If $p, q$ are both prime numbers. and $p+q, p-q$ are prime numbers too, prove that $p^2 - q$ is a prime number too.

[hide="Big Hint"]
The only possible values for $(p, q)$ are $(5, 2)$.
[/hide]

[hide]
clearly exactly one between $p$ and $q$ has to be $=2$. If $p=2$, then $p-q$ can't be prime. So $q=2$. But since $p$ is prime,
$p\equiv 1 \vee 2 \mod3$, and one between $p-2$ and $p+2$ is $\equiv 0 \mod 3$. So the only possibility is $p-2=3 \implies p=5$, and $p^2-q=23$ that is prime[/hide]

[hide]I also find the number p, q.
Because p-q is a prime number so p-q>0=> p>q
And the smallest prime number is 2 => p>2

p,  (p+q) are prime numbers so p, (p-q), (p+q) can't divide by 2.
=> (p+q)-p=q divide by 2=> q=2
  If $\ (p-q)\equiv0\mod3$ then p-q=3 and p=3+2=5, p+q=7(right)
  If $\ (p-q)\equiv1\mod3$ then p-q=3k+1 and p=3k+3which is not a prime number.
   If $\ (p-q)\equiv2\mod3$ then p-q=3k+2 and p+q=3k+6which is not a prime number.
So we have q=2 and p=5 an $\ p^2-q=\ 5^2-2=23$ is a prime number.
Math is my life.[/hide] :lol:


:blush: This is JaYESH FRM mUMBAI STUDIN IN 9TH STD
CaN ANY1 PLZ... TELL ME HOW 2 GO ABT 4 jUNIOR SCIENCE OLMPIAD

StoP WrITIng in CaPS!
And don't worry, if you've taken the initiative to come to aops in 9th, you're already halfway through. Rest, just wait for te geniuses to come and guide you ;-)
Warning: A few idiots like me will inevitably SPAM.


search
combinatorics theorems
combinatorics

find the sum of all three digit natural number which contain at least one odd digit and at least one even digit.

the sum of all 3 digit numbers with only odd digits is $ 25\cdot 111\cdot (1\plus{}3\plus{}5\plus{}7\plus{}9)\equal{}69375$. the sum of all 3 digit numbers with only even digits is $ 25\cdot 111\cdot (0\plus{}2\plus{}4\plus{}6\plus{}8)\minus{}5\cdot 11(0\plus{}2\plus{}4\plus{}6\plus{}8)\equal{}54400$. the sum of all 3 digit numbers is $ 100\plus{}\cdots \plus{} 999\equal{}494550$. so the answer is $ 494550\minus{}69375\minus{}54400\equal{}370775$.

First of all, this is a shockingly poor title for your post.  Second, this is a ridiculous forum in which to place this question.  Third, don't you do even a cursory search for the problems you post?  It's already been up here twice in the past week:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314847
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314839
I guess it's not surprising your rating is 1.6 !

Haha, I've just decided there's no reason to leave the crotchetyness to old people.  You kids get off my lawn!  :)

[quote="JBL"]I guess it's not surprising your rating is 1.6 ![/quote]
thanks JBL


inequalities
real analysis
real analysis unsolved

Let $ a_n$ be sequence of positive numbers and $ \alpha\in R$.Show that there exist $ \beta\in R$ such that the following inequality
|                                                    $ \alpha n \minus{} m \minus{} \beta|< a_n$
holds for infinitely many pairs of $ (m,n)$, where $ m\in Z$ and $ n\in N$.

[quote="Ibrogimov"]Let $ a_n$ be sequence of positive numbers and $ \alpha\in R$.Show that there exist $ \beta\in R$ such that the following inequality
|                                                    $ \alpha n \minus{} m \minus{} \beta| < a_n$
holds for infinitely many pairs of $ (m,n)$, where $ m\in Z$ and $ n\in N$.[/quote]
where is it from ?

i'm not sure but I think it's true.  :maybe:  but it's hard to prove


calculus
vector
geometry
analytic geometry
AoPSwiki
quadratics
AMC

Hi!
Some of you AMPers may know me...
Anyways, on topic, I'm a sophomore who currently is stuck in the 4 to 6 point range on the AIME.
I was wondering if anyone knows how to help me get out of this range...
I'm strongest with the number theory problems and weakest with the geometry. Anyways, even more specifically, I have taken Pre Calculus yet I'm not good with manipulating vectors and complex numbers. Does anyone have a guide to complex numbers and vectors? Also, I fail at geometry, so does anyone have a list of geometry topics I should study. And finally, anyone have a list of common algebraic manipulations that show up on these examinations?

All Help is Highly Appreciated,
Shri

Hello. I am like you in terms of strong and weak points (my favourite is number theory and I sucked at geometry). After reading a bit of Challenging Problems in Geometry, I feel as though my geometry has improved immensely. However, if you choose to buy this book, a would strongly recommend that you read the appendix before starting the problems. Hope that I was helpful.

wussup, shri G?

Ok you may not like me after this suggestion, but this is what I do to counteract the fact that Geometry is not my strong suit either:
bash

That's right, you have 3 hours. But make sure that after trying for at least some minutes, your Comp Geo skillz get you nowhere.

If it seems simple enough (yes this is the quite difficult part of the strategy), plot it on the coordinate plane and BASH. But as a guideline, almost definitely do not pursue anything with degree 8 or such and save anything greater than degree 4 and less than degree 8 for the last several minutes.
stuff that expands to only quadratics/LINEARS are very often good targets for coordinate bashing unless the coefficients get HUGE (I made an AoPSwiki contribution in which a=1311 in the quadratic equation)

This may also apply for Combinatorics. List the cases out (AIME 9/10 call to me). Perhaps you'll see a pattern; perhaps you won't. But perhaps the number of cases is not too large (aka the answers to AIME 9/10 which were 346 and 350 or something).

You definitely do not need elegant solutions for ANY of the AIME problems. Perhaps, even, I had the correct answer for 3-4 problems for which I would have failed if I had to write a proof.

can anyone tell me  a few good geometry books,so that i may improve my geometry?

Thanks 'stevenmeow' haha.
I'm trying to get my AIME score higher...currently I'm stuck in the low to mid range, scoring no lower than a 5 and no higher than a 7. Honestly, I mess up about two solutions per test, so I should really be scoring between 7 and 9, but I havent been getting any geo or combo problems...
Thats right, I took the 1985 AIME, got a 7, only answered the first ten and missed all three geo in it!
Argh...
I really need to work on accuracy. Also, any one know the topics I should study for #s 11-15 on the AIME. If I can get to where I can tackle two of those per test and get 7 or 8 of the first ten, I can finally make USAMO.
Thanks guys!


induction
combinatorics proposed
combinatorics

Seems easy.

If we "change backwardly" from 2005, the only possible sequence is: 2005 -> 1003 -> 502. So 502 and 1003 are the only positive integers similar to 2005.

The problem statement means that in the definition of similarity we are allowed to change both $ a$ and $ b$.

Im really interested in a solution to this problem. thanks for some help.

I think that we can prove next fact :
If $ a \sim b$ and $b \sim c$ then $a \sim c$ ( $a \sim b$ <=> $a$ and $b$ are similar )

If we can prove this fact, then easy, by induction, see that $n \sim k$ for any $n \in \mathbb{N}$ and $k \leq n$.

@Ovchinnikov Denis

How is transitivity supposed to help, im not able to get much out of it. if you could post your solution involving transitivity (involving, but not proving transitivity), i could probably extend that logic somehow

btw. transitivity of a relation R means that aRb and bRc $\implies$ aRc

Well if you can prove transitivity (and that is a big IF) then an induction on the set {$1,2,...,k$} for $k\in Z_{>0}$ would show that all positive integers are similar to one another. Base case is that $1$ and $2$ are similar to one another ($2=3*1-1$). Then we note that if it is true that all of  {$1,2,...,k$} are similar to each other and if $k+1$ is odd then $\frac{k+2}{2} \in$ {$1,2,...,k$} so $k+1$ is similar to {$1,2,...,k$}, if $k+1$ is even then it is similar to $\frac{3(k+1)}{2}$ as $2*\frac{3(k+1)}{2}-1=3k+2$ and $3(k+1)-1=3k+2$ and since $\frac{3(k+1)}{2}$ is a positive integer less than $k+1$ it is in {$1,2,...,k$} so $k+1$ is similar to {$1,2,...,k$}.

I think you need to prove another lemma: If a is similar to b and c then bis similar to c.

[quote=lasha]1. The transformation $ n \to 2n \minus{} 1$ or $ n \to 3n \minus{} 1$, where $ n$ is a positive integer, is called the 'change' of $ n$. Numbers $ a$ and $ b$ are called 'similar', if there exists such positive integer, that can be got by finite number of 'changes' from both $ a$ and $ b$. Find all positive integers 'similar' to $ 2005$ and less than $ 2005$.[/quote]
We will define $a \sim b$ as $a$ is similar to $b$
The first [color=#f00][b]Claim 1:-[/b][/color] The numbers which are divisible by 6 and those which are even and $\equiv 2 \pmod 3$ can't be similar to a number smaller than it.
[b][color=#ff9a00]Proof:-[/color][/b] This is obvious since if it could be represented as $a~b,a>b$ then it would be congruent to either $1 \pmod 2$ or $2 \pmod 3$ which contradicts this claim.
[color=#00f][b]Claim:2 (Transtivity)[/b][/color] If $a \sim b$ and $b \sim c$ then $a \sim c$
[color=#9a00ff][b]Proof:-[/b][/color] Since $a \to ........ \to c$ and $b \to ................ \to c$ after finitely many steps,we can put the two sequences together and get $a \to .......... \to b \to ..........\to c$ showing $a \sim c$
[b][color=#0f0]Claim 3:-[/color][/b] The transformations are all increasing i.e the transformations aren't reflexive.
[b][color=#0ff]Proof:-[/color][/b] Since $2n-1 \ge n$ and it is same for $3n-1$ this is true(except for only 1 $n$ i.e $n=1$)
By the above two claims it is easy to see that $2005 \to 1003 \to 503$ is the desired sequence and the answer is $\boxed{2}$
Also some of the above posts claimed that the answer is $n$.,Consider $n=6k$,it can't be similar to any number smaller than it,since it is neither congruent to $-1 \pmod 2$ or $-1 \pmod 3$ so in this case it would be $0 \neq 6k$ a clear contradiction.


(3rd Iberoamerican 1988)

Let k^3 = 2 and let x, y, z be any rational numbers such that x + y k + z k^2 is non-zero.
Show that there are rational numbers u, v, w
such that (x + y k + z k^2)(u + v k + w k^2) = 1.

First of all, it's not hard to show that $x+yk+zk^2=0\iff x=y=z=0$. We have $(x+yk+zk^2)(x+y\epsilon k+z\epsilon^2 k^2)(x+y\epsilon ^2 k+z\epsilon k^2)=t\in \mathbb Q^*$, where $\epsilon$ is a root of the eqn $x^2+x+1=0$.

This is because $x+yk+yk^2\ne 0\Rightarrow (x,y,z)\ne (0,0,0)\Rightarrow x+y\epsilon k+z\epsilon^2 k^2,x+y\epsilon ^2 k+z\epsilon k^2\ne 0$ (we show that $1,\epsilon k,\epsilon^2 k^2$ and $1,\epsilon^2 k,\epsilon k^2$ are independent over $\mathbb Q$ in the same way we showed it for $1,k,k^2$) and because, in general, $(a+b+c)(a+\epsilon b+\epsilon ^2 c)(a+\epsilon^2 b+\epsilon c)=a^3+b^3+c^3-3abc$.

On the other hand, $(x+y\epsilon k+z\epsilon^2 k^2)(x+y\epsilon ^2 k+z\epsilon k^2)=a+bk+ck^2,\ a,b,c\in \mathbb Q$, and we take $(u,v,w)=(\frac at,\frac bt,\frac ct)$.


Circles $A$ and $B$ have radii of 500 and 1000, respectively, and $AB=1337$.  Point $C$ is on $\overline{AB}$, but not on point $A$ or $B$.  Circle $C$ with radius $r$ is drawn such that it runs through the two points of intersection of circles $A$ and $B$.  How many integral values are there for $r$?

[hide]
Let circles A and B's intersection points be C and D.

$\frac{CD}{2}= \sqrt{1000^{2}-(1000-\frac{1500-1337}{2})^{2}}\approx 395.4$
So the minimum $r$ is 396 and can take values upto 999. So there's 999 - 396 + 1 = 604 values of $r$

[/hide]


geometry
incenter
trigonometry
analytic geometry
geometry proposed

Given triangle $ABC$ centroid $G$ and incenter $I$ and the point $J(\sin\frac{A}{2},\sin\frac{B}{2},\sin\frac{C}{2})$(in barycentric coordinates) prove that ${\angle JIG\ge 90^{0}}$

But there isn't solution in your link  :maybe: and my friend do you think I posted this problem on Hyacinthos, too :D


Let x, and y be binary n-tuples, where n is even. Suppose the number of 1's in both x and y are even(not necessarily the same). Show that x and y agree on an even number of places.

(0,1,0,1,1,1) would be an example of a binary 6-tuple.

[hide]
Two pairwise terms of the two $ n$-tuples agree if their sum is even. The even number of $ 1$s imply that the sum of the pairwise sums of the n-tuples is even, so an even number of the pairwise sums are odd, but since $ n$ is even, an even number of the pairwise sums are even.
[/hide]


analytic geometry
calculus
integration
number theory
greatest common divisor
combinatorics proposed
combinatorics

Prove that its not possible to draw a closed poligonal line in the plane such that all sides measure 1 unit, all vertices have rational coordinates, and there is an odd number of vertices.

1990, according to Kalva (problem 27):
[url]http://www.kalva.demon.co.uk/short/sh90.html[/url]

[quote="pascual2004"]Ok, I knew there was a solution but I want people to try it, its beautiful![/quote]
It is absolutely true remark. We should solve problems instead of searching for appropriate links.

I have just remembered another good problem:

Are there infinitely many points on unit circle s.t. all pairwise distances beetwen them are rational?

This is easy. :)

Since all vertices have rational coordinates, if such a polygon were to exist, we may translate it within the Cartesian plane so that one of the vertices is the origin.  Denote this vertex by $P_0$.  Now, we may multiply the coordinates of all the vertices by a suitable constant $k$ such that all the coordinates are integral, and their GCD is 1.

