Let be given a real number x and a clause A:
$[x[xn]]+1=[x^2n]\forall n=1,2,3,4,...$
1)$x=\frac{\sqrt{5}+1}{2}$,prove that A is right
2)A is right prove that 1<x<2

We have $[x^2n]=[x[xn]+x\{xn\}]=[x[xn]]+[x\{x[xn]\}+x\{xn\}]$. Therefore A equavalent:
(3) $1\le x\{xn\}+\{x[xn]\}<2$.
It give $A\Longrightarrow 1<x<2, x- irrational.$
Let $x=\frac{P_k}{Q_k}+\epsilon _k, |\epsilon _k|<\frac{1}{Q_kQ_{k+1}}$
From (3) we have (1): $x=\frac{1+\sqrt 5}{2}$.
If $x=\frac{1+\sqrt 5}{2}=y+1$, then $x\{xn\}+\{x[xn]\}=\{xn\}+y\{xn\}+\{y(xn-\{xn\})\}=1+\{xn\}.$
Therefore A equavalent 1).


Can you solve the interesting logic puzzle in the attached file?

Enjoy!

yup i got it :D
it was fun

[size=92][color=red]moved to puzzles/brainteasers[/color][/size]


An $m\times n\times p$ rectangular box has half the volume of an $(m+2)\times(n+2)\times(p+2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$  What is the largest possible value of $p$?

Anyone have a solution? This is one HARD problem...

[quote="du0_hell"]
since $m,n$ is an integer, $\frac{p+2}{p-2}$ have to be an integer also
$\frac{p+2}{p-2}=1+\frac{4}{p-2}$ meaning $\frac{4}{p-2}$ have to be an integer
thus the maximum value of $p$ that satisfies is 6[/quote]

I dont think this is true

Unfortunately, that is not the correct answer (according to Kalva) :(

seamusboyle is correct: the conditions $x = y\cdot z$, $x, z$ integers, do not suffice to show that $y$ is an integer.

Is this problem really that hard?
We have
$ (1 \plus{} \frac {2}{p})(1 \plus{} \frac {2}{m} \plus{} \frac {2}{n} \plus{} \frac {4}{mn}) \equal{} 2$
Since both factors are greater than 1, each factor is less than 2. We want to maximize the second factor and minimize the first, thus maximizing p.
We have $ 1 \plus{} \frac {2}{m} \plus{} \frac {2}{n} \plus{} \frac {4}{mn} < 2$, or $ (m \minus{} 2)(n \minus{} 2) > 8$.
So we try m=5,n=5 and m=3, n=11. Subbing in gives p=98 or 130, so 130 is the answer.

[quote="Ihatepie"]Is my solution rigorous enough? (If not, can someone help me complete it?)

[hide="Solution"]From what we are given, we know $ 2mnp \equal{} (m \plus{} 2)(n \plus{} 2)(p \plus{} 2)$
Rearranging, we get $ \frac {(m \plus{} 2)(n \plus{} 2)}{2mn} \equal{} \frac {p}{p \plus{} 2}$
Because $ p$ is always less than $ p \plus{} 2$ and all of $ m$, $ n$, and $ p$ are integers, $ 2mn$ has to be greater than $ (m \plus{} 2)(n \plus{} 2)$ by a positive integer, let's call it $ x$. We have $ 2mn \minus{} x \equal{} mn \plus{} 2m \plus{} 2n \plus{} 4$ or $ (m \minus{} 2)(n \minus{} 2) \equal{} 8 \plus{} x$. To maximize $ p$, we have to minimize $ x$, so we set $ x$ at its minimum, or $ 1$. We now have $ (m \minus{} 2)(n \minus{} 2) \equal{} 9$. This makes $ (m,n)$ either $ (5,5) \equal{} \frac {49}{50} \equal{} \frac {98}{100}$, $ p \equal{} 100$, or $ (3,11) \equal{} \frac {65}{66} \equal{} \frac {130}{132}$, $ p \equal{} 130$.  Because the latter $ p$ value is bigger, our answer is
$ 130$. [/hide]

The main part I'm not sure about is why minimizing $ x$ maximizes $ p$.[/quote]

Solution looks fine...

As for your question, if you minimize $ x$, then clearly the fraction $ \frac{(m\plus{}2)(n\plus{}2)}{2mn}$ increases, and it's not hard to show that $ \frac{p}{p\plus{}2}$ increases as $ p$ increases... So you need $ p$ maximal for $ x$ minimal.

Sorry for the bump :oops:

I was solving this problem and read the first solution ([url=https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14]of this page[/url], not this thread) and I'm slightly confused about this sentence:
[quote]
\[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]
[b]Clearly, we want to minimize the denominator[/b], so we test $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$.
[/quote]
Is it really clear that we want to minimize this denominator?

Thanks!


The number $ \sqrt {2}$ is equal to:

$ \textbf{(A)}\ \text{a rational fraction} \qquad\textbf{(B)}\ \text{a finite decimal} \qquad\textbf{(C)}\ 1.41421$
$ \textbf{(D)}\ \text{an infinite repeating decimal} \qquad\textbf{(E)}\ \text{an infinite non \minus{} repeating decimal}$

$ \sqrt{2}$ is an irrational number, so $ \boxed{E}$.


Find closed-form representations for the following:
\[ \sum_{k\equal{}0}^{n\minus{}1} k2^k\]
\[ \sum_{k\equal{}0}^{n\minus{}1} k(k\minus{}1)2^k\]

Consider
\begin{eqnarray*}
\displaystyle\sum_{k = 0}^{n-1} k \, x^k &=& x\displaystyle\sum_{k = 0}^{n-1} k \, x^{k-1}
\\&=&  x\displaystyle\sum_{k = 0}^{n-1} \dfrac{d}{dx}(x^k)
\\&=&  x\dfrac{d}{dx}\left(\displaystyle\sum_{k = 0}^{n-1} x^k\right)
\\&=&  x\dfrac{d}{dx}\dfrac{1 - x^n}{1-x}
\\&=&  x\dfrac{(1-x)(-nx^{n-1}) - (1 - x^n)(-1)}{(1-x)^2}
\\&=&  \dfrac{x - nx^{n} + (n-1)x^{n+1}}{(1-x)^2}
\end{eqnarray*}

Substitute in $ x = 2$.

HI;

You can use the summation calculus which has a summation by parts formula:

$ \sum_{k \equal{} a}^{b}f(k)\Delta g(k) \equal{} f(b \plus{} 1)g(b \plus{} 1) \minus{} f(a)g(a) \minus{} \sum_{k \equal{} a}^{b}g(k \plus{} 1)\Delta f(k)$

$ f(k) \equal{} k$
$ \Delta g(k) \equal{} 2^k$
$ \Delta f(k) \equal{} 1$
$ g(k) \equal{} 2^k$

Just plugging in the top formula:

$ n \ 2^n \minus{} 0 \minus{} \sum_{k \equal{} a}^{b}2^{k \plus{} 1}$
$ n \ 2^n \minus{} (2^{b \plus{} 2} \minus{} 2^{a \plus{} 1})$

$ \sum_{k \equal{} 0}^{n \minus{} 1}k2^{k} \equal{} n \ 2^n \minus{} 2^{n \plus{} 1} \plus{} 2$

Hi;

Just plug right into the Summation by parts formula that I used to get the first sum. You may have to use repeated summation by parts.

[hide="Elimination"]$ S \equal{} \sum_{k\equal{}0}^{n\minus{}1}(k^2 \minus{} k) 2^{k}$

Perform an index shift:
$ S \equal{} \sum _{k\equal{}\minus{}1}^{n\minus{}2} (k^2 \plus{} k) 2^{k\plus{}1} \equal{} \sum _{k\equal{}0}^{n\minus{}2} (k^2 \plus{} k) 2^{k\plus{}1} \equal{} 2 \sum _{k\equal{}0}^{n\minus{}2} (k^2 \plus{} k) 2^{k}$

Subtracting the latter from twice the former:

$ 2S \minus{} S \equal{}   2 \sum_{k\equal{}0}^{n\minus{}1}(k^2 \minus{} k) 2^{k} \minus{} 2 \sum _{k\equal{}0}^{n\minus{}2} (k^2 \plus{} k) 2^{k} \equal{} (n\minus{}1)(n\minus{}2) 2^{n\minus{}1} \minus{} 4 \sum_{k\equal{}0}^{n\minus{}2} k 2^k$

But you know what that remaining sum is from the first problem.[/hide]

It seems that, if $ n \geq c$:
$ \sum_{k\equal{}0}^n {k \choose c} 2^k \equal{} \minus{}(\minus{}2)^c \plus{} 2^{n\plus{}1} \sum_{k\equal{}0}^c {n\minus{}k \choose c\minus{}k} (\minus{}1)^k$
The RHS is a closed form if $ n$ is variable and $ c$ fixed ...

Your questions correspond to the cases $ c\equal{}1,2$.

I give combinatorial proofs [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url].


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Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?

$\text{(A)}\ 72 \qquad \text{(B)}\ 73 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 75\qquad \text{(E)}\ 76$

Search! I've noticed that a lot of the problems you just posted have been posted already, recently. :)

[quote="joml88"][quote="beta"]TOPIC LOCKED!!!!!

Jusk kidding  :D

BTW I don't understand all this topic locking business for posting problems that was posted before. I mean how are you supposed to know if a problem had been posted before? I mean why should you go searching the forums every time you want to post a problem?

Besides what's wrong with a problem being posted a few times?[/quote]

Yeah, all that searching also makes it ten times harder to put problems in the contest database...and that's going along slow as it is.[/quote]Agreed.   :)

I don't like my first solution. Here's a new one.  :roll:

[hide="Cauchy!"]Recognizing that the equation is a circle, we complete the square to get $(x-7)^2 + (y-3)^2 = 64$. Though we could solve this geometrically, applying the Cauchy-Schwarz inequality is much more effective.

From Cauchy-Schwarz, $((x-7)^2 + (y-3)^2)(3^2+4^2) \geq (3(x-7) + 4(y-3))^2$.

Because $(x-7)^2 + (y-3)^2 = 64$ and $3^2+4^2=5^2$, we can substitute and take the square root of both sides to get $\sqrt{8^25^2} \geq 3x-21+4y-12$.

Simplifying, $73 \geq 3x-4y$, so the largest possible value is $73$. (Can you find the minimum value?)  [/hide]

on the last equation, you mean $73 \geq 3x+4y$ instead of minus, right?

Nice solution, 4everwise!  I've never seen Cauchy used on a geometry problem before, but it's much quicker than the coordinate geometry solution.

Hey,

Redirected from: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198417

[quote="4everwise"]
The maximum (and minimum) values occur when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. (Do you see why?) Because the tangent line has slope $ \minus{} \frac34$, the radius has slope $ \frac 43$.
[/quote]

I do not quite understand how one determines that the maximum and minimum values occur at points tangent to the circle, how would one prove this?

Also, how would one solve this problem without resorting to Cauchy's inequality?

[hide="So far this is what I have,"]
First off, recognize that the solutions to,
$ max(3x \plus{} 4y)$

Must satisfy
$ {{x}^{2}} \plus{} {{y}^{2}} \equal{} {14x \plus{} 6y \plus{} 6}$

In addition, note the above can be rewritten as,
$ {{(x \minus{} 7)}^{2}} \plus{} {{(y \minus{} 3)}^{2}} \equal{} {{8}^{2}}$

Which fits the equation of a circle with,
r = 8
(h,k) = (7, 3)

So letting,
$ 3x \plus{} 4y \equal{} C$

Solving for y,
$ {y} \equal{} {{\frac { \minus{} 3x}{4}} \plus{} {\frac {C}{4}}}$

Now substituting this in to,
$ {{(x \minus{} 7)}^{2}} \plus{} {{(y \minus{} 3)}^{2}} \equal{} {{8}^{2}}$

Yields,
$ {{\left(x \minus{} 7\right)}^{2}} \plus{} {{\left({\left({{\frac { \minus{} 3x}{4}} \plus{} {\frac {C}{4}}}\right)} \minus{} 3\right)}^{2}} \equal{} {{8}^{2}}$

From here is the next step to set this equation equal to zero and solve using the quadratic equation?

But how would one solve this with the quadratic since there are two variables: x and C.
[/hide]
Thanks,

-APSStudent

[quote="APSStudent"]I do not quite understand how one determines that the maximum and minimum values occur at points tangent to the circle, how would one prove this?[/quote]

Consider the family of lines $ 3x \plus{} 4y \equal{} k$ with varying $ k$.  We want to find the maximum value of $ k$ such that the intersection of this line with the given circle is a point; as we vary $ k$ (which moves the line up and down) the intersection of the line with the circle is two points until it becomes one point (where the line is tangent to the circle), and then it stops intersecting at all.

Hey,

[quote="t0rajir0u"]
Consider the family of lines $ 3x \plus{} 4y \equal{} k$ with varying $ k$.  We want to find the maximum value of $ k$ such that the intersection of this line with the given circle is a point; as we vary $ k$ (which moves the line up and down) the intersection of the line with the circle is two points until it becomes one point (where the line is tangent to the circle), and then it stops intersecting at all.
[/quote]
Thanks for the reply, t0rajir0u.
[hide="So, if I understand correctly since,"]
$ Ax \plus{} By \equal{} C$

Can be rewritten as,
$ {y} \equal{} {{\frac{\minus{}3x}{4}} \plus{} {\frac{C}{4}}}$

Which when rewriting in slope-intercept form is,
$ {y} \equal{} {{{m}x} \plus{} {b}}$

I can see that,
$ {b} \equal{} {\frac{C}{4}}$

So that,
$ {C} \equal{} {{4}{b}}$

Which is a directly proportional relationship--showing that as $ C$ increases so does $ b$.

So, I can see from the above that given some,
$ {{(x \minus{} h)}^{2}} \plus{} {{(y \minus{} k)}^{2}} \equal{} {{r}^{2}}$

the,
$ max(Ax\plus{}By)$

Occurs at the $ max(b)$. Which can only occur when the line $ Ax\plus{}By$ makes only one intersection (ie. is tangent to) with the circle. In addition, the $ min(C)$ also occurs at the $ min(b)$ which can also only occur at one intersection (ie. is tangent to) with the circle.

