inequalities
calculus
derivative
logarithms
function
inequalities solved

I proved it too, with differentiation:

WLOG $y\geq x$

Let $F(y)=x^x+y^y-x^y-y^x$

Then $\frac {dF}{dy}=y^y(\ln y +1)-\frac {y^xx}{y}-x^y\ln x$

Now divide into cases where $y\geq 1$ and $y<1$.

In all cases we have $F'(y)\geq 0$ with equality iff $y=x$.  This leads to $F(y)\geq 0$, which is the desired result.

He made a mistake!!!!!
I think:
when
x >= y ,x>= 1
1>=x >= y >= (e^-1)
it is quiet easy (function:x^a-x^b)
when
1 >= x >= y ;
y <= (e^-1)
We have to try another way.

[quote="blahblahblah"]Myth, thank you for pointing out my all time mathematical low - getting a derivative wrong when calculating in Maple.

Anyways, it's clear now.[/quote]

 :lol: Hehe, I wouldn't claim that as my all time mathematical low. Do you beat this: 2+2=5? Here we got Bush arithmetic!!! :rotfl: :rotfl:  :D

[quote="zxingtan"]I have proved $n=2$
first we can assume $x>=y$
let $f(t)=t^x-t^y$      now we should prove $f(x)>=f(y)$
so $f'(t)=ln t(t^x-t^y)$
it is obvious that $f'(t)>=0$
so $f(x)>=f(y)$[/quote]

 :blush:

[quote="TonyCui"]It looks very easy.
But it is very hard.
It takes me two weeks.
I just give a special form:
(n=2)
x^x+y^y >= x^y+y^x(x,y>0)
(anlysis)

[i]Edited by Myth[/i][/quote]

can you show it?

I think you can Finsh it!
Gook Luck  :blush:

[quote="huhuhu"]I prove that:
(x/y)^(x-y)>=1,x>0,y>0
 so:x^x*y^y>=x^y*y^x.
lemma:a>=b>=c>=d>0,ad>=bc ,than:a+d>=b+c
so we have:x^x+y^y>=x^y+y^x[/quote]
I think you are creavity!

I have something wrong in my solution,but I have a new one.
let x>=y,it is easy when x >=1,so we have 1>x>=y and x^x*y^y>=x^y*y^x
lemma:if:(x^x)^(2^n)+(y^y)^(2^n)>=(x^y)^(2^n)+(y^x)^(2^n),(x^x)*(y^y)>=(x^y)*(y^x),then:(x^x)^(2^(n-1))+(y^y)^(2^(n-1))>=(x^y)^(2^(n-1))+(y^x)^(2^(n-1)).
the lemma is easy to prove.
when n is very very big,we have (x^x)^(2^n)+(y^y)^(2^n)=(x^y)^(2^n)+(y^x)^(2^n)-->0.
under the lemma,we have:x^x+y^y>=x^y+y^x
Is that OK?

My second solution is similer with yours.
I used to want it be a secret.HaHa!

I think your solution doesn't have any mistake.I will not chack it.
By the way.I will tell you.Only this problem cost me four days.
I only find two ways to prove it.
I think you are creavity.
You are very strong! :D

Clear it have been contributed.
It is not the last edition.(The last eaidion I have forgot it)
Say sorry to Friends who is not from China.
I will translate it soon.
It USE CCJ instead of CJK.

In this edition I only use one way to prove
But In the finally edition I use two way to prove it
I'm on my father's computer, so I can't find it.
Sorry!!!!!!!!


ratio
geometric series
geometric sequence

There is actually a pretty neat trick for this... You'll like it when you see it.

What is the arithmatic mean of all the vlaues of $2^x$ where $x$ is a whole number and $0\le x\le 11$?

Seems a bit more difficult than the first problem...doesn't it?

[hide]

Lets do this the long way and begin by adding up the numbers.

$2^0 \Rightarrow 1$ Sum: 1
$2^1 \Rightarrow 2$ Sum: 3
$2^2 \Rightarrow 4$ Sum: 7
$2^3 \Rightarrow 8$ Sum: 15
$2^4 \Rightarrow 16$ Sum: 31
...

Is it just me, or is there a pattern of sum sort? The sum is always one less then the next number in the series!
Let's proof this shall we?

How can we begin?
Why don't we use the formula for the sum of a geometric series!
$\frac{a-ar^n}{1-r}$
Where $a$ is the first term, $r$ is the rate and $n$ is the number of the final term.

Pluging in the definate values in any case gives us a first number of 1 and a rate of two yeilding:
$\frac{1-1 \cdot 2^n}{1-2} \Rightarrow \frac{1-2^n}{-1} \Rightarrow 2^n-1$

Please notice that zero is supposedly the first number...so we count from zero up...making one the second number two the third and so on.

Using that little trick we can find the sum of the first eleven powers of two as $2^12-1 \Rightarrow 4096-1 \Rightarrow 4095.

Then we divide by the number of terms which is from 0 to 11 inclusive giving us twelve...so the answer in this case is  341.25

That little trick comes in handy for speed and is definately better than having to add up all the values :lol:.

So remember the sum is the next power of 2 minus 1.
[/hide]

[quote="MCrawford"]What is the arithmetic mean of all the values of $2^x$ where $x$ is a whole number and $0 \le x \le 4$?  Express your answer as a decimal.[/quote]
[hide="solution"]
(1+2+4+8+16)/5
31/5
[b]6.2[/b]
[/hide]

Farazy, the entire formula is $\frac {p^{n+1} - 1}{p-1}$ for the sum $p^0 + p^1 + p^2 + ... + p^n$. Just some extra info, unless of course you allready knew it  :lol:.
The proof is basically the same thing except $p$ is substituted in for $2$.
[hide]Sum of finite geometric sequence: $a \frac {1-r^n}{1-r}$. If we want to find the sum of the geometric sequenece $p^0 + p^1 + p^2 + ... + p^n$, then, $p$ is the common ratio, $1$ is the first term, and the number of terms is $n+1$. Plugging in these values we get $\frac {1-p^{n+1}}{1-p} = \frac {p^{n+1}-1}{p-1}$.[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

let $ G$ be both Noetherian and artinian abelian group ,prove that $ G$ is finite.

modules which are artinian and noetherian have a (finite) composition series. thus it's enough to show that abelian simple groups are finite (and in fact, cyclic of prime order). but this is very clear.

What about non-abelian groups?
Does there exist an infinite simple group, in which every proper subgroup is contained in some finite maximal subgroup?


Q is the midpoint of PR. $ \overline{PR}$ is 12 cm longer than $ \overline{PQ}$. What is the number of centimeters in QR?

[asy]draw((-1.2,0)--(1.2,0),Arrows); unitsize(1inch);
pair P,Q,R; P = (-1,0); R = (1,0);
dot(P);  dot(Q); dot(R);
label("P",P,S); label("Q",Q,S); label("R",R,S);[/asy]

Since Q is the midpoint of PR, we find that PQ is 12 cm. This is the same as QR, so our answer is $ 12$.


The number of inches in both the length and the width of a rectangle are prime numbers. The area of the rectangle is 391 square inches. What is the number of inches in the perimeter of the rectangle?

[hide]I just know this one

$17*23$

[/hide]-jorian

[quote="ckck"]Is there any quick way to do this other than guess and check? :?:[/quote]
Okay, a couple comments.
1) There's no systematic formula or equation that gives a prime number so this is somewhat convoluted. Writing an equation isn't that bad.
2) [hide]17 is the 7th prime and 2, 3, 5, 7, 11, and 17 can each be checked under 10 seconds with certain tricks[/hide]

[quote="Ravi B"]Because 391 is 400 - 9, we can factor 391 using difference of squares.[/quote]
That is very clever... It gives 20-3, or 17, one of the factors. Never thought of that! Very Cool!  :lol:

[quote="Ravi B"]Glad you liked it.  Here's a related problem from the national countdown round about 3 years ago.  The two students took a while to solve it, but now you can solve it quickly.

What is the largest prime factor of 221?[/quote]
[hide]17*13
PS: When I saw this the first time in practice I checked all the primes up to 11 then I buzzed in and said 221 because I thought it was prime :D[/hide]


vector
superior algebra
superior algebra solved

given a subspace spanned by three vectors, how do i find a a vector in r3 that is orthogonal to it?

specifically, the subspace is spanned by { (1, -1, 3), (5, -4, -4), (7, -6, 2) }.

any help regarding the methodology of solving this problem would be greatly appreciated.

thanks.

s.

If $S$ is a subspace generated by $a_1,\ldots a_m$, then solve $\langle x,a_i\rangle=0\ \forall i=\overline{1,m}$. This would be the most obvious methdology.
In this case, let $S$ be the subspace generated by $(1,-1,3),(5,-4,-4),(7,-6,2)$. $(7,-6,2)=(5,-4,-4)+2(1,-1,3)$, so $S$ is the same as the subspace generated by $(1,-1,3),(5,-4,-4)$. Now solving $\left\{\begin{array}{cc}a-b+3c=0\\5a-4b-4c=0\end{array}\right.$ yields $a=16c,b=19c$. So for examaple $(16,19,1)$ is orthogonal to $S$.

We are to find the normal vector of the plane which is determined by $A(1,-1,3),B(5,-4,-4),C(7,-6,2)$.
Since $\vec{AB}=(4,-3,-7),\vec{AC}=(6,-5,-1)$, we get $\vec{AB}X\vec{AC}=(-32,-38,-2)\parallel (16,19,1)$


function
quadratics
algebra
polynomial
geometry
geometric transformation
functional equation

For $ x,y \in {R}$. We have
$ f(x)\plus{}f(y)\equal{}f(x\plus{}y)\minus{}xy\minus{}1$
$ f(1)\equal{}1$, find $ n \neq 1$ such that $ f(n)\equal{}n$

Plugging in $ x\equal{}1$ we get $ y\plus{}2\plus{}f(y)\equal{}f(y\plus{}1)$

Hence, $ f(0)\equal{}\minus{}1$, $ f(\minus{}1)\equal{}\minus{}2$, and $ f(\minus{}2)\equal{}\minus{}2$. Hence, we have $ n\equal{}\minus{}2$

[hide="A more complete solution"]
Let $ y\to 1$ to get $ f(x)\plus{}f(1)\equal{}f(x\plus{}1)\minus{}x\minus{}1\implies f(x\plus{}1)\equal{}f(x)\plus{}x\plus{}2$, or $ f(x)\equal{}f(x\minus{}1)\plus{}x\plus{}1$.
$ f(1)\equal{}f(0)\plus{}2\implies f(0)\plus{}2\equal{}1\implies f(0)\equal{}\minus{}1$.
$ f(2)\equal{}f(1)\plus{}2\plus{}1\equal{}4$
$ f(3)\equal{}f(2)\plus{}3\plus{}1\equal{}8$
$ f(4)\equal{}f(3)\plus{}4\plus{}1\equal{}13$
$ f(n)\equal{}f(n\minus{}1)\plus{}n\plus{}1\equal{}f(n\minus{}2)\plus{}2n\plus{}1\equal{}f(n\minus{}3)\plus{}3n\equal{}\cdots$

At this point, I noticed that the coefficient of $ n$ is the value being subtracted from $ n$ in the recursive definition, and the trailing constant follows a pattern like $ 1\plus{}1\minus{}1\plus{}1\minus{}2\plus{}1\minus{}3\plus{}\cdots$, or $ (1\minus{}0)\plus{}(1\minus{}1)\plus{}(1\minus{}2)\plus{}(1\minus{}3)\plus{}\cdots$.

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

Now, $ \sum_{k\equal{}0}^{n\minus{}1}(1\minus{}k)\equal{}\sum_{k\equal{}0}^{n\minus{}1}1\minus{}\sum_{k\equal{}0}^{n\minus{}1}k\equal{}n\minus{}\frac{n(n\minus{}1)}{2}\equal{}\frac{n(3\minus{}n)}{2}$.

So we have $ f(n)\equal{}f(0)\plus{}n^2\plus{}\frac{n(3\minus{}n)}{2}\equal{}\frac{n^2\plus{}3n\minus{}2}{2}$

It is easy to check that
$ f(x)\plus{}f(y)\equal{}\frac{x^2\plus{}3x\minus{}2\plus{}y^2\plus{}3y\minus{}2}{2}\equal{}\frac{x^2\plus{}y^2\plus{}3x\plus{}3y\minus{}4}{2}$,
$ f(x\plus{}y)\minus{}xy\minus{}1\equal{}\frac{x^2\plus{}2xy\plus{}y^2\plus{}3x\plus{}3y\minus{}2\minus{}2xy\minus{}2}{2}\equal{}\frac{x^2\plus{}y^2\plus{}3x\plus{}3y\minus{}4}{2}$
and so our condition is met.

So we need to solve the equation $ n^2\plus{}3n\minus{}2\equal{}2n\implies n^2\plus{}n\minus{}2\equal{}0\implies n\equal{}\frac{\minus{}1\pm3}{2}\equal{}1, \boxed{\minus{}2}$
[/hide]

You would have saved time if you had written the functional equation as $ f(x) \minus{} f(x\minus{}1) \equal{} \nabla_1 [f](x) \equal{} x\plus{}1$, and noted that a function over the integers with linear finite differences must be a quadratic. You can quickly find the particular solution by comparing coefficients.

For example, in this case, the functional equation is solved by any function of the form $ x^2/2 \plus{} C x \plus{} 1$

Including the initial condition, we find that $ C \equal{} \minus{}\frac{1}{2}$

As [b]TZF[/b] mentioned, $ f(x) \minus{} f(x \minus{} 1) \equal{} x \plus{} 1$ implies that f is a quadratic, a reasonably quick way of finding the function is as follows,

$ \sum_{x\equal{}2}^{y} \left(f(x) \minus{} f(x \minus{} 1)\right) \equal{} \sum_{x\equal{}2}^{y} \left(x \plus{} 1\right)$

$ f(y) \minus{} 1 \equal{} \frac{y(y\plus{}1)}{2} \minus{} 1 \plus{} (y \minus{} 1)$

Which holds for all y in $ \mathbb{R}$ by the identity theorem.

That's not at all the case, and I'm not sure what theorem you're using.  We aren't given that $ f$ is a polynomial, so the only thing we know about $ f$ is its values at a translation of the integers.

