geometry
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Hi,

USACO Shaping Regions is another problem which I have been stuck on for a long time. I initially tried looking at the grid row by row (as suggested by somebody on the topcoder.com forums)but this times out for the 11th case. So I was wondering how to split the rectangles (the USACO site says that we should split the rectangles when an overlap occurs).

I've figured out the coordinates of where 2 rectangles will intersect, I just can't figure out how to make the different cases for the overlapping

Thanks in advance

Well look at the given hint very carefully. (it is already discussed in this forum)

when a rectangle intersects another than u can cut a rectangle into a few rectangular pieces.

--------
| +-+ |
| | - | |
| +-+ |
---------

Look at the picture (not good :mad: ) there is a rectangle inside bigger one. so can u break the bigger rectangle into some rectangles?

yes i understand that but how many different cases will there be for the overlaps

i am having trouble making the cases

thanks

well i assumed all the possible cases. But in official soln they did it very simply. i can't remember though. But well keep trying! if u can't then i may help u by reading that soln again! Or well, weeblie may help u as well!

You don't have to make one for all cases...

Just make the "generic" case with one rectangle completely inside theother (i.e. split to 9, or more correctly, 8 rects), and check if the "splitted" rectangles are valid or not (like "left x coord" < "right x coord", or something like that :P).

Check the geometry section of this article for how to split up a rectangle
http://www.topcoder.com/index?t=features&c=feat_010505

ok so let's say I write the general case and then split the inner rectangle (after eliminating the im-possible splitted rectangles, as weeblie said by checking left coord<right coord and upper coord>down coord (is this right?) )

now let's take the case where a part of the overlapping rectangle is OUTSIDE the rectangle under it.  once we split this overlapping rectangle we will get three rectangles outside the rectangle under the overlapping rectangle.

now do i pass these 3 rectangles to the intersect and split function recursively or what? becasue how do i take care of these rectangles if they overlap some other rectangle?

i hope what i am saying is clear, i'm going nuts thinking about this!!!

if possible can u also post the official soln so that i understand what u guys mean

thanks

Well, i am not claer about what u r saying! but well there are four solns of this problem at the official site. one is brute force "board coloring" with code. One with intersecting rectangles with out code. one with the faster way just keeping the boundary points in account with code. and another code without description. so sorry i can't send those! again my code used all the possible cases :blush: so it will not help u either. so either u think about some perticular test cases and tell us so that we can understand what u wanted to mean or u may think ur self with a particualr case.  :)

can you give a basic idea of the 'faster' solution where we just have to keep the boundary points in account?

thanks
p.s. - i finally solved usaco contact (using binary search)

Copied and paste:

There is another solution to this problem, which runs in O(n*n*log n) time, but is quite tricky. First, we add one big white rectangle at the bottom - the paper. Then we make two arrays: one containing all vertical edges of the rectangles, and the other the horizontal ones. For each edge we have its coordinates and remember, whether it's the left or right edge (top or bottom). We sort these edges from left to right and from top to bottom. Then we go from left to right (sweep), jumping to every x-coordinate of vertical edges. At each step we update the set of rectangles seen. We also want to update the amount of each color seen so far. So for each x we go from top to bottom, for each y updating the set of rectagles at a point (in the structure described below) and choosing the top one, so that we can update the amounts of colors seen.

The structure to hold the set of rectangles at a point should allow adding a rectangle (number from 1..1000), deleting a rectangle, and finding the top one. We can implement these operations in O(log n) time if we use a heap. To make adding and deleting run in O(log n) we must also have for each rectangle its position in the heap.

So the total time spent at each point is O(log n). Thus the algorithm works in O(n*n*log n) time.

hmm... that seems even more complicated
i guess i'll just go back to trying the intersection idea.

if it's not asking for too much can somebody mail me the code and description of the intersecting rectangles at sas_junk@yahoo.co.uk

I am sorry, I've been trying my best to solve this but I'm just not getting it

thanks

could you PM me your solution too?

I've had the same problem.

For this one, I basically just sat and drew out a LOT of rectangles and figured out every 'type' of overlapping possible (there are 16 ;) ).  Then I went back threw them all and wrote down their relationships (which will all be unique) between their upper right and lower left coordinates in each instance.  Then it just comes down to lots of if statements.  Yes, it takes a lot of manual labor, but if you use the info correctly, you wont time out.  Good luck!

actually the idea is same. but there is a problem for this kinda problem and that is it si very tought o debug it! :huh:

In few words you have ta have an if for each way than rectangles can overlap:
EG.
[code]else if(rec[c].lx>rec[i].lx&&rec[c].ly>rec[i].ly&&rec[c].lx<rec[i].ux&&rec[c].ly<rec[i].uy&&rec[c].ux>=rec[i].ux&&rec[c].uy>=rec[i].uy){
                     rec.push_back();
                     rec[rec.size()-1].lx=rec[i].lx;
                     rec[rec.size()-1].ly=rec[i].ly;
                     rec[rec.size()-1].ux=rec[c].lx;                            //               ----------
                     rec[rec.size()-1].uy=rec[i].uy;                           //     --------|           |
                       rec[rec.size()-1].color=rec[i].color;                  //    |          ----------
                     rec.push_back();                                        //     |_______|
                     rec[rec.size()-1].lx=rec[c].lx;                            //
                     rec[rec.size()-1].ly=rec[i].ly;                            // LOOK AT '='
                     rec[rec.size()-1].ux=rec[i].ux;                         //IT PACKS SOME CASES IN ONE ;-)
                     rec[rec.size()-1].uy=rec[c].ly;
                     rec[rec.size()-1].color=rec[i].color;
                }[/code]

i=rectangle under
c=rectangle above

This is only of other 16 cases. Scrap cases on the paper.
Keep coordinates of new rectangles and at the end simply compute their area.

And that's everything. Keep it simple ....


Solve for value of x:

$6^x + 3^x - 2^x=0$

Testing x=-1 yields a result, 3^x+6^x is increasing faster than 2^x, so x=-1 is the only solution.

Yes, you're right, $\displaystyle{\frac{1}{6}+\frac{1}{3}-\frac{1}{2}=0}$ , but is there a way to do this without just trying numbers?

I'm not sure, but I don't think logs really help.

I thought about factoring (like $(1-3^x)(1+2^x)=1$), but that doesn't help either... :|

$x$ must be negative since $2<3,6$
so $x=-a$ for some positive $a$
then
$2^{a}+1=3^{a}$
thus $x=-1$


Let $(x_1,y_1),(x_2,y_2),...,(x_n,y_n)$ be all distinct pairs of integers which satisfy the equation $(x-5)(x-77)=3^y$
Find the value of $\sum^{n}_{k=1}x_k$

Is the answer 82? If it is, then I'll post what I did... if not, then  :blush:  :blush:  :blush:

EDIT: Drat!! Seamus beat me to it...
at least we have the same answer :)


Suppose someone has the common cold.  Some symptoms, such as having a runny nose, are very common.  Other symptoms, such as having a sore back, are uncommon and rare.  Which is more likely: someone with the cold only having a sore back, or someone with the cold have both a sore back and a runny nose?

Are they independent?

If so, this essentially becomes: A person has a cold; is it likely or unlikely that he will have a runny nose?  Thus the probability of having the sore back with the runny nose is greater.

Yes, they are independent.

I agree that the probability of having the sore back with the runny nose is greater.  However, according to the book where i got the question, there is a greater probability of having just a sore back.  Could the book be wrong?

I think its the first one, but i cant really explain it mathematically..

maybe you are giving us probabilities that are too general or maybe the answer is really "have a sore back" with no mention of runny nose

does your book give the probability of having a runny nose?
$ \frac{2}{5}$ is considered common by some u know

The book doesn't give the exact probabilities, but it is safe to assume that the probability of having a runny nose is greater than 1/2 and the probability of having a sore back is less than 1/2.

If it says a sore back, does it not include/exclude the runny nose?  If so, having a sore back would have a greater probability than having a sore back plus another condition, since the probability of the condition is not 1.


Prove that the trisectors of an angle of a triangle can never trisect the opposite side.

[hide="solution"]In $\triangle{ABC}$, let points $D$ and $E$ be on $BC$ such that $AD$ and $AE$ trisect $\angle{BAC}$.

Assume the angle trisectors also trisect $BC$. From the givens, $AD$ bisects $\angle BAE$. Because $BD=DE$, from the angle bisector theorem, $AB=AE$. Then $\triangle ABD=\triangle AED$, and it follows that $AD\perp BC$. Similarly, it can be shown that $\triangle AEC=\triangle AED$, and it follows that $AE\perp BC$. But then $\triangle ABC$ has two distinct altitudes from $A$, a contradiction.[/hide]

[hide]proof by contradiction:  let the triangle be ABC with D and E on base BC.  we have <BAD=<DAE=<EAC and BD=DE=EC.  by the angle bisector theorem, AB/1=AE/1, so AB=AE.  we can do the same thing to prove AD=AC.  now <ABE=<AEB=x and <ACD=<ADC=y.  also, call the trisection of the top angle z.  using the sum of angles in ABD, we have x+180-y+z=180, so x-y+z=0.  we can do the same with ACE to get y-x+z=0.  adding these equations together gives 2z=0, so z=0:
[b]QED[/b] :D
[/hide]


integration
calculus
probability
expected value
probability and stats

Suppose I choose two points from a (1-dimensional) Gaussian distribution with mean 0 and standard deviation $\sigma$. What will be the expected value of the distance between the two points?

I am thinking it is $\frac{1}{2\pi\sigma^{2}}\int_{-\infty}^\infty \int_{-\infty}^\infty e^{(-x_{1}^{2}-x_{2}^{2})/2\sigma^{2}}|x_{1}-x_{2}|\textrm{d}x_{1}\textrm{d}x_{2}$, but I would like to simplify this (into one integral) in terms of $d = |x_{1}-x_{2}|$.

Here's the simplification: the difference of two independent normals has the same distribution as the sum, and is a normal with twice the variance.
You're looking at $\frac1{\sqrt{2\pi}}\cdot\frac1{\sqrt{2}\sigma}\int_{-\infty}^{\infty}e^{-x^{2}/(4\sigma^{2})}\cdot |x|\,dx= \frac1{\sigma\sqrt{\pi}}\int_{0}^{\infty}xe^{-x^{2}/(4\sigma^{2})}\,dx$
$=\frac1{\sigma\sqrt{\pi}}\left[-2\sigma^{2}e^{-x^{2}/(4\sigma^{2})}\right]_{0}^{\infty}=\frac{2\sigma}{\sqrt{\pi}}$


A triangle $\bigtriangleup ABC$ is given and let $AD$ be, its median and $E,$ $F,$ two point on $AC,$ $AB$ respectively,
such that $AF = 5(BF)$ and $CE = 3(AE).$

We denote as $L,$ the symmetric point of $A,$ with respect to the midpoint $N$ of $EF$ and the points $P\equiv BE\cap CF,$ $K\equiv BC\cap AP.$

Prove that $LM = 2(KL),$ where $M\equiv AD\cap KL.$

Kostas Vittas.

PS. The general problem states that for any variable triangle $\bigtriangleup ABC,$ whose its side segments $AB,$ $AC$ are constant,
then for $\frac{AE}{CE}= p$ and $\frac{AF}{BF}=q$ $\Longrightarrow$ $\frac{KL}{LM}= constant.$

[color=darkblue][b]Nice problem, Vittasko ![/b] Generally, $\boxed{\ \left\{\begin{array}{c}\overline{EA}=p\cdot\overline{EC}\\\ \overline{FA}=q\cdot\overline{FB}\end{array}\right\|\implies\frac{\overline{LK}}{\overline{LM}}=\frac{1+pq}{p+q}\ }$.

[b]Thank you for it ![/b] I"ll come back soon.[/color]

$\{\begin{array}{c}\boxed{\ \triangle ABC\ }\\\\ D\in BC\ ,\ \overline{BD}+\overline{DC}=0\\\\ E\in AC\ ,\ \overline{EA}=p\cdot \overline{EC}\\\\ F\in AB\ ,\ \overline{FA}=q\cdot \overline{FB}\end{array}$ $\|\begin{array}{c}LE\parallel AB\ ,\ LF\parallel AC\\\\ P\in BE\cap CF\\\\ K\in AP\cap BC\\\\ M\in AD\cap KL\end{array}\}$ $\implies$ $\boxed{\ \frac{\overline{LK}}{\overline{LM}}=\frac{1+pq}{p+q}\ }$.

[color=darkblue][b][u]Proof.[/u][/b] I"ll use the vectorial system with the origin $A$, i.e. for any point $X$ of plane, $\overrightarrow{AX}=X$. Thus, $2D=B+C$ and

$\{\begin{array}{c}(1-p)E=-pC\\\\ (1-q)F=-qB\end{array}$. Apply the Ceva's theorem : $\frac{\overline{KB}}{\overline{KC}}\cdot\frac{\overline{EC}}{\overline{EA}}\cdot\frac{\overline{FA}}{\overline{FB}}=-1$ $\implies$ $\frac{\overline{KB}}{\overline{KC}}=-\frac{p}{q}$ $\implies$ $\boxed{(p+q)K=qB+pC}$.

Observe that $L=E+F\implies$ $\boxed{(1-p)(1-q)L=-q(1-p)B-p(1-q)C}\ \ (*)$ and exists $\lambda\in\mathcal R$ so that $M=\lambda (B+C)$.

$M\in KL$ $\Longleftrightarrow$ $\|\begin{array}{ccc}q & p & p+q\\\\--q(1-p) &-p(1-q) & (1-p)(1-q)\\\\ \lambda & \lambda & 1\end{array}\|=0$ $\Longleftrightarrow$ $\lambda=\frac{pq}{1+pq}$, i.e. $\boxed{(1+pq)M=pq(B+C)}$.

