i assume n2 is n^2.

[hide]

(n-10)^2 = n^2-20n+100, which is less than n^2-19n+99.

(n-9)^2 = n^2 - 18n + 81, which is greater than  n^2-19n+99 if n>18.  This means that for n>18, n^2-19n+99 is in between two consecutive perfect squares, therefore it cant be a perfect square.  So n<=18, and we check these values.  I get 4 of them.  [hide][/hide][/hide]

[hide]

n^2-19n+99 = k^2, but the equation is difficult to check by hand for numbers. So using a bit of "cleverosity", I factored it to (n-11)(n-9)+n=k^2, which is much easier.

Then I set up the inequalities like Cat and solved to get a range.

(n-9)^2 :le: (n-11)(n-9)+n :le: (n-10)^2

Solved for the = cases to get 1:le:n:le:18

Since n^2-19n+99 is a quadratic, with a vertex at -b/2a = 9.5, which is half way between 1 and 18, we only need to check numbers between 1 and 9, and then add 9 to each of the sol'ns to get the other solutions.

n=1 : 81 <- square

n=2 : 65

n=3 : 51

...

n=9 : 9

So the sol'ns are at n=1,9,10, and 18. Sum of these numbers: 38.[/hide]

Quote:Find the sum of all positive integers n for which n^2 - 19n + 99 is a perfect square.

[hide]Let n^2 - 19n + 99 = m^2.

Then n^2 - 19n + 361/4 + 35/4 = m^2

multiplying both sides by 4 conveniently yields

4n^2 - 78n + 361 + 35 = 4m^2

(2n-19)^2 + 35 = (2m)^2

(2m)^2 - (2n-19)^2 = 35

(2m + (2n-19))(2m - (2n-19) = 35

(2m + 2n - 19)(2m - 2n + 19) = 35.

The only pairs of factors of 35 are 1 and 35, 5 and 7, 7 and 5, and 35 and 1.

Adding these pairs together yields either 36 or 12, so

(2m + 2n - 19) + (2m - 2n + 19) = 12 or 36.

4m = 12 or 36

m = 3, 9

n^2 - 19n + 99 = 9

n^2 - 19n + 90 = 0

(n - 9)(n-10) = 0

n = 9 or 10

or

n^2 - 19n + 99 = 81

n^2 - 19n + 18 = 0

(n-18)(n-1) = 0

n = 1 or 18

So the solutions are 1, 9, 10, and 18. Sum = 038

I favor this method because there is no guessing-and-checking involved.  You just make sure everything is an integer, then use the factors of the number on the right side. [/hide]

indeed, but then you sum it, but correct anyways

oh, sorry, i just glanced at the end, but didnt notice it

confuted wrote:Very nice mathfanatic. I like that too... but how did you choose
[hide]n^2 - 19n + 361/4 + 35/4 = m^2[/hide]? Surely you didn't just see that, did you?

No, I was using the technique of completing the square.  Basically, if you have a problem like

x:^2: + 4x + 2 = 0, solve for x

You can say "well, x:^2: + 4x + 4 is a perfect square, so I'll add 2 to both sides and get:"

x:^2: + 4x + 4 = 2

Then, rewrite as a perfect square:

(x+2):^2: = 2

So,

x+2 = +- sqrt(2)

x = -2 +- sqrt(2).

(This is actually the method used to derive the quadratic formula).

How does one go about doing this for a general problem, like

ax:^2: + bx + c = 0?

Well, recall that a perfect square (mx+n)^2 expands to (mx)^2 + 2mnx + n^2, so a=m^2, b=2mn, and c = n:^2:.

So n = b/(2m) = b/(2*:sqrt:(a)).  Since c = n:^2:, c = b:^2:/4a

Thus in order to make ax:^2: + bx + c into a perfect square, we need to make c equal to b:^2:/4a.  In the example of x:^2: + 4x + 2 = 0, we needed to make the 2 into a 4 because 4:^2:/4(1) = 4.

Now, referring directly to the last example (spoiler):

[hide]

We had n^2 - 19n + 99 = m^2.  We want to make the n^2 - 19n + whatever into a perfect square because...well, just because.  So we find that c must equal (-19)^2/4*1 = 361/4.  So we separate 99 into 361/4 and 35/4.  Then we complete the square to get (n-19/2)^2 + 35/4 = m:^2:.  Of course, because I hate fractions (in fact, my math teacher had this thing called "Kill All Denominators", or KAD for short), I multiplied by 4 first.  It comes to the same thing.[/hide]


limit
function
inequalities
number theory unsolved
number theory

Edit for some more generality.

Define $f(k)$ to be the maximum number of distinct ways an integer $1 < f(k) < k$ can be written as the sum of two squares.

1.  Find $f(2006)$.

2.  Prove $\lim_{k \to \infty} f(k) = \infty$.

3.  Find an algebraic function $g(x)$ such that $\lim_{k \to \infty} \frac{ f(k) }{g(k) } = 1$, or prove that none exist.

Well, the only integer we have to look at is $2005$, cause we can add some $1^2$ to every other number.
In general, this can be calculated recursivly.

Oh, sorry - two nonzero squares.  *edits problem condition*  :blush:

Distinct = order and sign is not taken into account¿

First lets define $A_2(n)$ as the number of representations of $n$ as $n=a^2+b^2$ with $a,b \in \mathbb{Z}$ and $B_2(n)$ as the number of distinct representations as sum of nonzero squares.
$A_2$ is not exactly what we want since it's possible that $a,b \leq 0$, but thats (neglecting the cases $ab=0$ and $a=b$) proportional to the number we look for (by the factor $8$).
Now we have:
1. case  $n=c^2$: $8 B_2(n)=A_2(n)-4$
2. case  $n=2 c^2$: $8 B_2(n)=A_2(n)+4$
3. case  all other: $8 B_2(n)=A_2(n)$

Now $A_2(n)= 4 \prod_{ \substack{ p \in \mathbb{P} \\ p \equiv 1 \mod 4}} (v_p(n) +1)$ and thus:

1. case  $n=c^2$: $2 B_2(n)+1=\prod_{ \substack{ p \in \mathbb{P} \\ p \equiv 1 \mod 4}} (v_p(n) +1)$
2. case  $n=2 c^2$: $2 B_2(n)-1=\prod_{ \substack{ p \in \mathbb{P} \\ p \equiv 1 \mod 4}} (v_p(n) +1)$
3. case  all other: $2 B_2(n)=\prod_{ \substack{ p \in \mathbb{P} \\ p \equiv 1 \mod 4}} (v_p(n) +1)$

So e.g. $B_2(13 \cdot 5^k) = k+1$ which proves 2).

To solve 3) you would have to use ugly inequalities (the result will be something using $ln(n)$ etc.) to asymptotically evaluate the supremum of $\prod_{ \substack{ p \in \mathbb{P} \\ p \equiv 1 \mod 4}} (v_p(n) +1)$ with small enough $v_p(n)$.


geometry
symmetry
incenter
exterior angle
geometry unsolved

[quote="kprepaf"]let ABC be a triangle acute-angles.

let P , Q and R be respectively random points taken respectively on sides [AB] , [AC] and [BC] .

1) find the position of the three points P , Q and R on the triangles that gives birth to a triangle PQR with minimal perimeter.

2) prove that the remaining triangles (the remaining areas on ABC) are similar to each other and all similar to their father ABC.[/quote]
Very easy!
1) P,Q,R are the foot of alitudes.
2) You should say that the state is true with P,Q,R in 1). It is extremely obvious.

[quote="kprepaf"][quote]Very easy!
1) P,Q,R are the foot of alitudes.
2) You should say that the state is true with P,Q,R in 1). It is extremely obvious.[/quote]

you should prove 'em , we all know the result  :roll: or am i dumb  :|[/quote]

Let R,Q are fixed, denote Q' is the symmetry point of Q w.r.t BC. When Q',P,R are on a line , RP+PQ is min , BC is the bisector of exterior angle of angle QPR
So when RP+PQ+RQ is min , A,B,C are excenter of triangle PQR ,the incenter of triangle PQR is the orthocenter of triangle ABC
So the problem is solved

[quote="plane geometry"]
Let R,Q are fixed, denote Q' is the symmetry point of Q w.r.t BC. When Q',P,R are on a line , RP+PQ is min , BC is the bisector of exterior angle of angle QPR
So when RP+PQ+RQ is min , A,B,C are excenter of triangle PQR ,the incenter of triangle PQR is the orthocenter of triangle ABC
So the problem is solved[/quote]

thanks , i'm still looking at you reply ..
By the way , the triangle that fits condition 1) is called : orthic triangle (probably due to " orthocenter ")
and it has some interesting applications in practice as i once heard , so if someone can talk a bit about that , it'll be helpful  :roll:


What do the following graphs look like? Which values need special consideration? y = (x^2)(x-5)^3(/5) and y = (x^2)(x-5)^(2/5)

[quote="AndrewTom"]Hi. I spotted that there had been a reply to this (from JoeBlow, I think) as I was rushing out last night but it seems to have disappeared.[/quote]
Hello Andrew
Ya i too have seen his post.But where it is now? :wink:


AMC
AIME
trigonometry
ratio
rotation
analytic geometry
percent

Hello,

I think you know the concept of this type of thread.  I start off with a problem.  Someone solves it, posts a [i]solution[/i] (not just an answer), and then proceeds to post the next problem.

Some things to remember:

$\bullet$ Keep the problems numbered.
$\bullet$ [b]Keep the problems at Getting Started level.[/b]  The last Problem Solving Marathon exceeded the Getting Started level.  Remember, proofs are almost always unacceptable for Getting Started.  The problems should stay around AMC level.  Only the very easiest AIME problems are acceptable (like #1 on the AIME...sometimes #1 is too hard).  Only [b]very basic[/b] trig is allowed.
$\bullet$ Keep the thread moving by posting a problem after you have solved a problem.

Ok.  Here we go:
[color=blue][size=150]
Problem 1[/size][/color]

How many digits are in the number $4^{17}\cdot 5^{30}.$

[hide]$2^{34} \cdot 5^{30}=10^{30} \cdot 16$

That's 32 digits.
[/hide]

[size=150]Problem 2[/size]

The product $N$ of three positive integers is 6 times their sum. One of the integers is the sum of the other two. Find the sum of all possible values of $N$.

It's AIME #1, but it's really easy. :)

[quote="chess64"][hide]$2^{34} \cdot 5^{30}=10^{30} \cdot 16$

That's 46 digits.
[/hide]

[size=150]Problem 2[/size]

The product $N$ of three positive integers is 6 times their sum. One of the integers is the sum of the other two. Find the sum of all possible values of $N$.

It's AIME #1, but it's really easy. :)[/quote]

[hide]
$N=abc=6(a+b+c)$

  Lets say $c=a+b$

$ab(a+b)=6[ 2(a+b)]$

$ab(a+b)=12(a+b)$

$ab=12$

Now $a$ and $b$ can be 1,12 or 2,6 or 3,4  (it dosent matter which ones a, and which ones b)

  Thus the possible values for $N$ are:

$1\cdot12\cdot13=156$

$2\cdot6\cdot8=96$

$3\cdot4\cdot7=84$

  Thus the sum of the possible values for $N$ are $156+96+84=\boxed{336}$ [/hide]

EDIT: Isn't number one 32? :?

[quote="G-UNIT"][quote="chess64"][hide]$2^{34} \cdot 5^{30}=10^{30} \cdot 16$

That's 46 digits.
[/hide]

[size=150]Problem 2[/size]

The product $N$ of three positive integers is 6 times their sum. One of the integers is the sum of the other two. Find the sum of all possible values of $N$.

It's AIME #1, but it's really easy. :)[/quote]

[hide]
$N=abc=6(a+b+c)$

  Lets say $c=a+b$

$ab(a+b)=6[ 2(a+b)]$

$ab(a+b)=12(a+b)$

$ab=12$

Now $a$ and $b$ can be 1,12 or 2,6 or 3,4  (it dosent matter which ones a, and which ones b)

  Thus the possible values for $N$ are:

$1\cdot12\cdot13=156$

$2\cdot6\cdot8=96$

$3\cdot4\cdot7=84$

  Thus the sum of the possible values for $N$ are $156+96+84=\boxed{336}$ [/hide][/quote]
Oh! I had almost solved my fist AIME problem, execpt I made an algebra error.

[quote="236factorial"]While we're waiting, what happened to marathon 2? Should this be stickied to prevent future losses?[/quote]

Actually, joml88 moved it to Intermediate because it was getting too hard. :D

That means I think I should start going to the intermediate forum more. :)

[quote="236factorial"][quote="chess64"]Actually, joml88 moved it to Intermediate because it was getting too hard. :D

That means I think I should start going to the intermediate forum more. :)[/quote]

Yeah, I definetly saw an increase in difficulty in the other marathon (as with all the marathons).[/quote]
Someone just post a problem.

At getting started level:  :D  (sorry G-UNIT, but Penguinintegral is impatient)

[size=200]Problem 3[/size]

I will quit AoPS in 300 days. Now it is August 12. What month will I quit?

[quote="aidan"]Problem 3:

300 days is approximately equal to 10 months. Since August 12th is close to the middle of the month, we can safely add 10 months to August and hence you will quit in June (or will you?). Now this is really getting started. Let's keep it near to this level...

[b][size=150]Problem 4[/size][/b]

A large watermelon weighs $20$ kg, with $98$% of its weight being water. It is left to stand in the sun, and some of the water evaporates, such that now, only $95$% of its weight is water. What does the watermelon now weigh?[/quote]

I could do this calculating the ratios easily.........but is there a more elegant way?[/quote]

[quote="236factorial"]What is $\sin{900^{\text{o}}}$?[/quote]

[hide]
$\sin {(900-720)}=\sin{180}=\sin {(2 \cdot 90)} = 2 \sin {90} \cos {90} = 0$
[/hide]

[size=150]Problem 6[/size]

Four positive integers $a, b, c,$ and $d$ have a product of $8!$ and satisfy
$ab + a + b = 524$,
$bc + b + c = 146$, and
$cd + c + d = 104.$

What is $a - d$?

[quote="236factorial"][hide].98-x=.95(1-x)
.03=.05x
x=3/5

3/5*20=12 kg[/hide]

[size=200]Problem 5[/size]

What is $\sin{900^{\text{o}}}$?[/quote]
I think there is a flaw in his solution, can someone confirm this?
sin900=sin540=sin180=0

Problem 6
A set S of points in the xy plane is symetic about the origin, both axes, and the line x=y. If the point 2,3 is in S, what is the minimum number of points in S?

