geometry
circumcircle
3D geometry
trigonometry
ratio
number theory
relatively prime

[b] DEADLINE IS MONDAY JANUARY 24th, 2005 [/b]

PPPC
Round 2
Power Question (1)
Value: 8 points

Consider the integers from $1$ to $10,000,000,000$. Which are there more of: those in which the digit 1 occurs or those in which it does not occur?

PPPC
Round 2
Power Question (2)
Value: 10 Points

(a) What is the smallest $n>0$ for which $n^2-n+41$ is not prime? (3)

(b) What is the smallest $n>0$ for which $n^2-79n+1601$ is not prime? (4)

(c) What is the remainder when $5^{8190}$ is divided by $8191$? (3)

PPPC
Round 2
Power Question (3)
Value: 10 points

Consider $a,b,c>0$

Prove $27abc \leq (a+b+c)^3 \leq 9(a^3+b^3+c^3)$

PPPC
Round 2
Power Question (4)
Value: 12 points

Prove that $7^n+3n-1$ is divisible by 9.

PPPC
Round 2
Problem (1)
Value: 3 points

What is the value of $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$?

PPPC
Round 2
Problem (2)
Value: 4 points

Express $\frac{(4-5i)^2}{(2-3i)^2}$ in the form $a+bi$.

PPPC
Round 2
Problem (3)
Value: 4 points

What is the correspondence (# of elements in A compared to # of elements in B) between the set of positive integers, $1, 2, 3, ..., n$ and the set of even integers, $2, 4, 6, ..., 2n$ (both sets are infinite)?

PPPC
Round 2
Problem (4)
Value: 5 points

Consider $\triangle ABC$ with sides of length $13$, $14$, and $15$. What is the distance between the centers of the incircle and circumcircle of $\triangle ABC$?

PPPC
Round 2
Problem (5)
Value: 3 points

How many integers $0 \leq x \leq 1776$ are relatively prime to $1776$.

PPPC
Round 2
Problem (6)
Value: 4 points

Bob has red, blue, green, yellow, purple, and orange paint. If he wants to paint each face of his Rubik's cube a different color, in how many different ways can he do this if two colorings are considered the same when one can be obtained from the other by rotating the cube?

PPPC
Round 2
Problem (7)
Value: 5 points

Find all $x$ which satisfy $(4(3x+4))^{\frac{1}{3}}-(3(4x-7))^{\frac{1}{3}}=1$

PPPC
Round 2
Problem (8)
Value: 6 points

Find all $x$ which satisfy $16 \cos \frac{x}{7} \cdot \cos \frac{2x}{7} \cdot \cos \frac{4x}{7} \cdot \cos \frac{8x}{7}=1$

PPPC
Round 2
Problem (9)
Value: 4 points

What is the 1967th digit in the sequence of positive integers $1, 2, 3, 4, 5, ...$?

PPPC
Round 2
Problem (10)
Value: 2 points

How many zeroes are at the end of $10,000 !$ ?

I've decided to add 2 more power questions this round...the deadline will be Sunday, January 16th, 2005.

What kind of answer are you looking for in problem 3? An expression, text?

A ratio, of the number of elements of one to the number of elements in the other. Does that answer your question?

One other thing, this round I would like questions submitted in sets of at least 4 so they do not flood my inbox. Also, as stated previously, each contestant will have 2 chances to answer each question correctly after which his answers will not be considered. This does not apply to the power questions where partial credit may be awarded two solutions containing certain steps that may lead toward a solution of the problem. Thanx! :lol:

If you sent me contest answers last night could you please resend them to me because I accidentally cleared all of my inbox messages since I thought them all from last round. So far I only have answers from JGeneson. Sorry about this guys! ;)

NOTE: I have decided to make a change this round. Please submit ALL power questions and problems at once. You will still have 2 submissions for both Power 2:1-2:4 and Problems 2:1-2:10. Hope to have many new contestants this round! Happy solving!!! :lol:  :lol:  :lol:

do u mean by solve all 14 and sent or....

Solve the power and send them as one PM with title Power Questions and solve the Problems and send them as one PM with title Problems.

sry... i just broke that rule.... :(

i just sent my out...

Not a problem. Just resend them with the others when u finish.

Erm, I broke that rule too :P ...

I'll do that, though.

Oh!

For number 10, do you mean how many 0s [b]total[/b]? That would be just insane...

:(  :(  :(  Due to enormous amounts of hw and projects this round will be delayed for a few weeks. I am sorry to all for this delay!!!

Sorry guys...i'm dead with hw and  swamped with competitions...i'll restart soon!


function
real analysis
integration
LaTeX
calculus
absolute value
real analysis unsolved

Proof that when f is a complex Lebesgue-integrable function on (0,a) (a>0) and
g:=Int_(x,a) f(t)/t dm(t)
then:
a) g(x) is defined for all x in (x,a)
b) g: (0,a)->C continuous
c) g Lebesgue measurable on (0,a)
d) Int_(0,a)g(x)dm(x)=Int_(0,a)f(x)dm(x)

[Moderator edit: always put spaces after your colons ( : ) unless you intend to use an emoticon.]

More moderator edits. I think this is what you mean:

Prove that when $f$ is a complex Lebesgue-integrable function on $(0,a),\,(a>0)$ and

$g: =\int_x^a\frac{f(t)}t\,dm(t),$ then:

a) $g(x)$ is defined for all $x\in(0,a).$ [Note: $(0,a),$ not $(x,a).$]

b) $g: (0,a)\mapsto\mathbb{C}$ is continuous

c) $g$ is Lebesgue integrable on $(0,a).$

d) $\int_0^ag(x)\,dm(x)=\int_0^af(x)\,dm(x).$

Claudia: the best reason for learning a little $\LaTeX$ if you're going to post here is that you can get more people to look at your problems that way.

Parts a) and b) are pretty much routine. I'll focus on parts c) and d).

It's obvious that there will be iterated integrals present, which in turn leads to the assumption that the proof will involve either Fubini's Theorem or Tonelli's Theorem. In fact, we'll use both.

First note that $|g(x)|\le\int_x^a\frac{|f(t)|}t\,dt$. Hence,

$\int_0^a|g(x)|\,dx\le\int_0^a\int_x^a\frac{|f(t)|}t\,dt\,dx$

$=\int_0^a\frac{|f(t)|}t\,\int_0^tdx\,dt=\int_0^a|f(t)|dt<\infty$

since by assumption $f$ is integrable. The change of order of integration was justified by Tonelli's Theorem.

Now that we know that the double integral is absolutely convergent, we can repeat the exact same steps without the absolute value, this time justifying the change of order of integration using Fubini's Theorem. The result is part d).


Let a,b,c be positive real number, prove that,
\[ \frac{1\plus{}a\plus{}b}{a\plus{}ab}\plus{}\frac{1\plus{}b\plus{}c}{b\plus{}bc}\plus{}\frac{1\plus{}c\plus{}a}{c\plus{}ca} \geq \frac{6}{\sqrt{1\plus{}abc}}\]

[quote="mathpk"]Let a,b,c be positive real number, prove that,
\[ \frac {1 \plus{} a \plus{} b}{a \plus{} ab} \plus{} \frac {1 \plus{} b \plus{} c}{b \plus{} bc} \plus{} \frac {1 \plus{} c \plus{} a}{c \plus{} ca} \geq \frac {6}{\sqrt {1 \plus{} abc}}
\]
[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^\plus{}.$ :wink:

Then $ \frac{1\plus{}c\plus{}a}{c(1\plus{}a)} \equal{} \frac{1}{c}\plus{}\frac{1}{1\plus{}a} \rightarrow \infty$. :wink:


Let the 2x2 matrix

$ A \equal{} \left[\begin {array}{cc} a & b \\
c & d \end {array} \right]$

How can I found a necessary and sufficient condition in order the matrix A to be similar to the following matrix

$ B \equal{} \left[\begin {array}{cc} 2 & 1 \\
0 & 2 \end {array} \right]$

I would be grateful if someone show me a step-by-step solution.

Thanks in Advance

$ a\plus{}d\equal{}4,ad\minus{}bc\equal{}4$ and ($ b\not\equal{}0$ or $ c\not\equal{}0$).

[quote="loup blanc"]$ a \plus{} d \equal{} 4,ad \minus{} bc \equal{} 4$ and ($ b\not \equal{} 0$ or $ c\not \equal{} 0$).[/quote]

Could you explain me how extract these results

Similar matrices share many properties, including trace, determinant, and eigenvalues. Also, if B is diagonalizable and A is similar to B, then A is also diagonalizable. Observe that your matrix B is not diagonalizable, so your matrix A cannot also be diagonalizable.


Solve the equation $ 4\sin(\pi x) \equal{} 4x^{2} \minus{} 4x \plus{} 5$

Rewrite and observe that

$ 0\ge 4(\sin \pi x \minus{} 1) \equal{} (2x \minus{} 1)^2\ge 0$ so it is necessary that $ x \equal{} \frac {1}{2}$ which checks as a solution.


symmetry
number theory unsolved
number theory

Find all pairs m, n of positive integers such that (m^2 - n) divides (m + n^2) and (n^2 - m) divides (m^2 + n).

I would think this could certainly be easy olympiad level.

Solution:
[hide]
The conditions may be written as:
A) $m+n^{2}=k_{1}(m^{2}-n)$
B) $m^{2}+n=k_{2}(n^{2}-m)$

There is total symmetry in the two conditions, so there are two cases: m=n or m>n.

Case 1: m=n

Then we get $\frac{n(n+1)}{n(n-1)}=k$ and the only solutions are clearly $n=m=2,3$.

Case 2: m>n

Then m=n+a for a positive integer a. This gives, from the two conditions:

A) $\frac{n^{2}+n+a}{n^{2}+(2a-1)n+a^{2}}=k_{1}$ and

B) $\frac{n^{2}+3n+a^{2}}{n^{2}-n-a}=k_{2}$.

From (A) for all values $a$ the denominator is clearly larger than the denominator unless a=1 in which case the whole fraction is 1 for all n. So we need only investigate a=1 for (B).

Then (B) gives $\frac{n^{2}+3n+1}{n^{2}-n-1}=k_{2}=1+\frac{4n+2}{n^{2}-n-1}$

So this is only an integer if $\frac{4n+2}{n^{2}-n-1}$ is an integer. Since for n>5 the denominator is greater, we need only investigate n=(1,2,3,4,5). Only n=1,2 give integers. Then (n,m)=(1,2);(2,3).

Thus all solutions (n,m) are $\boxed{(n,m)=(1,2);(2,1);(2,2);(2,3);(3,2);(3,3)}$.[/hide]


search
combinatorics unsolved
combinatorics

[color=indigo]A boy plays the following game:[/color]

[color=darkblue]He has 3 stiks: [i]left[/i], [i]middle[/i] and [i]right[/i].
Firstly, on the left stick are placed $ n$ rings, numbered from $ 1$ to $ n$ in random order.
He can take the upper ring from [i]left [/i]or [i]middle[/i] stick and move it to next one. From the [i]right[/i] stick ring can not be moved.
The aim of the game is to move all rings from the [i]left[/i] stick to [i]right[/i], such that, finnaly, rings are placed on the right stick in correct order (counting from down). For example for $ n \equal{} 3$ and configuration $ 3 \minus{} 1 \minus{} 2$ he can play like this:

$ \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline 3 & \phantom{1} & \phantom{1} \\
1 & \ & \ \\
2 & \ & \ \\
\hline \end{tabular} \Longrightarrow \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline & & \\
1 & & \\
2 & 3 & \phantom{1} \\
\hline \end{tabular}\Longrightarrow \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline & & \phantom{1} \\
& 1 & \\
2 & 3 & \\
\hline \end{tabular}\Longrightarrow \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline & & \\
& & \\
2 & 3 & 1 \\
\hline \end{tabular}\Longrightarrow$

$ \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline \phantom{1} & & \\
& 2 & \\
& 3 & 1 \\
\hline \end{tabular}\Longrightarrow \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline \phantom{1} & & \\
& & 2 \\
& 3 & 1 \\
\hline \end{tabular}\Longrightarrow \begin{tabular}{\parallel{}l \parallel{} l \parallel{} l \parallel{}} \hline \phantom{1} & & 3 \\
& & 2 \\
& & 1 \\
\hline \end{tabular}$[/color]

[color=darkred]For given $ n$, how many configurations do exist, such that boy could win the game?[/color]

[color=green][b]Example:[/b] for $ n \equal{} 3$ answer is $ 5$.
The following configurations could lead to victory: $ 123$, $ 132$, $ 213$, $ 312$, $ 321$.
With configuration $ 231$ boy will never win. [/color]

This is essentially stack-sortability.  The set of these permutations can also be characterized as the set of 231-avoiding permutations, and they're enumerated by the Catalan numbers.  To see the Catalan recursion, proceed as follows: consider the position of the largest element $ n \plus{} 1$ in the permutation.  Then we can write our permutation as $ u(n \plus{} 1)v$ for some words $ u, v$.  It's not hard to see that in order for the overall configuration to be winning, $ u$ must contain the values $ 1, 2, \ldots, i$ for some $ i$ (possibly 0 or $ n$) while $ v$ contains $ i \plus{} 1, i \plus{} 2, \ldots, n$.  Moreover, $ u$ and $ v$ must themselves be winning permutations.  Moreover, if our permutation has this structure, then it is certainly a winning permutation.  Thus, if $ W_n$ is the number of winning permutations of length $ n$, we have that $ W_{n \plus{} 1} \equal{} \sum_{i \equal{} 0}^n W_i W_{n \minus{} i}$, with $ W_0 \equal{} 1$.  This verifies our claims.

Previous posts with non-zero relevence to this one: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199042 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162478

Well I don't understand JBL's post fully. The word in the very first sentence "stack-stability" goes over my head. :blush:

Here's how I solved it if someone is interested:

The answer to this problem is equal to the reverse problem:

[color=indigo]A boy plays the following game:[/color]

[color=darkblue]He has 3 stiks: [i]left[/i], [i]middle[/i] and [i]right[/i].
Firstly, on the right stick are placed $ n$ rings, numbered from $ 1$ to $ n$ in increasing order from bottom.
He can take the upper ring from [i]right [/i]or [i]middle[/i] stick and move it to previous one. From the [i]left[/i] stick ring can not be moved.
The game ends when all rings are moved from the [i]right[/i] stick to [i]left[/i]. How many different endgames can occur?[/color]

In each move there are two choices: whether to move from the right stick or from the middle stick. Call the former move1 and the later move2

Each number goes through exactly one move1 and exactly one move2. Hence there are $ n$ move1 s and $ n$ move2 s.