You're right. I should've said if we consider x an integer, then the function f on the integers must be a quadratic polynomial, in which case we know that the equation I wrote above holds for y in the integers.


search
inequalities
inequalities proposed

For $ x,y,z$ positive reals

$ \frac {2(yz \plus{} zx \plus{} xy)(x^2 \plus{} y^2 \plus{} z^2)}{x \plus{} y \plus{} z} \geq yz\sqrt {2y^2 \plus{} 2z^2} \plus{} zx\sqrt {2z^2 \plus{} 2x^2} \plus{} xy\sqrt {2x^2 \plus{} 2y^2}.$

Look at

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1614159#1614159

for wonderful Ji Chen's proof. :wink:

A related inequality is

$ yz\sqrt {2y^2 \plus{} 2z^2} \plus{} zx\sqrt {2z^2 \plus{} 2x^2} \plus{} xy\sqrt {2x^2 \plus{} 2y^2}$

$ \leq2(\sqrt{2}\minus{}1)(x^3 \plus{} y^3 \plus{} z^3) \plus{}(3\minus{}2\sqrt{2})[x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y)] \plus{} 6(\sqrt{2}\minus{}1)xyz$

holds for nonnegative real numbers $ x,y,z.$


On Thursday (Feb 8), we will be hosting a Math Jam to discuss the harder problems from the AMC 10A and 12A.  As we have done in the past, we will present full solutions to problems 21-25 on both contests, and then (time permitted) do other solutions upon request.

The Math Jam will be tomorrow (Thursday) at 7:00 PM Eastern / 4:00 PM Pacific.  [b]Note the early start time![/b]

To join the Math Jam, you must be logged in to your Art of Problem Solving account.  Then, up to 15 minutes before the Jam starts, simply click on the "Classroom" link on the left side of the screen.  Participation is free!

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for the complete Math Jam schedule.[/url]

I won't be able to join the Math Jam tomorrow due to ARML tryouts; when will a transcript of the Jam be available?

[quote="DPatrick"]All of our Math Jams are transcripted.  Highlight on the "Math Jams" tab above, then click the "Transcripts" option.[/quote]

I know; I'm just wondering [i]when[/i] the transcript will be online. Is it instantaneous, or does it take a while to be processed?

Thanks.  :)

Not instantaneous, but probably later that evening.  Friday morning at the latest.

I'm excited for this. This will be the first MathJam i am attending! :D

Are there going to be any game mathcounts math jams?

Also, just wondering, but the floor is constant for both AIMEs, right?

[quote="bpms"]Are there going to be any game mathcounts math jams?[/quote]We have not planned any, but that may change.

[quote]Also, just wondering, but the floor is constant for both AIMEs, right?[/quote]Not necessarily.  Often it turns out that one version of the AIME is much harder than the other.  In that case, adjustments can be made.

The transcript for the Math Jam is now available.  [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=161]Click here.[/url]

does anyone have the link to the list of colleges administering the amc 12?

[quote="buzzer11"]Will transcripts ever be deleted?[/quote]I wouldn't say "ever" (ever is a long time), but we haven't deleted any Math Jam transcripts so far, and don't expect to.

If you ever decide to delete some, are you going to inform us?

[quote="DPatrick"][url]http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml[/url][/quote]

thanks for the link, but I can't find any schools that are close to DC.


ratio
geometry
parallelogram
geometry solved

$ABC$ is a given triangle. $ABDE$ and $ACFG$ are the squares drawn outside of the triangle. The points $M$ and $N$ are the midpoints of $DG$ and $EF$ respectively. Find $MN: BC$

Nice little problem. A slight generalization:

[color=blue][b]Problem.[/b] On the sides CA and AB of a triangle ABC, parallelograms CAGF and ABDE are erected. Let M and N be the midpoints of the segments DG and EF. Show that $\overrightarrow{MN}=\frac{\overrightarrow{BC}}{2}$.[/color]

[i]Solution.[/i] Since CAGF and ABDE are parallelograms, $\overrightarrow{DE}=\overrightarrow{BA}$ and $\overrightarrow{GF}=\overrightarrow{AC}$. Since M is the midpoint of the segment DG, we have $\overrightarrow{AM}=\frac{\overrightarrow{AD}+\overrightarrow{AG}}{2}$; similarly, $\overrightarrow{AN}=\frac{\overrightarrow{AE}+\overrightarrow{AF}}{2}$. Thus,

$\overrightarrow{MN}=\overrightarrow{AN}-\overrightarrow{AM}=\frac{\overrightarrow{AE}+\overrightarrow{AF}}{2}-\frac{\overrightarrow{AD}+\overrightarrow{AG}}{2}=\frac{\left(\overrightarrow{AE}-\overrightarrow{AD}\right)+\left(\overrightarrow{AF}-\overrightarrow{AG}\right)}{2}$
$=\frac{\overrightarrow{DE}+\overrightarrow{GF}}{2}=\frac{\overrightarrow{BA}+\overrightarrow{AC}}{2}=\frac{\overrightarrow{BC}}{2}$,

what solves the problem.

  Darij


There's a mistake I think - instead of $x_n , y_n , z_n$ there should be $a_n , b_n , c_n$


Let $ ABC$ be an equilateral triangle, and $ P,Q,R$ three points in its interior satisfying
\[ \measuredangle PCA \equal{} \measuredangle CAR \equal{} 15^{\circ},\ \measuredangle RBC \equal{} \measuredangle BCQ \equal{} 20^{\circ},\  \measuredangle QAB \equal{} \measuredangle ABP \equal{} 25^{\circ}.\] Compute the angles of triangle $ PQR$.

This is a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=107969 . I have started the new topic just for the resources section.

  Darij from the Pentacracy


ratio
limit
geometry
rectangle
quadratics
algebra
quadratic formula

what exactly is the golden ratio and what is the purpose of it
and how is it derived ?

Let [tex]F_k[/tex] denote the kth Fibionacci (probably not spelled right) number.  The golden ratio phi is defined as

[tex]
\begin{eqnarray*}
\displaystyle\phi & = & \lim_{k \to \infty} \frac{F_k}{F_{k-1}}\\
 \displaystyle      & = & \lim_{k \to \infty} \frac{F_{k-2} + F_{k-1}}{F_{k-1}}\\
  \displaystyle     & = & 1 + \lim_{k \to \infty} \frac{F_{k-2}}{F_{k-1}}\\
   \displaystyle    & = & 1 + \frac{1}{\phi}
\end{eqnarray*}
[/tex]

Solving for [tex]\phi[/tex] gives [tex]\phi = \frac{1+\sqrt{5}}{2}[/tex]  Note that we want the positive solution since the limit is a ration of two positive numbers.  As for the use, it is used in a formula for the kth Fibionacci number, and the Greeks thought that the best looking rectangles had a side ratio 1:[tex]\phi[/tex]

All I know is the golden ratio is exactly one more then its recipricoal and it is:
(1+ :rt5: )/2
You  can get it by:
x=(1/x)+1
x :^2:-x-1=0
and then you do the quadratic formula and get:
x=(1 :pm:  :rt5: ) /2

The Golden Ratio is traditionally derived by answering the question:

What is the side of a rectangle whose other side is 1, such that if a square of side 1 is removed from the rectangle the remaining rectangle is similar to the original?


linear algebra
matrix
vector
linear algebra unsolved

a)Let $A$ be an $n*n$ symmetric real matrix with$a_{ii}=1$ for all $i$ and $|a_{ij}|\leq n^{\frac{-1}{2}}$ for all $j\neq i$.prove that $A$ has rank at least $\frac{n}{2}$.
b)Let $A$ be an $n*n$ real matrix of rank $d$,let $k$ be a positive integer,and let $B$ be an $n*n$ matrix with  $b_{ij}=a_{ij}^k$.prove that the rank of $B$ is at most ${k+d\choose k}$.
enjoy :)

Obs: deleted after saying a great non-sense!

After the great non-sense I said before, here is what I've done for a). Consider any complex matrix. I want to show that if all columns have length equal to 1 relatively to the norm imposed by the hermitian product $x\cdot y=C(y_1)x_1+...+C(y_n)x_n$ where $C(y_i)$ is the conjugate of $y_i$, then the rank of A is at least $\sum_{i}{|a_{ii}|^2}$.  It will obviously follow that for any matrix A we have $%Error. "rankA" is a bad command.
\geq\sum_{i}{\frac{|a_{ii}|^2}{{|a_{i1}^2|}+...+|a_{in}^2|}}$ and from here we will have $rankA\geq\frac{n}{\frac{n-1}{n}+1}$, from where $rankA\geq\frac{n^2}{2n-1}>n/2$.
    Now, the proof of the first claim. Let $c_i$ the column vectors of $A$ and suppose that they all have lenght $ l(c_i)=1$. Let $ (e_1,...,e_n)$ the canonical base in $C^n$ and let $r=rankA$. By Gram-Schmidt theorem we can find $r$ ortonormals vectors $v_1,...,v_r$ with the same vectorial space generated as the $c_i$. Then we have $c_i=\sum_{t=1,r}{(c_i\cdot v_t)v_t}$ and thus we have $|a_{ii}^2|=|c_i\cdot e_i|^2\leq\sum_{j=1,r}{|c_i\cdot v_j|^2}\cdot\sum_{j=1,r}{|v_j\cdot e_i|^2}$ (Cauchy-Schwarz inequality). Now, sum this over i and observe that the last sum is exactly r.

The second problem is really wierd, since again I have a better estimate, but it is surely wrong, as it turned out to be with the first part of the problem.
     Suppose thus that $rankA=d$ and regard its columns. We may assume the first d among them are a basis in the vectorial subspace generated by the columns. Then there exist $ x_{11},...,x_{1d},...,x_{n-d,1},...,x_{n-d,d}$ such that $c_{d+1}=x_{11}c_1+...+x_{1d}c_d,..., c_n=x_{n-d,1}c_1+...+x_{n-d,d}c_d$. Let $A=(a_{ij})$. If we manage to prove that the last $n-d$ columns of the matrix $a_{ij}^k$ generate a subspace of dimension at most ${k+d\choose k}-d$, then we are done, since then all columns will generate a subspace of dimension at most ${k+d\choose k}$.
     But using the multinomial formula we have that the columns $C_{i+d}$ of the matrix $a_{ij}^k$ verify $ C_i=^{t}( (x_{i1}a_{11}+...+x_{id}a_{1d})^k,...,(x_{i1}a_{n1}+...+x_{id}a_{nd})^k)$=$\sum_{j_1+...+j_d=k, j_s\geq 0}{\frac{k!}{j_1!...j_d!}\cdot x_{i1}^{j_1}...x_{id}^{j_d}C_{j_1,...,j_d}}$ where $C_{j_1,...,j_d}=^{t}(a_{11}^{j_1}...a_{1d}^{j_d},..., a_{n1}^{j_1}...a_{nd}^{j_d})$. Thus the vectorial space generated by $C_i$ is in the vectorial space generated by $C_{j_1,...,j_d}$, which has dimension at most the number of solutions of $j_1+...+j_d=k$, which is smaller than ${k+d\choose k}-d$ for k at least 2.
   Could anyone please show me the mistake?


Fifty ping-pong-balls are numbered 1, 2, 3, ..., 49, 50. A ball is
chosen at random. What is the probability that the number on
the selected ball is a multiple of 5 or 7 or both? Express your
answer as a common fraction.

There are $ \left\lfloor \frac{50}{5}\right\rfloor \equal{}10$ multiples of $ 5$ from $ 1$ to $ 50$ and $ \left\lfloor \frac{50}{7} \right\rfloor \equal{}7$ multiples of $ 7$, so there are $ 10\plus{}7\equal{}17$ numbers that are multiples of $ 5$ or $ 7$... not really.

We've counted the multiples of $ 5$ [b]and[/b] $ 7$ twice, but we only want to count them once.  A multiple of $ 5$ and $ 7$ is a multiple of $ 35$, and there is $ \left\lfloor \frac{50}{35}\right\rfloor \equal{}1$ multiple of $ 35$, so there are $ 17\minus{}1\equal{}16$ numbers that are multiples of $ 5$ or $ 7$, or both.

Finally, the probability is $ \frac{16}{50}\equal{}\boxed{\frac{8}{25}}$


I'm going to a summer class in UC Berkeley, and I've been assigned some homework pages. I'm taking a geometry class so i can skip to Algebra II in High-Shool. I'm 14 years old.

I'm having trouble on this problem. It's about angles, and I can't quite get it. I can only guess the angles. [u]I KNOW HOW TO DRAW IT OUT[/u], i just don't know how you can find the angles with given measures. Please help if you can.

Thank You

36. line OC bisects angle AOB, line OD bisects angle AOC, line OE bisects angle AOD, line OF bisects angle AOE, and line OG bisects angle FOC.

a. If (measure of) angle BOF = 120, then (measure of) angle DOE= ____ ?

b. If (measure of) angle COG = 35, then (measure of) angle EOG = ____ ?

Here is an image of the problem:
[url]Http://img382.imageshack.us/my.php?image=06152007101832pmrn3.jpg[/url]

Due to the difficulty, this problem belongs in the "Getting Started" forums. I suspect that most problems for an introductory geometry class would go in either the "Getting Started" or the "Intermediate Topics" forum.


Hello there,

I am interested in a effective solution for a "permutation" i'm not entirely sure what type of permutation it is, but i do believe it is s repetative one.

There is 3 different objects:

1)      x
2)      o
3)      v

And there are six different digits eg: ? ? ? ? ? ?
                                                    1 2 3 4 5 6

Example:

1)     x x x x x x
2)     x x x x x o
3)     x x x x o o
4)     x x x o o o
5)     x x o o o o
6)     x o o o o o
7)     o o o o o o
8)     o o o o o v
9)     o o o o v v
10)   o o o v v v
11)   o o v v v v
12)   o v v v v v
13)   v v v v v v
14)   v v v v v x
15)   v v v v x x
16)   v v v x x x
17)   v v x x x x
18)   v x x x x x

20)   x o x o x o
21)   o x o x o x
22)   v o v o v o
23)   o v o v o v

24)   x v o x v o
25)   o v x o v x
26)   v x o v x o
27)   x o v x o v

28)   o x v v x x
29)   o x v v v x
30)   o x v v v v

I figured that if the permutation is repeatable then the result should be 216 different permutations but does anybody have a quick formula to tell me how many there are just to be sure for a second opinion? and secondly possibly a print out of each single permutation result.

Thanks.

[quote="Geehoon"]you have three options for each digits.

$3^{6}=729$[/quote]

Thankyou, you have been a great help.


given 1000 points in a place contain 800 red points,200 blue points.Prove that:There exists a line partitionned a place into 2 half such that each haft contains 400 red points,100 blue points

Sorry for my poor presentation.

If no 3 pts are collinear, let L(x) be the line with inclination x with the horizontal line which partition red pts into harves and the minimum distance from the line to red pts in one side of the line is equal to the minimum distance from the line to red pts in another side. Let f(x) be the number of blue pts on the left side of L(x). Note that f(x) and f(x+pi) refer to different side, so f(x) = 200-f(x+pi). Since f(x) is "continuous" (| lim x + ->y f(x) - lim x - ->y f(x) | <= 1), therefore there exist y such that f(y) = 100.