From the relation $(*)$ obtain : $(1-p)(1-q)L=pq(B+C)-(qB+pC)$, i.e. $\boxed{(1+pq)M=(1-p)(1-q)L+(p+q)K}$,

where $(1-p)(1-q)+(p+q)=(1+pq)$. In conclusion, $\boxed{\boxed{\frac{\overline{LK}}{\overline{LM}}=\frac{1+pq}{p+q}}}$.[/color]

[quote="Virgil Nicula"][color=darkblue][b]Nice problem, Vittasko ![/b] Generally, $\boxed{\ \left\{\begin{array}{c}\overline{EA}=p\cdot\overline{EC}\\\ \overline{FA}=q\cdot\overline{FB}\end{array}\right\|\implies\frac{\overline{LK}}{\overline{LM}}=\frac{1+pq}{p+q}\ }$.
 [/color][/quote]

Very good. It is exactly the wanted conclusion and it is not necessary the notation of the constant side segments $AB,$ $AC,$
when the ratios $p,$ $q,$ are known.

The solution I have in mind, is based on [b][size=100]Davidd's[/size][/b] problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=149143
and I will post it here next week.

Best regards, Kostas Vittas.


Find $ n$ such that $ 2^3\cdot3^2\cdot n\equal{}10!$.

We can factor $ 10!$ into $ 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\equal{}2^8\cdot3^4\cdot5^2\cdot7$.
Dividing by $ 2^3\cdot3^2$, we have $ 2^5\cdot3^2\cdot5^2\cdot7\equal{}\boxed{50400}$.
Notice that $ 2^3\cdot3^2\equal{}8\cdot9$, so $ \frac{10!}{8\cdot9}\equal{}10\cdot7!\equal{}\boxed{50400}$.

so 10! means 1*2*3*4*5*6*7*8*9*10?

just wonderin thanks! :D

Yes. In general for a positive integer $ n$, $ n!=n(n-1)(n-2)\dots\(2)(1)$. (Note that $ 0!=1$.)


inequalities
quadratics
inequalities proposed

If a,b,c>0 and abc=1 then sum a/b^2(c+1) >= 3/2. :)

[quote="hardsoul"]Really easy:

Put $a=y/x,b=z/y,c=x/z,x+y+z=1,f(t)=\frac{t^3}{1-t}$ obviously $f-$  is increasing, $\frac{1}{x}-$ is decreasing  so by rearrengement we have $\sum_{cyc}\frac{f(x)}{y^2}\ge \sum_{cyc}\frac{f(x)}{x^2}=\sum_{cyc}\frac{x}{y+z}\ge \frac{3}{2}$.[/quote]

i obtain $\sum \frac{f(x)}{zy}$. is $\sum \frac{f(x)}{zy} \geq \sum \frac{f(x)}{y^2}$ true?

anyway i have the brute force soln:

Let $x,y,z$ be positive reals such that $a=\dfrac{x}{y}$, $b=\dfrac{y}{z}$ and $c=\dfrac{z}{x}$.

$\Leftrightarrow \sum\limits_{cyc} \dfrac{\dfrac{x}{y}}{\dfrac{y^2}{z^2} \left( \dfrac{z}{x}+1 \right)} \geq \dfrac{3}{2}$

$\Leftrightarrow \sum\limits_{cyc} \dfrac{\dfrac{x}{y}}{\dfrac{y^2}{z^2} \cdot \dfrac{z+x}{x}} \geq \dfrac{3}{2}$

$\Leftrightarrow \sum\limits_{cyc} \dfrac{x}{y} \cdot \dfrac{z^2}{y^2} \cdot \dfrac{x}{z+x} \geq \dfrac{3}{2}$

$\Leftrightarrow \sum\limits_{cyc} \dfrac{x^2z^2}{y^3(z+x)} \geq \dfrac{3}{2}$

$\Leftrightarrow \sum\limits_{cyc} \dfrac{x^5z^5(x+y)(y+z)}{x^3y^3z^3(x+y)(y+z)(z+x)} \geq \dfrac{3}{2}$

$\Leftrightarrow 2 \sum\limits_{cyc} x^5z^5(xy+xz+y^2+yz) \geq 3x^3y^3z^3(2xyz+x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)$

$\Leftrightarrow 2 \sum\limits_{cyc} x^6yz^5 + 2 \sum\limits_{cyc} x^6z^6 + 2 \sum\limits_{cyc} x^5y^2z^5 + 2 \sum\limits_{cyc} x^5yz^6 \geq 6x^4y^4z^4 + 3x^3y^3z^3 \sum\limits_{sym} x^2y$

$\Leftrightarrow 2 \sum\limits_{sym} x^6yz^5 + \sum\limits_{sym} x^6z^6 + \sum\limits_{sym} x^5y^2z^5 \geq \sum\limits_{sym} x^4y^4z^4 + 3\sum\limits_{sym} x^5y^4z^3$

Since $(6,6,0) \succ (4,4,4)$, $(6,5,1) \succ (5,4,3)$ and $(5,5,2) \succ (5,4,3)$, by Muirhead, the inequality above is true.

After $\sum_{cyc} \dfrac{x^2z^2}{y^3(z+x)} \geq \dfrac{3}{2} $ i continued using Cauchy:
$\sum_{cyc} \dfrac{x^2z^2}{y^3(z+x)}\cdot \sum_{cyc} y(x+z)\geq \left(\sum_{cyc} \dfrac{xy}{z}\right)^2$
Now we will prove that
$\sum_{cyc} \dfrac{x^2z^2}{y^3(z+x)}\geq \dfrac{\left(\sum_{cyc} \dfrac{xy}{z}\right)^2}{\sum_{cyc} y(x+z)}\geq\dfrac{3}{2}$
It's equivalent to:
$(x^2y^2+y^2z^2+z^2x^2)^2\geq 3x^2y^2z^2(xy+yz+zx)$
An it flows from Muirhead, cause $(4,4,0),(4,2,2) \succ (3,3,2)$.

anyway, is there any solution not using Muirhead?

Hmm, it depends...
In my solution at the end you can just use ineqs:
$(x^2y^2+y^2z^2+z^2x^2)^2\geq \frac{(xy+yz+zx)^4}{9}$
$\frac{(xy+yz+zx)^3}{9}\geq 3x^2y^2z^2$
They are just ineqs between arithetic, geometric and quadratic means ;)

If $a,b,c>0$ and $abc=1$ then $\sum \dfrac{a(c+1)}{b^2} \geq 6$.

and yes, it is more trivial than the first one.

This is my problem from the RMNO 2005 shortlist.

[quote="ciprian"]If $a,b,c>0$ and $abc=1$ then $sum a/b^2(c+1) >= 3/2. :)$[/quote]

I have for this one two solutions. One of them is already posted,but the second one is direct.
 It follows from the substitution $a=x^2/yz$.

[tex]\displaystyle \sum \dfrac{a(c+1)}{b^2} \geq 6[/tex]
It is equivalent to:
[tex]\displaystyle \sum \dfrac{ac}{b^2} + \sum \dfrac{a}{b^2} \geq 6[/tex]

The rest follows by applying AM-GM inequality to each sum separately and then adding them up concludes the proof.
Isn't it?

I have a nice solution:

Put $ a = \frac{x}{y} , b= \frac{y}{z}, c=\frac{z}{x} $.
We need prove:
$\frac{x^3}{yz(y+z)} + \frac{y^3}{xz(x+z)} + \frac{z^3}{xy(x+y)} \geq \frac{3}{2}$
We have: $x^3 + y^3 \geq xy(x+y)$
=> $\frac{x^3}{yz(y+z)} + \frac{y^3}{xz(x+z)} + \frac{z^3}{xy(x+y)} \geq  \frac{x^3}{y^3+z^3} + \frac{y^3}{x^3+z^3} + \frac{z^3}{x^3+y^3} \geq \frac{3}{2}$.
This is Nesbit inequalities.


Vieta
inequalities
algebra
binomial theorem

Find the pairs of the constants $ (a,\ b,\ c,\ d)$ such that all roots of $ x^6\minus{}6x^5\plus{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}1\equal{}0$ are positive.

[hide]
Let $ x_1, x_2, \dots, x_6$ be the roots of the equation. Vieta's gives us

$ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6 \equal{} 6 \implies \frac {x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \plus{} x_6}{6} \equal{} 1$
$ x_1x_2x_3x_4x_5x_6 \equal{} 1 \implies \sqrt [6]{x_1x_2x_3x_4x_5x_6} \equal{} 1$

However, by the AM-GM inequality we have that the arithmetic and the geometric mean of 6 positive numbers are only equal if the numbers themselves are equal. From here on it's easy, as we have $ x_1 \equal{} x_2 \equal{} \dots \equal{} x_6 \equal{} 1$.

To get the coefficients, a simple way would be to use the binomial theorem to get

$ x^6 \minus{} \binom61x^5 \plus{} \binom62x^4 \minus{} \binom63x^3 \plus{} \binom64x^2 \minus{} \binom65x \plus{} \binom66 \equal{} 0$

or

$ x^6 \minus{} 6x^5 \plus{} 15x^4 \minus{} 20x^3 \plus{} 15x^2 \minus{} 6x \plus{} 1 \equal{} 0$

Therefore, the only solution is $ (a, b, c, d) \equal{} (15, \minus{} 20, 15, \minus{} 6)$
[/hide]


On the blogs page, there are two page navigation lines at the top, but no page navigation thingies at the bottom.

Is this a bug?

Also, is it only on the MathLinks version? I believe so.

It's most likely caused by scrolling to fast down.
Do you use the middle mouse button, arrow keys, or the scrollbar?

It happens without scrolling, too.  :wink:

I think that it's an issue of a misplaced block of code; the second page navigation bar should be at the bottom of the page.

No... I think it's a messed up Java Script code, IE can Misinterpret things.
If you like perfection, I would recommend Firefox, it'll probalbly show it right, If you enjoy overall appearance, stick with IE. (Unless you already use Firefox?) :huh:

I use Google Chrome; does it work on Firefox?

I haven't tried Google Chrome yet, but as Google's mainly search, I don't know about its software abilities...  I wouldn't download a different browser just for that bug though, especially if it could be the webpage, I just gave recommendation for those if you ever considered switching.  Just don't download things rashly, it slows down computer an awful lot. (If you do  to much :wink:)

one of them is supposed to be on top of your blog, but if you don't have one, it duplicates...

it looks different on my computer because i have a blog, but that thing has always been messed up anyway...

[quote="levans"]I'm not really sure if this is a bug.

Basically, there is one line of navigation, then a listing of the blogs you own, then the navigation line again. If you have no blogs, they'll probably appear directly next to each other. Although I agree this doesn't make much sense. I'll look into having the second line moved to the bottom.[/quote]

I notice the word [i]blogs[/i]. Am I too confused or did I find somewhere that you can only have one blog? If you can have multiple blogs, how do you make more?

[quote="r15s11z55y89w21"]I notice the word [i]blogs[/i]. Am I too confused or did I find somewhere that you can only have one blog? If you can have multiple blogs, how do you make more?[/quote]
Only one active blog. Old deleted blogs also show up.

Besides, would you want to manage several blogs?  This ways better, also, it wouldcause spam cause people would spam on their blogs and if they were deleted by the admins, they use their old one.


trigonometry
real analysis
real analysis unsolved

IS it ((PI^2)/16)-1/2?i just multiplied and divided the whole thing by 4,and then use that 1/(1^2)+1/(3^2)+1/(5^2)+...=(Pi^2)/8 :D

Using complex variables I get the sum is equal to: $ \frac{\pi^{2}}{16}\minus{}\frac{1}{2}$ as well...

If $ S\equal{}\sum_{n\equal{}1}^{\infty}\frac{1}{(2n\minus{}1)^{2}(2n\plus{}1)^{2}}$, then we get $ 2S\plus{}1\minus{}\frac{\pi^{2}}{16}\minus{}\frac{\pi^{2}}{16}\equal{}0$.

To get that I used:  $ f(z)\equal{}\frac{\pi\cot\pi z}{(2z\minus{}1)^{2}(2z\plus{}1)^{2}}$ over the square contour with vertices at $ (N\plus{}\frac{1}{2})(\pm 1\pm i)$ and caculating the residues at $ z\equal{}\frac{1}{2}$ and $ z\equal{}\minus{}\frac{1}{2}$ to be equal to $ \minus{}\frac{\pi^{2}}{16}$.

complex variables is overkill, i myself used trig, but anyway, both methods are cool. :D
so, how about the sums

$ \frac{1}{1^{2}\cdot 3^{2}\cdot 5^{2}}\plus{}\frac{1}{3^{2}\cdot 5^{2}\cdot 7^{2}}\plus{}\frac{1}{5^{2}\cdot 7^{2}\cdot 9^{2}}\plus{}\frac{1}{7^{2}\cdot 9^{2}\cdot 11^{2}}\plus{}\frac{1}{9^{2}\cdot 11^{2}\cdot 13^{2}}\plus{}\ldots$

and

$ \frac{1}{1^{2}\cdot 3^{2}\cdot 5^{2}\cdot 7^{2}}\plus{}\frac{1}{3^{2}\cdot 5^{2}\cdot 7^{2}\cdot 9^{2}}\plus{}\frac{1}{5^{2}\cdot 7^{2}\cdot 9^{2}\cdot 11^{2}}\plus{}\frac{1}{7^{2}\cdot 9^{2}\cdot 11^{2}\cdot 13^{2}}\plus{}\ldots$

can you see a pattern emerging?

