring but I guess everything has their season.

I was wondering how many people on AoPS took this contest.  Post here if you did.  You can also share your results if you like.  I guess I ll share mine:

I got a 53 and was 21st in the 11 th grade division.  My school got 19th overall.

I know now that I could ve done a few things differently to improve my score.  For one, I shouldn't have tried to go through each problem in order.  I realized too late that they were grouped by section and that there were some that I can get really fast near the end... I could probably do some of the counting/prob or NT problems faster than the basic arithmetic ones.   But it doesn't matter any more until next year.

Thanks for the competition.

joml88, you're right that you probably could achieve a higher score by skipping problems that are easy but time-consuming.  The initial problems (the test starts with basic arithmetic) are essential for students who do not have experience or expertise with later subjects, but participants who can answer later problems would probably be better served to skip some of the initial problems.  If one were shooting for a perfect score, the question to ask of each question would be "Can I do this in 18 seconds?".  Surprisingly, many basic questions can take longer than this, while an easy question in a more advanced topic can often take only a few seconds.

Rep123max, because we (the company) intend to sell a book of our tests in a few years, the tests are copyrighted and cannot be publicly distributed.  That said, if people wish to discuss the problems they found most interesting, that is not a problem.  We do have a short sample test on our website, too.

In general, the test covers basic questions in topics from arithmetic to pre-calculus.  The first question asked the sum of three three-digit numbers, and the first half of the test was mostly arithmetic, Algebra I, and geometry.  The second half covers Algebra II, Trigonometry, Number Theory, Sequences & Series, Probability & Combinatorics, Statistics, Sets, and "Potpourri" (the final problem was a simple cryparithm).

I only received a 38. Meh... And I got the first two problems wrong. :P

Anyways, better start preppin' for more difficult competitions!

I thought the test was awesome- the format of the contest was definitely very different from what we were used to earlier. It was my first time taking the Fall Startup Event, and I believe that the biggest failing point was strategy- I didn't skip anywhere near enough questions that I knew how to do but could only do with some time expenditure. On some questions, like those that you know you could figure out but which you don't know how to do immediately, I would recommend skipping and possibly coming back.

In terms of results, I wound up with a 63 and my school did reasonably well, 11th I think?

I only ended up attempting seventy some odd problems and they were clustered toward the beginning, which might have been a poor distribution. Anyways, it was hard but a lot of fun!

In my opinion, the two main factors in the Fall Start-Up are how fast you can do the problems and how many "stupid mistakes" you make.  This year, the second was nice to me. ;)  I got 72, I think, so I placed 4th among sophomores (I was pretty happy, since that's 17 up from last year's score for me) and my school placed 4th.  I recall going through almost every problem in the beginning, then remembering, "Hey, I can skip problems!" and starting to skip on the third (or later) page or so.  There was one problem in particular that I KNOW I blew a substantial amount of time on, but overall, I'm happy with my results!

About distributing the previous year's problems: It's okay if it's just within our school, right?  Say, in the math club?

My friend beat me with an 81 (Evan Zhao). :(

On top of that, our teacher didn't read that we had to have one score from each grade, so we only had 3 scores, which added up to 216. If we had the 9th grade score, we probably would have won. Oh well.

I was a bit too careless (76). I'll try harder next year.

i do not understand, i got a 59 and my name is not on the list ;)

anyway would have been 14th in the 12 grade

pretty fun contest.  :D
I got a 54, which ranged from 12th-15th for sophomores.  Could have done better, but it was still a fun contest.

[quote="joml88"]Why are you guys posting in the [b]2004[/b] Fall Startup thread?[/quote]

good point,  :D

didn't realize that before


[tex]b_n= \left(\frac{1}{n+1}\right){ 2n \choose n}[/tex]

Show that [tex]b_{n+1} = \frac{2(2n+1)}{(n+2)}*b_n[/tex] for [tex]n \geq 0[/tex]

This is where I've come
[tex]\left(\frac{1}{n+2}\right){2n+2 \choose n+1} = \frac{4n+2}{(n+2)(n+1)}*{2n \choose n}[/tex]

I don't know where to go from here. Well, I'm not even sure what I've done is correct.

Thank you for any contribution.

You have \[b_n= \left(\frac{1}{n+1}\right){ 2n \choose n} =\frac{(2n)!}{(n+1)!n!}\]So, you obtain, after some simplification that \[\frac{b_{n+1}}{b_n}=\frac{(2(n+1))! \cdot n!}{(2n)!(n+2)!}=\frac{2(2n+1)}{n+2}\]

You allready have the answer.

The $n+1$ at the bottom and the $n \choose 2n$ is equal to $b_n$ and $4n+2 = 2(2n + 1)$


topology
geometry
3D geometry
real analysis
real analysis unsolved

Hilbert's space $ \mathcal{L}_{2}$ is subset of direct product $ \mathbb{R}^{\mathbb{N}}=\prod_{i\in \mathbb{N}}\mathbb{R}$.
Prove that the topology $ \mathcal{U}$ on $ \mathcal{L}_{2}$ which is defined by norm $ \parallel x\parallel =\sqrt{\sum_{i=1}^{\infty}(x_{i})^{2}}$ is different from relative topology which is induced by topology $ \mathcal{V}$ of product $ \mathbb{R}^{\mathbb{N}}$. Prove that on [b]Hilbert's square[/b] $ I^{\infty}=\prod_{i\in\mathbb{N}}[0,\frac{1}{i}]\subseteq \mathcal{L}_{2}\subseteq \mathbb{R}^{\mathbb{N}}$ topology $ \mathcal{U}$ and $ \mathcal{V}$ are equivalent.

The set $ \prod_{i\in\mathbb N}[0,1]$ is compact in the product topology, but not in the norm topology.
For the second part, first prove that the Hilbert cube is compact in the norm topology. Then use the general topological fact: a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. [url=http://www.mathreference.com/top-cs,haus.html]To put it another way[/url],
[quote]A compact Hausdorff space walks a fine line. Add more open sets and lose compactness. Take away open sets and the space is no longer Hausdorff.[/quote]


An oval circumscribes a football field (300 feet by 160 feet). The oval also goes through a flag post 100 feet away from two of the corners of the football field. Two circles, each centered at the two foci and has the radius of 100 yards, intersect each other at points P and Q. PQ intersects the rectangle at points P' and Q'. Find P'Q'.

[color=cyan]Wouldn't it be more reasonable to put a question that requires knowledge of high school mathematics on a forum not devoted to middle school-level stuff?[/color]

Actually all you need to do is to draw a diagram, a rough one, and you will get the answer. You could solve this in under a minute if you know what foci meant.

Foci: Two points inside an oval such that the distance between any point on the oval and the sum of the distances to these two points equal another sum of the distances to these two points from another point on the oval, no matter what.
...simply put, if you tie a string around the focus (singular form of "foci") and tie that string to a pencil, then t tie it to the other focus and try to write a circle, you get an oval.

Algebraically, If d1 and d2 dontes the distance from a point element of the oval to the foci and d3 and d4 denotes the distance from another point element of the oval, then you get d1+d2=d3+d4


$ abc\equal{}1$
Prove that:
$ \frac{ab}{a^5\plus{}b^5\plus{}ab}\plus{}\frac{bc}{b^5\plus{}c^5\plus{}bc}\plus{}\frac{ca}{c^5\plus{}a^5\plus{}ca}\leq1$

[quote="Xaenir"]$ abc \equal{} 1$
Prove that:
$ \frac {ab}{a^5 \plus{} b^5 \plus{} ab} \plus{} \frac {bc}{b^5 \plus{} c^5 \plus{} bc} \plus{} \frac {ca}{c^5 \plus{} a^5 \plus{} ca}\leq1$[/quote]
I'll work on only one fraction for now.
$ \frac {ab}{a^5 \plus{} b^5 \plus{} ab}$

$ \frac {abc^2}{a^5c^2 \plus{} b^5c^2 \plus{} abc^2*1}$

$ \frac {c}{a^5c \plus{} b^5c \plus{} a^2b^2c^2}$
By the rearrangement inequality,

$ a^5c^2 \plus{} b^5c^2 \plus{} a^2b^2c^3\ge (a\plus{}b\plus{}c)(a^2b^2c^2)$
So, by summing cyclically on the fractions we have, we get our LHS is less than or equal to
$ \frac{a\plus{}b\plus{}c}{(a\plus{}b\plus{}c)(a^2b^2c^2)}\equal{}1$

the step with the rearrangement is kinda iffy, I'm not sure if that worked or not.

this is my solution, I think it's nice
we have $ a^5\plus{}b^5\geq ab(a^3\plus{}b^3)$
and $ a^3\plus{}b^3\geq ab(a\plus{}b)$
therefore $ a^5\plus{}b^5\plus{}ab \geq ab(ab(a\plus{}b)\plus{}abc)\equal{} a^2b^2(a\plus{}b\plus{}c)$
then $ \sum \frac{ab}{a^5\plus{}b^5\plus{}ab} \leq \sum \frac{1}{ab(a\plus{}b\plus{}c)} \equal{}\sum \frac{c}{a\plus{}b\plus{}c}\equal{}1$

this ,or similar like this one was in IMO Shorlist


Can someone help me understand the Shoelace Theorem? I just can't seem to memorize it...

ALSO PLEASE POST SOME EXAMPLE PROBLEMS I COULD WORK ON!

[quote="Thunder365"]The Shoelace Theorem can give you the area for any n-sided polygon with points on the coordinate plane. The wiki article seemed a little confusing, so here is my explanation.

[hide="Easy Explanation of Shoelace Theorem"]
1.)Write all the coordinates going in order by clockwise or counterclockwise, and copy the first one at the end. Here is my example:
$ (2,1)$
$ (3,2)$
$ ( \minus{} 2,3)$
$ ( \minus{} 2, \minus{} 2)$
$ (2,1)$ [b]*Note how this one is copied from the top.[/b]
2)Draw diagonal lines connecting a y coordinate to the x coordinate that is below and to the left of it. Multiply.
..........$ X,Y$
..........$ (2,1)$
....$ 3 \equal{} (3,2)$Product of 1 and 3
$ \minus{} 4 \equal{} ( \minus{} 2,3)$Product of 2 and -2
$ \minus{} 6 \equal{} ( \minus{} 2, \minus{} 2)$Product of 3 and -2
...$ \minus{} 4 \equal{} (2,1)$Product of -2 and 2
I didn't draw the lines because I can't.
3) Now do the same thing on the other side by multiplying each x coordinate with the y coordinate below and to the right of it.
$ X,Y$
$ (2,1)$
$ (3,2) \equal{} 4$
$ ( \minus{} 2,3) \equal{} 9$
$ ( \minus{} 2, \minus{} 2) \equal{} 4$
$ (2,1) \equal{} \minus{} 2$
4) Your lines should now look like a laced shoe. Add both columns separately. The left column is $ \minus{} 11$, and the right one is $ 15$.
5) Take the absolute value of the difference between the two columns, which is $ | \minus{} 26|$, and divide by $ 2$, which gives $ \boxed{\boxed{\boxed{13}}}$ in this example[/hide]

To convince yourself that this method works, try this method with easy #'s like $ (1,1), ( \minus{} 1,1), ( \minus{} 1, \minus{} 1),$ and $ (1, \minus{} 1)$, which we know is a square with side 2 and area of 4.[/quote]

[quote="PowerOfPi"]Problem:
A quadrilateral has an area of $ 60$. The four vertices have coordinates of $ ( \minus{} 4,5), ( \minus{} 4, \minus{} 6), (x, \minus{} 3), (x,1).$ Find the positive value of $ x$ using the Shoelace Theorem.[/quote]

Also, there is some discussion about this theorem in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282647]Tip of the Week.[/url]

Is the answer to the example 4?

I used the shoelace thereom with the x and y coordinates and got -9x-24 and 6x+36.  The diference between the two I think is 15x+60(absolute value)  I divided it by two and subtracted 30 to get 15/2x=30.  I then got x=4.  Am I wrong?

You are right. However, an interesting thing about the shoelace theorem is that you have to put the vertices in clockwise order, otherwise sometimes with variables you can end up getting multiple answers. Assuming the clockwise order of the vertices is the order that was given in the question, that is correct.

