function
inequalities
combinatorics proposed
combinatorics

[quote="k2c901_1"]Every square on a $n\times n$ chessboard is colored with red, blue, or green. Each red square has at least one green square adjacent to it, each green square has at least one blue square adjacent to it, and each blue square has at least one red square adjacent to it. Let $R$ be the number of red squares. Prove that $\displaystyle \frac{n^2}{11} \le R \le \frac{2n^2}{3}$.[/quote]Is diagonal adjacency counted?

I'll assume diagonal adjacency is not counted.

Let X be the set of all cells in the chessboard. Then we know #X = $n^{2}$.
Let R, G and B be the set of all RED cells, the set of all GREEN cells, and the set of all BLUE cells. you know #R+#G+#B = #X
For x in X, let N(x) = {y in X | y is adjacent to x }. I know #N(x) <= 4.

There exists a function F : X -> X satisfying the following three conditions :
(1) If r $\in$ R then F(r) $\in$ G $\cap$ N(r).
(2) If g $\in$ G then F(g) $\in$ B $\cap$ N(g).
(3) If b $\in$ B then F(b) $\in$ R $\cap$ N(b).

F is fixed from now on.

R_1 = { r in R | r, F(r) and F(F(r)) are either on the same row or on the same column }
R_2 = R \ R_1

For example, if r $\in$ R_1 and g = F(r) and b = F(F(r)) then the arrangement of r, g and b must be one of the following 4 forms.
[code]
r       b
g       g       r g b       b g r
b       r
[/code]

If r $\in$ R_2 then the arrangement must be one of the following 8 forms.
[code]
r g     b   r       r
  b   r g   g b   b g   ...  and the rest 4 cases.
[/code]

We'll say a function f : Y -> Z is [b]at-most-d to one[/b] if for each z $\in$ Z we have $\# f^{-1}(z)$ <= d. Remember that then we have #Y <= d #Z.

Claim:  the six important inequalities.

#R_1 <=   #G
#R_2 <= 2 #G
#R_1 <= 3 #B
#R_2 <= 4 #B   (they are 1,2,3,4. beautiful)

#R <= 3 #G
#R <= 7 #B

Proof: 5th ineq follows from 1st and 2nd. 6th ineq follows from 3rd and 4th. Now we only have to prove the first 4 ineqs.
Let $(F\mid R_{1})$ be the restriction of F on R_1. Considering possible rgb arrangements, we can show that (F|R_1) is injective. Hence the 1st ineq.
Also we can show the function $(F\mid R_{2})$ is at-most-2 to one, considering rgb arrangements. Hence the 2nd ineq.
$(F^{2}\mid R_{1})$ is at-most-3 to one (Considering that one of N(b) is red, thus blocking one possible pre-image of F^2. Hence at-most-3. ) Hence the 3rd ineq.
$(F^{2}\mid R_{2})$ is at-most-4 to one. Hence the 4th ineq.
Done the six inequalities.

Apply the proof of 5th and 6th inequalities to cyclic permutations of R,G and B, we have another four ineqs.

#G <= 3 #B
#G <= 7 #R

#B <= 3 #R
#B <= 7 #G

We've collected enough ineqs to do the lower and the upper bound of #R.

The lower bound.
11 #R = #R + 7 #R + 3 #R >= #R + #G + #B = #X

The upper bound.
Recall the four inequalities which I called 'beautiful'.
To get the upper bound, we do   2 (1st ineq) + (2nd ineq) + (4th ineq).
2 #R <= 2 #R_1 + #R_2 + #R_2 <=  2 #G + 2 #G + 4 #B = 4 (#G+#B) = 4 ( #X - #R)
Hence, 3 #R <= 2 #X .

Done the 1/11 <= #R/#X <= 2/3. sloved it.

I'm wondering if there is a nice at-most-2 to 1 function from R to G $\cup$ B. That would make a delicious proof of #R/#X <= 2/3. Let's try to find it.

- [edit]
Remark: what's done to get the lower bound. If we apply it to get an upper bound, we get #R/#X <= 21/31. The motivation for partitioning R into R_1 and R_2 is to lower this upper bound from 21/31 to 2/3.


linear algebra
matrix
geometry
algebra
linear equation

Lets say I just have the determinant of a matrix. With just that, what would be the formula to determine the matrix contents (ie which numbers it contains, and in which row and column) or if there is a way to change the matrix into a formula. Any help would be appreciated.

There is no such formula or method; you can't possibly get $ n^2$ independent bits of information from one.

Another reason is interpreting the determinant as an area, there is no way to give the surface to which the area applies. An infinite number of shapes may have a given area.

Well is the any formula that you could run against a matrix to be able to quickly "decode" it back to the matrix?

What technique will work best for your purposes depends on specifically what you're trying to do (and what you mean by "quickly":  asymptotically or by hand?), so could you provide more details?

Those pieces of information aren't enough to uniquely recover the matrix unless you at least make the simplifying assumption that the matrix you want is $ 3 \times 3$ and has integer coefficients, and even then it's far from obvious that the solution is unique.

One small point: the statement that its rank is 3, together with its nonzero determinant, do fix its size as $ 3\times 3.$ But beyond that, there are four items of information, which shouldn't uniquely determine 9 elements.


Pascal\u0027s Triangle
binomial coefficients
number theory unsolved
number theory

For what value of n does  $ (a + b)^n $  yield only even coefficients for  $a^rb^{(n-r)} $   for  $ 0 < r < n $

I guess this is equivalent to which rows of Pascal's Triangle have all even entries apart from the obvious 1's on the ends. I think the answer is $ 2^n  $ for  n greater than or equal to 1. Can this be proved?

Also,
Can you generalise this to a result about the coefficients taken  $ mod  p$ for  $ p $ a prime?
Ie. Even coefficients in the rows of Pascal's Traingle after taking the values mod p

[url=http://planetmath.org/encyclopedia/LucassTheorem.html]Lucas' Theorem[/url] should answer this for general primes $p$: the binomial coefficients $\binom nk$ are divisible by $p$ for all $k\in\overline{0,n}$ iff $n$ is a power of $p$.

but what about the case where the coefficients aren't divisible by P, but leave an even number mod P. That would still count. I don't get the link. I'm not sure it's what I'm after.

I understood that you wanted "even residues mod $p$", but looking for the case when the binomial coefficients are divisible by $p$ instead simply seems like a more natural problem, and it's a generalization of your first question: when are the coefficients even? The answer is: when $n$ is a power of $2$. At least, this answers that question.

"even residue mod p" doesn't even make sense unless $2|p$; after all, $7\equiv12\mod 5$.

Of course :), but you can talk about residues whose representatives among $0,1,\ldots,p-1$ are even. Maybe that's what he meant. Anyway, if this is indeed what he meant, it still looks like a totally useless track to pursue.


In C^3 over the complex let:
e1 = (1, 0, -i ), e2 = (1+i, 1 - i, 1), e3 = (i, i, i )
Show that the vectors e1, e2, e3 are a basis for C^3. Express the vector (a, b, c) as a linear combiantion of the basis vectors.

Show that $ x(1,0,\minus{}i)\plus{}y(1\plus{}i,1\minus{}i,1)\plus{}z(i,i,i)\equal{}(0,0,0)\Rightarrow x \equal{} y \equal{} z \equal{} 0$ which is very easy. For second part simply solve sistem $ x(1,0,\minus{}i)\plus{}y(1\plus{}i,1\minus{}i,1)\plus{}z(i,i,i)\equal{}(a,b,c)$.


calculus
integration
complex analysis
complex analysis unsolved

Calculate integral on segment $[-i, i]$ on imaginary ax, by taking branch such $\frac{\sqrt{1+z^{2}}}{z}\rightarrow-1$ when $z \rightarrow \infty$.
\[\int \frac{\sqrt{1+z^{2}}}{z^{2}-1}\]

If I let $z=ix$ and neglect the change of contour, I get the equivalent integral:

$(-i) \int_{-1}^{1}\frac{\sqrt{1-x^{2}}}{1+x^{2}}= (-i) (-\arcsin{x}+\sqrt 2 \arctan{\( \frac{\sqrt 2 x}{\sqrt{1-x^{2}}}\) })$ evaluated between $-1$ and $1$ which gives:  $(-i) (-\pi+\sqrt 2 \pi)$.


Does anyone else think there should be a forum for high school olympiad problems?  In particular, I'm thinking of USAMO/IMO/CMO/whatever other national olympiad.  Advanced and Intermediate can sometimes go past high school...

[quote="probability1.01"]Advanced and Intermediate can sometimes go past high school...[/quote]

Are you saying that there are challenge problems on Intermediate forum that are harder than USAMO problems? That shouldn't be. Advanced is difficulty level unbounded above, and thus includes Putnam and the like, but I sure wouldn't call USAMO only Intermediate.

What does everyone else think about the "natural" levels of various problems in the curent three-way division among Getting Started, Intermediate, and Advanced? There seem to still be too many too difficult problems in the lower two forums.

I agree that the Advanced is the proper forum for high school Olympiad problems (with the exception of the rare very easy one that might be only Intermediate).

[quote="tokenadult"]What does everyone else think about the "natural" levels of various problems in the curent three-way division among Getting Started, Intermediate, and Advanced? There seem to still be too many too difficult problems in the lower two forums.[/quote]

Too many difficult problems? :?

I think the beginners should be the more simple questions, like the mock AMC tests and so on, perhaps a little higher.

Intermediate should be the olympiad questions from smaller countries, or any problem that can be solved by an untrained but intelligent mind

Advanced has all above that, includen USAMO but that is already quite hard, even for the advanced section. Most of the advanced section is much easier than USAMO.

Keep in mind the audience of the site - we have students who are just beginning, as in ages 11-14.  The Getting Started is for them.

The Intermediate is for students who progressed beyond the basics but are not yet at the olympiad level.  In other words, the problems you're describing as for the previous level.

The Other Problems section of the Advanced problems is for the beginning olympiad problems (and those that don't fit the other subjects).


inequalities
trigonometry
function
inequalities unsolved

Let $x,y,z \in [0,1)$ with  $x+y+z=1$  Prove that

$\sqrt{\frac{xy}{z+xy}}+\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{y+zx}} \leq \frac{3}{2}$

It's equivalent to $\sum\sqrt{\frac {xy}{(z+x)(y+z)}}\le \frac 32$

I think similar things have been solved here (I remember struggling on this a while back.

You are right Grobber, but since I do not remember which topic it was and since the proof is short I will give it once again :
Using your remark, it now suffices to use AM/GM to get
$\sqrt { \frac {xy} {(x+z)(y+z)} } \leq  \frac 1 2 ( \frac x {x+z} + \frac y {y+z} )$.
We get similar inequalities for the other terms, and summing you get the result.

Pierre.

Thanks for this remark, Grobber! Indeed, now we have a very simple proof:

Consider the triangle ABC with the sidelengths a = y + z, b = z + x and c = x + y. Then, the semiperimeter s of triangle ABC equals $s=\frac{1}{2}\left( a+b+c\right) =x+y+z$, and we have s - a = x, s - b = y and s - c = z. If A, B, C are the angles of triangle ABC, then

$\displaystyle \sin \frac{A}{2}=\sqrt{\frac{\left( s-b\right) \left( s-c\right) }{bc}}=\sqrt{\frac{yz}{\left( z+x\right) \left( x+y\right) }}=\sqrt{\frac{yz}{x\left( x+y+z\right) +yz}}=\sqrt{\frac{yz}{x+yz}}$.

Similarly,

$\displaystyle \sin \frac{B}{2}=\sqrt{\frac{zx}{y+zx}}$

and

$\displaystyle \sin \frac{C}{2}=\sqrt{\frac{xy}{z+xy}}$.

Hence, instead of proving

$\displaystyle \sqrt{\frac{xy}{z+xy}}+\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{y+zx}} \leq \frac{3}{2}$,

it is enough to show

$\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}\leq \frac{3}{2}$.

But this trivially follows from Jensen, since the function f(x) = sin x is concave on the interval $\left[ 0;\;\frac{\pi }{2}\right]$, and hence we have

$\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}\leq 3\cdot \sin \left( \frac{1}{3}\cdot \left( \frac{A}{2}+\frac{B}{2}+\frac{C}{2}\right) \right) =3\cdot \sin \frac{\pi }{6}=\frac{3}{2}$.

This proves everything. Okay, Pierre, your proof is much simpler, but you know I'm a triangle geometer...

  Darij

:D
I'm not quite agree with you Darij. The proof I gave (I'm sure that someone else gave it here too so it is not 'my' proof) is shorter but geometic proofs of inequalities are often quite difficult to find and elegant, in my point of view.

Pierre.

[quote="pbornsztein"]but geometic proofs of inequalities are often quite difficult to find and elegant[/quote]

Actually, for me the geometric interpretation was really straightforward. I am one of those who always have in mind triangles and the a = y + z, b = z + x, c = x + y ("Ravi") substitution when proving inequalities for three numbers...

  Darij

Both solutions are very nice! I enjoy reading both. Thanks! Cheers!  :D  :D  :D


A square region is cut into polygonal regions by two straight lines. What is the maximum number of sies any one of the non-overlapping polygonal regions could have?

When you make this, try not to make a line intersect at a corner.  :wink:

I thought I got 7 but I am not sure if that is possible.

No, $ \boxed{6}$ is the highest possible number of regions.

