[quote="uldivad9"]Trial and error works, usually that works very fast for these types of problems. The answer is
[hide]$ 3 \plus{} 2\sqrt2$.[/hide][/quote]

Thanks, but could you please explain your process of guess and check?

Blech. Okay.

Note that the '17' must be the square of an integer plus some other non-square. Look at the squares:
If the first square is 1, the non-square is 16, but 16 is a square
If the first square is 4, the non-square is 13 and the $ b\sqrt{c}$ is $ \sqrt13$, but that doesn't work.
If the first square is 9, the non-square is 8 and the $ b\sqrt{c}$ is $ 2\sqrt2$, and this works, so we are done.

$ (a\plus{}b\sqrt{2})^2\equal{}17\plus{}12\sqrt{2}$
$ a^2\plus{}2b^2\equal{}17$
$ 2ab\sqrt{2}\equal{}12\sqrt{2}$
$ ab\equal{}6$

Substitute into the first equation to get the desired values.

To be more specific, $ 17 = a^2 + 2b^2$ and $ 12 = 2ab$ where $ a$ and $ b$ are integers.

[hide="Derivation"]
$ \sqrt {17 + 12\sqrt {2}} = a + b\sqrt {c}$
$ 17 + 12\sqrt {2} = (a + b\sqrt {c})^2 \leftarrow$ Square both sides
$ 17 + 12\sqrt {2} = a^2 + 2ab\sqrt {c} + b^2c = (a^2 + b^2c) + 2ab\sqrt {c} \leftarrow$ Expand and group

By grouping of terms without square roots in them, we can see that the term without square roots contributes to the $ 17$ and the term with the square root contributes to the $ 12\sqrt {2}$. At this point, we can see that the only choice for $ c$ is $ 2$. Thus we can rewrite as follows:

\begin{align*} a^2 + 2b^2 & = 17 \\
2ab\sqrt {2} & = 12\sqrt {2} \end{align*}

Simplifying:

\begin{align*} a^2 + 2b^2 & = 17 \\
ab & = 6 \end{align*}

This last equation greatly reduces the trial and error we have to do.
Thus the only combinations we have to check are $ (1, 6), (2, 3), (3, 2), (6, 1)$. We can readily see that $ (3, 2)$ works because $ 3^2 + 2 \cdot 2^2 = 9 + 8 = 17$.

Therefore, our answer is $ 3 + 2\sqrt {2}$.

The general form
$ \sqrt {x + y\sqrt {z}} = a + b\sqrt {c}$
uses similar methods as above to reduce down to this system of equations:
\begin{align*} a^2 + cb^2 & = x \\
ab & = y \\
c & = z \end{align*}

provided that the original expression is reducible.
[/hide]

[quote="12markkram34"]Can someone please tell me how to solve this:

30. The number sqrt(17 + 12 * sqrt(2)) is an example of a number expressed in "embedded radical form" because of the sqrt(2) under the radical. What is the equivalent value of this number expressed in simplest radical form a + b * sqrt(c) where a and b are integers and c is a positive integer containing no perfect square factors greater than 1?

Thanks.
-12markkram34[/quote]

$ \sqrt{17+12\sqrt{2}}=a+b\sqrt{c}$.

We can assume that $ c=2$.

$ \sqrt{17+12\sqrt{2}}=a+b\sqrt{2}$.

Squaring,

$ a^2+2ab\sqrt{2}+2b^2=(a^2+2b^2)+(2ab\sqrt{2})=17+12\sqrt{2}$.

So, $ ab=6$ and $ a^2+2b^2=17$.

So,

$ a=\frac{6}{b} \implies \frac{36}{b^2}+2b^2=17 \implies
2b^4-17b^2+36=0 \implies (2b^2-9)(b^2-4) \implies b=\pm 2, \pm \frac{3\sqrt{2}}{2}$.

However, $ b$ is an integer, so $ b=\pm2$, making

$ a=3$.

Also, $ 3-2\sqrt{2}$ is neagtive, which the original obviously

is not. So,

$ \boxed{3+2\sqrt{2}}$.

It's long, but it works.

It takes a lot of practice to get used to it, but it does the job

very well.


algebra
polynomial
geometry
factorial
modular arithmetic
function
binomial theorem

If $ a^{2k} \plus{} a^kb^k \plus{} b^{2k}$ can be factorized (in $ \mathbb{Q}$), what do we know about $ k$?

I've already proved that if $ k$ [b]isn't [/b]a power of 3, then the polynomial [b]can[/b] be factorized...
I don't know how to prove  that if $ k$ [b]is[/b] a power of 3, then the polynomial [b]cannot[/b] be factorized...

It's reducible iff $ x^{2k}\plus{}x^k\plus{}1$ is reducible ie. iff  $ (x\plus{}1)^{2k}\plus{}(x\plus{}1)^k\plus{}1$ is reducible. For $ k$ a power of three all coeficients are divisible by 3, the constant coeficient is 3 so we can aply Eisenstein rule.

Sorry... It's the first time I post something in this zone (Olympic area), so I don't know what Eisenstein's rule is... Can you explain in a few words? (I looked it up in Wikipedia, but didn't understand anything!  :( )

Thank you! :D

Sorry for bad Czech -> English translation. I meant Eisenstein's criterion: http://en.wikipedia.org/wiki/Eisenstein%27s_criterion You probably found some other theorems by Eisenstein.

Application of this criterion is easy as long as we can prove its requirements for p=3. The first coefficient is 1, so first requirement is satisfied. Last coefficient is 3, which yields the third requirement. Most difficult is the second one.
By binomial theorem we can evaluate the coefficients. We have to show that for 0<t<k $ 3|{2k\choose t}$, for 0<t<k $ 3|{2k\choose t}\plus{}{k\choose t}$ and that $ 3|{2k\choose k}\plus{}1$.
First two parts can be done just by evaluating the powers of 3 which divide the factorials involved in the binomial coefficients. The third part is a bit harder. Let's cancel the highest possible power of three in $ \frac{2k!}{(k!)^2}$ (using the previous technique we can show that after that neither numerator nor denominator is divisible by 3) and evaluate the result mod 3. The denominator is a square, therefore it is 1 mod 3. The numerator is a product containing 1+2k/3+2k/9+...+2=k twos (mod 3), so it is $ 2^k\equiv 2\pmod 3$. The whole fraction is $ 2/1\equiv 2\pmod{3}$, thus $ 3|{2k\choose k}\plus{}1$.

[quote="Kondr"]Sorry for bad Czech -> English translation. I meant Eisenstein's criterion: http://en.wikipedia.org/wiki/Eisenstein%27s_criterion You probably found some other theorems by Eisenstein.[/quote]

Yeah, actually I found something strange about power series in algebraic function which are trascendental! :D

Thank you very much for the patience and for the help! I've learnt something very useful :)


All I (now) know about the modulus is that $a\equiv b \pmod c$ means when $a$ is divided by $c$ you get a remainder of $b$. Is that right? And could anyone post some sample questions involving it for me to look at/ask stupid questions about?

People have asked this so many times. Search it!  :)

Yes, mods are just pretty much remainders okay. Say you have $12/equiv/2(mod 5)$, that means when $12$ is divided by $5$ it gives a remainder of $2$. Thus that is also equivalent to $17$ because when divided by $5$ it gives a remainder of $2$.

[quote="Holier Than Thou"]Aw, I'm too lazy.... well people can respond if they want to, if not I'll get around to it sometime.[/quote]

Hmm, if you really want to find out, searching is the best way. It's better than what you can get from one or two posts.

If you're too lazy, well... nothing I can say about that.


function
algebra
domain
limit
calculus
calculus computations

Let's say I have two functions $ g(x)_n$ and $ f(x)_n$. I know that $ g(x)_n\sim f(x)_n$ for a certain x=c, and can prove it by taking the limit as n approaches infinity of the quotient of the two functions and seeing that it equals one. My question is, if the domain of the functions is all reals, how can I prove that x=c is the only x that makes the statement $ f(x)_n\sim g(x)_n$ true?

The truth of that statement depends on your choice of functions, so... what are your functions?

$ \lim_{x \minus{} > c} [\sum^\infty_{n \equal{} 0} \frac {1}{2^{2n \plus{} 1}}\sum^n_{k \equal{} 0} {2n\choose2k}(2k \plus{} 1)^{ \minus{} x}\sim \sum^\infty_{n \equal{} 0} \frac {1}{2^{2n \plus{} 3}}\sum^n_{k \equal{} 0} {2n \plus{} 2\choose2k}(2k \plus{} 1)^{ \minus{} x}]$

This is not exactly it, but hopefully you can show me how I can prove that x=c is the only x in the domain that makes the above statement true.


vector
function
real analysis
real analysis unsolved

Prove that in an infinite Hilbert space there exists a linearly independent set of vectors of continuum cardinality.

It suffices to prove this for separable Hilbert spaces, and since all such spaces are isometrically isomorphic, so, in particular, isomorphic as vector spaces, in the algebraic sense, it suffices to prove it for any separable Hilbert space. We can take our space to be $L^2([0,1])$, and in this case we find, for instance, the system of functions $f_a,\ a\in(0,1]$, where $f_a$ is the characteristic function of $[0,a]$. Obviously, these are linearly independent.


Please help me with this problem:
  Let $P \in \mathbb{C}[X]$, $deg P=2$, $P(0)=1$ and $|P(z)| \le |P(1)|, \forall |z|\le 1.$ Prove that $P \in \mathbb{Z}[X]$.
   Please move it to right section. I couldn't find any section fit for a complex numbers problem. Thank you for your effort.

i dont think the problem is true.
let $P(z)=\frac{1}{2}z^2+\frac{1}{2}z+1$
then $|P(z)|\leq 2=P(1)$
yet P has non-integer coeffs.


Asymptote
analytic geometry
function
conics
ellipse
parameterization

how do you make ticks on the coordinate axis?? (To mark off like 1, 2, 3, 4, ...)

thanks for any help.

In the olympiad package, there is a command called pathticks. [url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Macros_and_Packages[/url]

However, I am having trouble making it work  :(

For all such cases, there are 2 good advices

1) [b]Read the manual.;)[/b] There are commands for ticks described in the section about the graph package.

2) [b]Spend some time learning the basic programming and do it yourself. :)[/b]. Indeed, what you need is just two crossing lines with small perpendicular lines and numbers at the appropriate positions. So, just write two loops and you'll be done:
[asy]size(300); pen p=fontsize(7); draw((-10,0)--(10,0)^^(0,-10)--(0,10)); for(int k=-9;k<10;++k) { if (k!=0) {draw((k,-0.2)--(k,0)); label(string(k),(k,-0.2),S,p);} } for(int k=-9;k<10;++k) { if (k!=0) {draw((-0.2,k)--(0,k)); label(string(k),(-0.2,k),W,p);} }[/asy]

Well, being a weird person I like to do things the way they are meant to be (meaning if they implemented it I won't program my own).

But I figured out how to do it shortly after the post - I DID read my manual :P

Oh and something extra:

I given up trying to have asymptote auto-graph step functions for me...so is there any command similar to dot(), except it draws an empty circle instead??? I tried circle() but because my x-axis and y-axis scale are different I got an ellipse....

[quote="pkerichang"]Well, being a weird person I like to do things the way they are meant to be (meaning if they implemented it I won't program my own).[/quote]
I belong to the opposite school (meaning I would rather do by myself everything I can do without too much effort; this way I'll be at least sure about how it all works and shall be able to modify everything I don't like :P). Anyway, I owe you an apology for completely overlooking this post :blush:. I should have answered long ago but I noticed it only today :)

As to your extra question, if you get an ellipse using [b]circle()[/b], you should be able to get a circle using [b]ellipse()[/b] with appropriate parameters though I would rather keep the x and y scales the same and graph an appropriate multiple of the function instead :).

[quote="fedja"]1) [b]Read the manual.;)[/b] There are commands for ticks described in the section about the graph package.
[/quote]I've installed the graph_pi package which makes things simple

[asy]import graph_pi; size(200,0); graphicrules(yunit=1cm, xmin=-1, xmax=4, ymin=-1, ymax=3); grid(); cartesianaxis();[/asy]

How do I use the graph_pi package? Where can I find documentation for this package? For example, how do I get rid of the half-steps in Vornicu's graph image? And also, label the tick marks?

[quote="123456789"]How do I use the graph_pi package?[/quote]
You may find examples [url=http://piprim.tuxfamily.org/asymptote/modules/graph_pi/index.html]HERE[/url]

[quote="123456789"]Where can I find documentation for this package?[/quote]
There isn't documentation because I think the examples are sufficient to understand, my English is very poor and I have no time to do it.
Any help is welcome.

[quote="123456789"]For example, how do I get rid of the half-steps in Vornicu's graph image? And also, label the tick marks?[/quote]
[code]
import graph_pi;
size(200,0);
graphicrules(xmin=-1, xmax=4, ymin=-1, ymax=3);

grid(xstep=0,ystep=0);

cartesianaxis(xticks=Ticks(Label(Fill(white)), NoZero, ptick=invisible),
              yticks=Ticks(Label(Fill(white)), NoZero, ptick=invisible));

labelx("$O$",0,SW);
[/code]
As the package [b]graph_pi[/b] seems not be update in this site I put the result is [url=http://piprim.tuxfamily.org/asymptote/modules/graph_pi/index.html#fig0003]here[/url] (figure 0003).

Regards.

[quote="fedja"]Where can I download graph_pi from? I tried [url]http://piprim.tuxfamily.org/asymptote/modules/graph_pi.asy[/url] but their version has a bug and I'm too lazy to go through the code to remove it.[/quote]
I updated the package (it's may be useful to refresh the page) and you must see the line:
[b]Last modified: Fri Mar 28 15:57:10 CET 2008[/b]
However, can you send me an example which show the bug, the error message that you get and the version of Asymptote used please ?

I'm still using 1.33 (should upgrade, probably, but am too lazy). The message was:
/graph_pi.asy: 194.29: cannot cast expression to 'marker'
/graph_pi.asy: 215.29: cannot cast expression to 'marker'
There is no example to show. import graph_pi; already produces this error and it is the same with the newer version.
I downloaded base_pi from the same page, by the way.

[quote="fedja"]I'm still using 1.33[/quote]
Here is the problem... :|

We will soon update the Asymptote version on the site too.

Edit: Asymptote upgraded to 1.42 and also graph_pi upgraded.

[quote="i_like_pie"]Ph. Ivaldi - You should update [url=http://svnweb.tuxfamily.org/filedetails.php?repname=piprim/asymptote&path=%2Ftrunk%2Fgraph_pi.asy&rev=0&sc=0]this[/url] as well. :)[/quote]
It's done  :oops:
Thanks.


Let $ r_1$, $ r_2$, $ r_3$, and $ r_4$ be the roots to the equation $ ax^4 \minus{} 3x \plus{} 1 \equal{} 0$, for $ a \in {1, \minus{} 1, 3, \minus{} 3}$. Find the sum of all possible values of $ r_1^4 \plus{} r_2^4 \plus{} r_3^4 \plus{} r_4^4$.

Hint: Most of the info you get is useless.