[quote="mathwizarddude"]I'm not the author of the problem as it is from oneplusone, but I'll try to rephrase it according to my interpretation: Suppose there is a particle at point T, and a line $l$ at a distance of $10$ from T. Another particle is located at point $P\in l$, and $PT=d$. Suppose the particle at $T$ start traveling toward $l$ with speed $v$ (in a direction not necessarily perpendicular to $l$), and the particle at $P$ start traveling along line $l$ with speed $10v$. Find the minimum value of $d$ such that the two particles will meet on the line.[/quote]

I am sorry but the question is really hard to phrase properly. Tom is at $T$, and a line $l$ distance 10 from $T$. $P$ is on $l$ such that $PT=d$. Then the question is basically asking the minimum $d$ such that there exists a point $A$ on $l$ such that $\frac{PA}{AT}\geq 10$.

Note that $\frac{PA}{AT}=\frac{\sin\angle ATP}{\sin\angle APT}$ which is maximized when $\angle ATP=90$. Let $B$ be the foot of perpendicular from $T$ to $l$. Then by similar triangles $\frac{PA}{AT}=\frac{PT}{TB}\geq 10$, so $d=100$.

The new problem, does irreducible means that the polynomial cannot be expressed as a product of 2 non constant polynomials with integer coefficients?

[quote="oneplusone"]
The new problem, does irreducible means that the polynomial cannot be expressed as a product of 2 non constant polynomials with integer coefficients?[/quote]

Yes, a polynomial that can't be expressed as product of $n$ (and clearly $2$) non-constant polynomials is irreducible.
The coefficients depends on the problem, i.e. if the problem says $p(x)$ is irreducible over $\mathbb Z[x]$ it means prove that $p(x)$ can't be expressed as product of two non-constant polynomials with integer coefficients.

[quote="gouthamphilomath"]I request oneplusone to post the solution to that problem. Meanwhile, I am posting a...

[b][u]New problem:[/u][/b]

Prove that $p(x) = (x-1)^2(x-2)^2...(x-2010)^2+1$ is irreducible over $\mathbb{Z}$[/quote]

Solution:
Suppose $p(x)=f(x)g(x)$ can be reduced. Since $p(x)>0$, $f(x),g(x)$ must not have a root, so they are always positive or negative. WLOG let both of them be positive for all $x$. For $x=1,...,2010, f(x)g(x)=1$ thus $f(x)=g(x)=1$. consider the function $f(x)-1$ with 2010 roots from 1 to 2010. Thus $f(x)=(x-1)(x-2)...(x-2010)F(x)$ and similarly $g(x)=(x-1)...(x-2010)G(x)$. But since their product has a highest power of 4020, the sum of their highest power is 4020, thus $F(x)=a,G(x)=b$ are constants. Now we can just expand, and it is easy to check that their product cannot form $p(x)$.

New problem:
2 people are playing a game on an 8x8 chessboard with a knight. The knight moves in an L-shape. The first person puts the knight anywhere he wants. Then the players take turns (second player first) to move the knight in a square which the knight have not landed on yet. The loser is the person who cannot make a move. Who will win?

[quote="mathwizarddude"][quote="oneplusone"][quote="mathwizarddude"]I'm not the author of the problem as it is from oneplusone, but I'll try to rephrase it according to my interpretation: Suppose there is a particle at point T, and a line $l$ at a distance of $10$ from T. Another particle is located at point $P\in l$, and $PT=d$. Suppose the particle at $T$ start traveling toward $l$ with speed $v$ (in a direction not necessarily perpendicular to $l$), and the particle at $P$ start traveling along line $l$ with speed $10v$. Find the minimum value of $d$ such that the two particles will meet on the line.[/quote]

I am sorry but the question is really hard to phrase properly. Tom is at $T$, and a line $l$ distance 10 from $T$. $P$ is on $l$ such that $PT=d$. Then the question is basically asking the minimum $d$ such that there exists a point $A$ on $l$ such that $\frac{PA}{AT}\geq 10$.

Note that $\frac{PA}{AT}=\frac{\sin\angle ATP}{\sin\angle APT}$ which is maximized when $\angle ATP=90$. Let $B$ be the foot of perpendicular from $T$ to $l$. Then by similar triangles $\frac{PA}{AT}=\frac{PT}{TB}\geq 10$, so $d=100$.[/quote]
Why do we need $PA/AT$ maximized (also to maximize $PA/AT$, just having $\angle ATP=\pi/2$ would not suffice; we must also have $\angle APT\to0$)?
[quote="oneplusone"]Solution:
Suppose $p(x)=f(x)g(x)$ can be reduced. Since $p(x)>0$, $f(x),g(x)$ must not have a root, so they are always positive or negative. WLOG let both of them be positive for all $x$. For $x=1,...,2010, f(x)g(x)=1$ thus $f(x)=g(x)=1$. consider the function $f(x)-1$ with 2010 roots from 1 to 2010. Thus $f(x)=(x-1)(x-2)...(x-2010)F(x)$ and similarly $g(x)=(x-1)...(x-2010)G(x)$. But since their product has a highest power of 4020, the sum of their highest power is 4020, thus $F(x)=a,G(x)=b$ are constants. Now we can just expand, and it is easy to check that their product cannot form $p(x)$.[/quote]
Can't we also have $f(x)=g(x)=-1$ for some $x\in\{1,2,...,2010\}$? Also why is it WLOG to assume both $f$ and $g$ to be positive for all x? (Can't we also have both of them be negative for some x?)[/quote]

We need to maximize $\frac{PA}{AT}$ to make sure the maximum is at least 10. $\angle APT$ is fixed, we cannot change that.

If $f(x)=g(x)=-1$, we replace $f(x)$ by $-f(x)$ and replace $g(x)$ by $-g(x)$ and nothing else changes. If it can be negative, then by Intermediate Value Theorem there will be a root, so $p(x)$ will have a root.

Well I Think That Whatever You All Are Discussing
Can Be Stopped
While I'll Post A New Problem

Find The Function

$ f : N \to N $

$2f(x^2 + y^2) = f(x)^2 + f(y)^2 $

[hide="Solution Maaybe"]By Observtion $f(s)=1$ is oviously a solution.  Since the problem said find the funciton instead f find all functions it can be assumed there is only one answer.[/hide]

Someone post new problem. That is If I am correct.

well that is NOT the only solution
$f(x)=1 , 0 , 2x$ are also solutions
and sorry about "only the function"

I think there should be a real solution for this functional equation since we can just randomly choose numbers that work. For example f(x)=-1 works and neither of you have pointed it out.

$f(x) = -1 $ Doesn't work as RHS Has $f(x)^2$

If it does not work, I think you still get the point that we should not state answers but post solutions.

Are you sure the function $f:\mathbb{N}\to \mathbb{N}$? Is so, then $f(x)=0$ should not even be a solution.

And I think you all missed this question...

[quote="oneplusone"]
New problem:
2 people are playing a game on an 8x8 chessboard with a knight. The knight moves in an L-shape. The first person puts the knight anywhere he wants. Then the players take turns (second player first) to move the knight in a square which the knight have not landed on yet. The loser is the person who cannot make a move. Who will win?[/quote]


Which of the following describes the graph of the
equation (x + y)2 = x2 + y2 ?

A. the empty set
B one point
C. two lines
D. a circle
E. none of these

Are those supposed to be exponentiation or multiplication?  I'm assuming exponentiation, so $ (x\plus{}y)^{2}\equal{}x^{2}\plus{}2xy\plus{}y^{2}\equal{}x^{2}\plus{}y^{2}\Rightarrow 2xy\equal{}0$.  Then either $ x\equal{}0$ or $ y\equal{}0$.


[quote="Myth"]Is there some short commands, which can be used in Latex equation on forum?
I mean we very often write $\mathbb{N}$ or $\mathbb{Z}$, or something else. I think, it will be great thing if we will be allowed to use their short variants, '\N' for example.[/quote]

I think it will be the best if we propose a list of heavy used "long and tedious" commands and the shortcut which should be defined for it. Let's start. It is your turn to start, Myth.  ;)

[quote="Myth"]I think, it will be vain, since Valentin will say "I don't want to implement it" (at least, I am sure for 80% that it will be his answer).[/quote]

It is just command subsitution.  :)


analytic geometry
graphing lines
slope
logarithms
calculus
calculus computations

How many tangent lines can be drawn from the point $(a,\ b)$ to the curve $y=e^{x}$?

$y\;=\;e^{x}\;\;\; .............\Rightarrow\;\;\; \frac{dy}{dx}\;=\;e^{x}$

let us pick a general point on $e^{x}$, call it $(p,q)$

slope of tangent at that point is therefore $e^{p}$, and the coordinates are given by $(p,e^{p})$.

the equation of tangent is given by: $y\;-\;e^{p}\;=\;e^{p}\;(x-p)$

so, if the tangent passes through a point $(a,b)$, we will have: $b\;-\;e^{p}\;=\;e^{p}\;(a-p)$

now, we essentially [b][u]need[/u][/b] to solve the equation for $p$.
$b\;=\;e^{p}\;(1+a-p)\;\;\; .............\Rightarrow\;\;\; \ln b\;=\;p\;+\;\ln{(1+a-p)}$

i can analyze 3 distinct cases by looking at the equation:
$\textbf{(i)}\;\;\;\quad if\;\;\;b\leq 0\;\;\;\Rightarrow\;\;\;we\;\;will\;\;always\;\;get\;\;1\;\;tangent\;\;line\;\;only,$
$regardless\;\;of\;\;whatever\;\;value\;\;a\;\;might\;\;be$

$\textbf{(ii)}\;\;\;\quad if\;\;\;b>0,\;\; and \;\;a\;\;is\;\;chosen\;\;such\;\;that\;\;the\;\;point\;\;lies\;\;to\;\;the$
$\textbf{right}\;\;of\;\;e^{x},\;\;i.e.\;\;b<e^{a},\;\;then\;\;we\;\;will\;\;always\;\;get\;\;2\;\;tangent\;\;lines$

$\textbf{(iii)}\;\;\;\quad if\;\;\;b>0,\;\; and \;\;a\;\;is\;\;chosen\;\;such\;\;that\;\;the\;\;point\;\;lies\;\;to\;\;the$
$\textbf{left}\;\;of\;\;e^{x},\;\;i.e.\;\;b>e^{a},\;\;then\;\;we\;\;will\;\;have\;no\;\;tangent\;\;lines\;\;at\;\;all$


geometry
topology
analytic geometry
advanced fields
advanced fields unsolved

The problem is indeed an exercise in spivak's 'a comprehensive introduction to differential geometry' vol 1, chap 2, exercise 24.
I can do 24(a) and a half part of (b), but got stuck on the other half....
I first gave the result of 24(a), you can just use it:
If M is a $ C^\infty$ manifold, a set $ M_1$ in M can be made into a k-dimensional submanifold of M if and only if around each point in $ M_1$ there is a coordinate system $ (x,U)$ on M such that $ M_1 \cap U\equal{}\{p: \quad x^{k\plus{}1}(p)\equal{}\cdots\equal{}x^n(p)\equal{}0\}$.

Now here is the second part of the question which I got stuck:

If the subset $ M_1$ can be made into a closed submanifold, then the coordinate systems mentioned above exist around every point of M (not just $ M_1$).

My idea is that suppose M is connected, then let $ W\equal{} \bigcup U$, where U is the coordinates mentioned above, then W is an also a submanifold in M containing $ M_1$, W is open obviously, I want to show W is also closed so that W=M, then we are done.  But I got stuck here, maybe my idea was wrong.....

Please note that in Spivak's book, a closed submanifold $ M_1$ in M is defined as:
(a) an immersed submanifold whose inclusion map is also an embedding, and:
(b) the subset $ M_1$ is a closed subset of M.

Namely, a submanifold is just an embedding submanifold, and a closed submanifold is an embedding submanifold with itself is a closed set in the whole space.

Thank you in advance! Hope some guys who are familar with Spivak's book can help me~

I would call that a (closed) embedded submanifold; leaving that keyword off confuses the issue with mere immersed submanifolds. I did learn from a different textbook.

Here's the thing: coordinate systems are local, and aren't defined on all of $ M$. For points not on $ M_1$, just choose $ U$ not intersecting $ M_1$, and coordinates that take values in some half-space missing that subset.

If you require coordinates to cover all of $ \mathbb{R}^n$ (a very impractical restriction if you're going to compute with those coordinates), you have to glue together a path from $ M_1$ to $ p$, and this necessitates the extra assumption that $ M$ is connected. For that method, your approach is at least partially right. For the next step, I think you have to have to do the actual gluing at least once. Take one such $ U$ and a point $ p$ with a charted neighborhood $ V$ intersecting $ U$ but not $ M_1$, prune $ U$ back to $ U'$ intersecting $ U$ simply, and then get coordinates on $ U'\cap V$.

[quote="frankmelody"]If M is a $ C^\infty$ manifold, a set $ M_1$ in M can be made into a k-dimensional submanifold of M if and only if around each point in $ M_1$ there is a coordinate system $ (x,U)$ on M such that $ M_1 \cap U \equal{} \{p: \quad x^{k \plus{} 1}(p) \equal{} \cdots \equal{} x^n(p) \equal{} 0\}$.[/quote]Note that this implies that $ M_1$ is locally closed in $ M$.
[quote]If the subset $ M_1$ can be made into a closed submanifold, then the coordinate systems mentioned above exist around every point of M (not just $ M_1$).[/quote]Take some $ p\notin M_1$. What if you choose a chart $ (U,x)$ with $ p\in U$ such that $ U$ does not meet $ M_1$? By the assumption that $ M_1$ be closed this choice is possible.
[quote]My idea is that suppose M is connected, then let $ W \equal{} \bigcup U$, where U is the coordinates mentioned above, then W is an also a submanifold in M containing $ M_1$, W is open obviously, I want to show W is also closed so that W=M, then we are done.[/quote]This cannot be proven in this situation - due to the counterexample as follows: Take the closed submanifold $ \mathbb R\times\{0\}$ in $ \mathbb R^2$ and $ W\equal{}U\equal{}\mathbb R\times [\minus{}1,1]$ with the obvious chart. Then $ W$ is not $ \mathbb R^2$!

Note that it is not possible to expect things (here: open subsets) locally defined around every point of a proper closed subset to have such a "range of coverage" of all of $ M$. Sometimes some permanence properties - like analyticity - can do this job.

I hope that these comments can be helpful. :)

... correction of typo: The counterexample should certainly read as follows:[quote="Third Edition"]Take the closed submanifold $ \mathbb R\times\{0\}$ in $ \mathbb R^2$ and $ W \equal{} U \equal{} \mathbb R\times (\minus{} 1,1)$ with the obvious chart.[/quote]


$\frac{dy}{dx}=sin(y)$
Does this equation has a closed form solution?


