parameterization
vector
calculus
calculus computations

1. Why the arc from (1,0,0) to (1,0,2pi) is described by the parameter interval 0 <= (less and equal)t<=2pi

2. the point (1,0,0) corresponds to the parameter vaule t=0

any helps or any hints is great appreciated!  Thanks!

1. Bumping a thread, especially bumping a thread on the same day you posted it, is considered rude. Please don't do that.

2. Your question is completely meaningless as stated, and will remain meaningless until you provide the actual parametric equations. In fact, it is impossible to determine from what you said what the curve is.

There is nothing remotely like "the" parametric representation of a curve. Every curve can be parameterized in a great profusion of possible ways, over a great profusion of possible intervals for the parameter. It is true that in certain particular situations, one can have and remember a "usual" or "convenient" way to parameterize that curve - but there's nothing unique about any of that.

this is the whole question and some partial solution:

Question:  find the length of the arc of the circular helix with vector equatio r(t)=costi+sintj+tk from the point (1,0,0) to the point (1,0,2pi)

solution:
r'(t)=-sinti+costj+k
/r'(t)/=sqrt root of 2

then it says

The arc from (1,0,0) to (1,0,2pi) is described by the parameter interval 0<=t<=2pi.

//I understand the first two part, but I don't understand the part where it convert from the points to the parameter interval.
Can anyone explain it to me?

did anyone understand my question or is this too simple for every here.   :mad:   :mad:

It's just saying that when $t$ ranges from $0$ to $2 \pi$, it describes the curve from $(1,0,0)$ to $(1,0,2 \pi)$ so you should find the arc length for $t$ in that interval.

but how do you find the arc length for t in that interval, can you be a little specific?


The numbers $1, 2, 3$, $\ldots$, $n^2$ are arranged in an $n\times n$ array, so that the numbers in each row increase from left to right, and the numbers in each column increase from top to bottom. Let $a_{ij}$ be the number in position $i, j$. Let $b_j$ be the number of possible values for $a_{jj}$. Show that \[ b_1 + b_2 + \cdots + b_n = \frac{ n(n^2-3n+5) }{3} . \]

it is clear that $\forall\ i,\ i^2\leq a_{ii}\leq (n^2-(n-i+1)^2+1)$. adding all these as feasible possibilities gives the answer desired, so it suffices to prove that all these choices are feasible,i.e., for each such $a_{ii}$ as above there exists a grid with $a_{ii}$ as its $(i,i)^{th}$ entry. To see this consider a grid of the form
$\begin{array}{cc} A&B\\C&D\end{array}$,
where $A$ is $i\times i$, $B$ is $(i-1)\times (n-i)$, $C$ is $(n-i)\times (i-1)$ and $D$ is $(n-i+1)\times (n-i+1)$(of course, they are not disjoint; $A$ and $D$ share the entry $a_{ii}$, the other pairs are disjoint). Fill in this grid by filling rowwise and filling in the order $A,B,C,D$(i.e., $A$ first and then $B$ and then $C$ and so on). this is clearly a 'good' filling.
it is now easy to see that in this grid, one can just 'bump' out the element $a_{ii}$ which occurs in one of $B$ or $C$(if it not already in $A$), by moving elements rowwise(though i am not rigorous here i think the idea is quite clear) and get a suitable grid with $a_{ii}$ in the $(i,i)^{th}$ entry.

Actually the real problem is,
The numbers $1, 2, 3$, $\ldots$, $n^2$ are arranged in an $n\times n$ array, so that the numbers in each row increase from left to right, and the numbers in each column increase from top to bottom. Let $a_{ij}$ be the number in position $i, j$. Let $b_j$ be the number of possible values for $a_{jj}$. Show that\[ b_1 + b_2 + \cdots + b_n \leq \frac{ n(n^2-3n+5) }{3} . \]


function
geometry
analytic geometry
geometry unsolved

Let $ E$ be the set of all triangles whose only points with integer coordinates (in the Cartesian coordinate system in space), in its interior or on its sides, are its three vertices, and let $ f$ be the function of area of a triangle. Determine the set of values $ f(E)$ of $ f.$

In the plane:  [url=http://en.wikipedia.org/wiki/Pick%27s_theorem]Pick's theorem[/url] immediately gives the only possible value $ 0 \plus{} \frac{3}{2} \minus{} 1 \equal{} \frac{1}{2}$.  (Possibly the problem is asking for a proof of Pick's theorem.)

Does the problem take place in $ \mathbb{R}^3$?  (The wording is not clear.)

In $3$-dimensional space, clearly $f(E)$ is unbounded; just consider triangles $A(n,0,0), B(n,1,0), C(0,0,1)$, with $AB=1$, $AC = \sqrt{n^2+1}$, $BC = \sqrt{n^2+2}$.


If Alan, Joe, and 24 other people with different names randomly line up in a single line, then what is the probability that Alan is standing next to Joe?

Well I'm normally bad at these types of problems, but I'll try anyway.

There are $ 26$ in total so we can arrange them in $ 26!$ ways.

If Joe and Alan are standing near each other than there is $ 24! \cdot 25 \equal{} 25!$ ways they are standing next to each other. So it would be $ \frac{25!}{26!} \equal{} \frac{1}{26}$

This seems wrong...

Correct me if I am wrong.
If you make Joe and Alen 1 Unit There are 25! Ways of them standing together. but then you multiply it by 2 because Joe and Alen can switch places. Therefore there is (25!*2)/26! ways of them standing next to each other. 2/26=1/13.
(The * is a multipliication sign. sorry it looks wierd)
Once Again, correct me if I am wrong[/img]

[hide="The answer actually is"]
$ \frac{1}{13}$.
[/hide]


Solve this equation:
1, $ \sqrt{2\minus{}x^2}\plus{}\sqrt{2\minus{}\frac{1}{x^2}}\equal{}4\minus{}(x\plus{}\frac{1}{x})$
2, $ \sqrt {x^2 \plus{} x} \plus{} \sqrt {1 \plus{} \frac {1}{x^2}} \equal{} \sqrt {x \plus{} 3}$

[hide="1. answer"]
Just plugging in values $ x \equal{} 1$[/hide]

I don't think there is a nice solution to the second one. edit: nevermind, forgot to check $ \minus{}1$

Number 1
Note that $ \sqrt {2 \minus{} a^2} \le 2 \minus{} a$ with equality when a=1, since this is equivalent to $ 2(a \minus{} 1)^2 \ge 0$.
Rewriting the right side as $ (2 \minus{} x) \plus{} (2 \minus{} 1/x)$ and applying our lemma twice gives $ a \equal{} 1$ is the only solution.

For number 2, squaring gives $ x^2 \plus{} \frac {1}{x^2} \plus{} 2\sqrt {x^2 \plus{} x \plus{} 1 \plus{} \frac {1}{x}} \equal{} 2$
But we know $ x^2 \plus{} \frac {1}{x^2} \ge 2$ by AM-GM, with equality when $ x^4 \minus{} 1 \equal{} 0$
The only possible reals are 1,-1. Subbing both in shows only -1 works.


find a geometrical solution for the following system
X+Y=50
X+Z=60
Y+Z=70

triangle with sides 20,30,40?

I think your questions should be more specific than 'find a geometric solution'... ;)

And I don't see the link with calculus...

20 , 30 and 40 cannot be solution of the system and sides of triangle because of the triangle inequality.
To be more specific :x y and z are parts of the sides of triangle .What are those parts.What are their values.

a triangle with sides 50-60-70 then, and the place they are cut by the incircle? :? I mean there are so many possible solutions...


Hi,

Can we find a continuous map $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ \{x,f(x),f(f(x)),\cdots\}$ is dense in $ \mathbb{R}$ for some $ x \in \mathbb{R}$ ?

Thanks,
blang.


AMC
AMC 12
AMC 12 B
probability
trigonometry
calculus
integration

7) There is a piggy back with 100 coins (nickels, dimes, and quarters) whose total value is $8.35. It does not necessarily have coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

21) A particle moves 8 cm in a straight line from A to B, then turns at an angle  :theta:  chosen at random from (0, :pi: ) and moves 5 cm to C in a straight line. Find the probability that AC < 7.

24) a, b, c are positive integers such that a<b<c. The system of equations 2x+y=2003 and y=|x-a|+|x-b|+|x-c| has exactly one solution. Find the min value of c.

25) Three points are placed randomly and independently on a circle. Find the probability that all three pairwise distances between the points are less than the radius of the circle.

Fierytycoon wrote:7) There is a piggy back with 100 coins (nickels, dimes, and quarters) whose total value is $8.35. It does not necessarily have coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

[hide]Ignoring the restriction on the number of coins:

835=167 nickels

835=33 quarters and a dime, or 33 quarters and two nickels

835=83 dimes and a nickel

So we have 1 :le: d :le: 83 without the restriction on the number of coins

1 dime = 2 nickels, so from '835=83 dimes and a nickel' we can subtract a dime and add two nickels for a net gain of 1 coin... we have 84 coins now, so we need 16 more, so we need to perform that operation 16 times, subtracting 16 dimes... That gives a maximum number of 67 dimes.

Now let's take 167 nickels and turn some of it into quarters... we can remove 5 nickels and add a quarter to subtract a total of 4 coins. If we do this 16 times we'll have 103 coins -> 16 quarters and 87 nickels, which becomes 16q, 85n, 1d -> 16q,83n,2d -> 16q, 81n,3d, so the minimum number of dimes is 3.

67-3=64

[/hide]

That problem seemed a bit harder/longer than an AMC #7... there must be an easier way

21. We want to find the limiting value, i.e. the random value for that angle thing that will make AC=7. Anyway, this angle thing will be the supplement of angle ABC, so we find that by the law of cosines: cos ABC=(8^2+5^2-7^2)/(2*8*5)=1/2 => ABC=pi/3 => weird angle thing is 2pi/3. so AC<7 is weird angle thing is >2pi/3, i.e. 1/3 chance of that.

24. i can't see a way to do this other than taking cases which i don't feel like doing right now.

25. All three points must lie withini 1/6 of the circumference. So it doesn't matter where the first point is placed. The second point will fall within pi/3 of this first point with probability 1/3. Depending on where the first two points are placed relative to each other, the probability that the last point falls in the right place varies linearly from 1/3 to 1/6. We take the average (y'know, like average value of an integral. makes things simpler than actually taking a frickin integral) and get 1/4. So we multiply 1/4 with 1/3 to get 1/12.

[quote="Fierytycoon"]
7) There is a piggy back with 100 coins (nickels, dimes, and quarters) whose total value is $8.35. It does not necessarily have coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

25) Three points are placed randomly and independently on a circle. Find the probability that all three pairwise distances between the points are less than the radius of the circle.[/quote]

[b]7.[/b]
n, d, q :ge: 0
n + d + q = 100
5n + 10d + 25q = 835  :arrow:  n + 2d + 5q = 167

n + d + q = 100
n + 2d + 5q = 167

d + 4q = 67
d = -1 + 4n, q = 17 - n; n :ge: 1
3 :le: d :le: 67
67 - 3 = [b]64[/b]

[b]25.[/b]
[b]0[/b].  To prove this, let the radius of the circle be 1, making it a unit circle.  If the points are placed less than one unit from each other, then the sum of the distances between the three points around the circle is 3, but that isn't possible since 3 is less than 2:pi:.  Therefore, it is impossible.  Since all circles are similar, it has been proven that it is impossible.

7) Yeah, that's pretty much how I did it. The solution isn't very nice though :(

21) Ok. (I screwed up on this one cause I got to the part where cos(theta)=1/2 when AC=7, and thought that since cos(theta) varies from 1 to -1 when x varies from 0 to pi, the probability was (1/2)/2=1/4)

24) By taking cases, you mean letting each |x-k| be x-k and then k-x?

25) Mystic, why is "All three points must lie withini 1/6 of the circumference. " true?
Warped, the points can be in the interior of the circle too, not just the circumference.

The problem says the points are [i]on[/i] the circle though, not inside of it.

[quote]The problem says the points are on the circle though, not inside of it.[/quote]

Yeah, but "on the circle" generally means on any part of the circle I think (the AMC people used "on" too). In any case, if the points could only lie on the circumference, your reasoning would be correct. However, the answer isn't 0.

[quote="WarpedKlown1335"]If the points are placed less than one unit from each other, then the sum of the distances between the three points around the circle is 3, but that isn't possible since 3 is less than 2:pi:. [/quote]

The shortest distance between any two points is a straight line. An arc necessarily is longer than the straight-line path, and therefore the sum of the distances between the three points will <b>always</b> be less than 2:pi:. (I think, correct me if I'm wrong.)

I got the same answer as Mystic.

[quote="Fierytycoon"]Yeah, but "on the circle" generally means on any part of the circle[/quote]

A circle is the locus of points a fixed distance from a specified center, so "circle" includes only its circumference (no interior points).

I also agree with mystic's answer for 25... is it just me, or are these incredibly hard for AMC problems?

meh. i thought these were easy. except for that cases one which i don't want to do (yeah, that's what i meant fyty) and that coin problem which i hate. i hate coin problems. hrmph. neal gave it to me a while ago.

Hmm, ok, about taking that average for the probability for the third point in #25, is this a common technique for nondiscrete probability problems?

