Consider a convex polygon $A_{1}A_{2}\ldots A_{n}$ and a point $M$ inside it. The lines $A_{i}M$ intersect the perimeter of the polygon second time in the points $B_{i}$. The polygon is called balanced if all sides of the polygon contain exactly one of points $B_{i}$ (strictly inside). Find all balanced polygons.

[Note: The problem originally asked for which $n$ all convex polygons of $n$ sides are balanced. A misunderstanding made this version of the problem appear at the contest]

[quote="prowler"]Instead of answering $n=3$, you can answer that "triangle" is a balanced polygon. In any case the problem is beautiful ![/quote]

What about the solution?

i have one solution (i think)

the solution is very long, but the basic ideas are:

1) to prove that no 2k-convex polygon $A_{1}A_{2}...A_{2}k$ is balanced, see that, if M is in the interior of $A_{1}...A_{k+1}$, there are k+1 segments $MA_{i}$ to intersect  k sides of the polygon. so, two of them will intersect the same side.

2) for n = 3, 5 or 7, the polygons are balanced. for $n\geq9$ i think that no polygons is balanced (i THINK that i have exemples)

(PS: i understand the problem in the correct way?)

[quote="prowler"]The problem is from AMM:[/quote]

Well, I'm not sure if you missed a page or not in your file, because the two pages don't seem continued.

Yes, they do not seem to be continued.
But it is like it was published in AMM.
They just sketched the solution, the rest is on you.


Prove If  a,b,c>0 then
$ \frac{a}{b\plus{}c^3}\plus{}\frac{b}{c\plus{}a^3}\plus{}\frac{c}{a\plus{}b^3}$:geq$ \frac{3\sqrt[2]{3}}{\sqrt[2]{3}\plus{}\sqrt[2]{a^2\plus{}b^2\plus{}c^2}}$


calculus
integration
number theory unsolved
number theory

Q.Find all positive integeral solutions of the equation:--

    $7^x + 1 = 3^y + 5^z$

[quote="VANCHANH123"]I find :(1,1,1) ?[/quote]
IF IT IS THE ONLY ANSWER( I DON'T THINK SO) THEN YOU HAVE TO PROVE IT :mad:


function
algebra
polynomial
advanced fields
advanced fields theorems

It' s possible to find  rigurous definitions  of  following notions:
[color=green] [b] Special functions [/b] [/color] , or
[color=green] [b] Classical  polynomials  [/b] [/color]  ?

http://www.math2.org/math/integrals/specialfuns.htm

http://en.wikipedia.org/wiki/Orthogonal_polynomials

I don't know if that's exactly what you were looking for though..

[quote="mathmanman"]I don't know if that's exactly what you were looking for though..[/quote]
Indeed, on these  sites there are'nt rigurous definitions. However, thanks for interest.

Maybe this link is a bit more interesting : http://www.math.niu.edu/~rusin/known-math/index/33-XX.html .

[quote="mathmanman"]Mayb... more interesting : http://www.math.niu.edu/~rusin/known-math/index/33-XX.html .[/quote]
Many thanks, but please  insert [b]yours own definition(s).[/b]
 After reading /inspecting the sites, I have'nt seen  any convenable (rigurous,"descriptive")  definitions.
Regards,

Perhaps a "good definition",  (after [b]H. Bateman, 1882 - 1946=a reputed  specialist in the field of Special functions[/b]) ,is :
[quote="Harry Bateman"]
[b][color=blue]Classical polynomial (sequence) [/color][/b]=[b] any polynomial (sequence) which has received individual attention in at least one research publication .[/b] [/quote]


parameterization
function
probability
conditional probability
probability and stats

Consider a Bernoulli process with parameter p and define
X = time to first success
Y = 1 if the first two trials have the same outcomes and 0 if they have different outcomes
Z = time to first failure

Find the joint frequency function for (X, Y, Z) and the (joint) conditional distribution of (X,Z) given Y = y for y = 0 and 1.

My attempt:

Here are the possible values:

If x = 1, then y could be 1 or 0 (we don't know precisely which), and $ z \geq 2$

If x = 2, then y = 0 since the second trial is the first success and z = 1.

For $ x \geq 3$, y = 1 since the first two trials are both failures, and z = 1.

I think these are correct, unless I made an oversight. But what do I need to do for the rest of the problem (I tried googling)? Thanks in advance.

Apparently the first question asks you to find the probability that $ X\equal{}x,Y\equal{}y,Z\equal{}z$ in terms of $ x,y,z$. It is just a (somewhat boring) case consideration.

If $ y\equal{}0$, then you either have $ X\equal{}1,Z\equal{}2$, or vice versa. The probability of each is $ p(1\minus{}p)$.

If $ y\equal{}1$, then one possibility is that we have two successes in the beginning. Then automatically $ X\equal{}1$, and $ Z\equal{}z>2$ means that we have $ z\minus{}1$ successes followed by a failure. I leave finding the corresponding probability to you. The other possibility that we start with two failures is just a mirror image of this one.

So, the final answer will be something like $ f(1,0,2)\equal{}f(2,0,1)\equal{}p(1\minus{}p)$, $ f(1,1,t)\equal{}\dots$, $ f(t,1,1)\equal{}\dots$ for $ t>2$ and all other values are $ 0$.

I prefer to leave the conditioning to you (unless you don't understand what the conditional probability is).


linear algebra
matrix
vector
superior algebra
superior algebra unsolved

$ p$ is a prime number and $ n$ is a positive number.Find $ |GL_{n}(\mathbb F_p)|$. :)

[hide="Hint"]
In how many ways can you choose the first row of an invertible matrix? The second row must be chosen to be linearly independent of the first row - in how many ways can this be done? The third row? ...the last? For each row, consider how many linear combinations there are of the previous rows, and choose the new row to be different from all of these.
[/hide]

[hide="Solution"]
To build an invertible matrix over $ \mathbb{F}_p$, we choose the first row $ v_1$ to be any non-zero vector. This can be done in $ p^n \minus{} 1$ ways. $ v_1$ spans a 1-dimensional subspace of size $ p$ (all vectors of the form $ c_1 v_1$ with $ c_1 \in \mathbb{F}_p$), so the next row $ v_2$ must be chosen to be different from all of these. This leaves $ p^n \minus{} p$ options for $ v_2$.

The 2-dimensional subspace spanned by $ \{v_1, v_2\}$ has $ p^2$ elements since each such vector has a unique representation as $ c_1 v_1 \plus{} c_2 v_2$. Therefore $ v_3$ can be (and has to be) one of the $ p^n \minus{} p^2$ vectors that are not in this 2-d subspace.

Continuing on, we see that the number of such matrices is

$ |GL_n(\mathbb{F}_p)| \equal{} (p^n \minus{} 1)(p^n \minus{} p)(p^n \minus{} p^2)\cdots(p^n \minus{} p^{n \minus{} 1}) \equal{}$
$ \equal{} p^{n^2}(1 \minus{} \frac{1}{p})(1 \minus{} \frac{1}{p^2}) \cdots (1 \minus{} \frac{1}{p^n})$
[/hide]

...one additional note - the same approach and the same formula apply for ${ |GL_n(\mathbb{F}_q})|$ for $ q$ a prime power.


A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?

Are we allowed to take $1$ as one of the smallest divisors of the integer? I suppose we can.

yep :lol:, I am sure that 2 has to be one of the smallest. If 2 is not there, that means there cnanot be any other evens and hence the sum of the squares will be an even, but 2 is not a factor, contradiction. So i think that both 1 and 2 need to be there.

Nice argument for 2.  For 3 and 4: [hide]If we want to have 3 as well, we either must have {1, 2, 3, 4}, {1, 2, 3, 5} or {1,2, 3, 6} [because we certainly get 6 once we have 2 and 3], but none of these work.  So no 3.  4 dies by a mod argument: since all squares are 0 or 1 mod 4 and we have at least one of each, we can't get them to add up to 0.[/hide]

[hide]

[quote="JBL"]If we want to have 3 as well, we either must have {1, 2, 3, 4}, {1, 2, 3, 5} or {1,2, 3, 6} [because we certainly get 6 once we have 2 and 3], but none of these work.  So no 3. 0.[/quote]

whats wrong with {1,2,3,7} or {1,2,3,8}, etc?

[/hide]

here's my argument for 3

[hide]

if one of the numbers is $3$, then we know that the fourth number must be even, say, $2n$

then we have:

$1^{2}+2^{2}+3^{2}+(2n)^{2}= 4n^{2}+14 = 2(2n^{2}+7)$

this must now be divisible by $2n$, so we have:

$\frac{2(2n^{2}+7)}{2n}= 2n+\frac{7}{n}$

so $n=1$ or $7$, but when $n=1$ we have $2$ which we already have, so $n=7$

now we have $1^{2}+2^{2}+3^{2}+14^{2}= 210$ but this doesn't work because $14$ can't be a smallest divisor since $7$ is even smaller than that

[/hide]

[quote="gotztahbeazn"][hide]whats wrong with {1,2,3,7} or {1,2,3,8}, etc?[/hide][/quote]

If you have 2 and 3 already, you must have a factor 6.  We're taking the 4 smallest factors.  Thus, if we have 1, 2 and 3, the next smallest must be no larger than 6.

err i think i got it

[hide]

we know that $1$ and $2$ are two of the smallest positive divisors, and we know that one of the numbers has to be odd and the other has to be even, so we experiment with $5$

now we have for some natural number $n$:

$1^{2}+2^{2}+5^{2}+(2n)^{2}= 4n^{2}+30 = 2(2n^{2}+15)$

dividing this by $n$ gives:

$\frac{2(2n^{2}+15)}{2n}= 2n+\frac{15}{n}$

hence $n=1,3,5,$ or $15$ but if $n=1$ we have $2$ which is already accounted for, if $n=3$ then we have $6$ but that would make $3$ an even smaller divisor, and if we have $15$ then again $3$ would be an even smaller divisor, so $n=5$ in this case which gives us $(1,2,5,10)$, and checking:

$1^{2}+2^{2}+5^{2}+10^{2}= 130$ which works!

now $130 = 2 \cdot 5 \cdot 13$ and so $\fbox{13}$ is our answer i think

unfortunately for me, i just got lucky that the next odd after $3$ worked... so does anyone know how to do this without using guess and check almost?

[/hide]

Well , if you want to peek:

[hide]We know there is $1$, and $2$  but not $4$
so the choices are $(1,2,p,2p)$ or $(1,2,p,q)$  (p,q odd primes)

second  could not work, as some of squares is odd (and our number is even)
So we  get:
$n=1^{2}+2^{2}+p^{2}+4p^{2}=5(1+p^{2})$ obviously this is 0 mod 5  so p has to be 5. :) [/hide]


trigonometry
geometry
incenter
geometry unsolved

For any triangle $ ABC$ prove that :

\[ \sin^2 A\plus{}\sin^2 B\plus{}\sin^2 C\le 2\plus{}16\sin^2 \frac{A}{2}\sin^2 \frac{B}{2}\sin^2 \frac{C}{2}\]

[quote="Pain rinnegan"]For any triangle $ ABC$ prove that :
\[ \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C\le 2 \plus{} 16\sin^2 \frac {A}{2}\sin^2 \frac {B}{2}\sin^2 \frac {C}{2}\]
[/quote]

1. $ \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C \equal{} \frac {s^2 \minus{} r^2 \minus{} 4Rr}{2R^2}$

2. $ \sin^2 \frac {A}{2}\sin^2 \frac {B}{2}\sin^2 \frac {C}{2} \equal{} \frac {r^2}{16R^2}$

$ \Leftrightarrow s^2 \le 4R^2 \plus{} 4Rr\plus{}3r^2$ -which is true : $ IH^2\equal{}4R^2\plus{}4Rr\plus{}3r^2\minus{}s^2$ ( $ I$ -incenter , $ H$ -orthocenter ).


In what case do you use the word "Reason" compared with the word "reason"?

Any Comments would appreciated,

Thanks in advance.

Huh? That's the same word. As usual, just capitalize it if it starts a sentence.