However, how I can determine from the beginning which of the two intersections when the line is tangent to the circle will yield the $ max(C)$ or $ min(C)$?.
[/hide]
[hide="Additionally, where do I go from here?"]
When I get to the point where I have,
$ {y} \equal{} {{\frac { \minus{} 3x}{4}} \plus{} {\frac {C}{4}}}$

Then substitute that in to,
$ {{(x \minus{} 7)}^{2}} \plus{} {{(y \minus{} 3)}^{2}} \equal{} {{8}^{2}}$

I get,
$ {{\left(x \minus{} 7\right)}^{2}} \plus{} {{\left({\left({{\frac { \minus{} 3x}{4}} \plus{} {\frac {C}{4}}}\right)} \minus{} 3\right)}^{2}} \equal{} {{8}^{2}}$

However, I am not sure how to proceed from here since I have two unknowns: $ x$ and $ C$.

So, how do I proceed from here?
[/hide]
Thanks,

-APSStudent

[hide="Where to go"]From the equation of the circle...

The lines of the form $ 3x\plus{}4y\equal{}k$ have slope $ \minus{}\frac{3}{4}$.  You want the point tangent to the circle with a slope of $ \minus{}\frac{3}{4}$.  Letting $ O$ be the center of the circle and $ P$ being the point of tangency, the slope of $ \overline{OP}$ is $ \frac{4}{3}$.[/hide]

What do the functions $ max(3x \plus{} 4y)$ mean?  A line doesn't have a maximum.  Thanks!

$ 3x \plus{} 4y$ does not actually represent a line (there is no equals sign there); that expression is shorthand for "the maximum value of $ 3x \plus{} 4y$ subject to the constraint $ x^2 \plus{} y^2 \equal{} 14x \plus{} 6y \plus{} 6$."

[hide]As previously noted, $ \sqrt{(x\minus{}7)^2\plus{}(y\minus{}3)^2}\equal{}8$, and $ 3x\plus{}4y\minus{}k\equal{}0$, where $ k$ is the max/min
So, the distance from the line $ 3x\plus{}4y\minus{}k\equal{}0$ to $ (7,3)$ is $ \dfrac{|3\cdot7\plus{}4\cdot3\minus{}k|}{\sqrt{3^2\plus{}4^2}}\equal{}8$
Thus, $ k\equal{}\minus{}7$ is the min and $ k\equal{}73$ is the max.
[/hide]

[quote="Xantos C. Guin"][hide]As previously noted, $ \sqrt {(x \minus{} 7)^2 \plus{} (y \minus{} 3)^2} \equal{} 8$, and $ 3x \plus{} 4y \minus{} k \equal{} 0$, where $ k$ is the max/min
So, the distance from the line $ 3x \plus{} 4y \minus{} k \equal{} 0$ to $ (7,3)$ is $ \dfrac{|3\cdot7 \plus{} 4\cdot3 \minus{} k|}{\sqrt {3^2 \plus{} 4^2}} \equal{} 8$
Thus, $ k \equal{} \minus{} 7$ is the min and $ k \equal{} 73$ is the max.
[/hide][/quote]

How did you get $ \dfrac{|3\cdot7\plus{}4\cdot3\minus{}k|}{\sqrt{3^2\plus{}4^2}}\equal{}8$?  Thanks!

The distance between the line $ Ax \plus{} By \plus{} C \equal{} 0$ and a point $ (x_0,y_0)$ is $ \frac {|Ax_0 \plus{} By_0 \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$.

Since $ \sqrt {(x \minus{} 7)^2 \plus{} (y \minus{} 3)^2} \equal{} 8$, the distance between $ 3x \plus{} 4y \minus{} k$ and $ (7,3)$ is $ 8$.

What was the motivation for using Cauchy in 4everwises Sol?

[quote=4everwise][hide="Partial Solution"]Some inspection of the given equation tells us that it is a circle. Completing the square gives us $(x-7)^2+(y-3)^2=8^2$, which is a circle of radius $8$ centered at $(7,3)$. In order to satisfy the conditions specified, we know that $3x+4y$ must intersect the circle at at least one point.

The maximum (and minimum) values occur when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. (Do you see why?) Because the tangent line has slope $-\frac34$, the radius has slope $\frac 43$. [/hide]

Sorry, don't feel like finishing it now. Am feeling tired for some reason.  :(   :|[/quote]

[hide=finishing the partial solution (no Cauchy)]

The radius has slope $\frac 43$, therefore it is 6.4 above and 4.8 to the right of the center (this can be seen by building a 3-4-5 triangle, scaled up by 1.6). Therefore, the point is $(11.8, 9.4)$ so $3x+4y=35.4+37.6=73$

Also, the minimum possible value of $3x+4y$ is $2(3*7+4*3)-73=-7$.

[/hide]

For maximum (and minimum) b=3x+4y , that is four times the
"vertical intercept" of the stright line, it must be tangent
therefore touching a single point, in each case, of the circle.

So, the quadratic equation, for instance in $x$, obtained by
substitution of $y={{b-3x}\over 4}$ in the conditioning equation
(that of the circle) must have a null discriminant.
That leads us to the quadrstic equation in $b$:
$b^2-66b-511=0$ with roots $-7$ for the minimum and $73$, the
desired one.

from where u have found this..if mean from any previous year question or others

i think this was an AMC question but i might be wrong


The random variable $ N$ takes the values 0,1,2,..............

Independently of each other, each of N balls is placed either in urn A with probability p,  or in urn B with probability 1-p ,resulting in $ N_{A}$ balls in urn A and $ N_{B}$ =$ N$-$ N_{A}$ balls in urn B.

Prove that the random variables $ N_{A}$ , $ N_{B}$ are independent if and only if $ N$  has a Poisson distribution.

I would try the following if $ N$ has a poisson distribution then
Show that $ E[e^{i.u.N_A +i.v.N_B}]=E[e^{i.u.N_A}].E[e^{i.v.N_B}]$ for all $ (u,v)\in \C^2$
by condionning over $ N$
for the necessity part I don't see it for the moment.

Hoipe it helps.


7. Two factors have a product of $ 3\cdot9^{2x}\minus{}6\cdot27^x$ One of the factors is $ 3^x \minus{}2$ What is the oter factor?

I get the basic idea of making it into 3's and start. But I get stuck.
[hide="Answer"]$ 3^{3x\plus{}1}$ or $ 3\cdot3^{3x}$ or $ 3\cdot 27^x$[/hide]


[quote="AdithyaGanesh"]I do not have Wolfram's Mathematica, so would it be okay if I asked someone with Mathematica to perform something for me?[/quote]

I am not sure what you need mathematica for, but if you asked someone else to use mathematica to perform something for you, then you would be giving a part of your answer away, which in a way is discussing the problem/solution.

Since you're asking your friend to perform it; however, it seems like a simple task. So I say use this: wolframalpha.com. It's a site built by Wolfram Research and can perform some "outrageous" calculations mathematica is capable of.

You could always get the trial version of Mathematica and do it.

There's also the illegal way...

Remember, though, all questions were designed to be solvable without the use of programming or a super-calculator, or whatever.

If you just need to check something, I second the use of wolframalpha.com.

Okay, thanks guys.  If all problems are designed for solving without crazy calculation, then I'll keep trying.


i need the book (ALGEBRAIC INEQUALITIES old and new methods)but i cant buy it because i am in Iran please someone send it i will send its price .
 here is my Email: mgeo377@gmil.com
 thanks a lot :)

[quote="great math"]I think you may need a basic inequality book but it's not easy but not too hard![/quote]
 thank you very much my friend :wink:


1. Use a combinatorial argument to show that $ \forall \ \{n,m,k\} \in \mathbb{Z^\plus{}}$ with $ \{n,m\} \ge k$

$ \sum_{j\equal{}0}^k\binom{n}{j}\binom{m}{k\minus{}j} \equal{} \binom{n\plus{}m}{k}$

2. We are given $ \mbox{n}$ points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when n = 6, there are 15 intersections.

Thanks!

I just started combinatorics so please don't leave out any necessary steps! Thanks again :)

1. Hint: both sides count the number of ways to choose $ k$ objects out of $ m\plus{}n$ objects - the right side directly, the left by counting different cases and adding them up. (what are the cases?)

2. Hint: how many of the $ n$ points on the circle are involved in an intersection of two chords?

Ah right thanks I got Q 1 now. :)

For Q 2 I'm still a bit stuck, I found out that every 4 points makes 1 intersection but then...?

well that's it.. every selection of 4 points gives you an intersection, and every intersection comes from a selection of 4 points, so...

oh lol, how thick-headed am I  :|

$ \binom{n}{4}$ intersections!

Thanks pleurestique!


real analysis
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I have a question now that my analysis course is over that comes from our book (baby Rudin).  I never got a satisfactory answer, and I'm curious about it.
[quote]
If $f$ is defined on $E$, the $graph$ of $f$ is the set of points $(x,f(x))$, for $x\in E$. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact.[/quote]
I have the forward direction.  The backward direction seems like it needs some more work.

someone posted the same question few days back. please use the search function to locate the post


Let $ X,Y$ be i.i.d. random variables, with $ E(X)<\infty$.
Show that $ E\left(\frac{X}{X\plus{}Y}\right)\equal{}\frac{X\plus{}Y}{2}$ a.s..

The property is indeed very strange: it is false. First, of all the left hand side is a number; the right hand side is a r.v. Secondly, even in the trivial case when $ X \equal{} Y \equal{} const$, the property doesn't hold (unless $ const \equal{} 1/2$).

Did you mean the following property?
\[ E(X| X \plus{} Y) \equal{} (X \plus{} Y)/2
\]
[hide]
This property is trivial since $ E(X| X \plus{} Y)  \equal{} E(Y| X \plus{} Y)$ by symmetry; and $ E(X| X \plus{} Y)  \plus{} E(Y| X \plus{} Y) \equal{} E(X\plus{}Y| X\plus{}Y) \equal{} X\plus{}Y.$
[/hide]


function
integration
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Take a look at http://www.mathlinks.ro/viewtopic.php?t=214686
I am posting it here because I want a rigorous Proof that $ \boxed{6}$ is true.
Also contradict my reasoning if you think my answers aren't appropriate.

[hide="My answers"]
$ \boxed{TFFFFT}$[/hide]
[hide="Intuitive reasoning"]
If the population is distributed uniformly, then the result equals the actual population. Now suppose the population is clustered at some places and light at some places. Where the density is high, everyone gives high populations. And since there are many people who give opinions, The high value is exaggerated when taking into the mean. On the other hand, where the density is low, the people there gives low values, but since there are so few of them, their opinion hardly contributes to the arithmetic mean. So the result gets higher than the actual population.

Example: suppose all the population is uniformly distributed in to half the area. Then the result gets doubled. I'm pretty sure there's no inverse scenario.[/hide]
[hide="My approach to a proof"]
Suppose $ f : \mathbb L\to \mathbb R^ \plus{}$ be the population density function of the country where $ \mathbb L$ is the set of all points on the country.

$ \int_{l\in \mathbb L} dl \equal{}$ total area. Suppose it is $ A$.

$ \int_{l\in \mathbb L} f(l) \ dl \equal{}$ total population. Suppose it is $ P$.

$ f(l)\cdot A \equal{}$ the opinion on the country's population of a citizen on point $ l$.

Note that when taking the arithmetic mean the opinion of every individual point must be weighted with the density on that point because there are that many people on that point to give an opinion.

$ \frac {\int_{l\in \mathbb L} (f(l)\cdot A)\cdot f(l) \ dl}{P} \equal{}$ the arithmetic mean of the opinions $ \equal{}$ result, Denote it to be $ R$

We need to prove that
$ R\geq P\iff \boxed{\big(\int_{l\in \mathbb L} dl\big)\cdot \big(\int_{l\in \mathbb L} (f(l))^2 \ dl \big)\geq \big(\int_{l\in \mathbb L} f(l) \ dl\big)^2}$

How to prove it?[/hide]
[hide="To Mods"]This isn't a double post. Because I'm asking for proof here. If you still want to delete one of the two posts, then delete the one in "Classroom maths" forum[/hide]

No one interested?

I think that the inequality $ \boxed{\big(\int_{l\in \mathbb L} dl\big)\cdot \big(\int_{l\in \mathbb L} (f(l))^2 \ dl \big)\geq \big(\int_{l\in \mathbb L} f(l) \ dl\big)^2}$ could be common and well known if it is true (I very much believe it is true, it looks somewhat similar to cauchy schwarz inequality). I don't know anything about inequalities with integrals. So someone please help me.

[quote="rofler"]It shoould be the integral form of cauchy schwartz, look it up on wikipedia.[/quote]
 :thumbup:
[url=http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#L2]Cauchy schwarz integral form[/url]
$ g(x)\equiv 1\implies$ the inequality. Thus the proof is complete.
Thanks a lot rofler.


geometry
geometric transformation
homothety
geometry unsolved

If $ L_b,L_c$ are defined then vertice $ A$ can be constructed by the intersection of them.

Now consider the homothety $ H_{A}^{\frac{2}{3}}: G\mapsto M$ then $ M$ is the midpoint of $ BC$.

Consider the symmetric with centre $ M$, transform $ A\mapsto A'$ then $ ABA'C$ is a parallelogram. Thus $ A'B//L_c$ and $ A'C//L_b$, following the construction of $ B$ and $ C$.

Therefore, $ \triangle ABC$ is constructed.


How many four-digit positive integers have at least one digit that is a 2 or a 3?

Let's find the number of four-digit positive integers that have no 2's or 3's. There are $ 7$ ways to choose the thousands digit, $ 8$ for the hundreds, $ 8$ for the tens, and $ 8$ for the units, giving a total of $ 7 \cdot 8^3 \equal{} 3584$.

There are $ 9000$ four-digit positive integers in total, so our answer is $ 9000 \minus{} 3584 \equal{} \boxed{5416}$.