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 10
MATHCOUNTS

who thinks that AIME is a lot easier than the AMC 12???
cuz i HATE the time limit on the AMC's...i'd rather have 15 HARDER problems on 3 hours than 10 easy problems, 10 okay problems, and 5 hard problems in 1 hour 15 minutes...with an average of 3 minutes per problem. Most of the "hard" problems require some sort of pre-learned trick. On the AIME, we have time to actually think things out and find our own ways to do problems.

I disagree that the AIME is easiest. I think the USAMO is [u]much[/u] easier. (come on, we get 1.5 hours per question!)

 :P Seriously though, the questions on the AMC 12 level are much easier than those on the AIME level. AMC 12 problems (especially the first 20 or so) are much more straight forward, while AIME questions tend to invoke much more thought that you will probably end up needing 12+ minutes per question. Indeed, if you work on the first 20 something questions at a decent rate, then you should have plenty of time to think out the last several questions.

[quote="hunter34"]AIME is easier because the answers are from 1-999.[/quote]

umm thats 1000 choices (its 000-999)...

and technically all answers can be applied, they would just make you transform it.

Anyways, AMC its from A-E, 5 choices..

Ill take my chance with the 5 choices...even if you get counted off for guessing because the reward is much higher than with 1000 choices since most of the answers can be narrowed down.

And for the AMC, who cares about time?  When you say 10 easy, 10 okay, 5 hard, compared to the AIME thats 20 easy, 4 okay, 1 hard (and the first 10 questions can be done in like 15 minutes so its really 15 questions for 60 minutes) and all AIME problems would go from hard-impossible...

Plus with 3 hours, most of the questions, no matter how long you think, youll never get it.

EDIT: wait azjps just said (basically) the last part of what I said...but anyways thats two people...sorry i didnt real azjps popst after reading that thing about USAMO.

well maybe it's just cuz i don't know any of the "concepts" you guys are talking about, and I'm okay with reasoning everything out on the AIME =P

WHAT? AMC12 harder? Wow...I don't quite see how that's possible
For me, I can get quite decent scores on AMC12s (130s I'm hoping 140+ on the real thing coming up) and I don't find the time limit a problem at all. AIME is different, I find that there is NOT ENOUGH TIME :( Well I sorta suck so that's no wonder but...

There is some overlap in difficulty. The very easiest AIME problems (usually the first few on the exam) are often easier than the hardest back-page AMC 12 problems. But that's just an overlap. It's not the whole test in either case.

The first page of the AMC 12 (or AMC 10) is going to contain problems that should be accessible to just about anyone, including people with no problem solving background. It is in the interest of the test writers not to totally intimidate those who come into the test relatively casually. It's really better if we can have people walking away thinking "At least I got [i]something[/i] right."

That motivation is not there on the AIME. Those taking the AIME have already proven themselves. The AIME seeks to separate those student on the basis of some somewhat more difficult challenges. And more difficult challenges do require more time. That's built in.

[quote="Narcissa"]hmmm
maybe i mean it a different way
i enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]
Yes I agree on that, not necessarily the formula oriented part but the more fun part. That doesn't have anything to do with difficulty though...

[quote="Pakman2012"]Here is whats going on guys.

With each test, there are different expectations for success. Like that is literally all there is to it.

AMC - 130 is considered a pretty good score. which is like.. 21ish questions right, or roughly 80% of the questions.

AIME - 8/9 is considered a pretty decent score. It is usually more than enough to pass. Thats roughly thats like between 50 and 60%, which is alot better than AMC.

Omigosh.

It definitely helps that the first few AIME questions from the last few years have been super approachable by even those with limited mathematical experience. I personally enjoy doing AIME problems alot more than AMC because AMC problems generally wind up being alot more nitpicky/tricky than AIME problems.

I never really like the comparison of the last 5 of AMC to first 5 of AIME or whatever. The first 5 of the AIME are generally pretty nice and straight forward, whereas the last 5 of an AMC 12 are generally as unapproachable as possible. It has become kind of a standard to do like the first 5-6ish AIME problems in the first hour, 3-4ish in the second hour, and 1-2ish in the last hour, along with a nice nap somewhere in there.

I dunno. AMC is just a qualifying test, as is the AIME. Just do well enough to pass well and move on to the next test.[/quote]
Hmmmm...interesting, I don't find the last 5 of the AMC12 nearly as hard as some medium AIME questions (thanks in part to multiple choice, I can save time and work by checking the range of the answers they give :D I usually get 3+ right), but the first 7 or so AIME questions are generally pretty straightforward. The second half is a nightmare for me though...
To me AMC12 is not that different from AMC10 which is not that much different from mathcounts and mathcounts is super basic stuff...

[quote="Narcissa"]hmmm
maybe i mean it a different way
i enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]

Well that's different.

Do you like the USAMO even better?

[quote="dnkywin"]Well, I did some past AIMEs and AMCs, and their difficulty really can vary from year to year. (I got 12 on 2003 AIME, 8 on 2004 AIME.) :ewpu:[/quote]
That's true, although my scores hover around 6 :(

[quote="dnkywin"]Well, I did some past AIMEs and AMCs, and their difficulty really can vary from year to year. (I got 12 on 2003 AIME, 8 on 2004 AIME.) :ewpu:[/quote]

I feel like these competitions all tend to get little more difficult each year. The AMC actually seems to stay relatively stable, though. (Is it true that the alternate, AMC12-B, is harder than the AMC12-A?)

for me, thinking under pressure = hard. how bout that :P

[quote]Do you like the USAMO even better?[/quote]no, cuz i can't prove for life.

[quote]When you say you enjoy the a lot more, you are probably only talking about problems 1-4ish because after that, you wont enjoy them at all.[/quote]i dont know, like the first 10?

[quote="azjps"]
AMC 12 problems (especially the first 20 or so) are much more straight forward, while AIME questions tend to invoke much more thought that you will probably end up needing 12+ minutes per question. Indeed, if you work on the first 20 something questions at a decent rate, then you should have plenty of time to think out the last several questions.[/quote]

Okay, I take this back .. on my most recent practice test I spent 10 minutes working on #4, and in the end I only got #2 and 4 wrong. That's just weird  :rotfl:

[quote="firecricket91"][quote="Narcissa"]hmmm
maybe i mean it a different way
i enjoy doing AIME problems a LOT more than AMC12 problems, probably cuz I get to think more freely and because I'm not as pressed for time. Additionally AMC12 problems are more formula-oriented and AIME problems are less so, I think.[/quote]

what do you mean formula-oriented.

No AMC or AIME problem should be strictly formula oriented..that should be for school math.

They are equally "formula-oriented" or lack there of.

When you say you enjoy the a lot more, you are probably only talking about problems 1-4ish because after that, you wont enjoy them at all.[/quote]

True, no contest can be done by school knowledge alone. But the first 15 or so problems of the AMC 12 can be made shorter by knowing shortcuts, while AIME has less apparent problems that seem impossible if you don't know any more than theorems. AIME might be easier due to the time constraint, though I actually find AMC 10 to be more time-pressuring than the 12, so neither the 12 or the AIME really give time problems (except for finding a careless mistake at the last minute). And on the average, AIME is slightly harder. They give you just enough easy problems so that you need to work hard in order to reach your goal.


[color=darkred]Let $ABC$ be a triangle. The its incircle $C(I)$ touches the lines $AB$, $AC$ in the points $E$, $F$ respectively. The line $EF$ meets the circle $w$ with the diameter $[BC]$ in the points $X$, $Y$ so that the line $AI$ separates the points $X$, $F$. The same circle $w$ cuts the lines $AB$, $AC$ in the points $M$, $N$ respectively. Denote the intersections $K\in MX\cap NY$ and $L\in NX\cap MY$. Prove that $K\in AI$ and $L\in HI$, where the point $H$ is the orthocenter of $\triangle ABC$.[/color]

Here's my main ideas. ( I'm busy now so I can't post the full solution, sorry :( )
- Prove that $BY$ and $CX$ are pass through $I$ - only calculating with Menelaus's theorem.
Use this lemma: ( I think it's well-known )
Tak 6 points $A,B,C,X,Y,Z$ in an given circle. Then the intersections of $AY$ and $BX$, $BZ$ and $CY$, $CX$ and $AZ$ are collinear.
+ With $M,Y,C,N,X,B$: We have $K,I,A$ are collinear
+ With $M,X,B,N,Y,C$: We have $L,I,H$ are collinear

Here's my solution. I'm sorry to post a new post...but I can't edit my above post now  :blush:
[b]Lemma:[/b]
Take on a given circle six points $A_{1},A_{2},A_{3}$ and $B_{1},B_{2},B_{3}$. Then the intersections of $A_{1}B_{2}$ and $A_{2}B_{1}$ - called as $C_{3}$ , $A_{2}B_{3}$ and $A_{3}B_{2}$ - called as $C_{1}$ , $A_{1}B_{3}$ and $A_{3}B_{1}$ - called as $C_{2}$ are collinear.
I think it's well-known, so I only post my main ideas to prove it:
- Prove that: If $A,B,C,D,M$ lie on the circle $(O)$, then $(M.ABCD)$ is constant when $M$ vary
- Let $A_{2}B_{1}$ intersects $A_{1}B_{3}$ at  $M$, $A_{2}B_{3}$ intersects $A_{3}B_{1}$ at $N$
Because $(A_{1}.B_{1}B_{2}B_{3}A_{2})=(A_{3}.B_{1}B_{2}B_{3}A_{2})$ so $(B_{1}C_{3}MA_{2})=(NC_{1}B_{3}A_{2})$
But we've known that $C_{2}B_{1},C_{2}C_{1},C_{2}M,C_{2}B$ intersects $A_{2}B_{3}$ at $N,C'_{1}, B_{3}, A_{2}$, respectively, then $(B_{1}C_{3}MA_{2})=(NC'_{1}B_{3}A_{2})$. It means that $C_{1}\equiv C'_{1}$, and $C_{1},C_{2},C_{3}$ are collinear.
[hide="Figure"][img]http://i108.photobucket.com/albums/n9/PDatK40SP/735656425.jpg[/img][/hide]
To Mods: I'm quite busy now so I post my figure I've created before from Photobucket, and don't like ImageShack, it will not be cleared  :wink:
[b]Back to the problem:[/b]
$BI$ intersects $EF$ at $Y'$. We have $\hat{Y'FC}= 180^{o}-\hat{EFA}= 90^{o}+\frac{1}{2}\hat{BAC}= \hat{BIC}$ so $I,Y',F,C$ are concyclic. Then, $\hat{BY'C}= \hat{IY'C}= \hat{IFC}= 90^{o}$, it means that $Y'$ lies on $w$, or $Y \equiv Y'$. So $BY$ passes through $I$, and similarly, $CX$ passes through $I$
Use the lemma with $M,Y,C$ and $N,X,B$ we have $K,I,A$ are collinear
Use the lemma with $M,X,B$ and $N,Y,C$ we have $L,I,H$ are collinear
P/s to Mods: Could you delete my above post for me please ? Thank you  :)

[quote="PDatK40SP"]
$BI$ intersects $EF$ at $Y'$. We have $\hat{Y'FC}= 180^{o}-\hat{EFA}= 90^{o}+\frac{1}{2}\hat{BAC}= \hat{BIC}$ so $I,Y',F,C$ are concyclic. Then, $\hat{BY'C}= \hat{IY'C}= \hat{IFC}= 90^{o}$, it means that $Y'$ lies on $w$, or $Y \equiv Y'$. So $BY$ passes through $I$, and similarly, $CX$ passes through $I$
[/quote]

i think you might mixed up some notations, but the proof is ok.

another proof for the collinearity of $B,I,Y$ etc you can see here the lemma i used to solve megus's problem : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132338[/url]


induction
algebra
polynomial
algebra solved

Let F(n) be the n-th Fibonnacci number, where F(1) = F(2) = 1, F(n+2) = F(n+1) + F(n). Let p(x) be a polynomial of degree 990 such that p(k) = F(k) for k = 992, 993, ... , 1982. Show that p(1983) = F(1983) - 1.

Unfortunately, on http://www.kalva.demon.co.uk/short/sh83.html
u cannot find the solution
:)
:)
:D

I never heard of "Newton interpolation" ...


Hi, my brother is going to be writing the MOEMS competition (Division E) this year. I wanted to know what the average mark in MOEMS is.

Why don't you put a poll.

BTW, I think the average mark is 3. The people in my class normally get like 1

[quote="solafidefarms"]No. The average mark is *not* 3. From last years' statistics, something like 40% of the test-takers on each round got none right. The average was about 2.[/quote]

It depends on the year. But are you serious that 40% get 0 :P

i know that when i did it i always got all correct (25/25)
-jorian

the average in my class was a 18, 18/5= 3.6 (I was also in an advanced math class though)

[quote="satyam"][quote]No. The average mark is *not* 3. From last years' statistics, something like 40% of the test-takers on each round got none right. The average was about 2.[/quote]

Where did you get the statistics from? I tried to look for it over the web, but couldn't find anything useful.[/quote]

The MOEMS people send out statistics for the last round with each new set of tests to the teachers. My teacher would read the statistics to us to encourage us. That's why I'm not quite sure on the numbers.


let $ a,b,c>0$. Prove that
$ a\sqrt{\frac{b}{c}}\plus{}b\sqrt{\frac{c}{a}}\plus{}c\sqrt{\frac{a}{b}}\geq \sqrt{3(ab\plus{}bc\plus{}ca)}$

[quote="VIMF"]let $ a,b,c > 0$. Prove that
$ a\sqrt {\frac {b}{c}} \plus{} b\sqrt {\frac {c}{a}} \plus{} c\sqrt {\frac {a}{b}}\geq \sqrt {3(ab \plus{} bc \plus{} ca)}$[/quote]
It is equivalent to the old and easy problem:

If $ x,$ $ y,$ $ z$ are positive real numbers, then
$ \frac{x^2}{y}\plus{}\frac{y^2}{z}\plus{}\frac{z^2}{x} \ge \sqrt{3(x^2\plus{}y^2\plus{}z^2)}.$


Set $ \sin x+\sin y=\cos x+\cos y=\frac{1}{2}$ , So $ \sin 2x=?$

$ Sin(y)=1/2-Sin(x)$
$ Cos(y)=1/2-Cos(x)$
by squaring both equations and adding them we get:
$ Sin^{2}(y)+Cos^{2}(y)=1/4+Sin^{2}(x)-Sin(x)+1/4+Cos^{2}(x)-Cos(x)$
applying that $ Cos^{2}(a)+Sin^{2}(a)=1$
$ 1=1/2+1-Sin(x)-Cos(x)$
$ 1/2=Sin(x)+Cos(x)$
Squaring again and from $ Sin(2x)=2Sin(x)Cos(x)$ we get:
$ 1/2=Sin^{2}(x)+Sin(2x)+Cos^{2}(x)$
then $ Sin(2x)=-1/2$

LaTeX hint: \sin and \cos give $ \sin$ and $ \cos$, which looks much nicer than what you were typing.