[quote="Textangle"]Let $ r_1$, $ r_2$, $ r_3$, and $ r_4$ be the roots to the equation $ x^4 \minus{} 3x \plus{} 1 \equal{} 0$, for $ a \in {1, \minus{} 1, 3, \minus{} 3}$. Find the sum of all possible values of $ r_1^4 \plus{} r_2^4 \plus{} r_3^4 \plus{} r_4^4$.[/quote]
where does a come in?

Oh no... there was an a but I erased it... It's fixed now.

Summing up $ a(r_i)^4 \minus{} 3(r_i) \plus{} 1 \equal{} 0$ for each of the roots, we obtain $ a\left(\sum r_i^4\right) \equal{} 3\left(\sum r_i\right) \minus{} 4 \equal{} \minus{}4$ (as $ \sum r_i \equal{} 0$ by Vieta's), so $ \sum r_i^4 \equal{} \frac {\minus{}4}a$. The answer easily follows, and it also suggests the (not much) easier way.

I was just about to post that, azjps. lol


as you can see in the image bellow, a messege appeared that said I had 4 smiles when I had just three. No, I didn't work with photoshop (or any other program)!

that's a bug, right? It should be corrected...

If the admins would have nothing better to do than enabling you to post twice that much smileys...

[quote="mustafa"][quote="Ignite168"]Because :? is hidden in :?: and :o is hidden in :oops:

[/quote]

Wait a second, if that is true, how did you just use 6 smileys in 1 post?[/quote]
I'm pretty sure that the number of smileys you can have in a post is dependent on the amount of text in that post; the more text you have, the more the number of smileys you can have is. I think there is eventually an upper bound though (maybe ~10 smileys?). For example, in this post I can have several smileys:

 :D  :lol:  :blush:  :maybe:  :)  :(

(Generally it's not a good idea to post a large amount of smileys though, especially if the post is something like a single " :) ".)

[quote="mustafa"][quote="Ignite168"]Because :? is hidden in :?: and :o is hidden in :oops:

[/quote]

Wait a second, if that is true, how did you just use 6 smileys in 1 post?[/quote]

didn't you notce the excesive amount of color tags in his post when you quoted him?

It's not relative to the amount of text, it's actually relative to the number of words in your post.


quadratics
analytic geometry
graphing lines
slope

[quote="SplashD"][quote="bpms"]$x+y=3$
$x^{2}+y^{2}=5$
What is the smallest possible value of $\frac{x}{y}$?[/quote]

[hide]$x=1$
$y=2$

answer: $\boxed{\frac12}$

is that right?[/hide][/quote]
That is right.[hide="spoiler"]It wasn't intended to be hard :D[/hide]

[hide]common sense shows that x= 1 or 2 and that y= 2 or 1.

the smallest value is 1/2 or 1 half.

[/hide]

[hide]This system is so simple you can see what x and y are right away--1 and 2. So we let x=1 and y=2 to have the smaller fraction $\frac{1}{2}$.[/hide]

[hide="solution"]
y=3-x
Rewrite and factor:
$(x-1)(x-2)$

By solving for x and y, it is observable that the smallest value for $\frac{x}{y}$ is $\frac{1}{2}$.
[/hide]

[hide]Well, there are only 2 solutions to this system

(1,2) and (2,1).

So minimum value of y/x is $\frac{1}{2}$[/hide]

You guys all did trial and error though, can you show there are only 2 solutions?

[quote="akalra1"][hide]bpms, this problem was kinda easy

$x+y=3$
$x^{2}+y^{2}=5 \longrightarrow$
$x=1$
$y=2$
$\text{OR}$
$y=1$
$x=1$
$\text{so minimum value of}\frac xy = \boxed{\frac12}$[/hide][/quote]
This problem was supposed to be easy  :wink:

There are two solutions, [i]Ignite168.[/i]

One, is when:
$x=1;\ y=2$

Another is when:
$x=2;\ y=1$

We couldn't use the second one, because bpms wanted the [b]smallest possible value of $\frac{x}{y}$[/b], which was $ \frac{1}{2}$, not $ \frac{2}{1}$.

[quote="anirudh"]There are two solutions, [i]Ignite168.[/i]

One, is when:
$x=1;\ y=2$

Another is when:
$x=2;\ y=1$

We couldn't use the second one, because bpms wanted the [b]smallest possible value of $\frac{x}{y}$[/b], which was $\frac{1}{2}$, not $\frac{2}{1}$.[/quote]
He said there are two and you didn't show it. I can, but I made this problem so that doesn't count.

guess and check saves the day again!! duh da da dum! :P

Well, 1+2=3 and 1+4=5. You find the smallest value: .5

I used algebra and THEN trial and error. :P  and got (23,47)

[hide]
If $x=1$ and $y=2$ (or the other way around), the fraction would be $\frac{1}{2}$.[/hide]

An easy way to see that there are only 2 solutions is to make a mental graph. x^2+y^2=5 is a circle around the origin and x+y=3 is a negative-sloped line that goes through it in 2 places: only 2 solutions.

[quote="bpms"][quote="jclarkemathL314159"]I can show that there are only two solutions, i think...

[hide]
$x+y=3$
$x^{2}+y^{2}=5$

Square the first equation to get $x^{2}+y^{2}+2xy=9$. Thus $2xy=4$ and $xy=2$

We now have $x+y=3$ and $xy=2$, $x$ and $y$ are the roots of the quadratic $a^{2}-3a+2=0$ (since these are the sum and product of the roots). We just solve the quadratic to get $a=1,2$. So the solutions of the system are (1,2) and (2,1), and the smallest possible value of $\frac{x}{y}$ is $\frac{1}{2}$.[/hide]

That was a solution without brute force. If you didn't look at it, try solving this one. Obviously, trial and error is not the way to go with much larger numbers.

$x+y=70$
$x^{2}+y^{2}=2738$.[/quote]
I intended for the problem to be simple and obvious[/quote]

I was providing a "rigorous" algebraic proof that there were only two solutions. Now I see that k8reindeer's proof is much nicer.

[quote="bpms"]$x+y=3$
$x^{2}+y^{2}=5$
What is the smallest possible value of $\frac{x}{y}$?[/quote]

[hide]
x=1,2 y= 2,1 (im lame I just guessed and got it on the first try. I DO know how to solve it though...)

x/y could be 2/1 or 1/2.
of course 1/2 is smaller[/hide]


AMC USA(J)MO USAMO Vieta inequalities calculus derivative

[quote="Jakey"]USAMO 1983:
Prove that the zeros of $ x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} d \plus{} e \equal{} 0$ cannot all be real if $ 2a^2 < 5b$

Solution:
[hide] By vieta's theorem, $ \minus{} a \equal{} m \plus{} n \plus{} q \plus{} r \plus{} s$, where $ m,n,q,r,s$ are the roots. Also, $ b$ refers to the permutations of $ m,n,q,r,$ and $ s$ taken 2 at a time.

Then, we know that by AM-GM, $ m^2 \plus{} n^2 \geq 2mn$. If we do this repeatedly such that we have achieved all elements of $ b$ for the right hand side, the sum of all the inequalities would equal to $ 4m^2 \plus{} 4n^2 \plus{} ... \plus{} 4s^2 \geq 2(mn \plus{} mq \plus{} mr \plus{} ... \plus{} rs)$. Then, we could rewrite the whole thing by dividing everything by 2 and adding $ 4b$ to both sides. Ultimately, we would reach $ 2a^2\geq5b$. Because of this, there is a contradiction and the roots must be real.

Is there anything wrong? or can you suggest something better? thanks!

[/hide][/quote]

That's how I did it. I think you phrased the last sentence wrong; what you proved is that [hide]if the roots were all real, then $ 2a^2\geq5b$, so if $ 2a^2 < 5b$ then the roots cannot all be real. [/hide] Everything else looked fine though, you did the exact same stuff I did.

[quote="Jakey"]USAMO 1983:
Prove that the zeros of $ x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} d \plus{} e \equal{} 0$ cannot all be real if $ 2a^2 < 5b$[/quote]

Here is my solution:

Let $ p,\ q,\ r,\ s,\ t$ be roots of the equation, we can write $ x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} d x\plus{} e \equal{} (x \minus{} p)(x \minus{} q)(x \minus{} r)(x \minus{} s)(x \minus{} t)$.

Comparison of the coefficients of $ x^5,\ x^4$ gives $ p \plus{} q \plus{} r \plus{} s \plus{} t \equal{} \minus{} a, \ pq \plus{} pr \plus{} ps \plus{} pt \plus{} qr \plus{} qs \plus{} qt \plus{} rs \plus{} rt \plus{} st \equal{} b$.

$ \therefore 2a^2 \minus{} 5b \equal{} 2(p \plus{} q \plus{} r \plus{} s \plus{} t)^2 \minus{} 5(pq \plus{} pr \plus{} ps \plus{} pt \plus{} qr \plus{} qs \plus{} qt \plus{} rs \plus{} rt \plus{} st)$

$ \equal{} 2\{p^2 \plus{} q^2 \plus{} r^2 \plus{} s^2 \plus{} t^2 \plus{} 2(pq \plus{} pr \plus{} ps \plus{} pt \plus{} qr \plus{} qs \plus{} qt \plus{} rs \plus{} rt \plus{} st)\}$

$ \minus{} 5(pq \plus{} pr \plus{} ps \plus{} pt \plus{} qr \plus{} qs \plus{} qt \plus{} rs \plus{} rt \plus{} st)$

$ \equal{} 2(p^2 \plus{} q^2 \plus{} r^2 \plus{} s^2 \plus{} t^2) \minus{} (pq \plus{} pr \plus{} ps \plus{} pt \plus{} qr \plus{} qs \plus{} qt \plus{} rs \plus{} rt \plus{} st)$

$ \equal{} \frac {1}{2}\{(p \minus{} q)^2 \plus{} (p \minus{} r)^2 \plus{} (p \minus{} s)^2 \plus{} (p \minus{} t)^2 \plus{} (q \minus{} r)^2$

$ \plus{} (q \minus{} s)^2 \plus{} (q \minus{} t)^2 \plus{} (r \minus{} s)^2 \plus{} (r \minus{} t)^2 \plus{} (r \minus{} s)^2 \}< 0$

but for real numbers $ p,\ q,\ r,\ s,\ t$, that can't be true.

To be honest, I also thought the same idea.

Maybe we can generize as follows:

Prove that the roots of $ x^n\plus{}a_1x^{n\minus{}1} \plus{}a_2x^{n\minus{}2}\plus{}\cdots \plus{}a_{n\minus{}1}x\plus{}a_n\equal{}0\ (n\geq 2)$ can't all be real if $ (n\minus{}1)a_1^2<2na_2$.

Is this true?

hey kunny. basically, the backbone of our solutions is the same. but, i have a problem with yours... what if thr problem asked you to find a condition such that the roots can't all be real. This would mean that there's no mention about $ 2a^2<5b$, right? So, basically, my solution just revolved around the given first (the 5 roots) before arriving at the given inequality.

anyways, that's just my comment. but, i know it's correct. just gathering some probable solutions.:)

THANKS!!!:D

[quote="Ravi B"]Here is a calculus approach.  Suppose that $ x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e$ has all real roots.  Then by Rolle's theorem, the same is true of its derivative, its second derivative, and its third derivative.  Its third derivative is the quadratic $ 60x^2 \plus{} 24ax \plus{} 6b$.  Because the quadratic has all real roots, its discriminant is nonnegative.  So $ (24a)^2 \minus{} 4(60)(6b)$ is nonnegative.  Hence $ 2a^2 \ge 5b$.[/quote]

hi RaviB, thanks for offering a calculus approach. but, this is olympiad style so i'll stick to high school maths for the final solution.:) THANKS for replying to my post.:)

[quote="kunny"]Prove that the roots of $ x^n \plus{} a_1x^{n \minus{} 1} \plus{} a_2x^{n \minus{} 2} \plus{} \cdots \plus{} a_{n \minus{} 1}x \plus{} a_n \equal{} 0\ (n\geq 2)$ can't all be real if $ (n \minus{} 1)a_1^2 < 2na_2$.[/quote]
Yes, this generalization is true.  For variety's sake, here's another proof.  By Vieta's theorem, the arithmetic mean of the roots is $ \minus{}\frac{a_1}{n}$ and the quadratic mean of the roots is $ \frac{a_1^2 \minus{} 2a_2}{n}$.  If all the roots are real, we can apply the QM-AM inequality to get the contrapositive of your statement.


Let $ p_1 \equal{} 2, p_2 \equal{} 3, ...$ be the sequence of prime numbers. Prove or disprove the following statement: $ \exists N,\forall n\geq N, p_n > 3n$

All primes greater than 3 are either of form $ 6m \minus{} 1$ or $ 6m \plus{} 1$ otherwise they would be a multiple of $ 2$ or $ 3$.
It can be checked that $ p_9 \equal{} 29$ therefore $ p_{9 \plus{} 2k} \ge 29 \plus{} 6k > 3(9 \plus{} 2k)$ and $ p_{10 \plus{} 2k} \ge 31 \plus{} 6k > 3(10\plus{}2k)$ for $ k \ge 0$
Therefore the assertion is true with $ N \equal{} 9$


geometry
geometric transformation
rotation

The letters C, O, U, N and T are randomly placed around a circle. One such arrangement is shown here. If we always start with the C and continue to read the rest of the letters in order clockwise, in how many different orders can the letters appear?

[asy]draw(circle((0,0),1),dashed);
label("U",dir(90));
label("C",dir(90+360/5));
label("O",dir(90+2*360/5));
label("N",dir(90+3*360/5));
label("T",dir(90+4*360/5));[/asy]

COUNT is the same as TCOUN, they are just rotations of the circle COUNT. To eliminate this problem with rotations, we'll fix a point, say C. Then the other points, OUNT, can be arranged in $ 4!=\boxed{24} $ ways.

An alternate explanation would be the number of ways to place the letters of $\text{COUNT}$ without regard to rotation is $5!$.  However, there are $5$ rotations for each combination, thus we must divide by $5$ to get $\frac{5!}{5}=4!=\boxed{24}$.

In my aops book, you have to do (n-1)!/2. 4!/2 = 12

[quote=DRTFROG2]It never said c cannot move
Pls help me with this[/quote]

C can move, but we can turn the circle and it stays the same. To stop this, we keep C in one spot, and that removes all the rotations.

[quote=Darksid][quote="akhil0422"]That is simply $ 4!\equal{}\boxed{24}$.[/quote]
ya that was simple 4![/quote]

It was a little more complicated than that but yes it ended up being 4!
Because 5!/5=4!=24

[hide=solution] Since we always read the letters clockwise, this is really the same as counting the number of linear permutations of the 5 letters, given that C must go first. So, there are $1 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 4! = \boxed{24}$ orders.[/hide]

[hide=Post #10 by Yiyj1][url=aops.com/community/user/584706][b]Yiyj1[/b][/url] · 23 minutes ago [url=aops.com/community/p22991508](view)[/url][color=transparent]helo[/color]
[quote=Darksid][quote="akhil0422"]That is simply $ 4!\equal{}\boxed{24}$.[/quote]
ya that was simple 4![/quote]

Please hide your solutions.