[quote=tfz][hide]2006 AMC10A[hide]#21[hide]There are 9000 four-digit positive integers. For those without a 2 or 3, the first digit could be one of the seven numbers 1, 4, 5, 6, 7, 8, or 9, and each of the other digits could be one of the eight numbers 0, 1, 4, 5, 6, 7, 8, or 9. So there are \[
9000- 7\cdot 8\cdot 8\cdot 8=\boxed{5416}
\] four-digit numbers with at least one digit that is a 2 or a 3.[/hide][/hide][/hide][/quote]

lol that is actual answer

I can't believe that early AMCs were that easy, like this probably can only be on a modern amc like p10>


Solve the following questions correct to 3 decimal places. A college has 600 students. A survey has determined that
73 read French
41 read German
44 read Spanish
19 read French and German
18 read French and Spanish
9 read German and Spanish
6 read all three languages
What is the probability a randomly selected student reads exactly 2 of these 3 languages?
What is the probability a randomly selected student reads at least 1 of the 3 languages?
What is the probability a randomly selected student reads exactly 1 of the 3 languages?
What is the probability a randomly selected student does not read any of the 3 languages?

Any help appreciated...

[hide="Solution"]
Draw Venn diagrams http://en.wikipedia.org/wiki/Venn_Diagram
Only French: 42
Only Spanish: 23
Only German: 19
Only one language: 84

Similar:
Only two languages: 28
Three languages: 6
None: 482

1) $ \frac{28}{600}\equal{}0.047$
2) $ \frac{118}{600}\equal{}0.197$
3) $ \frac{84}{600}\equal{}0.14$
4) $ \frac{482}{600}\equal{}0.803$
[/hide]


Fie ABCD un patrat de latura 1.
 Sa se demonstreze ca pentru orice punct P din plan are loc inegalitatea
 $PA\cdot PC + PB\cdot PD \ge 1$

Am gasit o solutie urata de tot.
Putem fixa originea sistemului de referinta in $O$, centrul patratului, si axele diagonalele.
Atunci $A(-\frac{\sqrt{2}}{2},0),C(\frac{\sqrt{2}}{2},0),B(0,-\frac{\sqrt{2}}{2}),D(0,\frac{\sqrt{2}}{2})$.
Si punctul $P(x,y)$. Voi studia functia:
${{f(x,y)=PA\cdot PC+PB\cdot PD=\sqrt{x^2+(y-\frac{\sqrt{2}}{2} )^2}\cdot \sqrt{x^2+ (y+\frac{\sqrt{2}}{2} )^2}+\sqrt{y^2+ (x-\frac{\sqrt{2}}{2}} ) ^2}+\sqrt{y^2+ (x+\frac{\sqrt{2}}{2}} ) ^2}$
Am cautat punctele de extrem pt  functia $f$ rezolvand sistemul  : $ \{ \begin{array}{c} \frac{\partial f}{\partial x}=0\\\frac{\partial f}{\partial y}=0 \end{array}$. Solutia sistemulul este $(0,0)$ .Asadar $\min_{x,y\in R}(f(x,y))=1$.Deci $f(x,y)\geq 1$.(q.e.d)

Foarte frumoasa problema.Iata solutia mea,daca P e in interiorul patratului:
Proiectam punctul $P$ pe laturile patratului astfel incat $M \in [AB],N\in [BC],S\in [DC],T\in[AD].$ Astfel se formeaza dreptunghiurile $AMPT,PMBN,PNCS,DSPT$,unde diagonalele sunt congruente rezulta ca se formeaza patrulaterul $MNST$,care are laturile egale cu $PA,PB,PC,PD$.Aplicam inegalitatea lui Ptolemeu pentru patrulaterul $MNST$,unde avem:$MT\cdot NS+MN\cdot ST \leq SM\cdot TN\iff PA\cdot PC+PB\cdot PD \leq 1$
QED

Eu am gasit singur problema, pornind de la Ptolemeu si vroiam sa vad daca se gasesc si alte solutii..nu stiu sigur insa daca e sau nu originala poate a mai aparut pe undeva

Vom cauta minimul expresiei. Fie $O$ centrul patratului. Din teorema medianei in $PAC$ si $PBD$ avem ca:
$\ 4PO^{2}=2(PA^{2}+PC^{2})-AC^{2}=2(PB^{2}+PD^{2})-BD^{2}$
Atunci avem ca $2PO^{2}+1=PA^{2}+PC^{2}=PB^{2}+PD^{2}$, de unde avem ca:
$2PO^{2}+1\geq\ PA\cdot\ PC\ (1)$
$2PO^{2}+1\geq\ PB\cdot\ PD\ (2)$
Insumand relatiile $(1)$ si $(2)$ avem ca $2PO^{2}+1\geq\ PA\cdot\ PC+PB\cdot\ PD$
Pentru un $PA\cdot\ PC+PB\cdot\ PD$ minim trebuie sa facem ca limita sa superioara sa fie minima. Aceasta se intampla pentru $PO=0$, deci $P=O$, deci pentru $PA\cdot\ PC+PB\cdot\ PD=1/2+1/2=1$ q.e.d.

problema este cunoscuta:Fi abcd un dreptunghi cu laturile a si b atunci PA*PC+PB*PD>=a*b.Este o problema a profului meu pe care am vazut-o printr-o gazeta!


inequalities
trigonometry
geometry
national olympiad

Here's the problems for the China MO 2005:
[url=http://www.mathlinks.ro/Forum/resources/files/china/65/China-National_Olympiad-2005-65-37.pdf]PDF Version[/url]

\[ \text{ Day I }  \] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24387]Problem 1 Discussion[/url]
Suppose  $\theta_{i}\in(-\frac{\pi}{2},\frac{\pi}{2}), i = 1,2,3,4$. Prove that, there exist $x\in \mathbb{R}$, satisfying two inequalities\[ \cos^2\theta_1\cos^2\theta_2-(\sin\theta\sin\theta_2-x)^2\geq0,  \]\[ \cos^2\theta_3\cos^2\theta_4-(\sin\theta_3\sin\theta_4-x)^2\geq0  \]
if and only if\[ \sum^4_{i=1}\sin^2\theta_i\leq2(1+\prod^4_{i=1}\sin\theta_i + \prod^4_{i=1}\cos\theta_i).  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24378]Problem 2 Discussion[/url]
A circle meets the three sides $BC,CA,AB$ of a triangle $ABC$ at points $D_1,D_2;E_1,E_2; F_1,F_2$ respectively. Furthermore, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that the lines $AL,BM,CN$ are concurrent.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24388]Problem 3 Discussion[/url]
As in the graph, a pond is divided into 2n (n $\geq$ 5) parts. Two parts are called neighborhood if they have a common side or arc. Thus every part has three neighborhoods. Now there are 4n+1 frogs at the pond. If there are three or more frogs at one part, then three of the frogs of the part will jump to the three neighborhoods repsectively. Prove that for some time later, the frogs at the pond will uniformily distribute. That is, for any part either there are frogs at the part or there are frogs at the each of its neighborhoods.
[img]http://www.mathlinks.ro/Forum/files/china2005_2_214.gif[/img]
\[ \text{ Day II }  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24689]Problem 4 Discussion[/url]
4.The sequence {$a_n$} is defined by: $a_1=\frac{21}{16}$, and for $n\ge2$,\[ 2a_n-3a_{n-1}=\frac{3}{2^{n+1}}.  \]Let $m$ be an integer with $m\ge2$. Prove that: for $n\le m$, we have\[ \left(a_n+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left(m-\left(\frac{2}{3}\right)^{{\frac{n(m-1)}{m}}}\right)<\frac{m^2-1}{m-n+1}.  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24690]Problem 5 Discussion[/url]
5.There are 5 points in a rectangle(including its boundary) with area 1, no three of them are in the same line. Find the minimum number of triangles with the area not more than $\frac{1}{4}$, vertex of which are three of the five points.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24692]Problem 6 Discussion[/url]
6.Find all nonnegative integer solutions $(x,y,z,w)$ of the equation\[ 2^x\cdot3^y-5^z\cdot7^w=1.  \]

[i]Edited by Myth and orl, contributed by mecrazywong, siuhochung and jackhui.[/i]

[quote="qxy"]Wow, I think mathlinks is so quick, first day of CMO has just passed for one day!
I think if i wait for the problems in the Chinese magazine, I should wait for them at least one month.
So great! :D[/quote]

If I wait for them in Crux (a Canadian journal), I would need to wait for at least 4 years...

(and I might have never seen it, I add as a footnote)

[quote="billzhao"]Where are the day 2 problems?[/quote]

Hopefully it is not the same as with vess who never posted the second day problems of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=19727#p19727]Bulgarian TST 2004.[/url]  ;)

[quote="yptsoi"][quote="billzhao"]Where are the day 2 problems?[/quote]
I think the second day problems will be posted tonight since the hk team will be back, and it is easier for them to find a computer in Hong Kong than in China.  :)[/quote]We're a-waiting for them ..

PS Last time I checked, Hong Kong was a part of China :D :D ...

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24689]Problem 4 Discussion[/url]

4.The sequence {$a_n$} is defined by: $a_1=\frac{21}{16}$, and for $n\ge2$,\[2a_n-3a_{n-1}=\frac{3}{2^{n+1}}.\]Let $m$ be an integer with $m\ge2$. Prove that: for $n\le m$, we have\[(a_n+\frac{3}{2^{n+3}})^{\frac{1}{m}}(m-(\frac{2}{3})^{{\frac{n(m-1)}{m}}})<\frac{m^2-1}{m-n+1}.\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24690]Problem 5 Discussion[/url]

5.There are 5 points in a rectangle(including its boundary) with area 1, no three of them are in the same line. Find the minimum number of triangles with the area not more than $\frac{1}{4}$, vertex of which are three of the five points.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24692]Problem 6 Discussion[/url]

6.Find all nonnegative integer solutions $(x,y,z,w)$ of the equation\[2^x\cdot3^y-5^z\cdot7^w=1.\]

[i]Edited by Myth and orl[/i]

This is the Day 2 problems. Sorry for posting them late but I really didn't get any chance to use a computer after Day 2 competition :(
Generally they're much harder than Day 1(at least that's what I think). This time, like last year, the committee marked the paper extremely strict! They deducted me much points for a minor mistake in q.5 :(
It's really a surprise that Russia just got 5 silvers 1 bronze. Anyway, it's much better than Hong Kong.

P.S. Valentin, the time we were given is 4.5 hrs, not 4 hrs.

We were not able to get to a computer in last two days...   :(
Here is some stat:
23 golds.
4 perfect scores.
HK team: 8 bronze.
Macau team: 4 bronze.
Russian team (I guess it's 2nd team): 5 silver 1 bronze.

I've added the links and the problems in the first post also. The PDF file is now complete. Thanks mecrazywong,  siuhochung and jackhui for telling us the problems :)

[quote="mecrazywong"]

P.S. Valentin, the time we were given is 4.5 hrs, not 4 hrs.[/quote]

This might be also changed in the .pdf-file.  :)

Yes but usually 30 minutes are given for students to study the problems and ask the committee questions about the problems :D

Congratulations wpolly!!!  :D  :D  :D

When/where is the team selection camp going to be held?

And could someone explain a bit about how CMO works? Is it like a mini-IMO? How do the markings take place and what are the "teams", for example.

[quote="wpolly"]23 golds
47 silvers
and everyone else got a bronze :?
The gold and silver cut-offs are 36 and 25, respectively.
I am one of the 4 perfect score :D  :D  :D[/quote] WOW! Congrats!!! :D

[quote="wpolly"]
I am one of the 4 perfect score :D  :D  :D[/quote]

Congrats from me as well. But somehow the perfect scorer wanted to posted the Chinese TSTs 2004 once...  :?

[quote="wpolly"]23 golds
47 silvers
and everyone else got a bronze :?
The gold and silver cut-offs are 36 and 25, respectively.
I am one of the 4 perfect score :D  :D  :D[/quote]
wpolly, not everyone got a bronze. The cut-off of bronze is 12(4) points, but only a few(less than 10) cannot get a prize. They tends to give everyone a prize, but unfortunately there aren't enough bronze medal.
Anyway, congratulation! Perfect Scorer wpolly!

I have just noticed a typo of problem 1:
The first ineq should be $\cos^2\theta_1\cos^2\theta_2-(\sin^2\theta_1\sin^2\theta_2-x)^2\ge0$.

[quote="wpolly"]23 golds
47 silvers
and everyone else got a bronze :?
The gold and silver cut-offs are 36 and 25, respectively.
I am one of the 4 perfect score :D  :D  :D[/quote]
Congratulations!

[quote="iura"]Well, congratulations Wpolly!
Good luck to you at IMO TST!

By the way, why does the CMO level decrease from year to year?
I remember in $2002$ there were $4$ problems harder than any problem at CMO $2005$.
It seemess  that from $2002$ on the level is monotonically decreasing.. and this is bad since we lose one of the hardest olympiads in the world...[/quote]
Really?
Have you seen the problems of 2004 CMO? They are the hardest of every year! So, this year they set the problems easier, and I suppose the Day 2 problems are not so easy as day 1, rite?

[quote="iura"]I don't agree with you, Mecrazywong. I saw the problems of CMO $2004$, and I think only the inequality was super-hard.
Comparing it with CMO 2002, I would say that Q5,Q2,Q6 are approximately at the same level as this inequality.
So CMO 2002 is the hardest ever IMHO[/quote]
I'm nothing good in inequality. I don't know which one is harder.
For q1 and q3 and q4 of cmo 2002, I would surely say that they are easy to most people, but for cmo 2004, there isn't any such problems. Moreover, the cut-offs of cmo 2002 are also much higher than 2004.

How many students compete in the CMO?

How many are from "main part" of China not including HK, Macau, Russia, etc.?
(Places where the mathematics curriculum and education system are governed by committees in Beijing rather than some local variations.)

What is the procedure for Russia to compete?  Is it a different group of students
than the Russian IMO team?

[quote="wpolly"]I am one of the 4 perfect score :D  :D  :D[/quote]
Congratulations to you! You are so great!

[quote="fleeting_guest"]What is the procedure for Russia to compete?  Is it a different group of students
than the Russian IMO team?[/quote]
I could tell you some information about the Chinese Team in ARMO, the province(of China) which its team got the most marks in CMO in the last year could send its team to ARMO in the next year. So I suppose that the Russia Team might be similar to the Chinese Team.

Russian team at CMO - the students, which take the first 6 places in russian summer school.

[quote="siuhochung"]wpolly, your name is 王晨?[/quote]

yes ,he is my friend!

[quote="wpolly"]I am one of the 4 perfect score :D  :D  :D[/quote]

Congratulations. Well done.