[quote]Take a look at the answer choices for #25 to see that you don't really have to do very much to get it right.[/quote]

Simon, what do you mean by this?

well i just thought of it. it seems logical, especially after you cover that stuff about average value of an integral.

Well, it seems obvious even without covering that integral stuff, heh.

Anyways, would someone mind briefly explaining #24? I'm trying cases of the absolute values but I'm not really getting anywhere.

[quote="Fierytycoon"][quote="ComplexZeta"]Take a look at the answer choices for #25 to see that you don't really have to do very much to get it right.[/quote]

Simon, what do you mean by this?[/quote]

You can get crude bounds for this problem and see that the answer must be within the bounds. Only one answer is.

"crude bounds"? Well, I think 1/9 is pretty close to 1/12.

Quote:24) a, b, c are positive integers such that a<b<c. The system of equations 2x+y=2003 and y=|x-a|+|x-b|+|x-c| has exactly one solution. Find the min value of c.

Here is my solution:[hide]Let f(x) = 2x + |x - a| + |x - b| + |x - c|.  Note that f(x) :ge: 2x + |x - a| + (b - x) + (c - x) = b + c + |x - a| :ge: b + c.  Furthermore, the inequality is an equality if and only if x = a.  Also, note that f tends to :inf: as x tends to :pm::inf:.  Also, f is continuous.  Putting all those facts together, we see that the equation f(x) = 2003 has exactly one solution if and only if b + c = 2003.

So the problem becomes much simpler.  We are just asked for the min value of c subject to the condition b + c = 2003.  Well, 2c > b + c = 2003, so c :ge: 1002.  Furthermore, we can choose c = 1002 and b = 1001 to see that 1002 can be achieved.  So the answer is 1002.[/hide]

Ravi B:

Ok, the solution makes sense. However, just a question: how did you ever think of doing what you did as described in the first paragraph of your previous post?

my thoughts:

Well firstly letting f(x) be that thing is easy enough because you want that to be 2003..
then you can just play with that for a while.. you'll probably work out that x=a is the minimum, or feel like you want that 2x to cancel out, or ..

I thought of it because it was similar to the following well-known result.  Let p, q, r, s, and t be real constants.  Then the minimum of |x-p| + |x-q| + |x-r| + |x-s| + |x-t| occurs (uniquely) when x is the median of p, q, r, s, and t.  That's basically our situation, with the five constants being 0, 0, a, b, and c.  The proof of that well-known result is similar to what I did above.

Bah, I never seem to have heard of all these "well known results" everyone speaks of.  :D

In any case, thanks for the help everyone.

Look at Kiran Kedlaya's inequality notes, 1st section I think. For arbitrary \sum_{i=1}^n x-x_i, we have minimum at median if n is odd. if n is even, we have minimum on median interval

for example |x-1|+|x-3|+|x-5|+|x-15| has minimum on [3,5]
|x-1|+|x-3|+|x-5| has minimum at x=3.

etc.

Ah, I found it. Except it was just a problem (without solution). Thanks though.


modular arithmetic
number theory
prime numbers
combinatorics proposed
combinatorics

Prove that there is an arrangement $ a_1,a_2,\cdots$ of the members of $ \mathbb{N}$ (each member shows up in the sequence exactly once) such that $ a_i | a_{i\plus{}1}\plus{}a_{i\plus{}2}$ for any $ i\geq 1$.

[hide]We do the proof by construction. Start with the $ 1,2$. Assume that we have constructed the sequence $ 1,2,\ldots,m,l$ such that each pair of consecutive terms are relatively prime. Let $ k$ be the least positive integer which not in the sequence $ 1,2,\ldots,m,l$. Our aim now is to include $ k$ in the sequence.

Claim: There exist a prime number $ p$ which is not in the sequence $ 1,2,\ldots,m,l$ such that $ m|l\plus{}p$ and $ \gcd(l,p)\equal{}\gcd(k,p)\equal{}1$.

Proof: Note that $ m|l\plus{}p\implies p\equal{}xm\minus{}l$ for some $ x\in\mathbb{N}$. By Dirichlet's Theorem, there exist infinitely many primes of the form $ xm\minus{}l$. Since the sequence $ 1,2,\ldots,m,l$ contain finitely many terms and there exist finitely many $ n$ such that $ \gcd(l,n)\ne1,\gcd(k,n)\ne1$, then there exists a prime $ p$ such that $ m|l\plus{}p$ and $ \gcd(l,p)\equal{}\gcd(k,p)\equal{}1$. Claim proved.

So there exist prime numbers $ p,q$ which can be included to the sequence so that the sequence becomes $ 1,2,\ldots,m,l,p,q$. Let $ a$ be the integer which is not in the sequence yet such that $ p|q\plus{}a$, $ q|a\plus{}k$ and $ \gcd(a,k)\equal{}1$. This is possible because, by Chinese Remainder Theorem, there exist infinitely many $ a$ such that $ a\equiv\minus{}q\pmod{p}$ and $ a\equiv\minus{}k\pmod{q}$, and the general solution is of the form $ a\equiv n\pmod{pq}$ or $ a\equal{}xpq\plus{}n$. Clearly, there exists an integer $ x$ such that $ a$ is not in the sequence yet and $ \gcd(a,k)\equal{}1$. Thus the sequence becomes $ 1,2,\ldots,m,l,p,q,a,k$.

So the sequence can be continued so that each integer appears exactly once. Q.E.D.[/hide]


Ring Theory
superior algebra
superior algebra unsolved

We were given a 5 step problem on our homework set relating to primary decompositions, I have worked out a solution to most of them but I can't seem to get he last part; maybe I've made some mistakes or something. In any way, input is highly appreciated =)

Let $ k$ be a field and $ R \equal{} k[x,xy,y^2,y^3]$ We are given that $ P\equal{}(xy,y^2,y^3)$ and $ Q\equal{}P^2\equal{}(x^2y^2,xy^3,y^4,y^5)$. Prove the following:

(a) $ (0)$ is not in $ Ass_R(R/Q)$ (where Ass denotes the set of associated primes)
My answer: We know associated primes contain $ Ann(R/Q)$ which is $ Q$, so $ (0)$ is not an associated prime

(b) Prove that $ P$ is minimal over $ Ann_R(R/Q)$
We know $ P \equal{} Rad(Q)$ (the radical ideal of $ Q$) so $ P$ is the intersection of all primes containing $ Q$, hence it is the smallest such prime containing $ Q \equal{} Ann(R/Q)$.

(c) Find $ m\in R/Q$ such that $ Ann(m)\equal{}P$.
$ xy$ by inspection

(d) Find the $ P$-primary part of $ R/Q$
$ (y^2, y^3)$ again by inspection (I think (c) and (d) are right...)

(e) Prove that $ Ass(R/Q) \equal{} \{P,M_1...M_r\}$ for some maximal ideals $ M_1... M_r$ in $ R$ with $ r \geq 1$. For this part you may assume without proof that all prime ideals of $ R$ strictly containing $ P$ are maximal.
Example: $ M \equal{} (x,xy,y^2,y^3)$ is maximal.

(f) finish finding a primary decomposition of $ Q$
now what?

I have an idea: I'm thinking $ (x^2,xy,y^4,y^5)\cap (y^2,y^3)$ but Im not sure how to show that this is indeed a primary decomposition (in particular, I don't know how to show the first term in the intersection is primary)


analytic geometry
geometry
rectangle
floor function
ceiling function
geometry unsolved

For $ s,t \in \mathbb{N}$, consider the set $ M\equal{}\{ (x,y) \in \mathbb{N} ^2 | 1 \le x \le s, 1 \le y \le t \}$. Find the number of rhombi with the vertices in $ M$ and the diagonals parallel to the coordinate axes.

[hide]For each rhombi with vertices in $ M$ and the diagonals parallel to the coordinate axes, the rectangle whose midpoints are the vertices of the rhombi, and whose sides are parallel to the coordinate axis, also has vertices in $ M$.

For each rectangle with sides parallel to the coordinate axis, if the rectangle's length and width are both even, then the rhombi formed by the midpoints of the sides will have its vertices in $ M$ and the diagonals parallel to the coordinate axes.

Thus, we need to count the number of rectangles with sides of even lengths that are parallel to the coordinate axes.

There are $ \dbinom{\lfloor\dfrac{s}{2}\rfloor}{2}\plus{}\dbinom{\lceil\dfrac{s}{2}\rceil}{2}$ ways to choose the $ x$ coordinates of the vertical sides.

There are $ \dbinom{\lfloor\dfrac{t}{2}\rfloor}{2}\plus{}\dbinom{\lceil\dfrac{t}{2}\rceil}{2}$ ways to choose the $ y$ coordinates of the horizontal sides.

Therefore, the answer is: $ \left[\dbinom{\lfloor\dfrac{s}{2}\rfloor}{2}\plus{}\dbinom{\lceil\dfrac{s}{2}\rceil}{2}\right]\cdot\left[\dbinom{\lfloor\dfrac{t}{2}\rfloor}{2}\plus{}\dbinom{\lceil\dfrac{t}{2}\rceil}{2}\right]$[/hide]


abstract algebra
superior algebra
superior algebra unsolved

An ideal $ I$ of $ A$ is radical iff $ r(I)=I$.

1. Let $ I_1$ and $ I_2$ be primary ideals of $ A$ such that $ r(I_2)\subsetneq r(I_1)$ but $ I_2$ is not $ \subseteq I_2$. Show that $ I_1\cap I_2$ is not primary, but $ xy\in I_1\cap I_2$ implies either $ x^n\in I_1\cap I_2$ or $ y^m\in I_1\cap I_2$.

2. Let $ J$ be an $ I$-primary ideal of $ A$. Show that $ \frac{a}{1}$ is zero in $ A_I$ implies $ a\in J$. Deduce that the kernel of $ \phi: A\rightarrow A_I$, $ a\mapsto\frac{a}{1}$ is contained in the intersection of all $ I$-primary ideals of $ A$.

Notation: $ S=A\setminus I$ and write $ A_I$ for $ S^{-1}A$.

You have typos in both of these problems.

1) "but $ I_2$ is not $ \subset I_2$"

2) "in $ A_I$ implies $ a$" (this one I'm guessing is supposed to read $ a\equal{}0$).

1. Let $ I_1$ and $ I_2$ be primary ideals of $ A$ such that $ r(I_2)\subsetneq r(I_2)$ but $ I_2\nsubseteq I_1$. Show that $ I_1\cap I_2$ is not primary, but $ xy\in I_1\cap I_2$ implies either $ x^n\in I_2\cap I_2$ or $ y^m\in I_1\cap I_2$.

2. Let $ J$ be an $ I$-prmary ideal of $ A$. Show that $ \frac{a}{1}$ is zero implies $ a\in J$. Deduce that the kernel of $ \phi: A\rightarrow A_I$, $ a\mapsto \frac{a}{1}$ is contained in the intersection of all $ I$-primary ideals of $ A$.

Notation: $ S\equal{}A\setminus I$ and write $ A_I$ for $ S^{\minus{}1}A$.

This is the correct questions... Sorry... can anyone helps ?

There are still typos.

For (1), pick an $ x\in I_2\minus{}I_1$, and a $ y\in I_1\minus{}r(I_2)$, and consider $ xy\in I_1\cap I_2$.  For the second part of (1), note that $ xy\in I_1\cap I_2$ implies either $ x$ or $ y$ is in $ r(I_2)$.

If $ \frac{a}{1}$ is zero in $ A_I$, then there exists $ b\in A\minus{}I$ such that $ ab\equal{}0$.  Show how to conclude $ a\in J$.

Steve


inequalities
quadratics
inequalities unsolved

Let a,b,c be the sides of a triangle and and let x,y,z be real numbers.
Prove that

    $\sum (b/c+c/b-1)x^2  \geq  \sum (3-(b+c)/a)yz$

It may be possible to me post the 'official' solution, unfortunately your reference is incorrect. Problem 1051 appeared exactly where you noticed, but the statement seems very far from your problem....

Pierre.

Sorry, the right reference is:

Elem. Math. (1973) pag.141-146 "Two non-negative quadratic form"


trigonometry
geometry
area of a triangle
inequalities unsolved
inequalities

$x,y,z$ are real numbers,
prove $(x^{2}+xy+y^{2})(y^{2}+yz+z^{2})(z^{2}+zx+x^{2})\geq (xy+yz+zx)^{3}$

I had tried to explain it by Cosine Law(in case x,y,z>0), then xy+xz+yz is the area of a triangle but failed.

[quote="hanschinaman"]$x,y,z$ are real numbers,
prove $(x^{2}+xy+y^{2})(y^{2}+yz+z^{2})(z^{2}+zx+x^{2})\geq (xy+yz+zx)^{3}$
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142609


a)prove that there exists a,b \in N s.t. a4+b4=70613
b)find a,b,c\in N s.t. a^3+b^3+c^3=99100

for the second one: the idea is to take a=99^33*4 b=99^33*3 c=99^33*2 and the sum becomes 99^99*(64+27+8)=99^100

for the first one take a=706^3*5 b=706^3*3, and it works!


calculus
integration
trigonometry
function
calculus computations

Evaluate $ \frac{1}{\displaystyle \int _0^{\frac{\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}$

[color=darkred]Why [b]2006 and 2008[/b] ? More better $ 2n$ and $ 2(n \plus{} 1)$ ... or $ p$ and $ p\plus{} 2$ ... or ...[/color]

Of cource we can genelize the problem as what you say.

