calculus
integration
trigonometry
real analysis
real analysis unsolved

Can you evaluate the following integral
\[ \int\limits_0^{\frac{\pi }
{2}} {\frac{{dx}}
{{3 \plus{} \sin ^2 x}}} .
\]
by using method of residue  :D .

Here is a possible solution:

Let $ I$ be the value of your integral. Then, using the substitution $ x\equal{}\pi\minus{}y$, we get that

$ I\equal{}\frac{1}{2}\int\limits_{0}^{\pi}\frac{dy}{3\plus{}\sin^{2}y}\equal{}\frac{1}{2}\int\limits_{0}^{\pi}\frac{2dy}{7\minus{}\cos(2y)}$.

The substitution $ 2y\equal{}t$ shows that

$ I\equal{}\frac{1}{2}\int\limits_{0}^{2\pi}\frac{dt}{7\minus{}\cos(t)}$.

Now this is an integral that can be solved by residue theory via the classical substitution $ \sin(t)\equal{}\frac{u^{2}\minus{}1}{2iu}$.

Besides complex variables, one may also use the Weierstrass substitution $ x\equal{}2\arctan t.$ Then $ dx\equal{}\frac{dt}{t^2\plus{}1}$ and $ \sin t\equal{}\frac{2t}{t^2\plus{}1}.$

Then $ \frac 1{3\plus{}\sin^2x}\equal{}\frac{(1\plus{}t^2)^2}{3(1\plus{}t^2)^2\plus{}4t^2}\equal{}
\frac{(1\plus{}t^2)^2}{3t^4\plus{}10t^2\plus{}3}.$

$ \int_0^{\frac{\pi}2}\frac{dx}{3\plus{}\sin^2x}\equal{}
\int_0^{\infty}\frac{t^2\plus{}1}{3t^4\plus{}10t^2\plus{}3}\,dt$

$ \equal{}\int_0^{\infty}\frac{t^2\plus{}1}{(3t^2\plus{}1)(t^2\plus{}3)}\,dt$

$ \equal{}\frac14\int_0^{\infty}\frac1{3t^2\plus{}1}\plus{}\frac1{t^2\plus{}3}\,dt$

Using that $ \int_0^{\infty}\frac{dt}{t^2\plus{}a^2}\equal{}\frac{\pi}{2a},$ we get that the original integral is

$ \frac14\left(\frac13\cdot\frac{\sqrt{3}\pi}2\plus{}\frac{\pi}{2\sqrt{3}}\right)
\equal{}\frac{\pi}{4\sqrt{3}}.$

If I write $ \frac{1}{3\plus{}\sin^2x}\equal{}\sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^k\sin^{2k}x}{3^{k\plus{}1}},$ and then interchange integral and sum, I get

$ \int_0^{\frac{\pi}2}\frac{dx}{3\plus{}\sin^2x}\equal{}\frac{\pi}2\sum_{k\equal{}0}^{\infty}
\frac{(\minus{}1)^k}{3^{k\plus{}1}}\cdot\frac{(2k\minus{}1)(2k\minus{}3)\cdots3\cdot1}{(2k)(2k\minus{}2)\cdots4\cdot2}$

$ \equal{}\frac{\pi}2\sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^k(2k)!}{2^{2k}3^{k\plus{}1}(k!)^2}.$

[quote="Kent Merryfield"]If I write $ \frac {1}{3 \plus{} \sin^2x} \equal{} \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k\sin^{2k}x}{3^{k \plus{} 1}},$ and then interchange integral and sum, I get

$ \int_0^{\frac {\pi}2}\frac {dx}{3 \plus{} \sin^2x} \equal{} \frac {\pi}2\sum_{k \equal{} 0}^{\infty} \frac {( \minus{} 1)^k}{3^{k \plus{} 1}}\cdot\frac {(2k \minus{} 1)(2k \minus{} 3)\cdots3\cdot1}{(2k)(2k \minus{} 2)\cdots4\cdot2}$

$ \equal{} \frac {\pi}2\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k(2k)!}{2^{2k}3^{k \plus{} 1}(k!)^2}.$[/quote]
Last sum can be calculated. Because in classroom we finded closed form for $ I\equal{}\sum \frac{(2k)!}{(k!)^{2}}x^{k}$ it's I don't remember exact answer, but I cans say that it's calculated by standart method.

To use the formula milin wrote, note that we should first factor $ \frac13$ out of the sum and then let $ x\equal{}\minus{}\frac1{12}.$

We then get $ \frac{\pi}{6}\cdot\frac{1}{\sqrt{1\plus{}\frac13}}\equal{}
\frac{\pi\sqrt{3}}{12}\equal{}\frac{\pi}{4\sqrt{3}}.$

The various calculations are in agreement.


in a forest the distance between all 2 trees is less than the difference of the height of them. find the minimum value of the distance of a rope such that we can embed the forest with that rope . the height of all trees is less than 100 meters

The distance of a rope can be infinitely small...

[quote="JBL"]Presumably it means "find the shortest rope that works for all such forests."[/quote]
yes you are right
hint: the answer is 200 but why ?.........   :?:

[quote="PhilG"]"a rope such that we can embed the forest with that rope"

I dont understand this phrase. However. The tallest tree is less than 100m tall, all oher trees must be within 100m of this tree. So the forest is inside a circle of diameter 200m. Is that what was wanted?[/quote]
yes ,but the trees of the forest can not fill the circle (think!) then there is an other shape (not a circle) with that conditions
find the minimum value of the rope. ..
sorry my english is not good
a rope which we can embed the forest with
i think this is better

[quote="PhilG"]Do you mean that the rope must go round the forest to enclose it?
Then $ 200\pi$ would do it.[/quote]
i know the rope with distance 10000000000000 do it to !
but i mean :[size=150]find the shortest![/size]
 :lol:

[quote="sobe86"]Visit all the trees in order of height from shortest to highest, by the assumption of the question, the total length of this will be less than or equal to 100m.  Then return to the shortest, this will be less than or equal to 100m.  Total less than or equal to 200m.  200m is only necessary if all the trees are in a straight line of length 100m.

That OK?[/quote]
can you explain more please ? :?:

[quote="Elixir10"]Let $ n$ denote the number of trees. We find that all the trees have distinct heights. Denote the heights of the trees as $ h_1, h_2, h_3,... h_n$, where $ h_1 > h_2 > h_3 > ... > h_n$.

Then, denote the distance between the trees with heights $ h_i$ and $ h_{i \plus{} 1}$ as $ d_i$, so

$ d_1 < h_1 \minus{} h_2$
$ d_2 < h_2 \minus{} h_3$
$ \vdots$
$ d_{n \minus{} 1} < h_{n \minus{} 1} \minus{} h_n$

Hence $ d_1 \plus{} d_2 \plus{} ... \plus{} d_n < h_1 \minus{} h_n < 100$.

This means that if the rope were to go around the trees in order from tallest to shortest, it would only need to be 100 m; then, going back from the shortest tree to the tallest tree would be even shorter, because all that is needed is to go through a straight path, which is less than or equal to the first path going from tallest to shortest. Hence the minimum rope is 100 m + 100 m = 200 m.[/quote]
why can we [size=150]embed[/size] the forest with that rope?


First and most, this round was relatively harder than ones past and it was not easy to get a perfect score. The problem was not impossible but it was not easy for anyone to get it easy.

So, to have relatively similar amount of qualifers on middle school and high school, if you have tried, you'll be qualified.

Only one person has qualified in normal way (which means perfect score), and that is:

[size=150]Jimmy[/size]

Being only one to qualify, I give you:

[hide="Prize_to_see"]
 :first: [/hide]

Rest of qualifiers:

h_s_potter_2002
allrighty
white_horse_king88
Zyloch
sky9073
Pkothari1013
ln(dx/dy)

This is also thread discussing the solution. Please share your answer with others here.

2. Brute-forced (found all intelligent numbers, added them up, etc) and got 3866. (I think it's wrong somehow)
5. a^2 - b^2 = (a + b) (a - b) = 103. Since 103 is prime, a+b = 103 and a-b = 1. The answer is 1.

That nice - for 5 I apparently didn't pay attention to the fact that it was asking for a - b. Ah - that would have been nice to read the problem.

[quote="sky9073"]2. Brute-forced (found all intelligent numbers, added them up, etc) and got 3866. (I think it's wrong somehow)
[/quote]
Don't worry, it's right.

If anyone has any questions on the problems, I'd be glad to help.

I can't believe I forgot to try the middle school questions.  I hate myself so much now.

Why didn't I qualify? You said all my answers were correct!

[quote="Silverfalcon"]You'll see the list of people who qualified after some dates.

Your answers are correct.

You're qualified.

Your name's going to be posted.
[/quote]

Sorry about it.

I probably deleted some of pm without knowing it.

I edited so thanks for pointing out. :)

Just curious, what exactly were the answers to the questions? I'm not too sure which one I got wrong, so just want to know lol

I'm sorry I've seen to late this forum  :( . The exercises are not so hard  :P

[quote="Slizzel"]I'm sorry I've seen to late this forum  :([/quote]
Welcome .. you know what they say ... "better late then never" :P :)

[quote="Slizzel"]The exercises are not so hard  :P[/quote]Then move to the next forum. If you find all high-school forums not so hard, you may want to consider the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Forums[/url] ;)

For number 5 I don't understand why a+b has to be 103 and a-b has to be 1. Why can't the two number be radicals for fractions and still multiply to 103?

It doesn't matter a+b being 1 and a-b being 103 (or reverse) because when you add them, the b's are going to get cancelled anyway.

Remember that the answer has to be in integers.

I'm probably missing something obvious, but where did it ever say the answers had to be integers?

I asked, I think it was posted in the forum after the questions.  But think about it, if the answer didn't have to be in integers, then there are infinitely many solutions.

Thats exactly what I was wondering that there were infinitely many solutoins, but silverfalcon said these problems were hard so i figured it was just me  :blush:  So I ended up not entering. Eh, better luck next time.


I'm a soph and got 122 on AMC 12 and probably 6 on AIME this year..

My goal for next year is to do well in AIME and USAMO

Would WOOT be a good class for me??

Thanks,

Assuming you will be continuing your mathematics studies this summer from your current point, WOOT would indeed be a good choice for you.  We'll have details on the site about 2006-7 WOOT within a few weeks.

[quote="1234567890"]Is AoPS going to shut down the Olympiad classes and replace them with WOOT? I think you were mentioning that as a possibility before.[/quote]

This is highly likely.  At this point, we expect to end the Olympiad classes and expand/alter WOOT.  The olympiad classes will in part be folded into WOOT.


Given $ a, b, c \geq\ 0$. Prove that: $ a^5 \plus{} b^5 \plus{} c^5 \plus{} 4(a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3) \geq\ 5(a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2)$

@[hide]Materazzi: Have fun, my brother :) [/hide]


Here are the 3 problems out of the 25 for my homework that I can not figure out. Any help would be greatly appreciated.

1.Evaluate to four decimal place using a calculator e2.9

2.Find the equation of the hyperbola satisfying the given condition: Vertices at (9.0) and (-9,0); foci at (11,0) and (-11,0)

3.Find the focus and directrix of the parabola x2 = 12y

[hide="3"]
If you put a parabola in the form $y = a(x-h)^{2}+k$, the focus of the parabola is $\left(h, K+\frac{1}{4a}\right)$

and the directrix is
$y = k-\frac{1}{4a}$

For your problem we have (I'm assuming this is what you meant)
$x^{2}= 12y$.
$y = \frac{1}{12}(x-0)^{2}+0$
We know a few things. The vertex is always at $(H, K)$, in this case it's at $(0,0)$ and as previously stated, the focus is at $(H, K+\frac{1}{4a})$

The h value for the focus is at 0, and the $K+\frac{1}{4a}$ value is $0+\cfrac{1}{4 \cdot \frac{1}{12}}$
Thus, the focus is at $(0, \frac{12}{4})$, or more simply, $(0,3)$.

Try to use what I've given you to find the equation of the directrix. This form for parabolas is called "Standard Form"; and can be found by completing the square.
[/hide]

Ok this makes since so would the answer to the directix be y=-3?

Correct. This form is really quite helpful on tests, as it gives much information for such a small equation.

Yes, but I'm not so good at those, and wouldn't be confident in the answer. I might confuse you more.

Why dont you give it a shot and we can compare answers?

Ok, here it is:
[hide="2"]
Because parabolas can be written in the form:
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1$
(This parabola opens towards the X-Axis)
OR

$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}= 1$
(This parabola opens towards the Y axis)

Where a and b are coords of the vertices.