$ \sum_{k\equal{}0}^{\infty}\frac{1}{(2k\plus{}1)^{2}\cdots (2k\plus{}2n\plus{}1)^{2}}\equal{}\frac{1}{n!^{2}}\frac{{2n\choose n}}{2^{2n}}\left(\frac{\pi^{2}}{8}\minus{}\frac{1}4\sum_{k\equal{}1}^{n}\frac{(3k\minus{}1)\, 2^{4k}}{k^{3}\,{2k\choose k}^{3}}\right)$ could be the pattern.

Try to prove it!  :wink:

Or try this one: $ \sum_{k\equal{}1}^{\infty}\frac{1}{k^{2}\cdots (k\plus{}n)^{2}}\equal{}\frac{1}{n!^{2}}{2n\choose n}\left(\frac{\pi^{2}}{6}\minus{}3\sum_{k\equal{}1}^{n}\frac{1}{k^{2}{2k\choose k}}\right)$


I cannot prove this -->   $ \frac {1}{2} \minus{} \frac {1}{3} \plus{} \frac {1}{4} \minus{} ... \minus{} \frac {1}{99} \plus{} \frac {1}{100} > \frac {1}{5}$ :blush:  :blush:


find the real  number [b]m[/b]
\[ \therefore
\]
the common point D  of the lines
\[ \left( {m \minus{} 6} \right)x \plus{} \left( {m \minus{} 1} \right)y \plus{} 13 \equal{} 0
\]
 and
\[ \left( {2m \plus{} 1} \right)x \plus{} 3my \minus{} 9 \equal{} 0\]
\[ \therefore
\]

D
\[ \in
\]
of the line
\[ 2x \plus{} y \minus{} 5 \equal{} 0
\]

Just treat the three equations as a system and solve for the 3 variables.

I solved the equations as a system but i didn't come to the right solution.
is it any other way to  solve the problem or only with system of equations  ?

i also have to say that i'm not so good in solving systems of equations  :blush:

[quote="grn_trtle"]So he is asking for a number $ m$ that makes the three lines concur?[/quote]
Yes. Interpret it however you want. Basically, just solve a system of three equations.

There's two solution sets:
\[ \{m,x,y\}\equal{}\{\frac{14}{5},\frac{55}{17},\minus{}\frac{25}{17}\}, \{2,3,\minus{}1\}.\]

[quote="123456789"][quote="grn_trtle"]So he is asking for a number $ m$ that makes the three lines concur?[/quote]
Yes. Interpret it however you want. Basically, just solve a system of three equations.

There's two solution sets:
\[ \{m,x,y\} \equal{} \{\frac {14}{5},\frac {55}{17}, \minus{} \frac {25}{17}\}, \{2,3, \minus{} 1\}.
\]
[/quote]

thanks a lot "123456789"


Ba Salam

Yek soale khob

Agar $ x,y,z\ge 0$ va $ x^2\plus{}y^2\plus{}z^2\equal{}xyz$ Sabet konid
\[ xy\plus{}yz\plus{}zx\ge 2(x\plus{}y\plus{}z)\plus{}9\]

ba estefade az namosavie tavafoghi be rahati natije mishavad:\[ x \plus{} y \plus{} z \ge 9\]

hamchenin ba couchi darim:\[ \begin{array}{l} (x^2  \plus{} y^2  \plus{} z^2 )(y^2  \plus{} z^2  \plus{} x^2 ) \ge (xy \plus{} yz \plus{} zx)^2  \\ \Rightarrow xy \plus{} yz \plus{} zx \le xyz \\  \end{array}\]

hal ba estefade az namosavi rooberoo ke esbatash sade ast darim:\[ xy \plus{} yz \plus{} zx \ge \sqrt {3xyz(x \plus{} y \plus{} z)} \]

\[ \Rightarrow xy \plus{} yz \plus{} zx \ge \sqrt {3(xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z)} \]
\[ \Rightarrow xy \plus{} yz \plus{} zx \ge 3(x \plus{} y \plus{} z)\]

pas khahim dasht:\[ xy \plus{} yz \plus{} zx \ge 2(x \plus{} y \plus{} z) \plus{} (x \plus{} y \plus{} z) \Rightarrow xy \plus{} yz \plus{} zx \ge 2(x \plus{} y \plus{} z) \plus{} 9\]

rahe hale man(albate fek konam balaei behtare):
ba couchi darim:
$ (xy\plus{}yz\plus{}zx)(z\plus{}x\plus{}y)\geq 9xyz$
$ xy\plus{}yz\plus{}zx\geq 2(x\plus{}y\plus{}z)\plus{}9
\\\Leftrightarrow\frac{9xyz}{z\plus{}x\plus{}y}\geq 2(x\plus{}y\plus{}z)\plus{}9$
ba tavajoh be in namosavi $ {x^2}\plus{}{y^2}\plus{}{z^2}\geq xy\plus{}yz\plus{}zx$
va inke $ {x^2}\plus{}{y^2}\plus{}{z^2}\equal{}xyz$ khahim dasht: $ {x^2}\plus{}{y^2}\plus{}{z^2}\geq 3(x\plus{}y\plus{}z)$
baraye esbate in ham az couchi komak migirim va inke:
 $ {x^2}\plus{}{y^2}\plus{}{z^2}\geq xy\plus{}yz\plus{}zx$
 pas $ {x^2}\plus{}{y^2}\plus{}{z^2}\geq 3(x\plus{}y\plus{}z) \Leftrightarrow {(xy\plus{}yz\plus{}zx)^2}\geq 3xyz(x\plus{}y\plus{}z)  \Leftrightarrow {x^2y^2\plus{}y^2z^2\plus{}z^2y^2}\geq {x^2yz\plus{}y^2zx\plus{}z^2yx}$
ke inam baz ba hamon namosavi balaei hale.
manam 1 soal mizaram:
agar $ x\plus{}y\plus{}z\equal{}3$ bashad sabet konid: $ (\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})\geq{xy\plus{}yz\plus{}zx}$

$ (\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})\geq{xy\plus{}yz\plus{}zx}\leftrightarrow x^2\plus{}y^2\plus{}z^2\plus{}2(\sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z})\ge 9$

Vali midanim $ \sqrt{x}(\sqrt{x}\plus{}2)(\sqrt{x}\minus{}1)^2\ge 0$ Pas $ x^2\plus{}2\sqrt{x}\ge 3x$
va $ 3x\plus{}3y\plus{}3z\equal{}9$ :wink:


The Texas forum has been split into a public forum, for the general population, and a private forum, for Texans interested in ARML.

Anyone interested in joining the foremetioned private Texas forum should private message me or post in this thread. Thanks!

[quote="ashwinrk_jain"]Is there any charges for joining[/quote]
No, of course not. We have the same kind of arrangement for the Southern California ARML team, with a private, hidden forum. It's still part of AoPS. All you have to do is convince the moderator that you belong. He can admit you.

(Thanks to Max300):

For those Texans who are already in the group, you may have to log out and then log back in to access the private forum. Sorry for the troubles! :(

silly question...but... could someone briefly tell me what ARML is?

also, where's the public forum

[quote="ARML Contests"]
ARML is like no other mathematics contest. After months of planning and preparations, tryouts, and practice sessions, busloads of students stream onto three college campuses, turning sedate institutions into beehives of excitement and anticipation. New friendships are made; old ones are renewed. Superb mathematics students from across the country are drawn together by their love of mathematics, eager to measure their abilities against other talented students as well as against a collection of truly challenging non-routine problems. ARML is distinguished by the fact that it creates communities of mathematics students and teachers. ARML's format provides for a variety of problem-solving situations and the fine problems that distinguish ARML provoke and promote mathematical insight and inventiveness.[/quote]
Interested yet? Check out:
http://www.arml.com/
http://www.artofproblemsolving.com/Forum/index.php?f=348

By the way, this is the Texas public forum. The private forum is located at
http://www.artofproblemsolving.com/Forum/index.php?f=178

I've added you to the group, so you should be able to see it now.

Oh yeah, I remember your telling me at camp. You've been added to the usergroup, so you should be able to see the forum now.

[quote="sapphyre571"]ok I was invited to ARML this year
I just couldnt go[/quote]

Just wondering...who are you?

The private forum has been dead for over a month now....

ARML is over and people are at summer camps, but I guess I can post some problems to keep you busy. ;)

[quote="fireemblem13"]lol... ur one of the first ppl from aops that i know in person. i dont know many ppl....
[quote]
Just wondering...who are you?

The private forum has been dead for over a month now....[/quote]
who are you... just wondering[/quote]

My name's Akash. I go to Klein High Skew.

How can I access the private forum? (Like do I have to be on the ARML team before or something?)
EDIT: Actually, I just read your first post. =) Can I join the private forum? (I'm pretty sure you know who I am... =P)

I've added you to the usergroup, Maxima.

Can you add me? I was unable to attend ARML this year but I did in 2010 and was in 2011 Tx nats team for mathcounts.

Yeah sure.  You should be able to see it now.

Where's the private forum, or is this the private forum?

You can view the private forum at:
http://www.artofproblemsolving.com/Forum/viewforum.php?f=178

Can you add me? I went to the 2011 ARML.

Sure! You should be able to access the forum now.

Has this forum been taken down? I get the error message "Information You are not authorized to read this forum."

i think texas died a while ago

nothing has been posted in the past 2+ years in the private forum and nothing posted in this forum for the past 2 months.

This whole forum barely has any activity. There is sometimes no point in posting anything.

I'm not actually sure if that forum exists anymore, as it is no longer in the listing of states.


Hi. I have this "easy" problem with rank of a matrix. In MatLab I have this:
[b]A =

     1     1     2     2
     2     3     7     6
     1     2     5     4
     1     0     1     2
     2     1     0     1

>> rank(A)

ans =

     3[/b]
But I dont understand how the rank is 3, cuz all the columns are linear independet, right? shouldnt it be 4?
sorry for the noob question, but i still didnt get how to know the rank of a matrix... thks in advance

StinkyCat

I think you're falling into a trap common to students in a first linear algebra course - you haven't yet fully internalized the definition of independent. Yes, you can tell at a glance if a set of two vectors is independent. All you have to check there is whether they point in the same direction (are scalar multiples of each other). But once you're past two vectors, you can't tell at a glance. These 4 columns do point in 4 different directions, but that most emphatically does not tell you that they're independent. And you can't tell without computation.

I did Gaussian elimination in that matrix by hand (with several pauses for arithmetic debugging) and arrived at a row echelon form with three pivots (and two zero rows accross the bottom.) The rank of this matrix [b]is[/b] 3. MATLAB is telling you the truth.

That row elimination also tells me what the solutions are to the homogeneous equation $ Ax\equal{}0$. We get scalar multiples of $ [1\ \minus{}1\quad 1\ \minus{}1]^T.$

That means that if we write the four columns of this matrix as $ a,b,c,d,$ we have $ a\minus{}b\plus{}c\minus{}d\equal{}0.$ Now tell me: what is the precise mathematical definition for a set of vectors to be independent?


calculus
integration
geometry
3D geometry
calculus computations

Can somone draw the volume of the solid using disks:

Intergral of -x+1 from 0 to 1

Thanx

And also Im planning to draw the figure of the disks on the poster board. Im planning to cut 15 circles too stack on top of each other and decrease the radius all the way close to zero. My graph is 10 cm apart from 0 to 1. So can some one list the radiuses of the 15 circles.

[quote="DVO"]$\int_0^1-x+1dx$ ...  That's an area, not a volume.

Did you try to figure out what those radiuses should be?  How'd that go?[/quote]

If I'm correct to find the volume it should be 2pi times the integral above... I cant remember at the moment

That's not quite right, it's a little more complicated than that.

This hasn't been stated, but I assume the volume we are discussing is the volume of the solid that you would get if you took the graph of $-x+1$ from $0$ to $1$ and revolved it around the x-axis.

That volume could be approximated using cylinders.  It might be a good exercise to actually draw the picture and compute the approximation that you get when using, say, five cylinders.

The approximation that you get in this way will look just like a Riemann sum.  If we repeat this approximation process, using more and more cylinders, then the Riemann sums we come up with will approach a certain integral.  Thinking about it this way, you can figure out what that integral must be.

Of course, the solid in question is just a cone.  One might argue that this is a lot of trouble to go through just to get the volume of a cone.  (Then again, I personally don't know how to derive the formula for the volume of a cone without using Calculus.)

[quote="DVO"]Of course, the solid in question is just a cone.  One might argue that this is a lot of trouble to go through just to get the volume of a cone.  (Then again, I personally don't know how to derive the formula for the volume of a cone without using Calculus.)[/quote]

Deriving that "pointy things are a third" doesn't require calculus, in the sense of knowing how to take an integral, although it does rely on one idea from calculus: if two solids have at each height an equal cross-section, they have equal volumes.  Then you just find one pointy thing whose volume you can calculate, and the rest is straightforward.  You still need to be able to general the value of $pi\cdot r^2$ for a circle, of course, but there was a not-exactly calculus proof of that known since antiquity.


function
real analysis
real analysis solved

Prove that there aren't functions f from [0,1] in the reals such that the points of discontinuity of f are exacly the irrational ones.

More generally, it is a classical result (but too long to prove here) that
(1) The set of points of discontinuity of a real function is a countable union of closed sets
or equivalentely that :
The set of points of continuity of a real function is a countable intersection of open sets.

It follows that  there is no function from [0,1] to |R such that f is exactly discontinuous on the irrational numbers :
Suppose, for a contradiction, that such a function does exist. Then [0,1]-Q is a countable union of closed sets, each of which with void interior (since [0,1]-Q has void interior). Since [0,1] \cap Q is countable it is also a countable union of closed sets with void interior. Then, [0,1] = Q \cup ([0,1]-Q) is a countable union of closed sets with void interior. From Baire's theorem, it follows that [0,1] has void interior. A contradiction.