[quote="kunny"]Evaluate $ \frac {1}{\displaystyle \int _0^{\frac {\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}$[/quote]

the rule is to always work [b]down [/b]with the [i]multiple[/i]-angle arguments of trig functions (and avoid working with the [i]power[/i]-angle arguments)

Let $ I\equal{}\int\cos^{2006}\,x\cdot\sin(2008\,x)\;\textbf dx$
$ \equal{}\int\cos^{2006}\,x\;\left[\sin\,x\cdot\cos(2007\,x)\;\plus{}\;\cos\,x\cdot\sin(2007\,x)\right]\;\textbf dx$
$ \equal{}\int\cos(2007\,x)\left[\cos^{2006}\,x\cdot\sin\,x\right]\;\plus{}\;(\cos^{2007}\,x)\left[\sin(2007\,x)\right]\;\textbf dx$
$ \equal{}\minus{}\frac{1}{2007}\;\int\cos(2007\,x)\;\textbf d\left(\cos^{2007}\,x\right)\;\plus{}\;(\cos^{2007}\,x)\;\textbf d\left(\cos(2007\,x)\right)$

since
$ \textbf d(u\,v)\equal{}u\,\textbf dv\plus{}v\,\textbf du$

$ I\equal{}\minus{}\frac{1}{2007}\;\int\;\textbf d\left(\cos(2007\,x)\cdot\left(\cos^{2007}\,x\right)\right)$

and so, the answer to the original question is the current year, $ 2007$.

Your answer is correct. My solution is exactly same as yours, misan. if this problem comes the next year's Japanese university entrance exam, as pointed by Virgil Nicula, the problem would generalized with hint.

It seemed that I had posted the similar problem with answer 2005 or 2006, I'm sure that someone could solved with complex number,
maybe $ e^{ix} \equal{} \cos x \plus{} i\sin x$, but I can't find his post.

[url=http://www.mathlinks.ro/viewtopic.php?t=70994]You can see here![/url]

[color=darkblue][b]The Misan's drawing up is fine ![/b]

I edited more in detail such (for the my high school - students) :

$ I_n\equiv\int_0^{\frac {\pi}{2}}\cos^nx\sin (n + 2)x\ \mathrm {dx} = A + B$, where

$ A\equiv\int_0^{\frac {\pi}{2}}\cos^nx\sin x\cos (n + 1)x\ \mathrm {dx}$ and $ B\equiv\int_0^{\frac {\pi}{2}}\cos^{n + 1}x\sin (n + 1)x\ \mathrm {dx}\ .$

$ \boxed {\ \begin{array}{ccc} u(x) = \cos (n + 1)x & \Longrightarrow & u'(x) = - (n + 1)\sin (n + 1)x\ . \\
 \\
v'(x) = \cos^nx\sin x & \Longrightarrow & v(x) = - \frac {1}{n + 1}\cdot\cos^{n + 1}x\ .\end{array}\ }\|$ $ \implies$

$ A = [ - \frac {1}{n + 1}\cdot \cos^{n + 1}x\cos (n + 1)x]_0^{\frac {\pi}{2}} - B$ $ \implies$

$ A + B = [ - \frac {1}{n + 1}\cdot \cos^{n + 1}x\cos (n + 1)x]_0^{\frac {\pi}{2}}$ $ \implies$ $ \boxed {\ I_n = \frac {1}{n + 1}\ }\ .$

In the nice [b]Kunny[/b]'s exercise, $ I_{2006}=\frac {1}{2007}\ .$[/color]


search
induction
combinatorics proposed
combinatorics

Given a set $\mathcal M$ with $n\ge 3$ points with no three collinear, show that we can find three points $A,B,C\in\mathcal M$ s.t. every line determined by two points in $\mathcal M\setminus\{A,B,C\}$ cuts at most one of the segments $|AB|,|AC|$.

The problem is simpler than I expected it to be. I spent most of the evening trying to search for something in convex covering of the set, but I think the only solution is by induction.
So assume we proved that for n and we try to prove it for n+1. Let's consider our angle BAC with no line passing through both [AB] and [AC], and let X be the (n+1)-th point just added. For the induction to make difficulties it is needed to exist a line through X and another point D which cuts both [AB] and [AC]. We distinguish 2 important cases: when X and D are on the same side of both [AB] and [AC], and when they are not. In the second case it is easy to check that we can always choose XAD to be the needed angle. In the first case, let's assume that X and D are closer to the side AC. If X or C are on the same side of BC as A than than XAB or DAB satisfy. Else XCA satisfy. These are easy to see on a drawing.

Thanks! Now think about this: take one of the pairs formed by a point $A$ and a line $BC$ s.t. the distance $d(A,BC)$ is minimal, and we are done :).

Actually, we still have some explaining to do. For example, if we choose the points in this way, $\angle ABC,\angle ACB$ must be acute, otherwise we can take $B,AC$ or $C,AB$ instead of $A,BC$. Now the conclusion should be easy to derive.

yeah...you're right. Why didn't I think of this?  :blush:
But at least I also have a solution, which values just as many points as yours.


Let a,b,c is real numbers such that a+b+c=0 .Prove that :
TEX]\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2} \leq\frac{-15}{2}[/TEX]

hello, do you meant
Let $ a,b,c$ are real numbers such that $ a\plus{}b\plus{}c\equal{}0$ .Prove that :
$ \frac{ab}{c^2}\plus{}\frac{bc}{a^2}\plus{}\frac{ca}{b^2} \leq\frac{\minus{}15}{2}$
Sonnhard.

Substitute $ c\equal{}\minus{}a\minus{}b$. Then

$ \frac{bc}{a^2}\plus{}\frac{ca}{b^2}\plus{}\frac{ab}{c^2}\plus{}\frac{15}{4}\equal{} \frac{(a \minus{} b)^2 (2 a \plus{} b)^2 (a \plus{} 2 b)^2}{ 4 a^2 b^2 (a \plus{} b)^2}\equal{}\frac{(b\minus{}c)^2(c\minus{}a)^2(a\minus{}b)^2}{4a^2b^2c^2}\ge 0$

I want to prove:
$ P \equal{} \frac{ab}{c^2} \plus{} \frac{bc}{a^2} \plus{} \frac{ca}{b^2} \plus{} \frac{15}{4} \le 0$

Then $ P \equal{} \frac{4a^3b^3 \plus{} 4b^3c^3 \plus{} 4c^3a^3 \plus{} 15a^2b^2c^2}{4a^2b^2c^2}$.

Let's take a look at the expression $ A \equal{} 4a^3b^3 \plus{} 4b^3c^3 \plus{} 4c^3a^3 \plus{} 15a^2b^2c^2$.
Setting $ a\equal{}b\equal{}x$ gives $ c\equal{}\minus{}2x$ and thus $ A \equal{} 4x^6 \minus{} 2 \cdot 4 \cdot 8 x^6 \plus{} 15 \cdot 4 x^6 \equal{} 0$. Thus $ a\minus{}b \mid A$. And since $ A$ is symmetric we get $ (a\minus{}b)^2 \mid A$. In the same way $ (b\minus{}c)^2 \mid A$ and $ (c\minus{}a)^2 \mid A$.

Now $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \mid A$. $ A$ have degree $ 6$, and no matter how we substitute with $ a\plus{}b\plus{}c\equal{}0$ it will still have degree $ 6$. Then $ A \equal{} k \cdot (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$. Setting $ a\equal{}1$, $ b\equal{}2$, $ c\equal{}\minus{}3$ we get $ A \equal{} k \cdot (\minus{}1)^2 \cdot 5^2 \cdot (\minus{}4)^2 \equal{} k \cdot 400$. And $ A \equal{} 4a^3b^3 \plus{} 4b^3c^3 \plus{} 4c^3a^3 \plus{} 15a^2b^2c^2 \equal{} \minus{}400$. Thus $ k \cdot 400 \equal{} \minus{}400 \iff k \equal{} \minus{}1$. And thus $ A \equal{} \minus{}(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$. So $ P \equal{}  \frac{\minus{}(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{4a^2b^2c^2} \le 0$ for all $ a,b,c \neq 0$ such that $ a\plus{}b\plus{}c\equal{}0$. So

$ \frac{ab}{c^2} \plus{} \frac{bc}{a^2} \plus{} \frac{ca}{b^2} \le \frac{\minus{}15}{4}$


Princeton
college
floor function
algebra
polynomial
number theory
modular arithmetic

Do denominators need to be rationalized?

Also what would be the factors of zero if we go by the positive integer definition?

Denominators should be rationalized and fractions should be put in lowest form. You shouldn't have a 0 in the denominator, and if an answer is 0 (i.e. in the numerator), just write 0. So I have to admit I don't understand what you're asking about 0.

Note: we're not going to be horribly tyrranical about answer format. If there's a question where it's somehow unclear about what format it should be in, then we will accept ones that could reasonably defended as "simplest" form. For example, if we don't tell you how to write an answer and it is
$\frac{\sqrt{5}+1}{2}$
and you write
$\frac{\sqrt{5}}{2}+\frac{1}{2}$
I don't see how we could fairly count your answer wrong.

Does this make our policy clear? If not, please ask whatever other questions you have.

Okay so I guess what I was asking was in terms of, say, $\frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$ for the first question.

For the second question, it's like, is for example say $36$ a factor of $0$?

[quote="JSteinhardt"]Okay so I guess what I was asking was in terms of, say, $\frac{1}{\sqrt{3}}$ or $\frac{\sqrt{3}}{3}$ for the first question.[/quote]
Preferably $\frac{\sqrt{3}}{3}$. But as I said, we will try to specify the answer format when it isn't obvious (such as an integer), and in case we don't, we will most likely accept anything that's not too outlandish (i.e., the grader is sure by looking at it and not thinking too hard that it's right).

[quote="JSteinhardt"]For the second question, it's like, is for example say $36$ a factor of $0$?[/quote]
That is in fact a good question. I don't know whether it is conventional in professional mathematics to count any natural number as a divisor of 0, but we'll try to avoid the question entirely (i.e., we'll specify positive integers when 0 would present an unenlightening but confusing extra case).

Okay cool thanks for the clarifications.

I am reasonably sure that the definition fluctuates whimsically among professional mathematicians, which is why I wanted to find out which one Princeton was using, but I guess avoiding the problem altogether is probably best.

Actually the problem came up from this question:

For how many integers $n$, $0 \leq n < 144$, is $144$ a factor of $n^{2}-36$?

It seemed extremely innocent until some people answered $11$ and the rest answered $12$, which caused somewhat of a mess and lead me to find out what the policy would be at Princeton just in case the same thing happened.

That would be easily avoided if 36 were replaced by some non-square... :P So as you can see, we should be able to circumvent the problem by not having any questions where it would come up.

Well the problem is a bit messier if $36$ isn't a square.

However it could've also been avoided by being like find all $n$ such that $n^{2}= 144\lfloor\frac{n^{2}-36}{144}\rfloor+36$.

Why would any integer [i]not[/i] be counted as a divisor of zero?  It fits the definition :  for any integer $n$, there exists an integer $m$ (which coincidentially happens to be zero all the time) such that $mn = 0$.

neelnal :  Factor. :wink:

[quote="Boy Soprano II"]Why would any integer [i]not[/i] be counted as a divisor of zero?[/quote]
Mainly because saying something "is a divisor of 0" is kinda meaningless. Also, then we can no longer say in general that $a | b \implies a \le b$. It's just a question of "where do we start talking about 0 as a special case." No deep mathematical meaning.

[quote="Xevarion"]Mainly because saying something "is a divisor of 0" is kinda meaningless. Also, then we can no longer say in general that $a | b \implies a \le b$. It's just a question of "where do we start talking about 0 as a special case." No deep mathematical meaning.[/quote]
But you can't do this anyway unless you restrict yourself to positive integers.  And most expressions regarding zero are kind of meaningless.  We count zero as an even number, after all, don't we?  And isn't it congruent to zero mod every integer?  I understand about treating zero as a special case, but I think that "every integer divides zero" is special enough, and more consistent, too.

Boy Soprano: Actually it was something more like $6$ divides $n$, and $12$ does not divide $n$. Or you can Modular Arithmetic and CRT but that is much uglier.

Also, for polynomial answers, will it be fine to write, say,
(x+1)^2 instead of x^2+2x+1?

It should be fine. Again, we'll try to make it unambiguous, and if it's unclear, we'll take anything that's not too weird and is right. :P

I don't know what you guys are talking about...
If I'm in the grading room at the time, I will absolutely NOT accept $\frac{1}{\sqrt{3}}$.  Denominators MUST be rationalized!

Everything (except 0) is a divisor of 0.

wait seriously?

also $a | b \Leftrightarrow \exists k \in \mathbb{Z}, b = ka$; it seems like $0 | 0$


B and C are distinct digits in the 6-digit number N:  N = 1B3, 45C.  If N is divisible by 15, how many different numbers could N be?

A.  3	B.  4	C.  5	D.  6	E.  7

[hide=" click to reveal hidden content"] $ 7:$

$ 5 \mid N \implies C=0$ or $ 5$.

$ C=0 \implies B= 2, 5$ or $ 8$, since $ 3%Error. "midN" is a bad command.
$

$ C=5 \implies B= 0, 3, 6$ or $ 9$, since $ 3 \mid N$.