Find the number of ways of distributing [b]r[/b] candies

among [b]n[/b] children so that at least [b]m[/b] of the children

receive at least one candy each

Solution.We shall all through assume $ r\geq m$

[i] Candies are identical and children are distinct.

$ \sum_{k \equal{} m}^{min(n,r)} {r \minus{} 1\choose {k \minus{} 1}}{n\choose k}$

[ii] Candies are distinct and children are distinct.

$ n^r \minus{} \sum_{k \equal{} 1}^{k \equal{} m \minus{} 1} ( \minus{} 1)^{m \minus{} 1 \minus{} k}{n\choose k}k^r$

[iii] Candies are distinct and children are identical.

$ \bigg( \sum_{h \equal{} m}^{h \equal{} min(n,r)} \sum_{k \equal{} 0}^{k \equal{} h \minus{} 1} \frac {( \minus{} 1)^k}{r!} {h\choose k} (h \minus{} k)^r \bigg) \minus{} \bigg( \sum_{h \equal{} 1}^{h \equal{} m \minus{} 1} \sum_{k \equal{} 0}^{k \equal{} h \minus{} 1} \frac {( \minus{} 1)^k}{r!} {h\choose k} (h \minus{} k)^r \bigg)$

[iv] Candies are identical and children are identical.

$ \sum_{h \equal{} m}^{h \equal{} min(n,r)} P(r,h)$
where $ P(r,h)$ denotes the number of partitions of r into h parts.


modular arithmetic
number theory
relatively prime
algebra proposed
algebra

Prove that for all positive integers $n$, $n\geq 2$, the
number:
\[\frac{3^{n}-2^{n}}{n}\]
cannot be an integer number.

Laurentiu Panaitopol, TST 1987

I just have a question before I post my probable soln... If we prove this result ( and also in similar other questions) for primes, is it automatically or through some reasoning, true for all nos..???

If $p$ is prime, it is quite easy.
But if not, how to think?

My idea:
represented $3^{n}$ by $(2+1)^{n}$ . then use residue to consider the coefficients in every terms.

Let $p$ be the smallest prime factor of $n$.  We will prove the stronger claim $3^n \not\equiv 2^n \pmod{p}$.

If $p$ is 2 or 3, then one side is a multiple of $p$ and the other is not, so we are done.  So assume $p > 3$.

Let $x$ be a solution to $2x \equiv 3 \pmod{p}$.  Let $d$ be the order of $x$ modulo $p$.  By Fermat's Little Theorem, $d \le p - 1$.  Furthermore $d$ is bigger than 1 because $2 \not\equiv 3 \pmod{p}$.  Because all the prime factors of $n$ are bigger than $d$, $n$ must be relatively prime to $d$, and in particular $n$ cannot be a multiple of $d$.  Thus $x^n \not\equiv 1 \pmod{p}$.  Multiplying by $2^n$, we get $3^n \not\equiv 2^n \pmod{p}$.


$ 361\equal{}19^2$, so $ 19*(19\plus{}12)\equal{}589\plus{}36\equal{}625$. I'm just not sure how to do that fast in your head....


function
inequalities
inequalities theorems

Let $ x_{1},x_{2},\cdots,x_{n}$ and $ y_{1},y_{2},\cdots,y_{n}$ are two sets of nonnegative quantities. Now, if we have all the following conditions fulfilled:

$ x_{1} + x_{2} + \cdots + x_{n}\leq y_{1} + y_{2} + \cdots + y_{n},$

$ \sum_{i < j}x_{i}x_{j}\leq\sum_{i < j}y_{i}y_{j},$

$ \vdots$

$ x_{1}x_{2}\cdots x_{n}\leq y_{1}y_{2}\cdots y_{n},$

(generally the $ r$-th elementary symmetric functions $ E_{r}(x)\leq E_{r}(y)$ for any natural number $ r$ with $ 1 \leq r \leq n$)

then $ \sum_{i = 1}^{n} x_{i}^p\leq \sum_{i = 1}^{n} y_{i}^p$ for $ 0 < p < 1$.

[hide="Proof."] Let $ \sum_{i = 1}^n{a_{i}^p} = F(E_1,E_2,\cdots,E_n),(0 < p < 1)$,

where $ E_{r} = \sum_{1\leq i_{1} < i_{2} < \cdots < i_{r}\leq n}{a_{i_{1}}a_{i_{2}}\cdots a_{i_{r}}},(1\leq r\leq n)$.

Then $ \frac {\partial a_{i}}{\partial E_{r}} = \frac {( - 1)^{r - 1}a_{i}^{n - r}}{\prod_{j\neq i}(a_{i} - a_{j})},(1\leq i\leq n)$,

$ \frac {\partial F}{\partial E_{r}} = \sum_{i = 1}^n{pa_{i}^{p - 1}\frac {\partial a_{i}}{\partial E_{r}} = p( - 1)^{r - 1}\sum_{i = 1}^n\frac {a_{i}^{p + n - r - 1}}{\prod_{j\neq i}(a_{i} - a_{j})} > 0}$,

where $ 0 < p < 1\Longrightarrow n - r - 1 < p + n - r - 1 < n - r$,

and the last inequality see: http://www.mathlinks.ro/Forum/viewtopic.php?t=194070[/hide]

[quote="Ji Chen"]

where $ E_{r} \equal{} \sum_{1\leq i_{1} < i_{2} < \cdots < i_{r}\leq n}{a_{i_{1}}a_{i_{2}}\cdots a_{i_{r}}},(1\leq r\leq n)$.

Then $ \frac {\partial a_{i}}{\partial E_{r}} \equal{} \frac {( \minus{} 1)^{r \minus{} 1}a_{i}^{n \minus{} r}}{\prod_{j\neq i}(a_{i} \minus{} a_{j})},(1\leq i\leq n)$,

[/quote]
I am the beginer of the inequality. Can you explain it?

I am the beginer of the inequality, who will help me? :maybe:


Angela, whose weight ($ F_g$) is $ 400$ newtons, sits on a tire swing in her backyard.  Linda pulls Angela horizontally to the right so that the rope of the swing makes a $ 30^{\circ}$ angle with the horizontal.  If Angela is at rest in this position, what is the tension ($ T$) in the rope and what force ($ P$) does Linda exert on Angela?

Thanks in advance!

Hint: Linda must be pretty strong. With the rope closer to horizontal than vertical, the force Linda exerts, $ P,$ will be greater than Angela's weight.

(Further hint: consider horizontal and vertical components.)

I still don't get it.  I think that there are three forces exerted on Angela, $ T$, $ P$, and $ F_g$.  However, I don't know how to get the vector equation and then use it to solve the problem.

The gravitational force on Angela is a purely vertical downward force of $ 400$ Newtons. Linda exerts a force on Angela that is purely horizontal, of magnitude $ P.$ The rope exerts a force $ T$ on Angela. If we resolve this into components, we get an upward vertical force of $ T\sin 30^{\circ}$ and a horizontal force of $ T\cos 30^{\circ}.$

The horizontal forces must balance and the vertical forces must balance. This gives you two equations for the two unknowns $ T$ and $ P.$

[hide="Solution"][b]Horizontal Components[/b]
$ P \minus{} T\cos 30^{\circ} \equal{} 0$

[b]Vertical Components[/b]
$ T\sin 30^{\circ} \minus{} F_g \equal{} 0$

here $ F_g \equal{} 400 N$

Here no need for considering Torque
solving,we get,$ T\equal{}800$ and $ P\equal{}400\sqrt{3}$

I may be wrong as i am pathetic in calculations :oops:

[/hide]

That is correct. As I said, Linda must be pretty strong.


How do you calculate the force of gravity inside a planet? Inside the planet Earth, for instance, there is some mass above us and around us.

integrate
like um
let ds be a bit of mass-calculate the mass of ds from the density of the sphere, and calculate the force f=g*m1*m2/r^2. then integrate all fs and you get the overall force.
like you do for charge distributions

could you give a more specific example please :blush: ??????


How many divisors of 8! are larger than 7!?

I have no clue how to do this... any hints?[/hide]

[hide="hint"] Try breaking it up into "factor pairs".  [/hide]
If you have thought about it and still can't solve it, then here is the solution:
[hide="solution"] We can look at a factor pair in the notation of (x,y) where x<y.  The first few will be
\[(1,8!) \]

\[(2, \frac{8!}{2}) \]

\[(3, \frac{8!}{3}) \]

\[(4, \frac{8!}{4}) \]

\[(5, \frac{8!}{5}) \]

\[(6, \frac{8!}{6}) \]

\[(7, \frac{8!}{7}) \]
Now, notice that when the "x" term becomes greater than or equal to 8, the "y" term will become less than or equal to $8!/8=7!$.  Thus, we can only take these $7$ factor pairs into account, which contain 7 factors greater than 7!.  [/hide]

[hide]um, well 7! = 7*6*5*4*3*2 sooooo what's larger than that is...

8! or when 8 replaces either 7, 6, 5, 4, 3, 2 (ie 8*7*5*4*3*2). Count up[b] 7[/b]. :D[/hide]


What is the mechanism of Amadori rearrangement?

This is a nice link:

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

Is the Amadori Rearrangement just a curious observation or is it useful in synthesis?

Its obviously used in the formation of glucosazone  :wink:

Well fair enough but how applicable is it in synthesis?

Why are the powerpoint and the description in Wikipedia different?

No, this rearrangement is not used in synthesis.


geometry
incenter
circumcircle
geometry unsolved

Prove that the incenter, circumcenter and the point of intersection of the diagonals of the quadrilateral are
    collinear.


Let x, y be positive numbers, 's' the least of { x, (y+1/x), 1/y}. What is the greatest possible values of 's' and what are the corresponding values of x and y?

please give a detail solution.
thank you... :)

Is the middle term $ \frac{y\plus{}1}{x}$, or $ y\plus{}\frac{1}{x}$?

[hide]
Let x=1/a y=b

$ {\frac{1}{a},\frac{1}{b},a\plus{}b}$

If $ \frac{1}{a}$ is the smallest one, we have

$ \frac{1}{a}\le\frac{1}{b}$ so $ b \le a$ and $ \frac{1}{a}\le a\plus{}b\le2a$ so $ \frac{1}{\sqrt{2}}\le a$ and $ \frac{1}{a}\le\sqrt{2}$ w/ equality happening when a=b and a equals $ \frac{1}{\sqrt{2}}$ so consequently they both equal it.

Now, if $ a\plus{}b$ is the smallest number and is bigger than $ \sqrt{2}$ then WLOG $ a\ge\frac{\sqrt{2}}{2}$ and $ b\le\frac{\sqrt{2}}{2}$ so $ \frac{1}{b}\ge\sqrt{2}$ w/ equality iff $ a\equal{}b\equal{}\frac{\sqrt{2}}{2}\equal{}\frac{1}{\sqrt{2}}$

Therefore, we have $ x\equal{}\sqrt{2}$ and $ y\equal{}\frac{1}{\sqrt{2}}$ with the minimal value of s being $ \sqrt{2}$
[/hide]

[quote="Brut3Forc3"]Is the middle term $ \frac {y \plus{} 1}{x}$, or $ y \plus{} \frac {1}{x}$?[/quote]

it is the second expression


Alright, so I have algebra homework on the second day of school(yeah, I know). I'm in algebra 1 and have no clue what my teachers are talking about. So here are a few questions; you DON'T have to answer them I just need to know how to solve them, if you want to answer, be my guest.

[hide]Evaluate each expression if x = -3, y = 6, z = -4

1. -(x + (-y) + z)(letters italicized) =

2. 3x - 2y + z =

3. 2yz - 3x =

4. 4x - y - 3xz(letters italicized)

5.(below)

 xz

 ----

 -2y[/hide]

After, if you can help me with Combining Like Terms, please do so if you have the time.

Why weren't you in school?

All you have to do to evaluate the expressions is:
1. Replace the variables(letters) with the value that is assigned to them($ x \equal{}\minus{}3, y \equal{} 6, z \equal{}\minus{}4$).
2. Use order of operations to compute each equation.

well, my teacher taught it this way combining like terms. so we have something like $ 4x\plus{}5x^{2}\minus{}3\plus{}16x^{2}\plus{}14x^{3}\minus{}2x$. so what you do is you give every monomail(look it up if u dont no wht it is) an item/animal name. so lets just say constants are penguins, $ x$'s are pandas, $ x^{2}$ are duckys, and $ x^{3}$ are firetrucks. first we look at the penguins. all we have are negative 3 penguins. so we leave it be. then we have the pandas. we have 4 pandas minus 2 pandas equals 2 pandas. do that with duckies. 5 duckies plus 16 duckies equals 21 duckies. and we leave the firetrucks alone because theres only one set of them. so the answer is $ 14x^{3}\plus{}21x^{2}\plus{}2x\minus{}3$. once you get used to it, you wont need duckiess or pandas or firetrucks to help you.


The fraction $\frac{2}{15}$ can be expressed as $\frac{1}{x} + \frac{1}{y}$, where $x$ and $y$ are positive integers with $x > y$, in 2 ways, $\frac{1}{30} + \frac{1}{10}$ and $\frac{1}{120} + \frac{1}{8}$.

Find the ordered pair of positive integers $(x, y)$ with $x > y$, such that $\frac{2}{17} = \frac{1}{x} + \frac{1}{y}$.

If anyone could show me how to solve this problem with a full solution, it'd be greatly appreciated

[hide]Combine the fractions first:

$\frac{x + y}{xy} = \frac{2}{17} \rightarrow x + y = \frac{2xy}{17}$

Since we're dealing with integers, either x or y has to be a multiple of 17.  If either is 17, the other is also 17, which violates the condition $x > y$.  The only integer solution is (153,9).[/hide]

nice solution, and thanks - but how did you get the numbers 153 and 9? Is there some way to obtain those numbers? If so, can you show me how to get that?

Well, I kind of saw it by intuition, lol.  But I'll try to explain the reasoning behind it.  Let's say $x$ is a multiple of 17.  It can then be represented as $17k$, where $k$ is an integer greater than or equal to 1.  So:

$17k + y = 2ky \rightarrow 17k = (2k - 1)y$

Remember, $k$ and $y$ are integers.  So either $2k-1$ or $y$ is divisble by 17.  It can't be $y$.  If it was, you'll end up with the case we already covered.  Say $y = 17a$, then:

$k = (2k-1)a \rightarrow a \equiv 0 \pmod{k}$

$a = k \rightarrow 17k = (2k - 1)17k \rightarrow k = 1$

So that means $2k-1$ has to be divisble by 17.

