quadratics
calculus
Vieta
algebra
polynomial
complex numbers
Rational Root Theorem

I am most of the times unsure about this and the problems related to it.

Are there any comprehensive set of rules we can apply to find the nature of roots of any degree given a equation?

Any degree equation?  If the degree is 0, 1, or 2, I would assume you know how to find the roots.  If the degree is 3, there is a formula to find the roots, but it's a bit too complicated to flat-out memorize.  If the degree is 4, there is a really complicated method.  If the degree is 5 or greater, Abel and Galois showed that it is impossible to find a general solution for such an equation.

Well i was talking about the nature of the roots than the root themselves i.e, using the discriminant as we do in the case of a quadratic equation

Yes, you can use:

1.) [url=http://en.wikipedia.org/wiki/Rational_root_theorem]The Rational Root Theorem[/url] - which, along with the [url=http://en.wikipedia.org/wiki/Factor_Theorem]Factor Theorem[/url], Remainder Theorem, and Synthetic Division, every high school student should know!

2.) [url=http://en.wikipedia.org/wiki/Descartes_rule_of_signs]Descartes' Rule of Signs[/url]

3.) [url=http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem]The Complex Conjugate Root Theorem[/url] and the Real Conjugate Root Theorem.

4.) [url=http://en.wikipedia.org/wiki/Discriminant]The Discriminant[/url] (at least, in theory, but it's too complicated to memorize for degrees greater than 2 and it's never necessary for high school contests - so don't bother)

5.) [url=http://en.wikipedia.org/wiki/Intermediate_value_theorem]The Intermediate Value Theorem[/url]

6.) Some ideas from Calculus

I'd like to add [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_Formulas]Vieta's Formulas[/url] on to that list because it does give a very useful relations between the roots and the coefficients of any polynomial.

From the title, I'm assuming you just want to know this for a quadratic...
Mostly, you'll just need the discriminant. (But the Complex Conjugate, as well, if dealing with imaginaries...)

You should know that the quadratic formula gives the roots of the equations of the form $ ax^2\plus{}bx\plus{}c\equal{}0$, with $ a\neq0$:
$ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$

Here, $ b^2\minus{}4ac$ is the discriminant.

If $ b^2\minus{}4ac<0$, there are no real roots, and two complex conjugate roots. (i.e. The roots are of the form $ a\plus{}bi$ and $ a\minus{}bi$)

If $ b^2\minus{}4ac\equal{}0$, then there are, technically, two real roots, but they have the same value, and are often referred to as the double root, or sometimes, just as the single root.

If $ b^2\minus{}4ac>0$, then there are two real roots.

NOTE: In any polynomial, if a root is even (has a double root, quadruple root, etc.), then the x-axis will be tangent to the graph at the root. If the root is odd (has a single root, triple root, etc.), then the graph will "pass through" the x-axis.

I think you mean if the multiplicity of the root is even/odd?

because your wording is misleading ;-)

polynomials don't always have rational solutions that you can find with paper and pencil. I have had this question for a LONG time and it got answered by algebra book :)

Yes, I am talking about the multiplicity, but often, they are referred to as even and odd, just like functions.

The most comprehensive answer to this question is provided by [url=http://en.wikipedia.org/wiki/Galois_theory]Galois theory[/url], which is roughly a study of how symmetric the roots are.  It is Galois theory that proves that you cannot generally solve a polynomial of degree $ 5$ or higher by root extraction, but it is also Galois theory that allows you to extract as much information as possible about the roots even if you can't find them explicitly.

I'm not sure Galois theory is accessible to high school students.  Nevertheless, Galois and Abel were geniuses.


Consider an ordered set of six consecutive integers in increasing order. To create a new set of six integers, the first, third, and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased by two. The median of this new set is four more than the median of the first set. What is the sum of the median of the first set and the median of the second set?

[hide="answer(i think)"]
$ X$ is the smallest number of the 6 consecutive numbers.
The median of the first Set is:

$ \dfrac{(x \plus{} 2) \plus{} (x \plus{} 3)}{2}$

The median of the second Set is:

$ \dfrac{(x \plus{} 2) \plus{} (x \plus{} 3)}{2} \plus{} 4$

The Sum I guess is:
$ 2(\dfrac{(x \plus{} 2) \plus{} (x \plus{} 3)}{2}) \plus{} 4$
$ \equal{} \boxed{2x \plus{} 9}$; Where X is the smallest consecutive interger of the first set.

If I am wrong please tell me.
[/hide]

Come on guys...
[hide="Solution"]Let $ x\minus{}2,x\minus{}1,x,x\plus{}1,x\plus{}2,x\plus{}3$ be the set of integers.
The new set is $ 2x\minus{}4,x\plus{}1,2x,x\plus{}3,2x\plus{}4,x\plus{}5$.
The median of the first set is $ \frac{2x\plus{}1}{2}$
The median of the second set should be $ \frac{2x\plus{}9}{2}$
The sum is then $ 2x\plus{}5$.
Assuming that $ 2x\minus{}4>x\plus{}1$, $ x>5$
If $ 2x\minus{}4>x\plus{}3$, $ x>7$
If $ 2x\minus{}4>x\plus{}5$, $ x>9$
If this holds true, then the new set is $ x\plus{}1,x\plus{}3,x\plus{}5,2x\minus{}4,2x,2x\plus{}4$
The median of this set is $ \frac {3x\plus{}1}{2}\equal{}\frac{2x\plus{}9}{2}$
This implies that $ x\equal{}8,$ but then $ x\not>9$

We then try $ x>7$
The new set is then $ x\plus{}1,x\plus{}3,2x\minus{}4,x\plus{}5,2x,2x\plus{}4$
Notice that this set's median is the same as the other one.
We once again find that $ x\equal{}8$ and this satisfies the equation.
The sum is thus $ 21$

To check this: $ 6,7,8,9,10,11$- Median is $ 8.5$
$ 9,11,12,13,16,20$- Median is $ 12.5$
Holds true[/hide]

[quote="pascal_1623"][quote="mihail911"]im pretty sure the answer is numerical and i really don't know how you got that answer.
picking a set of 6 arbitrary integers doesn't work because this case obviously isn't general. i assigned variables but after the changes i could not tell which integer was bigger than which. anybody else wanna give a crack at it?[/quote]

all you have to do is put one of the consecutive integers as x, then you have x+1, etc.
just like mathcrazed did.[/quote]
that's exactly what i did but it didn't work out  :P

[hide="My Solution"]$ \text{Original Set: } \left\{x, x \plus{} 1, x \plus{} 2, x \plus{} 3, x \plus{} 4, x \plus{} 5\right\}$
$ \text{Modified Set: } \left\{2x, 2x \plus{} 4, 2x \plus{} 8, x \plus{} 3, x \plus{} 5, x \plus{} 7\right\}$

Because these are consecutive integers, the median is the average.  This avoids the situation of deciding whether $ x$ is positive, negative, or zero, which makes ordering the terms difficult.

Therefore, the median of the original set is $ \frac{6x \plus{} 15}{6}$ and the median of the modified set is $ \frac{9x \plus{} 7}{6}$.

The equation is now $ \frac{6x \plus{} 15}{6} \plus{} 4 \equal{} \frac{9x \plus{} 27}{6}$.

Solving, I get $ x\equal{}4$.

Thus, the original median is $ 6.5$ and the modified median is $ 10.5$.

The sum of these is $ \boxed{17}$.[/hide]

Is this right or wrong?  Thanks.

I have proved Cube's claim, and it worked, and he didn't make any mistakes, but neither did mathcrazed...

[quote="alkjash"][quote="Cube"][hide="My Solution"]$ \text{Original Set: } \left\{x, x \plus{} 1, x \plus{} 2, x \plus{} 3, x \plus{} 4, x \plus{} 5\right\}$
$ \text{Modified Set: } \left\{2x, 2x \plus{} 4, 2x \plus{} 8, x \plus{} 3, x \plus{} 5, x \plus{} 7\right\}$

Because these are consecutive integers, the median is the average.  This avoids the situation of deciding whether $ x$ is positive, negative, or zero, which makes ordering the terms difficult.

Therefore, the median of the original set is $ \frac {6x \plus{} 15}{6}$ and the median of the modified set is $ \frac {9x \plus{} 7}{6}$.

The equation is now $ \frac {6x \plus{} 15}{6} \plus{} 4 \equal{} \frac {9x \plus{} 27}{6}$.

Solving, I get $ x \equal{} 4$.

Thus, the original median is $ 6.5$ and the modified median is $ 10.5$.

The sum of these is $ \boxed{17}$.[/hide]

Is this right or wrong?  Thanks.[/quote]

Actually, when there is an even number of terms, the median of the set is the average of the middle 2 terms, which basically destroys your entire argument.[/quote]

I'm not sure if the number of terms matter, as the integers are consecutive.

For example, the average of $ \left\{1,2,3,4,5,6\right\}$ is $ 3.5$.  This is the same as the average of the two middle terms $ \frac{3 \plus{} 4}{2} \equal{} 3.5$.

I'm not too sure about this...

The first median doesn't change with this, since the numbers are consecutive.
On the other hand, the median and mean of the second set are different.

[hide="my solution"]
Set the original set = $ \{x, x\plus{}1, ... x\plus{}5\}$.
Now, we use casework based on the value of x:
x > 7:
The new set ordered is $ \{x\plus{}3, x\plus{}5, x\plus{}7, 2x, 2x\plus{}4, 2x\plus{}8\}$
Setting the new median equal to four more than the original median:
$ \frac{3x\plus{}7}{2} \equal{} \frac{2x\plus{}3}{2}\plus{}4$
So x = 6, which contradicts the fact that x>7.
However, for x = 6, 7, only the middle two terms are reversed; the median doesn't change, so x=6 works.
Now, $ 2x\plus{}5\plus{}4 \equal{} \boxed{21}$.[/hide]

hmm[hide]i get the same thing as everyone else with the sets
if we take a look at the first one, the median is x+2.5
then the median of the second one must be x+6.5
it is obvious x+3 is less than the mdian and 2x+8 is greater than the median
2x+4>x+7 and x+5<2x are both true if x>3.
if we assume that is true and solve,we get 3/2x+7/2 is the median
set that equal to x+6.5
x+6.5=1.5x+3.5
.5x=3
x=6
6 is greater than 3. then just substitue back in [/hide]

[quote="cognos599"]where did you get this problem?[/quote]
Look at the title. It says "National Sprint #22 2006".

[quote="resurrection"]The median of this set is $ \frac {3x \plus{} 1}{2} \equal{} \frac {2x \plus{} 9}{2}$[/quote]

How did you get the above?  Thanks!

For reference, the official answer is $ \boxed{21}$.


ARML practice is from 1-4 this sunday in the training building at florissant valley community college.  [room105/107]

meh just thought I'd post this.  Who's going? [I know at least 5 people from CA are going to say they are  :roll: ]

[quote="AIME15"]I'm going![/quote]

Hey maybe I'll go now!

@AIME: ohmygodrlynowai?

[quote="math154"]Nobody posted this because nobody's going.[/quote]
 :roll: clearly aime is.

But if I go i will be stalked by Truffles! Ahhhh!


analytic geometry
Pythagorean Theorem
geometry
arithmetic sequence

Starting at the origin, a penguin makes 40 moves. The nth move is a distance n^2/2 units. Its moves are successively due E, N, W, S, E, N ... . How far from the origin does it end up?

[quote="chess64"][hide]The first 4 moves are: $(0,0) \rightarrow (\frac{1}{2},0) \rightarrow (\frac{1}{2},2) \rightarrow (-4,2) \rightarrow (-4,-6)$.