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[color=#5b7083][aops]x[/aops] 10[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@Yiyj1][img]https://avatar.artofproblemsolving.com/avatar_584706.png[/img]
AoPS User[/tip]
Those users have not been online since 2011 and 2009.
So, they cannot edit it. Please also refrain from double posting, and simply edit the first post to include the second.

[hide=answer]Like the past answers, I believe it is $4!=24$ orders.[/hide]


I was wondering. [b]Can you take an AP Exam if you don't take the class?[/b] Because I am going to take IB Chemistry so I can take the AP Exam right? [b]Also, can you retake AP Exams the next year?[/b] I feel like doing AP Spanish in 10th. I probably get a 4 and not a 5, but I definitely going to get a five when I'm done with IB Spanish.

[quote="churchilljrhigh"]Can you take an AP Exam if you don't take the class?[/quote]

Anyone who signs up for them can take the AP exams, and there is no requirement to take a particular course beforehand.

http://www.collegeboard.com/student/testing/ap/about_faq.html

Hope this helps!

The answer to both questions is yes. But the exams are very expensive, and you may have to pay for it yourself in those situations. You would also need your school (or another school) to order the exam in enough time for you to take it.

But if you take the IB class, why not just take the IB exam?

It is $82.00 an exam if you sign up early (as in, the next week and a little more). After that, it is $107.00 a piece. A bunch of people complain about the prices for just a couple of exams, but face it; these are going to be the cheapest college credits you're ever going to earn.

I think the AP system offers benefits over the IB program (although admittedly I don't know much about IB) because the AP system is separate to each exam whereas the IB is one whole thing. Somebody clarify this (or tell me if I'm flat-out wrong).

As token amd gauss said, you don't need to take the class. You can also retake any AP exam, but I am unsure about scores overwriting other scores.

AP tests are always during about the first two weeks of May, so if you want to take them, you'd better pay soon. If you're going to take several tests, it's probably wise to start studying soon. I'm taking seven AP tests in May (if you count Physics C as two), so I'm starting this weekend :).

[quote="rlee"]these are going to be the cheapest college credits you're ever going to earn.[/quote]

If you go to a college that will trade them in for direct credit.

$85.00 an exam so you'd better be prepped. But yes, collegeboard.com has instructions on registering for APs that your school doesn't offer etc etc. Check with the school, they'll prolly have something.

[quote="churchilljrhigh"][b]Also, can you retake AP Exams the next year?[/b] I feel like doing AP Spanish in 10th. I probably get a 4 and not a 5, but I definitely going to get a five when I'm done with IB Spanish.[/quote]

Maybe I am not familiar enough, but I can't see a major benefit in taking an IB course and the AP exam.  You could probably retake an AP exam (as long as you fork over the 85 bucks!) but I don't think there's an option to cancel an AP score.  Also, if you're pretty sure you want to take both AP and IB spanish, and you think if you take AP Spanish next year you won't get a 5...why not slow down and just take AP Spanish after IB Spanish and get a 5 without taking the test twice?

BECAUSE I'm in Spanish 3. In Massachusetts there is no such thing as Spanish 4  why should I take Spanish 3 again.


Printing a book costs 2 cents per page plus $ \$2$ for the cover. How many pages does a book have if its total cost is twice as much as the total cost of a 100-page book?

A 100-page book costs $ 0.02 \times 100 \plus{} 2 \times 1 \equal{} 4$ dollars. So, we want to find the number of pages a book has that costs 8 dollars.

The cover is $ \$2$, so the cost of the pages alone is $ 8 \minus{} 2 \equal{} 6$ dollars. Therefore, there are $ 6 \div 0.02 \equal{} \boxed{300}$ pages.


The roots of $ f_1(x) \equal{} x^n$ are all at the origin on the complex plane.

The roots of $ f_2(x) \equal{} x^n \plus{} r^n$ form a circle of radius r, with its center at the origin of the complex plane.

The roots of $ f_3(x) \equal{} (x\minus{}b)^n \plus{} r^n$ form a circle of radius r, with its center at b.

How would the shape change if, instead of introducing constants, you introduce a term with x?

$ f_4(x) \equal{} (x\minus{}b)^n \plus{} x \plus{} r^n$

?

Then they'll no longer lie on any nice curve from "elementary geometry".

Hi, you told me about Rouche's theorem in the other thread and I think it is related to this problem. On Wikipedia, it says that Rouche's theorem can be applied like this:

Suppose you want to know something about the roots of:

$ f(z) \equal{} z^5 \plus{} 3z^3 \plus{} 7$

When $ f(z) \equal{} 0$,
$ z^5 \equal{} \minus{} 3z^3 \minus{} 7$.
$ |z^5| \equal{} | \minus{} 3z^3 \minus{} 7|$

Suppose I want to know the number of roots inside a circle at the origin of radius 2, then I would compute and observe that when the above is evaluated at $ z \equal{} 2$:
$ 32 > 31$

If 2 is a root, the two sides would equal. However, in this case, the side that grows faster is already greater than the side that grows slower, therefore, $ f(z)$ has no root outside the circle of radius 2.

I know this is not the full power of Rouche's Theorem, but is this explanation correct?

$ g(x) \equal{} (x \plus{} 1)^{n} \plus{} x \plus{} r^{n}$

First, I shift the axis by defining $ z \equal{} x \plus{} 1$, $ x \equal{} z \minus{} 1$, then:

$ g(z) \equal{} z^n \plus{} z \minus{} 1 \plus{} r^n$

To solve for the roots, set $ g(z) \equal{} 0$ and rearrange:

$ z^n \equal{} \minus{} z \plus{} 1 \minus{} r^n$
$ | z^n| \equal{} | z \minus{} 1 \plus{} r^n |$

Suppose I want to know whether $ g(z)$ would have roots within the unit circle:
$ 1 > r^n$
$ 1 > | r |$
If $ r$ is less than 1, the roots of $ g(z)$ will be within the unit circle. I have just checked it and it is not correct. For instance:
$ z^2 \plus{} z \minus{} 1 \plus{} 0.1^2$, where $ r\equal{}0.1$, and $ n\equal{}2$, has roots {-1.6, 0.6}.

?

It is not "evaluation at the point $ z\equal{}2$, but the inequality $ |z^5|>|\minus{}3z^3\minus{}7|$ on the [b]entire circle[/b] of radius 2 that is needed for applying Rouche! Do you see the difference? ;)

Is it only on the boundary of the circle?

$ f(z) \equal{} z^{5} \plus{} 3z^{3} \plus{} 7$
Check for $ | z^5 | > | 3z^3 \plus{} 7|$:

Suppose $ z \equal{} j2$
$ |z^5| \equal{} 32$
$ |3z^3 \plus{} 7| \equal{} | \minus{} 24j \plus{} 7| \equal{} 25$ It still holds.

$ g(z) \equal{} z^{n} \plus{} z \minus{} 1 \plus{} r^{n}$, suppose n = 2.
$ |z^2| > |z \minus{} 1 \plus{} r^2|$ ?

At $ z \equal{} 1$: $ 1 > r^2$, so $ 1 > r$.
At $ z \equal{} j$: $ 1 > \sqrt {(r^2 \minus{} 1)^2 \plus{} 1}$ so $ 1 > r$.
At $ z \equal{} \minus{}1$:
$ 1 > r^2\minus{}2$ if $ r^2 > 2$, so $ \sqrt{3}>r$, but $ r^2 > 2$ implies $ r>1$, we can't have that.
$ 1> 2\minus{}r^2$ if $ r^2 < 2$, so $ r> 1$. Impossible to have $ 1>r$ and $ r>1$.

Yes, only on the boundary (sorry for using an ambiguous word "circle"; I should have said "circumference" but was too lazy to type such a long word). Indeed, you cannot use Rouche for $ g(z)$ and the unit disk (but you already know that there may be zeroes outside the unit disk, so it shouldn't come as a surprise).


Consider a sample of size n from EXP(θ; η).
Compare MME and MLE for θ and η through Mean Squared Errors.

To find the mme and mle do you have to set θ to 1 or something??
Thanks.

Erm... Your abbreviations may be standard, of course, but I do not recognize them. What do $ Exp(\theta;\eta)$, MME, MLE, and "through mean squared errors" mean? The only thing that can be farther from reality than the assumption that everybody knows all the abbreviations and techniques you just learned in your statistics class is the assumption that somebody will spend his or her time trying to figure out what they might possibly mean ;). On the other hand, if you explain clearly what you are talking about, you may expect a lot of people to try to help you :).

MME is the method of moment estimator.
MLE is the Maximum likelihood estimator.
Exp(θ,η) is a 2 parameter exponential.
im not really sure what the mean squared error part is about though:(

you fix some parameters $ (\theta;\eta)$ and then compute the estimator $ \hat{\theta}, \hat{\eta}$ based on the observation of $ n$ independant observations. What have you done until now ?


[quote="Abdek"]Let $ a,b,c$ be a positive real numbers such that $ abc \equal{} 1$

show that :

$ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \ge \frac {8(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1)}$[/quote]
Because $ (a^2\plus{}bc)(b\plus{}c)\geq2\sqrt{bc(a^2\plus{}c^2)(a^2\plus{}b^2)}.$

prove that
$ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \geq \frac {(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}{(a \plus{} b \plus{} c)(bc \plus{} ca \plus{} ab) \minus{} 1}$

BQ

[quote="arqady"][quote="Abdek"]Let $ a,b,c$ be a positive real numbers such that $ abc \equal{} 1$

show that :

$ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \ge \frac {8(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1)}$[/quote]
Because $ (a^2 \plus{} bc)(b \plus{} c)\geq2\sqrt {bc(a^2 \plus{} c^2)(a^2 \plus{} b^2)}.$[/quote]

That's it !!  :lol:

[quote="arqady"][quote="Abdek"]Let $ a,b,c$ be a positive real numbers such that $ abc \equal{} 1$

show that :

$ (a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \ge \frac {8(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{((a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1)}$[/quote]
Because $ (a^2 \plus{} bc)(b \plus{} c)\geq2\sqrt {bc(a^2 \plus{} c^2)(a^2 \plus{} b^2)}.$[/quote]

nice...
 :)


For each pair of real numbers $ a \ne b$, define the operation $ \star$ as
\[ (a \star b) \equal{} \frac{a \plus{} b}{a \minus{} b}.\]
What is the value of $ ((1 \star 2) \star 3)$?


quadratics
inequalities
inequalities proposed

$a^2+b^2+c^2+d^2=1+5abcd$, a, b, c, d - real numbers.

find the minimum of $a^2+b^2+c^2+d^2$

We have (2nd ineq using AG):
$1=a^2+b^2+c^2+d^2-5abcd\leq a^2+b^2+c^2+d^2+|5abcd| \leq a^2+b^2+c^2+d^2+\frac{5}{16}(a^2+b^2+c^2+d^2)^2=x+\frac{5x^2}{16}$
with $x=a^2+b^2+c^2+d^2$ which we are trying to minamalize.
So we have $1\leq x+\frac{5x^2}{6}$
Solving this quadratic inequality we obtain (using also positiveness of $x$):
$a^2+b^2+c^2+d^2=x\geq \frac{4}{5}$
It attains minimum when $a=b=c=\frac{1}{\sqrt{5}}, d=-\frac{1}{\sqrt{5}}$


complex numbers
complex analysis
complex analysis unsolved

Prove that if $ \Re f(z)$ or $ \Im f(z)$ in an isolate singularity from up or down be bounded , then that singualarity is removable.

Assume that the real part of $ f$ is bounded from above. Then consider $ \exp\circ f$ which is bounded in an open neighbourhood of the isolated singularity, hence the singularity is removable. Since $ \exp$ is a local biholomorphism, $ f$ has a removable singularity, too. The other cases can easily be reduced to the case just proven.


inequalities
search
inequalities theorems

Hello,

Can anyone post Muirhead's Inequality here? I'm having trouble understanding the notation that [url=http://planetmath.org/encyclopedia/MuirheadsInequality.html]PlanetMath[/url] uses.

Thanks!

Thanks for your response.

A question about your notation: how does one expand the symmetric sums? For example, let [tex]x_1 = x[/tex], [tex]x_2 = y[/tex], [tex]x_3 = z[/tex], [tex]a_1 = 2[/tex], [tex]a_2 = 3[/tex], [tex]a_3 = 2[/tex], [tex]b_1 = 1[/tex], [tex]b_2 = 3[/tex], and [tex]b_3 = 3[/tex]. How would one write out the sums on the left and right sides of the equation?

Thanks in advance!

Bruno,
you have left out the condition that $a_1 \ge a_2 \ge \dots \ge a_n$ and $b_1 \ge b_2 \ge \dots \ge b_n$.

JSRosen3,
shuffling your numbers around, your example would now be $(a_1, a_2, a_3) = (3, 3, 1)$ and $(b_1, b_2, b_3) = (3, 2, 2)$, and the resulting inequality would be
\[2x^3 y^3 z + 2x^3 y z^3 + 2x y^3 z^3 \ge 2x^3 y^2 z^2 + 2x^2 y^3 z^2 + 2x^2 y^2 z^3.\]

I have a question too: Do $a$-s and $b$-s need to be positive integers?

Because i read an article where it said that they must be positive, but for example $(1,-1)\succ(0,0)$ and $ \frac ab + \frac ba \geq 2$, so Muirhead is true(i've come up with a couple of other cases too, so i thought: hey, why couldt it hold for all integers?)

Thanx

Does anyone have a proof of the Muirhead inequality?

[quote="billzhao"]The whole point of symmetric sum is that you sum over all permutation regardless of the expression. So the correct form is $\sum_\mathrm{sym} x^2 = 2x^2+2y^2+2z^2$. You will see the reason behind this once you use Muirhead.[/quote]

What does cyclic sum mean? (  $\sum_\mathrm{cyc}$  )

[quote="perfect_radio"]What does cyclic sum mean? (  $\sum_\mathrm{cyc}$  )[/quote]

Suppose we have 3 variables $x, y, z$ (which is the case most of the time). Then
\[
\sum_\mathrm{cyc}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b)
\]
For example, $\sum_\mathrm{cyc} x^2y = x^2y+y^2z+z^2x$ and $\sum_\mathrm{cyc} x^3 = x^3+y^3+z^3$.

[quote="jaka"]So i cant have for example $(1,0,0) \succ (\frac 13, \frac 13, \frac 13)$, which is AM/GM?[/quote]

why not?

[quote="billzhao"]No, Muirhead does not require exponents to be positive (this is evident since we can multiply both sides by $x_1x_2\cdots x_n$ enough times to make the exponents positive). The exponents don't even have to be integers.

However, the variables must be nonnegative, that is a must.

Another comment, the number one mistake I see with symmetric sum notation is people saying things like $\sum_\mathrm{sym} x^2 = x^2+y^2+z^2$. THAT IS NOT TRUE. The whole point of symmetric sum is that you sum over all permutation regardless of the expression. So the correct form is $\sum_\mathrm{sym} x^2 = 2x^2+2y^2+2z^2$. You will see the reason behind this once you use Muirhead.[/quote]

I admit that I am an abuser of this notation (hence I have generally dropped both the sym and cyc subscripts in my posts.  However, it's usually enough to count terms and note that in the manner that I usually do it,

$\sum x^ay^bz^c$

contains $6$ terms if $a\not=b\not=c$
contains $3$ terms if $a=b\not=c$ or one of the permutations thereof
contains $1$ term if $a=b=c$ (I usually just write $x^2y^2z^2$ or whatever instead

If there are $4+$ variables, I write out the proper symsums, just to avoid mistakes.