$2k - 1 = 17 \rightarrow k = 9$

$y = k = 9$

If $2k-1$ was a greater multiple of 17, either $k$ or $y$ would not be an integer, which is not permissible.  So there is only that one solution.

Hmm...I'm not sure you showed that both x and y couldn't be divisible by 17.  You made the statement that if both were 17, it is extraneous. Here is my work for the case in which both are divisible by 17, if this is redundant then oh well.
let $x = 17a$ and $y=17b$, where $a > b$
Furthermore, let $a = b+k$, where $k > 0$
$a+b = 2ab$
$2b+k = 2b^2+2bk$
$2b^2+b(2k-2)-k = 0$
$b = \frac{2-2k\pm\sqrt{4k^2+4}}{4}$
$b = \frac{1-k\pm\sqrt{k^2+1}}{2}$
As $b$ is an integer, and $k^2+1$ is not a perfect square for $k>0$
17 cannot both divide $x$ and $y$.
I'm certain there is a more streamlined approach.


integration
function
Functional Analysis
real analysis
real analysis unsolved

$f(x)$ is continuous on $[0,1]$. $\int_0^1 f(x) dx = \int_0^1 x f(x) dx = 0$. $\int_0^1 f(x)^2 dx = 1$. Find the maximum value of $\int_0^1 x^2f(x)dx$.

Let $e_1(x),e_2(x),e_3(x)$ be the functions we get by applying the Gram-Schmidt orthogonalization process to $1,x,x^2$ in the Hilbert space $L^2([0,1])$ with the usual norm $\langle f,g\rangle=\int^1_0fg\ \mbox dx$. Explicitly, we have $e_1(x)=1,e_2(x)=2\sqrt 3x-\sqrt 3$, and $e_3(x)=6\sqrt 5x^2-6\sqrt 5x+\sqrt 5$ (I used Mathematica).

We have $1=\|f\|^2\ge\langle f,e_1\rangle^2+\langle f,e_2\rangle^2+\langle f,e_3\rangle^2=\langle f,e_3\rangle^2$ (for all the functions we're interested in, i.e. those satisfying the given hypotheses), which means that the maximum of $\langle f,e_3\rangle$ is $1$ and it can be reached when $f=e_3$. Since it's clear that $\langle f,e_3\rangle=6\sqrt 5\langle f,x^2\rangle$ for all functions satisfying the given hypotheses, it means that the maximum we're looking for is $\frac 1{6\sqrt 5}$.

Is the answer correct, liyi?


group theory
number theory
prime factorization
superior algebra
superior algebra unsolved

Hello people!

I've quite a big problem with the following one:

Let G be a cyclic group of prime order p. Is AutG cyclic?

I've got to the point where I see that AutG is isomorphic to the multiplicative group $P_{p}=\{1,...,p-1\}$, so the question is equivalent to asking if $P_{p}$ is cyclic. But I'm quite blind beyond that.

The integers mod $p$ form a field. It is a standard theorem that any finite multiplicative group in a field is cyclic.

Thank you! I didn't know that theorem but now I know where to look for a proof.

A cool question is, what about finite cyclic groups in general : when has $\mathbb{Z}_m,+)$ a cylic automorphism group.

It is a well-known result, that $Aut(\mathbb{Z}/n) \cong (\mathbb{Z}/n)^*$ is cyclic if and only if $n=1,2,4,p^k$ or $2 p^k$ for a prime $p \geq 3$ and $k \geq 1$. One can even show:

If $n = \prod p^{v_p}$ is the prime factorization of $n$, then
\[ (\mathbb{Z}/n)^* \cong \prod (\mathbb{Z}/p^{v_p})^*  \]
And these factors are
$(\mathbb{Z}/2)^* \cong {1}$
$k > 1 \Rightarrow (\mathbb{Z}/2^k)^* = \span(-1) \oplus \span(5) \cong \mathbb{Z}/2 \times \mathbb{Z}/2^{k-1}$
$p > 2 \Rightarrow (\mathbb{Z}/p^k)^* \cong \mathbb{Z}/\varphi(p^k)$


trigonometry
geometry
3D geometry
maximization
IMO
IMO 1979

We consider a point $P$ in a plane $p$ and a point $Q \not\in p$. Determine all the points $R$ from $p$ for which \[ \frac{QP+PR}{QR}  \] is maximum.

hmm, do you mean that the fraction is maximal out of the 3 combinations given P,Q,R or it's maximal considering all points R? If it's the latter, then wouldn't R have to be points infinitely close to Q, so then the fraction would be infinitely large? I'm confused...

[quote="al.M.V."]hmm, do you mean that the fraction is maximal out of the 3 combinations given P,Q,R or it's maximal considering all points R?[/quote]

Maximal considering all points R on the plane p. The points P and Q are fixed.

[quote="al.M.V."]If it's the latter, then wouldn't R have to be points infinitely close to Q, so then the fraction would be infinitely large? I'm confused...[/quote]

The point R cannot come infinitely close to Q, since the point Q doesn't lie on the plane p (this condition was formerly invisible because of a LaTeX error in Orl's post; now I corrected it), while the point R should lie on p.

Anyway, here is my solution of the problem:

Let T be the orthogonal projection of the point Q on the plane p. Then, the line PT is the orthogonal projection of the line PQ on the plane p, and thus forms the least angle with the line PQ among all lines through the point P which lie in the plane p; hence, $\measuredangle RPQ\geq\measuredangle TPQ$, and equality holds if and only if the point R lies on the ray PT (the only exception is when $PQ\perp p$; in this case, P = T, so the ray PT is undefined, and equality holds for all points R in the plane p, since we always have < RPQ = < TPQ = 90°).

From $\measuredangle RPQ\geq\measuredangle TPQ$, it follows that $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$; also, since the angles < RPQ and < TPQ are < 180°, their half-angles $\frac{\measuredangle RPQ}{2}$ and $\frac{\measuredangle TPQ}{2}$ are < 90°, so that from $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ we can conclude that $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$. Equality holds, as in the above, if and only if the point R lies on the ray PT (and, respectively, for all points R in the plane p if $PQ\perp p$).

On the other hand, we obviously have $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ with equality if and only if $\frac{\measuredangle QRP-\measuredangle PQR}{2}=0^{\circ}$, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base QR, i. e. if and only if PR = PQ, i. e. if and only if the point R lies on the sphere with center P and radius PQ.

Now, applying the Mollweide theorem in triangle QPR, we get

$\frac{QP+PR}{QR}=\frac{\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}}{\sin\frac{\measuredangle RPQ}{2}}$
$\leq\frac{1}{\sin\frac{\measuredangle RPQ}{2}}$                   (since $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$)
$\leq\frac{1}{\sin\frac{\measuredangle TPQ}{2}}$                   (since $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$),

and equality holds here if and only if equality holds in both of the inequalities $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ and $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ that we have used, i. e. if and only if the point R lies both on the ray PT (this condition should be ignored if $PQ\perp p$) and on the sphere with center P and radius PQ.

Hence, the point R for which the ratio $\frac{QP+PR}{QR}$ is maximum is the point of intersection of the ray PT with the sphere with center P and radius PQ (or, respectively, it can be any arbitrary point on the intersection of the plane p with the sphere with center P and radius PQ if $PQ\perp p$).

  Darij

Calculus is not supposed to be used, but let me 'borrow' it anyway:

Assume $R'$ is the point we are looking for instead of $R$. Let $H$ be the foot of $Q$ on plane $[i][b]p[/b][/i]$. Draw ray $P$[i]y[/i] that passes through $H$. Let [i]a[/i] = $PQ$, [i]b[/i] = $QR'$, [i]c[/i] = $PH$, [i]x[/i] = $PR'$, α = ∠$QPR'$, β = ∠$QPH$ and $[i]r[/i]$ = ($QP + PR'$)/$QR'$ = ([i]a[/i] + [i]x[/i])/[i]b[/i].
The ratio [i]r[/i] is maximum when [i]r[/i]² = ([i]a[/i] + [i]x[/i])²/[i]b[/i]² is also maximum. However, the law of cosines gives us ([i]a[/i] + [i]x[/i])²/[i]b[/i]² = ([i]a[/i]² + [i]x[/i]² + 2[i]ax[/i])/([i]a[/i]² + [i]x[/i]² – 2[i]ax[/i]×cosα).

Observe that angle α is independent of variable [i]x[/i], and for the ratio [i]r²[/i] to be maximum, cosα has to be maximum which occurs when α = β. Furthermore cosβ = [i]c[/i]/[i]a[/i], and ([i]a[/i]² + [i]x[/i]² + 2[i]ax[/i])/([i]a[/i]² + [i]x[/i]² – 2[i]ax[/i]×cosβ) = ([i]a[/i]² + [i]x[/i]² + 2[i]ax[/i])/([i]a[/i]² + [i]x[/i]² – 2[i]cx[/i]).

Note that [i]c[/i] is also a constant, and the only variable in ratio [i]r²[/i] is [i]x[/i]. To get its maximum, take the derivative of the ratio [i]r[/i] with respect to [i]x[/i] and set it to zero to get [i]x[/i] = [i]a[/i], and point $R$ is on ray $P$[i]y[/i] with $PR = PQ$.

[quote=darij grinberg]
Anyway, here is my solution of the problem:

Let T be the orthogonal projection of the point Q on the plane p. Then, the line PT is the orthogonal projection of the line PQ on the plane p, and thus forms the least angle with the line PQ among all lines through the point P which lie in the plane p; hence, $\measuredangle RPQ\geq\measuredangle TPQ$, and equality holds if and only if [b]the point R lies on the ray PT[/b] (the only exception is when $PQ\perp p$; in this case, P = T, so the ray PT is undefined, and equality holds for all points R in the plane p, since we always have < RPQ = < TPQ = 90°).[/quote]

then:

[quote]Now, applying the Mollweide theorem in triangle QPR, we get:
$\frac{QP+PR}{QR}=\frac{cos \frac{\angle QRP -  \angle PQR}{2}}{sin \frac{\angle RPQ}{2}}$
[/quote]

Because R lies on the ray of PT, so $\angle RPQ = \angle TPQ$. If T=P, $\angle RPQ=90^O$. Also in both cases $\angle QPR$ is fixed. So:
$\frac{QP+PR}{QR}$ is max, if and only if $\angle QRP =  \angle PQR \iff PR = PQ$.

The answer:
* R is on the ray of PT, where T is  the orthogonal projection of the point Q on the plane p, and |PR|=|PQ|
*Special case: T=P --> R is on the circumference of the circle on plane p with center P and radius |PQ|

Here is different approach to prove that $R$ lies on line $PT$, when $T\ne P$ (I took $T$ as projection of $Q$ onto the plane $p$ as others).
Let $f(X)=\frac{QP+PX}{QX}$. Assume $R$ doesnt lie on $PT$. Take point $R_0$ on ray $PT$ such that $R_0$ doesn't lie on segment $PT$ and

$PR-PT<R_0T<RT$. We can choose such $R_0$ because $PR-PT<RT$ from triangle inequality on $\triangle PRT$. $$QR^2=QT^2+TR^2>QT^2+TR_0^2=QR_0^2 \implies QR>QR_0$$ $$PR_0=TR_0+PT>PR \implies PR_0>PR$$ So we get $f(R_0)>f(R)$ which is contradiction. So $R$ should lie on line $PT$.

Note: Since the rest of my sol is almost same with others, I don't post it.


geometry
calculus
integration
analytic geometry
trigonometry
calculus computations

Compute $ \int \int_D \,dx\,dy$, where $ D \equal{} \{(x,y) \in \mathbb{R} ^2: x^2 \plus{} y^2 \minus{} 2x \geq 0,\,\, x^2\plus{}y^2 \leq 4, \,\, y \leq \sqrt{3} x \}$.

In polar coordinates:

$ x^2 \plus{} y^2 \minus{} 2x\ge0$ becomes $ r^2 \minus{} 2r\cos\theta\ge0.$ Adopt the convention that $ r\ge 0$ always (which is the most useful convention for integration, since you don't want to double-count) and this become $ r\ge2\cos\theta.$ (Really, $ r\ge\max(2\cos\theta,0).$)

$ x^2 \plus{} y^2\le 4$ becomes $ r\le 2$.

$ y\le\sqrt {3}x$ under the $ r\ge 0$ assumption could be written in several ways, including $ \minus{} \frac {2\pi}3\le\theta\le\frac {\pi}3.$

Doing the integral in polar coordinates is the right thing to do. I'll let someone else finish it.

Or we could do this by pure geometry with no integration at all: two circles, both of the same radius, where what you want is inside one and outside the other, cut by a line at a nice angle -- draw the picture and see what you can do with that. (Hint for the picture: both $ (0,0)$ and $ (1,\sqrt {3}\,)$ are points of triple intersection.)


YAAYYYY!!!!!!  :D

[url]http://www.bubblebox.com/play/puzzle/975.htm[/url]

yeah i've played this game i think i got to like level 8 and stopped  :D

i got to that first level with the three flaggies and the i stopped.

level 8 I just draw shapes and watch them fall


[b]Simplify the following: (remember to use order of operation A.K.A.

PEMDAS)

x^6-x+x^2/x*5x-4x^3x+x^4[/b]

Is this what you mean:
$x^6-x+x^2%Error. "divx" is a bad command.
*5x-4x^3+x^4$?

[quote="ragnarok23"]I think he means:
$\frac{x^6-x+x^2}{x}\times$ $5x-4x^3+x+x^4$[/quote]
ok then:
[hide]$x^5+x^4-4x^3+5x^2+2x-1$[/hide]


inequalities
function
trigonometry
calculus
derivative
calculus computations

Prove that $ \cos (\sin x)\geq 1\minus{}\frac{1}{2}x^2$ for any real numbers $ x.$

Let $ \mathrm{f}(x) \equal{} \cos{(\sin{x})}$ and $ \mathrm{g}(x) \equal{} 1 \minus{} \frac {1}{2}x^2$.

Now

$ \mathrm{f}'(x) \equal{} \minus{} \sin{(\sin{x})}\cos{x}$

$ \mathrm{f}''(x) \equal{} \minus{} \cos{(\sin{x})}\cos^2{x} \plus{} \sin{(\sin{x})}\sin{x}$

Among the points where $ \mathrm{f}'(x)$ vanishes are $ x \equal{} 0$ and $ x \equal{} \frac {\pi}{2}$. Since $ \mathrm{f}''(0) < 0$ and $ \mathrm{f}''\left(\frac {\pi}{2}\right) > 0$, $ \mathrm{f}(x)$ has a maximum at $ x \equal{} 0$ and a minimum at $ x \equal{} \frac {\pi}{2}$. Thus $ \cos{1}\leq \mathrm{f}(x) \leq1$ for $ 0\leq x\leq\frac {\pi}{2}$.