Similarly, every four moves, the penguin travels: $(0,0) \rightarrow \left(\frac{(4k+1)^2}{2},0\right) \rightarrow \left(\frac{(4k+1)^2}{2},\frac{(4k+2)^2}{2}\right) \rightarrow \left(\frac{(4k+1)^2 - (4k+3)^2}{2},\frac{(4k+2)^2}{2})\right) \rightarrow \left(\frac{(4k+1)^2 - (4k+3)^2}{2},\frac{(4k+2)^2-(4k+4)^2}{2}\right)$
(for some nonnegative integer $k$)

Simplifying the last expression, we get \[ \left(-4(2k+1), -2(4k+3)\right).  \]

Sorry, I just realized I defined $k$ wrong. :blush: You actually end up at \[ \left(-4(2k-1), -2(4k-1)\right).  \]

Hence the x-coordinate is the sum \[ \sum_{k=1}^{10} -4(2k-1) = -4\sum_{k=1}^{10} (2k-1) = -4\left[\sum_{k=1}^{10} (2k) -\sum_{k=1}^{10} 1\right] = -400,  \] and the y-coordinate is the sum \[ \sum_{k=1}^{10} -2(4k-1) = -2\sum_{k=1}^{10} (4k-1) = -2\left[\sum_{k=1}^{10} (4k) -\sum_{k=1}^{10} 1\right] = -420.  \]

Thus after 40 moves, you're at $(-400,-420)$ which by Pythagorean theorem is $580$ units from the origin. (This could be difficult, but you could use $20^2+21^2=29^2$.)[/hide][/quote]
Your solution scares small children. Try something more elegant :roll:

[hide="Elegant Solution"]Vertical distance:

$\frac{4^2-2^2}{2}+\frac{8^2-6^2}{2}+\ldots +\frac{40^2-38^2}{2}$
$\frac{2(4+2)}{2}+\frac{2(8+6)}{2}+\ldots +\frac{2(40+38)}{2}$
$2+4+6+8+\ldots +38+40=20(21)$

Horizontal distance:

$\frac{3^2-1^2}{2}+\frac{7^2-5^2}{2}+\ldots +\frac{39^2-37^2}{2}$
$\frac{2(3+1)}{2}+\frac{2(7+5)}{2}+\ldots +\frac{2(39+37)}{2}$
$1+3+5+7+\ldots +37+39=20(20)$

Let $20x=$ the distance of the penguin to the origin.  Then we have a right triangle with legs 20(20) and 20(21) and hypotenuse $20x$.  Consider the similar right triangle with all sides $\frac{1}{20}$ of the original triangle's sides.  This right triangle has legs 20 and 21 and hypotenuse $x$.  Solving for $x$,

$x^2=20^2+21^2$
$x^2=400+441=841=29^2$
$x=29$

Then the distance from the penguin to the origin is $20x=\fbox{580}$[/hide]

[hide="More elegant...?"]
Every odd-numbered move only changes the x-coordinate.  Summing up the changes to the x-coordinate, we have the total amount of movement in the x-axis is the difference between the eastward and westward moves, which is $\frac{1}{2} \left(1 - 3^2 + 5^2 - 7^2 +- ... + 37^2 - 39^2 \right) = \frac{1}{2} \left( (37+39)(37-39) + (33+35)(33-35) + ... + (1+3)(1-3) \right) = -(1 + 3 + 5 + ... + 39) = -20^2 = -400$.

Repeating the same strategy for the movement in the y-axis, we have $\frac{1}{2} \left(2 - 4^2 + 6^2 - 8^2 +-... + 38^2 - 40^2 \right) = \frac{1}{2} \left( (38+40)(38-40) + ... + (2+4)(2-4) \right) = -(2 + 4 + ... + 40) = -2(1 + 2 + ... + 20) = -2 \frac{20(21)}{2} = -420$.

Using $20^2 + 21^2 = 29^2$, we get the total distance as $\boxed{580}$.
[/hide]

Edit:  Ah, matt276eagles beat me to it  :lol:

Hey, mine isn't hard, it's just not elegant. It gets the job done, and it's not a lot of work (i wouldn't call it brute force).

[hide="Answer"]After the first four moves, the penguin has moved a total of $4$ units west and $6$ south. Over the next four moves, the penguin moves a further $12$ units west and $14$ south. After the next three, we have $20$ units west and $22$ south. Notice that we have an arithmetic progression with $w_1=4$ and $s_1=6$, and $d=8$ for both. Thus, over $40$ moves, we have a distance of $10w_1+45d=400$ west and $10s_1+45d=420$ south, giving a total distance of $\sqrt{400^2+420^2}=\boxed{580}$ units from the origin.[/hide]


geometry
geometric transformation
reflection
geometry solved

Let $ABCD$ be a convex quadrilateral, and write $\alpha=\angle DAB$; $\beta=\angle ADB$; $\gamma=\angle ACB$; $\delta= \angle DBC$; and $\epsilon=\angle DBA$. Assuming that $\alpha<\pi/2$, $\beta+\gamma=\pi /2$, and $\delta+2\epsilon=\pi$, prove that
\[(DB+BC)^2=AD^2+AC^2\]

[color=red][Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=30569 .][/color]

Reflect D across AB to get a point E. Then it can be shown easily E,B,C are collinear, AC=AD, CE=DB+BC and $\angle EAC=90^\circ$. The conclusion follows.


vector
calculus
search
geometry
geometric transformation
rotation
function

I am rising junior in HS, and will be taking math classes at a well known university near my school next year. My interest is in Number Theory, so I want to take a Number Theory class ASAP. However, upon meeting with the proffesors it is clear that the first "intro" NT class is too basic. The next NT class however requires a background of  Linear Algebra, and Abstract Algebra. I want to take Linear first semester, and Abstract second semester of this year. I have not yet taken vector calculus yet, and  a prof. has suggested I take this before abstract algebra just for tradition's sake. Do you think it won't be good if I skip vector calc during HS (obviously I would take it in university)? The prof. also warns of getting too advanced in HS (the second NT course is very advanced) and of "burnout".
What do u guys think I should do?

[quote="mathemonster"]a prof. has suggested I take this before abstract algebra just for tradition's sake. [/quote]
I've never seen the point of this.  I still haven't taken a multivariable class and personally I think people should learn linear algebra as soon as possible.

There's something to be said for knowing your limits and making sure your foundations are solid.  For example, I know several people who didn't take group theory seriously enough and are now taking very advanced courses even though their basic group theory is actually weak, so they run into problems sometimes.  (I am, to some extent, one of those people!)  But if you feel confident there's no reason you shouldn't be taking advanced courses.  (I assume you've read the Calculus Trap, though.)

Multivariable calculus is essential for some applications; you really need it for physics. If you stick to math, it isn't really a content prerequisite to anything for a while.
Take it eventually, but your priorities right now should be finding classes that are interesting with good instructors. Committing to a three-course sequence is risky, so figure out a backup plan in case it doesn't work out like you're hoping.

yes im just going to try and take linear algebra first semester and see how that goes. Do you guys think there is any merit to taking "honors" linear algebra rather than regular linear algebra?

That completely depends on the school and the professors.  My guess is that "honors" will be more proof-oriented, so you probably want to spring for that.  To be honest, you'll probably be bored to death in "regular."

Yea maybe. The prof said it really depends on the year and who is teaching the class.

Aside from the interest factor, you really will learn linear algebra better in the honors class, I'd think. You'd also be surrounded by more motivated students, and the class size would probably be smaller. I know I understood more about basic linear algebra doing proofs than by tediously row-reducing matrices for exercises and calculating eigenvalues by hand.

Why don't you search the internet course catalog for the two class options, and if you have a question, post them here? You can also ask the professors/otherwise find out what books they'd be using.

Suppose wanted to take linear algebra at umm... Brown (randomly selected college with a normal and honors lin-alg class). The following course descriptions (taken from Brown) are probably similar to whatever college you plan to take it at:

[code]MATH 0520 - Linear Algebra
Vector spaces, linear transformations, matrices, systems of linear equations, bases, projections, rotations, determinants, and inner products. Applications may include differential equations, difference equations, least squares approximations, and models in economics and in biological and physical sciences. MATH 0520 or 0540 is a prerequisite for all 1000-level courses in Mathematics except MATH 1260 or 1610. Recommended prerequisite: MATH 0180, 0200, or 0350.[/code]
[code]MATH 0540 - Honors Linear Algebra
Linear algebra for students of greater aptitude and motivation, especially mathematics and science concentrators with a good mathematical preparation. Matrices, linear equations, determinants, and eigenvalues; vector spaces and linear transformations; inner products; Hermitian, orthogonal, and unitary matrices; and Jordan normal forms. Provides a more extensive treatment of the topics in MATH 0520. Recommended prerequisites: MATH 0180, 0200, or 0350.[/code]

You can see that the focus is totally different, and the normal class covers less material. My stance has always been that if you understand linear algebra (or any other basic math class that applies to science/engineering) properly, then its applications should seem very intuitive when you learn/see them, and not take long to learn. It is much more useful, I think, for someone with a mature enough approach to learning math, to learn and understand the core concepts properly than to focus on trying to apply the core concepts in a first course.

It's strange that you should pick my alma mater, so naturally I feel compelled to comment. I took the honors version, and I graded and tutored for the regular section. I will talk about the regular class first.

Nearly all of the homework was computation. The occasional proof question was very simple, informal, and mostly algebraic manipulation. I had a very easy time grading and tutoring because students were only expected to demonstrate their mastery of the step-by-step instructions laid out for them at the beginning of each chapter. The largest obstacle they needed to overcome was to occasionally manipulate functions or polynomials as vectors, but the core of the class was grounded in two- or three-dimensional Euclidean space.

The honors class was a completely different beast. Class time was spent going over definitions and proofs. We had problem sets consisting of proving very general statements about very general vector spaces. There was maybe one computation in each set, and that one was rarely the focus of the set. Like TZF said, there was more material covered. That class kicked my butt, but that material and conceptual understanding was much more useful for me in advanced courses than the computational practice I would have gotten from the regular section.

FYI, the three recommended prerequisites are all different flavors of third semester calculus. I think it's recommended because many of the first examples that demonstrate an application of a theorem are from multivariable calculus. However, I know people who have taken the honors course without the calculus and done very well. It's not sequential: linear algebra does not necessitate calculus in the way fractions depend on division.

[quote="mathemonster"]However, upon meeting with the proffesors it is clear that the first "intro" NT class is too basic. [/quote]
Are you sure? I'm pretty familiar with the course like that that we have. We give it a junior-level number and have our "transition" or "intro to proof/discrete structures" course a prerequisite - and we have a lot of our math majors struggle with it. (We then use this course as one of two possible prerequisites for our abstract algebra course.)

A high school student coming into this course with a contest-culture background would indeed have several huge advantages over most of our students in that course. You've been using the language of modular arithmetic for some time; some of our students are just being introduced to that. You already know Fermat's Theorem and Euler's Theorem, which are two of the central theorems of the course. You've seen a proof or two (so I wouldn't worry about that prerequisite.) But have you really seen all of it? And have you seen the systematic development?

Here's an example: Suppose $ a$ is relatively prime to $ b$ and $ a$ divides $ bc.$ Then $ a$ divides $ c.$ Question: right now, without consulting any sources, can you write a convincing proof of this theorem? (And if you try using prime factorizations, I'll rule your argument invalid due to circularity.)

Proof:
* $ (a,b) \equal{} 1 \implies \exists x,y \in \mathbb{Z}$ s.t. $ ax \plus{} by \equal{} 1 \implies acx \plus{} bcy \equal{} c.$$ a|bc \implies \exists k \in \mathbb{Z}$ s.t. $ ak \equal{} bc \implies acx \plus{} aky \equal{} c \implies a(cx \plus{} ky) \equal{} c. cx \plus{} ky \in \mathbb{Z}$ by closure of $ \plus{}$ and $ \times$ in $ \mathbb{Z}, \therefore a|c.$
QED

Note: * Is the LDE in $ \mathbb{Z}$ which follows from $ \mathbb{Z}$ being a Euclidean Domain.

I can prove QR, Mobius Inversion, Minkowski's Thm etc.

So yea the class would be review. The Totient Theorem, and FLT are easy to prove, does your course prove reciprocity?

@Bictor717
Thanks for your reply. I think I should take the honors based on your comment, but I didn't know there were vector calc prereqs are you sure about that? I haven't had it but my friend took it before vector calc while he was in HS and got an A so it clearly can't be a core part of the Homeworks.

[quote="Kent Merryfield"][quote="mathemonster"]However, upon meeting with the proffesors it is clear that the first "intro" NT class is too basic. [/quote]
Are you sure? I'm pretty familiar with the course like that that we have. We give it a junior-level number and have our "transition" or "intro to proof/discrete structures" course a prerequisite - and we have a lot of our math majors struggle with it. (We then use this course as one of two possible prerequisites for our abstract algebra course.)[/quote]

elementary number theory was one of the easiest two upper-division courses I took as an undergrad (the other was some basic abstract algebra course), solely because of my contest background.  in fact, i turned out to be a pretty poor algebraist and a less than stellar number theorist, so it's not like i did well in these courses because of any natural aptitude for the subject.

Judging from his sig, I'm assuming mathemonster attended the PROMYS program. Based on what I've read and heard about PROMYS, I believe they cover a lot of the material from an introductory number theory course.

The prerequisites I was talking about were merely "recommended" in the course descriptions that TZF provided and, at Brown, could be easily navigated around. I didn't say that vector calculus was necessary for any homework assignments. In a linear algebra class, the professor might, for example, relate a basis transformation matrix to a change of coordinates in an iterated integral. While this wouldn't be a large part of a lesson, knowing what the calculus thing means might make the understanding of the linear algebra material easier, but it's entirely possible to learn the linear algebra without calculus.

In fact, the example [b]Bictor717[/b] gave about coordinate transformations is a good example of a reason why it is better to have an understanding of basic linear algebra before taking classes such as vector calculus and differential equations that use linear algebra.


calculus
integration
analytic geometry
trigonometry
calculus computations

$\int_{-\infty}^\infty e^{\frac{-t^{2}}{2}}\,dt$
i found the solution to be $\sqrt{2\pi}$ in mathematica, but im not sure the steps to integrate this type of problem

A very famous problem, with a very famous solution.

Integrate $\iint_{\mathbb{R}^{2}}e^{-\frac{x^{2}+y^{2}}2}\,dx\,dy$ by two methods:

Using rectangular coordinates, we find that it's the square of the integral you want to find.