It seems, that all 23 gold medallists are from the mainland China  :?  Thanks [b]qxy[/b] for posting that list in the Chinese forum.


conics
parabola
geometry proposed
geometry

Dear everyone.

For the following problem which I got from my little and young angel, I got only one and very long solution 3 days ago.
But I long for beautiful, purely geometric and short solutions. May I ask you a favor? I will wait for your advice.
And I wonder if this problem is well known. Thank you for reading my post.

Please read the following.

Draw two parallel lines (parallel to AB, CA respectively) through an arbitrary point P on BC and let the intersection points be D, E respectively.
For the midpoint M of BC, let the line parallel to AM (passing through P) meet DE at K. Find the locus of the point K.  :)

Since $\tfrac{BE}{EA}=\tfrac{BP}{PC}=\tfrac{AD}{DC},$ then the series $E$ and $D$ with base lines $BA,AC$ are projective and clearly not perspective $\Longrightarrow$ lines $DE$ envelope a projective conic $\mathcal{P}$ tangent to $AB,AC$ at $B,C$ (images of the common point A in the respective series). Since these series are clearly similar, then $\mathcal{P}$ is a parabola (the A- [url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Artzt_paraboles.html]Artzt parabola[/url] of ABC).

Let $K^*$ be the tangency point of $DE$ with $\mathcal{P}.$ By degenerate Brianchon theorem, the lines $AK^*,BD,CE$ concur at $S,$ thus the pencil $A(B,C,S,Q)$ is harmonic, where $Q \equiv BC \cap DE$ $\Longrightarrow$ $(E,D,K^*,Q)=-1$ $\Longrightarrow$ $P(E,D,K^*,C)=-1$ $\Longrightarrow$ pencil formed by parallels through $A$ to $PE,PD,PK^*,BC$ is also harmonic $\Longrightarrow$ $PK^* \parallel AM$ $\Longrightarrow$ $K \equiv K^*$ $\Longrightarrow$ $K$ runs on $\mathcal{P}.$


Sea $n$ un entero positivo. Pruebe que alguno de los numeros $n$ ó $3n$ contiene algun digito del conjunto ${1,2,9}$ en su representecion decimal.

bueno trataremos de encontrar un contra ejemplo al problema luego veremos q esto es imposible

bueno sabemos q $n$ no puede empezar en $1$ o $2$ o $9$ para buscar el contra ejemplo , luego vemos que $n$ podria empezar en $3$ o $4$ o $5$ pero vemos q $3n$ empezaria necesariamente en uno , si es que $n$ empezara con un $6$ , $3n$ empezaria en $1$ o $2$ , luego si $n$ empieza en $7$ o $8$ , $3n$ empezara en $2$ , luego como  se analizaron par todos los $n$ posibles (respecto a su primera cifra) vemos que no hay contradiccion alguna por ello es valido el problema . espero q se entienda la solucion  :(


inequalities
function
real analysis
induction
calculus
derivative
algebra theorems

Given that $ \frac {f(x_1) +\ f(x_2)}{2} \geq f( \frac{x_1 + x_2}{2} ) $ , prove that

\[ \sum _{i=1} ^ n \frac{f(x_{i})}{n} \geq f( \frac {\sum _{i=1} ^n x_i}{n} ) \]

I know that this is a property of concave up functions. But can somebody prove it from the basics. I.e. Can somebody first prove the concavity/convexity and establish the result without using any theorm beyond Lagrange's Mean Value Theorm. I am preparing for IIT - JEE and this question is from there ( for special case of $ n =\ 3 $ ). There , one is expected to solve without using any theorm from advanced maths

[hide]Administrators - this question might be from calculus. I'd be happy to shift it there if you want. Can you plz guide me about it[/hide]

Here's the "special case n=3":

adding up $\frac {f(x_1)+f(x_2)}{2}\ge f\left( \frac{x_1+x_2}{2}\right)$ and $\frac {f(x_3)+f(x_4)}{2}\ge f\left( \frac{x_3+x_4}{2}\right)$ yield

\[\frac {f(x_1)+f(x_2)+f(x_3)+f(x_4)}{4}\ge \frac {f\left( \frac{x_1+x_2}{2}\right)+f\left( \frac{x_3+x_4}{2}\right)}{2}\ge f\left(\frac {x_1+x_2+x_3+x_4}{4}\right)\].
Now, let $x_4=\frac {x_1+x_2+x_3}{3}$. Plugging this in the previous inequality immediately gives
\[\frac {f(x_1)+f(x_2)+f(x_3)}{3}\ge f\left(\frac {x_1+x_2+x_3}{3}\right)
\],
as desired. The same trick works in the general case...Try it!

[quote="Rafal"]there is a simple inductive proof of the inequality but I'm not sure if it's so important to mention it here.[/quote]

Pleases do!!!!
ALso can you correlate this property and concavity/convexity of functions via a rigorous proof?!!!

just look what enescu did ;),
1. First prove (what's easy) that for $n=2^m$ it's true.
2. Then prove that if it's true for $k+1$ then it's true for $k$ (putting $x_{k+1}=\frac{x_1+...x_k}{k}$).

I understood the part of proving the inequality. But what about the relation of such functions to convexity????? :(  :oops:

This is obviously the Jensen inequality for convex/concave functions...

exactly.

Basically I am asking for the proof of jensen's Inequality and its converse(if true)

If it's not true , I am also asking for a proof of that

Rushil
Exactly ,what do you want? :?
Rafal had showed a basic proof with induction.

Basically I am asking for the proof that if $ f'(x) > 0 $ , prove that
$ \frac{ f (x_1) + f(x_2) }{2} \geq f ( \frac{x_1 + x_2}{2} ) $.

The general case can easily be obtained as you have shown above.. ,once we establish the inequality for $ n = 2$. My question is this!! To relate the inequality for $ n=2$ with concavity without using graphs!! A rigorous proof plz!

A function satisfying the above midpoint inequality is not necessarily convex.

As for a proof, one can be found here:

http://www-math.mit.edu/~rbm/18.100-F02.HMW/HMW7s.pdf

(it contains the solution to question 5.14 out of Rudin)

Well, I don't have a document with a proof of it, and I'm not going to take the time to write it down.  The pdf that I gave you contains the proof that if $f$ is twice differentiable, then it is convex iff $f''(x)>0$.

Honestly, it's not that hard (induction on the number of $\lambda_i$'s), and in general, you should stop asking people incessantly for hints and whatnot.  You write a huge number of posts with no real content, and it shouldn't really continue.  Rafal gave you more than enough information to prove the result you're looking for here, and I have written the proof for

$\frac {f(x)+f(y)}{2}\geq f\left(\frac {x+y}{2}\right)\implies$

$\frac {\sum_i f(x_i)}{n}\geq f\left(\frac {\sum_i x_i}{n}\right)$

on this forum on at least three different occasions.

I think I get it now. I am sorry if I posted useless posts. I wasn't clear about the definitons previously , and that is why I was confused.

The definiton of convexity IS that $ \frac{ f(x_1) + f( x_2) }{2} \geq f ( \frac{x_1 + x_2}{2} ) $ as aspecial case and a general form also!!!Ok now ,everything is foine and all the proofs are ok.

Obviously , the general case is easy once we establish tge above

Again, Thanks!!! Please correct me if I am wrong!!!

[quote="Rushil"]i'm basically asking for a proof of Jensen's inequality!!![/quote]

I don't understand what you want. If you want the proof of Jensen's, Rafal has written the outline of solution to you. If you want to know something about convexity, you have to know that $\frac{f(x)+f(y)}{2} \ge f(\frac{x+y}{2})$ means f is convex only when f is continuous. Same for the second derivative test. Moreover, $\frac{f(x)+f(y)}{2} \ge f(\frac{x+y}{2})$  is more primitive, because the definition of convexity is that $af(x)+(1-a)f(y) \ge f(xa+(1-a)y)$ with $0<a<1$. So I think it's strange to ask one to prove from $f''(x) \ge 0$ to $\frac{f(x)+f(y)}{2} \ge f(\frac{x+y}{2})$

[quote="Rushil+Edits"]Basically I am asking for the proof that if $ f''(x) > 0$ , prove that
$ \frac { f (x_1) \plus{} f(x_2) }{2} \geq f ( \frac {x_1 \plus{} x_2}{2} )$.

The general case can easily be obtained as you have shown above.. ,once we establish the inequality for $ n \equal{} 2$. My question is this!! To relate the inequality for $ n \equal{} 2$ with concavity without using graphs!! A rigorous proof plz![/quote]

$ f''(x) > 0 \rightarrow f'$ is is increasing.
let $ x_{1} < x_{2} \rightarrow x_{1} < \frac {x_{1} \plus{} x_{2}}{2} < x_{2}$
so $ f(x_{2}) \minus{} f(\frac {x_{1} \plus{} x_{2}}{2}) \equal{} \frac {f'(c_{1}) (x_{2} \minus{} x_{1})}{2}$ for some $ c_{1} \in (\frac {x_{1} \plus{} x_{2}}{2},x_{2})$.
also $ f(\frac {x_{1} \plus{} x_{2}}{2}) \minus{} f(x_{1}) \equal{} \frac {f'(c_{2}) (x_{2} \minus{} x_{1})}{2}$ for some $ c_{2} \in (x_{1},\frac {x_{1} \plus{} x_{2}}{2})$.
clearly $ f'(c_{1}) > f'(c_{2})$ hence
considering the case $ x_{1} \equal{} x_{2}$
we have
$ \boxed{\frac { f (x_1) \plus{} f(x_2) }{2} \geq f ( \frac {x_1 \plus{} x_2}{2} )}$


Given that $ \bigtriangleup BCE$ and the parallelogram $ ABCD$ have equal area, what is the length of $ AB$ in centimeters?
[asy]size(200); import geometry_dev; defaultpen(linewidth(0.8));
pair A = (0,0), B = (4,0), C = (4 + sqrt(7),3), D = (sqrt(7),3), E = (12,0), F = (4 + sqrt(7),0);
draw(A--B--C--D--cycle);
draw(C--E--B);
draw(Label("3 cm"),C--F,dashed); markrightangle(C,F,E);
draw(Label("8 cm", align=S), (B.x+0.5,B.y - 0.5)--(E.x-0.5,E.y - 0.5),Arrows);
label("$A$",A,SW); label("$B$",B,S); label("$C$",C,N); label("$D$",D,NW); label("$E$",E,S);[/asy]

$ BE\cdot h/2\equal{}h\cdot AB$
$ AB\equal{}BE/2\equal{}4$


if    a/b = c/d       a ≠ 0 , b ≠ 0 , c ≠ 0 , and  d ≠ 0  which of the following is true ?

a)  a/d = c/b      b) ac = bd       c)  a/c = b/d       d)  b/a = c/d

hello, the answer c) is true $ \frac{a}{c}\equal{}\frac{b}{d}$.
Sonnhard.

[hide="Solution"] $ \frac{a}{b}\equal{}\frac{c}{d} \implies ad\equal{}bc \implies \frac{a}{c}\equal{}\frac{b}{d}$.[/hide]


Natural numbers [b]n[/b], [b]m[/b] satisfy equation:
[b]3n^2 + n = 4m^2 + m[/b].
Prove that [b]n-m=p^2[/b],  p is an integer number

Posted many times. Search for Iran 1997 or something like that.

Factorise it and youre done otherwise use Pells equation...

Bomb


Frankly, I have read over the TOU and I have obeyed every single rule in it. There was not a single one that I could have possibly broken, it was yesterday's topic about "butt vs boobs" and which you prefer on a women. I feel like I have been abused, and not treated fairly because of made a topic that this forum usually does not see.

[quote="Kuku_King"]...made a topic that this forum usually does not see.[/quote]
I wonder why?

As I said in that topic, it is offensive to women. If you think it's not, then I suggest you just rethink.

how is it offensive?

What if a guy came and said "Hey which puppy looks cuter, I want to get one."

Would that also be deleted? How about if some one said "I want to own a farm, which cow looks better?" Is that also going to be deleted?

This is just prejudice because I made a topic that this topic doesn't usually recieve.

Now you're comparing women to puppies and cows? Good luck...

Go complain to the admins if you're not willing to see any sense.

Thats your counter?

Are you saying that now animals aren't living beings and don't have feelings? More prejudice. I have not broken any rule in TOU, and I brought examples of how I am being prejudiced.

Ok what about if some one said "Which girl looks better?" Would THAT be deleted?

Yes, that's my counter. No need to start flaming because of one topic.

For all you know, the topic could've been deleted because it sparked some anger, and not because of the original idea.

If you're not willing to at least try to understand why it was deleted, then what's the point of making this topic in the first place?

Because...there is no reason for it to be deleted...when it doesn't break the TOU.

However, some people may consider that offensive, which, in my opinion, is a good enough reason to have that topic removed.

I agree with i_like_pie, and lingomaniac88.
That topic was not only offensive, but also inappropriate.
How is it inappropriate?
You were discussing about females' private part and wanting to know which one we like better.
What do you call that?

It doesn't matter if it's in the TOS or not, the admins can do whatever they want on their website. No explanation is necessary.

If you don't feel it is fair, you are welcome to show your displeasure by leaving the website.


There are two mistakes you have to fix. First, you multiplied some of the terms by 16x instead of 16. Second, I'm not sure if by 3x+2/16 you meant 3x+1/8 or (3x+2)/16. If it was the former, you didn't multiply 3x by 16. If it was the latter, it should've become -3x-2, because you are subtracting the [i]quantity[/i] of 3x+2, not just 3x. If you fix that you should get the right answer.

[quote="yeoviltown_fc"]2x+1/7 - 3x+2/16 = 1

times by 16 to give:

16x(2x)+16x(1)/7 - 3x+2 = 16

32x +16/7 - 3x+2 = 16

times by 7 to give:

32x+16 - 21x+14=112

11x +30 = 112
11x= 82
[b]x=82/11

Where have I gone wrong, as i seem to think the answer is simply x=10[/b][/quote]

Ah I see it now. Thanks for your help.