(Other similar codes: \max, \min, \log, \ln, \tan, \cot, \lim, etc.)

possible typo in the last 2 lines ... should be 1/4 (with answer something like -3/4)


So when did/does your break start?

Extra credit to anyone who realizes why the break couldn't've started on Monday :)

The third option is to specify whether or not you have a half day or not, also if you're school doesn't have Friday off [size=59]that's too bad :twisted: [/size] I mean please say so :)

Also wondering if there's a school who doesn't celebrate Thanksgiving Day in the US (I mean I'm guessing some religious schools don't but who knows?)

Hehe.  It could have started on Monday.  They have [b]Sunday[/b] school at church, you know.   :wink: </jk>

Anyway, we have a half day on wednesday, and then this big potluck thing, and then we're home free.

we had school all day today, and have Thursday and Friday off...my teachers, however, felt the need to ruin break by giving too much homework...grrrr...

Wednesday for me! Well... it actually started Tues. because we went to Classics Day at OU  and didn't go to school.

Who's the lucky duck who got to have a whole week off? :)

Over here middle schoolers get their break by Wed...break for high school starts Thurs.

darn thanksgiving is over.:(
school will start on monday....
pumpkin pie all gone.....:cry:

Not to mention the fact that there's no major break until Christmas...*sigh* Did you know that most people suicide on New Year's Day (not Chinese :)) out of federal holidays? OK that was random...*sigh*

@above, please refrain from necroposting. This thread had not been posted in for nineteen years.

Other. I live in Canada, where Thanksgiving is in October.

my break is November 23rd to 25th

appreciate how my school gives 3 days of vacation.

[quote=MathWinner121]I do not celebrate thanksgiving[/quote]

bro this post is older than twice your age stop bumping these threads

[quote=mzarony0]My thanksgiving break's in October lol

I'm in Canada[/quote]

Hello, fellow Canadian!!! I really love how Thanksgiving is in October, because the weather is not gruesomely snowy.


geometry
circumcircle
trigonometry
incenter
analytic geometry
IMO Shortlist

Let $ Q$ be the centre of the inscribed circle of a triangle $ ABC.$ Prove that for any point $ P,$
\[ a(PA)^2 \plus{} b(PB)^2 \plus{} c(PC)^2 \equal{} a(QA)^2 \plus{} b(QB)^2 \plus{} c(QC)^2 \plus{} (a \plus{} b \plus{} c)(QP)^2,
\]
where $ a \equal{} BC, b \equal{} CA$ and $ c \equal{} AB.$

.Let $w=C(I,r)$ be the incircle of the triangle $ABC$.
Then there are the following wellknown relations:

$\blacksquare \ 1^{\circ}.\ a\cdot IA^2+b\cdot IB^2+c\cdot IC^2=abc.$
$\blacksquare \ 2^{\circ}.\ \left(\forall\right) P,\ a\cdot PA^2+b\cdot PB^2+c\cdot PC^2=(a+b+c)\cdot PI^2+abc.$

[u]Examples.[/u]
$Ex\ 1.\ \ P: =I\Longrightarrow 1^{\circ}.$
$Ex\ 2.\ \ P: =O\Longrightarrow 2sR^2=2s\cdot OI^2+4Rsr\Longrightarrow OI^2=R^2-2Rr.$

Let $ABC$ be a fixed triangle, a fixed point $F(u,v,w)$ and the mobile point $M: =M(x,y,z)$
(with barycentrical coordinates). I note the power $p(X)$ of the point $X$ w.r.t. the circumcircle
of the triangle $ABC$. Then for $\left(\forall\right) M$ there are the following relations:

$1^{\circ}.\ p(M)=-(yza^2+zxb^2+xyc^2).$
$2^{\circ}.\ x\cdot MA^2+y\cdot MB^2+z\cdot MC^2=-p(M).$
$3^{\circ}.\ u\cdot MA^2+v\cdot MB^2+w\cdot MC^2=MF^2-p(F).$
$4^{\circ}.\ FM^2=-[(y-v)(z-w)a^2+(z-w)(x-u)b^2+(x-u)(y-v)c^2]\ge0.$

The [u]remarcable points[/u] for a triangle $ABC: \ A(1,0,0),\ B(0,1,0),\ C(0,0,1);$

$G(\frac 13,\frac 13,\frac 13),\ I\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right),\ N\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right), I_a\left(\frac{-a}{2(s-a)},\frac{b}{2(s-a)},\frac{c}{2(s-a)}\right);$

$H(\cot B\cot C,\coth\cot A,\cot A\ cot B),\ O\left(\frac{R^2}{2S} \sin 2A,\frac{R^2}{2S} \sin 2B,\frac{R^2}{2S}\sin 2C\right);$

$\Gamma\left(\frac{(s-b)(s-c)}{r(4R+r)},\frac{(s-c)(s-a)}{r(4R+r},\frac{(s-a)(s-b)}{r(4R+r)}\right)$ (the [u]Gergogne's point[/u]);

$L\left(\frac{a^2}{a^2+b^2+c^2},\frac{b^2}{a^2+b^2+c^2},\frac{c^2}{a^2+b^2+c^2}\right)$ (the [u]Lemoine's point[/u] or the [u]simedian center[/u]).

[u]Remark.[/u] You observe that: $x+y+z=1,\ \frac{[MBC]}{|x|}=\frac{[AMC]}{|y|}=\frac{[ABM]}{|z|}$,
where $[XYZ]$ is the area of the triangle $ABC.$

Here is a solution with vectors.  The claim is clearly true for $ P\equal{}Q$.  No we shall show that given a point $ P$ satisfying the property, another point $ R$ will satisfy the property as well.  That is, we must prove

$ a(PA)^{2}\plus{}b(PB)^{2}\plus{}c(PC)^{2}\equal{} a(QA)^{2}\plus{}b(QB)^{2}\plus{}c(QC)^{2}\plus{}(a\plus{}b\plus{}c)(QP)^{2}$

$ \Rightarrow a(RA)^{2}\plus{}b(RB)^{2}\plus{}c(RC)^{2}\equal{} a(QA)^{2}\plus{}b(QB)^{2}\plus{}c(QC)^{2}\plus{}(a\plus{}b\plus{}c)(QR)^{2}$

So we must prove the respective differences are the same:

$ \sum a(PA^2\minus{}RA^2)\equal{}(a\plus{}b\plus{}c)(QP^2\minus{}QR^2)$

We use vectors:

$ \sum a(PA^2\minus{}RA^2)\equal{}(a\plus{}b\plus{}c)(QP^2\minus{}QR^2)$

$ \Leftrightarrow \sum a(PA\minus{}RA)(PA\plus{}RA)\equal{}(a\plus{}b\plus{}c)(QP\minus{}QR)(QP\plus{}QR)$

$ \Leftrightarrow \sum a(P\minus{}R)(P\plus{}R\plus{}2A)\equal{}(a\plus{}b\plus{}c)(P\minus{}R)(P\plus{}R\plus{}2Q)$

$ \Leftrightarrow \sum a(P\plus{}R\plus{}2A)\equal{}(a\plus{}b\plus{}c)(P\plus{}R\plus{}2Q)$

$ \Leftrightarrow \sum a(2A)\equal{}(a\plus{}b\plus{}c)(2Q)$

Now we simply have the vector definition of Q, the incenter of the triangle.

Place masses $a, b, c$ on vertices $A, B, C$. It is clear that the centre of gravity is $Q$. By parallel axis theorem, the solution follows.

[b]My solution uses $I$ for incenter instead of $Q$:[/b]

Let $I$ be the origin to all the vectors I will work with. First off, note that $\vec{I}=0$ if we take the origin as $I.$

We know that $\vec{PA}=\vec{IP}+\vec{IA}.$

The same holds true for the rest as well. So, the left hand side of our equation becomes $$a(\vec{IP}+\vec{IA})^2+b(\vec{IP}+\vec{IB})^2+c(\vec{IC}+\vec{IP})^2.$$
We expand that to get $$(a \cdot PA^2 + b \cdot PB^2 + c \cdot PC^2 = a \cdot IA^2 + b \cdot IB^2 + c \cdot IC^2 + (a + b + c) \cdot IP^2)+2\vec{IP}(a\vec{IA}+b\vec{IB}+c\vec{IC}).$$
All that remains is showing that $2\vec{PI}(a\vec{IA}+b\vec{IB}+c\vec{IC})=0.$

First off, since we took $I$ as the origin, we know that $a\vec{IA}+b\vec{IB}+c\vec{IC}$ is the same as $a\vec{A}+b\vec{B}+c\vec{C}.$

Now, this reminds us of the barycentric coordinates of incenter of $\triangle ABC.$ Namely, we know that $$\dfrac{a}{a+b+c}\vec{A}+\dfrac{b}{a+b+c}\vec{B}+\dfrac{c}{a+b+c}\vec{C}=\vec{I}=0.$$Multiplying through by $a+b+c$ gives $a\vec{A}+b\vec{B}+c\vec{C}=0$ as desired.

So, we have shown that $a \cdot PA^2 + b \cdot PB^2 + c \cdot PC^2 = a \cdot IA^2 + b \cdot IB^2 + c \cdot IC^2 + (a + b + c) \cdot IP^2.$

[hide=Solution]
Let $P$ be the origin, and let $I=Q$. Then, we need to show $aA\cdot A+bB\cdot B+cC\cdot C=aIA^2+bIB^2+cIC^2+(a+b+c)I\cdot I$. We have
\begin{align*}
IA^2&=\frac1{(a+b+c)^2}(bB+cC-(b+c)A)\cdot(bB+cC-(b+c)A)\\
&=\frac1{(a+b+c)^2}((b+c)cb^2+(b+c)bc^2-bca^2)\\
&=\frac1{(a+b+c)^2}{bc(b+c)^2-bca^2}\\
&=\frac{bc(b+c+a)(b+c-a)}{(a+b+c)^2}\\
&=\frac{bc(b+c-a)}{a+b+c}.\end{align*}
Therefore, $aIA^2+bIB^2+cIC^2=abc$. Now, we have
\begin{align*}
(a+b+c)I\cdot I&=\frac1{a+b+c}(aA+bB+cC)\cdot(aA+bB+cC)\\
&=\frac1{a+b+c}(a^2A\cdot A+b^2B\cdot B+c^2C\cdot C+2abA\cdot B+2bcB\cdot C+2caC\cdot A)\\
&=\frac1{a+b+c}(A\cdot A(a^2+ab+ac)+B\cdot B(ba+b^2+bc)+C\cdot C(ca+cb+c^2)-abc(a+b+c))\\
&=aA\cdot A+bB\cdot B+cC\cdot C-abc.\end{align*}
Therefore, we get $aA\cdot A+bB\cdot B+cC\cdot C=aIA^2+bIB^2+cIC^2+(a+b+c)I\cdot I$.


geometry
inequalities proposed
inequalities

Let $a,b,c$ be the side lengths of a triangle of area $S$. Prove that
\[2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})\ge 4\sqrt 3S \]

See here:
stronger than Klamkin:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31598[/url]


How many distinct integers are in the solution set of $ \left|\frac{1}{3}m \plus{} 2\right| \leq 1$?

$ \left|\frac {1}{3}m \plus{} 2\right|\leq 1$

$ \frac {1}{3}m \plus{} 2 \leq 1$ or $ \frac {1}{3}m \plus{} 2 \geq \minus{} 1$

$ \frac {1}{3}m \plus{} 2 \leq 1$
$ m \leq \minus{} 3$

$ \frac {1}{3}m \plus{} 2 \geq \minus{} 1$
$ m \geq \minus{} 9$

$ \minus{} 9 \leq m \leq \minus{} 3$ so there are $ \boxed{7}$ solutions.

EDIT: Noo beaten again.

Darn much easier way:
The interval is 2 so 2/(1/3)+1(because it's $\le$)=7


S is a set of 100 stones. f(S) is the set of integers n such that we can find n stones in the collection weighing half the total weight of the set. What is the maximum possible number of integers in f(S)?

:?: So where do you define the "weight"?


Do either one of these help you qualify for the ARML???

How do you qualify for the USAMO???

For the second question:  see the Frequently Asked Questions, USAMO #5
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

The AMC has essentially no relationship with ARML. Your state/school/region coaches may use AMC scores to help pick teams, there is no official qualification mechanism.

For the first question: ARML team selection is governed locally. It all depends on what information the coaches in your state collect and act on. That information might include AMC/AIME scores, and it might not. Ask about it locally.

[Or essentially what worthawholebean said.]

In addition to above posts, I recommend you read this Wiki article that thoroughly covers the selection process.

http://www.artofproblemsolving.com/Wiki/index.php/How_to_join_an_ARML_team


Please solve the following problem taken from Greece 95

Equation    ax^2 + (c-b)x  + (e-d) = 0  has two real roots  each of them > 1
Prove that equation  ax^4 + bx^3 + cx^2 + dx + e + 0 has at least one real root.

 Thanks for Your help. :)


Let $n$ be a positive integer. The set $\{1,2,\ldots,3n\}$ is partitioned in three subsets $A,B,C$, each one containing $n$ elements.
Prove that there is $x \in A$, $y\in B$ and $z \in C$ such that one of these integers is the sum of the two others.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=521113]All problems of the France TST 2006[/url]

recycling,reclycling,... see this http://www.mathlinks.ro/Forum/viewtopic.php?t=3781


algebra
polynomial
advanced fields
advanced fields solved

Let S the subset of $ \mathbb{C}^4$ given by the points $ \{(n,2^n,3^n,6^n)\}$ for n in $ \mathbb{N}$. Find generators for the ideal $ I(S)$ of polynomials in $ \mathbb{C}[X,Y,Z,W]$ that are zero in $ S$, and find the dimension of the associated affine variety $ V(I(S))$ (the Zariski closure of $ S$).

$ W\minus{}YZ$ belongs to $ I(S)$.
Let $ P(X,Y,Z,W)\in I(S)$. By using that $ W\minus{}YZ\in I(S)$ we can assume that $ P\in\mathbb C[X,Y,Z]$ (just divide by W-YZ and take the residue).