Do you think the shoelace thereom will ever be covered in Alcumus?  I want to feel like I can truly master it.

[quote="Thunder365"]However, an interesting thing about the shoelace theorem is that you have to put the vertices in clockwise order, otherwise sometimes with variables you can end up getting multiple answers.[/quote]
I must comment on this too. The points only have to be in cyclic order, not necessarily clockwise.

[hide="@maybach"]
I suppose the Shoelace Theorem is part of geometry, and I don't think geometry Alcumus is coming out anytime soon.[/hide]


linear algebra
matrix
probability
combinatorics unsolved
combinatorics

In an $m \times n$ matrix with real numbers, we color at least $p$ of the greatest numbers in each column ($p \leq m$) and at least $q$ of the greatest numbers in each row ($q \leq n$).

Show that at least $pq$ elements of the matrix are colored twice.

This is Arthur Engel, [i]Problem-Solving Strategies[/i], Chapter 8, exercise 26.

[b]EDIT:[/b] Please ignore the posting below. The generalization proposed is incorrect. - Darij

[hide="SPAM"]What is more interesting: Is the following generalization true?

Let $ m$ and $ n$ be positive integers. For each $ 1\leq i\leq m$ and $ 1\leq j\leq n$, consider a real $ u\left( i,j\right)$ and two nonnegative reals $ a\left( i,j\right)$ and $ b\left( i,j\right)$ such that $ \sum_{i=1}^{m}a\left( i,j\right) =1$ for every $ j$ and $ \sum_{j=1}^{n}b\left( i,j\right) =1$ for every $ i$. Assume that:

- all $ mn$ reals $ u\left( i,j\right)$ are pairwisely distinct;

- for any $ i$ and any two $ j_{1}$ and $ j_{2}$ satisfying $ u\left( i,j_{1}\right) <u\left( i,j_{2}\right)$, we have $ b\left( i,j_{1}\right) \leq b\left( i,j_{2}\right)$;

- for any $ j$ and any two $ i_{1}$ and $ i_{2}$ satisfying $ u\left( i_{1},j\right) <u\left( i_{2},j\right)$, we have $ a\left( i_{1},j\right) \leq a\left( i_{2},j\right)$.

In other words, if we consider the matrix having the entry $ u\left( i,j\right)$ in the $ i$-th row and the $ j$-th column, then the reals $ a\left( i,j\right)$ determine a probability distribution on every column of the matrix, and this probability distribution is, so to speak, "monotonically increasing with respect to the $ u\left( i,j\right)$ ". Similarly for the $ b\left( i,j\right)$ instead of $ a\left( i,j\right)$ and rows instead of columns.

Now, my conjecture is that $ \sum_{i=1}^{m}\sum_{j=1}^{n}a\left( i,j\right) b\left( i,j\right) \geq 1$. Is this true? This would yield Engel's problem since, denoting by $ u\left( i,j\right)$ the entry in the $ i$-th row and the $ j$-th column of the matrix, and setting

$ a\left( i,j\right) =\left\{\begin{array}[c]{c}\frac{1}{a},\text{ if }u\left( i,j\right) \text{ is one of the }a\text{ largest numbers in its column;}\\ 0,\text{ if not}\end{array}\right.$;
$ b\left( i,j\right) =\left\{\begin{array}[c]{c}\frac{1}{b},\text{ if }u\left( i,j\right) \text{ is one of the }b\text{ largest numbers in its row;}\\ 0,\text{ if not}\end{array}\right.$,

we would immediately get the problem's statement from the conjectured generalization.

Anyone willing to check whether the generalization holds?[/hide]

  Darij


Hi, I have some problems in finding the value of the sum $\sum_{k=0}^{mn-1}{(-1)^{[k/n]+[k/m]}}$. In fact, I need to show that if $n>m$ are odd and $n$ is not a multiple of m then the sum is equal to 1. I've done the other cases, but this one... Help. :(

This one is not so easy... On the integer line, mark points 0, n, 2n, 3n, ... , m, 2m, ... , mn. So, segment (0,mn) is divided by this points into smaller segments. Mark them black and white (alternating). So, if k is in the black segment, in your sum you have summand +1, if it's white then you have -1. You only need to prove that total lenght of black segments is greater by 1 than the total lenght of white segments. Indeed, it is the Balkan MO 2004 problem 4 in different form! But you need that m and n are relatively prime. I checked n=15, m=9 and the result is 9, not 1.

Yes, this one with coprime numbers is solved in http://www.mathlinks.ro/Forum/viewtopic.php?t=5332

Yes, the one with coprime numbers I knew from Balkan olympiad, but I am really amazed, since this problem was given in an oral examination in France at a school that is not exactly the best. I wonder if the examinator was ill or not.   :D

Don't know if that can help you but I have found (I hope there is no errors):
\[\sum_{k=0}^{mn-1}(-1)^{[k/n]+[k/m]}=\sum_{i=0}^{m-1} (-1)^i \sum_{k=0}^{n-1} (-1)^{[(in+k)/m]}-1\]

Oral Mines Ponts. Ok, I've finally found a short smart solution. I will post it soon. Still, the examinator remains idiot! No professor from Louis le Grand solved it.

You have found the solution for what?  :?
I have computed that sum for $n=15$, $m=9$ and obtained 9. So what we are to prove?  :?

I have found the solution for the general case. The answer is simpler than expected. Let's consider $ f(m,n)$ our ugly sum. Let us write $ m=du, n=dv$ with $(u,v)=1$. Then using the fact that $ [(j+kd)/(du)]=[k/u]$ for all j between 0 and d-1, we easily find that $ f(m,n)=d\cdot\sum_{k=0}^{duv-1}{(-1)^{[k/u]+[k/v])}}$. Next, write this sum as sum between 0 and $ uv-1$, then sum between $uv$ and $2uv-1$ and so on and you will find that $ f(m,n)=d(1+(-1)^{u+v}+...+(-1)^{(d-1)(u+v)})f(u,v)$ . Next, it is easy to see that in fact if $u+v$ is odd we have $ f(u,v)=0$ and thus let us consider u+v even, that is u and v are odd. Thus $f(m,n)=d^2 f(u,v)$. Furthermore, par simmetry we may assume that $u$ is smaller than v.  Next, if we group our terms of the sum, we find that [code]$ f(u,v)=\sum_{j=0}^{u-1}\sum_{k=0}^{v-1}{(-1)^{j+[(vj+k)/u]}}$. Since $u$ is odd, we have $(-1)^x=(-1)^{ux}$ and thus we can write $ f(u,v)=\sum_{j=0}^{u-1}\sum_{k=0}^{v-1}{(-1)^{ju+u[(vj+k)/u]}}$. Now, since $ u[(vj+k)/u]=vj+k-(vj+k)(mod u)$ and since $(-1)^{(u+v)j}=1$ and $ (vj+k)(mod u)$ are the numbers 0,1,...,u-1 for all k fixed and j going from 0,1,...,u-1, we find $f(u,v)=1$. Thus, the conclusion.


function
limit
floor function
algebra
binomial theorem
calculus
calculus computations

Find $\lim_{n \rightarrow+\infty}\sqrt{n}.(1-\max_{1\leq k \leq n}(\sqrt{k}-[\sqrt{k}]))$

(D.Mitin)

[x] is floor function

Claim:

$\max_{1\le k\le n}\left(\sqrt{k}-\left\lfloor\sqrt{k}\right\rfloor\right)=1-\frac1{2\sqrt{n}}+o(n^{-1/2})$

According to that claim, the limit asked for is $\frac12.$

If $n=m^{2}-1$ for some positive integer $m$ then surely the maximum is achieved for $k=n=m^{2}-1.$ In that case, by the binomial theorem,

$\sqrt{k}=\sqrt{m^{2}-1}=m-\frac1{2m}+o(m^{-1})$

$\sqrt{k}-\left\lfloor\sqrt{k}\right\rfloor=1-\frac1{2m}+o(m^{-1})$

Now suppose $m^{2}-1\le n\le (m+1)^{2}-2.$ The maximum should still be the same. If we use $k=(m+1)^{2}-2,$ then we would get $\sqrt{k}=m+1-\frac2{m+1}+o(m^{-1})$ and

$\sqrt{k}-\left\lfloor\sqrt{k}\right\rfloor=1-\frac2{m+1}+o(m^{-1})$

For sufficently large $m,$ that is smaller than our previous estimate.

Hence for $m^{2}-1\le n\le (m+1)^{2}-2,$ we have that

$\max_{1\le k\le n}\left(\sqrt{k}-\left\lfloor\sqrt{k}\right\rfloor\right) = 1-\frac1{2m}+o(m^{-1})$

We can then manipulate $m^{2}-1\le n\le (m+1)^{2}-2$ into

$\sqrt{n+1}\le m\le \sqrt{n+2}-1.$

From this we see that

$1-\frac1{2m}+o(m^{-1})=1-\frac1{2\sqrt{n}}+o(n^{-1/2})$


A number is randomly selected from the even integers 2 through 100. What is the probability that the number is a multiple of 6? Express your answer as a common fraction


logarithms
integration
probability
function
calculus
limit
number theory

$ \log f(x) \equal{} \int_2^x\log (1 \minus{} \frac {f(t)}{t})\ dt\ \ \forall x > 2$
given that $ (i)f(2) \equal{} 1,(ii)f(2) \equal{} a$ for some a $ 0<a<1$ find $ f(x)\ \ \forall x > 2$
Thank you.

$ \ln f(x) \equal{} \int_{2}^{x}{\ln \left(1 \minus{} \frac {f(t)}{t}\right)dt}$
Here should be $ f(x) > 0$ and $ f(x) < x$. Then, $ \ln \left(1 \minus{} \frac {f(t)}{t}\right) < 0$ and $ \int_{2}^{x}{\ln \left(1 \minus{} \frac {f(t)}{t}\right)dt} < 0$. So $ f(x) < 1$.
Finally $ 0 < f(x) < 1$  $ \forall x > 2$.

Alsow $ \ln f(x) \equal{} \int_{2}^{x}{\ln \left(1 \minus{} \frac {f(t)}{t}\right)dt} > \int_{2}^{x}{\ln \left(1 \minus{} \frac {1}{t}\right)dt} \equal{} \ln {\frac {4(x \minus{} 1)^{x \minus{} 1}}{x^x}}$

So: $ f(x) > \frac {4(x \minus{} 1)^{x \minus{} 1}}{x^x} \equal{} \frac {4(1 \minus{} \frac {1}{x})^{x}}{x \minus{} 1}\to \frac {4}{e(x \minus{} 1)}$

Now: $ \frac {f'(x)}{f(x)} \equal{} \ln \left(1 \minus{} \frac {f(x)}{x}\right)$ and we have $ f'(x) < 0$. This is all I can say...

Thank you very much for your response.

Although I dont have the answer yet I find the hint that this is not an easy problem from your post.That is why I feel the need to say what is the aim of this problem(I created it). I created it from number theory. I understood that $ 0 < f(x) < 1$  $ \forall x > 2$ because I designed it to be a probability function - to be more precise the probability of $ x$ being a prime. I know that you think it does not make sense since $ x$ is real.

Some random integer has probability $ (1 \minus{} \frac {1}{p})$ of not being devisible by the prime $ p$.Thus the probability of $ n$ being a prime is $ \prod (1 \minus{} \frac {1}{p_i})$ where $ p_i$ are the primes $ \leq n$.In that case I have to know the primes $ \leq n$. I can't know all of them for arbitrary $ n$. But I have the individual probabilities of them being prime. I thought rather than knowing them all I might consider the thread of reals where no number is completely prime or composite, but weighted by the probability of them being prime I can take them all in the product.The rest you can easily understand.However earlier i assigned the probability of $ 2$ being prime to be $ 1$. But I am not sure about it. So I have edited. Nor do I have where to start the integration(in this case 2).This is NOT any kind of proof or anything. I just want to do some experiments.

I plan to integrate the function to find the function that approximates the number of primes $ \leq n$.