Remark that it is also true that the converse of (1) holds, that is : For any given set A which is a countable union of closed sets, there exists a real function f such that f is exactly discontinuous on A.

Pierre.


Find all $a,b,c\in Z$ such that $a^2+b^2+c^2 = a^2b^2$

this is AHSIMC short answer 30b

please, no answers. If a mod sees this, lock this until tuesday.

(warat_suk, I realize you probably didn't mean to cheat, especially seeing as you live in thailand or something, but it IS on a current contest)

[i]warut_suk[/i] most probably didn't cheat, for this is actually some past-year olympiad problem. Since this is a current contest problem, I shall not mention the country and year.


when I try to build, (in TeXnicCenter), it says:
[code] [DocOpen('%bm.pdf')][FileOpen('%bm.pdf')][/code]
[code] Cannot execute command.[/code]
this is in a dialog box.

I thing that the command in is the fancyhdr.sty file, is that messed up or something?

Is anyone else having this problem?

I think this problem will be fixed once you start writing up your solutions. I had problem too when I tried to view the unaltered solutions.tex file. Does your LaTeX-Result say: 0 Error(s), 0 Warning(s), 0 Bad Box(es), 0 Page(s)?
If so, that's probably the problem. Try typing something between the begin and end document tags and see if you can view it now.

This has been discussed before eg at http://www.artofproblemsolving.com/Forum/topic-42360.html

To summarize:
In TexnicCenter check the paths in the Profiles (press Alt+F7) are correct.
Click on LaTeX=>PDF and on LaTeX=>PS=>PDF then on Viewer tab and make sure that Path of executable is pointing to the correct path for Adobe Reader.


In a triangle $ ABC$ the points $ D,E,Z,H,\Theta$ are the midpoints of the segments $ BC,AD,BD,ED,EZ,$ respectively. If $ I$ is the point of intersection of $ BE$ and $ AC$, and $ K$ is the point of intersection of $ H \Theta$ and $ AC$, prove that:
(a) $AK\equal{}3CK$
(b) $ HK\equal{}3H\Theta$
(c) $ BE\equal{}3EI$
(d) the area of $ ABC$ is $ 32$ times that of $ E \Theta H$.

# a) AK=3CK

since, O is mp. of EZ and H is mp. of ED.
HK parallel BC

t(AHK) similar t(ADC)

AK/AC=AH/AD

let, AD=x
AH= x/2+x/4=3x/4
thus, AK/AC=3/4
4AK=3AC
4AK=3AK+3CK
AK=3CK

b) HK=3HO

already, AH/AD=3/4
HK/DC=3?4
4HK=3DC
4HK/2=3BD/2
2HK=3ZD
HK=3HO

similarly c) ...

d){EOH}= EZH/2
       = EZD/4
       = EBD/8
       = ABD/16
       = ABC/32 #

this problem is easy and can be solved be tales theorm :lol:

Later posted here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=446963


Here's a nice problem.  There is an odd number of people in a plane.  Their mutual distances are different.  Everyone shoots his nearest neighbour.  Prove that a) at least one person survives; b) No one is hit by more than 5 bullets; c) the path of the bullets do not cross; d) the set of segments formed by the bullet paths does not contain a closed polygon.

Masoud Zargar

Does everyone shoot at the same time, or do they shoot one at a time?

By induction.  Suppose its true for n people (n odd).  Add two people.  Now the two people with shortest distance (say, A B) shoot each other.  We revert to the "n" hypothesis, except that its possible for some people to shoot A or B.  These shots are essentially "wasted shots."  Since in hypothesis atleast one person survives, if we remove some guns, then atleast one person still survives.

Suppose ABCDEF shoot O.  Consider angles AOB, BOC, ...., EOF, FOA.  Atleast one is $\le 60$ deg.  By "larger side larger angle" one of the AB's are smaller than the OA's.  Contradiction.

Suppose A shoots B and C shoots D, with paths AB and CD crossing at O.  We have a quad ACBD, with B closest to A and D closest to C.  This means the main diagonal AB is shorter than AD or AC, and CD is shorter than AC or BC.    So AC is longer than both of the main diagonals, contradiction.

Suppose we have polygon $A_1, A_2, \dots, A_n$, with $A_1$ shooting $A_2$, ... until $A_n$ shoots $A_1$.  Consider circle with center $A_1$ and radius $A_1A_2$.  It cannot have any points $A_i$ inside.  But for $A_n$ to shoot $A_1$, $A_1$ must be closest to $A_n$, where the distance $A_1A_n$ must be less than $A_{n-1}A_n$, which must be less than .... $A_2A_1$.  It implies $A_n$ is in the circle.  Contradiction.


geometry
circumcircle
trigonometry
geometry proposed

In $ \triangle ABC ,\hat{ABC}\equal{}60$  and $ \hat{ACB}\equal{}70$ .Point $ D$ is on the line segment $ BC$ such that $ \hat{BAD}\equal{}20$ .Prove that $ AB\plus{}BD\equal{}AD\plus{}DC$

Hi!
I dont have time to post all the solution; but I can write the main idea.
Find the circumcenter using orthocenter!

Nice problem with a simple solution!

Applying sine Theorem, we have:
$ AB\plus{}BD\equal{}\frac{\sqrt{3}sin50^o}{sin20^o}BD$  (1)
$ AD\plus{}DC\equal{}\frac{\sqrt{3}sin50^o}{sin20^o}BD$  (2)
The conclusion follows from (1) and (2).

[quote="armpist"]Is it doable without trigonometry?

  Thanks.
  M.T.[/quote]

I don't think it is doable without trigonometry.
Sincerely.

Take $E$ on $(AB$ and $F$ on $(AC$, $E,F$ external to triangle $\triangle ABC$, thus triangles $\triangle BED, \triangle DFC$ are isosceles and the perpendicular bisectors of $DE, DF$ are external angle bisectors of $\triangle ABC$ and the circumcenter of $\triangle DEF$ is the $A-$excenter of $\triangle ABC$, and $AE,AF$ are tangent to the circle $\odot DEF$, thus being done.

Best regards,
sunken rock


i'm from kansas and i got 12th at mathcounts nats in 2005. our state doesn't solely consisted of farmers and rednecks, you know.

Hello!
I was beginning to think that there wasn't anybody else from Kansas on this site.
I do see what you mean, pianoforte. Just look at the other three posts.  :)

I was wondering why there was a KS forum. Not every state has a forum, I don't think, so it seemed odd that there's a KS forum when nobody ever posts here.

[quote="maryam rashin"]cheghad tv mibinin?tv ro bishtar mipasandin ya cinema ro?iranian or american or indian or...?
che bazigari ro doos darin?chera?
man montazere javabam [/quote]

what language is that in??? :rotfl:


function
number theory unsolved
number theory

To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now


Which one of these letters is the odd one out?

$\xi
I
\chi
Y
\psi
\pi
\phi
\theta$

sorry the latex got messed up.  They are supposed to be in a straight line.

Yes, but even then, I still think it might be the first one since the others are *sort of* symmetric (vertical axis of almost-symmetry) and the first one isn't.

i guess your answer works but theres another answer.

I have no idea. All I can see is that two of them are uppercase English letters, while the others are lowercase Greek letters.

i'll give you guys until Wednesday, then i'll post the answer

$\theta$ (theta).

It's the only one not pronounced with an "I" sound at the end


inequalities
number theory
number theory unsolved

Hello guys. I am new in number theory, so I will not wonder if you kill this problem very fast.So here it goes :Solve the following equation in natural numbers: $x^{2}+y^{2}=1997(x-y)$.
I am quite sure that this can be done using inequalities, but unfortuanately  I don't know how exactly.

You can write it as $(x+y)^2+(x-y-1997)^2=1997^2$. So we have pythagorean triple here. But 1997 is a prime number, so it can be wriiten as a sum of two perfect squares in the unique way: 1997=34^2+29^2. Thus we have two possibilities:
1) $x+y=34^2-29^2$ and $x-y-1997=\pm2\cdot 34\cdot 29$;
2) $x+y=2\cdot 34\cdot 29$ and $x-y-1997=\pm(34^2-29^2)$.

A strange problem...


inequalities
LaTeX
algebra unsolved
algebra

[quote="stevenwang1992"]I  am  a high school student in China.I have an  inequality .I want to have it upgraded but I meet some problems.
The equality is as follows:
       Let F=(a/a+b)^x   +(b/b+c)^x  +(c/c+a)^x   and a,b,c  are  positive.
thenF>=3*(2^x)    when x=-1,-2,-3,-4,-5-----------
      F<=3/2            when x=1
      F>=3/(2^x)     when x=1,2,3,4,5-------------
  I don't know whether I have done it correctly .
And if it's correct Why when x=1 make    F  have maximum  but  not  minimun![/quote]

Hello [b]stevenwang1992[/b]!

For $ x\equal{}1$ inequality
\[ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}a}\leq \frac{3}{2}\]

does not hold for $ x\equal{}\frac 3 4, y\equal{}\frac 1 2, z\equal{}1$.

Maybe this is a misprint, and you wanted to write not $ \left(\frac{a}{a\plus{}b}\right)^x\plus{}\left(\frac{b}{b\plus{}c}\right)^x\plus{}\left(\frac{c}{c\plus{}a}\right)^x$ (as you wrote), but $ \left(\frac{a}{b\plus{}c}\right)^x\plus{}\left(\frac{b}{a\plus{}c}\right)^x\plus{}\left(\frac{c}{a\plus{}b}\right)^x$ ?

I do not know the solution for this problem. There are 2 suggestions for you:

1. Use inequalities forums insted of algebra forum, see http://www.mathlinks.ro/index.php?f=32

2. Learn Latex notation so that your formulas will look glad. This is easy. Example: you need to write $ \$$\frac{a}{b+c}$ \$$ to receive the result $ \frac{a}{b\plus{}c}$. Hint: click at any Latex formula at this forum and you will see, how it was done.

Sergey Markelov

First I want to say that I haven't miswritten my equality.
The fact is that when  x=1 the inequality has neither a minimum nor a maximum.
When x=1      (a/a+b)^x +(b/b+c)^x +(c/c+a)^x   a one power inequality.   Its  algebraic  structure is different from
x=2,3,4,5,6.........
so it has   neither a minimum nor a maximum.
then For  inequality (a/b+c)^x+(b/c+a)^ x+(c/a+b)^x  we can use the  POWER  INEQUALTY and CAUCHY INEQUATY to solve it


geometry
3D geometry
number theory unsolved
number theory

let $ (a_n)$ be a sequence defined by $ a_1\equal{}1$ and $ a_{n\plus{}1} \equal{} \frac{1}{n}(1\plus{}a_{1}^2 \plus{}a_{2}^2\plus{}...\plus{}a_{n}^2)$

Is $ a_n \in \mathbb{N}$ for all $ n$?

The sequence defined by $ x_0 \equal{} 1$ and $ x_n \equal{} \frac {1\plus{}x_0^2 \plus{} \cdots \plus{} x_{n\minus{}1}^2 }n$, has been created by Göbel (not Gödel). Lenkstra found that $ x_{43}$ is not an integer.

If you replace squares by cubes, you will have to wait until $ x_{89}$.

Many other similar sequences (powers, initiale value...) have been investigated.

Pierre.


algebra
functional equation
algebra unsolved

solve the following functional equation
$f : \mathbb{R} \rightarrow \mathbb{R}$
$f(x^2+y+f(y))=2y+f(x)^2$
where $x,y$ are any reals

http://www.mathlinks.ro/Forum/viewtopic.php?t=68972
http://www.mathlinks.ro/Forum/viewtopic.php?t=38603
http://www.mathlinks.ro/Forum/viewtopic.php?t=22929

  Darij

Yes, this is quite hard, it was on Bulgarian TST in 2003.
But it has very beautiful solution :)


geometry
3D geometry
superior algebra
superior algebra unsolved

Suppose $ \alpha_1, ..., \alpha_n$ are elements of an extension field of $ F$ such that $ (\alpha_1)^3 \in F$ and $ (\alpha_{i\plus{}1})^3 \in F(\alpha_1, ..., \alpha_i)$ $ (i < n)$.

Show that $ [F(\alpha_1, ..., \alpha_n): F]$ is either $ 3^k$ or $ 2 \times 3^k$ for some $ k$.

I think it amounts to show that some element of $ F(\alpha_1)$ has three distinct cube roots in $ F(\alpha_1)$, viewing from the previous problem, where $ [F(\alpha_1): F] \equal{} 2, 3$. I am not sure how to show this, or if this is even true.

Let $ i\equal{}min\{k|[F(\alpha_1,\cdots,\alpha_k): F(\alpha_1,\cdots,\alpha_{k\minus{}1})]\equal{}2\}$. We use the Zetax-notation in your previous problem:$ \alpha_i\zeta$or $ \alpha_i\zeta^2\in{F(\alpha_1,\cdots,\alpha_{i\minus{}1})}$; thus $ \zeta$ and $ \zeta^2$ are in $ {F(\alpha_1,\cdots,\alpha_i)}$ and, by your previous problem, there doesn't exist any extension field of degree 2 for $ k>i$.  QED


calculus
integration
logarithms
trigonometry
calculus computations

Evaluate \[ \int_0^{ \plus{} \infty } {\frac{{x^{\varphi  \minus{} 1}  \minus{} x^{\lambda  \minus{} 1} }}
{{\left( {1 \plus{} x} \right)\ln x}}dx}\]  \[ 0 < \varphi  < \lambda  < 1\]

Good luck.  :lol:

We have,
$ I \equal{} \int_{0}^{\plus{}\infty }{\frac{{x^{\varphi\minus{}1}\minus{}x^{\lambda\minus{}1}}}{{\left({1\plus{}x}\right)\ln x}}dx}$

$ \equal{}\int_{0}^{\plus{}\infty } \left (\int_{\lambda\minus{}1}^{\varphi\minus{}1} x^{t} dt \right) \frac{1}{1 \plus{}x} dx$

$ \equal{}\int_{\lambda\minus{}1}^{\varphi\minus{}1} \int_{0}^{\plus{}\infty } \frac{x^t}{1 \plus{}x} dx dt$

Well, I got a relatively long calculation so I will give only the answer, so if I'm wrong I will not waste time writing it all. I got:

$ \ln \tan \frac {\varphi \pi}{2} \minus{} \ln \tan \frac {\lambda \pi}{2}$

(Should be correct.. A quick numerical check in Mathematica confirmed it..)