We have $ P(n,2^n,3^n)\equal{}0$ for all $ n$. Look at $ P$ as a polynomial in $ Y,Z$ with coefficients in $ \mathbb C[X]$. Let $ d$ be the degree of $ P$. If we assume for a contradiction that $ P\neq 0$, then $ d$ must be a well defined integer. Let $ k$ be the greatest integer such that $ Z^kY^{d\minus{}k}$ appears in $ P$ with nonzero coefficient.
It should now be clear that if we consider $ Q\equal{}P/(Z^kY^{d\minus{}k})$ then $ lim_{n\to\infty}Q(n,2^n,3^n)$ is nonzero which contradicts $ P\in I(S)$.

Therefore $ I(S)\equal{}(W\minus{}YZ)$. The dimension of the closure is 3.


Prove that in $ \triangle\ \mathrm{ABC}$ we have $ \boxed {\ \mathrm N\Gamma\ \parallel\ \mathrm{BC}\ \Longleftrightarrow\ a(b \plus{} c) \equal{} b^2 \plus{} c^2\ \Longleftrightarrow\ \mathrm F\in (\mathrm{AM})\ }$

where  $ \mathrm M$ - [u]midpoint[/u] of $ [\mathrm{BC}]$ and $ \mathrm N$ - [b]Nagel[/b]'s , $ \Gamma$ - [b]Gergonne[/b]'s , $ \mathrm F$ - [b]Feuerbach[/b]'s [u]points[/u] .

Seems it so difficult ?! I don't think. This problem was proposed for
 the RMG - contest (shortlist !) by [b]Evgeniy Kulanin, Rusia[/b]. I have
a short metrical proof of it. Maybe someone has a synthetical proof ...

I think it is nice,

Let $ AN\cap BC\equal{}A_1,A\Gamma\cap BC\equal{}A_2$ thus \[ N\Gamma\parallel BC\Leftrightarrow \frac{AN}{AA_1}\equal{}\frac{A\Gamma}{AA_2}\Leftrightarrow \frac{p\minus{}b\plus{}p\minus{}c}{p\minus{}a\plus{}p\minus{}b\plus{}p\minus{}c}\equal{}\frac{1/(p\minus{}b)\minus{}1/(p\minus{}c)}{1/(p\minus{}a)\plus{}1/(p\minus{}b)\plus{}1/(p\minus{}c)}\Leftrightarrow\frac{a}{p}\equal{}\frac{a(p\minus{}a)}{\sum (p\minus{}b)(p\minus{}c)}\Leftrightarrow a(b\plus{}c)\equal{}b^2\plus{}c^2\]

We have $ F(a\minus{}a\cos(B\minus{}C),..,..)$ therefore $ F\in AM\Leftrightarrow b\minus{}b\cos(C\minus{}A)\equal{}c\minus{}c\cos(A\minus{}B)\Leftrightarrow a(b\plus{}c)\equal{}b^2\plus{}c^2$

So combine them we have nice relation $ \boxed {\ \mathrm N\Gamma\ \parallel\ \mathrm{BC}\ \Longleftrightarrow\ \mathrm F\in (\mathrm{AM})\ }$

[b]Thanks, [u]Encyclopedia[/u] ! [/b] My proof is [url=http://mateforum.ro/viewtopic.php?t=3471][color=red][b]here[/b][/color][/url].

Thank you for nice relation in triangle :), it is quite hard for me to understand solution in Romanian, but now I understood :oops: !

You are right, [b]Encyclopedia[/b] ! In at most two or three words ... But no all know barycentrical coordinates. For them is my proof. Thank you.


geometry
circumcircle
geometric transformation
cyclic quadrilateral
geometry unsolved

The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $O$.  The circumcircles of triangles $AOB$ and $COD$ intersect again at $K$.  Point $L$ is such that the triangles $BLC$ and $AKD$ are similar and equally oriented.  Prove that if the quadrilateral $BLCK$ is convex, then it is tangent [has an incircle].

Why no one answered :( Is this to hard? Or because no sketch from me?

Need better translation as my graphic does not agree with the problem.

Let $P \equiv OK \cap BC.$  $\angle BKP= \angle BAO= \angle CDO= \angle CKP$ $\Longrightarrow$ $KP$ bisects $\angle BKC.$ Let the angle bisectors of $\angle KBL$ and $\angle KCL$ cut $LC,LB$ at $U,V.$ Then

$\angle KCL = \angle KCB + \angle ADK = \angle ACB - \angle KCO + \angle ADB + \angle ODK$

$\Longrightarrow \angle KCL= 2\angle BCA=\angle 2\angle KCV$

Similarly, $\angle KBU=\angle CBD.$ Hence $CA,CV$ and $BD,BU$ are isogonals WRT $\angle KCB$ and $\angle KBC,$ respectively. If $I \equiv BU \cap CV,$ then $O,I$ are isogonal conjugates WRT $\triangle KBC,$ but since $KO$ bisects $\angle BKC,$ it follows that $I \in KP$ $\Longrightarrow$ Internal angle bisectors of the convex $BLCK$ concur at $I,$ i.e. $BLCK$ has an incircle centered at $I.$

To prove that $BLCK$ has an incircle, we have to prove that $KB-KC=BL-LC$ by Pitot's Theorem.
By Ptolemy in $AKOB$ and $DKOC$, we will have
$KB-KC=\frac{OB.AK+KO.AB}{OA}-\frac{KD.OC+KO.DC}{DO} $

First, we have $OB/OA=OC/DO$ by Power of point $O$
and $AB/OA=DC/DO$ : will simplify the expression:
$KB-KC=\frac{OB}{OA}(AK-KD)$
Now, $\triangle BLC \sim \triangle AKD$ gives us $BC/AD=BL/AK=LC/KD=\frac{BL-LC}{AK-KD}$
So
$KB-KC=\frac{OB}{OA}\cdot \frac{AD(BL-LC)}{BC}=BL-LC$

Straightforward length computation.

We need to show that $BK - KC = BL - LC$, so since $BL - LC = \frac{BC}{AD} (AK - KD)$ from the similar triangle condition, we want to show that
$$AD \cdot KC + BC \cdot KA = AD \cdot KB + BC \cdot KD.$$
Since $\triangle AKC \sim \triangle BKD$, we have that $KC = \frac{AC}{BD} \cdot KD, KA = \frac{AC}{BD} \cdot KB,$ so that plugging this into the left side of the above means that we wish to show
$$AD \cdot \frac{AC}{BD} \cdot KD + BC \cdot \frac{AC}{BD} \cdot KB = AD \cdot KB + BC \cdot KD,$$
which is equivalent to
$$(AC \cdot AD - BC \cdot BD) \cdot KD = (DA \cdot DB - CA \cdot CB) \cdot KB.$$
Since $\triangle KBA \sim \triangle KDC$, we have that $\frac{KB}{KD} = \frac{AB}{CD},$ so we only need to show that
$$(AC \cdot AD - BC \cdot BD) \cdot CD = (DA \cdot DB - CA \cdot CB) \cdot AB.$$
However, if we observe that $[\triangle ABC] = \frac{AB \cdot BC \cdot CA}{4R}$, where $[]$ denotes area and $R$ is the circumradius of $ABCD$, and three other cyclically symmetric expressions, plugging this into the above expression easily finishes.
$\square$

[b]S[/b]olved with [b]mueller.25[/b], [b]starchan[/b], [b]Siddharth03[/b], [b]AdhityaMV[/b]

Let $I'$ be the isogonal conjugate of $O$ in $\triangle ADK$. Let $I$ be the corresponding point in $\triangle BCL$ to $I'$ in $\triangle ADK$. Since $\angle AKO = \angle ABO = \angle DCO = \angle DKO$, $K$ lies on the angle bisector, meaning so does $I'$, its isogonal conjugate. As a consequence, $LI$ bisects $\angle BLC$.

Next, note that $\angle LBK = \angle LBC + \angle KBC = \angle KAD + \angle KBC = \angle KAO + \angle CAD + \angle KBC = 2 \angle DBC$.

And $\angle IBL = \angle I'AK = \angle OAD = \angle DBC$, so $BI$ bisects $\angle KBL$. Similarly, $CI$ bisects $\angle KCL$, implying that $I$ is in fact the incenter of quadrilateral $BLCK$ (as it is the intersection of three angle bisectors), so we are done. $\blacksquare$


Putnam
rotation
analytic geometry
geometry
geometric transformation
college contests
Putnam complex

Let $n$ be a positive integer, $n \ge 2$, and put $\theta=\frac{2\pi}{n}$. Define points $P_k=(k,0)$ in the [i]xy[/i]-plane, for $k=1,2,\dots,n$. Let $R_k$ be the map that rotates the plane counterclockwise by the angle $\theta$ about the point $P_k$. Let $R$ denote the map obtained by applying in order, $R_1$, then $R_2$, ..., then $R_n$. For an arbitrary point $(x,y)$, find and simplify the coordinates of $R(x,y)$.

[u][b]The author of this posting is : Kent Merryfield[/b][/u]
____________________________________________________________________

Work in the complex numbers.

Let $\omega = e^{i\theta} = e^{2\pi i/n}$. Multiplication by $\omega$ rotates counterclockwise by $\theta$. The mapping $R_k$ rotates around $k$. To do that, let $z^{\prime} = R_kz$. Then $z^{\prime}-k = \omega (z-k)$, which becomes $z^{\prime} = R_kz = \omega z - \omega k + k$.

Then $R_2R_1z = \omega \left(\omega z - \omega + 1\right) - 2\omega + 2 = \omega^2 z - \omega^2 - \omega + 2$.

We claim that

$R_kR_{k-1}\dots R_2R_1z = \omega^k z - \omega^k - \omega^{k-1} - \cdots - \omega + k$.

This claim can be verified by a straightforward induction.

Hence,

$R_nR_{n-1}\dots R_2R_1z = \omega^n z - \omega^n - \omega^{n-1} - \cdots - \omega + n$.

But $\omega^n + \omega^{n-1} + \cdots + \omega$ is a geometric sum that adds to zero. Also, $\omega^n = 1$. Hence,

\[R_nR_{n-1}\dots R_2R_1z = z+n\].

The overall operation is translation by $n$. The image of $\left(x,y\right)$ is $\left(x+n,y\right)$.

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

A are the morning problems, B are the afternoon problems.

[u][b]The author of this posting is : bugzpooder[/b][/u]
____________________________________________________________________

[quote="Kent Merryfield"]The overall operation is translation by $n$. The image of $\left(x,y\right)$ is $\left(x+n,y\right)$.[/quote]

are you sure this is right? i got (x-n, y)

[u][b]The author of this posting is : bugzpooder[/b][/u]
____________________________________________________________________

unless i understood the problem wrong, (2,0) gets mapped to (0,0), then a rotation about itself gives (0,0).
(x,y) gets mapped to from rotation about (1,0), then we rotate about (0,0) to get (2-x,-y) (x-2,y)

[u][b]The author of this posting is : ComplexZeta[/b][/u]
____________________________________________________________________

I wrote a rotation of $\theta$ about $z_0$ as $z\mapsto \left(z-z_0\right)e^{i\theta} + z_0$. Then it's easier to do computations.


[quote="zaya_yc"]$ a,b,c\geq 0$

$ \frac {1 \plus{} a}{1 \plus{} b} \plus{} \frac {1 \plus{} b}{1 \plus{} c} \plus{} \frac {1 \plus{} c}{1 \plus{} a}\leq 3 \plus{} a^{2} \plus{} b^{2} \plus{} c^{2}$[/quote]
Let $ c \equal{} \max\{a,b,c\}.$ Then
$ \frac {1 \plus{} a}{1 \plus{} b} \plus{} \frac {1 \plus{} b}{1 \plus{} c} \plus{} \frac {1 \plus{} c}{1 \plus{} a}\leq 3 \plus{} a^{2} \plus{} b^{2} \plus{} c^{2}\Leftrightarrow$
$ a^2 \plus{} b^2 \plus{} c^2\geq\frac {(a \minus{} b)^2}{(1 \plus{} a)(1 \plus{} b)} \plus{} \frac {(c \minus{} a)(c \minus{} b)}{(1 \plus{} a)(1 \plus{} c)},$ which true because
 $ a^2 \plus{} b^2\geq(a \minus{} b)^2\geq\frac {(a \minus{} b)^2}{(1 \plus{} a)(1 \plus{} b)}$ and $ c^2\geq(c \minus{} a)(c \minus{} b)\geq\frac {(c \minus{} a)(c \minus{} b)}{(1 \plus{} a)(1 \plus{} c)}.$


Prove that we cannot arrange the numbers from 1 to 81 on a $9 \times 9$ chessboard so that on every line the sum of the first 4 numbers is equal to the sum of the last 5 numbers.

Consider all 9 rows. The rectangle containing the first four numbers of each row has the same sum as the rectangle containing the last five numbers of each row. Since these cover the entire square, the sum of all numbers in the square must be even. But $ 1\plus{}2\plus{}\cdots\plus{}81 \equal{} 3321$, which is odd. Contradiction.

Another question: What if the numbers were $ 0, 1,\ldots, 80$ instead?

[hide="0 to 80"]
I came up with millions of ways for this to work.

Let all the rows come in the form $ a\plus{}1,a\plus{}3,a\plus{}6,a\plus{}8,a,a\plus{}2,a\plus{}4,a\plus{}5,a\plus{}7$, where $ a$ is a multiple of 9. Of course, this is only one of the very many examples that work.
[/hide]

:maybe:
The way I understood that, $ a \equal{} 9$ gives $ 10, 12, 15, 17, 9, 11, 13, 14, 16$ as a row, and that doesn't work. Also, I think you have to consider the columns as well (the original post specified all lines).

Can you clarify?

Wow. I must have been extremely tired when I wrote that, because it's completely not true unless $ a\equal{}0$. If the original post means all lines, does it mean columns from top to bottom? The diagonal from bottom left to top right?


Can anyone provide me a proof of Descartes' rule of Signs?


calculus
integration
calculus computations

Calculate $ \int \frac {x^2}{{(xsinx \plus{} cosx)}^{2}} dx$

P.S. I'm sorry it is already posted.

Instead if it is not posted - you may try to find:

$ \int_{0}^{\frac {\pi}{4}} arctg \sqrt {\frac {cos2x}{2cos^2x}} dx$

[quote="borislav_mirchev"]Calculate $ \int \frac {x^2}{{(xsinx \plus{} cosx)}^{2}} dx$[/quote]

[url=http://www.mathlinks.ro/viewtopic.php?p=1153094#1153094]See 226[/url]

kunny, it is a problem from a russian contest. The second problem is from the same source.
You may try it.


geometry
symmetry
area of a triangle
Heron\u0027s formula
angle bisector
geometry unsolved

Two isosceles triangles, which sides are x, x, a and x, x, b respectively, have the same area; a≠b. Find x.