Another discontinuous approach is: http://www.mathlinks.ro/viewtopic.php?t=192125

I am sorry if it appears nonsense to you. Then you could just solve the problem or prove it is wrong. I would be happy with that.

I am not much interested about the proof because I wont understand anything of it.Being 14 years old I know VERY LITTLE calculus. For example I didnt understand how $ \int_{2}^{x}{\ln \left(1 \minus{} \frac {1}{t}\right)dt} \equal{} \ln {\frac {4(x \minus{} 1)^{x \minus{} 1}}{x^x}}$. But I would very much like to see the answer.

The problem with using this to estimate the number of primes is that there's no such thing as a "random integer", uniformly distributed. For each $ n$, there is a uniform distribution on $ \{1,2,\dots,n\}$- but the events of being divisible by $ p_k$ aren't exactly independent anymore.

If you work with the ideas enough, you eventually come to analytic number theory, and the Prime Number Theorem. It's an advanced subject, which draws together a [i]lot[/i] of different areas. If you're just starting calculus, the proof is guaranteed to be over your head.

Thank you sir.
I never heard of the Prime Number Theorem before. So I was trying these things. I hope I shall understand these things when I grow up and learn enough mathematics.

\[ \ln \left(1 \minus{} \frac {1}{t}\right) \equal{} \ln \left(\frac {t \minus{} 1}{t}\right) \equal{} \ln \left(t \minus{} 1\right) \minus{} \ln t
\]

\[ \int {\ln \left(t \minus{} 1\right) dt} \equal{} (t \minus{} 1)\left(\ln \left(t \minus{} 1\right) \minus{} 1\right) \equal{} \ln \left(t \minus{} 1\right)^{t \minus{} 1} \minus{} t \plus{} 1
\]

\[ \int {\ln t dt} \equal{} t\left(\ln t \minus{} 1\right) \equal{} \ln t^t \minus{} t
\]

\[ \int {\ln \left(1 \minus{} \frac {1}{t}\right)dt} \equal{} \int{\ln \left(t \minus{} 1\right)dt} \minus{} \int{\ln t dt} \equal{} \ln \left(t \minus{} 1\right)^{t \minus{} 1} \minus{} t \plus{} 1 \minus{} \ln t^t \plus{} t \plus{} C \equal{}
\]

\[ \equal{} \ln \left(t \minus{} 1\right)^{t \minus{} 1} \plus{} 1 \minus{} \ln t^t \plus{} C
\]

\[ \int_{2}^{x} {\ln \left(1 \minus{} \frac {1}{t}\right)dt} \equal{} \ln \left(x \minus{} 1\right)^{x \minus{} 1} \plus{} 1 \minus{} \ln x^x \minus{} 1 \plus{} \ln 2^2 \equal{} \ln \frac {4(x \minus{} 1)^{x \minus{} 1}}{x^x}
\]
To your condition $ (ii)$:
Look, if we take $ \lim_{x\to 2 \plus{} }{f(x)} \equal{} a < 1$, but we have: $ f(x) > \frac {4(x \minus{} 1)^{x \minus{} 1}}{x^x}$ and if we take $ x\to 2 \plus{}$ here we will get $ \frac {4(x \minus{} 1)^{x \minus{} 1}}{x^x}\to 1$, so I think $ a$ could not be $ < 1$. This means that if $ f(2) \equal{} a < 1$ then does not exist such function $ f(x)$.

Now, we got the equation:
\[ f'(x) \equal{} f(x)\ln\left(1 \minus{} \frac {f(x)}{x}\right)
\]
We know, that for small $ t$ we have $ \ln (1 \plus{} t) \equal{} t \plus{} O(t^2)$. So for big $ x$ we can approximatlely write:
\[ f'(x) \approx \minus{} f(x)\cdot \frac {f(x)}{x}
\]
Let $ f(x) \equal{} f$, we will have:
\[ \frac {df}{f^2} \approx \minus{} \frac {dx}{x}
\]

\[ \int \frac {df}{f^2} \approx \minus{} \int\frac {dx}{x}
\]

\[ \minus{} \frac {1}{f} \approx \minus{} \ln x \minus{} \ln C
\]
So, we have:
\[ \boxed{f(x) \approx \frac {1}{\ln x \plus{} \ln C}}
\]
$ C$ is some constant... This only can show the speed of changing of the function $ f(x)$ for big values of $ x$.

I solved this differential equation by computer by RGK4 with condition $ f(2) \equal{} 1$ and found, that $ f(1000000) \approx 0.06996593$, so we can find the value of $ C$.
\[ C \equal{} \frac {1}{1000000}e^{\frac {1}{0.06996593}} \approx 1.611491539\approx 1.61149
\]
Finally for big $ x$ we have:
\[ \boxed{f(x) \approx \frac {1}{\ln x \plus{} 0.47716}}
\]
Even for $ x \equal{} 3$ our formula gives $ 0.635$ and the real answer is $ 0.668$. So I think we could use it with large exactness.  :jump:

[quote="Yuriy Solovyov"]I solved this differential equation by computer by RGK4 with condition $ f(2) = 1$ and found, that $ f(1000000) \approx 0.06996593$, so we can find the value of $ C$.
\[ C = \frac {1}{1000000}e^{\frac {1}{0.06996593}} \approx 1.611491539\approx 1.61149
\]
Finally for big $ x$ we have:
\[ \boxed{f(x) \approx \frac {1}{\ln x + 0.47716}}
\]
Even for $ x = 3$ our formula gives $ 0.635$ and the real answer is $ 0.668$. So I think we could use it with large exactness.  :jump:[/quote]

Thank you again sir Solovyov for putting so much effort into my problem.

Do you mean $ \frac {1}{\ln(3)}$ by "real answer is $ 0.668$".I found here( http://en.wikipedia.org/wiki/Prime_number_theorem ) that the original Prime Number Theorem states that the function should be $ \frac {1}{\ln(x)}$. $ C$ will become more and more negligable when $ x$ becomes larger and larger. I cant believe I got soooo close to getting and proving The Prime Number theorem. I think my thoughts can be modified and transformed into a proof of The Prime Number Theorem.

I request mathlinkers to think about it too. (Dont get the impression that I am not thinking about it leaving all work to others. :D )

So could anyone integrate the function ${ \frac {1}{\ln x + 0.47716}}$? then I can check how much close it is to $ \pi (x)$

Those functions don't have elementary antiderivatives. We can estimate $ \int_a^b \frac1{\ln x\plus{}c}\,dx\equal{}e^{\minus{}c}\int_{ae^c}^{be^c}\frac1{\ln t}\,dt$, but we can't evaluate it exactly because the antiderivative $ \text{Li}$ of $ \frac1{\ln t}$ is not an elementary function.

Oh! That's bad news.
I just took a closer look at the solution. Though I dont understand most of it I realise there are some approximations. But cant we calculate it more efficiently :?:. I am asking for it because my aim is to integrate the function(may be not algebrically but numerically?) and that would multiply the error greatly. And we again cannot integrate the function but approximate it. And it will be best if we can directly calculate the antiderivative without getting $ f(x)$. And why could not we just calculate $ \ln C$ by $ \lim_{x\rightarrow 2^ \plus{} } f(x)$ that gives $ \ln C \equal{} 1 \minus{} \ln 2\approx 0.3068528194$. And I am again asking what you meant by "the real answer"?

Please dont mind if I write something wrong because I am new to calculus.

Despite all the approximations $ \int_2^x f(x)\ dx$ is (according to mathematica) approximately $ 96.0\%$ and $ 97.0\%$ accurate to $ \pi(x)$ at $ x \equal{} 100000$ and $ x \equal{} 10000000$ respectively if we use $ \ln C \equal{} 0.47716$ and approximately $ 97.5\%$ and $ 98\%$ accurate respectively if we use $ \ln C \equal{} 0.3068528194$

An asymptotic calculation of $ \text{Li}$:
$ \int\frac1{\ln(x)}\,dx\equal{}\frac{x}{\ln x}\plus{}\int\frac1{(\ln x)^2}\,dx\equal{}\frac{x}{\ln x}\plus{}\frac{x}{\ln^2 x}\plus{}\int\frac{2}{\ln^3 x}\,dx$
$ \equal{}\cdots\equal{}\sum_{k\equal{}0}^{n\minus{}1} \frac{k!x}{\ln^{k\plus{}1}x}\plus{}\int\frac{n!}{\ln^{n\plus{}1}x}\,dx$

If you apply this to definite integrals, the series diverges. Taking too many terms is fatal. What you do get are accurate estimates of how fast it grows at $ \infty$. The leading term $ \frac{N}{\ln N}$ is a decent estimate, but the full integral is better.

I don't try to prove that $ \int_{2}^{x}{f(x)dx} \approx \pi (x)$.
I only sayed, that my formula $ f(x)\approx \frac {1}{\ln x \plus{} 0.47716}$ is true for big values of $ x$. I've solved differentianl equation with this condition.
We can't calculate $ \ln C \equal{} 1 \minus{} \ln 2$ because we can't write $ \ln(1 \minus{} t)\approx \minus{} t$. This is true for small $ t$. But in our case $ t \equal{} \frac {1}{2}$ and it's not small.
$ \ln\left(1 \minus{} \frac {1}{2}\right) \equal{} \minus{} 0.69314\ne \minus{} 0.5$, but for $ x \equal{} 10000$ for example:
$ \ln\left(1 \minus{} \frac {1}{10000}\right) \equal{} \minus{} 0.0001000050003\approx \minus{} 0.0001$

"Real answer" is the value, that I got by solving this equation using RGK4. This is the method of numeral decision of differential equations.
For example:
$ f(3) \equal{} 0.668304318318980518$
$ f(4) \equal{} 0.551605603440041392$
$ f(10000000) \equal{} 0.0602581313654575804$

Now let's clculate this value by my formula:
$ f(3) \equal{} \frac {1}{\ln 3 \plus{} 0.4771601719} \equal{} 0.6346093898$
$ f(4) \equal{} \frac {1}{\ln 4 \plus{} 0.4771601719} \equal{} 0.5366377243$
$ f(10000000) \equal{} \frac {1}{\ln 10000000 \plus{} 0.4771601719} \equal{} 0.06025818528$

AS we can see my formula for big $ x$ approximate real solution. Now let's calculate this using your constant $ C$:
$ f(3) \equal{} \frac {1}{\ln 3 \plus{} 0.3068528194} \equal{} 0.7115082362$
$ f(4) \equal{} \frac {1}{\ln 4 \plus{} 0.3068528194} \equal{} 0.5906161093$
$ f(10000000) \equal{} \frac {1}{\ln 10000000 \plus{} 0.3068528194} \equal{} 0.06088299162$

As we can see for big $ x$ my formula is much closer to real answer. :) And we can use it only for big $ x$ if we want an exact result.


calculus
derivative
inequalities
inequalities proposed

Why should we solve it without using derivatives?
Jensen solves this and the other two inequalities easily.

[quote="shoki"]for $ a,b,c,d > 0$ with

$ \sum a^2 \equal{} 1$

prove that

 $ \sum \frac {1}{1 \minus{} a} \ge 8$

sorry if it has been posted.
since i want to see a solution different of mine plz solve it without using derivatives and things like that.thx for your help :)[/quote]
The inequality is equivalent to :

$ \sum \frac{a}{1\minus{}a} \geq 4$

From Caushy-Shcwarz inequality we have :

$ \sum \frac{a}{1\minus{}a}. \left( \sum a \minus{} 1 \right) \geq \left( \sum a \right)^2$

It suffices to prove that :

$ \left( \sum a \right)^2 \geq 4 \sum a \minus{} 4$

Which is equivalent to :

$ \left( \sum a \minus{}2 \right)^2 \geq 0$

Done !