So, looking that the parabola was vertices on the X axis, and is centered at the origin, we can see that we have to use the first form:
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1$

We know the location of our vertices, so we substitute:
$\frac{x^{2}}{-9^{2}}-\frac{y^{2}}{9^{2}}= 1$

$\frac{x^{2}}{81}-\frac{y^{2}}{81}= 1$

This is considered the standard form of hyperbolas.
[/hide]

[hide="#2"][quote="Mark9182"]Do you have any clue about number 2?[/quote]

i posted one in the other thread (not sure)
2.Find the equation of the hyperbola satisfying the given condition: Vertices at (9.0) and (-9,0); foci at (11,0) and (-11,0)

find the other vertices
121-81=40
so the others have a length of square root of 40.
Luckily it is centered at 0,0 so the equation is:

$x^{2}$/81 - $y^{2}$/40=1
[/hide]


Let us have a set $\{a_n\}_{n\geq 0}$, such that $a_0=a_1=5$ and
\[ a_n=\frac{a_{n-1}+a_{n+1}}{98} \]
Prove that $\frac{a_n+1}{6}$ is a perfect square $\forall n\in\mathbb{N}\cup\{0\}$.

Masoud Zargar

if my calculations are not flawed then we have:

$2a_n = (5-2\sqrt6)(49+20\sqrt6)^n + (5+2\sqrt6)(49-20\sqrt6)^n$

We just need to prove: $\frac{(5-2\sqrt6)(49+20\sqrt6)^n + (5+2\sqrt6)(49-20\sqrt6)^n + 2}{12}$ is a square.

Anyone any ideas?


geometry
incenter
3D geometry
sphere
tetrahedron
geometry unsolved

1. Let I and G be respectively the incenter and the centroid of a triangle ABC. Let $R_{1}, R_{2}, R_{3}$ be the circumradii respectively of the triangles IBC, ICA, IAB and let $R'_{1}, R'_{2}, R'_{3}$ be the  circumradii respectively of the triangles GBC, GCA, GAB. Prove that $R'_{1}+R'_{2}+R'_{3}\geq R_{1}+R_{2}+R_{3}.$

2. Let $AA_{1}, BB_{1}, CC_{1}$ be the inner angled bisectors of a triangle ABC; let $A_{2}, B_{2}, C_{2}$ be the touching points of the incircle of triangle ABC winth the sides BC, CA, AB respectively. Let $S, S_{1}, S_{2}$ be the areas of the triangles $ABC, A_{1}B_{1},C_{1}, A_{2}, B_{2}, C_{2}$ respectively. Prove that $\frac{3}{S_{1}}-\frac{2}{S_{2}}\leq \frac{4}{S}.$

3. Let I and r be respectively the center and the radius of the sphere inscribed in a
tetrahedron ABCD. Let $r_{a}, r_{b}, r_{c}, r_{d}$ be the radii of the sphere inscribed respectively in a tetrahedron IBCD, IACD, IABD, IABC. Prove that $\frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}+\frac{1}{r_{d}}+\leq \frac{4+\sqrt{6}}{r}$

4. Let Sxyz be a trihedral angle with $\widehat{xSy}=121^{o}, \widehat{xSz}=59^{o}.$ A is a point on Sx, OA=a. On the ray bisecting the angle zSy, take the point B such that $SB=a.\sqrt{3}.$ Calculate the measures of the angles of triangle SAB.

These problems are from current Maths & Youths issues, please don't answer them.

  darij


A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one? \[ \text {(A) } 10 \qquad \text {(B) } 11 \qquad \text {(C) } 18 \qquad \text {(D) } 19 \qquad \text {(E) } 20  \]

[hide="Solution"]
If you start with $n^2$ tiles, then after two operations you will have $(n-1)^2$ tiles left. After $2k$ operations you will have $(n-k)^2$ tiles left. If $n-k=1 \Leftrightarrow k=n-1$, then $2(n-1)$ operations will have been performed. In this case, $n=10$, so $2(10-1)=\boxed {18}$ operations are required to reduce the number of tiles to 1.
[/hide]

[hide="Answer"]We first remove $10$ tiles, and are left with $90$. Then $9$ tiles, left with $81$; $9$, left with $72$; $8$, left with $64$; $8$ left with $56$; etc. Continue the pattern, and we have $18$ repetitions. $\boxed{C}$[/hide]

[hide]there are 10 perfect squares from 1 to 100. Removing 10 gives us 90 left. THe removing 9 perfect squares (1-81) gives us 81. Removing 9 again gives us 72. Then we remove 8 to get 64. Remove 8 and have 56 left. Then you will have 49 left. (hey wait a second, these are perfect squares). Continuing to 1, it will take 18 steps. [/hide]


I have a problem

I have 2 aces and I want to know the probability of getting at least one more ace drawing 3 cards.

I have 50 cards left and 2 of them are aces

2/50 + 2/49 + 2/48 = 0.122148......

On the other hand I have 2 aces and 48 non-aces left
There are 19,600 combinations of 50 cards taken 3 at a time.
The 2 aces can combine in 2 ways taken 1 at a time
The 48 non-aces can combine in 1128 ways taken 2 at a time

2*1128 = 2256

2256/19600 = 0.1151.....

Why do I get 2 different numbers?

Thanks in advance

Suppose the problem asked instead, "what's the probability of getting at least one more ace if I draw 19 more cards?"  If we follow your first solution to the given problem, what answer do we get?

More dramatically, suppose we replace "19" with "49" -- following your first solution, what do we get?  Following your second solution, what do we get?

[hide="What has gone wrong?"]In your first solution, there's a dependency problem -- after you pick the first card, the probability that the second card gives you what you want depends on what the first card was.  In your second solution, you didn't consider all possible cases.  Either solution can be made to work, but actually the quickest solution to this problem is probably to consider the complement and find the probability that none of the three cards you choose is an ace.[/hide]

Thank you very much JBL

First there is a mistake in my post
There are 125,000 different 3 card hands in a 50 card deck
The 19,600 came from a different problem

I took your advice and what I came up with is
There are 110,592 3 card hands in the 48 non-ace deck
110,592/125,000 = 0.884736
1-0.884736 = 0.115264

I think that is correct, sites that give poker odds say 12%

I am still confused about the first method but I will keep working on it

Thanks JBL

[quote="kaaib"]There are 125,000 different 3 card hands in a 50 card deck
The 19,600 came from a different problem[/quote]  No, you were better-off the first time: there are 125,000 [i]ordered[/i] triples of cards from a 50-card deck [i]allowing repetitions[/i], and these include (for example) 50 triples that consists of three copies of the same card.  (The 19,600 is for unordered triples without repetition; order doesn't matter (as long as you do it the same in all calculations), but repetition does.)  Again, for your last calculation, what does it give you if you replace "draw three cards" with "draw 49 cards"?

The fact that your answer rounds to 12% doesn't indicate that it's any better than either of your first two answers ;)

I have seen the error of my ways
I am using a online calculator and it defaults to "with repetitions"

Now I get 0.11755----

I have a 2/50 shot at the first ace
If I don't get it I got a 2/49 shot
If I don't get it I got a 2/48 shot

If I continue down I get to 2/2
All the other probabilities add to this to give me a probability >1

I'm really confused

You've finally written down a correct answer :).  You might have a more productive learning experience if you didn't rely on this online calculator and instead figured out the various counting answers and probability answers by hand.  (It's often better in this context to not do all the arithmetic out at once, but instead to leave things in factored form -- instead of writing 125,000, writing it as $ 50 \cdot 50 \cdot 50$ makes it clearer where it came from.  Then you can put the whole arithmetic expression into a normal calculator at the end.)

[quote="kaaib"]I have a 2/50 shot at the first ace
If I don't get it I got a 2/49 shot
If I don't get it I got a 2/48 shot

If I continue down I get to 2/2
All the other probabilities add to this to give me a probability >1

I'm really confused[/quote]
This is what I was pointing out in part of my first comment (for example, in the hidden text).  The key here is that the phrase "If I don't get it" indicates that you should be using a conditional probability.  In particular, the probability "I don't get the first ace but I get the second one" is not $ \frac{2}{49}$ but instead is $ \frac{48}{50} \cdot \frac{2}{49}$ -- you need to consider both the fact that you got the ace on the second try [i]and[/i] the fact that you didn't get the ace on the first try.

Many thanks JBL

It has been years since  I studied
I got out one of my old books and it all came back  to me
I can't believe that I forgot about the conditional probabilities
I was over simplifying the problem
When playing poker online there is only 30 secs to make a choice
I was looking for a quick answer
I now have a chart

You have been a great help
Thank you very much


continued fraction
number theory unsolved
number theory

I deal with the one of titu's book 'Diophantine Equations' and I saw some strange expression using bar symbols.
For two examples

ex1)
There exist i,j,p \in \{0,1,2,...,c_{1}\}, i\not=j, p\not=0, such that
\displaystyle \frac {p-1} {c_{1}}\prec l_{i}-.................\cdots
because there are c_{1} intervals of the form \left(\frac{p-1}{c_{1},\frac{p}{c_{1}}\right), p=\bar{1,c_{1}}

ex2)
l_{k}=[k\surd{D}+1], k=\bar{0,c_{1}}

If you know, please teach me your knowledge

I believe that notation is used to denote a repeating (simple) [url=http://mathworld.wolfram.com/ContinuedFraction.html]continued fraction[/url] expansion.


Solve:-
1)$\sqrt{3x^2-4x+34}+\sqrt{3x^2-4x-11}=9$

2)find square root of:-
$\frac{3}{2}(x-1)+\sqrt{2x^2-7x-4}$

Both of them are beautiful
 :whistling:


Has anyone on the A Team gotten their calculator yet? I still haven't. Didn't Archie say he was going to send them to us?

Yeah, he said they'll be sent, and no, I haven't gotten one yet either. They'll come eventually... :)

Calvin did you get your \$250 bucks from the state math conest yet? (Intentional mispelling, if you have seen my trophy, you will know.)


Given a square $ABCD$ with side $1$, and a square inside $ABCD$ with side $x$, find (in terms of $x$) the radio $r$ of the circle tangent to two sides of $ABCD$ and touches the square with side $x$. (See picture).

Since the circle is tangent to two sides of the squares then the diagonal of $ABCD$ passes through the center of the circle and the point of contact between the circle and the smaller square. Then $r\sqrt{2} + r + x\sqrt{2} = \sqrt{2}$.


function
induction
algebra
functional equation
algebra proposed

Let $ f: A \rightarrow \mathbb{R}$, $ (A \subseteq \mathbb{R})$, be a function so that $ f(x\plus{}y)\equal{}f(x) f(y)\minus{} f(xy)\plus{}1$, for every $ x,y \in A$.

If $ f: A \rightarrow \mathbb{R}$, $ (\mathbb{N} \subseteq A \subseteq \mathbb{R})$, is a solution of given functional equation then prove:

$ f(n)\equal{} \frac{c^{n\plus{}1} \minus{}1}{c\minus{}1} ,$  for every $ n \in \mathbb{N}$ when $ c \neq 1 ,$  and

$ f(n)\equal{} n\plus{}1,$  for every $ n \in \mathbb{N}$ when  $ c \equal{} 1,$ where $ (c\equal{}f(1)\minus{}1)$.

a)	Solve given functional equation for $ A\equal{} \mathbb{N}$

b)	Assuming $ A\equal{} \mathbb{Q}$, find all functions $ f$ which satisfy given equation and $ f(1997) \neq f(1998)$.

Let $ f(1) \equal{} c \plus{} 1 \equal{} \frac {c^2 \minus{} 1}{c \plus{} 1}$. By induction (x=n,y=1) $ f(n \plus{} 1) \equal{} c\frac {c^{n \plus{} 1} \minus{} 1}{c \minus{} 1} \plus{} 1 \equal{} \frac {c^{n \plus{} 2} \minus{} c \plus{} c \minus{} 1}{c \minus{} 1} \equal{} \frac {c^{n \plus{} 2} \minus{} 1}{c \minus{} 1}$. But n=m=2 give $ 2f(4)\equal{}f(2)^2\plus{}1$ or
$ 2(c^4\plus{}c^3\plus{}c^2\plus{}c\plus{}1)\equal{}(c^2\plus{}c\plus{}1)^2\plus{}1$ or $ c^4\equal{}c^2$ or $ c\equal{}\minus{}1,0,1$.
1. $ c\equal{}1$ give $ f(x)\equal{}x\plus{}1$ solution.
2, $ c\equal{}0$ give $ f(x)\equiv 1$ - solution.
3. $ c\equal{}\minus{}1$ $ f(x)\equal{}0$ if $ x\minus{}odd$, $ f(x)\equal{}1$ if $ x\minus{}even$.


