geometry
number theory proposed
number theory

Find, with proof, all triples of integers $ (a,b,c)$ such that $ a, b$ and $ c$ are the lengths of the sides of a right angled triangle whose area is $ a \plus{} b \plus{} c$

W.L.O.G. $ a\leq b < c$, by Pythagorean's theorem $ a^2 \plus{} b^2 \equal{} c^2$, we can set $ a \equal{} m^2 \minus{} n^2, \ b \equal{} 2mn,\ c \equal{} m^2 \plus{} n^2$ for positive integers $ m,\ n\ (m > n)$.
Then the area : $ \frac {1}{2}ab \equal{} a \plus{} b \plus{} c\Longleftrightarrow \frac {1}{2}(m^2 \minus{} n^2)\cdot 2mn \equal{} (m^2 \minus{} n^2) \plus{} 2mn \plus{} (m^2 \plus{} n^2)$

$ \Longleftrightarrow m(m \plus{} n)(m \minus{} n) \cdot n \equal{} 2m(m \plus{} n) > 0$

$ \therefore (m \minus{} n)n \equal{} 2\Longleftrightarrow (m \minus{} n,\ n) \equal{} (1,\ 2),\ (2,\ 1)$, yielding $ (m,\ n) \equal{} (3,\ 2), (3,\ 1)$.

Ans. $ (a,\ b,\ c) \equal{} (5,\ 12, \ 13),\ (6,\ 8,\ 10),\ (8,\ 6,\ 10),\ (12,\ 5,\ 13)$.

$ K \equal{} rs \equal{} r \cdot \frac{a \plus{} b \plus{} c}{2} \equal{} a \plus{} b \plus{} c \Leftrightarrow r \equal{} 2$.  Kunny is using the identity $ r \equal{} n(m\minus{}n)$ (a rather particular identity for right triangles, but one that I find useful for problems of this type anyway).  Using this technique you can easily calculate the Pythagorean triangles of any given inradius.

That's right. Especially for a right angled triangle with side lengths $ a,\ b,\ c\ (a\leq b < c)\ \boxed{r \equal{} \frac {a \plus{} b \minus{} c}{2}}$ holds.


floor function
trigonometry
linear algebra
matrix
algebra
polynomial
calculus

Prove

$ \prod_{k\equal{}1}^{\lfloor \frac{n\minus{}1}{2}\rfloor} \tan\left(\frac{k\pi}{n}\right)\equal{}\left\{ \begin{matrix} \sqrt{n} & \text{for} & n \;\; \text{odd} \\ \\ 1 & \text{for} & n\;\; \text{even} \end{matrix} \right.$

consider the polynomial $ p(x) \equal{} \frac {x^n \minus{} 1}{x \minus{} 1} \equal{} \prod_{k \equal{} 1;n \minus{} 1}(x \minus{} e^{i2k\pi/n})$ ,and evaluate $ p(1)$ and $ p(\minus{}1)$

another way:
$ \sin n\alpha \plus{}i\cos n\alpha \equal{} (\sin \alpha \plus{}\cos \alpha)^{n}\equal{} \cos^{n}\alpha(\tan \alpha \plus{}i)^{n}\equal{}\cos^{n}\alpha (\sum_{k\equal{}0}^{n}C_{n}^{k}\tan^{n\minus{}k}i^{k})$
hence you have :
$ \sin n\alpha\equal{}\cos ^{n}\alpha P_{m}(\tan \alpha)$ where you already know all coeficient of polynom $ P_{m}(x)$
hence  $ \alpha \equal{} \frac{k\pi}{n}$ will be root
or for polynom you have :
$ x_{k}\equal{}\tan \frac{k\pi}{n}$ and final use Viete theorem  you are done


function
floor function
inequalities
algebra unsolved
algebra

Consider the function $f$ defined by

$f(n)=\frac{1}{n}{\sum^{n}_{i=1}}{\lfloor \frac{n}{i}\rfloor}$

for all positive integers $n$. Prove that:
(a)  $f(n+1)>f(n)$ for infinitely many $n$
(b)  $f(n+1)<f(n)$ for infinitely many $n$

For each positive integer $n$, denote $S_{n}= \sum_{i=1}^{n}\left[\frac{n}{i}\right]$, then $f(n)=\frac{1}{n}S_{n}$.

Assume that $f(k) > f(k+1), i.e., S_{k+1}<(1+1/k)S_{k}$ for only finite many positive integer k, then there is an positive integer $n$ such that for all positive integer $m$, we have \[S_{n+m+1}\geq \left(1+\frac{1}{n+m}\right)S_{n+m}\geq\cdots\geq \left(1+\frac{1}{n+m}\right)\cdots \left(1+\frac{1}{n}\right) S_{n}.\] Let $a=1+\frac{1}{n}>1$, then $S_{n+m+1}\geq a^{m+1}S_{n}\geq a^{m+1}\quad (*)$. However for each positive integer k, $S_{k}\geq k+(k-1)+\cdots+1=k(k+1)/2$, it implies that $S_{n+m+1}/a^{m+1}$ reaches to $0$ when $m$ tends to $\infty$. A contradiction with (*).

Assume that $f(k) < f(k+1)$ for only finite many positive integer k. then there is an positive integer $n$ such that $f(m)\leq f(n)$ for all $m\geq n$. Let $b=f(n)$, then $S_{m}\leq bm\quad(**)$.  Put $j=[m/2]$, we have \[S_{m}= \left(\left[\frac{m}{1}\right]+\cdots+\left[\frac{m}{j}\right]\right)+\left(\left[\frac{m}{j+1}\right]+\cdots+\left[\frac{m}{m}\right]\right)\\ \geq m\left(\frac{1}{1}+\cdots+\frac{1}{j}+\right)-j+(m-j) \geq m\left(\frac{1}{1}+\cdots+\frac{1}{j}\right).\] For each $x\geq 0, ln(1+x)\geq x$, then the above inequality implies
$S_{m}/m \geq ln(\frac{2}{1})+ln(\frac{3}{2})+\cdots ln(\frac{j+1}{j})=ln(j+1) =ln([m/2]+1)$. Hence $S_{m}/m$ reaches to $\infty$ if $m$ tends to $\infty$. A contradiction with (**).

For each positive integer $n$, denote $S_{n}= \sum_{i=1}^{n}\left[\frac{n}{i}\right]$, then $f(n)=\frac{1}{n}S_{n}$.

Assume that $f(k) < f(k+1)$ for only finite many positive integer k. then there is an positive integer $n$ such that $f(m)\leq f(n)$ for all $m\geq n$. Let $b=f(n)$, then $S_{m}\leq bm\quad(*)$.  Put $j=[m/2]$, we have \[S_{m}= \left(\left[\frac{m}{1}\right]+\cdots+\left[\frac{m}{j}\right]\right)+\left(\left[\frac{m}{j+1}\right]+\cdots+\left[\frac{m}{m}\right]\right)\\ \geq m\left(\frac{1}{1}+\cdots+\frac{1}{j}+\right)-j+(m-j) \geq m\left(\frac{1}{1}+\cdots+\frac{1}{j}\right)\quad (**).\] For each $x\geq 0, ln(1+x)\leq x$, then the above inequality implies
$S_{m}/m \geq ln(\frac{2}{1})+ln(\frac{3}{2})+\cdots ln(\frac{j+1}{j})=ln(j+1) =ln([m/2]+1)$. Hence $S_{m}/m$ reaches to $\infty$ if $m$ tends to $\infty$. A contradiction with (*).

Now we prove that there are infinitely many $k$ such that $f(k) > f(k+1)$. For each prime number $p> 8$, from (**), we get $S_{p}>2p$. Note that $\left[\frac{p-1}{i}\right]=\left[\frac{p}{i}\right]$ for all $i=2,\cdots,p-1$. It implies that $S_{p}=S_{p-1}+2$. Hence \[f(p)-f(p-1)=\frac{S_{p}}{p}-\frac{S_{p-1}}{p-1}=\frac{S_{p}}{p}-\frac{S_{p}-2}{p-1}=-\frac{S_{p}-2p}{p(p-1)}<0.\] Because there are infinite primes, thus the assertion holds true. QED


Let $ z$ and $ z\plus{}1$ be $ n$th roots of unity, where $ n$ is a positive integral multiple of 7. Compute the smallest possible numerical value for $ n\minus{}z^3.$

[quote="pi^2/6"]Let $ z$ and $ z \plus{} 1$ be $ n$th roots of unity, where $ n$ is a positive integral multiple of 7. Compute the smallest possible numerical value for $ n \minus{} z^3.$[/quote]
[hide="Is this right?"]
If z and z+1 are roots of unity, they must be either cis60 and cis120, or cis -60 and cis-120. For n to be an integral multiple of 7, while have roots at the 60 and 120 degree positions, n must be >=42. 42 is the minimum, so we take that to be n.
z^3= either cis(120x3) or cis(-120x3), both of which are equal to 1, so z^3=1.
n-z^3=41[/hide]

[quote="serialk11r"][hide="Is this right?"]
If z and z+1 are roots of unity, they must be either cis60 and cis120, or cis -60 and cis-120. For n to be an integral multiple of 7, while have roots at the 60 and 120 degree positions, n must be >=42. 42 is the minimum, so we take that to be n.
z^3= either cis(120x3) or cis(-120x3), both of which are equal to 1, so z^3=1.
n-z^3=41[/hide][/quote]
[hide="Almost"]z=cis 60 or cis-60, so z^3=-1, and the total answer is 43.[/hide]

What? How is z cis60? If z=cis60=1/2+root3/2, then z+1=3/2+root3/2 which is not a root of unity...


function
limit
real analysis
topology
absolute value
advanced fields
advanced fields unsolved

Let $f_{n} : \mathbb{R} \to  \mathbb{R}$ be a sequence of continous functions that converge at every point. Prove that there exist an interval $I \subset \mathbb{R}$  and a number $M$ such that $|f_{n}(x)| \leq M$  for every $x \in I$ and every $n$

Any suggestion?
Thanks.

Do you know Baire's theorem ? If so, it is an easy consequence of that theorem.

[quote="niski"]Hello! Well, I don't know is it required to know Baire theorem (since I when I saw this question I did not know in which context it is included). Anyway, I know that Baire theorem states that if a metric space is complete it is nonmeager in intself...how you would work out this solution using this theorem?
Thanks![/quote]

Note, that since $\lim_{n \to \infty}f_n(x)$ exists for every real $x$, for each fixed $x$, the sequence $\{f_n(x) \}_n$ - Cauchy and,thus, for every $x$ and every $n$ there exists $p$ such that for all $q \geq p$ we have:
$|f_q(x) - f_p(x)| < \frac{1}{n}$, which means that $x \in S_{n,p} = \bigcap \{ y, |f_q(y) - f_p(y)| < \frac{1}{n} \forall q \geq p \}$ , so for all $n$ $\bigcup_{p} S_{n,p} = \mathbb{R}$ and each of $S_{n,p}$ is closed. Now, you may apply Baire Category Theorem and the left is for you ...

[quote="alekk"]$F_M=\{x : |f_n(x)| \leq M, \forall n > 0\}$ is closed. $\bigcup_{M \in \mathbb{N}} F_M = \mathbb{R}$ so ..[/quote]
Oh, stupid me, we can do it easier.
Niski, follow Alekk - his idea is much more appropriate in this problem.

[quote="eugene"][quote="alekk"]$F_M=\{x : |f_n(x)| \leq M, \forall n > 0\}$ is closed. $\bigcup_{M \in \mathbb{N}} F_M = \mathbb{R}$ so ..[/quote]
Oh, stupid me, we can do it easier.
Nilski, follow Alekk - his idea is much more appropriate in this problem.[/quote]

I saw the Baire theorem statment in Rudin Book and it said that
"If $X$ is a complete metric space, the intersection of every countable collection of dense open subsets of $X$ is dense in $X$"

So if we are working in $\mathbb{R}$, the theorem means that if we pick any countable collection of dense open subsets of $\mathbb{R}$ and consider its intersection, then  this intersection will be again dense (and open) in $\mathbb{R}$.

So in this case, why you guys are considering $\mathbb{R}$ as union of closed subsets?

Ok. Using the other statement of the theorem, then I know that at least one
$F_{M}$ contains a non empty open subset. This $F_{M}$ is the candidate for $I$ ? why  $F_{M}$ must be an interval?
Thanks

in open sets you can always find intervals !

Rigth! So the only thing left to prove is that $F_{M}$ is closed. Let's try
Fix $M$ and let $(x_{m})$ be a convergent sequence of elements of $F_{M}$  with $\bar{x}$ as limit. We have to prove that $\bar{x} \in F_{M}$.
Now we have that $|f_{n}(x_{m})| \leq M, \forall n > 0$ and $\forall m > 0$ so, because the absolute value is continous and $f_{n}$ is contiunous for every $n > 0$ then
$|f_{n}(\bar{x})| = \lim_{m \to  \infty} |f_{n}(x_{m})| \leq M$..
but I think something must be wrong since I did not used the hypothesis that $f_{n}$ is a sequence of continuous functions [b]that converges at every point[/b]...
any suggestions?

Thanks again!

I think it is correct.
Or you write $F_M = \bigcap_n f_n^{-1}([-M ; M])$ and because of the continuity of $f_n$, $f_n^{-1}([-M ; M])$ is closed.