Using polar coordinates ... (You finish it from here.)

This is a very classical example of change of variable.

[hide="Solution"]
Let $I = \int_{-\infty}^{\infty}e^{-t^{2}/2}\, dt$. Then

$I^{2}= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^{2}+y^{2})/2}\, dx dy$

Now use change of variable $x = r \cos\theta$ and $y = r \sin\theta$. This transform carries $[0,\infty)\times[0,2\pi]$ into the whole $\mathbb{R}^{2}$ plane, and $dxdy = rdrd\theta$. So

$I^{2}= \int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^{2}/2}\, rdr d\theta = 2\pi \int_{0}^{\infty}e^{-r^{2}/2}\, rdr$

Subsitute $t = r^{2}/2$, then $dt = rdr$ and

$I^{2}= 2\pi \int_{0}^{\infty}e^{-t}\, dt = 2\pi$.[/hide]

This is also the Fourier transform of the Gaussian evaluated at 1.
But then you need to know how the Fourier transform works.


So I have to choose between these two universities and I am having a hard time doing so.
Here are some questions that I hope will be answered.
1. Which university has a better math department?
2. Which university has a bigger reputations, is academically better etc.?
3. Which one would you choose?
4. Advantages or disadvantages each university has over the other.

I am inclined to choose Tufts. Bare in mind however (especially when answering three) that there is another problem. The financial aid package at Brandeis is 45000 dollars while the one at Tufts is 42000, leaving a difference of around 5000 dollars (cost of attending is bigger at Tufts) between the two. Paying this sum would not be impossible but not exactly pleasant either.

Thanks a lot.

I wouldn't say that either has a world-famous math department but I don't think either one is terrible either.  I applied to Brandeis but not Tufts (got in to Brandeis).  Brandeis is in more of a secluded area while Tufts is in the city.  Which one of those do you like more?  And if aid matters, then choose Brandeis.  You can't go wrong with either.

I don't know much about either, but I have once taken I combinatorics course with Ira Gessel, who is at Brandeis, during PROMYS (a summer math program). He is an excellent professor who really knows his stuff, but, more importantly, who can also explain the stuff really well. He was also very involved and enthusiastic with research projects, which means he might be a good professor if you're interested in undergrad research in combo. However, this is just a minor thing and you should probably pick the school based on location, campus, activities, gut feeling, etc. if money doesn't matter. I don't think you can go wrong with either one. :)


For some reason, when I try to submit a new thread or post, It posts twice!
Help!!!!

P.S. This was a double of the [u]Avatar[/u] post.

This happens to me too. Make sure you press the submit button only once. If you just reply to the topic, QUICKLY delete the extra(s). If it is not the last post, you can't delete it. You have to do it quickly because if someone sees it, they might post a reply complaining, and not let you delete it. [b]If someone posts a double post, PM him/her and let them delete it, don't reply[/b].


algebra
partial fractions
calculus
calculus computations

I have tried to split in partial fractions but I fail:

$ \frac{\minus{}4p\plus{}6}{(p\minus{}1)^2}$ who can we split this? Thanks.

$ \frac { \minus{} 4p \plus{} 6}{(p \minus{} 1)^2} \equal{} \frac { \minus{} 4}{p \minus{} 1} \plus{} \frac {2}{(p \minus{} 1)^2}$ is the best you're going to get.

Yes but how can you come their? Who can you find this answer?

I need to split $ \minus{}4p \plus{} 6$ into multiples of powers of $ p \minus{} 1$.  So, start from the highest power: I have a -4 in front of a $ p$, so I need to take $ \minus{}4(p \minus{} 1)$.  This gives $ \frac{\minus{}4p \plus{} 6}{(p \minus{} 1)^2} \equal{} \frac{\minus{}4(p \minus{} 1) \plus{} 2}{(p \minus{} 1)^2} \equal{} \frac{\minus{}4}{p \minus{} 1} \plus{} \frac{2}{(p \minus{} 1)^2}$.


factorial
calculus
derivative
trigonometry
real analysis
real analysis solved

Amusingly, a very close problem can be found in SPRA (page 70, problem 4.4.6):
Find $\sum_{n\geq 0}\frac {n!^2}{(2n+1)!}z^n$ ($|z|<4$) :D

Define $f(x)=\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}x^{2n-1}.$
We will derive and solve a differential equation for $f$, then evaluate the solution at 1.

We have $4xf'(x)=\sum_{n=1}^\infty
\frac{(n!)^2}{(2n)!}\cdot4(2n-1)\cdot x^{2n-1}=\sum_{n=1}^\infty 2n
\frac{((n-1)!)^2}{(2n-2)!}x^{2n-1}$
and $(x^3f(x))'=\sum_{n=2}^\infty \frac{((n-1)!)^2}{(2n-2)!}\cdot2n\cdot x^{2n-1},$ so
$4xf'(x)=(x^3f(x))'+2x$, accounting for the first term. Evaluating the derivative and dividing by $x,$ we have $(4-x^2)f'(x)=3xf(x)+2.$ Also, we have $f(0)=0$ as an initial condition.

This is a first-order linear differential equation. The homogeneous equation $(4-x^2)y'=3xy$ has $y=(4-x^2)^{-\frac32}$ as a solution, so we set $f(x)=(4-x^2)^{-\frac32}g(x)$ and get $g'(x)=2\sqrt{4-x^2}.$
Integrating this, $g(x)=x\sqrt{4-x^2}+4\sin^{-1}\left(\frac x2\right)+C.$ Our initial condition gives $C=0,$ so $f(x)=(4-x^2)^{-\frac32}\left( x\sqrt{4-x^2}+4\sin^{-1}\left(\frac x2\right)\right)=\frac{x}{4-x^2}+\frac{4\sin^{-1}\left(\frac x2\right)}{\sqrt{4-x^2}^3}$ and $f(1)=\frac1{3\sqrt3}\cdot\left(\sqrt3+4\cdot\frac{\pi}{6}\right) =\frac13+\frac{2\pi}{9\sqrt3}.$


Hi everybody... I'm Mattia, from Italy...
I partecipated at last IMO.
I published my photo at http://www.esnips.com/web/IMO2008
(beacause of an error the first is the last shot and the last photo is the first shot  :oops: )

I suggest that if somebody posts photos in internet could write here the address!!!
In particular i'm searching photos of the party after the excursion at the Retiro Park, because at the party italian team received the prize for the Gymchana Contest

Goodbye everybody and sorry for my bad english

There are lots of photos in Facebook but i don't know if there are the photos you're searching for.


Here's the problem:

On side $ BC$ of $ \triangle ABC$ point $ A_1$ is taken so that $ BA_1: A_1C \equal{} 2: 1$.  What is the ratio in which median $ CC_1$ divides segment $ AA_1$?

This from Prasolov's [i]Problems in Planimetry[/i] in the section about similar triangles.  I used an analytic argument to show that $ AD: DA_1 \equal{} 3: 1$.  Does anyone know how to solve this problem using similar triangles?  I'm trying to improve my skills in that area.  :)

Does Ceva's theorem count? I see it more having to do with area ratios than anything, though, than similar triangles.

Draw $ \overline{BB_1}$ through $ D$. By Ceva's theorem, $ AB_1: B_1C\equal{}2: 1$; hence, $ [ADB_1]: [CDB_1]\equal{}2: 1$. Then, $ [ADC_1]\equal{}[BDC_1]$ and $ [ACC_1]\equal{}BCC_1]$, so $ [ADC]\equal{}[BDC]$. We also have that $ [BDA_1]: [CDA_1]\equal{}2: 1$, so $ [CDA_1]\equal{}[CDB_1]$.

Thus, $ AD: DA_1\equal{}[CDA]: [CDA_1]\equal{}\boxed{3: 1}$.

I think I know of such a solution.

$ Draw A_1E \parallel{} BC_1$ to intersect $ CC_1$ at E.
$ \Delta CEA_1\sim \Delta CC_1B.$
So,$ EA_1/BC_1 \equal{} CA_1/CB \equal{} 1/3 \equal{} EA_1/AC_1$

$ \Delta DEA_1$ and $ \Delta AC_1D$ are similar.
So,$ DA_1/DA$
=$ EA_1/AC_1 \equal{} 1/3$


algebra
function
domain
calculus
calculus computations

It looks like the domain is empty, because the expression under the square root is always negative. If you had $ \sqrt{1\minus{}x^2\minus{}y^2}$ instead, that would be more interesting.

If we do computations in $ \bf C$, the multi-function does evaluate to a single real number when $ x \in \bf Z$, $ y \minus{} \frac{1}{2} \in \bf Z$.  So one could argue for the corresponding set as the domain.  Either way, it's not a very interesting question.


Prove that for every $ x \in [0, \pi/2)$, the following inequality holds:

$ (\sin \ x \plus{} 3x)(x \ \cos \ x \plus{} 4 \ \sin \ x)(\sin \ x \plus{} \sin \ 2x) \ge 30x \sin \ 2x \ \sin \ x$.

If you cancel a factor of sin(x) and then expand, you can rearrange it into something that's trivially positive... but i'm not sure of a cleaner way :(


this is an amc 12 question, but its easy stuff to people who plan on going to nats.

If a, b, and c are positive real numbers such that a(b+c)=152, b(c+a)=162, and c(a+b)=170, then abc is?

happy solving. if you need any hints, just ask. or wait for a solution.

I changed the naming because "sigh" doesn't describe the problem at all.

lol. i was wondering why it was changed

and btw, i may have exaggerated about it being easy cheese for nats people. it may for some. it may not for some others. dont make it discourage you!

After spending a needless 35 minutes onthis problem, I finally got it.

[hide]

720

[/hide]

okay, ill show the faster way that only took me around 1 minute *laughs evilly for 35 minutes, oh the irony*

Solution:[hide]just add the equations to get 2(bc+ab+ca)=484, so bc+ab+ca=242. if yall subtract this from the equations given previously you receive ca=80, bc=90, and ab=72, multiplying you get a:^2:b:^2:c:^2=80 x 90 x 72=518400=720:^2:, and you take the postive root of that to get abc=720. voila![/hide]

nice question... i took around 5 minutes to think of a good way to solve it, and i was wasting time till i realized that the (b+C) was really  a(b+c).... then it was easy

[hide]720 since after analyzing carefully you get ab - bc = 10, ca-ba=8, cb-ab=18. Solve for ab, bc, and ca, multiply them together and you get a^2 * b^2 * c^2  which is (abc)^2. square root 518400 and you get 720[/hide]

[hide]I got ba=72, ca=80, and bc=90(using system of equations).  I multiplied and took the square root.
 :sqrt: 518400 = 720[/hide]

[quote="Monkey"]This is what I did:

[hide] First, add all the equations together to get 2ab+2bc+2ac=484, and divide by 2 to get ab+bc+ac=242. Then, replace bc+ac with c(b+a)=170, so ab=72. Repeat with the other two pairs of sums. Afterwards, you know that ab=72, bc=90, and ac=80. Multiply all of them together to get (abc)^2=518400, and take the square root to get abc=720.

Answer: 720[/hide][/quote]

yeah... :) thats exactly what i did


Just curious to see what kind of pianos everyone has ;)
For me, Yamaha U3 (upright) :(

i got a baldwin upright

any1 here have a shulze pollman?

Nice piano. Yamaha, upright. Dude, I have a friend that is such a... He has a Steinway, but he never uses it.

Ummm wow no one has a grand? I have one and it's hardly used (my brother plays sometimes).

wat kinda grand?  a good upright can be much better than a bad grand

I've got an annoying Russian one. :P  I can't read Russian so I don't know it's called. :rotfl:

Not a piano, but I've got a Casio LK-55 (Keyboard)

Hmmm those pianos at mop were from the 1920's :P.

[quote="kookamunga"]I have a baby grand. The only marking I could find on it was Khabe on the front... I've never even heard of that before.[/quote]

Is it possible that it's knabe, which is a well-known brand?

[url]http://www.knabepianos.com/[/url]

I have a Diapason. My initial complaint about it was that it was off-key by about one semitone (I was able to verify this using my Casio keyboard, a Yamaha Grand Piano, and some computer software). Unfortunately, I have seen others of the same model and it seems that all Diapasons are off-key.

A 100 year old stienway that belonged to my great grandmother. It is an amazing piano, and I love it. it is Baby Grand

an at least 100 year old upright piano that says it's restored by simpson&son
somehow it sounds like a grand. seriously

i have a yamaha and a.... electronic keyboard :P

A wonderful   Baby Grand Stienway that is about 100 years old...belonged to my Great Grandmother...my parents had it refinished...it is very un-stiff for a stienway.
My Teacher has a nine foot long piano made for a famous pianiist who was really short.