[quote="spang"]What does $\succ$ mean?[/quote]

majorization

[quote="DPopov"]Is Muirhead the same thing as "Bunching" ?[/quote]
Yes.

Ah, thanks. I have another question; if you have something that holds by Muirhead can you conclude that the cyclic form of that inequality will also hold or does it depend on the situation?

In general, Muirhead does not help in proving a cyclic inequality.

That's the symbol used for a symmetric sum ;) (http://www.mathlinks.ro/Forum/viewtopic.php?highlight=symmetric+sum&t=6697)

Must $a_{1},\ldots,a_{n}$ and $b_{1}, b_{2},...,b_{n}$ be integer numbers?

Sorry for this stupid question, but how do you pronounce "Muirhead"?
Obviously I'm somehow a bit dumb to speak out the "Muir" comfortably.....

Everyone I know pronounce "Muir" like "Mir" in German (or Russian, Polish and probably many more languages  :)  )

I guess we can ask vinoth, since stephen muirhead is on australian team  :D

Every nice inequalities paper contains good stuff about Muirhead's inequality. Here you have some well-known names: Mildorf's [i]Olympiad inequalities[/i], Kedlaya's $A<B$, and Lee's [i]Topics in inequalities[/i]. You may search for them in Google, if you don't know them. :)

[quote="Armo"][quote="perfect_radio"]Does anyone have a proof of the Muirhead inequality?

[/quote]
It can be easily done by Karamata.[/quote]

How exactly ? I found only complicated proofs , but "easy"  :o
i'm very stupid , so help please  :(

[quote="Dragos Calinescu"]Hey can anyone please tell me something, cause i am a bit confused: let's say $ \sum_{sym}a^3\ge\sum_{sym}a^2b$ but $ \sum_{cyc}a^3\ge\sum_{cyc}a^2b$ isn't always true, am i right?[/quote]

Bad example - $ \sum_{cyc} a^3 \equal{} \sum_{cyc} \frac{2}{3}a^3 \plus{} \frac{1}{3}b^3 \ge \sum_{cyc} a^2b$ by AM-GM. But yeah, you have the general idea - symmetric inequalities don't always imply the corresponding cyclic inequality.

For instance, $ \sum_{sym} a^3b^2d \ge \sum_{sym} a^2bc^2d$ by Muirhead, but $ \sum_{cyc} a^3b^2d \ge \sum_{cyc} a^2bc^2d$ isn't true when we plug in $ a \equal{} 1000, b \equal{} 1, c \equal{} 1000, d \equal{} 1$.

Zeb, is right. Even for 3 variables. The Muirhead condition is not the same as the cyclic condition.
Look towards the end of
http://www.mathlinks.ro/viewtopic.php?search_id=1813538618&t=103761

[quote="jaka"]Can anybody please post a link to a proof of the Muirhead's theorem...please!  :)[/quote]

Wikipedia is a very useful site, you know. You can get everything(practically) there...... :lol:

http://en.wikipedia.org/wiki/Muirhead%27s_inequality

Also see the AoPS Wiki:

http://www.artofproblemsolving.com/Wiki/index.php/Muirhead%27s_Inequality.

If you still don't think you have got it/understood it, we members are here for you!!!!!!! :)

Correct me if I'm wrong, but jaka asked for a proof, and the linked wiki pages do not offer one.

But K.Y. Li does:

see pages 2 and 4: http://www.math.ust.hk/excalibur/v11_n1.pdf

[quote="great math"]Is that right to apply Muiheard in the following situation?
\[ \sum_{cyc}a^3b^2 \ge \sum_{cyc}a^3bc\]
[/quote]
If you ask the question as you do, the response is as simple as this: you CANNOT apply Muirhead because Muirhead is only about symmetric sums... (see above)  :!: Muirhead is just a certain symmetrical application of AM-GM. As shown in the thread mentioned above, there are necessary and sufficient conditions showing [i]when [/i]AM-GM can be applied in an asymmetrical way to prove cyclic sum inequalities of this type. BTW, it cannot for the one you give, which means it is wrong.)

Often this theory will use with Schur Inequality


If 0<a<b<c<d are real numbers such that a+b+c+d \leq 1 prove that

a 2 +3b 2 +5c 2 +7d 2  \leq 1

The problem statement should be a > b > c > d > 0.

We will prove it (WLOG) for a + b + c + d = 1.

1 - a2 - 3b2 - 5c2 - 7d2
= (a + b + c + d)2 - a2 - 3b2 - 5c2 - 7d2
= 2(ab + bc + cd + da + ac + bd - b2 - 2c2 - 3d2)
= 2[d(a - d) + d(b - d) + d(c - d) + c(a - c) + c(b - c) + b(a - b)]
> 0.

You are right, Arne.
Sorry for my mistake.


Prove that for every natural number $N$ there exist a natural number $M$ such that $\overline{MN} \cdot N = \overline{NM}$. (May be there are some zeros at the beginning of $M$.)

For example, if $N=46$, then $M=0100021743857360295716$. So, $10002174385736029571646 \cdot 46 = 460100021743857360295716$.

Suppose $N$ is a $n$-digit number, $M$ is $m$-digit number. Then
\[(10^nM+N)N=10^mN+M.\]
It follows that $M$ is divisible by $N$, so $M=Ns$, and we obtain $10^nNs+N=10^m+s$, i.e.
\[s=\frac{10^m-N}{10^nN-1}.\]
Therefore we are to find $m\in\mathbb{N}$ s.t. $10^nN-1\mid 10^m-N$. It is equivalent to $10^nN-1\mid 10^{m+n}-1$, and we are done :D

[quote="Myth"]Remike! Почитай http://www.mathlinks.ro/Forum/viewtopic.php?t=21455
?егоже называть так задачи. Про?ви фантазию, чтобы по названию можно было догадать??, о чём топик.[/quote]
Извин?ю?ь, ? не читал ?той ветки. Вроде переименовал.

[quote="Myth"]Remike! Почитай http://www.mathlinks.ro/Forum/viewtopic.php?t=21455
?егоже называть так задачи. Про?ви фантазию, чтобы по названию можно было догадать??, о чём топик.[/quote]

It seems that this number theory subforum has changed into the Russian national local community. But I guess I am the only non-native Russian speaker who has noticed it.  :D


Find all trios of natural numbers $ (p,q,n)$ satisfying $ p,q$ are pimes and $ p(p\plus{}1)\plus{}q(q\plus{}1)\equal{}n(n\plus{}1)$

$ p(p\plus{}1)\plus{}q(q\plus{}1)\equal{}n(n\plus{}1) \Rightarrow p(p\plus{}1)\equal{}(n\minus{}q)(n\plus{}q\plus{}1)$

If $ p|n\minus{}q \Rightarrow n\minus{}q \ge p$ and $ n\plus{}q\plus{}1>p\plus{}1 \Rightarrow (n\minus{}q)(n\minus{}q\plus{}1)>p(p\plus{}1)$

It follows that $ p|n\plus{}q\plus{}1 \Rightarrow n\plus{}q\plus{}1\equal{}kp$

$ p(p\plus{}1)\equal{}(n\minus{}q)kp \Rightarrow p\plus{}1\equal{}nk\minus{}kq \Rightarrow p\equal{}nk\minus{}kq\minus{}1$

$ n\plus{}q\plus{}1\equal{}kp\equal{}k(nk\minus{}kq\minus{}1) \Rightarrow k^2(n\minus{}q)\minus{}k\minus{}(n\plus{}q\plus{}1)\equal{}0$

$ \Delta \equal{}1\plus{}4(n\minus{}q)(n\plus{}q\plus{}1)\equal{}(2n\minus{}2q\plus{}1)^2$

$ \Rightarrow k\equal{}\frac {1\plus{}2n\minus{}2q\plus{}1} {n\minus{}q}\equal{}\frac {2(n\minus{}q\plus{}1)} {n\minus{}q}$

But $ k \in \mathbb{N} \Rightarrow n\minus{}q|2(n\minus{}q\plus{}1)$

From $ n\minus{}q|2(n\minus{}q)$ we deduce that $ n\minus{}q|2 \Rightarrow n\minus{}q \in \{ 1,2 \}$ because $ n>q$

Similar $ n\minus{}p \in \{ 1,2 \}$

$ \bullet$ 1st case: $ n\minus{}q\equal{}1$ and $ n\minus{}p\equal{}1$

$ \Rightarrow p\equal{}q \Rightarrow 2p(p\plus{}1)\equal{}(p\plus{}1)(p\plus{}2) \Rightarrow p\equal{}q\equal{}2$ and $ n\equal{}3$

$ \bullet$ 2nd case: $ n\minus{}q\equal{}1$ an $ n\minus{}p\equal{}2$

$ \Rightarrow p(p\plus{}1)\plus{}(p\plus{}1)(p\plus{}2)\equal{}(p\plus{}2)(p\plus{}3) \Rightarrow p(p\plus{}1)\equal{}2(p\plus{}2) \Rightarrow p^2\minus{}p\minus{}4\equal{}0$

$ \Delta \equal{}1\minus{}4(\minus{}4)\equal{}17 \Rightarrow \sqrt {\Delta} \equal{}\sqrt {17}$ $ \not\in \mathbb{N}$

$ \Rightarrow (p,q,n) \in \emptyset$

$ \bullet$ 3rd case: $ n\minus{}q\equal{}2$ and $ n\minus{}p\equal{}1$

Similar with 2nd case

$ \bullet$ 4th case: $ n\minus{}q\equal{}2$ and $ n\minus{}p\equal{}2$

$ 2p(p\plus{}1)\equal{}(p\plus{}2)(p\plus{}3) \Rightarrow p^2\minus{}3p\minus{}6\equal{}0$

$ \Delta\equal{}9\minus{}4(\minus{}6)\equal{}9\plus{}24\equal{}33 \Rightarrow \sqrt {\Delta}\equal{}\sqrt {33} \not\in \mathbb{N}$

$ \Rightarrow (p,q,n) \in \emptyset$

Finally, $ (p,q,n)\equal{}(2,2,3)$


I m given these rules for division:

a)$ t|t$
b)$ 1|t$
c)$ t|0$
d)$ t|s\wedge s|r \rightarrow t|r$
e)$ s \neq 0 \wedge t|s \rightarrow t \le s$
f)$ t|s \wedge s|t \rightarrow s \equal{} t$
g)$ t|s\rightarrow t|(r\cdot s)$
h)$ t|s \wedge t|r \rightarrow t|(s \plus{} r)$

note: $ \wedge$ = "and" , $ \rightarrow$ = implies, $ s'$= successor of $ s$

how would I show $ (t|s\wedge t|s')\rightarrow t \equal{} 1$?

note:$ s'$= successor of $ s$ i.e. 4 =successor of 3

What would be a rigorous way to show this in the natural numbers( i don't have subtraction, only addition and multiplication)

The reply has already been given in your other thread; all of those rules only allow you to deduce that if $ t | s$ then $ t$ divides something bigger than $ s$; none of them allow you to work the other way, namely, to conclude that $ t | 1$.

what if you were also given $ t|s \leftrightarrow (\exists c)(s \equal{} t\cdot c )$?

As has also been stated in the other thread, this is the standard definition of divisibility, and all well-known properties can be deduced from it.

is reasoning by contradiction the only way, or is there a better way?

Without the use of subtraction, I believe the proof by contradiction mentioned in the other thread is the best way.

Prove$ (t|s\wedge t|s')\rightarrow t \equal{}$

Need to show for two cases $ s \equal{} 0$ and $ s\neq0$

$ s \equal{} 0$

we simply have the case $ t|1\rightarrow t \equal{} 1$

by hypothesis, we have $ t|1$
by proposition b) we have $ 1|t$
then by prop. f) we have $ t|1 \wedge 1|t \rightarrow t \equal{} 1$

$ s\neq0$

let $ s' \equal{} s \plus{} 1$
we'll try to prove a contradiction
by hypothesis we have$ t|s$ and $ t|s'$ and assume $ t\neq1$
then by prop e)$ s \neq 0 \wedge t|s \rightarrow t \le s$
thus we have two cases: i) $ t \equal{} s$ and ii)$ t < s$
case i)
if $ t \equal{} s$ then we know that $ t|s \equal{} 1$ but $ t\nmid s' \equal{} s \plus{} 1$.
how do we know this? proceed by induction.
base case:( since $ t\neq 1$) we start with t=2
clearly, $ 2|2$ and for $ n \in\mathbb{N} , 2|2n$ but it's clear that $ 2\nmid 2n \plus{} 1$
step:Assume true for $ k$ where $ k \equal{} mn$ for $ m,n\in\mathbb{N}$

this is where I'm stuck. I just dont see how i can show definitively for $ k \plus{} 1$

the idea of this proof is to show that for $ t \equal{} s$, we can't have $ t \equal{} s$ since $ t\nmid s'$ and thus doesn't satisfy our hypothesis. and for $ t < s$, show that becuase $ t < s$ and $ t < s'$ for some arbitrary $ s$, then work backwards inductively ( method of infinte descent?) such that we get $ t < 1$ which contradicts the fact that $ t\in\mathbb{N}$. So I sorry that I m so bad at both cases.

any help is very much appreciated.

Thanks

Please continue in http://www.mathlinks.ro/Forum/viewtopic.php?t=173409 .


geometry
analytic geometry
trapezoid
trigonometry
geometry unsolved

In a triangle ABC with points of tangency of incircle K, L and M on BC, CA and AB respectively are constructed points P, Q and R, which are the cutting KL, LM and MK with straight lines parallel to LM, MK and KL by A, B and C respectively. Let be X, Y and Z the intersections of PM, QK y RL with the incircle of ABC respectively. Proves that the lines AX, BY and CZ are concurrent.