Now consider a vertex $P_1$ which is adjacent to $P_0$.  Since all coordinates are integral, $k^2 = P_0P_1^2$ is an integer.

For any two vertices $P_a (a_1, a_2)$ and $P_b (b_1, b_2)$, denote $D(P_a, P_b) \equiv (a_1 - b_1 + a_2 - b_2)$ (mod 2).

If $k^2 = P_0P_1^2$ is odd, this implies that $D(P_0, P_1) \equiv 1$ (mod 2).  this in turn implies that $D(P_i, P_{i+1}) \equiv 1$ (mod 2) for all the vertices of the polygon (including the one that goes back to P_0).  But since there are an odd number of vertices, $\sum_{all vertices} D(P_i, P_{i+1}) \equiv 1$ (mod 2), a contradiction.

If $k^2 = P_0P_1^2$ is even, this implies that $D(P_i, P_{i+1}) \equiv 0$ (mod 2) for all the vertices of the polygon. Considering mod 4, we see that either the horizontal and vertical distances between adjacent vertices are both odd or both even.  If they are both odd, we get a contradiction since there are an odd number of vertices.  If they are even, we get a contradiction because of the assumption that GCD = 1.

[quote="pascual2004"]BTW, sorry jhaussman, I am sorry, i think you just wanted to tell the source, thanks.(there is no sol in kalva)[/quote]
jhaussman! Accept my apologies.

[quote="Myth"]Valiowk! I see that you and I are one-year forum members! Happy birthday!    ;)  :lol:  :D  :P[/quote]

Happy anniversary to you too, Myth! :D :lol: //considers letting fireworks off in the thread but decides it's probably too dangerous ;)

[quote="Myth"]Are there infinitely many points on unit circle s.t. all pairwise distances beetwen them are rational?[/quote]

As for this one, I think this works:

Assume the radius of the circle is $\frac 12$ instead (it's better for what follows). Perform an inversion with one of the points as a pole. Assume the power of the inversion is $1$. It's now easy to see that it's enough to find infinitely many points $A_n$ on the line which is the image of the circle s.t., wrt the point of tangency between the line and the circle, $x_n=t+r_n$, where $x_n$ are the coordinates of $A_n$ and $r_n$ are rationals, and, furthermore, we should have $(t+r_n)^2+1=a_n^2$, where $a_n$ are, again, rational (these are the distances from the pole to the images of the points). This is easy to do: take $t=0$ and $r_n=\frac {x_n^2-1}{2x_n}$.

So the answer is err.. "yes"? :) Think it's Ok, Myth?

Great work!
Original solution is completely different.
Suppose cos(a) and sin(a) are rationals (for example, cos(a)=3/5 and sin(a)=4/5). Consider points (cos(2na),sin(2na)), n=1,2,3,... For any two such points distance between them is sin(ma) for some m, and it is clear sin(ma) is rational.

No worries, like pascual2004 says, I just wanted to give the reference for completeness.


modular arithmetic
MATHCOUNTS
Euler
function
number theory

When the three numbers 1238, 1596, 2491 are divided by a positive integer [b][i]m[/i][/b], the remainders are all equal to a positive integer [b][i]n[/i][/b].  Find [b][i]m[/i][/b] + [b][i]n[/i][/b].

This means that

$ 1238\equiv1596\equiv2491\equiv n\pmod m$, such that $ 1\leq n<m$, or equivalently,

$ 1238\equal{}am\plus{}n$, $ 1596\equal{}bm\plus{}n$, and $ 2491\equal{}cm\plus{}n$, for integers $ 0\leq a,b,c$.

Subtracting equations, we get

$ 358\equal{}2\cdot179\equal{}(b\minus{}a)m$ and $ 895\equal{}5\cdot179\equal{}(c\minus{}b)m$, so

$ m\equal{}\{1,179\}$.

However, if $ m\equal{}1$, then $ n\equal{}0$, but that is not possible (we must have $ n\geq1$).

Thus, $ m\equal{}179$, and $ n\equal{}164$, and our final answer is $ 179\plus{}164\equal{}\boxed{343}$.

There must be an easier solution.

Great solution, it was good. I just think that there might be an alternative method of solving. I think i have seen a problem similar to this one on a MathCounts test. Do MathCounts problems require knowledge of modular arithmetic, or is there another solution that is easier but harder to find?

I agree that my solution is somewhat unsatisfying, but this is the most obvious solution to me. In solutions from MC, I usually see guess and check instead of modular arithmetic. However, there doesn't seem to be a good guess and check solution for this one...

It would be useful to know modular arithmetic for MC.

I agree. Some important properties for MC are

Addition, Subtraction, Multiplication, Exponentiation (easy)
Division (somewhat trickier; if $ ax\equiv b\pmod m$, then $ x\equiv \frac ba\pmod{\frac m{\gcd(a,m)}}$)
$ a\equiv b\pmod m\Longleftrightarrow c \equal{} \frac {a \minus{} b}m\Longleftrightarrow a \equal{} mc \plus{} b$ for some integer $ c$.

If you want, learn [url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Fermats Little Theorem[/url], [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_theorem]Eulers Totient Theorem[/url], and maybe the [url=http://www.artofproblemsolving.com/Wiki/index.php/Carmichael_function]Carmichael Function[/url].

You guys are right. Knowing modular arithmetic not only is a good method of solving problems like these, but it is also three times as fast.


Let $ \frac {yb \plus{} (1 \minus{} y)c}{a} \equal{} \frac {yc \plus{} (1 \minus{} y)a}{b} \equal{} \frac {ya \plus{} (1 \minus{} y)b}{c} \equal{} k$.
Then, we get $ yb \plus{} (1 \minus{} y)c \equal{} ka\cdots\boxed{1}, yc \plus{} (1 \minus{} y)a \equal{} kb\cdots\boxed{2}, ya \plus{} (1 \minus{} y)b \equal{} kc\cdots\boxed{3}$.
Add each side of $ \boxed{1},\boxed{2},\boxed{3}$, we get $ k \equal{} 1\quad or\quad a \plus{} b \plus{} c \equal{} 0$.
  If $ k \equal{} 1$, it's easy to show $ y \equal{} \minus{} 1$(If a,b,c are all different), to product each sides of $ \boxed{1},\boxed{2},\boxed{3}$.  We got $ y \equal{} \minus{} 1$, so $ 2c \equal{} a \plus{} b, 2a \equal{} b \plus{} c, 2b \equal{} c \plus{} a$. The next work is easy.
  If $ a \plus{} b \plus{} c \equal{} 0$, we get $ \frac {y(a \plus{} 2b) \minus{} a \minus{} b}{a} \equal{} \frac {y(b \plus{} 2c) \minus{} b \minus{} c}{b} \equal{} \frac {y(c \plus{} 2a) \minus{} c \minus{} a}{c}$. After arangement of $ \frac {y(a \plus{} 2b) \minus{} a \minus{} b}{a} \equal{} \frac {y(c \plus{} 2a) \minus{} c \minus{} a}{c}$, we get $ 2y(a^2 \minus{} bc) \equal{} a^2 \minus{} bc$. If $ a^2 \equal{} bc$, , $ ab \plus{} bc \plus{} ca \equal{} 0(\because a^2 \plus{} ab \plus{} ac \equal{} 0)$, so $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$. It contradics to assumption. So, $ y \equal{} \frac {1}{2}$  if $ a \plus{} b \plus{} c \equal{} 0$. The next work is easy.


find the number which divided by 7 gives a remainder of 2, and by 5 gives a remainder of 3...

I'm confused in how to do this with modulor arithmetic.. i think i saw this problem another day somewhere but i can't find it..

[hide="HINT"]
Chinese Remainder Theorem (CRT)
[/hide]

I hope you meant the smallest number...

[hide="elementary logic"]It is $3\pmod5$, so it ends in a 3 or 8.
Let's list a couple of numbers are like this.

3, 8, 13, 18, 23, 28, 33, 38, 43, 48, ...

We look for a number that is $2\pmod7$, so if we subtract two from each of the terms in the list above, we can search for multiples of 7:

1, 6, 11, 16, 21, 26, 31, 36, 41, 46, ...

21 seems to be divisible by 7, so this works.
This number corresponds to 23 in the first list, and it is the answer.

[/hide]

In other words, the solution is $x\equiv 23\pmod{35}$.

$m\equiv{n\pmod{k}}$ means that when m is divided by k, it leaves a remainder of n.

[quote="DanK"]Could someone explain to me what modular arithmetic is?  All the basics of what it means and everything...?[/quote]

Use search function... you'll find lots of results!!!!!!!!!...

Ok, so now I understand what it is and how it works, but I still don't see how you use it for this problem.


modular arithmetic
Diophantine Equations

Prove that the product of five consecutive positive integers is never a perfect square.

$ (x\minus{}2)(x\minus{}1)x(x\plus{}1)(x\plus{}2)\equal{}a^2\Longrightarrow x^2\minus{}1\equal{}u@,x^2\minus{}4\equal{}v^2$ or $ x^2\minus{}1\equal{}3u^2,x^2\minus{}4\equal{}3v^2$ - had not solution.

A typical terse, and incomplete "solution" from Rust. We want $ (x\minus{}2)(x\minus{}1)x(x\plus{}1)(x\plus{}2) = x(x^2-1)(x^2-4) = a^2$.
We have $\gcd(x,x^2-1) = 1$, $\gcd(x,x^2-4) \mid 4$, $\gcd(x^2-1,x^2-4) \mid 3$.

If $\gcd(x,x^2-4) =1$ and $\gcd(x^2-1,x^2-4) = 1$, it follows $x^2-1$, $x^2-4$ must be perfect squares, impossible.

If $\gcd(x,x^2-4) =1$ and $\gcd(x^2-1,x^2-4) = 3$, it follows $x\not \equiv 0 \pmod{3}$, and so $3(x^2-1)$, $3(x^2-4)$ must be perfect squares, impossible.

If $\gcd(x,x^2-4) > 1$ it follows $x=2y$, and so $x^2-4 = 4(y^2-1)$. Since then $x^2-1$ is odd, it follows we need have $y=2z$, and so $x = 4z$, $x^2-4 = 4(4z^2-1)$. If now $\gcd(x^2-1,x^2-4) = 1$, it follows $z$, $4z^2-1$ must be perfect squares, impossible. Finally, if  $\gcd(x^2-1,x^2-4) = 3$, another analysis must be done, but I run out of patience.

Of course, by a result of Erdös and Selfridge, no product of consecutive positive integers is ever a perfect power.

Whoever is Reading this Post, I advise that I am usually online from Monday to Friday at 6: 30. Now, this problem is a Good introdution to study the Erdos Selfridge theorem.  One solution basically consists in writing each of the five numbers as the product of a square and a square free integer, and notice that there are only 4 possibilities for the values of the square free numbers; since we have five numbers, by the Pigeonhole principle, two share the same "square free part". To finish the problem, it is suffice to consider each of the 4 cases and note that the diference of two numbers is bounded above by 4; we find that there is only one case to consider: squarefree part equal 1; and numbers 1 and 4, i.e., the case 5!, which is not a square.


abstract algebra
advanced fields
advanced fields unsolved

Let $ K$ be a number field,and let $ O_K$ be the ring of the integers of $ K$,and $ a,b$ be two nonzero ideal of $ O_K$,
then must the abel groups $ O_K/a$ and $ b/ab$ be isomorphic?

Yes, that's right, since $ b$ is a projective $ \mathcal O_K$-module of rank 1 and the quotient ring $ \mathcal O_K/a$ has only finitely many prime ideals. In other words prove the following

[b]Lemma:[/b] Let $ A$ be a commutative ring with finitely many prime ideals and $ P$ a finitely generated $ A$-module projective of rank 1. Then $ P$ is a free $ A$-module (of rank 1).


A multiple  dwelling has 50 letter boxes. IF 101 pieces of mail are correctly delivered and boxed, show that there is atelast one letter box with 3 or more pieces of mail. ( sorry if this is too easy- I wasnt  sure whre to post it)

Someone move this to basics (if you want), so I can moderate it  :lol:

[hide="hint"]Pigeon hole principle. [/hide]

[quote="nonie"]Wait I thought I put this in Mathcounts?[/quote]

I guess this had some proof, so it was moved to getting started... it's probably too hard for the basics and mathcounts  :P

hmmmm, this is odd... an elementary pigeon hole problem in getting started..... i'd like to hear the justification behind this........

101 = 2*50 +1, think if we wanted to have the least in each box, we would have 2 in each of the 50 boxes, and then 1 left over, and since all the boxes have 2 in them, adding 1 letter will make it so there are 3 letters in one box, this is generally the pigeon hole principle
if you have kn+1 objects and you put them into n boxes, one box must have at least k+1 objects

[quote="mathnerd314"]This was in the AoPS text, wasn't it? Deja vu.

If all you have to do is know the pigeonhole principle and refer to it, then it's pretty easy. But if you have to prove the pigeonhole principle, you'd need to use some algebra, and of course proving.[/quote]

Actually I got it from a different book.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Does there exist an infinite group with only a finite number of subgroups.
[hide="Answer from the book"]no, but i dunno why. if anyone knows , please tell me :) [/hide]

Denote the group by $G$. Consider two cases:
1. There is an element of infinite order.
[hide]
Then the group contains a subgroup isomorphic to $\mathbb Z$. Since $\mathbb Z$ has infinitely many subgroups, $G$ has infinitely many subgroups as well.
[/hide]

2. Each element has a finite order.
[hide]
Pick an element $a_{1}$; let $A_{1}$ be the subgroup generated by it. Pick another element $a_{2}$ from $G\setminus A_{1}$; let $A_{2}$ be the subgroup generated by $a_{2}$. Then pick $a_{3}\in G\setminus(A_{1}\cup A_{2})$ e.t.c.

All subgroups $A_{1}, A_{2}, \dots$ are distinct.
[/hide]

Consider $\mathbb{Z}$. It is clear that this is an infinite group. What are all subgroups of $\mathbb{Z}$? Since $\mathbb{Z}$ is cyclic; all subgroups have to be cyclic too. These are $k \mathbb{Z}$ for $k \in \mathbb{Z}$. It is clear that the  subgroups generated by these subsets are different subgroups. Since there are infinitely many of them, we can conclude that there are infinitely many subgroups of $\mathbb{Z}$

Now, the general case. For some $x \in G$ that has infinite order, we can write in general that:

$<x>$ has all the distinct subgroups  $<x^{1}>$, $<x^{2}>$, $<x^{3}>$, ... So those are infinitely many subgroups of $G$.

edit: too late for case 1

Ok,Nice proof and nice problem.