[quote="rachid"][quote="shoki"]for $ a,b,c,d > 0$ with

$ \sum a^2 \equal{} 1$

prove that

 $ \sum \frac {1}{1 \minus{} a} \ge 8$

sorry if it has been posted.
since i want to see a solution different of mine plz solve it without using derivatives and things like that.thx for your help :)[/quote]
The inequality is equivalent to :

$ \sum \frac {a}{1 \minus{} a} \geq 4$

From Caushy-Shcwarz inequality we have :

$ \sum \frac {a}{1 \minus{} a}. \left( \sum a \minus{} 1 \right) \geq \left( \sum a \right)^2$

It suffices to prove that :

$ \left( \sum a \right)^2 \geq 4 \sum a \minus{} 4$

Which is equivalent to :

$ \left( \sum a \minus{} 2 \right)^2 \geq 0$

Done ![/quote]

seems that sth is going wrong with rachid's solution,can anyone find a proof for this?

[quote="shoki"]seems that sth is going wrong with rachid's solution,can anyone find a proof for this?[/quote]
[quote="FelixD"]$ \sqrt {a} \ne a$.[/quote]
Dear shoki and FelixD,my solution is correct  :wink:

BTW $ \sum a \minus{} 1 \equal{} a\plus{}b\plus{}c\plus{}d\minus{}1 \equal{} a(1\minus{}a) \plus{} b(1\minus{}b) \plus{} c(1\minus{}c) \plus{} d(1\minus{}d)$  :lol:

Rachid's solution is correct. Note that $ \sum a(1\minus{}a)\equal{}\sum (a\minus{}a^2)\equal{}(\sum a)\minus{}1.$

ah sorry rachid,i'm new in inequality....
thx again for your solution.

I thought he means $ \sum (a\minus{}1)$.
But again
@shoki: why shouldn't we use derivatives?

in fact we can prove that :

$ \frac{1}{1\minus{}a} \ge 4a^2\plus{}1$

$ \leftrightarrow 0 \ge \minus{}a(2a\minus{}1)^2$

by using this inequality for each fraction we are done!


1.Let x,y be positive real numbers satisfy: $ x^3\plus{}y^3\equal{}x\minus{}y$
Prove that: $ x^2\plus{}y^2\leq 1$
2.$ 0 \leq a,b,c\leq1$ and $ a\plus{}b\plus{}c\equal{}2$.Prove that: $ a\plus{}4abc\leq2$

[quote="dulldrake"]1.Let x,y be positive real numbers satisfy: $ x^3 \plus{} y^3 \equal{} x \minus{} y$
Prove that: $ x^2 \plus{} y^2\leq 1$
2.$ 0 \leq a,b,c\leq1$ and $ a \plus{} b \plus{} c \equal{} 2$.Prove that: $ a \plus{} 4abc\leq2$[/quote]
IT VERY EASY

[quote="dulldrake"]
2.$ 0 \leq a,b,c\leq1$ and $ a \plus{} b \plus{} c \equal{} 2$.Prove that: $ a \plus{} 4abc\leq2$[/quote]
The following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}2.$ Prove that
\[ a \plus{} 4abc\leq2\]

Rewrite the inequality as follows: $ b\plus{}c\ge 4abc\equal{}4(2\minus{}b\minus{}c)(bc)$

If $ s\equal{}b\plus{}c$ and $ p\equal{}bc$, then $ 4\ge s^2\ge 4p\ge 0$...Fix $ s$, the inequality is strongest when $ p$ as large as possible:
$ s\minus{}4(2\minus{}s)p\ge s\minus{}(2\minus{}s)s^2\equal{}s(1\minus{}2s\plus{}s^2)\equal{}s(s\minus{}1)^2\ge 0$.

Equality iff $ s^2\equal{}4p$ and $ s\equal{}0$ or $ s\equal{}1$... $ (a,b,c)\equal{}(2,0,0)$ or $ \left(1,\frac{1}{2},\frac{1}{2}\right)$.

For number 1, we have
$ x^2 \minus{} y^2 \equal{} (x^3 \plus{} y^3)(x \plus{} y) \ge (x^2 \plus{} y^2)^2$
So we can assume
$ x^2 \minus{} y^2 > 1$
Or else the given would be true.
But then that means $ x > 1$, but from the given, $ y^3 \plus{} y \equal{} x \minus{} x^3$, and $ x \minus{} x^3$ is negative for $ x > 1$, contradiction.
What is the easier proof?

Here is my solution for 2
We can assume $ a >\frac{2}{1\plus{}4bc}$, or the problem is true. Then,
$ 2\equal{}a\plus{}b\plus{}c >\frac{2}{1\plus{}4bc}\plus{}b\plus{}c$
$ 2(1\plus{}4bc) >2\plus{} (b\plus{}c)(4bc\plus{}1)$
$ 8bc >(4b^2c\plus{}c)\plus{}(4bc^2\plus{}b) \ge 8bc$ by AM-GM, contradiction

[quote="dgreenb801"]I have a long solution for 1.
What is the easier proof?[/quote]
The easier proof is here:
$ \frac {x^3 \plus{} y^3}{x \minus{} y} \equal{} 1$
Now it's sufficient to prove that: $ x^3 \plus{} y^3 > \equal{} (x \minus{} y)(x^2 \plus{} y^2)$, which is equivalent to: $ 2y^3\plus{}x^2y \minus{} xy^2\geq0$, which is obviously true since $ x>y$.
The equality doesn't occur.


A $ 10$ unit by $ 10$ unit square is disassembled into unit squares. Two separate squares are then built using all of these unit squares. What is the edge length of the smaller
of these two squares?

A  unit by  unit square is disassembled into unit squares. Two separate squares are then built using all of these unit squares. What is the edge length of the smaller
of these two squares?

the only 2 integer squares that can fit into 100 unit squares are 64 and 36, and 36=6 x 6 which the answer is 6.


What is the positive difference between the sum of the first 100 positive odd integers and the sum of the first 100 positive even integers?

The formula for the sum of the first $ n$ odd numbers is $ n^2$, and the formula for the sum of the first $ n$ even numbers is $ n(n\plus{}1)$. So we have:

$ |100^2 \minus{} (100)(101)| \implies |10000 \minus{} 10100| \implies \boxed{100}$

The sum of the first 100 even integers is $ 2(1\plus{}2\plus{}\cdots\plus{}100)\equal{}2\plus{}4\plus{}\cdots\plus{}200\equal{}100(101)\equal{}10100$.  The sum of the first 100 odd numbers is simply the difference between the first 200 numbers and the first 200 even numbers.  So that is $ \frac{200(201)}{2}\minus{}10100\equal{}100(201)\minus{}10100\equal{}10000$.  Then $ \left|10100\minus{}10000\right|\equal{}100$.

Alternatively, $ (2\minus{}1)\plus{}(4\minus{}3)\plus{}(6\minus{}5)\plus{}\cdots\plus{}(200\minus{}199)\equal{}100$.

I just had this problem, and apparently the answer is 72. Is this is a bug or am I missing something?

Between us, it looks like we have come up with 2-3 ways to reach the same answer of 100.  It's probably a bug.[/hide]


Suppose that 3 lines are given, having length 1, a , and b.

(a) Describe a compass-and straight edge construction for a line segment of length $a \times b$ (multiplication)

(b) Describe a compass-and straight edge construction for a line segment of length $\sqrt{a}$ (multiplication)

How would I go about doing this?

this really isnt calculus but....

 consider the line OA with length a.draw a line with length 1 at O making any angle of your choice with OA.label its end point M. connect M and A. measure out b units on the line OM and construct a line parallel to MA. the intercept of this line on the line OA gives youa*b. similiarly for sqrt a.


Ten cars containing a total of $ 32$ people passed through a checkpoint. If none of these cars contained more than $ 4$ people, what is the greatest possible number of these cars that could have contained exactly $ 2$ people?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

The official solution on the website showed a guess and check method. How would you do this without guess and check? Does it involve the Pigeonhole Principle? (I am not very familiar with it nor am I familiar with how to use it.)

first assume that every car carries exactly 2 people.
then we will need 16 cars.
now there are only 10 cars and none of these cars containes more than 4 people
we can just combine 6 cars with other 6

in the end
we have 4 cars with 2 people and 6 cars with 4 people

[hide="Solution"]If all $ 10$ cars had $ 2$ people in each of them, there would be a total of $ 20$ people. There are, however, $ 32$ people, so those extra $ 12$ people also need to be somehow arranged amongst the $ 10$ cars, while keeping the number of cars with exactly $ 2$ people at a maximum.

To accomplish this, those $ 12$ people are split up into pairs, and a pair is placed in each of $ 6$ of the cars.  Thus, those $ 6$ cars now each have $ 4$ people in them, while the remaining $ 4$ cars each have $ 2$ people in them. There is no other way to arrange the people amongst the cars such that the number of cars with exactly $ 2$ people in them would be greater, so the answer is $ {\text{(D)}\ 4}$.[/hide]


ARML
geometry
analytic geometry
algebra
polynomial

1.
Compute $1^2 - 2^2 + 3^2 - 4^2 + \cdots - 1998^2 + 1999^2$.

2.
Find all real values of $m$ such that both roots of the equation
\[ x^2 - 2mx + m^2 - 1 = 0 \]
are greater than -2 but less than +4.

3.
An experiment is performed in which each trial consists of tossing an ordinary six-sided die repeatedly and adding the numbers that come up; in each trial, as soon as the total exceeds 15, we stop tossing the die.  For this experiment, what final total is expected to occur most often?

4.
Equilateral triangle $ABC$, with side 2, is in plane $P$.  Segments $\overline{BX}$ and $\overline{CY}$ are drawn perpendicular to plane $P$, both on the same side of the plane.  If $BX=1$ and $CY=3$, compute the area of triangle $AXY$.

5.
In triangle $ABC$, $a=11$, $b=13$, and the median to side $c$ is 10.  Compute the area of the triangle.

6.
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3 = 0$.  If the maximum and minimum values of $x^2+y^2$ are $M$ and $m$ respectively, compute the numerical value of $M-m$.

7.
The sum
\[ \frac{1}{1!9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!1!} \]
can be written in the form $\frac{2^a}{b!}$, where $a$ and $b$ are positive integers.  Find the ordered pair $(a, b)$.

8.
The equations $75x^4+ax^3+bx^2+cx+12=0$ and $12x^5+dx^4+ex^3+fx^2+gx+75=0$ have a common rational root $k$ which is nonintegral and negative.  The number $k$ is also a root of $3hx^3+hx^2+jx+40h=0$.  The coefficients in the equation are integers, and $h\not=0$.  Compute the other two roots of this last equation.

9.
The roots of the equation $x^5-40x^4+Px^3+Qx^2+Rx+s=0$ are in geometric progression.  The sum of their reciprocals is 10.  Compute the numerical value of $|S|$.

10.
The polar coordinates of points $A$, $B$, and $C$ are $A(2, 70^\circ )$, $B(2\sqrt{3}, 40^\circ )$, and $C(3, 10^\circ )$.  Compute the area of triangle $ABC$.

Can you tell me which of these (in spoiler tags) are correct before I put solutions?

01. [hide]1999000[/hide]

02. [hide](-2,3)[/hide]

03. [hide]16[/hide]

04. [hide]sqrt10[/hide]

05. [hide]66[/hide]

06. [hide]8[/hide]

07. [hide](1,10)[/hide]

08.

09.

10. [hide]sqrt3[/hide]

I'm not very good with polynomials.

#8: Do you know the rational roots theorem?

#9: What relationships do you know between the roots and the coefficients of a polynomial?

#2 You're half right.  Check your inequalities.

#7 You're half right.  What kind of expression are relating this to?

I know both those things for polynomials, but  I guess I need more practice applying them. Now I get [hide]+/- 2i sqrt10[/hide] for #8. I'll look at #9 some more.

I made careless mistakes on #2 and #7. I guess that's why it helps to write up your solutions as you go.. I'm still too tired right now. Now I get #2: [hide](-2,4)[/hide] and #4: [hide](9,10)[/hide]. Are the rest correct?

On a side note, I keep having dreams/nightmares about math. Is that bad? In particular, last night's dream was your response to this topic (before I saw the real one), and you were just posting nonsense in caps and bold. I got 4 of them right, but the number of problems kept changing. I think it's anxiety.