I thought what you said. My acquaintance says, the difference usage of "Reason" and "reason" is that the former is used when we would say something objectively and the latter, subjectively. How do you think about his opinion?

I think he is mistaken about the usual English pattern of special treatment of words. For American editors, the definitive book on this subject is the Chicago Manual of Style.

http://www.amazon.com/Chicago-Manual-Style-University-Press/dp/0226104036/


Divisibility
number theory
binary representation
minimization
IMO Shortlist

Let $ n$ be a composite natural number and $ p$ a proper divisor of $ n.$ Find the binary representation of the smallest natural number $ N$ such that

\[ \frac{(1 \plus{} 2^p \plus{} 2^{n\minus{}p})N \minus{} 1}{2^n}\]

is an integer.

I can't understand why it is smallest? Who can explain?

It's the smallest because there exists a unique $N$ which is less than $2^n$, and both of [b] tjhance[/b]'s examples above are $<2^n.$

My motivation for the examples given above is the fact that

$\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots,$

for $0 < x < 1.$ If we plug in $2^p + 2^{n-p}$ for $x$ in the equation above, we can obtain the examples given above.


I just went to state MC. Didn't make it to nats. Very unhappy. Anyway, I want to know What kind of high school competitions there are, and what I should do to prepare for them.  Thanks!

Well the main ones are the AMC and the ARML.

Are you new to competitions? I only learned about them in 8th grade, and although I made it to nats, I was very mad.  :mad:


[quote="Halbalba"]Is there an algebraic way to solve this equation?

cos(6x) = cos (x)

When I apply some addition theorema and substitute z = cos(x) I get

32z^6 - 48z^4 + 18z^2 - z - 1 = 0

One obvious solution is x = 0 i.e. z = 1, so the equaition shrinks to

32z^5 + 32z^4 - 16z^3 - 16z^2 + 2z + 1 = 0

and then Im lost.[/quote]

You can write $ 32z^5 \plus{} 32z^4 \minus{} 16z^3 \minus{} 16z^2 \plus{} 2z \plus{} 1\equal{}(4z^2\plus{}2z\minus{}1)(8z^3\plus{}4z^2\minus{}4z\minus{}1)$

which gives you the two roots of $ 4z^2\plus{}2z\minus{}1$ : $ \frac{\sqrt 5\minus{}1}4\equal{}\cos(\frac{2\pi}5)$ and $ \frac{\minus{}\sqrt 5\minus{}1}4\equal{}\cos(\frac{4\pi}5)$

And you just have to use Cardan for solving $ 8z^3\plus{}4z^2\minus{}4z\minus{}1\equal{}0$

But, to be honest, the direct resolution is much more simple than the algebraic one and gives the results $ \frac{2k\pi}5$ and $ \frac{2k\pi}7$

Nice.

@Patrick: what do you mean by "direct" resolution?

@Rust: how do you get there?

I think rush is illustrating what pco means.

Also, rust's identity is quite well known, here is the idea:
Let $ X\equal{}e^{ia},Y\equal{}e^{ib}$
\[ 2\sin a \sin b\equal{} \frac{(X\minus{}X^{\minus{}1})(Y\minus{}Y^{\minus{}1})}{2}
\\\equal{} \frac{ XY \plus{} (XY)^{\minus{}1} \minus{} [XY^{\minus{}1} \plus{} YX^{\minus{}1}]}{2}
\\\equal{} \cos (a\plus{}b)\minus{}\cos(a\minus{}b)\]

[quote="Halbalba"]Nice.

@Patrick: what do you mean by "direct" resolution?

[/quote]

I mean $ \cos a\equal{}\cos b$ $ \iff$ $ a\equal{}\pm b\plus{}2k\pi$

Here : $ 6x\equal{}\pm x\plus{}2k\pi$ and so $ x\equal{}\frac{2k\pi}5$ and $ x\equal{}\frac{2k\pi}7$

Thanks, now I get it  :blush:

Just didnt think of all the symmetries.


[hide]Just rewrite as a qaudratic equation in $x$.
Then you get the discriminant is equal to $-3y^{2}+8y-8$, which is always negative. So the quadratic equation in $x$ has no zero's, and it's main coëfficient is positive so the inequality is true[/hide]


Ring Theory
superior algebra
superior algebra unsolved

Let $ P$ and $ Q$ be prime ideals of a ring $ R$.  Is it true that $ P\cap Q$ is a prime ideal of $ R$?

I have a feeling that this is not true, but I can't think of an explicit counterexample.  If we take $ ab\in P\cap Q$ then we must have $ a$ or $ b$ in $ P$ and $ a$ or $ b$ in $ Q$.  An example where $ a\not\in P$ and $ b\not\in Q$ would prove that this statement is false.  Does anyone know of such ideals?  Thanks.

$ (2)\cap(3)\equal{}(6)$ is not prime in $ \mathbb Z$. owk

What if P were prime ideal in ring of integers of number field R. Is possible to prove P intersect Z is prime ideal in Z?

Yes. In more general manner, if $ R \subset S$ are rings (with the same unity) and $ P$ is a prime ideal of $ S$, then $ R \cap S$ is a prime ideal of $ R$. It's in general wrong for maximal ideals.


Trivial : $a=b=c=x \Longrightarrow P= 3x\sqrt{1+x^2} \Longrightarrow 0 < P < +\infty$
 :cool: Manifestly, there are missing conditions !! :cool:

For the maximum (all the sums are cyclic):
From BCS we get
$\sum\dfrac{a^2}{1+a^2}\sum(1+b^2)(1+c^2)\geq P^2$ (1)
$\sum\dfrac{a^2}{1+a^2}\leq 3- \sum\dfrac{1}{1+a^2}$ Using BCS or Jensen we easily obtain $\sum\dfrac{1}{1+a^2}\geq \dfrac{9}{4}$ Thus
$\sum\dfrac{a^2}{1+a^2}\leq \dfrac{3}{4}$ (2)
Moreover
$\sum(1+b^2)(1+c^2)=5+\sum b^2c^2\leq5+\dfrac{(\sum a^2)^2}{3}=\dfrac{16}{3}$ (3)
From (1), (2), (3) we have $P\leq2$. Equality for $a=b=c=\dfrac{\sqrt{3}}{3}$

The maximum, if i am not mistaken, was posed as the third problem in the Mediterranean mathematical olympiad in 2002.

[/quote]

The maximum, if i am not mistaken, was posed as the third problem in the Mediterranean mathematical olympiad in 2002.[/quote]

You mean the minimum, right? Because you proved the maximum. But I have not the problems of the mediterrenean 2002 so I can't know


quadratics function trigonometry conics algebra

well, one obvious, brute-forcing solution is to use substitution. I really don't see any other way

[quote]
If x = 1, then
a(1)^{2}+b(1)+c =-8
a+b+c =-8 ... #1

If x = 2, then
a(2)^{2}+b(2)+c = 4
4a+2b+c = 4 ... #2

If x = 3, then
a(3)^{2}+b(3)+c = 24
9a+3b+c = 24 ... #3

Subtract #1 from #2 and you get
3a+b = 12 ... #4

Again subtract #1 from #3 and you get
8a+2b = 32
4a+b = 16 ... #5

Now you can subtract #4 from #5
(4a+b)-(3a+b) = 16-12
a = 4
[/quote]

That's what I started doing, but I didn't take it that far.  :huh:

[quote="happyme"]The table shown represents the relationship between $x$ and $y$ in a quadratic equation of the form $ax^{2}+bx+c$, where $a, b$ and $c$ are integers. What is the value of $a$?

TABLE:

$x$:    1   2   3    4   5     6        7
$y$:  -8   4   24  49  80    117    160

respectively[/quote]

The data in this table does not represent a quadratic relationship.  [I suspect a minor typographical error.]  A solution method that attempted to use the bad data would have a very difficult time.

[quote="ProtestanT"][quote="happyme"]The table shown represents the relationship between $x$ and $y$ in a quadratic equation of the form $ax^{2}+bx+c$, where $a, b$ and $c$ are integers. What is the value of $a$?

TABLE:

$x$:    1   2   3    4   5     6        7
$y$:  -8   4   24  49  80    117    160

respectively[/quote]
[hide]
use the standard form for a quadratic equation $y=n(x-h)^{2}+k$
as you can see, $a=n$
so $y=a(x-h)^{2}+k$
$-8=a(1-h)^{2}+k$
$a(h-1)^{2}=-8-k$
$4=a(2-h)^{2}+k$
$a(h-2)^{2}=4-k$
$24=a(3-h)^{2}+k$
$a(h-3)^{2}=24-k$
Subtract $a(h-1)^{2}=-8-k$ from $a(h-2)^{2}=4-k$ to get
$a(-2h+3)=12$
$-2ah=12-3a$
$2ah=3a-12$
$h=\frac{3a-12}{2a}$
Subtract $a(h-2)^{2}=4-k$ from $a(h-3)^{2}=24-k$ to get
$a(-2h+5)=20$
$a(\frac{-3a+12}{a}+5)=20$
$-3a+12+5a=20$
$2a=8$
$a=4$
[/hide][/quote]I thought it was already in standard form.  Isn't your form called vertex form?

No. My form is Standard form. The form $y=ax^{2}+bx+c$ is General form. (refered to as the Quadratic Function)
From Algebra and Trignometry Fifth Edition:
[quote="Pg 257"]
Standard Form of a Quadratic Function
The quadratic function
$f(x)=a(x-h)^{2}+k, a\ne0$
is in [b]standard form.[/b] The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h,k)$. If $a>0$, the parabola opens upward, and if $a<0$, the parabola opens downward.
[/quote]

[quote="Pg 254"]Definition of Quadratic Function
Let $a$, $b$, and $c$ be real numbers with $a\ne0$. The function
$f(x)=ax^{2}+bx+c$
is called a [b]quadratic function.[/b][/quote]

[quote="ProtestanT"]No. My form is Standard form. The form $y=ax^{2}+bx+c$ is General form.[/quote]
Quoting from my textbook:
"A quadratic function is a function that can be written in the [b]standard form[/b] $y=ax^{2}+bx+c$..."
"Another useful form of a quadratic function is the [b]vertex form[/b], $y=a(x-h)^{2}+k$."

what textbook do you have?
notice how mine said the other one is the Standard Form.

Just like standard forms of other Conics all identify their Centers (like $\frac{x-h}{a^{2}}+\frac{y-k}{b^{2}}=1$; $\frac{x-h}{a^{2}}-\frac{y-k}{b^{2}}=1$; etc),
but the General form of Conics is $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

As above stated, the #s are incorrect. If they had been, and considering that this is a target, you could always just plot the points and find the quadratic reegresion, b/c graphing calcs are allowed.

Yeah, that should be a 5 in the $y$ row instead of a 4.

And if one changes the data set like that:

[hide]
$a = 3$
$b = 4$
$c =-15$
[/hide]

[quote="Quevvy"][quote="ProtestanT"][quote="Quevvy"]I was checking the answer with my calculator, and $a=4$ doesn't work for all the numbers. Then you would get $b=0$ and $c=-12$ (from undefined117's work) and that doesn't fit all the points.[/quote]
plot the points and do a QuadReg. See if it fits everything.[/quote]
If you do a QuadReg, you get $a=3, b=4.071428571, c=-15.42857143$ and that fits all of the points.[/quote]
then a is $3$. the point is that there is a quadratic for which all the points fit.


Given $a+b\geq1$
 Prove that ${a^{4}+b^{4}}\geq\frac{1}{8}$

we can suppose that $a\leq b$
so by using the tchebyshev inequality  we have
 $a^{4}+b^{4}\geq\frac{1}{2}(a^{2}+b^{2})^{2}$
                                 $\geq\frac{1}{2}(\frac{a+b}{2})^{2}$
                                 $\geq\frac{1}{8}$

[quote="Nekruzjon_eko"][quote="robpal"]We can take $a+b=1$ now, using power-mean we have
\[\sqrt[4]{\frac{a^{4}+b^{4}}{2}}\geq \frac{a+b}{2}\]

\[a^{4}+b^{4}\geq \frac{1}{8}\]
[/quote]
If $a+b>1$ then?[/quote]
$\sqrt[4]{\frac{a^{4}+b^{4}}{2}}\geqslant \frac{a+b}{2}\geqslant \frac12$, so in fact $a^{4}+b^{4}\geqslant \frac14$.
Solution for the case $a+b=1$ was sufficent to the problem, because if the inequality is proved for $a+b=1$, we can simply rescale the variables to any given sum (of course satisfying $a+b\geqslant 1$) enlarging the left hand side.


geometry
geometric transformation
reflection
geometry proposed

(V.Protasov, 8) For a given pair of circles, construct two concentric circles such that both are tangent to the given two. What is the number of solutions, depending on location of the circles?