Therefore $ E: 7$.

[/hide]


abstract algebra
algebra
polynomial
function
domain
vector
group theory

Find the kernel and range of each of the linear operators on P3:
L(p(x)) = xp'(x)
L(p(x)) = p(x) - p(x)'
L(p(x)) = p(0)x + p(1)

My prof did these examples in class but didn't work through them, so I know that the first is ker(L)= P1, L(P3) = Span(x^2, x) etc. But I have no clue how to solve any of these step by step???

Also if you have any idea how to show that L: V->W is one-to-one if and only if       ker(L) = {0v}, I'm guessing it has to do with if L(v1)=L(v2) then v1=v2 ???

Your examples:
The first has a dimension-1 kernel; the constants. The image is all polynomials in your space which are divisible by $x$.
The second is bijective- trivial kernel, image is the whole space. The inverse operation $L^{-1}$ is $L^{-1}(p)=p+p'+p''+\dots$ (this sum terminates for any polynomial).
The third has a 2-dimensional range (the constant and linear polynomials, and its kernel is the set of all polynomials divisible by $x(x-1)$.

A general algebraic fact: A group homomorphism has trivial kernel if and only if it is injective.
The proof: If $f$ is injective, $f^{-1}(e)$ can contain at most one element ($e$ is the group identity, in either domain or codomain). Since $f(e)=e$ always, $f^{-1}(e)=\{e\}$ and we are done.
Conversely, if $f^{-1}(e)=\{e\}$, and $f(x)=f(y)$, we have $f(x)(f(y))^{-1}=e$, so $f(x)f(y^{-1})=f(xy^{-1})=e$ and $xy^{-1}=e$. Multiplying on the right by $y$, $x=y$ and $f$ is injective.

Since nearly all algebraic structures, including vector spaces, are groups with extra properties, this holds in those structures as well. In a vector space, we would state this using addition instead of "multiplication".
The idea of this proof is also important; if something's true at the origin, we can translate it to anywhere in the space.


solve for the x

it's 2 right triangles.

HINT:

try to solve this by using similar triangels....

ok....

let y be the distance between those vertical lines , and let c be the distance between first vertical line and x

now we have

603/y=x/(y-c)

951/y=x/c

we get

xy=951c=603y-603c

1554c=603y

c=(201/518)y

x=(201/518)951


geometry incenter circumcircle inradius inequalities

consider a triangle ABC.Let I be its incenter and let M,M,P be the mid points of the triangle's sides.Prove that
               IM^2+IN^2+IP^2>_r(R+r)
   where R and r are the circumradius and inradius respectively.
please prove it.

[hide]In this proof I'll use the identity $ ab \plus{} bc \plus{} ca \equal{} r^2 \plus{} s^2 \plus{} 4Rr$, which is proven [url=http://www.mathlinks.ro/viewtopic.php?p=1498406#1498406]here[/url].

Let $ M$ be the midpoint of $ AB$, and $ I_c$ the foot of the perpendicular from $ I$ to $ AB$. Then

$ I_cM \equal{} |AI_c \minus{} AM| \equal{} \left|s \minus{} a \minus{} {c\over 2}\right| \equal{} \left|{a \plus{} b\over 2} \minus{} a\right| \equal{} \left|{a \minus{} b\over 2}\right|$

Hence $ IM^2 \equal{} r^2 \plus{} \left({a \minus{} b\over 2}\right)^2$, and similarly for the other two.

The desired sum $ \sigma$ is therefore

$ \sigma \equal{} 3r^2 \plus{} {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca\over 2}$

$ \sigma \equal{} 3r^2 \plus{} {(a \plus{} b \plus{} c)^2 \minus{} 3(ab \plus{} bc \plus{} ca)\over 2}$

$ \sigma \equal{} 3r^2 \plus{} 2s^2 \minus{} {3\over 2}(r^2 \plus{} s^2 \plus{} 4Rr)$

$ \sigma \equal{} {3r^2\over 2} \plus{} {s^2\over 2} \minus{} 6Rr$

Hence we have to prove

$ {3r^2\over 2} \plus{} {s^2\over 2} \minus{} 6Rr\geqslant r^2 \plus{} Rr\iff r^2 \plus{} s^2\geqslant 14Rr\iff ab \plus{} bc \plus{} ca\geqslant 18Rr$

Using $ R \equal{} {abc\over 4P}$ and $ r \equal{} {P\over s}$ where $ P$ is the area of the triangle, the last inequality becomes

$ ab \plus{} bc \plus{} ca\geqslant {9abc\over 2s} \equal{} {9abc\over a \plus{} b \plus{} c} \equal{} \frac {9}{{1\over ab} \plus{} {1\over bc} \plus{} {1\over ca}}$

but that's just AM-HM, so the initial inequality is proven. The equality holds for $ ab\equal{}bc\equal{}ca$ i.e. for an equilateral triangle.[/hide]


geometry
incenter
trigonometry
angle bisector
IMO Shortlist

Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG \equal{} DE$. Prove that $ CA \equal{} CB$.

[i]Original formulation:[/i]

Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF \equal{} DE,$ prove that $ AC \equal{} BC.$

[quote="Fermat -Euler"]Let $ABC$ be a triangle, and let the angle bisectors of its angles $CAB$ and $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $F$, respectively. The lines $AD$ and $BF$ meet the line through the point $C$ parallel to $AB$ at the points $E$ and $G$ respectively, and we have $FG = DE$.
Prove that $CA = CB$.[/quote]

Let a=BC, b=CA, c=AB.
We have BF/AD=(BF/FG)/(AD/DE)=(c/a)/(c/b)=b/a, but it's easy to see from the formulae of the type la=2bc*cos(A/2)/(b+c) that the order of the bisectors of a triangle is inverse to that of the sides (if a>=b>=c then la<=lb<=lc). If b>a then BF/AD=b/a>1, which is false, so b<=a => contradiction. We do the same for b<a, so the only possible situation is a=b, Q.E.D.

[hide="another solution"]WLOG assume $ a\ge b$ and hence $ \angle A\ge \angle B$. Let $ H$ be on segment $ FG$ such that $ DH\parallel AB$ (easy to see that such point exist because $ a\ge b$) and $ I$ be on $ CG$ such that $ DI\parallel BG$. We have $ DI\equal{}HG\ge FG\equal{}ED$, so $ \angle DEI\ge \angle EID$. But $ \angle DEI\equal{}\angle DAB\equal{}\frac12\angle A\ge\frac12\angle B\equal{}\angle GBA\equal{}\angle BGE\equal{}\angle DIE$. So equalities hold everywhere and we get $ CA\equal{}CB$.

Q.E.D. :D[/hide]

Let $ I$ be the incenter, and let $ CI$ intersect $ AB$ at $ P$. Since $ GE \parallel AB$, we have $ \triangle ABD \sim \triangle DEC$ and $ \triangle ABF \sim \triangle FCG$. Thus,
$ \frac {BD}{DC} \equal{} \frac {AD}{DE}$ and $ \frac {AF}{FC} \equal{} \frac {BF}{FG}$. Since $ DE \equal{} FG$, dividing the two we have
$ \frac {BD}{DC} \cdot \frac {CF}{FA} \equal{} \frac {AD}{BF}$. But by Ceva, $ \frac {BD}{DC} \cdot \frac {CF}{FA} \equal{} \frac {PB}{PA} \equal{} \frac {BC}{CA}$ (angle bisector theorem). So $ \frac {BC}{CA} \equal{} \frac {AD}{BF}$. But if $ BC \neq CA$, then this would mean that  lengths of the angle bisectors are in the same order as the lengths of the sides they intersect, a contradiction by the generalization of the Steiner-Lehmus Theorem in the link [url]http://www.cems.uvm.edu/~cooke/sl.pdf[/url]. Thus, $ BC \equal{} CA$.

Never seen such an easy geometry in the ISLs atleast.

Wlog let $a \ge b$.Note that $CF=\frac{ab}{a+c},\angle{GCA}=A,\angle{FGC}=\frac{B}{2}$

So application of sine rule in $\triangle{CFG}$ yeids

$\frac{GF}{sinA}=\frac{ab}{(a+c)sin\frac{B}{2}}$

Similarly sine rue in $\triangle{CDE}$ yeilds

$\frac{ED}{sinB}=\frac{ab}{b+c}$

Dividing the two relations we get

$\frac{(a+c)b}{(b+c)a}=\frac{sin\frac{A}{2}}{sin\frac{B}{2}} \ge 1$

by our assumption.

This gives

$\frac{(a+c)b}{(b+c)a} \ge 1$

$\Rightarrow ab+bc \ge ab+ac \Rightarrow b \ge a$

So we obtain $b=a$

[quote=WolfusA]OK, maybe some logical order. Everyone up, who made assumption that $a\geq b$ and didn't check the case $a<b$ has only a half of solution. By the way you can't have an assumption that $a=b$, so you can't make the case $a\geq b$, because you have to prove, that $a=b$. How can the proof be correct, when you prove something supposing it. Correct solution is when you show, that both inequalities $a>b$ and $a<b$ are contradiction, so there must be $a=b$.[/quote]

Your first complaint is incorrect because you can make the assumption $a\ge b$ because the two cases are exactly symmetric, hence the "Wlog" (just flip your diagram). In addition, your second complaint is incorrect because supposing $a\ge b$ is not the same thing as $a=b$. Obviously at least one of $a\ge b, a\le b$ is true, and since they are symmetric you may assume one of them. You should learn some logic yourself before criticizing others.

[quote=WolfusA]I don't mean that WLOG part is incorrect.
So if you like so this equality in assumption check yourself that this $p \vee q \implies p$ is tautology. And put here $p: a=b$ and $q: a>b$. Anyway the second case (after WLOG) is $a<b$ (not $b\geq a)$.[/quote]

I'm not sure exactly what you're saying (I'm pretty sure the first thing you said isn't a sentence), but $p \vee q \implies p$ is not a tautology. $a\ge b$ does not always imply $a=b$, that is absurd. Also, the point of WLOG is that there are no cases, we only have to check one of the cases by symmetry. The reason I said $b\ge a$ rather than $b>a$ is to preserve the symmetry, so that the WLOG is valid.

Sorry, but I made a mistake with this "tautology". I withdraw from accusations.

If $AC \neq BC$, WLOG, assume that $\angle CAB > \angle CBA$, then we have \[BC > AC, BG > AE.\] Since $GF=DE$, we have $FB=BG-GF >AE-GF=AE- DE=DA$, so \[GF=\frac{FB \cdot GC}{AB}=\frac{FB \cdot BC}{AB} > \frac{DA \cdot AC}{AB}=\frac{DA \cdot CE}{AB}=DE\]  which contradicts $GF=DE$, thus, $AC=BC$. $\blacksquare$


linear algebra
matrix
algebra
polynomial

Hey,

can anyone help with the following question please?

A square matrix A is called nipotent if A^n = 0 for some n >=1. Find all nilpotent diagonalizable matrices.

I arranged the terms as such: A^n = PD^nP-1
But I only found one: the zero matrix. Is there more and how should i go about finding it?

Thanks!

There's only the zero matrix. All other nilpotent matrices are not diagonalizable.

consider the minimal polynomial, you will get the result


From the conditional we have :
$ 2na_n = (2n - 3)a_{n - 1}$
$ \Leftrightarrow 2(n - 1)a_{n - 1} - 2na_n = a_{n - 1}$
And so
$ \sum_{i = 1}^n a_i = \sum_{i = 1}^n ({2ia_i - 2(i + 1)a_{i + 1}) = 2a_1 - 2(n + 1)a_{n + 1}}$
From the condion easy to check that $ a_n > 0$ so
$ 1 - 2(n + 1)a_{n + 1} < 1$
It follow that :
$ \sum_{i = 1}^n a_i< 1,\forall n \in N$

$ \sum_{n\equal{}1}^{\infty} a_nx^n\equal{}1\minus{}\sqrt{1\minus{}x}$, therefore $ \sum_{n\equal{}1}^{\infty}a_n\equal{}1.$


[quote="ch1n353ch3s54a1l"]For integers $k$ and $n$ such that $1\le k<n$, let $_{n}C_{k}=\frac{n!}{k!(n-k)!}$. Then $(\frac{n-2k-1}{k+1})_{n}C_{k}$ is an integer

(A) for all $k$ and $n$
(B) for all even values of $k$ and $n$, but not for all $k$ and $n$
(C) for all odd values of $k$ and $n$, but not for all $k$ and $n$
(D) if $k=1$ or $n-1$, but not for all odd values of $k$ and $n$
(E) if $n$ is divisible by $k$, but not for all even values of $k$ and $n$[/quote]

[hide]
I think the answer is $\boxed{A}$

$(\frac{n-2k-1}{k+1})(_{n}C_{k})\rightarrow (\frac{n-2k-1}{k+1}) (\frac{n!}{k!(n-k)!}) \rightarrow (n-2k-1)(\frac{n!}{(k+1)!(n-k)!})$

$(n-2k-1)(\frac{n!}{(k+1)!(n-k)!}) \rightarrow \frac{n-2k-1}{n+1}\cdot \frac{(n+1)!}{(k+1)!(n-k)!}) \rightarrow (\frac{n-2k-1}{n+1}) (_{n+1}C_{k+1}) \rightarrow ( 1+\frac{-2k-2}{n+1}) (_{n+1}C_{k+1})$

If we distribute, we get $(_{n+1}C_{k+1})+\frac{-2k-2}{n+1}\cdot \frac{(n+1)!}{(k+1)!(n-k)!}$

If we factor out $\frac{n+1}{k+1}$ from $\frac{(n+1)!}{(k+1)!(n-k)!}$ and multiply it with $\frac{-2k-2}{n+1}$, we get -2. So, we have $(_{n+1}C_{k+1})+(-2)\frac{(n!)}{(k!)(n-k)!}\rightarrow (_{n+1}C_{k+1})-(2) (_{n}C_{k})$

[/hide]

EDIT #35421415123124 : Man, I edited this post a lot of times because I screwed up on basic arithmetic in the beginning, which screwed up the whole solution. I didn't quite double check, but I think this is right.