If we write the sequence of moves: it will be something like 1211122122
The middle stick can't have negative number of rings. So:
given any subsequence of this starting from the left side, the number of 2 s will not exceed the number of 1 s.
These are called dyck words of length $ 2n$. The number of such dyck words are $ C_n$, the $ n$-th Catalan number.
So there can be $ C_n$ different games.
(Also two different games can not lead to same endgame, because, suppose the first $ k$ moves are same for two games, and the next choice of move is different. Suppose after the first $ k$ moves there is the ring $ u$ on top of middle stick and ring $ v$ on top of right stick. If move1 is chosen $ u$ will end up above $ v$ at last in the left stick. And if move2 is chosen then $ v$ will end up above $ u$ at last in the left stick. Thus the 2 endgames are different)

So the answer is $ \boxed{\binom{2n}{n} \minus{} \binom{2n}{n \minus{} 1}}$, which is the formula for the $ n$-th Catalan number.

[quote="Akashnil"]Well I don't understand JBL's post fully. The word in the very first sentence "stack-stability" goes over my head.[/quote]  A stack is a just a "first-in, last-out" storage object, like the sticks in this problem.  "Stack-sortability" is the process of sorting using a stack, which is essentially what we're doing in this problem.  The first two sentences of my post are for culture -- the rest of the first paragraph is a (very terse) solution.  What part of it is not clear?

Yours is a nice bijective solution.

Hmm... what was troubling me was that you considered $ u,v$ as words. Thought they were variables. I hadn't seen something like that. And the stack word made me feel it was too advanced. Didn't think much. Now I understand it well. nice.
Thanks for that about stack-stability. I think I heard it sometimes when my elder brother were doing a programming course

Is this game tower of Hanoi?

[url=http://en.wikipedia.org/wiki/Tower_of_Hanoi]Tower of Hanoi[/url] - Wikipedia
[url=http://mathworld.wolfram.com/TowerofHanoi.html]Tower of Hanoi[/url] - MathWorld
[url=http://demonstrations.wolfram.com/TowersOfHanoi/]Towers of Hanoi[/url] - Towers of Hanoi

No: in Towers of Hanoi, the rings on each stick must be in increasing order at all times.  The way this is phrased certainly evokes Towers of Hanoi, but it looks to me like the similarities are pretty superficial.

Here are some threads on the forum dealing with Towers of Hanoi-related problems:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167584
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=248253722&t=139613
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=248253722&t=2732
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=248253722&t=32680


Calculate the exact value of $\sin(1^\circ)$.

[hide="What I got"]
After various applications of the double-angle, half-angle, and triple-angle identities, I arrived at
\[-\frac{\left(1+i \sqrt{3}\right) \sqrt[3]{\sqrt{-8-\sqrt{3}-\sqrt{15}-\sqrt{2 \left(5-\sqrt{5}\right)}}-\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{2 \left(5-\sqrt{5}\right)}}}}{2^{8/3}}\]
\[-\frac{1-i \sqrt{3}}{2 \sqrt[3]{2 \left(\sqrt{-8-\sqrt{3}-\sqrt{15}-\sqrt{2 \left(5-\sqrt{5}\right)}}-\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{2 \left(5-\sqrt{5}\right)}}\right)}}\]
This involves complex numbers, which can be confusing, but they all cancel out ($\sin(1^\circ)$ is obviously a real number, after all)

Try showing the work, and for even more fun, show how the complex numbers cancel each other out.  :lol:
[/hide]

Reduce 30 degrees to 15 by 1/2 angle formula.
Reduce 15 to 5 by 1/3 angle forumla.
Now Im not sure if a 1/5 reduction formula is expressible in radicals, can it be?

[quote="mets501"][quote="sylow_theory"]
Now Im not sure if a 1/5 reduction formula is expressible in radicals, can it be?[/quote]
I don't think so, but I couldn't say for sure.  I used only double-angle (and 1/2-angle) and triple-angle (and 1/3-angle) formulas, which are clearly expressible in radicals.[/quote]

Yeah, but how do you get to 1 degree? Its not possible.(Unless you do it infinite number of times).

[quote="sylow_theory"]Yeah, but how do you get to 1 degree? Its not possible.(Unless you do it infinite number of times).[/quote]
We know the following:
$\sin 30^\circ=\frac{1}{2}$
$\cos 30^\circ=\frac{\sqrt{3}}{2}$
$\sin 36^\circ=\sqrt{\frac{5-\sqrt{5}}{8}}$
$\cos 36^\circ=\sqrt{\frac{3+\sqrt{5}}{8}}$
(the latter 2 can be derived, but take them for granted for now)
Now can you show it?


Suppose 20% of the eighth-grade students participate in Math Olympics and 95% of these students get the mathematics teacher of their choice next year. Only 50% of the eighth-grade students who do not participate in Math Olympics get the mathematics teacher of their choice next year. What percent of eighth-grade students get the mathematics teacher of their choice next year?

[hide]95% of the 20% who participate get their choice, this is 19% of all 8th graders participate and get their choice.
50% of the 80% who do not participate get their choice, so that's 40%...
So, the total is 59% who get their choice teacher.[/hide]


[quote="shobber"][b]Randy:[/b] "Hi Rachel, that's an interesting quadratic equation you have written down.  What are its roots?''

[b]Rachel:[/b] "The roots are two positive integers.  One of the roots is my age, and the other root is the age of my younger brother, Jimmy.''

[b]Randy:[/b] "That is very neat!  Let me see if I can figure out how old you and Jimmy are.  That shouldn't be too difficult since all of your coefficients are integers.  By the way, I notice that the sum of the three coefficients is a prime number.''

[b]Rachel:[/b] "Interesting.  Now figure out how old I am.''

[b]Randy:[/b] "Instead, I will guess your age and substitute it for $x$ in your quadratic equation $\dots$ darn, that gives me $-55$, and not $0$.''

[b]Rachel:[/b] "Oh, leave me alone!''

(1) Prove that Jimmy is two years old.

(2) Determine Rachel's age.[/quote]

[hide]
First, let the polynomial be $P(x) = ax^2 + bx + c$ and let Rachel and Jimmy's ages be represented by $r$ and $j$. we know that $r + j = -b/a$ and $rj = c/a$ so we know that $a + b + c = p = a -a(r + j) + arj = a(rj - r - j + 1) = a(r-1)(j-1) = p$

Since we know that $r > j$ jimmy must be two years old in order for the sum to be prime. Also then, $a = 1$ to maintain primeness.

So now the polynomial becomes $P(x) = x^2 - (r + 2)x + 1$ so rachel can be ages:$6,8,12,14,18,..$ let Randy's guess be $g$
then we have $g^2 - (r+2)g + 1 = 55 \implies \frac{g^2 - 54 -2g}{g} = r$ so that $g - 2 - 54/g = r$ Rachel's age is an integer so $54/g$ must be an integer $54 = 2(3)^3$ and if we test the divisors of $54$ for $g$ we find that the only tenable age for rachel is $24 = 27 - 2 - 2$[/hide]

[b]Theorem:[/b]
Let $W(x)\in \mathbb{Z}[x]$. If $a$, $b$ are distinct integers, then:  $a-b|W(a)-W(b)$

Let $x_1$ be smaller root of our equation. And $x_2$ the other root. By the theorem we have:
$1-x_1|f(1)-f(x_1)\rightarrow x_1-1|f(1)$. Analogously $x_2-1|f(1)$. Since $f(1)$ is a prime we have few possibilities:
$x_1-1=1$ or $x_1-1=f(1)$ and $x_2-1=1$ or $x_2-1=f(1)$. Since $x_2>x_1$ we have $x_1=2$ and $x_2=f(1)+1$
By Viete'a theorem : $2+x_2=-\frac{b}{a}$, $2x_2=\frac{c}{a}$, $b=-a(2+x_2)$, $c=2ax_2$
$f(1)=a+b+c=a-a(2+x_2)+2ax_2=a-2a-ax_2+2ax_2=-a+ax_2=a(x_2-1)$. Since $f(1)$ is a prime $a=1$.
So our function looks like that: $f(x)=x^2+bx+c$ , $f(2)=0$
$4+2b+c=0\Rightarrow c=-4-2b$ $x_2=f(1)+1$ We know. that $f(y)=-55 \quad y\geq 3$
$(y-2)(y-x_2)=-55$ $y=3, 5, 13$By analysing this cases we get $x_2=18$, and $f(x)=x^2-20x+36$, $f(7)=-55$, $f(1)=17$, so this function satisfies all conditions.And the answers: $a) \boxed{2}$ $b) \boxed{18}$

[b]Agolsme[/b]
Your result is wrong, because if $g$ is Randy's guess, then $f(g)=-55$ not $55$.

[hide="Solution"]

Let $f(x) = a(x-s)(x-t)$ be Rachel's quadratic, where $s$ and $t$ are the ages of Jimmy and Rachel, respectively $(s < t)$. Since the sum of coefficients is equal to $f(1)$,
$f(1) = a(1-s)(1-t) = a(st-(s+t)+1) = p$ for some prime number $p$.
Thus either $a = 1$, and $(st-(s+t)+1) = p$, or $a = p$, and $(st-(s+t)+1) = 1$. If $a = p$, and $(st-(s+t)+1) = 1$, then $(s-1)(t-1) = 1$, and since $s,t$ are positive integers, $s=t=2$, contradiction since $s<t$.
Thus, $a = 1$, and $(st-(s+t)+1) = p$, or $(s-1)(t-1) = p$. Since $p$ is a prime, and $s<t$, we must have $s-1 = 1$ and $t-1 = p$, or $ s = 2$. Therefore, Jimmy is $2$ years old.

Now we have: $f(x) = (x-2)(x-t) = x^2 - 2x - tx + 2t = t(2-x)+x^2-2x$.
For some integer $m$ that Randy guesses, $f(m) = -55$.
Thus, \[ \begin{aligned}
f(m) = t(2-m)+m^2-2m = -55 &\Rightarrow t(2-m) = -55+2m-m^2 \\ &\Rightarrow t = \frac{m^2-2m+55}{m-2} = m + \frac{55}{m-2} \end{aligned} \]

Since $t$ is an integer (Rachel's age), $m-2|55$, so $m = 3,7,13,57$. Testing the cases:

If $m = 3, t = 58$, and $f(x) = (x-2)(x-58) = x^2-60x + 116$. However, $f(1) = 1-60+116 = 57 = 3\times 19$, not a prime. Thus $m \not= 3$.
If $m = 7$ or $m = 13, t = 18$, and $f(x) = (x-2)(x-18) = x^2-20x + 36$. Also, since $f(1) = 1-20+36 = 17$ which is a prime, then Rachel is $18$ years old (which satisfies all conditions).
If $m = 57, t = 58$, and $f(x) = (x-2)(x-58) = x^2-60x + 116$. However, $f(1) = 1-60+116 = 57 = 3\times 19$, not a prime. Thus $m \not= 57$.

Therefore, Rachel is $18$ years old.[/hide]

[hide="'Solution"]

WLOG $P(x)=c(x-a)(x-b)$, where $a,b$ are positive integers. Note that the polynomial is the same as $cx^2+c(-a-b)x+acb$ which has sum $c(1-a-b-ab)$ or $c(1-a)(1-b)$.    We must have one of $1-a, 1-b$ being $-1$ since if they both weren't then the sum of the coefficients wouldn't be $1$ (and we can't have it being $1$ since that would mean that the roots would be $0$).  Therefore, in all cases, one of the roots is $2$.

Now, to find the other root.  Let the root that is $2$ WLOG be $a$, so $P(x)=c(x-2)(x-b)$.  The sum of the coefficients is prime implies that $-c(1-b)$ is prime.  Let $1-b=-m$ where $m\le 0$.  Therefore, we get $cm$ is prime.  Since $m=0$ is out of option, we must have $m\ge 1, c\ge 1$.  If $m=1$, then we get $1-b=-1$ so $b=2$.  Therefore, we must have for some $k$, that $c(x-2)^2=-55$.  However, this implies $x=3, c=-55$ or $x=1, c=-55$, therefore in both cases we get the sum of the coefficients being $55$ which is not prime.

Therefore $c=1$, and we have $m$ being prime.  $P(x)=(x-2)(x-b)$.  The choices for the two factors are $(-5, 11)$, $(5, -11)$, $(-55, 1)$, $(55, -1)$, (11, -5), (-11, 5)

If $x-2=-5$, then $x=-3$, so we get $-3-b=11$, absurd. If $x-2=5$, then $x=7$, so $7-b=-11$, $b=18$.  If $x-2=-55$, then $x=-53$, so $-53-b=1$ absurd.  If $x-2=55$, then $x=57$, so $57-b=-1$, or $b=58$. If $x-2=11$, then $x=13$, so $13-b=-5$ again giving $b=18$.  If $x-2=-11$ then $x=-9$, then $-9-b=5$ again absurd.

The fact that we haven't used on this last part is that the sum of the coefficients of $P(x)$ is prime, or $2b-b-2+1$ is prime or $b-1$ is prime.  The first one gives $p=57$ the second gives $p=17$ which is prime, while the first isn't, so $b=\boxed{18}$ as well.

As a quick check:  $(x-2)(x-18)=x^2-20x+36$ which has sum of coefficients $17$ (prime), and for $x=13$, we get $11(-5)=-55$ and $x=7$ we get $5(-11)=-55$.    [/hide]

Its a rather old thread and I am really sorry for reviving it but couldn't prevent adding up my solution to this sweet and easy problem.

Let $P(x)$ denote the polynomial. Then we set $ P(x)= ax^2+bx+c$.Let $\alpha,\beta$ be its roots. Then,
$\alpha + \beta = \dfrac{-b}{a}$ and $\alpha\beta = \dfrac{c}{a}$. Since $\alpha,\beta\in\mathbb{Z} \implies
a|b,a|c$.From here we obtain that $a=1$. Also, $a-a(\alpha+\beta)+a(\alpha\beta)\in P$
This Factors into $a(\alpha-1)(\beta-1)=p$. Wlog, Let $\alpha$ be greater than $\beta$. Hence to maintain the primeness, we must have $\beta=2$. This concludes the first part.
Now we rewrite $P(x)$ as $P(x)= x^2-(2+\alpha)x+2\alpha$. Since $P(k)=-55, k\in \mathbb{Z}$,
We have $P(k)= k^2-k(2+\alpha)+2\alpha=-55$. $\implies (k-2)(k+\alpha)=-55.$
Solving this, we obtain that $\alpha = 18$.


function
algebra
domain
trigonometry
topology
calculus
calculus computations

Find a continuous function defined on $ \mathbb{R}$ such that the image of a closed set is not closed. =)

That just means it's not a closed map, which is of course not necessary for continuity.  Define $ \pi_{1}: \mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ to be the projection onto the first coordinate.  You can check that this is an open map, but not a closed map.  The subset $ C \equal{} \{x\times y\mid xy \equal{} 1\}$ is a closed subset of $ \mathbb{R}\times\mathbb{R}$, but $ \pi_{1}(C) \equal{} \mathbb{R}\backslash\{0\}$ is not closed in $ \mathbb{R}$.