/bump

Somebody help please. crastybow and I have some ideas but a lot are intuitive such as the radius of the smaller circle is on the perpindicular bisector of the line segment connecting the radii of the 2 given circles and a bunch more but before I go into too much I was kind of wondering if anybody else had a solution

If given circles have equal radii, then there are infinite number of solutions. Otherwise, the number of solutions is equal to the number of intersection points of the given circles. In the latter case, the common center of sought-for circles must be the reflection of an intersection point of the given circles across the midpoint of the segment joining their centers.


Does anybody know the other way to solve problem 17 at "Are You Ready" in AIME/Olympiad counting & probability? Thanks.

I believe that this is the problem:
Find two proofs that for every positive integer n, the following equality holds:
[tex]\displaystyle{{n\choose0}-{n\choose1}+{n\choose2}-{n\choose3}+...+(-1)^n{n\choose{n}}=0}[/tex]

[quote="polymorphic"]one way: let x = -1 in the binomial expansion of (1+x)^n.
another way: same number of even subsets as odd subsets[/quote]

First seems like the most intuitive, but the second is only obvious to me when n is odd... is there an obvious reason it's true when n is even?

what do you mean by same number of even subsets as odd subsets?

I think even and odd subsets were intended to mean subsets with an even number of elements and those with an odd number of elements, respectively.

[quote="zabelman"][quote]First seems like the most intuitive, but the second is only obvious to me when n is odd... is there an obvious reason it's true when n is even?[/quote]

If the numbers are a_1, ..., a_n, then the correspondence is with a subset of the first n-1 numbers paired with that same subset with a_n thrown in. One is even, one is odd, and it's 1-1.[/quote]

Oh yes of course, the shiny coin.  thanks


Suppose that in any given year, the population of a certain endangered species is reduced by 25%. If the population is now 7500, in how many years will the population be 4000?

Well, I know that the formula for this is P(t) = Ce^(kt).

I just can't figure out the value of k.

i did 7500 x .25 = 1875... 7500 - 1875 = 5625...
so i set up 7500x = 5625 and got .75.
so, essentially.. the population is growing at .75 per year.
so i used .75 as k and:

4000 = 7500e^(.75t)
4000/7500 = e^(.75t)
8/15 = e^(.75t)
ln(8/15) = .75t lne
ln(8/15) = .75t
t = [ln(8/15)] / .75
t = -.8381448792 years

This can't be correct because time isn't negative, and if you do it logically, it would be around 2. something years:

7500 x .75 = 5625 ----> 1 year
5625 x .75 = 4218.75 ----> 2 yrs
4218.75 x .75 = 3164.0625 ----> 3 yrs

So it's somewhere between 2 and 3 years... closer to 2 years.

and if i set up 4218.75x = 4000
x = .9481481481

so i know after 2 years.. i can multiply by .9481481481 and i get 4000..
but i don't know how to convert that into years.. since that is a percentage.

BLEH.. brain is about to explode.. don't know what to do.. lol

i don't need any more help with this one.. thanks anyway


Please do tell the procedure to delete my AoPS account. Thanks in advance.

Try to search:

http://www.mathlinks.ro/viewtopic.php?search_id=1438112909&t=161324

You can't delete it.


calculus
integration
analytic geometry
pigeonhole principle
Putnam

Let there be given nine lattice points (points with integral co-ordinates) in three dimensional Euclidean space. Show that there is a lattice point on the interior of one of the line segments joining two of these points.)

There are $ 2^{3}=8$ different possibilities for the parities of the coordinates, so by Pigeonhole at least two of the nine points must have the same parity in all three coordinate values. If we have two points $ (x_{1},y_{1},z_{1})$ and $ (x_{2},y_{2},z_{2})$ where corresponding values are of the same parity, then their midpoint will have integral coordinates, so we are done.

[quote="samath"]...What is the interior of a line segment?[/quote]

The region inside the boundaries.  Just like you consider the interior of a 2-dimensional connected region to be the set of points "inside the boundary" , you define the interior of a line segment to be the part between the endpoints.  Speaking of which, I wonder how the "interior" of a region is defined?  Probably the set of points not on the boundary such that no matter which direction you go in that dimension, you will reach a non-boundary point before you reach a boundary point.

Ah.. a Putnam problem from  baby bloomers generation ...(1970's) :)

[quote="mdk"]wow, I didn't even know that was a Putnam problem[/quote]
It was A1.


A standard deck of 52 cards is shuffled and the cards are dealt face up one at a time until an ace appears. Show that the probability of getting the first ace on or before the ninth card is greater than 50%.

[quote="jmerry"]That's not a valid argument, since it doesn't include an accurate probability.

The probability of getting no aces in the first nine cards is $\frac{48\cdot 47\cdot 46\cdot 45\cdot 44\cdot 43\cdot 42\cdot 41\cdot 40}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44}=\frac{43\cdot 42\cdot 41\cdot 40}{52\cdot 51\cdot 50\cdot 49}=\frac{43\cdot 41\cdot 2}{17\cdot 13\cdot 7\cdot 5}$
$=\frac{82\cdot 43}{85\cdot 91}<\frac{43}{91}<\frac12$

The probability of getting the first ace somewhere in those nine cards is 1 minus the probability of getting no aces.[/quote]

What you did is basically what I did only I just said probability of getting an ace < some fraction < 1/2.

[quote="neelnal"]4/52+4/51+4/50+...+4/45+4/44>9(4/52)>1/2[/quote]
That's not the correct probability. Using your method, the probability of drawing an ace in the first 13 cards is greater than 1, which is definitely false.

[quote="neelnal"]4/52+4/51+4/50+...+4/45+4/44[/quote]

A common probability mistake.

It is true that the probability of getting an ace as your first card is $\frac{4}{52}$.  However, the probability of getting an ace within the first two cards is [b]not[/b] the sum of $\frac{4}{52}$ and $\frac{4}{51}$.  In fact, it is

$1-(1-\frac{4}{51})(1-\frac{4}{52})$

Which is the complement of the probability that you will not draw any aces in the first two cards.  The reason your simple addition is not valid is that it does not account correctly for the case where you draw two aces.


linear algebra
matrix
geometry
3D geometry
linear algebra unsolved

1/ let $ A\in GL_{n}(R)$ such that $ A.A^{t}.A\equal{}A^{t}.A.A^{t}$

prove that $ A$ is symetric matrix

$ A^{t}$ is the transpose of $ A$

2/ let $ A\in M_{n}(R)$ such that $ A.A^{t}.A\equal{}A^{t}.A.A^{t}$

Does  $ A$  stay symetric matrix ?

An idea for beginning.
1) We use the polar decomposition $ A\equal{}SO$ where $ S$ is SPD and $ O$ is orthogonal. The relation can be rewritten $ S^3O\equal{}O^TS^3$. We may assume that $ S\equal{}diag(\lambda_1,\cdots,\lambda_n)$ with $ \lambda_i>0$. Let $ O\equal{}[u_{ij}]$; then for all $ ij$: $ \lambda_i^3u_{ij}\equal{}u_{ji}\lambda_j^3$. It remains to prove that $ O$ is a diagonal matrix.
After I don't know.

1) $ (AA^T)^3 \equal{} (AA^TA)(A^TAA^T) \equal{} (AA^TA)^2 \equal{} (A^TAA^T)^2 \equal{} (A^TA)^3,$
but $ AA^T, A^TA \in \mathcal{S}_n(\mathbb{R})$ so they're diagonalizable and the above implies (I still need to check this) $ AA^T \equal{} A^TA.$

Finally, $ AA^TA^T \equal{} A^TAA^T \equal{} AA^TA$ and $ (A \in \mathbf{GL}_n(\mathbb{R}) \implies A^T \in \mathbf{GL}_n(\mathbb{R}))$ yield $ A^T \equal{} A,$ we're done.

Welldone mathmanman.
The last piece is easy: a SPD matrix $ B$ admits an unique square root which is SPD. By the same method, we can prove that $ B$ admits an unique cube root which is SPD.

Of course.  :blush: Thank you [b]loup blanc[/b].

As for 2), I guess it follows from the fact that $ \mathbf{GL}_n(\mathbb{R})$ is dense in $ \mathcal{M}_n(\mathbb{R}).$

Now, the fact that $ \mathbb{R}$ was the base field was essential for our proof, what if the field is arbitrary?

[quote="mathmanman"]
As for 2), I guess it follows from the fact that $ \mathbf{GL}_n(\mathbb{R})$ is dense in $ \mathcal{M}_n(\mathbb{R}).$

Now, the fact that $ \mathbb{R}$ was the base field was essential for our proof, what if the field is arbitrary?[/quote]
You wanna use a continuity argument, right? Suppose $ A$ satisfies the given condition. Define $ \tilde A: \equal{}A\plus{}\varepsilon I$, for some small $ \varepsilon$. Does $ \tilde A$ necessarily satisfy the problem's condition?(Maybe I'm missing sth. Its 1 am you know!)

Four remarks:
a) The result remains true in $ \mathbb{C}$ if we replace $ A^T$ by $ A^*$.
b) If we keep the transpose then the result is false in $ \mathbb{C}$: take $ A\equal{}\begin{pmatrix}1&j\\1&\minus{}j\end{pmatrix}$ where $ j\equal{}exp^{2i\pi/3}$.
c) 2) is not solved (cf. The Captain's post).
d) I can conclude with the polar decomposition (cf. my last post) but the proof is more complicate than that of mathmanman:
we return to the matrix $ S^3O\equal{}O^TS^3$ that we note $ B$. $ B$ is a symmetric matrix which admits as polar decomposition $ B\equal{}S^3O$; moreover $ TU$ is the polar decomposition of a symmetric matrix iff $ TU\equal{}UT$ and $ U\equal{}U^T$ (easy). Therefore $ S^3O\equal{}OS^3$ and $ O\equal{}O^T$.
$ A$ is symmetric iff $ SO\equal{}OS$ and $ O\equal{}O^T$; we can conclude because $ S^3O\equal{}OS^3$ and $ S$ SPD imply that $ SO\equal{}OS$ (easy).

I think that my method works for any real matrix $ A$.
Let $ A\equal{}SO$ a $ A$-polar decomposition. We can assume that $ S\equal{}diag(0_p,T_q)$ where $ T_q$ is SPD. The condition is $ S^3O$ is a symmetric matrix.Let $ O\equal{}\begin{pmatrix}B&C\\D&E\end{pmatrix}$.  $ S^3O\equal{}\begin{pmatrix}0&0\\T^3D&T^3E\end{pmatrix}$ then $ T^3D\equal{}0,T^3E\equal{}E^TT^3$;therefore $ D\equal{}0$ and $ E$ is an orthogonal matrix. Then $ T^3E$ is the polar decomposition of a symmetric invertible matrix and $ E\equal{}E^T$, $ T^3E\equal{}ET^3$. Moreover $ TE\equal{}ET$ because the $ T^3$-commutant is the same that the $ T$-commutant. $ A\equal{}\begin{pmatrix}0&0\\0&TE\end{pmatrix}$; $ TE$ is symmetric and $ A$ also.      QED

 $ T^E\equal{}E^TT$ because the $ T^3$-commutant is the same that the $ T$-commutant.


superior algebra
superior algebra solved

Let us define an ordered field as a field $F$ together with a relation $<$ such that:
(i) If $x$ and $y \in F$, then exactly one of $x < y$, $y < x$ and $x=y$ is true.
(ii) Sums and products of positive (i.e., $> 0$) elements are positive.