Also, since $ \mathrm{f}\left(\frac {\pi}{2} \plus{} x\right) \equal{} \mathrm{f}\left(\frac {\pi}{2} \minus{} x\right)$ and $ \mathrm{f}(x \plus{} \pi) \equal{} \mathrm{f}(x)$, f is symmetrical about $ x \equal{} \frac {\pi}{2}$ and has period $ \pi$; hence we have determined the whole of the function f just from $ 0\leq x\leq\frac {\pi}{2}$ i.e. $ \cos{1}\leq \mathrm{f}(x) \leq1$ for all $ x$.

When $ x\in\left[0,\frac {\pi}{2}\right]$, $ \sin{x}\leq{x}$. $ \therefore\ \sin{(\sin{x})}\leq\sin{x}$
$ \Rightarrow\ \sin{(\sin{x})}\cos{x}\leq\sin{x}\cos{x} \equal{} \frac {1}{2}\sin{2x}\leq\frac {1}{2}\,(2x) \equal{} x$
$ \Rightarrow\ \mathrm{f}'(x) \equal{} \minus{} \sin{(\sin{x})}\cos{x}\geq \minus{} x \equal{} \mathrm{g}'(x)$

Hence, $ \forall\,x\in\left[0,\frac {\pi}{2}\right]$, $ \mathrm{f}(x) \minus{} \mathrm{f}(0)\geq \mathrm{g}(x) \minus{} \mathrm{g}(0)$, i.e. $ \mathrm{f}(x)\minus{}1\geq \mathrm{g}(x)\minus{}1$, i.e. $ \mathrm{f}(x)\geq \mathrm{g}(x)$.

For $ x > \frac {\pi}{2}$, $ \mathrm{g}(x)$ is negative whereas $ \mathrm{f}(x)$ is always positive, so $ \mathrm{f}(x)\geq \mathrm{g}(x)$ again. Thus $ \mathrm{f}(x)\geq \mathrm{g}(x)$ for all $ x\geq0$. Which means, since f and g are both even functions, that $ \mathrm{f}(x)\geq \mathrm{g}(x)$ for all $ x\in\mathbb{R}$.

[quote="Jane_Bennet"]Thus $ \cos{1}\leq \mathrm{f}(x) \leq1$ for $ 0\leq x\leq\frac {\pi}{2}$.[/quote]

This doesn't follow: having [i]a[/i] maximum at $ x \equal{} 0$ doesn't imply that [i]the[/i] maximum of the function is at $ x \equal{} 0$, and likewise for the minimum.  On the other hand, this is easy to argue without going to far as the second derivative test: on $ [0, \frac{\pi}{2}]$ we have that $ \sin$ is monotonically increasing from $ 0$ to $ 1$, while on $ [0, 1]$ we have that $ \cos$ is monotonically decreasing, so the composition is decreasing over $ [0, \frac{\pi}{2}]$.

This is same as proving 1-cos(sinx)<=(x^2)/2

1-cos(sinx)=2sin((sinx)/2)^2

for x belonging to[0.pi/2],    (sinx)/2<=x/2

sin((sinx)/2)^2 <=(sin(x/2))^2 <=(x/2)^2

2sin((sinx)/2)^2 <=(x^2)/2

1-cos(sinx)<=(x^2)/2 for x belonging to[0.pi/2]

Now,cos(sinx) lies in the interval [cos1,1] whereas 1-(x^2)/2 becomes -ve for values of           x >=pi/2

Since cos(sinx) is an even function the same can be proved for -ve values of x as well


inequalities geometry incenter trigonometry

In a triangle ABC let I be the incentre

Prove that (a^2 +b^2 +c^2)/2 >AI^2 +BI^2+CI^2

it's easy if we use the cos theorem:
[tex]a^{2}=b^{2}+c^{2}-2bc \cos A[/tex]

in this problem:
[tex]a^{2}=BI^{2}+CI^{2}-2BI \cdot CI \cos BIC[/tex]
[tex]b^{2}=AI^{2}+CI^{2}-2AI \cdot CI \cos AIC[/tex]
[tex]c^{2}=AI^{2}+BI^{2}-2AI \cdot BI \cos AIB[/tex]

and now we need to see that [tex] AIB, AIC, BIC > \frac{ \pi}{2} [/tex] so the [tex] \cos [/tex] are negatives and we get
[tex]a^{2}+b^{2}+c^{2}=2(AI^{2}+BI^{2}+CI^{2}) + 2(AI \cdot BI \cdot \cos AIB + AI \cdot CI \cdot \cos AIC + BI \cdot CI \cdot \cos BIC) [/tex]
[tex]\frac{a^{2}+b^{2}+c^{2}}{2}>AI^{2}+BI^{2}+CI^{2} [/tex]


induction
combinatorics proposed
combinatorics

Find all permutation  $ (a_1, a_2,..., a_n)$  of $ (1, 2, 3,..., n)$ such that there is exist only one number such that $ a_i > a_{i\plus{}1}$

There are exactly n-1 such permutations.

In the permutation $ \{1,2,\dots,n\}$ we can allow for exactly one pair to exchange position and they have be consecutive.

There's more than just $ n\minus{}1$.  For example, for $ n\equal{}4$, there are  $ 2134, 3124, 4123, 1324, 1423, 2314, 2413, 3412, 1243, 1342, 2341$, making $ 11$ in all by my count.

My suggestion:

If you have a permutation in the desired class, you can think of it as the union of two increasing sets of numbers, $ a_1<a_2, \dots <a_i$ and $ a_{i\plus{}1}<a_{i\plus{}2}<\dots<a_n$.  Conversely, you can try and form permutations by first choosing some subset $ A$ to be the first group ($ a_1$ through $ a_i$), and forming the permutation by listing everything in $ A$ in increasing order, followed by everything not in $ A$ (again in increasing order).

This [b]almost[/b] works, but there are some choices of $ A$ which will fail and give you a permutation you don't want to count.  Which ones are those, and how many are there?

We can solve this problem by induction and i will post this way when I free time

There are exactly $ 2^n-n-1$ such permutations.

Consider the permutation $ \{1,\dots,n\}$. Pick a non empty subset of $ \{1,\dots,n-1\}$ that is not $ \{1,2,\dots,k\}$. Delete exactly those elements from the permutation $ \{1,\dots,n\}$ and call the resulting permutation $ P_1$. Arrange the elements of the subset in ascending order and attach it either at the beginning or end of $ P_1$. A subset of form $ \{1,2,\dots,k\}$ can only be attached to the right of $ P_1$.
\[{ 2(2^{n-1}-n})+n-1\]

example: n=4


function
trigonometry
algebra unsolved
algebra

find all function $ f:\mathbb{R}^{\plus{}}\rightarrow\mathbb{R}$ that for all $ x,y\in\mathbb{R}^{\plus{}}$ we have:
$ f(x\plus{}y)\equal{}f(x^{2}\plus{}y^{2})$

Only $ f(x)\equiv const$. Consider $ x\equal{}rcos\phi, y\equal{}r\sin\phi$, it give $ f(r^{2})\equal{}f(z), r\le z\le z\sqrt 2$.


algorithm
number theory
Euclidean algorithm
6th edition

How did you solve it? I tried proving there`s  no solution by checking modula with so many numbers, but it didn`t work.

My solution uses a lemma similar to jin's one :

[hide]
Lemma :

If $n$ is of the form $4k-1$, with $n \geq 0$, then the equation
\[ a^2 \equiv -1 \]
has got no solution modulo $n$.

(it is easily proven by taking a prime $p \equiv -1$ (mod.4), and taking the $(p-1)/2$-th power)

I then distinguish the cases, x odd and x even.

1) x odd
then $y^3 \equiv x^2 +4-27 \equiv 2$ (mod.4), which is impossible for a cube.

2) x even
then $y^3 \equiv 1$, so $y \equiv 1$.
Thus $x=2x'$, $y=4y'+1$, and
\[ 4(x'^2+1) = (4y'+4)(y^2-3y+9) \]
By simplifying by 4, we get $x'^2 \equiv -1$ (mod.$y^2-3y+9$), which is impossible since $y^2-3y+9 \equiv 3$ (mod.4), and is positive.[/hide]

[quote="jin"]This is my solution in .pdf,and I am lazy,because I copied it from my solution.[/quote]
Could you please explain how you drew the last conclusion about y^2-3y+9 in detail?

Well, it indeed isn't as trivial as it first seems :
$x^2+2^2 \equiv 0$ (mod. $y^2-3y+9$),
so, since there is some prime $p \equiv 3$ (mod.4) which divides $y^2-3y+9$, we get
$x^2+2^2 \equiv 0$ (mod. $p$),
thus $p$ divides $2$, which is impossible.

[quote="perfect_radio"][quote="cadge_nottosh"]I think I should go into a corner and hang my head in shame.
I used unique factorisation in $\mathbb{Z}[\sqrt{23}]$.
  :oops:
Jack[/quote]

Well, was your solution very long or what?[/quote]

Even if it was long, it would be interesting to see it. Could you please post it?

Not especially long, I just don't like deploying facts which I don't know how to prove.

But I don't know a way of proving the fact (as the Euclidean algorithm does not work in $\mathbb{Z}\sqrt{23}$).
If anyone can help me with this it would be much appreciated.
It is true though: see [url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A003172]Sloane's A003172[/url]

Hope it is right!

Jack

I think I have explain it clearly,well,y^3 + 3^3 is divisible by 4,so y^3≡1(mod4),and then y≡1(mod4),
y^2≡1(mod4),3y≡3(mod4),so y^2-3y+9 ≡3 (mod 4),by the conclusion upon and there is a prime p≡3(mod4), divides y^2-3y+9,so it divides x^2+2^2,ant by the lemma,it divides 2,a contradiction[/quote]


I have to do this problem for high school geometry.

A line is tangent to two intersecting circles at P and Q.  The common chord is extended to meet segment PQ at T.  Prove that T is the midpoint of segment PQ.

Can anyone help me?

You need to prove that the lengths of $ TP$ and $ TQ$ are equal.
Call the points where the circles intersect (the ends of the common chord) $ R$ and $ S$. $ R$, $ S$, and $ T$ are colinear by construction. You must have seen a theorem linking the length of the segments formed by lines from an exterior point that intersect a circle. For one circle, that would involve the lengths of $ TR$, $ TS$ and $ TQ$. For the other circle, $ TR$, $ TS$ and $ TP$ would be involved.

Let the two circles intersect at $ X$ and $ Y$ with $ X$ closer to segment $ PQ$.

By power of a point, $ TP^2\equal{}TX*TY\equal{}TQ^2$. Thus $ TP\equal{}TQ$ and we are done.

Note: $ XY$ is called the radical axis with respect to the two circles and it is known that for any point on the radical axis, it has equal powers to both circles.

[quote="resurrection"]Let the two circles intersect at $ X$ and $ Y$ with $ X$ closer to segment $ PQ$.

By power of a point, $ TP^2 \equal{} TX*TY \equal{} TQ^2$. Thus $ TP \equal{} TQ$ and we are done.

Note: $ XY$ is called the radical axis with respect to the two circles and it is known that for any point on the radical axis, it has equal powers to both circles.[/quote]

The radical axis isnt just defined for Common tangents but for any tangents of equal lengths intersecting it.

We can even do this using general equation of the circle.


modular arithmetic
number theory proposed
number theory

Let $p$ be a prime number.
a) Define $A_k=\{(a,b,c)\mid 1\leq a<b<c\leq p-1,\ abc=k\pmod{p}\}$. Prove that $|A_1|\geq |A_k|$ for all $k$.
b) Suppose $p=4k+1$. Define $A_k=\{(a,b,c,d)\mid 1\leq a<b<c<d\leq p-1,\ abcd=k\pmod{p}\}$. Prove that there are $i$ and $j$ s.t. $|A_i|>|A_1|>|A_j|$.

If no one gets by Monday could you please post the solution. Unless of course you don't know either.

Who is Karpov?

Sorry about repost I accidently hit esc and I think it reposted when I refreshed. BTW I only know of a chess playing Karpov.

D. Karpov is a teacher from Sankt-Peterburg.

For the first one:

We count the sets $\{(a,b),c\}$ with $a\ne b$ s.t. $abc=k$, and we find that in order to prove the assertion in the problem, we must show that there are at most as many such sets with $a=c$ or $b=c$ for $1$ as there are for $k$. Call this $(*)$. If we count the pairs $(a,b)$ with $a^2b=k$, we find that $(*)$ is equivalent to the fact that there are at least as many $a^3=1$ as there are $a^3=k$, and this is clear: if $p=3u+1$, then $x^3=1$ has three solutions, and $x^3=k$ has at most $3$ solutions, while if $p=3u+2$, $x^3=k$ has one solution for all $k$.

I'm pretty sure I might have made some counting error though :).


Let $ a,b,c \geq 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[ \frac{a}{4\minus{}b}\plus{}\frac{b}{4\minus{}c}\plus{}\frac{c}{4\minus{}a} \leq 1\]

[quote="king-of-inequalities"]Let $ a,b,c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {a}{4 \minus{} b} \plus{} \frac {b}{4 \minus{} c} \plus{} \frac {c}{4 \minus{} a} \leq 1
\]
[/quote]
It's equivalent to the following known inequality.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}1.$ Prove that:
\[ a^2b\plus{}b^2c\plus{}c^2a\plus{}abc\leq\frac{4}{27}\]

[quote="Rose_joker"]After expanding it remains to prove that
$ 4\geq a^2c \plus{} b^2a \plus{} c^2b \plus{} abc$
let $ b$ be the middle one among $ a,b,c$we have
$ a(c \minus{} b)(a \minus{} b)\leq 0$
$ a^2c \plus{} ab^2 \plus{} c^2b \plus{} abc\leq a^2b \plus{} abc \plus{} c^2b \plus{} abc \equal{} b(a \plus{} c)^2$
but from am-gm we have
$ 2b(a \plus{} c)^2\leq (\frac {2b \plus{} (a \plus{} c) \plus{} (a \plus{} c)}{3})^3 \equal{} 8$
so
$ a^2c \plus{} ab^2 \plus{} c^2b \plus{} abc\leq 4$
 :)[/quote]
yes it's strong than canada's MO one.


Does the sum of the reciprocals of Fibonacci numbers converges?