32x +16/7 - 3x+2 = 16

times by 7 to give:

32x+16 - 21x[b]+14[/b]=112        I relised that you have to subtract the 14 and not add it basically

11x= 110
11x= 10
[b]x=110/11 =10! YAY ![/b]


Three of the six faces of a cube are selected at random.  What is the probability that they all share a common vertex? Express your answer as a common fraction.

the answer is 4/5 because there is a total of 6+5+4+3+2+1 faces or 10 and there are 8 vertexes... and 8/10 simplifies to 4/5

Please hide your answers :!:
[hide]
There are $\binom 63$ ways to choose the faces and each vertex has 1 set of 3 faces that come together on it.
So the answer is $\frac{8}{20}=\boxed{\frac 25}$[/hide]

Or you could do this:

[hide="Other method"]First face chosen does not matter. For the second face, 4 out of 5 work. For the third face, 2 out of 4 work. Therefore, $1\cdot \frac{4}{5}\cdot \frac{2}{4}=\boxed{\frac{2}{5}}$.[/hide]


abstract algebra
superior algebra
superior algebra unsolved

Hi all,

I have a problem with Boolean algebra. How do I tackle the following problem?

Prove in Boolean rings that if

a+b=a+c

and

a.b=a.c

Then b=c.

Thanks!

What is your definition of a Boolean ring? I'm used to there being subtraction in such a ring.

[quote="enoughsaid05"]Hi all,

Prove in Boolean rings that if

$ a\plus{}b\equal{}a\plus{}c$

and

$ a \cdot b\equal{}a \cdot c$

Then $ b\equal{}c.$

Thanks![/quote]

Let R be a Boolean ring. Then (R,+) is an abelian group. Hence, a+b=a+c implies b=c and the proof terminates.


For all $k\in\mathbb{N}$, the number $k!$ can be expressed in the form $2^{A_{1}}3^{A_{2}}5^{A_{3}}\dots{p_{n}}^{A_{n}}\dots$, where $p_{n}$ is the $n^\text{th}$ prime and all $A_{n}$ are integers greater than or equal to zero.

1. Prove that $A_{1}\ge A_{2}\ge A_{3}\ge\dots$.
2. Is there a formula for finding out the number of zeroes that $k!$ ends with when written in base $B$?

[hide]2: There is a formula only if $B$ is a compostive number (not sure about proof yet). For example, with $B=10$, a $0$ shows up at the end of $k$ factorial for every pair of $2$ and $5$ in the ppf of $k!$. However, since the $5$'s are much less common, they ultimately determine how many of these pairs we have. Formula is $\lfloor \frac{k}{5}\rfloor+\lfloor \frac{k}{5^{2}}\rfloor+\ldots$ (counts the number of multiples of $5$'s, the number of multiples of $5^{2}$, etc.
Another example is $B=6$. Since $B=6=2*3$, a $0$ occurs for every pair of a $2$ and a $3$. $3$'s are rarer, so we would see how many factors of $3$ $k!$ has in its ppf.[/hide]

Drunner, I think your solution is flawed because what if base B is prime factorized at 2^2*3 would that still work unless there is something I am missing.[/hide]

[quote="noneoftheabove"]Drunner, I think your solution is flawed because what if base B is prime factorized at 2^2*3 would that still work unless there is something I am missing.[/quote]
Indeed -- that solution will work nicely in the prime case and in selected composite cases (where there is one "big" prime factor), but it is not quite so nice in general.  Of course, that can be solved with a quick application of the $\min$ function.

yeah in the case where $B=2^{2}*3=12$ things get more complicated. you get a $0$ from a pair of $4$ and $3$ or a pair of $6$ and $2$. i believe that if $a$ is the number of pairs of $4$ and $3$ and $b$ is the number of pairs of $6$ and $2$ then $a=b$. that means you could just count the number of pairs of $4$ and $3$ and not even worry about the pairs of $6$ and $2$.
i'm not 100% sure about these cases though

[quote="lingomaniac88"]I have a slightly easier question, but is kinda based on the first one I posted.

Is there a formula for finding out the number of zeroes that $k!$ ends with when written in base $B$, where $B=p_{f}p_{g}$ and $p_{n}$ is the $n^\text{th}$ prime number?[/quote]

If a number has $k$ zeros at the end in base $B$, then it is divisible by $B^{k}$ and not $B^{k+1}$. So count the number of times $p_{f}$ and $p_{g}$ appear in the product $k!$ and take the minimum (obviously the larger prime), which gives you $k$.


Let $ a,\,b,\,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}1.$ Prove that
\[ \frac a{1\plus{}a}\plus{}\frac b{1\plus{}b}\plus{}\frac c{1\plus{}c}\ \geqslant\ \frac1{1\plus{}a^2\plus{}b^2\plus{}c^2}\]

this is a direct result of Cauchy-Schwarz (or Titu's lemma, but that's just Cauchy-Schwarz)  :lol:

\[ \frac{a}{1\plus{}a}\plus{}\frac{b}{1\plus{}b}\plus{}\frac{c}{1\plus{}c}\ \equal{} \frac{a^2}{a\plus{}a^2}\plus{}\frac{b^2}{b\plus{}b^2}\plus{}\frac{c^2}{c\plus{}c^2}\ \geq\ \frac{1}{1\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}}\]

\[ \frac {a}{1\plus{}a}\plus{}\frac {b}{1\plus{}b}\plus{}\frac {c}{1\plus{}c}\ \geqslant\ \frac{1}{1\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}}\]
\[ f(x)\equal{}\frac{1}{x}\]
\[ af(1\plus{}a)\plus{}bf(1\plus{}b)\plus{}cf(1\plus{}c) \geqslant\ f[a(1\plus{}a)\plus{}b(1\plus{}b)\plus{}c(1\plus{}c)]\]
jensen

Thanks. :)

[quote="exodd"]\[ f(x) \equal{} \frac {1}{x}\]

\[ af(1 \plus{} a) \plus{} bf(1 \plus{} b) \plus{} cf(1 \plus{} c) \geqslant\ f[a(1 \plus{} a) \plus{} b(1 \plus{} b) \plus{} c(1 \plus{} c)]\]
jensen[/quote]
What about $ f(x)\equal{}\frac1{1\plus{}x}\,?$ That’s also convex. ;)

[quote="itiselizabeth"]Thanks. :)

[quote="exodd"]
\[ f(x) \equal{} \frac {1}{x}\]

\[ af(1 \plus{} a) \plus{} bf(1 \plus{} b) \plus{} cf(1 \plus{} c) \geqslant\ f[a(1 \plus{} a) \plus{} b(1 \plus{} b) \plus{} c(1 \plus{} c)]\]
jensen[/quote]
What about $ f(x) \equal{} \frac1{1 \plus{} x}\,?$ That’s also convex. ;)[/quote]

yes, that works as well.


[u]Definition[/u]: For any $ x\in\mathbb{R}$ and integer $ k\geq 0$ we define $ \binom{x}{0}\equal{}1$ and $ \binom{x}{k}\equal{}\frac{x(x\minus{}1) \cdots (x\minus{}k\plus{}1)}{k!}, \ \ k \geq 1.$

Prove that for any integers $ m,n,k$ where $ k\geq 0$ and $ n\geq 1$  we have:
$ \binom{\frac{m}{n}}{k} \in \mathbb{Z}[1/n].$

That isn't a proof is it? In the real numbers, is it obvious $ a\plus{}(\minus{}b) \equal{} \equal{}0$ if and only if $ \minus{}a \equal{} \minus{}b$? Obviously, but still prove it :P


Given that $a,b,x,y$ are all positive integers. Find all possible cases of $a,b$ that $ax^2+x = by^2+y$ implies $x-y$ is a perfect square.

Pascual2005 , I didn't understand :blush: What do you mean :oops:

He means it will work if $b-a$ is a perfect square  :)
But what about other cases?

I realy want to say what warut_suk saied and I think it 's just true for the cases that
b-a is a perfect squre :lol:  :clap:
because I 've seen a question like this:
If 3*x^2+x=4*y*2 then x-y is a perfect squre.
And I know the answer. :yup:

Well, how to prove it that $b-a$ not a perfect square won't work? Is it obvious? Why?


modular arithmetic
IMO
IMO Shortlist
number theory
Perfect Squares
IMO 1996
quadratic reciprocity

The positive integers $ a$ and $ b$ are such that the numbers  $ 15a \plus{} 16b$ and $ 16a \minus{} 15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?

Lets denote $15a+16b=x^2$, $16a-15b=y^2$ so we have $x^4+y^4=(15^2+16^2)(a^2+b^2)$,
since $x^4=(15b+16b)^2$,$y^4=(16a-15b)^2$ summing up this two we get desired.
So $x^4+y^4=481(a^2+b^2)$ also $481=13\times 37$.
We note that $m^4\equiv -1 (mod13)$(*) for m, natural number, but $m^12\equiv1(mod13)$ [by F.L.T] therefore (*) leads to contradiction.
Analogously $m^4\equiv-1(mod37)$ gives contradiciton since $m^36\equiv1 (mod37)$.
So $x^4+y^4 \equiv0 (mod13)$ states that $x\equiv y\equiv 0 (mod13)$, neither x and y doesnt leave 0 on division by 13. And the last one is impossible, since $m^4+n^4 \equiv 0 (mod13)$ and $m^4$ doesnt leave $-1(mod13)$, if
$m^4+n^4\equiv0 (mod13)$ and $n$ is not congruent to $0$ in $mod13$; noting that 13- is prime, we have a natural k, such that $kn\equiv 1(mod13)$, but from here $(mk)^4\equiv -(nk)^4 \equiv -1$. Thus $x\equiv y\equiv0 mod13$ and analogously $x\equiv y\equiv 0mod37$.
Resulting we have $x,y\geq481$ and minimal value of $x^2,y^2$ is $481^2$

Answer: $481^2$

davron

Here's a different solution.

Put $ 15a \plus{} 16b \equal{} x^{2}$ and $ 16a \minus{} 15b \equal{} y^{2}$ as [b]davron[/b] did. Then $ 240a \plus{} 256b \equal{} 16y^{2}$ and $ 240a \minus{} 225b \equal{} 15y^{2}$; subtracting gives $ 481b \equal{} 16x^{2} \minus{} 15y^{2}$. Also, $ 225a \plus{} 240b \equal{} 15x^{2}$ and $ 256a \minus{} 240b \equal{} 16y^{2}$; adding gives $ 481a \equal{} 15x^{2} \plus{} 16y^{2}$.

Note that $ 481 \equal{} 13\cdot37$. So we have $ 37|(16x^{2} \minus{} 15y^{2}) \minus{} (15x^{2} \plus{} 16y^{2}) \equal{} x^{2} \minus{} 31y^{2}$, so $ x^{2}\equiv31y^{2}\pmod{37}$. Assume that $ y$ is not divisible by 37, so it is relatively prime to 37. Then, $ x$ isn't divisible by 37 either. We have that $ 31y^{2}$ is a quadratic residue modulo 37, but $ \Big(\dfrac{31y^{2}}{37}\Big) \equal{} \Big(\dfrac{31}{37}\Big)\cdot\Big(\dfrac{y^{2}}{37}\Big) \equal{} \Big(\dfrac{6}{31}\Big)\cdot( \minus{} 1)^{15\cdot18}\cdot1 \equal{} \Big(\dfrac{2}{31}\Big)\cdot\Big(\dfrac{3}{31}\Big) \equal{} ( \minus{} 1)^{15\cdot16}\cdot\Big(\dfrac{1}{3}\Big)\cdot( \minus{} 1)^{1\cdot15} \equal{} \minus{} 1$, contradiction. So $ 37|x,y$.

We also have $ 13|x^{2} \minus{} 31y^{2}$, so $ x^{2}\equiv5y^{2}\pmod{13}$. Assume that $ y$ is not divisible by 13, so it is relatively prime to 13. Then, $ x$ isn't divisible by 13 either. We have that $ 5y^{2}$ is a quadratic residue modulo 13, but $ \Big(\dfrac{5y^{2}}{13}\Big) \equal{} \Big(\dfrac{5}{13}\Big)\cdot\Big(\dfrac{y^{2}}{13}\Big) \equal{} \Big(\dfrac{3}{5}\Big)\cdot( \minus{} 1)^{2\cdot6}\cdot1 \equal{} \Big(\dfrac{2}{3}\Big)\cdot( \minus{} 1)^{1\cdot2}\cdot1 \equal{} \minus{} 1$, contradiction. So $ 13|x,y$.

Now, $ 481|x,y$, and since 481 is square-free, $ 481^{2}|x^{2},y^{2}$, and $ x^{2}$ and $ y^{2}$ are at least $ 481^{2}$. To get equality, take $ a\equal{}481\cdot31, b\equal{}481$.

Yet another solution.

The minimum possible value may be attained by setting $a = 13 \cdot 31\cdot 37$ and $b = 13 \cdot 37$, which causes $15a + 16b = 16a - 15b =481^2$. We prove this below.

Let $15a + 16b = m^2$ and $16a - 15b = n^2$ for integral $m,n$. Solving for $a$ and $b$ in terms of $n$, we get:

\[
a=\frac{16m^2 - 15n^2}{13\cdot 37},b=\frac{15m^2 + 16n^2}{13\cdot 37}
\]

Since $a$ is integral, we must have that $16m^2 \equiv 15n^2 \bmod 13\Rightarrow m^2 \equiv 5n^2 \bmod 13$. The quadratic residues, modulo $13$ are $0$, $\pm 1$, $\pm3$, $\pm 4$. Multiplying these residues by $5$ yields $0$, $\pm 2$, $\pm 5$, $\pm 7$, modulo $13$, suggesting that $13 \mid m, n$.

Further, we must have that $16m^2 \equiv 15n^2 \bmod 37 \Rightarrow m^2 \equiv 31n^2 \bmod 37$. The quadratic residues, modulo$37$ are $0$, $\pm 1$, $\pm 3$, $\pm 4$, $\pm 7$, $9$, $\pm 10$, $\pm 11$, $\pm 12$, $\pm 16$. Multiplying these residues by $31$ yields $0$, $\pm 2$, $\pm 5$, $\pm 6$, $\pm 8$, $\pm 13$, $\pm 14$, $\pm 15$, $\pm 18$, $20$, modulo $37$, suggesting that $37 \mid m, n$.