I have a bit of trouble with this... can anyone help plz? (I know it should be fairly simple)

thx

by heron's formula you have that
$\frac{1}{4}\sqrt{(2x+a)(2x-a)a^2}=\frac{1}{4}\sqrt{(2x+b)(2x-b)b^2} \\\Rightarrow 4x^2a^2-a^4=4x^2b^2-b^4 \\\Rightarrow (a^2-b^2)(4x^2-(a^2+b^2))=0$,
and since $a^2-b^2\neq 0$ we must have that
$x=\frac{1}{2}\sqrt{a^2+b^2}$

Here is my solution:

   You can avoid  use of Heron's formula.

   Just cut the triangle along the symmetry line and

   turn 180 around the midpoint of the base one of the halves

  so that half of the old base becomes

  the altitude in the new triangle

   and  the old altitude becomes half of the new base.

   This transformation does not change the area.

   Now you see what the value of the side X must be.

   Thank you.

   M.T.

Let's call $ ADB$ and $ ADC$ the two triangles, having $ BD\equal{}AD\equal{}CD$. In order the altitudes from A to the bases $ BD\equal{}CD$ we need $ AD$ as angle bisector of $ \angle ADC$, in which case we get $ AB \equal{} AC$, and this is not the case, or $ B, D$ and $ C$ are collinear and, in this case $ AB\perp AC$, hence $ 4x^2\equal{}BC^2\equal{}AB^2\plus{}AC^2\equal{}a^2\plus{}b^2$, done.

Best regards,
sunken rock


salam
ye soale sede
$2n+1$nafar adam hastand ke har $n$ nafari hade ahgal yek dooste moshtarak darand
sabet konid yek nafar hast ke ba hame doost ast.

rahnama i:say konid khoshe n+1 rasi besazid tozih khastid begid.

soale russia grade 9 salaye 2001 ta 2004
in rahe khushe giri yeki az sari tarin rahashe harchand ye rahe algorithmic khodam daram barash,avalin bar.


email
function
set theory
real analysis
real analysis solved

Hi every body . I have some question about counting operation and set theory.

1\ If $X_1,X_2,...$ be countable sets then $\bigcup_{n=1}^{\infty}X_{n}$ is countable set.

2\[0;1] has the same potency with (0;1)

I want to know about set theory but I haven't any book or article a bout this problem .
Could you lend me some ebook about this  problem?
Thanks .
my email address k09tranhoan@gmail.com

I'm so sorry . My English is so poor.

(1)for example:
X1:x11  x12  x13  x14  ......
x2:x21  x22  x23  x24  ......
x3:x31  x32  x33  x44  .......
..... .... .....   .....  ....  ..... ..

you can acquire a new sequence
X:x11, x12, x21, x13, x22, x31, x14,x23,x32,x41....
then you can  delete some repeated value.
X will be a countable set

(2)
you can take a countable set  A from (0,1)
B=AU{0,1} is also a countable set
we can make a function f
A:{xi},B:{yi},f(xi)=yi
if x in (0,1),but x is not in A,then f(x)=x
so [0;1] has the same potency with (0;1)

You can read some books about real variable function.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Calculete integral : $ \int_{0}^{2\pi}\left(\frac {\sin 5\phi}{\sin\left(\frac {\phi}{2}\right)}\right)^{6}d\phi$

[hide="Answer"]$ 110504\pi$[/hide]


On Monday, 30 girls and 0 boys in the Crescent Valley Band missed practice, resulting in a $ 2 : 5$ ratio of girls to boys. On Tuesday, 30 girls and 48 boys missed practice, resulting in a $ 2 : 1$ ratio of girls to boys. How many students are in the band?

48 more boys missed the band practice, and the ratio reduced by 4.

so ratio 1 is 12 students.

ratio 5, which is 'no boy missed practice', is 12*5=60.

and ratio 2, which is '30 girls missed practice', is 12*2=24.

so the answer is 60+24+30=114. (boys = 60, girls = 54)

answer : 114


Given a parallelogram $ABCD$, join $A$ to the midpoints $E$ and $F$ of the opposite sides $BC$ and $CD$. $AE$ and $AF$ intersect the diagonal $BD$ in $M$ and $N$. Prove that $M$ and $N$ divide $BD$ into three equal parts.

We can show that $\Delta BEM$ is similar to $\Delta DAM$, because all angles are equal.
$\frac{BE}{AD}=\frac{BM}{DM}=\frac{1}{2}$, so:
\[ DN+NM = 2 BM (1) \]

In a similar way, we prove that $\Delta DNF$ is similar to $\Delta BNA$, because all angles are equal.
$\frac{DF}{AB}=\frac{DN}{BN}=\frac{1}{2}$, so:
\[ MN+BM = 2 DN (2) \]

subtracting $(1)$ from $(2)$ gives:
\[ DN-BM=2BM-2DN \iff 3 (BM-DN) = 0 \iff BM=DN \]

Plugging this into $(1)$, finally tells us that \[ DN=MN=BM \]

Cool.

There is another solution - think of centroids!

Oh, yeah, I see. $BD$ and $AE$ are two medians of $\Delta ABC$, and intersect with the ratio $\frac{1}{2}$. Then the same in $\Delta ADC$, sum up and get the answer.

[i]Now it is too late , but tommorow I ' ll try to prove or disprove the converse:[/i]

 << If  we take the midpoints $E$  and $F$ of sides $CB, CD$ of a quadratilateral $ABCD$  and the diagonal $BD$  is divided into three equal parts from the two segments $AE$  and $AF$ , then $ABCD$ is a parallelogram. >>

[u]Babis[/u]

[i](Morning) Yes ! It is true![/i]
 [b]A hint.[/b]

 $AMCN$must be a parallelogramm , so  $AC$ and  $NM$(and also $BD$ ! ) bisect each other .

 [u]Babis[/u]


This 0ne stumped me. If there are 52 well shuffled deck of cards in a standard deck and I deal 13 cards to each of the four players (including me) then what is the probability that I as one of the players can get
1) All four aces (the other cards don't matter)
2) All the 13 cards are from the same suite
3) All the 13 cards go from Ace, 2, 3 ... Jack, Queen, King sequentially irrespective of the suite from which they are

[quote="kyyuanmathcount"]Umm... this may be for the test forum or the MATHCOUNTS or Getting Started.

Move please!!!![/quote] This is Getting Started.  :|


MATHCOUNTS
geometry
analytic geometry
congruent triangles

How many square units are in the area of the convex quadrilateral with vertices (0,0), (3,0), (2,2), and (0,3)

I know the answer but not the solution thanks

Answer
[hide]6 square units[/hide]

[hide]
There are a couple of ways of going about this, you could use the Shoelace Theorem and then you'd have
0  0
3  0
2  2
0  3
0  0
(0 + 6 + 6 + 0) - (0 + 0 + 0 + 0) = 12
Divide by 2, to get $6$, your answer.

You could divide the figure into a square, and two triangles. One triangle has the coordinates (2,2) (0,2) and (0,3). Height is 1, base is 2, area of that is 1. The next triangle has the coordinates of (2,2) (3,0) (2,0), height is 2, base is 1, area of that is 1. The square has coordinates (0,0) (0,2) (2,0) and (2,2). The sidelength is 2, square that to get 4. Lastly, add 4, 1, and 1 up to get 6.[/hide]

Shoe string method:

http://artpad.art.com/?ixhawi10rspo

Note: I know it's not flawless, but hopefully you'll get it.

By the way, I recommend you view it at "fast" speed.

You can also divide it into two congruent triangles by drawing the line segment from (0,0) to (2,2). Each one has a base of 3 and a height of 2, so each one has an area of 3*2/2 = 3. Double it and you get [b]6[/b].

shoe string method???

never heard of it....can you give me a link on it? oh and preferably not with flash...

i can't install the latest version, so it never works :oops:

Here, in PDF:

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf


search
algorithm
logarithms
floor function

Suppose there is a stack of paper with words that needs to be sorted alphabetically, and there are three people in the room. What is the most efficient way to sort the stack?

My thought:

1) Divide the big stack into three or more smaller stacks
2) Each person grabs a smaller stack
3) For each paper, put it into a pile that starts with that alphabet
4) When a person is done with their stack get another stack or take a portion from another person's stack
5) When all papers are in piles (there are at most 26 of them), each person gets a pile and sort it alphabetically
6) When all 26 piles are sorted, simply stack them up to get the final stack

Is this the most efficient method?

Now there is an additional constraint:

Instead of a room, the three people are sitting in a subway with quite many people moving. They can't lay the paper around to make piles. They can only hold on to their piles. Holding a stack right, everyone could only keep one pile (So there can only be three piles at most). In addition, to avoid papercut, paper can only transfered to another person by placing it on the top of their stack. Pulling a paper from another person is not allowed (even if that page is on top) because that is the easiest way to get papercut. Similar to the original situation, the three people might not work at the same speed.

What is the most efficient way to do the same job?

This isn't exactly what I've done, but I do have experience with sorting papers- typically, putting student homework papers in alphabetical order by name.
As a one-person job with some room to work, I typically go with a merge sort. Sort piles of up to about five or six, and then merge those.

What are the key properties of a stack of papers as a data structure? Finding a location is about as costly as a search, which is a binary search or maybe a little better. Inserting something or removing something once you know where to do it is cheap. The storage space cost that matters is the number of stacks, not the size of those stacks.

The difficulty in the second case is finding any algorithm that sorts at all. Here's what I can do:
Insertion/bubble sort for three piles:
1. Person A starts with the whole stack, and passes the top paper to C.
2. Look at the top sheet of the pile. If it belongs later than what C is holding,
2a. C passes their paper to B face down.
2b. A passes their top paper to C.
Otherwise, 2c. A passes their top paper to B face down.
3. Repeat step 2 until A is holding nothing.
4. C passes their paper to B face down.
5. B flips their stack and passes it back to A.
6. Repeat steps 1-5 until everything is sorted. This takes up to $ n\minus{}1$ passes if there are $ n$ papers.

"Finding a location is about as costly as a search" [above]
"The storage space cost that matters is the number of stacks." [above]

With these two in mind, I thought of an alternate method of sorting in the first, kind of like a recursive sorting algorithm.

1) Divide the pile into three arbitrary stacks and give them to the three people
2) For each person: for each paper: put the paper in one of three stacks: "A-I","J-R",or "S-Z".
3) For each stack:
*Divide the stack into three arbitrary piles and distribute
*Sort the stack into "first third, second third, and third third" eg: "A-C", "D-F", "H-J"
*Divide and sort into each letter
*Sort each letter-stack, perhaps with similar methods if the stacks are large
4) Put the sorted stacks on top of each other.

This method reduces the amount of time spent searching for a location and perhaps the storage space as well, as there are always 3 active stacks instead of 26.

Hi, I don't understand the algorithm jmerry said. Could you do a simulation?

Suppose a stack is like this: Top [ 9 6 7 4 1 3 2 8 5 0 ] Bottom

Initially:
A: Top [ 9 6 7 4 1 3 2 8 5 0 ] Bottom
B: Top [ ] Bottom
C: Top [ ] Bottom

A looks at the first page. It is 9 (Suppose they don't know that 9 is the max number).
A gives that to C

A: Top [ 6 7 4 1 3 2 8 5 0 ] Bottom
B: Top [ ] Bottom
C: Top [ 9 ] Bottom

Now A sees the page with a 6 on it. It passes it to B faced down:

A: Top [ 7 4 1 3 2 8 5 0 ] Bottom
B: Top [ 6 ] Bottom
C: Top [ 9 ] Bottom

Next, A sees page 7, it is still less than the page C has, so A gives B faced down:

A: Top [ 4 1 3 2 8 5 0 ] Bottom
B: Top [ 7 6 ] Bottom
C: Top [ 9 ] Bottom

A sees page 4. Now what happens?

On Jmerry's algorithm:

We have
A: Top [ 4 1 3 2 8 5 0 ] Bottom
B: Top [ 7 6 ] Bottom
C: Top [ 9 ] Bottom

We will simply continue doing what we have done so far, so A gives "4" to B face down (and all other ones as well) - (3)

A: Top [ ] Bottom
B: Top [ 0 5 8 2 3 1 4 7 6 ] Bottom
C: Top [ 9 ] Bottom

C then passes to B face down (4):

A: Top [ ] Bottom
B: Top [ 9 0 5 8 2 3 1 4 7 6 ] Bottom
C: Top [ ] Bottom

B flips the stack and passes to A (5):

A: Top [ 6 7 4 1 3 2 8 5 0 9 ] Bottom
B, C are empty.
Repeat steps 1-5:

A: Top [ 6 4 1 2 3 7 5 0 8 9 ] Bottom
B, C are empty
Note how the "8" has moved to its correct location now.

And again and again, and so on.

It takes way more than n-1 passes that jmerry was saying. Now it looks like n log n or something.

I think that by "passes," he meant the number of times 1-5 would be repeated.  While this algorithm does take a fair amount of steps, I think it's fairly efficient with respect to the restrictions.

But do you think that it is a waste of processing from B and C? They are not doing anything. They are just holding the stacks. Wouldn't it be better if they are also doing some sorting?

This isn't about minimizing the total number of comparison, it is about sorting is quickly. Right now it looks like B and C are idling most of the time.

What about this algorithm:

The stack of paper have names (like the original prompt). Instead of labeling the people A,B,C, they are now called person 1, 2, 3

STEP 1) Initially, the stack is separated into three, and each person gets one.

STEP 2) In the first round, each person look through their own stack and extracts a stack that contains A-I. By extract I meant when you see a paper in the range of A-I, put it at the back of the stack you are holding. You go through each paper in your own stack. When you see a name you have already seen, you know you are done.

STEP 3) When all three are done, they combine the A-I stacks for Person 1, Person 1 starts to sort this stack on his own

STEP 4) While Person 1 is sorting, Person 2 and 3 do similarly to get a stack that is from J-R and another S-Z.

STEP 5) When they get those stacks, they sort them independently.

In this algorithm they only pass stacks for three rounds. The first round happens when the A-I stack is formed. The second round happens when the J-R and S-Z stacks are formed. The last pass is just combining the three sorted stacks.

The algorithm I gave was an $ n^2$ algorithm. Yes, it's not efficient with processing time- but you can't really process a stack beyond truly simple bits with one person that can only hold one stack. Even that extraction operation- do you have three hands? You need one to hold the top half of the stack as you look, one to move the paper, and one to hold the bottom. Only two hands? Oops; you just lost your place.
One person, alone, with no room to lay stacks down, [i]can't[/i] sort a stack.