But, complete the proposed :D

My solution:

Define: \[ J\left( \theta  \right) \equal{} \int_0^{ \plus{} \infty } {\frac{{x^{\theta  \minus{} 1} }}
{{\left( {1 \plus{} x} \right)\ln x}}dx}\]

Where \[ \theta  \in \left( {0,1} \right)\]

\[ \underbrace  \Rightarrow _{Leibniz}J'\left( \theta  \right) \equal{} \int_0^{ \plus{} \infty } {\frac{{x^{\theta  \minus{} 1} }}
{{\left( {1 \plus{} x} \right)}}dx}  \equal{} \Gamma \left( \theta  \right)\Gamma \left( {1 \minus{} \theta } \right) \equal{} \pi \csc \left( {\pi \theta } \right)\]

Now, separating variables and integrating:

\[ \int_\lambda ^\varphi  {dJ}  \equal{} \int_\lambda ^\varphi  {\pi \csc \left( {\pi \theta } \right)d\theta }\]

\[ \Rightarrow J\left( \varphi  \right) \minus{} J\left( \lambda  \right) \equal{} \int_{\pi \lambda }^{\pi \varphi } {\csc \left( u \right)du \equal{} \ln \left( {\frac{{\csc \left( {\pi \lambda } \right) \plus{} \cot \left( {\pi \lambda } \right)}}
{{\csc \left( {\pi \varphi } \right) \plus{} \cot \left( {\pi \varphi } \right)}}} \right)}\]

But, \[ J\left( \varphi  \right) \minus{} J\left( \lambda  \right) \equal{} \int_0^{ \plus{} \infty } {\frac{{x^{\varphi  \minus{} 1}  \minus{} x^{\lambda  \minus{} 1} }}
{{\left( {1 \plus{} x} \right)\ln x}}dx}\]

\[ \therefore \int_0^{ \plus{} \infty } {\frac{{x^{\varphi  \minus{} 1}  \minus{} x^{\lambda  \minus{} 1} }}
{{\left( {1 \plus{} x} \right)\ln x}}dx}  \equal{} \ln \left( {\frac{{\csc \left( {\pi \lambda } \right) \plus{} \cot \left( {\pi \lambda } \right)}}
{{\csc \left( {\pi \varphi } \right) \plus{} \cot \left( {\pi \varphi } \right)}}} \right)\]

----
Another solution?  :roll:


Scholarships info is now being sent to every applicant.
Watch your email box.

Yeah, I got it!!!
I got enough scholarship to go there.
So see you guys there.
If any of you want to contact me and to be my friend, please feel free to PM me now.

I am accepted with full scholarship, too!!
It will be lovely to attend Mathcamp this summer, it will be a great experience!!

Thank You MathCamp

"Due to your outstanding performance on the USAMTS, Canada/USA Mathcamp is pleased to award you a $200 merit scholarship to Mathcamp 2006."

Yay. But...how is 65 total an outstanding performance, exactly?

[quote="bookaholic"]"Due to your outstanding performance on the USAMTS, Canada/USA Mathcamp is pleased to award you a \$200 merit scholarship to Mathcamp 2006."  Yay. But...how is 65 total an outstanding performance, exactly?[/quote]

The USAMTS scholarships go to anyone who is among the top 30 boys or the top 30 girls on this year's USAMTS.

[quote="Alexz"]Thanks Mathcamp!!!
I will be given full scholarship![/quote]

Good! so, I presume you will be coming to mathcamp only. Hope to see you there.
Are you travelling from Singapore ?  Then we could travel together, as I will also be going via Singapore. Let me know your travel plans if the idea is OK with you.

[quote]travel together[/quote] Speaking of which, if anyone's planning to fly through Tokyo, it'd be lovely to travel together. :)

- Y. '05

As long as we're on the topic, how about people flying through San Francisco? (Or San Jose, I guess.)


quadratics
superior algebra
superior algebra unsolved

Show without appealing to the Pepin's Test that 3 is a primitive root modulo Fermat primes.

If we assume 3 to not be a primitive root mod p why must 3 be congruent to a square? Can someone show this?

Lemma: Let $ p \equal{} 2^n \plus{} 1$ be a prime with $ n\geq 1$. Then the set of primitive roots modulo $ p$ is the set of quadratic non-residues modulo $ p$.

Now since $ p \equal{} 2^n \plus{} 1$ and $ n\geq 3$, $ p\equiv 1 \mod 4$. Also, $ p\equiv 2 \mod 3$. For if $ p\geq 1
\mod 3$, then $ 2^n \plus{} 1\equiv 1 \mod 3$, get a contradiction. As $ p\equiv 1 \mod 4$ and
$ p\equiv 2 \mod 3$, we have $ p\equiv 5 \mod 12$. By the quadratic reciprocity law, $ (3/p) \equal{}\minus{}1$. By lemma, $ 3$ is a primitive root modulo $ p$ for all $ p$.


Let $ x_1,$ $ x_2,$ $ x_3,...$ are real numbers such that $ x_{n \plus{} 2} \equal{} \frac {x_{n \plus{} 1}x_n \plus{} x_n^2}{x_n \minus{} x_{n \plus{} 1}}$ for all $ n\in\mathbb N$ and $ x_1 \equal{} x_{2008}.$
 Prove that $ x_2\neq x_{2007}.$

If $ x_1 \equal{} 0$, then $ x_2\not \equal{} 0, \ x_3 \equal{} x_5 \equal{} .. \equal{} x_{2007} \equal{} 0\not \equal{} x_2.$
If $ x_2 \equal{} 0$, then $ x_1\not \equal{} 0, \ x_4 \equal{} x_6 \equal{} ... \equal{} x_{2008} \equal{} 0\not \equal{} x_1$.
Therefore from $ x_1 \equal{} x_{2008},x_2 \equal{} x_{2007}$ we get $ x_i\not \equal{} 0, \ \forall i\le 2008$ and $ x_n \equal{} x_1\frac {x_n}{x_1} \equal{} x_1y_1*...y_{n \minus{} 1}$, were $ y_k \equal{} \frac {x_{k \plus{} 1}}{x_k}$.
Sequence $ y_n$ defined by $ y_{n \plus{} 1} \equal{} f(y_n)$, were $ f(x) \equal{} \frac {1 \plus{} x}{x(1 \minus{} x)} \equal{} \frac {1}{f( \minus{} \frac 1x )} \equal{} \frac {1}{f(\frac {1 \plus{} x}{x \minus{} 1})} \equal{} f(\frac {1 \minus{} x}{1 \plus{} x}).$
From $ x_1 \equal{} x_{2008}\to y_1*y_2...y_{2007} \equal{} 1$, from $ x_2 \equal{} x_{2007}\to y_2*...y_{2006} \equal{} 1$ or $ y_1y_{2007} \equal{} 1$ or .
Let $ g(x) \equal{} xf(x)$, then $ g(g(x)) \equal{} \minus{} \frac 1x , g(g(g(g(x))))\equiv x$.
But $ g(x) \equal{} \frac {x_{n \plus{} 2}}{x_n}$ if $ x \equal{} \frac {x_{n \plus{} 1}}{x_n}$ or $ g(y_n) \equal{} y_ny_{n \plus{} 1}$.
Equations $ x_1 \equal{} x_{2008}$ and $ x_1 \equal{} x_{2007}$ are polinomial equations for $ x \equal{} \frac {x_2}{x_1}$. I think calculations (give above) may be used for prove, that these equations had not common solutions.

[hide="The hint"] There is a proof in one line. [/hide]


USAMTS
search
LaTeX
MATHCOUNTS
videos
summer program
Mathcamp

The USA Mathematics Talent Search (USAMTS) is a free mathematics competition open to all United States middle and high school students. The USAMTS is primarily funded by the National Security Agency, which has funded the program since 1992.

Starting with the 2004-5 school year, the USAMTS will be administered by the Art of Problem Solving Foundation.

The USAMTS consists of four rounds of five problems each, and allows students a full month to work out their solutions. Carefully written justifications are required for each problem. The problems range in difficulty from being within the reach of most high school students to challenging the best students in the nation.  Read more about the USAMTS [url=http://www.usamts.org]here[/url].

Cool! I was planning on 2004-2005 being A+MATH's first year in USAMTS, and now it looks like there have been some significant new participants joining the USAMTS management team. I'll be sure to tell parents on other online forums about this.

Thanks for spreading the word.

If he hasn't mastered it already, you may also want to get A+MATH [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]started on LaTeX[/url].  We'd like to see as many USAMTS students as possible use LaTeX, both because it's an important skill to develop and because LaTeX is much easier to read than handwriting!  We are building the LaTeX pages linked above in order to help students learn LaTeX, and as an incentive in the USAMTS, we will only be publishing student solutions which are typeset in LaTeX (the most nicely polished/typeset solutions will be published as part of the official solutions).

We are the USAMTS now.  The NSA is not administering the contest anymore, though they will be funding it and doing some of the grading.  Initially, the contest will be run very similarly to past years, though we will be able to do some things with the internet that they couldn't (primarily due to restrictions on the use of government web pages).  For example:

1) We're building the LaTeX pages and strongly encourage students to submit their work electronically.
2) We'll have a leaderboard and stats page.
3) We'll have USAMTS Math Jams after each round.
4) We'll have all reporting done online (you'll get your scores and grader comments on this site).

I'm sure we'll come up with others, and are open to suggestions (with the caveat that we certainly won't be able to do everything we'd like to!)

This is great news!

Maybe this way the USAMTS site will be updated once every season or so. Will the Honorable Mention/Medal prize qualifying scores remain the same?

We haven't set the prize criteria yet.

The site will be updated every round with scores, stats, solutions, etc.

Wow, this is very interesting. Having AoPS taking over the USATMS.

And there is my usual question, will Canadians be allowed to participate now?

Is AoPS going to notify the winners of this year's USAMTS?

Unfortunately, we only have funding for the US.  We will be seeking funding to expand the program as soon as our application for the appropriate non-profit status is complete.  If you know of Canadian foundations that might have an interest in funding the program, please let us know!

If we already have a control number with USAMTS, and we've participated in previous years, we still have to register again (through AoPS) correct?

jimmy

Congratulations on getting the USAMTS contract.  I have every confidence you folks will do a great job!

I happen to love TeX (and LaTeX), and I have personally used TeX for nearly 2 decades.  However, I am deeply concerned about your policy regarding publishing only (La)TeX solutions.

With USAMTS, I think students should concentrate primarily on the mathematics of their solutions.  Students should concentrate secondarily on their ability to clearly express their solutions to others.  Forcing a particular mode of expression is completely irrelevant to the purpose of the USAMTS!

Most schools have MS Word.  MS Word's "Equation" add-on does an adequate job typesetting most mathematics.  Some schools have MathType or other typesetting software.  Very few schools have any version of TeX or LaTeX, and I suspect most schools will be resistant to installing and maintaining another software package, even if it is "free".  In fact, I would be willing to give odds that among every student and employee in our local school district (~18,000 people), not a single student or employee has ever produced a (La)TeX document using school facilities.

My 7th and 8th grade students have a difficult time just "typing" their solutions.  Teaching them to use LaTeX would completely overwhelm them, and rob them of the  limited time I have for coaching math.

Budding young math students may also need WYSIWYG capabilities not generally available in TeX packages.

Over the years, several brilliant solutions by 7th, 8th, and 9th graders have been posted on the USAMTS Web site.  I fear these solutions would never have been presented if a policy of "LaTeX-only solutions" had existed.

Currently, I think many students strive for a solution worthy of a "5", a solution worthy of a commendation, and a solution worthy of publication.  For those who can't reasonably use LaTeX, this restriction eliminates one of the students' incentives for doing a good job.  For those students who find LaTeX is tedious, you may find you have lost those students entirely.

I understand that LaTeX is essential to the professional mathematician (and probably to the collegiate mathematician).  Virtually all mathematics journals accept or require LaTeX submissions.

However, the majority of students who excel at USAMTS (or MATHCOUNTS or AMC or etc.) do not go on to careers in mathematics!

In my entire professional career as an electical engineer and a software engineer, I have met hundreds of people who use and write mathematics on a daily basis.  However, I have met only a handful of engineers who use LaTeX at all.

Please!  I beg you to reconsider your policy of only publishing LaTeX solutions.

Well, there is a very easy tutorial about installing LaTeX and using it.  Seeing as to how summer break is approaching, your students will have plenty of time to get acquainted with the language before the first round.  I'm willing to bet that most LaTeX users taught the language themselves even without a step-by-step tutorial.

rcv:

Where does it say that only Latex solutions are allowed? I thought they only encouraged Latex usage. Thanks.

I think he was referring to the fact that nifty solutions that are posted are to be submitted in LaTeX.

We will consider possibly including other solutions, but it is unlikely that we will choose to do so.  Here are some of our reasons for this decision and responses to some of your concerns:

We are preparing a pretty thorough tutorial for using LaTeX, and our message board does suggest there are lots of students who like using it.

We would hope that the possibility of having a solution published is not a major reason for students to participate in USAMTS - this is a very minor reward compared to others.