Letting $m = 13 \cdot 37 \cdot m_1, n = 13 \cdot 37 \cdot n_1$ and substituting, we get that $a = 13\cdot 37 \cdot (15m_1^2 + 16n_1^2)$ and $b = 13\cdot 37 \cdot (16m_1^2 - 15n_1^2)$. Substituting once more, our we get that $15a + 16b = 13^237^2m_1^2$ and $16a - 15b = 13^237^2n_1^2$. Setting $m_1 = n_1 = 1$ yields the desired result.

[hide = Solution]
The answer is $481^2$. Let $15a + 16b = x^2$ and $16a - 15b = y^2$. Then solving for $(a, b)$ gives \[(a, b) = \left(\frac{15x^2 + 16y^2}{15^2 + 16^2}, \frac{16x^2 - 15y^2}{15^2 + 16^2}\right).\] Note that $15^2 + 16^2 = 481 = 13\cdot 37$, so we must have \[13 \cdot 37 \mid 16x^2 - 15y^2.\] But $16$ is a square, and $15 \equiv 2$ is not a square mod $13$ because $13 \not\equiv \pm 1 \pmod{8}$, so $13$ must divide both $x$ and $y$. Similarly, $15$ is not a square mod $37$ by Quadratic Reciprocity: \[\left(\frac{15}{37}\right) = \left(\frac{3}{37}\right)\left(\frac{5}{37}\right) = \left(\frac{37}{3}\right)\left(\frac{37}{5}\right) = 1\cdot (-1) = -1,\] so $37$ must divide both $x$ and $y$ as well. So then $x$ and $y$ are both multiples of $481$, meaning $x^2$ and $y^2$ are at least $481$. On the other hand, taking $x = y = 481$ gives a positive integer solution for $a$ and $b$, so $481^2$ is attainable.
[/hide]

Check [url=https://artofproblemsolving.com/community/c146h150496p25272692]here[/url] for my solution and some others.


The homologation of functional groups is an important tool in organic synthesis. In this and the next problems we could explore some examples. Let's begin with carboxilic acids.

Devise a method for converting a carboxilic acid to its next higher homolog:

$RCO_{2}H \longrightarrow RCH_{2}CO_{2}H$

Carcul

Arndt-Eistert Synthesis
via wolff  rearrangement

That's right. The Arndt-Eistert synthesis is the standard procedure for the homologation of acids, and it is made up of three steps:

1) Conversion of the acid to the acyl halide (with thionyl chloride for example).

2) Reaction of the acyl halide with an excess of diazomethane to produce a diazoketone.

3) Conversion of the diazoketone into the homologated acid with H2O/Ag2O, photolysis, or simply heating. This step makes use of the [i]Wolff Rearrangement[/i], where a diazoketone rearranges into a ketene.

How about problem 7?


Let $F(x)\in Z[x]$, and $F(1),F(2),...,F(n)$ is all not divisible by $n$. Is it necessary that $F(x)$ has no integer roots?

Suppose $x$ is an integer root of $f$.  Then $f(x) = 0$ mod n.  However, consider the function modulo n.  Then because $x^k$ mod n = $r^k$ mod n, where r is the residue of x, it follows that F(r) is not zero mod n.  Contradiction.  So there are no integer roots of f.


A $100\times 100$ chessboard is cut into dominoes ($1\times 2$ rectangles). Two persons play the following game: At each turn, a player glues together two adjacent cells (which were formerly separated by a cut-edge). A player loses if, after his turn, the $100\times 100$ chessboard becomes connected, i. e. between any two cells there exists a way which doesn't intersect any cut-edge. Which player has a winning strategy - the starting player or his opponent?

[quote="Summerburn"]Is the board cut in an arbitrary way, or is it cut in the "normal" way?[/quote]
What do you mean by "normal" way?

[i] Second person has the winning strategy.[/i]
[b]LEMMA :[/b] the maximum number of moves that are needed for the second person to punch all dominoes all arround the board so that raising one corner of the board makes all the border dominoes raised(call [b]A[/b]), is less than needed moves for one to lose(call [b]B[/b]).
$\blacktriangledown$ let second person starts punching all squares which are in the side of BIG square with all of their neighbours, it takes no more than 8*99 moves , summing up it with [i]first person's moves[/i] it will get no more than 2*8*99 moves, for which any square belongs to exactly 1 dominoe , so the number of all dominoes punched together is no more than 2*2*8*99 which is clearly less than the whole number of dominoes;proved the Lemma $\square$.
[b]LEMMA :[/b] any connected set of dominoes with no square in its inner part not belonging to it, has an even number of segments as its sides and has an even number of small square's sides on its border.
$\blacktriangledown$ first part of Lemma caused by surrounding any horizontal segment by two vertical segments and vice versa so that  the number of horizontal segments is equal to the number of vertical segments, the second part came from shadowing this shape  so that the shape become a line parallel to $x$ axes,it's obviouse that the thickness of this line is twice more than the thickness of the line which makes the border of the shape, same arguement for $y$ axes will get the same result $\square$

let the second person punchs all dominoes in the border of the BIG square, and then play illegantly; assume at one point, second player has no move to do i.e for any 2 squares which he punchs them,the BIG square becomes a connected shape. it means that the BIG square is now a collection of two connected shape,in which any two squares belong to one figure are punched to eachother and any two squares which do not belong to a same figure, are not connected, hence the number of all punchs which has been not uses untill now is even by the second Lemma, on the other hand the number of punchs needed to punch all the adjacent squares to eachothers, is even, following that the number of punchs used till now is even and so it's now the first player's turn. Contradiction.$\square$
:)

Sorry, but what does it mean to punch all dominoes or raise a corner? Thanks!

Does anyone have a solution to this problem or an explanation of the solution above? Thank you so much for all your help!

We will prove for $n \times n$ chess board for sufficient large $2 \mid n$ that the second person has a winning strategy.

First, consider a tile $X$ in $2n \times 2m$ (in here, we refer to a tile if there is no cut-edge in that tile), each cell in tile $X$ can glue/connect to some different cells in a different tile, the sum of connection of all the cells in tile $X$ is called [i]connected number[/i]. We call a tile [i]odd tile[/i] if it has odd connected number, otherwise [i]even tile[/i]. For example, in the square $4 \times 4$ below: the red tile is [i]odd tile[/i] because it has connected number of $3$, yellow tile is [i]odd tile[/i] with connected number of $9$, purple tile is [i]even tile[/i] with connected number of $4$.
[asy]
size(4cm);
for (int i=0; i<=4; ++i) { draw((0,i)--(4,i)^^(i,0)--(i,4)); }
fill((0,1)--(0,2)--(1,2)--(1,3)--(3,3)--(3,2)--(2,2)--(2,1)--cycle,linewidth(2)+yellow);
fill((2,0)--(3,0)--(3,2)--(2,2)--cycle, blue);
fill((0,0)--(2,0)--(2,1)--(0,1)--cycle, red);
fill((1,3)--(3,3)--(3,4)--(1,4)--cycle, purple);
fill((0,2)--(1,2)--(1,4)--(0,4)--cycle, green);
fill((3,0)--(4,0)--(4,4)--(3,4)--cycle, grey);
[/asy]
We start with $1 \times 2$ dominos. It not hard to observe that odd tile $1\times 2$ can only appear at the border of the board (for example red tile, green, blue), all tiles not at the border has connected number of $6$, so it's all [i]even tiles[/i]. Not all $1 \times 2$ tiles at the border are odd tiles (e.g. purple tiles), we notice that odd tiles can be either at the conner of the board (green, red tiles) or connect to the border with its shorter edge (blue tile). Hence, it not hard to see that there are even number of odd tiles on the chess board (because the length of the chessboard are even).

On the other hand, we can prove that if we connect two odd tiles we get an even tile (grey tile obtained by connecting two $1\times 2$ dominos), if we connect one odd tile to one even tile we get odd tile (yellow tile).

Back to the problem, let two people be $A,B$ and $A$ starts. We observe that if we need to glue all the cells together, we need an even number of connections $\tfrac 32 n^2- 2n$ to make (e.g. for $4\times 4$ chessboard, we need $16$ connections). Hence, if after some turn, all there left are two odd tiles then the next turn must belong to $B$, which means $B$ loses. If there are two even tiles left then $A$ loses.

From this, we think of a strategy of winning to $B$: at every turn, he must connect all odd tiles together to create an even tile. By doing this, eventually, there will be two even tiles left, which means it's $A$'s turn to connect these two tiles, means $A$ loses. Since all the odd tiles are on the borders, the maximum number of connection $B$ have to make is $4n-8$. $A$ can stop $B$ from doing this if he can create two odd tiles in at most $4n-8$ moves. This is impossible because there are at least $\tfrac 12 (n-4)^2$ even tiles $1 \times 2$ which need at least $(n-4)^2$ connections to make (each tile has minimum connected number of $4$ and there are $\tfrac 12 (n-4)^2$ tiles). For sufficient large $n$ then $(n-4)^2>4n-8$. Thus, $A$ can't stop $B$ from creating two even tiles.

Thus, $B$ has a winning strategy.

[b]Remark.[/b] $B$ can also win in $2m \times 2n$ chessboard for any $m,n \in \mathbb{N}$.

Wow, thank you so much for this proof! I really appreciate all your help.

[quote=MathPanda1]No, no it is a great explanation, my brain just froze completely. Thank you so much for your amazing proof! May I ask, how did you come up with such a clever proof? Thanks![/quote]

I try for $4 \times 4$. When I was doing that, I realised the [i]odd, even[/i] tiles.  :blush:

Interesting, thank you for all your help!

[quote=shinichiman][quote=MathPanda1]Oh sorry, I suddenly realized that I don't completely understand your proof  :blush:  Why is it that if after some turn, all there left are two odd tiles then the next turn must belong to $B$, which means $B$ loses and if there are two even tiles left then $A$ loses? Thank you so much for all your help![/quote]

Because there are total of even connections (which is $\tfrac 32 n^2-2n$) so if there are two odd tiles left, that means there are odd number of connections (that connect those two) left, that means it's $B$'s turn. $B$ must connect those two odd tiles which connects the chessboard so $B$ loses.

It's ok, it's because my explanation is not that good.[/quote]
Noope. You may have some connections which were omitted since the squares they connect (which are now part of a tile) were already connected by another path. Your solution is wroong.


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$  Prove that
1)
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [5]{\frac {a^2b \plus{} b^2c \plus{} c^2a}{3}}
\]
2)
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^3b \plus{} b^3c \plus{} c^3a}{3}}
\]
[hide="remark"] With the same conditions $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [4]{\frac {a^2b \plus{} b^2c \plus{} c^2a}{3}}$ and
$ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3b \plus{} b^3c \plus{} c^3a}{3}}$ are wrongs already.[/hide]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$  Prove that
1)
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [5]{\frac {a^2b \plus{} b^2c \plus{} c^2a}{3}}
\]
[hide="remark"] With the same conditions $ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [4]{\frac {a^2b \plus{} b^2c \plus{} c^2a}{3}}$ and
$ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10]{\frac {a^3b \plus{} b^3c \plus{} c^3a}{3}}$ are wrongs already.[/hide][/quote]
It is equivalent to: $ (a\plus{}b\plus{}c)^5 \ge 27(a^2b\plus{}b^2c\plus{}c^2a)(ab\plus{}bc\plus{}ca)$
and,we have an nice solution by can_hang2007 :)


function
advanced fields
advanced fields unsolved

Let $A$ and $b$ be two sets with size m and n respectively. show that the number of onto functions from A to B is $^{n}C_{0}n^{m}-^{n}C_{1}(n-1)^{m}+^{n}C_{2}(n-2)^{m}-^{n}C_{3}(n-3)^{m}......^{n}C_{n}(n-n)^{m}$.

[hide]
Hint: use the priciple of inclusion and exclusion.[/hide]

There's another way, which is to develop a recursive identity for this function of $m$ and $n.$ Try it.


A set of balls contains $ n$ balls which are labeled with numbers $ 1,2,3,\ldots,n.$ We are given $ k > 1$ such sets. We want to colour the balls with two colours, black and white in such a way, that

(a) the balls labeled with the same number are of the same colour,

(b) any subset of $ k\plus{}1$ balls with (not necessarily different) labels $ a_{1},a_{2},\ldots,a_{k\plus{}1}$ satisfying the condition $ a_{1}\plus{}a_{2}\plus{}\ldots\plus{}a_{k}\equal{} a_{k\plus{}1}$, contains at least one ball of each colour.

Find, depending on $ k$ the greatest possible number $ n$ which admits such a colouring.

The answer is $ n\equal{}k^{2}\plus{}k\minus{}2$.

===============================================
Lower bound: $ n\equal{}k^{2}\plus{}k\minus{}2$ is possible.

Use color white for the numbers $ 1,\ldots,k\minus{}1$ and for $ k^{2},\ldots,k^{2}\plus{}k\minus{}2$ .
Use color black for the numbers $ k,\ldots,k^{2}\minus{}1$.

If there was a monochromatic black solution to the equation, then
$ a_{1},\ldots,a_{k}$ all are $ \ge k$. Hence $ a_{k\plus{}1}\ge k^{2}$; contradiction.
If there was a monochromatic white solution to the equation, then
$ a_{k\plus{}1}\ge k^{2}$; hence one of $ a_{1},\ldots,a_{k}$ must be strictly greater
than $ k\minus{}1$. Hence one of $ a_{1},\ldots,a_{k}$ must be at least $ k^{2}$, and the
others are all at least 1. Hence their sum is at least $ k^{2}\plus{}k\minus{}1$; contradiction.

===============================================
Upper bound: $ n\equal{}k^{2}\plus{}k\minus{}1$ is not possible.

For the sake of contradiction assume that there is a solution with $ n\equal{}k^{2}\plus{}k\minus{}1$.
Assume w.l.o.g. that 1 is colored white.
Then $ k$ must be black.
Then $ k^{2}\equal{}k\plus{}\cdots\plus{}k$ must be white.
Then $ x\equal{}k^{2}\plus{}k\minus{}1\equal{}k^{2}\plus{}(1\plus{}1\plus{}\cdots\plus{}1)$ must be black.
And $ y\equal{}k^{2}\minus{}k\plus{}1$ must be black, since $ y\plus{}(1\plus{}\cdots\plus{}1)\equal{}k^{2}$.