Show that in an ordered field squares of non-zero elements are always $> 0$.

Seems obvious since $\forall u \in F-\{0\}, u>0$  or $-u>0$.

I can't show it :( I can show that we cannot have both $u > 0$ and $-u > 0$, but I don't know how to show that we can't have both $u < 0$ and $-u < 0$.

You're right, I wrote without thinking. This is not immediate. I'll give it a look tomorrow ...

Hello,

If we take $K=F_{2}$ and define the relation $<$ by : $0<1$, then $(K,<)$ is an ordered field, but $1=1^{2}$ is not positive.

Olivier.

We're missing an axiom. As stated, any order in which $0$ is the largest element works, and these examples don't deserve to be called ordered fields.
The standard is to start with the set of positive elements and derive an order from that; the trichotomy axiom would be that either $x$ is zero, $x$ is positive, or $-x$ is positive.

Ok, thanks. This makes me pretty angry. I wasted more than one hour of my time trying to prove this false proposition (it is an excercise on a course). Very annoying time-waster. I guess I should've realized that the statement is false, though.

I realized it as soon as I started to think about it this morning: under the shower lol.


algebra
polynomial
function
probability
search
expected value
functional equation

The difficulty level ranges a lot i guess. I came up with most of the problems this afternoon, and I'm surely inspired by other problems i've seen before. :)

1) Let $ p(x)$ be a polynomial with positive real coefficients and $ deg(p) \ge 2$. Let $ x_1, ... , x_n \in \mathbb{R}_ \plus{}$ such that $ p(x_1) \plus{} ... \plus{} p(x_n) \equal{} k \in \mathbb{R}$. Find the minimum of $ x_1 \plus{} ... \plus{} x_n$.
[b]EDIT:[/b] You should find the maximum of $ x_1 \plus{} ... \plus{} x_n$. :)

2) For each $ k \in \mathbb{R}_ \plus{}$ find the minimum of $ \sqrt {y^2 \plus{} k^2 \minus{} \sqrt {3}yk} \plus{} \sqrt {y^2 \plus{} x^2 \minus{} \sqrt {3}yx} \plus{} \sqrt {k^2 \plus{} x^2 \minus{} \sqrt {3}kx}$ when $ x,y \in \mathbb{R}_ \plus{}$.

3) Let $ a,b,c,x,y,z \in \mathbb{R}_ \plus{}$ such that:
$ \begin{cases} abx \equal{} c(y \plus{} z) \\
bcy \equal{} a(z \plus{} x) \\
caz \equal{} b(x \plus{} y) \\
\end{cases}$
Prove that
i) $ a^s \plus{} b^s \plus{} c^s$ has no limit when $ s \in \mathbb{R}_ \plus{}$
ii) $ a^s \plus{} b^s \plus{} c^s \ge a^r \plus{} b^r \plus{} c^r$ when $ s > r$ and $ s,r \in \mathbb{R}$
iii) When we choose $ a,b,c$ such that $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$ has reached minimum. Then $ abc$ has reached minimum as well.

4) For each $ k,n \in \mathbb{N}$ such that $ k \equiv \minus{} 1 \bmod n$ find infinitly many solutions to: $ a^k \plus{} b^k \equal{} c^n$ where $ a,b,c \in \mathbb{N}$

5) Show in as many ways as possible: $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \le \frac {8}{27}$ when $ a \plus{} b \plus{} c \equal{} 1$ and $ a,b,c \in \mathbb{R}_ \plus{}$

6) Given $ x_1,...x_n,y_1,...,y_n,z_1,...,z_n \in \mathbb{R}_ \plus{}$ such that:
$ \begin{cases} x_1 \plus{} ... \plus{} x_n \equal{} a \\
{y_1}^2 \plus{} ... \plus{} {y_n}^2 \equal{} b \\
{z_1}^3 \plus{} ... \plus{} {z_n}^3 \equal{} c \\
\end{cases}$
Then find the maximum of $ (x_1y_1z_1)^{\frac {6}{11}} \plus{} ... \plus{} (x_ny_nz_n)^{\frac {6}{11}}$

7) Find the average number of times you have to toss a coin to get head.

8) Find all functions $ f: \mathbb{R} \backslash \{0\} \to \mathbb{R} \backslash \{0\}$ such that:
$ f(x \plus{} y) \equal{} xyf(x^3 \plus{} f(y)), \quad \forall x,y \in \mathbb{R} \backslash \{0\}$

Good luck :D

In the future, please post unrelated questions in separate threads.

[hide="1, remark"] I am not sure if this problem is solvable without the use of [url=http://en.wikipedia.org/wiki/Lagrange_multiplier]Lagrange multipliers[/url]. [/hide]
[hide="2, hint"] Law of Cosines. [/hide]
[hide="4, generalization"] Generalizes to $ \gcd(k, n) \equal{} 1$.  Find integers $ p, q$ such that $ \minus{} pk \plus{} qn \equal{} 1$.  Pick integers $ x, y$ and let $ a \equal{} x(x^k \plus{} y^k)^p, b \equal{} y(x^k \plus{} y^k)^p$.  Then

$ a^k \plus{} b^k \equal{} (x^k \plus{} y^k)^{pk \plus{} 1} \equal{} (x^k \plus{} y^k)^{qn}$

so we can let $ c \equal{} (x^k \plus{} y^k)^q$. [/hide]
[hide="7, remark"] Generalize to an $ n$-sided die.  What about an arbitrary probability distribution $ (p_n)$ with $ \sum_{n \equal{} 1}^{\infty} p_n \equal{} 1$? [/hide]

[quote="t0rajir0u"]In the future, please post unrelated questions in separate threads.

[hide="1"] This problem does not appear to be solvable without the use of [url=http://en.wikipedia.org/wiki/Lagrange_multiplier]Lagrange multipliers[/url] (and you have only to look at the equality case to see why).  If $ P(x)$ is a monomial, this is an application of Power Mean, but for a generic polynomial things get much trickier. [/hide]
[hide="2, hint"] Law of Cosines. [/hide]
[hide="4, generalization"] Generalizes to $ \gcd(k, n) \equal{} 1$.  Find integers $ p, q$ such that $ \minus{} pk \plus{} qn \equal{} 1$.  Pick integers $ x, y$ and let $ a \equal{} x(x^k \plus{} y^k)^p, b \equal{} y(x^k \plus{} y^k)^p$.  Then

$ a^k \plus{} b^k \equal{} (x^k \plus{} y^k)^{pk \plus{} 1} \equal{} (x^k \plus{} y^k)^{qn}$

so we can let $ c \equal{} (x^k \plus{} y^k)^q$. [/hide]
[hide="7, remark"] Generalize to an $ n$-sided die.  What about an arbitrary probability distribution $ (p_n)$ with $ \sum_{n \equal{} 1}^{\infty} p_n \equal{} 1$? [/hide][/quote]
Ok. I'll do that from now on :)

What do you mean with "What about an arbitrary probability distribution $ (p_n)$ with $ \sum_{n \equal{} 1}^{\infty} p_n \equal{} 1$?". What is "arbitrary probability distribution" ?

Sorry about number 1. It should be maximum. I'll edit it :)

hello, nice problems, thank you very much.
Sonnhard.

The answer to #7 is 2, and a substantial generalization (larger alphabets, waiting for longer patterns) was recently discussed [url=http://www.mathlinks.ro/viewtopic.php?search_id=1738585740&t=228842]here[/url].

[quote="t0rajir0u"]An arbitrary probability distribution can be thought of as a die with infinitely many sides where the probability of getting side $ 1$ is $ p_1$, the probability of getting side $ 2$ is $ p_2$, etc.  Such a distribution has to satisfy two rules:

- The sum of the probabilities is $ 1$, and
- Each probability is greater than or equal to zero.

In particular, it is impossible for each probability to be equal.  An example of such a distribution is $ p_n \equal{} \frac {1}{n(n \plus{} 1)}$ or $ p_n \equal{} \frac {1}{2^n}$.  Now, the question:  what is the expected number of times you need to roll this infinite die before you get side $ i$ for some $ i$?[/quote]
After solving the case $ p_n \equal{} \frac {1}{2^n}$ it seemed like the expected number of times is $ \frac {1}{p_n}$. I would appreciate if you looked through my proof, and point any mistakes out to me :)
There is $ P_i$ chance to it on the first roll. And then there is $ (1 \minus{} P_i)P_i$ for the second roll, and $ (1 \minus{} P_i)^2P_i$ chance for the third, etc.
So we set $ a_k \equal{} k \cdot P_i \cdot (1 \minus{} P_i)^{k \minus{} 1}$ And we want to compute $ a_1 \plus{} ... \plus{} a_k$ for $ k \to \infty$.
We note that $ a_k \minus{} a_{k \plus{} 1} \equal{} kP_i(1 \minus{} P_i)^{k \minus{} 1}(k \minus{} (k \plus{} 1)(1 \minus{} P_i))$ $ \equal{} kP_i(1 \minus{} P_i)^{k \minus{} 1}(kP_i \minus{} (1 \minus{} P_i)) \equal{} P_ia_k \minus{} P_i(1 \minus{} P_i)^k$.
Thus $ a_1 \minus{} a_{k \plus{} 1} \equal{} (a_1 \minus{} a_2) \plus{} ... \plus{} (a_k \minus{} a_{k \plus{} 1})$ $ \equal{} P_i(a_1 \plus{} ... \plus{} a_k) \minus{} P_i( (1 \minus{} P_i) \plus{} .... \plus{} (1 \minus{} P_i)^k )$ $ \iff$

$ a_1 \plus{} ... \plus{} a_k \equal{} \frac {1}{P_i}(a_1 \minus{} a_{k \plus{} 1}) \plus{} (1 \minus{} P_i) \plus{} .... \plus{} (1 \minus{} P_i)^k$ $ \equal{} \frac {1}{P_i}(a_1 \minus{} a_{k \plus{} 1}) \plus{} \frac {1 \minus{} (1 \minus{} P_i)^{k\plus{}1}}{P_i} \minus{} 1$
Since $ a_{k \plus{} 1} \to 0$ when $ k \to \infty$ we know that:
$ a_1 \plus{} ... \plus{} a_k \to \frac {1}{P_i}a_1 \plus{} \frac {1}{P_i} \minus{} 1$ when $ k \to \infty$. But $ \frac {1}{P_i}a_1 \equal{} 1$ so:

$ a_1 \plus{} ... \plus{} a_k \to \frac {1}{P_i}$ when $ k \to \infty$. So the expected number of times you need to roll the dice is exactly $ \frac {1}{P_i}$ :D QED

[quote="Mathias_DK"]
8) Find all functions $ f: \mathbb{R} \backslash \{0\} \to \mathbb{R} \backslash \{0\}$ such that:
$ f(x \plus{} y) \equal{} xyf(x^3 \plus{} f(y)), \quad \forall x,y \in \mathbb{R} \backslash \{0\}$[/quote]
[hide="Solution"]fix $ y$ at any real value.
Then consider the following cubic equation in $ x$ where $ y$, and $ f(y)$ are merely coefficients:
$ x \plus{} y \equal{} x^3 \plus{} f(y)$
It being a cubic with real coefficients must possess at least one real root.
moreover none of the roots of it is zero unless $ y \equal{} f(y)$
even in that case $ x$ has a non-zero real root (namely 1)
Thus whatever the value of $ y$ there is at least one non-zero real $ x_0$ satisfying:
$ x \plus{} y \equal{} x^3 \plus{} f(y)$
Plug in that $ x_0$ for $ x$ in the original functional equation
$ f(x_0 \plus{} y) \equal{} x_0yf(x_0^3 \plus{} f(y))$
But $ f(x_0 \plus{} y) \equal{} f(x_0^3 \plus{} f(y))\neq 0$
Hence $ 1\equal{}x_0y\implies x_0 \equal{} \frac {1}{y}$
But that $ x_0$ also satisfies:
$ x \plus{} y \equal{} x^3 \plus{} f(y)$
$ \iff \frac {1}{y} \plus{} y \equal{} \frac {1}{y^3} \plus{} f(y)$
$ f(y) \equal{} y \plus{} \frac {1}{y} \minus{} \frac {1}{y^3}$
Remember that we had fixed $ y$ at some arbitrary value at the very beginning.
Hence $ f(y) \equal{} y \plus{} \frac {1}{y} \minus{} \frac {1}{y^3}$ holds for all real $ y\neq 0$
However it doesn't work in the main equation for all $ x,y$
contradiction.
Hence there is no function satisfying the functional equation[/hide]
P.S. did you randomly create the problem? :P

[quote="Mathias_DK"]4) For each $ k,n \in \mathbb{N}$ such that $ k \equiv \minus{} 1 \bmod n$ find infinitly many solutions to: $ a^k \plus{} b^k \equal{} c^n$ where $ a,b,c \in \mathbb{N}$[/quote]

$ (a,b,c) \equal{} \left((1 \plus{} t^k),t(1 \plus{} t^k),(1 \plus{} t^k)^{\frac {k \plus{} 1}{n}}\right)$

 :P  :D

anyways, nice problems. :)

Akashnil: Great minds think alike:) It was almost the exact same way I solved 4 and 8 ^^
Yes. The problem was pretty much randomly created. I thought i would have some function to equal $ xy$ times the same function again, and then I just saw if it created a good function. (In this case it didn't)


function
vector
real analysis
algebra
functional equation
linear algebra

Show that there exists a function $ f: \mathbb{R} \rightarrow \mathbb{R}$ that is additive ($ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$ for all $ x,y \in \mathbb{R}$) but not linear (I'm assuming linear in the linear transformation sense here?).