The USAMTS is not a school program - most students do it on their own, so the availability of LaTeX in schools is not a concern for us.  Besides, we are not preventing non-LaTeX students from participating at all.

We believe most, if not very nearly all, of the kids who participate in the USAMTS are capable of learning and using LaTeX if they decide to do so.  It's certainly far less complicated than the video games the kids master.

There is also a serious time saving for us - even if we permitted the publication of other solutions, we would probably never do so, as we would always elect to publish the elegant solutions in LaTeX rather than the hand-written ones we would have to typeset ourselves.  Better for us to be honest with the students and let them know that published solutions will come from the ranks of students using LaTeX than to have that be an unmentioned policy.  Please keep in mind, that while we have a contract from NSA to do this work, we still expect to volunteer quite a bit of time ourselves.  (In fact, all our work done on the USAMTS so far is volunteer work that will not be funded by the NSA.)

I've already had one of my students send me 'his first LaTeX document' since we published this policy - the benefits of giving avid students the incentive to learn LaTeX are, we believe, outweighed by the costs of losing the less interested students who opt out of the program since they won't be published.

Also, please note that students who don't use LaTeX are still just as eligible for 5's or commended's.

I can't do almost any USATMS problem :(. How do I get better?

USAMTS problems aren't like the "one trick and then compute like hell" mathcounts problems. These problems require many insights and are very difficult problems. They may take days, even weeks to solve. The problems may range from early AIME to easy/mid USAMO so a good way to get better is to do AIME problems which is what I have been doing. I used to be in a similar situation as you, but after doing lots and lots of very hard problems I find myself able  to solve many USAMTS problems.

USAMTS problems are a lot like ARML power round problems albeit a bit trickier and harder. @countyguy dont expect to solve olympiad level problems in 5 mins!! You have to spend more time!!

A strategy that I sometimes use is that I carry around a piece of paper with a problem that I find hard to solve around school. Between brakes/class changes/ free time in class I think about that problem. When I get home, I find that I have made significant progress.

[quote="Debdut"]Since USAMTS is online.Will it be possible in future that it is spread to Asian countries because many of Indian AOPSers want it.[/quote]

Check out IMTS. It's international, while the USAMTS is restricted the the USA.

I think the problems are often common (don't take my word for it, though.)

Oh. Thanks tc1729. Well, then sorry, Debdut. I don't know what to tell you, except for there are several online olympiad-level math contests. Try Google.

But to answer your original question, I don't think they'll extend USAMTS to foreigners anytime soon.

EDIT: Fixed username. Derp. Thanks cire_il.

UGH I hate duplicate avatars.

If you look closely, the user above was actually tc1729.

What is the USAMTS like? Can you give me some sample problems from previous years? What types of questions do they have on the USAMTS? And can you give me some strategies?

Problems from previous years' tests can be found [url=http://www.usamts.org/Problems/U_Problems.php]here.[/url]

Since the test is untimed, I don't really know if there is a lot of strategy involved.

You can use programs to solve problems, BTW.

@above, But you still have to justify why your program is correct.

Everyone, please check the official rules online at USAMTS for stuff like this; while AoPS is a great community, we aren't perfect, and the rules may change as well, so the official ones are always your best (and safest) bet.

[quote="ssilwa"]
A strategy that I sometimes use is that I carry around a piece of paper with a problem that I find hard to solve around school. Between brakes/class changes/ free time in class I think about that problem. When I get home, I find that I have made significant progress.[/quote]

Wow, thats exactly what i do!! It's a really effective method.

Only American students can receive prizes, but I'm pretty sure that everyone can still try the problems/submit.

So if I solve USAMTS problems, I send my solution to them, do they consider my solution ?


There are 4 couples lining in a line . How many ways are there such that no couple is standing adjacent to each other ?


If a,b,c>0 proove that:

\[ \frac{a^4}{bc}+\frac{b^4}{ac}+\frac{c^4}{ab} \geq \frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \]

That's trivial.We only use Cauchy and then Muirhead.

We can use the RE-ARANGEMENT INEQUALITY
Probably it works with CBS too...

OK, you're right.We can use this or Cauchy&Muirhead. :D


analytic geometry
real analysis
real analysis unsolved

Does exist a set of points on the unit circle of coordinates rationals such that this set is dense in the unit circle and that distance for any two points is rational ?

It's been discussed before. Check [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=rational+distances&t=44864]this[/url] out, and also the link in that thread.


There is a list of seven numbers. The average of the first four numbers is 5 and the average of the last four numbers is 8. If the average of all seven numbers is 46/7, what is the common number to both sets of four numbers?

[hide]Let the sum of the first 3 numbers be a, and the sum of the last 3 be b. Let the 4th number be x.

a + x = 20
b + x = 32

a + x + b = 46

Add the first two, and take away the last.
x = 6[/hide]


Two problems  :)

1.
For the real numbers $a,b,c$ we know that
$a<b+c$, $b<a+c$ and $c<a+b$
Prove that they are all positive.

2.
For the real numbers $a,b,c$ we know that
$\mid a \mid<\mid b+c \mid$, $\mid b \mid<\mid a+c \mid$
 and $\mid c \mid<\mid a+b \mid$
Prove that they have the same sign.

Have fun!! :)

[hide="problem 1"]
Add two of the three togather. [/hide]

Correct!  :)

Go on to the next problem!!! :)

I'd just square both sides of each ineq. and do what shobber did solving the first problem. Then put common factors outside the brackets...

For the first, can we use that $a,b,c $ represent sides of  a triangle!!


Let ABC be a triangle and P be an interior point of ABC.If PA=sqrt(2),
PB=sqrt(5), PC=sqrt(10) then find the maximum value of perimeter of ABC triangle.

something tells me... law of cosines.. but I'm a bit rusty

Let ABC be a triangle and P be an interior point.Let PA,PB and PC distances be constant.Find the locus of P when the perimeter of ABC maximum.
I think the locus must be intersection of bisectors but I can not prove it.


Aceasta problema a aparut in nr. 9/1988 al revistei mentionate si are enuntul:
"Fie ABC un triunghi oarecare. Simetricele dreptei BC faţă de dreptele AB şi AC se intersectează în punctul K. Demonstraţi, că dreapta AK trece prin centrul cercului circumscris triunghiului ABC."

Inainte de a vizualiza solutia din ATTACHMENT, incercati sa gasiti solutia singuri!
Cautati apoi si o solutie sintetica.

[quote="mihai miculita"][color=darkred][b]Fie $ ABC$ un triunghi oarecare. Simetricele dreptei $ BC$ faţă de dreptele $ AB$ şi $ AC$ se intersectează în punctul $ K$.
Demonstraţi că dreapta $ AK$ trece prin centrul $ O$ al cercului circumscris triunghiului $ ABC$.[/b][/color][/quote]
[color=darkblue][b]Demonstratie.[/b] Notam punctul $ L$ pentru care $ \{\begin{array}{c}LB\perp AB\\ \ LC\perp AB\end{array}$. Ne vom concentra atentia asupra triunghiului $ BKC$. Apar doua cazuri dupa cum dreapta $ BC$ [u]nu separa[/u] ([b]separa[/b]) punctele $ A$, $ K$. Semidreapta $ [KA$ este bisectoarea unghiului $ \widehat{BKC}$, $ L\in KA$, punctul $ A$ este [u]incentrul[/u] ([b]$ K$- exincentrul[/b]) si punctul $ L$ este $ K$- [u]exincentrul[/u] ([b]incentrul[/b]) pentru $ \triangle BKC$. Cercul circumscris triunghiului $ ABC$ are diametrul $ AL$, adica centrul $ O$ al acestuia este mijlocul segmentului $ [AL]$. In concluzie, $ O\in\overline{KAL}$. De fapt problema este o exprimare echivalenta a unei proprietati cunoscute :[/color]

[color=darkred][b]" In triunghiul $ ABC$ punctele $ B$, $ C$ apartin cercului de diametru $ [II_{a}]$, adica $ A'B = A'C = A'I = A'I_{a}$ ".[/b][/color]


For $p,q\in\mathbb{N}$ where $p>q$ prove that

$\sum^{\infty}_{k=1}\frac{(pq)^k}{(p^k-q^k)(p^{k+1}-q^{k+1})}=\frac{q}{(p-q)^2}$

Happy New Year 2006  :D

:first:  :coolspeak:

you method is nice by using substitution


geometry
vector
linear algebra
matrix
calculus
linear algebra unsolved

hello all, i am taking a linear algebra course this semester and am not doing well in it, the only area of difficulty i am having is the proofs, any pointers as to how to tackle them ? Here were some specifics i was having trouble with:

1.Show if S is an element of T and T is finite and Linearly Independent then S is Lin. Independent
A:Since S is an element of T any of the properties that belong to T also belong to S, therefor S would be lin. indep.

2. Prove that if T and S are subspaces of a vector space V then S intersects T is a subspace of V.
A: don't know where to even begin with this

3. For what values of t is the set S={(t,t,t),(t,t,0),(t,0,1)} Linearly Independent
A: for this I put t should be like this: (1,0,0),(0,1,0),(0,0,1) b/c if you do this c1v1...c3v3=0,0,0 , c1=c2=c3=0

4.Prove that if S={v1,...vn} and T={v2,..vn} and span(S)=span(T) then v1 is a linear combination of v2...vn

5. Prove that if S={v1,v2,v3} are Linearly dependent and v1,v2,v3 not equal to 0 and v1 is not a multiple of v2,and v2 is not a multiple v3,and v1 is not a multiple of v3, then the span of {v1,v2} =span S

6. Prove that if A is an non matrix then the set of nxn matrices B where AB=0 is a subspace of the vector space of nxn matrices.

When i see these on the test I do not know where to start It seems that I am only good and the algebra type problems where robot like calculations are involved . when it comes time to "think" i do not know how to approach them or where to start any help would be appreciated on this . thanks you

Somehow, I don't think that Linear Algebra is a "Getting Started" topic.

OK! then where was i supposed to put up my post ?

Your problems look like Pre-Olympiad Problems, but I'm only in middle school, so I'm not familiar with the difficulty. At least I don't know how to do them.

probably in the intermediate or pre olympiad section.

1.  I think you mean that $S\subseteq T$ - if not, clarify.

2.  Say $a,b\in S\cap T$.  Then $ca+b\in S$ (because it is a subspace), $ca+b\in T$, so $ca+b\in S\cap T$.

3.  $t\not=0$

4.  $v_1\in\text{span}(S)$, so that $v_1\in\text{span}(T)$, so that $v_1=\sum_i c_iv_i$ for $i=2,3,\ldots,n$, which is the desired result.

5.  $v_1$ and $v_2$ are linearly independent, and $v_3=c_1v_1+c_2v_2$ for some nontrivial $c_i$'s, or else we would havel linear independence.  The result (almost) follows.

6.  Say $AB=0$, $AC=0$, and $x$ is a scalar.  Then $A(xB+C)=x(AB)+AC=x0+0=0$

Well, Linear Algebra is a course that is offered after Calculus CD at the high school I'm entering so... I think it belongs in a category where Calculus is accepted.

The appropriate forum for this is "Matrix-Linear Algebra".


geometry
perimeter
inequalities
AMC
AMC 10
AMC 10 A
LaTeX

how can i delete this post now??

sorry i cant figure out how to add on, only to edit  :P

By the Angle Bisector Theorem, $ \frac{GF}{FI}\equal{}\frac{GH}{HI}$, or $ \frac{GF}{3}\equal{}\frac{GH}{8}$, so $ GF\equal{}\frac{3GH}{8}$. Since GF is an integer, let GH=16 to achieve the minimum perimeter (if GH=8, then GF=3, and the triangle becomes a straight line).  Then, GF=6 (and the triangle inequality holds for this triangle), so the minimum perimeter is $ 11\plus{}16\plus{}6\equal{}\boxed{33}$.

If you guys haven't learned about the angle bisector theorem, then yeah, that's pretty tough.  If you have though, it's not that hard.

[quote="carlospena69"]whats the triangle inequality[/quote]
(I believe it was edited)

The triangle inequality states the for a triangle with side lengths $ a$, $ b$, and $ c$, we have
\[ a\plus{}b >c \\
a\plus{}c>b \\
b\plus{}c >a\]

In other words, the sum of any two sides of a triangle is always greater than the third.

Sorry for reviving. But is it just me, or is this problem identical to a problem on AMC10A this year?

Is there a way to solve this problem without Angle Bisector Theorem?

Original Problem: Find the smallest possible perimeter of Triangle ABC, where BD is an angle bisector, and AD=3, and CD=8.

[size=84][color=red]Forum Mod: Hi \$LaTeX\$. This is technically not reviving, so even if what you said wasn't so enlightening, it's not all that bad. [/color][/size]


search function LaTeX MATHCOUNTS AMC USA(J)MO USAMO

Is it possible to use modular arithmetic to prove the Chicken McNuggets Thereom?

Yes, but it's not that simple.
Look at this for more info. http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mcnugget+chicken&t=129763

what is the chicken mcnuggets theorem!?!?

use the search function at the top of your screen...type in "chicken mcnuggets theorem"

$ ab-a-b$

this equation (the mcnuggets theroem) finds the max number that cannot be achieved by adding any ammoung of $ a$ and $ b$ together.

EDIT: corrected $ \text{\LaTeX}$ error.

I think you made a typo. It's $ ab-a-b$. Also, it only works on numbers without common factors, other than 1.

[quote="kimmystar94"]y is it called the McNuggets theorem???[/quote]
McDonalds used to only sell McNuggets in packs of 4 and 7 (correct me if I'm wrong, which I probably am), which are relatively prime.

yea stargirl i think so, but basically formulas are bad for you.

EDIT: reason already posted above.

Didnt The All-Time Greatest MATHCOUNTS problems have a solution plus a proof to the problem? I dont have the book with me now...anyone able to copy and write the proof from the book down?