[color=darkblue]Hi every One, I have two questions about IQ. May you help me please.[/color]

[color=darkblue][u][b]Question 1[/b][/u][/color]
[color=darkred]Let these number: $ \minus{} 1$  ;  $ 1$  ; $ \minus{} 2$  ; $ \minus{} 1$  ; $ 0$  ; $ ?$

Find number in position "$ ?$"[/color]

[color=darkblue][u][b]Question 2[/b][/u][/color]
[color=darkred]Let these number: $ 0$  ; $ \minus{} 1$  ; $ 0$  ; $ 0$  ; $ 1$  ; $ ?$

Find number in position "$ ?$"[/color]

Possible "simple" dependencies:

1) $ x_{n\plus{}1}\equal{}x_nx_{n\plus{}2}\minus{}1$
2) $ x_{n\plus{}2}\equal{}x_n\plus{}x_{n\plus{}1}\plus{}1$

Frankly speaking, both questions do not contain enough information to be 100% sure about any solution. Whoever created this IQ test did a very poor job. Your coloring problem was far better :).

what do you mean by find the number in position?


Solve the following equation for $ x,y,z\in \mathbb R$

$ x^4\plus{}y^4\plus{}z^4\minus{} 4xyz\plus{}1\equal{} 0$

Using AM-GM we have: x^4+y^4+z^4+1>=4/xyz/>=4xyz(/x/ means absolute value of x)
The equality occurs when /x/=/y/=/z/=1, which means (x,y,z)={(1,1,1),(1,-1,-1),(1,1,-1),(-1,-1,1)...)(there are 8 solutions to this equation)(Q.E.D)

[quote="ghjk"]Using AM-GM we have: $ x^4+y^4+z^4+1\ge4|xyz|\ge4xyz$ (|x| means absolute value of x)
The equality occurs when $ |x|=y|=|z|=1$, which means $ (x,y,z)={(1,1,1),(1,-1,-1),(1,1,-1),(-1,-1,1)...}$(there are 8 solutions to this equation)(Q.E.D)[/quote]

$ \boxed{(x,y,z)=\{(1,1,1),(1,-1,-1),(1,1,-1),(-1,-1,1) \cdots\}}$
Sorry ,your solutions may not all suit for the equation.

 :oops:   :D

[quote="vinskman"][quote="ghjk"]Using AM-GM we have: $ x^4 + y^4 + z^4 + 1\ge4|xyz|\ge4xyz$ (|x| means absolute value of x)
The equality occurs when $ |x| = y| = |z| = 1$, which means $ (x,y,z) = {(1,1,1),(1, - 1, - 1),(1,1, - 1),( - 1, - 1,1)...}$(there are 8 solutions to this equation)(Q.E.D)[/quote]

$ \boxed{(x,y,z) = \{(1,1,1),(1, - 1, - 1),(1,1, - 1),( - 1, - 1,1) \cdots\}}$
Sorry ,your solutions may not all suit for the equation.

 :oops:   :D[/quote]
Oopps you're right! I forget to add the condition that xyz>=0, which means there are at least 2 same signed variables. Now probably we're done!


Let X be any real number. Prove that among the numbers: X, 2X, ... , (n-1)X there is one that differs from an integer by at most 1/n.

Hi,

My attempt :

[hide]

Divide the distance between two integers into n blocks of 1/n width each.

Now, if any of X, 2X, ... , (n-1)X lies in the 1st or last block between some two integers, then we are done.

Assume none of them lie in the first or last blocks between some two integers.

Then, each of them will have to be in one of (n-2) blocks. (they might be between different integer pairs)

Let, iX and jX lie in the same block, but possibly between different integer pairs. (i > j)

Then, iX - jX = some integer + fractional part,

where the fractional part is less at most 1/n.

Thus, among the given integers (i-j)X lies at a distance of at most 1/n from an integer.

[/hide]

*****************************************************************************************

Sambit

[quote="player"]Sorry...this dosn't look like a valid solutions.

it needs a case 3: if the numbers are not in the two end blocks and no two numbers are in the same block.

This is a harder case to show, and it needs to be shown for your above proof to be valid.

I might be wrong  :o[/quote]

There's $n-1$ numbers and $n-2$ non-end blocks.

Case 3 never happens.


limit
logarithms
calculus
calculus computations

evaluate this limit:

$ \lim_{n\to\infty}\sqrt[n]{1^{2}\plus{}2^{2}\plus{}...\plus{}n^{2}}$

thanx.

We actually have a closed formula for the sum of the squares of the first n natural numbers: $ \sum_{k \equal{} 1}^n k^2 \equal{} \frac{n(n\plus{}1)(2n\plus{}1)}{6}$. Therefore, your limit L is

$ L \equal{} \lim \left(\frac{n(n\plus{}1)(2n\plus{}1)}{6}\right)^{1/n} \equal{} 1$.

Alternatively,

$ L \equal{} \lim \frac{\sum_{k \equal{} 1}^{n\plus{}1} k^2 }{\sum_{k \equal{} 1}^n k^2 } \equal{} \lim \frac{1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \plus{} (n\plus{}1)^2}{1^2 \plus{} 2^2 \plus{} ... \plus{} n^2}$

$ \equal{} \lim \frac{1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \plus{} (n\plus{}1)^2}{(n\plus{}1)^3} \cdot \left(\frac{1^2 \plus{} 2^2 \plus{} ... \plus{} n^2}{n^3} \right)^{\minus{}1} \cdot \frac{(n\plus{}1)^3}{n^3}$

$ \equal{} \frac13 \cdot \left(\frac13 \right)^{\minus{}1} \cdot 1 \equal{} 1$. $ \Diamond$

In general:

$ \lim _ {n \to \infty} \sqrt [n] {\sum_{i \equal{} 1}^n i^p} \equal{} \lim _ {n \to \infty} \sqrt [n] {\sum_{i \equal{} 0}^{p\plus{}1} c_i\,n^i} \equal{} \lim _ {n \to \infty} \sqrt [n] {c_{p\plus{}1}\,n^{p\plus{}1}} \equal{}$

$ \equal{} \exp\left(\lim _ {n \to \infty} \frac {\ln c_{p\plus{}1} \plus{} (p\plus{}1) \ln n}{n}\right) \equal{} ^\text{H} \exp\left(\lim _ {n \to \infty} \frac {p\plus{}1}{n}\right) \equal{} 1$

The $ \equal{} ^\text{H}$ denotes the usage of L'Hospital's theorem.


How many five-digit zip codes exist such that each of the first three digits is greater than 4?

There are 5 digits greater than four(5,6,7,8, and 9) and 10 total digits, so the number of zip codes is $ 5 \cdot 5 \cdot 5 \cdot 10 \cdot 10  \equal{} 5^3 \cdot 10^2 \equal{} 12500$.

There are 5 possible digits for the first 3 so there are 125 ways the last 2 can be any digit from 0-9 which is 10 so its 125*100=12500.


How many positive integers have exactly three proper divisiors, each of which is less than 50?  (A proper divisor of a positive integer $ n$ is a positive integer divisor of $ n$ other than $ n$ itself.

Why isn't the answer 15C3 -   15 prime numbers choose three to be the divisors.

If you have three prime numbers a, b, c, then the proper divisors of abc are:
1, a, b, c, ab, bc, ac; there are 7 divisors there.

You can find the actual solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=365485]here[/url].

I hope that helps  :)


geometry
geometric transformation
vector
algebra solved
algebra

We have a square ABCD with AB=1 and a figure F wich has area included in it.Prove that, if the distance betwen every two points of F is less than 0.001 than the area of F is less than 0.34

Indication:We make two translations of F by vectors wich make an angle of 60
and demonstrate that F,F1,F2 where F1,F2 are the images dont have common points and are included in another square(somehow concentric with the initial square) .

???
Why are you giving so precise hints? The problem turns to be trivial...

It is very easy. We can consider two parallel translations T_1 and T_2, defined by vectors u and v such, that |u|=|v|=0.001, and \angle(u,v)= \pi /3. Than T_1(F) \cap F=T_2(F) \cap F=T_1(F) \cap T_2(F)=\emptyset.
So S(F)\leq (1.001^2)/3\leq 0.34.

More interesting to prove that the area of F is less than 0.287


geometry
circumcircle
trigonometry
angle bisector
geometry proposed

Given is $ \triangle ABC$ such that $ \angle A \equal{} 57^{\circ}$, $ \angle B \equal{} 61^{\circ}$ and $ \angle C \equal{} 62^{\circ}$. Which segment is longer: the angle bisector through $ A$ or the median through $ B$?

[i]Author: N.Beluhov[/i]

Internal bisector of $ \angle A$ cuts $ BC$ at $ D$ and the circumcircle $ (O)$ again at $ X.$ $ A', B'$ are midpoints of $ BC, CA.$ Since $ \triangle ABC$ is approximately equilateral, to the first order of magnitude,

$ \sin \widehat {XAA'} \equal{} \frac {A'X}{AA'} \sin \widehat{A'XA} \approx \frac {_1}{^3} \sin (\widehat C \minus{} \widehat B) \equal{} \frac {_1}{^3} \sin \frac {\pi}{180}$ $ \Longrightarrow$ $ \angle {XAA'} \approx \frac {\pi}{540}$ $ \Longrightarrow$

$ \frac {AD}{AA'} \approx \cos \widehat{XAA'} \approx 1 \minus{} \frac {\pi^2}{2 \cdot 540^2}.$

$ \frac {BB'}{AA'} \approx \frac {1 \plus{} \cos \widehat B}{1 \plus{} \cos \widehat A} \approx 1 \plus{} \frac {\sin \widehat B}{1 \plus{} \cos \widehat B} \cdot (\widehat A \minus{} \widehat B) \approx 1 \minus{} \frac {\sin{\frac {\pi}{3}}}{1 \plus{} \cos \frac {\pi} {3}} \cdot \frac {4 \pi}{180} \equal{} 1 \minus{} \frac {4\pi \sqrt 3}{540}$

$ 1 \minus{} \frac {AD}{AA'} \ll 1 \minus{} \frac {BB'}{AA'}$ $ \Longleftrightarrow$ $ \frac {\pi^2}{2 \cdot 540^2} \ll \frac {4\pi \sqrt 3}{540}$ $ \Longleftrightarrow$ $ \pi \ll 8 \cdot 540\sqrt 3$ $ \Longrightarrow$ $ \boxed{AD > BB'}$.
___________________________________________________

I am really curious, what passes as a solution in a mathematical competition ?


calculus
integration
limit
calculus computations

Let: $ u_n \equal{} \int_{0}^{1} \frac{dx}{1\plus{}x^n}$

Then prove that $ \lim u_n \equal{} 1$

Just pass the limit. Monotone convergence, dominated convergence both do it


Find maximum number of points of intersection of $ n$  circles in a plane

Each pair of circles can intersect in at most two points, and it's easy to see that such arrangements exist for all $ n$, e.g., circles of radius 1 centered at $ (a_1, 0), (a_2, 0), \ldots, (a_n, 0)$ with $ 0 < a_1 < \ldots < a_n < 1$.  (We can always choose $ a_i$ given $ a_1, \ldots, a_{i \minus{} 1}$ to not duplicate any intersection points because there are infinitely many choices of $ a_i$, finitely many points to avoid, and we hit each point with a unique "wrong" choice.)

The answer is $ n(n\minus{}1)$.But i do not understand the logic behind it.

Every pair of circles can be made to intersect in two points.  There are $ \binom{n}{2} \equal{} \frac{n(n\minus{}1)}{2}$ pairs of circles.  Thus ....


This is a post explaining how to add new members to your local community in case you should forget.

Go to the Usergroups button (at the top of this forum page).  Within your usergroups should be your local community.  At the bottom of that page is a field that allows you to add new members.  Type or paste the username of the user into that field and press the submit button.  Don't press Enter or Return or else the user won't be added.

Just wondering, if people living in the US put their state in their location field, did you just add them to that state's community?

Yes.  The goal was to get the communities going as quickly as possible.


http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14909

 :(

Let the given ratio be $ \frac{PA}{PB}\equal{}m$

Let $ C$ be a point on the circle $ (O)$ such that $ CP\perp PO$. Then $ CP^2 \equal{} PA\cdot PB$

$ m\equal{}\frac{PA}{PB} \equal{} \frac{PA^2}{PA\cdot PB} \Rightarrow PA \equal{} CP\sqrt{m}$

[size=150]Construction[/size]
On the extension of $ CP$ we construct a segment $ PF$ with length $ m\cdot CP$
We draw the circle $ w$ with diameter $ CF$. Let $ H\in PO \cap w$
Then $ PA\equal{}PH$

____________

[color=brown][u][b]Note[/b][/u]

Let the line $ PO$ intersect the circle $ (O,R)$ at the points $ S,T$ with $ SP\leq TP$

Then the minimum and the maximum possible ratio is respectively $ \frac{SP}{TP}$ and $ \frac{TP}{SP}$
So we can only construct the figure iff $ \frac{SP}{TP} \leq m \leq\frac{TP}{SP}$[/color]


[quote="M4RI0"]Prove that a circle centered at point $(\sqrt{2},\sqrt{3})$ in the cartesian plane passes through at most one point with integer coordinates.