EDIT: I think I got it now

[quote="Zellex"][quote="ch1n353ch3s54a1l"]For integers $k$ and $n$ such that $1\le k<n$, let $_{n}C_{k}=\frac{n!}{k!(n-k)!}$. Then $(\frac{n-2k-1}{k+1})_{n}C_{k}$ is an integer

(A) for all $k$ and $n$
(B) for all even values of $k$ and $n$, but not for all $k$ and $n$
(C) for all odd values of $k$ and $n$, but not for all $k$ and $n$
(D) if $k=1$ or $n-1$, but not for all odd values of $k$ and $n$
(E) if $n$ is divisible by $k$, but not for all even values of $k$ and $n$[/quote]

[hide]
Since $_{n}C_{k+1}$ is always an integer, we ignore it, for it satisfies the condition.

Thus, we want $\frac{n-2k-1}{n-k}$ to be an integer.

$\frac{n-2k-1}{n-k}\rightarrow 1+\frac{-k-1}{n-k}$

[/hide]

[/quote]

How do we know that if $\frac{n-2k-1}{n-k}$ is not an integer that it can't be made an integer by multiplying it by $_{n}C_{k+1}$? :ninja:

How about

\begin{eqnarray*}\frac{n-2k-1}{k+1}\binom{n}{k}&=& \left(\frac{n-k}{k+1}-1\right) \binom{n}{k}\\ &=& \frac{n-k}{k+1}\binom{n}{k}-\binom{n}{k}\\ &=& \binom{n}{k+1}-\binom{n}{k}\end{eqnarray*}


trigonometry
analytic geometry
ratio
geometry
angle bisector

Well i suppose you could use the fact that it's a curve to show that that is true, but...how would you prove it...this i have to think about
Interesting problem btw!

hmm. this problem is awfully familier. i've seen it before. there's a theory....

Let $O$ be the coordinate origin of the xy-coordinate system and $(O, r = 1)$ the unit circle intersecting the positive x-axis at a point $U$, i.e., OU = 1. Let $t$ be the tangent to the unit circle at the point $U$, i.e., $t \perp OU \equiv$ x-axis. WLOG, assume that $x_2 \ge x_1$. Let $P_1, P_2$ be the intersection of rays $OP_1, OP_2$ with the tangent $t$, such that $\measuredangle UOP_1 = x_1$, $\measuredangle UOP_2 = x_2$. Then

$UP_1 = \tan{x_1} \le \tan{x_2} = UP_2$

$OP_1 = \sqrt{OU^2 + UP_1^2} = \sqrt{1 + \tan^2{x_1}} \le$

$\le \sqrt{1 + \tan^2{x_2}} = \sqrt{OU^2 + UP_2^2} = OP_2$

Let $OM$ be the median of the triangle $\triangle OP_1P_2$ from the vertex $O$ to the side $P_1P_2$ and $OB$ the bisector of the angle $\measuredangle P_1OP_2$. $M$ is the midpoint of the segment $P_1P_2$ and

$P_1M = MP_2$

$UM = \frac{UP_1 + UP_2}{2} = \frac{\tan{x_1} + \tan{x_2}}{2}$

$\measuredangle UOB = \frac{x_1 + x_2}{2},\ \ UB = \tan{\frac{x_1 + x_2}{2}}$

Since the angle bisector $OB$ cuts the side $P_1P_2$ in the ratio of the adjacent sides $OP_1, OP_2$,

$\frac{P_1B}{BP_2} = \frac{OP_1}{OP_2} \le 1,\ \ P_1B \le BP_2\ \ \Rightarrow\ \ P_1B \le P_1M$

$\tan{\frac{x_1 + x_2}{2}} = UB = UP_1 + P_1B \le UP_1 + P_1M = UM = \frac{\tan{x_1} + \tan{x_2}}{2}$


MATHCOUNTS
algorithm
number theory
divisibility tests

Is there a way to determine if a number in base n is a perfect square in base 10 without actually converting to base 10? For example how do you dteremine if $ 1111_4$ is a perfect square in base 10?

The property of being a perfect square is independent of what base you choose to express a number in.  It's just a matter of where it's more convenient to perform your calculation.

I don't think I quite get what your saying. so there is no way to tell that 10000 base 3 is a perfect square without performing the equivalent of a square root in base 3 or converting to base 10 and find out. speaking of that, how do you perform a square root in base three?

How would you perform a square root in base A? (A = 10, avoiding 10 here to emphasize that the base we choose doesn't matter and that "10" is not a special number).

10000 base 3 = $ 3^4$, and that is indeed a perfect square. In fact, for any base $ x$, 10000 base $ x$ is a perfect square because that number is equal to $ x^4$, which is a perfect square.

To determine whether $ 1111_4$ is a perfect square, express it as $ 4^3 \plus{} 4^2 \plus{} 4 \plus{} 1 \equal{} (4 \plus{} 1)(4^2 \plus{} 1)$, which is not a perfect square. The only way to compute this efficiently (without resorting to number theory tricks) is to calculate this number in base 10 and "compare" it to your "list of known perfect squares". In this case, $ 5 \cdot 17 \equal{} 85$ is not a perfect square.

Actually, the only values for which $ x^3 \plus{} x^2 \plus{} x \plus{} 1$ is a perfect square is $ x \equal{} 1$ or $ 7$, but base 1 does not make sense, so $ 1111$ is a perfect square only in base 7.

I'm honestly curious - how do you, personally, tell if a number is a square if you are given it in base $ 10$, and why doesn't your method extend to other bases?

I usually just factor it and look at its prime factors. That's about thte only way I know how to do it. Or if I have a calculator handy I just type it in. Why do you ask, is there a different way to do this? I'm not sure how to apply this method to other bases. Do you just do base number division?

For smaller numbers, I can usually "tell" whether it is a perfect square. If it is a larget number ($ > 10000$), I resort to tricks such as, the last digit cannot be 2, 3, 7, 8, or if it is divisible by some prime number $ k$, it must be divisible by $ k^2$.

I guess for other bases, you need to be familiar with the properties of perfect squares before you can do those tricks.

[quote="ommfgrotfl"]I usually just factor it and look at its prime factors.[/quote]
Do you know how to divide in other bases?  It's the same.  Everything depends on how good you are at doing arithmetic in bases other than $ 10$.

[quote="ommfgrotfl"]Why do you ask, is there a different way to do this? [/quote]
If you went through Mathcounts, you probably memorized the first 26 or 33 perfect squares.  What you can do is lop off two digits from the end of the number until you reach a number that's "close" to a perfect square and then square a few numbers that look right to check.  (Of course, you should check last digits and remainders first.)

For example, if you wanted to figure out whether $ 16777216$ was a perfect square, lop off the last six digits to get $ 16$, which tells you that the square root is greater than $ 4000$.  Lopping off the last four digits gets you $ 1677 < 41^2$, which tells you that the square root is less than, but close to, $ 4100$.  Checking the last digit tells you it needs to end in six, so guess $ 4096$, which is correct.

In other bases, you should be familiar with what last digits are possible and what divisibility tests are easy to do.  The general principle is that if you are in base $ b$ the divisibility tests with respect to any divisor of $ b, b \plus{} 1, b \minus{} 1$ are the easy ones; this generalizes what happens in base $ 10$.  But since the above algorithm requires some familiarity with perfect squares to work properly, you're better off converting to base $ 10$.  There's not much of a point learning how to do this in every base.


Suppose that the matrix of a linear map $f: \mathbb{C}^3 \rightarrow \mathbb{C}^3$ (with respect to a well-chosen basis) is the following Jordan matrix: \[ \begin{pmatrix} 2 && 0 && 0 \\ 1 && 2 && 0 \\ 0 && 0 && 2 \end{pmatrix}.  \] We could also regard $\mathbb{C}^3$ as a REAL vector space (of dimension 6). Which basis of $\mathbb{C}^3$ (as a vector space over $\mathbb{R}$) could we choose such that the matrix $f$ is in its Jordan form with respect to this basis? And which Jordan matrix do we get then?


quadratics
calculus
inequalities unsolved
inequalities

[color=darkblue]1) Let $ x,y\in R$. Find minimum of $ A \equal{} x^2 \plus{} 5y^2 \plus{} 4xy \minus{} 6x \plus{} 5y \minus{} 9$

2) Let $ x,y\in R$. Find maximum of $ B \equal{} \minus{} x^2 \minus{} 10y^2 \plus{} 6xy \plus{} 4x \minus{} 3y \plus{} 2$

3) Let $ x,y\in R$. Find maximum of $ C \equal{} \minus{} x^2 \minus{} 4y^2 \plus{} 2xy \plus{} 2x \plus{} 10y \minus{} 3$[/color]

[quote="ibrahimsenturk"][url=http://imageshack.us][img]http://img103.imageshack.us/img103/437/yerlminkk8.gif[/img][/url]
[URL=http://g.imageshack.us/g.php?h=103&i=yerlminkk8.gif][IMG]http://img103.imageshack.us/img[/quote]

[color=darkblue]Thank you very much. But I think we can done it by other way?[/color]

We want to find the minimum of $ (x \plus{} 2y \minus{} 3)^2 \plus{} y^2 \plus{} 17y \minus{} 18$, so we minimize the quadratic $ y^2 \plus{} 17y \minus{} 18$ and set x so x+2y-3=0. The minimum of that quadratic is when $ y \equal{} \minus{} 17/2$, hence $ x \equal{} 20$. No calculus is needed. The others can be solved similarly.

[color=darkblue]Thank you very much.[/color]


function
induction
algebra unsolved
algebra

Hello, I am not too a "math student", but I would like to have of the assistance for this exercise.

A function f is defined over the set of all positive integers and satisfies.
 f(1) = 1996.
f(1)+f(2)+..+f(n) = n²f(n) for all n>1

Calculate the exact value of f(1996)

Thanks.

Set f(1) = a.  We can re-write the given equation as $f(n) = \frac{f(1) + f(2) + \ldots + f(n - 1)}{n^2 - 1}$.  Calculate the values f(1), f(2), f(3), f(4), f(5) in terms of a.  The pattern should be clear at that point: $f(n) = \frac{2a}{n(n+1)}$.  We will prove it by induction.  Assume it true for n = k so $f(k) = \frac{2a}{k(k+1)}$.  $f(k+1) = \frac{f(1) + \ldots + f(k - 1)+f(k)}{(k+1)^2-1}=\frac{(k^2-1)f(k) + f(k)}{k(k+2)}$ by the recursion.  Thus, by the inductive hypothesis, $f(k+1) = \frac{k}{k+2}f(k) = \frac{2a}{(k+1)(k+2)}$, so by induction the theorem holds and $f(1996) = \frac{2\cdot1996}{1996\cdot1997} = \frac{2}{1997}$.

*Edited for a minor clerical error at the end.*


[quote="infinity"]Hello, I have a problem with this problem could you please help me out?

When a certain positive integer N is divided by a positive integer d, the remainder is 7. If 2n + 3 is divided by d, the remainder is 1. Find all possible values of d.

I have only one solution so far and it is D=16 and N=23...
is there any more and what are they?

Thanks heaps[/quote]
[hide]All variables are natural numbers.
n = d*a+7  <=>  2n = 2*d*a + 14
2n + 3 = d*b + 1  <=>  2n = d*b - 2

So: 2*d*a + 14 = d*b - 2  <=>  d*(b - 2*a) = 16

So the possibilitys for d are the divisors of 16: d = 1, d = 2, d = 4, d = 8, d =16.

You should check if alle these d's are indeed solutions, but this wont't be too difficult.
For example: d = 1 => b = 2*a + 16. Suppose a = 1 and b = 18 => n = 8. Indeed: 8 = 1*1+7 and 2*8 + 3 = 19 = 1*18 + 1. [/hide]

[quote="infinity"]Thankyou for answering my question....

but there seems to be a problem here, you said that 1, 2, and 4 would work. They don't! However 8 and 16 do work. 1,2 and 4 do not work because if d=2 then n=9, this wont work because 9:2 equals 4 remainder 1, not remainder 7. Similarly for 1 and 4. Anyway i will work out the other numbers by myself...thanks for the help![/quote]

Oops, ofcourse you can't have a remainder 7 if you divide by 1, 2 or 4. Stupid of me.  :o  ;)

Happens often to me. I think, "ok I have it", but I forget the last trivial step...


abstract algebra
superior algebra
superior algebra solved

We saw today a quite ugly and technical proof of the following result: over a local ring, all projective modules finitely generated are free. Is there a nice solution?  :maybe:

You can find a proof in Matsumura. If your proof is shortest tell me a reference or just write me some.