[i]Please [/i] use the Search feature. The Chicken McNugget Theorem has been posted several times.

i agree. there have been WAY too many posts on chicken mcnugget theorem..seriously people, this is what the search function is for..don't revive old topics..i think a mod should lock this topic

[quote="minicon"]mihail911 is right.  I searched and found 20 results.
[size=200][color=red]ATTN: MOD  PLEASE LOCK THIS TOPIC[/color][/size][/quote]

in AoPS 1, it says it was not a well-known theorem. Which means we have to prove it on the USAMO

No one wants to do that, so they are trying to make it well-known

please dont rate this as spam like all my other posts >_>


Guys, what happened to toonami? It used to be awesome (.hack//sign, bowboy bebop). But now all I see is horrible stuff.

wtf
is toonami
the
saturday or whatever segment on cartoon network
hmm what
dude they used to show rave master
man rave master kicks ass
OH YEAH they also had BOBOBO BOBOBOBO
god that was hilarious

Dude toonami always failed, so I stopped watching it 5 years ago.

wtf you clearly have not watched/read rave master and bobobo bobobobo

Of course. I remember pissing my pants in laughter with boboboboobobo.

I still miss .hack//sign.

wtf i was in korea back then
watching sc

I don't even remember toonami but it seems familiar

meh. this topic is totally pointless.
oh well w/e.
toonami had OK shows i guess there were a lot of pokemon movies i think.
bobobobobobobobobo was the best show ever though.  :rotfl:


if integers $ (X,Y)$ satisfy the equation $ 13x^2\equal{}y^8\plus{}8$.
prove that $ 10 | Y$or $ X\equiv1(mod 10)$

You should post this in High School Basics .

$ 2|y$ or $ R.H.S.\equal{}1(mod \;\; 8) \Rightarrow x$ is odd $ \Rightarrow L.H.S.\equal{}5(mod \;\; 8)$, contradiction.

$ 5|y$ or $ R.H.S.\equal{}4(mod \;\; 5)$ by Fermat's little theorem $ \Rightarrow x^2\equal{}3(mod \;\; 5)$ for some integers $ x$ , which is easy to check to be impossible.

So $ 10|y$


algebra
system of equations
combinatorics proposed
combinatorics

A dog run behind a fox. They have distance 60 steps of the fox. When the fox make 9 steps the dog makes 6 steps. but 3 steps of the dog=7 steps of the fox.After how many steps of dog the dog will catch the fox?

maybe this is not in the right section, intermediate section is more apropiate! I eman is an algebra problem that can be  easily solved with a simple system of equations! ;)

6 steps of dog have lenght like 14 steps of fox so when dog make 6 steps they have lengjt of 14 foxes steps, for that time fox makes only 9 steps so dog is closer for 14-9=5 steps  :). In the beginning dog was 60 foxes steps away from fox and in every 9 steps of the fox the dog is 5 steps closer, so hi will catch fox when fox makes $\frac{60}5 \cdot 9=108$ steps

Number is   \huge{108}

 :)  :)  :)  :)  :)  :)  :)  :)  :)

He's asking dogsteps m@re.
Every move consists of 6 dogsteps, and he gets $\frac{6\cdot7}{3}-9 = 5$ steps closer per move, so we need $\frac{60\cdot 6}5 = 72$ steps.

However this problem comes a bit late to me, I'd have enjoyed it more when I was in primary school. ;)

oooooops my mistake

Well i enjoyed this kind of problems a few years ago but it feels good to remember....


We wish to place natural numbers around a circle such that the following property is satisfied: the absolute values of the differences of each pair of neighboring numbers are all different.
a) Is it possible to place the numbers from 1 to 2009 satisfying this property
b) Is it possible to suppress one of the numbers from 1 to 2009 in such a way that the remaining 2008 numbers can be placed satisfying the property

a) It is not possible, there are $ 2009$ different pairs of neighboring numbers and the absolute differences ranges from $ 1$ to $ 2008$.

for b),$1507$ is also valid
we get $1,2009,2,2008,...,502,1508,503,1506,504,...,1006,1004,1005$
are these two the only possible numbers to be ignored in order to satisfy the condition?


Let T_1 and T_5 be the time of the first and fifth arrivals in a Poisson process with rate R.

a) Find the conditional density of T_1 given that there are 10 arrivals in the time interval (0,1).
b) Find the conditional density of T_5 given that there are 10 arrivals in the time interval (0,1).
c) Recognize the answers to a) and b) as named densities and find the parameters.


the energy efficiency of the aerobic celular respiration reaction is??????
i have different answers and i dont have a clue which one is the correct!!
is it 40% which is about equal to most energy conversion rates?
is it: about 40% which is more efficient than most energy conversion rates???
is it: about the 25% which is about equal to most energy conversion rates?????
or it is: about 25% which is less efficient than most energy conversion rates??

????????????  :(
ttthhhhhxxxx

http://www.rwc.uc.edu/koehler/biophys/8e.html

The second one


${A=\{x|x\in \mathbb{Z}\land 0\leq x\leq 9\}}$
${\left(n\left|\gamma_{n}\right|s|\lambda_{s}\right)\in A}$

If the postulate this valid

${0<1<2<3<4<5<6<7<8<9}$

Proove that

${k\in \mathbb{Z}}$
${\sum_{i=0}^{k}\gamma_{i}\geq \sum_{i=0}^{k}\lambda_{i}}$

Is valid if and only if

${0\leq h\leq k\land h\in \mathbb{Z}}$
${\gamma_{h}\geq \lambda_{h}}$

Proove this using the postulate

[hide]

If this is wrong, disproove, does not give only a counterexample,

proove or disproove
[hide]
Give the correct solution of the terms[/hide]

[/hide]

First of all, this is nearly incoherent.  Second of all, if you assert that $X$ always happens and I give a counter-example, that [i]is[/i] a disproof.

[quote]Give the correct solution ...[/quote]

read

and have a 10^i  on the sum

i forgot

hehe


Jason counts up from 1 to 9 skipping the even numbers, and then immediately counts down without skipping any of the numbers, and then back up to 9 skipping the evens, and so on, alternately counting up and down
\[ (1, 3,5,7,9,8,7,6,5,4,3,2,1,3,5,\ldots ).\] What is
the $ 1000^{\text{th}}$ integer in his list?

First, see how many numbers are in the sequence going one time through. 1,3,5,7,9,8,7,6,5,4,3,2,1, this is 14 numbers. We are trying to find the 1000th number, so we do 1000/14, and we come up with 71 with a remainder of 6. This means that we can go through the sequence 71 times, and then the 6th number of the sequence is the 1000th number. We find that the 6th number is 8, so ans=8 This may not be correct, but I hope it is. :)

I just realized a mistake. There are actually only 12 numbers in the sequence. :oops: Lol, I can't count. So 1000/12 is 83 remainder 4. So we go through the sequence 83 times, and the 4th number, 7=ans. Sorry about that.


The following holds thanks to AM-GM: \[\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)}= \frac{mn}{(m+n+1)(m+1)(n+1)}\leq \frac{\sqrt[6]{mn}}{12}.\] But $m,n\in\mathbb{N}\ \Rightarrow\ mn\geq 1$, so $\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)}\leq \frac{1}{12}$ with equality iff $m=n=1$.

In fact, $f(2,2)=\frac{4}{45}$ and is the maximum. This can be proven easily when looking at the equality case. (Yes, Kurt, yes, always the same ;))

$\frac{m}{2}+\frac{m}{2}+\frac{n}{2}+\frac{n}{2}+1\ge 5\sqrt[5]{\frac{m^{2}n^{2}}{16}}$,
$\frac{m}{2}+\frac{m}{2}+1\ge 3\sqrt[3]{\frac{m^{2}}{4}}$,
$\frac{n}{2}+\frac{n}{2}+1\ge 3\sqrt[3]{\frac{n^{2}}{4}}$.

So if $mn\ge4$ it is proven that $f(m,n)\le\frac{4}{45}$. For $mn<4$ it is trivial, since that means WLOG $m=1$ and $n\le3$.

Excuse me, [b]Peter VDD[/b], can you elaborate some more? Plugging those inequalities into $\frac{mn}{(m+1)(n+1)(m+n+1)}$ I get quite an ugly expression, something like $\frac{2^{32/15}}{45(mn)^{1/15}}$...

EDIT: Oh, I see, now you split that into $mn\geqslant 4$ and $mn<4$... Got it.


Define a space $ X$ to be $ T_1$ if given any $ x$ in $ X$, the singleton set $ \{x\}$ is a closed set.

Define space $ X$ is said to be limit point compact if every infinite subset of $ X$ has a limit point in $ X$.

Define a space $ X$ to be countably compact if every countable open cover has a finite subcover.

Show that for any $ T_1$ space $ X$, $ X$ is limit point compact if and only if $ X$ is countably compact.

$ \Rightarrow$

Assume that $ X$ is limit point compact. Let $ \{U_i\}_{i\equal{}1}^{\infty}$ be an open cover. If there is no finite subcover then for each $ n \geq 1$, there exists $ x_n \notin \cup_{i\equal{}1}^n U_i$. Because each $ x_n$ can occur only finitely many times, $ \cup_{i\equal{}1}\{x_n\}$ is an infinite set, so it must have a limit point, say $ x$. $ x$ must be in $ U_{i_0}$ for some $ i_0$, but then $ U_{i_0}$ provides a neighborhood of $ x$ that intersects only finitely many of the $ x_n$. By T1, we can remove these $ x_n$ from $ U_{i_0}$ and it will still be open. The resulting set will not intersect any of the $ x_n$, contradicting the fact that $ x$ is a limit point of $ \cup_{i\equal{}1}\{x_n\}$.

$ \Leftarrow$

Assume that $ X$ is countable compact. Let $ S \equal{} \{x_n\}_{i\equal{}1}^{\infty}$ be an infinite set. Assume $ S$ has no limit points. Then $ S$ is closed and each $ x_n$ has a neighborhood $ V_n$ that is disjoint from all the other $ x_n$. $ X\minus{}S$ together with all the $ V_n$ then form an open cover of $ X$. But $ V_n$ is the only set in that collection that contains $ x_n$, so there cannot be a finite subcover, which is a contradiction.

oh.. i realised my mistake...

anyone has the example of a non-$ T_1$ space which is countable compact but not limit compact?

You won't find one. I didn't use T1 in my proof of $ \Leftarrow$.

My apologies. I was too tired. :blush:
I meant a non-$ T_1$ space which is [b]limit compact[/b] but [b]not countable compact[/b].

Let $ S \equal{} \{a,b\}$ be a set with two elements and the indiscrete topology (i.e. $ \{\{a,b\},\emptyset\}$). Let $ T \equal{} \{t_1,t_2,t_3,...\}$ be a countably infinite set with the discrete topology (i.e. every subset is open).

Consider the product space $ S \times T$. Any nonempty set (in particular, any infinite set) in this space must contain either $ (a,t_i)$ or $ (b,t_i)$ for some $ t_i$. In the first case $ (b,t_i)$ is a limit point. In the second case $ (a,t_i)$ is a limit point. So, every nonempty set (in particular, every infinite set) has a limit point.

This space is not countably compact because the open cover $ \left\{\{(a,t_1),(b,t_1)\},\{(a,t_2),(b,t_2)\},\{(a,t_3),(b,t_3)\},...\right\}$ clearly has no finite subcover.


Find the smallest constant $ k>0$ such that $ \frac{ab}{a\plus{}b\plus{}2c}\plus{}\frac{bc}{b\plus{}c\plus{}2a}\plus{}\frac{ca}{c\plus{}a\plus{}2b} \le k(a\plus{}b\plus{}c)$ for every $ a,b,c>0$.

[hide]If we plug in $ a\equal{}b\equal{}c\equal{}1$, we get $ k\geq \frac{1}{4}$. To show that this value of $ k$ works, multiplying by $ 4(ab\plus{}bc\plus{}ca)$ gives
\[ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}\sum_\text{cyc} \frac{4ab(ab\plus{}bc\plus{}ca)}{a\plus{}b\plus{}2c}\]
\[ \equal{}\sum_\text{cyc} \frac{ab((a\plus{}b\plus{}c)(a\plus{}b\plus{}2c)\minus{}4(ab\plus{}bc\plus{}ca))}{a\plus{}b\plus{}2c}\]
\[ \equal{}\sum_\text{cyc} \frac{ab((a\minus{}b)^2\plus{}c((c\minus{}a)\minus{}(b\minus{}c)))}{a\plus{}b\plus{}2c}\]
\[ \equal{}\sum_\text{cyc} \frac{ab(a\minus{}b)^2}{a\plus{}b\plus{}2c}\plus{}abc\left( \frac{a\minus{}b}{b\plus{}c\plus{}2a}\minus{}\frac{a\minus{}b}{a\plus{}c\plus{}2b}\right)\]
\[ \equal{}\sum_\text{cyc} \frac{ab(a\minus{}b)^2}{a\plus{}b\plus{}2c}\minus{}\frac{abc(a\minus{}b)^2}{(a\plus{}c\plus{}2b)(b\plus{}c\plus{}2a)}\]
\[ \equal{}\sum_\text{cyc} ab(a\minus{}b)^2\left( \frac{1}{a\plus{}b\plus{}2c}\minus{}\frac{c}{(a\plus{}c\plus{}2b)(b\plus{}c\plus{}2a)}\right)\]
After clearing denominators, it's equivalent to showing
\[ 0\leq \sum_\text{cyc} ab(2a^2\plus{}5ab\plus{}2b^2\plus{}2ac\plus{}2bc\minus{}c^2)(a\minus{}b)^2\]
WLOG let $ a\geq b\geq c$, then the $ (a\minus{}b)^2,(a\minus{}c)^2$ terms are nonnegative. As $ (a\minus{}c)^2\geq (b\minus{}c)^2$, it remains to show
\[ 0\leq a(2a^2\plus{}5ac\plus{}2c^2\plus{}2ab\plus{}2bc\minus{}b^2)\plus{}b(2b^2\plus{}5bc\plus{}2c^2\plus{}2ab\plus{}2ac\minus{}a^2)\]
But this is evident, as it is at least $ a(2ab\minus{}b^2)\plus{}b(2ab\minus{}a^2)\equal{}ab(a\plus{}b)\geq 0$.[/hide]

wow, so much computations  :o i don't think they are necessary... you can try using that $ \frac{1}{x\plus{}y}\leq \frac{1}{4}\left(\frac{1}{x}\plus{}\frac{1}{y}\right)$.
so you get $ \sum\frac{ab}{a\plus{}b\plus{}2c}\leq \frac{1}{4}\sum\frac{ab}{a\plus{}c}\plus{}\frac{ab}{b\plus{}c}\equal{}\frac{1}{4}$... i think this inequality appears also in old&new inequalities, and the given reference is Gazeta Matematica, [Viorel Vajaitu, Alexandru Zaharescu]...


hi

I just cant seem to solve this problem and need help.