That is a rather involved solution :P
Is there any typical function such as $ f(x) \equal{} \frac{sin(x)}{x}$? Not necessarily that one.

JRav, I fail to see how that's a function on $ \mathbb{R}$.

$ e^x$ is continuous on $ \mathbb{R}$, but the image of $ \mathbb{R}$ is $ (0, \infty)$.  The reason this example exists is that $ \mathbb{R}$ is not compact, i.e. the reason this problem isn't totally trivial is because it is true that the image of compact sets are compact, so any counterexample to this problem needs to start with a closed subset of $ \mathbb{R}$ that isn't compact, i.e. bounded (by Heine-Borel).

I don't fully understand how $ e^x$ is a cts. function from a closed set though :S Isn't $ \mathbb{R}$ open ?

Edit: OOOOOh i was playing around with the definition again. And the compliment of $ \mathbb{R}$ is the empty set which is both open and closed. Hence $ \mathbb{R}$ is closed as well. Thanks a lot again :):)

Edit 2: Typed that as you typed your post =) It is kind of amusing. My analysis professor is possibly the worst person to be learning from. She doesn't know the material herself and hardly knows the definitions of things when asked. She told us the real line is strictly open teehehe.

It is open.  It's also closed.  ("Sets are not doors," as they say.)  Are you familiar with the definitions of open and closed?

Open and closed are relative definitions.  If $ Y\equal{}[0,1]\cap (2,3)$, then $ [0,1]$ is open relative to $ Y$ but not $ \mathbb{R}$.

Which naturally makes sense the more I think of it. It depends on the space in which the objects are being dealt with.
Thanks =)

[quote="t0rajir0u"]JRav, I fail to see how that's a function on $ \mathbb{R}$.[/quote]
I've corrected it.  I neglected to add in that extra "$ \times\mathbb{R}$" for the domain.  I'm pretty sure that this now works.

My understanding is that "defined on $ \mathbb{R}$" means that the domain, not the codomain, is $ \mathbb{R}$, but I'm not sure what the intent of the original problem was.

It's not supposed to be defined on $ \mathbb{R}$, it's supposed to be defined on $ \mathbb{R}\times\mathbb{R}$, which is now corrected.  It just maps any ordered pair $ (x,y)\mapsto x$, so I don't see why that isn't a function.

The original question asks for a function [i]defined on $ \bf R$[/i].

I seemed to have neglected that part :oops: .  But going back to what Fermat's Little Turtle posted, consider $ f(x)\equal{}\sin\left(\frac{1}{x}\right)$.  Obviously, $ [\minus{}1,1]$ is closed in $ \mathbb{R}$.  Now $ f^{\minus{}1}([\minus{}1,1])\equal{}\mathbb{R}\backslash\{0\}$, which is not closed in $ \mathbb{R}$.


quadratics
number theory
least common multiple
algebra

x^4-x^3-x^2-1=0                                                                                2.proove thata^2+b^2is greater than or equal to 2ab                                                                                3.given that p and q are constants proove that 1\2(logp+logq)=log(p-q)divided by3.andlog(p+q)divided bysquare root of 13.                                                                                4.find the number of ways in which at least 3 out 5 people can be sited in a row.                                                                                5.proove that the permutation of n taking 3 at a time divided by the combination of  n taking 2 at time is equal to 6                                                                                6.do you know any good formular for binomial expansion.                                                                             7.if a and b are roots of a quadratic equation,we know that a+b=-b\a and ab=c\a.what is a-b and a\b

1. x^4-x^3-x^2-1=0
2. prove that a^2+b^2 is greater than or equal to 2ab
3. given that p and q are constants proove that 1\2(logp+logq)=log(p-q) divided by3. antilog (p+q) divided   by square root of 13.
4. find the number of ways in which at least 3 out 5 people can be sited in a row.
5. proove that the permutation of n taking 3 at a time divided by the combination of n taking 2 at time is   equal to 6
6. do you know any good formular for binomial expansion.
7.if a and b are roots of a quadratic equation,we know that a+b=-b\a and ab=c\a.what is a-b and a\b

Answers
1. $ x^4 \minus{} x^3 \minus{} x^2 \minus{} 1 \equal{} (x\plus{}1).(x\minus{}1).(x^2\plus{}1) \minus{} (x\plus{}1).x^2 \equal{} (x\plus{}1).(x^3\minus{}2.x^2\plus{}x\minus{}1)$
2. Consider $ a^2\plus{}b^2\minus{}2.a.b \equal{} (a\minus{}b)^2 >\equal{} 0. Hence a^2\plus{}b^2 >\equal{} 2.a.b$
4.What do you mean?
5. I think the question must be
    proove that the permutation of n taking 3 at a time divided by the combination of n taking 3 at time is    equal to 6
   Answer is n.(n-1).(n-2)/(n.(n-1).(n-2)/6) = 6
6. Again what formula for binomial expansion do you want?
7. $ a\minus{}b \equal{} (a\plus{}b)^2 \minus{} 4.a.b$
    for a/b put $ a \equal{} b.k$ and solve for k

1)factorise completely.(5)find n.(6)a very good  one. any you know except pascals triangle and combination sys..

1. I believe that is the complete factorization.
5. $ \\\begin{eqnarray}
\frac{(_n P_3)}{(C_{2}^n)}=6
&\leftrightarrow& \frac{n(n-1)(n-2)}{n(n-1)/2}=6\\
&\leftrightarrow& n-2=3\\
&\leftrightarrow& n=5
\end{eqnarray}$
6.$ (x+y)^n=\sum_{i=1}^n {(C_{i}^n)(x^i)(y^n-i)}$(<--is that what you are talking about??)

Or, for #2, even though the proof madness gave is basically the same as the proof for the two variable AM-GM, you could also use AM-GM on $ a^2+b^2$:
$ \frac{a^2+b^2}{2}\ge{\sqrt{a^2b^2}\rightarrow\\a^2+b^2\ge{2ab}}$.

#2(explain your proof.                                                                                #6explain the formular.                                                                                new set.#1(n+3)!/n!=(n+1)(n+2)(n+3).proove                                                                         #2 n!/r!(n-r)!+n!/(r-1)!(n-r+1)!=(n+1)!/r!(n-r+1)! proove also.

To prove nCr + nC(r-1) = (n+1)Cr just take common lcm and simplify or take a combinatorial route  :)

for No.4, is it 5P3+5P4+5P5=300?
and if the order doesnt matter that will be 5C3+5C4+5C5=16


calculus
derivative
logarithms
calculus computations

I am learning the derivative of log, but first I try to understand the inverse properties of log. My question is: Why e^ln(x)=x?
btw, is the derivative of lna^x equal to lna+x/a? But my book wrote it equaled to itself?

[quote="PunPride"]x = e^y

y = ln(x)

therefore,

x = e^ln(x)

does that help?[/quote]

wait, I mean that when you just look at e^ln(x), how can you tell it's equal to x?

[quote="PunPride"]because you would know that y = ln(x), so ln(x) is just plugged in where y should originally be, which is a power of e (e^y). so when this is plugged in you know it is equal to x because x = e^y or x = e^y...

x = e^ln(x) and x = e^y are the same... just that instead of y , you now have ln(x) because y = ln(x) when you solve for x=e^y.

you have to remember that x=e^y is the same as y=ln(x) because ln(x) just means Log(base e)(x)[/quote]

OH, then I got it. Since y=ln(x), then it's the same thing. Thank you.

Here is one more problem, is the derivative of lna^x equal to lna+x/a? But my book wrote it equaled to itself?

[quote="PunPride"][quote="akech"]If you meant to write: $ \ln(a^{x})$, then we have $ \ln(a^{x}) \equal{} x\ln(a)$ so when you take the derivative then you treat $ \ln(a)$ as a constant so that the derivative will be $ \ln(a) * 1 \equal{} \ln(a)$[/quote]

why is ln(a) treated as a constant? and where did the x go?[/quote]

What is derivative of x when you differentiating with respect to x?

if your answer is 1, then you are right if not then check your book. What is the derivative of kx where k is a constant? it is k, right? the reason why ln(a) is treated as a constant is because it does not depend on x.

[quote="MIA6"]Thanks, akech. I got it.[/quote]

Akuna matata;-
Akech


In how many ways can $ 1440$ be written as the product of two numbers?

thank you...

$ 1440\equal{}2^5\cdot 5\cdot 3^2$.
therefore,the answer is the half of $ 6\cdot 2\cdot 3$,or $ \boxed{18}$.

no, the answer is not $ 18$.
its given that answer is $ 26$
and please give the answer with reason

thank you...

If real numbers - $ \infty$
If integers - $ 36$
If positive integers - $ 18$


Find all naturall solutions of $ x^2\minus{}51y^2\equal{}13$.

the problem is easy. just check two sides of the equation (mod 5).first note that x^2=0,1 or -1 (mod 5), 51=1(mod 5)[so 51y^2=0,1,-1(mod 5) again] and that 13=3 (mod 5). and our conclusion follows...

i think you are wrong, because $ 64\minus{}51\equal{}13$


calculus
derivative
real analysis
real analysis unsolved

Let $ f$ be twice continuously differentiable, with compact support. ($ f \in C^2_c(\mathbb{R})$)

Show that $ \frac {f(x \plus{} h) \minus{} f(x)}{h} \rightarrow \frac {d f}{d x}(x)$ uniformly on $ \mathbb{R}$ as $ h \rightarrow 0$.

Taylor's theorem with the Lagrange form of the remainder can be manipulated to show that
\[ \frac{f(x\plus{}h)\minus{}f(x)}h\minus{}f'(x)\equal{}\frac12f''(\xi)h\]
for some $ \xi$ between $ x$ and $ x\plus{}h.$

Since $ f''$ is continuous and compactly supported, it is bounded. Let $ M\equal{}\max_{x\in\mathbb{R}}(|f''(x)|).$ Then we have that
\[ \left|\frac{f(x\plus{}h)\minus{}f(x)}h\minus{}f'(x)\right|\le\frac{M|h|}2\]
That goes to zero as $ h\to0,$ and since that bound depends only on $ h$ and not on $ x,$ the convergence is uniform.

==============================================

A followup question: the stated hypotheses here seem very strong. Can we get away with weakening any of them?

About Kent's question

is the existence and boundedness of $ f''$ enough?

Oh, we can do better than that. $ f\in C_c^1$ suffices, as does $ f\in C^1$ with $ f'(x)\to0$ as $ |x|\to\infty.$ We don't need a second derivative at all.

WOW, Thanks a lot !

But I can't prove the theorem if $ f \in C^1_c$..

Help me, please..

Note that $ f\in C_c^1$ implies that $ f'$ is uniformly continuous. Given $ \epsilon>0$ find $ \delta>0$ such that $ |x\minus{}y|<\delta$ implies that $ |f'(x)\minus{}f'(y)|<\epsilon.$

By the Mean Value Theorem, for each $ x$ and $ h$ there exists $ \xi$ between $ x$ and $ x\plus{}h$ such that $ \frac{f(x\plus{}h)\minus{}f(x)}h\equal{}f'(\xi).$ Now suppose $ |h|<\delta,$ which implies that $ |\xi\minus{}x|<\delta.$
\[ \left|\frac{f(x\plus{}h)\minus{}f(x)}h\minus{}f'(x)\right|\equal{}|f'(\xi)\minus{}f'(x)|<\epsilon\]
and we are done.


If you start writing all the natural numbers in order, what will be the 3001st digit you will write?

[hide]In the first 9 natural numbers, you will write 9 digits.
$10-99\Rightarrow180$ digits.
Since $3001>3(999)>\text{Number of digits total in all natural numbers less than 1000}$, you know the digit will land in the thousands.
$100-999\Rightarrow2700$
Now, $x$ is how far the number is from 999
$9+180+2700+4x=3001$
$2889+4x=3001$
$4x=112$
$x=28$
Thus, the number is $999+28=1027$.  Because $x$ is an integer, it will be the 4th digit in 1027, which is $\boxed{7}$[/hide]

[hide]There are 9 one-digit numbers.
There are 90 two-digit numbers.
There are 900 two-digit numbers.
3001-9-2*90-3*900=112
112/4=28 r0. The 28th four-digit number ends with 7[/hide]


modular arithmetic
induction
quadratics
function
number theory unsolved
number theory

Let $ n$ and $ k$ be positive integers such as either $ n$ is odd or both $ n$ and $ k$ are even. Prove that exists integers $ a$ and $ b$ such as $ GCD(a,n) \equal{} GCD(b,n) \equal{} 1$ and $ k \equal{} a \plus{} b$

Let $p_1,p_2,\ldots,p_t$ be the prime factors of $n$. If there is an $x$ s.t. $p_i\not |k-x,\ p_i\not |x$, then we can take $a=k-x,b=x$. If $n$ is odd, we can do this because there are at least $3$ residues $\pmod{p_i},\ \forall i$, and we can select $x\not\in\{0,k\}\pmod{p_i}$ (we are sure it exists from the Chinese Remainder Theorem). On the other hand, if $n$ is even, but so is $k$, we select $x\equiv 1\pmod 2$, and for the rest of the primes $p_i$ we do what we did above.

So you see, if $n$ is odd, $k$ can be anything we want, which makes me proud of having translated the problem correctly :).

Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=32278]hjbrasch:[/url]

Induction on the number of prime divisors of $ n$ (not elegant, but it works).

As for the induction start assume $ n \equal{} p^{t}$. If $ p \equal{} 2$ holds, set $ a \equal{} 1,b \equal{} k \minus{} 1$ and remember that $ k$ must be even then. If $ p > 2$ choose any residue $ a$ different from $ 0,k\bmod{p}$ and set $ b \equal{} k \minus{} a$.

As for the induction step write $ n \equal{} p^{t}n'$ with $ p$ an odd prime and $ n'$ not divisble by $ p$. Since $ n,n'$ have the same parity, we can assume the existence of integers $ a' \plus{} b' \equal{} k$, which are both coprime to $ n'$. Consider the quadratic function $ x\rightarrow f(x) \equal{} (a' \plus{} xn')(b' \minus{} xn')$, which is not identical to zero modulo $ p$, since $ n'^{2}$ is not divisible by $ p$. Therefore, there exists $ x$ such that $ f(x)$ is not divisible by $ p$ and we can set $ a \equal{} a' \plus{} xn', b \equal{} b' \minus{} xn'$ to complete the proof.


integration
logarithms
trigonometry
calculus
limit
real analysis
real analysis unsolved

hello, for the second integral i have found $ \int_{0}^{1}\frac{\ln(x^2\minus{}x\plus{}1)}{x\minus{}1}\,dx\equal{}\frac{\pi^2}{18}$.
Sonnhard.

The first one is a bit complicated (my solution, anyway), so I'm just gone give an outline and the whole answer tomorrow morning (it's late here).