You can also look here http://www.mathlinks.ro/Forum/viewtopic.php?t=122188&highlight=module+projective


given that $ \frac{3a^2\plus{}4}{4a\plus{}3}\equal{}k$
the question is, is $ k$ an interger/does there exist any $ k$ that is integral?

[quote="21 7 15"]given that $ \frac {3a^2 \plus{} 4}{4a \plus{} 3} \equal{} k$
the question is, is $ k$ an interger/does there exist any $ k$ that is integral?[/quote]

When $ a\equal{}1$, $ k\equal{}1$.  Therefore, yes, there is a value of $ k$ that is integral.  (Not very mathematical, it was the first thing that struck me.)

its actually, $ \frac{3a^2\plus{}2}{4a\plus{}3}\equal{}k$
this is the question now is $ K$ integral :roll:

[hide="Full Solution"]$ \frac{3a^2\plus{}4}{4a\plus{}3} \equal{} \frac{1}{16}\left(12a\minus{}9\plus{}\frac{91}{4a\plus{}3}\right)$.

Therefore, it is a necessary (yet not necessarily sufficient) condition that $ 4a\plus{}3$ divides $ 91$. The only divisors of $ 91$ are $ 1$, $ 7$, $ 13$, and $ 91$, which correspond to $ a$ values of $ \minus{}\frac{1}{2}$, $ 1$, $ \frac{5}{2}$, and $ 22$. We care only for the values where $ a$ is an integer.

$ a \equal{} 1$ yields $ k\equal{}1$.

$ a \equal{} 22$ yields $ k \equal{} 16$.

The above two solutions are the only two solutions, since any other value of $ a$ yields a nonintegral value for $ \frac{91}{4a\plus{}3}$, which correlates to a noninteger value for $ k$.

[/hide]

in my second post i made one change to the problem

[hide="solution to modified problem"]
I use the same methods as darkdieuguerre did in his earlier post on the unmodified problem.

$ \frac{3a^{2}\plus{}2}{4a\plus{}3}\equal{}\frac{3}{4}a\minus{}\frac{9}{16}\plus{}\frac{59}{16(4a\plus{}3)}\equal{}\frac{1}{16}\left ( 12a\minus{}9\plus{}\frac{59}{4a\plus{}3} \right )$

So, it is necessary for $ 59$ to divide $ 4a\plus{}3$. As $ 59$ is prime, it's only factors are $ 1$ and $ 59$. The corresponding values of $ a$ are $ \frac{\minus{}1}{2}$ and $ 14$.

We need to check our solutions, because (as darkdieuguerre pointed out) dividing $ 59$ is not sufficient to ensure that $ k$ is an integer.
When $ a\equal{}14$ (the only integer solution), we get,
$ k\equal{}\frac{3*14^{2}\plus{}2}{4*14\plus{}3}\equal{}10$

So, the only solution is $ \boxed{a\equal{}14, k\equal{}10}$.
[/hide]


I know that, when I was on a Unix system, there was a latex2html command that generated webpages from one's documents.  I see that there are pages on converting LaTeX to pdf documents, but nothing on converting the docs to webpages (and - hey- Word with Equation Editor can convert to html)

So... does Miktex include this feature?  If not, does latex2html exist for windows somewhere?  And if so, I recommend adding info on that to the LaTeX pages.  HTML is far more accessible than pdf's.

Here's a link I found via google:
[url]http://www.mayer.dial.pipex.com/l2h.htm[/url].

I agree that LaTeX looks better in PDF than in html.

There are a number of programs that will convert to HTML. You'll find comparisons of their advantages and disadvantages at http://www.mayer.dial.pipex.com/samples/comparisons.htm and some examples of what they produce at http://www.mayer.dial.pipex.com/samples/example.htm

These programs are not part of MiKTeX (even though Tex4ht looks as if it is, but it's only a few files); they all have Windows versions. Click on their names in the comparisons page to find their source.

[quote="rrusczyk"]We explain how we get LaTeX on our site [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_HelpFAQ.php#mysite]here[/url], but that won't address converting a whole TeX document to html.  Anyone know how to do this?  (Though I do think that for longish documents, 2-3 pages or more, PDF is more reader-friendly)[/quote]
Please feel free to copy and use whatever you want from my pages listed above and from the home page at http://www.mayer.dial.pipex.com/tex.htm


modular arithmetic
search
function
combinatorics proposed
combinatorics

An even number of people have a discussion sitting at a circular table. After a break they sit down again in a different order. Show that there must be two people with the same number of people sitting between them before and after the break.

Let $ n$ be a positive integer, $ 2n$ be the number of people sitting around that circular table, and number the people as well as their chairs with the integers from $ 1$ to $ 2n,$ so that two consecutive integers represent two consecutive chairs.

Let $ b_i$ be the number of the chair of the $ i \minus{}$th person before the break and $ a_i$ the one after.
From now on, $ a_i$ and $ b_i$ will be considered as integers $ \mod 2n.$
Let $ d_i \equal{} a_i \minus{} b_i.$

Assume the $ d_i$ are pairwise distinct.
Since there are $ 2n$ of them and $ 2n$ possible remainders $ \mod 2n,$ any value is taken exactly once, so that :
$ \sum_{i \equal{} 1}^{2n} d_i \equiv \sum_{i \equal{} 1}^{2n} i \equiv n \pmod {2n}.$
But, similarly, $ \sum_{i \equal{} 1}^{2n} b_i \equiv n \equiv \sum_{i \equal{} 1}^{2n} a_i \pmod {2n},$ which gives a contradiction to the above congruence.

Thus, we may find two distinct indices $ j$ and $ k$ such that $ d_j \equiv d_k \pmod {2n},$ meaning that the people with respective numbers $ j$ and $ k$ are the sought people.

EDIT : ok, after using the search function, I see that this problem comes from the [url=http://www.mathlinks.ro/viewtopic.php?t=60414]ISL 1988[/url], and that my solution is the same as the one presented on [url=http://web.archive.org/web/20041210120507/www.kalva.demon.co.uk/short/soln/sh8831.html]kalva.[/url]


$ \begin{cases} \frac{xy}{x\plus{}y}\equal{}1\minus{}z\\\frac{yz}{y\plus{}z}\equal{}2\minus{}x\\\frac{zx}{z\plus{}x}\equal{}2\minus{}y \end {cases}$

We have $ xy\plus{}yz\plus{}zx\equal{}x\plus{}y\equal{}2(y\plus{}z)\equal{}2(x\plus{}z)$. The equality of the last two gives $ x\equal{}y$, then equating the second and third gives $ z\equal{}0$. Finally, plug into the last two original equations to get $ x\equal{}y\equal{}2$, which also satisfies the first equation. Therefore the only solution is $ \boxed{(2,2,0)}$.


Suppose the sequence $a_n$ satisfies
$a_n=a_1a_2\cdots a_{n-1}+1$ for every $n\ge2$ and $a_1=1$
Find the least $M$ for which
$\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_m}<M$ for every $m$.


Am o carte mai veche cu probleme de la olimpiadele de matematica in care apare aceasta sectiune. Am vazut ca in [url=http://www.mathlinks.ro/Forum/resources.php]Resources[/url] nu se gasesc probleme atat de vechi (1968-1974). Le voi posta deocamdata pe cele din 1968, sperand ca acestea se refera la TST :blush:

[b]Pb. 1[/b] Sa se afle numerele naturale $a$ si $b$ stiind ca suma numerelor naturale mai mari decat $a$ si mai mici decat $b$ este $10000$.

[b]Pb. 2[/b] Sa se construiasca grafic triunghiul dreptunghic $ABC$, cunoscand ipotenuza $BC=a$ si mai stiind ca $\frac{4}{BC}=\frac{1}{AE}+\frac{1}{AC}$ (G. D. Simionescu)

[b]Pb. 3[/b] Sa se afle cu cate zerouri se termina numarul $N=1958!$ (Ionel Atanasiu)

[b]Pb. 4[/b] Din numarul $123456789101112\cdots9899100$ sa se scoata $20$ de cifre astfel ca numarul ramas sa fie cat mai mare posibil. (mi se pare foarte familiara, dar nu-mi amintesc de unde o stiu)

[b]Pb. 5[/b] Fie $25$ de puncte in spatiu astfel ca: trei puncte sunt coliniare, dintre cele $25$ nu exista $5$ puncte coplanare, iar afara de grupele de cate $4$ puncte care contin cele trei puncte coliniare nu exista $4$ puncte coplanare. Cate plane sunt determinate de cele $25$ de puncte? (I. Stanescu)

[b]Pb. 6[/b] Sa se arate ca daca $a+b+c=\frac{\pi}{2}$ atunci \[\sin{a}+\sqrt{2}(\sin{b}+\sin{c}) \le 2\] In ce caz avem egalitate? (C. Ionescu-Tiu)

[b]Pb. 7[/b] Sa se afle restul impartirii prin $35$ a numarului \[N=3^{12n+10}\cdot 10^{6m+3}+300\cdot 72^{72p+30}\]
$m$, $n$, $p$ fiind numere naturale. (E. Rusu)

Apropo de dezvoltarea colectiei de probleme de pe acest site, cred ca am toate problemele date la ONM din primul an pana acum (la nivel de liceu). Le voi posta si pe acelea, nu promit ca ma voi misca prea repede. :D

Eu ma cam indoiesc ca problemele de mai sus sunt de la TST.  :wink:

[quote="icx"] [b]Pb. 4[/b] Din numarul $123456789101112\cdots9899100$ sa se scoata $20$ de cifre astfel ca numarul ramas sa fie cat mai mare posibil. [u](mi se pare foarte familiara, dar nu-mi amintesc de unde o stiu) [/u][/quote]

poate din " Probleme de teoria nr si combinatorica pt juniori"-autori: L. Panaitopol si D. Serbanescu

Da mychrom, de acolo este. Iar in acea carte ce crezi ca scrie la sursa? Baraj 1968 :P Deci problemele chiar sunt de la TST.

O sa le fac poze si le voi posta intr-o arhiva (desi n-a mers ultima data cand am incercat asta). Mychrom, iti pun aici si problemele de la ONM de la prima editie pana in 2003 la cls a 9-a. In seara asta sau cel tarziu maine; sunt foarte, foarte presat de timp. :blush:

De [url=http://www.wikiupload.com/download_page.php?id=156265]aici[/url] se pot downloada subiectele de la faza nationala ( cls a 9-a; fisierele sunt jpg, am pozat paginile, pur si simplu ) din ultimii aproximativ 50 de ani. :) Acum am vazut si eu ca nu pot atasa arhive. Happy happy joy joy mychrom, a? :D

PS: Nu ma puneti sa pozez si solutiile ca mor :lol:

[quote="icx"] Happy happy joy joy mychrom, a? :D

[/quote]

 :rotfl: dupa capacitate... :(

Am pus testele de selectie din perioada 1969-1974 [url=http://www.wikiupload.com/download_page.php?id=157755]aici[/url] :) Imi pare rau ca nu pot face pdf-uri din ele ca sa usurez si munca altora :blush:


1. Prove that between any two non equal real numbers there is a rational number.
2. Prove that between any two non equal real numbers, there is an irrational number

[hide="1"] Write out the base-$b$ representation of both reals.  They are distinct; hence, their digits differ.  A terminating $b$-imal must therefore exist between them (simply terminate the digits of the smaller real at some arbitrary point after the digit at which the two reals in question differ, and add one to the last digit).  QED. [/hide]
[hide="2"] It is a well-known lemma that a non-terminating $b$-imal is irrational iff its $b$-imal representation has no period.  From the same logic as the first, such a number must exist between two distinct reals.  QED. [/hide]


geometry
circumcircle
angle bisector
geometry proposed

[color=darkblue][b][u]Enunciation.[/u][/b] Let $ABCD$ be a quadrilateral inscribed in the circle $w$ so that $E\in AB\cap CD$ and $B\in (AE)$. Denote : the midpoints $M$, $N$ of the sides $[AB]$, $[CD]$ respectively ; the circumcircle $w'$ of the triangle $MEN$ ; the intersections $L$, $K$ of the lines $AB$, $CD$ respectively with the common chord of the circles $w$, $w'$. Suppose that exists at least a common point $X$ of the circles with the diameters $[AB]$, $[CD]$. Prove that $2\cdot m(\widehat{BXC})\in \left\{\ m(\widehat{LXK})\ ,\ 360^{\circ}-m(\widehat{LXK})\ \right\}$.

[b][u]Remark.[/u][/b] This problem is a fine aplication of the harmonical division.[/color]

hmmm Where is grobber?  :D

Solution:
Let $O$ be the circle of $w$.
And note that circle $w'$ has the diameter $EO$, hence the common of $w$ and $w'$ is the polar of $E$ wrt $w$ hence
$(ALBE)=(DKCE)=-1$

Now look at $X (ALBE)$ and $XA \perp XB$ hence $XB$ is the angle bisector of $LXE$.