[b]1).[/b]
$ \mbox{Ussing barycentric coordinates if: } p = \frac {a + b + c}{2} \mbox{, then have: } \\
A(1;0;0); B(0;1;0); C(0;0;1); K(0;p - a;p - b); L(p - c;0;p - a); M(p - b;p - a;0); \\
\mbox{and equation of incircle is: } \\
a^2yz + b^2xz + c^2xy - (x + y + z).[(p - a)^2.x + (p - b)^2.y + (p - c)^2.z] = 0.$   (1)
[b]2).[/b]
$ \mbox{The equation of line KL is: } \\
| \begin{array}{ccc} x & y & z \\
0 & {p - c} & {p - b} \\
{p - c} & 0 & {p - a} \end{array}| = 0\Leftrightarrow (p - a).x + (p - b).y - (p - c).z = 0.$     (2)
$ \mbox{Similary, obtain equation of line LM: } - (p - a).x + (p - b).y + (p - c).z = 0.$     (3)
[b]3).[/b]
$ \mbox{The point of infinity of line ML is: } (b - c; - b;c)\Rightarrow \mbox{ the equation of line AP is: } \\
cy + bz = 0.$  (4)
[b]4).[/b] $ \mbox{The coordinates of point P is solution of system: }$
$ \{\begin{array}{c} (p - a).x + (p - b).y - (p - c).z = 0 \\
cy + bz = 0 \end{array}  \Rightarrow \\
\Rightarrow P[b(p - b) + c(p - c); - b(p - a);c(p - a)].$
[b]5).[/b]
$ \mbox{The equation of line MP is: } |\begin{array}{ccc} x & y & z \\
{p - b} & {p - a} & 0 \\
{b(p - b) + c(p - c)} & { - b(p - a)} & {c(p - a)} \end{array}| = 0 \\
\Leftrightarrow c(p - a).x - c(p - b).y - [2b(p - b) + c(p - c)].z = 0\Leftrightarrow \\
\Leftrightarrow x = \frac {c(p - b).y + [2b(p - b) + c(p - c)].z}{ c(p - a)}.$  (5)
[b]6).[/b]$ \mbox{Solving system: }$
$ \{\begin{array}{c} a^2yz + b^2xz + c^2xy - (x + y + z).[(p - a)^2.x + (p - b)^2.y + (p - c)^2.z] = 0 \\
x = \frac {c(p - b).y + [2b(p - b) + c(p - c)].z}{ c(p - a)}\end{array}\Rightarrow\\
\Rightarrow \frac{4b^2(p-b)^2}{c^2}.z^2-\frac{a^2(p-a)+b^2(p-b)+c^2(p-c)}{p-a}.yz=0\Rightarrow\\
\Rightarrow (z=0) \lor ( \frac{y}{z}=\frac{4b^2(p-a)(p-b)^2}{c^2.[a^2(p-a)+b^2(p-b)+c^2(p-c)]}).\\
\mbox{a). value z=0 corespond to the point M, and }\\
\frac{y}{z}=\frac{4b^2(p-a)(p-b)^2}{c^2.[a^2(p-a)+b^2(p-b)+c^2(p-c)]}
\mbox{ -is equation of line AX.}$
[i]something is not right!, must again calculations ...[/i]
[b]I will continue later...[/b]

It's easy to see that AXLP is cyclic quadrilateral.
Let $ MK\cap AP\equal{}\{J\}$. By [b]Miquel[/b]'s theorem we get AXMJ is cyclic.
Then L,X,J are collinear.
JPLM is a trapezoid so KX passes through the midpoint of ML. Similarly for MZ, LY
We obtain $ KX, MZ, LY$ are concurrent.
Therefore $ AX, BY, CZ$ are concurrent.

I begin by similar way than livetolove, proving by anges that J, L and X are colinears and that AMKP is cyclic.
Then <XAL=<XPL=<MPK=<MAK, and then AX is the isogonal of AK, then the point is the Isogonal of Gergonne.
I was wondering if this is a result known... do you know? :|

[quote="livetolove212"][quote="Johan Gunardi"][quote="livetolove212"]
We obtain $ KX, MZ, LY$ are concurrent.
Therefore $ AX, BY, CZ$ are concurrent.[/quote]How does the concurrency of $ KX,MZ,LY$ implies the concurrency of $ AX,BY,CZ$?[/quote]
[hide]Use [b]Ceva sine[/b] theorem[/hide][/quote]I still don't understand. Please write in detail.

Applying [b]Ceva-sine[/b] theorem for triangle $ AML$ we get $ \frac{sin \angle MAX}{sin \angle LAX}\equal{}\frac{sin^2 \angle AMX}{sin^2 \angle XML}\equal{}\frac{sin^2 \angle MKX}{sin^2 \angle XKL}$.
Similar for $ BY, CZ$ we are done.


$m,n \in \mathbb N^{*}$ , $n\geq m$

Να δειχθεί ότι

$\left(1+\frac{1}{n}\right)^{n-2m}\leq \frac{n}{2m}$  .

p.s. Κι αυτή δικιά μου! Ελπίζω να μην είναι ήδη γνωστή.  :wink:
p.s.2 Όσο και αν φαίνεται περίεργο δεν έχω λύση γι'αυτή. Απλώς ήταν μια έμπνευση της στιγμής και επειδή την απέδειξα για διάφορες σχέσεις μεταξύ των m και n (π.χ. 2m>=n, κ.λ.π.), υποθέτω ότι ισχύει γενικά. Αν όχι, ας βρεθεί αντιπαράδειγμα.  :P

Στέλιος

8eto $n=2m+k$ opote me kapies praksules ginete
$2m(2m+k+1)^{k}\leq (2m+k)^{k+1}$ i isodinama
$(1+(k+1)/(2m))^{1/(k+1)}\leq (1+k/(2m))^{1/k}$
omos i akolu8ia $a_{n}=(1+n/(2m))^{1/n}$
ine f8inusa kai tini sto 1........kai ara i ipo8esi isxiii!!!  :coolspeak:
pantos den mporo na katalavo pos ir8e i idea...etsi sto ksekarfoto..????

 :D

Kalispera RAKAR,

Poli wraia h skepsi sou na katalikseis stin isodinami sxesi pou egrapses. To monadiko pou den pianw amesws einai gia pio logo i $a_{n}$ pou pairneis einai fthinousa. Mipws mporeis na to eksigiseis perissotero?

(Prepei episis na pareis ksexwrista tis periptwseis k=0,-1 dioti gia na pareis tin isodinamia pou grafeis prepei na tis eksaireseis kai na tis meletiseis meta mones tous).

Alexandros

Y.G. Stelio elkistikotati askisi Mpravo! :)

Λοιπόν, την ανισότητα την απέδειξα! Και όχι μόνο αυτό αλλά απέδειξα και μια γενίκευση της. Το μόνο που χρειάζεται είναι να κάνεις μερικές πράξεις φέρνοντάς την σε μια ισοδύναμη μορφή και μετά να δουλέψεις με κατάλληλες συναρτήσεις (παίρνοντας παραγώγους, αποδεικνύοντας μονοτονίες, κ.λ.π., η λύση είναι αρκετά μεγάλη όμως...). Παρ'όλα αυτά αν κάποιος έχει καμιά καλή ιδέα για κανέναν πιο κομψό τρόπο ας την πει.  :wink:

Επίσης, Ράλλη κι εγώ δεν κατάλαβα πως είναι τόσο προφανές ότι η ακολουθία $a_{n}$ είναι φθίνουσα. Αν και ισχύει εφόσον η άσκηση είναι σωστή, θέλει μια απόδειξη.

[b][u]Παρατηρήσεις πάνω στην ανισότητα[/u][/b] (...τις οποίες ανακάλυψα εκ των υστέρων)

$1)$Η ισότητα ισχύει αν και μόνο αν $n=2m$ ή $n=2m-1$.

$2)$Για $n\leq 2m$ η ανισότητα αποδεικνύεται απευθείας χρησιμοποιώντας την ανισότητα Αριθμητικού-Γεωμετρικού μέσου (πώς; ). :wink:

$3)$H ανισότητα μπορεί να επεκταθεί και στους πραγματικούς αριθμούς, αν βγάλουμε την συνθήκη $(n\geq m)$ και στη θέση της βάλουμε μία από τις συνθήκες: ($n\geq 2m>0$) ή ($-1>n\geq 2m$) ή ($2m-1\geq n>0$) ή ($-1>2m-1\geq n$)  (...βέβαια τώρα όπου $2m$ μπορούμε να βάλουμε και σκέτο $m$ και στις συνθήκες και στην ανισότητα).

Στέλιος

gia su cretanman!
ontos xriazete na paro  idiki periptosi gia k=0 kai -1(gia na min exo provlimata me ta rizika..)ala  kai s aytes tis periptosis vgeni poly eykola...
stelio,
oso gia tin akolu8ia $a_{n}$ oti ine f8inusa,den prokipti giati i askisi ine sosti ala prokipti i or8otita tis askisis apo to gegonos oti ayti ine f8inusa.....
yaxno na vro pu to ixa diavasi(itan ena palio biblio algebras tu patera mu)pu elege oti to orio tis $(an+b)^{1/n}$ me a,b 8etikus pragmatikus, ine to 1 otan to n tini sto apiro....(sti sigekrimeni periptosi a=1/2mkai b=1..)
aplos den 8imame an ixe kanena periorismo...gia ayto molis to vro 8a tin grayo analitika tin apodiksi tis monotonias tis $a_{n}$......

 :D

Gia tin sigekrimeni den einai diskolo na deikseis oti to orio einai to $1$ afou isxiei

$1<a_{n}=\sqrt[n]{\frac{n}{2m}+1}=\sqrt[n]{\frac{n+2m}{2m}}\stackrel{n\geq m}{\leq}\sqrt[n]{\frac{n+2n}{2m}}<\sqrt[n]{3n}\to 1$ kathws $n\to\infty$

(Einai gnwsto oti $\lim{\sqrt[n]{n}}=1$ kai $\lim{\sqrt[n]{a}}=1$ an $a>0$)

Pws voithaei auto stin euresi tis monotonias tis $a_{n}$ ?.

Alexandros

lipon...den vrika to vivlio pu eyaxna (krima gt 8imame oti ixe mia komyi apodiksula...) ala vrika mia dikia mu apodiksi oxi toso geniki (stin morfi an+b diladi....)ala apodiknii ayto pu 8elo....
cretanman ixa kani edit,ala prolaves kai estiles gia ayto to stelno se kenurgio mnm...
lipon esto$f(x)=(ax+1)^{1/x}$me a 8etiko pragmatiko, kai x>0
i sinartisi ayti exi proti paragogo
(ax+1)^{1/x} epi [-In (ax+1)/(x^{2})+a/(x(ax+1))]
arki ndo oti ine i deyteri agili arnitiki gt tote 8a ine f'(x)<0 kai ara f8inusa...
esto lipon deyteri agili <0 <=>ax/(ax+1)<In(ax+1) (1)
arki lipon na apodiksume tin (1)...
omos (x-1)/x<=Inx<=x-1 (2) (8eorite gnosti,eleos min apodikso kai ayti pos vgeni......pernis paragogus monotonies klpklp....rutina..  :P  )
e,vazontas stin (2) opu x to ax+1,to aristero melos tis (2) dini tin (1)....telos.....sorry ala den ta pao kala me tin latex.....
cretanman to orio tis akolu8ias den voi8ai pu8ena stin eyresi tis monotonias etso opos to ipa pio pano,aplos to anefera etsi gt 8imomun ayto to 8eorimataki s ayto to palio vivlio pu leo...

 :D

Για χάρη της συνέπειας παραθέτω την ανισότητα στη τελική της μορφή.

$m,n \in \mathbb Z^{*}$ με $n \neq-1$

Ν.δ.ό.   $\left(1+\frac{1}{n}\right)^{m-n}\geq \frac{m}{n}$

(με ισότητα αν και μόνο αν $n=m$ ή $n=m-1$) .

Στέλιος

vrika mia orea apodiksi se ena vivlio kai tin metafero aytoleksi...
i sinartisi $f(x)=Inx$ ine kili se olo to pedio orismu tis ka8os $f''(x)<0$
to grafima tis sinartisis perna apo ta simia $(k,Ink),(k+1,In(k+1))$
i ey8ia py pernai apo ayta ta simia exi eksisosi $\frac{y-Ink}{x-k}=In(k+1)-Ink$i isodinama  $y=Ink+(x-k)In(1+\frac{1}{k})$ omos logo tis kirtotitas tis sinartisis 8a ine $f(x)\geq y$ gia ta x pu anikun sto $[k,k+1]$
 diladi apologari8mizontas ,pernume
$\frac{x}{k}\geq (1+\frac{1}{k})^{x-k}$ to zitumeno...

 :D

[quote="RAKAR"]
i sinartisi $f(x)=Inx$ ine kili se olo to pedio orismu tis ka8os $f''(x)<0$
to grafima tis sinartisis perna apo ta simia $(k,Ink),(k+1,In(k+1))$
i ey8ia py pernai apo ayta ta simia exi eksisosi $\frac{y-Ink}{x-k}=In(k+1)-Ink$i isodinama  $y=Ink+(x-k)In(1+\frac{1}{k})$
[/quote]
Μπορείς να δουλέψεις κι έτσι αν $x,k>0$. Στην άσκησή όμως ισχύει $m,n \in \mathbb Z^{*}$. Αντιμετωπίζεις δηλαδή ένα μέρος της.

[quote="RAKAR"] diladi apologari8mizontas ,pernume
$\frac{x}{k}\geq (1+\frac{1}{k})^{x-k}$ to zitumeno...
[/quote]
Καταλήγεις σε λάθος συμπέρασμα (το ζητούμενο είναι άλλο...).
Η τελευταία ανισότητα ισχύει μόνο για τα $x$ που ανήκουν στο διάστημα $[k,k+1]$ (για τα υπόλοιπα ισχύει η αντίστροφη).


number theory
prime numbers
number theory proposed

Natural numbers $a, b$ and $c$ are pairwise distinct and satisfy \[a | b+c+bc, b | c+a+ca, c | a+b+ab.\]
Prove that at least one of the numbers $a, b, c$ is not prime.

Assume that $a,b,c$ is prime numbers . we have $a|a+b+c+ab+ac+bc$ , $b|a+b+c+ab+ac+bc$ , $c|a+b+c+ab+ac+bc$ , $(a,b)=1 , (a,c)=1 , (b,c)=1$
$\longrightarrow$ $abc|ab+bc+ca+a+b+c$ and $a<b<c$ . now we have : $abc \leq ab+bc+ca+a+b+c < 3bc+3c$ $\longrightarrow$
$ab < 3b+1$ $\longrightarrow$ $a<4$
$1)a=2$ now we have $2|b+c+bc$ , but $b,c$ are odd numbers . Contradiction !
$2)a=3 \longrightarrow 3bc\leq 3b+3c+bc+3+b+c$ $\longrightarrow$ $2bc\leq 6b+6c+3 < 12c+3$ $\longrightarrow$ $b<6 \longrightarrow$ $b=5$
now we have $c|3+5+15$ $\longrightarrow$ $c|23$ $\longrightarrow$ $c=23$ $\longrightarrow 3|5+23+115$ $\longrightarrow 3|143$ . Contradiction .

[quote="khashi70"]
$2)a=3 \longrightarrow 3bc\leq 3b+3c+bc+3+b+c$ $\longrightarrow$ $2bc\leq 6b+6c+3 < 12c+3$ $\longrightarrow$ $b<6 \longrightarrow$ $b=5$
now we have $c|3+5+15$ $\longrightarrow$ $c|23$ $\longrightarrow$ $c=23$ $\longrightarrow 3|5+23+115$ $\longrightarrow 3|143$ . Contradiction .[/quote]
Isnt it easier: \[3bc\leq 3b+3c+bc+3+b+c \iff 2bc\leq 4b+4c+3 \implies b<4\] contradiction?

Assume all three numbers are primes. Clearly none of them can be $ 2$ otherwise it would divide an odd number. So we have:
$ a|(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1,b|(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1,c|(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1$ and since $ a,b,c$ are distinct primes: $ abc|(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1$.
Now note that:
$ 1<\frac{(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1}{abc}<\frac{(a\plus{}1)(b\plus{}1)(c\plus{}1)}{abc}\equal{}\frac{a\plus{}1}{a}\times \frac{b\plus{}1}{b}\times \frac{c\plus{}1}{c} \leq \frac{4}{3} \times \frac{6}{5} \times \frac{8}{7} < 2$
and thus $ abc\nmid (a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1\Rightarrow$ at least one of $ a,b,c$ is not a prime.