Now I have another question:
Does there exist an infinite group with all subgroups is finite :?:

I'm not totally sure, but I guess that the union of roots of 1 of order a power of 3 should work.

[quote="rustam"][quote="Yury"]Denote the group by $G$. Consider two cases:
1. There is an element of infinite order.
[hide]
.... isomorphic to $\mathbb Z$.
[/hide][/quote]
Sorry is there any other proof not using the isomorphism, because i did not reach that  chapter and i think this exercise should be solved in the very elementary way  :blush:[/quote]
Read yannick's post.

[quote="-oo-"]Yes, it works. If $p$ is a prime, denote the $p$-sylow subgroup of $\mathbb{Q}/ \mathbb{Z}$ by $\mathbb{Z}(p^\infty)$. It has the presentation
$\mathbb{Z}(p^\infty) = < x_{1}, x_{2}, .... : p x_{1}= 0 , p x_{2}= x_{1}, p x_{3}= x_{2}, ... >$
from which the property can be seen.[/quote]

*bump*

This is some old thread. But can someone please elaborate on the

'Yes, it works. If $p$ is a prime, denote the $p$-sylow subgroup of $\mathbb{Q}/ \mathbb{Z}$ by $\mathbb{Z}(p^\infty)$.'

part :) I don't know what $\mathbb{Z}(p^\infty)$ stands for.

It is called "Prüfer group", see http://en.wikipedia.org/wiki/Pr%C3%BCfer_group

$\mathbb{Z}(p^\infty)$ can be represented, as -oo- wrote, as the p-sylow group of $\mathbb Q / \mathbb Z$. It is the subgroup of fractions (mod 1) whose denominator is a power of $p$.

Or it is the group of p^n-th roots of unity, where n runs through the natural numbers.

LiMa


Let $ a$, $ b$, $ c$ be positive reals. Prove that:
$ \sqrt{\frac{a^{3}}{a^{3}\plus{}\left(b\plus{}c\right)^{3}}}\plus{}\sqrt{\frac{b^{3}}{b^{3}\plus{}\left(c\plus{}a\right)^{3}}}\plus{}\sqrt{\frac{c^{3}}{c^{3}\plus{}\left(a\plus{}b\right)^{3}}}\ge 1$.

yes, hardsoul
this solution is similar IMO-01---very cleary
and now, solve new pro:
PRO: let be given possitive real numbers $a,b,c,d$
pro the inequality: $\sum{\frac{a}{\sqrt[3]{a^3+63bcd}}}\ge 1$
okie?

[quote="manlio"]This is a problem of 2004 Polish Mathematical Olympiads.

You can solve it using

$\displaystyle{\frac{a}{\sqrt[3]{a^3+63bcd}}}\ge \frac {a^p}{a^p+b^p+c^p+d^p}$

It is not hard to find $p$ in my opinion.

Try ;)[/quote]

yes, $p=\frac{21}{16}$
okie?

:D  :D  :D    i've check my work

and i think $p=\frac{21}{16}$.exactly!!!!

[quote="bulba"]For Reyes's inequality:
 Take a+b+c=1 And use Jensen's inequality[/quote]

How?


$ x>1$
$ 4x^5\minus{}7\equal{}k^2$
$ x^5 \in \mathbb Q \Rightarrow (x^5)^3 \in \mathbb Q$
similarly $ x^{13} \in \mathbb Q$
$ \frac{x^{15}}{x^{13}} \in \mathbb Q \Rightarrow x^2 \in \mathbb Q$
similarly $ x \in \mathbb Q$
if $ x\equal{}\frac{a}{b}$ with $ (a,b)\equal{}1$ then
$ 4a^5\equal{}(k^2\plus{}7) b^5 \Rightarrow 4| k^2\plus{}7$ otherwise both $ a$ and $ b$ are even $ \Rightarrow k\equal{}2m\plus{}1$
$ 4a^5\equal{}(4m^2\plus{}4m\plus{}1\plus{}7) b^5 \Rightarrow a^5\equal{}(m^2\plus{}m\plus{}2) b^5$
$ b\equal{}1$ otherwise $ b|a \Rightarrow x$ is a natural number
Summing up:
$ x^5\equal{}(m^2\plus{}m\plus{}2)$
$ x^{13}\equal{}(n^2\plus{}n\plus{}2)$ with $ m,n,x$ natural numbers
$ x\equal{}2$ is a solution
Pending to analize for $ x \ge 3$

[quote=mszew]Pending to analize for $ x \ge 3$[/quote]
/Bump..

[quote=ZetaX]Find all real numbers $ x$ such that $ 4x^5 \minus{} 7$ and $ 4x^{13} \minus{} 7$ are both perfect squares.[/quote]

Let $4x^5-7=a^2,4x^{13}-7=b^2$ for nonnegative integers $a,b$. We have $x\neq0$. Thus $x=\frac{x^{26}}{x^{25}}=\frac{\left(\frac{b^2+7}{4}\right)^2}{\left(\frac{a^2+7}{4}\right)^5}\in\mathbb{Q}$. Since $4x^5-7$ is a nonnegative integer, $x$ is a positive integer.

By looking at $4x^5-7=a^2$ modulo $2$, we see that $a$ is odd. Thus we have in the PID $\mathbb{Z}\left[\tfrac{1+\sqrt{-7}}{2}\right]$ the factorization $\left(\tfrac{a+\sqrt{-7}}{2}\right)\left(\tfrac{a-\sqrt{-7}}{2}\right)=x^5$. Note that $\tfrac{a+\sqrt{-7}}{2}$ and $\tfrac{a-\sqrt{-7}}{2}$ are coprime. Thus there are $m,n\in\mathbb{Z}$ with
\begin{align*}
\tfrac{a+\sqrt{-7}}{2}=\left(n+m\tfrac{1+\sqrt{-7}}{2}\right)^5=\frac{1}{2^5}\left(\left((2n+m)^5-70\cdot(2n+m)^3m^2+245\cdot(2n+m)m^4\right)+\left(5(2n+m)^4m-70(2n+m)^2m^3+49m^5\right)\sqrt{-7}\right)
\end{align*}
So we have
\begin{align*}
16a=(2n+m)^5-70\cdot(2n+m)^3m^2+245\cdot(2n+m)m^4\\
16=5(2n+m)^4m-70(2n+m)^2m^3+49m^5
\end{align*}
In particular we get from the last equation $m\mid 16$ and $16\equiv49m^5\equiv-m\pmod{5}$. Thus $m\in\{-16,-1,4\}$. But $m=-16$ does not work $\pmod{2^5}$ and $m=4$ does not work $\pmod{7}$. So $m=-1$ and
\begin{align*}
16=-5(2n-1)^4+70(2n-1)^2-49\\
(2n-1)^4-14(2n-1)^2+13=0\\
\left((2n-1)^2-7\right)=36\\
(2n-1)^2\in\{1,13\}\\
(2n-1)^2=1\\
n\in\{0,1\}
\end{align*}
But $n=0,m=-1$ contradicts $a\geq0$. So $n=1,m=-1$ implying $a=11$ and $x=2$.

Since $4\cdot2^5-7=11^2$ and $4\cdot 2^{13}-7=181$, the only real number such that $4x^5-7$ and $4x^{13}-7$ are both perfect squares is $x=2$.


inequalities
search
inequalities unsolved

Let $\ a,b,c$ be positive real numbers ,
 prove that   $\frac{a}{\sqrt{a^{2}+8bc}}$+$\frac{b}{\sqrt{b^{2}+8ac}}$+$\frac{c}{\sqrt{c^{2}+8ba}}$ $\geq 1$

[quote="2006"]Let $ \ a,b,c$ be positive real numbers, prove that

$ \frac {a}{\sqrt {a^{2} \plus{} 8bc}} \plus{} \frac {b}{\sqrt {b^{2} \plus{} 8ca}} \plus{} \frac {c}{\sqrt {c^{2} \plus{} 8ab}}\geq 1.$[/quote]$ \left[5\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 4(bc \plus{} ca \plus{} ab)\right]^2 \minus{} \left(a^2 \plus{} 8bc\right)(5a \plus{} 2b \plus{} 2c)^2$

$ \equal{} \frac {(b \plus{} c)(b \plus{} c \minus{} 2a)^2[20a \plus{} 17(b \plus{} c)] \plus{} (b \minus{} c)^2\left[130(2a^2\plus{}bc) \plus{} 208a(b \plus{} c) \plus{} 83(b^2\plus{}c^2)\right]}{4}$

$ \geq0\Longrightarrow$

$ \sum{\frac {a}{\sqrt {a^{2} \plus{} 8bc}}}\geq\sum{\frac {a(5a \plus{} 2b \plus{} 2c)}{5\left(a^2 \plus{} b^2 \plus{} c^2\right) \plus{} 4(bc \plus{} ca \plus{} ab)}} \equal{} 1.$

This problem is introduced as Example 2.4.3 on page 63 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.readbuy.cn/product/product_800815.html]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

See also : http://www.mathlinks.ro/viewtopic.php?t=15294

Just search for "IMO 2001" and you get heaps of search results. This problem has been posted at least a hundred times before. ;)
For example:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=440931849&t=17451
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=440931849&t=15294
The official solution uses UCT theorem as stated and described in the second link by [b]scorpius119[/b].
 :)


symmetry
geometry
circumcircle
trapezoid
perpendicular bisector
geometry unsolved

In triangle $ABC$ denote the point of tangency of inscribed circle to side $AB$ by $E$. Let $\xi$ be the perpendicular bisector of $AB$ and $D$ the image of point $C$ in symmetry by the line $\xi$. A circle is tangent to lines $AC$, $BC$ and circumcircle of triangle $ABC$, at points $X$, $Y$ and $F$, respectively. Prove that points $D$, $E$ and $F$ are collinear.

Very Nice Problem!  :D

My solution:

We know that ABCD is one isóceles trapezoid (with one easy angle chasing)
and that CF and CK are isogonals (where K is the point such that the excircle of ABC, opposite to C, touch AB) by http://www.mathlinks.ro/Forum/viewtopic.php?t=136055

so, <FCB = <KCA.

but, by simetry, E is the point such that the excircle of ADB, opposite to D, touch AB. we have that the triangles ABC and ABD are congruent, so, <KCA = <EDB.

now, we have <EDB = <FCB, and that A, D, C, B and F are in the same circle.

so, the problem is solved!  :wink:

(sorry for my bad² english)

[quote="cvix"]Great solution! :)[/quote]

thanks!

where this problem are from?


How can it be proved that any number can be written as the sum of at most three triangular numbers?

hello, see here
GE Andrews, ETPHKA! num = Д + Л + Д, J. Number Theory, 23(1986) 285-293. Paul T. Bateman, Adolf Hildebrand &: George B. Purdy, Expressing a positive integer ...
Sonnhard.


conics
parabola
ellipse
hyperbola
algebra proposed
algebra

I solved a conic section problem with my teacher yesterday but since I've checked over the problem I think I may have made a mistake would you please check my work?

Question: Identify the conic section represented by each equation. If it is a parabola, give the vertex. If it is a circle, give the center and radius. If it is an ellipse or a hyperbola, give the center and foci.

1.) 4x(squared) + y (squared) - 16X - 6y + 9 = 0

I came up with (x-2)squared + (y-3)squared/4 = 1

but when i was with my teacher i came up with (X-2)squared/4 + (y-3)squared/16

Can you tell me which one is right? or they could both be wrong but I think I made some addition or subraction problems on the first one.
Thanks for your time.

These problems should not be posted here.  The answer you obtained with your teacher is correct.


find the product of all odd integers less than $10000$. You can leave your answer in factoral form.

Product of all even integers less than 10000 is:
$5000! \cdot 2^{5000}$

Denote $P$ as the product of all odd integers less than 10000:
$5000! \cdot 2^{5000} \cdot P = 10000!$

$P =\boxed{ \frac{10000!}{5000!2^{5000}}}$


For certain real values of a,b,c, and d, the equation x^4 + ax^3 + bx^2 + cx + d = 0 has for non real roots. The product of two of these roots is 13+i and the sum of the other two is 3+4i. find b.

help, i dont undersntad the official solution

thanks

Oh, this was a good question.  First step: what do you know about complex roots of polynomials?

yes

i think i understand it now.

JBL, i knew they came in conjugate pairs, i just didnt really know what to do after getting a=-6

A bit shorter of a way... just simplifies the steps through some nice algebra

[hide]

If the roots are p, q, conj(p), conj(q), then

the sum of two of these is 3+4i, obviously these cant be conjugates of each other, so lets say

p+q = 13+i.

therefore conj(pq) = 13+i => pq = 13-i

for notation, lets let r = conj(p), s = conj(q), then

b = pq + pr + ps + qr + qs + rs = (p+q)(r+s) + pq + rs

= (3+4i)(3-4i) + 13+i + 13-i = 9 + 16 + 26 = 51.

[/hide]


I wonder if people make less careless mistakes on more difficult problems.

Please participate in the poll.

The ranges now include all possible responses.

Hoping for a normal distribution.  :lol:

x=0. No mistakes, just not realizing that I already knew how to do the problems and losing points because of it.

That's basically the same story with me. I guess not counting 14 where I misread the problem.


To solve the famous Vietnam ineq

[quote]For $a,b,c$ positive real numbers such that

$21ab+2bc+8ca \leq 12$

find the least value of

$P=\frac 1a +\frac 2b +\frac 3c $ [/quote]

you can use AM-GM for three times

[quote]$21ab+\frac{28}{15}\cdot{\frac{1}{a}}+\frac{112}{25}\cdot{\frac{1}{b}}\ge{\frac{84}{5}}$

$2bc+\frc{144}{75}\cdot{\frac{1}{b}}+\frac{18}{5}\cdot{\frac{1}{c}}\ge{\frac{36}{5}}$

$8ab+6\cdot{\frac{1}{c}}+\frac{4}{3}\cdot{\frac{1}{a}}\ge{12}$[/quote]

and then, after adding, you have

$\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\ge{\frac{15}{2}}$

My ?????????????????? is:

When we use AM-GM, how do you choose the coefficients $K,A,B,C$ ?