Lock this.  :roll:


function
calculus
derivative
limit
real analysis
real analysis solved

Let f be a two times derivable function on an interval that contains 0, whose second derivative is continous. Find the limit of the expression $\frac{f(x+y)-f(x)-f(y)+f(0)}{xy} $ when $ (x,y)$ tends to $(0,0)$. This is really a super nice problem with a super nice solution.

Let $L$ be the finite limit of $f''$ in 0 : $ \displaystyle L = \lim_{x \to 0} f''_{(x)}$

Let $m,M$ be such that : if $|y| \leq |x|$ then $m(x,y) = y, M(x,y)=x$ , else $m(x,y)=x , M(x,y)=y$ . Manifestly $|\frac{m}{M}| \leq 1$

Denote $g$ be $g(x,y) = \frac{f(x+y)-f(x)-f(y)+f(0)}{xy}$ . Manifestly $g(x,y) = g(m,M)$

By using the Taylor developpement, we define $u,v,w$ such that :

$f(m+M)- f(M)= mf'(M) + \frac {m^2f''(u)}{2}$ , where $|u-M| \leq |m|$

$f(m) - f(0) = mf'(0) + \frac {m^2f''(v)}{2}$ , where $|v| \leq |m|$

$f'(M) - f'(0) = Mf''(w)$ , where $|w| \leq |M|$

The initial expression becomes:

$h(x,y)=h(m,M) = f''(w) + (f''(u)-f''(v)) \frac {m}{2M} $

It's easy to see that $(x,y) \rightarrow 0 ===> (m,M) \rightarrow 0 ===>  (u,v,w) \rightarrow 0  ===> h(x,y)$ converges and $h(x,y) \rightarrow L$

:cool:   need you the continuity of $ f''$ ?! :cool:

I think you only need to know that it has a limit in $0$.

:D  Really , Grober ??!!!   Very interesting ! :D


Let an [i]L-tile[/i] be a 2x2 sqaure tile with one of its 1x1 tiles removed from it. Is it possible to completely tile a 5x9 rectangle with L-tiles?

[hide="A side note..."]This problem had a bunch of people stumped for quite a while. It was hilarious when they found out the simple solution.[/hide]

... nvm i lied xD i cant count

Nice solution to
|||
vvv

[quote="Castor496"][hide]I think this is right. :maybe: You need to arrange 15 L-tiles like this. [img]http://img248.imageshack.us/img248/7089/ltilesbq4.jpg[/img][/hide]

 :?:[/quote]
Correct. People couldn't get it at first because the answer was "Yes", but the arrangement wasn't completely obvious at first glance, so they assumed using the axioms of problem solving that the answer was "No".


You don't need a calculator (although the thinking is much higher than classroom level, PenguinIntegral, you can move this please if you want).

[hide]900=30^2, so it's higher than 30. 1600=40^2, so it's less than fourty. The only number that gives a unit digit of 6 when squared are 4 and 6, so just try both of them and you will get the answer of 36. [/hide]

[quote="anirudh"][quote="joejia"]Calculate: $\sqrt{1296}$.[/quote]

36[/quote]

Did you just use a caculator? If not, you should tell us how you got to that answer. If so, you should hide your answer.

[quote="joejia"]Calculate: $\sqrt{1296}$.[/quote]
[hide]
30x30 = 900
40 x40 =1600

so the answer is more than 30, but less than 40.

35x35 = 1225
36x36 = 1296

teh answer is 1296.
[/hide]

[hide]
By using prime factorization, you get $\sqrt{(3^4 * 2^4)} = 3^2 * 2^2 = \framebox{36}$ [/hide]

[hide]
i would use guess and check
lets try 30
that equals 900
how about 35
that equals 1225
so we are close...
lets try 36
that equals 1296
[/hide]

correct, that was intended for anirudh.

by the way, you have 500 posts.  :)

oh, sorry. he's offline anyway.

yay! 3 stars. :)

The way I do it is

Draw a diagram much like we do with long division. In fact, the whole method is very much like long division. We mark off every other digit of the number under the square root sign, starting at the decimal point. Every two digits of that number produces one digit of the answer.

We estimate the first digit of the answer: greater than 4 and less than 5, so we choose 4. Now we put that 4 to the left of the square root sign. We multiply 4x4 and get 16, which we put under the 19. We subtract and get 3, and we bring down two digits (00 in this case), making 300.

Here's the part that is different from long division (where the divisor stays the same): We double the four on the left, making 8, and put that to the left of the 300. Now we estimate what times 80+what is just less than 300. 4x84 is greater than 300, so 3x83 is what we want. 3 is the next digit of the answer, and we put that 3 to the right of the 8 (making 83 (or 80+what)) on the left side. 3x83=249, which we put under the 300 and subtract, getting 51. We bring down 00 from above.

Now we get 86 (twice 43) on the left side. And we repeat the operation, getting 865x5 and 8708x8 and 87169x9. I did that last one to see which way to round the previous answer. So the answer is 4.359. That process may look daunting. But it is fairly fast.

From this website: [url]http://www.jimloy.com/arith/sqrt.htm[/url]

That's what our teacher taught us. Except this thread was split to the high school getting started forum  :P

that's the same website i went too. :) but really, would you ever be put in a situation where you would have to do that? if you need to find the square root for a math problem and the number is not a fairly small perfect square, there is a 99.9% chance you did something wrong with the problem.

My teacher says this is a good way to use on a "calculator without the square root button." Well, every calculator, even four-function ones, have a square root button, so this is sort of not useful.  ;)

[quote="h78reg"][quote]My teacher says this is a good way to use on a "calculator without the square root button." Well, every calculator, even four-function ones, have a square root button, so this is sort of not useful. [/quote]

if a calculator doesn't have a square root button, just raise the number to the 1/2 power.[/quote]

Usually, if the calculator doesn't have a square root button, it won't have a power button either.  ;)

then the Board of Education must fund better calculators. :D that really is sad if a school's calculators can't do square roots.

Here's the stupid part. The calculators HAVE SQUARE ROOT FUNCTIONS!!! I really don't know why we learn stupid things in school  :roll:

Why do we even need that algorithm? It never comes up in mathcounts....

[quote]
Here's the stupid part. The calculators HAVE SQUARE ROOT FUNCTIONS!!! I really don't know why we learn stupid things in school
[/quote]

same here. my school gives us TI-34's and TI-83's and still teaches this to us. as for mathcounts, i don't think anyone knows about it in my school. the school isn't really problem solving-friendly.........

I don't know......it's just another pointless thing we learn in school. Like that dumber other way to do subtraction I can't even remember how anymore.

well, i think its somewhat worth knowing. after all, theoretically, because of calculators we don't need to know how to add, subtract, multiply, or divide for that matter.

Yeah it's not as bad as some of the other stuff....

your answer is correct. I just had to quote it and it ruined it all for you.  :P
Similarly, you can square 27, add 27, stick a 25, and divide it by 100.

[quote="236factorial"]okay anirudh, what is $27.5^2$?  ;)[/quote]

quick way to square anything ending in 5 (or .5 or .05.....)
if the number's decimal representation is $ab5$
multiplying $ab$ by $ab+1$ and slapping a 25 on the end gives the square of $ab5$
example:
$755^{2}=$
$75\times76=75^{2}+75= (7\times8)$with a 25 slapped on the end $(5625)$add 75= and you get$5700,$ slap on a 25 to the end of 5700 and it gives
$\fbox{570025}$ (the correct answer)
I hope you guys can understand that

Yay shortcuts!

[quote="anirudh"][quote="236factorial"]okay anirudh, what is $27.5^2$?  ;)[/quote]

[hide="Dont click this"]
[hide="Hidden text"]
[hide="Hidden text"]
27.5^2=27.5*27.5 which by hand comes out to be 756.25

Also, for the root 1296, here's another way to do it:

30^2=900
40^2=1600

Therefore it has to be between 30 and 40.

Now: 35^2=1225, which is close.

go up to 36^2 to get 1296.
The answer is then 36

[/hide]
[/hide]
[/hide]
 :mad:[/quote]

This kind of things are stupid, and do not make MathLinkers improve. Please don't do it anymore.

sapphyre571, careful with the decimal points.

Besides, avoid reviving old old topics. That has gotten almost out of control lately on this forum.

[quote="236factorial"]Besides, avoid reviving old old topics. That has gotten almost out of control lately on this forum.[/quote]

Yes trully. Thanks for locking the other one.

$1296=900+2\cdot 30 \cdot 6 + 36=(30+6)^2$ [hide]But on the other hand if you have difficulties finding $\sqrt{1296}$ I dought you will see the factorising of $a^2+2ab+b^2$[/hide] :D

[quote="pohoatza"]lol... :rotfl: 36 :)[/quote]

Please don't start spamming here too

This topic has been discussed enough.
The answer has been given, as have several methods of obtaining the answer.
Locked.


I googled, but no answer found.
When did Japanese start rebuilding Hiroshima and Nagasaki and when did new residents check in? What kind of technique did they use to clean up the nuclear pollution? In my mind, it's a tough task even nowadays.

they rebuilt those cities? who would have thought. i would have abonded them to avoid radiation damage, but that's just me.

apparently they were rebuilt.

these atomic bombs were not nearly as powerful as the ones we have today, so not as much of the horrific radiation was released.  i believe hiroshima had a yield of like 15 kilotons; the largest bombs today have yields of like 15 megatons.

as to when, i do not know.  sry

well the aftermath was huge: many people caught leukemia... but they were, i believe, rebuilt when the whole country was regeared with army by United States, in protecting the Eastern Asian line against the Soviet Union, who pushed N. Korea to attack S. Korea. AKA Korean War.

[quote="stupidkid"]i thought the radiation is still there, so rebuilding would create a lot of problem.[/quote]They were actually rebuilt. Anyway, the area that would still have recidues of radiation would be very very small (not more than a few blocks), and even there the radiation would be very small, most of the really dangerouse radiation disipated within the next 100 hours, what would not be the case with one of the bombs nowdays! mainly because they are more than a thousand times more powerful. I recomend [url=http://www.csi.ad.jp/ABOMB/data.html]this page[/url] with interesting information about it!


Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
   $ \frac {1}{a \sqrt{2a \plus{} b \plus{} c}} \plus{} \frac {1}{b \sqrt{2b \plus{} c \plus{} a}} \plus{} \frac {1}{c \sqrt{2c \plus{} a \plus{} b}} \geq \frac {3}{2}$

It's obviously trues by Am-Gm and CS  :)

[quote="nguoivn"]It's obviously trues by Am-Gm and CS  :)[/quote]
hi nguoi vn,you can post  your solution!

[quote="2424"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
   $ \frac {1}{a \sqrt {2a \plus{} b \plus{} c}} \plus{} \frac {1}{b \sqrt {2b \plus{} c \plus{} a}} \plus{} \frac {1}{c \sqrt {2c \plus{} a \plus{} b}} \geq \frac {3}{2}$[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\frac {1}{a \sqrt {2a \plus{} b \plus{} c}}\right)^2\sum_{cyc}\frac{2a\plus{}b\plus{}c}{a}\geq\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)^3.$
Hence, it remains to prove that $ 4(ab\plus{}ac\plus{}bc)^3\geq9\sum_{cyc}ab^2c^2(2a\plus{}b\plus{}c),$ which is obvious.

[quote="2424"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
   $ \frac {1}{a \sqrt {2a \plus{} b \plus{} c}} \plus{} \frac {1}{b \sqrt {2b \plus{} c \plus{} a}} \plus{} \frac {1}{c \sqrt {2c \plus{} a \plus{} b}} \geq \frac {3}{2}$[/quote]
By Chebyshev we have :
$ LHS \ge \frac{1}{3}(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})(\frac {1}{\sqrt {2a \plus{} b \plus{} c}} \plus{} \frac {1}{\sqrt {2b \plus{} c \plus{} a}} \plus{} \frac {1}{\sqrt {2c \plus{} a \plus{} b}} ) \ge$ $ \frac{\sqrt{3}}{2}(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\frac{1}{\sqrt{a\plus{}b\plus{}c}} \ge \frac{3}{2}$   ;(  $ q^2 \ge 3p$ )

[quote="arqady"][quote="2424"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Prove that
   $ \frac {1}{a \sqrt {2a \plus{} b \plus{} c}} \plus{} \frac {1}{b \sqrt {2b \plus{} c \plus{} a}} \plus{} \frac {1}{c \sqrt {2c \plus{} a \plus{} b}} \geq \frac {3}{2}$[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\frac {1}{a \sqrt {2a \plus{} b \plus{} c}}\right)^2\sum_{cyc}\frac {2a \plus{} b \plus{} c}{a}\geq\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^3.$
Hence, it remains to prove that $ 4(ab \plus{} ac \plus{} bc)^3\geq9\sum_{cyc}ab^2c^2(2a \plus{} b \plus{} c),$ which is obvious.[/quote]

  :lol:
$ \sum_{cyc}\frac {2a \plus{} b \plus{} c}{a} \equal{} 3 \plus{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$
We have:
$ 4(ab \plus{} bc \plus{} ca)^3 \ge 12(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \ge 9(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \plus{} 27\equal{}9[(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}3]$
So:
$ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})^3 \ge \frac{9}{4}\sum_{cyc}\frac {2a \plus{} b \plus{} c}{a}$


For a sequence $ \{x_n\}_{n\geq 0}$, for $ k \in \mathbb{N}, r \in \{0,1,...,k \minus{} 1\}$   define

$ S_{k,r} \equal{} \liminf_{n \rightarrow \infty} \frac {\sum_{0 \leq ki \plus{} r \leq n} x_{ki \plus{} r}}{\sum_{0 \leq i \leq n} x_i}$

Does there exist a [b]nonconstant[/b] sequence $ \{x_n\}{n\geq 0}$ such that

(i) $ \liminf_{n \rightarrow \infty} \frac {1}{\sum_{0 \leq i \leq n} x_i} \equal{} 0$

(ii) $ \forall$ $ k \in \mathbb{N}, r \in \{0,1,...,k \minus{} 1\}$  , $ S_{k,r} \equal{} \frac {1}{k}$

$ x_i\equal{}const\not \equal{}0$ satisfyed 1) and 2).