[quote="Dojo"]Woohoo! Go cheap uprights! I have a Young Chang. Fortunately (or unfortunately?), unlike other instruments, you generally use an instrument that is provided for you when you perform, so I'm not too concerned.[/quote]
I think that depends on several factors, at least for me it does. If you have less performing experience, the fact that you are using a different instrument can really throw things off. If you have gotten more comfortable, then often times the piano on stage is better than what you have at home, which can really help. Of course if you have a very nice piano at home then you have a good chance of having to go play on a worse piano on stage, which would suck.

I own a Kawai (grand piano, about 8 years old), and the tone (when it is in tune :D) is amazing!
I sometimes dislike having a nice piano at home, because when I participate in competitions, half the time, the piano is not of a very good quality. If the piano is out of tune, that's worse, since it constantly bothers me when I am playing. I've gotten used to lower-quality pianos after accompanying my choir on an extremely out-of-tune upright piano. :)

[quote="Dojo"]Woohoo! Go cheap uprights! I have a Young Chang. [/quote]

Strange. I also have a young chang (My two cousins on my dad's side- their last names are Young and Chang). It typically fails, what with all the buzzing and breaking and the pedal not working and it being half a note out of tune.

[quote="louiethechewy123"]
Strange. I also have a young chang (My two cousins on my dad's side- their last names are Young and Chang). It typically fails, what with all the buzzing and breaking and the pedal not working and it being half a note out of tune.[/quote]
You're korean, right? I believe it is a korean brand. Anyway, mine works perfectly fine. I don't have any buzzing/breaking/pedal failure/intonation. And if you piano is out of tune, you should call a piano tuner...


Let be a triangle $ ABC$ such that $ (\sin \ B\plus{}\sin \ C)^2\equal{}\cos \ 2A\plus{}\frac{7}{2}$ . Prove that the triangle is equilateral .

If all angles are equal then...

$ (2\sin(A))^2 \equal{} \cos(2A) \plus{} 7/2$

$ 4\sin^2(A) \equal{} 2\sin(A)\cos(A) \plus{} 7/2$

$ 4\sin^2(A) \equal{} 1\minus{}2\sin^2(A) \plus{} 7/2$

$ \dfrac{6\sin^2(A) \equal{} 9/2}{6}$

$ \sin^2(A) \equal{} 3/4$

$ \sqrt{\sin^2(A)} \equal{} \sqrt{3/4}$

$ \sin(A) \equal{} \pm\dfrac{\sqrt{3}}{2}$

$ \arcsin(\dfrac{\sqrt{3}}{2}) \equal{} A$

$ 60^\circ \equal{} A \equal{} B \equal{} C$

Is that what you asked for???

I asked for $ (\sin \ B \plus{} \sin \ C)^2 \equal{} \cos \ 2A \plus{} \frac {7}{2}\Rightarrow \widehat{A}\equal{}\widehat{B}\equal{}\widehat{C}$ , but you proved $ \widehat{A}\equal{}\widehat{B}\equal{}\widehat{C}\Rightarrow (\sin \ B \plus{} \sin \ C)^2 \equal{} \cos \ 2A \plus{} \frac {7}{2}$


induction
real analysis
real analysis unsolved

[quote="ilyasu1"]
case 1:
...

Then take our closed uncountable set to be $[a,b] \cap A$.

[/quote]

Why is that set closed in $\mathbb R$? Believe me, it's not that simple.

Here's something to ponder: if you assume the Continuum Hypothesis is wrong (since it's independent of the Zermelo-Fraenkel+Choice system of axioms, you can either accept it or refute it), then the answer is clearly "yes": take any uncountable set with cardinal smaller than the continuum. It's well-known that every closed uncountable subset of the real line has continuum cardinality, so the set described above works.

Grobber made a couple of very nice observations. Here is one more: the set of closed subsets of $\mathbb R$ has cardinality continuum. So, the problem can be generalized as follows: given a set $X$ of cardinality [i]continuum[/i] and a [i]continuum[/i] of its subsets $F_\alpha$ each of cardinality [i]continuum[/i], find a subset $Y\subset X$ of cardinality [i]continuum[/i] such that $F_\alpha\setminus Y\neq\varnothing$ for all $\alpha$. Of course, the particular cardinality (continuum) seems not so important here; what probably matters is that it is infinite and the same everywhere. So, the first reasonable step towards the solution may be to check the statement with [i]continuum[/i] replaced by [i]infinite countable[/i]. ;)

Of course, this approach will, probably, require using Axiom of Choice and not everyone is willing to accept it. It would be extremely interesting to get an example using neither the Continuum Hypothesis, nor the Axiom of Choice. BTW, Harazi, do you know a solution yourself? (Now, when the "Unsolved" and "Proposed" sections are united, it may be a good idea to somehow indicate such things in the post :))

For most problems I post I do have solutions, but for this one I really did not manage to find one. Sorry. :blush:  (I agree it is a little bit embarrasing to have only a mixture of proposed and unsolved problems section; after all, why?).

Wow! Thanks, fedja! :) It's much cleaner to think about it in these terms. Did you turn this beautiful idea into a proof?

I think I can prove your assertion for any infinite cardinal instead of [i]continuum[/i], but I'll have to check it and recheck it, since I usually mess up such proofs :).

[quote="grobber"] Did you turn this beautiful idea into a proof? [/quote]
At least, I can do it for the countable infinite case. Since it involves only some fairly simple induction, I believe one can do it using transfinite induction for every cardinality but I haven't checked the details yet...

[quote="grobber"]I think I can prove your assertion for any infinite cardinal instead of [i]continuum[/i], but I'll have to check it and recheck it, since I usually mess up such proofs :).[/quote]
Great! Go ahead and post the solution when you are ready.  :)

Ok, here goes nothing :).

Assume we have some infinite cardinal $\alpha$, a set $X$ with $\alpha$ elements, and a collection $(A_x)_{x\in X}$ of  $\alpha$ subsets of $X$, each one having $\alpha$ elements. We want to prove that there is a subset of $X$ with $\alpha$ elements which does not contain any of the sets $A_x$.

Well-order $X$ by using the least ordinal (I'll identify this ordinal with $X$) of cardinality $\alpha$. We can now use an inductive process to refute an element from each $A_x$ and keep another element from any $A_x$ in order to construct a set with the required properties: we simply consider the set formed by all the kept elements:

We can clearly do the above for the first set $A_{x_0}$ ($x_0$ is the smallest element in $X$). Now suppose we have refuted and kept an element from each set $A_x,\ x<v$ for some $v\in X$. The set of refuted elements can be put in a one-to-one correspondence with the set $\{x\in X|x<v\}\ (*)$, and this set has cardinality $<\alpha$, since it's a segment of $X$ (by segment I mean precisely a set of the form $(*)$), and $X$ is the smallest ordinal with cardinality $\alpha$. The same goes for the elements kept up until now, so the union of the two sets has cardinality $<\alpha$, which means that it's a subset of an initial segment of $X$, $\{x\in X|x<t\}$. $|A_v|=\alpha$, so there is an element in $A_v$ which is larger than $t$. Refute this element, and keep the element in $A_v$ which comes right after the one we have refuted ($A_v$ is also naturally well-ordered by the order imposed on it by that on $X$).

Does it have any chance of being correct? :)

[quote="grobber"]
Does it have any chance of being correct? :)[/quote]
Basically it's fine :) though not completely formal and the choice of the refuted and kept elements in $A_v$ is done in a very fancy way (with some obvious typos). On the other hand, all one needs here is to choose 2 distinct elements in the set $A_v$ of cardinality $\alpha$ not belonging to some set of cardinality strictly less than $\alpha$, which can always be done. Anyway, I would consider it a valid solution, if I were reading it in some math. paper. Of course, any criticism is welcome and I will gladly join Grobber in the effort to clarify any detail anybody finds confusing.

I still wonder if we can find a construction for the initial Harazi's problem avoiding the use of Axiom of Choice (though I do not have the slightest idea of how this may be accomplished :().

[quote="fedja"]
the choice of the refuted and kept elements in $A_v$ is done in a very fancy way (with some obvious typos). On the other hand, all one needs here is to choose 2 distinct elements in the set $A_v$ of cardinality $\alpha$ not belonging to some set of cardinality strictly less than $\alpha$, which can always be done. [/quote]

Yes, it's pretty clumsy, actually :). Just to give an example, there's no need for the two elements we choose to be consecutive. I did find one of those typos though :).

[quote="harazi"]Is there a non-numerables set of real numbers such that any subset which is closed in the set of real numbers is numerable?[/quote]

Now we know that the answer is 'yes'. Are there any Borel sets with this property?

The answer is "No". Hint: there are no Suslin sets with this property (for the proof, it is convenient to represent each Suslin set as the result of applying the A-operation to closed intervals).


Check the properties of the element Manganese on [url=http://www.infoplease.com/periodictable.php?id=25]this web page[/url]. Do you see anything strange?

[hide=":D"]it was discovered in sweden and it's name is italian   [/hide]

The name is latin, not italian. Anyway, that's not what I was referring to. Look closer. Don't you see something odd?

Can oxidation number of metals take on negative values? I thought they must be positive. Anyway, I have never heard of $ \minus{}1$ oxidation state for $ Mn$.

Yes, that's what I was referring to. It seems strange that metals can achieve negative oxidation numbers, but indeed they can under special conditions. For example, when sodium metal is dissolved in ethylenediamine (and in other amines or ethers) there is the formation of both $ Na^{\plus{}}$ and $ Na^{\minus{}}$ ions. Also, some metal carbonyls, $ M(CO)_{n}$ (where the metal has zero oxidation state), can be reduced to metal carbonyl anions, in which the metal has formally a negative oxidation state.


Prove for given $n$, there is an integer $N(n)$ such that any collection of $N\geq N(n)$ points in the plane, no three on aline, has a subset of $n$ points forming a convex n-gon.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=47979 , http://www.mathlinks.ro/Forum/viewtopic.php?t=26701 , http://www.mathlinks.ro/Forum/viewtopic.php?t=21981 .

  Darij


inequalities
function
induction
algebra
polynomial
floor function
inequalities unsolved

Prove that:/{n*square root2} - {n*square root3}/>1/(20n^3)(n is positive integer!)
PS:/ / means absolute value

$ \text{Prove that: }\sqrt [n]{3}\minus{}\sqrt [n]{2} > \frac {1}{20n^3}?$

Thx for your help but i mean n times square root of 2 similar to 3. And you also miss aboslute value!

Do you mean $ | n \sqrt{2} \minus{} n \sqrt{3} | > \frac{1}{20n^3}$?  But the LHS is increasing and the RHS is decreasing, so this is trivial.  Perhaps you meant

$ | n \sqrt{2} \minus{} m \sqrt{3} | > \frac{1}{20n^3}$

or something along those lines?

Yeah but u lack this sign:{ }.This mean that an odd integer function like [ ] is the greatest integer function! That's why the topic is:Strange and hard!

[quote="t0rajir0u"]Do you mean $ | n \sqrt {2} \minus{} n \sqrt {3} | > \frac {1}{20n^3}$?  But the LHS is increasing and the RHS is decreasing, so this is trivial.  Perhaps you meant

$ | n \sqrt {2} \minus{} m \sqrt {3} | > \frac {1}{20n^3}$

or something along those lines?[/quote]
If it means $ | n \sqrt {2} \minus{} n \sqrt {3} | > \frac {1}{20n^3}$
<=>$ |\sqrt {3} \minus{} \sqrt {2}| > \frac {1}{20n^2}$
But with $ n \in Z,n \neq 0$ we have:
$ |\sqrt {3} \minus{} \sqrt {2}|  > \frac {1}{20} > \frac {1}{20n^2}$
 :P

I will type it again:
Pove that:/ {n times square root2} - {n times square root 3}/>1/(20n^3)(n is positive integer)
PS:{ } means the odd integer function and it equals: {x}=x-[x]

[quote="mihai miculita"]$ \text{Prove that: }\sqrt [n]{3} \minus{} \sqrt [n]{2} > \frac {1}{20n^3}?$[/quote]
It is obviosly
\[ \sqrt [n]{3} \minus{} \sqrt [n]{2} \equal{} exp(\frac {ln2}{n})(exp(\frac {ln 3 \minus{} ln 2}{n}) \minus{} 1) > \frac {ln(3/2)}{n} > \frac {2}{5n} > \frac {1}{20n^3}.
\]

[quote="mihai miculita"]$ \text{Prove that: }{|}\{n.\sqrt {2}\} \minus{} \{n.\sqrt {3}\}{|} > \frac {1}{20.n^3}.$

($ \text{If }x\in\mathbb{R} \text{ then: [x] \equal{} k} \in\mathbb{Z}\Longleftrightarrow k\le{x} < k \plus{} 1 \text{; and x \equal{} [x] \plus{} \{x\}}.$)[/quote]

It's a main problem I want you to help me! Please don't solve anything except it!  :(

It seems that I have no clue to start with.