1) Rewrite $ \cos x \equal{} \frac {1 \minus{} \tan^2\frac {x}{2}}{1 \plus{} \tan^2\frac {x}{2}}$

2) Make the substitution $ \tan\frac {x}{2} \equal{} z$

3) Use $ \sin x \equal{} \frac {2 \tan\frac {x}{2}}{1 \plus{} \tan^2\frac {x}{2}}$ to get rid of the "substitution residue"

4) Now you have $ I \equal{} 2 \int ^ 1 _ 0 \frac {1}{1 \minus{} z^2}\ln \left(\frac {3 \plus{} z^2}{1 \plus{} 3z^2}\right)\,dz$

5) Define $ F(\alpha) \equal{} 2 \int ^ 1 _ 0 \frac {1}{1 \minus{} z^2}\ln \left(\frac {\alpha \plus{} z^2}{1 \plus{} \alpha \, z^2}\right)\,dz$

6) Differentiate $ F$ with respect to $ \alpha$ and calculate the integral with respect to $ z$

7) Now you have $ F'(\alpha) \equal{} \frac {\pi}{\sqrt {\alpha}\left(\alpha \plus{} 1\right)}$

8) Integrate this from $ 1$ to $ 3$ (the integral is equal to $ 2 \pi \tan^{ \minus{} 1}(\sqrt {\alpha})$ )and use the fact that the integral definition of $ F$ gives $ F(1) \equal{} 0$

9) Use the fact that $ I \equal{} F(3) \equal{} \frac {\pi^2}{6}$

First one (a bit less complicated approach than the one proposed in my last post):

Define $ F(\alpha) = \int ^{\frac{\pi}{2}}_0 \frac{1}{\cos x} \ln(\frac{\alpha + \cos x}{\alpha - \cos x})\, dx$

Then, we have:

$ F'(\alpha) = \int ^{\frac{\pi}{2}}_0 \frac{1}{\cos x} \frac{\alpha - \cos x}{\alpha + \cos x} (\frac{1}{\alpha - \cos x}-\frac{\alpha + \cos x}{(\alpha - \cos x)^2})\, dx$

$ = \int ^{\frac{\pi}{2}}_0 \frac{1}{\cos x} \frac{1}{\alpha + \cos x} (1-\frac{\alpha + \cos x}{\alpha - \cos x})\, dx$

$ = \int ^{\frac{\pi}{2}}_0 \frac{1}{\cos x} \frac{1}{\alpha + \cos x} (\frac{\alpha - \cos x - \alpha - \cos x}{\alpha - \cos x})\, dx$

$ = 2 \int ^{\frac{\pi}{2}}_0  \frac{1}{\alpha + \cos x} \frac{1 }{\cos x - \alpha }\, dx$

$ = 2 \int ^{\frac{\pi}{2}}_0  \frac{1}{\cos^2 x - \alpha^2} \, dx$

Now, we know that:

$ \sin x = \frac{2 \tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}$

$ \cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$

From this we can make the substitution $ \tan \frac{x}{2} = z$ and this leads to:

$ \cos x = \frac{1-z^2}{1+z^2}$

$ dx = \frac{2}{1+z^2}\,dz$

For the ease of calculating, we can now look at the indefinite integral:

$ I_{\alpha} (x) = 2 \int   \frac{1}{\cos^2 x - \alpha^2} \, dx$

Then with the substitution, we have:

$ I_{\alpha}(2 \tan^{-1} z)  = J_{\alpha} (z) = 2 \int   \frac{1}{(\frac{1-z^2}{1+z^2})^2 - \alpha^2} \frac{2}{1+z^2}\,dz$

Using some simple manipulation, we can see that:

$ 2\frac{1}{(\frac{1-z^2}{1+z^2})^2 - \alpha^2} \frac{2}{1+z^2} = -\frac{2}{\alpha (\alpha-1)}\frac{1}{z^2 + \frac{\alpha+1}{\alpha-1}} -\frac{2}{\alpha (\alpha+1)}\frac{1}{z^2 + \frac{\alpha-1}{\alpha+1}}$

Plugging this in the inegral, we can see that it's a classic integral of the form:

$ \int \frac{1}{z^2+a} \,dz = \frac{1}{\sqrt{a}} \tan^{-1} (\frac{z}{\sqrt{a}}) + C$

This gives us:

${{ J_{\alpha} (z) = - \frac{2}{\alpha \sqrt{\alpha^2 - 1}} \tan^{-1} (z \sqrt{\frac{\alpha - 1}{\alpha+1}}})- \frac{2}{\alpha \sqrt{\alpha^2-1}}\tan^{-1} (z \sqrt{\frac{\alpha+1}{\alpha-1}}}) + C$

${{ = - \frac{2}{\alpha \sqrt{\alpha^2 - 1}} (\tan^{-1} (z \sqrt{\frac{\alpha - 1}{\alpha+1}}})+\tan^{-1} (z \sqrt{\frac{\alpha+1}{\alpha-1}}}) )+ C$

Now:

$ F'(\alpha) = J_{\alpha} (\tan \frac{x}{2}) |^\frac{\pi}{2} _ {x=0}$

${{ =  - \frac{2}{\alpha \sqrt{\alpha^2 - 1}} (\tan^{-1} (\sqrt{\frac{\alpha - 1}{\alpha+1}}})+\tan^{-1} (\sqrt{\frac{\alpha+1}{\alpha-1}}}) )$

Now, use the fact that for $ x > 0$ we have:

$ \tan^{-1} x + \tan{-1}\frac{1}{x} = \frac{\pi}{2}$

and this gives you:

$ F'(\alpha) = - \frac{\pi}{\alpha \sqrt{\alpha^2 - 1}}$

From the integral definition of $ F$ we can see that:

$ \lim_{\alpha \to \infty} F(\alpha) = 0$

so this gives us:

$ \text{Result} = F(2) = F(2) - F(\infty) = \int ^\infty _ 2 \frac{\pi}{\alpha \sqrt{\alpha^2 - 1}}\, d\alpha =$

$ = - \pi \tan ^{-1} \frac{1}{\sqrt{x^2-1}}|^\infty_2 = \pi \tan ^{-1} \frac{1}{\sqrt{3}} = \frac{\pi^2}{6}$

Summation:

$ \boxed{\boxed{\int ^{\frac{\pi}{2}}_0 \frac{1}{\cos x} \ln(\frac{2+ \cos x}{2- \cos x})\, dx = \frac{\pi^2}{6}}}$

Hi,

I've another way to compute the first integral but it needs some classic tools (see wikipedia for example) :

1) the Wallis integral and the developpement of arcsin are
\[ W_{2n}: \equal{} \int_{0}^{\pi/2} \cos(x)^{2n} dx \equal{} \frac {(2n)!}{n!^2 4^n} \times \frac {\pi}{2}\]

\[ \forall x \in [ \minus{} 1,1] \qquad \arcsin(x) \equal{} \sum_{n\geq 0} \frac {(2n)!}{4^n (n!)^2 (2n \plus{} 1)} x^{2n \plus{} 1}\]
2) And so
\[ \frac {\pi \arcsin(x)}{2x} \equal{} \sum_{n\geq 0} \frac {W_{2n} }{(2n \plus{} 1)}x^{2n}\]
Let's go now.  By derivate we have  for all $ t\in ] \minus{} 1,1[$ and $ x\in ]0,\pi/2[$
\[ \frac {d}{dt} \ln( 2 \plus{} t ) \minus{} \ln(2 \minus{} t) \equal{} \frac {1}{2 \plus{} t} \plus{} \frac {1}{2 \minus{} t} \equal{} \frac {4}{4 \minus{} t^2} \equal{} \frac {1}{1 \minus{} (t/2)^2} \equal{} \sum_{n\geq 0} \frac {t^{2n}}{4^n}\]

\[ \frac {\ln(2 \plus{} t) \minus{} \ln(2 \minus{} t) }{t} \equal{} \sum_{n\geq 0} \frac {t^{2n}}{4^n (2n \plus{} 1)}\]

\[ \frac {1}{\cos(x)} \ln \left(\frac {2 \plus{} \cos(x)}{2 \minus{} \cos(x)} \right) \equal{} \sum_{n\geq 0} \frac {\cos(x)^{2n}}{4^n (2n \plus{} 1)}\]
There's absolute convergence on $ [0,\pi/2]$, so
\[ \int_{0}^{\pi /2} \frac {1}{\cos(x)} \ln \left(\frac {2 \plus{} \cos(x)}{2 \minus{} \cos(x)} \right) dx \equal{} \sum_{n\geq 0} \frac {W_{2n}}{4^n (2n \plus{} 1)} \equal{} \pi \arcsin(\frac {1}{2}) \equal{} \frac {\pi^2}{6}\]

Second one..

Define:

$ F(\alpha) \equal{} \int ^1 _ 0 \frac {\ln \left(\frac {x (x \minus{} 1)}{\alpha} \plus{} 1\right)}{x \minus{} 1} \, dx$

Then:

$ F'(\alpha) \equal{} \minus{} \frac {1}{\alpha} \int ^1 _ 0 \frac {x}{x^2 \minus{} x \plus{} \alpha}\,dx \equal{} \minus{} \frac {1}{2\alpha} \int ^1 _ 0 \left(\frac {2x \minus{} 1}{x^2 \minus{} x \plus{} \alpha} \plus{} \frac {1}{x^2 \minus{} x \plus{} \alpha}\right)\,dx$

$ \equal{} \minus{} \frac {1}{2\alpha} \ln \left(x^2 \minus{} x \plus{} \alpha\right) \Big|^1_0 \minus{} \frac {1}{2\alpha} \int ^1_0\frac {1}{x^2 \minus{} x \plus{} \alpha} \,dx$

The left part equals $ 0$ so we have:

$ F'(\alpha) \equal{} \minus{} \frac {1}{2\alpha} \int ^1_0\frac {1}{x^2 \minus{} x \plus{} \alpha} \,dx \equal{} \minus{} \frac {1}{2\alpha} \int ^1_0\frac {1}{\left(x \minus{} \frac {1}{2}\right)^2 \plus{} \alpha \minus{} \frac {1}{4}} \,dx$

$ \equal{} \minus{} \frac {1}{2\alpha} \int ^{\frac {1}{2}}_{ \minus{} \frac {1}{2}}\frac {1}{z^2 \plus{} \alpha \minus{} \frac {1}{4}} \,dz$

$ \equal{} \minus{} \frac {1}{\alpha} \int ^{\frac {1}{2}}_0\frac {1}{z^2 \plus{} \alpha \minus{} \frac {1}{4}} \,dz$

$ \equal{} \minus{} \frac {1}{\alpha} \frac {1}{\sqrt {\alpha \minus{} \frac {1}{4}}} \tan^{ \minus{} 1} \left(\frac {1}{2\sqrt {\alpha \minus{} \frac {1}{4}}}\right)$

Since from the integral definition of $ F$ we have $ \lim _ {\alpha \to \infty}F(\alpha) \equal{} 0$ we can write:

$ \text{Result} \equal{} F(1) \equal{} \minus{} \left(F(\infty) \minus{} F(1)\right) \equal{} \int ^\infty _ 1 \frac {1}{\alpha} \frac {1}{\sqrt {\alpha \minus{} \frac {1}{4}}} \tan^{ \minus{} 1} \left(\frac {1}{2\sqrt {\alpha \minus{} \frac {1}{4}}}\right) \,d\alpha$

$ \equal{} \minus{} 2 \left(\tan^{ \minus{} 1}\left(\frac {1}{\sqrt {4\alpha \minus{} 1}}\right)\right)^2\Big|^\infty_{\alpha \equal{} 1} \equal{} 2 \left(\tan^{ \minus{} 1}\left(\frac {1}{\sqrt {3}}\right)\right)^2 \equal{} \frac {\pi^2}{18}$

Summation:

$ \boxed{\boxed{\int ^1 _ 0 \frac {\ln \left(x^2 \minus{} x \plus{} 1\right)}{x \minus{} 1} \, dx \equal{} \frac {\pi^2}{18} }}$

$ I \equal{} \int_0^{\frac{\pi}{2}} \frac{1}{\cos x} \ln \left(\frac{2 \plus{} \cos x}{2 \minus{} \cos x}\right)\,dx \equal{} \minus{} \int_0^{\frac{\pi}{2}} \sec x \ln \left(\frac{1 \minus{} \frac12 \cos x}{1 \plus{} \frac12\cos x}\right)\,dx$.

Now we using the result in post #4 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=240804]here[/url], with $ a \equal{} \frac12$ and $ b \equal{} \minus{} \frac12$ we get

$ I \equal{} \minus{} \frac12 \left[\arccos^2\left(\frac12\right) \minus{} \arccos^2\left(\minus{} \frac12\right)\right] \equal{} \minus{} \frac12 \left[\left(\frac{\pi}{3}\right)^2 \minus{} \left(\frac{2 \pi}{3}\right)^2 \right] \equal{} \frac{\pi^2}{6}$.


A graph has n points and 100 edges. A move is to pick a point, remove all its edges and join it to any points which it was not joined to immediately before the move. What is the smallest number of moves required to get a graph which has two points with no path between them?

problem required to find the minimum of moves that could afford the condition(graph which has two points with no path between them)
So answer is 2.
Solution:
it's easy to conclude that we always can find point such that degree of this point will equal 2(if we can't find such point  - we would find point's degree which equal 1 which is easier variant).
then we  make sequentially move with 2 points that joint to point that we had already found. after this 2 moves finded point will have no connection so we have graph which has two points with no path between them.
1 move is not enough(example: cycle of 100 point)/

We cannot always find a vertice with degree 2. Consider a complete graph.

There can be anywhere from 15 to 101 points (assuming already a path between any two points) but I don't see how that really helps.


calculus
integration
trigonometry
logarithms
calculus computations

Wrong forum. Send it to  :arrow:  [url=http://www.mathlinks.ro/index.php?f=296][color=red]calc[/color][/url] forum please.

[quote="FOURRIER"]$ \int{\pi/2}\frac{cosx}{\sqrt{1+sin^{2}(x)}}$[/quote]
$ \int_{0}^{\pi/2}\frac{\cos x}{\sqrt{1+\sin^{2}x}}\,dx$

Let's set $ u=\sin x\implies du=\cos x\,dx$, the integral becomes to

$ \int_{0}^{\pi/2}\frac{\cos x}{\sqrt{1+\sin^{2}x}}\,dx=\int_{0}^{1}\frac{1}{\sqrt{1+u^{2}}}\,du$

A new change of variables defined by $ v=u+\sqrt{1+u^{2}}$ yields

\begin{eqnarray*}\int_{0}^{\pi /2}{\frac{{\cos }}{{\sqrt{1+\sin^{2}x}}}\,dx}&=&\int_{1}^{1+\sqrt 2 }{\frac{1}{v}\,dv}\\ &=&{\ln| v|}|_{1}^{1+\sqrt 2 }\\ &=&\ln({1+\sqrt 2 })\,\blacksquare\end{eqnarray*}


This is a 4 by 4 magic square. You are to use the odd numbers from 1 to 31 $(1, 3, 5, 7, \dots, 31)$. Use each number once and only once. Each Column, each row, and each of the two diagonals must add up to 64.