When I have

Problem Set: \printproblemset  Problem Number: \theqnumber  Page: \thepage

How do I get spaces between "\printproblemset" and "Problem"?  Because right now, I'm getting something like (if problemset = 1)

Problem Set: 1Problem [more stuff]

Try \ followed by a space so you get
Problem Set: \printproblemset \ Problem Number: \theqnumber Page: \thepage


Let $ABC$ be an acute triangle and let $D,\,E,$ and $F$ be the feet of the altitudes from $A,\,B,$ and $C$ to sides $BC,\,CA,$ and $AB$ respectively. Let $P,\,Q,$ and $R$ be the feet of the  perpendiculars from $A,\,B,$ and $C$ to $EF,\, FD,$ and $DE$ respectively. Prove that
\[2(PQ+QR+RP) \ge DE+EF+FD\]

Hint: Triangle inequality w/ Ptolemy's inequality.

[quote="April"]Let $ABC$ be an acute triangle and let $D,\,E,$ and $F$ be the feet of the altitudes from $A,\,B,$ and $C$ to sides $BC,\,CA,$ and $AB$ respectively. Let $P,\,Q,$ and $R$ be the feet of the  perpendiculars from $A,\,B,$ and $C$ to $EF,\, FD,$ and $DE$ respectively. Prove that
\[2(PQ+QR+RP) \ge DE+EF+FD \]
[/quote]
I wonder if there exists a triangle $\triangle XYZ$ such that $X,Y,Z$ lie on $BC,CA,AB$ respectively,and $\triangle XYZ \sim \triangle PQR$ s.t $\frac{XY}{PQ}=2$... :maybe:
then the problem will follow immediatly...


i will need 9 people to join
there will be 2 teams
the war begins when one person from one team insults another person from another team (dont take these insults seriously, its just a game). Then the other team has a choice: either the insulted person can insult back with a counter-insult, or a teammate of the insulted person can compliment the insulted person to prove the insult wrong. Once a team has no good move to make (cannot form an insult or compliment), the person currently insulted is "dead". Then the other team starts off the next round. when a team loses all its members, it loses the war. Signups are first come first serve, need nine people!
Edit: when you sign up, you must specify if you are on my team or the other team. 4 members can join my team, while 5 members can join the team opposing me. my team gets to strike first, but dont worry, this war will be done twice where the other team strikes first. if each team wins one war, the final result is a tie. if one team wins both wars, that team is the final winner.

yeah why not ill join the team without mathelete...

Yeah, I'll join too... As an insult, I won't join mathelete's team either.  :)

okay:

team 1:
mathelete
team 2:
chompy
latex
spaceguy524

still need 4 members on team 1 and 2 members on team 2

okay:
Team 1:
mathelete
gogators
Team 2:
chompy
latex
spaceguy524

mathelete, gogators

chompy, latex, spaceguy524, westiepaw

just need one more on team 2 and that team will be complete.

Join the available team. (Mathelete's team)

[quote="abcak"]Join Mathlete's team[/quote]

I think you're a bit too late... The teams are

mathelete, gogators, EggyLv.999, meewhee009, and KING 1

hjoon0125, chompy, latex, spaceguy524, westiepaw

Dang, i checked the Games Forum too late :wink:

My sis could've help me with this :lol:

Okay:
mathelete, gogators, EggyLv.999, mwehee009, KING 1

hjoon0125, chompy, latex, spaceguy524, westiepaw

war will start this friday, cause i am too busy. i will let mwehee009 (chosen by random.org) fire the first insult.
remember: it is just a game, dont take the insults seriously. also, dont use profanity in the "insults".

I SO wish i could join.

Could you just please, please, PLEASE have 5 people on each team?

[quote="polkadotman1"]
Could you just please, please, PLEASE have 5 people on each team?[/quote]

There are 5 people on each team.

Go try to PM mathlete. Maybe you can convince him. :)

Oh YES, I fire first insult. :evilgrin:
Noting from the rules;
whoever I insult... do not, I repeat DO NOT take it seriously, I'm nice but when told to, I can be mean. So yes... early sorry!

@spaceguy524
[quote="mathelete"]Okay:
[b]war will start this friday[/b], cause i am too busy. i will let mwehee009 (chosen by random.org) fire the first insult.
remember: it is just a game, dont take the insults seriously. also, dont use profanity in the "insults".[/quote]

i'm not in this game but am anxious to see how fail you bunch of losers are at coming up with insults(don't take this seriously either)

Can we start?
Okay, I will.

Heyyy... Latex...I need you...........I want you............To get out of my face.

(IDMI. (guess what that is))

*sniff*, *sniff*

u left me out!!!!!!!!!!!!!!!!

mathelete, you don't deserve to be an AoPSer.

(again, don't take insults seriously)

[quote="mathelete"][quote="spaceguy524"]@LaTeX: You are very smart. :) :) :)[/quote]
really? well, you cannot tell if somebody is smart unless you are also smart, which you are not!!![/quote]

Spaceguy not smart? Are you kidding!?  :o  Spaceguy made Mathcounts, which is more than you could do!

meewhee is so forgetful, she... um... what was I talking about?

xD

I hope meewhee forgets.

then i can joinz :evilgrin:  :evilgrin:  :evilgrin:

oh crap sry i thought this game didn't start yet. Silly me.

Since everyone forgot about this, then am i automatically in?

This is slow....and yes most are forgetting.
polkadot can take over for me :D

that is, if he is able to do so!!! can you now???
i did not think so.

*sigh*

i hav 1 more post than you mathlete!!!! yay!!!!!

and i hav the right to take over chompy if he lets me and he does.

dead topic.

maybe too inappropriate.

lock this topic up mods

too late, you should have called the mods sooner. i guess you are really slow. and yay i have one less post than you, take that!

You're one to talk. You've been losing the game for quite some time. You couldn't win the game if my life depended on it.
Also, when was the last time you cooked cookies? You haven't cooked cookies since your ma helped you. I can do it [i]all by myself![/b]

and because you do it all by yourself, they turn out bad. while mine turn out really good since my mother helps me. also, the first insult does not count since you can only do one.


group theory
abstract algebra
superior algebra
superior algebra solved

Every finite group is isomorphic to a subgroup of some $A_n$.

Let $G$ act on two copies of itself. Each element acts as a permutation, and each such permutation is obviously even.

Therefore, if $G$ has $k$ elements, $G$ is isomorphic to a subgroup of $A_{2k}$.

Or:
every group of order $n$ is subgroup of $S_n$ (Cayley's theorem)
Consider this subgroup as subgroup of $S_{n+2}$ and 'adjungate' the cycle $(n+1 , n+2)$ to all odd permutations, then you get it as subgroup of $A_{n+2}$

[quote="jmerry"]Let $ G$ act on two copies of itself. Each element acts as a permutation, and each such permutation is obviously even.

Therefore, if $ G$ has $ k$ elements, $ G$ is isomorphic to a subgroup of $ A_{2k}$.[/quote]

what do u mean act on two copies of itself , Could you clarify the action only ?

Did u mean $ G$ is $ G \minus{} Set$ via $ g.x \equal{} g. \minus{}(x)$ , where is $ g.\minus{}$ is a permutation on $ G$, if yes , how do u sure this is even permutation ?

 :)

$ G$ acts fauthfully on $ G$ by left multiplication, i.e. we have a monomorphism $ G \to Sym(G)$. this yields a monomorphism $ G \to Sym(G) \times Sym(G) \to Sym(G \dot{\cup} G)$. the images have the form $ \sigma \circ \sigma'$, where $ \sigma,\sigma \in Sym(G)$ are actually the same but regarded on different copies of $ G$. in particular, they have the same sign and thus the sign of their product is $ 1$.


AMC
AIME
geometry
geometric transformation
homothety
rectangle
parameterization

The solutions by this Web Site's instructors are posted in the Math Jams transcripts.  The solutions are, of course, very good!

How about posting novel solutions to see some different methods?

I'll start the ball with a different solution to #10.

In our Olympiad Geometry class we teach homothecy.  Students asked if it could be used on the AIME.  I answered 'rarely, though we often wrote Mandelbrot problems for homothecy.'

Then you give this cool solution, in which the homothecy of the 'desired' area to the half-rectangle solves the problem quickly.

8-)

2004 AIME-I #15

I do not consider the following to be elegant.  But it is different.

I am hopeful someone will have a truly elegant solution they will be willing to share.  This problem is so out of the ordinary for AIME, there *must* be an elegant solution that is within the norm for AIME problems.

As MCrawford suggested during the Math Jam, false starts are a pitfall for this problem.  I considered trying to count the cases directly.  I considered taking the 9's complement of the decimal input parameter (and adapting the function).  I considered making a tree.  I considered making a 20-stage finite state machine.  I considered making a single-stage finite state machine.  While thinking about this, I thought about Turing's tape.  This eventually led to the following.  [You will be thinking "only from the mind of a software engineer".]  Cheers!  ...

As the sequence x1, x2, x3, …, xn is generated, let’s create a string to represent which of the three operations the function performs with each step.  Let c(x) be a string representing the operation(s) of function f(x) in generating d(x).

We’ll use “I” to indicate the function incremented the parameter.  f(x)=x+1
We’ll use “/” to indicate the function shifted the parameter by a factor of 10.  f(x)=x/10
And we’ll use “Z” to indicate the function has found the terminal result.  f(x)=1

Every pattern ends with "Z".  Except for the coding of x=1, every pattern ends with “/Z”.

If the input parameter, x1, contains trailing zeros, the first steps to be performed by the function are to shift the parameter by factors of 10.  We insert a shift symbol (“/”) at the beginning of the output string for each trailing zero in the input parameter.

Leading and trailing zeros have now been stripped from our decimal input parameter.  If what remains is 1, we're done, and the coding is some consecutive sequence of zero or more “/” symbols, followed by a “Z”.

Otherwise, there are digits to be “coded”.  We append an initial “I” to the output string.  Then consider each digit of the input parameter, from right-to-left, and append the output string with the appropriate code from the following table:
[code]
	0	IIIIIIIII/
	1	IIIIIIII/
	2	IIIIIII/
	3	IIIIII/
	4	IIIII/
	5	IIII/
	6	III/
	7	II/
	8	I/
	9	/
[/code]

The value of d(x) is the length of the string, c(x) we have just determined.

Try some examples, and convince yourself the coding correctly represents the operation of function, f.
[code]
        c(1)        Z                    d(1)=1
        c(10)       /Z                   d(10)=2
        c(100)      //Z                  d(100)=3
        c(9)        I/Z                  d(9)=3
        c(90)       /I/Z                 d(90)=4
        c(87)       III/I/Z              d(87)=7
        c(349)      I/IIIII/IIIIII/Z     d(349)=16
        c(34900)    //I/IIIII/IIIIII/Z   d(34900)=18
[/code]

Our goal is to determine how many valid strings have a length of 20.  All strings are of the following form, where “x” can represent a “/” symbol or a “I” symbol.

[code]
	c(x) = “xxxxxxxxxxxxxxxxxx/Z”
[/code]

The problem is now a combinatorics problem…  “How many ways can we place “I” and “/” symbols in a string of length 18, without violating certain conditions?”

I identify the exceptions we must consider:[list=1]
[*]A low-order zero digit is never coded as “IIIIIIIII/”.  [The function would use shifts to code low-order zeros.]
[*]A high-order zero digit is never coded as “IIIIIIIII/”.  [The function does not code high-order zero digits.  It "stops".]
[*]A stand-alone one digit is never coded as “IIIIIIII/”.  [The function never performs in increment if the input parameter is a power of 10.]
[*]The longest run of "I"s is produced by the code for zero.  No longer run of "I"s can be produced by the function.[/list]

Next I consider how these exceptions translate into conditions for the combinatorics problem:[list]
[*]We can never allow a run of 10 or more consecutive “I” symbols in the string.  This one condition deals with two issues.  First, there is no digit that generates a code of 10 or more consecutive “I” symbols.  Second, the low-order digit in the coded string can never be “0”.  Since the code for “0” is nine consecutive “I”s, and since we always begin the coding of digits with an “I”, the prohibition against 10 consecutive “I” symbols solves both problems.

[*]The high-order digit in the coded string can never be “0”.  Testing for this one condition also deals with the other special case…

The string cannot contain the code for “1”, unless it also contains the code for at least one other digit.  The only case this affects is “/////////IIIIIIIII/Z”, which is a false coding for “1,000,000,000”.[/list]

I enumerate the possibilities.  The left column shows various patterns of the strings of length 20.  The right column describes why they are a special case.  Each “x” in the pattern strings indicates a position that can be filled by an increment (“I”) or a shift (“/”).  The total number of choices within each pattern is just 2^n, where n is the number of “x” symbols in the pattern.  I have carefully organize the patterns so there is no duplication between the impossible choices.

[code]
        xxxxxxxxxxxxxxxxxx/Z    2^18 total choices.  But some of these are impossible…

         xxxxxxxx/IIIIIIIII/Z    256 impossible choices.  x cannot code a high-order zero-digit, and x=1,000,000,000 cannot be coded in this manner.