I think you will still have to pay a reduced amount, but it might vary a little bit depending on whether it's an AoPS course, EPGY course, university course, etc.

assuming it's an aops course, the following quote from the Online Math Classes page may be of help

[quote]Drop Option. Students who drop the course within 10 days after the first class will receive a full refund. This gives students the chance to attend two full classes and take a look at the MidTerm challenge set before committing to a 12 week class.[/quote]

[quote="sisira gorthala"]I am not sure about taking the course.  Is it possible for me to audit the course without paying?

I would like to continue based on how much I like the course.[/quote]It is not possible to audit without paying first.  However, as mentioned above, what you can do is enroll in the course and pay the fee.  This will allow you to attend the first 2 class sessions, and if you are not happy with the course at that point, you can drop the course and receive a full refund of the course fee.

Please note that in our AMC courses, there is no Midterm Challenge set.  This is one reason why these courses are less expensive than our regular subject courses.

Thanks!  So far it seems like a pretty intense course.


topology
function
complex analysis
advanced fields
advanced fields unsolved

Well, this looks trivial and I have to confess my laziness but here it goes: how do you prove that the complementary of a simple closed curve $\gamma$ in the complex plane has exactly two connected components ?

I know how to prove the existence of the unbounded component, but cannot see right away how to show the existence and unicity of the second one ...

This is the Jordan curve theorem.
It is indeed very hard, giving a proof here should be a difficult task.
Hatcher's book (algebraic topology) has a proof, also for the general case.

Isn't that actually fiercely disputed?  I heard that it took until 2005 before they finally came up with a computerfree proof? :maybe:

Dou you mix it up with something else, e.g. the four-color-theorem¿
The only thing I found that sounds related was:
"Recently, a proof checker was used by a Japanese-Polish team to create a "computer-checked" proof of the theorem (Grabowski 2005)." [MathWorld]

Okay, the guy who wrote the problem I tried to do is not very clever then. This was as follows: given a simple curve $\gamma$ in the complex plane, and two functions $f$ and $g$ analytic on and inside $\gamma^{*}$ such that $f=g$ on $\gamma^{*}$ (the star is the image of), then $f=g$ inside the curve.

Apart from the fact that this is trivial complex analysis,  refering to the "inside" of a simple curve seemed not that obvious to me.

Thanks for the name of the theorem, by the way.


nice!

10\10

BTW

How do I start the newfolder game I get it opened and everything.

Wh9ich file i klick?

Mafioso, I'm not saying you're bad, and I'm definitely not saying I'm better, but that hidden code art is pretty squished together. You might want to try spacing the letters out. I wasn't really able to tell what that meant until I saw that it was supposed to be "Mafia".

[quote="la mafioso"]

Rate it out of 10 please?
[/quote]

How about a 1?

Mafioso -- Extract the folder out of its zip (keep the zip)
There should be an html file which has the link.

Thanks by the way :D

[quote="math92"][quote="la mafioso"]K, watching mafia is boring, so for anyone waiting for mods to make a decision, waiting for accusations and defenses to be given, or just to plain old have fun, try some of it!

It's very simple. Draw stuff but make sure it's not inappropriate.

If later requested, I might make a guide.

Here's some of my special hidden content art.

[hide="?___#___X___??X##_??X##___X___"]
[hide="??_##__?_#__?_______X_____?_#_"]
[hide="?_X_#__?X#__??X##___X_____?X#_"]

Yay! It says mafia! ain't it cooool?

[/hide]
[/hide]
[/hide]

Rate it out of 10 please?

Then do yours and your ffriends and enemies will rate.[/quote]
what is your picture??? :huh:[/quote]

Its not really a picture; it just says mafia.
I rate it a 1 because its REALLY squished together and I couldnt read it until i knew what it said.

I definitely agree, it is too squashed up to tell what it really says until you know what it is supposed to say.

i like Darknights.  I give it a 10.
 __   _ ._|_ |_ (_|   )
| | |(_|   |_ | |   |   /_

that's the best i got

If you put your drawings in <code> tags, like Hexaditidom did, they will be in a monospace font, so they won't look warped.

Yeah, hate to be a spoil sport, but
http://www.network-science.de/ascii/


calculus
integration
limit
trigonometry
search
function
real analysis

Check out the links over here: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=966433094&t=134465

[quote="suweijie"]Suppose that $f\in C[0,2\pi]$.Show that$\lim_{n\rightarrow+\infty}\int_{0}^{2\pi}f(x)|\sin nx| dx=\frac{2}{\pi}\int_{0}^{2\pi}f(x) dx$. :lol:[/quote]
there is many idea to solve it.
I represent you one (which was discused in my classroom)
at the first you see
$\lim_{n\rightarrow+\infty}\int_{0}^{2\pi}\sin nx dx=0$ or
$\lim_{n\rightarrow+\infty}\int_{0}^{2\pi}C_{n}\sin nx dx=0$
where $C_{n}$ is some constant.
then you write
$\lim_{n\rightarrow+\infty}\int_{0}^{2\pi}G(x)\sin nx dx>\lim_{n\rightarrow+\infty}\int_{0}^{2\pi}f(x)\sin nx dx>\lim_{n\rightarrow+\infty}\int_{0}^{2\pi}H(x)\sin nx dx=0$
where $G(x)$ and $H(x)$ is "*****"-constant functions from $f(x)$.
I don't know how it  "*****" called in english.


Ok, I have tried to do this problem but seem to have no idea how to get the answer indicated. The question is:

The lengths of the sides of isosceles triangle ABC are $ 3x\plus{}62$, $ 72\plus{}30$ and $ 52\plus{}50$ feet. What is the least possible number of feet in the perimeter of triangle ABC?

[hide="My Solution"]Well, as long as the third side isn't 0....so...if integer, 1 I guess...meaning $ 102\plus{}102\plus{}1\equal{}205$... right?[/hide]

[hide="Correct Answer"]232[/hide]

Someone plz help!!!!

I'm pretty sure you made a typo, or there's a typo in your test something.  It should say:

The lengths of the sides of isosceles triangle ABC are $ 3x \plus{} 62$, $ 7x \plus{} 30$ and $ 5x \plus{} 50$ feet. What is the least possible number of feet in the perimeter of triangle ABC?

Solving this question should give you the correct answer.

Suppose $ 3x \plus{} 62$ and $ 5x \plus{} 50$ were the equal sides (this minimizes the perimeter.) We have the equation $ x \equal{} 6$. Therefore, the perimeter would be 232.


We are given a convex quadrilateral $ ABCD$ with sides $ AB \equal{} 9$, $ BC \equal{} 8$, $ CD \equal{} 1$, and $ DA \equal{} 2$. What is the largest possible area of this quadrilateral?

[hide]Maximum area of a quadrilateral with given side measures occurs when it is cyclic. By Brahmagupta's formula:

$ A_{\textbf{max}} \equal{} \sqrt{9 \cdot 8 \cdot 1 \cdot 2} \equal{} 12$, where I have used the fact that $ AB\plus{}CD\equal{}BC\plus{}DA \iff s\minus{}a\equal{}c, \; etc.$ to expedite computation -- if a quadrilateral has an incircle and is also cyclic, its area is the squareroot of the product of the sidelengths.[/hide]

[hide]Another approach:
We have the formula of the area of a convex quadrilateral:
square root[(p-a)(p-b)(p-c)(p-d)-abcd.cos^2(B+D)/2] with p is a half-perimeter of the convex quadrilateral
Applying this formula to the problem above, we easily see that the maximum occurs when abcd.cos^2(B+D)/2=0,which means B+D=180. Therefore, the quadrilateral will be cyclic when it reaches its maximum area: square root(p-a)(p-b)(p-c)(p-d)=12(Q.E.D)[/hide]

The max area of a quadrilateral is when its diagonals are perpendicular, because the area of a quad is $ (d_1)\cdot(d_2)\cdot(sin A)$/2, where A is the angle between the diagonals (do you see why?). So we draw the diagonals, they intersect at O, by Pythag $ OA^2 \plus{} OB^2 \equal{} 81$, $ OB^2 \plus{} OC^2 \equal{} 64$, $ OC^2 \plus{} OD^2 \equal{} 1$ $ OD^2 \plus{} OA^2 \equal{} 4$. Solve the system. (I didn't remember Brahmagupta's, that's why I did it this way).

Whoops, my method reaches a contradiction, $ OA^2\plus{}OD^2\equal{}4\equal{}81\minus{}64\plus{}1\equal{}18$.


I need help with this question. Can someone help me?

[b][u]The question is:[/u][/b]

A martial arts Instructor devised a way to train the martial arts academy students in the art of mass combat. These battles are divided into 2 rounds. Rounds A and B.

[b][u]Round A :[/u][/b] The students are allowed to fight against anyone and as many opponents they choose. (i.e. 1 Student can go against 5, or 3 can go against 7) However, the students cannot change opponents.

[b][u]Round B :[/u][/b] The students must not spar with the opponents encounted in the previous round and must spar with all those they didn't fight in the first round.

This is to ensure that the weakest student will have a chance to fight the strongest of the lot and that the rest will be trained in fighting as a part of a group or against many. Each student is tagged numerically and their sensei draws up a table that lists the opponents of each student.

[b][u]Question:[/u] In one particular battle, thier sensei discovered that the table he gets in the 2nd round could be made identical to that in the 1st round simply by re-labelling the students. If all the students are now divided into groups of 4, what is the possible numbers of students that will be left over? [/b]

I need to know how many and why?   :?:
And I also need a good and reasonable explination so that I may study from it.
I would be very grateful to anyone who could help. Thank you.  :D

[quote="K81o7"]Question: Does a given student participate in two fights total, one against all of the first set of people he chooses, and the second against all of the others, or does he fight each individually?

If individually, this problem is simpler...because you know the total number of fights, and can work from there...[/quote]

one against all of the first set of people he chooses, and the second against all others he did not battle yet. But they can also work in numbers, not just individually.

[quote="spraycan18"] If all the students are now divided into groups of 4, what is the possible numbers of students that will be left over? :D[/quote]

  what do you meant by this question?

I'm guessing that means what possibilities are there for the number of students mod 4.

But if I understood the way you had the listing work, it is necessary that in the two rounds in question, the number of pairs of students is the same?

Well, because $\binom{n}{2}$ is even for any n, it seems that any residue is possible...

What is the problem source?
(And ${n\choose 2}$ is not always even... take $n=3$)

if you guys think it's not hard [b]solve it![/b]
please.  :D

[quote="ychwyo"][quote="K81o7"]
Well, because $\binom{n}{2}$ is even for any n...[/quote]

Really?  :wink:[/quote]

Er yeah...what was I thinking there...

$\binom{n}{2}$ is even iff n is 0, 1 mod 4.  My bad  :D

I didn't mean to say that...

That's interesting, what you just said implies that ${n^{2}\choose 2}$ is always even..


[quote="brunojoyal"]Given a set $ S$ having $ n$ elements, find the number of unordered pairs of disjoint subsets of $ S$.[/quote]

[hide="Hint."]Consider three sets: (1)set A subset of S, (2)set B disjoint from A & subset of S,(3)the rest of S i.e. S-A-B.
each element can go into any one of the three.[/hide]

Well here is my solution.

If $ S$ is a set, let $ D(S)$ be the set of unordered pairs of disjoint nonempty subsets of $ S$.
Let $ |S|\equal{}n$.

$ 2|D(S)| \equal{} {n \choose 1} (2^{n\minus{}1}\minus{}1) \plus{} {n \choose 2} (2^{n\minus{}2}\minus{}1) \plus{} ... \plus{} {n \choose {n\minus{}1}} (2\minus{}1)$
$ \equal{} (2\plus{}1)^n \minus{} 2^n \minus{}1 \minus{} \sum_{j\equal{}1}^{n\minus{}1}{n \choose j} \equal{} 3^n\minus{}2^n\minus{}1 \minus{} (2^n\minus{}2) \equal{} 3^n\minus{}2^{n\plus{}1}\plus{}1$

Hence $ |D(S)| \equal{} \frac{3^n\plus{}1}{2}\minus{}2^n$.


calculus
integration
geometry
inequalities
limit
logarithms
calculus computations

Find the area of the figure such that the points $(x,\ y)$ satisfies the inequality $\lim_{n\to\infty}(x^{2n}+y^{2n})^{\frac{1}{n}}\geq \frac{3}{2}x^{2}+\frac{3}{2}y^{2}-1.$

[quote="amcavoy"]I tried logarithms but didn't get anywhere.  Will you post a hint?[/quote]

Compare $x^{2}$ with $y^{2}.$


limit
function
calculus
calculus computations

Assume that $ \lim_{n\to\infty}{x_n\equal{}c}$ s.t. for $ \forall n\in\mathbb{N}_\plus{}$, $ |x_n\minus{}A|\geq\delta_0$.