1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31
1,2,3,4,5,6 ,7  ,8  ,9  ,10,11,12,13,14,15,16

31---
-15,19-
-21,17-
---1

I GIVE UP, WAS I ON THE RIGHT TRACK?
-JORIAN

[quote="bpms"]I think this should be moved to puzzles and brainteasers...[/quote]

Nobody goes to puzzle and brainteasers :(

i tried for a long while and couldn't get it
sry,
jorian

It's not hard

Since all numbers are of the form 2n-1, where n ranges from 1 to 16, we can simply:
1) make a magic square with 1~16
2) substitute with 2n-1

the well known construction of 4*4 magic square is:

[color=red] 1[/color][color=blue] 1615[/color][color=red] 2[/color]
[color=blue]14[/color][color=red] 3[/color] [color=red] 4[/color][color=blue]13[/color]
[color=blue]12[/color][color=red] 5[/color] [color=red] 6[/color][color=blue]11[/color]
[color=red] 7[/color][color=blue] 10 9[/color][color=red] 8[/color]

but why am I doing your sister's homework?

The solution for 4*4 magic square is:
 1  15 14  4
12  6   7   9
 8  10 11  5
13  3   2  16
then follow lightrhee's method

[quote="jhredsox"]thats wrong, the diagonals don't add up to 64!
-jorian[/quote]
oops :blush:  Forgot that this is only for rows and columns.


S22

A bee can fly or walk only in a straight line between any two corners on the inside of a cubic box of edge length 1. The bee managed to move so that it visited every corner of the box without passing through the same point twice in the air or on the wall of the box. What is the largest possible length of such a path?

My answer is 6 times square root of 2+square root of 3

:) But i think you would have to read the question carefully to get this correct =P


trigonometry
inequalities proposed
inequalities

Let $ a$, $ b$, $ c$ be positive real numbers such that $ bc \equal{} 1 \plus{} a\left(b \plus{} c\right)$.
Find the maximum of expression:
$ P \equal{} \frac {6a}{\sqrt {1 \plus{} a^{2}}} \plus{} \frac {4}{\sqrt {1 \plus{} b^{2}}} \plus{} \frac {3}{\sqrt {1 \plus{} c^{2}}}$

ie. $3\cos \alpha+4\cos \beta-6\cos(\alpha+\beta)$ :wink:

Thank you Albanian Eagle, but I don't like trigonometric. Can you propose to me  another solution for this problem? :D

Albanian Eagle

Can you post ypur solution in details please?
I`m very interested

By substituting $tan \alpha$ for b, $tan \beta$for c, and therefore $a=-cot(\alpha+\beta)$.

E ZAIZAI, I THINK YOU DON'T LIKE TRIGONOMETRY BECAUSE YOU DON'T KNOW TRIGONOMETRY . HO`HO`

HOC HANH` NGU NHU BO` MA LEN DAY ON` AO` HA CON

If she works on such problems, then logically, she surely knows trigonometry.  :wink:

I have no idea for this problem, but Sanl, please post some more meaning, i'm Vietnamese and i can understand what you posted. I think you can't stand in this forum more time.

Finally, to hold down abide a pending questions  :(
to Sanl: maybe you are a as mad as a hatter man  :rotfl:

[quote="cefer"][b]@boxedexe[/b]: zaizai-hoang isn't a girl  :D[/quote]
 :rotfl: Sorry about that Well, my statement still holds true. :)

Anyway, even using trigonometry I can't solve the problem. Can somebody post the solution?  :(


[quote="ashwinrk_jain"]its a quite simple problem.
$-1<x<5$[/quote]

can you post your solution?

[quote="shobber"]If $\lim_{n \to \infty}\frac{n \cdot 3^{n}}{n \cdot (x-2)^{n}+n \cdot 3^{n+1}-3^{n}}= \frac{1}{3}$, then find the range of $x$.[/quote]

[hide="Full-written solution"]
Clearly, \[\begin{aligned}\frac{n \cdot 3^{n}}{n \cdot (x-2)^{n}+n \cdot 3^{n+1}-3^{n}}& = \frac{n}{n\cdot\left(\frac{x-2}{3}\right)^{n}+3n-1}\\ & = \frac{1}{\left(\frac{x-2}{3}\right)^{n}+3-\frac{1}{n}}\end{aligned}\] Therefore, \[\begin{aligned}\lim_{n \to \infty}\left(\left(\frac{x-2}{3}\right)^{n}+3\right) & = \lim_{n \to \infty}\left(\left(\frac{x-2}{3}\right)^{n}+3-\frac{1}{n}\right) = \frac{1}{\lim_{n \to \infty}\left(\frac{1}{\left(\frac{x-2}{3}\right)^{n}+3-\frac{1}{n}}\right)}\\ & = \frac{1}{\lim_{n\to\infty}\left(\frac{n \cdot 3^{n}}{n \cdot (x-2)^{n}+n \cdot 3^{n+1}-3^{n}}\right)}= 3 \end{aligned}\] So we should have that \[0 = \lim_{n \to \infty}\left(\left(\frac{x-2}{3}\right)^{n}+3\right)-3 = \lim_{n \to \infty}\left(\frac{x-2}{3}\right)^{n},\] which is only possible if $\left|\frac{x-2}{3}\right| < 1$, i.e. $|x-2| < 3$, i.e. $x\in\left]-1,5\right[$.[/hide]


modular arithmetic
number theory
greatest common divisor
induction
number theory unsolved

We define the sequence $(a_n)_{n=1}^\infty$ s.t. $a_1=a_2=1$ ,$a_3=m$ , $a_{n+1}=\frac {k+a_na_{n-1}}{a_{n-2}}$ and $gcd(m,k)=1$. Prove that $a_n\in\mathbb{Z}$ iff $k\equiv {-1}(mod m)$

Iterating, we get $a_6 = \frac{k+(m^2+km+k)(m+k)}{m}$. Note that $k+(m^2+km+k)(m+k) \equiv k^2 + k = k(k+1) \pmod{m}$. Thus for $a_6 \in \mathbb{Z}$, we must have $k(k+1) \equiv 0 \pmod{m}$. Since $\gcd (m,k) = 1$, it follows that $(k+1) \equiv 0 \pmod{m}$ and we are done.

:oops_sign:
That's only for the forward direction... Haven't solved the backward direction.  :blush:

Let's complete the solution of this nice problem. The direction "all $a_n$ are integers" $\Longrightarrow$ "$k\equiv -1\mod m$" was shown by Dblues in post #2; so it remains to prove the other direction. Actually, this other direction doesn't need the condition gcd (m; k) = 1. Let's formulate this direction as a theorem:

[color=blue][b]Theorem 1.[/b] Let m and k be two positive integers such that $k\equiv -1\mod m$, and let $\left(a_n\right)_n$ be a sequence defined by $a_1=a_2=1$ and $a_3=m$ and by the recurrence relation $a_{n+1}=\frac{k+a_na_{n-1}}{a_{n-2}}$ for any integer $n\geq 3$. Then, for any natural number n, the number $a_n$ is an integer.[/color]

[i]Proof.[/i] Since $k\equiv -1\mod m$, we have $m\mid k+1$. Thus, there exists a positive integer q such that k + 1 = mq. Consider this q.

Now, we will show:

[color=blue][b]Lemma 2.[/b] For any positive integer n, we have

$a_{2n-1}+a_{2n+1}=\left(m+1\right)a_{2n}$;
$a_{2n}+a_{2n+2}=\left(q+1\right)a_{2n+1}$.[/color]

[i]Proof of Lemma 2.[/i] Let u be an integer with $u\geq 1$. Then, by the recursive definition of our sequence, $a_{u+3}=\frac{k+a_{u+2}a_{u+1}}{a_u}$. In other words, $a_ua_{u+3}=k+a_{u+2}a_{u+1}$, so that $a_ua_{u+3}-a_{u+2}a_{u+1}=k$. Applying this to the number u + 1 instead of u, we get $a_{u+1}a_{u+4}-a_{u+3}a_{u+2}=k$. Thus,

$\left(a_u+a_{u+2}\right)a_{u+3}-\left(a_{u+2}+a_{u+4}\right)a_{u+1}$
$=\underbrace{\left(a_ua_{u+3}-a_{u+2}a_{u+1}\right)}_{=k}-\underbrace{\left(a_{u+1}a_{u+4}-a_{u+3}a_{u+2}\right)}_{=k}$
$=k-k=0$.

Thus, we obtain the equation $\left(a_u+a_{u+2}\right)a_{u+3}=\left(a_{u+2}+a_{u+4}\right)a_{u+1}$. This equation will be called the [b]crucial equation[/b].

Now, we will prove Lemma 2 by induction over the number n.

[i]Induction base:[/i] For n = 1, the equation $a_{2n-1}+a_{2n+1}=\left(m+1\right)a_{2n}$ is pretty clear, since it comes down to $a_1+a_3=\left(m+1\right)a_2$, what is trivial for $a_1=a_2=1$ and $a_3=m$. The equation $a_{2n}+a_{2n+2}=\left(q+1\right)a_{2n+1}$ is also easy to check: it comes down to $a_2+a_4=\left(q+1\right)a_3$, but this is true (since the recursion of our sequence yields $a_4=\frac{k+a_3a_2}{a_1}=\frac{k+m\cdot 1}{1}=k+m$, so that $\underbrace{a_2}_{=1}+\underbrace{a_4}_{=k+m} = 1+k+m=\underbrace{\left(k+1\right)}_{=mq}+m=mq+m=\left(q+1\right)\underbrace{m}_{=a_3}=\left(q+1\right)a_3$). So Lemma 2 is proven for n = 1.

[i]Induction step:[/i] Assume that Lemma 2 is proven for some integer $n\geq 1$. This means that we have the equations

$a_{2n-1}+a_{2n+1}=\left(m+1\right)a_{2n}$;
$a_{2n}+a_{2n+2}=\left(q+1\right)a_{2n+1}$.

Now we have to prove that Lemma 2 also holds for n + 1; this means that we have to prove the equations

$a_{2n+1}+a_{2n+3}=\left(m+1\right)a_{2n+2}$;
$a_{2n+2}+a_{2n+4}=\left(q+1\right)a_{2n+3}$.

In fact, by the crucial equation, applied to u = 2n, we have $\left(a_{2n}+a_{2n+2}\right)a_{2n+3}=\left(a_{2n+2}+a_{2n+4}\right)a_{2n+1}$. But since $a_{2n}+a_{2n+2}=\left(q+1\right)a_{2n+1}$, this becomes $\left(q+1\right)a_{2n+1}a_{2n+3}=\left(a_{2n+2}+a_{2n+4}\right)a_{2n+1}$. Division by $a_{2n+1}$ yields $\left(q+1\right)a_{2n+3}=a_{2n+2}+a_{2n+4}$, so one of the required equations is proven.

Now, by the crucial equation, applied to u = 2n - 1, we have $\left(a_{2n-1}+a_{2n+1}\right)a_{2n+2}=\left(a_{2n+1}+a_{2n+3}\right)a_{2n}$. Since $a_{2n-1}+a_{2n+1}=\left(m+1\right)a_{2n}$, this transforms into $\left(m+1\right)a_{2n}a_{2n+2}=\left(a_{2n+1}+a_{2n+3}\right)a_{2n}$. Dividing this by $a_{2n}$, we get $\left(m+1\right)a_{2n+2}=a_{2n+1}+a_{2n+3}$, so the second required equation is also proven.

Thus, Lemma 2 is proven for n + 1. Hence, the induction step is complete, and Lemma 2 is established.

Now, the assertion of Theorem 1 becomes trivial: For any integer $n\geq 1$, the equation $a_{2n-1}+a_{2n+1}=\left(m+1\right)a_{2n}$ shows that if $a_{2n-1}$ and $a_{2n}$ are integers, then $a_{2n+1}$ is an integer, too, and the equation $a_{2n}+a_{2n+2}=\left(q+1\right)a_{2n+1}$ shows that if $a_{2n}$ and $a_{2n+1}$ are integers, then $a_{2n+2}$ is an integer, too. Thus, starting with the fact that the numbers $a_1$ and $a_2$ are integers (since they are = 1), we can recursively show that all numbers $a_n$ are integers, and Theorem 1 is proven. This solves the problem.

[[b]EDIT:[/b] A more detailed writeup of the above proof of Theorem 1 can now be found in [url=http://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf]my [i]Mathematical Problem Solving[/i] notes[/url] (Exercise 4.10.9 in the current numbering; if the numbering changes, search for "The core of our argument will be the following two observations").]

  darij

Nice solution [i]darig[/i]! I like the style in which you present your solutions. It's rigorous, clear, and easy to follow.  :lol:
Oh, just a comment, the condition $\gcd (m, k) = 1$ is already implied from the condition $k \equiv -1 \pmod{m}$, so the gcd condition is not necessary but still true. :P


[hide="stuck here"]
In order for 2d-1 to be a square it has to equal a square of an odd number (2z+1)^2

d=2z^2+2z+1

13d-1 needs to be a square and equals 12(mod13)

so 13d-1=(13x+5)  since 5 and -5 are the numbers you square to get the quadratic residue of 12 in modular 13

so d=13y^2+10y+2

and d=20x^2+8x+1

all now I have to do is prove that the three quadratics cannot be equal for any positive integer as x,y,z but I can't

[/hide]

[quote="Inspired By Nature"]Let d be any positive integer not equal to 2,5, or 13. Show that we can find distinct $(a,b)$ in the set $(2,5,13,d)$ such that $ab-1$ isn't a perfect square.[/quote]

[hide]The quadratic residues of $\pmod{16}$ are $0,1,4,9$

Since $2d-1$ has to be an odd perfect square, $d\equiv 1,5,9,13\pmod{16}$

If $d\equiv 5\pmod{16}$, then $5d-1\equiv 8\pmod{16}$ which is not a perfect square.
If $d\equiv 9\pmod{16}$, then $5d-1\equiv 12\pmod{16}$ which is not a perfect square.
Therefore, $d\equiv 1,13\pmod{16}$

But if $d\equiv 1\pmod{16}$, $13d-1\equiv 12\pmod{16}$ and therefore is not a perfect square.
And if $d\equiv 13\pmod{16}$, $13d-1\equiv 8\pmod{16}$ and therefore is not a perfect square.

Thus we can always find distinct $(a,b)$ in the set $(2,5,13,d)$ such that $ab-1$ isn't a perfect square where $d$ is any postive integer.[/hide]

[quote="SplashD"]The quadratic residues of $\pmod{16}$ are $0,1,4,9$

Since $2d-1$ has to be an odd perfect square, $d\equiv 1,5\pmod{16}$[/quote] Isn't it $d\equiv 1,9\pmod{16}$?

EDIT: Never mind.