[quote]$8ac + KC\frac 1c +KA\frac 1a \geq.......$

$2bc + KB\frac 1b + K(3-C)\frac 1c \geq..........$

$21ab + K(1-A)\frac 1a +K(2-B) \frac 1b \geq.....$[/quote]

so that when we collect all terms we have

$8ac +2bc+21ab +K(\frac 1a +\frac  2b + \frac 3c ) \geq H$

and finally, using the condition,

$(\frac 1a +\frac  2b + \frac 3c ) \geq \frac {H-12}{K}$

Please, can you think of this question?

Its easier after you guess the equality case - then the constants are "obvious" because of the equality condition of AM-GM.

How do you guess the equality case for this ineq?

you haven't mistake-malino :D
but can you tell me ??????? max: $\frac{H-12}{K}$.okie?


function
complex analysis
geometry
geometric transformation
reflection
complex analysis unsolved

If $ f$ is an analytic function that statisfies : If $ z$ is a real number, then $ f(z)$ is a real number, and if $ z$ is in the form $ ib$, where $ i\equal{}\sqrt {\minus{}1}$, then $ f(z)$ is in the form $ ic$, where $ b$ and $ c$ is a real number. Prove that $ f(\minus{}z)\equal{}\minus{}f(z)$. Thanks in advance.

Since $ f$ is real-valued on the real axis, its Taylor coefficients at $ 0$ are real. The same applies to $ if(iz)$. Therefore,

I think this may be another approach --

I think $ \overline{f(\overline{z})}$ is holomorphic? By the given condition, it agrees with $ f$ on the real line and hence everywhere. Similarly we can reflect over the imaginary axis in this way and combining the two reflections gives the desired behavior of $ f$ with respect to the antipodal map.


Are you sure about the statement?

$xy=n(x+y)^2\geq 4nxy$ so there's no solution

[quote="lomos_lupin"]A little but funny problem  :) :

[color=blue]Problem[/color]Find all of the answers of the following equation in natural numbers.

$,x,y,n \in \mathbb{N} , \frac{xy}{(x+y)^2}=n$[/quote]
Mybe $\frac{(x+y)^2}{xy}=n$ :roll:

that would be easy too:

Suppose that $x>y$ and call $x_0 = \frac{x}{(x,y)}$ and $y_0 = \frac{y}{(x,y)}$. Now as $x_0 >1$ there's prime $p|x_0$. It divides right side of $x_0 ^2+y_0 ^2=(n-1)x_0y_0$ but does not divide the left side. Therefore $x=y$ and $n=3$ is the only solution.

[quote="lomos_lupin"]It was TOO obvious ,wasnt it?
but Tiks just made it a little more like a problem  :D[/quote]
I think this two ones is not olimpiad problems :lol: ,mybe these can be proposet only  in schoole's  olimpiads :lol: (for grad 7 :D )


function
inequalities
induction
analytic geometry
graphing lines
slope
algebra

Find all functions $ f: \mathbb{Z}\rightarrow \mathbb{Z}$ such that $ f(f(m))\equal{}m\plus{}1$.

[hide]Let $ f(a)\equal{}b$ for some a and b
If $ b<a$ and $ a\minus{}b\equal{}k$
Then
$ f(b)\equal{}f(f(a))\equal{}a\plus{}1$ and $ f(a\plus{}1)\equal{}f(f(b))\equal{}b\plus{}1$
$ f(b\plus{}1)\equal{}f(f(a\plus{}1))\equal{}a\plus{}2$ and $ f(a\plus{}2)\equal{}f(f(b\plus{}1))\equal{}b\plus{}2$
Continually taking repeated functions and building up, eventually we'd get:
$ f(b\plus{}k)\equal{}f(f(a\plus{}k))\equal{}a\plus{}k\plus{}1\equal{}f(a)\equal{}b$
So
$ 2a\minus{}b\plus{}1\equal{}b\\
b\equal{}\frac{2a\plus{}1}{2}$
and we get
$ f(a)\equal{}\frac{2a\plus{}1}{2}$
$ f(a\plus{}x)\equal{}\frac{2a\plus{}1}{2}\plus{}x$

There has to be some $ a$ for which $ f(a)\le a$ because otherwise
$ a\plus{}1\equal{}f(f(a))>f(a)>a$ but the inequalities are strict and everything is an integer, so $ a\plus{}1>a\plus{}2$, contradiction.
And no $ a$ could have $ f(a)\equal{}a$:
$ f(f(a))\equal{}a\plus{}1\equal{}a$, contradiction there.
Unless I made a mistake, that means there is no such function because $ f(a)<a$ implies
$ f(a)\equal{}\frac{2a\plus{}1}{2}$ but $ \frac{2a\plus{}1}{2}$ isn't a integer, but there also has to be some a such that $ f(a)<a$[/hide]

Let $ P(m)$ be the mentioned property of our function:
$ P(0)$ gives us $ f(f(0))\equal{}1$
Let us have $ f(0)\equal{}a$
Then:
$ f(f(a))\equal{}a\plus{}1$
so $ f(1)\equal{}a\plus{}1$
Let us prove by induction that $ \forall_{m\in Z_{\plus{}}} \ f(m)\equal{}a\plus{}m$
We check for m=1 OK
Proof:
$ f(m\plus{}1)\equal{}f(f(f(m)))\equal{}f(f(a\plus{}m))\equal{}(a\plus{}m)\plus{}1\equal{}a\plus{}(m\plus{}1)$
So now we easily check:
$ a\plus{}(a\plus{}m)\equal{}m\plus{}1$
$ a\equal{}\frac{1}{2}$
Now we know $ f(0)\equal{}\frac{1}{2}$
so it is clear that our function is $ f(x)\equal{}x\plus{}\frac{1}{2}$ :wink:

[quote="polskimisiek"]Let $ P(m)$ be the mentioned property of our function:
$ P(0)$ gives us $ f(f(0)) \equal{} 1$
Let us have $ f(0) \equal{} a$
Then:
$ f(f(a)) \equal{} a \plus{} 1$
so $ f(1) \equal{} a \plus{} 1$
Let us prove by induction that $ \forall_{m\in Z_{ \plus{} }} \ f(m) \equal{} a \plus{} m$
We check for m=1 OK
Proof:
$ f(m \plus{} 1) \equal{} f(f(f(m))) \equal{} f(f(a \plus{} m)) \equal{} (a \plus{} m) \plus{} 1 \equal{} a \plus{} (m \plus{} 1)$
So now we easily check:
$ a \plus{} (a \plus{} m) \equal{} m \plus{} 1$
$ a \equal{} \frac {1}{2}$
Now we know $ f(0) \equal{} \frac {1}{2}$
so it is clear that our function is $ f(x) \equal{} x \plus{} \frac {1}{2}$ :wink:[/quote]

$ f(x)\equal{}x\plus{}\frac{1}{2}$ doesn't satisfy $ f: \mathbb{Z} \rightarrow \mathbb{Z}$

This doesn't seem right...
[hide]Take the function $ f$ of both sides:
$ f(f(f(m))) \equal{} f(m \plus{} 1)$
$ \Rightarrow f(m) \plus{} 1 \equal{} f(m \plus{} 1)$
And again:
$ f(f(m) \plus{} 1) \equal{} f(f(m \plus{} 1))$
$ \Rightarrow f(f(m) \plus{} 1) \equal{} m \plus{} 2$
And again:
$ f(f(f(m) \plus{} 1)) \equal{} f(m \plus{} 2)$
$ f(m) \plus{} 2 \equal{} f(m) \plus{} 2$

So we can see that after repeating this we have $ f(m \plus{} n) \equal{} f(m) \plus{} n$ for all positive integers $ n$ which implies $ f(n)$ is linear. Since it has a slope $ \frac {f(m \plus{} n) \minus{} f(m)}{n} \equal{} 1$, it is in the form $ f(m) \equal{} m \plus{} c$, for some constant $ c$. Plugging it into to the original equation, we have $ f(f(m) \equal{} f(m \plus{} c) \equal{} m \plus{} 2c \equal{} m \plus{} 1\Rightarrow c \equal{} 1/2$. However, the function $ f(m) \equal{} m \plus{} 1/2$ (which has a domain of the integers from the original conditions) has a range that has no integers, so there are no such functions $ f$.  :) [/hide]

Edit: I didn't see polskimisiek or Simon's OR caffieneboy's  :oops:   post, but $ f(x) \equal{} x \plus{} 1/2$ is not $ \mathbb{Z}\rightarrow \mathbb{Z}$  :wink:

Aah, yes, but the conclusion follows :wink:

Are functions such as $ f(x) \equal{} x \plus{} x$ $ mod$ $ 2$ acceptable?

By empty function you mean $ f(m) \equal{} 0$? We can plug this back in and we have $ f(f(m)) \equal{} f(0) \equal{} 0 \equal{} m \plus{} 1$ for all $ m$, which doesn't work  :wink:

Edit: @begoner: Plug in that function for even $ m$, and it doesnt work either

[quote="caffeineboy"]Actually, I think we all missed something. [/quote]
Why do you think so? There are simply no such functions.

Begoner, your function doesn't satisfy the condition :wink:

[quote="polskimisiek"]Begoner, your function doesn't satisfy the condition :wink:[/quote]

Yeah, I know, I was just wondering if weirder functions were acceptable.  This one almost works, for example (if $ x$ is a positive number, it returns $ \minus{} x$; if it's negative, it returns $ \minus{} x \plus{} 1$; if it's $ 0$, it doesn't work).

$ f(x) \equal{} \minus{} x \plus{} \frac {x^2 \minus{} x|x|}{2x^2}$.

Well of course, weird functions are always acceptable if they are sufficient. Mostly, when solving functional equation we won't have any weird stuff like that (I've just seen it once). And almost is the right word for this :wink:

Yeah usually unless there is an obvious weird function after attacking the function, in a contest our solution would suffice. I really wish they would say continuous in problems like these though...

Sorry, continuous? It is $ f: \mathbb{N} \rightarrow \mathbb{N}$. It can't be continuous in the normal sense of the word. I think you mean polynomial, which isn't really that great since we lose loads of functions

For any $ x$ and $ y$ such that $ x \neq y$, $ f(x) \neq f(y)$.  Otherwise $ f(f(x)) \equal{} f(f(y))$ and $ x \plus{} 1 \equal{} y \plus{} 1$, a contradiction.

Say $ f(x) \equal{} a \plus{} 1$ for some integral values of $ x$ and $ a$.  Then $ f(a) \equal{} x$ because $ f(f(a)) \equal{} a \plus{} 1$ and there can be no other value $ b$ (except for $ f(a)$) such that $ f(b) \equal{} a \plus{} 1$.  Continuing in this manner, $ f(x \minus{} 1) \equal{} a$, $ f(a \minus{} 1) \equal{} x \minus{} 1$ and $ f(a \minus{} k) \equal{} x \minus{} k$.  Then $ f(a \minus{} (a \minus{} x))$ $ \equal{}$ $ x \minus{} (a \minus{} x)$; thus, $ f(x) \equal{} 2x \minus{} a$.  However, it was previously determined that $ f(x) \equal{} a \plus{} 1$, which is a contradiction.  Therefore, no function that meets the conditions exists.

[quote="Anaxerzia"]I meant for the general case :). For this case a modular function could possibly work...[/quote]
It won't. The solutions above show that no function, no matter how weird, can work.


Let $a,b$ are positive integers, prove that $(a+1/2)^{n}+(b+1/2)^{n}$ is an integer only for finitely many positive  integers.

Does it work if we assume that it's true for all integers, than show a contradiction?

Because $a = b = 1$ is a contradiction if $n > 2$

You need to prove that no matter which ordered pair of positive integers you use, that result will hold. Proving it for $(1,1)$ doesn't say anything about $(2,2)$ (unless you use induction).

hmm im not sure i quite understand this problem. If $ n \in \mathbb{Z}^{+}$, then doesn't there only exists solutions when $ n=1$? In fact, there are infinitely many solutions when $ n=1$. Well here's what I have so far.
[hide]
Let $ n=k>1$ and $ s^{k}= a^{k}+b^{k}$. So we have

\begin{eqnarray*}\left(a+\diplaystyle\frac{1}{2}\right)^{k}+\left(b+\diplaystyle\frac{1}{2}\right)^{k}&=& \frac{1}{2^{k}}\left((2s)^{k}+(2s)^{k-1}\binom{k}{1}+(2s)^{k-2}\binom{k}{2}+\cdots+(2s)\binom{k}{k-1}+1\right) \\ \end{eqnarray*}

All the terms in the parentheses on the RHS are even except the $ 1$. Therefore, their sum cannot be divisible by $ 2^{k}$ and there can be no integer value.

[/hide]

[quote="paladin8"]$ \left(\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\in \mathbb{Z}$.[/quote]
 :huh:
$ \frac{9}{4}+\frac{1}{4}= \frac{10}{4}= \frac{5}{2}\not\in \bf Z$.  In general, $ a+\frac{1}{2}= \frac{2a+1}{2}$, and $ (2a+1)^{2}\equiv 1 \pmod 4$, so when $ n = 2$ we [i]always[/i] end up with half an odd integer.  However, when $ a = 0$ and $ b = 3$ we have $ \left(\frac{1}{2}\right)^{3}+\left(\frac{7}{2}\right)^{3}= 43 \in \bf Z$, so there are [i]some[/i] solutions with $ n > 1$.

The problem might be clearer like this:

Show that, for [i]any[/i] pair $ a, b$ of integers, there are only finitely many values of $ n$ such that $ (a+1/2)^{n}+(b+1/2)^{n}$ is an integer.

[i]Edit: The first part of this post is in response to a post which paladin8 deleted while I was typing.[/i]

Yeah, sorry about that. I'm tired from work... lol. :P

[quote="NeverOddOrEven"]hmm where did i go wrong in my solution? it's clearly wrong since JBL showed that there are some solutions when $ k>1$ but I can't seem to find where.[/quote]
[quote="NeverOddOrEven"][hide="Emphasis added"]\begin{eqnarray*}\left(a+\diplaystyle\frac{1}{2}\right)^{k}+\left(b+\diplaystyle\frac{1}{2}\right)^{k}&=& \frac{1}{2^{k}}\left((2s)^{k}+(2s)^{k-1}\binom{k}{1}+(2s)^{k-2}\binom{k}{2}+\cdots+(2s)\binom{k}{k-1}+\boxed{1}\right) \\ \end{eqnarray*}[/hide][/quote]
$ 1+1 \neq 1$.  ;)

anybody have a way to characterize the largest power of $ 2$ in denominator of the reduced fraction? it seems to decrease in general, but it does not form a strictly decreasing sequence for $ (a,b)=(0,3)$, for example.

oh haha thanks cincodemayo  :blush:

hmm so how about this?