Prove that:$ |c\minus{}A|\geq\delta_0$

Suppose that $ |c\minus{}A| < \delta_0$. Then, as $ \lim x_n \equal{} c$, we can take $ x_n$ as close as we want to $ c$, and so there must be an order $ p$ such that $ |x_n \minus{} A| < \delta_0$ for $ n > p$. But this contradicts the hypothesis that $ |x_n \minus{} A| \geq \delta_0$ for all n. So we must have $ |c\minus{}A|\geq\delta_{0}$. $ \Diamond$

The function $ f(x) \equal{} |x\minus{}A|$ is continuous on $ \mathbb{R}$, therefore from $ f(x_n) \geq \delta_0$ follows $ f(c) \equal{} \lim_{n \to \infty} f(x_n) \geq \delta_0$.


x-box is still rocking.  the ps2 are wayyy toooo oldddd.  the new final fantasy games are only for pc now.

I think you should make an option of gaming pcs, considering the graphics cards in gaming pcs are better than those used in x-box and ps2.

I say PS3 is gonna be the best. Isn't FFXII going to be for PS2?

[quote="DarkKnight"]Isn't FFXII going to be for PS2?[/quote]

Yes. I most certainly wouldn't mind if they released it for PC too though. *mutters*

[quote="djimenez"][quote="alkachi"]what about next generation??[/quote]
WOW Alkachi, two topics about Game Consoles, you sure love videogames![/quote]

yes i love videogames  :lol:

[quote="juicybooty911"]i wonder what it would be like if they combine playstation and x-box? and maybe use pc graphic cards[/quote]
and what about competition ???? ;)

Nintendo, since it's the manufacturer of all game consoles I own.


I'll do this with generating functions, since I like that method, even though its an application of a far-more-powerful-than-necessary method. To wit:

[hide] First note that I'm shifting the index back by one.
$\frac{(A(x)-b_0)}{x} = \frac{4}{1-4x} - 3A(x)$
$(\frac{1}{x}+3)A(x) = \frac{4}{1-4x} + \frac{1}{x}$
Multiplying through by $x$ and then isolating $A(x)$,
$A(x) = \frac{4x + 1 - 4x}{(1-4x)(1+3x)}$
Then partial fractions:
$A(x) = \frac{4}{7(1-4x)} + \frac{3}{7(1+3x)}$
The power series for this function has the coefficients of $b_n$. This yields (changing the index back) $b_n = \frac{4^n}{7} - \frac{(-3)^n}{7}$.[/hide]

I think Kunny's sign on the power of 3 is off; check the first few terms. Otherwise, though, I think his solution is nicer than mine.

Oh, thanks :) ,but  :blush: .I will edit.

kunny


Let x and y be positive integers such that xy divides $ x^{2}\plus{}y^{2}\plus{}1.$Show that$ \frac{x^{2}\plus{}y^{2}\plus{}1}{xy}\equal{}3$

That's NOT pre-olympiad level....

The solution i know uses vieta jumping......

@yildobekir: Look up Vieta Jumping. Some of the problems you've posted are classic for being solvable by Vieta Jumping.

[hide="Practicing my Vieta jumping"]
Suppose $ x^2\plus{}y^2\plus{}1\equal{}kxy$. Suppose a particular solution $ (x,y)\equal{}(a,b)$ is such that this is true and $ a\plus{}b$ is minimal. We assume that $ a\ne b$. Without loss of generality, $ a>b>0$.

Consider the quadratic in $ x$: $ x^2\minus{}bkx\plus{}1\plus{}b^2\equal{}0$. We already know that $ x\equal{}a$ is a particular root of the quadratic. By Vieta's formulas, the sum of the roots is $ bk$, so the second root is $ bk\minus{}a$. Also, the product of the roots is $ 1\plus{}b^2$, so the second root must be $ \frac {1\plus{}b^2}{a}$.

Because $ bk\minus{}a$ is clearly an integer, we must have $ \frac {1\plus{}b^2}{a}$ as an integer, and it must be positive given that $ a$ and $ b$ are positive, so this second root is also a solution to the quadratic.

\[ a>b\implies a\geq b\plus{}1\implies a^2\geq b^2\plus{}2b\plus{}1\implies a^2>a^2\minus{}2b\geq b^2\plus{}1.\]

Therefore, $ \frac{b^2\plus{}1}a<a$, and $ \frac{b^2\plus{}1}{a}\plus{}b<a\plus{}b$, so we have contradicted our assumption that $ a\plus{}b$ is minimal. Therefore, our assumption that $ a\ne b$ must have been false, and so $ a\equal{}b$.

Then $ \frac{x^2\plus{}y^2\plus{}1}{xy}\equal{}\frac{2x^2\plus{}1}{x^2}\equal{}2\plus{}\frac{1}{x^2}$.

$ x^2$ is necessarily $ 1$, so we have $ 2\plus{}1\equal{}3$ $ \Box$.
[/hide]


I'm starting to amass portable applications on my USB drive, and I was just wondering, what are some must-have portable applications? :)

hmmm

i only have N

if anyone can post a game up here, that we can download from here, i guess taht would be nice.

unless we cannot share  :(

Sorry - we're not allowed to fileshare. However, you're free to check out portableapps.com, where they have tons of free games - just not the illegal types :P

[quote="7h3.D3m0n.117"]I have Portable Pidgin, Firefox, GIMP, along with some classic games like Counter Strike :)[/quote]

Pidgin doesn't work for me!! It never loads my buddy list. :(

i cant.

but, there are always school library computers.......

ARGH!!!

I CANNOT GO TO portableapps.com/games!

darn it.

darn

darn

@#%! :cursing:

[quote="buzzer11"]Pidgin doesn't work for me!! It never loads my buddy list. :([/quote]

Try reinstalling it. Then make sure you've entered your correct username/password combo.

I also have UltraSurf on my flash drive, letting me bypass the school filters and granting my access to Facebook, Youtube, etc. I'm also interested to see if there are any copies of other FPSes out there which aren't based of Source engines (aka better graphics - although Counter Strike Source's graphics are not bad), such as Halo or Call of Duty.

BZFlag.
It's multiplayer shooting game.
No blood. Just tanks.
There is a chat feature
Cursing is punished.

[quote="1=2"]BZFlag.
It's multiplayer shooting game.
[b]No blood.[/b] Just tanks.
There is a chat feature
[b]Cursing is punished.[/b][/quote]

Sigh...I'll pass on that, thanks. Guess one shortfall of PC shooters is that most of it is dominated by 12 year olds who think cursing every second is cool.


Let $ a,b,c$ be positive real numbers such that $ a^6\plus{}b^6\plus{}c^6\equal{}3$. Prove that
\[ a^7b^2\plus{}b^7c^2\plus{}c^7a^2\le 3\]

[quote="Tenoreoz"]From Cauchy's on the LHS, we have $ (a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2)^2 \le (a^7 \plus{} b^7 \plus{} c^7)(a^2 \plus{} b^2 \plus{} c^2)$. By AM-GM on the resulting RHS, $ (a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2)^2 \le 9(abc)^3$. Using AM-GM on the original equality gives $ 1\ge abc$. Therefore, $ (a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2)^2 \le 9(1)^3 \implies a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2 \le 3$.[/quote]
Are you sure or are you joking?

just an idea...very obvious and common...
$ \sum a^{7}b^{2}\equal{}(a^{6}\plus{}b^{6}\plus{}c^{6})(ab^{2}\plus{}bc^{2}\plus{}ca^{2})\minus{}(\sum ab^{8}\plus{}\sum a^{6}bc^2)\equal{}3(ab^{2}\plus{}bc^{2}\plus{}ca^{2})\minus{}(\sum ab^{8}\plus{}\sum a^{6}bc^2)<3(ab^{2}\plus{}bc^{2}\plus{}ca^{2})$
If we can prove that $ ab^{2}\plus{}bc^{2}\plus{}ca^{2}<1$, then i think we are done but i doubt if its true and this does does not consider when equality exists... :maybe:

[quote="hell_ever"]
If we can prove that $ ab^{2} \plus{} bc^{2} \plus{} ca^{2} < 1$, then i think we are done but i doubt if its true and this does does not consider when equality exists... :maybe:[/quote]

Take $ a\equal{}b\equal{}c\equal{}1$ and it's wrong. :wink:

[quote="nayel"]Let $ a,b,c$ be positive real numbers such that $ a^6 \plus{} b^6 \plus{} c^6 \equal{} 3$. Prove that
\[ a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2\le 3
\]
[/quote]
Nice inequality, nayel! :lol:
$ a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2\le 3\Leftrightarrow\left(a^6 \plus{} b^6 \plus{} c^6\right)^3\geq3\left(a^7b^2 \plus{} b^7c^2 \plus{} c^7a^2\right)^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a^{18} \plus{} 3a^{12}b^6 \plus{} 3a^{12}c^6 \plus{} 2a^6b^6c^6 \minus{} 3a^{14}b^4 \minus{} 6a^9c^7b^2\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a^{18} \plus{} 2a^{12}b^6 \minus{} 3a^{14}b^4\right) \plus{} \sum_{cyc}\left(c^{12}a^6 \plus{} 3a^{12}c^6 \plus{} 2a^6b^6c^6 \minus{} 6a^9c^7b^2\right)\geq0,$
which true by AM-GM.
Try it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176217


Round 2 of the USAMTS is now available at http://www.usamts.org.  The postmark deadline for solutions is November 19, 2007.

I'm not sure where to post this  :blush: but for #3, the Remarkable Array question, do I need to prove that I found the smallest largest element in the fourth row, or can I just 'find it'?

A little late...it's due tomorrow you know....
But anyways, we can't really say anything. Just read the problem and clearly state any assumptions you make in your solution I guess. (one of the mods said that in another post)

I know the round's due tommorrow  :lol:

Well, my question is rather general, not entirely specific to number 3, so it's not like I'm trying to get hints on this problem  :huh: If we are asked to 'find something', I assume that a rigorous proof is not looked for. I simply wanted to know whether I would be penalized for not having proved that I had found the minimum value...

The website says, "Remember to fully justify all your answers--don't just send numerical answers." Just giving a numerical answer can generally only recieve a 1 (if it's right).

Do I have to draw a figure for the problem 5?? Can I give a solution without a figure??Please reply me as soon ayou can because the due is tomorrow! :(

You can submit a solution without a diagram, as to whether or not that will affect your solution ...

Wow guys - that's why you don't procrastinate.

Wait what  :huh:  ... its due tommorrow  ?! Time to start


I'm doing AIME problems and AMC12/AHSME problems right now, but I find that my score isn't increasing as fast as I want it to improve. Would making the jump to olympiad problems improve my score faster? And how would someone who really hasn't had much olympiad experience make the jump from pre-olympiad AIME type problems to USAMO type olympiad problems?

[quote="Ichverstehenicht"]I started olympiad at my school this year, and only qualified for AIME, which is heller hard. Are there any good resources besides old problems (which by the way I would like a copy of the questions from last years AMC 12 and AIME because my math teacher took it for the summer. This would be appreciated greatly) that can help me improve to at least qualify for the newly expanded USAMO?[/quote]

I see you are still challenged by AIME problems. Then go here and buy all those books, study it hard and you qualify for USAMO easily:

- http://www.mathlinks.ro/Books/AoPS_B_Texts.php

Of course, you need some more training problems:

- http://www.mathlinks.ro/Forum/resources.php?c=182

But the best is personal instruction by taking classes, that is go here:

- http://www.mathlinks.ro/Classes/AoPS_C_ClassesS.php

Guided by top-notch instructors who are offering those classes for an incredible low price. Go ahead and subscribe.  :)

@Ichverstehenicht: Dann wirst du es locker verstehen. Glaub' mir.  :D

this is the wrong forum to discuss this, there is an entire forum about the American Math Competitions, and topic like these (and many other "how to's")have been discussed many many many times!

[quote="Karth"]"Dann wirst du es locker verstehen. Glaub' mir."? You mean that I will barely understand the books?  :huh:[/quote]

Ok in English for you what I said above. "Then you will understand it easily. Believe me."

Are these books specifically realted to the USAMO only?

No, actually. They help truly very much for the AMC and AIME and maybe a little on the USAMO, but for real help with the USAMO, I think Engel's book Problem Solving Strategies would really help.  :)


Hello:

How many ways are there to arrange the word STATISTICS so that no identical letters are side by side(such as TT, SS, etc.)?.

This is rather tough. I can easily find the number of total arrangements. 50400.

I thought about 'binding' them together and counting them as one 'letter' and then discounting those,

but I think it may be more involved than that.

[quote="Z-coli"]Instead of trying to count the number of ways that the letters from STATISTICS can be arranged so that no two are next to each other, how about you count the number of ways the letters from STATISTICS can be arranged so that every two are together, and then subtract this number from the total number of arrangements, to determine what you wanted to find.  To do this, it will be neccessary to count SS as 1 letter, TT as 1 letter, and so on.[/quote]

Umm, I think that's what he meant when he said

[quote="fredoniahead"]

I thought about 'binding' them together and counting them as one 'letter' and then discounting those,
[/quote]

I think that should work, but you should be careful because there are three pairs of double letters, so I think the answer should be Total number of combos - number with SS's-number with TT's-number with II + number with both SS and TT +number with both SS and II + number with TT and II - number with TT, II, and SS.

not 100% sure though.