[quote="JBL"][quote="SplashD"]The quadratic residues of $\pmod{16}$ are $0,1,4,9$

Since $2d-1$ has to be an odd perfect square, $d\equiv 1,5\pmod{16}$[/quote]

$2d-1 \equiv 1, 9 \pmod{16}\implies 2d \equiv 2, 10 \pmod{16}\implies d \equiv 1, 5 \pmod{\bf{8}}$ -- I don't know what this does to your solution.

The easiest way to do this problem is using [hide]mod 4[/hide][/quote]

Yeah, that was my mistake. I fixed it now.

I remember trying mod 4 and it didn't work...

[quote="raul_v20"]Well guys can neone pls explain me

 wat are quadratic residues...

 n frm dat how do u arrive at (mod16)
why use (mod16)..[/quote]

1.  $a$ is a quadratic residue $\bmod m$ iff there exists $b$ such that $b^{2}\equiv a \bmod m$.

2.  $16$ is a square, and $2 | \phi(16) = 8$, so $\bmod 16$ does not have very many quadratic residues.  If we can show that $ad-1$ is not congruent to a quadratic residue $\bmod 16$, then it cannot be a square.

Wasn't this an IMO problem some time ago?  I think it's labelled as one in my art of problem solving book...

That's right. [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=366557#p366557]IMO 1986[/url] to be precise.

It was an IMO 1986 problem.

[hide="So long..."]We have to show that atleast one of $2d-1$, $5d-1$, $13d-1$ is not a perfect square. Let's try proof by contradiction and assume all three are perfect squares:
$2d-1=x^{2}$ $2d=x^{2}+1$
$5d-1=y^{2}$ $5d=y^{2}+1$
$13d-1=z^{2}$ $13d=z^{2}+1$

This forces $x$ to be odd. Let $x=2p+1$
$x^{2}= 4p^{2}+4p+1 = 4p(p+1)+1$

Atleast one of $p$, $p+1$ is even, so $4p(p+1)$ contains a factor of 8. Thus,
$x^{2}\equiv1\pmod8$
$x^{2}+1\equiv2\pmod8$
$2d\equiv2\pmod8$
$d\equiv1\pmod{\frac{8}{(8,2)}}$
$d\equiv1\pmod4$
[b]So $d$ is also odd.[/b]
$5d-1\equiv4\pmod4$
$13d-1\equiv12\pmod4$
So $y^{2}$ and $z^{2}$ are multiples of $4$. So $y$ and $z$ are multiples of $2$, i.e. they are even. Let
$y=2q$
$z=2r$
Since $z^{2}-y^{2}=8d$, $(r+q)(r-q)=2d$. This implies that $r$ and $q$ are either both odd or both even (or else their product would be odd). Since they are both even, their product is divisible by 4:
$2d\equiv0\pmod4$
$d\equiv0\pmod2$
[b]Thus, d is also even. [/b]

WAIT!!! Look at the bold terms. They contradict each other, so the temporary hypothesis was false, and the statement is proven. That was a long proof, and might've left some of you snoring already :sleep2:

QED[/hide]

[quote="vishalarul"]It was an IMO 1986 problem.

[hide="So long..."]We have to show that atleast one of $2d-1$, $5d-1$, $13d-1$ is not a perfect square. Let's try proof by contradiction and assume all three are perfect squares:
$2d-1=x^{2}$ $2d=x^{2}+1$
$5d-1=y^{2}$ $5d=y^{2}+1$
$13d-1=z^{2}$ $13d=z^{2}+1$

This forces $x$ to be odd. Let $x=2p+1$
$x^{2}= 4p^{2}+4p+1 = 4p(p+1)+1$

Atleast one of $p$, $p+1$ is even, so $4p(p+1)$ contains a factor of 8. Thus,
$x^{2}\equiv1\pmod8$
$x^{2}+1\equiv2\pmod8$
$2d\equiv2\pmod8$
$d\equiv1\pmod\frac{8}{(8,2)}$
$d\equiv1\pmod4$
[b]So $d$ is also odd.[/b]
$5d-1\equiv4\pmod4$
$13d-1\equiv12\pmod4$
So $y^{2}$ and $z^{2}$ are multiples of $4$. So $y$ and $z$ are multiples of $2$, i.e. they are even. Let
$y=2q$
$z=2r$
Since $z^{2}-y^{2}=8d$, $(r+q)(r-q)=2d$. This implies that $r$ and $q$ are either both odd or both even (or else their product would be odd). Since they are both even, their product is divisible by 4:
$2d\equiv0\pmod4$
$d\equiv0\pmod2$
[b]Thus, d is also even. [/b]
WAIT!!! Look at the bold terms. They contradict each other, so the temporary hypothesis was false, and the statement is proven. That was a long proof, and might've left some of you snoring already :sleep2:
QED[/hide][/quote]
As long as the proof maybe, its awesome and very easy to understand. Nice job vishalarul. :lol:
P.S. [quote="vishalarul"]That was a long proof, and might've left some of you snoring already :sleep2: [/quote]
I really like your conclusion, lol.

Actually, I looked in the answer key, and used the main ideas.  :blush: So I really don't deserve the credit.

well, its an IMO question, so you still get some of the credit. :)


inequalities
trigonometry
inequalities proposed

[quote="Tiger100"]Show that in any acute triangle the following inequality is truly:

$ CosA.CosB.CosC \le (1 \minus{} CosA)(1 \minus{} CosB)(1 \minus{} CosC)$[/quote]
It's true for all triangle.  :wink:

[quote="Tiger100"]Show that in any acute triangle the following inequality is truly:

$ CosA.CosB.CosC \le (1 \minus{} CosA)(1 \minus{} CosB)(1 \minus{} CosC)$[/quote]

$ \Leftrightarrow 8\sin^2(A/2)\sin^2(B/2)\sin^2(C/2) \ge \cos(A)\cos(B)\cos(C)$

$ \sin(A/2)\sin(B/2)\sin(C/2) \equal{} \frac {r}{4R}$

$ \cos(A)\cos(B)\cos(C) \equal{} \frac {s^2 \minus{} (2R \plus{} r)^2}{4R^2}$

So we must prove: $ \frac {r^2}{2R^2} \ge \frac {s^2 \minus{} (2R \plus{} r)^2}{4R^2} \Leftrightarrow 3r^2 \plus{} 4Rr \plus{} 4R^2 \ge s^2$ - known inequality , holdes for all triangles

[quote="2424"][quote="Tiger100"]Show that in any acute triangle the following inequality is truly:

$ CosA.CosB.CosC \le (1 \minus{} CosA)(1 \minus{} CosB)(1 \minus{} CosC)$[/quote]
We have ineq than tronger this ineq!
$ CosA.cosB.cosC \leq \frac {8}{27}sin^2Asin^2B.sin^2C$
With all triangle ABC!
It's easy by Am-Gm! :P[/quote]
we only have to prove this in case ABC is not an obtuse triangle
so we can assume that
$ cosA\equal{}\frac{bc}{(b\plus{}a)(c\plus{}a)}$,..........
the ineq becomes:
$ 8(a\plus{}b\plus{}c)^3\ge 27(a\plus{}b)(b\plus{}c)(c\plus{}a)$
which is true by AMGM

[quote="2424"][quote="Tiger100"]Show that in any acute triangle the following inequality is truly:

$ CosA.CosB.CosC \le (1 - CosA)(1 - CosB)(1 - CosC)$[/quote]
We have ineq than tronger this ineq!
$ CosA.cosB.cosC \leq \frac {8}{27}sin^2Asin^2B.sin^2C$
With all triangle ABC!
It's easy by Am-Gm! :P[/quote]

It is holdes for all triangles :

$ \sin(A)\sin(B)\sin(C) = \frac {rs}{2R^2}$

$ \cos(A)\cos(B)\cos(C) = \frac {s^2 - (2R + r)^2}{4R^2}$

So we must prove: $ s^2 \le \frac { (2R + r)^2}{1 - \frac {8}{27}\frac {r^2}{R^2}}$

$ s^2 \le 2R^2 + 10Rr - r^2 + 2\sqrt {R}(R - 2r)^{\frac {3}{2}}$

$ t = \frac {r}{R} \le \frac {1}{2}$

$ \Leftrightarrow \frac { (2 + t)^2}{1 - \frac {8}{27}t^2} - 2 - 10t + t^2 - 2(1 - 2t)^{\frac {3}{2}} \ge 0$ - which is true for $ 0 \le t \le \frac {1}{2}$

Equality holdes for : $ (A = B = C)$  or $ (A \rightarrow \frac {\pi}{2}$ and $ B \rightarrow \frac {\pi}{2})$


The problem reads $\sum_{k=0}^{r}{r\choose6}{12-r\choose 6-k}$, but since those numbers aren't always positive, and by the solution, I believe it should read $\sum_{k=0}^{r}{r\choose k}{12-r\choose 6-k}$.

Yeah that's right. On those files we didn't correct the errors that were found during/after the competition. Do you think I should make an "errata" document that lists all the errors on the tests and in solutions?

Ok I'll do that as soon as I make sure I'm going to pass all of my classes. (Note: this might mean "after exams")


A customer ordered 15 pieces of gourmet chocolate. The order
can be packaged in small boxes that contain 1, 2 or 4 pieces
of chocolate. Any box that is used must be full. How many
different combinations of boxes can be used for the customer's
15 chocolate pieces? One such combination to be included is
to use seven 2-piece boxes and one 1-piece box.

You can have up to 3 4-piece boxes.
If you have 3 4-piece boxes, you can have 0-1 2-piece boxes, for 2 cases total.
If you have 2 4-piece boxes, you can have 0-3 2-piece boxes, for 4 cases total.
If you have 1 4-piece box, you can have 0-5 2-piece boxes, for 6 cases total.
If you have 0 4-piece boxes, you can have 0-7 2-piece boxes, for 8 cases total.

Add up all the cases to get 20 cases total.


function
integration
inequalities
calculus
floor function
real analysis
real analysis unsolved

suppose that $ f: \mathbb{R}\to \mathbb{R}$ is a positive function such that $ \int_{\mathbb{R}}f(x) dx < \infty$. Is it true that $ \sum_{n \geq 1}\frac{f(x \sqrt{n})}{\sqrt{n}}< \infty$ for almost every $ x \in \mathbb R$

$ \sum_{n\geq 1}{\frac{f(x\sqrt{n})}{\sqrt{n}}}\leq \int_{1}^{\infty}{\frac{f(x\sqrt{y})}{\sqrt{y}}}dy=\{ x\sqrt{y}=t;dy=\frac{2t}{x^{2}}dt\}=\frac{2}{x}\int_{x}^{\infty}{f(t)}dt<\infty$  :roll:

You tacitly assumed that $ f$ is monotone on half lines - that it decreases for $ x>0$ and increases for $ f<0.$ You made that assumption when you wrote the inequality between the sum and the integral.

What are you going to do if $ f$ is not monotone?

One note (not in itself an answer to the original question):

Let $ g(x)=\sum_{n=1}^{\infty}\frac{f(x\sqrt{n})}{\sqrt{n}}.$ Then $ \int_{\mathbb{R}}g(x)\,dx=\infty.$

Just interchange the limit and sum (justified by the Monotone Convergence Theorem). After making the natural change of variables, we have

$ \int_{\mathbb{R}}g(x)\,dx= \sum_{n=1}^{\infty}\frac1n\int_{\mathbb{R}}f(y)\,dy$

Had $ g$ turned out to be integrable, that would have proved the claim, but the failure of this argument doesn't prove anything: $ g$ might be finite almost everywhere but still fail to be integrable.

An argument based on Kent's idea: we show that $ h(x)\cdot\sum_{n=1}^{\infty}\frac{f(x\sqrt{n})}{\sqrt{n}}$ is integrable for some suitable nonnegative function $ h$.

Since the sum behaves badly near $ x=0$, we want $ h(0)=0$. For this to be meaningful, $ h$ shouldn't be zero anywhere else.
Now, $ \int_{\mathbb{R}}h(x)\sum_{n=1}^{\infty}\frac{f(x\sqrt{n})}{\sqrt{n}}\, dx=\sum_{n=1}^{\infty}\int_{\mathbb{R}}\frac1n h\left(\frac{y}{\sqrt{n}}\right)f(y)\,dy$

If $ \sum_{n=1}^{\infty}\frac1n h\left(\frac{y}{\sqrt{n}}\right)$ is bounded, this integral will converge.
After some experimentation, I found $ h(x)=\frac{x^{2}}{1+x^{4}}$ works;
$ \frac1n h\left(\frac{y}{\sqrt{n}}\right)=\frac{y^{2}}{n^{2}+y^{4}}$, which is less than $ \frac1{y^{2}}$ when $ n<y^{2}$ and is less than $ y^{2}\frac1{n^{2}}$ when $ n>y^{2}$. The sum of the $ \lfloor y^{2}\rfloor$ terms in the first case is less than $ 1$, and the sum of the terms in the second case is less than $ \frac{y^{2}}{\lfloor y^{2}\rfloor}\approx 1$ (For $ |y|<1$, it's less than $ \zeta(2)$) by an integral estimate.
With a little more care, we could show that the whole sum is less than 2.

Restated, my claim is that $ \frac{x^{2}}{1+x^{4}}\sum_{n=1}^{\infty}\frac{f(x\sqrt{n})}{\sqrt{n}}$ is an integrable function of $ x$, and therefore the sum is finite a.e.


Prove that for all $ n\in\mathbb{N}$
$ (2n + 1)^3 - 2$ can be expressed as a sum of $ 3n-1$ perfect squares, such that each perfect square is greater than 1.

[hide]
notice that
$ (2(n \plus{} 1) \plus{} 1)^3 \minus{} 2$
$ \equal{} (2n \plus{} 1)^3 \minus{} 2 \plus{} (2n \plus{} 1)^2 \plus{} (2n \plus{} 3)^2 \plus{} (4n \plus{} 4)^2$

thus,it can be proven by induction.
[/hide]

How couldn't I solve it..  :oops:

Thank you!


If $ f$ is differentiable at $ a$, then $ f$ is continuous at $ a$.

Before reading the proof of this theorem, I approached it myself and did the following:

[quote]We can assume from the givens that $ f'(a) \equal{} \lim_{x \to a} \frac {f(x) \minus{} f(a)}{x \minus{} a}$ exists. With some algebraic manipulation, we get:

$ f'(a) \equal{} \frac {\lim\limits_{x \to a} [f(x) \minus{} f(a)]}{\lim\limits_{x \to a} (x \minus{} a) }$

$ f'(a) \equal{} \frac {\left( \lim\limits_{x \to a} f(x) \right) \minus{} \left( \lim\limits_{x \to a} f(a) \right) }{\lim\limits_{x \to a} (x \minus{} a)}$

$ f'(a) \equal{} \frac {\left( \lim\limits_{x \to a} f(x) \right) \minus{} f(a) }{\lim\limits_{x \to a} (x \minus{} a)}$

Since $ \lim_{x \to a} (x \minus{} a)$ exists, it can be considered a constant and thus you can multiply both sides of the equations by said expression.