[quote="Cicatriz"]
On a side note, I keep having dreams/nightmares about math. Is that bad? In particular, last night's dream was your response to this topic (before I saw the real one), and you were just posting nonsense in caps and bold. I got 4 of them right, but the number of problems kept changing. I think it's anxiety.  :?[/quote]

My hope is that students find math enjoyable once they get into it.  We all have anxiety over many things, but it's good to find ways to enjoy everything we do for what it is.

Cicatriz wrote:I made careless mistakes on #2 and #7. I guess that's why it helps to write up your solutions as you go.. I'm still too tired right now. Now I get #2: [hide](-2,4)[/hide] and #4: [hide](9,10)[/hide]. Are the rest correct?

Check #2 again.

[quote="MCrawford"]My hope is that students find math enjoyable once they get into it.  We all have anxiety over many things, but it's good to find ways to enjoy everything we do for what it is.[/quote]

I do enjoy it... I just generally have anxiety about almost anything.  And posting my answers right before I went to sleep is partly to blame. :maybe:


 I hope you will enjoy...  :)

 Prove that if $2n+1$ and $3n+1$ are both perfect squares then $5n+3$ is a composite number. For any n element of $Z$

Let $n=0$ , we get that $(2n+1)=1$ and $(3n+1)=1$ are both squares, and $(5n+3)=3$ is not composite, so I think it must be, $n \in \mathbb{Z}^+$.

So, we will assume that $n \in \mathbb{Z}^+$.

Let $2n+1=k^2$ and $3n+1=m^2$.
From this equations we get that $1=3k^2-2m^2 \ldots (*)$ and $n=m^2-k^2$.
Then $5n+3=5(m^2-k^2)+3(3k^2-2m^2)=4k^2-m^2=(2k+m)(2k-m)$;
Suposse that $2k-m=1$ then $m=2k-1$, reemplacing this in $(*)$ we obtain $3k^2=1+2(2k-1)^2$ that have solutions $k=1$ and $k=\frac{5}{3}$, since $k$ is integer $k=1$, and with this $n=0$ (=><=), so $2k-m >1$ and $(2k+m)(2k-m)$ is composite.

$Tipe$

Very nice solution  :)  there is another proof which is more more easier please try to find that.
I think this problem looks like Germanys Selection Tests Number Theory problem  ;) ...

Davron is my name meams Live a life ... :lol:

Please everybody try to find another solution for this problem !!!

It is nice and i know that you will like that please all Math Killers my friends !!!  :)
May all you wishes come true !!!
Solve it
Zetax i know that you are a good at playing with Numbers try it please ...

Davron

Somebody has a solution ???  :?

I have a lot of curiosity!!  :lol:

$Tipe$

Is there anyone interested with this problem ?

Davron

Whats up men dont you wanna solve this or its too easy for you...? I cant understand  :oops:  Ok my hint use just one digit number to get the result. This time i hope that you will answer alooo everyone please.

Davron

Well, my solution is pretty much the same as Tipe's one..

Just notice that : $5n+3 = (2n+1)+(3n+1)+1$
But! $2n+1 = a^2$ and $3n+1 = b^2$, so : $5n+3 = a^2 + b^2 + 1$, but, like Tipe did, we can get that $3a^2 - 2b^2 = 1$.
So : $5n+3 = 4a^2 - b^2 = (2a+b)(2a-b)$..

:)

Ok thanks, but your proof is not finished yet...

Davron

I try using mod 10 but I only get that $n$ is congruent to 0 or 5 mod 10... if $n$ is congruent 5 mod 10 : (5n+3) is congruent 8 mod 10... composite !
I dont know if it's the correct way.  :?

$Tipe$

No its not my way, is mine is more nice...  :)

Davron

So.... What is your solution Davron  :?  ?

$Tipe$


algorithm
search
algebra
polynomial
graph theory

I have a problem that I'm trying to create a good algorithm to solve.  I wasn't sure where to post this, so I guessed.  Here goes.

You have some arbitrary number of units.  All of these units are defective and are caught by specific tests.  Each test has a time associated with it.  You need to determine a very good subset of these tests that will cover all units.  For example, you have the following tests:

Test 1:  10 units of time
Test 2:  8 units of time
Test 3:  3 units of time

and the following units:

Unit 1:  Fails tests 1,2
Unit 2:  Fails tests 3
Unit 3:  Fails tests 1,3

In this case, tests 2 and 3 cover all units for a total time unit cost of 11.

How would you proceed for a case of several thousand units and tests?  This clearly falls outside the realm of being able to brute force an optimal result by considering all permutations.  Currently I am using a greedy weighted algorithm as follows:

Algorithm A
1a)  determine weight for each test based on the formula (1/time unit cost)*(defective units caught)
2a)  select highest weight test, add this test to our final test set
3a)  remove all units failing that test from consideration
4a)  repeat 1-3 unitl all units have been covered

I found that I could improve this algorithm as follows:
Algorithm B
1b)  look for any units failing one test
2b)  add that test to final test set
3b)  remove all units failing that test from consideration
4b)  repeat 1b -3b until all single failures have been removed
5b)  begin Algorithm A with new reduced unit set

Here are the current assumptions I'm working under.
1)  There will be approximately 10000 units
2)  There will be 2-5000 tests
3)  Each unit will fail at most 50 tests
4)  Completely random failing tests (although likely this is a distribution)

I'm not looking for a definitive answer to this question, as I don't believe one exists :)  I am interested in possible similar problems and algorithms for solving them or just general input.

Thanks in advance :)

[quote="tj_thompson"]
Algorithm A
1a)  determine weight for each test based on the formula (1/time unit cost)*(defective units caught)
2a)  select highest weight test, add this test to our final test set
3a)  remove all units failing that test from consideration
4a)  repeat 1-3 unitl all units have been covered
[/quote]

I don't think that's enough

Watch this example:

Test 1: Units 1 Cost 1
Test 2: Units 4,5 Cost 4
Test 3: Units 2,3,4,5 Cost 10

The weights are 1, 0.5 and 0.4 respectively (they are sorted by the weight)

If we take them on-by-one, we have a cost $ 1+4+10=15$

But if we skip the Test-2, we cover all the units and we have a cost $ 1+10=11$

This is quite a famous (hard) problem - I suggest you google for Weighted Set Cover, you may find something useful.

Pontios --  You are correct, you can devise sets that would result in a non optimal coverage solution.  Unfortunately, the only way to guarantee optimality is to brute force search all permutations.  This becomes impossible with test sets larger than about 30 though as this is O(n!) (this could be reduced some by clever optimization, but it will be extremely processing intensive in any case).  So the real problem is determining the best approximation algorithm.

TripleM -- You are also correct, it does appear to be a famous problem.

Looks like the problem is a weighted vertex minimum cover problem dealing with a bipartite graph.

Now, I know that a vertex minimum cover problem is NP-hard.  If you are dealing with a bipartite graph, this can be solved in polynomial time due to Konig's theorem however.

Unfortunately, I think adding weighted vertices to the problem moves it back to NP-hard.  Haven't thought enough about this to decide for sure, and I'd welcome input here :)

Thanks for your replies, this is actually a real problem I'm trying to solve and will have to implement in one way or another.  I appreciate the input.


geometry
geometric transformation
reflection
function
algorithm
algebra
polynomial

How do was go about solving this problem...lets talk about it here...

 Suppose that you are organizing housing accommodations for a group of four hundred university students. Space is limited and only one hundred of the students will receive places in the dormitory. To complicate matters, the Dean has provided you with a list of pairs of incompatible students, and requested that no pair from this list appear in your final choice. This is an example of what computer scientists call an NP-problem, since it is easy to check if a given choice of one hundred students proposed by a coworker is satisfactory (i.e., no pair from taken from your coworker's list also appears on the list from the Dean's office), however the task of generating such a list from scratch seems to be so hard as to be completely impractical. Indeed, the total number of ways of choosing one hundred students from the four hundred applicants is greater than the number of atoms in the known universe! Thus no future civilization could ever hope to build a supercomputer capable of solving the problem by brute force; that is, by checking every possible combination of 100 students. However, this apparent difficulty may only reflect the lack of ingenuity of your programmer. In fact, one of the outstanding problems in computer science is determining whether questions exist whose answer can be quickly checked, but which require an impossibly long time to solve by any direct procedure. Problems like the one listed above certainly seem to be of this kind, but so far no one has managed to prove that any of them really are so hard as they appear, i.e., that there really is no feasible way to generate an answer with the help of a computer. Stephen Cook and Leonid Levin formulated the P (i.e., easy to find) versus NP (i.e., easy to check) problem independently in 1971.

-----------

Immunocal Distribution Center
San Francisco, CA
http://www.immunotec.com/buy

Theoretical computer science is, indeed, very interesting.  However it has little to do with the American Mathematics Competition, the subject of this forum.  A better place to discuss topics such as these would be the [url=http://www.artofproblemsolving.com/Forum/index.php?f=331]Computer Science and Informatics[/url] forum.

you r allowed to write about p vs np here.

Isn't that the exact same problem from the Clay Math Institute?
Just curious, lets say someone proves that P does not equal NP, then would there be any value in that result?

$P \neq NP$ would also be very useful. It will tell us that is is pointless trying to solve some problems. It will also prove that one-way functions DO exist and it will be a big breakthru in cryptography.

Most NP-Hard problems involve subset checking. For example, the clique problem asks if there is a complete subgraph of k nodes within a graph G.
A graph of n nodes has 2^n possible subsets of size ranging from 0-n. We may think of these subsets as the proper input to the problem except that these subsets happen to be compressed into a graph of n nodes. Think of 2^n files with sizes 0Kb to nKb. We compress these 2^n files into a .zip file and wish to see if there's a file of size k Kb in that .zip file with some particular property (maybe composed of all 1s), without having to extract all the files of size k Kb (there's n choose k such files).
I find it interesting to convert NP-Complete problems into more material problems. For example, the following problem is NP-Complete:
. You're given a grid composed of black and white boxes (think crossword puzzle), and are asked to determine the size of the biggest square composed entirely of black boxes by first reordering the rows and then the columns.

imho this problem is not enough studied with pen and paper but all
these people are searching for the answers where they are not,
a computer is really the dumbest thing ever,it is the classical yeti
with no brains but big with allot of muscles in the sense that it can do
allot of stuff fast but it will never think.
The closest thing to solving this problem would be genetic algorithms,
neural networks and stuff that is really not developed enough as to say
that it is really usefull.For this matter i think that instead of trying to make
a really stupid backtracking algorithm with allot of constraints to give us
a solution to as above was said a wide solution space problem on a machine
with allot of cpu,ram,stackstuff and...
The solution should be thought on paper not on machines,because machines
will always be very stupid except for the case in wich neural networks and
genetic algorithms get studied better.

I once met a man at a software competition at wich i too took part that made
a scheduling algorithm based on dynamic programming for any school in wich
he only had to introduce the professors and their preferences meaning what
hours they need to be off and what hours can they be in,in wich days they
want have lessons or courses and it gave a solution in polynomial time.
Meaning that for an average highschool it gave a solution in about 30seconds.
Now that was a real piece of work.I think that would be similar to what you are
telling here,this problem a81

I don't know how familiar you are with this problem, but this is a theoretical problem and it's studied with pen, paper and everything else. It's a very general problem.
Also, computers can be made to act more intelligent or perform tasks more intelligentely than many multicelular organisms and indeed alot of people. And it's not the machine, but the algorithm that the machine is running that is stupid or not, and that is perhaps inherited from the programmer. Computers are only tools, like pens, we don't blame pens for not being able to come up with sophisticated theorems, why do i hear "stupid" computers so often? Unless of course you are unreasonably expecting too much.