[color=#ff9a00]solution[/color] [hide]Assume all of them are in fact prime numbers so $a|a+b+c+ab+ac+bc$ same goes for $b$ and $c$. Because they are pairwise distinct $\longrightarrow $ $abc|a+b+c+ab+bc+ac$ so $\frac{a+b+c+ab+bc+ac}{abc}$is an element in $\mathbb{N}$ assume WLOG $a=2$ it's obvious that this isn't possible since $a+b+c+ab+bc+ac$ is odd and $abc$ is even so take WLOG ${a,b,c}={3,5,7}$  $NOTE$:([color=#f00]WE ARE DOING THIS TO FIND THE MAXIMUM OF THE EXPRESSION[/color]) so when doing this the value of the expression is $\frac{86}{105}$ this is a clear contradiction since $\frac{86}{105}$ is less than $1$ so we are done [/hide]


Vasc said nice and easy inequality for $a+b++c=3$
\[\frac{a}{b^{2}+2c}+\frac{b}{c^{2}+2a}+\frac{c}{a^{2}+2b}\ge 1. \]

Why he said it is easy. :maybe:

Dear Manlio. It follows the Cauchy-Schwars inmmediately.

Notice that

\[\sum_{cyc}ab^{2}+2\sum_{cyc}ab \le 9.\]


logarithms
function
integration
limit
calculus
calculus computations

For B, consider the function $f(x) = \sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}}$ and note that $f(1) = \sum_{n=1}^{\infty}\frac{1}{n^{2}}= \frac{\pi^{2}}{6}$ and f(0) = 0. We also have, $f(x) =-\int \frac{\ln(1-x)}{x}\,dx$. We are thus looking for $f(\frac{1}{2})$
Now differentiate f(x): $f'(x) = \sum_{n=1}^{\infty}\frac{x^{n-1}}{n}=-\frac{\ln(1-x)}{x}$.
Remember now that $\sum_{n=1}^{\infty}x^{n-1}= \frac{1}{1-x}$. Substituting x by 1-x and then integrating we have:

$\sum_{n=1}^{\infty}(1-x)^{n-1}= \frac{1}{x}\Rightarrow  \sum_{n=1}^{\infty}-\frac{(1-x)^{n}}{n}= \ln x \Rightarrow  \sum_{n=1}^{\infty}-\frac{(1-x)^{n-1}}{n}= \frac{\ln x}{1-x}\Rightarrow  \sum_{n=1}^{\infty}\frac{(1-x)^{n}}{n^{2}}= \int \frac{\ln x}{1-x}\,dx$.

So we have:

$f(1-x) = \sum_{n=1}^{\infty}\frac{(1-x)^{n}}{n^{2}}= \int \frac{\ln x}{1-x}\,dx$.

So: $f'(x)+f'(1-x) =-\frac{\ln(1-x)}{x}+\frac{\ln x}{1-x}=-[\ln x\ln(1-x)]'$.
Next, let's integrate this expression:

$\int^{x}_{1}[f'(x)+f'(1-x)]\,dx =-\int^{x}_{1}[\ln x\ln(1-x)]'\,dx$
$\Rightarrow  f(x)-\frac{\pi^{2}}{6}+f(1-x)-0 =-\ln x\ln(1-x)+\lim_{x\to1}\ln x\ln(1-x)$

This limit is zero, and so

$f(x)+f(1-x) = \frac{\pi^{2}}{6}-\ln x\ln(1-x)$.

Finally, make $x = \frac{1}{2}$ and we get:

$f(\frac{1}{2})+f(\frac{1}{2}) = \frac{\pi^{2}}{6}-\ln(\frac{1}{2})\ln(\frac{1}{2}) \Rightarrow  f(\frac{1}{2}) = \frac{1}{2}\left(\frac{\pi^{2}}{6}-(\ln 2)^{2}\right)$


Let $G$ be the centroid of triangle $ABC$. Through $G$ draw a line parallel to $BC$ and intersecting the sides $AB$ and $AC$ at $P$ and $Q$ respectively. Let $BQ$ intersect $GC$ at $E$ and $CP$ intersect $GB$ at $F$. If $D$ is midpoint of $BC$, prove that triangles $ABC$ and $DEF$ are similar

everything stays in proportion, so three pairs corresponding sides are all parallel, so similarity follows. I have a trivial algebraic solution... :D

First Question of 1991 Aian Pacific Mathematical Oympiad
here is a link to a solution:
[url]http://www.kalva.demon.co.uk/apmo/apmo91.html[/url]


let x+y+z=1 and x,y,z nonnegative
Prove that
$ (1\minus{}x^2)^2\plus{}(1\minus{}y^2)^2\plus{}(1\minus{}z^2)^2\le(1\plus{}x)(1\plus{}y)(1\plus{}z)$
When do we have equality?

[b]My solution;[/b]
The inequality is equivalent to
$ \begin{array}{l}
 1 \plus{} x^4  \plus{} y^4  \plus{} z^4  \le 2(x^2  \plus{} y^2  \plus{} z^2 ) \plus{} xyz \plus{} (xy \plus{} yz \plus{} zx) \\
  \\
 \end{array}$
or \[ \begin{array}{l}
 2xyz \plus{} x^4  \plus{} y^4  \plus{} z^4  \le x^3  \plus{} y^3  \plus{} z^3  \\
  \\
 \end{array}\]
\[ \Leftrightarrow 2xyz \le x^3 (y \plus{} z) \plus{} y^3 (x \plus{} y) \plus{} z^3 (x \plus{} y)\]
Which is true from the AM-GM inequality


inequalities
geometry
incenter
trigonometry
function
inequalities proposed

Let $ABC$ be an acute triangle $ABC$ with the incenter $I\Longrightarrow$

${{\blacksquare \ 1^{\circ}}.\ \overline {\underline {\left| \ \sum \frac{1}{\sin \frac A2}\ge 6\ \sqrt{\frac{R}{2r}}\ \right| }}\ge 6.\ \ \ \blacksquare \ 2^{\circ}}.\ \underline {\overline {\left| \ \sum IA\ge 3\sqrt{2Rr}\ \right| }}\ge 6r.$

[u]Remark.[/u] The above inequalities are equivalently.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=63801

Is easy to see that te first implies the second:

$\sum{IA} = \sum{\frac{r}{sin(A/2)}}$.

to prove the first:

we know that $r/R = cosA+cosB+cosC - 1 = 2(1 - \sum{sin^2(A/2)})$.

so, simplifying, we have to prove that:

$(1-sin^2(A/2)-sin^2(B/2)-sin^2(C/2))(\frac{1}{sin(A/2)}+\frac{1}{sin(B/2)}+\frac{1}{sin(C/2)})^2\geq9$.

the function $f(x) = sin^2x$ is concave in $[0,\pi/2]$ and the function $g(x) = 1/sinx$ is convex in $[0,\pi/2]$.

so, use jensen and the problem is solved.  :)

$f(x)=\sin ^2x\Longrightarrow f'(x)=2\sin x\cos x=\sin 2x\Longrightarrow f''(x)=2\cos 2x.$
Why the function $f$ is concave in $\left[ 0,\frac{\pi}{2}\right]$ ???
 :spider:  :gathering:  :spider:

Nice emoticons Virgil :D
I see in Brazil for convex function not must be $f\"(x)\geq 0$ :D


Prove $n$ is congruent to 1 modulo 4 if and only if $n$ is congruent to 1 modulo 8 or congruent to 5 modulo 8.

[hide]Consider $mod 8$. We have that $n$ is written as $8k + m$.
$8k + m \equiv m$ $mod 4$ but on the other hand we have $8k + m \equiv 1$.
So it is $m \equiv 1$.Hence $m = 1$ or $m = 5$[/hide]


Suppose $n\in\mathbb{N}$. Prove that $\sqrt[n]{n}<1+\sqrt{\frac{2}{n}}$

We are to prove in using binominal theorem.

How??Are you sure it can be proved with BT? :?:


Consider the set of all five-digit numbers whose decimal representation is a permutation of the digits $1, 2, 3, 4, 5$. Prove that this set can be divided into two groups, in such a way that the sum of the squares of the numbers in each group is the same.

Denote $ \sigma$ the bijection on $ \{1,...,5\}$ given by $ \sigma(i) \equal{} 6 \minus{} i$, denote $ S \equal{} \{n\in\mathbb{N}|n > 33333\text{ and }n\text{ is a digit permutation of }12345\}$. Denote the numbers in $ S$ as $ [a,b,c,d,e]$ and consider the differences
\[ {T_{abcde} \equal{} [a,b,c,d,e]^2 \minus{} [a_{\sigma},b_{\sigma},c_{\sigma},d_{\sigma},e_{\sigma}]^2 \equal{} 66666\cdot [(a \minus{} a_\sigma),(b \minus{} b_\sigma),(c \minus{} c_\sigma),(d \minus{} d_\sigma),(e \minus{} e_\sigma)].}
\]
Then there is to prove that $ S \equal{} S_1\overset\cdot{\smash\cup} S_2$ with $ \sum_{[a,b,c,d,e]\in S_1}T_{[a,b,c,d,e]} \equal{} \sum_{[a,b,c,d,e]\in S_2}T_{[a,b,c,d,e]}$.

Now note that $ x \minus{} x_\sigma\in\{ \minus{} 4, \minus{} 2,0,2,4\}$, and consider the differences $ T_{[p,q,r,s,t]} \minus{} T_{[p_\tau,q_\tau,r_\tau,s_\tau,t_\tau]}$, where $ \tau \equal{} (12)(3)(4)(5)$, with $ 2$ appearing left of $ 1$ in $ [p,q,r,s,t]\in S$ (note $ p_\tau \equal{} p$) -- and note that each element of $ S$ is uniquely represented in this way.
The differences are one of $ 9, 90, 99, 900, 990, 999$, appearing respectively $ 6,6,6,4,4,4$ times.

Since $ 999 \minus{} 990 \plus{} 9 \equal{} 0$, $ 999 \minus{} 900 \plus{} 99 \equal{} 0$, $ 990 \minus{} 900 \plus{} 90 \equal{} 0$ and $ 99 \minus{} 90 \minus{} 9 \equal{} 0$, we can indeed split $ S$ in two disjoint sets with equal $ T$-sums, and we're done.


[url=http://imageshack.us][img]http://img218.imageshack.us/img218/4427/problemva2.jpg[/img][/url]
Shot at 2007-08-04

Try this problem without induction...can you manipulate it algebraically?

Yeah, I usually prove it directly:

$ (n+1)\cdot (n+2)\cdots (2n)=\frac{(2n)!}{n!}=\frac{[(2n)\cdot (2n-2)\cdots 2][(2n-1)\cdot (2n-3)\cdots 3\cdot 1]}{n\cdot (n-1)\cdots 2\cdot 1}$

$ =2^{n}\cdot [(2n-1)\cdots 3\cdot 1]$.

The forum is equipped with LaTeX.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690
.  Also, please try to give your posts names that carry some indication of their content.  Thanks.

Thanks, Z. I wasn't trying to take a jab at Will Hunting. :)

and...

I will make sure to make it clearer in the future.


limit
function
integration
floor function
complex numbers
real analysis
real analysis unsolved

Hello,

Does anyone sees why is this true

$\{a_n\}_{n\in\mathbb Z}$ is a secuence of real or complex numbers such that $\lim_{|n|\rightarrow\infty}a_n=0$ and $\Delta^2a_n=a_n-2a_{n-1}+a_{n-2}\geq0$. Then

\[\lim_{n\rightarrow+\infty}n(a_n-a_{n+1}) = 0\]

I know that $(a_n-a_{n+1})$ is monotonically decreasing, but after that, I don't see the limit!

You can use the fact that if $\displaystyle \sum_{n=1}^{\infty}a_n$ converges and $\displaystyle lim_{n\to\infty}a_n=0$,$a_n$ monotone then $a_n=o(\frac{1}{n})$.
After that you can apply it to the series $\displaystyle \sum_{n=1}^{\infty}(a_n-a_{n+1})$

[quote="grobber"]The following problem has been discussed before, I believe:

Given a decreasing function $f:[0,\infty)\to[0,\infty)$ s.t. $\lim_{x\to\infty}\int^x_0 f(t)dt$ is finite, show that $xf(x)\to 0$ as $x\to\infty$.[/quote]
According to Cauchy's Convergence Principle, for $\epsilon > 0$, there exists $G>0$ such that $\int_\frac{x}{2}^x f(t)dt<\frac{\epsilon}{2}$ for $x>\frac{x}{2}>G$.

In fact, $f(x)$ is decreasing, hence
$0\leq \frac{1}{2}xf(x) = \int_\frac{x}{2}^x f(x)dt\leq \int_\frac{x}{2}^x f(t)dt<\frac{\epsilon}{2}$
which implies that $xf(x)<\epsilon$ when $x>2G$

Thanks to all of you, but you are all using the fact that $\sum a_n$ converges, but that is not assured by the fact that $\lim_{n\rightarrow\infty}a_n=0$.

[quote="zhaobin"]can I ask what do you mean by $\Delta^2a_n=a_n-2a_{n-1}+a_{n-2}\geq0 $? :)[/quote]$\Delta:\mathbb{C}^{\mathbb N}\rightarrow\mathbb{C}^{\mathbb N}$ is an operator defined on the space of real or complex sequences by $\Delta\{a_n\}=\{a_n-a_{n-1}\}$. Usually one consider the equation on the elements instead of the full sequence, and one writes $\Delta a_n = a_n - a_{n-1}$. By $\Delta^2a_n=\Delta(\Delta a_n)$, that happens to be exactly $a_n-2a_{n-1}+a_{n-2}$.

[quote="djimenez"]Thanks to all of you, but you are all using the fact that $\sum a_n$ converges[/quote]

That's not true, actually :). If you read eugene's post, you'll see that he applies the result to the sequence $(a_n-a_{n+1})_n$, and the series $\sum(a_n-a_{n+1})$ does converge. If you want to apply the problem I mentioned (which is pretty much equivalent to what eugene said anyway :)), take $f(x)=a_n-a_{n+1}$ whenever $\lfloor x\rfloor=n$.

:oops: You're right, I misread! I appologize. By the way, thanx y'all!


articles geometry AMC AIME trigonometry calculus integration

Talk.

Scores, places, and answer keys will be posted soon. I'm bit busy and have to eat but thanks to everyone who participated.

[hide="Answer keys"]

1. C
2. B
3. C
4. C
5. E
6. A
7. A
8. B
9. C
10. D
11. A
12. E
13. B
14. E
15. B
16. C
17. B
18. A
19. C
20. D (One problem in Math magazine. Got it from one of Berkeley's article and I think the person said this problem was discussed in Geometry Revisited(not sure?). Again, not easy problem as it looks.)
21. B (Last problem in one of HMMT's)
22. D (Old AHSME in 80's)
23. A (IMO)
24. B (IMTS/USAMTS Problem)
25. E (Old AIME Problem #15)
[/hide]

Here are my answers. It seemed like the later ones were quite a bit harder than usual; the last 10 were almost as hard as the last 5 of the AMC12B06 in my opinion.  Hopefully I got more than 100..............