Now, can you do better than my algorithm? Sure. If you start by splitting into $ k$ pieces of approximately equal size and then restack, you reduce the work by almost a factor of $ k$. Also, person C can do more with the "buffer"; if they keep some of the stack in unaltered order (face down), that can split in half and do a bubble pass on the other half at the same time. Do that right (with different cuts for each run), and it's an $ n\log n$ algorithm.
Here's one version of it:
As before, except add a case in step 2:
If A's top sheet is in a certain set (depending on pass), A passes that to C, who swaps it to the bottom of their stack.
What is that set? Here's one version, assuming approximately uniform distribution among letters:
First pass: A-M. Second pass: A-F and N-S. Third pass: A-C and G-I and N-P and T-V. Fourth pass: A, D, G, J-K, N, Q, T, W-X. Fifth pass: B, E, H, J, L, O, R, U, W, Y. In later passes, don't bother with anything; you probably couldn't remember anyway.
Then, in step 5, pass back from both B and C with a merge. A will take them face down, and then flip the stack up for the next pass.

That gives you the first-letter sort in five passes, while incrementally improving the rest. The number of passes required after that is comparable to the number of papers in the most numerous letter.

I see what you mean. Practically you could sort the stack in your hands by using your index finger as a bookmark. This matches the original assumption that a person could sort a stack in their hands, if they want to.

Could we talk more in terms of real people and real paper? I wasn't trying to dress up a computer problem. I was really talking about people sorting paper. I was wondering what algorithm could human use to sort it effectively.

I would probably get the three people to split the papers into three (hopefully) equal sized piles and then sort, then mergesort those piles one at a time.

If the stack got huge, then just doing it with your fingers would be ghastly.

If you have enough people, maybe you can just do it for each letter of the alphabet (and maybe combine some) sort each letter, then merge the piles.

In general, the most efficient use of time (to try to maximize the number of people active at any time as well as minimize the number of steps used by these people) for the first (unbound) scenario would likely be the mergesort algorithm with $ n$ divisions, where $ n$ is the number of people who are participating.

When the space-papercut restriction is added, however, the most efficient use of time appears to be something based on a mergesort with $ \lfloor \frac{n}{2} \rfloor$ divisions, where $ n$ is again the number of participants.

[quote="Twinbee"]Could we talk more in terms of real people and real paper?[/quote]
As I said back in my first post, under real-world conditions with effectively unlimited space, I've used mergesort, with a base of piles of about five or six each, sorted ad hoc. The problem with space-limited algorithms is that the more you try to do with a pile without putting it down, the more likely you are to drop part of the pile on the floor, where it will partially scatter and you'll lose some of your sorting. It's hard to quantify, but I'd be conservative about such things.

I didn't realize that when you merge the stacks,
you don't necessarily have to look at all three stacks at once.
I always felt that took me a long time.


Support
graph theory
number theory open
number theory

Which is the minimum $ n$ that have the following properties,

$ a_1,a_2,\cdots,a_n$ is a permutation of $ 1,2,\cdots,n$

$ a_1 \plus{} a_2$, $ a_2 \plus{} a_3$, $ \cdots$, $ a_{n \minus{} 1} \plus{} a_n$ and $ a_n \plus{} a_1$ are all squares?

Here's a quick argument showing $ n \geq 31$. I don't know if $ n\equal{}31$ actually supports such a sequence, and I can't see a clever way to attack this - just brute force Hamilton circuit search.

Presumably $ n\equal{}1$ is too degenerate to matter, so $ n \geq 2$. The smallest integer which pairs up with 2 to yield a square is 7, so $ n \geq 7$. But we must have another mate for 7 other than 2, which forces $ n \geq 9$. Now 9 forces 16 and 16 forces 20. But if $ n \geq 20$ we must take care of 18, which pairs up with 7 to yield 25 but has no other potential partner until we get to 31.

Luckily, with $ n\equal{}31$ we no longer have any isolated numbers: the graph on $ \{1,\ldots,31\}$ where edges connect two numbers if they add up to a square has minimum degree 2, which is a trivial necessary condition for a Hamiltonian cycle. Checking if this graph actually has a Hamiltonian cycle is tedious, but if it doesn't one can hope that $ n\equal{}32$ or $ n\equal{}33$ will do better.

OK, so the answer is 32. As I mentioned before 31 is a lower bound. I was able to find the following near miss for $ n\equal{}31$:

5 31 18 7 29 20 16 9 27 22 3 13 12 4 21 28 8 17 19 30 6 10 26 23 2 14 11 25 24 1 15

That's a near miss because the first and last elements don't add up to a square, as required. I did this by hand because I was too lazy to program at this stage :blush: but after failing to find a full cycle I gave up a and turned to my trusty computer for help. It confirmed that there is no solution for $ n\equal{}31$, so unless my code is buggy there really isn't one.

However, for $ n\equal{}32$ there is a unique full cycle:

1 8 28 21 4 32 17 19 30 6 3 13 12 24 25 11 5 31 18 7 29 20 16 9 27 22 14 2 23 26 10 15


inequalities
symmetry
inequalities proposed

[quote="bibibibi"]Hi, apply Jensen:
$\sum \frac{1}{\sqrt{a^{2}+bc}}\leq 3\sqrt{\frac{3}{a^{2}+b^{2}+c^{2}+ab+bc+ca}}$[/quote]

I think you've got the sign reversed here

By AM-QM, we have
\[\sum_{cyc}\frac{1}{\sqrt{a^{2}+bc}}\leq 3\sqrt{\frac{ \sum_{cyc}\frac{1}{a^{2}+bc}}{3}}\]
so it's enough to prove that
\[\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab}\leq \frac{3(a+b+c)^{2}}{2(ab+bc+ca)^{2}}\]
which can be solved easyly but tediously by expanding, which will leave
\[\sum_{sym}a^{4}b^{3}c+4\sum_{cyc}a^{4}b^{2}c^{2}+3\sum_{cyc}a^{3}b^{3}c^{2}\leq 3\sum_{cyc}a^{6}bc+\sum_{sym}a^{5}b^{3}+4\sum_{cyc}a^{4}b^{4}. \]

Any more beautiful solutions? I'd like to see one.

There is one more solution, maybe. It's by SOS. The following inequality also holds:

\[{1 \over \sqrt{a^{2}+b^{2}}}+{1 \over \sqrt{b^{2}+c^{2}}}+{1 \over \sqrt{c^{2}+a^{2}}}\leq{3 \over \sqrt{2}}\cdot{a+b+c \over ab+bc+ca}. \]

[quote="Lovasz"]There is one more solution, maybe. It's by SOS.[/quote]
What about posting this SOS solution.

[quote="Lovasz"]
The following inequality also holds:
\[{1 \over \sqrt{a^{2}+b^{2}}}+{1 \over \sqrt{b^{2}+c^{2}}}+{1 \over \sqrt{c^{2}+a^{2}}}\leq{3 \over \sqrt{2}}\cdot{a+b+c \over ab+bc+ca}. \]
[/quote]
This is a trivial consequence of the easy inequality
\[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le\frac{3(a+b+c)}{2(ab+bc+ca)}\]
which was already mentioned at http://www.mathlinks.ro/Forum/viewtopic.php?t=111171

[quote="ductrung"][quote="Lovasz"]There is one more solution, maybe. It's by SOS.[/quote]
What about posting this SOS solution.[/quote]
I just rewrite it ina possible form:
\[\sum_{sym}{a^{2}+bc \over b^{2}+ca}+3 \leq{3(a+b+c)^{2}\over ab+bc+ca}. \]

[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[\frac{1}{\sqrt{a^{2}+bc}}+\frac{1}{\sqrt{b^{2}+ca}}+\frac{1}{\sqrt{c^{2}+ab}}\le\frac{3}{\sqrt{2}}\!\cdot\!\frac{a+b+c}{ab+bc+ca}\]
[/quote]
To prove this, we only need to prove that
\[\underset{cyc}{\sum}\frac{1}{\sqrt{a^{2}+bc}}\le \underset{cyc}{\sum}\frac{\sqrt{2}}{a+b}(1)\\ \underset{cyc}{\sum}\frac{\sqrt{2}}{a+b}\le \frac{3}{\sqrt{2}}.\frac{a+b+c}{ab+bc+ca}(2) \]
To prove (1), we only need to use Chebyshev. To prove (2), we only need AM - GM. :)

[quote="toanhocmuonmau"][quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{1}{\sqrt{a^{2}+bc}}+\frac{1}{\sqrt{b^{2}+ca}}+\frac{1}{\sqrt{c^{2}+ab}}\le\frac{3}{\sqrt{2}}\!\cdot\!\frac{a+b+c}{ab+bc+ca}\]
[/quote]
To prove this, we only need to prove that
\[ \underset{cyc}{\sum}\frac{1}{\sqrt{a^{2}+bc}}\le \underset{cyc}{\sum}\frac{\sqrt{2}}{a+b}(1)\\ \underset{cyc}{\sum}\frac{\sqrt{2}}{a+b}\le \frac{3}{\sqrt{2}}.\frac{a+b+c}{ab+bc+ca}(2) \]
To prove (1), we only need to use Chebyshev. To prove (2), we only need AM - GM. :)[/quote]

Dear Ductrung, is your solution similar to this one or more elementary?

This solution is OK but very hard to think  :(

Thank you very much.

[quote="manlio"]Dear Ductrung, is your solution similar to this one or more elementary?[/quote]
Hi Manlio,
I posted this version just for fun. The actual inequality I proved at that time is the one mentioned by Yimin Ge above, which now is just an easy corollary of [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=131247[/url].
My solution back then is very simple: split it into two inequalities
\[ \frac{a(b+c)}{a^{2}+bc}+\frac{b(c+a)}{b^{2}+ca}+\frac{c(a+b)}{c^{2}+ab}\le\frac{(a+b+c)^{2}}{ab+bc+ca}\]
and
\[ \frac{bc}{a^{2}+bc}+\frac{ca}{b^{2}+ca}+\frac{ab}{c^{2}+ab}\le\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\]
It is VERY easy to prove the above two inequalities. I am busy at the moment but will post my solutions soon (if nobody does that until then :))

[quote="ductrung"]My solution back then is very simple: split it into two inequalities
\[ \frac{a(b+c)}{a^{2}+bc}+\frac{b(c+a)}{b^{2}+ca}+\frac{c(a+b)}{c^{2}+ab}\le\frac{(a+b+c)^{2}}{ab+bc+ca}\]
and
\[ \frac{bc}{a^{2}+bc}+\frac{ca}{b^{2}+ca}+\frac{ab}{c^{2}+ab}\le\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\]
It is VERY easy to prove the above two inequalities. I am busy at the moment but will post my solutions soon (if nobody does that until then :))[/quote]

Very nice these two inequalities but I cannot solve them in concise manner :oops: using only Cauchy, AM-GM and Shur.

P.S. I like a lot all inequalities which can be solved in concise manner using only these three elementary inequalities

[hide="Hint for Manlio"]
\[ \frac{a(b+c)}{a^{2}+bc}+\frac{b(c+a)}{b^{2}+ca}+\frac{c(a+b)}{c^{2}+ab}\leq 3\leq \frac{(a+b+c)^{2}}{ab+bc+ca}\]
and
\[ \frac{bc}{a^{2}+bc}+\frac{ca}{b^{2}+ca}+\frac{ab}{c^{2}+ab}\leq\frac32\leq \frac{(a+b+c)^{2}}{2(ab+bc+ca)}\]
[/hide]

[hide=";)"]Vornicu-Schur for number 1...[/hide]

  Darij

[quote="manlio"]Thank you very much, Darij. :wink:

I will try.[/quote]

Dear Darij. I think both your inequalities are wrong?!

OMG, I should learn calculating...  :blush:

In fact, the first one is at least correct for a, b, c sides of a triangle.

  Darij

[quote="manlio"]
\[ \frac{bc}{a^{2}+bc}+\frac{ca}{b^{2}+ca}+\frac{ab}{c^{2}+ab}\le\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\]
Sol:

Write it as
$ \sum \frac{a^{2}}{a^{2}+bc}+\frac{(\sum a)^{2}}{2\sum ab}\geq 3$

Now use Cauchy

$ \sum \frac{a^{2}}{a^{2}+bc}\geq \frac{(\sum a)^{2}}{\sum ab+\sum a^{2}}$

to reduce our ineq at

$ \frac{(\sum a)^{2}}{\sum ab+\sum a^{2}}+\frac{(\sum a)^{2}}{2\sum ab}\geq 3$

Clearing denominators we are left with

$ 3(\sum ab)(\sum a)^{2}+(\sum a^{2})(\sum a)^{2}\geq 6(\sum ab)(\sum a^{2})+6(\sum ab)^{2}$

Now use $ (\sum a)^{2}\geq 3\sum ab$ and all is OK (I hope  :maybe: )

Dear Ductrung is your proof the same?

Thank you very much.