I tried to prove that any circle with center at $(0,0)$ has at most one point with coordinates $(a-\sqrt{2},b-\sqrt{3})$;$a,b \in \mathbb{Z}$.
So that when we translate the center to $(\sqrt{2},\sqrt{3})$ we have what we wanted to show.[/quote]
I think the approach below is probably a simpler one, though there may be other equally simple ways of solving the problem.

Let us assume, for the sake of contradiction, there is a circle with center $O(\sqrt{2}, \sqrt{3})$ such that it passes through two points with integer coordinates. Let these two points be $A(a,b)$ and $B(c,d),$ where $a,b,c,d \in \mathbb{Z}.$ Note that the line $AB$ is a chord of the circle.

Let the midpoint of $AB$ be $C,$ where $C\equiv (\frac{a+c}{2}, \frac{b+d}{2}).$ Let $L$ denote the line passing through $C$ and perpendicular to $AB.$ Now, slope of line $AB$ equals $\frac{b-d}{a-c}\Rightarrow$ Slope of the line $L$ equals $\frac{a-c}{d-b}.$

So, the equation of line $L$ is given by

$y-\frac{b+d}{2}={(\frac{a-c}{d-b})}(x-\frac{a+c}{2})$

Since $L$, passes through the point $O$, we must have
$\sqrt{3}-m = n(\sqrt{2}-p),$ where $m=\frac{b+d}{2}, n=\frac{a-c}{d-b}$ and $p=\frac{a+c}{2}.$ Note that $m,n$ and $p$ are all rational.

So, we get
$\sqrt{3}-n\sqrt{2}= m-np$

$\Rightarrow 3+2n^{2}-2n\sqrt{6}={(m-np)}^{2}$

$\Rightarrow \sqrt{6}$ is rational, which is clearly a contradiction.

Hence no such circle exists.


[quote="paganinio"]does this message board have quick reply?
sorry if this has been asked before[/quote]If it would have had one, since you seem to know how quick reply works, you would have obviously seen it :)
I am not planning on installing such a feature because it's the spammer's paradise ;)

[quote="paganinio"]I understand,, but to be honest, this forum isn't very convenient
to post a reply, first you have to click the button, and wait until it loads, then type and press Alt+S, then wait another page which asks you to choose between going to the thread and going to the forum, and finally you finish a post.[/quote]However it seems that a lot of people post a lot ;) I find it really easy to post something when you want it to, because you also have to hit preview to make sure your post is displayed correctly with $\LaTeX$ (if you're not confident enough in your $\LaTeX$ skills :D).

Now I realize a lack of quick reply is really effective in preventing spammers.


I was just wondering what the amount of math books people have read are correlated to how well they have done on math competitions.
For example, you could say i did aops v1 and made AIME
or i did engels and made usamo. that being said arent aops books the best?

I think it varies.  You are sure to find people who read every AoPS books and made it no where, just as you are sure to find people who read nothing and made MOP.  For me, I read (skimmed) AoPS V2 and scored 4-5 on the AIME.

Remember, just because you owned an copy of AoPS, it doesn't make you the best mathlete out there. Most of the skill in accomplishing such a position is due to the student him/herself, and his work habits.


[quote="pbornsztein"]Such a partition does not exist. You will find more details at :

http://mathcentral.uregina.ca/MP/previous2003/feb04sol.html

Pierre.[/quote]
Kudos Pierre!!! It makes me happy when I see your name on somewhere else besides MathLinks :D  :D

I remember many years ago one mathematician told me about that problem. Unforunately, I don't remember details. Maybe I am wrong, but a large number of mathematicians (in France?) tried to solve it for a period of time.

[quote="orl"]Wanted is a partition of the unit square in an odd number of area-equal non-overlapping triangles.[/quote] I think that the general idea is the following. Suppose we can dissect a unit square into $n$ triangles. Consider a rational lattice with step $\frac{1}{n}$ (or $\frac{1}{ns}$, where $s$ is odd). The trick to prove is that it is possible to deform the dissection in such a way that new dissection points will be in that lattice.

May be someone will enlighten us?  :?


Find all $ f$  $ R \rightarrow R$
$ f(x\plus{}f(y)\plus{}yf(x))\equal{}y\plus{}f(x)\plus{}xf(y)$

[quote="hxy09"]Find all functions $ f:$  $ \mathbb{R} \rightarrow \mathbb{R}$ such that for every couple of reals $ x$ and $ y$ we will have
$ f(x \plus{} f(y) \plus{} yf(x)) \equal{} y \plus{} f(x) \plus{} xf(y)$.[/quote]

Donate $ f(0) \equal{} a$.

$ x \equal{} 0$ $ \Rightarrow f(ya \plus{} f(y)) \equal{} y \plus{} a$. (1)
$ y \equal{} 0$ $ \Rightarrow f(x \plus{} a) \equal{} f(x) \plus{} ax\Leftrightarrow f(x) \equal{} f(x \plus{} a) \minus{} ax$. (2)

(1)+(2) gives $ f(f(x \plus{} a)) \equal{} x \plus{} a\Leftrightarrow f(f(u)) \equal{} u$ $ \Rightarrow$ we get that $ f(x)$ is a bijection (obviously $ f(\mathbb{R}) \equal{} \mathbb{R}$ and also $ f(a) \equal{} f(b)\Rightarrow a \equal{} f(f(a)) \equal{} f(f(b)) \equal{} b$). From this, we also get $ f(f(0))\equal{}f(a)\equal{}0\Rightarrow f(a\plus{}f(a)\plus{}af(a))\equal{}a\plus{}f(a)\plus{}af(a)\Leftrightarrow f(a)\equal{}a\equal{}0$.

Donate $ f(1)\equal{}b$ ($ f(b)\equal{}1$).

$ x\equal{}1$ $ \Rightarrow f(by\plus{}f(y)\plus{}1)\equal{}b\plus{}y\plus{}f(y)$
$ y\equal{}b$ $ \Rightarrow f(x\plus{}1\plus{}bf(x))\equal{}b\plus{}f(x)\plus{}x\equal{}f(bx\plus{}f(x)\plus{}1)$ $ \Rightarrow bf(x)\plus{}x\equal{}f(x)\plus{}bx$ which is equivalent to $ b(f(x)\minus{}x)\equal{}f(x)\minus{}x$ for every $ x\in \mathbb{R}$.

Therefore we will have either $ f(x)\equal{}x$ for every $ x\in \mathbb{R}$ or $ b\equal{}1$ $ \Rightarrow f(1\plus{}x\plus{}f(x))\equal{}1\plus{}x\plus{}f(x)\Leftrightarrow f(A)\equal{}A$, since $ g(x)\equal{}1\plus{}x\plus{}f(x)$ is a bijection unless $ f(x)\equal{}\minus{}x$ for some $ x$s, which can be refuted.  :)

[quote="BG Yoda"]$ \Rightarrow f(1 \plus{} x \plus{} f(x)) \equal{} 1 \plus{} x \plus{} f(x)\Leftrightarrow f(A) \equal{} A$, since $ g(x) \equal{} 1 \plus{} x \plus{} f(x)$ is a bijection unless $ f(x) \equal{} \minus{} x$ for some $ x$s, which can be refuted.  :)[/quote]
Why it can be refuted?I think $ g(x)$ doesn't run the whole $ R$ can't obtain $ f(x) \equal{} \minus{} x$ for some $ x$s :(


Find all natural numbers $x,y,z$ such that $\frac{97}{19}=5+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$.

[hide]let x=a, y=b, z=c (i like to use a, b and c  :) )

We can rewrite it as : 97/19 = (5 abc +5a+5c+bc+1)/(abc+a+c)
97 and 19 are prime numbers:

5abc+5a+5c+bc+1= 0 mod(97), and abc+a+c=0 mod(19)

bc+1 can be 2 , then 5 (abc+a+c)=95 ==> abc+a+c=19=0 mod 19
but bc+1 cant be 7,12,... because abc+a+c has to be 0 mod 19.

so bc=1 , b=1 and c=1
and a+a+1=0 mod 19 so a=9

so the solution is (x,y,z)=(9,1,1)[/hide]


rotation
geometry
perimeter
calculus
calculus computations

A rotating beacon is located 1 kilometer off a straight shoreline. If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be moving to a viewer who is 1/2 kilometer down the shoreline?

I'm not sure I understand your question.  Which part of the beam of light?  Why do you say "[i]appear[/i] to be moving"?

Still, I'll go ahead and write an answer.

The distance from the person to the beacon is $ \sqrt{2}$ km.  That is the radius of a circle whose perimeter is $ 2\pi \sqrt{2}$ km.  The point on the beam of light that we are watching goes around this circle $ 3$ times per minute.  That's a distance of $ 6 \pi \sqrt{2}$ km in a minute, giving a speed of $ 6 \pi \sqrt{2}$ km/min $ \equal{} 360 \pi \sqrt{2}$ km/hour.

I am sorry I forget 1/2 in front of kilometer.
The question said "If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light [i]appear to be moving [/i]to a viewer who is 1/2 kilometer down the shoreline?" means what is the speed of the light move at the moment viwer see it. (I think so it is what it means.)

Who wrote this question? I don't think it's phrased very well.  Different points on the beam move with different speeds.  I assume we are focusing our attention on the point on the beam whose distance from the beacon is equal to the distance of the person from the beacon.

In any case, I think I understand what the question is trying to ask.

The $ \frac12$ doesn't change much.  It just means that instead of $ \sqrt{2}$ km, the distance from the person to the beacon is $ \sqrt{1^2 \plus{} (\frac12)^2} \equal{} \sqrt{\frac54}$.  The speed becomes $ 360\pi \sqrt{\frac54}$ km/hr.


Let $a, b, c>0$. Prove that the following inequality holds:

               \[ \frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.  \]

 [b] Remark. [/b] The inequality from above is sharper than the one proposed by Hungthkn. Here it is

                         \[ \[ \frac{a^{2}+bc}{(b+c)^{2}}+\frac{b^{2}+ca}{(c+a)^{2}}+\frac{c^{2}+ab}{(a+b)^{2}}\geq\frac{3}{2}. \]

It's easy by Schur Extension Inequality.

Yes, the inequality is equivalent to the assertion
\[ \sum \frac{1}{(b+c)^2}(a-b)(a-c)\geq 0  \]


trigonometry calculus derivative function

Find $ \min_{z \in \mathbb{C} - \mathbb{R} } \frac {Im{(z^5)}}{Im^5{(z)}}$ . And find $ z \in \mathbb{C} - \mathbb{R}$ for which the minimum case occurs .

[hide]
Minimum is zero.

Consider in $ re^{i\theta}$ form

$ \frac {Im(z^5)}{Im^5(z)} \equal{} \frac {r^5\sin 5 \theta}{r^5 \sin^5 \theta} \equal{} \frac {\sin 5 \theta}{\sin^5 \theta}$

If $ \arg z \equal{} \frac {\pi}{5}$ then the minimum is obtained.

The required elements lie on a straight line, angle $ \frac {\pi}{5}$ starting at the origin (almost) and going out in the first quadrant[/hide]

[b]Edit: Whoops, ignored negative cases[/b]

[hide="Are you sure ?"] This was the tricky part . The minimum is $ \minus{} 4$ . One solution is : $ \arg{z} \equal{} \frac {\pi}{4}$ .
[/hide]

We get that $ \frac {\Im(z^5)}{\Im^5(z)} \equal{} \frac {r^5\sin 5 \theta}{r^5 \sin^5 \theta} \equal{} \frac {\sin 5 \theta}{\sin^5 \theta}$ from $ r\operatorname{cis}{\theta}$ form.  Taking the derivative and simplifying, we get that the derivative is $ \frac {5\sin{4\theta}}{\sin^6{\theta}}$.  The minimum occurs when this is 0.  Since $ z$ is not real, $ \theta\neq0$, so this is 0 when $ \theta \equal{} \frac {\pi}{4}$, which gives $ \boxed{ \minus{} 4}$.

Is there a nice way to minimize that quantity without derivatives?