C
B
C
C
E

B
A
B
C
D

A
E
D
E
B

B
_
A
C
C

_
_
_
_
_

o crap. I messed up 11.

Didn't do harmonic mean properly, and got E ( i put $(p+w)(p-w)=p^2+w^2$, woops).

These were my answers
1. C
2. B
3 C
4 C
5 E
6 B
7A
8 B
9 E
10 D
11 A
12 E
13 D
14 E
15 B
16
17 B
18
19 C
20
21
22
23 A
24
25

I thought they were a lot harder than the AMC 2006B

I only worked for 45 minutes, but I looked briefly at the last five, and they were considerably harder than normal. #24 for example, reminded me of an IMO problem from the '60s, though it wasn't as hard. Well, here are my answers:

1. C
2. B
3. C
4. C
5. E
6. B
7. A
8. B
9. C
10. D
11. A
12. E
13. D
14. E
15. Didn't really understand wording of problem...
16. -
17. B
18. A
19. C
20-25. -

I disagree with Xantos/Penguin about #13 and I disagree with barce about #9. Everything else is in agreement with everyone else, I think.

#13 was a graph points, and punch trig on calc problem for me.
i got abc.debf=324.0625 making a+b+c+d+e+f=20.
Although I might have been wrong.

My answers were:
1. C
2. B
3. C
4. C
5. E
6. B
7. A
8. B
9. E
10. D
12. E
13.  D
14. E
15. B
16. C (can someone explain this one? I didn't get an integral answer, so I rounded up.)
17. B
18. A
19. C

20-25 ??

Answer keys are posted. I'll try to grade them soon but if it's not done anytime soon, you can just look at the key and do it by yourself.

I'll give solution to #18 because it's pretty nice problem. It's actually derivation of 2004 ARML Individual Problem.

[hide="Solution for 18"]

Remember that $\frac{x^m}{x^n} = x^{m-n}$. So:

$\displaystyle \sum^{2006}_{j=1} (i^{2004-a_j}) = \sum^{2006}_{j=1} \frac{i^{2004}}{i^{a_j}} = \sum^{2006}_{j=1} \frac{1}{i^{a_j}}$

The value of $i^n$ changes in every four terms. So, writing Fibonacci sequence as base 4, we get:

$1,1,2,3,1,0 | 1,1,2,3,1,0 | ...$

So, every 6 terms, there is a cyle. The value is 334 cycles with 2 1's. This is:

$\left(\frac{1}{\sqrt{-1}} + \frac{1}{\sqrt{-1}} + \frac{1}{-1} + \frac{1}{-\sqrt{-1}} + \frac{1}{1} + \frac{1}{\sqrt{-1}} + \frac{1}{1}\right)334 + \left(\frac{1}{\sqrt{-1}} + \frac{1}{\sqrt{-1}}\right) = (-2i)334 + (-2i) = -670i$

Then, $|\mathbb{X}| = |-670| = 670$ so answer is $6 \cdot 7 \cdot 0 = 0$. [/hide]

[quote="solafidefarms"]Here are my answers. It seemed like the later ones were quite a bit harder than usual; the last 10 were almost as hard as the last 5 of the AMC12B06 in my opinion.  Hopefully I got more than 100..............[/quote]

This was my intention. First 15 were what I thought "You need to get this." The main reason was that 14 right and 11 blank is the minimum score you need to get to qualify. I'm guessing that in next year's AMC, the writers will try to put some harder problems after 15 to see who are really "smart."

I had hard time finding last 5 problems. The last response I had was that the test had too easy last problems (and it did) so I wanted to find good hard problems. Luckily, it worked I guess.

can someone explain 16-19? yeah...   :oops:

I just posted 18. I'll do #19.

Another very interesting ARML problem (I'm not sure where it's from.. It's from packet I had)

[hide="Solution for 19"]

For $N$ to be maximized, we need to have all piles in similar numbers (not same!). Since $\frac{100}{7} > 14,$ we can set pile as:

$11,12,13,14,15,16,17$

$15+16+17 = 48$ so make $16 \rightarrow 18,$ then $15+17+18 = 50.$ So, $N \geq 3.$ Now, we'll prove that $N < 4$.

Assume $N = 4.$ Then your top 3 piles have sum of 49 or less. Also, your last 4 piles must have sum of 51 or more. Let $x$ be the smallest number in top pile and $y$ be the largest one in bottom pile. We know $x > y$.

Since $16+17+18 = 51 > 49, x \leq 15$. Since $11+12+13+14 = 50 < 51, y \geq 15.$ So, $15 \leq y < x \leq 15$ but there is no such value. Contradiction. So, $\fbox{N = 3}$. [/hide]

wow, i completely forgot about submitting these answers! oh well, my answers anyway

[hide="mine"]
1. C
2. B
3. C
4. C
5. E
6. B
7. A
8. B
9. C
10. D
11. A
12. E
13. B
14. E
15. B
16. - (um, i didn't end up with an integer)
17. A
18. A
19. C
20. -
21. B and D (don't they both work? b and c can take on different values)
22. C
23. A (DEFAULT CHOICE!!!)
24. B
25. -
[/hide]

[color=red][size=59]edit: ouch.[/size][/color]

SilverFalcon- I think that your answerkey is wrong for number 6.  Everyone else has B but is says A on yours.  Thanks for the help.

All of us missed #6.
Please post a solution.

Yay! I'm the only one who submitted it on time and got #13 right!

Unofficial score:130.5

I don't know if #6 is wrong. Is it?

I don't remember very well. But it was just using sequence formula of $S = \frac{a-ar^n}{1-r}$ and putting in the numbers and canceling the stuffs.

I know how to solve these but I'm looking for elegant solutions other than ones I know for:

#20
#24
#25

Anyone mind to share?

I'm having trouble with the wording of number 15. When it says point must be in the fourth quadrant, does that mean the randomly selected point can be anywhere in the fourth quadrant? So, to get the probability of choosing something in an infinitely large space, you get the area of that space? I get how $\frac{\pi-2}{4}$ is derived [area of space], but that counts as the probability?

are you sure about number 13?

if the key's right, i got 17 right, 2 wrong, 6 blank

99

Uh.... I don't think I'm able to grade the solutions. I need to get my English essays done..  :?

So, just grade by yourself. I apologize for this. But I can answer any question you have regarding problems and its solutions so if there are any, ask here.

[hide="Solution for 17"]

Note that rearranging, we get $y^2 - x = \frac{6-x}{x}$. Then solving for $y$ and taking square roots:

$y = \sqrt{\frac{6-x}{x} + x}$

The left fraction can further be simplified by separating.

$y = \sqrt{\frac{6}{x} - 1 + x}$

Now, since $x$ is a positive integer, it must be a divisor of 6. Trying 1,2,3,6, we see that only 2,3 works. Now, both coordinates give $y = 2$ so only $(3,2)$ satisfy the given condition and the desired probability is $\frac{1}{2}$.[/hide]

Nevermind, my score was a 117.  I didn't know how the scores work, but yeah, stupid mistake on the cos a one, i added all the digits, instead of noticing that it was a+b+c+d+e+f

[hide="24 without oversimplifying"]
The area of quadrilateral $ABCD$ can be found by splitting it along either diagonal into two triangles. Using $[ABC]=\frac 12 ab\sin C$, the area ratios by shrinking these triangles can be found. When $\triangle ABC$ shrinks into $\triangle PBQ$, $\triangle BCD$ into $\triangle QCR$, $\triangle CDA$ into $\triangle RDS$, and $\triangle DAB$ into $\triangle SAP$, one side is multiplied by $\frac{1}{k+1}$, another is multiplied by $\frac{k}{k+1}$, and the angle between the two is unchanged. So the area of each triangle gets multiplied by $\frac{k}{(k+1)^2}$.

Adding this area for all four triangles, the area of the four triangles surrounding $PQRS$ is $\frac{2k}{(k+1)^2}$ times the area of $ABCD$. But $[PQRS]$ is $\frac{13}{25}$ of $[ABCD]$ so $\frac{2k}{(k+1)^2}=\frac{12}{25}$.

Cross-multiply to get $25k=6k^2+12k+6\Rightarrow 0=6k^2-13k+6=(3k-2)(2k-3)$. So $k$ is either 2/3 or 3/2 $\Rightarrow \boxed{B}$.
[/hide]

[quote="krustyteklown"]hmm, just looked at #23; why isn't it infinitely many?

isn't that just schur's inequality...[/quote]

It's supposed to be a less than sign on #23. I'll post my real solution (obviously after taking it, since it was easy to see that one solution would generate infinitely many, hence the (DEFAULT CHOICE!!!)). And I'm not familiar with Schur, so someone else who isn't might still be able to read this solution.

[hide]From the triangle inequality, $a+b-c>0$ and the other cases are similar. So one way to attack this is to expand out $(a+b-c)(b+c-a)(c+a-b)$ and see what happens. Since this is symmetric, the only terms that need to be looked at are $a^3$, $a^2b$, and $abc$.

The $a^3$ coefficient is obviously -1. For the $a^2b$ coefficient, there are 3 ways to choose the position of $b$. One results in all positives, another has 1 negative, and the third has a product of two negatives, for a positive. The total coefficient of $a^2b$ is 1. Finally, the $abc$ coefficient can be found by looking at the sum of the coefficients, which is 1. Since the sum of the coefficients of the terms already accounted for is $3(-1)+6(1)=3$, the $abc$ coefficient is -2.

Amazingly, this accounts for almost all of the left side of the original inequality! Multiplying the original inequality by -1 (reversing the sign) and adding $abc$ gives $(a+b-c)(b+c-a)(c+a-b)>abc$.

Let $x=a+b-c$, $y=b+c-a$, $z=c+a-b$ so $x$, $y$, and $z$ are positive real numbers. Also $a=\frac{x+z}{2}$, $b=\frac{x+y}{2}$, and $c=\frac{y+z}{2}$. So $xyz>\frac{x+y}{2}\cdot \frac{x+z}{2}\cdot \frac{y+z}{2}$.

But by AM-GM, $\sqrt{xy}\leq \frac{x+y}{2}$, $\sqrt{xz}\leq \frac{x+z}{2}$, $\sqrt{yz}\leq \frac{y+z}{2}$. Multiplying these inequalities gives $xyz\leq \frac{x+y}{2}\cdot \frac{x+z}{2}\cdot \frac{y+z}{2}$. So the original inequality has no solutions $\Rightarrow \boxed{A}$.
[/hide]

[quote="vvr1590"]how do you do number 9?[/quote]

$\triangle CDE$ formed by the angle $\angle BEC$ creates a right triangle. The triangle is a $45-45-90$ right triangle [since a diagonal of a square will cut the angle in half]. $\triangle EAB$ is an isosceles triangle, with $\angle BAE$ being $150 degrees$ [60+90]. This means that $\angle AEB$ has a measure of $15 degrees$. $90 degrees - \angle AEB - \angle CED= 90-15-45=\boxed{30}$

[hide="Problem 20"]
This is an old and well-known chestnut problem, and it does appear in Geometry Revisited, among other places.  Here's an interesting article about it:
[url=http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf]mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf[/url]

One of my favourite proofs appears in one of Honsberger's books, and uses the 18-gon centred at $A$, one of whose sides is $BC$.
[/hide]

[hide="Problem 24"]
If $AP: PB = k$, then
\[ \frac{AP}{AB} = \frac{k}{k + 1} \quad \text{and} \quad \frac{PB}{AB} = \frac{1}{k + 1}. \]
Similarly,
\[ \frac{DS}{DA} = \frac{k}{k + 1} \quad \text{and} \quad \frac{SA}{DA} = \frac{1}{k + 1}. \]
Therefore,
\[ \frac{[SAP]}{[DAB]} = \frac{SA}{DA} \cdot \frac{AP}{AB} = \frac{k}{(k + 1)^2} \ \Rightarrow \ [SAP] = \frac{k}{(k + 1)^2} [DAB]. \]

Similarly,
\begin{eqnarray*} [PBQ] &=& \frac{k}{(k + 1)^2} [ABC], \\ \: [QCR] &=& \frac{k}{(k + 1)^2} [BCD], \\ \: [RDS] &=& \frac{k}{(k + 1)^2} [CDA] \\ \Rightarrow \ [SAP] + [PBQ] + [QCR] + [RDS] &=& \frac{k}{(k + 1)^2} ([DAB] + [ABC] + [BCD] + [CDA]) \\ &=& \frac{2k}{(k + 1)^2} [ABCD] \\ \Rightarrow \ \frac{2k}{(k + 1)^2} &=& 1 - 0.52 = 0.48. \end{eqnarray*}

Solving, we get the two solutions $k = 3/2$ and $k = 3/2$.  (B)
[/hide]

Yes, that's the same source I got for that problem. I solved with the one method about extra construction with $k$ because I like pure geometry best.  :D

[quote=pi_kachu]how do you do 16?

16. In the right triangleY U O,∠Y U Ois trisected and the angle trisec-tors intersectY OatT1andT2. If the order of points inY OisY, T1, T2, O,and it is given thatY U= 54 andOU= 47, find the remainder when(Y U+T2U)(T1U+OU) is divided by 35.[/quote]

As it's a right-angled triangle, the right angle is trisected into $30^{\circ}$ angles. It is therefore easy to solve using trig or Cartesian coordinates.

Can someone provide a link to this mock? Thank you! :P

When and where did this mock contest take place?


I don't understand.  $z$ is arbitrary, so does the third equation even matter?

Must is

[quote="Beat"]Given real numbers x, y such that
$a+b = 6$,
$ax+by = 10$,
$ax^{2}+by^{2}= 24$,
$ax^{3}+by^{3}= 62$,
determine $ax^{4}+by^{4}$.[/quote]

Then denote $u_{n}=ax^{n}+by^{n}$, we have $u_{n+2}=cu_{n+1}+du_{n}(n=0,1,2,...)$. Now, find $c,d$ :D


Nazaretoon raje be emtahan rooz aval computer chi bood ?
Sakh bood ya asoon ? Chand ta hal kardin ? pashid biyayd begid  :)

bacheha salam!

guya in ahamzaman soalaye entekhabe teame imo 09 emun bude! ke fght yeki azin emtehan 100 shode[azarmsa]
yeki 75[pooya vahidi]
baghie am beyne 50 ta 60!