        IIIIIIIIIIxxxxxxxx/Z    256 impossible choices.  No code has a run of 10 “I”s.
        /IIIIIIIIIIxxxxxxx/Z    128 impossible choices.  Ditto.
        x/IIIIIIIIIIxxxxxx/Z    128 impossible choices.  Ditto.
        xx/IIIIIIIIIIxxxxx/Z    128 impossible choices.  Ditto.
        xxx/IIIIIIIIIIxxxx/Z    128 impossible choices.  Ditto.
        xxxx/IIIIIIIIIIxxx/Z    128 impossible choices.  Ditto.
        xxxxx/IIIIIIIIIIxx/Z    128 impossible choices.  Ditto.
        xxxxxx/IIIIIIIIIIx/Z    128 impossible choices.  Ditto.
        xxxxxxx/IIIIIIIIII/Z    128 impossible choices.  Ditto.
[/code]

I have 2^18=262144 total choices.  Subtract 1536 impossible choices.  We are left with 260608 valid choices.

Prime factor:  260608 = 2*2*2*2*2*2*2*2*2*509.

Sum the distinct prime factors:  2+509=511.

WOW! That's amazing. I probably will never come up something like this(I probably will never solve this problem)

(Worked w/ Fergie to get this)

So yeah the basic idea is to work backwards. Start at 1 then multiply by 10 (denoted by B) or subtract 1 (denoted by A) at each step. Then you get a 19 letter word of A's and B's, with the following conditions.

1. The word starts with a B (you must go to 10 first)
2. The word does not start with BAAAAAAAAA (B, 9 A's) (otherwise you hit 1. But 20 is the first spot where you hit 1, so no good).

3. The word does not have AAAAAAAAAA (10 A's), Since you cant do 100 -> 99 -> ... -> 91 -> 90, as 90 can only go to 9, not 91.

This is now a simpler counting problem.

Consider only the remainder of the word after the B (18 letters). Then there are 2^18 possibilities. But some do not work. They fall into two nonoverlapping categories.

1. Starts with 9 A's but has no 10 consecutive A's.
2. Has 10 consecutive A's.

1) The word is of form AAAAAAAAABXXXXXXXX, where each X can be anything without violating any conditions. This makes 2^8 words of this form.

2) Either the word starts with 10 A's (2^8 words), or has a BAAAAAAAAAA (B, 10A's) in the word. There are 8 ways to choose where this goes in the word (pick the number of letters before it from 0 to 7), and 2^7 ways to pick the other letters (7 spots remain), making 8 * 2^7 = 2 ^ 10 words.

Therefore, there are 2^18 - 2 * 2^8 - 2^10 = 2^18 - 3 * 2^9 = 509 * 2^9 words. The sum of its prime divisors is 509 + 2 = 511 as desired.

#14.

Take an overhead view and use power of a point to find the distance (on the ground) between tangent point and the unicorn is 4 :sqrt: 5. Then unwind the rope into a right triangle ABC, with right angle C, and BC = 4, AB = 20. Mark point D on AC s.t. DC = 4:sqrt: 5 and draw a perpendicular through D to AC letting this intersect AB at E. Then find the length AE and you have your answer.

#6 -- multiply by 1+0x and pair stuff up:

(1+0x)(1-x) = 1 - x - 0*1*x^2
(1+2x)(1-3x) = 1 - x - 2*3*x^2
(1+4x)(1-5x) = 1 - x - 4*5*x^2
.
.
.
(1+14x)(1-15x) = 1 - x - 14*15*x^2

The the only way to get an x^2 term is either to take 2 -x terms and all the rest 1's -- there are 8 C 2 = 28 ways to do that, so that contributes 28 to the product, or to take a 2k(2k+1)x^2 term from one factor and all the rest 1, which contributes
-0*1 + 2*3 + 4*5 + ... + 14*15 = -616 (add it all up or derive a formula) to make 28-616 = -588.

I had an alternate solution to #15 that wasn't particularly slick but was a different approach.  So, instead of looking at x, you look at the tens complement of x, that is, 10^n - x, where 10^n is the least power of 10 >= x (or n is one less than the number of digits of x), and where leading zeros are added such that the tens complement of x has the same number of digits as x.  Given leading zeros, the tens complement of x uniquely defines x, and the tens complement of x can be any nonnegative integer (with leading zeros) that does not begin with 9 (this is the part I missed on the test).

Then if you look at what f does to the tens complement of x, either it decreases the last digit by 1, or if the last digit is 0, it removes the last digit.  Hence d(x) is equal to the sum over all the digits in d(x) (including leading 0's) of one plus the digit -- this is a sum of a sequence of numbers between 1 and 10 such that the first number is not 10 (because the first digit of a tens complement is not 9).

So what we want to count is the number of sequences of numbers between 1 and 10 with the first number not 10 that sum to 19.  This is now fairly doable.  If we didn't have the restrictions, the answer would be 2^18 -- if there are 19 stars in a row, there are 18 spots between them in which to put a bar or not put a bar, and then the number of stars between successive bars gives such a sequence.  After that, it's just subtracting stuff off, which is fairly routine and not so bad.

[quote="Nukular"]#15.  So yeah the basic idea is to work backwards.[/quote]

Very nice!  Combinatorically, this is equivalent to my solution.  But your solution provides two major advantages:

With your solution, it wasn't necessary to go through the intermediate "coding" step.

Second, by working backwards, you had only two special cases to deal with.  I had four special cases to deal with.  Until I noticed I could combine my special cases into just two, I had several dozen exception patterns.

I also like your other alternate solution and those by Alison and masamune!

#9
 U_2 is a 3-4-5 triangle. The segment has to be drawn from the vertex of the rectangle to one of the opposite side. The longer leg of U_2 has to be 6 or 7.  The greater angle of U_2 is  :cong: to one of the angle of V_2.

U_1 is a 3-4-5 triangle. It must have the greater angle  :cong: one of the angle of V_1. So the segment must be drawn from the hypotenuse to the longer leg that's parallel to the shorter leg.

Suppose longer leg iof U_2is 7, then the shorter leg has to be 21/4, so shorter base of V_2 has to be 3/4 and the longer base is 6. (3/4)/6=x/3 where x is the shorter leg of U_1. x=3/8
Suppose it'6, then the shorter leg is 9/2, so the shorter base is 5/2. (5/2)/7=x/3, so x=15/14.
We are going to have smaller area if x is smaller, so x=3/8

The longer leg is 1/2, 1/2*(3/8)*1/2=3/32. 3+32=35

Nukular knows me from many places, most notably PROMYS '03, since he was my roommate.

For AIME problem #3 --

   I'm not sure how elegant this is, but .....

   In each of the 12 quadrilaterals, draw a single diagonal line segment. By assuming a little bending on these segments, you now have a convex polyhedron with 48 triangular faces, but the same 26 vertices. The effect of these auxilary line segments is the creation of an addition 12 space diagonals. In the original quadrilaterals, the vertices not connected by the auxilary form space diagonals in the new convex polyhedron.

   So, at this point, it's just of matter of computing the number of space diagonals in a convex polyhedron (48 triangles, 26 vertices) then subtracting 12 for the additional ones you created.

Spoiler:

[hide] 3*48 = 144 vertices, with redundancies.

If k triangles meet at a single vertex, then the number of space diagonals that can be formed from that vertex is (25 - k).

Assume for a moment, that k can be either 5 or 6, and x is the number of vertices with k=5.  The redundancies handled by:

    5x + 6 (26-x) = 144     which yields  x = 12

   So, 12 vertices with 5, and 14 vertices with 6.

   This yields    12 * (25 -5) + 14 * (25 - 6) = 506.

    Dividing by 2 (since each space diagonal counted twice), yields 253.

   Subtracting 12 (created by splitting quadrilaterals) and you get 241.

  Note:

  Other combinations like k either 5 or 7 produces

      19 5-type vertices and 7 7-type which yield..

      19 * (25 -5) + 7 * (25 - 7) = 506

   k either 4 or 6 produces

      6 4-type vertices and 20 6-type which yield..

      6 * (25 -4) + 20 * (25 - 6) = 506

[/hide]

For #3, what I did was

the total number of segments joined by any vertices is 26C2

that include the 60 edges, which we don't want

it also include all the diagonals of Quads that we don't want
there are 2 diagonals for each quad, so total 12*2=24
26C2-60-24=241

beta's way is by far the best I have seen (it's the way I did it; it surprised me how easy the solution was to find.)

And rcv, wow. That's just really, REALLY amazing. The odd thing being that I started thinking about Turing too during that problem, but dismissed it as nothing more than my ADD. (It was really just a shadow of an idea; wondering if I could use computers to solve it.)

[quote="masamune2387"]For the coefficient problem...

All I did was find all possible pairwise products between -1, 2, -3 ..., -15.

I did this by (-1 + 2 - 3 + 4... - 15)^2 -  :Sigma: x = 1 to 15, x^2, and divided the whole thing by 2. I got -588, so |-588| = 588.[/quote]

That's amazing.  Nice job.

Thanks Magnara :D

Gotta love working with generating functions.

#4,
note that the midpt. of the hypotenuse is equidistant from all vertices of a right triangle, so the locus of points that's 1 any of the vertices is 4 quarter of a circle with radius 1. The enclosed region by the locus is 4- :pi:  :approx: 86.

Beta-
  Awesome solutions for 3 and 4.  I came up with the all triangles approach for #3, because I was having trouble visualizing the quads and triangles together.  Yours is so much slicker.

For #8, it's like stepping around a clock with a thousand numbers, without hitting the same spot twice, before coming back to you starting point (and you can't step by 1). Stepping by more than 500, is like stepping by less than 500 in the opposite direction.

 So, it's all the integers > 1 and < 500 that are relatively prime to 1000 = (2*5)^3

  So, (499) -  (250 multiples of 2) - (50 odd multiples of 5) = 199

Yep. For that star problem:

phi(1000) = 4*25*4

there are 400 relatively prime numbers. Get rid of 1 and 999, so there are 398, and they pair up, so divide by two. The answer is then 199.

[quote="beta"]#4. note that the midpt. of the hypotenuse is equidistant from all vertices of a right triangle[/quote]
Nice, simple explanation, Beta.  For those who are visual, I've attached a picture.


calculus
logarithms
probability
number theory
least common multiple
ratio
function

Prove that \[ x^a y^b \leq ax + yb \] whenever $x,y > 0$ and $a+b=1$. [edit]  of course, $a,b > 0$ [/edit]

Can anybody show me a proof without calculus?

(My proof would go as follows: we prove it first for $a,b \in \mathbb{Q}$ using normal AM-GM; after that, using a continuity argument - :(, we prove it for $a,b \in \mathbb{R} - \mathbb{Q}$)

I'm really, really, really interested in a more elementary proof ...

Just A Hint: Prove rational situation first.
                 Show the irrational in continued fraction.

[quote="Arne"]That is exactly what perfect_radio wrote, Skyward_sea.

I don't have a more elementary prove but I do have a very short one -

since ln is concave we have $\ln{\left(x^ay^b\right)} = a \ln{x} + b \ln{y} \leq \ln{\left(ax + by\right)}$ :)[/quote]

of course, your proof is correct. but showing that ln is concave implies some calculus :(, so, indeed, it isn't what i'm looking for

[quote="Arne"]That is exactly what perfect_radio wrote, Skyward_sea.

I don't have a more elementary prove but I do have a very short one -

since ln is concave we have $\ln{\left(x^ay^b\right)} = a \ln{x} + b \ln{y} \leq \ln{\left(ax + by\right)}$ :)[/quote]

Sorry I said "continued fraction"(like $\sqrt2=1+\frac{1}{2+\frac{1}{2+...}}$), I thought it was different from "a continuity argument "

Ummmm...

x=1
y=1/2

a=-1
b=2

That would mean that
1/4 < 0, which is false.
Are we assuming a,b > 0?

Yes, $a$ and $b$ (or more generally, $x_1,x_2,...,x_n$ with sum $1$) define a probability distribution, so they must be positive.

FWIW - Here is an interesting  proof when  $a$ and $b$ are  rational and positive,  using only elementary math.

First since  $(\sqrt x_1-\sqrt x_2)^2  \ge 0$ gives
$\frac {x_1+x_2}  2 \ge  (x_1 x_2)^{(1/2)}$
Similarly  $\frac {x_3+x_4}  2 \ge  (x_3 x_4)^{(1/2)}$
which gives:
$\frac {x_1+x_2+x_3+x_4}  4 \ge \frac {(x_1 x_2)^{(1/2)}+ (x_3 x_4)^{(1/2)}} 2 \ge (x_1 x_2 x_3 x_4)^{(1/4)}$

Keep doing it for $k$ times and if  $n=2^k$ then: (Eq 1):
$\frac {x_1+x_2+x_3+x_4+ \dots +x_n}  n   \ge (x_1 x_2 x_3 x_4 \dots x_n)^{(1/n)}$

Now if $a$ and $b$ are rational and $m$ is LCM of the denominators  then one can write:
$a = p/m $ and $b = q/m$ where $p$ and $q$ are integers and $p+q=m$

Now select  $k$ large  enough so that $n=2^k \ge  m$ and select:

First $p$ ....   $x_1 x_2 \dots  x_p = x$
Next $q$ ....  $x_{p+1} x_{p+2} \dots  x_{p+q} = y $
and the rest $(n - p - q)=n-m $ ...  $x_{p+q+1} \dots x_n =  (ax+by) $

Now eq 1 gives:

${\frac  {p x + q y + (n -m) (ax+by)} n } \ge  (x^p y^q (ax+by)^{n-m})^{(1/n)}$
which gives: $ax + by \ge (x^a y^b)$

If $a$ and $b$ are not rational, one can use the  method suggested (approximate as continued fraction).
 step 1 show that it is true for x=y
step 2 show that if x is not equal to y  the difference between LHS and RHS is posiitive, and  use the limit for LHS and RHS... etc..

how can you expect a calculus theorem to be proven properly without calculus? :?