-job- let's clarify i can't see where i said computers were stupid.
i said "stupid backtracking algorithm".
also i said that $\text{computers=high speed jackasses}$ and i am convinced of this
until someone gets to solve in an elegant fashion problems with it
i am a programmer,coder what do you want to say and i know some maths too and i
am not as unfamiliar as you think with these problems

I thought you were using the word "stupid" excessively, and my computer got offended... :P
In my opinion, if someone gets to solve problems in an elegant fashion with a computer, than the computer is not smart either, just the same way it's not dumb. Only when the computer is programming itself, maybe with a learning system, would i start to call computers dumb (inefficient learning system), or smart (efficient learning system). Semantics...
I'd rather talk about the P NP problem, any strategies?

deeeeeep studies of neural networks and genetic algorithms thats all

But the P = NP problem is as much about solving a very hard problem (the NP-Complete problems) in polynomial time as it is about finding a contradiction that shows that those problems can't be solved in polynomial time. And given that it is unlikely that P = NP, then finding a contradiction is probably the way to go (assuming that the P = NP problem is not unsolvable which is not out of the question). So are you hoping that these neural networks and genetic algorithms will one day get so smart that they will be able to find a contradiction and prove P != NP, or what are you proposing?

its all about a specific problem
it is not natural to consider a large set of problems at hand
and also what do you do if the cases for wich the problems works and the contradictions
are little relative to the search space of the problem ? this is a much bigger problem wich
can be solved with only one element that is LUCK!

but how did fermat discover the theorem that keeps his name in history?
it was a big part of intuition involved and euler had an unhumanly big intuition when
he generalized fermat on his little theorem.
That is what we should built into the neural networks the intuition
There is a book written by Hadamard title: "Psichology of invention in the mathematical field"
I am reading this now and that is my point of inspiration,but i think you agree

my conclusion is that LUCK is not LUCK its more scientificaly called intuition and this
intuition is based on some rules so it can algorithmically be described
But intuition resides in experience

I'm sure Fermat had plenty of intuitions before the correct one. It's trial and error for the most part.


inequalities
induction
real analysis
real analysis unsolved

How to solve this (preferably with mathematical induction)?

$\sqrt[n]{n}-1 \leq \sqrt{\frac{2}{n-1}}$

I know it may seem very easy but I really can't solve it.

yup, to be more specific,

expand $(1+\sqrt{\frac{2}{n-1}})^{n}$


Solve the equation $x^{3}-3x=\sqrt{x+2}$

One solution $x_{1}=2$. But this equation have 2 real solution. Second $-1<x_{2}<0$.

Given $x=2cos 2a$ then $cos 6a=cos a$and ....


induction
linear algebra
linear algebra unsolved

let $ n \in N^*,A,B \in M_{n}(R)$ and $ E_{A,B}\equal{}${$ X \in M_n(R),AX\plus{}XA\equal{}B$}.
1)prove that $ E_{A,B}$is non empty if and only if ,$ \forall C \in E_{A,0},tr(BC)\equal{}0$
2)prove that ,if $ AB\plus{}BA\equal{}0$ and $ E_{A,B}$ is non empty ,then $ B$ is nilpotent

Have somebody a solution for this
b ) $ AB\plus{}BA\equal{}O_{n}$ and exist $ X$ s.t $ AX\plus{}XA\equal{}B$ then B is  nilpotent ?

$ AB\equal{}\minus{}BA$ gives $ AB^k\equal{}(\minus{}1)^kB^kA$ by induction for all $ k\in\mathbb{N}$.
Then $ B\equal{}AX\plus{}XA$ gives
$ B^{2k}\equal{}AXB^{2k\minus{}1}\plus{}XAB^{2k\minus{}1}\equal{}AXB^{2k\minus{}1}\minus{}XB^{2k\minus{}1}A\equal{}AY_k\minus{}Y_kA$
for $ Y_k: \equal{}XB^{2k\minus{}1}$ for all $ k\in\mathbb{N}$.
And then $ \text{Tr}\, B^{2k}\equal{}\text{Tr} (AY_k\minus{}Y_kA)\equal{}\text{Tr} ( AY_k)\minus{}\text{Tr}(  Y_kA)\equal{}0$ for all $ k\in\mathbb{N}$,
which gives $ B^2\in\mathcal{M}_n(\mathbb{R})$ nilpotent, and so is $ B$.
(This is well known and works only in characteristic $ 0$)


calculus
integration
trigonometry
calculus computations

$ \int_{0}^{\inf}\frac {|sinx|}{x^p}dx$

For what value of p(>0) does the integral converge/diverge?

Hi.. $ \int_ {0} ^ {\infty}\frac {|sinx|}{x^{p}}dx \le \int_ {0} ^ {\infty}\frac {1}{x^{p}}dx$. Thus when p $ \g > 1$ . The integral converges.. :maybe:

And it diverges for $ p\le1$ because you essentially can't do any better than that comparison.

Limits are from 0 to inf not from 1 to inf.

Which means that it diverges for $ p\ge 2.$ as well.

If you are comparing it with $ \frac{1}{x^p}$ then it will diverge for all p>0.

No, we make a smarter comparison than that. It compares to $ \frac1{x^p}$ for large $ x$ (say, $ x>1$), while for $ x$ near zero, we use that $ \sin x\approx x$ to compare to $ \frac1{x^{p\minus{}1}}.$ You will get convergence for $ 1<p<2.$


[i] Problem
  Let a,b,c be positive real numbers such that $ abc \equal{} 1$
     Pove that

  $ \frac {a^2b^2}{a^2b^2 \plus{} a^7 \plus{} b^7} \plus{} \frac {b^2c^2}{b^2c^2 \plus{} b^7 \plus{} c^7} \plus{} \frac {c^2a^2}{c^2a^2 \plus{} a^7 \plus{} c^7} \leq 1$

    Love An forever[/i]

[quote="nkht-tk14"]
  Let a,b,c be positive real numbers such that $ abc \equal{} 1$
     Pove that

  $ \frac {a^2b^2}{a^2b^2 \plus{} a^7 \plus{} b^7} \plus{} \frac {b^2c^2}{b^2c^2 \plus{} b^7 \plus{} c^7} \plus{} \frac {c^2a^2}{c^2a^2 \plus{} a^7 \plus{} c^7} \leq 1$

   [/quote]
$ \sum_{cyc}\frac {a^2b^2}{a^2b^2 \plus{} a^7 \plus{} b^7}\leq\sum_{cyc}\frac {a^3b^3c}{a^3b^3c \plus{} a^4b^3 \plus{} a^3b^4} \equal{} \sum_{cyc}\frac {c}{a \plus{} b \plus{} c} \equal{} 1.$


Hi,

Since I'm a moderator at getting started, the most important thing for me to do is post the challenging problem for getting started and sending out anyone who is beyond the getting started to intermediate (so we can have more room for newcomers). So, I have thought of one plan. This plan requires all the people's help so please try to cooperate  :) .

Here's the plan:

I'm going to collect the problems from users of this site in level of getting started so approximately AMC-10 or AMC-11 or MATHCOUNTS levels. What I'm going to do after I collected problems is that I'll organize in the level of difficulty and post it on this forum. Then users who are past the getting started and users who are on getting started can come and try to solve the problems.

This is not a contest where you have to solve many questions to be the BEST. That is not what mathematics is about. The mathematics, at least in my point of view, to get better at it, not necessarily better than others. Of course, it wouldn't be bad if you're the best but getting better at it is still something big.  ;)

So, if you have good questions, please mail them to me by:

g89unit@gmail.com

Remember, more problem I receive from users, better it gets.

Also, the people from intermediate level can send the problems so I can set them out for getting started.

If I see the problem is too hard, I'll not use it.

So, please mail me the problems!!

:)

[quote="tokenadult"]AMC 10 is intermediate level.[/quote]

Even AMC 10 #1-8 or 10? They seem pretty doable for Getting Started, barring the exceptions each year. I'm pretty sure that they'd be sort of good problems for Getting Started.

To tokenadult,

I don't fully agreed that AMC-10 is on intermediate, at least in this forum. I noticed that people in the intermediate level has great intelligency than what it is to be expected on the forum. Like I remember posting one problem that I had from my class this year and found that like three people mentioning it's getting started.

So, maybe AMC-12 is intermediate. But AMC-10, well, I'll go with like 80% getting started and 20% intermediate. I heard some young students in middle school taking AMC-12 so I guess that part is depending on where you look at it.

Plus, I'll judge the problems when I receive it. So, please send me some!

[quote="tokenadult"]AMC 10 is intermediate level.[/quote]

Does that make the AMC 12 and AIME pre-olympiad?

Well, all three levels that precede Olympiads are pre-Olympiad, aren't they? So then the question becomes what contests are typically done by people "getting started" in problem-solving, which are done by people "in between" (that is, "intermediate") in problem-solving, and which are done by people just before they get into Olympiad level ("pre-Olympiad"). I'm noticing bracket creep downward here again, that is harder problems being put in lower forums, such that undergraduate math majors are answering problems in Getting Started. Undergraduates are POST-Olympiad, it seems to me, because the IMO is restricted to people who haven't matriculated at an undergraduate program.

Well, I'm an undergraduate math major.

I don't know if I've ever answered a question in there, unless it was clearly not for that section, but I do try to help out where I can, as do posters like myth, etc, from what I can see.

It's pretty hard on determining the levels of the problems and people. Like AHSME, the level of it had been in intermediate because it was on high school students covering from 9-12 grades (not necessarily they're the only ones who took that test but mostly) and at least in my opinion, they're considered as intermediate.

Yet, according to tokenadult, part of AHSME which is currently known as AMC-12, should be then in pre-Olympiad with AIME and well, I don't know about that. I mean AMC-12 does involve some calculus concepts but I don't really know and be sure of that it's considered as "pre-Olympiad" topic.

For blahblahblah, you should be... hmm.. I don't know.. Somewhere in this forum that's higher than where I'm standing. :)

Silverfalcon

P.S. Please send me the problems for this. I see many people are discussing on it but practically, I have received no problems for this and it's very sad :(. After all, it's for getting started people who will be future Olympiad people..

I post seriously in Intermediate, Pre-Olympiad, Olympiad, and Advanced forums, depending on the nature of the question.


With a rational denominator, the expression $ \frac {\sqrt {2}}{\sqrt {2} \plus{} \sqrt {3} \minus{} \sqrt {5}}$ is equivalent to:

$ \textbf{(A)}\ \frac {3 \plus{} \sqrt {6} \plus{} \sqrt {15}}{6} \qquad\textbf{(B)}\ \frac {\sqrt {6} \minus{} 2 \plus{} \sqrt {10}}{6} \qquad\textbf{(C)}\ \frac {2 \plus{} \sqrt {6} \plus{} \sqrt {10}}{10}$
$ \textbf{(D)}\ \frac {2 \plus{} \sqrt {6} \minus{} \sqrt {10}}{6} \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
$ \frac{\sqrt{2}}{(\sqrt{2}\plus{}\sqrt{3})\minus{}\sqrt{5}} \cdot \frac{(\sqrt{2}\plus{}\sqrt{3})\plus{}\sqrt{5}}{(\sqrt{2}\plus{}\sqrt{3})\plus{}\sqrt{5}}\equal{}\frac{2\plus{}\sqrt{6}\plus{}\sqrt{10}}{2\sqrt{6}}$ so $ \frac{2\plus{}\sqrt{6}\plus{}\sqrt{10}}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}}\equal{}\frac{2\sqrt{6}\plus{}6\plus{}2\sqrt{15}}{2 \cdot 6}\equal{}\frac{3\plus{}\sqrt{6}\plus{}\sqrt{15}}{6}$ or $ \boxed{\textbf{(A)}}$.
[/hide]

[hide]$ \frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}} = \frac {\sqrt {2}}{(\sqrt {2} + \sqrt {3}) - \sqrt {5}}$. Multiply this by the conjugative radical of $ \frac {(\sqrt {2} + \sqrt {3}) + \sqrt {5}}{(\sqrt {2} + \sqrt {3}) + \sqrt {5}}$ to clear the radical.

Simplifying this you get:
$ \frac {2 + 2\sqrt {3} + 2\sqrt {5}}{(2 + \sqrt {3})(2 + \sqrt {3}) + (2 + \sqrt {3})(\sqrt {5}) - (\sqrt {5})(2 + \sqrt {3}) - 5}$

Simplifying this you get:
$ \frac {2 + 2\sqrt {3} + 2\sqrt {5}}{2 + 4\sqrt {3}}$.

Multiply by the conjugative radical $ 2 - 4\sqrt {3}$.