@harazi: can you help me solve this problem?? I think it can be proved by mathematical induction because I see a nearly similar problem in Russia competition:
Prove that:sum of {square root k)=<(n^2-1)/2(n is natuaral number and k is positive integer form 1 to n^2)

[quote="mihai miculita"]$ \text{Prove that: }{|}\{n.\sqrt {2}\} - \{n.\sqrt {3}\}{|} > \frac {1}{20.n^3}.$

($ \text{If }x\in\mathbb{R} \text{ then: [x] = k} \in\mathbb{Z}\Longleftrightarrow k\le{x} < k + 1 \text{; and x = [x] + \{x\}}.$)[/quote]

We shall prove a somewhat stronger assertion: for each $ n\in\mathbb{N}$ it holds $ \left\Vert n(\sqrt 3-\sqrt 2)\right\Vert>\frac 1{20n^3}$, where $ \Vert\cdot\Vert$ denotes distance to the nearest integer. It is clear why this assertion is stronger that the one being asked for, and it is also clear that it immediately follows if we prove that $ \bigg|(\sqrt 3-\sqrt2)-\frac m n\bigg|>\frac 1{20n^4}$ for each $ n\in\mathbb{N}$ and $ m\in\mathbb{Z}$, therefore we shall prove the latter.

Notice that $ \sqrt 3-\sqrt 2$ is a root of irreducible polynomial $ P(x)=x^4-10x^2+1$. We are now determining the maximal value of $ |P'(x)|=|4x^3-20x|$. Because of $ P''(x)=12x^2-20$, it holds $ P''(x)=0\Leftrightarrow x=\pm\sqrt{\frac 5 3}$, what is necessary, and it is easily checked that it is also a sufficient condition for the required maximum, that is, $ \max_{x\in\mathbb{R}}|P'(x)|=\left|4\left(\sqrt{\frac 5 3}\right)^3-20\sqrt{\frac 5 3}\right|\approx 17.21<20$. Further, since $ \left|P\left(\frac m n\right)\right|=\frac{\left|m^4-10m^2n^2+n^4\right|}{n^4}\geqslant\frac 1{n^4}$ (because of the irreducibility of $ P(x)$), the mean value theorem finally gives ${{ \frac 1{n^4}}\leqslant\bigg|\underbrace{P(\sqrt 3-\sqrt 2)}_{=0}-P\left(\frac m n\right)\bigg|=|P'(\xi)|\bigg|(\sqrt 3-\sqrt2)-\frac m n\bigg|<20\bigg|(\sqrt 3-\sqrt2)-\frac m n\bigg|}$, what was to be proven.

I would like to say something more. The proof above was not made by myself specifically for this occasion, but it actually is an adaptation of a proof of Liouville’s approximation theorem. There is also a similar, but much stronger Roth’s theorem, from which follows that for all $ \varepsilon >0$ inequality $ \left|\left\{n\sqrt{3}\right\}-\left\{n\sqrt{2}\right\}\right|>{1\over n^{1+\varepsilon}}$ holds for each $ n\in\mathbb{N}$ but finitely many. Unfortunately, there is no way to determine how much exceptions could possibly be, and we therefore were obliged to use weaker, but more concrete methods.


Prove the Shaphiro inequality for bound 5n/12.

Sorry, is Shapiro.
The proof is not hard,just add to all fractions [b]5/4[/b] and split into 2 fractions each and apply AM-GM


Let say the natural number n good if there exists such integers: a1 (sub 1),a2,...,an that a1+a2+...+an=a1*a2*...*an=n. Find all good numbers[/i]

Clearly, if $ n$ is odd, then all numbers $ a_1,a_2,\dots,a_n$ gotta be odd. Thus, we can, by default, say that $ a_i\equiv 1$ or $ \minus{} 1$ mod $ 4$. Let there be $ k$ numbers among $ a_i$,which give $ 1$ mod $ 4$.

Therefore, $ k \minus{} (n \minus{} k)\equiv ( \minus{} 1)^{n \minus{} k}\equiv n$ mod $ 4$, which means $ k \minus{} n$ is even, or $ n\equiv 1$ mod $ 4$.

Now let $ n \equal{} 4m \plus{} 1$, we're allowed to choose $ a_1 \equal{} a_2 \equal{} \dots \equal{} a_{2m} \equal{} 1$, $ a_{2m \plus{} 1} \equal{} \dots \equal{} a_{4m} \equal{} \minus{} 1$ and $ a_{4m \plus{} 1} \equal{} 4m \plus{} 1$. Apparently, these numbers satisfy the statement.

So the only case left is when $ n$ is even. There has to be an even number of even numbers among $ a_i$, consequently, there has to be at least two even numbers.(if there are none, the product of $ a_i$ is odd). So $ n$ is divisible by $ 4$, let $ n \equal{} 4k$.

if $ k$ is odd:

the collection: $ 2k, \minus{} 1, \minus{} 1,\dots, \minus{} 1,1,\dots,1,\minus{}2$ satsifies thoroughly both statements.(there are $ k \minus{} 2$ of $ \minus{} 1$, $ 3k$ of $ 1$ and $ 1$ of $ \minus{} 2$).

If $ k$ is even:

the collection: $ 2k, \minus{} 1, \minus{} 1,\dots, \minus{} 1,1,\dots,1, 2$ satsifies thoroughly both statements.(there are $ k$ of $ \minus{} 1$, $ 3k \minus{} 2$ of $ 1$ and $ 1$ of $ 2$).

Hopefully, I didn't mess up smth again  :)


problem 1:Given $ a,b,c \ge 0$ and a+b+c=4.Prove that
$ 2a \plus{} ab \plus{} abc \le 9$
problem 2:Given a,b,c>0.prove that :
$ \frac {(a^3 \plus{} b^3)(b^3 \plus{} c^3)(c^3 \plus{} a^3)}{abc} \ge \sqrt {8(a^4 \plus{} b^2c^2)(b^4 \plus{} c^2a^2)(c^4 \plus{} a^2b^2)}$
problem 3:given $ a,b,c \ge 0$.
$ \frac {ab}{(a \plus{} b)^2} \plus{} \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \plus{} \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{k(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2} \le \frac {3}{4}$
Prove this ineq for$ k \equal{} 12$,$ k \equal{} 4$
problem 4:Given $ a,b,c \ge 0$.prove that:
a)$ \sqrt {\frac {a^2 \plus{} bc}{b^2 \minus{} bc \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} ca}{c^2 \minus{} ca \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} ab}{a^2 \minus{} ab \plus{} b^2}} \ge 3$
b)$ \sqrt {\frac {b^2 \minus{} bc \plus{} c^2}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {a^2 \minus{} ab \plus{} b^2}{c^2 \plus{} ab}} \plus{} \sqrt {\frac {c^2 \minus{} ca \plus{} a^2}{b^2 \plus{} ac}} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \ge 4$
problem 5:Given $ a,b,c \ge 0$ and a+b+c=3
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge 3 \plus{} \frac {9(ab \plus{} bc \plus{} ca \minus{} 3)^2}{4}$
problem 6:Given $ a,b,c \ge 0$ $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$ and $ k \ge 3$Prove that:
$ a \plus{} b \plus{} kc \ge 4$
problem 7 :Prove for all real a,b,c
$ \frac {a[(k \minus{} 1)a \plus{} kb]}{(a \plus{} b)^2} \plus{} \frac {b[(k \minus{} 1)b \plus{} kc]}{(a \plus{} b)^2} \plus{} \frac {c[(k \minus{} 1)c \plus{} ka]}{(a \plus{} b)^2} \le \frac {k^2 \plus{} 4k \minus{} 4}{4} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

If k=-1 we get:
$ \sum\frac {a(2a \plus{} b}{(a \plus{} b)^2} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {7}{4}$
if k=0 We get
$ \frac {a^2}{(a \plus{} b)^2} \plus{} \frac {b^2}{(b \plus{} c)^2} \plus{} \frac {c^2}{(c \plus{} a)^2} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 1$
If k=1 ,we get an old problem of canhang2007
$ \frac {ab}{(a \plus{} b)^2} \plus{} \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \le \frac {1}{4} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$(Vo Quoc Ba Can)
If k=2 we get:
$ \sum\frac {a(a \plus{} 2b)}{(a \plus{} b)^2} \le 2 \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
problem 8: Given $ k > 0$ and a=b+c=3.prove that:
$ \frac {a^2}{ka^2 \plus{} b \plus{} c} \plus{} \frac {b^2}{kb^2 \plus{} c \plus{} a} \plus{} \frac {c^2}{kc^2 \plus{} a \plus{} b} \le \frac {3}{k \plus{} 2}$
Problem 9:a+b+c=1 and $ a \plus{} b \le 0$.prove
$ \sum\sqrt {a^2 \minus{} a \plus{} 1} \ge 3$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.

[quote="secrets"]
problem 8: Given $ k > 0$ and a+b+c=3.prove that:
$ \frac {a^2}{ka^2 \plus{} b \plus{} c} \plus{} \frac {b^2}{kb^2 \plus{} c \plus{} a} \plus{} \frac {c^2}{kc^2 \plus{} a \plus{} b} \le \frac {3}{k \plus{} 2}$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.[/quote]
This inequality is false :wink:
Try $ k\equal{}\frac{1}{2}$,we have:
$ \frac{a^2}{ka^2\plus{}b\plus{}c} \ge \frac{1}{k\plus{}2} \plus{}\frac{5}{(k\plus{}2)^2}(a\minus{}1)$
It is equivalent to:
$ \frac{(a\minus{}1)^2(9\minus{}3k\minus{}5ka)}{(k\plus{}2)^2(ka^2\minus{}a\plus{}3)} \ge 0$
Which is obvious true :)
Then, we have:$ LHS \ge RHS$

Are you sure?
[hide]10 phút nữa em sẽ check lại bài này,anh làm những bài còn lại đi ạ[/hide]

[quote="secrets"]problem 1:Given $ a,b,c \ge 0$ and a+b+c=4.Prove that
$ 2a \plus{} ab \plus{} abc \le 9$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.[/quote]
We have$ ab\plus{}abc\equal{}ab(c\plus{}1) \le \frac{(a\plus{}3b\plus{}3c\plus{}3)^3}{243} \equal{}\frac{(15\minus{}2a)^3}{243}$
So,$ LHS \le \frac{486a\plus{}(15\minus{}2a)^3}{243} \equal{}\frac{(a\minus{}3)^2(8a\minus{}132)}{243} \plus{}9 \le 9$
The eqality holds when$ (a,b,c)\equal{}(3;1;0)$ :)

[quote="secrets"]
problem 3:given $ a,b,c \ge 0$.
$ \frac {ab}{(a \plus{} b)^2} \plus{} \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \plus{} \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{k(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2} \le \frac {3}{4}$
Prove this ineq for$ k \equal{} 12$,$ k \equal{} 4$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.[/quote]
By AM-GM inequality we have:
$ \frac{3}{4} \minus{}\sum \ \frac{ab}{(a\plus{}b)^2} \equal{}\frac{1}{4}(\sum \ \frac{(a\minus{}b)^2}{(a\plus{}b)^2}) \ge \frac{3}{4} \sqrt[3]{\frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}} \ge \frac{3}{4} \frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}$
So, this inequality is true with $ k \le \frac{3}{4}$ :)

[quote="secrets"]I also checking for problem,k>0,I thinks it is true[/quote]
$ k\equal{}\frac{1}{2}$ is wrong,my friend
This inequality is true with $ k \ge 3$ :)

[quote="secrets"]
problem 2:Given a,b,c>0.prove that :
$ \frac {(a^3 \plus{} b^3)(b^3 \plus{} c^3)(c^3 \plus{} a^3)}{abc} \ge \sqrt {8(a^4 \plus{} b^2c^2)(b^4 \plus{} c^2a^2)(c^4 \plus{} a^2b^2)}$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.[/quote]
Suppose $ abc\equal{}1$
By AM-GM inequality ,we have:
$ (a^3\plus{}b^3)(b^3\plus{}c^3)\equal{}(b^6\plus{}a^3c^3)\plus{}b^3(a^3\plus{}c^3) \ge 2\sqrt{(b^6\plus{}a^3c^3)(b^3(a^3\plus{}c^3)}$
Similar, we have:
$ LHS^6 \ge 8^4(a^9\plus{}1)^2(b^9\plus{}1)^2(c^9\plus{}1)^2$
So,Iwill must prove that:
$ 8(a^9\plus{}1)^2(b^9\plus{}1)^2(c^9\plus{}1)^2 \ge (a^6\plus{}1)^3(b^6\plus{}1)^3(c^6\plus{}1)^3$
By HOLDER inequality,we have:
$ (1\plus{}1)(a^9\plus{}1)(a^9\plus{}1) \ge (a^6\plus{}1)^3$
Then ,we have Q.E.D :)

[quote="secrets"]
Problem 9:a+b+c=1 and $ a \plus{} b \le 0$.prove
$ \sum\sqrt {a^2 \minus{} a \plus{} 1} \ge 3$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.[/quote]
Since $ a\plus{}b\plus{}c\equal{}3$ and $ a\plus{}b\le 0$, then $ c\ge 1$,so$ \sqrt{c^2\minus{}c\plus{}1} \ge 1$
$ \sqrt{(\frac{1}{2}\minus{}a)^2\plus{}\frac{3}{4}} \plus{}\sqrt{(\frac{1}{2}\minus{}b)^2\plus{}\frac{3}{4}} \ge \sqrt{(1\minus{}a\minus{}b)^2\plus{}3} \ge 2$
Then, we have Q.E.D :)