I mean, the reason they ever started with calculus was to avoid long basic proofs like that...

Peter, did you follow the proof at all?

the majority of it is just a proof of AM-GM, by appropriate choice of the numbers, we can prove it for rationals, and by continuity it holds for all reals.  Of course there is a faster proof using calculus, but you need to know that $f''(x)\geq 0$ implies convexity, which is not particularly simple.

Here is the continuity part.

Say that $G(\lambda)=\lambda a+(1-\lambda)b-a^\lambda b^{1-\lambda}\geq 0$ for all $\lambda\in\mathbb{Q}\cap (0,1)$.  Then if there is some $t\in\mathbb{R\setminus Q}\cap (0,1)$ for which $G(t)<0$, by continuity (since half of the equation is linear in $\lambda$ and the other part is continuous), there exists an interval $(t-\epsilon,t+\epsilon)$ in which $G(t)<0$.  But any such interval contains rationals, a contradiction.

$x^ay^b\leq ax+by\ (a,b,x,y>0,a+b=1)$

$\Longleftrightarrow \frac{a\log x+b\log y}{b+a}\leq \log \left(\frac{ax+by}{b+a}\right)$

You can imagine [color=blue]Internal Ratio[/color]. :)

kunny

Well in my opinion showing the graph of ln() is enough to deduce convexity if you don't know the details... but of course proofs without calculus are "possible", as shown above. It just seems a little pointless to me. (but then again, me = a person knowing calculus)

Well, why don't you prove that a function with $f''(x)>0$ is convex then?

Of course I know, blahblahblah. I am just lazy, rather than , I would say the very Seeing is Believing!

I wish I could draw a graph with Tex.


A man of weight $w$ is in an elevator of weight $w$.The elevator accelerates vertically up at a rate $a$ and at a certain instant has a speed $V$.

(a) What is the  apparent weight of the man?

(b) The man climbs a vertical ladder within the elevator at a speed $v$ relative to the elevator.What is the man's rate of expenditure of energy?

a
$m|(g-2a)|$

for b, i'm not sure what is does the word expenditure means but if it means change, then
$\frac{w}{2g}(v)^2$

Your answer is incorrect, amirhtulsa :(

The rate of expenditure of energy means power output.

kunny

for the first one, in terms of $w$
$w(1-\frac{a}{g})$??

[hide="a"]$W = w(1+\frac{a}{g})$[/hide]

[hide="b"]$P = Wv = vw(1+\frac{a}{g})$
The velocity V is doomed to be irrelevant, I think.[/hide]

The first part.
[hide]
$N-mg=ma$
So
$N=ma+mg$[/hide]

For the second one, I still don't quite understand the meaning. But if it means power, then i think that:
[hide]$P=(v+V)(ma+mg)$[/hide]

Shobber's a) part is the same as mine, but in b) I still think that V should NOT play any role in any of those problems.

Recall the Einstein's (or whoever it was) law of relativity of two systems moving with constant speed against each other. How can you even FIND OUT how fast is the elevator moving? (I definitely think it shouldn't be possible by measuring your power climbing a ladder.)

Do you really think that the stewardess in an airplane, who is pushing the drink&meals cart with velocity 0.3 m/s = 1 ft/s has to exert 1000 times larger power when the plane is in the air at cruising speed 300 m/s, than when it is sitting on the ground? (Not talking of a stewardess pushing the cart in opposite direction;-)

Do you have the answers to this question Kunny?

Yes. Here are the answers.

(a) $w\left(1+\frac{a}{g}\right)$

(b) $w\left(1+\frac{a}{g}\right)(V+v)$

I think your B) answer is not right. Maybe I'm wrong, but still, look at my previous post in this forum.

From your formula for (b), when the man climbs with a zero speed and the acceleration is zero and the elevator is moving at a high speed, the man has a big power output. Which doesn't seem right to me at all.


probability
geometry
perpendicular bisector

Ok, so you are given the following diagram and asked to find the length of two segments one connecting the center of the circle to the midpoint of the triangle's base and the other connecting a perpendicular bisector on the base to the circle . [geogebra]19c921643b05462da5a71f324ff7a16724611df9[/geogebra] This is what I did: I made the segment connecting the center of the circle to the midpoint of the base of the triangle = 2root3. Then I subtracted it from the radius: root16 - root12=root4. This means that the remaining segment is 2. Subtracting that from the radius gave me 2. Which means root12=2. What I am I doing wrong? Please help.

I haven't seen the applet yet, but $ \sqrt{16} \minus{} \sqrt{12} \neq \sqrt{16\minus{}12}$.

oops  :blush: this is worse than when I was a poker probability did not work out because 4^2 is not eight. Thank you.


What is the slope of any line parallel to the graph of the line determined by the equation $ \frac{2}{3} x \plus{} \frac{2}{5} y \equal{}2$?

But then they ask us for the slope, so the answer is -5/3.


1/ Let $ a,b,c>0$ such that $ abc\equal{}1$. Prove that
$ a\plus{}b\plus{}c \ge \frac2{a\plus{}b}\plus{}\frac2{b\plus{}c}\plus{}\frac2{c\plus{}a}$

2/ Let $ a,b,c,d>0$ such that $ abcd\equal{}1$. Prove that
$ a\plus{}b\plus{}c\plus{}d \ge \frac3{a\plus{}b\plus{}c}\plus{}\frac3{b\plus{}c\plus{}d}\plus{}\frac3{c\plus{}d\plus{}a}\plus{}\frac3{d\plus{}a\plus{}b}$

3/ Let $ a_i>0$ such that $ \Pi a_i\equal{}1$. Prove that
$ a_1\plus{}a_2\plus{}...\plus{}a_n\ge \frac{n\minus{}1}{a_1\plus{}a_2\plus{}...\plus{}a_{n\minus{}1}} \plus{}...\plus{}\frac{n\minus{}1}{a_n\plus{}a_1\plus{}...\plus{}a_{n\minus{}2}}$

[quote="akai"]1/ Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
$ a \plus{} b \plus{} c \ge \frac2{a \plus{} b} \plus{} \frac2{b \plus{} c} \plus{} \frac2{c \plus{} a}$
[/quote]
$ a\plus{}b\plus{}c\geq\sqrt a\plus{}\sqrt b\plus{}\sqrt c\geq \frac2{a \plus{} b} \plus{} \frac2{b \plus{} c} \plus{} \frac2{c \plus{} a}.$

[quote="akai"]
2/ Let $ a,b,c,d > 0$ such that $ abcd \equal{} 1$. Prove that
$ a \plus{} b \plus{} c \plus{} d \ge \frac3{a \plus{} b \plus{} c} \plus{} \frac3{b \plus{} c \plus{} d} \plus{} \frac3{c \plus{} d \plus{} a} \plus{} \frac3{d \plus{} a \plus{} b}$
[/quote]

Rinse repeating in a similar way to arqady

$ \frac {a \plus{} b \plus{} c}{3} \ge \sqrt [3]{abc} \Longleftrightarrow \frac {3}{a \plus{} b \plus{} c} \leq \frac {1}{\sqrt [3]{abc}} \equal{} \sqrt [3]{\frac {1}{abc}} \leq d$

[quote]3/ Let $ a_i > 0$ such that $ \Pi a_i \equal{} 1$. Prove that
$ a_1 \plus{} a_2 \plus{} ... \plus{} a_n\ge \frac {n \minus{} 1}{a_1 \plus{} a_2 \plus{} ... \plus{} a_{n \minus{} 1}} \plus{} ... \plus{} \frac {n \minus{} 1}{a_n \plus{} a_1 \plus{} ... \plus{} a_{n \minus{} 2}}$[/quote]

For the more general one, we may also rinse repeat

$ \frac {{\displaystyle \sum_{k \in S \minus{} \{a_m\}}k}}{n \minus{} 1} \geq \sqrt [n \minus{} 1]{\prod_{k \in S \minus{} \{a_m\}}} \Longleftrightarrow \frac {n \minus{} 1}{{\displaystyle \sum_{k \in S \minus{} \{a_m\}}k}} \leq \sqrt [n \minus{} 1]{a_m} \leq a_m$

Where $ S \equal{} \{ a_1,a_2,a_3, \ldots, a_n\}$

rachid, of cause!
$ \sqrt a\equal{}\frac{1}{\sqrt{bc}}\geq\frac{2}{b\plus{}c}$ and $ a\plus{}b\plus{}c\geq\sqrt a\plus{}\sqrt b\plus{}\sqrt c\Leftrightarrow\sum_{cyc}(x^6\minus{}x^4yz)\geq0,$ which is true by Muirhead, where $ a\equal{}x^6,$ $ b\equal{}y^6$ and $ c\equal{}z^6,$ $ \{x,y,z\}\subset\mathbb R^\plus{}.$

[quote="arqady"]rachid, of cause!
$ \sqrt a \equal{} \frac {1}{\sqrt {bc}}\geq\frac {2}{b \plus{} c}$ and $ a \plus{} b \plus{} c\geq\sqrt a \plus{} \sqrt b \plus{} \sqrt c\Leftrightarrow\sum_{cyc}(x^6 \minus{} x^4yz)\geq0,$ which is true by Muirhead, where $ a \equal{} x^6,$ $ b \equal{} y^6$ and $ c \equal{} z^6,$ $ \{x,y,z\}\subset\mathbb R^ \plus{} .$[/quote]
 $ a \plus{} 1    \geq 2\sqrt{a }$
  hence    we have  :
 $ a \plus{} b\plus{}c  \plus{}3  \geq 2(\sqrt{a}  \plus{}\sqrt{b} \plus{}\sqrt{c})  \equal{} (\sqrt{a}  \plus{}\sqrt{b} \plus{}\sqrt{c})\plus{}(\sqrt{a}  \plus{}\sqrt{b} \plus{}\sqrt{c})  \geq  (\sqrt{a}  \plus{}\sqrt{b} \plus{}\sqrt{c})  \plus{} 3\sqrt[6]{abc}$
    or  $ a\plus{}b\plus{}c \geq  \sqrt{a}  \plus{}\sqrt{b} \plus{}\sqrt{c}$
 i think  this solution is better

I predict that the following result also true:

$ a\plus{}b\plus{}c\plus{}6 \ge \frac6{a\plus{}b}\plus{}\frac6{b\plus{}c}\plus{}\frac6{c\plus{}a}$

where $ a,b,c>0$ and $ abc\equal{}1$.

Thank for nice solution and nine problem.

[quote="dduclam"]I predict that the following result also true:

$ a \plus{} b \plus{} c \plus{} 6 \ge \frac6{a \plus{} b} \plus{} \frac6{b \plus{} c} \plus{} \frac6{c \plus{} a}$

where $ a,b,c > 0$ and $ abc \equal{} 1$.[/quote]

No body have idea for it? :wink:

[b]*[/b]A my new result:

$ a \plus{} b \plus{} c  \ge \frac2{a \plus{} b} \plus{} \frac2{b \plus{} c} \plus{} \frac2{c \plus{} a}$

where $ a,b,c \ge 0$ and $ ab\plus{}bc\plus{}ca \equal{} 3$

It's nice but easy  :)

[quote="dduclam"]

$ a \plus{} b \plus{} c \ge \frac2{a \plus{} b} \plus{} \frac2{b \plus{} c} \plus{} \frac2{c \plus{} a}$

where $ a,b,c \ge 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$

[/quote]
It is equivalent to $ \sum_{cyc}(a^3b \plus{} a^3c \minus{} 2a^2bc)\geq0.$
[quote="dduclam"]I predict that the following result also true:

$ a \plus{} b \plus{} c \plus{} 6 \ge \frac6{a \plus{} b} \plus{} \frac6{b \plus{} c} \plus{} \frac6{c \plus{} a}$

where $ a,b,c > 0$ and $ abc \equal{} 1$.[/quote]
Yes, it is true. The Mixing Variables method kills it very easy.

[quote="dduclam"]

[b]*[/b]A my new result:

$ a \plus{} b \plus{} c \ge \frac2{a \plus{} b} \plus{} \frac2{b \plus{} c} \plus{} \frac2{c \plus{} a}$

where $ a,b,c \ge 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$

It's nice but easy  :)[/quote]

My solution:

$ a \plus{} b \plus{} c \ge \frac2{a \plus{} b} \plus{} \frac2{b \plus{} c} \plus{} \frac2{c \plus{} a}$

<=> $ 3(a\plus{}b\plus{}c)\ge \frac{2(ab \plus{} (a\plus{}b)c)}{a\plus{}b} \plus{}\frac{2(bc \plus{} (b\plus{}c)a)}{b\plus{}c}\plus{}\frac{2(ca \plus{} (c\plus{}a)b)}{c\plus{}a}$

<=> $ a\plus{}b\plus{}c\ge \sum\frac{2ab}{a\plus{}b}$ also true because  $ (a\plus{}b)^2\ge 4ab$
 :)


real analysis
function
real analysis unsolved

Show that if $ (f_n)_n$ is a sequence in $ L^1[0, 1]$ is such that $ \sum_{n\equal{}1}^{\infty}  \parallel{} f_n \parallel{}_1 < \infty$ then $ \sum_{n\equal{}1}^{\infty} | f_n(s) | < \infty$ for almost every $ s \in [0, 1]$.