Are you sure this works? $ \binom{n\minus{}k\plus{}1}{k}$ doesn't mean the number of ways to arrange $ k$ objects in $ n$ places such that no two are next to each other.

For erdogankerem123's proposed method, it would be very tedious; resembling a brute force solution, because you would also have to discount cases where the two IIs were together, for example, from the two SSs and TTs were together.

[quote="Temperal"]For erdogankerem123's proposed method, it would be very tedious; resembling a brute force solution[/quote]

Yeah, I thought it would be tedious, which is why I didn't actually work it out  :D .

[quote="Temperal"]
 because you would also have to discount cases where the two IIs were together, for example, from the two SSs and TTs were together.[/quote]

Didn't I already discount that?

Thanks everyone for your interest and input. Much appreciated. These are tough counting problems. I have arrived at and been given various solutions.

[quote="Ognjen09"]When you are making arrangement you have 10 "places" where you put letters S, T, A, I and C. In order to make wished

arrangement you can choose 3 places from 10 such that no two are next to each other for letters S in $ {8 \choose 3}$ ways, 7 places

left, $ {5 \choose 3}$ places for letters T, 6 places left, $ {5 \choose 2}$ for letters I, and 2 places left for A and I.

At the end you have $ {8 \choose 3}\cdot{5 \choose 3}^{2}\cdot2$ arrangements.[/quote]

I'm quite sure this isn't right because you get a different answer if you choose the two spaces for the I's first.

Yes, you are right, my solution is incorrect. I wanted to solve the problem in a short way and I made a mistake.


Hey guys,

I really want to make MOP next year, but it's just so hard to memorize all those darn formulas for USAMO.

Should I just make formula sheets and start memorizing from the beginning? Or do you think it's better to memorize parts of different theorems every day?

No. You are approaching the situation incorrectly. If you understand the formulas, you can quickly rederive them. You have to focus on proof writing and creativity.

Yes, cardinal rule: Don't memorize, understand and be able to derive them. It'll serve you much better in the long run.

[quote="themonster"]Dude, trigonometry1.01 just doesn't have the same kind of irony/trickiness as probability1.01.  Get a life.[/quote]
What are you talking about?

First of all, we are NOT the same person.

Second of all, sorry for my misconception about USAMO prep, but you don't have to make fun of me for it.

eh? this is pretty much how I prepare for USAMO every year

the only problem is I lost some of my formula sheets so my score went down this year

I was pretty unhappy

I have seen Arnov's formula sheets. He wouldn't let me photocopy them, so now I'm trying to write my own.

I suggest, trigonometry1.01, that you read Richard Rusczyk's "What is Problem Solving?"

It can be found in the "Articles" section of this website.

However, I have a fairly strong suspicion that you, in fact, already know that formula sheets can only get you so far, and that you have deliberately designed this username to start ridiculous threads.

Time will tell.


group theory
abstract algebra
limit
modular arithmetic
superior algebra
superior algebra unsolved

Let $A_{n}$ be the number of groups with at most $n$ elements.

Let $B_{n}$ be the number of groups with at most $n$ elements that contain at least one proper nontrivial normal subgroup.

Is it true that $\lim_{n\to\infty}\frac{B_{n}}{A_{n}}= 1$?

Just by looking at the list of simple finite groups, one can easily answer this question. Now, the question is: is there a proof avoiding this tremendously hard result (classification of finite simple groups)?

I wonder if there is a relatively accessible proof that there are only finitely many sporadic simple groups. Probably not.

What do you mean? There is a classification of finite simple groups: they are some sporadic groups (I forgot their number), some Lie groups and cyclic of prime order and finally alternating groups. The number of these is always $O(n^{x})$ with $x<1$.

Well, post them. I'm afraid I don't know them very well and I want to see this kind of things. :wink:

Are you kidding? I don't think it's so difficult to go and read this in any advanced book on the theory of groups. I don't know how to tell you, but first I don't know all of them, secondly some are monstrously large or complicated. You know, there are some mathematical results which are not that easy as you think: this one is among the most difficult mathematical result ever obtained.

Actually, I wonder if there is number-theoretic proof of the original question. Because my professor conjectured what I said based on the following:

A group has a (nontrivial proper -- i'll leave those off after this) normal subgroup if (and only if?) there is only one p-Sylow for some prime p dividing the size of the group. So, it is necessary that there be multiple p-Sylows for each p in order for the group to have no normal subgroup.

This means for each prime $p$ such that $p \mid |G|$ we must be able to find some $n_{p}$ so that $n_{p}\mid |G|$ and $n_{p}\equiv 1 \pmod p$. (I hope I stated that condition right.)

My professor thinks this is generally difficult to do, whence his guess. (I don't know why he didn't think of the classification of groups. But after all, he is a number theorist, not an algebraist.)

I don't think the only if part works. Anyway, this is indeed the most natural approach, but it does not seem easy to conclude from here, because indeed you will probably find not too many possibilities for its cardinal (even though I don't think that's easy), but for each possibility you have no control on the number of finite simple groups having that order. Probably I'm idiot and I do not see why...

[quote="Cezar Lupu"]Well, post them[/quote]
[url=http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups]Sure thing[/url]


ABCDEF is a regular hexagon with M and N the midpoints of CD and EF, respectively. FC intersects AM at X and FC intersects BN at Y . Find XY, if a side of the hexagon has length 6.

[hide="Solution"]
Extend $ AX$ and $ BC$ to meet at $ G$. Then, $ BG \parallel AD$, so $ \triangle AMD \sim \triangle CMG$, but $ CM \equal{} MD$, so $ \triangle AMD \cong \triangle CMG$. Also, since $ CF \parallel AB$, $ \triangle GCX \sim \triangle GBA$. Thus, $ \frac {CX}{AB} \equal{} \frac {CG}{BG} \equal{} \frac {CG}{BC \plus{} CG} \equal{} \frac {AD}{BC \plus{} AD} \equal{} \frac {2}{3}$, since $ AD \equal{} 2BC$. So $ CX \equal{} 4 \equal{} FY$ by symmetry, but $ CF \equal{} 12$ so $ XY \equal{} 4$.[/hide]
 [geogebra]afb7c3ac59b0f16a685d33211b8de28bb10c0978[/geogebra]


[quote="Art of Owna"](-1,3,-2)

Add all three equations to get 2(xy+xz+yz)=-14, so we get xy+xz+yz=-7, subtratiing the three original equations from this equation, we get xy=-3, xz=2, yz=-6, multiplying these three equations, we get (xyz)^2=36, so xyz=6, now substituting xy, xz, and yz into this equation, we get x=-1, y=3, z=-2[/quote]
[hide]$(xyz)^{2}=36$ implies $xyz=\pm 6$ not $xyz=6$[/hide]

Ya, good catch.. so not only does (-1,3,-2) work, but (1,-3,2) also works


inequalities
search
inequalities unsolved

[quote="Gohan"]Im trying to solve this problem for some time...
Let $x,y,z$ be positive real numbers and $x+y+z=1$
Prove that:
$xy+yz+zx-3xyz\leq \frac{1}{4}$
While trying to solve it I came to this one
$6xyz\geq x^3+y^3+z^3$
how to continue ??? :blush:[/quote]
You have done it wrongly I think.
Homogenization gives
$(xy+yz+zx)(x+y+z)-3xyz \le \frac{1}{4} (x+y+z)^3$
(The following are symmetric sum)
$4 \sum x^2 y \le \frac{1}{2} \sum x^3 + 3\sum x^2 y + \sum xyz$
$\sum x^2 y \le \frac{1}{2} \sum x^3 + \sum xyz$
But then by schur we have $\frac{1}{2} (\sum x^3 + \sum xyz) \ge \sum x^2 y$
So it's true because $\frac{1}{2} \sum xyz \ge 0$.

[quote="nttu"]It's an ineq from HongKong MO 1999  :)[/quote]
yes?? I haven't seen this problem..  :blush:

wrong :( huh... cant do anything right :oops:
Good solving :D
Anyway here is mine :)
The left side of the ineq is:
$xy+xz+yz-3xyz$
$=xy(1-z)+yz(1-x)+zx(1-y)$
$=xy(x+y)+yz(y+z)+zx(z+x)$
$\leq (\frac{x+y}{2})^2(x+y)+(\frac{y+z}{2})^2(y+z)+(\frac{x+z}{2})^2(x+z)$
Now the whole ineq becomes:
$(x+y)^3+(y+z)^3+(x+z)^3\leq 1$
$(x+y)^3+(y+z)^3+(x+z)^3\leq (x+y+z)^3$
$6xyz \geq x^3+y^3+z^3$
And this is where i'm stuck for ever  :(  :(  :(

It's normal to be stuck forever, since the inequality you want to prove is not always true. Actually, it is always reversed.

[quote="siuhochung"]
But then by schur we have $\frac{1}{2} (\sum x^3 + \sum xyz) \ge \sum x^2 y$[/quote]
Schur ?!?!?!? .. havent heard of it .. could you tell me what is schur's ineq ?

Gohan,

Schur's inequality is a well-known one that is used quite often:

If $a,b,c\geq0$ and $n \in N$ then $a^n(a-b)(a-c)+b^n(b-a)(b-c)+c^n(c-a)(c-b)\geq0$

Also search [i]mathlinks[/i] forums for more information about Schur's ineq
and exercises  in which it is used!!!  :)

Isn't schur true for $ n \in \mathbb{R} $

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7942]www.mathlinks.ro/Forum/viewtopic.php?t=7942[/url]


Two distinct positive integers from 1 to 50 inclusive are chosen.  Let the sum of the integers equal $ S$ and the product equal $ P$.  What is the probability that $ P\plus{}S$ is one less than a multiple of 5?

[hide]Let the selected numbers be $ a$ and $ b$. Then we have $ a\plus{}b\plus{}ab\equal{}5k\minus{}1, k \in Z \Rightarrow (a\plus{}1)(b\plus{}1)\equal{}5k$.

Either $ (a\plus{}1)$ or $ (b\plus{}1)$ must be a multiple of $ 5$, so either $ a$ or $ b$ must be an element of $ {4,9,...,49}$. Thus, we have $ \frac{2\binom{10}{1}\binom{40}{1}}{2}\plus{}\binom{10}{2}$ ways to select $ a$ and $ b$, out of the $ \binom{50}{2}$ total ways, giving a probability of $ \boxed{\frac{89}{245}}$.[/hide]

A little simpler (for me at least):

xy + x + y = 4 mod 5

trying one of them = 4 mod 5:

4y + 4 + y = 4 mod 5
5y + 4 = 4 mod 5
5y = 0 mod 5 (always)

So when x = 4 mod 5, y can be any number
0 mod 5: 100 ways (10 ways to pick 0 mod 5, 10 ways to pick 4 mod 5)
1 mod 5: 100 ways
2 mod 5: 100 ways
3 mod 5: 100 ways
4 mod 5: 45 ways ( nCr(10, 2) both are = 4 mod 5)

Then of course 1225 possible picks (50 choose 2), 445/1225.

pumaking94:

Yes, if one number is congruent to 4 in mod 5, the other number can be anything.  But you didn't check that if one number isn't congruent to 4 mod 5, the other number has to be congruent to 4 mod 5...because if this weren't true, there'd be more possibilities for both numbers...but it turns out that if you let x or y be congruent to the numbers 0 through 3, the other number ends up always being congruent to 4.

EDIT:  Forget what I just said.  If you let one number be congruent to 4 in mod 5, you see the other number can be any number.  So if you first choose any number, and then a second one that isn't congruent to 4 in mod 5, then P+S won't be congruent to 4 in mod 5, because if the second number were congruent to 4 in mod 5 then P+S would be congruent to 4 in mod 5.

Why can't $(a+1)$ and $(b+1)$ both be multiples of 5?

Also, how did @mismatchtea get $\frac{2\binom{10}{1}\binom{40}{1}}{2}$? How come you can't just immediately do $\binom{10}{1}\binom{40}{1}$ since the order wouldn't matter if choosing an element from a set?

[hide] Nice mix of subjects. So we have $(a+1)(b+1)$ is a multiple of 5. Compute the probability $a$ is, $b$ is a multiple of $5$, and subtract off the probability both are multiples of 5 to remove over counting. [/hide]

This is a very cool problem, and both solutions are excellent. However, since I am still a bit of a rookie, I need help understanding this problem. Why does (a+1)(b+1) have to be a multiple of 5? I can't figure out what that has to do with anything. Thanks for the help!

Think about it this way. If the first number chosen is a and the second number chosen is b and z is an integer, ab+a+b=5z+4. Factoring gives a(b+1)+b=5z+4. Then adding 1 to each side allows you to factor it to become (a+1)(b+1)=5(z+1). Since z is an integer that means that (a+1)(b+1) is a multiple of 5.

oh no I didn't see that you can use SFFT

This is a cool problem, nice difficulty for a lvl 25 problem I think.