5x^7 - 21x^6 - 51x^5 + 141x^4 +264x^3 + 342x^2 -560x + 96 = 0

And the other one:

1 + (x^2/2) + (x^3/6) = 0

The methods they want you to use are newton's method an dthe bisection method.
horners method can also be used.

[quote="numero1"]hi

I just cant seem to solve this problem and need help.

$5x^7 - 21x^6 - 51x^5 + 141x^4 +264x^3 + 342x^2 -560x + 96 = 0$

And the other one:

$1 + \frac{x^2}{2} + \frac{x^3}{6} = 0$

The methods they want you to use are newton's method an dthe bisection method.
horners method can also be used.[/quote] :lol:


What common fraction is one-third of the way from $ \frac{1}{4}$ to $ \frac{1}{3}$ on a number line?


In a paralellogram $ ABCD$ , $ AD \parallel EF$ and the ratio of area $ I,II,II$ is $ 2: 3: 6$, then $ DF: FG: GC\equal{}$?

[hide="Solution"]
Assume that $ [\text{I}]\equal{}2$, $ [\text{II}]\equal{}3$, $ [\text{III}]\equal{}6$.

Draw $ AF$, which splits region $ \text{I}$ in half. Then $ [AFD]: [\text{II}]\equal{}1: 3$. They have the same heights, so the ratio of their areas is the ratio of their bases; therefore, $ DF: FG\equal{}1: 3$.

$ [\text{II}\plus{}\text{III}]\equal{}9$. Draw $ EC$, which splits $ EFCB$ in half, so that $ [ECB]\equal{}4.5$. Then $ [EGC]\equal{}6\minus{}4.5\equal{}1.5$. Again, $ \triangle EFG$ and $ \triangle EGC$ have the same heights, so the ratio of their areas is the ratio of their bases, which must be $ 3: 1.5$.

Therefore, your answer is $ \boxed{2: 6: 3}$.
[/hide]

[hide="Solution"]Let $ DF\equal{}x, FG\equal{}y$, and $ GC\equal{}z$.

Then, we will have $ AE\equal{}x$ and $ EB\equal{}y\plus{}z$.

WLOG, let $ I\equal{}2$, $ II\equal{}3$, and $ III\equal{}6.$

Then, we will have $ x\equal{}2, \frac{y}{2}\equal{}3, \frac{y\plus{}2z}{2}\equal{}6$.

Solving for $ x, y, z$, we have $ x: y: z\equal{}2: 6: 3$.

Answer: 2:6:3[/hide]


function
topology
real analysis
real analysis solved

$E$ is a measurable set. $f$ is defined on $E$. They satisfy that: for any given $\epsilon > 0$, there always exists a closed set $F\subseteq E$ such that $m(E\setminus F) < \epsilon$ and $f$ is continuous on $F$.
Prove that $f$ is measurable.

eh...I solve it myself now...

My solution goes like this:

Suppose $F_n$ be the closed set $F$ corresponding to the given [tex]\epsilon = \frac{1}{n}[/tex].

Let [tex]M = \bigcup\limits_{n=1}^\infty F_n[/tex] and $N=E\setminus M$.

\begin{eqnarray*}
mN &=& m(E-M) = m(E-\bigcup F_n) \\
&=& m(E\cap (\bigcup F_n)^c)\\
&=& m(E\cap (\bigcap F_n^c))\\
&=& m(\bigcap E\cap F_n^c)\\
&=& m(\bigcap E\setminus F_n)\\
&\leq& m(E\setminus F_n) < \frac{1}{n}
\end{eqnarray*}

So $mN = 0$

$\{x\in E:f(x)>a\} = \{x\in M: f(x)>a\} + \{x\in N: f(x)>a\}$

From $mN = 0$ we know that the measure of $\{x\in N: f(x)>a\}$ is zero.

And $\{x\in M: f(x)>a\} = \bigcup\limits_{n=1}^\infty \{x\in F_n: f(x)>a\}$

Since $F_n$ is closed and $f$ is continous, each $\{x\in F_n: f(x)>a\}$ is measurable, therefore $\{x\in M: f(x)>a\}$ then $\{x\in E:f(x)>a\}$ is measurable, hence $f$ is measurable on $E$.


We have given the following grid:

SAIPPUAKAUPPIAS
AIPPUAKAUPPIASA
IPPUAKAUPPIASAI
PPUAKAUPPIASAIP
PUAKAUPPIASAIPP
UAKAUPPIASAIPPU
AKAUPPIASAIPPUA
KAUPPIASAIPPUAK
AUPPIASAIPPUAKA
UPPIASAIPPUAKAU
PPIASAIPPUAKAUP
PIASAIPPUAKAUPP
IASAIPPUAKAUPPI
ASAIPPUAKAUPPIA
SAIPPUAKAUPPIAS

We can start at any S and every turn you can go one letter up, left, down, or right. You can go to the same position more than once. Compute how many ways you can read the Finnish palindrome "saippuakauppias" from the grid.

You sell soap?
[hide="Hint"]Pascal's for the first half[/hide]

[quote="mathgeniuse^ln(x)"][quote="puuhikki"]I don't see how to use Pascal's triangle here. What do you mean be first half?[/quote]

He means consider the top trianlge that spells whatever it is.

Then notice that for each row, the number of ways it can spell is $2^{r-1}$, where r is the row number.

So since there are 15 letters, there will be 2^14, for the top part and 2^14 for the bottom part

So the answer is 2^15=32768.[/quote]

close.  notice that it is a palindrome

[quote="mathgeniuse^ln(x)"]Okay, I am sorry that I didn't notice that. :wink:

Anyways, since it is a palindrome you multiply it by 2 whatever I had, so the answer is 65536[/quote]

umm no.  the answer is 33something something something. :maybe:

for some reason i keep seeing the word asian in there but theres no n :D

[quote="puuhikki"][quote="mathgeniuse^ln(x)"]So the answer is 2^15=32768.[/quote]
Unfortunately no. The correct answer is [hide]334544.[/hide] Try again.[/quote]

Oh, I didn't see that.

Where is this contest from?

I don't see how you can get 334544, a way large number.

The problem is from Suomen Kuvalehti, a Finnish magazine, from 1920s. I saw the solution from another forum:

The basic idea is to compute separatelly from every K-lettered square how many ways we can form a word KAUPPIAS. Is this number is $n$ for some K-square, we can form a word SAIPPUAKAUPPIAS in $n^{2}$ ways, where the specific letter K is in the middle of the word. This is because every SAIPPUAKAUPPIAS-route we have to go back to some KAUPPIAS-word beginning from the first S to the letter K. After that we have to go along some KAUPPIAS (which can be different from the first KAUPPIAS) beginning with K to the last S.

We have 16 K-squares, so the number in the first half is $\sum_{i=0}^{15}N_{i}^{2}$. The grid is symmetric to its middle point so we take the sum twice: $2\sum_{i=0}^{15}N_{i}^{2}$, as K-squares are numbered from up to down and left to right.

Now the problem is to compute every $N_{i}$. This can be done by noting that when we move in the grid one square either down or right, we move also one letter right on the word SAIPPUAKAUPPIAS. Respective, when moving up or left, we move one letter left in the word SAIPPUAKAUPPIAS.

It can be seen that for example by moving from the third K-square up (corresponding th index $i=2$) we can form the KAUPPIAS-route only by moving always right_or_down or left_or_up but being always inside the grid. The number of different right/down routes is $2^{7}$ and the number of different left/up routes is $\binom{7}{i}=\binom{7}{2}$. This can be generalized in every K in the upper left part of the grid so we get the formula
\[2\sum_{i=0}^{7}\left (2^{7}+\binom{7}{i}\right )^{2}= 334544. \]

Very Hard :wink:

And, yeah, that is a neat solution.

But if Blah, wasn't a palindrome, then my first answer would've been correct

[quote="mathgeniuse^ln(x)"]Very Hard :wink: [/quote]
Yeah. I thought that the solution would be easier but I didn't realize its hardness before I saw the solution.


find the least positive integer s such that 625s+38 is a multiple of 4096

[quote="t0rajir0u"]Consult the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]Extended Euclidean algorithm[/url].[/quote]

umm how would that be useful to this problem? :?:

$ 4096 | 625s \plus{} 38 \Leftrightarrow$
$ \exists t : 625s \plus{} 38 \equal{} 4096t \Leftrightarrow$
$ 4096t \minus{} 625s \equal{} 38$

Now, if you can find integers $ x, y$ such that

$ 4096x \plus{} 625y \equal{} \gcd(4096, 625) \equal{} 1$

then

$ 4096(38x \plus{} 625k) \plus{} 625(38y \minus{} 4096k) \equal{} 38$

for any integer $ k$.  It remains to vary $ k$ to find the minimal value of $ s \equal{} 4096k \minus{} 38y$.

you can use mods to reduce the problem quite a bit although i think it may be the similar to using Euclidean's algorithm and this process is somewhat long and tedious.
[hide = solution using mods]
625s + 38 = 4096t
38 = 4096t = 346t mod 625
346t- 38 = 0 mod 625
346t - 38 = 625u
-38 = 625u = 279u mod 346
279u +38 = 0 mod 346
279u +38 = 346v
38 = 346v = 67v mod 279
67v - 38 = 279w
-38 = 11w mod 67
11w + 38 = 67x
5 = x mod 11
you want the minimum x to give you the minimum s so x = 5.
then back track with a calculator or something... and you get s = 1658.
[/hide][/hide]

woops... i dont think my hiding exactly worked..

The process you described is exactly equivalent to the extended Euclidean algorithm, and yes, it is tedious either way.


Let $ a>b,m>n$ be positive integers.Let $ gcd(m,n)\equal{}d$.Prove that $ gcd(a^m\minus{}b^m,a^n\minus{}b^n)\equal{}a^d\minus{}b^d$.

[quote="KDS"]Let $ a > b,m > n$ be positive integers.Let $ gcd(m,n) \equal{} d$.Prove that $ gcd(a^m \minus{} b^m,a^n \minus{} b^n) \equal{} a^d \minus{} b^d$.[/quote]

That's obviously wrong : Choose $ (a,b,m,n,d)\equal{}(4,2,3,2,1)$ and you get $ \gcd(4^3\minus{}2^3,4^2\minus{}2^2)\equal{}4\ne 4^1\minus{}2^1\equal{}2$

of course ,
KDS must have mentioned $ \gcd(a,b)\equal{}1$.
well, for a proof...
obviously we have $ a^d\minus{}b^d |gcd(a^{m}\minus{}b^{m},a^{n}\minus{}b^{n})$ so let's assume that there exists a prime $ p$ which isn't in $ a^d\minus{}b^d$ but it is in $ gcd(a^{m}\minus{}b^{m},a^{n}\minus{}b^{n})$.
just see that this prime can't be $ 2$.
so it must be an odd prime.
let's assume that $ k$ is the smallest number for which we have
$ p|a^k\minus{}b^k \implies v_p(a^m\minus{}b^m)\equal{}v_p(a^k\minus{}b^k)\plus{}v_p(\frac{m}{k})$.
(note that we can easily prove $ k|m,n$ by assuming $ k \nmid m,or,n$ and getting a contradiction)
and the same case is for $ a^n\minus{}b^n$.
but just see that we must have $ k|m,n \implies k|d$.
so this prime also exists in $ a^d\minus{}b^d$.
so we must assume that there exists a prime $ p$ but doesn't have the same power in $ a^d\minus{}b^d$ and
$ gcd(a^{m}\minus{}b^{m},a^{n}\minus{}b^{n})$.
but by the same method mentioned above we can see that
$ min(v_p(\frac{m}{k}),v_p(\frac{n}{k}))\equal{}v_p(\frac{d}{k})$.
if $ p\equal{}2$ we must use this formula $ v_2(a^l\minus{}b^l)\equal{}v_2(\frac{a^2\minus{}b^2}{2})\plus{}v_2(l)$
(with the assumption $ 2 |l$ if not then $ v_2(a^l\minus{}b^l)\equal{}v_2(a\minus{}b)$).
so we r done!

[quote="pco"][quote="shoki"]of course ,
KDS must have mentioned $ \gcd(a,b) \equal{} 1$. ...[/quote]

Yes, new kind of olympiad problems more and more frequent : give a wrong problem statement and wait for people modify it and solve the self-modified problem. :(

Pco,
who thinks people nowadays are unable to simply copy a less_than_two_lines text without error[/quote]
allez,ne soyez pas aussi dur... :wink:

The equality does hold in $ \mathbb{Z}[a, b]$, though.