This looks simple to prove.  However, I do not see how to prove this right now.  Any hints on how to proceed would be appreciated.

Hint: is the function $ g(x)\equal{}\sum_{n\equal{}1}^{\infty}|f_n(x)|$ integrable?


Is it possible for you guys to post PDFs of Transcripts instead of having to copy and paste into text editors?  I think it would help [b][i]Lots[/i][/b] of people.  Just a suggestion. :D

If you really want to use a PDF copy, then just copy and paste to MS word and then use [url=http://www.primopdf.com/]PrimoPDF[/url].

I support HTML copies, so ccy's transcript player can be used.

PrimoPDF doesn't work with some systems. http://cutepdf.com/ is a better choice.

There should be a php that echoes the transcripts based on a url var.

[quote="Nexmus"]Guys! The whole point of this is that converting to PDF takes too LONG!!! See, waiting for the transcript to load completely (without any squares with x's inside them) takes like 10 tries. Next, pasting into word takes 4 minutes, and finally converting takes 2 minutes. So, basically could you give us the pdf's ready to download in the "my classes" page?[/quote]

Those times are exaggerated.
On a standard connection, in Firefox, loading takes about 30 seconds.

On a normal Windows XP Dell, pasting takes maybe 10 seconds.
Converting takes maybe a minute, probably less.

Just in case you guys didn't notice, there's a new "Print" feature now, so it makes it printer friendly and all. While downloading a 1.7 GB file, I loaded a transcript (1.5 hrs) in about 10 secs. (SDSL here)

It doesn't take too long to convert after that. Maybe 40 secs?

Sorry guys, maybe it's just me. But, I'm in a town in the middle of nowhere with a connection sucks !@#$ and am about 1000 miles from the nearest server.

Note #1: ad above.

Note two: I support this too, but I believe there are problems ya'll aren't seeing. First, PDFs are easier to transfer between people. This means one dishonest person can pay for WOOT and pass *all* the class materials around quite easily. Second, it's not straightforward to convert it to PDF on the server side. BBCode is not perfectly easily translatable. Third, in a browser, the images may be cached so that multiple loads of one transcript only cause, say, 1.5 full page loads worth of traffic. In a PDF, someone clicks it 8 times, it get downloaded 8 times: guess what, you've tripled the load on the server on  the transcripts.


So I've compiled all the answers from my own test and from the various threads on here. I've got what I believe to be the definitive list, but I would like some confirmation on 15.

1. 372
2. 200
3. 578
4. 450
5. 888
6. 640
7. 553
8. 896
9. 259
10. 710
11. 179
12. 91
13. 640
14. 676
15. 389 (pretty sure)

[quote="Arnizzle"]How sure are you guys about 11? Because I was pretty sure I was right with 227.
I got 128pi + 96sqrt3[/quote]

100%.

All of the above ARE correct, according to the Math Jam.


The volume of the new shape is $ 2\cdot2\cdot1\equal{}\boxed{4}$ times the volume of the original shape.


I wrote a tournament test and I discovered that it was incredibly hard, much harder than I intended. In fact, people who got a 7 at ARML last year got a 4 on my ARMLish ciphering test. I've posted it at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=932033#932033 if you'd like to take a look at it, answers will be posted on Tuesday.

Where is the solution thread to the problems (if any)? Thanks!


function
real analysis
real analysis unsolved

Prove the existence of $ f,g: \mathbb{R}\rightarrow\mathbb{R}$ measurable such that $ f\circ g$ is not measurable.

I would use two Cantor-type sets $ C_1$ (with positive measure) and $ C_2$ (with zero measure). They can be arranged so that there is a continuous (even Lipschitz) bijection $ g\colon C_1\to C_2$ that can be extended to $ \mathbb R$. Choose a nonmeasurable set $ E\subset C_1$ and let $ f$ be the characteristic function of $ g(E)$.


Given that $ n$ is an integer and $ 0 < 4n <30$, what is the sum of all possible integer values of $ n$?

$ 0 < 4n < 30$ simplifies to $ 0 < n < 7 \frac{1}{2}$. Thus, positive integers $ 7$ and less fit the restrictions on $ n$. Adding these $ 7$ positive integers, $ 1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\equal{}\frac{7(7\plus{}1)}{2}\equal{}7(4)\equal{}\fbox{28}$.


Propose a synthesis of p-aminoacetophenone from aniline.

Quite very simple
1. Protect the NH2 by converting it into NHCOCH3 ucing acetic anhydride
2. acylate in the para posn using acetic anhydride
3. use NaOH to convert NHCOCH3 back into NH2

[quote="Madness"]Quite very simple[/quote]

Is that so? Humm, let's see:

[quote="Madness"]2. acylate in the para posn using acetic anhydride[/quote]

And what are the reaction conditions to do that? Aren't you forgeting something?


function
superior algebra
superior algebra unsolved

Find a bijection between the real line R and the unbounded interval (0, inf) = {x in R: x>0}.

Can the bijection be f: x-->x+1?

A function $ f: X\to Y$ is [i]surjective[/i] if for each $ y\in Y$ there is at least one $ x\in X$ such that $ f(x) \equal{} y$.  A function is [i]injective[/i] if for each $ y\in Y$ there is at most one $ x\in X$ such that $ f(x) \equal{} y$.  A function is a bijection if it is both injective and surjective.

Your function is a bijection from $ \mathbb{R}\to \mathbb{R}$, but its range is not $ (0,\infty)$, so it is not the bijection you're looking for.

Try $ f(x) \equal{} e^x$.


Let $R = gs - 4$.  What $S = 8$, $R = 16$.  What $S = 10$, $R$ is equal to:
\[ \text{(A)}\ 11 \qquad \text{(B)}\ 14 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 21 \qquad \text{(E) none of these} \]

[hide][color=green]the answer is 21, or D.[/color][/hide]

[hide]The first condition gives us that when S is 8,  R is 16, so when we substitute the values into the equation, we get
16 = 8g - 4,
g = 20/8 = 5/2, solving for g.
Now when S is 10, R is:
(5/2)(10) - 4,
or 21, which is [color=brown]D[/color][/hide]


inequalities
algorithm
inequalities proposed

[i]"If it is dumb and works, then it is not dumb."[/i] (Apocryphal Murphy law)

There seems to be a general method for killing inequalities like this one. I will demonstrate it at the example of your nice inequality (which surely has a much simpler proof).

Denote

$ f\left(a, b, c, d\right)$
$ \equal{} \left(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\right) \minus{} \left(a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 d^2 \plus{} d^2 a^2 \plus{} a^2 c^2 b^2 d^2\right)$.

We must show that $ f\left(a, b, c, d\right) \geq 0$ for all positive a, b, c, d. Now, since $ f\left(a, b, c, d\right)$ is obviously symmetric in a, b, c, d, we may WLOG assume that

  $ a\geq b\geq c\geq d$.

After some algebra,

  $ f\left(a, b, c, d\right) \minus{} f\left(a, b, c, c\right)$
$ \equal{} \left(c \minus{} d\right)\left(c \left(a^2 \plus{} b^2 \minus{} 2ab\right) \plus{} d\left(a^2 \plus{} b^2 \minus{} cd \minus{} d^2\right)\right)$.

Now, a well-known inequality states $ a^2 \plus{} b^2 \minus{} 2ab \geq 0$; also, from $ a\geq b\geq c\geq d$, we have $ c \minus{} d \geq 0$ and $ a^2 \plus{} b^2 \minus{} cd \minus{} d^2 \geq 0$. Thus,

  $ f\left(a, b, c, d\right) \minus{} f\left(a, b, c, c\right) \geq 0$,

and $ f\left(a, b, c, d\right) \geq f\left(a, b, c, c\right)$.

Another calculation gives

  $ f\left(a, b, c, c\right) \minus{} f\left(a, b, b, b\right)$
$ \equal{} \left(b \plus{} c\right)\left(b \minus{} c\right)\left(2a^2 \plus{} b^2 \minus{} 2ab \minus{} c^2\right)$.

Again, $ b \minus{} c \geq 0$ and $ 2a^2 \plus{} b^2 \minus{} 2ab \minus{} c^2 \geq 0$ because of $ a\geq b\geq c\geq d$. Therefore,

  $ f\left(a, b, c, c\right) \minus{} f\left(a, b, b, b\right) \geq 0$,

and $ f\left(a, b, c, c\right) \geq f\left(a, b, b, b\right)$.

Finally,

  $ f\left(a, b, b, b\right) \equal{} a \left(a \minus{} b\right)^2 \left(a \plus{} 2b\right)$,

what is obviously $ \geq 0$. Hence,

  $ f\left(a, b, c, d\right) \geq f\left(a, b, c, c\right) \geq f\left(a, b, b, b\right) \geq 0$,

qed..

Some comments about this solution:

Since $ f\left(a, a, a, a\right) \equal{} 0$, we could also write

  $ f\left(a, b, c, d\right) \geq f\left(a, b, c, c\right) \geq f\left(a, b, b, b\right)\geq f\left(a, a, a, a\right)$

instead of

  $ f\left(a, b, c, d\right) \geq f\left(a, b, c, c\right) \geq f\left(a, b, b, b\right)\geq 0$.

This would show the algorithm more clearly.

One thing more: If I were at the IMO, I wouldn't have written the full argument I wrote above, but I would just prove by calculation the identity

  $ f\left(a, b, c, d\right) \equal{} \left(c \minus{} d\right)\left(c \left(a^2 \plus{} b^2 \minus{} 2ab\right) \plus{} d\left(a^2 \plus{} b^2 \minus{} cd \minus{} d^2\right)\right)$
$ \plus{} \left(b \plus{} c\right)\left(b \minus{} c\right)\left(2a^2 \plus{} b^2 \minus{} 2ab \minus{} c^2\right) \plus{} a \left(a \minus{} b\right)^2 \left(a \plus{} 2b\right)$,

without giving any hints how I found this formula, just as if I had guessed it. Subsequently I would conclude that $ f\left(a, b, c, d\right) \geq 0$. The corrector surely would be fascinated about my ingeniousity and I'd get a special prize for an elegant solution. ;-)

[b]EDIT:[/b] The method of showing that $ f\left(a, b, c, d\right)\geq 0$ by subsequently passing from $ \left(a,b,c,d\right)$ to other quadruples such that the value of $ f$ becomes smaller and smaller on each step is the so-called [b]mixing variables[/b] technique of solving inequalities. Hence, in the above proof, I more or less discovered this method for myself. I am obviously not the first one to discover it; I could bet that the famous special prize solution of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55327]IMO 1983 problem 6[/url] was also found by this method.

  Darij

Dear Anh Cuong,

This is Turkevici's inequality... ;)

Let [tex]S=a^4+b^4+c^4+d^4+2abcd-(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)[/tex].

Then S is symmetric, so we can suppose that the biggest two numbers among [tex]a,b,c,d[/tex] is [tex]a,b[/tex].

Therefore

[tex]S=(ab+cd-c^2-d^2)^2+2cd(c-d)^2+(a-b)^2((a+b)^2-(c^2+d^2))\geq 0[/tex].

The above problem was used in KMO winter program, and In-seok Seo (2002 IMO) proposed the above solution :)


I have seen the term "subset ordering of sets" at http://en.wikipedia.org/wiki/Order_theory

What I can understand now is it is something related to the ordering of sets.

But I can't understand literally what "subset ordering of sets" means.
What is the subset, what are the sets, and how they relate to each other?

The page offers a definition: [quote]If a set A contains all the elements of a set B, then B is said to be smaller than or equal to A. [/quote]
That is, we consider $B$ to be smaller than (or equal to) $A$ precisely when $B$ is a subset of $A$. It is a rather natural ordering of sets.

Thanks, but I still not really understand the whole picture.
Can you please literally explain what is "subset ordering of sets"?
Very thanks=)

The concept can be broken down as follows.
1. Ordering
2. Ordering of sets
3. Subset ordering of sets.

1. Ordering is a relation between some (possibly quite abstract) objects that is [b]reflexive, antisymmetric, and transitive[/b] (these terms are explained by Wikipedia). We usually write $a\le b$ to indicate this kind of relation between objects $a$ and $b$. This need not have anything to do with usual inequalities.

2. A special case of part 1 arises when the objects that we are trying to order are subsets of the same set. For example, they could be intervals on the real axis, in which case we could order them by declaring that $I\le J$ when the length of interval $I$ is less than or equal to the length of interval $J$. According to this ordering, $[0,2]\le [1,6]$.

3. This is a special case of part 2. Again the objects to be ordered are subsets of the same set. But this time we consider a very special ordering: $A\le B$ if $A$ is contained in $B$. According to this ordering, $[0,2]\le [-1,4]$, but it is not true that $[0,2]\le [1,6]$.


geometry
trapezoid
similar triangles
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, P a point, P* the isogonal of P wrt ABC, Pb the perpendicular to AB at A, Pc the perpendicular to AC at A,
B" the meet point of Pb and BP, C" the meet point of Pc and CP
and A' the meet point of the perpendicular to BC through P* with BC.
Prove that AA' is perpendicular to B"C".
Sincerely
Jean-Louis

Hello!