[quote="secrets"]
problem 4:Given $ a,b,c \ge 0$.prove that:
a)$ \sqrt {\frac {a^2 \plus{} bc}{b^2 \minus{} bc \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} ca}{c^2 \minus{} ca \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} ab}{a^2 \minus{} ab \plus{} b^2}} \ge 3$
b)$ \sqrt {\frac {b^2 \minus{} bc \plus{} c^2}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {a^2 \minus{} ab \plus{} b^2}{c^2 \plus{} ab}} \plus{} \sqrt {\frac {c^2 \minus{} ca \plus{} a^2}{b^2 \plus{} ac}} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \ge 4$
I post it in maths.vn,but all people are busy,so I post it in mathlinks.[/quote]
b) See solution of can_hang 2007
a) Soppose$ c\equal{}min(a,b,c)$.
Then$ LHS \ge \frac{a}{b}\plus{}\frac{b}{a}\plus{}\sqrt{\frac{ab}{a^2\minus{}ab\plus{}b^2}} \equal{}\frac{(a\minus{}b)^2}{ab}\minus{}\frac{(a\minus{}b)^2}{\sqrt{ab}(\sqrt{ab}\plus{}\sqrt{a^2\minus{}ab\plus{}b^2})}\plus{}3 \ge 3$  :)

[quote="secrets"]Thanks,quykhtn,but all people,Do you like this problem? :([/quote]
I think many people like inequalities,secrets :)

[quote="nguoivn"]Yes, I also think your ineqs is very nice  :)
@secrets: [hide]Ki~ nang sang' tac' cua? em rat tot', tiep tuc phat huy nhe'. Bay gio` tui. anh cung~ it di lam` bdt roi`(de? time lam` 1 so viec khac:P) , da~ den' luc' nhuong`san choi cho cac ban. tre? khac nhu em, Quy', Materazzi,... Hi vong. cac' em se~ phat trien tot hon nua~ de? dua bdt Viet Nam len 1 tam` cao moi'. Good luck  :)  [/hide][/quote]
@ nguoivn [hide] cam' on anh nhieu- lam, rat tiec' khi anh it lam bdt nua? :( .em va cac ban. se co gang hon nua ,mong dc anh va moi. nguoi giup do :)  :)  :)
[/hide]

Please , I request you ONLY ONE thing, PLEASE do not post so many problems at once in a single topic, it puzzles the others enormously...... :blush:  :(

One cannot think about a single one and the discussion is not CONCENTRATED or CENTRALISED on a single inequality, that is why I did not find much interest in this topic..... :o

Please forgive me and post NEW TOPICS to discuss NEW PROBLEMS in every forum.

                                        Sincerely,
                                                potla.


$ \tan^2\frac {\pi}{16} \plus{} \tan^2\frac {2\pi}{16} \plus{} \ldots \plus{} \tan^2\frac {7\pi}{16} \equal{}$

$ \equal{} \frac {1 \minus{} \cos\frac {\pi}{8}}{1 \plus{} \cos\frac {\pi}{8}} \plus{} \frac {1 \minus{} \cos\frac {\pi}{4}}{1 \plus{} \cos\frac {\pi}{4}} \plus{} \frac {1 \minus{} \cos\frac {3\pi}{8}}{1 \plus{} \cos\frac {3\pi}{8}} \plus{} \frac {1 \minus{} \cos\frac {\pi}{2}}{1 \plus{} \cos\frac {\pi}{2}} \plus{} \frac {1 \minus{} \cos\frac {5\pi}{8}}{1 \minus{} \cos\frac {5\pi}{8}} \plus{} \frac{1\minus{}\cos\frac{3\pi}{4}}{1\plus{}\cos\frac{3\pi}{4}}\plus{}\frac {1 \minus{} \cos\frac {7\pi}{8}}{1 \plus{} \cos\frac {7\pi}{8}} \equal{}$

$ \equal{} \left(\frac {1 \minus{} \cos\frac {\pi}{8}}{1 \plus{} \cos\frac {\pi}{8}} \plus{} \frac {1 \plus{} \cos\frac {\pi}{8}}{1 \minus{} \cos\frac {\pi}{8}}\right) \plus{} \left(\frac {1 \minus{} \cos\frac {\pi}{4}}{1 \plus{} \cos\frac {\pi}{4}} \plus{} \frac {1 \plus{} \cos\frac {\pi}{4}}{1 \minus{} \cos\frac {\pi}{4}}\right) \plus{} \left(\frac {1 \minus{} \cos\frac {3\pi}{8}}{1 \plus{} \cos\frac {3\pi}{8}} \plus{} \frac {1 \plus{} \cos\frac {3\pi}{8}}{1 \minus{} \cos\frac {3\pi}{8}}\right) \plus{} 1 \equal{}$

$ \equal{} \frac {2 \plus{} 2\cos^2\frac {\pi}{8}}{\sin^2\frac {\pi}{8}} \plus{} \frac {2 \plus{} 2\cos^2\frac {\pi}{4}}{\sin^2\frac {\pi}{4}} \plus{} \frac {2 \plus{} 2\cos^2\frac {3\pi}{8}}{\sin^2\frac {3\pi}{8}} \plus{} 1 \equal{}$

$ \equal{} \frac {6 \plus{} 2\cos\frac {\pi}{4}}{1 \minus{} \cos\frac {\pi}{4}} \plus{} 6 \plus{} \frac {6 \plus{} 2\cos\frac {3\pi}{4}}{1 \minus{} \cos\frac {3\pi}{4}} \plus{} 1 \equal{} 35$


1/640 = x/10^y

i cross mutilplyed and got 10^y=640x

the problem is find the least vaules of x??

The point is to find the lease vaules of x i worked out and got 2^7 times 5^1 times x

now i need to get 5 ^ 7 power...???

If you have $10^y = 640 x$, you can factor to get $10^{y-1} = 2^6 x$. If $y=7$, we'll have the smallest solution with $x=5^6$. (This is because $2^k \cdot 5^k = 10^k$.)


Solve the equation: $ (x^2\minus{}1)^2\minus{}4(x\minus{}1)^2\equal{}12(x\plus{}1)^2$

$ (x\minus{}1)^2(x\plus{}1)^2\minus{}4(x\minus{}1)^2 \equal{} 12(x\plus{}1)^2$

Let $ a\equal{}x\minus{}1$ and $ b\equal{}x\plus{}1$, then

$ a^2b^2 \minus{} 4a^2\equal{} 12b^2$
$ a^2b^2 \minus{} 4a^2\minus{}12b^2\equal{}0$
$ (a^2\minus{}12)(b^2\minus{}4) \equal{} 48$
$ (a\minus{}2\sqrt{3})(a\plus{}2\sqrt{3})(b\minus{}2)(b\plus{}2)\equal{}48$

Continuing to solve by factors of $ 48$ will get some of the solutions but not probably not all.


Hey,

can anyone pleae help me on the followeing problems.

1. find all positive integers x,y s.t. x^2+7=2^y. there are only two soltions but how do we prov e that..

2. Fina all triples x,y,z of coprime postive integers s.t. x^3+y^3+z^3=6xyz. Only solution is 1 2 3 but how we prove that...

Thanks

BOMB

- Unifies to unity.

There are 5 solutions to the first equation, known as the Ramanujan-Nagell equation.  The largest solution is 2^15 - 7 = 181^2.


:!: [size=150]prove this  inegalité [/size]

there is no one give a solution :!:  :?:


analytic geometry
algebra unsolved
algebra

let $ (D)$ a line intersect with a curve with an equation:
$ y\equal{}x^4\minus{}6x^3\plus{}4x\minus{}1$
in 4 points which their cordinates are $ (x_1,y_1)$,$ (x_2,y_2)$,$ (x_3,y_3)$ and $ (x_4,y_4)$
prove that the sum : $ x_1(x_2\plus{}x_3\plus{}x_4)\plus{}x_2(x_3\plus{}x_4)\plus{}x_3x_4$  is indepent  from the place of $ (D)$

$ if \ (D): \ y \equal{} mx \plus{} n \Rightarrow X^4 \minus{} 6x^3 \plus{} 4x \minus{} 1 \equal{} mx \plus{} n \Rightarrow \\
\Rightarrow x^4 \minus{} 6x^3 \plus{} (4 \minus{} m).x \minus{} (n \plus{} 1) \equal{} 0 \Rightarrow \\
x_1x_2 \plus{} x_1x_3 \plus{} x_1x_4 \plus{} x_2x_3 \plus{} x_2x_4 \plus{} x_3x_4 \equal{} 0; (\forall)m,n\in \mathbb{R}.$

Let the equation of line $ D$ be $ y\equal{}ax\plus{}b$. If $ x_i$ is the abscissa of an intersection point, we have
\[ x_i^4\minus{}6x_i^3\plus{}(4\minus{}a)x_i\minus{}(1\plus{}b)\equal{}0.
\]
Therefore $ x_1(x_2\plus{}x_3\plus{}x_4)\plus{}x_2(x_3\plus{}x_4)\plus{}x_3x_4$ is the coefficient of $ x^2$, which is 0, and does not depend on $ D$.


There are two values of $ a$ for which the equation $ 4x^2 \plus{} ax \plus{} 8x \plus{} 9 \equal{} 0$ has only one solution for $ x$. What is the sum of these values of $ a$?

$ \textbf{(A)}\ \minus{}16\qquad
\textbf{(B)}\ \minus{}8\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ 8\qquad
\textbf{(E)}\ 20$

[quote="Silverfalcon"]There are two values of $a$ for which the equation $4x^2+ax+8x+9 = 0$ has only one solution for $x$. What is the sum of these values of $a$?

A. -16
B. -8
C. 0
D. 8
E. 20[/quote]

[hide]
For a general quadratic ($ax^2+bx+c=0$) to only have one solution, it's discriminant ($b^2-4ac$) must be 0.  $4ac$ in this case would be 144, so we need the value of b to be 144.  $(8+a)^2=144$ has solutions 4 and -20.  $4+(-20)=-16$
A
[/hide]

[hide="Answer"]$4x^2+(8+a)x+9=0$ must be a perfect square, so we have $(2x+3)^2=0$ or $(2x-3)^2=0$, which give $b=8+a=\pm 12$, so we have $a=4,-20$, and the sum of the values is $-16\Rightarrow \boxed{A}$.[/hide]

just do the discriminant test. it should be obvious from there.


geometry
circumcircle
incenter
geometric transformation
rotation
homothety
ratio

Let $P$ be an arbitrary point on the arc $AB$ (which doesn't contain the point $C$) of the circumcircle of triangle $ABC$. Let $I$ and $J$ be the incenters of triangles $APC$ and $BPC$. Prove that, when the point $P$ is varying, the circumcircle of triangle $PIJ$ intersects the circumcircle of triangle $ABC$ again at a fixed point $K$.

[color=red][[b]Moderator edit:[/b] This problem has also been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=19777 .][/color]

I think I have a proof, but it?s long, hard to follow and I didn?t prove everything I used (maybe I even made some mistakes). Here it goes anyway

 :D
Let T be the incenter of ABC (I am used to I being the incenter, but I is already taken). By an angle chase we can see that TI perpendicular to TJ. We can also see that I is on the circle ATC and J is on the circle BTC. I guessed what K must be. Let K be a point on the smaller arc AB s.t. KT is the bisector of angle AKB. We just use some more angle-chasing to prove some properties of K. For example triangles KAT and KTB are similar (with their vertices in this order). It is easy to see that angle IPJ is Pi/2-C/2, so inorder to prove what we want it is enough to prove angle IKJ=Pi/1-C/2 for any P. Consider the similitude (rotation composed with a homothety) of center K, angle Pi/2-C/2 and ratio k=KT/KB=KA/KT (because triangles KAT and KTB are similar).

The image of B is T and the image of  T is A. For the 2 circles BTC and ATC we have the pairs of points (B, T) on BTC and (T, A) on ATC such that the second pair is the image of the first one. This isn?t enough to say that circle ATC is the image of circle BTC, but we can also prove (I?m not going to do it here, it?s just calculations) that radius(circle ATC)/radius(circle BTC)=k. If we also prove this, combined with those 2 pairs of points, it?s enough to say that circle ATC is the image of circle BTC by the similitude we considered.

J is as we have defined it and let X be its image. T is the image of B so the angle between BJ and TX is the angle of the similitude=Pi/2-C/2. Angle BJT is Pi-C/2 because J is on the circle BTC, and from these 2 equalities we get angle JTX = Pi/2.  X is on circle ATC and because I is on the circle ATC and JTI is also Pi/2 (I wrote it before in this post, without a proof, it?s just angle-chasing), it means that I must be X, so I is the image of J by the similitude we considered, which has center K and angle Pi/2-C/2, so angle JKI = Pi/2-C/2, what we wanted to prove. Please tell me if you find any mistakes, and I hope you?re patient enough to read it all...
 :D Bye!