Same holds for $XC$ and $KXE$. conclusion follows.


geometry
geometric transformation
rotation
function
calculus
derivative
complex numbers

1. Let $ z_{k}, k \equal{} 1,2,\ldots, n$ be the complex numbers lying on the same side of a straight line passing through the origin.
 Show that $ z_{1} \plus{} z_{2} \plus{} \ldots \plus{} z_{n}\neq 0\,\,$ and $ \,\, \frac {1}{z_{1}} \plus{} \ldots \plus{} \frac {1}{z_{n}}\neq 0$

2. Let $ \mathbb{D} \equal{} \{z : |z| < 1\}$ .  Assume that $ f : \mathbb{D}\to\mathbb{D}$ is analytic and $ f(0) \equal{} 0$, $ f$ is not a rotation (i.e $ f(z)\neq ze^{i\phi}$), define $ f^{1}(z) \equal{} f(z), \ldots, f^{n \plus{} 1}(z) \equal{} f(f^{n}(z))$ . Prove that $ \sum_{n\geq 1} f^{n}(z)$ converges uniformly to analytic functions on compact subsets of $ \mathbb{D}$

thanks so much

A hint for number 2. Use Schwartz Lemma to bound the derivative of $ f$ at 0. (In other words, do you have reason to believe that it is strictly less than 1?) Now use the fact that the iterates give a power of the derivative. i.e. $ f'(0)\equal{}r\Rightarrow f_n'(0)\equal{}r^n$.

I used subscripts so that the prime was clearer.

Can you get it from these facts?

I am sorry, I still dont know why you are considering the derivative? from the definition $ f^{1}(z) \equal{} f(z),\ldots, f^{n\plus{}1}(z) \equal{} f(f^{n}(z))$, $ f^{n}(z)$ is not  nth derivative of f.

Yes, HilbertThm90 understands that; he's looking at the first derivative of the $ n$th function, not anything involving $ n$th derivatives.

It's still a bit too local for my tastes. Here's another way of looking at things:

On the closed disk $ D_r$ of radius $ r$ centered at zero, what can you say about the maximum of $ \left|\frac{f(z)}{z}\right|$? If $ |f(z)|\le K|z|$ on that disk, what can you say about $ f^n$ there?

To be a little more explicit. If $ f'(0)=r<1$, then the fact that $ f^n'(0)=r^n$ means that $ f^n'(0)\to 0$. But note that if it is strictly less than 1 at 0, then it is strictly less than 1 on a neighborhood of zero. Thus $ \{f^n'\}$ the sequence of derivatives converges uniformly on some neighborhood of 0 (and possibly by further argument on all of $ \mathbb{D}$?).

I'm sorry. I haven't solved the problem, maybe I shouldn't have been misleading. I'm just throwing out ideas. This is a common trick for these types of problems.

Ah. jmerry beat me to the reply, so this was probably an unnecessary post.


calculus
vector
real analysis
Stanford
college
trigonometry
MIT

I have several questions concerning some stuff..

I took AP Calculus BC last year, and I want to do an independent study in math during the summer, before I head off to college. I was wondering what are the best texts to study Multivariable Calculus (more here) and Linear Algebra from. (I guess those are things you do after Single Variable Calculus?)

I am a potential physics major, so I'm not looking for a book too rigorous on proofs. I want to read a book that's not so difficult to follow, but I'm torn b/c I kind of want to try books w/ some prestige (if that makes any sense)

I was also wondering what Vector Calculus is, is that same as Multivariable??

Anyway these are the texts I'm considering,
(If Vector Calculus is indeed Multivariable)

h.m. schey   Div, Grad, Curl, All That
Walter Rudin   Principles of Mathematical Analysis

Georgi E. Shilov   Linear Algebra
Gilbert Strang   Introduction to Linear Algebra

Can you please help me find a good MultiV book[/list]

[quote="chloe L"]I took AP Calculus BC last year, and I want to do an independent study in math during the summer, before I head off to college. I was wondering what are the best texts to study Multivariable Calculus (more here) and Linear Algebra from. (I guess those are things you do after Single Variable Calculus?)[/quote]

Tom Apostol's books [url=http://www.amazon.com/Calculus-Vol-One-Variable-Introduction-Algebra/dp/0471000051/]Calculus volume 1[/url] and [url=http://www.amazon.com/Calculus-Vol-Multi-Variable-Algebra-Applications/dp/0471000078/]Calculus volume 2[/url] are always good to look at.

[quote="chloe L"]I am a potential physics major, so I'm not looking for a book too rigorous on proofs. I want to read a book that's not so difficult to follow, but I'm torn b/c I kind of want to try books w/ some prestige (if that makes any sense)[/quote]

A potential physics major might want to look right away at [url=http://www.amazon.com/Road-Reality-Complete-Guide-Universe/dp/0679776311/]The Road to Reality[/url] and work through all the math exercises in that book. Then visit the [url=http://www.phys.uu.nl/~thooft/theorist.html]How to Become a Good Theoretical Physicist[/url] Web site (by a Nobel laureate who encourages students to take initiative in self-study) and eventually take a look at [url=http://www.amazon.com/Geometric-Algebra-Physicists-Chris-Doran/dp/0521480221/]Geometric Algebra for Physicists[/url] and similar books.

[quote="chloe L"]I was also wondering what Vector Calculus is, is that same as Multivariable??[/quote]

Yes, that is largely the same subject. Hubbard and Hubbard's book [url=http://matrixeditions.com/UnifiedApproach.html]Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach[/url] is a helpful introduction to this subject.

[quote="chloe L"]Anyway these are the texts I'm considering,
(If Vector Calculus is indeed Multivariable)

h.m. schey   Div, Grad, Curl, All That
Walter Rudin   Principles of Mathematical Analysis

Georgi E. Shilov   Linear Algebra
Gilbert Strang   Introduction to Linear Algebra

Can you please help me find a good MultiV book[/quote]

Schey's book is a book for physicists by a physicist. Some of the math terminology is old-fashioned, but commonly encountered among physicists. Rudin (called "baby Rudin" or "little Rudin" is a classic book on mathematics. Shilov's and Strang's books on linear algebra are both good.

Also see this forum's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]FAQ on Preparing for College Math Study[/url] for somewhat more information. Good luck.

I would work on the assumption that even with the self-study, you won't be skipping any courses. That is, you'll go ahead and take your college's courses in multivariable calculus, linear algebra and differential equations.

You don't need to be in a big hurry. There's time to learn things thoroughly, and to learn them in your courses.

what you've been studying really hard and you feel that you have mastered the material, then is it okay to skip?

I'd recommend trying to skip classes whenever possible.

I don't quite agree with ComplexZeta's recommendation to skip classes whenever possible. When someone takes an AP test an places into MV/linear algebra in their first year of college, their knowledge of single variable calculus is certainly sometimes not as ideal. I know this happened to me, when I did the 50H sequence at Stanford this year. I felt like the BC Calculus class I took was calculus for engineers. There was zero rigor in my BC course, whereas the 50H series was based entirely on rigor.

I guess what I'm saying is this: there's a difference between being able to skip a course and mastering the material in the course you skipped. Be very certain that you thoroughly know the material you're trying to skip.

I am also much more conservative about skipping courses than ComplexZeta.

It's probably OK to skip courses every so often, but if you do that, make sure you go back and review what you skipped or skimmed. Rarely do you understand something fully the first time through.

Well, first of all, I skipped most undergraduate math classes, and I think I turned out okay in the long run, so I have evidence that it can work :)

My thought is that, while it is necessary that you learn the material eventually, I don't see too much advantage in learning it in the order that someone else says you should. So I picked up most of the undergraduate stuff on the way while learning graduate mathematics, basically as it was needed.

Perhaps more importantly, it's a lot of fun to be learning a lot (and confused from time to time) than bored most of the time, at least for me. (I had so much fun and learned so much while failing graduate classes my freshman year of college.)

Chloe L, one question you might ask yourself is whether you liked your Calc BC textbook, or at least found it adequate?  If so, then you might look for the multivariable textbook by the same author(s).  (Textbook publishers are like movie studios.  All successful calc books MUST spawn sequels or prequels.)  At a minimum, you'll be familiar with the notation, style, etc.  It won't be a mind-blowing experience, but it's just multivariable calculus, after all.

Also, you did not ask for advice on skipping courses, but plenty of advice has been given.  Here's my two cents on self-studying a course such as multivariable:  Do as many problems as possible, ideally at least half (less, if they seem painfully easy).  If this experience gives you no serious problems, then it's probably safe to skip the course, but if there's a lot you cannot do on your own, then a formal course might be the way to go.

You should also keep in mind that those two courses cover a lot of ground.  As a very conservative estimate, I'd say there's more material there (typical first year of college) than  there was in the last two years of high school combined.

[quote="yenlee"]You should also keep in mind that those two courses cover a lot of ground.  As a very conservative estimate, I'd say there's more material there (typical first year of college) than  there was in the last two years of high school combined.[/quote]
That's not hard to do.  Seriously, high school math courses are not the greatest.  Pre-calculus is primarily a review of Algebra II and trig, so you are basically comparing it with AP Calc.  And anyway, the college courses are much better than the high school courses (I hope--If not I will be very bored :roll: ).

http://math.mit.edu/~tkemp/18.024/index.html
this was the theoretical multivariable calc course i took this year at MIT..
it was very good.you can do all the homeworks and exams...
we used hubbard and hubbard and apostol
and for differential forms and stokes thm. the prof wrote some notes that are in the site too.

Although Apostol's book is very good, it's probably too proof-oriented for you, and an easier book would serve better as an introduction to multivariable calculus.

If you are looking for a less-rigorous, easy to read introduction to multivariable calculus, I highly recommend the [url=http://www.math.gatech.edu/~cain/notes/calculus.html]online textbook[/url] from Georgia Tech by George Cain and James Herod.

This book is nice because it:
- is not too rigorous, but rather gives a nice introduction to the concepts (and might be good for a physics major)
- focuses on MV, with a sufficient amount of linear algebra
- introduces and explains the concepts very well in my opinion
- is free! (and online, for easy access anywhere)

I'm currently a high school sophomore who took BC Calculus this year (and hardly learned anything new). I'm currently learning multivariable calculus through that book. My goal is to do a bit of study over the summer (finish that book soon and look through the first Apostol book) and then complete the work for MIT's 18.024 Calculus with Theory II Course (mentioned in the post above) in junior year, using MIT's OpenCourseWare.

I would also recommend [i]Div, Grad, Curl, and all that[/i] by H. M. Schey to understand the said concepts and their relevance to physics.

Though everyone has been giving you advice on calculus, I'd like to point you out to some Linear Algebra.

As calc rulz mentioned, he is using MIT  OpenCourseWare to help learn multivariable. But, OCW also has a great set of lectures on Linear Algebra, even including video lectures. Plus, the book that you mentioned (Gilbert Strang Introduction to Linear Algebra) is used in this course! So, it worked out quite nicely for you. (Speaking of which, mind if you tell me where you plan on buying it?)

[url=http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/CourseHome/index.htm]MIT OpenCourseWare Link[/url]

[quote="tokenadult"]Rudin (called "baby Rudin" or "little Rudin" is a classic book on mathematics. [/quote]

This is true, but it's also a horrible text to learn multivariable calculus out of for various reasons.  The relevant chapters (9 and 10 iirc) seem like almost an afterthought and they will be near impossible to understand without going through the previous 8, which is a pretty formidable task in and of itself.


logarithms
counting
derangement
function
geometry
3D geometry
factorial

Using the digit 3 three times, and no other digits, (with any combination of operators), create 8.
You may not append it (as in [b]33[/b] + 3 ).

(square root is considered not to use the digit 2.)[/b]

[hide]$ 3!\plus{}3!/3$[/hide] is one answer...

[hide="Some more?"]
Depending upon how loosely you can interpret operator,

$ \left(3!/3\right)^3$

With logarithms, $ (\log_{\sqrt {3}} 3)^3$,

With derangements, $ !3 \plus{} 3 \plus{} 3$ or $ (!3)^{\sqrt {3 \times 3}}$
[/hide]

Im kinda confused because this question is very ambigious.

It kinda depends what sign we can use

For example, $ (3\minus{}3^{\minus{}})(3)\equal{}8$ would work

and if square roots is not considered to use the digit two, then putting it to a - power is not considered to use the digit -1 (since you said - signs can be used)

Yes, sorry it's ambiguous.

No, aside from using the square root function, if you are going to do an exponentiation, you must use ONLY the digit three.

Your example isn't really correct, because $ \minus{}$ does not inherently mean $ \minus{} 1$.

So far my favorite is $ (log_{\sqrt{3}} 3)^3$ :)

Is the subfactorial just the number of permutations of $ n$ objects that aren't the original arrangement? I don't think this is right, because then $ !n\equal{}n!\minus{}1$, which seems silly. Or is it when none appear in their place in the original arrangement? That sounds right...

In the case of $ n \equal{} 3$, $ !n \equal{} n! \minus{} 1$.

Wait, that's not right...

You could just randomly define an operation that takes in three 3s and returns 8... :P

Althought it's a little ambiguous, I think it's referring to [b]defined operators[/b].  :wink:

[quote="Temperal"]Is the subfactorial just the number of permutations of $ n$ objects that aren't the original arrangement? I don't think this is right, because then $ !n \equal{} n! \minus{} 1$, which seems silly. Or is it when none appear in their place in the original arrangement? That sounds right...[/quote]

The [url=http://www.artofproblemsolving.com/Wiki/index.php/Derangement]subfactorial[/url] is indeed the latter definition you mentioned.