[hide="Yes , there is a way without derivates"] Let $ z\equal{}a \plus{} bi$ , where $ a,b \in \mathbb{R}$ and $ b \neq 0$ .
then , $ Im{(z^5)}\equal{}5a^4\minus{}10a^2b^2 \plus{}b^5$ .
So , $ \frac{Im{(z^5)}}{Im^5{(z)}} \equal{}  5 \left(\frac{a}{b} \right)^4\minus{}10 \left(\frac{a}{b} \right)^2 \plus{} 1$ .
Let $ x \equal{} \left( \frac{a}{b} \right)^2$ .
We need to find the minimum value of the function $ f: [0;\infty) \to \mathbb{R}$ with $ f(x) \equal{} 5x^2 \minus{} 10x\plus{}1$ .
So $ f_{min}\equal{} f(1)\equal{}\minus{}4$ .
So the asked numbers are $ z\equal{}a(1 \pm i )$ . done ..[/hide]


function
algebra
domain
real analysis
geometry
3D geometry
integration

Let $ f$ be a real measurable function defined on a domain $ \Omega\subset\mathcal R^n$. For each real number $ t$, set $ \Phi_t \equal{} \{x\in\Omega: f(x) \equal{} t\}$. Prove that $ m(\Phi_t) \equal{} 0$  for almost every $ t$, where $ m$ denote the Lebesgue measure in $ \mathcal R^n$

Even stronger: $ m(\Phi_t)\equal{}0$ for all but countably many values of $ t$. You can't have uncountably many disjoint measurable sets of positive measure.

[quote="mlok"]Even stronger: $ m(\Phi_t) \equal{} 0$ for all but countably many values of $ t$. You can't have uncountably many disjoint measurable sets of positive measure.[/quote]
but does it hold when $ m(\Omega)\equal{}\plus{}\infty$?

[quote="fedja"]Yes, because a set of positive measure would have an intersection of positive measure with some large cube and $ \mathbb R^n$ can be covered by countably many cubes.[/quote]
Ah, thank you. I didn't regconize that simple argument. I used Fubini's theorem to conclude that $ \int_{ \minus{} \infty}^{ \plus{} \infty}\varphi(t)dt \equal{} 0$, where $ \varphi(t) \equal{} m(\Phi_t)$.


1. If 76 grams of lithium completely reacted with water, what would be the mass of hydrogen gas released?
(Relative atomic masses: H, 1.0; Li, 7.0; O, 16.0)

A. 5.4 g
B. 10.9 g
C. 19.0 g
D. 21.7 g

2. The proton numbers for elements X, Y, and Z are between 10 and 18. X has 1 electron less than a noble gas, Y has one electron more than a noble gas, while Z has 3 electrons less than a noble gas. Which of the following is a metal?

A. X
B. Y
C. Z
D. X and Y

3. In which compound does vanadium have an oxidation number of +4?

A. $NH_{4}VO_{2}$
B. $K_{4}[V(CN)_{6}]$
C. $VSO_{4}$
D. $VOSO_{4}$

4. 1.85 grams of copper (II) carbonate reacted with $4.00 \times 10^{-2}$ moles of hydrochloric acid to produce 0.06600 grams of carbon dioxide, 0.2700 grams of water and a certain mass of copper (II) chloride. If this reaction occurred to completion, what is the excess reagent and how many moles of it is left?
(Relative atomic masses: H, 1.0; C, 12.0; Cl, 35.5; Cu, 63.5)

A. Copper (II) carbonate, 0.015 moles
B. Copper (II) carbonate, 0.025 moles
C. Hydrochloric acid, 0.010 moles
D. Hydrochloric acid, 0.025 moles

5. The combustion of 54.9 grams of a sample of hydrocarbon produced 38.0 grams of water. What is the percentage of hydrogen (by mass) in the hydrocarbon?
(Relative atomic masses: H, 1.0; C, 12.0; O,16)

A. 2.11%
B. 4.22%
C. 7.69%
D. 69.2%

[hide="Number 1"]Lithium and Oxygen form $Li_2O$

$76$ grams of lithium forms $10\frac{6}{7}$ mol.

This amount can form $5\frac{3}{7}$ mol $Li_2O$.
Therefore, $5\frac{3}{7}$ mol $H_2O$ is needed.
Therefore, $5\frac{3}{7}$ mol $H_2$ is needed.

$5\frac{3}{7}\times 2\times 1\approx\boxed{B. 10.9 g }$[/hide]

[hide="Number 2"]X) one less than a noble gas is a halogen.
Y) one more than a noble gas is an alkali metal.
Z) 3 less than a noble gas is either a nonmetal or a metalloid.

Therefore $\boxed{B. Y}$[/hide]

[hide="Number 3"]$NH_4$ and $O_2$ have a total of 1+. V has a oxidation number of 1-.
$K_4$ and $(CN)_6$ have a total of 2-. V has a oxidation number of 2+.
$SO_4$ has a total of 2-. V has a oxidation number of 2+.
$SO_4$ and $O$ have a total of 4-. V has a oxidation number of 4+.

Therefore, $\boxed{D. VOSO_{4}}$[/hide]

By the way, there is an extra 0 in the number of moles of CO_2 in problem 4
Number 4:
[hide][b]C.[/b]  The weight of CuCO3 is 123.5, so the number of moles is 1.85/123.5 ~ 0.015 moles. Balancing the chemical equation, we find that there are two moles HCl for each mole of CuCO3. Thus 0.03 of the moles of HCl are used, leaving 0.010 moles left over. [/hide]
Number 5:
[hide][b]c.[/b] 38 g H2O is 38/18 = 2.111 moles. Thus there are 4.222 grams of H, and 4.222 / 54.9 ~ 0.0769. [/hide]


i didn't know, when i uninstalled mathematica, that i had to fax them a system transfer application, so i just went ahead and did it. yet, now i know this, and on the system transfer application, i need to write my old math id. however, i don't have that. is the math id crucial, or not?

i am sorry but i really don't understand what you are talking about......

i don't think anyone can really answer my question unless they have transferred mathematica from one computer to another....so, does anyone know?


OK so we have this assignment on the population of China.
QUESTION:
let N = number of years
     A = the number of people in N year = 2 000 000 000 people
     P = present value = 1 100 000 000 people
     I  = interest of people every year = 1.5%/a
I i arranged the question as a compound interest question;
A = P(1+I)^N
2 000 000 000 = 1 100 000 000 (1 + 0.015)^N

Is it possible to isolate N in that question?  I can't.

So i went to another option;  Graphing it.

I downloaded a graphing calculator program from http://www.graphcalc.com.  I can't figure it out.  I need  to find a formula that will give me a line that i can use to solve for N.

Is that possible?

Any help is appreciated,
Thanks...

Use logarithms.

$\log_a b=c$ means $a^{c}=b$.

Note that you can find $\log_{10} x$ with a calculator, and using the property $\log_a b=\frac{\log_{10} b}{\log_{10} a}$ you can get the answer.

[hide=" Without LaTex"]

Not sure about international standards but lg means logarithm in base 10..

2 000 000 000 = 1 100 000 000 (1 + 0.015)^N

N = lg( 2 / 1.1) / lg( 1.015 ) =~ 40.143

Was this what you were looking for or did I fail to understand the question?

[/hide]

Ok i found the answer VIA graphing calculator.  The answer is approx. 40 years.  I think i will include both the graph and the algebraic solution in my answer.

Thanks all!


how do I get the three dots for 'therefore'?

Ensure you have amssymb loaded then it's \therefore $\therefore$
Being a maths symbol it needs to be inside dollar signs.

It's a good idea always to have amsmath, amssymb and amsthm loaded as they are so useful. The [url=http://tug.ctan.org/tex-archive/info/symbols/comprehensive/]Comprehensive LaTeX Symbol List[/url] is always worth consulting as well.


inequalities
calculus
integration
function
real analysis
real analysis unsolved

If  f(x) is a non negative ,continuous,convex function on the closed  interval  [0,1] such  that   f(0)=0,show that
$2$\int_0^1 x^2f(x)\,dx+1/12<=($\int_0^1f(x)\,dx)^2$

Do you mean $2\int_0^1 x^2f(x)\,dx+1/12\leq (\int_0^1f(x)\,dx)^2$?

That inequality cannot hold for all functions that are non-negative, increasing, convex, and satisfy $f(0) = 0$.


AMC
USA(J)MO
USAMO
AIME
geometry
AIME II

Oh I made some silly mistakes and only got 115.5 on AMC 12A, I think the last 4 questions are really hard. And if I take both A and B, which score will be in my USAMO index? Do the scores on A and B count the same or not? And What's the usual USAMO index for those who take AMC 12?

Just out of curiosity, how many questions did you answer?

They will choose your higher score to calculate your USAMO index. The scores from A and B are equivalent.

I answered 24, I guessed 3 but all wrong, then made mistakes on the other 2, really messed up.

There is no "usual". See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1221302#1221302]this post[/url] for recent USAMO qualifying standards.

If you are taking the AMC 12, you should aim for:

125 AMC 12, 8 AIME (somewhat unlikely, since the AIME is much harder than the AMC 12)

or

135 AMC 12, 7 AIME (this is probably more realistic.)

Hope that helps!
 :)

Thanks blackbelt14253 and jmerry, it really helps.

//and the last 5 problems on the AMC can be as hard as a #10 on AIME.//

I have to disagree...

[quote="the future"]actually why would a 125/8 be unrealistic (aside from the fact that you cant get 125)?

I think its actually easier to get a 8 on the AIME than to get a 125 on the AMC 12 since AMC has more time pressure and the last 5 problems on the AMC can be as hard as a #10 on AIME.[/quote]

I agree, the time pressure killed me. AIME has much more time for each problems.

It's usually above 200 but not above 210. Last year AIME was easy and the floor was still 6 so I guess it depends.

Does once in the last four years qualify as usual? There's a lot of fluctuation, enough that I'd rather just give all the data than pick any single number.

Remember also that most of our data is pre-expansion.

Hey, I got 117 on the AMC [b]10[/b] and then got an 8 on the AIME last year.  Sometimes people just have bad/good days.  (I had them respectively).  Plus, in order to get an 8, you only need to get 8 questions out of 15, but to get a 135, you have to get 22 out of 25.  Way less room for error.

I really don't think 8 is that hard on AIME.  Last year I got 102 AMC 12A and 6 on AIME II, because of 2 really stupid mistakes (one in counting on #1, the other was in multiplication in a massively long geometry problem, somewhere around #11).  So I almost got an 8.  Anyway, this year I'm hoping for 10 because that will definitely qualify me for usamo (115.5 AMC 12 score, again more stupidity losing me points.)  Unfortunately, scores without stupidity are not taken into account.


superior algebra
superior algebra unsolved

Problem:
Let $ (\Theta_{i})_{i\in I}$ family of congruence of $ A$ ( $ A$ is an algebra. ) If $ \Theta\equal{}\cap_{i\in I}\Theta_{i}$ then $ A/\Theta$ a a subdirect product  of $ \Pi A/\Theta_{i}$
I know that $ A/\Theta$ and  $ \Pi A/\Theta_{i}$ are isomotphic. Why $ A/\Theta$ is a subdirect product  of $ \Pi A/\Theta_{i}$

Probably $ A/\Theta$ is only isomorphic to a subalgebra of $ \Pi A/\Theta_i$.   It is a pretty nice subalgebra though, and if you apply any of the projections to $ A/\Theta_i$, then when restricted to $ A/\Theta$ it will be a sujection.  Often that is the definition of subdirect product (sub-object of direct product where each projection remains surjective when restricted to the sub-object).

For instance if A=Z, T_1 = 6Z, T_2 = 10Z, then T = 30Z, and sure enough A/T = Z/30Z is a subalgebra of Z/6Z x Z/10Z, but it is not the whole direct product.


" Daca (gof)(x) = x, oricare ar fi x real si f, g sunt definite pe R cu valori in R, atunci f si g sunt inversabile". o = compunere

Afirmatia este clar falsa, dar nu gasesc un contraexemplu...

multumesc

$f(x)=e^x$ si $g(x)=0$ pentru $x\le 0$, $g(x)=\ln x$, pentru $x>0$.

Va multumesc foarte mult. Ce simplu era! Mi-ati scos un bolovan din cap, si pe drum tot la asta m-am gandit.

Numai bine si succese pe toate planurile


trigonometry limit calculus derivative LaTeX function inequalities

Hello, this is my first post. I'm in Calc AB or calc I, I suppose you would call it.

I'm having a bit of trouble finding the limit (as x-> 0) of xcscx. It's easy to do graphically, but I'm interested in how to do it using calculus.

First I did the product rule to get cscx - xcscxcotx

Then I broke it into sin and cos: 1/sinx - xcosx/sin^2(x)

To: (sinx-xcosx)/ (sin^2(x))

Then I could change the denominator to 1-cos^2(x), and multiply the whole thing by (1+cos^2(x)) but that doesn't work either 'cause (1-cos^4(x)) is still 0. I simply can't figure out how to get rid of 0 for the denominator. I've spent forever on this problem and nobody else I know in the class has figured it out either.