[hide="Going further in my previous solution"]
Let $ s^{n}= a^{n}+b^{n}$.

\begin{eqnarray*}\left(a+\diplaystyle\frac{1}{2}\right)^{n}+\left(b+\diplaystyle\frac{1}{2}\right)^{n}&=& \frac{1}{2^{n}}\left((2s)^{n}+(2s)^{n-1}\binom{n}{1}+(2s)^{n-2}\binom{n}{2}+\cdots+(2s)\binom{n}{n-1}+2\right) \\ &=& \frac{(2s+1)^{n}+1}{2^{n}}\\ \end{eqnarray*}
So we want to find when
\[ (2s+1)^{n}+1 \equiv 0 \pmod{2^{n}}\Rightarrow (2s+1)^{n}+1-a\cdot 2^{n}= 0\]
Since we now have a polynomial, does that tell us that there are only finitely many solutions? i'm not sure though.

[/hide]

[quote="NeverOddOrEven"]
Let $ s^{n}= a^{n}+b^{n}$.
[/quote]
Do you mean $ s_{n}= a^{n}+b^{n}$? It's a subscript, not a superscript. Your notation is confusing, and I think you even confused yourself here:

[quote="NeverOddOrEven"]$ \ldots = \frac{(2s+1)^{n}+1}{2^{n}}$[/quote]
You can't contract the expansion like that. Even in the smaller cases, $ (s_{1})^{2}= (a+b)^{2}= a^{2}+2ab+b^{2}\neq a^{2}+b^{2}= s_{2}$.

The expansion is really: \[ (2a+1)^{n}+(2b+1)^{n}= \sum_{k=0}^{n}\dbinom{n}{k}2^{k}(a^{k}+b^{k}) = \sum_{k=0}^{n}\dbinom{n}{k}2^{k}\cdot s_{k}\]


in parallelogram ABCD, angle ABC measures 75 degrees, AF is perpendicular to BC, DE=2AB, AE is perpendicular to AD, what is the measure of AED?

Is it $60^{\circ}$?

edit: nvm

All I can figure out is $\triangle AED$ is similar to $\triangle FEB$

here is a really simple solution
take $M$ the middle of $[DE] $
now $m(ABC)=75$so $m(DAC)=115$ so $m(ADB)+m(ABD)=75(1)$
but $m(DAE)=90$ so $m(ADE)+m(AED)=90(2)$ so $m(AED)-m(ABD)=15$
if $M $ middle of $[DE]$ then $AM=ME=MD=AB$ so $ m(AME)=m(ABD)$
and because $AM=ME$ $m(AEM)=m(MAE)$ so $m(AME)+m(AEM)+m(MAE)=180$ so
$m(ABD)+2m(AED)=180$ but $m(AED)-m(ABD)=15$ so $m(AED)=65$

bravo!  :D Why does AM=ME?

Edit: Nvm, i got it.

well AM=ME because DAE is a right angled triangle and M is the middle of DE (you prove this by taking a point F such that DAEF is a rectangle then M is the intersection of the diagonals :) )  (by the way  i forgot to say i thought it was a really nice problem :D )


Find all natural numbers [b]n [/b]for which the product of its digits is

$ n^2 \minus{} 13n \minus{} 25$ .

Kindly show the working,

Thank You!

[hide]First we'll eliminate all the digits that can't appear in $ n$.

Take $ N: \equal{} n^2 \minus{} 13n \minus{} 25$ modulo $ 2$. We have $ N\equiv n^2 \minus{} n \minus{} 1\equiv n(n \minus{} 1) \minus{} 1\pmod{2}$, but as $ n(n \minus{} 1)$ is always even, we get $ N\equiv 1\pmod{2}$. Therefore $ n$ can't contain either of the digits $ 0,2,4,6,8$.

Next, $ N\equiv n^2 \minus{} n \minus{} 1\pmod{3}$. Checking all the possibilites for $ n$, we find that $ n^2 \minus{} n \minus{} 1\not\equiv 0\pmod{3}$, hence $ n$ doesn't contain $ 3,9$ either.

Now $ N\equiv n^2 \plus{} n \plus{} 3\pmod{7}$. Checking $ n\equiv 0,\pm 1,\pm 2,\pm 3\pmod{7}$, we get $ N\not\equiv 0\pmod{7}$ as well.

So finally we conclude that $ n$ contains only $ 1$ and $ 5$ as digits, therefore $ N \equal{} 5^k$ for some non-negative $ k$.

Now $ n^2 \minus{} 13n \minus{} 25 \equal{} 5^k$. Assume $ k\geqslant 2$. Then $ n^2 \minus{} 13n\equiv 0\pmod{25}\iff n(n \minus{} 13)\equiv 0\pmod{25}$. Among the numbers $ n, n \minus{} 13$, at most one is divisible by $ 5$, hence we must have either $ n\equiv 0\pmod{25}$ or $ n \minus{} 13\equiv 0\pmod{25}$. But neither is possible, since the first possibility implies $ n$ ends in $ 00, 25, 50$ or $ 75$, and the second one implies $ n$ ends in $ 3$ or $ 8$ - both of which contradict the conclusion that $ n$ contains only $ 1$'s and $ 5$'s.

Therefore $ k$ is at most $ 1$.

Now we just have to solve two quadratics: $ n^2 \minus{} 13n \minus{} 25 \equal{} 1$ and $ n^2 \minus{} 13n \minus{} 25 \equal{} 5$. Only the latter yields a solution satisfying all the conditions, and that's $ n \equal{} 15$.[/hide]


There's a new student union building at UNLV, which is where the Saturday morning competition will be held. I've just found out that the food court in that building will be open until 3:00 on Friday. The food court includes the following: Jamba Juice, Starbucks, Sub Connection, Taco Bell, and Panda Express.

I don't know what else is in there - what kind of gathering places there might be, or whether there's a game room. But it seems likely that that location might be a good one for meeting people from other teams.

Who else will be there for lunch on Friday?  We're currently trying to plan our meals, and I was hoping to plan something for lunch that would encourage socializing with other teams, if there are any there.  In particular, one idea that occurred to me was to get together with another team or two and get Subway takeout (we have adults with cars who can pick it up) to eat in Tonopah as kind of a lunch social thing.  Anyone interested?

Also, Jamba is apparently closed for renovation. :(

Our best estimate is that we will arrive at about 1:00, and the very first thing we'll do is check in. At that point - 1:30  or so - we'll encourage anyone who's hungry to go over to that food court in the student Union. San Diego will probably arrive at about the same time we do. We're 3 teams and San Diego is 2 teams.

The only way what you're talking about would work for us would be if you made it a late lunch.

[quote="Sly Si"]I was hoping to plan something for lunch that would encourage socializing with other teams, if there are any there.[/quote]
Name tags, anyone? And if there are name tags should people put their real names or their AoPS screen
names? (For me, of course, the answer to that question is "Yes.")

On another subject: it appears that there are no short-handed teams, while several of us, notably including SFBA, NorCal, and us, appear to be bringing a few extra students as alternates. There might be a near-complete official alternate team.

From the housing list, I count exactly 16 alternates.  Not sure what they'll do about that; hopefully the 16th can just join the other 15 on a team, since they're not competing officially anyway.

And we won't be able to do a late lunch, as we're practicing during that time.

You might have overcounted. For instance, Colorado has no alternates.  I don't think there will be more than 15.

[quote="Paul Dreyer"]The locations
(all in Tonopah, where your team will check-in) are:

Alabama: 3S
Arizona: 2S
Colorado: 4N and 5N
New Mexico: 6N
Northern California: 3S and 4S
Philippines: 6C
San Diego: 2C
SF Bay: 3N and 4N
Southern California: 5S and 6S
Taiwan: 3C and 4C
Utah: 4C
Washington: 5C
[/quote]

Official weather forecast:

[quote]Tonight: Mostly clear. Lows in the mid 70s. Southwest wind around 10 mph.

Friday: Sunny. Highs around 101. Light wind becoming south around 10 mph in the afternoon.

Friday Night: Mostly clear. Lows in the mid 70s. Southwest wind 10 to 15 mph.

Saturday: Mostly sunny. Highs around 101. Light wind becoming southwest around 10 mph in the afternoon.

Saturday Night: Partly cloudy. Lows in the upper 70s. [/quote]
... but you knew it would be like that. Perfectly normal, maybe 3-4 degrees cooler than last year.

I know it's been awhile, but for my own planning purposes, does anyone remember when the tests ended Saturday morning last year?

I'm going to lock this because people (read: me) are getting confused because this thread discusses 2007 and not 2008.  Post in the new 2008 thread if you have questions or comments about the Las Vegas site in 2008.


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

Let $P$ and $Q$ be antipodal points on the circumcircle of triangle $ABC$, and let $P'$, $Q'$ be their isotomic conjugates, respectively.  Prove that $PQ' \perp P'Q$.  (Thus, they intersect on the circumcircle.)

[quote="prowler"]Sorry, but what is definition for "antipodal" points ?[/quote]
The distinct points lying on the circle and passing through the center.

Sorry, but what are the definitions for "isotomic conjugate"? (antipodal point - i don't understood; a point passes through some ...?)

Isotomic conjugate is the concurrency of the lines joining the vertex and the reflection of the foot of the corresponding Cevian in the midpoint of the corresponding side.

See solution here
https://artofproblemsolving.com/community/c6h1365992p28314736


function
trigonometry
algebra unsolved
algebra

Let [tex]h:R --> R[/tex] be an arbitrary function. The task: It's always possible to find some nondecreasing functions [tex]f,g:R-->R[/tex] such that for all real [tex]x[/tex] is satisfied that [tex]h(x) = f(x) - g(x)[/tex]?

If we have only computable number of [tex]x[/tex], it can be easily showed that the statement is valid, but I don't know whether for uncomputable number of [tex]x[/tex] it's valid too. Maybe it's well-known problem, but actually I don't know what to do with it. Thanks for a help.

Let $f$ be a real function.
Let $I = [a,b]$ be a closed interval, and $s = (a=x_0 , x_1, \cdots, x_{n-1}, x_n = b)$ be a subdivision of $I$ (where $x_i < x_{i+1}$ for all $i$).
Let $V(f,s) = \sum_{i=1}^n |f(x_i) - f(x_{i-1})|.$

We will say that the function $f$ has 'bounded variations' (I don't know the exact english wording, I've just translated it from the french 'à variations bornées') on the interval $I$ if the set of the numbers $V(f,s)$ when $s$ describes the set of all the subdivisions of $I$ is bounded. Its upper bound is $V(f)$ and called its total variation over $I$.

It is a classical result to prove that $f$ has bounded variation on $I$ if and only if $f$ is the difference of two non-decreasing functions on $I$.

Thus, to prove that the answer is 'no' to your question we only have to prove that there exists a real function with unbounded variations...

You may prove (I leave it to the reader) that if
$f_n(x) = \frac { \sin (10^nx)} {2^n}$
then $f (x) = \sum_{n=0}^{+\infty} f_n(x)$ is continuous on $\mathbb {R}$ but has unbounded variations on every closed interval with positive length.

Pierre.

Another simple continuous example is $f(x)=x\sin \frac{1}{x}$.


algebra
polynomial
linear algebra
linear algebra unsolved

Is the following statement true? [ If you say a statement is true you should explain why , if you say it is false then you shoud give a counter example]

d) All 3x3 matrices over REAL NUMBERS{|R} have an eigenvector in $|R^3$

What is the degree of caracteristic polynomial ? [hide] (odd ;))[/hide]


This is about the last question in AIME Math Jam on Aug. 30.

The question, Jennifer, (as i remember it) is:

Find a such that a and b are integers and x^2-x-1 is a factor of ax^17+bx^16+1.

Someone please correct me if I'm wrong.

Transcript would be available soon, so you can refer to that [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=12]here[/url] when they are available. (It's not yet available, FYI)

Special thanks to mathfanatic :)

Haha, OK.
I understand how Fibonacci and Binet's managed to slip itself into this question, but really, you don't need it.

Your average algebraic manipulation would do the trick.

Oy!  I just remembered that I couldn't make this Math Jam either.  Stupid DSL router.  Oh well.  Maybe I'll save all the transcripts onto my hard drive for fun.  Whoopee!

Anyway, I'm thinking reaaaly hard on this problem.

Does anyone need the solution to the original problem?


search
combinatorics proposed
combinatorics
Chip firing

Let $ G$ be a connected locally finite graph, i.e. a (possibly infinite) connected graph all of whose vertices have finite degree. For a vertex $ x$, denote its degree by $ d(x)$. On some of the vertices of $ G$ we place some tokens such that the total number of tokens is finite. Now we play the following game, in a sequence of steps: a step consists of finding a vertex $ x$ on which at least $ d(x)$ tokens are placed, removing $ d(x)$ tokens from $ x$, and placing one on each of $ x$'s neighbors. We then say that vertex $ x$ was "fired" (hence the name "chip-firing game"). This game could terminate, or it could go on forever. Prove the following:

(i) Given the graph $ G$ and a starting configuration of tokens, if one legal game as described above finishes, then every legal game finishes. If this is the case and all games do end in finitely many steps, then every vertex is fired the same number of times regardless of the game. In particular, all games end after the same number of steps, and in the same final configuration.

(ii) If there is a vertex which is never fired, then the game finishes.

(iii) If $ G$ is infinite, then there is a vertex which is never fired; hence, according to (ii), the game finishes.

This solves a few problems which have appeared on the forum. I can remember at least two, both shortlisted for the IMO: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=219679]one from 1996[/url], which concerns a chip-firing game on a path infinite in both directions, and [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1927943989&t=2519]one from 1994[/url], which masks a chip-firing game on a cycle (for this second problem a little more work is required in order to apply the results here, but only a little :)).