For the second inequality I tried to use Cauchy in this form

$ \frac{a(b+c)}{a^{2}+bc}= \frac{a(b+c)^{2}}{(b+c)(a^{2}+bc)}\leq \frac{ab^{2}}{b(a^{2}+c^{2})}+\frac{ac^{2}}{c(a^{2}+b^{2})}$

but unfortunately this trick doesn't work so I am  :(

Thank you very much for your patience with me.[/quote]

For your patience, I will try to prove nice problem of Duc ... although I'm really busy these days. Actually, I just prove the second result as Duc says. As Darij say, by generalized Schur theorem, we have certainly
\[ \sum \frac{a(b+c)}{a^{2}+bc}-3=\sum\frac{(a-b)(a-c)}{a^{2}+bc}\le 0, \]
if $ a,b,c$ are three side-lengths of a triangle. Otherwise, we assume that $ a\ge b+c\ge b\ge c$. The inequality becomes
\[ \frac{(a+b+c)^{2}}{ab+bc+ca}-3+\sum \frac{(a-b)(a-c)}{c^{2}+ab}\ge 0. \]
Now, since $ a\ge b\ge c$, we refer that
\[ (b-c)^{2}(ab+bc+ca) \le (b^{2}-bc)(ab+bc+ca) \le (b^{2}+ca)(c^{2}+ab). \]
Thus
\[ \sum \frac{(a-b)(a-c)}{c^{2}+ab}\ge \frac{(c-a)(c-b)}{c^{2}+ab}+\frac{(b-a)(b-c)}{c^{2}+ab}\]

\[ \ge \frac{(b-a)(c-b)}{c^{2}+ab}+\frac{(b-a)(b-c)}{c^{2}+ab}\]

\[ =-\frac{(a-b)(b-c)^{2}(a-b-c)}{(c^{2}+ab)(b^{2}+ac)}\ge-\frac{(b-c)^{2}(a-b)^{2}}{(c^{2}+ab)(b^{2}+ac)}\ge-\frac{(a-b)^{2}}{ab+bc+ca}. \]
The desired result follows by
\[ \frac{(a+b+c)^{2}}{ab+bc+ca}-3=\frac{(a-b)^{2}+(c-a)(c-b)}{ab+bc+ca}\ge \frac{(a-b)^{2}}{ab+bc+ca}. \]

Thank you very much, hungkhtn and ductrung for your very nice solutions. I am really very happy to study them. :)

[quote="ductrung"]For any positive real numbers $ x,y$ and $ z$,
\[ \frac {1}{\sqrt {x^{2} \plus{} yz}} \plus{} \frac {1}{\sqrt {y^{2} \plus{} zx}} \plus{} \frac {1}{\sqrt {z^{2} \plus{} xy}}\leq\frac {3(x \plus{} y \plus{} z)}{\sqrt {2}(yz \plus{} zx \plus{} xy)}.\][/quote]
[quote="Yimin Ge"]By AM-QM, we have
\[ \sum\frac {1}{\sqrt {x^{2} \plus{} yz}}\leq \sqrt {\sum\frac {3}{x^{2} \plus{} yz}}.\][/quote]
$ \frac {9(x \plus{} y \plus{} z)^2}{2(yz \plus{} zx \plus{} xy)^2} \minus{} \sum\frac {3}{x^{2} \plus{} yz}$

$ \equiv \frac {3F(x,y,z)}{2(x^{2} \plus{} yz)(y^{2} \plus{} zx)(z^{2} \plus{} xy)(yz \plus{} zx \plus{} xy)^2}$.

Without losing generality, we may assuming $ x \equal{} \min\{x,y,z\}$.

$ F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} t)$

$ \equal{} 48(s^2 \minus{} st \plus{} t^2)x^6 \plus{} (s \plus{} t)(82s^2 \minus{} 61st \plus{} 82t^2)x^5$

$ \plus{} (57s^4 \plus{} 91s^3t \plus{} 71s^2t^2 \plus{} 91st^3 \plus{} 57t^4)x^4$

$ \plus{} 2(s \plus{} t)(10s^4 \plus{} 22s^3t \plus{} 23s^2t^2 \plus{} 22st^3 \plus{} 10t^4)x^3$

$ \plus{} (3s^6 \plus{} 21s^5t \plus{} 32s^4t^2 \plus{} 73s^3t^3 \plus{} 32s^2t^4 \plus{} 21st^5 \plus{} 3t^6)x^2$

$ \plus{} st(s \plus{} t)(3s^4 \plus{} 20s^2t^2 \plus{} 3t^4)x \plus{} s^3t^3(s^2 \plus{} 4st \plus{} t^2)\geq 0$.

[quote="toanhocmuonmau"]
\[ \underset{cyc}{\sum}\frac {1}{\sqrt {a^{2} + bc}}\le \underset{cyc}{\sum}\frac {\sqrt {2}}{a + b}
\]
[/quote]
For along time, we haven't got solution for this problem yet. Now, I will post my solution
[b]Problem[/b] (Vo Quoc Ba Can) Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \underset{cyc}{\sum}\frac {1}{\sqrt {a^{2} + bc}}\le \underset{cyc}{\sum}\frac {\sqrt {2}}{a + b}.
\]
[i]Solution.[/i]
By the Cauchy schwarz Inequality, we have
\[{ (\sum \frac {1}{\sqrt {a^2 + bc}})^2 \le (\sum \frac {1}{(a + b)(a + c)})(\sum \frac {(a + b)(a + c)}{a^2 + bc})} = \frac {2\sum a}{(a + b)(b + c)(c + a)}(\sum \frac {a(b + c)}{a^2 + bc} + 3)
\]
It suffices to show that
\[ \frac {2\sum a}{(a + b)(b + c)(c + a)}(\sum \frac {a(b + c)}{a^2 + bc} + 3) \le 2( \sum \frac {1}{a + b})^2
\]

\[ \Leftrightarrow \sum \frac {a(b + c)}{a^2 + bc} + 3 \le \frac {( \sum a^2 + 3\sum ab)^2}{(a + b)(b + c)(c + a)\sum a}
\]

\[ \Leftrightarrow \sum \frac {a(b + c)}{a^2 + bc} - 3 \le \frac {( \sum a^4 - \sum a^2b^2}{(a + b)(b + c)(c + a)\sum a}
\]

\[ \Leftrightarrow \sum (a - b)(a - c)(\frac {1}{a^2 + bc} + \frac {1}{(b + c)(a + b + c)}) \ge 0
\]
Due to symmetry, we may assume $ a \ge b \ge c$, since $ a - c \ge \frac {a}{b}(b - c)$. It suffices to show that
\[ a(\frac {1}{a^2 + bc} + \frac {1}{(b + c)(a + b + c)}) \ge b(\frac {1}{b^2 + ca} + \frac {1}{(a + c)(a + b + c)})
\]

\[ \Leftrightarrow c(a^2 - b^2)[(a - b)^2 + ab + bc + ca] \ge 0
\]
which is trivial.
Equality holds if and only if $ a = b = c.$ :)


Find the minimal value of integer $ n$ that guarantees:
Among $ n$ sets, there exits at least three sets such that any of them does not include any other; or there exits at least three sets such that any two of them includes the other.

$\textbf{(A)}\ 4  \qquad\textbf{(B)}\ 5  \qquad\textbf{(C)}\ 6  \qquad\textbf{(D)}\ 7  \qquad\textbf{(E)}\ 8$

Let $ A \subset B$ , $ C \subset D$ ,  and $ A \cap C \equal{} \{\}$. Now on there are 2 subsets and 2 non-subsets. The set $ E$ can be a non-subset. At that situation $ A,C,E$ are the three sets such that any of them does not include any other. Other possibility is $ E \subset A$ (of course it is same for $ C$). So $ A$ and $ C$ will include $ E$. The answer is $ B$ which is $ 5$.

PS: The translation of this problem was very difficult because it is also diffucult to understand the Turkish text.

This is a special case of [url=http://mathworld.wolfram.com/DilworthsLemma.html]Dilworth's Lemma[/url]: every [[partially ordered set]] with $ mn \plus{} 1$ elements contains either a chain of length $ m \plus{} 1$ or an antichain of length $ n \plus{} 1$.  Here $ m \equal{} n \equal{} 2$, where the partial order is "containment as a subset."  (A chain is a sequence of elements that are pairwise comparable, here a sequence of three nested sets; an antichain is a sequence of elements that are pairwise incomparable, here a sequence of three elements such that none contains any other.)


Let $ a,b$ be potisive numbers(a<b) and p is a prime number.
      prove that:$ \large C_{pb}^{pa}$ $ \equiv$ $ \large C_{b}^{a}$($ mod$p)
                       $ \large C_{pb\plus{}r}^{pa}$ $ \equiv$ $ \large C_{pb}^{pa}$ ($ mod$p)
                         where $ r$ be natural number $ r$ $ \in$ (0,b).

Can you give the proof of $ Lucas theory$? :D and how applying it to solve this problem? :o

Write $ p,pa,pb,a,b,r$ in base p and apply this theory.


inequalities
quadratics
inequalities proposed

\[ \boxed {\ \{a,b,c\}\subset (0,\infty )\Longrightarrow\left(a+\frac ca\right)^2+\left(b+\frac cb\right)^2\ge \frac{\left[(a+b)^2+4c\right]^2}{2(a+b)^2}\ }\ .  \]

[b]Remark.[/b] Thanks, Argady ! I corrected.

\[ \boxed {\ \{a,b,c\}\subset (0,\infty )\Longrightarrow\left(a+\frac ca\right)^2+\left(b+\frac cb\right)^2\ge \frac{\left[(a+b)^2+4c\right]^2}{2(a+b)^2}\ }\ .  \]

Quadratic mean-AM: $(a+\frac{c}{a})^2+(b+\frac{c}{b})^2 \geq \frac{(a+b+\frac{c}{a}+\frac{c}{b})^2}{2}$

Also: $\frac{1}{a}+\frac{1}{b} \geq \frac{4}{a+b}$

Then we have: $(a+\frac{c}{a})^2+(b+\frac{c}{b})^2 \geq \frac{[(a+b) + \frac{4c}{a+b}]^2}{2}$ ; $QED$


geometry parallelogram perimeter trigonometry ratio

The perimeter of a parallelogram $ABCD$ is 40, and its altitudes are 4 and 7.  Compute $\sin A$.

[hide]
11/20

AD times one altitude = area = AB times the other altitude
AD:AB=4:7 (inverse of ratio between altitudes)
AD+AB+BC+CD = 2(AD+AB) = 40

let AD=4x, 22x=40, x = 20/11, AD = 80/11

SinA = 4 over 80/11 = 11/20

even if the diagram is constructed differently, sinA = sin (180-A), so it remains the same.

[/hide]


[quote="hasan4444"]If $ 2^{2x \minus{} 1} \minus{} 3^{x \plus{} \frac {1}{2}} \equal{} 3^{x \minus{} \frac {1}{2}} \minus{} 4^x$ then $ x \plus{} x^{ \minus{} 1} \equal{}$

$ \textbf{(A)}\ \frac {11}{6} \qquad \textbf{(B)}\ \frac {13}{6} \qquad \textbf{(C)}\ \frac {17}{6} \qquad \textbf{(D)}\ \frac {19}{6} \qquad \textbf{(E)}\ \frac {23}{6}$[/quote]

Let $ y\equal{}x\minus{}\frac{1}{2}$, then the equation
$ \Leftrightarrow 4^y\minus{}3(3^y)\equal{}3^y\minus{}2(4^y)$
$ \Leftrightarrow 3 \times 4^y\equal{}4 \times 3^y$
$ \Leftrightarrow (\frac{4}{3})^{y\minus{}1}\equal{}1$
$ \Leftrightarrow y\minus{}1\equal{}x\minus{}\frac{3}{2}\equal{}0$
$ \Rightarrow x\equal{}\frac{3}{2}$
$ \Rightarrow x\plus{}x^{\minus{}1}\equal{}\frac{13}{6}$

So the answer is $ \textbf{(B)}$


inequalities
integration
limit
triangle inequality
real analysis
real analysis unsolved

Let $ \mathcal{P}$ denote the set of partitions of $ [a,b]$.  Given a partition $ P\in \mathcal{P}$ of $ [a,b]$ into $ n$ subintervals with $ a\equal{}t_0<t_1<\dots <t_n\equal{}b$, let $ v(f,P) \equal{} \sum_{k \equal{} 1}^n |f(t_k) \minus{} f(t_{k \minus{} 1})|$.  By definition, $ T(f) \equal{} \sup_{P\in \mathcal{P}} v(f,P)$.  Furthermore, $ \int_a^b|f'| \equal{} \lim_{k\to \infty}v\left(f,P_k\right)$ for some sequence of partitions $ \{P_k\}$.  But $ \{P_k\}\subset \mathcal{P}$, so $ \int_a^b|f'|\leq T(f)$ by definition of supremum.

[quote="JoeBlow"] Furthermore, $ \int_a^b|f'| \equal{} \lim_{k\to \infty}v\left(f,P_k\right)$ for some sequence of partitions $ \{P_k\}$.  [/quote]
Can you justify this?

Use mean value theorem!
$ v(f,P_n) \equal{} \sum_{k \equal{} 1}^n |f(t_k) \minus{} f(t_{k \minus{} 1})| \equal{} \sum_{k \equal{} 1}^n |f'(c_k)|{}|t_k \minus{} t_{k \minus{} 1}|\rightarrow \int_a^b |f'|$, as $ n\rightarrow \infty$, where $ c_k\in (t_{k \minus{} 1},t_k)$.

Thank you Mr.Doe. Can you prove equality when $ f$ is absolutely continuous?

[quote="msaid"]Let $ f$ be of bounded variation on $ [a,b]$.

Show $ \int_a^b | f' | \leq T(f)$ where $ T(f)$ denotes the total variation of $ f$ on $ [a,b]$[/quote]

We know that $ f$ has bounded total variation on $ [a,b]$ iff $ f\equal{}g\minus{}h$  for $ g$ increasing and $ h$ decreasing.  Now, use the definition of total variation, use the triangle inequality somewhere, you can prove the problem


[quote="y-a-s-s-i-n-e"]$ a$ , $ b$ and $ c$ $ \in$ $ {\mathbb R}^ +$  , prove that :

$ \frac {1}{2}(\frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}) \geq \frac {a + b + c}{\sqrt [3]{abc}}$[/quote]

$ \leftrightarrow \frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}\ge \frac {2(a + b + c)}{\sqrt [3]{abc}}$

we have $ \sum \frac {a}{b} \ge \frac {a + b + c}{\sqrt [3]{abc}}(*)$

and $ \sum \frac {a}{c} \ge \frac {a + b + c}{\sqrt [3]{abc}}$

$  \frac {b + c}{a} + \frac {c + a}{b} + \frac {b + a}{c}\ge \frac {a + b + c}{\sqrt [3]{abc}} + \frac {a + b + c}{\sqrt [3]{abc}}\ge \frac {2(a + b + c)}{\sqrt [3]{abc}}$

we proven$ (*)$
we have
$ \frac {2a}{b} + \frac {b}{c}\ge 3\sqrt [3]{\frac {a^2}{bc}} = \frac {3a}{\sqrt [3]{abc}}$

analogue with another inequality

done!~ :)

[quote="y-a-s-s-i-n-e"]$ a$ , $ b$ and $ c$ $ \in$ $ {\mathbb R}^ \plus{}$  , prove that :

$ \frac {1}{2}(\frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {b \plus{} a}{c}) \geq \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$[/quote]
by Chebyshev : $ LHS \ge \frac{1}{3}(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}) (a\plus{}b\plus{}c)$ :)


hello,
I would like to have no number associated with Introduction an conclusion.
I want the same result as
\section*{Introduction}

Unfortunately, by doing this, "Introduction" doesn't appear in the \tableofcontents.

Is there a way to force it?
How do you cope with this problem?

Thanks a lot.

Before \chapter*{Introduction}, add
\addcontentsline{toc}{chapter}{Introduction}
Similarly,
\addcontentsline{toc}{chapter}{Conclusion}

SSN.

Thanks a lot!

Did you change your message? I saw something about titlesec !

Anyway, this solution works fine. I may try to use the titlesec package to improve my documents title and headers.