Thanks alot!

$\lim_{x\to 0} x\csc x=\lim_{x\to 0}\frac{x}{\sin x}=\lim_{x\to 0}\frac1{\frac{\sin x}{x}}=\frac1{\lim_{x\to 0}\frac{\sin x}{x}}$.
$\lim_{x\to0}\frac{\sin x}{x}$ is the derivative of $\sin$ at 0 by definition. We know that it's 1. Taking the reciprocal (if the limit exists and is nonzero, that works), the original limit is 1.

Thank you very much. I would just like to ask about taking the inverse of the limit. You say that works if the limit exists and is non-zero? Well, two questions: how would I know if it exists and is non-zero? and why does this work? I'm afraid I don't know too much about what a limit is [i]exactly[/i]. :blush: My teacher isn't especially helpful, and neither is the textbook. If you could try to explain why this is true, I would very much appreciate it. :)

It just occured to me that there is a much easier way. I'm such an idiot. Just do l'Hopital's rule after you simplify it to x/sinx. :blush: That gives you 1/cosx = 1

I'm still interested in that inverse limit thing though. Thanks.

L'Hopital's rule fells a little circular here- how did you find that derivative in the first place?

There is an essential theorem, which should be one of the first things you prove once you have a definition for limits: We can do arithmetic.
In the textbook I have (Stewart), this is stated as the "Limit Laws":
Suppose that $c$ is a constant and the limits $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist. Then
1. $\lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$
2. $\lim_{x\to a}[f(x)-g(x)]=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$
3. $\lim_{x\to a}[cf(x)]=c\cdot\lim_{x\to a}f(x)$
4. $\lim_{x\to a}[f(x)g(x)]=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)$
5. $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ if $\lim_{x\to a}g(x)\neq0$

This list is somewhat redundant, but the point is clear. This also works for limits of sequences and limits at $\infty$. If you've seen the actual definition of a limit, you should try to prove these; this is part of the foundation that lets you do everything else.

[quote="jmerry"]L'Hopital's rule fells a little circular here- how did you find that derivative in the first place?[/quote]
I need to learn LaTex, but...

lim      xcscx
x-> 0

lim x/sinx

lim 1/cosx

now you just plug in zero for x to get 1/1, which equal 1. How does L'Hopital's rule feel circular? As to the rest... (I know this sounds ignorant, but believe me I'm trying) I don't know how you would prove that we can do arithmetic. I understand that we can, and I understand those laws, but I'm unsure as to what you mean by I need to prove them. Prove them how? As I said, I'm at a bit of a loss as to what an actual limit is. (It's sad that I have an A). I start school tomorrow, perhaps I'll have a lengthy discussion with my teacher. I asked him once, but he said we should talk about it later.

Thank you for your help. :) It's greatly appreciated.

[quote]How does L'Hopital's rule feel circular?[/quote]

In order to use L'Hôpital, you must know the derivative of $\sin x$

How to we know the derivative of $\sin x$ ? Let's try the definition:

$\sin'x=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}$

$=\lim_{h\to0}\frac{\sin x\cos h+\cos x\sin h-\sin x}h$

$=\lim_{h\to 0}\left(\frac{\cos h-1}h\cdot\sin x+\frac{\sin h}h\cdot\cos x\right)$

In order to complete this we must be able to prove - and prove [i]without[/i] already knowing the derivatives of any trigonometric function - what the following two limits are:

(1) $\lim_{h\to0}\frac{\sin h}h=1$

(2) $\lim_{h\to0}\frac{\cos h-1}h=0.$

Now go look in any calculus book. You'll probably be able to find a geometric argument setting up inequalities that prove (1) by the "squeeze theorem," and then get (2) from (1) via algebra and trigonometric identities.

The trap a large number of calculus students fall into is thinking that L'Hôpital's rule, since it's a formula, is some kind of magic that creates knowledge. It's not magic and it doesn't create knowledge - it organizes the knowledge that we already have.

In particular, you had to know that $\lim_{x\to0}\frac{\sin x}x=1$ before you can know the derivative of the sine, so a method that relies on the derivative of the sine wouldn't be a way of computing that limit, apart from patching over your faulty memory.


Even though the UCs havent announced any acceptances yet, I was wondering if anyone here is planning/applied to UCB, UCSB, or UCSD.

[quote="ComplexZeta"]I attend UCSB. It's a good place if and only if you are in College of Creative Studies, in my opinion.[/quote] Concurred. But for those who may not be as diligent as Simon, watch out for the tempations and distractions in beautiful Santa Barbara  :D

Yeah!   One of my  best friend (and  old Class mate)  managed  to get to teach a course *every*  (well almost every) summer at UCB.. (He is a faculty at UCLA  but  he teaches/taught  at UCB in   summer sessions)  He once confided  in me,.. "They don''t know but  I would pay [i]them[/i] to  let me teach here [UCB] .." :)

(Oh well,  Can't blame him, when we chose graduate schools, I was in snow and sleet in NE of US and he  would always rub it in ( he went to UCSD)  -  "and BTW how's the weather there"  .
 :(


ok, there are nine seats around a circular table. There are also 3 women and 6 men. In how many different ways can the women and men sit around the table?

please help...

[quote="lukezli"]are women and men distinguishable? if so, the answer is 9![/quote]

Uh, the formula for the number of ways to seat people IN A CIRCLE is $ (n\minus{}1)!$

$ (9\minus{}1)!$
$ 8!$
$ 40,320$ ways


limit
logarithms
Euler
real analysis
real analysis unsolved

[b] IF we define the following two series :

$ P\equal{} \sum\limits_{k\equal{}1}^{\infty} {\frac{(\minus{}1)^{k\plus{}1}}{k}}$. Alternating harmonic series

$ T\equal{}1\plus{}\frac{1}{3}\minus{}\frac{1}{2}\plus{}\frac{1}{5}\plus{}\frac{1}{7}\minus{}\frac{1}{4}\plus{}\cdot\plus{}\cdot\minus{}\cdots$

Find the sum of T and deduce the sum of P.

 :) [/b]

$ P\equal{}\lim_{n\rightarrow \infty}\sum_{k\equal{}1}^{2n} \frac{(\minus{}1)^{k\plus{}1}}{k}\equal{}\lim_{n\rightarrow \infty}(\sum_{k\equal{}1}^{2n} \frac{1}{k}\minus{}\ln2n)\minus{}\lim_{n\rightarrow \infty}(2\sum_{k\equal{}1}^{n} \frac{1}{2k}\minus{}\ln n)\plus{}\ln2\equal{}\ln2$
(Because $ \lim_{n\rightarrow \infty}(\sum_{k\equal{}1}^n\frac{1}{k}\minus{}\ln n)\equal{}c$: Euler constant)

Great work tdl

the sum of P is true. but the sum of T isn't ${ \frac{\ln(8)}{2}}$ I think there is something wrong.

I just see it very nice problem and I would like to share it with fabulous people here.

Thx

I sure that I do it correctly, you can try by calculator with $ n$ enough large!

Let $ A_n \equal{} 1 \plus{} \frac{1}{3} \plus{} \cdots \plus{} \frac{1}{2n\minus{}1}$ and $ B_n \equal{} \frac{1}{2} \plus{} \frac{1}{4} \plus{} \cdots \plus{} \frac{1}{2n}$. Then

(1) $ A_{p_n} \minus{} B_{q_n}$ converges if and only if $ \lim_{n\to\infty} \frac{p_n}{q_n}$ converges to some positive number $ \alpha$ and
(2) $ \lim_{n\to\infty} (A_{p_n} \minus{} B_{q_n}) \equal{} \ln (2 \sqrt{\alpha} )$.

For $ T$, we can let $ p_n$ and $ q_n$ by

$ p_n \equal{} 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, \cdots$
$ q_n \equal{} 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, \cdots$

and this gives $ \alpha \equal{} 2$. Hence $ T \equal{} \ln (2 \sqrt{2})$.


geometry
geometric transformation
reflection
3D geometry
prism

A light ray strikes a 45-45-90 glass prism(n=1.5) perpendicular to the hypotenuese. We know, since the critical angle is $ 41.8$ and $ 41.8 < 45$, it is totally intenaly reflected. it is reflected twice. When it leaves the prism, will the light ray be parallel to the ray that came in?

Does the critical angle affect the result angle in any way?

Solution.

Light is inside glass and hits the glass-air boundary. The beam is incident at an angle of 45 degrees to the normal to this boundary. The law of light reflection states that the angle of incidence is equal to the angle of reflection, as measured from the normal to the reflection surface. So the angle of reflection is at 45 degrees to this normal, so the beam changes it's direction by a right angle. Due to the geometry of the prism, the beam strikes again at 45 degrees, and is again diverted by an angle of 90 degrees back out of the glass. Clearly this means that the beam leaves the glass at the same angle to the incident beam, since it has been changed by an angle of 180 degrees. Note that this whole argument is independent od the critical angle. We only need the critical angle to be less than 45 degrees.


articles
number theory
least common multiple

Despite the general notion that no mathematics knowledge is required to solve a classic sudoku puzzle, sudoku can be used as a teaching aid in mathematics.  I have contributed a sudoku puzzle to the August 2008 issue of NRICH.

The NRICH Project is based in both the University of Cambridge's Faculty of Education and the Centre for Mathematical Sciences.  It is also part of the family of activities in the Millennium Mathematics Project.

My sudoku creation ("Multiplication Equation Sudoku") illustrates the concepts of multiplication and equations.

Please refer to http://nrich.maths.org/public/viewer.php?obj_id=6224

its good  :clap:  :thumbup: :coolspeak:
ye s, it can be used for teaching. its a good idea .

Glad to get the first nod to my idea in this forum.  In my opinion, sudoku can assist and supplement the teaching and learning of mathematics.

In the classroom, the teacher can try out the idea by giving such puzzles to his top students.  Such students will find the puzzles challenging.  They will also find fun in the learning of mathematics.

Don't revive old threads mathelete.  And I doubt you even read the article....

And aman, if you happen to check out this thread again, I think that's great.

AIME15, I am interested to read the article and thread mentioned in your reply to mathelete.

Anyway, mathelete, you have asked a very interesting question.

Please read the article on KenKen in http://entertainment.timesonline.co.uk/tol/arts_and_entertainment/games_and_puzzles/article3599704.ece

Although KenKen is claimed to be an entirely different puzzle from Sudoku, we can still detect the long shadow of Sudoku in its background.

Since February 2006, I have been contributing Sudoku variants with mathematical characteristics to every issue of NRICH.  Do you think the editor of NRICH and her professor colleagues at Cambridge University are so senseless as to accept my sudoku contributions if they find nothing useful in my puzzles for the mathematical education of the students?

aman, where can I find KenKen puzzles?

The "article" I was referring to was simply the link you posted at first.  As for the thread, it's this one.

Searching on Google though, the Wikipedia page referring to Mathematics of Sudoku was very interesting.

http://en.wikipedia.org/wiki/Mathematics_of_Sudoku

Well, it looks like I cannot see anything that is as near as the tip of my nose.

You can find KenKen puzzles on the following page of TimesOnLine:

http://entertainment.timesonline.co.uk/tol/sitesearch.do?query=kenken&submitStatus=searchFormSubmitted&mode=simple&sectionId=1017&queryKeywords=kenken&x=49&y=15

By the way, you can also find my latest stuff (Addition Equation Sudoku) on the following page of the October 2008 NRICH website:

http://nrich.maths.org/public/viewer.php?rss=1&obj_id=6310&part=index

To be more exact and accurate, all puzzles (including sudoku), interactive mathematical games and even board games can help in the teaching and learning of mathematics.

It is obvious that NRICH knows that puzzles and interactive mathematical games allow students to learn and play at the same time.

Beginning this month (November 2008), NRICH uses innovative flash interactivity for my sudoku creation ("LCM Sudoku") in http://nrich.maths.org/public/viewer.php?obj_id=6311

The variant illustrates the concept of Least Common Multiples.

My first version of "LCM Sudoku", using numbers and unknown variables, was posted on http://nrich.maths.org/public/viewer.php?obj_id=6018

To be more specific, that variant should be called "Algebraic LCM Sudoku".

I have posted a more challenging version of "LCM Sudoku", which uses advanced sudoku techniques, this month (November 2008) on the site http://www.chinasudoku.com/

For students who like fairy tales, they can try to solve another of my sudoku puzzle called "Story Sudoku".

My sudoku creation for the December 2008 NRICH http://nrich.maths.org/public/viewer.php?obj_id=6368 is named "Bochap Sudoku" which illustrates the concepts of the four basic arithmetic operations.

Hi,

Do you know sudoku++?