$ f'(a) \left( \lim_{x \to a} [x \minus{} a] \right) \equal{} \left( \lim_{x \to a} f(x)\right) \minus{} f(a)$

Now it is permissible to evaluate $ \lim_{x \to a} (x \minus{} a) \equal{} 0$. And so $ \lim_{x \to a} f(x) \equal{} f(a)$ as required.[/quote]

However, every proof I look at insists on showing $ \lim_{x \to a} [f(x) \minus{} f(a)] \equal{} 0$. Is there a mistake in my proof attempt, or is this approach seldom used for being inefficient?

There's not really any notable difference there. It doesn't really matter whether we put the constant $ f(a)$ inside or outside the limit.

[quote="Prime factorization"]With some algebraic manipulation, we get:
$ f'(a) \equal{} \frac {\lim\limits_{x \to a} [f(x) \minus{} f(a)]}{\lim\limits_{x \to a} (x \minus{} a) }$
[/quote]
And this step is a nonsense ;)

[quote]And this step is a nonsense [/quote]

Because the quotient rule of limits does not hold if the denominator computes to 0... right. So that means my approach was essentially a dead end?

Yes, because of that. :) It seems like your approach won't be any different from the standard one after the mistake is corrected (there seems to be essentially one way to correct it and it leads straight to the standard proof).


inequalities
LaTeX
inequalities unsolved

For the third inequality see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=57683[/url]. And please write the problems in $\LaTeX$, becuase it is inconvenient to open an attachment every time trying to see what are you talking about.


Find the real solution(S) to the equation (x+y)^2=(x+1)(y-2)  :huh:

[hide="Attempt"]$ (x + y)^2 = (x + 1)(y - 2) \\
x^2 + 2xy + y^2 = xy + y - 2x - 2 \\
x^2 + y^2 - y + 2x + xy - 2 = 0$

Finding $ y$ bounds:

$ x^2 + x(2 + y) + (y^2 - y - 2) = 0$

$ \begin{array}{rl}\Delta_y = & (2 + y)^2 - 4(y^2 - y - 2) \\
= & 4 + 4y + y^2 - 4y^2 + 4y + 8 \\
= & - 3y^2 + 8y + 12\end{array}$

For there to be real solutions, $ \Delta_y \geq 0$:

$ - 3y^2 + 8y + 12\geq 0 \Leftrightarrow \boxed{ - 4 - 2\sqrt {13} \leq y \leq - 4 + 2\sqrt {13}}$

Finding $ x$ bounds:

$ y^2 + y(x - 1) + (x^2 + 2x - 2)$

$ \begin{array}{rl}\Delta_x = & (x - 1)^2 - 4(x^2 - 2x - 2) \\
= & x^2 - 2x + 1 - 4x^2 + 8x + 8 \\
= & - 3x^2 + 6x + 9 \end{array}$

For there to be real solutions, $ \Delta_x\geq0$

$ - x^2 + 2x + 3\geq0 \Leftrightarrow \boxed{ - 1\leq x \leq 3}$

So I guess that if either the $ x$ or $ y$ values are in the bounds above, there will be a real solution. Dunno if I'm missing anything.[/hide]


Δίνεται τρίγωνο ΑΒC. Να προσδιορίσετε όλες τις τιμές της παράστασης: $\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}.$

Σ αυτήν εδώ ρε Σιλουανέ πώς κατέληξες στο αποτέλεσμα????  Βοηθά η De Moivre???

Η αριστερή είναι προφανής.

Για τη δεξιά πρόσεξε ότι

$ 16E^2 \equal{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} a^4 \minus{} b^4 \minus{} c^4$

από ήρωνα και De Moivre. :wink:

το πρόβλημα βασικά δν ήταν να την αποδείξω - αυτό μου ήταν αρκετά έυκολο-. Ο λόγος που ρώτησα είναι γιατί αναρρωτιέμαι πώς κατέληξε ο Σιλουανός απευθείας  σ αυτή τη σχέση από τη δοθείσα. Ποιά ήταν η συλλογιστική του κ πώς ξέρει ότι δν μπορούμε να βελτιώσουμε άλλο τις ανισότητες :wink:

Δε βελτιώνεται άλλο γιατί μπορείς να κατασκευάσεις τρίγωνο με οσοδήποτε μικρό εμβαδό θέλεις αν είναι αυτή η ερώτησή σου.

[quote="meinstein"] Ποιά ήταν η συλλογιστική του κ πώς ξέρει ότι δν μπορούμε να βελτιώσουμε άλλο τις ανισότητες :wink:[/quote]

Με το που τη βλέπεις σκέφτεσαι ποιες γνωστές ανισότητες ισχύουν ...Το ένα μέρος για το 1 είναι προφανές . Το για να βρεις το up bound θεωρείς ή εκφυλισμένο το τρίγωνο ή τη μια πλευρά του να τεινει στο 0  :wink:  και μετά προσπαθείς να αποδείξεις την ανισότητα

thanks  ρε Σιλουανέ. Αυτό εννοούσα :wink:


algebra
polynomial
Divisibility
modular arithmetic
number theory
IMO Shortlist

Let $ a$ be the greatest positive root of the equation $ x^3 \minus{} 3 \cdot x^2 \plus{} 1 \equal{} 0.$ Show that $ \left[a^{1788} \right]$ and $ \left[a^{1988} \right]$ are both divisible by 17. Here $ [x]$ denotes the integer part of $ x.$

Here's a sketch of an answer.  Let $a > b > c$ be the three real roots of $f(x) = x^3 - 3x^2 + 1$ and note that $a > 1 > b > 0 > c > -1$.  Let $t_n = a^n + b^n + c^n$.  Write down the first three values of $t_n$ and note that it satisfies a recurrence relation with driving polynomial $f$: that is, $t_{n+3} = 3t_{n+2} - t_n$.  Reduce modulo 17, find the cycles and check that $t_{1788}$ and $t_{1988}$ are zero mod 17.  Finally observe that $t_N$ = $\lfloor a^N \rfloor$ for large even $N$ since $\vert b \vert, \vert c \vert < 1$.

[quote="piever"]Finding the cycles could be a bit boring...

Maybe it is more easy to prove that $ a^{n}+b^{n}+c^{n}\equiv 4^{n}+5^{n}+11^{n}\pmod{17}$  :wink:[/quote]

And then finding the cycles again for $ 4^{n}$, $ 5^{n}$ and $ 11^{n}$... But you're right, it is clearly visible that their cycles' lengths are at most 16, which is not evident at the $ t_{i}$ sequence. However the latter cycle's length is 16, so this is not a big problem.
Another thing is, i don't think your identity shines from $ t_{n}$ :P

I solved for a.
Let a-1=k, then we have
$ k^3 \equal{} 3k \plus{} 1$
Let $ k \equal{} 2cos m$, then we have
$ 2(4cos^3m \minus{} 3cosm) \equal{} 1$
$ cos(3m) \equal{} 1/2$
$ 3m \equal{} 60$ gives the largest root.
So $ a \equal{} 2cos20 \plus{} 1$
The other roots are $ 2cos100 \plus{} 1$ and $ 2 cos 140 \plus{} 1$
Is it possible to solve it this way?

How could someone better explain this problem and fully reverse the recurrence is not my thing but it is very clear.
Thanks in advance.

Greetings  :D


While cleaning out a garage, John found four old single-digit house
numbers, one 3, one 4, and two 5s. The number of different two-digit house numbers he can create using any two of them is:

a. 12 b. 5 c. 6 d. 7 e. NG

[hide=" click to reveal hidden content"]

$ 7$:

$ 55, 34, 43, 35, 53, 45, 54$[/hide]


Let $ a,b,c$ positive reals. Prove that

$ \sum\sqrt{\frac{a\plus{}b}{a\plus{}c}}\plus{}\sum\sqrt{\frac{a\plus{}c}{a\plus{}b}}\leq\sum\frac{b\plus{}c}{\sqrt{bc}}$

[quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \sum\sqrt {\frac {a \plus{} b}{a \plus{} c}} \plus{} \sum\sqrt {\frac {a \plus{} c}{a \plus{} b}}\leq\sum\frac {b \plus{} c}{\sqrt {bc}}$[/quote]
$ \sqrt {\frac {a \plus{} c}{a \plus{} b}} \plus{} \sqrt {\frac {a \plus{} b}{a \plus{} c}} \equal{} \frac{2a\plus{}b\plus{}c}{\sqrt{(a\plus{}b)(a\plus{}c)}} \le \frac{2a\plus{}b\plus{}c}{a\plus{}\sqrt{bc}}$ (by cauchy-schwartz)
$ \frac{2a\plus{}b\plus{}c}{a\plus{}\sqrt{bc}} \le \frac{b\plus{}c}{\sqrt{bc}} \iff$
$ 2a\sqrt{bc} \le ab\plus{}ac \iff a(\sqrt{b}\minus{}\sqrt{c})^2 \ge 0$. So $ \sqrt {\frac {a \plus{} c}{a \plus{} b}} \plus{} \sqrt {\frac {a \plus{} b}{a \plus{} c}} \le \frac{b\plus{}c}{\sqrt{bc}}$
Summing up we get the desired.

[quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \sum\sqrt {\frac {a \plus{} b}{a \plus{} c}} \plus{} \sum\sqrt {\frac {a \plus{} c}{a \plus{} b}}\leq\sum\frac {b \plus{} c}{\sqrt {bc}}$[/quote]

$ \sqrt{\frac{a\plus{}b}{a\plus{}c}}\plus{}\sqrt{\frac{a\plus{}c}{a\plus{}b}}\leq\frac{b\plus{}c}{\sqrt{bc}}$

$ \Longleftrightarrow \frac{a\plus{}b}{a\plus{}c}\plus{}\frac{a\plus{}c}{a\plus{}b}\plus{}2 \le \frac{b^2\plus{}c^2\plus{}2bc}{bc}$

$ \Longleftrightarrow \frac{a\plus{}b}{a\plus{}c}\plus{}\frac{a\plus{}c}{a\plus{}b}\le \frac{b}{c} \plus{} \frac{c}{b}$

$ \Longleftrightarrow \frac{{a \plus{} b}}{{a \plus{} c}} \minus{} \frac{b}{c} \le \frac{c}{b} \minus{} \frac{{a \plus{} c}}{{a \plus{} b}}$

$ \Longleftrightarrow \frac{{a\left( {c \minus{} b} \right)}}{{c\left( {a \plus{} c} \right)}} \le \frac{{a\left( {c \minus{} b} \right)}}{{b\left( {a \plus{} b} \right)}}$

$ \Longleftrightarrow 0 \le \frac{{c \minus{} b}}{{b\left( {a \plus{} b} \right)}} \minus{} \frac{{c \minus{} b}}{{c\left( {a \plus{} c} \right)}}$

$ \Longleftrightarrow 0 \le \frac{{\left( {b \minus{} c} \right)^2 \left( {a \plus{} b \plus{} c} \right)}}{{bc\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}.$

true, so we have

$ \sum\sqrt{\frac{a\plus{}b}{a\plus{}c}}\plus{}\sum\sqrt{\frac{a\plus{}c}{a\plus{}b}}\leq\sum\frac{b\plus{}c}{\sqrt{bc}}.$


geometry
cyclic quadrilateral
circumcircle
national olympiad

[url=http://www.mathlinks.ro/viewtopic.php?t=235743]Problem 1[/url]
A positive integer is [i]dapper[/i] if at least one of its multiples begins with $ 2008$. For example, $ 7$ is dapper because $ 200858$ is a multiple of $ 7$ and begins with $ 2008$. Observe that $ 200858 \equal{} 28694\times 7$.

Prove that every positive integer is dapper.

[url=http://www.mathlinks.ro/viewtopic.php?t=235745]Problem 2[/url]
Let $ S$ be a set of $ 6n$ points in a line. Choose randomly $ 4n$ of these points and paint them blue; the other $ 2n$ points are painted green. Prove that there exists a line segment that contains exactly $ 3n$ points from $ S$, $ 2n$ of them blue and $ n$ of them green.

[url=http://www.mathlinks.ro/viewtopic.php?t=235748]Problem 3[/url]
Let $ x,y,z$ real numbers such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$. Find the minimum value of
\[ {x \over x^2 \plus{} 1} \plus{} {y\over y^2 \plus{} 1} \plus{} {z\over z^2 \plus{} 1}\]

[url=http://www.mathlinks.ro/viewtopic.php?t=235749]Problem 4[/url]
Let $ ABCD$ be a cyclic quadrilateral and $ r$ and $ s$ the lines obtained reflecting $ AB$ with respect to the internal bisectors of $ \angle CAD$ and $ \angle CBD$, respectively. If $ P$ is the intersection of $ r$ and $ s$ and $ O$ is the center of the circumscribed circle of $ ABCD$, prove that $ OP$ is perpendicular to $ CD$.

[url=http://www.mathlinks.ro/viewtopic.php?t=235744]Problem 5[/url]
Prove that for all integers $ a > 1$ and $ b > 1$ there exists a função $ f$ from the positive integers to the positive integers such that $ f(a\cdot f(n)) \equal{} b\cdot n$ for all $ n$ positive integer.

[url=http://www.mathlinks.ro/viewtopic.php?t=235746]Problem 6[/url]
The venusian prophet Zabruberson sent to his pupils a $ 10000$-letter word, each letter being $ A$ or $ E$: the [i]Zabrubic word[/i]. Their pupils consider then that for $ 1 \leq k \leq 10000$, each word comprised of $ k$ consecutive letters of the Zabrubic word is a [i]prophetic word[/i] of length $ k$. It is known that there are at most $ 7$ prophetic words of lenght $ 3$. Find the maximum number of prophetic words of length $ 10$.

Thanks for posting those problems. What about [url=http://www.mathlinks.ro/resources.php?c=25&cid=58]Brazil NMO[/url] before 1990?


use the area of a triangle. or maybe i misunderstood the question.

[quote="Wumbate"]I was reading a solution to a problem today and in the solution it used something equivalent to this:
In triangle ABC, let a_1 = the altitude to side a, b_1 = the altitude to side b, and c_1 = the altitude to side c.   a*a_1=b*b_1=c*c_1
It worked in the problem, can anyone confirm this?[/quote]

Yep that's correct.  If n_1 is  the altitude to side n, then the area of that triangle is always n_1*n/2.  So a*a_1/2=b*b_1/2=c*c_1/2.  Multiplying by 2, a*a_1 = b*b_1 = c*c_1.  The area of one triangle is always the same, no matter how you calculate it.

[quote="Wumbate"]Good point.  I didn't even think of area, I was trying out the right triangles or making similar triangles.[/quote]

Area can be a very powerful tool.