Construct $ B"M\parallel{}AC$, $ M\in BC$ and $ \left\{S\right\}\equal{}AB"\cap BC$.
$ \frac{CM}{CS}\equal{}\frac{AB"}{AS}\equal{}\frac{AB\cdot tan\left(\angle ABP\right)}{AB\cdot tan\left(\angle B\right)}\equal{}\frac{tan\left(\angle A'BP^{*}\right)}{tan\left(\angle B\right)}\equal{}\frac{A'P^{*}}{A'B\cdot tan\left(\angle B\right)}\Rightarrow$

$ \Rightarrow CM\cdot A'B\equal{}\frac{A'P^{*}}{tan\left(\angle B\right)}\cdot CS\equal{}\frac{A'P^{*}}{tan\left(\angle B\right)}\cdot\frac{AC\cdot sin\left(90^{\circ}\minus{}\angle A\right)}{sin\left(90^{\circ}\minus{}\angle B\right)}\equal{}$

$ \equal{}\frac{A'P^{*}}{tan\left(\angle B\right)}\cdot\frac{AC\cdot cos\left(\angle A\right)}{cos\left(\angle B\right)}\Rightarrow CM\cdot A'B\equal{}A'P^{*}\cdot\frac{AC\cdot cos\left(\angle A\right)}{sin\left(\angle B\right)}$     (1)

Similarly construct $ C"N\parallel{}AB$, $ N\in BC$ and $ \left\{T\right\}\equal{}AC"\cap BC$
$ \Rightarrow BN\cdot A'C\equal{}A'P^{*}\cdot\frac{AB\cdot cos\left(\angle A\right)}{sin\left(\angle C\right)}$     (2)

(1), (2) $ \Rightarrow CM\cdot A'B\equal{}BN\cdot A'C\Rightarrow\frac{A'B}{BN}\equal{}\frac{A'C}{CM}$     (3)
$ \left\{V'\right\}\equal{}B"M\cap AA'$ and $ \left\{V"\right\}\equal{}C"N\cap AA'$
$ \frac{A'A}{AV"}\equal{}\frac{A'B}{BN}\equal{}\frac{A'C}{CM}\equal{}\frac{A'A}{AV'}\Rightarrow V'\equal{}V"\equal{}V$.
$ C"V\parallel{}AB$ and $ AB\bot AB"\Rightarrow C"V\bot AB"$ (4)
$ B"V\parallel{}AC$ and $ AC\bot AC"\Rightarrow B"V\bot AC"$ (5)
(4), (5) $ \Rightarrow$ the point $ V$ is the orthocenter of the triangle $ AB"C"\Rightarrow VA\bot B"C"\Rightarrow AA'\bot B"C"$

Best regards, Petrisor Neagoe :)

Let $ X,\ Z,\ Y$ be, the orthogonal projections of $ A,\ B'',\ C''$ respectively, on the side-segment $ BC,$ of the given triangle $ \bigtriangleup ABC.$

Also, le $ P'$ be ( instead of $ P^{*}$ ), the isogonal conjugate of $ P,$ wrt $ \bigtriangleup ABC.$

The lines through the points $ X,\ Y,\ Z$ and perpendicular to $ B''C'',\ AB'',\ AC''$ respectively, are concurrent at one point so be it $ Q,$ as well ( [b][size=100]Carnot theorem[/size][/b] ).

So, because of the similar triangles $ \bigtriangleup ABC,\ \bigtriangleup QYZ,$ with $ AB\parallel QY$ and $ AC\parallel QZ,$ it is enough to prove that $ AA'\parallel XX',$ which is equivalent to prove that $ \frac {A'B}{A'C} \equal{} \frac {XY}{XZ}.$

$ \bullet$ From the similar right triangles $ \bigtriangleup A'BP',\ \bigtriangleup ABB'',$ we have that $ \frac {A'B}{AB} \equal{} \frac {A'P'}{AB''}$ $ \Longrightarrow$ $ \frac {A'B}{A'P'} \equal{} \frac {AB}{AB''}\ \ ,(1)$

Similarly, from the similar right triangles $ \bigtriangleup A'CP',\ \bigtriangleup ACC'',$ we have that $ \frac {A'C}{A'P'} \equal{} \frac {AC}{AC''}\ \ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {A'B}{A'C} \equal{} \frac {AB}{AC}\cdot \frac {AC''}{AB''}\ \ ,(3)$

$ \bullet$ Based on the below [b][size=100]Lemma[/size][/b], we have $ \frac {AB}{AC}\cdot \frac {AC''}{AB''} \equal{} \frac {RC''}{RB''} \equal{} \frac {XY}{XZ}\ \ ,(4)$ $ ($ because of the trapezium $ YZB''C''$ $ ).$

Hence, from $ (3),\ (4)$ $ \Longrightarrow$ $ \frac {XY}{XZ} \equal{} \frac {A'B}{A'C}\ \ ,(5)$

From $ (5)$ and because of the similarity of the triangles $ \bigtriangleup ABC,\ \bigtriangleup QYZ,$ with $ AB\parallel QY$ and $ AC\parallel QZ,$ we conclude that $ AA'\parallel XX'$ $ \Longrightarrow$ $ AA'\perp B''C''$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given and let $ AD$ be, its $ A$-altitude. Let $ P$ be, an arbitrary point on $ AD,$ between $ A,\ D$ and we draw an arbitrary line through $ P,$ which intersects the lines through vertex $ A$ and perpendicular to $ AC,\ AB,$ at points $ E,\ F,$ respectively. Prove that $ \frac {PE}{PF} \equal{} \frac {AB}{AC}\cdot \frac {AE}{AF}.$[/color][/size][/b]

$ \bullet$ Because of the proof of the above [b][size=100]Lemma[/size][/b] is very easy, I post only the schema and I think it is enough.

Kostas Vittas.

Dear Kostas and Mathlinkers,
I found your approach (orthopole) very interesting. I ask me always the same question: can we obtain the result with a minimum of calculation?
Sincerely
Jean-Louis

I liked also the argument of [b][size=100]orthopole[/size][/b] dear [b][size=100]Jean Louis[/size][/b], but after the reading of the solution by [b][size=100]Petrisor[/size][/b], I think we can have a more easier proof of your proposed problem, without this argument and also without trigonometric relations.

So, we denote the points $ N'\equiv AC\cap NC'',\ M\equiv AB\cap MB''$ and applying the [b][size=100]Thales theorem[/size][/b] we have :

From $ AB\parallel NN'$ $ \Longrightarrow$ $ \frac {BN}{BC} \equal{} \frac {AN'}{AC}$ $ ,(1)$

From $ AC\parallel MM'$ $ \Longrightarrow$ $ \frac {CM}{CB} \equal{} \frac {AM'}{AB}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {BN}{CM} \equal{} \frac {AB}{AC}\cdot \frac {AN'}{AM'}$ $ ,(3)$

From the similar right triangles $ \bigtriangleup AN'C'',\ \bigtriangleup AM'B''$ we have that $ \frac {AN'}{AM'} \equal{} \frac {AC''}{AB''}$ $ ,(4)$

From $ (3),\ (4)$ $ \Longrightarrow$ $ \frac {BN}{CM} \equal{} \frac {AB}{AC}\cdot \frac {AC''}{AB''}$ $ ,(5)$

$ \bullet$ From the similarities of the triangles $ \bigtriangleup A'BP'\approx \bigtriangleup ABB''$ and $ \bigtriangleup A'CP'\approx \bigtriangleup ACC'',$ we have $ \frac {AB}{AC}\cdot \frac {AC''}{AB''} \equal{} \frac {A'B}{A'C}$ $ ,(6)$ ( please see the [b][size=100]post[/size][/b] #3# ).

Hence, from $ (5),\ (6)$ $ \Longrightarrow$ $ \frac {BN}{CM} \equal{} \frac {A'B}{A'C}$ $ \Longrightarrow$ $ \frac {A'B}{BN} \equal{} \frac {A'C}{CM}$ $ ,(7)$

From $ (7),$ based on the [b][size=100]Thales theorem[/size][/b], we conclude that the line segment $ AA'$ passes through the point $ V\equiv NC''\cap MB''.$

Because of now, the point $ A$ is the orthocenter of the triangle $ \bigtriangleup VB''C'',$ we conclude that $ VA\equiv AA'\perp B''C''$ and the proof is completed.

Kostas Vittas.

[quote="jayme"]Dear Mathlinkers,
after toying with my programm,
let A* be the meetpoint of BC" and CB".
Prove that A*, A and P* are collinear.
Sincerely
Jean-Louis[/quote]
We can prove this result dear [b][size=100]Jean Louis[/size][/b], as a direct application of the [b][size=100]Isogonic theorem[/size][/b].  :wink:

Kostas Vittas.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 133
Sincerely
Jean-Louis


geometry
3D geometry
tetrahedron
geometry proposed

Let a tetrahedron $ ABCD$. Find $ M,N$ in $ AB,CD$ so that all planes over $ M,N$ cut $ ABCD$ become $ 2$ constant volume parts.

If $ M$ midpoint of $ [AB]$ and $ N$ midpoint of $ [CD]$, then the 2 volume is =, for all planes $ \pi; M,N\in\pi.$


I'm really not sure how looks the correct statement of a problem, because the inequality is not true for example for $n=3$, $a=b=c=1/3$. But I think that if Cezar meant $n(n-1)$ instead of $n(n+1)$ I got it. Here is my proof:

Denote $a_i=\frac{1}{x_i+1}$. Then $a_i \leq 1$, $\sum a_i \geq n-1$ and $x_i=\frac{1-a_i}{a_i}$.

So we have to prove that:
$\sum \frac{a_i}{1-a_i} \geq n(n-1)$
But using the Cauchy-Shwarz, the inequality $\sum a_i^2 \geq \frac{(\sum a_i)^2}{n}$ and the condition gives us:
$\sum \frac{a_i}{1-a_i} = \sum \frac{a_i^2}{a_i-a_i^2} \geq \frac{(\sum a_i)^2}{\sum a_i - \sum a_i^2} \geq \frac{(\sum a_i)^2}{\sum a_i - \frac{(\sum a_i)^2}{n}} = \frac{\sum a_i}{1-\frac{\sum a_i}{n}} \geq \frac{n-1}{1-\frac{n-1}{n}}=n(n-1)$
QED.

nice solution TomciO ,(it was the same that mine :lol: )


algebra
difference of squares
special factorizations
AMC

[hide="Solution"]The given expression equals
$ \frac{9^{2008}\minus{}9^{2006}}{9^{2007}\minus{}9^{2005}}$
Factoring, we have
$ \frac{9^{2006}}{9^{2005}}\cdot \frac{9^2\minus{}1}{9^2\minus{}1}\equal{}9 \Rightarrow \boxed{E}$[/hide]


The world would definately be a different place... my question, for worse? Or better?

Is this even a question?

edit: is this meant to imitate the below "If America didn't go to Iraq...." post?

[quote="ursula"][quote="bubka"]Of course it is a spoof of that.
By the way, the US did not join the war out of pity for the jews - they did it because there submarines were being attacked. So it was not really a choice, you see.[/quote]

Indeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.

No country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war

Ursula[/quote]
Well the holocaust was by large unknown until the advance into germany started happening.

[quote="ursula"][quote="bubka"]Of course it is a spoof of that.
By the way, the US did not join the war out of pity for the jews - they did it because there submarines were being attacked. So it was not really a choice, you see.[/quote]

Indeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.

No country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war

Ursula[/quote]

Amendment of the Neutrality Act, Lend-Lease, "Four Freedoms", Atlantic Charter, [i]then[/i] Pearl Harbor.

hi there but people, FDR was against Germany even before WWII had started.

And...

Great Britain headed the war against the U-Boats, and decrypted various Enigma messages throughout world war II and was able to read coded messages within hours of transmission which is very good

also, the enigma could have 15 billion+ ways to scramble a single letter so...

[quote="ursula"]

Indeed, the US joins the WWII after Pearl Harbor and launches all the Pacific campaigns. However, the jews weren't actually in China or something. The US doesn't really enters, except for supplies, some regiments, you know, the war big time until Normandy and I believe the reason mainly is that they foresaw the inevitability of the soviet victory over germany and they had to keep a portion of the western world.

No country behaved even acceptably morally speaking in WWII. However, it was the USSR who turned the tide and I'd say won 70% of the war

Ursula[/quote]

ussr wouldn't be able to win a victory if usa hadn't helped them (remember who told the soviets that the japan has no intentions in attacking ussr so stalin sent his elite divisions to europe) and don't forget all the armory that was sent to moscow

ussr hasn't won 70% of the war

btw, remember that hitler and stalin made an agreement about poland and borders in eastern europe

[quote="deimos"]
ussr wouldn't be able to win a victory if usa hadn't helped them (remember who told the soviets that the japan has no intentions in attacking ussr so stalin sent his elite divisions to europe) and don't forget all the armory that was sent to moscow[/quote]

A lot of weapons were sent to the USSR, however three times more were sent to Britain. I don't think you are correct in thinking the USSR wouldn't be able to win a victory, without the US. Sorry but the men in the front were the soviet ones, and they were the ones using those weapons you mention which I don't think represent a major fraction of the weapons the USSR was using. Anyway, this is Lend Lease and I reckon the importance of it, that pretty much makes the 30% I'm conceding to the other allies. I don't know according to you who told the soviets japan had no intentions in attacking the ussr, to my knowledge this information was provided by Richard Sorge, prominent spy of the soviet secret services.