[hide="Short answer"] Yes; the Fibonacci numbers grow exponentially (specifically, on the order of $ \phi^{n}$). [/hide]
[hide="Long answer"] $ F_{n}\equal{}\frac{\phi^{n}\minus{}\varphi^{n}}{\phi\minus{}\varphi}$, where $ \phi,\varphi$ are the roots of $ x^{2}\equal{} x\plus{}1$; $ \phi$ is the golden ratio, and $ \varphi$ is... the other golden ratio.  It's easy to prove convergence from here, but the actual value of the summation is not known.  See http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant .[/hide]

So does this mean the sum of the reciprocals of Lucas numbers also converges?

Yes it converges. If you know a little about the theory of convergence of infinite series - try the Ratio Test.

The Lucas numbers are given by $ \varphi^{n}\plus{}\phi^{n}$, which is also exponential (in fact, every second-order linear recurrence can be written explicitly in a form similar to these.)

In addition, we have the identity $ L_{n}\equal{} F_{n\minus{}1}\plus{}F_{n\plus{}1}$, hence $ L_{n}\ge F_{n}$.

Oh I see, by the way, what is meant by second-order linear recurrence?

The Fibonacci and Lucas numbers are both given by the recurrence $ a_{n\plus{}2}\equal{} a_{n\plus{}1}\plus{}a_{n}$.  A second-order (goes back two terms) homogeneous (has all terms of the same degree (no constants)) linear (is linear in both terms) recurrence is any recurrence of the form

$ a_{n\plus{}2}\equal{} A\cdot a_{n\plus{}1}\plus{}B\cdot a_{n}$


Are you guys still sure that Espn2 is not broadcasting the national Mathcounts competition?

National Spelling Bee at the Hyatt in Washington DC is on ABC right now. Jonathan Horton from AZ was on there, eliminated in round 8.

who will win??? gah, the suspence is killing me.

It was Kerry Close.

I still can't believe that she got that one (german???) word right.

That was insane :lol:

Um... We're talking about MATHCOUNTS in a MATHCOUNTS forum.

No, we are supposed to be talking about Mathcounts in a mathcounts forum. ON the contrary, we are not.

Well Mathcounts is clearly more useful than spelling bees.

I wanted the Canadian girl to win (Finola Hackett).

It's all Ignite's fault.

And yes, ESPN2 is not showing Mathcounts. For sure...

Sorry, I am slow, why is it Ignition's fault again?

I still think Mathcounts is better, just because competitors get the same questions. (And it's math!)

I think I'm better at spelling than at math. At the spelling bee that decided who went to states, I got third place! After that, I could have gotten almost every single word that they threw at the other mathletes up there! Darn! They don't give ribbons to Third placers! And at mathcounts, I got 4th!

There's a topic in G&FF about the spelling bee, called "Ursprache".

"Every single word that they threw at the [b]mathletes[/b] up there"?!? :P

*mathnerd314 continues to mumble dejectedly about "dirndl"*


analytic geometry
geometry
quadratics
Vieta
algebra

The points S, T, U and V have coordinates $ (s, ms)$, $ (t, mt)$, $ (u, nu)$ and $ (v, nv)$ respectively. The lines SV and UT meet the line $ y\equal{}0$ at the points with coordinates $ (p, 0)$ and $ (q, 0)$ respectively. Show that

$ p\equal{} \frac{(m\minus{}n)sv}{ms\minus{}nv}$

and write down a similar expression for $ q$.

Give that S and T lie on the circle $ x^{2} \plus{} (y\minus{}c)^{2} \equal{} r^{2}$, find a quadratic equation satisfied by $ s$ and $ t$ and hence determine  $ st$ and $ s\plus{}t$ in terms of $ m, c,$ and $ r$.

Given that S, T, U and V lie on the above circle, show that $ p\plus{}q\equal{}0$.

Part 1: We use the 2-point equation of a line:

$ (y\minus{}ms)\equal{}\frac{ms\minus{}nv}{s\minus{}v}(x\minus{}s)$

$ x\equal{}\frac{s\minus{}v}{ms\minus{}nv}(y\minus{}ms)\plus{}s$

$ x\equal{}\frac{(s\minus{}v)y\minus{}ms^2\plus{}msv\plus{}ms^2\minus{}nsv}{ms\minus{}nv}$

$ x\equal{}\frac{(s\minus{}v)y\plus{}(m\minus{}n)sv}{ms\minus{}nv}$

Now we substitute $ p$ for $ x$ and $ 0$ for $ y$:

$ p\equal{}\frac{(m\minus{}n)sv}{ms\minus{}nv}$

Similarly, we obtain $ q\equal{}\frac{(m\minus{}n)tu}{mt\minus{}nu}$ if we use $ t$ and $ u$ instead of $ s$ and $ v$, respectively.

Part 2: Substituting yields:

$ (s)^2\plus{}(ms\minus{}c)^2\equal{}r^2$ and $ (t)^2\plus{}(mt\minus{}c)^2\equal{}t^2$

$ (s^2)\plus{}(m^2s^2\minus{}2mcs\plus{}c^2)\equal{}r^2$ and $ (t^2)\plus{}(m^2t^2\minus{}2mct\plus{}c^2)\equal{}r^2$

$ (m^2\plus{}1)s^2\minus{}(2mc)s\plus{}(c^2\minus{}r^2)\equal{}0$ and $ (m^2\plus{}1)t^2\minus{}(2mc)t\plus{}(c^2\minus{}r^2)\equal{}0$

Therefore, $ s$ and $ t$ both satisfy the quadratic equation $ (m^2\plus{}1)x^2\minus{}(2mc)x\plus{}(c^2\minus{}r^2)\equal{}0$.
By Vieta's Laws, $ s\plus{}t\equal{}\frac{2mc}{m^2\plus{}1}$ and $ st\equal{}\frac{c^2\minus{}r^2}{m^2\plus{}1}$.

Part 3:
I'll work on it later. I can't think about the problem clearly after I worked on it a lot. Where did you get this problem anyways? It's pretty intriguing.

Thanks. (Someone asked me: it's from an  examination).

This is STEP III question 1 2009, a UK university entrance exam for Cambridge and Warwick - which I myself took on Monday amd looking at the date you posed this question you still had time to technically cheat (or allow other people to cheat) (as are the other problems you have posted in thiis thread). Please wait for 24 hours after the exam (which was taked on the same Monday you posted as it is not fair to  other people who are required to obtain certain grades to actually GO to the university they want to).

DeanK22


Results I can remember:

* denotes Duke scholarship eligibility.

1. John Berman - Hoggard
2. Kevin Lang - Myers Park
3. Jeremy Hahn - Michaux Academy (Homeschool)
4. William Schlieper - Green Hope
5. Joe Keefer* - NCSSM
6. Daniel Vitek - Enloe
7. Bryce Taylor - Mount Tabor
8. Vivek Bhattacharya* - Enloe
9. Andrew Hertzberg - Chapel Hill
10. Ray Wang*
11. Damien Jiang - Cary Academy
12. (?) Fei Fei Wang - Chapel Hill
13. (?) Calvin Deng - Ligon
14.
15.
16.
17.
18. (?) Allen Yang - Ligon
19. (?) Eric McCabe - (?)
20.

I know the following individuals were somewhere in the top 20.
Brendan Fletcher

Edits: Thanks to not_trig.

I got 17th =)

[hide]
not 18th
[/hide]

I think Brendan [hide]fletcher[/hide] got 20th

Berman got a \$1000 check [color=red][EDITED BY MODERATOR][/color].

EDIT: Woops =)

Top 10 got books/plaques/AoPS scholarships

EDIT: there was a 4-way tie for 14th. I got 4th in the tiebreaker =(. Uber Fail at integer answer  :wallbash_red:  :cursing:  :furious:

P.S.You have an odd number of LaTeX tags $ in your document and BBCode is enabled. You might have forgotten to close a tag of a LaTeX formula, so please check your document. If you want to use the symbol$ to represent currency (dollars) either disable BBCode (not recommended) or
use the code $ \$$ to create a image of the dollar sign (so that your post can be safely quoted afterwards).

Remember that it's best to use the Preview function so you can spot mistakes in your LaTeX syntax.

Siyang Chen (NCSSM) placed ~18th, and Michael Chang (NCSSM) placed ~16th.

final ranks are up
i failed  :wallbash_red:  :help: :blush:
[url=http://courses.ncssm.edu/goebel/STATECON/results/TOP20-08.HTM]Website[/url]
Good job to berman, he pwnd everyone.

Congratulations John! Good job! Have a nice summer vacation with your parents using that check. If you do not think you need that AoPS scholarship, give it to somebody who needs it (for example, my son). :)

[quote="aspiringchesslete"]
my mom is currently trying to get me to do state problems, but maybe I should go improve my olympiad combo instead?
[/quote]
if you listened to your mother, you could rank next to John, at least you could get into top 10.

[quote="jb05"]Heh I was about to get really confused about why someone with a random username is acting like he knows what's going on, and then I realized who you are. Very tricky.

[/quote]

[quote="MysticTerminator"]uh so who is this[/quote]

It is much easier to figure out who he is than to solve an IMO#1 problem.
By the way, great job on this year's Putnam Competition.

[quote="yc"]
if you listened to your mother, you could rank next to John, at least you could get into top 10.
[/quote]

The state math contest does not matter at all, especially to someone already guaranteed a spot on the ARML team if he wants it.  Olympiad practice is much more important and (at least in my opinion) much more fun  :) .  Better yet, people could actually study some math outside of contests  :wink: .

Apparantly last year only about half of the AoPS scholarships were used  :maybe:

[quote="n^4+4"]Well I don't see what a senior is going to do with an AoPS scholarship... he should just give it to a middle schooler.  Also a lot of the people who are already at Olympiad level don't really need to take AoPS classes.  Even if they wanted to take something $ \$$150 doesn't go too far in terms of paying for WOOT for example.[/quote]

150/650 is about 23.1% o_O


Enuntul este urmatorul:
        Fie ABC un triunghi ascuţitunghic. Notăm cu M şi N mijloacele înălţimilor BB1 şi CC1, iar cu P si Q punctele
de intersectie ale perechilor de drepte: (AM;CC1) si respectiv (AN; BB1).  Arătaţi că:
a). punctele M, N, P şi Q sunt conciclice.
b). in cazul în care punctele B, C, P şi Q sunt conciclice, triunghiul ABC este isoscel.

De fiecare data fac recomandarea de a incerca mai intai sa rezolvati problema singuri!
si numai dupa aceea, sau daca dupa mai multe incercari nu reusiti s-o rezolvati sa va uitati la solutie.

O gasiti in ATTACHMENT!

[quote="mihai miculita"][color=darkred]Fie $ ABC$ un triunghi ascuţitunghic. Notăm $ \{\begin{array}{c}E\in AC\ ,\ BE\perp AC\\ \ F\in AB\ ,\ CF\perp AC\end{array}\ \ ,\ \ \{\begin{array}{c}M\in (BE)\ ,\ MB = ME\\ \ N\in (CF)\ ,\ NC = NF\end{array}\ \ ,\ \ \{\begin{array}{c}P\in AM\cap CF\\ \ Q\in AN\cap BE\end{array}$. Sa se arate :

$ 1\blacktriangleright$ patrulaterul $ MNPQ$ este ciclic.

$ 2\blacktriangleright$ Patrulaterul $ BCPQ$ este ciclic $ \Longleftrightarrow$ triunghiul $ ABC$ este $ A$- isoscel.[/color][/quote]
[color=darkblue][b]Demonstratie.[/b] Notam ortocentrul $ H\in BE\cap CF$ al triunghiului $ ABC$. Asadar $ BCEF$ este ciclic, adica $ HB\cdot HE = HC\cdot HF$.

$ \blacksquare\ 1.$ Se observa ca $ \widehat{ABE}\equiv\widehat{ACF}$, adica $ ABE\sim ACF$ $ \implies$ $ M\sim N$ (corespondente). Asadar, $ \widehat{MAB}\equiv\widehat{NAC}$, adica
semidreptele $ [AM$ si $ [AN$ sunt izogonale in unghiul $ \widehat{BAC}$. Deci $ m(\widehat{QMA}) = m(\widehat{MAB})+m(\widehat{ABE}) =$
$ m(\widehat{NAC})+m(\widehat{ACF}) =$ $ m(\widehat{ANP})$,adica $ \widehat{QMA}\equiv\widehat{ANP}$, ceea ce inseamna ca  patrulaterul $ MNPQ$ este ciclic.

$ \blacksquare\ 2.$ Patrulaterul $ MNPQ$ este ciclic $ \implies$ $ HM\cdot HQ = HN\cdot HP\ \ (*)$. Prin urmare, patrulaterul $ BCPQ$ este ciclic $ \Longleftrightarrow$
$ HB\cdot HQ = HC\cdot HP$. Tinand seama de relatia $ (*)$ obtinem : patrulaterul $ BCPQ$ este ciclic $ \Longleftrightarrow$ $ \frac{HM}{HB}=\frac{HN}{HC}$ $ \Longleftrightarrow$
$ MN\parallel BC$ $ \Longleftrightarrow$ $ \frac{MB}{HB}=\frac{NC}{HC}$ $ \Longleftrightarrow$ $ \frac{BE}{HB}=\frac{CF}{HC}$ $ \Longleftrightarrow$ $ \frac{HE}{HB}=\frac{HF}{HC}$ $ \Longleftrightarrow$ $ \frac{HE\cdot HB}{HB^{2}}=\frac{HF\cdot HC}{HC^{2}}$ $ \Longleftrightarrow$
$ HB = HC$ $ \Longleftrightarrow$ ortocentrul $ H$ apartine mediatoarei segmentului $ [BC]$ $ \Longleftrightarrow$ triunghiul $ ABC$ este $ A$- isoscel.

[b]Observatie.[/b] In mod asemanator se poate demonstra urmatoarea usoara extindere deoarece in demonstratia precedenta am folosit doar ca :
$ \blacktriangleright$ punctele $ M$, $ N$ sunt corespondente in asemanarea mentionata (nu neaparat mijloacele celor doua inaltimi) ;
$ \blacktriangleright$ patrulaterul $ BCEF$ este ciclic (nu neaparat de diametru $ [BC]$).
Extinderea este fireasca si se constituie intr-un model didactic. Dupa parerea mea, autorul problemei a fost superficial in compunerea acesteia.
Poate gresesc, insa nu-mi pot imagina o solutie care sa foloseaca in mod esential particularitatile din ipoteza mentionate mai sus.[/color]

[quote="Virgil Nicula"][color=darkred]Un cerc de coarda $ [BC]$ taie laturile $ [AB]$, $ [AC]$ ale triunghiului ascutitunghic $ ABC$ in punctele $ F$, $ E$ respectiv. Notam punctele

$ M\in (BE)$, $ N\in (CF)$ pentru care $ \frac{MB}{ME}=\frac{NC}{NF}$ si intersectiile $ P\in AN\cap BE$, $ Q\in AM\cap CF$. Sa se arate :

$ 1\blacktriangleright$ patrulaterul $ MNPQ$ este ciclic.