This problem is probably too well known but meh.
I have a solution for this but I just wanted to give this problem to other people to try as I think it's pretty nice.  I also wanted to double check that my assumption above is correct  :P.

This must be already solved here, but I couldn't find it either.

It is a classical result to take a Hamel basis of $ \mathbb R/\mathbb Q$ assuming that you mean that $ f$ is not $ \mathbb R$-linear.

The keyword here is [url=http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation]Cauchy's functional equation[/url] and it has been posted many, many times before (with that keyword).

I'll add that I probably wouldn't use the word "nice" in this context. "Pathological," perhaps. To prove that any such examples exist, you must use some form of the Axiom of Choice. (The usual path is to use AC to prove Zorn's Lemma, and then Zorn's Lemma to prove that any vector space - including $ \mathbb{R}$ considered as a vector space over $ \mathbb{Q}$ - must have a Hamel basis.) Using that, we find the following: if $ \mathfrak{c}\equal{}2^{\aleph_0}$ is the cardinality of $ \mathbb{R},$ then there are $ 2^{\mathfrak{c}}$ solutions to this functional equation, only $ \mathfrak{c}$ of which are ordinary real-linear functions.

As can be said of any "construction" that invokes the Axiom of Choice, the functions so "constructed" (the weird ones, the ones that aren't ordinary linear) are quite literally unimaginable. Their graphs are dense in the plane. They aren't Lebesgue measurable. There is nothing I can say that would effectively describe them.

Yes, pathological is a much more fitting word (although I tend to associate other pathological functions with the word nice as well, like everywhere continuous nowhere differentiable functions. I guess in a way I consider it synonymous with interesting, which I probably shouldn't :lol:).


vector
calculus
rotation
trigonometry
integration
geometry
geometric transformation

I have a vector calculus problem. I'm trying to find length of arc using vector calculus. I have limited experience here.

Consider a semi circle with the length of the arc being $\pi r$. The base (diameter) of the semi circle is $2r$. Imagine a point $P$ at the intersection of the circumfrence and the base. Calculate the length of arc from 0 to $\pi$ radians as the vector rotates on point $P$

Is this easy or hard to do?

[quote="vector1"]Is this easy or hard to do?[/quote]
This is easy to do once the problem is understood. Which is not easy, at least for me. But I'll guess that you want to find the arclength of semicircle using its vector equation $\vec r(t)=r\cos t\,\vec i+r\sin t\,\vec j$, where $0\le t\le \pi$. Then $L=\int_{0}^{\pi}|\vec r\,{}'(t)|\,dt=\int_{0}^{\pi}r\,dt=\pi r$. Note the difference between $r$ and $\vec r$.

I tried to find the arc length of a semi circle using vector calculus and got this integral. Could some genius out there solve the integral please.

$\int \sqrt{1+\frac{1}{4}(tan^{3}\theta-tan \theta)^{2}}d\theta$

BTW I will give the full details on how I arrived at this integral if anyone is interested.

A semicircle is given by $\vec{r}(t)=\cos t\vec{i}+\sin t\vec{j}$, where $r$ is the radius of your circle. So the arclength is $\int_{0}^\pi |\vec{r}'(t)|dt=\int_{0}^\pi |-r\sin(t)\vec{i}+r\cos(t)\vec{j}|dt=\int_{0}^\pi \sqrt{r^{2}\sin^{2}(t)+r^{2}\cos^{2}(t)}dt= \int_{0}^\pi \sqrt{r^{2}(\sin^{2}(t)+\cos^{2}(t)}dt=\int_{0}^\pi rdt=r\pi$

HilbertThm90

Please reread the first post of this thread.

Note that I wanted the vector to rotate on point P which is NOT on the axis of the semicircle, rather it is at the intersection of the base and the circumfrence.

I will have to show how I got this integral
$\int \sqrt{1+\frac{1}{4}(tan^{3}\theta-tan \theta)^{2}}d\theta$

and then maybe it will make more sense.

I will also clear up a little error. I said in the first post that the vector would have to rotate from 0 to pi radians but rotation from 0 to pi\2 radians is sufficient


function
search
topology
real analysis
real analysis unsolved

Find as much as possible about metric spaces X, with the property that any continuous function from $ X$ to $ \mathbb{R}$ are necessarily bounded?
[ For example, all compact spaces have this property ... but I'm not too sure if the converse is even true, and I don't think it's trivial at all ].

See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=633177683&t=146751

Incidentally I got this question on my general topology exam a year ago, but I had already done it on this forum  :P

I don't quite see what Kalle means (the relation between this post and that one), anyway, I think a characterization is countable compactness.

I don't see how the question in the link is not exactly the same as this one.

For metric spaces, countable compactness is compactness. You only need sequences to understand a metric topology.

:blush: I don't know what I was thinking - both Kalle and jmerry are right.

Anyway, this is probably why I had such a post, it seems that for a normal topological space we have every continuous real-valued function is bounded <=> it is countably compact.


algebra
polynomial
inequalities
function
algebra proposed

[quote="flip2004"]Let $ f(x)=Ax^2+Bx+C $ with real coefficients.[b] Prove or disprove that $ |f(x)|\le 1 \; ,\; \; \forall x \in [a,b] , $ if and
only if  $ |f(a)|\le 1 \; ,\; |f(b)|\le 1 $ and moreover [/b]
\[
- \; 1 + \sqrt{\left(1-f(a)\right)\left(1-f(b)\right)}\le
\frac{f(a)+f(b)}{2}\; -\; 2f\left(\frac{a+b}{2}\right)\le  1 - \sqrt{\left(1+f(a)\right)\left(1+f(b)\right)}\; .
\][/quote]
[b]Solution:[/b] Let $ G(x)=Px^2+Qx+r $ with real coefficients. let us prove the proposition:
[b]Lemma.[/b][i]The inequality $G(x)\ge 0 $ is verified for all $  x\in [a,b] $, if and only if
\[
(1)\; \; \;  G(a)\ge 0\; ,\; G(b)\ge 0 \; \; \;  \mbox{and} \; \;  \displaystyle  2
G\left(\frac{a+b}{2}\right)- \frac{ G(a)+G(b)}{2} \ge \displaystyle
\sqrt{G(a)G(b)} .
\]
[/i]
[b]Proof.[/b] [b] 1).[/b] Suppose (1) as verified. Denote \[ m=\displaystyle
\frac{\sqrt{G(b)}-\sqrt{G(a)}}{b-a} \; ,\; \;
 n= \frac{b\sqrt{G(a)}-a\sqrt{G(b)}}{b-a}\; ,\; \;  \mbox{and}
\]
 \[
 (1.1)\; \; \;  K\equiv K[G]:=\frac{2}{(b-a)^2}\left[  2G\left(\frac{a+b}{2}\right)- \frac{G(a)+G(b)}{2}-
 \sqrt{G(a)G(b)}\right].
\]
Then $ K\ge 0 $ and on the other hand holds the identity
\[
G(x)=(mx+n)^2+K\cdot (x-a)(b-x)
\]
which implies $\; \;  (2)\; \; \;  G(x)\ge 0\; ,\;  \forall x\in [a,b]. $
[b]2).[/b] Further assume (2) as  verified . Then $ G(a)\ge 0\;  ,\; G(b)\ge 0  $ and moreover $ G(x) $ may be written as (Luk\'acs Theorem)
\[(3)\; \; \;  G(x)=(m_1x+n_1)^2+ K_1(x-a)(b-x) \; \; \;  \mbox{with}\; \;  K\ge 0 . \]
If we select in (3)  $  x\in \left\{a , \;  \frac{a+b}{2}, \;  b\right\} $ then we find
\[(*)\; \; \ ; \;   \left\{\begin{array}{lcl}
\displaystyle  (m_1a+n_1)^2 &=& G(a)\\
\displaystyle (m_1b+n_1)^2 &=& G(b)
\end{array}\right. \; \;  \mbox{and}  \; \;  K_1=K ,
\]
$ K $ being as in (1.1) . Note that the system (*) furnish us $  m_1,n_1 $  and  we must have $ K\equiv K[G]\ge 0 , $ therefore inequalities (1) are verified.
Further apply the equivalence
\[
G(x)\ge 0 ,\; \forall x\in [a,b],\hsp \Longleftrightarrow \hsp
G(a)\ge 0,G(b)\ge 0 \; ,\; K[G]\ge 0 ,
\]
to the polynomial functions $ G_1(x):=1- f(x) $ and $G_2(x):=f(x)+1\;
. $ Indeed $ \left\{\begin{array}{l}G_1(x)\ge 0\\
G_2(x)\ge 0
\end{array}\right.
 ,\; \forall x\in
[a,b] , $ if and only if $ |f(a)|\le 1\; ,\; |f(b)|\le 1\; , $ and $  K[G_1]\ge 0\; ,\; K[G_2]\ge 0 .$ But the last two inequalities are the same with
\[\left\{
\begin{array}{l}
\displaystyle
1-2f\left(\frac{a+b}{2}\right)+\frac{f(a)+f(b)}{2}-\sqrt{\left(1-f(a)\right)\left(1-f(b)\right)}\ge
0\\
\\
\displaystyle
1+2f\left(\frac{a+b}{2}\right)-\frac{f(a)+f(b)}{2}-\sqrt{\left(1+f(a)\right)\left(1+f(b)\right)}\ge
0\\
\end{array}\right.
\]
which completes the solution.


Hi you guys,

I'm in 9th grade I've been fully immersed in studying for the AIME this past month and want to know whether I am doing it the right way:
-I've never taken a full-time practice test
-I just print out problems from AoPS wiki and solve them. I only look at solutions once I've solved it. When I don't solve it, i usually look at other problems and sometimes, days later, i come back to it and solve it.

Should I continue like this, or take full-time AIME's? Also, should I get ACoPS if I can consistently solve 1-5 and most of 6-10 (I'm bad at the counting)? I occasionally get 11's, 12's, or 13's  in my strong areas (geometry or algebra).

Also, should I be doing olympiad problems yet?

It would certainly help to do a timed full test since you'll know how to pace yourself.  But I never did timed tests and I did fine.

Don't look at about 4-6 of the AIME tests, save these for a full-time practice. Do one soon to evaluate how you will do under timed conditions and work on your weak areas. Continue do more AIME and problems that are in your weak areas. Do lots of  timed tests starting ~6 weeks from AIME. That's what works for me anyway. Also, I hear that easy Olympiad problems help with the last 5 AIME problems.

i would suggest studying up your weaknesses. doing problems is amazing and will help you more than anything else, but if you are really bad in some areas (you mentioned counting) it would help to get some practice on that separately from problems. i would suggest looking at alcumus - i have never been that great at counting and probability myself, but i have been progressing very rapidly using that thing. if you still struggle, you could think of buying a book about it. i dont actually own any, so i cant tell you for certain if theyll help a lot, though im sure you wont get worse from studying them.

and definitely save some aimes for timed studying closer to the test date.