[quote="deimos"]btw, remember that hitler and stalin made an agreement about poland and borders in eastern europe[/quote]

I remember deimos, however one thing has nothing to do with the other. I also said above that no country behaved acceptably morally speaking. I'm not saying Stalin did the best things of his life during WWII. I'm saying the USSR won the war, I'm saying that the D-Day the war was won and the allies knew better than not coming and keeping a portion of the western world. If they hadn't they would have learned russian in Paris.

Ursula

Edited: I was kind kind of pissed of at the world last night and was a bit rude.

I think it's an interesting question whether the United States would have entered the war in Europe, had Hitler not declared war after the U.S. declared war on Japan.

While I'm always up for a little left-wing revisionist history, the idea that the U.S. entered the war in Europe to stop the Soviet Union, only a week after the Germans gave up their assault on Moscow, is a strained interpretation at best.  I think a better case can be made that stopping the Soviet Union was one of the reasons we nuked Japan.

What appears to be a good timeline of the war is available [url=http://www.historyplace.com/worldwar2/timeline/ww2time.htm]here[/url].

[quote="ursula"][quote="Piggypi314"]
Amendment of the Neutrality Act, Lend-Lease, "Four Freedoms", Atlantic Charter, [i]then[/i] Pearl Harbor.[/quote]

I don't understand. Is this supposed to be an argument against the claim that the US entered WWII after Pearl Harbor? Because I don't get it, of the four things you mention, only the Lend Lease program influenced directly and significantly in the course of the war - it's actual significance might be a bit exaggerated but nonetheless -. And with a world scale war going on, surely you are not telling me that the US approving to loan stuff to the allies marks its entrance in the war. Four Freedoms, do you mean that stating some basic freedoms you think people should have is anyhow historically as important as the Pearl Harbor attack and the events that broke out? The US politics those days was of course, shaped by the international conflict but I think the statement remains, the US joins WWII after Pearl Harbor.

Ursula[/quote]

Have you read the actual speech?  The whole thing is like "Hello.  These are four wonderful freedoms.  WE MUST GO TO WAR to protect these wonderful freedoms.  WE WILL GO TO WAR AND WE WILL WIN."

Of course, the US officially joined after Pearl Harbor, meaning Congress voted and yadda yadda.  But you should read up on undeclared wars, for fun if not for anything else.

[quote="JBL"]
While I'm always up for a little left-wing revisionist history, the idea that the U.S. entered the war in Europe to stop the Soviet Union, only a week after the Germans gave up their assault on Moscow, is a strained interpretation at best.  [/quote]

I didn't knew the  germans abandoned the assault on Moscow on june, 1944 and I don't think so. Are you sure about this? Because, as it appears in the link you gave us, on january 1944, the ussr already was advancing into poland, out of their borders. The front line was very far from Moscow by then, the siege of Leningrad was over and there was also the ukrainian front who took up Ukraine, all this happened before Normandy. I'd say that the soviets in june 1944 were knocking at the doors of eastern europe with inexorable drive.

However, I understand you can't make a good case of my opinion and it can be bashed because of the lack of supporting evidence. But the fact that the allies, with the US being essential - without the US there is no D day - enter Europe a bit en retard I think is clear. And although I'm always up for a little left-wing revisionist history myself, I think I'm not trying to do so right now, in the sense I'm not praising left-wing side actions, well may be so, but my intention rather than giving credit to left-wing as opposed to right-wing is giving credit to the other pole that existed as opposed to the only pole that survived. I think history in our schools is taught to minimize the role of the USSR in WWII. The landing on Normandy is presented as one of the greatest victories of humanity, which I think it is, but there are three or four victories by the soviets at least comparable to Normandy of which you are told very little, ok, maybe it didn't happened in your school. If you don't set up to find out yourself, you learn almost nothing about Kursk or Stalingrad or Moscow or the siege of Leningrad or even Berlin.

Totally unrelatedly, this also happens with physics. The physics made in the USSR was truly amazing, however most of it remained unknown in the west until someone rediscovered it. Then they received all the credit and many times this is what we know today. Names like Feynman, Gell-Mann, Wilson, Weinberg, I could go on, are all well known even to the broad public. However, russians like Landau, Zel'dovich, Bogoliubov, Ginzburg, Kapitsa, etc are much less well known although their contributions are truly magnificent aswell, in many ocasions, as is the case with Wilson and Bogoliubov some contributions were very related. I had a professor who told me about a conference he assisted where the speaker was explaining his result and there was this russian guy who stood up to say that all he was saying actually was known 40 years ago in goo ol' UUSR and was discovered by Landau. Of course, you had to go to some russian journal and read it in russian to be aware. Of these soviet physicists you got to know what was so outstanding that the world [i]had to know[/i] but no more.

[quote="Piggypi314"]
Have you read the actual speech? The whole thing is like "Hello. These are four wonderful freedoms. WE MUST GO TO WAR to protect these wonderful freedoms. WE WILL GO TO WAR AND WE WILL WIN."
Of course, the US officially joined after Pearl Harbor, meaning Congress voted and yadda yadda. But you should read up on undeclared wars, for fun if not for anything else.
[/quote]

I've read it diagonally, I think I've read enough and Hello, you keep missing the point. Roosevelt saying 'we will go to war and we will win' [i]is not[/i] going to war and winning. Others really [i]were[/i] at war and had not been for Pearl Harbor one could ask when the US would have actually entered the war. And above all, why are you telling me this? I say US doesnn't enters the war until Pearl Harbor and I also recognize the importance of Lend Lease, are you saying there is something absolutely inaccurate in this statement? What undeclared war are you speaking of? What the US was doing at least cualitatively is known and was somewhat known at the time as Mr Roosevelt comes to remind us, and I'm saying that it certainly wasn't fighting a war. Right until Pearl Harbor.

Ursula

[quote="ursula"]I didn't knew the  germans abandoned the assault on Moscow on june, 1944 and I don't think so. Are you sure about this? Because, as it appears in the link you gave us, on january 1944, the ussr already was advancing into poland, out of their borders. The front line was very far from Moscow by then, the siege of Leningrad was over and there was also the ukrainian front who took up Ukraine, all this happened before Normandy. I'd say that the soviets in june 1944 were knocking at the doors of eastern europe with inexorable drive. [/quote]
Whoa, whoa, 194[b]4[/b]?  The United States entered the war in Europe shortly after Germany declared war at the end of 194[i]1[/i].  By the time of the D-Day landings and Normandy in 1944, there were American troops in Rome, among other places.  From my earlier link: [quote="[url=http://www.historyplace.com/worldwar2/timeline/ww2time.htm#1941]historyplace.com[/url]"]Dec 5, 1941 - German attack on Moscow is abandoned.
Dec 6, 1941 - Soviet Army launches a major counter-offensive around Moscow.
Dec 7, 1941 - Japanese bomb Pearl Harbor; Hitler issues the Night and Fog decree.
Dec 8, 1941 - United States and Britain declare war on Japan.
Dec 11, 1941 - Germany declares war on the United States.
...
Jan 26, 1942 - First American forces arrive in Great Britain.
[/quote]

[quote="JBL"]
Whoa, whoa, 194[b]4[/b]?  The United States entered the war in Europe shortly after Germany declared war at the end of 194[i]1[/i].  By the time of the D-Day landings and Normandy in 1944, there were American troops in Rome, among other places.  From my earlier link:
[/quote]

OK sorry, I misunderstood you, I already mentioned the presence of American troops before D-Day in Europe, I just thought you were referring to Normandy since I still think represents the turning point in the other allies side - the turning point for the soviets had ocurred long ago as we seem to agree now. However what I thought was clearly permeated by my ignorance on the subject, I just checked some numbers and I was wrong, I thought the campaigns in the Mediterraneum (italy and north africa) were largely conducted by the british and that's not true, the US contribution was at least comparable to the british one. So  I was wrong and will have to study my history again :) So bad you were right so quickly, :lol: it removes the fun and I stop learning
With no intention to keep the argument because it was incorrect I might nevertheless say that anyhow the greatest enemy by far of the war was Germany. Italians weren't serious about the war and was comparably easy to overpower them. And I still think the war with Germany was essentially won by the USSR.

Ursula

[quote="kstan013"][quote="ursula"]However, it was the USSR who turned the tide and I'd say won 70% of the war

Ursula[/quote]Really? Who stormed the beaches of Normandy? Who (I really hate to say this) dropped the nukes on Hiroshima and Nagasaki? The Soviets really didn't have that great of a strategy.[/quote]

Nukes and strategy? Oh, please... Don't be silly. If Soviets were in the position in which USA was, the nukes would never been used, you can be sure of that!
And remember that every history book states that the crucial turn in WWII was the break of Stalingrad siege!

BTW, did you know that the US army losses in WWII were smaller than the losses of Yugoslav army?

Remember one thing: Soviets lost 10,000,000 soldiers and 12,000,000 civilians in WWII. The greatest war toll in history.

[quote="hsiljak"][quote="solafidefarms"]Was it the US's subs or their merchant vessels that were being attacked?[/quote]

Merchant vessels? That was in WWI... USA entered the WWII after the Japanese attack on the Pearl Harbor, if they thought me good in my history classses...[/quote]
And the Black Tom explosion! :ninja:

[quote="hsiljak"][quote="kstan013"][quote="ursula"]However, it was the USSR who turned the tide and I'd say won 70% of the war

Ursula[/quote]Really? Who stormed the beaches of Normandy? Who (I really hate to say this) dropped the nukes on Hiroshima and Nagasaki? The Soviets really didn't have that great of a strategy.[/quote]

And remember that every history book states that the crucial turn in WWII was the break of Stalingrad siege!

BTW, did you know that the US army losses in WWII were smaller than the losses of Yugoslav army?

Remember one thing: Soviets lost 10,000,000 soldiers and 12,000,000 civilians in WWII. The greatest war toll in history.[/quote]1) [i]crucial[/i] turn? yes, definitely a turn, but I don't know if it was the crucial hit.
2) Maybe they fought better and were less prone to be killed
Another thing is who *ended* the war. The Soviets take more credit than they deserve after completely taking control of Eastern Europe in the 1950's to the 1980's, that was a complete act of ruthlessness.

If you are going to say the nukes or the invasion of normandy was the crucial turn then you defiantely have no idea what you are talking about

Also America is the country that takes more credit then it deserves the war would be over if they hadn't joined it would have just taken a while longer

ps the bombs were not needed, the fact was the Soviets were freeing China faster then France surrendered, and the generals of Japan said they would have continued fighting after the bombs and did until they saw that there was no hope in China

And about fighting better yeah everyone fights better when they can just bomb from across the ocean, they did not even have to send troops unlike yugoslaiva which had to fight an invasion

Only thing in Europe that I can say America actually did well and before the war was determined was Africa and Italy but in the first case it was mostly Britian still and in Italy the Italians were the ones who actually overthrew Mussolini, America was mostly the supplier,

Also your last point, America shot, bombed, and well attacked Soviet troops to delay them in Germany so the United states could take over the lands, if that is not ruthless then I have no clue what is

Finally the dividing of Germany was America's idea the soviets said that it should be kept whole and did not care for it at all

Finally about strategy America had none what so ever it was kill everthing on your path to Berlin and had no way of freeing China

They also told the Soviets to free Korea but then said no don't and came back there

If you have anything else about strategy to talk about go ahead but I believe you have no clue what you are saying

This of Course is not to commend Stalin's actions after the war in any way

That's the power of money right there. Where did the money come from? A great army. Where did the great army come from? Money. It's a never-ending cycle.

Yeah if America was in Europe I doubt you would be talking about the money cycle now

America did not have a great Army in comparison to the one Germany raised in its depression

The only Reason America had money was because it used its isolation to declare nuetrality and sell things

Yes now I agree its army is great but of course there is the fact it is all in Iraq

At the time its army was worse then almost all other powers armies

The money came from isolation and industry nothing to do with the army

It was not the power of money but of isolation

and you still have not answered any of my points

[quote="ursula"]I don't know according to you who told the soviets japan had no intentions in attacking the ussr, to my knowledge this information was provided by Richard Sorge, prominent spy of the soviet secret services.[/quote]
he was working for soviets?! :o sorry, i didn't know that :blush:

[quote="hsiljak"]And remember that every history book states that the crucial turn in WWII was the break of Stalingrad siege!
[/quote]
[quote="hsiljak"]Remember one thing: Soviets lost 10,000,000 soldiers and 12,000,000 civilians in WWII.[/quote]

soviets had a great lost of soldiers in stalingrad battle because stalin wanted a glorious victory, and he didn't care about soldiers life.
do you know what soviet general zukov [as far as i have heard] have said about that battle (concerning stalin and hitler)

[quote="deimos"]do you know what soviet general zukov [as far as i have heard] have said about that battle (concerning stalin and hitler)[/quote]

No, what?

BTW, Russians lost 800,000 men (both soldiers and civilians) in Stalingrad battle. That's less than 5% of all their losses.

[quote="anirudh"][quote="bubka"]The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.[/quote]

OKAY. SO IF SOME DUDE GOES AND DOES SOME GREAT FEAT, THEN HE IS EXEMPT FROM ANY BAD STUFF HE DOES?