$ 2\blacktriangleright$ Patrulaterul $ BCPQ$ este ciclic $ \Longleftrightarrow$ triunghiul $ ABC$ este $ A$- isoscel.[/color][/quote]

[color=darkred][b][i]Domnule Miculita[/i][/b], nu stiu de ce nu va pot citi ATTACHMENT - ul ! Mai este cineva in situatia mea ?! Poate P.C. - ul meu are toane ...[/color]

[quote="Virgil Nicula"]

[color=darkred][b][i]Domnule Miculita[/i][/b], nu stiu de ce nu va pot citi ATTACHMENT - ul ! Mai este cineva in situatia mea ?! Poate P.C. - ul meu are toane ...[/color][/quote]

Mie-mi merge.Se poate sa fi fost pc ul de vina.


i start a math blog. i want to join with you.

  welcome to math blog:

 http://math.haiyanh.com/

i am in canada now


inequalities
rotation
inequalities proposed

Let $ a,b,c$ be posible real numbers satisfying $ ab\plus{}bc\plus{}ca>abc$. Prove that $ (a\plus{}b\minus{}c\minus{}1)(b\plus{}c\minus{}a\minus{}1)(c\plus{}a\minus{}b\minus{}1)<8$

I tried a transformation, but could not finish my proof....
Let $ x\equal{}a\plus{}b\minus{}c$, $ y\equal{}b\plus{}c\minus{}a$, $ z\equal{}c\plus{}a\minus{}b$
Now, the inequality becomes equivalent to:
$ (x\minus{}1)(y\minus{}1)(z\minus{}1) < 8$
$ \Rightarrow xyz\minus{}x\minus{}y\minus{}z\plus{}xy\plus{}yz\plus{}zx\minus{}1 < 8$
$ \Rightarrow ab(a\plus{}b)\plus{}bc(b\plus{}c)\minus{}a^3\minus{}b^3\minus{}c^3\minus{}a\minus{}b\minus{}c\plus{}ab\plus{}bc\plus{}ca\minus{}a^2\minus{}b^2\minus{}c^2\minus{}1 < 8$
$ \Rightarrow b(a\plus{}c)(b\plus{}c)\minus{}a^3\minus{}b^3\minus{}c^3\minus{}a\minus{}b\minus{}c\plus{}ab\plus{}bc\plus{}ca\minus{}a^2\minus{}b^2\minus{}c^2\minus{}1 <b(a\plus{}c)(b\plus{}c)\minus{}4abc\minus{}a\minus{}b\minus{}c\minus{}1<8$

It is getting clumsy....can you give me a hint please, mr.danh?

[hide="Hint"]No complicated transformation in my proof. We shold make a use of the following problem (as lemma):
[quote="mr.danh"]
[b]Lemma:[/b]
Let $ x,y,z$ be positive real numbers satisfying $ xyz\ge 1$. Prove that:
$ \frac{1}{x^3\plus{}y^3\plus{}1}\plus{} \frac{1}{y^3\plus{}z^3\plus{}1}\plus{}\frac{1}{z^3\plus{}x^3\plus{}1}\le 1$[/quote][/hide]

[quote="mr.danh"][hide="Hint"]No complicated transformation in my proof. We shold make a use of the following problem (as lemma):
[quote="mr.danh"]
[b]Lemma:[/b]
Let $ x,y,z$ be positive real numbers satisfying $ xyz\ge 1$. Prove that:
$ \frac {1}{x^3 \plus{} y^3 \plus{} 1} \plus{} \frac {1}{y^3 \plus{} z^3 \plus{} 1} \plus{} \frac {1}{z^3 \plus{} x^3 \plus{} 1}\le 1$[/quote][/hide][/quote]
\[ \sum {\frac{1}{{a^3  \plus{} b^3  \plus{} 1}}}  \le \sum {\frac{1}{{ab\left( {a \plus{} b} \right) \plus{} 1}}}  \le \sum {\frac{c}{{a \plus{} b \plus{} c}}}  \equal{} 1
\]

Hey, Sunkern_sunflora, I have a hint for you, why don't you notate   $ a\plus{}b\minus{}c\minus{}1\equal{}x$    and change the given condition into $ x, y, z$? Have fun!

Hello great math. I have tried it, putting $ a\plus{}b\minus{}c\minus{}1\equal{}x, b\plus{}c\minus{}a\minus{}1\equal{}y, c\plus{}a\minus{}b\minus{}1\equal{}z$, but I could not tackle the relation $ ab\plus{}bc\plus{}ca>abc$ from it, will you help me please??
  :maybe:


How can I find how fast this sum diverges in terms of x?

$ \sum_{n \equal{} 0}^\infty \sum_{k \equal{} 0}^{n} [{2n\choose2k} {(2k \plus{} 1)}^{ \minus{} x}] / 2^{2n \plus{} 1}$

0<x<1

Let $ a_n\equal{}\dfrac{1}{2^{2n\plus{}1}}\sum_{k\equal{}0}^{n}(_{2k}^{2n})(2k\plus{}1)^{\minus{}x}$; experiments with Maple "show" that $ a_n\sim\dfrac{1}{4n^x}$; thus $ \sum_{n\equal{}0}^{p}a_n\sim\dfrac{n^{1\minus{}x}}{4(1\minus{}x)}$.

[quote="loup blanc"] thus $ \sum_{n \equal{} 0}^{p}a_n\sim\dfrac{n^{1 \minus{} x}}{4(1 \minus{} x)}$.[/quote]

Why is there an n term in the last equation when there is no sum before it? Or is the sum implied?

Obviously you must read:
    $ \sum_{n \equal{} 0}^{p}a_n\sim\dfrac{p^{1 \minus{} x}}{4(1 \minus{} x)}$
when $ p\rightarrow\infty$.


quadratics
algebra
function
domain
inequalities
complex numbers

Find all solutions for $ x$ to the equation
\[ \sqrt[4] {x} \plus{} \sqrt[4] {97 \minus{} x} \equal{} 5
\]
[hide="Hint"]
Don't take the fourth power of both sides  :wink:
[/hide]

[hide]I'm getting $ x\equal{}16, 81$.
I said that $ x\le97$, and the only perfect fourth powers less than or equal to 97 are 0, 1, 16, and 81.
$ 16\plus{}81\equal{}97$
$ 16\equal{}2^4$ and $ 81\equal{}3^4$
$ \sqrt[4]{2^4}\plus{}\sqrt[4]{3^4}\equal{}2\plus{}3\equal{}5$
[/hide]

Partial Solution:
[hide]
Let $ \sqrt [4]{x} \equal{} A$ and
$ \sqrt [4]{97 \minus{} x} \equal{} B$.
Then $ A \plus{} B \equal{} 5$ and
$ A^4 \plus{} B^4 \equal{} 97$.
Substituting $ 5 \minus{} A$ for $ B$ in the second equation and simplifying yields
$ A^4 \minus{} 10A^3 \plus{} 75A^2 \minus{} 250A \plus{} 264 \equal{} 0$
By observation, we can see that two possible values of $ A$ are $ 2$ and $ 3$.
Dividing out the factors $ A \minus{} 2$ and $ A \minus{} 3$ yields $ A^2 \minus{} 5A \plus{} 44 \equal{} 0$.
I really don't know how radicals work with complex numbers (i.e. which is the positive root and which is the negative), so I don't know if the complex solutions to this equation can be fourth roots. Anyways $ 16$ and $ 81$ are definitely solutions to the original equation and maybe the quadratic will yield some more...[/hide]

[hide="cheap R-rated solution"]
We will avoid non-positive real solutions by alleging that non-integer powers of such numbers are not well defined. :P Henceforth all our functions' domains will therefore be subsets of $ (\mathbb{R}^\plus{}_0)^n$, for appropriate values of $ n$.

Observe that the function $ x \mapsto x^{1/4}$ is concave, so by Karamata's inequality, the function $ (x,y) \mapsto x^{1/4} \plus{} y^{1/4}$ is Schur concave.  This implies that the function $ f : x \mapsto x^{1/4} \plus{} (97\minus{}x)^{1/4}$ is monotonically increasing on $ [0,97/2]$ and monotonically decreasing on $ [97/2, 97]$.  Therefore $ f$ is injective on each of those intervals, so there are at most two solutions total, one in each interval.  It is easy to verify that those two solutions are 16 and 81.
[/hide]
That contained no elegant algebra at all; only very strong tools.  However, knowing Karthik, maybe Schur convexity is what he had in mind. :P

[hide="What is Schur convexity?"]
Okay, first, you need to know what [[majorization]] is.  Let $ f$ be a function mapping some subset of $ \mathbb{R}^n$ to the reals.  We say that $ f$ is Schur convex if for all $ \mathbf{x,y}$ in the domain of $ f$, $ \mathbf{x} \succ \mathbf{y}$ implies $ f(\mathbf{x}) \ge f(\mathbf{y})$.  We say that $ f$ is strictly Schur convex if this inequality is strict except when $ \mathbf{x}$ and $ \mathbf{y}$ are permutations (i.e., reorderings) of each other.  (We say that $ f$ is Schur concave if $ \minus{}f$ is Schur convex.)  Karamata's inequality says that a symmetric sum of (strictly) convex real functions is (strictly) Schur convex.

If this interests you, or if my explanation was too dense, you can read about Schur convexity in good books on inequalities, such as [i]The Cauchy-Schwarz Masterclass[/i] by J. Michael Steele and [i]Inequalities[/i] by Hardy, Littlewood, and Pólya.  The first of these is a lot less dense than the second, just for the record.[/hide]

Haha, it's true that one way is Schur's convexity. Eric got the closest to my solution. If you have $ A \plus{} B \equal{} 5, A^4 \plus{} B^4 \equal{} 97,$ try to express both in terms of $ A\plus{}B$ and $ AB$ then solve.


quadratics
number theory unsolved
number theory

A subset $S$ of the set $M=\{1,2,.....,p-1\}$,where $p$ is a prime number of the kind
$12n+11$,is [i]essential[/i],if the product ${\Pi}_s$ of all elements of the subset
is not less than the product $\bar{{\Pi}_s}$ of all other elements of the set.The
[b]difference[/b] $\bigtriangleup_s=\Pi_s-\bar{{\Pi}_s}$ is called [i]the deviation[/i]
of the subset $S$.Define the least possible remainder of division by $p$ of the deviation of an essential subset,containing $\frac{p-1}{2}$ elements.

Let $\Pi_s=a$, from Willson $\bar{{\Pi}_s}=-1/a$. So we have to find min minimal
$a-(-1/a)=a+1/a (mod p)$. So let's have a look at the first three cases:

1)$a+1/a=0 (mod p)$ means
$a^2=-1 (mod p)$ but  for p=3 (mod 4) is -1 not quadratic residue.

2)$a+1/a=1 (mod p)$ means
$4a^2-4a=-4 (mod p)$
$(2a-1)^2=-3 (mod p)$, bud for p=11 (mod 12) is -3 not quadratic residue.

3)$a+1/a=2(mod p)$.  We can obtain this as
*$[(p-1).(p-2).....((p+3)/2)).((p+1)/2)]-[1.2.....((p-3)/2).((p-1)/2)]$ or
#$[(p-1).(p-2).....((p+3)/2)).((p-1)/2)]-[1.2.....((p-3)/2).((p+1)/2)]$.
From Willson the product of these two brackets in * is -1 and their values mod p differ only in sign, so one is 1 and the other is -1, as well as the brackets in #.
But the values of first  bracket in * and # differ in sign, so one of * and # works.


trigonometry
calculus
calculus computations

[img]http://static.flickr.com/103/305583784_fa7f9c2ffa_o.png[/img]

How can I solve this kind of problem?
Our professor didn't teach this kind of problem.
He just teach the definition but he did not gave an example.
Saw in the net but his teaching is different from the net though.

And also this one

- lnx expand to the power of  x-2=0 using TSE or taylor series.

- logx in the powers of x-3=0

How can I solve this one?

thanks guys.

[quote="mlok"]By definition $\sinh x=(e^{x}-e^{-x})/2$ and $\cosh x=(e^{x}+e^{-x})/2$. Hence $e^{x}=\sinh x+\cosh x$. Now if you can show that $e^{ix}=\cos x+i\sin x$, it will follow that $e^{ix-x}=\frac{e^{ix}}{e^{x}}=\frac{\cos x+i\sin x}{\sinh x+\cosh x}$ (you have a typo in the formula).[/quote]

So what's the answer on this one sir?


OK.  I decided to make a moderate difficulty thread.  Make sure they're not too easy and not too hard.  Make sure you post a new problem and a solution to the previous problem.
PROBLEM 1:
How many ways can Patricia misspell MISSPELL if she uses all of those 8 letters and none more but puts them in the incorrect order?

[b]PROBLEM 1:
How many ways can Patricia misspell MISSPELL if she uses all of those 8 letters and none more but puts them in the incorrect order?[/b]
[hide="Solution"]Number of ways to missplell MISSPELL = Number of different words formed from all the letters of MISSPELL - 1 = $ \frac{8!}{(2!)(2!)} \minus{} 1$ = $ 10080 \minus{} 1 \equal{} 10079$ [/hide]

PROBLEM 2:
How many non-negative positive integral solution are there to the equation  $ x \plus{} 2y \plus{} 3z \equal{} 100$?


cane some one prove this:
If there are $ m$ items of one kind,$ n$ items of another kind and so on,then the number of ways of choosing $ r$ items out of these items is
coefficient of $ x^{r}\in\ (1 \plus{} x \plus{} x^{2} \plus{} x^{3} \plus{} \cdots \plus{} x^{m})(1 \plus{} x \plus{} x^{2} \plus{} \cdots \plus{} x^{n})\cdots$

is there any way you can rephrase that problem better?

[hide]
Actually, the number of ways to choose those r items is $ \min{{r\minus{}n,r\minus{}m,m,n,r}}$. Or something like that.

[/hide]

[quote="CircleSquared"][hide]
Actually, the number of ways to choose those r items is $ \min{{r \minus{} n,r \minus{} m,m,n,r}}$. Or something like that.