If you do get a book, then I would suggest maybe the aops intermediates or volume 2. Note that getting a book and learning from it are different things. If you really want to improve by reading books, then work all the problems and do as much as you can without the solutions. I find that sometimes you have to read over the same thing a few times until you can really understand it and solve the challenge problems.


hey could u plz tell me wat are the most important topics to prepare for the rmo

thanx in advance
 :)

rt.........we all r in d same boat so we almost know as much as u kow........

u can get an idea after seeing previous years papers........
well do P&C...etc,,,,,,revise inequalities............rest is all grey mattr

btw........wats the last date for the form........
ive not taken the form yet :huh:

i think the most imp patr in PnC 4 RMO (coz dat was the only ques i did rite last yr :) )

and ofc we can't forget geometry :P

@arnav
last date in my region is 4 october and i know that till now 990 students have registered and most of them from city montessori school....and i hate it

@ritu and @arnav thanx but tell me about number theory and bout fermat and wilson theorem are these also important along with introductory combinaotrics

thanx :)

Copy down problems from the Pre-olympiad section of Mathlinks/AOPS in a book and start solving them right away without looking at the solutions. Those are the type of problems you must expect in RMO. There are no specific topics to study as such.

But there is a standard pattern since the past years. You always get one number theory problem, 2 geometry problems, one inequality/polynomial, one combinatorics/set theory and one sequence/system of equations/vague problem.

hey one thing is common most of rmo question or of rmo type are in incorporated in a book named
"CHALLENGES AND THRILLS OF PRECOLLEGE MATHEMATICS"
now i dunno y?

there r many such books...
"An excursion in Mathematics" is still better

challenges and thrills is just an appetiser :P  :P

which book should be bouht by me for rmo........??? :P

i have a lot of problems but wats d fun if i cant solve any,......!!! :lol:

i need a book which gives detailed solutions.......!!!

also 1 more and probably a foolish ques :D .............is kvpy allowed for 12th class students.......""????

ofc kvpy exam is given by twelthies also---but different stream :P
120 seats 4 11th and just abt 20 4 12th :rotfl:

well i wud suggest the book "problem primer 4 olymp(or sth)"
it has soln of many ques...

hey does "primer........" teaches fundamental priniciples needed for olympiad or is too high level?

it teaches high level thing in a fundamental way :rotfl:


How many distinct diagonals of a convex heptagon (7-sided polygon) can be drawn?

What was more natural for me was is that there are 7 choose 2
= 21 line segments total among the 7 vertices.  Subtract the
7 line legments around the perimeter that are not diagonals to
get 14.

In an $ n$-gon, for every vertex, there are $ n\minus{}3$ corresponding vertices that form a diagonal. This overcounts by $ 2$, so
\[ d \equal{} \frac{n\left(n\minus{}3\right)}{2}.\]For $ d\equal{}7$, this equals $ \boxed{14}$.

the number of diagnals in a polygon with $n$ sides is: $$\binom n 2-n$$

[size=200]Can somebody explain I don't understand :( [/size]

Think about how to get a diagonal. You need to chose 2 from the number of vertices, but subtract $n$ because you don't want to count the edges.
This is $\binom{n}{2}-n$ which if you expand it out, you get $\dfrac{n(n-3)}{2}$.
Hope this helps :D

[quote=sadhana]The formula for the amount of diagonals is d=n(n-3)/2. If we substitute n for 7, then we will get[u] 28[/u].[/quote]

you forgot to divide by 2 :)


Er..the polls section?

But seriously, we have
\[ \frac {1}{2^0} \plus{} \frac {2}{2^1} \plus{} \cdots \equal{} \left(\frac {1}{2^0} \plus{} \frac {1}{2^1} \plus{} \cdots\right)^2 \equal{} 4
\]
so our expression is less than $ 3$.

[Moderator: do not post mathematical problems into this section!]


analytic geometry
geometry
inradius
incenter

A point $(x,y)$ is to be chosen in the coordinate plane so that it is equally distant from the x-axis, the y-axis, and the line $x+y = 2$. Then $x$ is

A. $\sqrt{2} - 1$
B. $\frac{1}{2}$
C. $2 - \sqrt{2}$
D. 1
E. Not uniquely determined

[quote="minsoens"][hide]
The two axis and the line y=-x+2 make an isoceles right triangle.
Let r be radius of the inscribed circle and the distance from the point to each of the sides of the triangles.
The isoceles right triangle can be divided into three smaller triangles, with each triangle having the center of the circle as one of its vertices.
The total area of the isoceles right triangle is:
$\begin{eqnarray*} Area &=& \frac{1}{2}(2)(2) = 2\\ &=& (\frac{1}{2})(2)r+(\frac{1}{2})(2)r+(\frac{1}{2})(2\sqrt{2})r\\ &=& r+r+ 2\sqrt{2}r\\ &=& r(2+2\sqrt{2}) \end{eqnarray*} r = x = 2-\sqrt{2}$
[/hide][/quote]

I think you make mistake with the last step. $r=x=\sqrt2 - 1$

So the right answer is $\boxed{A}$

(1+sqr(2))r=sqr(2)
r=2-sqr(2).The answer is C.

[hide="Answer"]We are looking for the center of the circle inscribed in the triangle formed by these three lines. The x- and y-intercepts of $x+y=2$ are $y=2$ and $x=2$, so we have a triangle of area $2$, and the semiperimeter is $\frac{2+2+2\sqrt{2}}{2}=2+\sqrt{2}$, so we have $A=rs\Rightarrow 2=r(2+\sqrt{2})\Rightarrow r=\frac{2}{2+\sqrt{2}}=2-\sqrt{2}\Rightarrow \boxed{C}$.[/hide]

[quote="prototypex24"][quote]So the right answer is A[/quote]

I don't think so... using a different method

Area, $A=2$, Semiperimeter, $s=1+1+\sqrt{2}$

inradius, r
$=\frac{A}{s}$
$=\frac{2}{2+\sqrt{2}}$
$=2-\sqrt{2}$
 x=r so  $(C)$[/quote]

Opps  :blush: yeah, the right answer is C. What I meant was that minsoens made a little mistake at the last part because he has $A=r(2+2\sqrt{2})=2$ so I said the right answer in that case is A ($(\sqrt{2}-1)(2+2\sqrt{2})=2$).

[quote="236factorial"][hide]Such a point is the inradius, and the triangle given by the bounded area is a right triangle, we use the formula $\frac{a+b-c}{2}$ to get $2-\sqrt2$. [/hide][/quote]

 :huh: why is the formula for the inradius $\frac{a+b-c}{2}$... Never heard of it, can someone help explain? :P

[hide](x,y,) lies at the inradius. $A=2$, $s=2+\sqrt{2}$, $r=2-\sqrt{2}$, $\boxed{C}$.[/hide]

[quote="Chocoboom"]Can't a point be at $(2+\sqrt{2}, 2+\sqrt{2})$ as well?
So my answer is E. Not uniquely defined.[/quote]
[hide]That's true. The point can be either the incenter or any of the excenters, so it is [b]E[/b].[/hide]
This topic is a bit old, though.  :wink:  :P

[quote="Silverfalcon"]A point $(x,y)$ is to be chosen in the coordinate plane so that it is equally distant from the x-axis, the y-axis, and the line $x+y = 2$. Then $x$ is

A. $\sqrt{2}-1$
B. $\frac{1}{2}$
C. $2-\sqrt{2}$
D. 1
E. Not uniquely determined[/quote]

[hide]
$2-\sqrt{2}$ so the answer is C :lol: [/hide]

[quote="Chocoboom"]Can't a point be at $(2+\sqrt{2}, 2+\sqrt{2})$ as well?
So my answer is E. Not uniquely defined.[/quote]
Don't revive old topics...

[quote="mathwizarddude"]I don't see the reason for not reviving an old topic particularly when the previous answer is incorrect...[/quote]

Perhaps you see the reason for not reviving a 2.5 year old topic only to say that its not wrong to revive old topics under certain circumstances?

[quote]Perhaps you see the reason for not reviving a 2.5 year old topic only to say that its not wrong to revive old topics under certain circumstances?[/quote]

 :rotfl:  :rotfl:  :rotfl:


calculus
integration
trigonometry
calculus computations

hello
[hide="hint 1"]
first do rationalization and then use integration by parts
[/hide]

[quote="dulldrake"]
$ J \equal{} \int^\frac {1}{2}_0\sqrt {\frac {1 \minus{} \sqrt {x}}{1 \plus{} \sqrt {x}}}dx$

[/quote]

Also try:
$ x\equal{}\cos^22x
\\
dx\equal{}\minus{}2\cos2x\sin2x$


calculus
integration
logarithms
LaTeX
calculus computations

Argh, I missed class today and I have no idea how to do this.

$ \int_{1}^{e} \frac{(1 \plus{} \ln{x})^2}{x} dx$

Any help would be greatly appreciated. Thanks.  :D

i think you should compute the square and then intergrate by part using x=(lnx)' and  you should be fine

There's a simpler way, known as substitution.

In general, if we have a integral of the form $ \int f(g(x))g'(x)\, dx$ or similar, then we can make the substitution $ u \equal{} g(x) \implies du \equal{} g'(x)\,dx$ (we must replace the $ dx$, and the $ du$ in the given equality accomplishes that). After pairing the given substitutions, we get $ \int f(g(x))g'(x)\,dx \equal{} \int f(u)\,du$

And if it was a definite integral like $ \int_a^bf(g(x))g'(x)\, dx$, then doing the substitution will also change the limits, yielding $ \int_a^bf(g(x))g'(x)\, dx \equal{} \int_{g(a)}^{g(b)}f(u)\,du$.

[quote="math92"]Argh, I missed class today and I have no idea how to do this.

$ \int_{1}^{e} \frac {(1 \plus{} \ln{x})^2}{x} dx$

Any help would be greatly appreciated. Thanks.  :D[/quote]

$ \int_{1}^{e} \frac {(1 \plus{} \ln{x})^2}{x} dx$
$ \equal{} \int_{1}^{e} (1 \plus{} \ln{x})^2 \times \frac {1}{x}dx$
Notice $ \frac {1}{x} \equal{} \frac {d}{dx}(1 \plus{} \ln{x})$
It's the chain rule. So the integral is
$ \equal{} \frac {(1 \plus{} \ln{x})^3}{3}$
Plug in values
$ \frac {8}{3} \minus{} \frac {1}{3} \equal{} \frac {7}{3}$

I'm a latex noob so i don't know how to do the long | sign


Let $ p$ is prime number, $ {n}\in\mathbb{N}$, $ Z_{k}\equal{}cos{\frac {2 k \pi} {p}}\plus{}i\sin{\frac {2 k \pi} {p}}$ , where  $ k\equal{}0,1,2.....p\minus{}1.$
Find $ \sum_{k\equal{}0}^{p\minus{}1}{Z_{k}^{n}}$

$ \sum_{k\equal{}0}^{p\minus{}1}Z_k^n \equal{} \frac{e^{2i\pi n} \minus{} 1}{e^{\frac{2i\pi n}{p}} \minus{} 1} \equal{} 0$


search Harvard college Princeton Yale Columbia

Hello guys,

So I am going to be a senior next year and I am still considering the universities I will apply to.
Here is a brief resume. I would like to hear everyone's opinion and if possible to chance me, or say what I can improve.
International student
Member of the math team of my school, a few awards (1-3rd places); TOP 30 of my country, most probably member of the national team next year;
Member of the Unicef Club - visited many orphanages; donated and organized a lot of activities that stimulated the donations and gathering of money used for charity;
President and founder of the TechClub - tought a lot of students - about 30 - how to program and to create their own websites; to use those for business; also this club exists on the internet; explained also search optimisation and webmarketing
Not part of school - I am owner and creator of about 10 websites which are functioning and this is like my business because I earn from those money (not saying what amount) 3 of them are one of the most visited websites :))
Member of MUN and JA - participated in a Manager for a day (chosen) + many conferences and awards from there.
10 awards from literary contests (national ones)
Sorry, I forgot something to add - GPA 5.72/6.00; AP Courses 5 taken - 3 fives and 2 fours; SAT - 2350; SAT II Chemistry 800
SAT II Physics 800
Consider those universities if u chance me - Harvard; Princeton, Yale, Columbia, Cambridge :)
Thank you :))
Talk later :)

Your resume seems impressive, but then again, so are the resumes of everyone else applying.  I would say you have as a good a chance as anyone.

i think that's really good. u might have a good chance.
i would probably take the ACT, but don't take my word for it.
I'm only a freshman next year :D

they'll love the fact that you're from a different country. seems very good, i think you have a great chance of getting into an elite university.