A new sudoku variant.

To know more

http://joujoujou.over-blog.com/article-25839168.html


I have no clue what the OCTM and OHMIO are like. How many problems/difficulty/time is alloted?

The only thnig we did even close ot that was something with like two columns and forty questions with pathetically easy problems...but I got a 39 due to some really retarded mistake

I don't know so much about it either, but what I know, I learned from:
[url]http://www.ohioctm.org/OCTM_Contest.htm[/url]

Oh cool they have old tests on there

I did '08 and got a 38. Stupid mistake and I had no clue what Euler's line was :D

I never really did geometry, after all :wink:

[quote="#H34N1"]I had no clue what Euler's line was :D[/quote]

how did I not know what the euler line is :o

thats used in like 1/7 geomery problems =xo


Similar to: prove that \[4|n\ \Longleftrightarrow\ \exists a_{1},a_{2},\cdots,a_{n}\in\{-1,1\}: a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n}a_{1}= 0\]

I believe this is an invariance problem from Problem Solving Strategies; I think you got it from there.  Dorost nist? :P  You can work a few cases and see that the sum does not change mod 4...I think. :)

Masoud Zargar

but i didn't get it from Problem Solving Strategies(na dorost nist) :wink: !i see it both in number theor's book and combinatories one!!!

Very well then.  Sorry, I thought you got it from Problem Solving Strategies.  It seems to be known else where too.  Here is a solution.

Let $P=x_{1}x_{2}x_{3}x_{4}+x_{2}x_{3}x_{4}x_{5}+...+x_{n}x_{1}x_{2}x_{3}=0$.  Denote a replacement of $p\rightarrow q$ by $p\circ q$.

$x_{i}\circ-x_{i}$ does not change $P\text{ mod 4}$ since 4 terms in P change their sign.  If three of the four selected terms have the same sign, then a replacement $x_{i}\circ-x_{i}$ changes $P$ by $\pm 4$.  If two of these four selected terms are >0 and the other two <0, then a replacement  does not change $P$.  If all four have the same sign, then $P$ changes by $\pm 8$.  Initially, $P\equiv 0\pmod 4$, thus $P\equiv \pm 4\equiv\pm 8\equiv 0\pmod 4$ remains invaraint.  Since a move does not change the congruency modulo 4, after a finite number of steps, $P=n\equiv 0\pmod 4$.

Masoud Zargar


advanced fields
advanced fields unsolved

Suppose we colour Z x Z using an infinite number of colours. Must there be an infinite subset T of Z such that either T x T is monochromatic or every pair in T x T gets a different colour? If not, can you find a statement of this type that does hold (perhaps allowing some additional types of colouring)?

Any idea about this? It seems complicated.

Thank you in advance!

A counterexample is given by each diagonal line $ x\minus{}y\equal{}a$ having a different color (lets call it $ a$). Then already any $ T \subset \mathbb Z$ with $ |T|>1$ already doesn't fulfill the requirements: let $ s,t\in T, s \neq t$. Then $ (t,t)$ and $ (s,s)$ have the same, but $ (t,t)$ and $ (s,t)$ have different colors.


Let $ ABC$ is a triangle and let $ a\equal{}BC, b\equal{}CA, c\equal{}AB$.

Prove that: $ \sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{b\plus{}c\minus{}a}\plus{}\sqrt{c\plus{}a\minus{}b}\le \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

[quote="thanhnam2902"]Let $ ABC$ is a triangle and let $ a \equal{} BC, b \equal{} CA, c \equal{} AB$.

Prove that: $ \sqrt {a \plus{} b \minus{} c} \plus{} \sqrt {b \plus{} c \minus{} a} \plus{} \sqrt {c \plus{} a \minus{} b}\le \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$[/quote]

That's APMO problem:
http://www.mathlinks.ro/viewtopic.php?p=451720#451720


A train is formed exactly as shown by connecting regular hexagons and regular triangles alternately, beginning with a hexagon. The side length of each polygon is 1 cm. How many hexagons are in a train that has a perimeter of 426 cm?

[hide]x is number of hexagons. Every hexagon has four sides except for the two on the end have five. Every triangle has one side, and there is one less triangle than there are hexagons.
$4x+(x-1)+2=426$
Solve for x to get 85 hexagons.[/hide]

[hide]
The relationship between number of hexagons (x) and perimeter (y) is 5x+1=y.
So, if 5x+1=426
Then, 5x=425
          x=85
The answer is 85 hexagons.[/hide]


Let f(x,y):$N^{2}\rightarrow{N^{2}}$ satisfy:
i)f(x,y)=f(y,x)
ii)f(x,y)=f(x,y+1)+f(x+1,y)$Forall{x,y\in{N}}$
Let $f(n,0)=x_{n}$.Prove that $2^{n}x_{n}\geq{x_{0}}$

[quote="Rust"]I think $f: Z_{+}^{2}\to N$ (not $\to N^{2}$).
In these case it is proved by induction.[/quote]
I agree with u,but can u post your solution ? :)


number theory
greatest common divisor
least common multiple

For three positive integers [tex]a,b,c[/tex], show that [tex]\gcd(a,\mbox{lcm}(b,c))=\mbox{lcm}(\gcd(a,b),\gcd(a,c))[/tex].

By the FTA, let:

a = z1^a1*z2^a2*.....zn^an
b = z1^b1*z2^b2*.....zn^bn
c = z1^c1*z2^c2*.....zn^cn

So:

LCM(b,c) = z1^(max{b1,c1})*z1^(max{b2,c2})*.......zn^(max{bn,cn})
GCD(a,b) = z1^(min{a1,b1})*z1^(min{a2,b2})*.......zn^(min{an,bn})
GCD(a,c) = z1^(min{a1,c1})*z1^(min{a2,c2})*.......zn^(min{an,cn})

It follows that LCM(GCD(a,b) , GCD(a,c)) =

prod(zi^(max{min{ai,bi},min{ai,ci} })

Similarly, GCD(a, LCM(b,c)) =

prod(zi^(min{ai, max{bi,ci}})

It remains only to show that min{ai, max{bi,ci}} = max{min{ai,bi},min{ai,ci}}

I don't have a nice proof immediately, so I'll case it out.

Wolog, bi  :ge: ci.

Then our expression reduces to:

min{ai, bi} = max{min{ai,bi},min{ai,ci}}

If ai  :ge: bi  :ge: ci, we have bi = max{bi,ci} = bi as desired.
If bi  :ge: ai  :ge: ci, we have ai = max{ai,ci} = ai as desired.
If bi  :ge: ci  :ge: ai, we have ai = max{ai,ai} = ai as desired.

Hence min{ai, max{bi,ci}} = max{min{ai,bi},min{ai,ci}} as desired and it follows that GCD(a, LCM(b,c)) = LCM(gcd(a,b) , gcd(a,c)).

[quote="fuzzylogic"][quote="blahblahblah"]
It remains only to show that min{ai, max{bi,ci}} = max{min{ai,bi},min{ai,ci}}

I don't have a nice proof immediately, so I'll case it out.

Wolog, bi  :ge: ci.

Then our expression reduces to:

min{ai, bi} = max{min{ai,bi},min{ai,ci}}

If ai  :ge: bi  :ge: ci, we have bi = max{bi,ci} = bi as desired.
If bi  :ge: ai  :ge: ci, we have ai = max{ai,ci} = ai as desired.
If bi  :ge: ci  :ge: ai, we have ai = max{ai,ai} = ai as desired.

Hence min{ai, max{bi,ci}} = max{min{ai,bi},min{ai,ci}} as desired and it follows that GCD(a, LCM(b,c)) = LCM(gcd(a,b) , gcd(a,c)).[/quote]

That's a good proof. The cases at the end are not necessary since bi :ge: ci implies that min{ai,bi} :ge: min{ai,ci}.[/quote]

Thanks.  This min/max argument seems to work well for nearly every GCD/LCM problem.


find all $n$ such that $2^{n-1}+1$ is divisible  by $n$


Find all $ (m,n)\in N$ such that   $ m^{2}n^{2}\minus{}5m\plus{}5n$ is a perfect square.

It is trivially a square for $ m \equal{} n$.

Assume $ n > m$.  Then, $ m^{2}n^{2}\minus{}5m\plus{}5n > m^{2}n^{2}$, or in order to be a perfect square, $ m^{2}n^{2}\minus{}5m\plus{}5n\geq m^{2}n^{2}\plus{}2mn\plus{}1$, or $ (5\minus{}2m)n\geq5m\plus{}1$.  Then $ m < 5/2$ and it can only be $ m \equal{} 1,2$ (I assume $ m,n$ are nonzero).  If $ m \equal{} 1$, then $ n^{2}\plus{}5n\minus{}5$ must be a perfect square larger than $ n^{2}$, but smaller than $ n^{2}\plus{}6n\plus{}9$, or it must be equal to either $ n^{2}\plus{}2n\plus{}1$, yielding $ n \equal{} 2$, or to $ n^{2}\plus{}4n\plus{}4$, yielding $ n \equal{} 9$.  If $ m \equal{} 2$, then $ 4n^{2}\plus{}5n\minus{}10$ is a perfect square smaller than $ 4n^{2}\plus{}8n\plus{}4$ but larger than $ 4n^{2}$.  Or, it equals $ 4n^{2}\plus{}4n\plus{}1$ yielding $ n \equal{} 11$.  So, if $ n > m$, the only possible $ (m,n)$ solutions are $ (1,2)$, $ (1,9)$, $ (2,11)$.

Assume $ n < m$.  Then, $ m^{2}n^{2}\minus{}5m\plus{}5n < m^{2}n^{2}$, or in order to be a perfect square, $ m^{2}n^{2}\minus{}5m\plus{}5n\leq m^{2}n^{2}\minus{}2mn\plus{}1$, or $ (2m\plus{}5)n\leq5m\plus{}1$, or now $ n \equal{} 1,2$ are the only possible answers.  Thus, making $ n \equal{} 1$ results in $ m^{2}\minus{}5m\plus{}5$ being a perfect square less than $ m^{2}$ but larger than $ m^{2}\minus{}6m\plus{}9$, or it must be equal to either $ m^{2}\minus{}4m\plus{}4$ yielding $ m \equal{} n \equal{} 1$, or to $ m^{2}\minus{}2m\plus{}1$, yielding non-integer $ m$.  Finally, $ n \equal{} 2$ results in $ 4m^{2}\minus{}5m\plus{}10$ being a perfect square less than $ 4m^{2}\plus{}4m\plus{}1$ but larger than $ 4m^{2}\minus{}8m\plus{}4$, or it must be equal to either $ 4m^{2}$, yielding $ m \equal{} 2$, or to $ 4m^{2}\minus{}4m\plus{}1$ yielding $ m \equal{} 9$.  So, if $ m > n$, the only possible $ (m,n)$ solution is $ (9,2)$.

All other solutions are $ m \equal{} n$.

this problem I take from my friend
1)$ (mn)^{2}\minus{}4m\plus{}4n$ is a perfect square
This is my problem
For each $ a\in N$ so that there exist finite $ (m,n)$ such that
$ (mn)^{2}\minus{}am\plus{}an$ is a perfect square.


geometry
trapezoid
symmetry
parallelogram
rhombus
geometric transformation
reflection

Given two positive lengths $h$ and $m$. Construct a cyclic trapezium $ABCD$ with $AB$ parallel to $CD$, perpendicular distance $h$ between $AB$ and $CD$, and $AB + CD = m$.

Such trapezium is not unique; here is how to construct the whole family:

Draw a segment AB=m, C is an arbitrary point lying on AB. k is the line s.t. k//AB and d(k, AB)=h. Constrcut the perp. bisector of BC which intersect k at D. Let E be the midpoint of AC. Through A and C draw the lines parallel to DE. Suppose they intersect k at F and G. Then BCGF is the required cyclic trapezium.

There may be some mistakes, as I rarely solve any construction problem.

I have taken the liberty to edit the above posts since there was some confusion at first.

Of course, Mecrazywong's solution is correct. Here is another one:

First consider our required trapezium ABCD. A trapezium is cyclic if and only if it is isosceles; thus, our cyclic trapezium ABCD with AB || CD must satisfy BC = AD.

Now let X, Y, Z, W be the midpoints of the sides AB, BC, CD, DA of our trapezium ABCD. Then, the segment YW is the midline of the trapezium ABCD (since it passes through the midpoints Y and W of its legs BC and AD), and thus it is parallel to the bases AB and CD and its length is $YW=\frac{AB+CD}{2}=\frac{m}{2}$. On the other hand, the line ZX passes through the midpoints Z and X of the bases CD and AB of our trapezium ABCD, and thus it is the symmetry axis of this trapezium; hence, it is perpendicular to the midline YW and to the bases AB and CD; consequently, the length of the segment ZX equals to the distance h between the lines AB and CD.