Dear
If you know "spiral similarity "try to prove this one
with "spiral similarity".(It is not easy!! ,If you needed
a hint , ask)
It's really delicious!!! to solve it in this way.

what is  the "spiral similarity "? (is it a geometrical transformation? )

cheers! :D

Yeah
it's a transformation, it's the combination of a homothecy and a
rotation.It's really great, (I believe even greater than Inversion)
You can try Yaglom's "geometric transformations" volume 2.

                Bye

yeah,i think that's wot grobber means by similitude,duznt he?so wots the proof using spiral similarity amir is tokin abt,is it sumthing different?

Yeah, that's what I mean by 'similitude'. That's how we call it in Romanian. I didn't know it was called 'spiral similarity'. I'll edit the post if you want me to.. :D

Dears
First We must introduce two lemmas:

(1)if any 2 lines AB and CD intersect in E .then the center of the spiral similarity(K) that takes A to C and B to D(and actually AB to CDdirectly) is :(this spiral similarity has to have actually  the angle <E and the ration AB/CD)
       K is the second intersection of the circumcircle of ACE and BDE.
 if we need that this similitude takes inversely A to D and B to C(But this time AB to CD reversely) the center is the second intersection of ADE and BCE.
                  (Try to prove this lemma, it is not difficult.)
(2) if the second intersection af the angle bisector of A in any triangle ABC with the circumcircle is M then MI=MB=MC( where I is the incenter)

So we have in our case:

Let PI intersect the circumcircle of ABC in  L and PJ in M.then K is the center of the spiral similarity that takes IL to JM, with the angle:<B/2+<C/2.
the ratio is IL/JM.But we know that MJ=MBand and LI=LC.But as M and L are the midpoints of the arcs BC andAC they are constant so we have a spiral similarity that has to take M to L with the constant angle <B/2+<C/2
and the constant ratio MB/LC. So the center is constant.

(great ...no?)


[quote="sandra90"]I know that you have allready mentioned that [b] everyone [/b] below the age of 20 can participate. But my questions are :has an 8th grade student any chance? will there be 8th grade problems also? :blush:

thank you :)[/quote]You can check previous problems to estimate the difficulty level. Often the problems are very difficult, but once in a while a very easy problem appears intentionately or not :D

In general I'm afraid it would be very hard for an 8th grader to participate in the MathLinks Contest.

[quote="siuhochung"][quote="Valentin Vornicu"][quote="mecrazywong"]Why don't give different problem sets for different grades?(just like APMO or RMO) This will surely encourage more people to join the contests.[/quote]Because 1) I don't have time and 2) problems are difficult to make cheating difficult :)[/quote]

like wt grinberg has said in a thread, the questions are becoming unsolvable... (at least to me, i'm still too weak..  :blush: ) :([/quote]He, he :) You have to carefully check the contest. Sometimes the problems are not as hard as they seem ;) :D

[quote="harazi"]Vali, you said something about problems that can be made in 8-th grade? I think I can make an inequality for 8-th grade, if you really want!  ;)[/quote]Hehe, I can make one too :P :D There were some very easy, but also very few, problems on the MathLinks Contest. But one needs to pay attention to all the problems to discover the easy ones :D

Valentin, don't forget that some inequalities which can be made with 8-th grade methods are not solved by 12-th grade students, so do not hurry. I was serious! ;)

Yes harazi....pls make an inequality for the 8th grade... :lol:


Can you write numbers in the fields of a 5 * 5 grid so that in any 2 * 2 part of the grid the sum of the four numbers should be negative, while in any 3 * 3 part of the grid the sum of the nine numbers should be positive?

couldnt get any valid solutions on mc forum, so i put it here

I think you'll always be caught by the middle 3x3 square. Ex:

[code]3    3     3    3     3

3   -10    3   -10    3

3    3     3    3     3

3   -10    3   -10    3

3    3     3    3     3[/code]


I am having a little trouble switching this equation from parametric to cartesian form.

x=3 sin 2t
y=1.5 cos t
0  _<_  t  _<_  2pi       (though this limit doesn't really matter i don't think)

I am going to continue trying on my own.  When I finally give up, I hope someone smarter than I will have gotten this.

Thanks

since there are at most 4 times this crosses a vertical line but the best way i can think is to do the inverse of it, then you just simplify to get $\pm 3 \sin( 2 \cos^{-1}(\frac{2}{3}y) =x$, and that is in cartesian so...but if you want it in function form, you need to split it up more

the thing i don't like about that is that it has a trig function and an inverse trig function.

i am looking for something more like the equation of a circle:   x^2+y^2=4

where the trig would be eliminated.

also i dont believe you can call the graph a "function," it doesn't pass the vertical line test  ;)

so far i have gotten to 4(x^2+y^2)=(cos 2t - 1.125)(cos 2t + 1)

so close.

if you wanted the graph to be a "function", you would need to break it into 4 separate functions
as for a circle with out trig, $y = \pm \sqrt{r^{2}-x^{2}}$


inequalities
algebra
polynomial
function
real analysis
complex analysis
complex analysis unsolved

Let $ f (x) \equal{} (x \minus{} x_1)(x \minus{} x_2) \ldots (x \minus{} x_n)$ be a polynomial with real coefficients with distinct real roots $ x_i$.
Is it true that for $ k \equal{} 0, 1, \ldots, n \minus{} 1$ holds:
\[ (\minus{}1)^{k \minus{} 1} \sum_{i \equal{} 1}^n \frac{e^{x_i} x_i^k}{ f' (x_i) } > 0 .\]

It certainly smells fishy: the right hand side is $ e^t$ times the linear function of $ t$ when $ k\equal{}1$ and $ x_i\equal{}y_i\plus{}t$. The coefficient at $ t$ in the linear part is certainly not always zero, so there cannot be any sign-preservation property unless you restrict $ x_j$ from either above or below.

First of all, I am not sure is this correct :)
But, I forgot to write the condition $ 0 < x_i < n$.

Still fishy: take $ n\equal{}2$. Then the denominators are the same up to sign and the numerator is an increasing function of $ x$ for both $ k\equal{}0$ and $ k\equal{}1$, so the sign of the sum must be the same for $ k\equal{}0$ and $ k\equal{}1$ in this case.

[hide="Anticipating the next correction:"]
If you remove the alternating sign and still impose the condition $ x_i>0$, it becomes true. The residue theorem and Laurent series are your best friends then ;).
[/hide]

What about the sign of
\[ \sum_{i \equal{} 1}^n \frac { x_i^k }{ f' (x_i) \sqrt {x_i} }
\]

The residue theorem implies that $ \sum_{i}\frac {x_i^k}{f'(x_i)}$ is $ 0$ for $ k\equal{}0,\dots,n\minus{}2$ and $ 1$ for $ k\equal{}n\minus{}1$. Let now $ g$ be any smooth function and let $ P$ be the interpolating polynomial for $ f$ at the points $ x_1,\dots,x_n$ of degree $ n\minus{}1$. Then the sum $ \sum_i\frac {g(x_i)}{f'(x_i)}\equal{}\sum_i\frac {P(x_i)}{f'(x_i)}$ equals to the coefficient of $ P$ at $ x^{n\minus{}1}$. This coefficient is just $ \frac 1{(n\minus{}1)!}P^{(n\minus{}1)}$. Now, since $ g\minus{}P$ has $ n$ real roots, we conclude by Rolle's theorem that $ g^{(n\minus{}1)}\minus{}P^{(n\minus{}1)}$ has at least one real root. But $ P^{(n\minus{}1)}$ is constant. Thus, we can conclude that the sum in question equals $ \frac 1{(n\minus{}1)!}g^{(n\minus{}1)}(\xi)$ for some $ \xi$ in the interval containing all $ x_i$. If $ g^{(n\minus{}1)}$ preserves its sign, we can tell the sign of the sum. This is the case in all the above examples.


Find the sum of the odd integers between 10 and 50.

all odds are 11, 13, 15...47, 49.

the sum of 1+3+5=the number of terms^2.

in this case, there were 25 terms which gets you to 635, but you should subtract 1+3+5+7+9, because you didn't include them, which is 25. 625-25=600, the asnwer

ps. this should be in the moems forum. i saw this exact question in a math olympiad

[hide]using the sum of an arithmetic seires formula,

$\frac{(11+49)20}{2}=600$[/hide]


How many posts are needed to advance to the next level? (eg. 20 for P vs NP, 100 for Poincare Conjecture, and thats all I know)

See the first post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280499]here[/url].

Thank you.

Thats weird, it says that the topic is under "Frequently asked questions," but I dont see that topic here.

It's farther down in the topic list. It should be a sticky.


integration
limit
real analysis
real analysis unsolved

Hi,

How can i find out if the following series converges?

thnx :)
[/code]

use wellknown fact :
$ \frac{\pi}{2}\minus{}\arctan n \equal{}\arctan^{\minus{}1}n$
also read something from here http://en.wikipedia.org/wiki/Arctan
it's will be good for you  :)
therefore :
.

[hide]
\[ \frac{\pi}{2}\minus{}\arctan n\equal{}\int_{n}^{\infty}\frac{1}{x^{2}\plus{}1}\ dx\geq\int_{n}^{\infty}\frac{x^{2}\minus{}1}{x^{4}}\ dx\]
\[ \equal{}\int_{n}^{\infty}\frac{1}{x^{2}}\minus{}\frac{1}{x^{4}}\ dx \equal{}\frac{1}{n}\minus{}\frac{1}{3n^{3}}\]
It follows that the series diverges.
[/hide]

$ S \equal{}\sum_{n \equal{} 1}^{\plus{}\infty}\left(\frac{\pi}{2}\minus{}\arctan n\right)$.

Let's compare S with the harmonic series through a limit, which is already known to diverge:

$ \lim_{n\to\infty}\frac{\frac{\pi}{2}\minus{}\arctan n}{\frac{1}{n}}\equal{}\lim_{n\to\infty}\frac{\minus{}\frac{1}{1\plus{}n^{2}}}{\minus{}\frac{1}{n^{2}}}\equal{}\lim_{n\to\infty}\frac{n^{2}}{1\plus{}n^{2}}\equal{} 1$.

As the limit is a real number different from zero, then S also diverges.

You can also use the fact that, (as Extremal seems to have already suggested) as n > 0, then

$ \frac{\pi}{2}\minus{}\arctan n \equal{}\arctan\frac{1}{n}$,

and the series with this last general term diverges, because

$ \lim_{n\to\infty}\frac{\arctan\frac{1}{n}}{\frac{1}{n}}\equal{}\lim_{n\to\infty}\frac{\frac{1}{n}\plus{}O(n^{\minus{}3})}{\frac{1}{n}}\equal{}\lim_{n\to\infty}[1\plus{}O(n^{\minus{}2})] \equal{} 1$.


trigonometry calculus derivative function

Prove that :

$ A\equal{}(\sqrt{2\sin \ x\plus{}4}\minus{}\sqrt{2\sin \ x})\cdot \sqrt{\sin \ x\plus{}1\plus{}\sqrt{\sin^2x\plus{}2\sin \ x}}$ is constant $ (\forall)x\in \left[0,\frac{\pi}{2}\right]$

hello, your term $ A$ is $ 2$ for all $ x \in [0,\frac{\pi}{2}]$. You must differentiate $ A$ with respect to $ x$
and after simplifying this derivative must be vanish for all these $ x$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]You must differentiate $ A$ with respect to $ x$
and after simplifying this derivative must be vanish for all these $ x$.
[/quote]

Sorry for my stupidity , but i can't understand that  :(

hello, what do we get when we differentiating a constant?
Sonnhard.

Simplifying the derivative is a method (because $ A$ is a constant function $ \Leftrightarrow$ derivative of it is $ 0$).

But for this easy problem I don't think we need to use it.

We have $ A$ is a constant $ \Leftrightarrow (\sqrt{2\sin \ x\plus{}4}\minus{}\sqrt{2\sin \ x})^2 (\sqrt{\sin \ x\plus{}1\plus{}\sqrt{\sin^2x\plus{}2\sin \ x}})$ is a constant
$ \Leftrightarrow 4(\sqrt{\sin \ x\plus{}1\minus{}\sqrt{\sin^2x\plus{}2\sin \ x}})(\sqrt{\sin \ x\plus{}1\plus{}\sqrt{\sin^2x\plus{}2\sin \ x}})\equal{}4$ is a constant, which is obvious , done.