Furthermore 2 must be white, since $ y\plus{}(2\plus{}\cdots\plus{}2)\equal{}x$.
Since 1 and 2 are white, $ k,\ldots,2k$ must be black.
Since $ k,\ldots,2k$ are black, $ k^{2},\ldots,k^{2}\plus{}k\minus{}1$ must be white.
Hence $ x\equal{}k^{2}\plus{}k\minus{}1$ must be white.  Contradiction.
===============================================


[hide="Hint"] The order of the cards in the deck determines exactly which cards Sampson draws and which card Dirk draws. [/hide]

[hide]
Building on the hint:
After Sampson's 1st draw, there are 51 cards left.
Number them 1-51 by order.  Sampson will draw all the even numbered cards (25 of them) and Dirk will draw all the odd numbered cards(26 of them).
The ace of spades needs to be one of the even numbered cards for Sampson to draw it, so the probability is $\frac{25}{51}$[/hide]

[quote="mr. math"]Sampson and dirk alternately draw cards from a (face down) deck of ordinary playing cards. Sampson draws first and his first draw is the Ace of Diamonds. If they continue drawing alternately, what is the probablity that Sampson will be one to draw the Ace of Spades?

How would you do this problem?[/quote]

[hide]There are 51 cards left.  The following sequence gives the prbability that Sampson draws the Ace of Spades:

$\frac{50}{51}\frac{1}{50}+\frac{50}{51}\frac{49}{50}\frac{48}{49}\frac{1}{48}+\frac{50}{51}\frac{49}{50}\frac{48}{49}\frac{47}{48}\frac{46}{47}\frac{1}{46}+\cdots$

$\frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\cdots$

There should be 25 terms in the sequence so $\boxed{\frac{25}{51}}$[/hide]

[edit]:just realized circlesquared has a solution thats 10 times better.  why didnt i think of that?


function
rotation
calculus
calculus computations

Hi, i need help with a problem:

The function $ y \equal{} x^2$ is rotated about the y-axis and y lies in [0,h].
The rotated graph forms a tank with the volume  $ V \equal{} \frac{\pi}{2} h^2 $

This tank is filled with water at a constant speed $ 2 \frac{\ m^3}{s} $ .
How rapidly does the water height increase when it is 1 m?

thanks!

$ V \equal{}\frac{\pi}{2}h^{2}$
Differentiate with respect to t, to get:
$ \frac{\mathrm{d} V}{\mathrm{d} t}\equal{}\pi h \frac{\mathrm{d} h}{\mathrm{d} t}$.
Substituting in given values, we get $ \boxed{\frac{2}{\pi}}$.


How many 4-digit number after the number 2009 is divisible by 2,3,4,5,6,7,8 and 9 ?

[hide]They are $ 2520$, $ 5040$ and $ 7560$, so $ 3$.

The numbers have to be divisible by $ 5, 7, 8, 9$ and $ 5\times 7\times 8\times 9\equal{}2520$[/hide]


[quote="SnowStorm"]Source: AHSME 1973 #1

A chord which is the perpendicular bisector of a radius of length $12$ in a cicle, has length

$\text{(A)} \ 3\sqrt{3} \qquad \text{(B)} \ 27 \qquad \text{(C)} \ 6\sqrt{3} \qquad \text{(D)} \ 12\sqrt{3} \qquad \text{(E)} \ \mbox{none of these}$[/quote]
[hide]
$\boxed{\text{(D)} 12\sqrt{3}}$
The lengths of the triangle constructed by a half chord that meets at the circle and the radius has hypotenuse $12$ and leg $6$, so the half chord has length $\sqrt{108}$. Multiply by two and simplify to get $12\sqrt{3}.$[/hide]

[hide]Let the chord touch the circle at X and Y and intersect the radius at P.

Using the equation of a circle, $PX^2+6^2=144^2$, since if the center of the circle is O, OP=6.
Thus $PX=6\sqrt3$, and $XY=2PX=12\sqrt3$. [/hide]


function
trigonometry
induction
algebra unsolved
algebra

Find all nonconstant $ f: R\rightarrow R$, such that for any real x,

$ f(\sin x+\cos x)=f(\sin x)+f(\cos x)$.

[quote="easternlatincup"]Find all nonconstant $ f: R\rightarrow R$, such that for any real x,

$ f(\sin x + \cos x) = f(\sin x) + f(\cos x)$.[/quote]

Very strange problem.

[hide="My solution"]

The equation is $ E(x)$ : $ f(\sqrt 2\sin(x+\frac{\pi}{4}))=f(\sin(x))+f(\cos(x))$

Let $ A=(-\infty,-\sqrt 2)$ $ \cup\{-1\}$ $ \cup (0,1]$ $ \cup (+\sqrt 2,+\infty)$

Consider that $ f(x)$ is defined arbitrarily in $ A$. We'll show first that all other values may be computed from there and then that the resulting function fit the requirement all over $ \mathbb R$

1) [u][b]when defined over $ A$, $ f(x)$ has mandatory values over $ \mathbb R\backslash A$[/b][/u]
1.1) For $ x\in(1,\sqrt 2]$ :
Let $ u\in(0,\frac{\pi}{4}]$ such that $ x=\sqrt 2\sin(u+\frac{\pi}{4})$ ($ u$ is unique)
Then $ \sin(u)\in A$ and $ \cos(u)\in A$ and $ E(u)$ $ \implies$ $ f(x)=f(\sin(u))+f(\cos(u))$

Notice that with this method, we also established that $ E(x)$ is verified $ \forall x\in(0,\frac{\pi}{4}]$

1.2) For $ x\in(-1,0]$ :
let the sequence of values in $ [\frac{\pi}{2},\pi)$ defined as :
$ u_0=\frac{\pi}{2}$
$ \forall n\geq 0$ : $ u_{n+1}=$ the unique root in $ [\frac{\pi}{2},\pi)$ of the equation $ \sin(x)+\cos(x)=\cos(u_n)$
$ u_n$ is a strictly increasing sequence whose limit is $ \pi$.

We'll define $ f(x)$ over $ x\in(-1,0]$ thru induction :

At the beginning, we have $ f(x)$ defined over $ (\cos(u_0),1]$

Consider now that $ f(x)$ is defined all over $ (\cos(u_n),1]$
For any real $ x\in(\cos(u_{n+1}),\cos(u_n)]$, $ \exists ! u\in [u_n, u_{n+1})$ such that $ x=\cos(u)$
$ \sin(u)\in(0,1]$ and so $ f(\sin(u))$ is defined
Since $ \sin(y)+\cos(y)$ is a decreasing function over $ [\frac{\pi}{2},\pi)$, $ \sin(u)+\cos(u)>\sin(u_{n+1})+\cos(u_{n+1})$ $ =\cos(u_n)$ and so $ f(\sin(u)+\cos(u))$ is defined.
Then $ E(u)$ $ \implies$ $ f(x)=f(\cos(u))=f(\sin(u)+\cos(u))-f(\sin(u))$
And $ f(x)$ is now defined over $ (\cos(u_{n+1}),\cos(u_n)]$ and so over $ (\cos(u_{n+1}),1]$

And since $ u_n\to\pi$, this allow us to define $ f(x)$ all over $ (-1,0]$

Notice that with this method, we also established that $ E(x)$ is verified $ \forall x\in[\frac{pi}{2},\pi)$

1.3) For $ x\in[-\sqrt 2,-1)$ :
Let $ u\in(\pi,\frac{5\pi}{4}]$ such that $ x=\sqrt 2\sin(u+\frac{\pi}{4})$ ($ u$ is unique)
Then $ \sin(u)\in [-\frac{\sqrt 2}{2},0)$ so $ f(\sin(u))$ is defined.
Same, $ \cos(u)\in (-1,-\frac{\sqrt 2}{2}]$ so $ f(\cos(u))$ is defined.
Then $ E(u)$ $ \implies$ $ f(x)=f(\sin(u))+f(\cos(u))$

Notice that with this method, we also established that $ E(x)$ is verified $ \forall x\in(\pi,\frac{5\pi}{4}]$

2) [u][b]When $ f(x)$ is defined as in point 1 above, $ E(x)$ is true $ \forall x\in\mathbb R$[/b][/u]
First, we obviously have to verify $ E(x)$ $ \forall x\in[0,2\pi)$
Second, by construction, we already know that :

$ E(x)$ is true $ \forall x\in(0,\frac{\pi}{4}]$
$ E(x)$ is true $ \forall x\in[\frac{pi}{2},\pi)$
$ E(x)$ is true $ \forall x\in(\pi,\frac{5\pi}{4}]$

2.1) $ E(0)$ is true :
$ E(\frac{\pi}{2})$ is true and $ E(\frac{\pi}{2})$ and $ E(0)$ are the same assertion

2.2) $ E(x)$ is true $ \forall x\in(\frac{\pi}{4},\frac{pi}{2})$
$ E(x)$ is true $ \forall x\in(0,\frac{\pi}{4})$ and $ E(x)$ and $ E(\frac{\pi}{2}-x)$ are the same assertion.

2.3) $ E(\pi)$ is true :
$ E(0)$ is true $ \implies$ $ f(0)=0$ $ \implies$ $ E(\pi)$ is true.

2.4) $ E(x)$ is true $ \forall x\in(\frac{5\pi}{4},2\pi)$
$ E(x)$ is true $ \forall x\in(\frac{\pi}{2},\frac{5\pi}{4})$ and $ E(x)$ and $ E(\frac{5\pi}{2}-x)$ are the same assertion.

Q.E.D.

[/hide]


real analysis
calculus
integration
inequalities
function
real analysis unsolved

Show that if f is bounded almost everywhere on A and m(A) $ < \infty$ ,then f is lebesgue integrable.

Since $ m(A) < \infty$ and $ f$ is bounded a.e., $ \left| f\right|\leq K$, so $ \left|\int_{A}f\right|\leq\int_{A}\left|f\right|\leq Km(A) < \infty$, which means that $ f\in L^{1}(A)$.

Hi JRav,
your first inequality is a fault of reasoning.

Which inequality are you referring to?  If $ f$ is bounded a.e., then $ \left|f\right|\leq K$ everywhere except on a set of measure 0, but I don't care about that set.

Or, since $ f$ is bounded a.e., both $ f^{\plus{}}$ and $ f^{\minus{}}$ are finite a.e. (positive and negative parts) so they are integrable on a set of finite measure, so $ \left|f\right|\equal{}f^{\plus{}}\plus{}f^{\minus{}}$ is integrable, which means $ f$ is.

How does it follow that $ f$ is measurable at all?

Hi JRav,
In fact there are 2 faults.
1) As Third Edition writes, what about measurability of $ f$ ?
2) Now assume $ f$ is measurable. When you begin your proof, $ \int{f}$ doesn't exist.
In other words $ f\in{L^1}$ iff $ \int{|f|}$ (that exists because $ f$ is measurable) is finite.

There's also the issue of exactly how to read the phrase "bounded almost everywhere". It's not a phrase I'd use at all, because of the confusion in my mind as to what it means.

It appears that JRav and others have chosen to read this as "essentially bounded" - that is, there is a set $ Z$ of measure zero such that $ f$ is bounded on $ A\setminus Z.$ And maybe that it what the phrase is supposed to mean.

But how different does "bounded almost everywhere" sound than "finite almost everywhere"? And there you run into very simple examples. $ \frac1x$ on $ [0,1]$ is finite almost everywhere and measurable, but it's not integrable.

I think that if you mean "essentially bounded" then it's best to say "essentially bounded," or to give an equivalent definition.

Kent,

I bumped into the same interpretation problem. But in either way of your suggestions the claim is not true, since we cannot show that $ f$ is measurable. Take, e.g., the indicator function of a [b]non-measurable[/b] subset $ A'$ of $ A$. Vitali's set gives such a non-measurable example for $ A\equal{}[0,1]$.


Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process, the contents of the two bags are the same?

$ \textbf{(A) } \frac 1{10} \qquad \textbf{(B) } \frac 16 \qquad \textbf{(C) } \frac 15 \qquad \textbf{(D) } \frac 13 \qquad \textbf{(E) } \frac 12$

[hide]"Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. "

There are exactly $5 \times 6 = 30$ ways of doing this. Now say a ball of green colour is selected and placed in Bob's bag. Then there will be 2 green balls in Bob's bag which can be selected to be returned into Alice's bag. There are 5 colours, so by symmetry, there are $5 \times 2 = 10$ ways to place a ball of the same colour to the ball which was chosen the first time back into Alice's bag. So the probability is $\frac{10}{30} = \frac{1}{3}$,

hence $(D)$.[/hide]

[hide="Much faster way"]Observe that no matter what color Alice gives Bob, Bob has 6 balls in his bag and 2 are the color Alice gave him. The only way the bags will be the same after the exchange is if Bob returns to Alice the same color Alice gave Bob. Thus Bob must give one of the 2 balls of the same color to Alice. There is a 2/6=1/3=D chance of doing this.[/hide]

Billy


inequalities
induction
complex numbers
triangle inequality
inequalities proposed

We will prove a slightly more general theorem. Note the difference!

[color=blue][b]Theorem 1.[/b] Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be 2n [/color][color=red]complex[/color][color=blue] numbers such that $\left|a_j\right| \leq 1$ and $\left|b_j\right| \leq 1$ for every j = 1, 2, ..., n. Then, the following inequality holds:

$\left| \prod_{j=1}^n a_j - \prod_{j=1}^n b_j \right| \leq \sum_{j=1}^n \left|a_j-b_j\right|$.[/color]

[i]Proof of Theorem 1.[/i]

We will prove Theorem 1 by induction over n.

Actually, Theorem 1 is trivial in the case n = 1 (in fact, in this case, the assertion of Theorem 1 reduces to $\left|a_1-b_1\right|\leq\left|a_1-b_1\right|$), so we can use the case n = 1 as the base of our induction, but we will actually make use of the case n = 2 in the induction step, so we can't avoid having to prove Theorem 1 for n = 2, too.