[hide=my sol]Basically if the numbers are x and y, we have xy+x+y=4 (mod 5) => (x+1)(y+1)=0 (mod 5). Then, at least one of x or y must be -1 (mod 5). We see "at least", so we proceed to use complementary counting. We do 50*4/5c2=40c2=780, and then note that the total number of outcomes is 50c2=1225, to get (1225-780)/1225=89/245.[/hide]

[quote=PizzaPerson]89/295 i thought.

also, friend me!!!![/quote]

what? thats not how you simplify fractions my guy

[hide=Alcumus Solution]There are a total of $\binom{50}{2}=1225$ ways to choose the two positive integers. Call these integers $a$ and $b$. The problem asks what the probability is that: $$ab+a+b=n-1$$where $n$ is a multiple of 5. We can add one to each side of this equation and factor: $$ab+a+b+1=(a+1)(b+1)=n$$Now, we need to count the number of values of $a$ and $b$ such that $(a+1)(b+1)$ is a multiple of 5. This will happen if at least one of the factors is a multiple of 5, which will mean $a$ or $b$ is one less than a multiple of 5.

There are 10 integers from 1 to 50 inclusive that are 1 less than a multiple of 5: $4,9,14, \dots, 49$. So, the number of ways to choose $a$ and $b$ so the product is $\textit{not}$ a multiple of 5 is $\binom{40}{2}=780$. Therefore, there are $1225-780=445$ ways to choose $a$ and $b$ that do satisfy the requirement, which gives a probability of: $$\frac{445}{1225}=\boxed{\frac{89}{245}}$$[/hide]

I typically like solving the problem in a different way after getting it right (or wrong), and luckily, I used complementary counting around the ending area.
However, I don't understand why we can't count the number of ways to choose a and b by doing 2*10*50 = 1000. My reasoning for that was that there are 10 choices for $a$ for it to be a multiple of 5, another  50 choices for b since it can be anything and that there are 2 choices for which one gets to be the multiple of 5. Why is this wrong?

Oops i found out nvm

[quote=Mogmog8][hide=Solution]Note there are 10 numbers of each residue modulo 5. If our numbers are a,b, we know 5|ab+a+b+1=(a+1)(b+1) so a or b is congruent to 4 mod 5. Hence, successes are 40*10+10C2=445. The total is 50C2 so our answer is 89/245.[/hide][/quote]

That’s what I did as well. I didn’t think of complementary counting.


trigonometry
Putnam
calculus
calculus computations

:D

find the minimum value of

${ |\; \sin\; x + \cos\; x + \text{sec}\; x + \text{cosec}\; x + \tan\; x + \cot\; x}\; |$

where $ x\in \mathbb R$

[b]PS:[/b] do [b]not[/b] use any technique that goes beyond a 6th or a 7th grader.

It's an old Putnam problem, and reduces to finding the minimum of
\[ \frac{y^2\minus{}y\plus{}2}{y\minus{}1}\]
Maybe you learned calculus in the 6th grade?

what is the minimum value of $ \alpha^2 \plus{} \beta^2 \equal{} 1$

$ \mid \alpha \plus{} \beta \plus{} \frac{1}{\beta} \plus{} \frac{1}{\alpha}\plus{}  \frac{\alpha}{\beta}\plus{}\frac{\beta}{\alpha}\mid \ge 2$ by observation that

the min will be at $ \alpha \equal{} \minus{}\beta$ or $ x\equal{} \frac{(8k\plus{}3)\pi}{4}, \frac{(8k\minus{}1)\pi}{4}$??

[quote="JoeBlow"]It's an old Putnam problem, and reduces to finding the minimum of
\[ \frac {y^2 \minus{} y \plus{} 2}{y \minus{} 1}
\]
Maybe you learned calculus in the 6th grade?[/quote]
That's good.


calculus
integration
inequalities
geometry
search
calculus computations

In $ xyz$ space, find the volume of the solid expressed by the sytem of inequality:

$ 0\leqq x\leqq 1,\ 0\leqq y\leqq 1,\ 0\leqq z\leqq 1$

$ x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 1\geqq 0$

[hide="attempt"]$ A \equal{} \{(x,y,z): (x \minus{} y)^2 \plus{} z^2\ge 1\} \subset [0,1]^3$

$ \int_A 1 dV \equal{} \int_0^1 \left[\int_{(x \minus{} y)^2\ge 1 \minus{} z^2} 1 dA \right]dz$

so we want to find $ \int_{(x \minus{} y)^2\ge w^2} 1dA$ for $ w\in [0,1]$

or $ |x \minus{} y|\ge w$ on $ [0,1]^2$, which is $ \int_0^1 \max(1 \minus{} (x \plus{} w),0) \plus{} \max(x \minus{} w,0)dx \equal{}$
case 1: $ (w\le \frac {1}{2})$
$ \int_0^w 1 \minus{} (x \plus{} w) dx \plus{} \int_w^{1 \minus{} w} 1 \minus{} (x \plus{} w) \plus{} (x \minus{} w) dx \plus{} \int_{1 \minus{} w}^1 x \minus{} wdx \equal{} 1 \minus{} 3w^2$
case 2: $ (w > \frac {1}{2})$
$ \int_0^w 1 \minus{} (x \plus{} w)dx \plus{} \int_{1 \minus{} w}^1 x \minus{} w dx \equal{} 2w \minus{} 3w^2$

so $ \int_A dV \equal{} \int_0^1 1_{z\ge \frac {\sqrt {3}}{2}} \plus{} 2\sqrt {1 \minus{} z^2}1_{z < \frac {\sqrt {3}}{2}} \minus{} 3(1 \minus{} z^2) dz$
$ \equal{} 2\int_0^{\frac {\sqrt {3}}{2}} \sqrt {1 \minus{} z^2} dz \minus{} (1 \plus{} \frac {\sqrt {3}}{2})$
[/hide]
$ \equal{} \frac {\pi}{3} \minus{} (1 \plus{} \frac {\sqrt {3}}{4})$ which is.. negative...

I get $ \frac {10 \minus{} 3\pi}{6}$. If this is correct I will post my work.

Edit: [hide="Work"]$ x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 1\ge 0\Rightarrow (x \minus{} y)^2\ge 1 \minus{} z^2$. From this we deduce the two conditions on $ x$:

$ x\ge y \plus{} \sqrt {1 \minus{} z^2}$
or
$ x\le y \minus{} \sqrt {1 \minus{} z^2}$

Also, for any z, we see that $ (x,y)\in [0,1]^2$. These three conditions mean that for any slice in the $ x\!\minus{}\!y$ plane at height $ z$, the volume is two triangles. These triangles have an area:

$ A\equal{}2*\left[\frac{1}{2}\left(1\minus{}\sqrt{1\minus{}z^2}\right)^2\right]\equal{}2\minus{}z^2\minus{}2\sqrt{1\minus{}z^2}$

$ V \equal{} \displaystyle \int_0^1 A\cdot dz \equal{} \int_0^1 (2 \minus{} z^2 \minus{} 2\sqrt {1 \minus{} z^2})\cdot dz \equal{} \frac {10 \minus{} 3\pi}{6}$ [/hide]

That's correct answer.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=19113349&t=23106]Challenge[/url]


function
trigonometry
probability and stats

\[ 1 \ TASK\]
Prove whether function $ \phi(t)$:
\[ \phi(t)\equal{}\frac{e^{\minus{}t^{2}}}{1\plus{}\sin^{2}t}\]
is characteristic function of some distibution on line.
\[ 2 \ TASK\]
Prove whether function $ \phi(t)$:
\[ \phi(t)\equal{}\frac{\cos t}{1\plus{}t^{2}}\]
is characteristic function of some distibution on line.

Characteristic functions are nothing but the Fourier transforms of pdf of the random random variables, provided pdf exists for a r.v.(i.e. for absolutely continuous r.v's).

So try to find the inverse transforms and see whether you get some valid pdf.

I offer you another way of solving. Use the next fact -
if $ \xi_1,\xi_2$ are 2 independent r.v. with respectively c.f. $ \psi_1(t),\psi_2(t)$ then $ \xi_1+\xi_2$ will have c.f. equal to $ \psi_1(t) \cdot \psi_2 (t), t \in \mathbb{R}$.
I'll solve the second task:
$ \cos (t)$ is c.f. of the following r.v. $ \mathbb{P}[\xi_1=\pm 1]=0.5$
$ \frac 1 {1+t^2}$ is a c.f. of a r.v. with p.d.f. $ p(x)=e^{-|x|}, x \in \mathbb{R}$ (it will be our $ \xi_2$ - it's called two-sides exponential distribution)
So your function is a(product of the latter c.f.) c.f. of r.v. $ x=\xi_1+x_2$, where $ xi_1,\xi_2$ - are independent and described the latter, due to uniqueness of с.f.

The next task I'll solve a little bit later if there is still need in this.


trigonometry
geometry
parallelogram
trapezoid
geometry proposed

Circle $ O_1$ and $ O_2$ intersect at $ P,Q$,
$ A,C$ lie on circle $ O_1$,$ B,D$ lie on circle $ O_2$
$ A,P,B$ are collinear,so do $ C,P,D$
$ M,N$ are midpints of $ AD$ and $ BC$ let $ O$ be the midpoint of $ O_1O_2$
angle$ APC \equal{} \theta$ $ (0 < \theta < 90)$
denote $ h$ the distance from $ O$ to $ MN$,$ K$ is the midpoint of $ PQ$
Prove that $ h \equal{} OK\cos \theta$

Denote $ X, Y$ the midpoints of $ AB, CD$.
It's well-known that $ X,Y$ lie on $ (O,OP). MXNY$ is a parallelogram then $ P\equal{}MN\cap XY$ is the midpoint of $ XY$ and we have $ OP\perp XY$.
$ OH\equal{}OK.\cos \theta\equal{}OP.\cos POK.\cos \theta$
$ \Leftrightarrow OH\equal{}OX.\cos\theta.\cos POK\equal{}OP.\cos POK$
$ \Leftrightarrow \cos HOP\equal{}\frac{OH}{OP}\equal{}\cos POK \Leftrightarrow \angle POK\equal{}\angle HOP$
$ \Leftrightarrow \angle XPN\equal{}\angle QXP$
Denote $ BQ\cap (O_1)\equal{}\{W\}, BQ\cap (O)\equal{}\{V\}$ then $ V$ is the midpoint of $ BW$.
We get $ NV//CW$ and it's easy too see that $ CWBD$ is a trapezoid thus $ VY//CW$
Therefore $ Y,N,V$ are collinear.
$ \Rightarrow \angle NYP\equal{}\angle XQB$
By angle chasing we obtain $ \Delta XQY\sim \Delta BQD \Rightarrow \frac{XQ}{BQ}\equal{}\frac{XY}{BD}\equal{}\frac{XP}{YN}$
$ \Rightarrow \Delta NYP\sim \Delta BQX \Rightarrow \angle YPN\equal{}\angle QXP$ and we are done.

Very nice solution,thank you very much :)
I paste another two solutions below

P.S.I do much better in calculating than geometry :D


geometry
geometric transformation
rotation
geometry proposed

let ABCD a square and costruct on AB externaly an isosceles triangle $ \triangle ABS$ ($ SA\equal{}SB$) and costruct externaly BC a similar triangle isosceles $ \triangle BCT$ ($ TB\equal{}TC$, $ \angle BTC \equal{} \angle ASB$). Let O the center of ABCD and M the midpoint of BT, prove that $ MO \perp SC$.

Call X the projection of M onto OT and Y the one of S onto BC, see that MX/SY = OX/CY = 1/2, hence the triangles OMX and CSY are similar and <MOX = <SCY, done.

Best regards,
sunken rock

Dear Mathlinkers,
1.  OM // DT
2.  the triangles CDT and BCS are directly egal
3.  The rotation with center O and angle 90° transform the first one to the second
4. done..
Sincerely
Jean-Louis


[quote="neutrus"]During cold weather, a car with a 20 liter radiator containing 10% antifreeze must increase the solution to 25% antifreeze. How many liters of 10% antifreeze must be drained and replaced with 100% antifreeze? Express your answer as a mixed number.[/quote]

Using the given info, we know that the radiator originally contains 2 liters of antifreeze and 18 liters of other stuff. We know that we want the radiator to have 4 liters of antifreeze and 16 liters of other stuff. Thus, we must remove 2 liters of the other stuff and then fill the empty space with antifreeze. Since the original solution is 10% antifreeze and 90% other stuff, we will also remove 2/9 liters of antifreeze when removing the 2 liters of other stuff (as the liquids are throughly mixed together). Thus, 2 2/9 liters of the current mix will be drained and replaced with antifreeze.

2 2/9 liters

actually, you want 5 l of antifreeze

then you need to get rid of 3 l of 'other stuff'
so you drain 10/9 *3=$ \frac{10}{3}$ l of the original miture

hi, this is how u do it. anyway the answer is 10/3 or 3 and 1/3

oh yeah... srry says to express as a mixed number
i do that alot
$ \frac{10}{3}\equal{}3\frac13$

i think mark has a really good solution and look at the one listed in the intro algbera book....


limit
linear algebra
matrix
algebra
polynomial
function
linear algebra unsolved

Let $ A\in\mathbb{C}^{m\times m}$ and let $ \|\cdot\|$ be any norm. Show that
\[ \lim_{t\to \plus{} \infty}\|e^{tA}\| \equal{} 0 \Leftrightarrow \alpha(A)  < 0,
\]
where $ \alpha(A) \equal{} \max_{j \equal{} 1..m} Re(\lambda_j)$ is the spectral abscissa.