[hide="i"]
Consider two games $ A$ and $ B$, and suppose they are different. Take the first firing $ f$ in $ A$ that does not exist in $ B$. If $ A$ is contained entirely in $ B$, then just switch $ A$ and $ B$. (If the firing from this vertex is done $ n$ times in $ B$, then you are looking at the $ (n\plus{}1)^{th}$ firing from this vertex in $ A$). Since all the firings before $ f$ are also done in $ B$, then after these firings are done in $ B$, it must be possible for $ f$ to be done in game $ B$ as well, so $ B$ cannot terminate unless it has $ f$ as well, a contradiction. Thus, if one game terminates, all such games will terminate and will consist of the same firings.
[/hide]

[hide="ii"]
Let $ f(v)$ be the number of times that a vertex $ v$ is fired. I claim that if $ f(v)$ is finite, then $ f(v')$ is finite for any vertex $ v'$ adjacent to $ v$.

Proof: Suppose $ v'$ is fired an infinite number of times. Since $ v$ is only fired a finite number of times, an infinite number of chips will accumulate onto $ v$, but this is impossible since there are only a finite number of chips.

Since we have one vertex $ v_1$ not fired at all, $ f(v_1)$ is finite, and since $ G$ is connected, we can find that every vertex $ v$ has $ f(v)$ finite. If $ G$ is a finite graph, or if there are only finitely many vertices $ v$ with $ f(v)>0$, then this game must terminate, so assume that there are infinitely many $ v$ with $ f(v)>0$.

Now, take any vertex $ v$ with $ f(v)>0$, and consider the last time it is fired, and look at one of its chips which goes onto a neighboring vertex $ v'$. Now, if $ v'$ is never fired again, then the chip will remain on $ v'$. Otherwise, $ v'$ will fire some chip back to $ v$, which will remain at $ v$.

So, for each of the infinitely many vertices $ v$ with $ f(v)>0$, there is some chip which must stop moving, but this is impossible as there are only finitely many chips. Thus, there are only finitely many $ v$ with $ f(v)>0$, so the game must terminate.
[/hide]

[hide="iii"]
Suppose that every vertex is fired. Consider infinitely many pairs of vertices, such that the two vertices of each pair are connected and no two pairs share a vertex. Take one of these pairs $ (v_1,v_2)$. When $ v_1$ is fired, there will be a chip on $ v_2$, and when $ v_2$ is fired, there will be a chip on $ v_1$. So, when one of these is fired, there will always be at least one chip on a vertex of this pair. But since we are assuming all vertices will be fired at least once but there are only finitely many chips, this is impossible.
[/hide]

An omission on my part: the converse to (ii) also holds.

(ii') If the game finishes, then there is a vertex which is never fired.

(of course, this is only interesting when the graph is finite)

[hide]
Suppose a game terminates but every vertex is fired. Then, for each vertex, consider the last time the vertex is fired. Take the vertex whose final firing is the earliest, call this vertex $ v$. Then all other vertices are fired after $ v$ is fired, including the neighbors of $ v$. Each neighboring vertex fires a chip onto $ v$, so $ v$ is able to be fired again. Then, the game could not have been terminated at this point.
[/hide]


$9$ can be expresses using exactly three occurrences of $3$ very easily: $9=3+3+3$

Can we express $9$ using exactly three occurrences of $3$, but without using any of the symbols $+$, $\times$, $/$ (addition, multiplication, division)?

You can use any other arithmetic operation, symbol.

[quote="DarkArchon"]$\sqrt[3]{3^{3!}}$[/quote]

Good work.

I am sure there are many more ways to do this...

${\lfloor\sqrt{\sqrt{\sqrt{\sqrt{3^{33}}}}}\rfloor}$

or, while i'm at it:

${{\lfloor\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3!^{(3!^{3})}}}}}}}}}}}}\rfloor}$


function
binomial coefficients
calculus
calculus computations

Is it possible that $ \binom n{2j \plus{} 1} \equal{} 0$ whenever $ 2j \plus{} 1 > n$ ? I don't get it.

You can consider that a convention, but it's a convention that makes sense.

Let $ \binom{n}{k}\equal{}\frac{n!}{k!(n\minus{}k)!}$ and rewrite that as $ \frac{\Gamma(n\plus{}1)}{\Gamma(k\plus{}1)\Gamma(n\minus{}k\plus{}1)}$

Now, what happens if either $ k>n$ or $ k<0?$ If $ n$ and $ k$ are still integers, we note that the Gamma function has a simple pole at $ k\plus{}1$ or at $ n\minus{}k\plus{}1.$ Adopt the kind of complex arithmetic in which division by $ \infty$ gives $ 0$ (the kind of arithmetic that says that the reciprocal of a function with a pole is a function with a zero) and we'll say that that binomial coefficient is zero.

There aren't any ways to choose more than $ n$ objects from $ n$ objects.

That works, too. Note that there is one way to choose all $ n$ objects, so $ \binom{n}{n}\equal{}1.$ In colloquial English, we often say that we have "no choice" but we really do have one "choice". Having no choice (meaning we can't do it at all) is what t0rajir0u means.

One way to use this notation: suppose I'm working on a formula for the sum of every third binomial coefficient in a row. I can write that sum as

$ \sum_{k\equal{}0}^{\infty}\binom{n}{3k}.$

Even though I wrote the upper limit of summation as $ \infty,$ this is really a finite sum, with an upper limit that depends on $ n$ -- well, it depends on $ n$ in a way that I'm temporarily too lazy to write down. And sometimes it's a mild convenience to be able to be that lazy.

To throw my penny in, yet another way to see it, midway between regular binomial coefficients and the $ \Gamma$ function. For non-negative integers $ k$ consider the functions $ b_k : \mathbb{R} \to \mathbb{R}$ defined by $ b_k(x) \equal{} \frac {x(x\minus{}1)\cdots (x\minus{}k\plus{}1)} {k!}$; these expressions appear in the generalized Newton's binomial formula.
Now, for integer $ n \geq k$ notice that $ b_k(n) \equal{} \genfrac {(} {)} {0pt} {} {n} {k}$.
What happens for $ k > n$? One factor in the product $ n(n\minus{}1)\cdots (n\minus{}k\plus{}1)$ becomes zero, so $ b_k(n) \equal{} 0$, which adds justification to the "convention".


inequalities
Euler
inequalities unsolved

[quote="ityaa"]in a triangle,prove or disprove
$\frac{a}{b}+\frac{b}{a} \leq \frac{3R}{R+r}$[/quote]Consider a $9-40-41$ triangle.
Then \[ \frac{3R}{R+r}<3<\frac{40}{9}+\frac{9}{40}, \] so the inequality is not necessarily true.

Sorry, but I know that Bandila's inequality is the folowing:
 In each triangle, we have $\frac{R}{r}\geq\frac{b}{c}+\frac{c}{b}$,
and I have a proof of it.

yes, sorine, that`s bandila and ityaa`s ineq is the Euler ineq applied in Bandila and etc :|

Kosmin, do you have a proof of Bandila's Inequality?Please post it.

Hint:
  We can use the Napier's Formulas and we get the following relation:
$\ l_a^2=\frac{4bcp(p-a)}{(b+c)^2}$, $\ p=\frac{a+b+c}{2}$.Easiliy we get the inequality $\ l_a^2\leq\ p(p-a)$ ...  It's very easy.


How  many numbers from 1 to 100 :
a) divisible by 2 ;
b) divisible by 5 ;
c) divisible  by 2 and by 5 ;
d) not divisible by 2 and  by 5 ?

How  many numbers from 1 to 100 not  divisible by 2 or by 3 ?

For the last question:
Numbers divisible by 2: 50, 3:33, both 2 and 3: 16
100-50-33+16=33

Or: In every 6 consecutive numbers, there will be 2 that is not divisible by 2 or 3. In 1-96 there are 2x(96/6)=32, adding one more for 97 makes 33


A line passing through the points $ (2a\plus{}4,3a^2)$ and $ (3a\plus{}4,5a^2)$ has slope $ a\plus{}3$.  Find the value of $ a$.

The slope for a line is: $ \frac {\textbf{y2-y1}} {\textbf{x2-x1}}$ when an ordered pair is expressed as $ (\textbf{x1},\textbf{y1}), (\textbf{x2},\textbf{y2})$. so the slope of this line is...

$ \frac {\textbf{5a^2-3a^2}} {\textbf{(3a+4)-(2a+4)}}$ $ =$ $ \frac {2a^2} {a}$ $ =$ $ 2a$. We are also given that the slope is $ a+3$, so we set up an equation that states:

$ 2a$ $ =$ $ a+3$. We subtract $ a$ from each side, and get

$ \boxed{\boxed{\boxed{a=3}}}$


Where do you suggest i get a book that can teach me maxima and minima?

Of functions?  Any basic calculus book can teach you how to find max and min of single-variable functions, and a multivariable calc book and extend this to several variables.

if you want just basic i suggest "Calculus" by Michael Spivak or "Calculus" by Larson, Hostetler, Edwards.

Spivak for basic?  Not a chance.  Go with any basic calculus book.  You don't need Spivak's rigorous treatment.  Great book, but not if you are unfamiliar with calculus going in.


Can you post the full score of China Team?

Besides Xiaosheng Mu (CHN1), Dongyi Wei(CHN4) also got a perfect scores!

[quote="Zhang Fangyu"]Besides Xiaosheng Mu, Dongyi Wei also got a perfect scores![/quote]
My congratulations. What about perfect scorers of the other countries?


1.Entry by: evilhampster

Newsbreak

The President passed his physical airplane with flying cats this morning. Doctors gave him an incompetent bill of health, but advised him to fail at least twenty minutes a day and to eat less boring food.

At the Mars zoo, a five-hundred pound Giant Panda reached out and grabbed a woman's incredulous camera right out of her foot when she tried to take his picture eating spaghetti. The Giant Pand then flew the camera.

Sometimes it doesnt pay to diet. Popular comedian Bob, whos lost more than $ \pi^2$ pounds, was virtually caught with his runners down when his pants fell to the floor as he performed in front of an audience of five hundered enthusiastic hitchhikers.

2. Entry: by Aleph0

Newsbreak

The President passed his physical spam with flying spammers this morning. Doctors gave him an spammy bill of health, but advised him to spam at least twenty minutes a day and to eat less cold food.

At the Antarctica zoo, a five-hundred pound penguin reached out and grabbed a woman's freezing camera right out of her liver when she tried to take his picture eating liver. The penguin then ate the camera.

Sometimes it doesnt pay to diet. Popular comedian Mr. Green, whos lost more than $ \pi$ pounds, was virtually caught with his emoticons down when his pants fell to the floor as he performed in front of an audience of five hundered enthusiastic Mad Libs.

3. Entry by: dragon96

Newsbreak

The President passed his physical slug with flying slime this morning. Doctors gave him an slimy bill of health, but advised him to slime-ify at least twenty minutes a day and to eat less slimy food.

At the snail garden zoo, a five-hundred pound snail reached out and grabbed a woman's squishy camera right out of her mouth when she tried to take his picture eating snail. The snail then ate the camera.

Sometimes it doesnt pay to diet. Popular comedian eugene, whos lost more than 3.1415926535897932384626433832795028841971693993 pounds, was virtually caught with his gastropods down when his pants fell to the floor as he performed in front of an audience of five hundered enthusiastic slugs.

4. Entry by: cf249

Newsbreak

The President passed his physical pineapple with flying dandelions this morning. Doctors gave him an dilapidated bill of health, but advised him to pounce at least twenty minutes a day and to eat less hilarious food.

At the New York City zoo, a five-hundred pound antelope reached out and grabbed a woman's curious camera right out of her nostril when she tried to take his picture eating pizza. The antelope then performed the camera.

Sometimes it doesnt pay to diet. Popular comedian David, whos lost more than -249 pounds, was virtually caught with his rainbows down when his pants fell to the floor as he performed in front of an audience of five hundered enthusiastic elevators.

[quote="Dojo"]

3. Entry by: dragon96

Newsbreak

The President passed his physical slug with flying slime this morning. Doctors gave him an slimy bill of health, but advised him to slime-ify at least twenty minutes a day and to eat less slimy food.

At the snail garden zoo, a five-hundred pound snail reached out and grabbed a woman's squishy camera right out of her mouth when she tried to take his picture eating snail. The snail then ate the camera.

Sometimes it doesnt pay to diet. Popular comedian [b]eugene[/n], whos lost more than 3.1415926535897932384626433832795028841971693993 pounds, was virtually caught with his gastropods down when his pants fell to the floor as he performed in front of an audience of five hundered enthusiastic slugs.

[/quote]
My pants fell down.  :blush:.  I lost 3 pounds.  Yay!  What's a gastropod?

Honestly, I can't say mine is the best.

And thanks for spelling my username wrong!

Try looking in your science book. Note that the main thing I entered were about snails and slugs. You might wanna search in the Index. Or I'll just tell you. It's the species that snails and slugs belong in. Clams and spuids are related to them.

:winner_first:  cf249
:winner_second:  Dragon96
:winner_third: Aleph0


Prove that a maximal simple graph having no $r+1$-clique has an $r$-clique.


probability
function
inequalities
LaTeX
probability and stats

Hello, everyone :).

I have a problem I can't solve. Maybe you can help me... Here it is:

From interval $ [1,10]$ we choose a natural number $ a$ (all results equally probable) and then we choose a real number $ \zeta$ from interval $ [1,a]$ (uniform distribution for each $ a$). Give the distribution of the random variable $ \zeta$.

My problem is that I don't know what I should do with the case $ a \equal{} 1$ - for this $ a$ there is only one possible value of the variable $ \zeta$, while for every other $ a$ there are infinitely many of them.

I would be grateful for an explanation. Thanks! :)

Not all random variables are either discrete or (absolutely) continuous. Some are a mixture of the two, or something even more complicated.

To simplify this, let's look at the same thing on $ [1,2]$. Half of the time, we choose $ a \equal{} 1$ and get $ \zeta \equal{} 1$. The other half of the time, we choose $ a \equal{} 2$ and get $ \zeta$ uniform in $ [1,2]$.
The simplest description is probably just a combination of both sides of the terminology; density $ \frac12$ on $ [1,2]$, probability $ \frac12$ of being $ 1$. Alternately, we could look beyond functions and call the density $ \frac12\cdot \mathbf{1}_{[1,2]}(x) \plus{} \frac12\delta(x \minus{} 1)$; that's the indicator function for the set and the Dirac delta. Finally, we could just look at the cumulative distribution function: $ P(\zeta\le x) \equal{} \begin{cases}0 & x < 1 \\
\frac12x & 1\le x\le 2 \\
1 & x > 2\end{cases}$.