[quote="yaofan"]i would probably take the ACT, but don't take my word for it.[/quote]

ignore this suggestion, and don't retake any of the SATs.

Actually, I think you should probably take the Math II exam.  Most top universities like to see a math exam and it seems like you should have no problem getting an 800.

Thank you, guys; I will have in mind your comments :)
Yes, I will take Math II (i didn't have enough time :P ).
Do you think that I will be able to get into Yale early decision, having in mind the short resume I wrote + most probably I will get 800 ot SAT II Math.
I know you cannot determine whether I will get accepted early decision or not, but just want to know what you think :)

Thank you in advance,

Salvatore

the literary aspect is impressive for a probable IMO participant next year

I'll point you back to [b]JRav[/b]'s first post because it really gets to the point.

Your credentials, statistics, and awards are certainly very strong, so we can say that you will at least be competitive with other viable international candidates (international admissions are held to higher standards than domestic admissions).

However, that's all we can say -- we have no information about three of the most important factors in admissions: your essays, your letters of recommendation, and overall coherence of your application.

So, if all you wanted is some comfort from forum users that say your application doesn't suck, you've got it right here. If you want actual, useful advice, considering asking questions like, "might _______ be an interesting topic for an admissions officer to read about in an essay?" or, "I'm thinking between person 1 and person 2 to write me letters of recommendation, this is the scope in which I know each of them, which will help my application more?"

Of course, you should keep it to specific questions like that -- if you post your essay and ask people for feedback, chances are you won't get any.

If you really want feedback, try posting at http://www.collegeconfidential.com.  I must warn you though, many of the people on that site have resumes as good as, if not better than yours so it may make you quite insecure.  But it will show you what you're up against and you may get some good feedback.  Note that none of the admissions officers on that site (there are a few) will evaluate your chances, so all of the feedback is from other college students and high school students.


Another question :

Let $a_n$ be defined as the number closest to $\sqrt{n}$ (for example $a_3=2$ , $a_{10}=3$), find the greatest integer less than or equal to $A$ where \[A=\sum^{2001}_{i=1}\frac{1}{a_i}\]

Well, you can even compute A exactly, we have $A=\sum^{2001}_{i=1}\frac{1}{a_i}=\frac{1327}{15}=88+\frac{7}{15}$. The proof is absolutely the same as in http://www.mathlinks.ro/Forum/viewtopic.php?t=5553 , with the only difference that the upper bound of summation is 2001 rather than 2004 (so you have to replace 24 by 21).

  Darij


Let $ ABC$ be a triangle with circumcircle $ w$. Point $ D$ lies on side $ BC$ such that $ \angle BAD\equal{}\angle CAD$. Let $ I_A$ denote the excenter of triangle $ ABC$ opposite $ A$, and $ w_A$ denote the circle with $ AI_A$ as a diameter. $ w,w_A$ meet at $ P$ ($ P\neq A$). The circumcircle of triangle $ APD$ meet $ BC$ again at $ Q$. Prove that $ Q$ lies on the excircle of triangle $ ABC$ opposite $ A$.

I will change notation, as follows:
[quote="discredit"]Let $ \triangle ABC$ be a triangle with circumcircle $ \omega$.Let $ AD$ be an angle bisector of $ \widehat{BAC}$,where $ D\in [BC]$.Let $ \zeta_A(I_A)$ be $ A$-excircle and $ \omega_A(I_X)$ is the circle with diameter $ AI_A$.Circles $ \omega$ and $ \omega_A$ intersect at $ P$.If $ A_1$ is the tangency point of $ BC$ with $ \omega_A$, then prove that points $ A,D,A_1,P$ are concyclic[/quote]
[b]Proof:[/b]
-Suppose $ AP$ intersects $ l_A$ at point $ X$,where $ l_A$ is the tangent line to $ \omega_A$ at point $ I_A$.
-Let $ \zeta(O_A)$ be the circumcircle of $ \triangle BI_AC$.If $ I$ is incircle of $ \triangle ABC$,then $ IBI_AC$ is cyclic,with $ \widehat{IBI_A} \equal{} \widehat{ICI_A} \equal{} \frac {\pi}{2}$,thus $ O_A$ lies on $ AI_A$.But as we know $ I_X$ lies on $ AI_A$,too,so circles $ \zeta$ and $ \omega_A$ are tangent to each other.
-[b]Lemma:[/b]
   Let $ \omega_1,\omega_2,\omega_3$ be circles with radical axes $ r_{12},r_{13},r_{23}$,then $ r_{12},r_{13},r_{23}$ are concurrent.
   [b] Proof:[/b] is well-known and easy.
-Now using lemma for circles $ \zeta$,$ \omega_A$,$ \omega$,we conclude that $ B,C,X$ are collinear.
-Since $ \widehat{IA_1X} \equal{} \widehat{I_APX} \equal{} \frac {\pi}{2}$ we conclude that $ A_1PXI_A$ is cyclic,hence
$ \boxed{\widehat{PAD} \equal{} \widehat{PII_A} \equal{} \widehat{PI_AX} \equal{} \widehat{PA_1X} \equal{} \pi \minus{} \widehat{DA_1P}}$,here we use that $ l_A$ is tangent to $ \omega_A$ and $ B,C,X$ are collinear.The proof is complete.


When I protested problem #2, the USAMTS replied this:

Your grading protest for USAMTS Year 20, Round 1, Problem 2 has been approved. Your score for the problem has been adjusted to 5.

And the comments below said:
Your solution meets the criteria for 4 points.

And the USAMTS page said I have 4 points for Problem #2.

The thing is, I have no idea whether I should deserve a 5 or a 4... :maybe:

Where did you get that? my score for 5 got moved up but I didn't get any message

[quote="xpmath"]Don't worry, the same thing happened to me, except instead of a 3 I got a 2. Basically, since you're getting some extra points, they'll have to approve your protest. However, you said that you should get a 5, so the email first says that your score is adjusted to 5. However, they're really only changing it to 4.

At least, that's what I concluded :maybe:[/quote]

But the funny thing is, I said I should get a 4. :maybe:
And the e-mail said I should get a 5, and then the comment went back to 4.

This is so confusing.

There was a brief moment while we were changing some things that the score listed in the email was one too high. I think this affected 4 people. Your current listed score on your USAMTS page is the correct score.

Ah, that explains it. Thanks levans, I kept wondering why the scores were different...


function
integration
logarithms
calculus
calculus computations

Given that: $ x^2y'$+xy=1 and $ y(1)\equal{}2$. Find the function y

This is a first-order linear ODE; we can standardize our methods for dealing with such problems. First step: get a standard form by making the coefficient of $ y'$ $ 1.$

$ y'\plus{}\frac1xy\equal{}\frac1{x^2}$

The integrating factor is $ e^{\int \frac1x\,dx}\equal{}e^{\ln x}\equal{}x.$

$ xy'\plus{}y\equal{}\frac1x$

$ (xy)'\equal{}\frac1x$

$ xy\equal{}\ln x\plus{}C.$

Now is a good time to put in the initial condition:

$ 1\cdot 2\equal{}\ln 1\plus{}C$ so $ C\equal{}2.$

$ xy\equal{}\ln x\plus{}2$

$ \boxed{y\equal{}\frac{\ln x}x\plus{}\frac2x}.$

Note that this solution is only valid for $ x>0.$ The original differential equation has a singular point at $ x\equal{}0$ and the given initial condition lies somewhere to the right of that singular point. We cannot meaningfully extend the solution across the singularity.


inequalities
geometric inequality
geometry proposed
geometry

With given triangle $ABC$ an arbitrary point $M$ we can prove
$\cot\frac{A}{2}MA+\cot\frac{B}{2}MB+\cot\frac{C}{2}MC\ge a+b+c(*)$
Now if we want a similar problem for two triangle we will get it
With two triangle $ABC,A'B'C'$ an arbitrary point $M$ prove that
$\cot\frac{A'}{2}MA+\cot\frac{B'}{2}MB+\cot\frac{C'}{2}MC\ge a+b+c(**)$
P/s:$(*)$ is quite hard I posted it but nobody solve and I think $(**)$ is very hard, it is true, I found it but I lost solution for it, therefore it is my last post about geometric inequality on ML, I won't post more about  GI if I haven't found my solution or anybody helps me solve to it thanks !

[quote="gemath"]With two triangle $ABC,A'B'C'$ an arbitrary point $M$ prove that
$\cot\frac{A'}{2}MA+\cot\frac{B'}{2}MB+\cot\frac{C'}{2}MC\ge a+b+c(**)$[/quote]

Obviously wrong: Take M = A and play around with A', B', C'.

  Darij

Oh sorry I remembered wrongly It must be
$\frac{a'}{a}\cot\frac{A'}{2}MA+\frac{b'}{b}\cot\frac{B'}{2}MB+\frac{c'}{c}\cot\frac{C'}{2}MC\ge a'+b'+c'$
 :)

Yesterday is a good day of me, I found the nice solution for the above problem in my old drafts therefore I will continue to post the geometric inequalities, thanks for your interest.  :)


Let $ABC$ a triangle and $O$ it`s circumcentre.$AO\cap BC=M$,$BO\cap AC=N$,$CO\cap AB=P$.We know that $AO=p$,where $p$ is a prime number and the lenghts $OM$,$ON$ and $OP$ are positive integers.
Find the lenghts of the triangle ABC.

It's easy.

Let $OM=x,ON=y,OP=z$
$\frac{p}{p+x}+\frac{p}{p+y}+\frac{p}{p+z}=2$
$pf(x,y,z)=2(p+x)(p+y)(p+z)$
$x<p, y<p, z<p$, and p is a prime number
$(p,x+p)=1,(p,y+p)=1,(p,z+p)=1$
But $p|2(x+p)(y+p)(z+p)$
Thus $p|2$
$p=2$
$x=y=z=1$
lenghts of the triangle ABC is $6\sqrt{3}$

Very nice and easy solution.Actually when I saw the problem in the contest,I saw the solution,but it was very computational.But this one makes the problem seem very easy.Thnx.


This is a small thing that I think people could benefit from. Every page in the forum is titled "Art of Problem Solving Forum." More than often, I want to visit a thread or a subforum that I've been to already. I use Firefox, and when I type "artof " in the address bar, it shows me everywhere I've been along with the page title. Unfortunately, they all say Art of Problem Solving forum, and finding a page can get annoying.

I think thread page titles should be the name of the thread, like "AOPS Forum - Help with AIME question" and subforum page titles should be "AOPS Forum - College Playground" or something similar.

Just a small thing that I think would be easy to change.

Not for me:

[img]http://img.photobucket.com/albums/v489/amitface/screen.jpg[/img]

I guess the Mathlinks skin does it, whereas the AOPS one doesn't.

So my suggestion still stands.

I really do not understand your question. It would still not help you, no matter what the beginning would be. You would still have to type artofproblemsolving.com etc. ...  :maybe:

In FF, you can see the titles, and just go directly, instead of re-navigating.