Simplify to get:

$ \frac { - 20 - 4\sqrt {3} + 4\sqrt {5} - 8\sqrt {15}}{ - 140} = \frac { - 10 - 2\sqrt {3} + 2\sqrt {5} - 4\sqrt {15}}{\70}$

SO the answer is E.
[/hide]

Wow, brute force latex is that makes sense

Looking at AIME15's post: I think I made a mistake simplifying the first conjugative radical expression.

[quote="Nandubetax"][hide]Simplifying this you get:
$ \frac {2 \plus{} 2\sqrt {3} \plus{} 2\sqrt {5}}{(2 \plus{} \sqrt {3})(2 \plus{} \sqrt {3}) \plus{} (2 \plus{} \sqrt {3})(\sqrt {5}) \minus{} (\sqrt {5})(2 \plus{} \sqrt {3}) \minus{} 5}$
[/hide][/quote]The mistake is at this step.
[hide="It should be"]$ \frac {2 \plus{} \sqrt {6} \plus{} \sqrt {10}}{(\sqrt{2} \plus{} \sqrt {3})^2 \minus{} (\sqrt{5})^2}$[/hide]


algebra polynomial calculus integration function

Prove that for a polynomial $f(x)=a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+a_{3}x^{n-3}+...+a_{n}$, the polynomial whose roots are the recipricol of the roots of $f(x)$ is $g(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}$. (Assuming that all the $a$'s are integers)

Find a polynomial with integral coefficients whose roots are $m$ times the roots of $f(x)$.
Find a polynomial with integral coefficients whose roots are $m$ more than the roots of $f(x)$.

That wasn't very interesting.  :(  :D

Of course, we need $m$ to be an integer. :maybe:

Yeah, I had to check that property of functions like 4 times before I believed it.  Just thinking about it would've helped too.

Another nice conceptual question, what is the polynomial in terms of the function f, which has roots that are the reciprocals of the roots of f?

That doesn't really make sense as a question, but you get what I mean.

Pakman, vishalarul already gave that question, why are you repeating it without solving?

vishalarul, the first question does not require the condition that $a_{i}$ be integer.  In fact, it holds true in any field.  (You have also mislabeled your coefficients slightly; the last term of $g(x)$ should be $a_{0}$, not $a_{1}$.)

Hokkage, technically your first answer should [hide="really be"] $m^{n}f( \frac{x}{m})$ [/hide]

As for the answer to the [hide="first question..."] Let $f(x) = a_{0}(x-r_{1})(x-r_{2}) ... (x-r_{n})$.  Then

$g(x) = a_{0}(1-r_{1}x)(1-r_{2}x) ... (1-r_{n}x)$ [/hide]

Edit:  Here is a question a little more interesting.

[b]Follow-up.[/b]  Find a polynomial with integral coefficients whose roots are the square of the roots of $f(x)$.

Oh woops... I didn't even read that part. My bad.

[quote="jhcreinhardt"]Is it

[hide]$f(\sqrt{x}) \times f(\sqrt{-x})$[/hide][/quote]

$f(\pm\sqrt x)$ may not be a polynomial.

Does multiplying them cancel that out? :maybe:

jhcreinhardt, your method does not give integral (or, in fact, even real) coefficients.  For example, $f(x) = x$.

Amending it to $f( \sqrt{x}) \cdot f(-\sqrt{x})$ EDIT:  works, apparently.   :blush:

[hide="Wrong Solution (due to a misunderstanding)."]
I think that $f(X^{2})$ works. It includes both square roots. Also, the multiplicity of $0$ is doubled, so we can divide by the appropiate $X^{k}$ in order to maintain the multiplicity of $0$ in $f(X)$ and $\frac{f(X^{2})}{X^{k}}$ the same (since $0$ has a single square root).
[/hide]

That gives the polynomial whose roots are the [b]square roots[/b] of the roots of the original, not the [b]squares[/b].

[quote="t0rajir0u"]Amending it to $f( \sqrt{x}) \cdot f(-\sqrt{x})$ also fails (for example, $f(x) = x^{3}+x$), although it is closer ;)[/quote]

I think that your counterexample is wrong. I also think that this works. Why? Mainly because $f( \sqrt{x}) \cdot f(-\sqrt{x}) \in \mathbb Z[X]$.

But I may be wrong  :!:

$f(\sqrt x)f(\sqrt-x)$ always works [hide="because"]
Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$.
Now let $b$ be the sum of all the terms with even exponents, with all the exponents halved. Let $c$ be the sum of all the terms with odd exponents, with all the exponents halved.
$f(\sqrt x)f(\sqrt-x) = (b+c)(b-c) = b^{2}-c^{2}$
All the terms of $b^{2}$ have integral exponents because all the terms in $b$ have integral exponents. All the terms in $c^{2}$ have  integral exponents because every term in $c$ has an exponent of an odd number divided by 2, so every possible way to multiply 2 terms in $c$ will get you an integral exponent. Thus $b^{2}-c^{2}$ is a polynomial. Sorry if that explanation was somewhat unclear.[/hide]
Can anyone extend this to higher powers?

[quote="perfect_radio"]Mainly because $f( \sqrt{x}) \cdot f(-\sqrt{x}) \in \mathbb Z[X]$.[/quote]

I should be more careful with my algebra  :blush:


function
limit
integration
real analysis
calculus
real analysis unsolved

Suppose that $ \{g_n\}$ is a sequence of positive continuous functions on $ I \equal{} [0,1]$, that $ \mu$ is a positive Borel measure on $ I$, and that:
(i) $ \lim_{n \rightarrow \plus{} \infty} g_n(x) \equal{} 0$ a.e [$ m$],
(ii) $ \int_I g_n dm \equal{} 1$ for all $ n$,
(iii) $ \lim_{n \rightarrow \plus{} \infty} \int_I f g_n dm \equal{} \int _I f d \mu$ for every $ f \in C(I)$.
Does it follow that $ \mu \perp m$ ?

($ m$ stands for the Lebesgue measure on $ I$ and $ C(I)$ for the real-valued continuous functions on $ I$)

*Edit: thanks HilbertThm90! *

[hide="Edited"]I think you are missing an integral sign in (iii)[/hide]

Let $ E\subset I$ given by $ E \equal{} \{x: \lim g_n\neq 0\}$. By (i) we have that $ m(E) \equal{} 0$.

Claim: $ \mu$ is concentrated at $ E$. Suppose not. Thus we have $ A\subset I$ such that $ A\cap E \equal{} \emptyset$, but $ \mu(A) > 0$.

Note: $ \lim g_n \equal{} 0$ everywhere on $ A$ since it doesn't intersect $ E$.

Define $ d\mu_n \equal{} g_n dm$. Since for any measurable $ K\subset I$ the integral $ \mu_n(E) \equal{} \int_E g_n dm$ makes sense, we have each $ \mu_n$ as a legitimate measure. We want $ d\mu_n\to d\mu$. Let's check.

We have $ |fg_n|\leq \|f\|_\infty\|g_n\|_1 \equal{} \|f\|_\infty$ a finite number. So $ fg_n$ is dominated by the constant function $ \|f\|_\infty$ and we can move the limit inside by Lebesgue Dominated Convergence. $ \lim\int_I fg_ndm \equal{} \int_I f\cdot(\lim g_n dm) \equal{} \int_I fd\mu$ by (iii). Thus $ \lim g_ndm \equal{} \lim d\mu_n \equal{} d\mu$.

Here is the contradiction, though. $ 0 < \mu(A) \equal{} \lim\mu_n(A) \equal{} \lim\int_A g_ndm \equal{} 0$. That last integral is zero since $ g_n\to 0$ uniformly on $ A$ by "Note."

Thus $ \mu$ is concentrated on $ E$ which is a set of Lebesgue measure zero giving $ \mu\perp m$.

[quote="HilbertThm90"]
We have $ |fg_n|\leq \|f\|_\infty\|g_n\|_1 \equal{} \|f\|_\infty$ a finite number.[/quote]

Mh... what if $ f$ is the constant function $ 1$ and $ g_n$ is such that $ g_n(0) \equal{} 0$, $ g_n\left(\frac {1}{m}\right) \equal{} m$, $ g_n$ is $ 0$ on $ \left[\frac {2}{m},1\right]$, linear on $ \left[0,\frac {1}{m} \right]$ and $ \left[\frac {1}{m},\frac {2}{m}\right]$ (for large $ m$)?

By (ii) we have $ \|g_n\|_1\equal{}\int_I |g_n| dm\equal{}\int_I g_ndm\equal{}1$, since $ g_n$ is positive.

I also just noticed that I wrote "any measurable $ K\subset I$", then wrote "$ E$" in the next thing. That should have been $ K$.

I understand [quote="HilbertThm90"]
We have $ |fg_n|\leq \|f\|_\infty\|g_n\|_1 \equal{} \|f\|_\infty$ a finite number.[/quote]
and
[quote="HilbertThm90"]$ fg_n$ is dominated by the constant function $ \|f\|_\infty$[/quote]
as: for all $ x \in I$, $ |f(x) g_n (x)|\leq \|f\|_\infty$.

This is false: take $ f \equal{} 1$ and $ g_n$ as in my previous post. Then: $ |f(1/m)g_n(1/m)| \equal{} m > 1 \equal{} \|f\|_{\infty}$.

After further thoughts, I think that the answer is "no", but I can't find a counterexample  :| ...

*Edit: moreover, the weak convergence $ \mu_n \rightharpoonup \mu$ [i]does not[/i] imply that for all borel sets $ B$ of $ I$ one has $ \mu_n(B) \rightarrow \mu(B)$. (it is true if $ \mu(\partial B) \equal{} 0$: it's the portmanteau theorem)*

Wow. Sorry. I obviously was thinking $ \|fg_n\|_1\leq \|f\|_\infty\|g_n\|_1$ which is meaningless in that context.

I was going to point out the weak convergence in the first post. That was the thing that concerned me most. I didn't really know what followed from it, and was just assuming it worked nicely.  :blush:

You say you think the answer is no. Do you have any intuitions on why that might help in my looking for a counterexample? I'm still stumped. I sort of lean towards it being true...


Harvard college MIT Princeton Stanford email AMC

Hello guys,
I think that there are many people that are going to be seniors next year. I was wondering where are you going to apply next year (which universities)

It would be interesting to hear this. I will modify the thread title to make it more clear what you are asking.

If anyone out there (California resident, most likely) is considering CSU Long Beach, do contact me directly by PM or email.

Well I guess I might as well be the first to say:

6 Universities
1. Harvard
2. MIT
3. Princeton
4. UC Berkeley
5. U Chicago
6. BYU

[quote="1/(ln x)"]
6. BYU[/quote]

BYU being Brigham Young University in Utah?

I'll be glad to help out.

We have:

1. Harvey Mudd
2. MIT
3. University of Chicago
4. Caltech
5. Stanford
6. Princeton
7. Case Western Reserve ? Harvard ? I don't know ?

The magical 7 spot is up for grabs (obviously). I'm kind of looking for a good physics university with great research opportunities that isn't uber-competitive to get into, and so maybe Case Western Reserve? I'd appreciate hints. Harvard might be in there because if I get in (1/10 chances, I'll take 'em), I get in for practically free!

I think it would be a good idea if you also post a little information about your achievements and interests (something like a short CV) :maybe:

I dream of Caltech,MIT and Princeton and want to major in Computer Science  :blush:  but it 's impossible for me to get into these colleges  :rotfl:
Now I am  learning at a technological university and  really not such an intelligent student .Do you think I have some chances to get into these colleges I list above If I try all my best ? :oops:

Apart from the obvious Ivies and other superpowers, I might suggest applying to Boston University -- they have some nice scholarships based solely on merit, including several half-tuition grants and a handful of full-tuition grants.

I don't know about your situation, but for me, studying at BU for free instead of blowing 220k on an Ivy was  something I seriously considered.

I also find BU to be quite underrated in the rankings.

As TZF states, my friend, who I met as a foreign exchange student from China, currently attends Boston University.  And Like TZF stated, he is getting alot of aid in the form of merit scholarships.