Then again, there's always the possibility that the writers didn't intend to ask this question, and including the point mass case was an oversight.

Thank you for your answer.

Your first suggestion is what I myself thought about, but as to the Dirac delta - I'd prefer not to involve it here.

Could you correct the fragment about cumulative distribution function?

One possibility for the question they might have meant to ask: Let $ a$ be a real number chosen uniformly on $ [1,10]$ and then ...

If we do it that way, we won't need to worry about $ a\equal{}1,$ which would be an event of probability zero.

Well, unfortunately I'm pretty sure they meant a [b]natural[/b] number  :( .

Besides, the problem with natural $ a$ seems to be quite interesting now!  :D

Anyway, thank you for the answers! (feel free to give more suggestions ;) )

I just corrected that fragment (reversed an inequality sign), although unfortunately the $ \text{\LaTeX}$ picture isn't loading for me.

You're missing a \ in \begin: $ P(\zeta\le x) \equal{} \begin{cases}0 & x < 1 \\
\frac12x & 1\le x\le 2 \\
1 & x > 2\end{cases}$
[color=green]Fixed now.[/color]


[quote="bodan"]Assume that $a,b,c$ are positive real numbers such that \[a+b+c=1.\] Prove the inequality \[a^{2}+b^{2}+c^{2}+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+8abc \geq 2(ab+bc+ca).\] [/quote]

$(a^{2}+b^{2}+c^{2})(a+b+c)^{2}+\sum a^{2}b^{2}+8\sum a^{2}bc \geq 2(ab+bc+ca)(a+b+c)^{2}$
$\iff \sum a^{4}\geq \sum a^{2}b^{2}$, which is trivial by AM-GM.


I honestly. HATE logic. If I study on my own, I will NEVER understand it. SO I'm going to post 4 If=>Then statements. I need a tutorial sort of thing and detailed explanations so even a nutshell like me will be able to do it on my own...
ALSO: I do not need an explanation on why some answers are wrong. YOU may be able to tell, however I cannot! :[ I just need to know how to answer these hell-ish questions! Thanks in advance guys~  :)

For all the questions- The question is, WHAT MUST BE TRUE?

If Charlie caught a bug, then the bug was green. If the bug was green, then the bug was not on the sidewalk.

F. If there is a green bug, then Charlie will catch it.
G. If Charlie caught a bug, then the bug was not on the sidewalk.
H. If the bug was on the sidewalk, Charlie caught it.
J. Charlie catches only bugs.
K. If the bug was not green, Charlie caught it on the sidewalk.

2.) Todd goes to the movie if Movie man is playing. Jane goes to the movies if Todd goes to the movies.

A. Todd does not go to the movies if Movie man is Not playing.
B. If Jane goes to the movies, Movie Man is playing.
C. If Jane does not go to the movies, then Movie Man is NOT playing.
D. If Todd goes to the movies, then Movie man is playing.
E. If Movie man is Not playing, then Jane does not go to the movies.

3.) If Han practices her violin, Tammy will play basketball. If Marianne goes out to dinner, then Tammy will play basketball.

A. If Han does not practice her violin, then Tammy will not play basketball.
B. When Marianne goes out to dinner, Han does not practice her violin.
C. When Han practices her violin, Tammy goes out to dinner.
D. If Tammy does not play basketball, then Han did not practice her violin and Marianne did not go out to dinner.
E. Tammy will not play basketball if Marianne eats dinner at home.

4.) If it is cold in the morning, I will watch cartoons. If it is not cold in the morning, I will walk the dog.

A. If I do not watch cartoons, then I will walk the dog.
B. If I walk the dog, then I will not watch cartoons.
C.If It is cold in the morning, I will not walk the dog.
D. If I do not walk the dog, then I do not watch cartoons.
E. If I watch cartoons, then it is cold in the morning.

-This is very crucial for me because I have a major test coming up on November 6th, and I will be posting more questions as the test nears. I MUST do well!! *Ahem and you guys are gonna help  :lol:

dang i love logics

1)

Let C = Charlie caught a bug
G = the bug was green
S=the bug was not on the sidewalk

then $ C\implies G$ and $ G\implies S$
by syllogism, $ C \implies S$, which equals to If Charlie caught a bug, the bug was not on the sidewalk.

Answer : G

2)

Let T=todd goes to the movie
M = Movie man is playing
J = Jane goes to the movies

then $ M \implies T$ and $ T\implies J$
syllogism; $ M\implies J$
which equals to If Movie man is playing, Jane goes to the movies.
note that if $ A\implies B$ is true, $ \sim B\implies \sim A$ is also true.
($ \sim A$ means not A)
so <If Jane doesn't go to the movies, movie man is not playing> is also true.

answer : C

3)

Let H=han practices her violin
T=Tammy will play basketball
M=marianne goes out to dinner
then $ H\implies T$ and $ M \implies T$
by note, $ \sim T \implies\sim H$ and $ \sim T \implies \sim M$, which is If Tammy does not play basketball, then Han did not practice her violin and Marianne did not go out to dinner, and thats D.

answer : D

4)
Let C=it is cold in the morning
W=i watch cartoons
D=i walk dogs
then $ C\implies W$ and $ \sim C\implies D$
by note, we can change $ \sim C\implies D$ to $ \sim D\implies C$
syllogism. $ \sim D\implies W$
which is If i dont walk dogs, i watch cartoons.
note again: it becomes 'if i watch cartoons, i do walk dogs.'

answer : A

TIP: the most important thing is syllogism and that note thing

I understand what you're getting at Fantasy, but can you pretend I know ABSOLUTELY nothing about this and try again? I've gotten the same replies and I still cant do them!  :( Gawd I feel stupid... x.x

Edit: I mean the books I do these from give me explanations with contraposative and all this complicated stuff. However, you do it without the need of that, and Im not used to it!  :(

Also, what do you mean by, "by note we can change this..."

blah sry im not very good at teaching so blahblahblah

[quote="Forte"]I understand what you're getting at Fantasy, but can you pretend I know ABSOLUTELY nothing about this and try again?[/quote]

i also memorized all this stuff.
like if there is a statement $ A\implies B$, $ \sim B\implies \sim A$ is its opposition, $ \sim A\implies \sim B$ is its converse, etc...

[quote="Forte"]Also, what do you mean by, "by note we can change this..."[/quote]

on #2, i said "note that if $ A\implies B$ is true, $ \sim B\implies\sim A$ is also true".
the rule of a statement is if a statement is true, the statement's opposition os also true.
what i meant by it is to use that fact.

AHA, I get it now, I've tried a few on my own and got them right! I'll post anything else if needed. Thanks A LOT!  :lol:

By "opposition", I believe you meant contrapositive. (I don't know why they chose that, of all things, either.)

Syllogism basically means that if A implies B, and B implies C, then you can assume that A implies C. Note that this only works in one direction. If A implies C, and B implies C, then you cannot assume that A implies B.

It's a very useful tool to have. You don't even have to know what it means to solve a problem using it.


trigonometry
LaTeX
induction
absolute value

$ \frac{\cos (15^\circ \minus{}x)}{\cos 15^\circ}\minus{}\frac{\sin (15^\circ \minus{}x)}{\sin 15^\circ}\equal{}4\sin 15^\circ$

$ \Longleftrightarrow \sin 15^\circ \cos (15^\circ \minus{}x)\minus{}\cos 15^\circ \sin (15^\circ \minus{}x)\equal{}4\sin 15^\circ \cos 15^\circ \sin 15^\circ$

$ \Longleftrightarrow \sin [15^\circ \minus{}(15^\circ \minus{}x)]\equal{}2(2\sin 15^\circ \sin 15^\circ)\sin 15^\circ$

$ \Longleftrightarrow \sin [15^\circ \minus{}(15^\circ \minus{}x)]\equal{}2\sin 30^\circ \sin 15^\circ$

$ \Longleftrightarrow \sin x\equal{}\sin 15^\circ$

$ \therefore x\equal{}15^\circ \plus{}360^\circ *n,\ 165^\circ \plus{}360^\circ *n\ (n\equal{}0,\ \pm 1,\ \pm 2,\ \cdots).$

hello, we have
$ \cos(15^{\circ}\minus{}x)\sec(15^{\circ})\equal{}\left(\frac{1\plus{}\sqrt{3}}{2\sqrt{2}}\cos(x)\plus{}\frac{\sqrt{3}\minus{}1}{2\sqrt{2}}\sin(x)\right)\sqrt{2}(\sqrt{3}\minus{}1)$
and
$ \sin(15^{\circ}\minus{}x)\csc(15^{\circ})\equal{}\left(\frac{\sqrt{3}\minus{}1}{2\sqrt{2}}\cos(x)\minus{}\frac{1\plus{}\sqrt{3}}{2\sqrt{2}}\sin(x)\right)\sqrt{2}(1\plus{}\sqrt{3})$
adding both terms we get $ 4\sin(x)$.
Sonnhard.

[quote="kunny"]That's right. After all, what's the problem?

Maybe Solve the equation with respect to $ x$, right?[/quote]
im sorry for this mistake
the question was reright
sorry

[quote="almonkez"]hi all
here i will post questions in trigonometry from simple  to complex
the first one :
 :play_ball:[/quote]

my solution  :maybe:

[quote="almonkez"] Prove that $ \left(1\plus{}\cos\frac{\pi}8\right)\left(1\plus{}\cos\frac{3\pi}{8}\right)\left(1\plus{}\cos\frac{5\pi}{8}\right)\left(1\plus{}\cos\frac{7\pi}{8}\right)\equal{}\frac 18$[/quote]

$ LHS\equal{}\left(1\plus{}\cos\frac{\pi}8\right)\left(1\plus{}\sin\frac{\pi}{8}\right)\left(1\minus{}\sin\frac{\pi}{8}\right)\left(1\plus{}\cos\frac{\pi}{8}\right)\equal{}$ $ \left(1\minus{}\cos^2\frac{\pi}{8}\right)\left(1\minus{}\sin^2\frac{\pi}{8}\right)$

$ \Longleftrightarrow LHS\equal{}\left(1\minus{}\frac{2\plus{}\sqrt 2}{4}\right)\left(1\minus{}\frac{2\minus{}\sqrt 2}{4}\right)\equal{}\frac 18$

[quote="Mateescu Constantin"][quote="almonkez"] Prove that $ \left(1 \plus{} \cos\frac {\pi}8\right)\left(1 \plus{} \cos\frac {3\pi}{8}\right)\left(1 \plus{} \cos\frac {5\pi}{8}\right)\left(1 \plus{} \cos\frac {7\pi}{8}\right) \equal{} \frac 18$[/quote]

$ LHS \equal{} \left(1 \plus{} \cos\frac {\pi}8\right)\left(1 \plus{} \sin\frac {\pi}{8}\right)\left(1 \minus{} \sin\frac {\pi}{8}\right)\left(1 \plus{} \cos\frac {\pi}{8}\right) \equal{}$ $ \left(1 \minus{} \cos^2\frac {\pi}{8}\right)\left(1 \minus{} \sin^2\frac {\pi}{8}\right)$

$ \Longleftrightarrow LHS \equal{} \left(1 \minus{} \frac {2 \plus{} \sqrt 2}{4}\right)\left(1 \minus{} \frac {2 \minus{} \sqrt 2}{4}\right) \equal{} \frac 18$[/quote]
thanks :Mateescu Constantin
in the last step we can solve without combuting sin p/8 or cos p/8  :coolspeak:
almonkez

[quote="Mateescu Constantin"][quote="almonkez"] Prove that $ \left(1 \plus{} \cos\frac {\pi}8\right)\left(1 \plus{} \cos\frac {3\pi}{8}\right)\left(1 \plus{} \cos\frac {5\pi}{8}\right)\left(1 \plus{} \cos\frac {7\pi}{8}\right) \equal{} \frac 18$[/quote]

$ LHS \equal{} \left(1 \plus{} \cos\frac {\pi}8\right)\left(1 \plus{} \sin\frac {\pi}{8}\right)\left(1 \minus{} \sin\frac {\pi}{8}\right)\left(1 \plus{} \cos\frac {\pi}{8}\right) \equal{}$ $ \left(1 \minus{} \cos^2\frac {\pi}{8}\right)\left(1 \minus{} \sin^2\frac {\pi}{8}\right)$[/quote]

$ \equal{} \sin ^ 2 \frac {\pi}{8}\cos ^ 2 \frac {\pi}{8}\equal{} \left(\frac {1}{2}\sin 2 \cdot \frac {\pi}{8}\right)^2$

$ \equal{} \frac {1}{4}\cdot \left(\frac {1}{\sqrt {2}}\right)^2\equal{} \frac {1}{8}$

[quote="kunny"][quote="Mateescu Constantin"][quote="almonkez"] Prove that $ \left(1 \plus{} \cos\frac {\pi}8\right)\left(1 \plus{} \cos\frac {3\pi}{8}\right)\left(1 \plus{} \cos\frac {5\pi}{8}\right)\left(1 \plus{} \cos\frac {7\pi}{8}\right) \equal{} \frac 18$[/quote]

$ LHS \equal{} \left(1 \plus{} \cos\frac {\pi}8\right)\left(1 \plus{} \sin\frac {\pi}{8}\right)\left(1 \minus{} \sin\frac {\pi}{8}\right)\left(1 \plus{} \cos\frac {\pi}{8}\right) \equal{}$ $ \left(1 \minus{} \cos^2\frac {\pi}{8}\right)\left(1 \minus{} \sin^2\frac {\pi}{8}\right)$[/quote]

$ \equal{} \sin ^ 2 \frac {\pi}{8}\cos ^ 2 \frac {\pi}{8} \equal{} \left(\frac {1}{2}\sin 2 \cdot \frac {\pi}{8}\right)^2$

$ \equal{} \frac {1}{4}\cdot \left(\frac {1}{\sqrt {2}}\right)^2 \equal{} \frac {1}{8}$[/quote]
thanks alot " kunny "
this is what i mean  :thumbup:

[quote="almonkez"]$ \triangle ABC\implies \boxed{\cos\frac A2\cos\frac{B\minus{}C}{2}\plus{}\cos\frac B2\cos\frac{C\minus{}A}{2}\plus{}\cos\frac C2\cos\frac{A\minus{}B}{2}\equal{}\sin A\plus{}\sin B\plus{}\sin C}$[/quote]

[b][u]Proof.[/u][/b]

$ \cos\frac A2\cos\frac{B\minus{}C}{2}\equal{}\cos\frac A2\left(\cos\frac B2\cos\frac C2\plus{}\sin\frac B2\sin\frac C2\right)\equal{}$ $ \frac{p}{4R}\plus{}\cos\frac A2\sin\frac B2\sin\frac C2$

$ \Longleftrightarrow \cos\frac A2\cos\frac{B\minus{}C}{2}\equal{}\frac p{4R}\plus{}\sqrt{\frac{p(p\minus{}a)}{bc}\cdot\frac{(p\minus{}a)(p\minus{}c)}{ac}\cdot\frac{(p\minus{}a)(p\minus{}b)}{ab}}\equal{}$ $ \frac p{4R}\plus{}\frac{S(p\minus{}a)}{abc}$

$ \implies LHS\equal{}\frac{3p}{4R}\plus{}\frac{S[3p\minus{}(a\plus{}b\plus{}c)]}{abc}\equal{}\frac{3p}{4R}\plus{}\frac{Sp}{4RS}\equal{}\frac{3p}{4R}\plus{}\frac{p}{4R}$ $ \equal{}\frac pR\equal{}\sin A\plus{}\sin B\plus{}\sin C$ .