[img]http://myfiles.zapto.org/aops_titles.PNG[/img]

I could juts click on teh row to go to the topic/forum.

it's not a big deal. just thought i'd throw it out there

The title of this page (the very top line of the browser) in AoPS skin is "Art of Problem Solving Forum."  (See the first sceen shot.)  This is also (mildly) aggravating when I have many tabs open, since each of them is titled "Art of Problem Solving Forum" and I can't tell which is which.

If you look at the second screen shot, the title of the page (in mathlinks skin) is "The Art of Problem Solving Forum :: View Topic - Forum name in web page title."

In fact, it might be even nicer if the actual name of the topic or forum came earlier in the title.  This is what happens on the Wiki: when I visit [[Perfect square]], the title of the page is "Perfect square - AoPSWiki," so I can immediately find it from among my tabs.

Yes because when I have tabs open, it says
the Art of Problem Solving :: View...
and I can't view the title.  It would be more useful to put the title first then the Art of Problem Solving, as JBL said.

I might design some customizable options as to what you want the page title to be ( x = the site name, y = the page name, and combinations of the two). That way everyone will be happy.


Let $\cos{\theta}=\cos{\alpha}\cos{\beta}$.
Prove, that
$\tan{\frac{\theta+\alpha}{2}}\tan{\frac{\theta-\alpha}{2}}=\tan^2{\frac{\beta}{2}}$.

\begin{eqnarray*}\tan\frac{\theta+\alpha}{2}\tan\frac{\theta-\alpha}{2} &=& \frac{\sin\frac{\theta+\alpha}{2}\sin\frac{\theta-\alpha}{2}}{\cos\frac{\theta+\alpha}{2}\cos\frac{\theta-\alpha}{2}}\\ &=& \frac{(\sin{\theta\over 2}\cos{\alpha\over 2})^2-(\cos{\theta\over 2}\sin{\alpha\over 2})^2}{(\cos{\theta\over 2}\cos{\alpha\over 2})^2-(\sin{\theta\over 2}\sin{\alpha\over 2})^2}\\ &=& \frac{\frac{1-\cos\alpha\cos\beta}{2}\cdot\frac{1+\cos\alpha}{2}-\frac{1+\cos\alpha\cos\beta}{2}\cdot\frac{1-\cos\alpha}{2}}{\frac{1+\cos\alpha\cos\beta}{2}\cdot\frac{1+\cos\alpha}{2}-\frac{1-\cos\alpha\cos\beta}{2}\cdot\frac{1-\cos\alpha}{2}}\\ &=& \frac{\frac{\cos\alpha(1-\cos\beta)}{2}}{\frac{\cos\alpha(1+\cos\beta)}{2}}\\ &=& \frac{2\sin^2{\beta\over 2}}{2\cos^2{\beta\over 2}}\\ &=& \tan^2{\beta\over 2}\end{eqnarray*}


analytic geometry
rotation
trigonometry
geometry
rectangle
inequalities
triangle inequality

Here is a problem I "created" (probably it is well-known and very very old, but I didn't knew it).

Given two points in the plane and their coordinates $ (x_1,y_1)$ and $ (x_2,y_2)$, find the coordinate axes.

Coordinate axes with respect to what? How are we to represent them? We need more information.

I don't get it; if a point is at $ (x_1,y_1)$, that means it's $ x_1$ units away from the $ y$-axis and $ y_1$ units away from the $ x$-axis, so you know exactly where the axes are.

But how do you measure the length of 1 unit if you have only one point? That's why you need at the very least, two...

Is the plane rotated, or do you know which way x and which way y is?

[hide="Determining orientation of axes"]Draw a circle $ O$ with diameter AB, where A and B are the given points.  Draw a circle with radius $ y_2 \minus{} y_1$ or $ x_2 \minus{} x_1$, centered at $ B$ or $ A$.  Let the point where this circle meets $ O$ be $ P$.  then $ PA$ and $ PB$ are parallel to the axes.[/hide]

If I just give you two points and their coordinates, you can't figure out the length $ y_2 \minus{} y_1$ without first knowing the orientation of the axes.

I assume we're working in Cartesian coordinates.

$ \| x \minus{} y \| \equal{} \sqrt {(x_2 \minus{} x_1 )^2 \plus{} (y_2 \minus{} y_1)^2}$, which is computable, so you know what the length of one unit is.

Now construct a circle of radius $ \| x \|$ around point $ x$ and a circle of radius $ \| y \|$ around point $ y$, both of which are computable by the distance formula. The intersection of these two circles gives you up to two possible locations of the origin.

If there are two, pick the particular origin such that, if $ x_1 y_2 \minus{} x_2 y_1 > 0$ then $ y$ is counterclockwise of $ x$, and if $ x_1 y_2 \minus{} x_2 y_1 < 0$, then $ x$ is counterclockwise of $ y$.

Now, you can use trig and arctan to draw the particular coordinate axes. That is, a point $ (1,1)$ would be separated from either coordinate axis by an angle of $ \pi /4$ about the origin, for example.

It looks like a lot of people weren't sure this problem made sense in the absence of a coordinate system, so let me put it this way:

You are given two points $ P_1, P_2$.  Attached to those points are a pair of rectangles; each point is a corner of the respective rectangle.  You are free to rotate the rectangles about those points.  Let $ Q_1, Q_2$ denote the corners of the rectangles opposite $ P_1, P_2$ respectively.  Determine an orientation of both rectangles such that both $ Q_1 \equal{} Q_2$ and the corresponding sides adjacent to $ Q_1, Q_2$ are parallel.  (These determine the coordinate axes.)

In general, $ Q_1 \equal{} Q_2$ says nothing about the sides adjacent to $ Q_1, Q_2$, but the way the problem is set up we have the additional piece of information about the distance between $ P_1, P_2$ and the triangle inequality comes into play.


[quote="roadnottaken"]Did you mean for the left side of the equation to be equal to the denominator of the right side?[/quote]

No. The equation is shown how it should be.

Can we do something like $t=-x^{2}/t+2x^{2}$?

I get the polynomial
$x^{8}+8x^{7}+30x^{6}+72x^{5}+120x^{4}+144x^{3}+121x^{2}= 64x+16$

After some finagling, I think I can prove that (x+1) is a factor.
Which means one solution is -1?
EDIT: After doing some synthetic division, I get $(x+1)(x^{7}+7x^{6}+23x^{5}+49x^{4}+71^{3}+73x^{2}+48x+16)$,
where the second factor is irreducible over the integers.

Are you sure?  "Irreducible over the integers" is not the same thing as "Has no integer roots."  For example, $x^{4}+x^{2}+1$ has no integer roots, but

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

[quote="cincodemayo5590"]Solve for $x \in \mathbb{R}$:

$x^{4}+4x^{3}+6x^{2}+8x+4 = \frac{-x^{2}}{x^{4}+4x^{3}+8x^{2}+8x+4}$

[hide="hint"]
Find a suitable substitution.[/hide][/quote]

Hrmmm... how about just multiply the denominator out to get

$(x^{4}+4x^{3}+7x^{2}+8x+4)^{2}-x^{4}=-x^{2}$.

By the awesome method of guess-and-check factoring, we can rewrite this as:

$(x+1)^{2}(x+2)^{2}(x^{2}+x+2)^{2}= x^{4}-x^{2}$
$(x^{2}+3x+2)^{2}(x^{2}+x+2)^{2}= x^{4}-x^{2}$
$\left( x+3+\frac{2}{x}\right)^{2}\left( x+1+\frac{2}{x}\right)^{2}= 1-\frac{1}{x^{2}}$

Hrmm... well a "squared" randomly disappeared the first time, and I don't know if this helps any but I'm too lazy to try more stuff at the moment.

[quote="probability1.01"][quote="cincodemayo5590"]Solve for $x \in \mathbb{R}$:

$x^{4}+4x^{3}+6x^{2}+8x+4 = \frac{-x^{2}}{x^{4}+4x^{3}+8x^{2}+8x+4}$

[hide="hint"]
Find a suitable substitution.[/hide][/quote]

Hrmmm... how about just multiply the denominator out to get

$(x^{4}+4x^{3}+7x^{2}+8x+4)^{2}-x^{4}=-x^{2}$.

By the awesome method of guess-and-check factoring, we can rewrite this as:

$(x+1)^{2}(x+2)^{2}(x^{2}+x+2)^{2}= x^{4}-x^{2}$
$(x^{2}+3x+2)^{2}(x^{2}+x+2)^{2}= x^{4}-x^{2}$
$(x^{2}+2x+2)^{2}-x^{2}= x^{4}-x^{2}$
$(x^{2}+2x+2)^{2}= x^{4}$
$x^{2}= x^{2}+2x+2$ (since both are non-negative)
$x =-1$.[/quote]

Rofl, "the awesome method of guess-and-check factoring".  :D   RRT also works.

Yeah, I just noticed that after I posted  :blush: .

[quote="Altheman"]i am pretty sure that $-x^{2}$ should be a $-x^{4}$....[/quote]

[hide="Yes, it allows this similar substitution..."] Let $t = x^{4}+4x^{3}+7x^{2}+8x+4$.  Then

$t+x^{2}= \frac{-x^{4}}{t-x^{2}}\implies t^{2}-x^{4}=-x^{4}\implies t = 0$

And since $t = (x+1)^{4}+(x+1)(x+3)$ we can knock out one root immediately and find a few more, it wouldn't take very much effort :) [/hide]

[hide="Continuing my solution..."] The problem is equivalent to solving

$x^{4}+4x^{3}+7x^{2}+8x+4 = 0$

I showed that we can write this as $(x+1)^{4}+(x+1)(x+3)$, so, factoring out,

$(x+1) \left( x^{3}+3x^{2}+4x+4 \right) = 0$

RRT and observation tells us that $-2$ is a root of the cubic.  Factoring out,

$(x+1)(x+2)(x^{2}+x+2) = 0$

Since $x^{2}+x+2 > (x+\frac{1}{2})^{2}\ge 0$ the quadratic has no real roots.  Hence our only real solutions are $x = \boxed{-1,-2 }$. [/hide]

[quote="cincodemayo5590"]Oh, oops. It should be $-x^{4}$. Sorry guys; the problem is edited.  :)

[edit]: I can't change my first post! Grr. If anyone knows how I can, please tell me....  otherwise, just know that the problem should be this:

Solve for $x \in \mathbb{R}$:

$x^{4}+4x^{3}+6x^{2}+8x+4 = \frac{-x^{4}}{x^{4}+4x^{3}+8x^{2}+8x+4}$[/quote]

I have noticed that after a certain amount of time (denoted by $t$  :P ) has elapsed, you can no longer edit a post.

[quote="t0rajir0u"]...$(x+1)^{4}+(x+1)(x+3)$, so, factoring out,

$(x+1) \left( x^{3}+3x^{2}+4x+4 \right) = 0$

RRT and observation tells us that $-2$ is a root of the cubic.[/quote]

A bit easier for factoring if we substitute $t=x+1$, getting $t^{4}+t(t+2)=t(t^{3}+t+2)$. Now it's obvious that $t+1$ is a factor, which gives $t(t+1)(t^{2}-t+2)$ which is of course equal to $(x+1)(x+2)(x^{2}+x+2)$


A jet airplane departing on time, flying between two airports at an average speed of 540 mph arrives eight minutes late.  Departing on time and flying at an average speed of 480 mph it arrives fifty-three minutes late.  What is the number of miles between the two airports?

[hide="Solution"]Let x be the distance between the airports and y be the time the plane's supposed to take.

x/9=y+8
x/9-8=y
x/8=y+53
x/8-53=y
x/9-8=x/8-53
x/9=x/8-45
x=9x/8-9*45
9*45=x/8
x=360*9
x=3600-360
x=3240[/hide]

[hide="My Answer."]Use the formula $d=rt$, where $d$ is distance, $r$ is rate, and $t$ is time.  We have that $d=540(t+8)$ and $d=480(t+53)$.
Substitute:
$540(t+8)=480(t+53)\\
9(t+8)=8(t+53)\\
9t+72=8t+424\\
t=352\ \rm{minutes}$
Now, solve for the distance (we have to divide 360 by 6 to get minutes into hours):
$d=540(6)\\
d=3240$
Therefore, the distance is 3240.[/hide]