You still can win a scholarship when applying to Ivies (even if you are an international student)
:)

[quote="stuck"]The magical 7 spot is up for grabs (obviously). I'm kind of looking for a good physics university with great research opportunities that isn't uber-competitive to get into, and so maybe Case Western Reserve? I'd appreciate hints. Harvard might be in there because if I get in (1/10 chances, I'll take 'em), I get in for practically free![/quote]
Is this undergrad? Because if it is, then there are many great physics programs that you can get into, and the difference in caliber in undergrad physics education is small (although I've personally listened to Feynman (I know he's no longer alive) and Lewin, and therefore would put Caltech and MIT as the top two if I could ever get in).

Um, UC Santa Barbara is pretty good in Physics and easy enough to get into. Colorado at Boulder is pretty good at Physics for an "easy" school. Penn State is ranked well. Rutgers is supposed to be good. Stony Brook is supposed to be good.

I don't know about the research opportunities. Your best bet in anecdotal, from attending students. Try to email professors at Universities you are considering.

Oh, and look at this http://www.physicsforums.com/showthread.php?t=240792 for some background on going into Physics.

Thank you very much. That was incredibly helpful.

I've been eying Colorado-Boulder for a bit now. Some rankings give it one of the best quantum mechanics programs for grads in the nation, and that is my primary interest in physics. Is it a good idea to think that those grad programs will affect me as an undergrad? Can I take advantage of those? Or are grad-undergrad programs sufficiently apart such that benefiting from them is unlikely?

While Feynman is dead, a friend at MIT has told me that Walter Lewin has gone a bit senile and I don't think he is teaching anymore.

Just sayin'

As far as my math goes, I don't have MOP, but like, USAMO qualification since 9th grade, and 15 on USAMO last year. I took BC Calc in 10th grade, and have gotten a lot farther since then.

I am on/captain of my school math team, and I am editor-in-chief of my school newspaper, along with a few other minor activities.

My (weighted) GPA is 4.57ish out of 5.0, though junior year was significantly better than others (4.70 for that year). A decent number of people from my school get into Harvard and MIT each year, say 2-5, and the average GPA of those people is around 4.60. I have some math and science awards (and I'm doing independent study in physics with a teacher this year), and I also have some language awards in both French and Latin, along with all perfect scores on national french exam, one year for latin exam, and 19th in the us on linguistics olympiad.

Thank you for the help!

seems like you have a good shot.
for myself im aiming to have 2 years USAMO qualification  (10 + 11)
chess ranking, BC calc in 10th grade, 5 poetry contests won (so far), and to be recruitable in squash.

hi,
  my GPA is a decent 4.87
could you tell me the GPA required to get admission in the top universities.

 thanks

[quote="ghjk"][quote="paladin8"]I believe blahblahblah is correct regarding financial aid for international students at Stanford; from what I've heard, it's generally not that great. On the other hand, Stanford is awesome so you should come anyway.[/quote]
According to the official website of Stanford, it says that it will give the full tuition fee for every accepted student whose family tuition is below $ \$$60000.

http://www.stanford.edu/dept/finaid/undergrad/how/enhancements.html

Is it correct, paladin8? Since I seriously consider applying it next year(it's my dream though), I want to know more about its financial aid program. I won't be able to study there even when I'm accepted(just assumption) if they don't give me financial aid.I'm studying in US as a high schooler though.[/quote]

If you are a domestic student, that will be the case. According to http://www.stanford.edu/dept/finaid/undergrad/how/index.html, the policy for international students is different, i.e. more limited.

8 Universities
1. MIT
2. Caltech
3. Stanford
4. Harvard
5. CMU
6. Berkeley
7. Binghamton
8. Stony Brook

and what about your co-cirricular activities?

[quote="stuck"]I'll be glad to help out.

We have:

1. Harvey Mudd
2. MIT
3. University of Chicago
4. Caltech
5. Stanford
6. Princeton
7. Case Western Reserve ? Harvard ? I don't know ?
[/quote]

I'll add Carnegie Mellon to that list.

As for a short stats profile:

36 ACT
790/800/800 Chem/Physics/Math II
670/750/800 Reading/Writing/Math SAT
1/500 Class Rank ~4.0 unweighted

Math Team, Scholastic Bowl, WYSE, Science Olympiad, NHS, Newspaper, Peer Tutor, EPGY, Varsity Tennis, Soccer till this year. Probably more.
Research at the astrophysics department at Fermilab.

But my AMC Score was 97.5 and AIME was I think a 4.  :blush: I'm working on it. :wink: Plus my school is notably average. The rumor is no one has ever gone Ivy, though I know two instances of MIT admission (in its history).
Hopefully that helps someone out.

Dude my school sends 8-9 people ivy[b] every[/b] year

[b]stuck[/b], if you pick your schools wisely, write thoughtful essays, and give a good argument as to why you are interested in those schools in your interview, you are very easily ivy-league material.

Here's what I say you should be paranoid about (at least a little): [b]letters of recommendation[/b]
If your schools are notably average, and even the best students are finding difficulty getting good placement, chances are that the teachers that are giving them letters of rec aren't doing a good job. And they just might be the same set of teachers too, if they are AP teachers. If you think that this may be the case, here are some guidelines that you might have already heard:

1. Pick your teachers wisely! Pick teachers that you know well personally, and that know what you do apart from things in their class

2. Find some time when they are not busy, and sit down with them. Ask them if they are busy. Ask them if they will write you a good letter. Ask them what kinds of things they would have to say about you. Tell them a bit about why you are interested in said university, if it's not on the common app. If it is on the common app, tell them why you are interested in some general traits of a college, if there is anything in common between the good schools you are applying to (ie, diverse program, diverse student body, etc). If necessary, tell them a bit about yourself.

3. If possible, ask someone at Fermi lab to write you a rec. They will have less generic things to say about you, they probably won't be writing as many letters of rec, and they will carry more weight than a schoolteacher.

4. Spend some time talking to your guidance counselor! Aside from a list of activities you're involved in, tell them some stories about yourself, or interesting facts. Let the GC say something interesting about you that your teachers might not cover.

[b]@ mathmonster:[/b] Correct me if I'm wrong... aren't you in a private school?

yes I am, so therefore we have less kids per grade (85 ish)

Thanks very much for your advice, TZF! I think I will really benefit with your help. It's comforting to know that I've got a chance, I suppose. I should probably start my essays, though. :|

I'm on pretty good terms with my teachers, and have filled them in on all the random stuff I do. I even built a Theremin in their classes once (Hey, you gotta do something post-AP tests). So I feel as if those will be fine. My English teacher I met this year, which is admittedly risky, but I feel as if she really likes me, plus English is a really participation-based class. She knows better than most all of the random literature I read in my free time. And my guidance counselor also does student council, so I've worked with him for 4 years.

If I were to get a letter of recommendation from one of my Fermilab adviser, under which pretense would I send it? Should I send it as supplementary material?

i think stuck will go ivy no question...

Many of you have experienced application processes or at least will experience them in the future.
I know that there isn't any particular formula for getting into university, but aren't there criteria that can get u closer to getting into Ivys ?
Thank you!
P.S. This topic is a nice discussion :)

[b]@ stuck:[/b]

[b][i]Regarding teacher recs:[/i][/b] Even still. If you give it more importance, they will. If they really like you, and you just go and give them a form, then they're more likely to give you a generic sounding one. I mean, the generic one will be full of nice things, but it will just sound generic. A letter can sound generic even if it contains details about what you specifically do. I'm sure your letters will come out just fine, I'm just saying that if you know there's a pattern of people from your school getting underplaced, it is in your best interest to take steps to address factors that might be causing that.

[b][i]Regarding your Fermi-lab rec:[/i][/b] You should call every individual school's office of admissions and ask them about it.

1. Some schools really don't like getting extra letters (usually bigger schools). If you send them an extra letter, it will probably end up in the trash.
2. Some schools will take extra letters, but tend to advise students not to, because a lot of AP+extracurriculuar-piling students will just send in a flood of letters. So you might have to be assertive about the importance of that letter before they give you a constructive suggestion.
3. Some schools will very openly accept additional letters.

What should you pass him off as? In most cases, research avisor would be best. If you show that you've been involved in some substantial research in high school, and suggest that you will be proactive about doing undergraduate research while in college, that's a real turn-on for many top-tier schools that have extensive facilities.

[quote="TZF"][b]@ stuck:[/b]

[b][i]Regarding your Fermi-lab rec:[/i][/b] You should call every individual school's office of admissions and ask them about it.

1. Some schools really don't like getting extra letters (usually bigger schools). If you send them an extra letter, it will probably end up in the trash.
2. Some schools will take extra letters, but tend to advise students not to, because a lot of AP+extracurriculuar-piling students will just send in a flood of letters. So you might have to be assertive about the importance of that letter before they give you a constructive suggestion.
3. Some schools will very openly accept additional letters.
[/quote]

My son went through this last year. He actually had seven people to write recs for him, and he wanted to know how many were reasonable. He wrote to each of the schools he was considering (a list similar to other kids posting on this topic). Replies were all over the map - some said bring it on, and others said one extra would be ok.  What schools do not want are extra recs from people who have similar functions - like two science teachers, etc. But, the fermi-lab rec would definitely add extra info for the adcoms. It is a good idea to check with the schools, but I am betting they will all welcome this extra rec. (my son ended up using 4 recs)


[hide="If nothing strikes, do brute force"]$ \frac 1x\minus{}\frac {3x}{4}\plus{}\frac 2y\plus{}y \ge 2$
$ \Leftrightarrow \frac {1}{3\minus{}y}\plus{}3\frac {y\minus{}3}{4}\plus{}\frac 2y\plus{}y \ge 2$
$ \Leftrightarrow \frac {7y^3\minus{}38y^2\plus{}55y\minus{}24}{4y(3\minus{}y)} \le 0$
$ \Leftrightarrow \frac {(y\minus{}1)^2(7y\minus{}24)}{4y(3\minus{}y)} \le 0$

which is true. Equality occurs when $ x\equal{}2,y\equal{}1$[/hide]

[quote="dulldrake"]Let $ x,y$ be positive real number satisfy: $ x \plus{} y \leq 3$.Prove that:
$ \frac {y \plus{} 2x}{xy} \plus{} \frac {4y \minus{} 3x}{4} \geq 2$
when the equality occur?[/quote]

Hi.

[hide="solution"]
The LHS $ \equal{}\frac{1}{x}\minus{}\frac{3}{4}x\plus{}\frac{2}{y}\plus{}y$, this is strictly decreasing for $ x$,and now $ 0<x\le 3\minus{}y$ ,so it is minimumized when $ x\equal{}3\minus{}y$ and we have:

LHS $ \ge \frac{1}{3\minus{}y}\minus{}\frac{3}{4}(3\minus{}y)\plus{}\frac{2}{y}\plus{}y$

$ \equal{} \frac{1}{3\minus{}y}\plus{}\frac{2}{y}\plus{}\frac{7}{4}y\minus{}\frac{9}{4}$

$ \equal{}\frac{1}{3\minus{}y}\plus{}(\frac{1}{2y}\plus{}\frac{1}{2y}\plus{}\frac{1}{2y}\plus{}\frac{1}{2y})\plus{}(\frac{y}{2}\plus{}\frac{y}{2}\plus{}\frac{y}{2}\plus{}\frac{y}{2})\plus{}\frac{3\minus{}y}{4}\minus{}3$

$ \ge 10\sqrt[10]{\frac{1}{3\minus{}y}\cdot \frac{1}{2y}\cdot \frac{1}{2y}\cdot \frac{1}{2y}\cdot \frac{1}{2y}\cdot \frac{y}{2}\cdot \frac{y}{2}\cdot \frac{y}{2}\cdot \frac{y}{2}\cdot \frac{3\minus{}y}{4}}\minus{}3$　(Since AM-GM inequality)

$ \equal{} 10\sqrt[10]{\frac{1}{2^{10}}}\minus{}3\equal{}2$,

so it's done and the equality holds if and only if $ x\equal{}3\minus{}y$,and $ \frac{1}{3\minus{}y}\equal{}\frac{1}{2y}\equal{}\frac{y}{2}\equal{}\frac{3\minus{}y}{4}$,i.e $ \boxed{x\equal{}2,y\equal{}1}$.

[/hide]