[man nemidunam vaghan intore ya na! manam az yeki shenidam]

ba in owsaf entezar mire emtehane sakhti bashe!
behar hal age bache ha bian began chia neveshtan malum mishe
___________________________________________________________

http://shaazzz.blogfa.com

[color=green][b]1-Morabba'e makhfi[/b]
[size=150]ahmad yek zir jadval 3*3 az jadvale 6*6e moghabel entekhab karde ast.
ali ghasd darad ba tedadi por3sh mahale in m0raba ra kashf k0nad.o0 dar har n0bat mitavanad be 1i az khanehaye jadval eshare k0nad va az ahmad bep0rsad ke aya in khane dar m0raba m0rede nazare o0 gharar darad ya kheir.
kamtarin tedade p0r3sh haye lazem baraye an ke ali betavanad makane m0rabae ahmad ra kashf k0nad chand tas?
baraye esbate kh0d bayad 1razesh era'e dahid
va niz neshan dahid ba kamtar az in tedad s0al nemitavan hamishe be javab re30d[/size]

[b]2-Ghurbagheye pahlevan![/b]
[size=150]1 gho0rbagheye pahlevan ro0ye mehvare adade sahih gharar gerefte ast.o0 dar har jahesh mitavanad be andazeye tavani az 2be chap ya rast beparad.be 0nvane mesal o0 mitavanad ba 2jahesh az 11 be 18 beresad:ebteda ba 1jahesh az 11 be 19 va dar jaheshe badi az 19 be 18 miparad.neshan dahid 2adade sahihe a,b v0jo0d darand ke o0 nemitavanad ba kamtar az 100jahesh az a be b beresad[/size]

[b]3-4 malakhe jahande[/b]
[size=150]4malakh dar 4ra'3 1 m0raba bar ro0ye safhe gharar gerefteand.har bar 1i az malakh ha ba1 jahesh az ro0ye malakhe dgari miparad va dar n0ghteye m0ghabel(gharine makane ghablie kh0d nesbat be makane malakhe dgar)f0ro0d miayad.aya m0mken ast bad az m0dati malakh ha ro0ye 4 ra'3 1 m0raba b0z0rgtar gharar begirand?(m0rabae b0z0rgtar m0mken ast davaran yafte bashad
[/size]

[b]4-Derakhte rangarang[/b]
[size=150]manzo0r az derakht tedadi dayere ast ke ba parekhat hayi be ham m0ta3l sh0deand t0ri ke parekhat ha hamdgar ra ghat nemik0nand va beine har 2dayere daghighan 1 ma30r az tarighe pare khat haye va3l v0jo0d darad.maryam va mina bazie rang amizie zir ra bar ro0ye 1derakhte delkhah anjam midahand:
ba sh0ro0 az maryam,har kas dar n0bate kh0d 1i az dayerehayi ke hano0z rang nash0de ra ba 1i az 6range ghermez,abi,sabz,zard,banafsh ya narenji rang mik0nad vali bayad rangi ke entekhab mik0nad ba range hich k0dam az dayerehayi ke ba 1 parekhat be in dayere m0ta3l hastand va ta k0no0n rang nash0deand 1i nabashad.mina mikhahad bazi ra be b0n bast bekeshanad yani vaziati ijad k0nad ke hano0z dayereye rang nash0dei v0jo0d dashte bashad vali harekate dgari m0mken nabashad.neshan dahid maryam mitavanad ba 1raveshe bazie kho0b,mina ra nakam b0gzarad!
[/color][/size]

ba chakavak movafegham . rooze aval sakht bood vali rooz dovom asoon bood .
ba in vaziat fek konam kaf bein 105 ta 115 bashe .  :!:
Baghiye chikar kardan ?   :?:
pashin biyayn begin .

http://shaazzz.blogfa.com/comments/Default.aspx?blogid=shaazzz&postid=46&timezone=12642&p=4

chizio ke mikhay injast!
nemidunam chera kaC nemiad mathlinks!

malume!!!! chun in site MMMMMAAAAAATTTTTHHHHH  linke informatic link ke nist!!! ine :wink:

are khob!
vali fek nemikonam site motabar tari ham bashe!
hala khube riazi ham tarkibiat dare!


for which $x$, $\sum_{i=1}^{n}|x-x_{i}|$ is at minimum?
[hide="solutions for n=2,3"]for n=2 x=a or b, for n=3, x=b[/hide]


linear algebra
matrix
linear algebra unsolved

Calculate: \[ \det A_{n\times n}\equal{}\det \begin{pmatrix} 1&2&3&\cdots&n\\ x&1&2&\cdots&n\minus{}1\\ x&x&1&\cdots&n\minus{}2\\ .&.&.&\cdots&.\\ x&x&x&\cdots&1\end{pmatrix}\]

Answer: $ \minus{} ( \minus{} x)^n \plus{} (1 \minus{} x)^n$
Proof:
Subtract the first column from the 2nd. Subtract the 2nd and 1st columns from the 3rd. Continue in that manner until you get:
$ \det A_{n\times n} \equal{} \det \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\
x & 1 \minus{} x & 1 & \cdots & 1 \\
x & 0 & 1 \minus{} x & \cdots & 1 \\
. & . & . & \cdots & . \\
x & 0 & 0 & \cdots & 1 \minus{} x\end{pmatrix}$
From the linearity of the determinant, it equals:
$ x\det \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\
1 & 1 \minus{} x & 1 & \cdots & 1 \\
1 & 0 & 1 \minus{} x & \cdots & 1 \\
. & . & . & \cdots & . \\
1 & 0 & 0 & \cdots & 1 \minus{} x\end{pmatrix} \plus{} \det \begin{pmatrix} 1 \minus{} x & 1 & 1 & \cdots & 1 \\
0 & 1 \minus{} x & 1 & \cdots & 1 \\
0 & 0 & 1 \minus{} x & \cdots & 1 \\
. & . & . & \cdots & . \\
0 & 0 & 0 & \cdots & 1 \minus{} x\end{pmatrix}$
The later determinant is equal to $ (1 \minus{} x)^n$ (the matrix is diagonal).
In the first one, exchange the first and last column (hence the $ ( \minus{} 1)^{n \plus{} 1}$ term before $ x^n$), and it is left to calculate
$ \det \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\
1 \minus{} x & 1 & 1 & \cdots & 1 \\
0 & 1 \minus{} x & 1 & \cdots & 1 \\
. & . & . & \cdots & . \\
0 & 0 & 0 & \cdots & 1\end{pmatrix}$
Call this determinant $ a_n$.
Calculating by the first column gives us the recurrence $ a_{n} \equal{} a_{n \minus{} 1} \minus{} (1 \minus{} x)a_{n \minus{} 1} \equal{} xa_{n \minus{} 1} \implies a_{n} \equal{} x^{n\minus{}1}$, and the conclusion follows.


inequalities
LaTeX
inequalities proposed

Let $(f_n)_{n \in {\mathbb N}^*}$ be the fibbonaci sequence defined by $f_1=f_2=1, f_{n+2}=f_{n+1}+f_n$ for all integers $n \geq 1$.
Show that the following inequality is true for all $n \in {\mathbb N}^*$\$\{1\}$:$\prod_{k=1}^{n} {f_k}^{f_k} \geq \left ( \frac{2f_n + f_{n-1}-1}{n} \right )^{2f_n + f_{n-1} -1}$.

Jeane,your inequality follows imediately from the well known:

  ${x_1}^{x_1}.{x_2}{x_2}. .... .{x_n}{x_n}\geq\frac({x_1+x_2+...+x_n}{n})^{x_1+x_2+...+x_n}$ ;)

yes cezare, i think u are right, but even if i put glasses i can not understand your latex writing  :o
 :P


modular arithmetic
number theory unsolved
number theory

i) Prove that there exist infinite $ n$ such that $ n \mid 2^n \plus{} 2$

I suppose that when it says to prove for infinite $ n$ and you can't figure it out with primes you have to find a complex representation of $ n$ to put there and see what happens but I can't figure out that representation.

ii) Prove that if $ p$ is prime, $ a$ is a primary element of $ \mathbb{Z}_p$ for every $ x,y \in \mathbb{Z}$ then
$ a^x \equiv a^y \mod p \Leftrightarrow x \equiv y \mod (p \minus{} 1)$. Now for this one I have no idea

2) it's typically called a [i]primitive[/i] element. $ a$ being a primitive element in $ \mathbb{Z}_p^{\times}$ means that $ a$ generates this multiplicative group, i.e., $ a^{p \minus{} 1}\equiv 1\pmod{p}$ and no smaller positive power of $ a$  is  $ 1\pmod{p}$. you need to show that $ a^{x \minus{} y}\equiv 1\pmod{p}\Rightarrow p \minus{} 1|x \minus{} y$. here's a more general statement.

If $ d$ is the order of $ a$ modulo $ n$ -- in other words, $ d$ is the smallest positive integer for which $ a^d\equiv 1\pmod{n}$ -- and if $ a^h\equiv 1 \pmod{n}$, then $ d|h$.

Proof: Assume that $ d\not| h$. then we can write $ h \equal{} dq \plus{} r$, where $ 0 < r < d$. then $ 1\equiv a^h \equal{} a^{dq \plus{} r} \equal{} (a^d)^qa^r\equiv 1^qa^r\equiv a^r\pmod{n}$, i.e., $ a^r\equiv 1\pmod{n}$. but this contradicts the minimality of $ d$. thus our assumption $ d\not| h$ is false, so we must have $ d|h$.

Does it goes backwards also ?
Meaning $ x\equiv y\mod (p\minus{}1) \Rightarrow a^{x}\equiv a^{y}\mod p$

[quote="nickop"]Does it goes backwards also ?
Meaning $ x\equiv y\mod (p \minus{} 1) \Rightarrow a^{x}\equiv a^{y}\mod p$[/quote]

yes. if $ p\minus{}1|x\minus{}y$, write $ x\minus{}y\equal{}(p\minus{}1)k$. then $ a^{x\minus{}y}\equal{}(a^{p\minus{}1})^k\equiv 1^k\equal{}1 \pmod{p}$, and so $ a^x\equiv a^y\pmod{p}$


A really addicting game:

http://rebarb.homeip.net:8081/filedb/pafiledb.php?action=file&game=12
 my whole computers class  is addicted to it and it isnt blocked by our school's filters....yet

[size=75]Warning: If you read the comments there's a lot of swearing.[/size]

my best is 416 feet...i've only tried it 3 times so far.
EDIT:i just tried it again and i got 988 feet!!!!! :D  :D


superior algebra
superior algebra unsolved

Let $ G < S_n$ such that for all $ 1\le i,j\le n$ there is $ \sigma \in G$ that $ \sigma(i) \equal{} j$.
Prove that $ n|\ |G|$

Apply the Orbit-Stabilizer theorem to the set $ X\equal{}\{{1,2,...,n}\}$. Let $ x\in{G}$ Since $ G$ acts transitively on $ X$ we have that
$ |G|\equal{}|X\parallel{}G_x|\equal{}n|G_x|$

That kind of stuff is easily found with Google. For example, see [url=http://planetmath.org/encyclopedia/OrbitStabilizerTheorem.html]here[/url].


algebra
polynomial
least common multiple
number theory proposed
number theory

Give $ k $ is an integer number.
Does it exist infinitely number $ n $ satisfying $ n $ is divided by $ i , \forall i \leq \sqrt[k]{n} $

No! In general, there doesnt exist a non-zero polynomial with integer coefficients such that $ f(n) $ is a multiple of $ j $ for $j=1,[\sqrt[k]{n}] $ for infinitely many n. I will use here the fact that the least common multiple of $ 1,2,...,n $ is at least $ 2^{n-1} $. This problem was given in 1999 in Chech-Slovak Match and its not that difficult. Try to solve it, its very nice. From here it follows that $ |f(n)|\geq 2^{[\sqrt[k]{n}]-1} $ for infinitely many n, false.

Nice solution,you are really quick,Harazi :) .
Here is a different one:
We will prove that if $ n $ satisfies the condition then $ n<p_{2k}^{2k} $ where $ p_{2k} $ is the$  2k^{th} $ prime number.
Assume $ n \geq p_{2k}^{2k} $ .Take$  i_m $ is the greatest power that $ p_{m}^{i_m} <\sqrt[k]{n}$ .Then we have $ p_{1}^{i_1} \cdot. \cdot p_{2k}^{i_{2k}} $ divides $ n $,or$ n \geq p_{1}^{i_1} \cdot... \cdot p_{2k}^{i_{2k}} . $
However [tex]  p_{1}^{i_1} \cdot... \cdot p_{2k}^{i_{2k}} \geq \frac{n^2}{p_1 \cdot ... \cdot p_{2k}} [/tex].
So $ n \leq p_1 \cdot ... \cdot p_{2k} < p_{2k}^{2k} $ (unreasonableness) :D

That's so cool. I am quick just because I only know to solve the easy problems. The hard ones I never manage to solve.  :(


This is very easy.You shouldn't have posted in the Olympiad Section .Be more carefull next time
x=8

[quote="Liblikaneiu"]Please help me to find x.[/quote]
[hide]
$\frac{23^{8}(23-1)}{22}=23^{x}$
$x=8$[/hide]


Prove that there are infinitely many integer solutions to

xx = y3 + z3.

Sorry ... positive integer solutions of course.


number theory
prime factorization
number theory unsolved

Show that the coefficients of a binomial expansion $(a + b)^n$ where $n$
is a positive integer, are  all $odd$, if and only if $n$ is of the form $2^k - 1$ for
some positive integer $k$.

We have to show that all $\binom{n}{k}$ are odd, and thus, the product of the first $k$ integers has the same number of 2's in its prime factorization as the product of the first $k$ integers counting down from $n$.

$\binom{n}{1}$ is odd tells us n is odd.
$\binom{n}{2}$ is odd tells us $n-1 \equiv 2\mod{4}$, so that it has the same number of 2's as 2.
Then n-2 is odd
$\binom{n}{4}$ is odd tells us $n-3 \equiv 4\mod{8}$, so that it has the same number of 2's as 4.
$n-7 \equiv 8\mod{16}$
.
.
.

If $(n+1)-2^{i}$ is divisible by $2^i$ for $2^i \le n+1$, as above, then $n+1$ is a power of 2, yielding the desired result.


Let $P$ be a point on the side $AB$ of a square $ABCD$, and $Q$ a point on the side $BC$.  Let $H$ be the foot of the perpendicular from the point $B$ to the line $PC$.  Suppose that $BP = BQ$.
Prove that $QH$ is perpendicular to $HD$.

From $PB = BC$ and $\Delta PBC \sim \Delta BHC$, we have $\frac{PB}{BC} = \frac{BQ}{BC} = \frac{BH}{HC}$, from which $\Delta HBQ \sim \Delta HCD$ and $\measuredangle BHQ = \measuredangle CHD$.
Then $\measuredangle BHQ + \measuredangle QHC = 90^{\circ} = \measuredangle DHQ$.


I'm confused.
Evaluate the (-1)^(2/2)
I have 2 solutions that are not equal.
First. it's [sqrt(-1)]^2 =-1
Second. it's sqrt((-1)^2)=1
How can I explain what's it happening?
 :(

For a student has less than $15$ years old then when write $a^{\frac{p}{q}}$ you mean that $a>0$.  :wink:


function
topology
real analysis
real analysis unsolved

$(f_{n})$ a bounded (in the supremum norm) sequence of continuous functions from a compact Hausdorff space to $\mathbb{R}$ or $\mathbb{C}$. The sequence converges pointwise to $f$ continuous on the same space. Show that for each $\varepsilon$, there exists a convex combination $g = \sum \lambda_{n}f_{n}$ such that $\parallel f-g\parallel _{\infty}<\varepsilon$. Convex means the $\lambda_{i}$ sum to 1.

[quote="alekk"]already posted I think. This is because the sequence converges weakly towards $f$, and that closed convex sets are also weakly closed.[/quote]
Precisely. I am sorry if this was posted before.