My favorite example (draw this out, it'll be a lot more elegant):
Make a square ABCD, with side length (a+b).  Draw a point on each of AB, BC, CD, and DC, and label them M,N,O, and P respectively such that AM=BN=CO=DP=a, and MB=NC=OD=PA=b.  Connect those points to form another square.  The total square should now be a tilted square in the middle and four right triangles around it.  We'll define the length of the tilted inner square to be c.
The total area of the large square is obviously (a+b)^2, or c^2+4*1/2*a*b (the inner square plus the four triangles).
a^2+2ab+b^2=c^2+2ab
a^2+b^2=c^2
Yipee!


In plan sunt date $\ n\geq 2$ puncte.Doi jucatori unesc, pe rand cate o pereche de puncte care nu sunt inca unite
printr-un vector in unul dintre cele doua sensuri posibile.Daca dupa mutarea unui jucator, suma tuturor vectorilor desenati este nula, atunci castiga al doilea jucator;daca toate perechile de puncte au fost unite si suma nu este nula, castiga primul.Cine castiga la un joc corect?
 $\ Nazar\ Agahanov, \ Rusia$

Primul jucator castiga.

De fiecare data cand ii vine randul sa mute, primul jucator poate alege doua puncte care nu au fost inca unite, cu distanta $\ell$ maxima printre toate distantele intre doua puncte neunite inca, si orienteaza vectorul dintre ele astfel incat sa faca unghi de cel mult $90^\circ$ cu vectorul suma (care e nenul, din moment ce jocul nu s-a terminat inca) din momentul respectiv.

Dupa mutarea asta, vectorul suma o sa aiba lungime strict mai mare decat $\ell$, si deci al doilea jucator nu o sa poata face suma nula, pentru ca din definitia lui $\ell$ rezulta ca nu exista doua puncte neunite inca si care au distanta $>\ell$ intre ele.


complex analysis and diffrential equation is more popular~`


number theory
prime numbers
number theory proposed

Find all positive numbers $a,b,c$ such that:
1)$14c^{2}=4ab-5(a+b)c$;
2)$a-b=2p$, where $p$ is prime numbers;
3)Neither of numbers $9c+1$, $18c+4$ dont are perfect square!

Can solve anyone?!

I think that don't is very hard problems? ..
Ok I can help you:
$a-b=2p\Leftrightarrow\ a+b=(a-b)+2b$, Now put $a+b=2q$ :wink:


symmetry
vector
linear algebra
linear algebra theorems

Let $v$ be a fixed vector in $\mathbb{R}^3$, and let $\times$ denote the traditional cross product from calculus. Let $T$ be the linear operator $T(x) = (x \times v) \times v$.

(i) Show that this operation is symmetric. You may use properties of the scalar triple product, but no computation of its matrix.

(ii) Compute the matrix.

I'm not looking for an answer to this problem. I'm just wondering, doesn't "symmetry" (for a linear operator) require a given form? I thought symmetric meant $\langle Tv, w \rangle = \langle v, Tw \rangle$, but no form is given.

Yes, symmetric meant $\langle Tv, w \rangle = \langle v, Tw \rangle$. Given form is not required.

[quote="darktreb"]But then does that mean $\langle v, w \rangle$ is the traditional dot product? Or that it can be any form?[/quote]
can be any form...
remember that a form have to satisfies some rules to be a dot product...(and u can use this)

So it needs to work for positive definite symmetric bilinear forms?

the problem would be misstated if the form was not defined

but here is the subtly in my opinion : where does your crossproduct come from?
in order to have that, not only do you need an inner product (or thus a set of orthonormal basisses) , you also need to define a certain set (half of them) to be righthanded

so if the crossproduct is 'classical' making that of (a,b,c) and (d,e,f)
(bf-ec,cd-ef,ae-bd) you have to use as form the inner product a*d+b*e+c*f


A point is selected inside an equilateral triangle. From this point, perpendiculars are dropped to each side. Show that the sum of the lengths of these perpendiculars is equal to the altitude length.

[quote="236factorial"]I have a problem believing that that's even true[/quote]
i know, me too. it took me a while to believe it but if you do it by areas...  :D

in this question you  just have to equate area of the small triangles to the big triangle

[quote="riddler"][quote="Palytoxin"][quote="riddler"]piece of cake for me! [b]rubbishly[/b] easy question[/quote]

ook... mr.riddler here is your trophy smilie:  :showoff:
now you can move on to olympiad section without interfering us, ty[/quote]

you cannot tell me wha to do!  :mad: got it![/quote]Wow, calm down.  I wouldn't call a question rubbishly easy, because there are people that may not find the question rubbishly easy, that you just insulted.  Palytoxin was suggesting to you that if a question is clearly not your level, then don't aggravate others.

[quote="gauss202"][quote="Iversonfan2005"]Here's my solution, when you drop perpendiculars from each side of the point, you form 3 equal triangles in which you add the areas up, you'll have the area of the large equalateral triangles. Therefore, the perpendiculars which are the altitudes for the smaller triangles are $1/3$ of the altitude of the equalateral triangle. Thus by adding all three of these altitudes, you have the altitude of the large equalateral triangle.[/quote]

Is the altitude of [i]each[/i] of the smaller triangles necessarily 1/3 the altitude of the equalateral triangle?[/quote]

No, you're right, but it will always add up to the larger equilateral triangle. When I did it though, I made the point as the circumcenter to which all the perpendiculars from each vertice of the equalateral triangle intersect which would make that true because the ratio is 1:3. I'm sorry about that, but the three altitudes would always add up to the larger.

sorry for calling the sum easy i didnt look at the consequences but the only thing was i had done it before.


How many terms does the arithmetic sequence
$ 15,20,25,30,\dots ,95$ have?

The sequence follows the formula $ 15\plus{}5n$ for $ n\equal{}0...16$.
One for each n, so $ \boxed{17}$ total.

hello, an explicit formula for your sequence is given by
$ a_n\equal{}a_1\plus{}(n\minus{}1)\cdot d$ with the difference $ d\equal{}5$
So we have $ 95\equal{}15\plus{}(n\minus{}1)\cdot 5$. From this we get $ n\equal{}17$.
Sonnhard.

Another way:

Simply subtract 15 from 95 to get 80, and then divide by 5 since the common difference is 5, and finally add 1.

$ 95 \minus{} 15 \equal{} 80 \div 5 \equal{} 16 \plus{} 1 \equal{} \boxed{17}$

A 3rd way (very similar to RightAnnihilator)

We see the terms re $ 3\cdot{5},4\cdot{5}....19\cdot{5}$

So it is the numbers 3 through 19. 19-3+1=17


Graph $ G$ contains $ 2n\minus{}1$ vertices.It is known that if we delete an arbitary vertex in $ G$ we are still able to find $ K_n$
Prove that $ G$ contains $ K_{n\plus{}1}$

This is a very famous problem, with a quite difficult solution (if I find time I may post one). Notice where $ 2n\minus{}1$ plays: for $ 2n$ we easily find the counter-example of $ K_n \cup K_n$.

[quote="mavropnevma"]This is a very famous problem, with a quite difficult solution .[/quote]

I agree with you :) The offcial solution is strange and incredible.But I am still waiting for a natural one :lol:

Let $ V$ be the set of vertices of $ G$. Let $ C_1, \ldots, C_k$ be all complete subgraphs of $ G$ with $ n$ vertices. We can assume that $ C_1 \cap \ldots \cap C_k \equal{} \emptyset$, because otherwise there would exist a vertex $ v$ which is connencted to all vertices of $ G$ which are contained in at least one $ K_n$ and we would get a $ K_{n\plus{}1}$. Let $ A_i \equal{} C_1 \cap \ldots \cap C_i$ and $ B_i \equal{} V \backslash (C_1 \cup \ldots \cup C_i)$. We have $ |A_1| \equal{} n, |B_1| \equal{} n\minus{}1, |A_k| \equal{} 0$. Thus, we can find a $ i \in \{1, \ldots, k\minus{}1\}$ with $ |A_i| > |B_i|$ and $ |A_{i\plus{}1}| \leq |B_{i\plus{}1}|$.
Assertion: $ D : \equal{} A_i \cup (C_{i\plus{}1} \backslash B_i)$ induces a $ K_m$ with $ m \geq n\plus{}1$.
1. Obviously, $ D$ induces a {\it complete} Graph.
2. We have \[ |D| \equal{} |A_i| \plus{} |C_{i\plus{}1} \backslash B_i| \minus{} |A_i \cap (C_{i\plus{}1} \backslash B_i)| \equal{} |A_i| \plus{} n \minus{} |B_{i} \backslash B_{i\plus{}1}| \minus{} |A_{i\plus{}1}| \equal{} n \plus{} |A_i| \minus{} |B_i| \plus{} |B_{i\plus{}1}| \minus{} |A_{i\plus{}1}| > n\]
So we are done!


Hi, can you help me solve these problems using modular arithmetic?

1)Find three 5 digit numbers that are divisible by 11. Each number must have 5 different digits and can not use 0.

2)Use all 10 digits 0-9 once each to make a 10 digit number that is divisible by 11.

3) The 7 digit number 23N7N51 is divisible by 11. What is the value of the digit N?

4) Write a divisibility rule for 144.

[hide="hint"]
If a number $ abcde$ is divisible by $ 11$ then $ a\minus{}b\plus{}c\minus{}d\plus{}e \equiv 0 \mod 11$[/hide]

[hide="Hint for first three"]
In general, $ \overline{a_na_{n \minus{} 1}\ldots a_1a_0}\equal{}a_n10^n \plus{} a_{n \minus{} 1}10^{n \minus{} 1} \plus{} \cdots \plus{} a_110^1 \plus{} a_010^0$

$ \equiv a_n( \minus{} 1)^n \plus{} a_{n \minus{} 1}( \minus{} 1)^{n \minus{} 1} \plus{} \cdots \plus{} a_1( \minus{} 1)^1 \plus{} a_0( \minus{} 1)^0$

$ \equiv a_0 \minus{} a_1 \plus{} a_2 \minus{} a_3\pm\cdots\pmod{11}$.
[/hide]
[hide="Hint for the last one"]
If you can find $ a,b$ such that $ ab \equal{} 144$ and $ \gcd(a,b) \equal{} 1$, then by [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]CRT[/url], it is sufficient to find the remainders in $ \pmod a$ and $ \pmod b$ to find the remainder $ \pmod{144}$.
[/hide]

I highly recommend looking at the hints given by previous posters and making a genuine attempt at the problems before looking at these.

[hide="Solution to #1"]
$ N=\overline{abcde}$.
$ 10^n\equiv (-1)^n\pmod{11}$.
$ N\equiv -a+b-c+d-e=(a-b)+(c-d)+e\equiv 0\pmod{11}$.
$ a$ and $ c$ are interchangeable, and $ b$ and $ d$ are interchangeable.
$ \boxed{N=12375, 17325, 37125}$
[/hide]

[hide="Solution to #2"]
Let $ N=\overline{a_0a_1\cdots a_9}\equiv a_0-a_1+a_2-\cdots\pmod{11}$.
$ \equiv a_0+a_2+a_4+a_6+a_8-(a_1+a_3+a_5+a_7+a_9)\pmod{11}$.
If $ a_0+a_2+\cdots+a_8-(a_1+a_3+\cdots+a_9)=11$, and we know that $ a_0+a_1+\cdots+a_9=45$, we can add these to get $ 2(a_0+a_2+\cdots+a_8)=56\implies a_0+a_2+\cdots+a_8=28$.
A solution to this, by trial and error, is $ (1,3,7,8,9)$. Therefore, the remaining digits are $ (2,4,5,6,0)$, and we can simply concatenate this all together in an alternating fashion to get $ \boxed{N=1234756809}$.
[/hide]

[hide="Solution to #3"]
$ -2+3-N+7-N+5-1=2N-12\equiv 0\pmod{11}$.
$ \boxed{N=6}$.
[/hide]

[hide="Solution to #4"]
$ 144=12^2=(3\cdot2^2)^2=3^22^4$.
If $ 144|n$, $ 2^4|n$ and $ 3^2|n$.
The test for $ 9$ is simply the sum of the digits of $ n$, as proven by the fact that $ 10\equiv 1\pmod{9}$.
Divide the number by $ 9$ if possible, and then simply check if the number can be divided by $ 2$ four times.

For example, suppose $ n=12341232$.
$ 1+2+3+4+1+2+3+2=18\equiv 0\pmod{9}$.
Divided by $ 2$, the number ends in $ 16$.
Divided by $ 2$ again, it ends in $ 8$.
Proceeding similarly, it's obvious that we can divide it by $ 2^4$, so it passes the test.

(Does anyone have a better solution?)
[/hide]

[hide="number 4"]
$ 144 \equal{} 2^43^2$. So as [b]Grn trtle[/b] said, to check for divisibility by nine, sum up the digits and check their congruence $ \mod 9$

To check where a number is divisible by $ 16$, we see that $ 2^4|10^4$ so we only need to consider the last four digits.
$ \Rightarrow a_31000 \plus{} a_2100 \plus{} a_110 \plus{} a_0 \equiv 8a_3 \plus{} 4a_2 \minus{} 6a_2\plus{} a_0 \mod 16$

So to check for divisibility of $ 16$ we can just plug the last three digits into the above equation.

I suppose this can be slightly simplified (hard to tell if this makes it easier);
[b] 1 [/b] If the thousands digit is even plug the last three digits into the equation $ 4a_2 \minus{} 6a_2 \plus{}a_0$ and check for congruence $ \mod 16$

[b] 2 [/b] If the thousands digit is odd "..." ; $ 8 \plus{} 4a_2 \minus{} 6a_2\plus{} a_0$[/hide]


LaTeX
binomial coefficients
calculus
calculus computations

Written in a compact form

\[ \prod_{k\equal{}1}^{\infty }\prod_{i\equal{}1}^{k}\left(\frac{(2 i)^{\left(\begin{array}{c}k\\ 2 i\minus{}1\end{array}\right)}}{(2 i\plus{}1)^{\left(\begin{array}{c}k\\ 2 i\end{array}\right)}}\right)^{2^{\minus{}k}}\]

 :D

This formula grows very fast.

I tried Log in both sides. but. the result is more complex.

It would look much nicer if you used \binom{}{} for your binomial coefficients:

\[ \prod_{k\equal{}1}^{\infty }\prod_{i\equal{}1}^{k}\left(\frac{(2 i)^{\binom{k}{2 i\minus{}1}}}{(2 i\plus{}1)^{\binom{k}{2 i}}}\right)^{2^{\minus{}k}}\]

Sorry, I every use Mathematica do generate my latex formulaes

[quote]TeXForm[Unevaluated[Product[((2*i)^Binomial[k, 2*i - 1]/(2*i +                  1)^Binomial[k, 2*i])^2^(-k), {k, 1, Infinity}, {i, 1, k}]]][/quote]

And misan could give a hint. those gamma over 2i is a little bit annoying