So, we have $YW=\frac{m}{2}$, ZX = h and $ZX\perp YW$. Since X, Y, Z, W are the midpoints of the sides AB, BC, CD, DA of the quadrilateral ABCD, the quadrilateral XYZW is a parallelogram; since its diagonals ZX and YW are perpendicular to each other, it is even a rhombus. Now, we know the lengths of the diagonals of this rhombus ($YW=\frac{m}{2}$ and ZX = h), and it is very easy to construct a rhombus knowing the lengths of its diagonals (take any point in the plane as the center of the rhombus, take any two perpendicular lines through this point as the diagonals, and locate the vertices of the rhombus on these diagonals since you have given the lengths of the diagonals and you know that the center of a rhombus bisects the diagonals). Now, once the rhombus XYZW is constructed, denote by a and c the parallels to the line YW through the points X and Z; let an arbitrary line g through the point Y intersect these lines a and c at two points B and C, and let the reflection of this line g in the line ZX (of course, this reflection will pass through the point W, since the line g passes through the point Y) intersect these lines a and c at two points A and D. Then, the quadrilateral ABCD is our required trapezoid.

  Darij

Nice constructions!

I would like to post one more

Let $A$ be a fixed point, $Ax$ a fixed ray, and one fixed line $L$ parallel to $Ax$, such that $d(L,Ax)=h$.

We'll construct all the trapezoids of this family, such that $B \in Ax$. Also $AB>CD$.

Now, consider such a trapezium. The perpendicular lines from $C$ and $D$ intersect $AB$ at the points $C', D'$ respectively. Also, $M$ is the midpoint of $AB$.

$D'C'CD$ is a rectangle, so $D'C'=CD$ and $M$ is the midpoint of $D'C'$ ($ABCD$ is isosceles).

$AC' = AM+MC' = \dfrac{AB}{2} + \dfrac{C'D'}{2} = \dfrac{AB+CD}{2} \Rightarrow$
$AC' = \dfrac{m}{2}$

So $C'$ is a fixed point in $Ax$. As a result, $C$ is a fixed point in $L$.

So here is the construction:
$Ax$ is a fixed ray.
Bring a line $L \parallel Ax \ : \ d(L,Ax)=h$
Take a point $C' \in Ax$ such that $AC'=\dfrac{m}{2}$
Take the point $C \in L$ such that $CC' \perp AC'$. ($AC$ will be the fixed diagonal of $ABCD$)

If $B$ is an abritrary point on $C'x$, take the reflection $D$ of $C$ w.r.t. the perpendicular bisector of $AB$.
Then of course $D \in L$.

$ABCD$ is a trapezium ($AB \parallel CD$) with height $h$ and

$AB+CD=AD'+D'C'+C'B+CD \ \ \ \ \ \ \ \ (D$ as defined above $)$

$AB+CD=(AD'+C'B')+(D'C'+CD)$
$AB+CD=2 \cdot AD'+2 \cdot D'C'$
$AB+CD=2 \cdot AC'$
$AB+CD=m$


quadratics
number theory proposed
number theory

$ x,y\in N \plus{}$ and $ (x,y) \equal{} 1$
$ p$ is an odd prime
It is known that$ x\equiv 1(mod 4),x\equiv 2(modp),y\equiv 0(mod4),y\equiv \minus{} 1(modp)$ where $ y$ is a perfect square
$ u_0 \equal{} 0,u_1 \equal{} 1,u_{n \plus{} 2} \equal{} xu_{n \plus{} 1} \plus{} yu_n(n\ge0)$
Prove that there exist no positive integers$ m,z$ such that $ u_m \equal{} pz^2$

This problem was from DengYu----gold medalist of 2007 IMO :)
The main idea is using Jacobi Symbol(maybe I say nothing),I will post his solution on this weekend if no solution is presented here :)


Hi
I need some useful liks about Hex opening ideas or strategies ,.......

[quote="asdf2345"]It's usually a type of magical spell or curse, similar to the ones demonstrated in Harry Potter.[/quote]

I really hope that was a joke...

Anyways, I don't really have any links...

[quote="nutz_for2.718281828"]he means the GAME. Here is a link:

[url]http://gprime.net/game.php/hexxagon[/url][/quote]

...I don't think that's the most commonly known hex...

Here is what I assumed he meant...http://en.wikipedia.org/wiki/Hex_%28board_game%29

Although I am into Go, I'm not really into Hex, so I can't offer input on hex sites...


Multiply by $ abc$ then subtract the right side from the left and call it $ f(a,b,c)$.

Then $ f(a,b,c)\equal{}(a \minus{} b)^2 (b \minus{} c)^2 (c \minus{} a)^2\plus{}(a^2 \minus{} a b \plus{} b^2 \minus{} a c \minus{} b c \plus{} c^2) (a^3 b \plus{} a b^3 \plus{} a^3 c \minus{}
   2 a^2 b c \minus{} 2 a b^2 c \plus{} b^3 c \minus{} 2 a b c^2 \plus{} a c^3 \plus{} b c^3)$

which is true by muirhead. I wonder how you arrived at this inequality. It seems quite sharp.

Thanks Altheman!
 :) Your solution is very nice.To arrive this inequality,I created the simple one first and then I tried to make it stronger :lol: .I also wonder how can you lead to $ f(a,b,c)\equal{}(a \minus{} b)^2 (b \minus{} c)^2 (c \minus{} a)^2\plus{}(a^2 \minus{} a b \plus{} b^2 \minus{} a c \minus{} b c \plus{} c^2) (a^3 b \plus{} a b^3 \plus{} a^3 c \minus{}
2 a^2 b c \minus{} 2 a b^2 c \plus{} b^3 c \minus{} 2 a b c^2 \plus{} a c^3 \plus{} b c^3)$
It is really a huge calculation. :wink:

P\s:I'm also glad to see another proof. :)

Once you clear out the $ abc$, you get a six degree inequality. $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\ge 0$ is a very strong six degree inequality that cancels the $ a^4b^2$ terms, etc. Then I factored the equation by computer...it was very fortunate that it worked out the way it did...

:huh: I can't find the term $ a^4b^2$ after clear out abc.Cab you show me where is it? :)


The famous Yin-Yang symbol is constructed from a circle of area $ 16\pi$ square units. (The shaded and unshaded regions are congruent.) In terms of $ \pi$, how many units long is the perimeter of the shaded region?
[asy]draw(unitcircle);
fill((0,1)--(0,-1)..(-1,0)..cycle);
fill((0,0)..(.5,-.5)..(0,-1)..(-.5,-.5)..cycle);
unfill((0,0)..(.5,.5)..(0,1)..(-.5,.5)..cycle);
draw((0,0)--(0,1));
label("$r$",(0,.5),E);[/asy]

That's half perimeter of big, and whole perimeter of smaller. Radius of the big: 4, small: 2
$ 4\pi \plus{} 4\pi \equal{} 8\pi$


I'm just curious about how different regional teams prepare for ARML.

How does your region choose its ARML Team, and how many team practices did you have?

no, once before NYSML, once before ARML.

This year, we practiced some relay problems online,
but NYC A team only got 6/40 on relays. :oops:

We hold practices over the whole school year, approximately once a month, although we only concentrate on ARML material as opposed to other types of problems in about the last two months. Most of our team comes from the greater Los Angeles metropolitan area and is within an hour's drive of our practice location in Long Beach. There is a separate group that practices in Santa Barbara and there is also the San Diego Math Circle; we had a few people each from each of those locations. Although Bill Fitz-Gibbon has been sick, I expect him to also run practice sessions in North Hollywood next year. I do the team selection, based on AMC/AIME scores, the results of our local contest (CSULB Math Day at the Beach) and various other factors. We've had a "C" team in each of the last two years, and haven't completely filled it yet, so I haven't yet had to tell anyone they can't come.

As for relays - relays are hard. My congratulations to Lehigh Valley for basically winning it on the relays. I really have no idea how a team can consistently get 12 or more points on each relay.

We Washintonians have no pracitces whatsoever... although I think we will in the future, now realizing that we have a good team ahead of us. SoCal has an advantage in that there are many "math circles" and other places to practice.

Chicago has tryouts and 4 practices, then one more practice in Iowa.  That's because we're "Chicago" and not "Illinois", so everyone can drive an hour or so once every few weeks.

[quote="Sly Si"]Minnesota has a state math league that our coaches use, along with AMC scores, to pick 30 (or so) people and 6 to 8 younger "students in training" (alternates).  The trainees then aren't guaranteed to go to the meet, but inevitably several people picked for the team can't go, so some of them take those spots.

We have practices on three Saturdays in May, all morning and part of the afternoon.  They basically consist of taking past events, followed by a run-down of the problems and what the coaches noticed about our performance.  The Gold (A) team is chosen based on performance in these practice events, along with which trainees will go to Iowa.[/quote]

Minnesota is a pretty large state. Do you draw students from all over the state? And if so, how many of them can get together to practice three times in a month?

[quote="Midnight"]Indeed we do, Geehoon, but I think it would probably be better if we could mandate online practice...we blow at relays, and we need all of the practice we can get.[/quote]
it also helps if people practice with the actual team that they compete with, rather than whoever happens to be there... that would help people better know what they are up for- how the other people on their team perform, and if they can work well together

it's really a shame that the pressure of actual competition is impossible to simulate >_<

[quote="Mariag"]it also helps if people practice with the actual team that they compete with, rather than whoever happens to be there... that would help people better know what they are up for- how the other people on their team perform, and if they can work well together[/quote]
I partially disagree. That's why I did not take too much time to organize teams with the actual teammates.
As JBL said, it is a half team - a half individual round.
I don't see any problem with the practice with someone else other than your actual teammates, as long as they are not extremely slower/faster than your actual teammates.

[quote="MysticTerminator"][quote="gauss202"][quote="jb05"]We have NCHSSMC (North Carolina High School State Mathematics Contest) which has the top 7% from regional competitions. The top 20 are automatically offered a spot on the team (we have 2 teams plus 4 alternates so 34 people), and a lot of those people can't go. The next choices are from AMC, AIME, the next 20 on the NCHSSMC, and various other things. It was interesting that at the competition, I had heard of most of the A team but hadn't heard of anybody else. I think the alternates are just the leftover people who agree to go because they have nothing better to do. Oh, and if you want to look: [url]http://courses.ncssm.edu/goebel/statecon/state.htm[/url]. The site's down a lot.[/quote]

Do you guys hold any practices before the competition?[/quote]

heh, our contest has an ACRONYM? whoa. I'm impressed

no practices. well we do have practices. like ... 1. nobody goes to it. which is why we utterly pwnt on the first two rounds, not so much on the next two (yet we rather inexplicably are really really good on super relay. won 2 of the last 3 years)[/quote]

Yeah...I'm not sure what's up w/ our super relay when we can't seem to get more than 1 relay right throughout the entire team on 2 relays.  It's too bad the relays cost us so badly in placement.  7th still isn't too much to complain about.  I look forward to seeing what you can do next you since you're only losing 3 of us seniors.  Not like we're that incredibly essential anyway.

But yeah...we don't do practice.  We don't even know who's going to be A and B team until the night beforehand.  I don't know how much of a difference it would've made though.  Anyway...I'm expecting great things from North Carolina A next year.

[quote="gauss202"]Minnesota is a pretty large state. Do you draw students from all over the state? And if so, how many of them can get together to practice three times in a month?[/quote]

We do have some students from all over the state, but most come from the Twin Cities area.  That's where most of the top students in the state come from anyway, though.  A lot of the students from far away come to the practices; I think some might drive down the night before, since they run 9-3:30.  I think usually every student who goes to Iowa comes to at least one practice, and almost all come to two or three.  Our turnout is usually 25-30 students out of 36 to 38 selected.


Does anybody know of any math olympiad preperation programs in nyc?

The Math-M-Addicts class discusses many olympiad-level problems.  See
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=665185#665185[/url].

My daughter is in the class, and I highly recommend it to anyone in the NYC area hoping to eventually do well on olympiads.

Well I already went there this Saturday, but are there any programs on weekdays, after school?

Not that I know of.  I heard a rumor that NYU might be starting a math circle next year.  I don't know when it would be held.  When I found out details, I will post them in this forum.

For an online course, you can try the WOOT training class available at this site.  It's too late for this year, but you can sign up over the summer for next year.

I haven't heard any rumor about that, but I too can investigate that--I'm not sure who on the faculty would run that, though.

~KC

I personally have not heard of the type of after school programs you are looking for.  Though I am a big proponent of math students also learning CS.  Try your hands on the web based USACO (USA Computing Olympiad) training:

http://train.usaco.org/usacogate