So we will now prove Theorem 1 for the case n = 2. This means that we will prove the inequality

$\left|a_1a_2-b_1b_2\right|\leq\left|a_1-b_1\right|+\left|a_2-b_2\right|$

for any four complex numbers $a_1$, $a_2$, $b_1$, $b_2$ such that $\left|a_1\right| \leq 1$, $\left|a_2\right| \leq 1$, $\left|b_1\right| \leq 1$ and $\left|b_2\right| \leq 1$. This is quite easy:

Since $\left|a_1\right| \leq 1$, $\left|a_2\right| \leq 1$, $\left|b_1\right| \leq 1$ and $\left|b_2\right| \leq 1$, with the help of the triangle inequality we get $\left|a_1+b_1\right|\leq\left|a_1\right|+\left|b_1\right|\leq 1+1=2$ and similarly $\left|a_2+b_2\right|\leq 2$, so that

$\left|\left(a_1-b_1\right)\left(a_2+b_2\right)\right|+\left|\left(a_2-b_2\right)\left(a_1+b_1\right)\right|=\left|a_1-b_1\right|\left|a_2+b_2\right|+\left|a_2-b_2\right|\left|a_1+b_1\right|$
$\leq\left|a_1-b_1\right|\cdot 2+\left|a_2-b_2\right|\cdot 2=2\left(\left|a_1-b_1\right|+\left|a_2-b_2\right|\right)$.

Thus,

$2\left(\left|a_1-b_1\right|+\left|a_2-b_2\right|\right)$
$\geq \left|\left(a_1-b_1\right)\left(a_2+b_2\right)\right|+\left|\left(a_2-b_2\right)\left(a_1+b_1\right)\right|=\left|a_1a_2+a_1b_2-b_1a_2-b_1b_2\right|+\left|a_2a_1+a_2b_1-b_2a_1-b_2b_1\right|$
$=\left|\left(a_1a_2-b_1b_2\right)+\left(a_1b_2-b_1a_2\right)\right|+\left|\left(a_1a_2-b_1b_2\right)-\left(a_1b_2-b_1a_2\right)\right|$
$\geq \left|\left(\left(a_1a_2-b_1b_2\right)+\left(a_1b_2-b_1a_2\right)\right)+\left(\left(a_1a_2-b_1b_2\right)-\left(a_1b_2-b_1a_2\right)\right)\right|$       (by the triangle inequality)
$=\left|2\cdot\left(a_1a_2-b_1b_2\right)\right|=2\left|a_1a_2-b_1b_2\right|$.

Thus, we get

$2\left|a_1a_2-b_1b_2\right|\leq 2\left(\left|a_1-b_1\right|+\left|a_2-b_2\right|\right)$,

so that $\left|a_1a_2-b_1b_2\right|\leq\left|a_1-b_1\right|+\left|a_2-b_2\right|$, and Theorem 1 is proven for n = 2.

Now we have to make the induction step, i. e. to show that if Theorem 1 holds for some integer n (with $n\geq 1$), then Theorem 1 also holds for n + 1.

In fact, this is almost immediate:

For any 2 (n + 1) complex numbers $a_1$, $a_2$, ..., $a_{n+1}$, $b_1$, $b_2$, ..., $b_{n+1}$ such that $\left|a_j\right| \leq 1$ and $\left|b_j\right| \leq 1$ for every j = 1, 2, ..., n + 1, we have $\left|\prod_{j=1}^n a_j\right|=\prod_{j=1}^n\underbrace{\left|a_j\right|}_{\leq 1}\leq\prod_{j=1}^n 1=1$, and similarly $\left|\prod_{j=1}^n b_j\right|\leq 1$, so that we can apply Theorem 1 for the case n = 2 to the numbers $\prod_{j=1}^n a_j$, $a_{n+1}$, $\prod_{j=1}^n b_j$, $b_{n+1}$ instead of $a_1$, $a_2$, $b_1$, $b_2$, and get

$\left| \prod_{j=1}^n a_j \cdot a_{n+1} - \prod_{j=1}^n b_j \cdot b_{n+1} \right|\leq \left| \prod_{j=1}^n a_j - \prod_{j=1}^n b_j\right| + \left|a_{n+1}-b_{n+1}\right|$;

in other words,

$\left| \prod_{j=1}^{n+1} a_j - \prod_{j=1}^{n+1} b_j \right| \leq \left| \prod_{j=1}^n a_j - \prod_{j=1}^n b_j\right| + \left|a_{n+1}-b_{n+1}\right|$.

But since we assumed Theorem 1 to hold for n, we can apply it to the 2n complex numbers $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ and get $\left| \prod_{j=1}^n a_j - \prod_{j=1}^n b_j \right| \leq \sum_{j=1}^n \left|a_j-b_j\right|$. Thus,

$\left| \prod_{j=1}^{n+1} a_j - \prod_{j=1}^{n+1} b_j \right| \leq \underbrace{\left| \prod_{j=1}^n a_j - \prod_{j=1}^n b_j\right|}_{\leq \sum\limits_{j=1}^n \left|a_j-b_j\right|}  + \left|a_{n+1}-b_{n+1}\right|$
$\leq \sum_{j=1}^n \left|a_j-b_j\right| + \left|a_{n+1}-b_{n+1}\right|= \sum_{j=1}^{n+1} \left|a_j-b_j\right|$.

And thus, we have shown that Theorem 1 holds for n + 1. This completes the induction step, and thus, Theorem 1 is proven.

  Darij


limit
logarithms
calculus
calculus computations

[quote="countdownmath"]Compute $\lim_{x\to\infty}(x+\ln x)^{p}-x^{p}$, where $p\in (0,1)$.[/quote]
$0$

can you be more specific in terms of how you got $0$?
that would be great, thanks.

From $\lim_{x\to\infty}\frac{\ln x}{x^\epsilon}= 0$, we have that $\ln x < x^\epsilon$ holds for any fixed positive $\epsilon$ and for sufficiently large $x$. Choose $\epsilon$ to be sufficiently small so that $\epsilon+p < 1$. Then

\begin{eqnarray*}(x+\ln x)^{p}-x^{p}& = & x^{p}\left[ \left( 1+\frac{\ln x}{x}\right)^{p}-1 \right]\\ & \leq & x^{p}\left[ \left( 1+\frac{x^{\epsilon}}{x}\right)^{p}-1 \right]\\ & = & x^{p}\left[ ( 1+x^{\epsilon-1})^{p}-1 \right] \end{eqnarray*}

But since $x^{\epsilon-1}< 1$ for $x \gg 1$, $( 1+x^{\epsilon-1})^{p}\approx 1+px^{\epsilon-1}$,

$x^{p}\left[ ( 1+x^{\epsilon-1})^{p}-1 \right] \approx x^{p+\epsilon-1}$.

Then by $\lim_{x \to \infty}x^{p+\epsilon-1}= 0$ and squeezing theorem, we have $\lim_{x \to \infty}(x+\ln x)^{p}-x^{p}= 0$.

[quote="countdownmath"]i'm not sure i follow how you got the approximately equal signs...[/quote]
 easy seen $\frac{\ln{x}}{x}$ when $x\to oo$ equal $0$. it's mean that this expression is very very small. therefore  by taylor we can write  $(x-\ln{x})^{p}$=$x^{p}(1-\frac{\ln{x}}{x})^{p}$=$x^{p}(1-p\frac{\ln{x}}{x}+O((\frac{\ln{x}}{x})^{2})$=$x^{p}-p\frac{\ln{x}}{x^{1-p}}+O((\frac{\ln{x}}{x})^{2-p})$ when $x\to oo$ then $p\frac{\ln{x}}{x^{1-p}}\to 0$ and $O((\frac{\ln{x}}{x})^{2-p})\to 0$ and  therefore $x^{p}-p\frac{\ln{x}}{x^{1-p}}+O((\frac{\ln{x}}{x})^{2-p})$~$x^{p}$ when $x\to oo$. therefore in our case we also must $-x^{p}$ and equal 0.


For $n \in \mathbb{N},\; n \geq 2$ and $x \in \mathbb{R}$ define \[ P_n(x) := n^nx((x + 1)^{n + 1} - 1)^{n - 1} - (n + 1)^{n - 1}((x + 1)^n - 1)^n.\] Is it true that this polynomial can be written in the form \[ P_n(x) = \sum_{k = n + 2}^{n^2} c_{n,\;k} x^k \] where $ c_{n,\;k} > 0$ for $n + 2 \leq k \leq n^2$?

WOW, this must be VERY hard - unsolved after more than one year!

The number $c_{n,k}$ is just the coefficient of $x^k$ when $P_n(x)$ is expanded. Sorry, I thought that would be obvious enough :)


Let a quadrangle $ ABCD$ with $ AC\cap BD\equal{}O$. Denote $ M,N$ are centre of triangle $ AOB,COD$ and $ P,Q$ are orthorcentre of triangle $ AOD,BOC$. Prove that $ MN$ is perpendicular with $ PQ$.

It is an old Russian Math Olympiad problem and probably a theorem.


In a triangle $\triangle{ABC}$ with all its sides of different length, $D$ is on the side $AC$, such that $BD$ is the angle bisector of $\sphericalangle{ABC}$. Let $E$ and $F$, respectively, be the feet of the perpendicular drawn from $A$ and $C$ to the line $BD$ and let $M$ be the point on $BC$ such that $DM$ is perpendicular to $BC$. Show that $\sphericalangle{EMD}=\sphericalangle{DMF}$.

Let $F_1$ be the reflection of $F$ in $BC$. The conclusion we want to reach is equivalent to the fact that $F_1,M,E$ are collinear, so this will be the aim.

The triangles $BF_1C,BMD,BEA$ have right angles in $F_1,M,E$ respectively, and the angles $\angle F_1BC,\angle MBD,\angle EBA$ are all equal to half the angle $\angle ABC$. In particular, they are equal, so the three mentioned triangles are similar. This means that $F_1,M,E$ are the images of $C,D,A$ respectively through a spiral similarity centered at $B$. Since $C,D,A$ are collinear, so are $F_1,M,E$, exactly what we wanted to show.

A different solution: We will use directed angles modulo 180°. Consider not only the orthogonal projection M of the point D on the line BC, but also the orthogonal projection N of the point D on the line AB. Since the line BD is the angle bisector of the angle ABC, these orthogonal projections M and N are symmetric to each other with respect to this angle bisector BD, and thus we have < EMD = - < END. Since < AED = 90° and < AND = 90°, the points E and N lie on the circle with diameter AD, so that < END = < EAD. Since the lines AE and CF are both perpendicular to the line BD, they are parallel to each other, and thus < DAE = < DCF. Finally, since < CFD = 90° and < CMD = 90°, the points F and M lie on the circle with diameter CD, and therefore < DCF = < DMF. Hence, < EMD = - < END = - < EAD = < DAE = < DCF = < DMF, and we are done.

  darij

[quote="carlosbr"]In a triangle $\triangle{ABC}$ with all its sides of diferent length, $D$ is on the side $AC,$ such that $BD$ is the angle bisector of $\sphericalangle{ABC}.$ Let $E$ and $F,$ respectively, be the feet of the perpendicular drawn from $A$ and $C$ to the line $BD$ and let $M$ be the point on $BC$ such that $DM$ is perpendicular to $BC.$ Show that $\sphericalangle{EMD}=\sphericalangle{DMF}.$[/quote]
Let $ D'$ be the foot of the external bisector of $ \angle ABC.$ Since the division  $ (A,C,D,D')$ is harmonic, then its orthogonal projection onto the internal bisector of $ \angle ABC$ is harmonic as well $ \Longrightarrow$ $ (E,F,D,B) \equal{} \minus{} 1.$ From $ DM \perp BC$ we conclude that rays  $ MD$ and $ MB$ bisect $ \angle EMF$ internally and externally $ \Longrightarrow$ $ \angle EMD \equal{} \angle DMF.$

From $\triangle ABE \sim \triangle CBF$ and $\triangle ADE \sim \triangle CDF$ we get $(E, F, D, B) = -1$, further see above.

Best regards,
sunken rock

let N be the symmetric point of M in respect of BD,then
$\angle EMD=\angle END=\angle EAD=\angle DCF=\angle DMF$.

Let $P =BC\cap AE$. We can easily see that triangle $DEP$ is congruent to triangle $DEA$ and that $DEPM$ is a cyclic cuadrilateral hence $\angle DAE=\angle EPD=\angle EMD$. Now, since $FDMC$ is a cyclic cuadrilateral, which can be proven easily, $\angle DMF=\angle DCF$ and since $CF\parallel EA$, $\angle DCF=\angle DAE$ hence $\angle DMF=\angle DCF=\angle DAE=\angle EPD=\angle EMD$.

Since $\angle DFC =\angle DMC=90, DMCF$ is a cyclic quadrilateral and thus $\angle DCF=\angle DMF$. Also, $AE||FC$ so $\angle DCF=\angle DAE$. Let $M'$ be the foot of the perpendicular to $AB$ from $D$, note that $M'EDA$ is cyclic because $\angle DM'A= \angle AED$ and thus $\angle DAE= \angle DM'E$. Finally, since $BD$ is an angle bisector, $\Delta DEM' \equiv \Delta DEM$ and therefore $\angle DM'E= \angle EMD$ completing the proof.

Let $G\in AE\cap BC$ and let $P_{\infty}$ be the point at infinity of lines $AE$ and $FC$.
We have that $\Delta GBA$ is isosceles, since $\overline{BE}$ is both an altitude and an angle bisector. Then, $E$ is the midpoint of $AG$.
Hence, $-1=(A,G;E,P_{\infty})\stackrel{C}{=}(D,B;E,F)$.
Since $\angle DMB=90^\circ$, we have that $MD$ bisects angle $\angle FME$, as desired.

[hide="Sketch"]
Let $X=AE\cap BC$. Then by easy angle chasing quadrilaterals $EMXD$ and $DMFC$ are cyclic, which, coupled with $AE \parallel CF$, easily lead to solution.
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In a triangle $\triangle{ABC}$ with all its sides of different length, $D$ is on the side $AC$, such that $BD$ is the angle bisector of $\angle{ABC}$. Let $E$ and $F$, respectively, be the feet of the perpendicular drawn from $A$ and $C$ to the line $BD$ . Let $M$ any point the point on $BC$ . Prove that  $DM \perp BC$ iff  $\angle{EMD}=\angle{DMF}$.
[hide=.]this is the shortlist version of g2, the only difference is the iff condition, this comment was created for the [url=https://artofproblemsolving.com/community/c2468825_iberoamerican_shortlist__geometry]ibero geo shortlist[/url] link[/hide]