Also, Turing was not the only person behind it? True that he was the one who designed the Turing Bombe, but Welchman and Winterbotham were also very involved in it. :wink:
\[\ast\qquad\ast\qquad\ast\qquad \]
Also, the US won a predominant part of WWII, especially in Japan. The man who cracked Purple did not even know how purple was like and stuff.

Also, the US was helping UK for a long time before it entered the war.

Furthermore, would anyone in their right mind seriously join someone who had just bombed them in some place killing 3403 people?

Also, as of 1942, the US read a lot of enigma messages using the US Navy Cryptanalytic Bombe by Joe Desch. :wink:

Next Up: Who dropped the atomic bomb?[/quote]
The atomic bomb drop is regarded more of as a bad thing rather than a good one (okay, so granted maybe it stopped the Soviet Union from invading more territory, but that's not really that important). Japan was already losing the fight against the American pacific force, and if France and England would direct all their forces to French Indochina and India, in addition with the Soviet war machine turning east, then the war would've swiftly been won anyways. The Atomic bomb killed thousands of innocent civilians; its an example of terrorism on an enormous scale.

And do you consider homosexuality a crime punishable by death? What did Turing's sexual orientation ever do to anyone?

In any case, America did do something, of course. For example, if 1,000 pounds of food need to be outputted by a factory, and I outputted 100 while someone else outputted 400 and another outputted 550, then I could claim that we wouldn't have met the quota without me, but it doesn't make me a predominant player.

[quote="bpms"][quote="anirudh"][quote="bubka"]The man behind the cracking of the enigma was Alan Turin, and he was later persecuted by the state because of his 'illegal' homosexuality and driven to suicide. So Great Great Britain has nothing to boast of mind you.[/quote]

OKAY. SO IF SOME DUDE GOES AND DOES SOME GREAT FEAT, THEN HE IS EXEMPT FROM ANY BAD STUFF HE DOES?

Also, Turing was not the only person behind it? True that he was the one who designed the Turing Bombe, but Welchman and Winterbotham were also very involved in it. :wink:
[/quote]
Actually, turing designed the collosus mark I which cracked the enigma code. That was his major accomplishment.[/quote]

the Colossus was to recover key settings for the Lorenz :wink:

The Lorenz was used by Hitler and the German High Command

the bombe was what helped crack enigma :wink:


Is anybody aware of a definition of generalized Carmichael numbers using a modified Korselt criterion

n is called r-Carmichael number,
if n is squarefree and if for all primes p with $ n \equiv r \mod(p)$ then $ n \minus{} 1 \equiv r \mod(p \minus{} 1)$

The normal Carmichael numbers would be 0-Carmichael numbers in the above sense.
Are there any composite r-Carmichael numbers for $ a\not \equal{} 0$ ?

Is anybody aware of a definition of generalized Carmichael numbers using a modified Korselt criterion

n is called r-Carmichael number,
if n is squarefree, n-r composite and if for all primes p with  $ n\equiv r\mod(p)$ then $ n\minus{}1\equiv r\mod(p\minus{}1)$ then

The normal Carmichael numbers would be 0-Carmichael numbers in the above sense.
Are there any composite r-Carmichael numbers for  $ a\not\equal{}0$


function
linear algebra
matrix
logarithms
calculus
integration

For any positive integer $ n$, let \[f(n) \equal{} \begin{cases} \log_8{n}, & \text{if }\log_8{n}\text{ is rational,} \\
0, & \text{otherwise.} \end{cases}\] What is $ \sum_{n \equal{} 1}^{1997}{f(n)}$?

$ \textbf{(A)}\ \log_8{2047}\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ \frac {55}{3}\qquad \textbf{(D)}\ \frac {58}{3}\qquad \textbf{(E)}\ 585$

[hide="Click for solution"]
Since $ \log_8n\equal{}\frac{1}{3}(\log_2n)$, it follows that $ \log_8n$ is rational iff $ \log_2n$ is rational.  The nonzero numbers in the sum will therefore be all numbers of the form $ \log_8n$, where $ N$ is an integral power of $ 2$.  The highest power of $ 2$ that does not exceed $ 1997$ is $ 2^{10}$, so the sum is:

$ \log_81\plus{}\log_82\plus{}\log_82^2\plus{}\log_82^3\plus{}\ldots\plus{}\log_82^{10}\equal{}0\plus{}\frac{1}{3}\plus{}\frac{2}{3}\plus{}\frac{3}{3}\plus{}\ldots\plus{}\frac{10}{3}\equal{}\frac{55}{3}$, or $ \boxed{\textbf{(C)}}$.
[/hide]


my solution

Because \[x^{2}y^{2}(x^{2}+y^{2})=2x^{2}y^{2}(2-xy) \leq 2 \Longleftrightarrow (xy)^{3}-2(xy)^{2}+1 \leq 0\]  \[\Longleftrightarrow xy(xy-1)^{2}+(1-xy) \geq 0\] Since $x+y=2,~x+y \geq 2\sqrt{xy}\leftrightarrow 1-xy \geq 0$, therefore, $x^{2}y^{2}(x^{2}+y^{2}) \leq 2$.


[quote="Virgil Nicula"]Given are the natural number $ n\ge 2$ and the real positive numbers $ x,y,a,b$ . Prove that

$ \boxed {\ \frac {a^nx^{n \plus{} 1} \plus{} b^ny^{n \plus{} 1}}{(x \plus{} y)^{n \plus{} 1}}\ge\left(\frac {ab}{a \plus{} b}\right)^n\ }$ . Particularly, $ \boxed {\ \left\{\begin{array}{c} x \plus{} y \equal{} 1 \\
\ a \plus{} b \equal{} 1\end{array}\right|\ \Longrightarrow\ \frac {x^{n \plus{} 1}}{a^n} \plus{} \frac {y^{n \plus{} 1}}{b^n}\ge 1\ }$ .

See [url=http://www.mathlinks.ro/viewtopic.php?t=193348]here[/url][/quote]
By Holder inequality, we have:
$ \left(\frac{1}{a}\plus{}\frac{1}{b}\right)^n\left(a^nx^{n \plus{} 1} \plus{} b^ny^{n \plus{} 1}\right) \ge \left(x\plus{}y\right)^{n\plus{}1}$
So we are done!


Is there any easy solution without expanding to show :
$ \minus{}\sum \frac{a^3}{(a\minus{}b)(c\minus{}a)}\equal{}a\plus{}b\plus{}c$


Let k and n be positive integers:$\frac{n}{2}<k\leq\frac{2n}{3}$Find the least number m such that we can place each of m pawns on a squares of an n*n chessboard so that no column and row contains a block of k adjacent unoccupied squares

This kind of problem is well-known.For example,see IMO3-1999


[quote="jensen"][quote="mohamadhosein"]HADAFE SHOMA AZ IN ZENDEGI CHIYE? :?:[/quote]
mishe yek kami bishtar tozih bedi?[/quote]

manzooreshon ineh keh, har ensan e salem ye hadafi too zendegish dareh.

masalan hadafe to gold medal of imo e, amma omran be in arezoo beresi. ;)  :D

fekr konam soalam be andazeye kafi vazeh bashe.hammon chizi ke ashegh gofte.
baraye che hadafi darin talash mikonin?


[quote="misan"]show:

$\textbf{(i)}\;\;\;\prod_{k=2}^{\infty}\;\left( 1+\frac{2}{k}\right)^{(-1)^{k}k}\;=\;\boxed{\frac{6}{\pi e}}$

$\textbf{and\;\;\;(ii)}\;\;\;\prod_{k=1}^{\infty}\;\left( 1+\frac{2}{k}\right)^{(-1)^{k+1}k}\;=\;\boxed{\frac{\pi}{2 e}}$

 :D[/quote]
some idea:
Stirling forever :
by stirling we also can find asymptotic of this expression:
$(2n)!!$~...
and
$(2n+1)!!$~...
try to find it and your problem will be solved
it's easy


calculus
integration
MATHCOUNTS
modular arithmetic
search
function

How do you find the smallest integral x for simultaneous congruences?
(i.e. x congruent to 2 (mod 3) and also congruent to 3 (mod 4) I'm trying to find a systematic way.

I came upon this problem in problem 9 of Warm-Up 10 (2005-2006) MATHCOUNTS School Handbook:

A college class has exactly enough students to from eight equal rows. On Monday a student is absent, and the professor is able to seat the students into five equal rows. On Tuesday two students are absent, and the professor can seat the students into nine equal rows. What is the least possible number of students in the class?

I have:
x congruent to 0 (mod 8)
x-1 congruent to 0 (mod 5)
x-2 congruent to 0 (mod 9)

You can use certain techniques to limit it, but you can't actually find it without some testing (most of the time). For example, you know that it's a multiple of 8. Next, you know that it's one more than a multiple of 5. If this was an even multiple of 5, then the number would end in a 1, which is impossible, so you know that it's a multiple of 8 that ends in a 6. Finally, you know that it is two more than a multiple of 9. For all multiples of 9, the sum of the digits is a multiple of 9, so this must add to 11. (20 and higher are too high, because you need a three-digit number for that.)

So, in this case, you can actually get the answer right off. It ends in a 6, and the digits sum to 11, so the answer is [hide]56[/hide].

[quote="hillarryous"]How do you find the smallest integral x for simultaneous congruences?
(i.e. x congruent to 2 (mod 3) and also congruent to 3 (mod 4) I'm trying to find a systematic way.

[/quote]

You have

$x\equiv2\pmod3$

$x\equiv3\pmod4$.

You add one to both sides to get

$x+1\equiv0\pmod3$

$x+1\equiv0\pmod4$.

That must mean that $x+1\equiv0\pmod{12}$, $x\equiv11\pmod{12}$.

In general modular congruences can be solved with the Chinese Remainder Theorem.  I think it's a bit too advanced to put in this forum (and I'm too lazy to), but if you use the search function you can probably find some stuff about it.


Find root of this equation: $ 7^{x \minus{} 1} \equal{} 6.log_{7}(6x \minus{} 5) \plus{} 1$

[quote="azjps"][hide]
Let $ z \equal{} x \minus{} 1$, and the equation becomes $ \frac {7^z \minus{} 1}{6} \equal{} \log_7 (6z \plus{} 1)$. But the LHS and RHS are inverse functions (if $ f(z) \equal{} \frac {7^z \minus{} 1}{6}$, then $ f^{ \minus{} 1}(z) \equal{} \log_7 (6z \plus{} 1)$), so equality comes when $ f(z) \equal{} \frac {7^z \minus{} 1}{6} \equal{} z \Longrightarrow z \equal{} 0, 1$, or $ x \equal{} \boxed{1,2}$.
[/hide][/quote]

[hide]Oh ! It's nice. Thank you. :P [/hide]


geometry
geometric transformation
reflection
invariant
incenter
circumcircle
perpendicular bisector

[color=green]I tried to invent a transformation like isogonal and isotomic transformations  :blush: . I thought that if Isogonal keep invariant incenter and
isotomic keep invariant baricenter then there's a tranformation like
those that keep invariant the circuncenter (I was searching some result
like the concurrence). So I got this results  :( .

   Given $ \triangle ABC$ and given a point $ P$ in its plane, let be $ O$ the
circumcenter of $ \triangle ABC$ and $ P_{a}$, $ P_{b}$, $ P_{c}$ the simetrics of $ P$ respect the
perpendicular bisector of sides $ \overline{BC}$, $ \overline{AC}$, $ \overline{AB}$ respectly. Then

   $ \blacksquare$ $ O$ is the circumcenter of the triangle $ P_{a}P_{b}P_{c}$.
   $ \blacksquare$ $ P$ lies on circumcicle of $ P_{a}P_{b}P_{c}$.
   $ \blacksquare$ $ \triangle P_{a}P_{b}P_{c}\sim \triangle ABC$.
   $ \blacksquare$ Let $ M, N, P$ the midpoints of sides $ \overline{BC}$, $ \overline{AC}$, $ \overline{AB}$ and let be $ M'$, $ N'$, $ P'$
the midpoints of $ \overline{P_{b}P_{c}}$, $ \overline{P_{a}P_{c}}$, $ \overline{P_{a}P_{b}}$, then ,obviusly $ \triangle MNP$ is similar to
$ \triangle M'N'P'$. Also $ \triangle MNP$ and $ \triangle M'N'P'$ are in perspective.

   I also got some others results but I'd like to anybody helps me to
find out a very interesting result there.

   Thanks a lot[/color]

[quote="darij grinberg"]All of your results are true, and I found particularly the last one very interesting. I have given proofs of these results now in my paper "Isogonal conjugation with respect to a triangle", which can be found as an attachment at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=569871#569871]http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 post #6[/url] and on [url=http://de.geocities.com/darij_grinberg/]my website[/url] (Theorems 10 and 11).

  Darij[/quote]

Your proof is very nice, my proof is longer than yours  :blush:

[quote="hucht"]Here is another result

Given $A, B, C \in l$ and a point $P \not \in l$. Let $A'$, $B'$, $C'$ the circumcenters of $\triangle PBC$, $\triangle PCA$, $\triangle PAB$ respectively. Then:

$\bigstar$ The orthogonal proyections of $P$ on the sides of $\triangle A'B'C'$ are collinear.
$\bigstar$ $P$ lies on the circumcircle of $\triangle A'B'C'$.

Sould anybody post anothers??[/quote]

Why did you post this in this topic? Actually, it's a different problem, I don't see a direct relation... It's ok if I move it into a new topic?

  darij