I can't get this one smooth. To be honest I have no idea how to handle $ \|e^{tA}\|$... can you show me the light?

Shouldn't that be
\[ \lim_{t\to \plus{} \infty}\|e^{tA}\| \equal{} 0 \iff \alpha(A) < 0\ ?
\]

Kent is right.
$ \alpha(A)<0$ is necessary. This part is easy because we may assume $ A$ upper triangular and we see that $ spectrum(e^{tA})\equal{}\{e^{t\lambda_1},\cdots,e^{t\lambda_n}\}$. Necessarily the diagonal of $ e^{tA}$ must tend to $ 0$. QED
$ \alpha(A)<0$ is sufficient. This part is more difficult. We may assume $ A\equal{}\lambda{I}\plus{}J$
where $ J$ is a nilpotent Jordan matrix of order $ k\leq{n}$ and $ \lambda$ is an $ A$-eigenvalue.
$ e^{tA}\equal{}e^{t\lambda}(I\plus{}tJ\plus{}\cdots\plus{}{t^{k\minus{}1}J^{k\minus{}1}})$. $ e^{tA}$ tends to $ 0$ iff $ e^{t\lambda}t^{k\minus{}1}$ tends to $ 0$. We are done.

The following result is a special case of this:

Let $ P(z)\equal{}z^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}a_1z\plus{}a_0$ be a complex polynomial and let $ \alpha(P)$ be the maximum of the real parts of the roots of $ P.$ Then consider homogeneous differential equation
\[ y^{[n]}\plus{}a_{n\minus{}1}y^{[n\minus{}1]}\plus{}\cdots\plus{}a_1y'\plus{}a_0y\equal{}0.\]
This equation has the property that $ \lim_{t\to\infty}y(t)\equal{}0$ for all solutions $ y$ if and only if $ \alpha(P)<0.$

Kent is right indeed.

I'm sorry loup blanc, I don't think that I understand your solution...
[quote="loup blanc"]$ \alpha(A) < 0$ is necessary. This part is easy because we may assume $ A$ upper triangular and we see that $ spectrum(e^{tA}) \equal{} \{e^{t\lambda_1},\cdots,e^{t\lambda_n}\}$. Necessarily the diagonal of $ e^{tA}$ must tend to $ 0$. QED[/quote]I get neither statement:[list][*]How do we get rid of the unitary factors $ Q$ from Schur's lemma (I guess that's how you assume $ A$ upper triangular)?
[*]why is $ spectrum(e^{tA})\equal{}e^{t\cdot spectrum(A)}$?[/list]

[quote="loup blanc"]We may assume $ A \equal{} \lambda{I} \plus{} J$ where $ J$ is a nilpotent Jordan matrix of order $ k\leq{n}$ and $ \lambda$ is an $ A$-eigenvalue. [/quote]Aren't you assuming here that $ \lambda$ is the only eigenvalue of $ A$? Otherwise how can you assume $ A\equal{}\lambda I\plus{}nilpotent$?

[quote="loup blanc"]$ e^{tA} \equal{} e^{t\lambda}(I \plus{} tJ \plus{} \cdots \plus{} {t^{k \minus{} 1}J^{k \minus{} 1}})$. $ e^{tA}$ tends to $ 0$ iff $ e^{t\lambda}t^{k \minus{} 1}$ tends to $ 0$. We are done.[/quote]... what are you doing exactly in this step? I completely lost track here...

Please can you explain the above steps into some more detail?
Thanks!

* How do we get rid of the unitary factors from Schur's lemma (I guess that's how you assume upper triangular)?
---
If $ A \equal{} QDQ^{ \minus{} 1}$, then $ e^{tA} \equal{} Qe^{tD}Q^{ \minus{} 1}$ (by expanding into series), so $ \lim_{t \rightarrow \infty} \parallel{}e^{tA}\parallel{} \equal{} 0 \Leftrightarrow \lim_{t \rightarrow \infty} \parallel{}e^{tD}\parallel{} \equal{} 0$.

    * why is spectrum(e^{tA})=e^{t\cdot spectrum(A)}?
---
Put $ A$ into Jordan form. Then the diagonal entries can be readily computed.

I want to know how loup blanc's argument goes over for the other side too.

Hi Kent,
your particular case is the case of the companion matrix of your polynomial.
Hi Peter and Soarer,
I'll try to be clear.
1) Functions of matrices:
More generally if $ f$ is an analytic function which is defined over $ spectrum(A)$, then we may define $ f(A)$ and then $ spectrum(f(A))=f(spectrum(A))$. (see the books of Horn and Johnson about matrix analysis,chapter matrix functions or the last book of N. Higham: functions  of matrices). There exist several methods to define $ f(A)$ ; one among is the tiangularization of $ A$.
2) Jordan's form:(over an algebraically closed field).
There exists $ P$ invertible s.t. $ A=PDP^{-1}$ where $ D=diag(\lambda_1{I{}_i{}_1}+J{}_i{}_1,\cdots,\lambda_k{I{}_i{}_k}+J{}_i{}_k)$,$ \lambda_j$ are the $ A$-eigenvalues,$ J{}_i{}_k$ is the nilpotent Jordan matrix of order $ i_k$,ie the matrix $ [u_{ij}]$ with $ u_{i,i+1}=1$ the other elements being $ 0$ (beware, the $ \lambda_i$ are not necessarily distinct !). Now $ e^{tA}=Pdiag(E_1,\cdots,E_k)P^{-1}$ where $ E=e^{t(\lambda{I}+J)}=e^{t\lambda}e^{tJ}=e^{t\lambda}(I+tJ+\cdots+{t^{k-1}J^{k-1}/{(k-1)!}})$ (one mistake in the formula in my precedent post) . Here I assume $ E$ is of dimension $ k$ and then $ J^k=0$.
It remains to prove that $ \alpha(A)<0$ implies $ e^{t\lambda}t^s$ tends to $ 0$ when $ t$ tends to infinity, that is obvious ($ s\geq{0}$).
Remark 1: from this we can deduce: there exist $ a>0,\sigma>0$ s.t. for all $ t\geq{0}$, $ \parallel{}e^{tA}\parallel{}\leq{ae^{-\sigma{t}}}$ iff all solutions of $ x'=Ax$ tend to $ 0$ when $ t$ tends to infinity.This is very important in the theory of linear ODE.
Remark 2:more precisely $ \parallel{}e^{tA}\parallel{}=O(e^{t\alpha(A)}t^{n-1})$ if $ A$ is of order $ n$.
Is it clear ?


$ m,n\in \mathbb N$
find all $ (m,n)$ such that
$ 3m^2n\equal{}n^3\plus{}415$

$ n$ | $ 415$

$ n\equal{}1,5,83,415$

But we also have $ 3$ | $ 415\plus{}n^3$

Hence, by modular arithmetic, it is easy to see that $ 1,415$ are impossible.

$ n\equal{}5$ and $ m\equal{}6$ works as a solution.

$ n\equal{}83$ does not work, so $ (n,m)\equal{}(5,6)$ is the only solution.


If $ x<y$, find the ordered pair of real numbers $ (x,y)$ which satisfies $ x^{3}+y^{3}=400$ and $ x^{2}y+xy^{2}=200$.

$ (x+y)^{3}=(x^{3}+y^{3})+3(x^{2}y+xy^{2})=1000$

Therefore $ x+y=10$. It follows that $ xy=20$ from the fact that $ x^{2}y+xy^{2}=xy(x+y)$.

10 is the sum of the roots and 20 is the product of the roots, so...

$ x^{2}-10x+20=0$

$ x=\frac{10-\sqrt{10^{2}-4\cdot\ 1 \cdot20}}{2}=5-\sqrt5$.

And our answer is $ (5-\sqrt5,5+\sqrt5)$.


"Since you are now studying geometry and trigonometry, I will give you a problem. A ship sails the ocean. It left Boston with a cargo of wool. It grosses 200 tons. It is bound for Le Havre ... There are 12 passengers aboard. The wind is blowing East-North-East. The clock points to a quarter-past-three in the afternoon. It is the month of May. How old is the captain?"

well, isn't the problem like so:

"Since you are now studying geometry and trigonometry, I will give you a problem. You are the captain of a ship who sails the ocean. It left Boston with a cargo of wool. It grosses 200 tons. It is bound for Le Havre ... There are 12 passengers aboard. The wind is blowing East-North-East. The clock points to a quarter-past-three in the afternoon. It is the month of May. How old is the captain?"

because otherwise it is impossible.. at least in my opinion.  :?

No, I think there is no extra condition necessary. Actually, in a variant of the problem (On a ship there are 22 sheep and 15 horses, etc., etc. ..., the final question being again on the age of the captain) is solved by an astrological interpretation of the data.

  Darij

Actually, I think a literature interpretation of the data would be more suitable.
The captain should be 16, as in the book by Jules Verne :) :D.
I am probably wrong, but it should be something like this :)

[quote="darij grinberg"]I knew this one, but do you have the solution? (I believe it was an astrological one, but I've never seen it.)

  Darij[/quote]

i dont have a solution  :(
how would you astrologically interpret the data  :?  :?  :P


$a,b,c>0$and $a+b+c=1$
prove that:
 $2+9abc \geq 7(ab+bc+ca)$

A boring and well-known inequality.I have two proof for it:
First solution:Using another form of Schur's ineq for n=3:$\frac{9abc}{a+b+c}\geq 4(ab+bc+ca)-(a+b+c)^2$.The relation a+b+c=1 implies:$9abc+1\geq 4(ab+bc+ca)$
Conjugate of:$1=(a+b+c)^2\geq 3(ab+bc+ca)$->we're done.
Second solution:Let's prove it by using mixxing variables(it's easy)


A cube has eight vertices (corners) and twelve edges. A segment,
such as $ x$, which joins two vertices not jointed by an edge is
called a diagonal. Segment $ y$ is also a diagonal. How many
diagonals does a cube have?
[asy]/* AMC8 1998 #17 Problem */
pair A=(0,48), B=(0,0), C=(48,0), D=(48,48);
pair E=(24,72), F=(24,24), G=(72,24), H=(72,72);
pen d = linetype("8 8");

draw(A--D--C--B--cycle);
draw(D--H--G--C);
draw(A--E--H);

draw(B--F--E);
draw(F--G);
draw(H--A--G, d);
label("$x$", (40, 61), N);
label("$y$", (40, 35), N);[/asy]

There are $ 12$ diagonals congruent to segment $ x$ ($ 2$ on each face). There are $ 4$ diagonals congruent to segment $ y$ (pick any corner on the top, then connect to the opposite corner on the bottom). This makes $ 16$ diagonals total.

the four vertices on the top line up with the four on the bottom to make 4 space diagonals. Each face has 2 face diagonals, so there are 12+4=16 diagonals

This was already answered. A few months ago. With the same solution you posted.

meh.  you do it all the time?  why?  for like a post count?  no offense, but thats really dumb...


All triangles have the same value, and all circles have the same value. What is the sum of three circles?\\
$ \Delta \plus{} \bigcirc \plus{} \Delta \plus{} \bigcirc \plus{} \Delta\equal{} 21$\\
$ \bigcirc \plus{} \Delta\plus{}\bigcirc\plus{}\Delta\plus{}\bigcirc \equal{} 19$\\
$ \bigcirc \plus{} \bigcirc \plus{} \bigcirc \equal{}$ ?

We know that a triangle is 2 bigger than a circle by equations one and 2.  Thus, we can let circle=x and triangle=x+2 to find 5x+6=21 so 5x=15, x=3.  3 circles is 3*3=9

[b][size=150]Solution 1[/size][/b]
$ \{\begin{array}{l}
3\triangle+2\bigcirc = 21\\
2\triangle+3\bigcirc = 19\\
\end{array}
\Rightarrow
5\triangle+5\bigcirc = 40 \iff \triangle+\bigcirc = 8$.
So,
$ \{\begin{array}{l}
3\triangle+2\bigcirc = 21\\
2\triangle+3\bigcirc = 19\\
\end{array}
\iff
\{\begin{array}{l}
16+\triangle= 21\\
16+\bigcirc = 19\\
\end{array}
\iff
\{\begin{array}{l}
\triangle = 5\\
\bigcirc = 3\\
\end{array}$
Thus, $ \bigcirc  + \bigcirc  + \bigcirc  = 3+3+3 = \boxed{9}$.

[b][size=150]Solution 2[/size][/b]
$ \{\begin{array}{l}
3\triangle+2\bigcirc = 21 \cdots \ooalign{\hss1\hss\crcr\bigcirc}\\
2\triangle+3\bigcirc = 19 \cdots \ooalign{\hss2\hss\crcr\bigcirc}\\
\end{array}
\Rightarrow
\{\begin{array}{l}
3\times\ooalign{\hss1\hss\crcr\bigcirc}-2\times\ooalign{\hss2\hss\crcr\bigcirc}: 5\triangle=25 \\
2\times\ooalign{\hss1\hss\crcr\bigcirc}-3\times\ooalign{\hss2\hss\crcr\bigcirc}: -5\bigcirc=-15 \\
\end{array}
\iff
\{\begin{array}{l}
\triangle = 5\\
\bigcirc = 3\\
\end{array}$
Thus, $ \bigcirc  + \bigcirc  + \bigcirc  = 3+3+3 = \boxed{9}$.


advanced fields
advanced fields unsolved

Let $ E \subset M$ where $ M$ is measurable and $ m(M) < \infty$. Show that $ E$ is measurable if and only if $ m(M) \equal{} m^{*}(E) \plus{} m^{*}(M \minus{} E)$. The "only if" part is trivial. I want help with the other part. Thanks.