[quote="Mateescu Constantin"][quote="almonkez"]$ \triangle ABC\implies \boxed{\cos\frac A2\cos\frac {B \minus{} C}{2} \plus{} \cos\frac B2\cos\frac {C \minus{} A}{2} \plus{} \cos\frac C2\cos\frac {A \minus{} B}{2} \equal{} \sin A \plus{} \sin B \plus{} \sin C}$[/quote]

[b][u]Proof.[/u][/b]

$ \cos\frac A2\cos\frac {B \minus{} C}{2} \equal{} \cos\frac A2\left(\cos\frac B2\cos\frac C2 \plus{} \sin\frac B2\sin\frac C2\right) \equal{}$ $ \frac {p}{4R} \plus{} \cos\frac A2\sin\frac B2\sin\frac C2$

$ \Longleftrightarrow \cos\frac A2\cos\frac {B \minus{} C}{2} \equal{} \frac p{4R} \plus{} \sqrt {\frac {p(p \minus{} a)}{bc}\cdot\frac {(p \minus{} a)(p \minus{} c)}{ac}\cdot\frac {(p \minus{} a)(p \minus{} b)}{ab}} \equal{}$ $ \frac p{4R} \plus{} \frac {S(p \minus{} a)}{abc}$

$ \implies LHS \equal{} \frac {3p}{4R} \plus{} \frac {S[3p \minus{} (a \plus{} b \plus{} c)]}{abc} \equal{} \frac {3p}{4R} \plus{} \frac {Sp}{4RS} \equal{} \frac {3p}{4R} \plus{} \frac {p}{4R}$ $ \equal{} \frac pR \equal{} \sin A \plus{} \sin B \plus{} \sin C$ .[/quote]
verry good "  Mateescu Constantin " thanks fot quick solution
ther is another simple solution  :read:

almonkez

[quote="almonkez"]hi all
the next :  :thumbup:
[color=blue][b](7)[/b][/color]
[url=http://www.servimg.com/image_preview.php?i=326&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/1110.jpg[/img][/url]
with my best wishes[/quote]
is it hard? :maybe:

[quote="Mateescu Constantin"][quote="almonkez"]$ \triangle ABC\implies \boxed{\cos\frac A2\cos\frac {B - C}{2} + \cos\frac B2\cos\frac {C - A}{2} + \cos\frac C2\cos\frac {A - B}{2} = \sin A + \sin B + \sin C}$[/quote]

[b][u]Proof.[/u][/b]

$ \cos\frac A2\cos\frac {B - C}{2} = \cos\frac A2\left(\cos\frac B2\cos\frac C2 + \sin\frac B2\sin\frac C2\right) =$ $ \frac {p}{4R} + \cos\frac A2\sin\frac B2\sin\frac C2$

$ \Longleftrightarrow \cos\frac A2\cos\frac {B - C}{2} = \frac p{4R} + \sqrt {\frac {p(p - a)}{bc}\cdot\frac {(p - a)(p - c)}{ac}\cdot\frac {(p - a)(p - b)}{ab}} =$ $ \frac p{4R} + \frac {S(p - a)}{abc}$

$ \implies LHS = \frac {3p}{4R} + \frac {S[3p - (a + b + c)]}{abc} = \frac {3p}{4R} + \frac {Sp}{4RS} = \frac {3p}{4R} + \frac {p}{4R}$ $ = \frac pR = \sin A + \sin B + \sin C$ .[/quote]

This can be done in another simpler way:  Use
$ 2\cos\frac{A}{2}\cos\frac{B-C}{2}=2\sin\frac{B+C}{2}\cos\frac{B-C}{2}=\sin B+\sin C$

This gives
${ 2\cos\frac{A}{2}\cos\frac{B-C}{2}+2\cos\frac{B}{2}\cos\frac{C-A}{2}+2\cos\frac{C}{2}\cos\frac{A-B}{2}=2\sin A+2\sin B+2\sin C}$
Hence the proof.

[quote][quote="vidyamanohar]

This can be done in another simpler way:  Use
$ 2\cos\frac {A}{2}\cos\frac {B - C}{2} = 2\sin\frac {B + C}{2}\cos\frac {B - C}{2} = \sin B + \sin C$

This gives
${ 2\cos\frac {A}{2}\cos\frac {B - C}{2} + 2\cos\frac {B}{2}\cos\frac {C - A}{2} + 2\cos\frac {C}{2}\cos\frac {A - B}{2} = 2\sin A + 2\sin B + 2\sin C}$
Hence the proof.[/quote][/quote][/quote]
this is my solution my dear look at replay 46
thanks for help
mr mohamed

[quote="Machodog"][hide]$ \cot{30^\circ}\cot{40^\circ} = 1 + \sec{20^\circ}$

\begin{eqnarray*} RHS & = & (1 + \sec{20^\circ})(\frac {\sin{40^\circ}}{\sin{40^\circ}}) \\
& = & \frac {1}{\sin{40^\circ}} (\sin{40^\circ} + \frac {2\sin{20^\circ}\cos{20^\circ}}{\cos{20^\circ}}) \\
& = & \frac {1}{\sin{40^\circ}} (\sin{40^\circ} + 2\sin{20^\circ}) \\
\end{eqnarray*}
Note that:
\begin{eqnarray*} 2\sin{20^\circ} & = & 2\cos{70^\circ} \\
& = & 2\cos{(40^\circ + 30^\circ}) \\
& = & \sqrt {3}\cos{40^\circ} - \sin{40^\circ} \\
\end{eqnarray*}

Substituting back in we get:
$ RHS = \frac {\sqrt {3}\cos{40^\circ}}{\sin{40^\circ}} = \cot{30^\circ}\cot{40^\circ}$[/hide][/quote]
 :10:    :clap2:
thanks
still there another solution i will wait
mr mohamed
 :wink:

[quote="almonkez"]hi all
the next :  :thumbup:
[color=blue][b](7)[/b][/color]
[url=http://www.servimg.com/image_preview.php?i=326&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/1110.jpg[/img][/url]
with my best wishes[/quote]

hi here are my solution  :wink:
[url=http://www.servimg.com/image_preview.php?i=342&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk14.gif[/img][/url]
with my best wishes
mr mohamed

Congratulations Mr. Mohamed. Nice methods ;)

[hide]Using the sum to product formula $ \cos{A} \plus{} \cos{B} \equal{} 2\cos\left(\frac {A \plus{} B}{2}\right)\cos\left(\frac {A \minus{} B}{2}\right)$ we have:

$ LHS \equal{} 2\cos{36^\circ}(\cos{12^\circ} \plus{} \cos{132^\circ})$
$ \equal{} 2\cos{36^\circ}\cdot 2\cos{72^\circ}\cos{60^\circ}$
$ \equal{} 2\cos{72^\circ}\cos{36^\circ}$
$ \equal{} \frac {2\sin{18^\circ}\cos{18^\circ}\cos{36^\circ}}{\cos{18^\circ}}$
$ \equal{} \frac {\sin{36^\circ}\cos{36^\circ}}{\cos{18^\circ}}$
$ \equal{} \frac {\sin{72^\circ}}{2\cos{18^\circ}} \equal{} \frac {1}{2}$[/hide]

[b]New Problem:[/b]
Prove that in triangle ABC,

$ \frac{a\minus{}b}{a\plus{}b}\equal{}\tan{\frac{A\minus{}B}{2}}\tan{\frac{C}{2}}$

[quote="Machodog"][hide]Using the sum to product formula $ \cos{A} \plus{} \cos{B} \equal{} 2\cos\left(\frac {A \plus{} B}{2}\right)\cos\left(\frac {A \minus{} B}{2}\right)$ we have:

$ LHS \equal{} 2\cos{36^\circ}(\cos{12^\circ} \plus{} \cos{132^\circ})$
$ \equal{} 2\cos{36^\circ}\cdot 2\cos{72^\circ}\cos{60^\circ}$
$ \equal{} 2\cos{72^\circ}\cos{36^\circ}$
$ \equal{} \frac {2\sin{18^\circ}\cos{18^\circ}\cos{36^\circ}}{\cos{18^\circ}}$
$ \equal{} \frac {\sin{36^\circ}\cos{36^\circ}}{\cos{18^\circ}}$
$ \equal{} \frac {\sin{72^\circ}}{2\cos{18^\circ}} \equal{} \frac {1}{2}$[/hide]

[b]New Problem:[/b]
Prove that in triangle ABC,

$ \frac {a \minus{} b}{a \plus{} b} \equal{} \tan{\frac {A \minus{} B}{2}}\tan{\frac {C}{2}}$[/quote]
here is the  solution " Machodog "
[url=http://www.servimg.com/image_preview.php?i=409&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk16.gif[/img][/url]
 :wink:

[quote="mrmohamed"][size=150][color=blue][b]our new today[/b][/color][/size]
[url=http://www.servimg.com/image_preview.php?i=406&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/811.jpg[/img][/url]
with my best wishes
mr mohamed :wink:[/quote]
tomorrow i will post the solution if any one didnt post
mrmohamed : almonkez

[quote="mrmohamed"] Prove that: $ \cos 24^{\circ} \plus{} \cos 48^{\circ} \plus{} \cos 96^{\circ} \plus{} \cos 168^{\circ} \equal{} \frac 12$ . [/quote]

[b][u]Proof.[/u][/b]

I'll use the well-known relation $ \ \boxed{\cos a \plus{} \cos b \equal{} 2\cos\frac {a \plus{} b}{2}\cdot\cos\frac {a \minus{} b}{2},\ \forall\ a,b\in\mathbb{R}}$ .

$ \implies \left\| \begin{array}{cc} \cos 24^{\circ} \plus{} \cos 96^{\circ} \equal{} 2\cos 60^{\circ}\cdot \cos 36^{\circ} \\
 \\
\cos 48^{\circ} \plus{} \cos 168^{\circ} \equal{} 2\cos 108^{\circ}\cdot \cos 60^{\circ} \end{array}\right|\ \bigoplus\ \implies$

$ \implies LHS \equal{} 2\cos 60^{\circ}(\cos 36^{\circ} \plus{} \cos 108^{\circ}) \equal{} \cos 36^{\circ} \plus{} \cos 108^{\circ}$ $ \equal{} \frac {\sqrt 5 \plus{} 1}{4} \plus{} \frac { \minus{} \sqrt {5} \plus{} 1}{4} \equal{} \frac 12$ .

Hint:
(a^2 - b^2 - c^2) tan A
= - (b^2 + c^2 – a^2) tan A
= - (2 bc cos A) tan A
= - 2 bc sin A
= - 4 S
= - 2 ca sin B
= - (2 ca cos B) tan B
= - (c^2 + a^2 – b^2) tan B


There must be a mistake.  What did you really mean?  Perhaps a square in the LHS?

No I checked the problem again it's true

Fix $x,y,z$ meeting the conditions and consider what happens when we put $tx,ty,tz$.  Then we get

$8(x^3+y^3+z^3)\geq t^3\cdot(9(x^2+yz)(y^2+zx)(z^2+xy))$ for all $t$, which is absurd...

sorry you are tight ,I edited my post
now how to solve this problem

And it turns out that it's been solved here before.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%28x%5E2%2Byz%29%28y%5E2%2Bzx%29%28z%5E2%2Bxy%29&t=6507


In a group of ten persons , each person is asked to write the sum of the ages of all the other nine persons. If all the ten sums form the 9 element set {82,83,84,85,87,89,90,91,92}, find the individual ages of the persons, assuming them to be whole numbers(of years).

[hide]Well, if we add the sums of the 10 elements in the set including the repeated element $x_i$, we get $9x = 873 + x_i$, where $x$ is the sum of all the ages $x_1, x_2 \ldots x_{10}$.

The only $x_i$ which makes that RHS $0 \pmod{9}$ is $90$. So $9x = 973 \Rightarrow x = 97$. And subtraction yields the individual ages.[/hide]

[quote="scorpius119"][hide]Since the ten sums form a 9 element set, one sum is repeated. Let $a$ be the repeated sum of nine ages. If $S$ is the sum of all of the ages, then adding the 10 sums of nine ages gives $9S=82+83+84+85+87+89+90+91+92+a$. Since $S$ is an integer, so must $a$.

$82+83+84+85+87+89+90+91+92+a\equiv 1+2+3+4+6+8+0+1+2+a\equiv a\equiv 0\pmod{9}\Rightarrow a=90$

$9S=82+83+84+85+87+89+2*90+91+92=973\Rightarrow S=97$.

Subtracting the equations for the sums of nine ages from $S$, the ages are 5, 6, 7, 7, 8, 10, 12, 13, 14, and 15.[/hide][/quote]

I couldn't understand the problem, that is why I was getting number of solutions.
Here according to question 10 sums form 9 element set, which I thought meant that only 9 sums are taken.

So, thanks any way. Your first line of solution itself sufficed me. :lol:

I still don't understand the question.

How can $10$ sums form a $9$ element set?

Well you should know that {1,2,3,3} is equivalent to {1,2,3}. Thus 4 elements have formed a 3-element set.
So now do you understand.
This was the same problem with me :P