0, but thats calculus and this is the high school intermidiate level


modular arithmetic
algebra unsolved
algebra

[quote="jagdish"][b](1)Find the largest 4 Digit no. when divided by 3,5,7,9 It gave a Remainder 1,3,5,7 respectively.[/b][/quote]

$ N\equal{}1\pmod 3$ $ \iff$ $ N\equal{}3a\plus{}1$
$ N\equal{}3\pmod 5$ $ \iff$ $ 3a\plus{}1\equal{}3\pmod 5$ $ \iff$ $ a\equal{}4\pmod 5$ $ \iff$ $ N\equal{}15b\plus{}13$
$ N\equal{}5\pmod 7$ $ \iff$ $ 15b\plus{}13\equal{}5\pmod 7$ $ \iff$ $ b\equal{}6\pmod 7$ $ \iff$ $ N\equal{}105c\plus{}103$
$ N\equal{}7\pmod 9$ $ \iff$ $ 105c\plus{}103\equal{}7\pmod 9$ $ \iff$ $ 6c\equal{}3\pmod 9$ $ \iff$ $ c\equal{}2\pmod 3$ $ \iff$ $ N\equal{}315k\plus{}313$

And the greatest $ 4\minus{}$digits such number is obtained when $ k\equal{}\left[\frac{9999\minus{}313}{315}\right]$ $ \equal{}30$

And so the result is $ N\equal{}315\cdot 30\plus{}313\equal{}9763$

[quote="hsbhatt"]Note that N+2 is divisible by 3,5,7, and 9 so N+2 is divisible by LCM(3,5,7,9) = 315

So N = 313 + 315k satisfies the requirement. Rest is same as pco's post[/quote]

Shorter   :) , you're right


Find alll the natural number solutions such that  $ m>n>1$

and  $ m^n\equal{}n^m$

[hide]For any prime $ p$ and any integer $ x$, let $ v_p(x)$ be the largest power of $ p$ dividing $ x$. $ m^n \equal{} n^m$ implies that $ n v_p(m) \equal{} m v_p(n)$. Since $ m > n$, $ v_p(n) < v_p(m)$, or else $ m v_p(n) > n v_p(m)$. This means that $ n | m$, so we may set $ m \equal{} kn$ for some integer $ k$. Then $ (kn)^n \equal{} n^{kn}$, giving $ kn \equal{} n^k$, so $ n^{k\minus{}1} \equal{} k$. If $ n > 2$, then $ n^{k \minus{} 1} > 3^{k\minus{}1}$. It can easily be shown by induction that $ 3^{k\minus{}1} > k$ for all $ k > 1$, so we must have that $ n \equal{} 2$, trivially resulting in $ k \equal{} 2$ as well. Hence, our only solution is $ (2, 4)$.[/hide]

[hide="Or"]You could take the $ \ln$ of both sides to get $ \frac {\ln m}{m} \equal{} \frac {\ln n}{n}$ and since $ \frac {\ln x}{x}$ is decreasing for $ x > e$, we only need to check $ n \equal{} 2$. But $ (2,4)$ is clearly a solution, and it can be the only as for $ m > 4$, $ \frac {\ln m}{m}$ would just be lower. This has been posted many times.[/hide]


topology
advanced fields
advanced fields unsolved

Give $R^\omega$ in the box topology. Let $Q^\infty$ denote the subspace consisting of sequences of rationals that end in an infinite string of 0's.Which countability axioms ( first countable , second countable , Lindelof, separable )  does this space satisfy?

[quote="jmerry"]Stronger than all of them: $\mathbb{Q}^\infty$ is a space of countably many points. It doesn't matter what topology you put on that.[/quote]

What stronger means?? Does a space with countably many points always first countable???

[quote="grobber"]It's separable and Lindelof, but if a space is countable, this doesn't imply it has a countable basis. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=282817#282817]this problem[/url].

Edit:

I don't like to double-post, so I'm modifying this message :).

When I first saw that problem I qouted, I thought of the following example of a countable but not second-countable topological space:

Take the countable set which we are about to endow with a topology to be $\mathbb N$, and let the open sets be the empty set, and the elements of a [url=http://www.google.ro/url?sa=t&ct=res&cd=3&url=http%3A//mathworld.wolfram.com/Ultrafilter.html&ei=kwD_QurjBa2kiALxw5FG]free ultrafilter[/url] consisting of subsets of $\mathbb N$.

Now, given countably many infinite subsets of $\mathbb N$, it should be easy to construct a subset of $\mathbb N$ with the property that neither it nor its complement in $\mathbb N$ contain any of the (countably many) given subsets. Since either this set or its complement belong to the ultrafilter, we're done: given a countable collection of open sets, we have found an open set which is not the union of sets in our countable collection.

If we want a countable space which is not even first-countable, just add $-1$ to all the non-empty sets of the topology defined above. Then $-1$ does not have a countable collection of fundamental neighbourhoods.

I hope it's correct. I bet there are simpler examples, but this is what I first thought of :).[/quote]

I think in the countable space, first countability and second countability are equivalent.

It is, and I obviously didn't see it when I wrote that :D.


geometry
inradius
inequalities
inequalities proposed

[quote="darij grinberg"][b]a)[/b] Let ABC be a triangle with a right angle at C. Let r be the inradius and $m_a$, $m_b$, $m_c$ be the medians of triangle ABC. Prove that $\frac{r^{2}}{m_{a}^{2}+m_{b}^{2}}\leq \frac{1}{20}$.

[b]b)[/b] Find the actual maximum of $\frac{r^{2}}{m_{a}^{2}+m_{b}^{2}}$.

  Darij[/quote]
the first part.
$r= (a+b-c)/2 $ and
 $m_{a}^{2}+m_{b}^{2} = c+1/4 *(a+b)$.
the inequality becomes $ 5(a+b-c)<=c+1/4 *(a+b)=5c/4 $
because we know that $ a+b=c  $
so we have :  $ (a+b-c)<=c/4 $
this is not difficult to prove,
we get easily $ c/2 <=a+b<=3c/2  $
using $ c=\sqrt(a+b) $ we can find the desired inequality


superior algebra
superior algebra unsolved

is this possible to partetion $ \mathbb{R}^2$ into the infinitly many number of connected dense sets?

I have no idea what your question means. Please use better English.

I guess the problem is:
Can you partition the plane in infinitely many connected and dense subsets?


Heather and Isabelle play a two-person game which is won by the first person to accumulate 10 points. At each turn Heather gains a point with probability of $ \frac{2}{3}$ . If she doesn't get a point then Isabelle gets a point. Isabelle is now ahead 9 to 6. What is the probability that Isabelle will win? Express your answer as a common fraction.

Heather must gain 4 points in a row to win, so she has a $ \left(\frac{2}{3}\right)^4\equal{}\frac{64}{81}$ probability of winning. Therefore, Isabelle's probability is $ \frac{17}{81}$.


geometry
geometric transformation
rotation
ratio
reflection
cyclic quadrilateral
perpendicular bisector

Let $P$ be a point inside of a triangle $ABC$ such that $\measuredangle PAC = \measuredangle PBC$. The perpendiculars from the point $P$ to the lines $BC$ and $CA$ meet these lines at the points $L$ and $M$, respectively. Prove that $DL = DM$, where $D$ is the midpoint of the segment $AB$.

We compose some spiral similarities:

The one of center $P$ which takes $M\to A$ (call it $S_1$), then one of center $D$ which takes $A\to B$ ($S_2$), and the one of center $P$ which takes $B\to L$ ($S_3$). The result has to be a rotation, because the ratios of $S_1,S_3$ cancel out (this is where we need the angle equality in the hypothesis). All we need to do now is to show that the fixed point of this composition is $D$. This is, however, pretty easy: the image of $D$ after $S_1$ is a point $D_1$ s.t. $PDD_1$ and $PMA$ are similar. The reflection of $D_1$ in $D$ is $D_2$ s.t. $PDD_1$ and $PDD_2$ are congruent, so $PDD_2$ is similar to $PLB$, which means that the image of $D_2$ after $S_3$ is $D$, Q.E.D.

Nice solution, Grobber! What I wanted to add is that this prolem is equivalent to Problem 3 of the 2nd Round of the BWM 2003. (Those who can read German will find the [url=http://bundeswettbewerb-mathematik.de/aufgaben/pdf/loesungen/loes_03_2_e.pdf]proposed solutions on the BWM site[/url].) In fact, that BWM problem states:

[i]Let the diagonals AC and BD of a cyclic quadrilateral ABCD intersect at a point P. Call L and M the orthogonal projections of the point P on the sides AB and CD of the quadrilateral ABCD. Then, prove that the perpendicular bisector of the segment LM passes through the midpoints of the sides BC and DA of the quadrilateral ABCD.[/i]

In fact, why is this equivalent to Pascual's problem? Well, look at Pascual's problem; if the lines AP and BP meet the lines BC and CA, respectively, at the points A' and B', then the equality < PAC = < PBC rewrites as < A'AB' = < A'BB'. In other words, the points A, B, A', B' lie on one circle. Hence, we have a cyclic quadrilateral BA'B'A, whose diagonals BB' and A'A intersect at the point P. The points L and M are the projections of the point P on the sides BA' and B'A of this quadrilateral. Hence, applying the BWM 2003 problem above to this cyclic quadrilateral, we see that the perpendicular bisector of the segment LM passes through the midpoints of the sides AB and A'B' of our quadrilateral. But the midpoint of the side AB is the point D, and thus the perpendicular bisector of the segment LM passes through the point D, so that DL = DM. Proof complete.

[b]EDIT:[/b] See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=30926

for this problem.

  Darij

well if you take the midpoints of AP(say X)and of BP(say Y)it's easy to prove that
triangles DYL=MXD and the rest it's obvious.


Let $ n$ be a positive integer. We consider the set $ P$ of all partitions of $ n$ into a sum of positive integers (the order is irrelevant). For every partition $ \alpha$, let $ a_{k}(\alpha)$ be the number of summands in $ \alpha$ that are equal to $ k, k = 1,2,...,n.$ Prove that
$ \sum_{\alpha\in P}\frac{1}{1^{a_{1}(\alpha)}a_{1}(\alpha)!\cdot 2^{a_{2}(\alpha)}a_{2}(\alpha)!...n^{a_{n}(\alpha)}a_{n}(\alpha)!}=1.$

Although technically a combinatorics problem, this is really based as much on concepts you'd see in a first abstract algebra course as those you'd see in a combinatorics course.

For each permutation $ \sigma$ of $ \{1, 2, \dots n\}$, we can define a corresponding partition by first writing $ \sigma$ in [url=http://en.wikipedia.org/wiki/Cycle_notation]cycle notation[/url], then taking the sizes of the cycles.

Fact: Given a partition $ \alpha$, there are $ \frac{n!}{1^{a_{1}(\alpha)}a_{1}(\alpha)! 2^{a_{2}(\alpha)}a_{2}(\alpha)! \dots n^{a_{n}(\alpha)}a_{n}({\alpha})!}$ permutations with cycle structure $ \alpha$.

I'm not including a proof of the Fact here (as this is what I would consider a "standard" result), but it's a worthwhile exercise if you haven't seen it before.  I'd be willing to try and write out a proof if anyone wants to see it.

Since each permutation has exactly one cycle structure, and there are $ n!$ permutations total, we have

$ \sum_{P}\frac{n!}{1^{a_{1}(\alpha)}a_{1}(\alpha)! 2^{a_{2}(\alpha)}a_{2}(\alpha)!,\dots n^{a_{n}(\alpha)}a_{n}({\alpha})!}=n!$.

Dividing by $ n!$ yields the given result.


Hi, as a APCS student, I am wondering what the USACO is? thanks

The USACO is basically a programming contest, where you try to solve problems using a bunch of algorithms and stuff.  So basically, it's kind of like the free-response section of the APCS test, except a lot harder, and little things like not putting the semicolon at the end and stuff count, because you are actually running the program to see it it works.

To try the last contst of the year, go to [url=http://usaco.org]usaco.org[/url], and most things should be explained there.  And make sure you have a proctor for the last test if you live in the US (it can be anyone, even your parents or little brother or someone).  Most of the stuff should be explained at the website for having a proctor and stuff.

It's the [url=http://oldweb.uwp.edu/academic/mathematics/usaco/]USA Computing Olympiad[/url], which is a series of contests to select the US team for the International Olympiad in Informatics.

Their next contest in April 28 -- see the website linked to above for more details.


integration
calculus
trigonometry
calculus computations

In other words, how would you evaluate the following:
$\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[f(t)]^{2}dt$

where f(t) is the general expression of the Fourier series given by

$f(t) = \frac{a_{0}}{2}+\sum_{n=1}^{\infty}[a_{n}\cos(w_{n}t)+b_{n}\sin(w_{n}t)]$

where $w_{n}= \frac{2\pi n}{T}$.

[For the sake of this post, I will assume that $f$ is real-valued. We could cope with complex values if necessary. We'll also assume that everything converges and that integrals and sums can be interchanged - the justifications can follow later.]

Just multiply out $f(t)^{2}=f(t)\cdot f(t)$ into a double sum. Integrate that double sum. Use the orthogonality of the cosines and sines to simplify the double sum down to just the diagonal elements. The answer will turn out to be

$\frac1T\int_{t_{0}}^{t_{0}+T}[f(t)]^{2}\,dt=\frac{a_{0}^{2}}{4}+\frac{1}{2}\sum_{n=1}^{\infty}(a_{n}^{2}+b_{n}^{2})$

This is known as Parseval's identity.

Thank you very much Kent Merryfield and Euclid 3.14.  :)