Obviously, by the fact that $ m(M) < \infty$, we can find measurable sets $ A, B \subset M$, s.t. $ E\subset A, M\setminus E\subset B, m^*(E) \equal{} m(A), m^*(M\setminus E) \equal{} m(B)$. Hence,
$ m(M) \equal{} m^*(E) \plus{} m^*(M\setminus E) \equal{} m(A) \plus{} m(B) \equal{} m(A\cap B) \plus{} m(A\cup B)$$ \equal{} m(M) \plus{} m(A\cap B).$
Then we have $ 0 \equal{} m(A\cap B)\geq m^*(A\setminus E)$.
It follows that $ A\setminus E$ is measurable and $ E \equal{} A\setminus (A\setminus E)$ is measurable as well.


Find the ordered pair $ (x,y)$ if $ (x+y)^{x}=101x+10y$.

The answer is quite easy to guess, but how would you do this without guessing?

geuss AND check. lol

would it involve logotharims?

Sorry how do you even guess and check for this?

I guess you can note the last digit must also be the last digit of $ x$ since its $ 10y$...but I dont even know what the answer is from guessing and checking for 30 seconds..

EDIT: nvm its $ (3,4)$...but yea I dont think theres a direct way with just computation.

So there's no way to do this without guess and check?

And seriously, how is my question spam? This is getting annoying...


Any idea as to how to start preparing for math olympiads ? I'm presently going to the ninth grade , and the singapore mathematics Olympiad is well , the first Olympiad I'll be writing...Anyways , could someone tell me as to how I start preparing {for all kinds of Olympiads of my level } ?
Thanks a real lot in advance..

[quote="amirhtlusa"]i guess the best way to prepare is to take a look take at previous exams and see what you know and what you dont. then go learn stuff... :D
but remember this:
NEVER MEMORIZE ANYTHING, TRY TO PROVE EVERY THING NEW...
and USE what you learn...[/quote]

I will expand on what amirhtlusa said.

by USING what you learn, I think he means that take advantage of the conditions/formulas/theorems, etc. That you are given. If it is inequality, look for certain characteristics taht reminds you of other well-known theorems. Get the problem down to simplicity. Then you will succeed.

Memorizing is not good, as you will probably be ducted points for writing something memorized with being proved. of course if you memorized the proof itself, then you are good. ;)

I'd recommend getting a few books with large numbers of olympiad problems in them.  If they are in the book, they are usually organized by topic in some way and that can help you to see patterns and methods that apply to to many different problems.  I've got Mathematical Olympiad Challenges+Treasures and both are excellent.  For a beginner, a book like the Mathematical Olympiad Handbook might not be bad because it has problems  in it that are all fairly approachable.  Good luck!  :)


We have a ping-pong ball of mass $m$. In a height $h$ above the table we hit it with so that the direction of its initial velocity is horizontal. We spinned the ball, so it starts spinning around its axis perpendicular to the direction of the initial velocity. After falling down on the table, it bounces back verticaly (perpendicular to the plane of the table) without spinning, (the spinning stops just as it collides with the table). At most how much heat was "made" during the collision, if the collision is fully elastic and the coefficient of kinetic friction between the table and the ball is $\mu$?

Yes very good Djole, thats correct.  :clap2:
You can post the solution if you want :D

Ok, so the most heat will be generated when the same force does both the translational and the rotational slowing down -- I mean, when the translational speed $v$ and the rotational speed $R\omega$ at the bottom point of the ball are parallel and have the same direction; this happens when $\vec \omega$ is perpendicular both to $\vec v$ and $\vec g$. This force is $F = \mu N$, and $N = \Delta p_{\perp}/\Delta t = 2m\sqrt{2gh}/\Delta t$ (I believe this is the easier part). We also have $F = mv/\Delta t$ and $FR = 2mR^{2}\omega/3\Delta t$, which leads to $R\omega = 3v/2$ and $v = 2\mu\sqrt{2gh}.$ Now it is simple to get the released heat, which is the sum of the rotational and (horizontally) translational kinetic energies: \[Q = \frac{mv^{2}}{2}+\frac{(2/3)mR^{2}\omega^{2}}{2}= 10\mu^{2}mgh.\]


$ \forall _{n\in N} \ a_{n}\wedge b_{n}$ are real numbers, such that $ a_{n}\leqslant b_{n}$. Show that there exist $ sup\lbrace a_{n}\rbrace _{n\in N}$ and also $ (sup\lbrace a_{n}\rbrace _{n\in N}) \ \in \bigcap_{n \equal{} 1}^{infinity}[a_{n}, b_{n}]$.

this doesnt sound true..

$ a_n \equal{} 1\minus{}\frac{1}{n}, b_n \equal{} 1\minus{}\frac{1}{2n}$ ?

Yes, it's wrong. Maybe this was intended: for all $ m,n$, $ a_m\le b_n$?

Hmmm... The question is correct  $ a_{n}\leqslant b_{n}$  for $ n\in N$.

Hmmm.... The question is asked properly - but it is wrong -> the point is that now even I see the nonsense, but I cannot prove that the question is incorrect... Could you prove (mathematicaly)  that the question does not have sense? I have a problem with that...

Thank you

kenn4000's example is the proof you need. Just write out the details of how it doesn't work.

Hmmm... OK.

Let me modificate the question:

$ \forall _{n\in N} \ a_{n}\wedge b_{n}$ are real numbers, such that $ [a_{n}, b_{n}]\subset [a_{n\plus{}1}, b_{n\plus{}1}]$. Show that there exist $ sup\lbrace a_{n}\rbrace _{n\in N}$ and also $ (sup\lbrace a_{n}\rbrace _{n\in N}) \ \in \bigcap_{n \equal{} 1}^{infinity}[a_{n}, b_{n}]$.

The supremum you want in that version is $ a_1$, and the intersection is $ [a_1,b_1]$. It's still a silly question.

I think, the problem should be stated as follows:
Let $ P_n\equal{}[a_n,b_n]\in\mathbb{R}$ be a sequence of non-empty intervals such that: $ \forall n\in\mathbb{N} \quad P_{n\plus{}1}\subset P_n$

Prove, that:

1) $ \exists x\in\mathbb{R}\quad\forall n\in\mathbb{N} \quad x\in P_n$
2) $ sup \{a_n\}\equal{}inf \{b_n\}\equal{}x$


quadratics
geometry
circumcircle
probability
algebra
perpendicular bisector
angle bisector

Canadian MO 2001:

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=77723
[b]Randy:[/b] "Hi Rachel, that's an interesting quadratic equation you have written down.  What are its roots?''

[b]Rachel:[/b] "The roots are two positive integers.  One of the roots is my age, and the other root is the age of my younger brother, Jimmy.''

[b]Randy:[/b] "That is very neat!  Let me see if I can figure out how old you and Jimmy are.  That shouldn't be too difficult since all of your coefficients are integers.  By the way, I notice that the sum of the three coefficients is a prime number.''

[b]Rachel:[/b] "Interesting.  Now figure out how old I am.''

[b]Randy:[/b] "Instead, I will guess your age and substitute it for $x$ in your quadratic equation $\dots$ darn, that gives me $-55$, and not $0$.''

[b]Rachel:[/b] "Oh, leave me alone!''

(1)Prove that Jimmy is two years old.
(2)Determine Rachel's age.

Problem 2:

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?t=77724
Let $ABC$ be a triangle with $AC > AB$.  Let $P$ be the intersection point of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle{A}$.  Construct points $X$ on $AB$ (extended) and $Y$ on $AC$ such that $PX$ is perpendicular to $AB$ and $PY$ is perpendicular to $AC$.  Let $Z$ be the intersection point of $XY$ and $BC$.
Determine the value of $\frac{BZ}{ZC}$.

Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?t=77726
Let $n$ be a positive integer. Nancy is given a rectangular table in which each entry is a positive integer.  She is permitted to make either of the following two moves:
 (1)  select a row and multiply each entry in this row by $n$.
 (2)  select a column and subtract $n$ from each entry in this column.
Find all possible values of $n$ for which the following statement is true:

Given any rectangular table, it is possible for Nancy to perform a finite sequence of moves to create a table in which each entry is $0$.

Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?t=77727
Let $P_0$, $P_1$, $P_2$ be three points on the circumference of a circle with radius $1$, where $P_1P_2 = t < 2$. For each $i \ge 3$, define $P_i$ to be the centre of the circumcircle of $\triangle P_{i-1} P_{i-2} P_{i-3}$.
(1)  Prove that the points $P_1, P_5, P_9, P_{13},\cdots$ are collinear.
(2)  Let $x$ be the distance from $P_1$ to $P_{1001}$, and let $y$ be the distance from $P_{1001}$ to $P_{2001}$. Determine all values of $t$ for which $\sqrt[500]{x/y}$ is an integer.

I did not post problem 2 because I don't have the diagram. Here is the problem:

There is a board numbered $-10$ to $10$ as shown.  Each square is coloured either red or white, and the sum of the numbers on the red squares is $n$.  Maureen starts with a token on the square labeled $0$. She then tosses a fair coin ten times.  Every time she flips heads, she moves the token one square to the right. Every time she flips tails, she moves the token one square to the left. At the end of the ten flips, the probability that the token finishes on a red square is a rational number of the form $\frac{a}{b}$.  Given that $a+b=2001$, determine the largest possible value for $n$.


[quote="shobber"]Find the greatest integer $n$, such that there are $n+4$ points $A$, $B$, $C$, $D$, $X_1,\dots,~X_n$ in the plane with $AB\ne CD$ that satisfy the following condition: for each $i=1,2,\dots,n$ triangles $ABX_i$ and $CDX_i$ are equal.[/quote]

See here: [url]http://mks.mff.cuni.cz/kalva/apmo/asoln/asol015.html[/url]

Here is the solution from our yearbook containing all our training material from 2001. This solution is probably copied from the official solution.

One of the sides $AX_i$ or $BX_i$ is equal to $CD$, thus $X_i$ is on one of the circles of radius $CD$ and centre $A$ or $B$. In the same way $X_i$ is on one of the circles of radius $AB$ with centre $C$ or $D$. The intersection of these four circles has no more than 8 points so that $n\le 8$.

Suppose that circle $S_B$ with centre $B$ and radius $CD$ intersects circle $S_C$ with centre $C$ and radius $AB$ in two points $X_1$ and $X_2$ which satisfy the conditions of the problem. Then in triangles $ABX_1$ and $CDX_1$ we have $BX_1=CD$ and $CX_1=AB$. Since these triangles are congruent then $AX_1=DX_1$, therefore $X_1$ and $X_2$ are on the perpendicular bisector of $AD$. On the other hand $X_1X_2$ is perpendicular to segment $BC$. Then $BC\parallel AD$ and $AB$ and $CD$ are the diagonals or non-parallel sides of a trapezoid.

Suppose that $AB<CD$. Then $BX_1=CD>AB=CX_1$. It follows that the distance from $A$ to the perpendicular bisector of $BC$ must be less than the distance from $D$ to this line otherwise we obtain a contradiction to the condition $AB<CD$. Then for any point $X$ in the perpendicular bisector of $BC$ we have $AX<DX$ and it is not impossible to have $AX=CD$, $DX=AB$. Thus if the circle with centre $A$ and radius $CD$ intersects the circle with centre $D$ and radius $AB$, then the points of intersection of $S_B$ with $S_C$ satisfy the condition of congruence, then the points of intersection of $S_B$ with $S_D$ do not. Thus no more than half of the 8 points of intersection of these circles can satisfy the condition of congruence, i.e. $n\le 4$.

If $n=4$ we have the following example where $ABDX_3X_2C$ is a regular hexagon, and $X_1$ and $X_4$ are the intersections of $X_2X_3$ with respectively $AC$ and $BD$.
[asy] pair A; pair B; pair C; pair D; pair X1; pair X2; pair X3; pair X4;

D=(1,0); C=(-1,0); B=(0.5,sqrt(3)/2);
A=B+C;
X2=-B; X3=-A; X4=X3+D; X1=X2+C;

draw(A--X1--X4--B--A);
draw(X2--C--D--X3);
draw(circumcircle(A,B,C));
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,W);
label("$D$",D,E);
label("$X_1$",X1,S);
label("$X_2$",X2,S);
label("$X_3$",X3,S);
label("$X_4$",X4,S);
[/asy]

What motivated the solver to get that configuration? While it does have nice symmetric properties, the motivation seems a bit obscure besides that bit about AB, CD being a trapezoid but even that contradicts the diagram.


I'm going as a counselor  :)

see you there!


There are a few things that I have noticed about Alcumus being back that may be glitches:

-I asked Alcumus to choose a new problem for me, but it just gave me a screen with just a thin white box but no problem

-Sometimes, I refresh and get a problem that I recently answered.

-On my report, all my subjects are mastered, but on the solution menu, it sometimes says that I am in a subject that I have passed rather than mastered.

Wow, thats weird. Thats never happened to me once, even since the new (or recently fixed) alcumus. Sorry, if that is happening to you, I really do not know how to fix it.  :( You could try PMing a admin or programming person for alcumus.

Don't worry, its just that when you were one, aops was probably adding new problems and/or features.