[/hide][/quote]

Er, no.  He means a situation like this:

Say we have $ 4$ math books, $ 10$ physics books, and $ 7$ chemistry books, and we want to pick out $ 5$ books to read for the weekend.  The number of ways in which we can pick $ 5$ books is the coefficient of $ x^5$ in

$ (1 \plus{} x \plus{} ... \plus{} x^4)(1 \plus{} x \plus{} ... \plus{} x^{10})(1 \plus{} x \plus{} ... \plus{} x^7)$

This way of looking at combinatorics is very powerful.  The result, by the way, follows from the definition of polynomial multiplication.

The intuitive explanation is that when you multiply polynomials, you take all of the possible choices for a factor and add them up. So for every way to choose $ r$ items, there's a corresponding sequence of factors which multiply to $ x^r$. Similarly, for any sequence of factors, you can find a way to choose $ r$ items. Therefore the coefficient must be equal to the number of ways to choose $ r$ items.

This is also a really interesting concept when you don't force your factors to be polynomials, but can be formal power series instead. Then similar to your product your product, you could write:
$ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots)(1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots) \equal{} (1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots)^2 \equal{} \frac{1}{(1\minus{}x)^2}$. This will immediately give you the number of solutions to $ a\plus{}b\equal{}r$ in nonnegative integers by looking at the coefficient of the $ x^r$ term of the product power series.

That is, in my opinion, one of the coolest things in math.


geometry
3D geometry
modular arithmetic
number theory
number theory unsolved

Prove that there exists infinitely many positive integers which cannot be represented as sums of two cubes of integers, but can be represented as sums of two cubes of positive rational numbers.

For example, any prime of the form $ 9k\plus{}4$ can be represented as the sum of two rational cubes but not of two integer cubes (see H. Cohen "Number Theory", Vol. 1, Section 6.4.6).
It is clear that there are infinitely many such primes.

[hide]Consider the number
$ (\frac{27k^3\plus{}3n\plus{}1}{3k})^3\plus{}(\frac{27k^3\minus{}3n\minus{}1}{3k})^3\equal{}2(729k^6\plus{}3 \cdot (3n\plus{}1)^2) \equiv 6 \pmod 9$. But cubes can only be $ 0$,$ 1$, or $ \minus{}1$ mod $ 9$, so this can't be a sum of cubes.[/hide]


As I was going to St Ives,
I met a man with seven wives,
each wife had seven sacks,
every sack had seven cats,
every cat had seven kits,
Kits, cats, sacks, and wives,
how many were going to St. Ives?

Found this in my Pre-Calc book enjoy!  :D

I say 1. Since it doesn't say the people he met were going. :)

I guess its just one since it doesn't say all of them were going to that place.  But heres a variation with a bunch of math.  There are 7 girls in a room.  Each girl has 7 bags.  Each bag has 7 big kittens.  Each big kitten is accompanied by a small kitten.  How many legs are there in the room.  No tricks just add up the legs.  good luck!

[hide="Solution"]
10990
[/hide]

nice solution. :D

i think this was intended to include the other people. it said kit, cats, and wives,

$ 7^0$ men
$ 7^1$ wives and so on
$ 7^4$ kits
too lazy to use a calc
$ S \equal{} 7^0 \plus{} 7^1...7^4$
$ 7S \equal{} 7^1 \plus{} 7^2...7^5$
$ 6S \equal{} 7^5 \minus{} 7^0$
$ S \equal{} 2801$
eh, i guess i can include the "I" in the peom so 2802

if u include yourself and the man you met then 7^4+1+1=2401+1+1=2403.
if u only mean wive ,sack,..............then 2401. :D
you r really generous u speak of sacks as though they were human. :rotfl:

7 wives had 7 sacks = 49 sacks total
Each cat has seven kits = 49 kits per sack * 49 sacks= 2401 total kits
Each sac has 7 cats = 7 cats per sack. * 49 = 343 total cats
7 wives
2 men
 Add all these up

Grand total of 2802!

But I still think the right answer is 1. Since it said he met a man on the way to st. ives. He wasn't in St. Ives and said nothing of the destination of the other partys.

[quote="Cronus"]if u include yourself and the man you met then 7^4+1+1=2401+1+1=2403.
if u only mean wive ,sack,..............then 2401.[/quote]

$ 7^4$ is only the number of kits there are. The total number of people (excluding "I") is $ 7^0\plus{}7^1\plus{}7^2\plus{}7^3\plus{}7^4\equal{}2801$, as abacadaea explained. Including "I" yields $ \boxed{2802}$.

[quote="Inequality"]Is any of these right?[/quote]

I'm afraid not.  :(  For one, you said that each wife or sack had $ 343$ objects, which is not correct (in either case). You could say that each wife has herself, $ 7$ sacks, $ 49$ cats, and $ 343$ kits, for a total of $ 1\plus{}7\plus{}49\plus{}343\equal{}400$ objects per wife. Multiplying by $ 7$ gives you $ 2800$, and adding the man and "I" gives you $ \boxed{2802}$.

ooooooooooooooooooooooooooooooooops I forgot to add the powers.one of the following 4 is right:
a.1,b.2800,c.2801,d.2802.untill mathe monster frightens us out. :rotfl:

I agree with you venezuela! :P Just needed to write it all down.

yes the answer is intended to be $ \boxed{2802}$


I am a year 10 student in Singapore.
Is it possible for me to take part in both AMC 10 and AMC 12?
Cause,the knowledge involved in AMC 12 is still  what I have learned~

This is answered in Frequently Asked Question #12 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


inequalities LaTeX function algebra polynomial

prove for x, y, z in $ \mathbb{R}^{\plus{}}$ that

$ \sum_{\text{cyc}}\frac {(x \plus{} y)\sqrt {1 \plus{} 4x^2y^2z}}{xy\sqrt {z}}\ge 3$

What does the "cyc" mean. :oops: Cyclic?

1) Yup cyc is Cyclic. I asked a similar question earlier... apparently

$ \sum_{\text{cyc}}a^2 \plus{} b \plus{} c \equal{} a^2 \plus{} b \plus{} c \plus{} b^2 \plus{} a \plus{} c \plus{} c^2 \plus{} a \plus{} b$

2) Here's what I got...

[hide]
First, we note that we are only dealing with positives, which helps when looking at squares and inequalities.

By Cauchy, it suffices to prove that

$ \frac {(x \plus{} y)^2(1 \plus{} 4x^2y^2z)}{x^2y^2z} \plus{} \frac {(y \plus{} z)^2(1 \plus{} 4y^2z^2x)}{y^2z^2x} \plus{} \frac {(x \plus{} z)^2(1 \plus{} 4x^2z^2y)}{x^2z^2y}\ge 9$

So now let's expand our terms... For simplicity (and cause I suck at Latex) I'll just use the first term...

$ \frac{(x \plus{} y)^2}{x^2y^2z} \plus{} 4(x \plus{} y)^2$
$ \equal{} 4(x \plus{} y)^2 \plus{} \frac{1}{y^2z} \plus{} \frac{2}{xyz} \plus{} \frac{1}{x^2z}$

Which, adding the other terms, results in

$ 4(x \plus{} y)^2 \plus{} 4(y \plus{} z)^2 \plus{} 4(x \plus{} z)^2 \plus{} \frac{1}{y^2z} \plus{} \frac{1}{x^2z} \plus{} \frac{1}{y^2x} \plus{} \frac{1}{z^2x} \plus{} \frac{1}{x^2y} \plus{} \frac{1}{z^2y} \plus{} \frac{6}{xyz}\ge 9$

Now we assume the sum is less than 9, which means that the sum of any of the terms of the equation must be less than 9, as all terms are positive:

$ 4((x \plus{} y)^2 \plus{} (y \plus{} z)^2 \plus{} (x \plus{} z)^2)<9$
$ x \plus{} y \plus{} z<\frac{9}{4}$

Also,

$ \frac{6}{xyz}<9$
$ xyz>2/3$

which is impossible by AM-GM. Hence, contradiction so the sum must be greater than 9, which proves the original assumption. QED.[/hide]

Did you check the problem, Temperal? Although I haven't found out the counter-example  of this inequality, but it seems to be wrong since the equality does not occur when x=y=z. This inequality is symmetric but it's hard to define where the equality occurs. Quite strange :|

Yes, I just checked it, and found that it was strict, and not non-strict. Sorry.  :oops:

The reason that you don't always find equality when $ x \equal{} y \equal{} z$ in this problem is because the inequality itself is non-homogenous - that is, we can't make an assumption such as $ xyz \equal{} 1$ (which would have helped) or so on.  Set $ x \equal{} y \equal{} z \equal{} t$ and we have a function of $ t$ that is obviously non-constant.  Specifically, $ f(t) \equal{} \frac {12\sqrt {1 \plus{} 4t^5}}{\sqrt {t^3}}$.  So any symmetric equality case will have to occur when $ f(t) \equal{} 3$.  We then have $ 4\sqrt {1 \plus{} 4t^5} \equal{} \sqrt {1 \plus{} t^3}$.  Squaring, we get $ 16 \plus{} 64t^5 \equal{} 1 \plus{} t^3$, so $ t$ is a root of the polynomial $ 64t^5 \minus{} t^3 \plus{} 15 \equal{} 0$.  This has four complex solutions and one negative real solution, hence there is no [i]symmetric[/i] equality case.  There could still be a non-symmetric case, however (but as Temperal showed, there isn't).

Edit:@unlimited3K: Close.  A cyclic sum written out in that form would be $ \sum_{cyc} a^2 \plus{} b \plus{} c \equal{} a^2 \plus{} b \plus{} c \plus{} b^2 \plus{} c \plus{} a \plus{} c^2 \plus{} a \plus{} b$.  Look at the 4th-6th terms closely - you have to preserve the same order.  That is, the variables are already in the order (a,b,c), so in the second term they have to "shift" to the left one to become (b,c,a), not (b,a,c).  That's trivial now, because addition is commutative, but if you had such an expression in a non-commutative algebraic structure (i.e. multiplication over matrices) you would have to be more careful.

Here's my solution(not sure if it's right):
W.L.O.G, we can assume that x>=y>=z
We will prove: f(x,y,z)>=f(square root(xy),square root(xy),z).
To prove this, we have:(x+y)square root(1+4x^2y^2z)/[xysquare root(z)]>=2square root(xy)square root(1+4x^2y^2z)/[xysquare root(z)](by AM-GM)
we will prove the sum of the rest two terms of f(square root(xy),square root(xy),c)=<the sum of the rest two terms of f(x,y,z)
After clearing the common denominator(xyz), we need to prove: (z+x)square root(y+4z^2x^2y^2)+(z+y)square root(x+4x^2y^2z^2)>=2(z+square root(xy))(quartic root(xy+4z^2x^2y^3))
We prove following inequalities: z(square root(y+4z^2x^2y^2)+square root(x+4z^2y^2x^2)-2quartic root(1+4xyz^2.square root(xy))^2.xy)>=0(squares both sides and use Cauchy+ the assumption that a>=b>=c)
                                    x.square root(y+4z^2x^2y^2)+y.square root(x+4x^2z^2y^2)>=2square root(xy).quartic root[(1+4xyz^2.square root(xy))^2.xy](again, square both sides and using AM-GM+a>=b>=c)
Now, we just need to prove the inequality when x=y!
At that time, after clearing all denominators(x^2z), we need to prove: 2x.square root(z+4x^4z^2)+2(z+x).square root(x+4z^2x^4)>=3x^2z
Easily to see that after squaring both sides, we have: LHS>=4x^2z+8x^2z>9x^2z=RHS
Therefore,the inequality is proved! THe equality does not occur :roll:
@Temperal: It's not strict! Since when we assume that two variables are equal, we see that the equality still not occur. That shows the strict of this inequality(I don't think 3 is the best constant) :wink:
@Teki-Teki:Thank you for your detailed and clear explanation! I forgot to check the homogenization of this inequality.
PS: Please check my solution! Don't be panicked because of bad LATEX.


linear algebra
matrix
trigonometry
vector
complex numbers
linear algebra unsolved

For $n \geq 3, x_i,y_i \in R, i=1,...,n$ prove $\det (A) = 0$ where $A = (a_{ij}), a_{ij} = cos(x_i + y_j)$.
[i hope it is clear now  :blush: ]

Could you pleaxe correct the typos and fully explain the problem. Thanx!

it works only for $n\ge 3$.
and it's suffices to show for $n=3$. For a biggr $n$ you just decompose the determinant until you get only determinants of order three which are all 0, and the conclusion follows.
for $n=3$.
The matrix can be rewritten like this:
(you can rewrite this for any $n$ in general... )
$\left(\begin{array}{lll} \cos x_1 & -\sin x_1 & 0\\ \cos x_2 & -\sin x_2 & 0\\ \cos x_3 & -\sin x_3 & 0\\ \end{array}\right) \left(\begin{array}{lll} \cos y_1 & \cos y_2 & \cos y_3\\ \sin y_1 & \sin y_2 & \sin y_3\\ 0 & 0 & 0\\ \end{array}\right)$
and now it's clear.
I think this problem was given three or four years ago at the county olympiad, at the 10th grade... I might be wrong though...

The problem Kuba was talking about: (I think)

(a) Let $x_1,x_2,x_3,y_1,y_2,y_3 \in \mathbb R$, $a_{ij} = \sin \left( x_i - y_j \right)$, $i,j=1,2,3$ and $A= \left( a_{ij} \right) \in \mathcal M_3$.

Prove that $\det A = 0$.

(b) Consider the non-zero complex numbers $z_1,z_2,\ldots,z_{2n}$, $n \geq 3$, such that $\left| z_1 \right| = \left| z_2 \right| = \ldots = \left| z_{n+3} \right|$  and $\textrm{arg} \, z_1 \geq \ldots \geq \textrm{arg} \, z_{n+3}$. Also consider the numbers $b_{ij} = \left| z_i - z_{j+n} \right|$, for $i,j \in \left\{ 1,2,\ldots,n \right\}$ and $B = \left( b_{ij} \right) \in \mathcal M_n$. Prove that $\det B = 0$.

[i]RMO 2004, 11th Grade, District Round, Problem 2[/i]

Yes of course $n \geq 3$, stupid me :)
I didn't know about that contest, my solution involves also geometrical arguments and linear systems: one can construct a polygon $A_1...A_n$ which sides make angles $x_1,...,x_n$ regarding a fix line, $A_1A_2 + A_2A_3 + \cdots + A_nA_1 = 0$ as vectors, put this in complex, $\sum_{k=1}^n a_k(cos(x_k) + i * sin(x_k)) = 0$, multiply by $cos(y_k) + i * sin(y_k), k = 1,...,n$ and get a linear system with our determinat and a solution not null [$a_1,...,a_n$] so the determinant must be 0.