[hide]
That $ !3 \equal{} 2$ gives many more solutions like $ \sqrt{!3^{3\plus{}3}}$. Or, replace all of the $ !3$s with $ \varphi(3) \equal{} 2$.  :lol:
[/hide]

Wow, you guys are [b]good[/b] at this digit stuff...

Here's my solution! (extremely cheesy)
[hide]Flip one of the three's around so you get an E or something, then smash it with another three to make an eight.  Then just take the cube root.

In other words, $ \sqrt[3]{\xi3}\equal{}2$ ! Yes![/hide]

Sorry if this spammed up the thread...I had to get that out.

Er.. the problem asked for [b]8[/b], not 2.

3(3!+3!)square root covers it all then +!3

Here's another factorial solution
[hide]
It uses the double factorial.

$ (3/3\plus{}3)!!$
[/hide]

Or
[hide]
$ \left(\frac{3!}{3}\right)^3$[/hide]


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

find all n s.t. there exists integral domain consisting of exactly n elements.

Every finite integral domain is a field. Every finite field has $p^{k}$ elements for some prime $p$ and positive integer $k$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26475]Old thread[/url]


modular arithmetic
quadratics
number theory proposed
number theory

If  $p$ is a prime $p>2$ and  $n$ is a natural number with  $n\not\equiv 0 \pmod{p-1}$
then prove that
 $1^{n}+2^{n}+\ldots+(p-1)^{n}\equiv 0 \pmod{p}$

Noting that p is odd, we pair up the summands

$\sum_{i=1}^{\frac{p-1}{2}}(i^{n}+(p-i)^{n}) \equiv \sum_{i=1}^{\frac{p-1}{2}}(i^{n}+(-i)^{n}) \mod{p}$

If n is odd, this is trivially equal to zero, so let us suppose n=2m.

So $\sum_{i=1}^{\frac{p-1}{2}}(i^{2k}+(-i)^{2k}) \equiv \sum_{i=1}^{\frac{p-1}{2}}2 i^{2k}\mod{p}$.

It is obvious that p divides this expression iff p divides $(1^{k})^{2}+(2^{k})^{2}+....((\frac{p-1}{2})^{k})^{2}$

If k=1, this is merely the sum of all the quadratic residues modulo p.

Sorry, be back later. :(


quadratics
number theory unsolved
number theory

Prove that if p is an odd prime, then $\sum_{a=1}^{p-2} (\frac{a(a+1)}{p}) = -1$

where $(\frac{a}{p})$ = 1 if a is a quadratic residue modulo p, -1 if a is a quadratic non-residue modulo p, 0 if p|a.

Hope someone can enlighten me. Thanks in advanced.

it is also a part of a problem in VietnamTST1999

Thanks pleurestique very much for your quick response.

But anyway, could you explain the use of your sum's indices please? When you put a' in, even though it's in the range 1->p-1 but a is still from 1 to p-2, isn't it? And I don't quite understand the part with 'm' and then sum - (1/p)

Thanks again.

for the first step, changing the range to include $p-1$ doesn't change anything because the extra term is $(\frac{(p-1)p}{p})$ which is 0

for the final step, i switched to $m$ because i wanted to take a sum where the index goes all the way to $p$ (the sum that you see turns out to be 0) but $a'$ doesn't exist for $a = p$... so i had to change the variable...

the reason that sum is 0 is because as $m$ ranges through $\{1, ..., p\}$, so does $1+m (\mod p)$... and there are as many residues in $\{1,...,p\}$ as there are non-residues, giving just as many $-1$'s in the sum as $1$'s.


Im not quite sure how to prove this one.
Let ABC be an equilateral triangle and let P be a point on Arc AB of its circumscribed circle. Let AB and PC intersect at D.
Prove that 1/PD=1/PA+1/PB

Hint: [hide]You have 4 points on a circle.  Which means you have a ______ _____________.  Which means you can use ________ _______.[/hide]

hm ok, so this is problem 6 on the maml2 exam from this week....

part A was to prove that PA + PB = PC, which is useful in this proof, i'll start with that assumption

i didn't get this during the test :(, but here u go...

Prove that 1/PD = 1/PA + 1/PB, lets think about how we can attain this result:

(PA + PB) / (PA*PB) = 1/ PD (common denominator)

(PA*PB) / (PA+PB) = PD (flippage)

(PA*PB) / (PC) = PD (applying the solution to part A to the denominator)

PA * PB = PC * PD

PA / PC = PD / PB.... ok.. this was just rearranging the thign they asked us to prove.. so now i'm going to prove this fact, which quickly leads to the 1/PD = 1/PA + 1/PB that they asked for.....

triangle PDB  :sim: triangle ADC

PD / PB = AD / AC

triangle PAC  :sim: triangle ADC

PA / PC = AD / AC

by transitive....

PA / PC = PD / PB ...that's what i wanted to prove.. rearrange that and use PA+PB = PC to get 1/PD = 1/PA + 1/PB


inequalities
logarithms
real analysis
real analysis solved

Prove that if $a>0,b>0$ and $2a^x+3b^x+2^x+3^x\ge 7$ for any $x\in\mathbb{R}$ then $a^2b^3=\frac16$

Hi Kuba!
Please, read http://www.mathlinks.ro/viewtopic.php?p=140098 and name problems properly next time!

With regards,
Myth

P.S. Rename this topic too :)

Oh.. I'm sorry Myth.. I have't seen that discussion yet... I promiss to name the problems properly next time


I'm Irina, I am from Macedonia and I participated in the 43rd IMO in UK. I had a honourable mention. I've also participate in 3 BMOs and have 2 bronze medals. After the IMO, I changed my intrests and now I'm studying finance in the USA. I like my major, but I still feel kind of nostalgic about the days when I was competing in math, so I still check math sites on regular basis. ( Anyway, finace is all about numbers, and I try to keep in track with math by reading math books& sites).
Irina

that's cool! were are you studying in the states Irina?


Let $ G$ be a $ k$-regular graph with $ n$ vertices, where $ k>0$. A subset of the set of the vertices of $ G$ is called an [i]independent subset of $ G$[/i] if no two vertices from this subset are connected to each other. Prove that the size of any independent subset of $ G$ is $ \leq\frac{n}{2}$.

[i]Source:[/i] Raymond Greenlaw, Rossella Petreschi, [i]Cubic Graphs[/i], 1995, §8.4. Downloadable from http://citeseer.ist.psu.edu/greenlaw95cubic.html .

  darij

Cute :).

The graph has $\frac{nk}2$ edges. On the other hand, if an independent set $S$ has $t$ vertices, then there are $tk$ edges with vertices in $S$, since every vertex in $S$ is adjacent to $k$ edges, and edges adjacent to different elements of $S$ are distinct. This means that $tk\le\frac{nk}2$, and we're done.


Can someone please help me with this question?

A student is moving a box of books with mass 35 kg up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25deg with respect to the incline with 185N force, what is the acceleration up the ramp? Assume that coefficient of kinetic friction is 0.27.

So I drew a diagram, and I can't come up with any different answer.
Fn = 257.66
Fk = 69.5689
using the ramp as x,
sum of x components: 26.71
divide that by 35 kg to get .76

But the answer is .90

What did I do wrong?

Applying Newton's second law to the parallel and perpendicular directions relative to motion:

$ ma = F \cos 25-\mu R-mg \sin 12$
$ R+F \sin 25 = mg \cos 12$,

where R is the force exercted by the plane. Now substitute by the given data (using $ g = 9.8 m/s^{2}$) and we get $ R = 257.3 N$, $ a = 0.77 m/s^{2}$, which agrees with your reasoning. So perhaps the solution given by the book is wrong, or some data is wrong.

Ok, I guess the book can be wrong in some way. Thanks!


calculus
integration
trigonometry
function
complex analysis

1$ \underset{z\equal{}\infty}{Res}  (z\plus{}2)^2 \sin \frac{1}{z}$
2$ \underset{z\equal{}1}{Res}  z e^{\frac{1}{z\minus{}1}}$

1) You expand $ \sin(1/z)$ around $ z \equal{} \infty$ and obtain
$ (z^2 \plus{} 2z \plus{} 2)({1\over z} \minus{} {1\over 6}{1\over z^3} \plus{} O(1/z^5))$ and the residue is the opposite of the coefficient of the power
$ 1/z$ that is $ \minus{} 23/6$

2) Here you expand around $ z \equal{} 1$ obtaining $ (1 \plus{} (z \minus{} 1)) (1 \plus{} {1\over z \minus{} 1} \plus{} {1\over2}{(z \minus{} 1)^2} \plus{} {1\over6}{(z \minus{} 1)^3} \plus{} ....)$
and take the coefficient of the power $ 1/z$ that is $ 3/2$

Find and build the region of convergence of the Laurent series:
$ \sum_{n\equal{}\minus{}\infty}^{\infty} 2^{|n|} z^n$

Lay in a Laurent series in a punctured neighborhood of z = i the function:
$ f(z)\equal{}\frac{1}{z^2\plus{}1}$


i was the school winner this year, and i got a pin. next year, if i take it at a different location, and i am the school winner, will i get a bronze medal, or another pin?

The AMCDirector answered this question:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=770077#770077[/url].


function
real analysis
topology
real analysis unsolved

this is from folland
Let $ f: [0,1]\to\[0,1]$ be the Cantor function and let $ g(x)=f(x)+x$
a.$ g$ is a bijection from $ [0,1]$ to $ [0,2]$ and $ h=g^{-1}$ is continuous from$ [0,2]$ to $ [0,1]$
b.If $ C$ is the Cantor set $ m(g(C))=1$
c.We know(ex 29 ch 1) tha $ g(C)$ contains a Lebesgue nonmeasurable set $ A$. Let $ B=g^{-1}(A)$. Then $ B$ is Lebesgue measurable but not Borel
d.There exist a Lebesgue measurable function $ F$ and a continuous function $ G$ on $ \mathbb_R$ s.t. $ F\circ G$ is not Lebesgue measurable

Let $ \Delta$ be the Cantor set and $ f: [0,1]\rightarrow [0,1]$ the Cantor function.  Now let
$ J_{\frac{1}{2}}=\left(\frac{1}{3},\frac{2}{3}\right)$
$ J_{\frac{1}{4}}=\left(\frac{1}{9},\frac{2}{9}\right)$, $ J_{\frac{3}{4}}=\left(\frac{7}{9},\frac{8}{9}\right)$
$ J_{\frac{1}{8}}=\left(\frac{1}{27},\frac{2}{27}\right)$, $ J_{\frac{3}{8}}=\left(\frac{7}{27},\frac{8}{27}\right)$, $ J_{\frac{5}{8}}=\left(\frac{1}{19},\frac{20}{27}\right)$, $ J_{\frac{7}{8}}=\left(\frac{25}{27},\frac{26}{27}\right)$
and so on.  That is, the $ J_{r}$'s constitute the removed middle thirds.  Now note that $ \Delta=[0,1]\setminus\bigcup_{r}J_{r}$, $ f$ is constant on each $ J_{r}$ ($ f(x)=r$ for each $ x\in J_{r}$), and $ f$ is an increasing continuous function.

Now define $ g(x)=x+f(x)$.  Clearly, $ g$ is strictly increasing, continuous, and $ g(0)=0$ and $ g(1)=2$.  Since $ g$ is continuous, one-one, onto, and a closed map, this implies that $ f^{-1}$ is continuous), $ g$ is a homeomorphism of $ [0,1]$ to $ [0,2]$.  Since $ g(x)=x+r$ on each $ J_{r}$, $ g$ maps $ J_{r}$ into another open interval with the same length as $ J_{r}$.  Then we have that $ m(g(J_{r}))=m(J_{r})$.  Hence we have that
\begin{align*}
m(g(\Delta)) & = m([0,2]\setminus\bigcup_{r}g(J_{r}))\\
& = m([0,2])-m\left(\bigcup_{r}g(J_{r})\right)\\
& = 2-\sum_{r}m(g(J_{r}))\\
& = 2-\sum_{r}m(J_{r})\\
& = 2-1=1
\end{align*}
Since $ m(g(\Delta))>0$, problem 18 says that there exists a nonmeasurable set $ B\subseteq g(\Delta)$.  Let $ A=g^{-1}(B)$.  Since $ A\subseteq\Delta$ and $ m(A)=0$, $ A$ is a measurable set.  Now since $ g$ is a homeomorphism, if $ A$ were a Borel set, then $ g(A)=B$ would be a Borel set.  But $ B$ is nonmeasurable so this is not possible.  So $ A$ is measurable, but not Borel.

Let $ G=g^{-1}(x)$, where $ g$ is as above and $ F=\chi_{E}$, where $ E=g^{-1}(N)=G(N)$ and $ N$ is a nonmeasurable set in $ \Delta$.  Then $ \chi_{E}$ is measurable on $ [0,1]$ since $ m(E)=0$.  But $ F\circ G$ is the characteristic function of $ N$ and so can't be measurable.


for $ ABCD$ is an cycle quadraliteral. E be the intersection of AD and BC (C between B and E) AC and BD intersect at F. Let M be the midpoint of the side CD, and N be a point on the circumcircle of the triangle ABM such that $ AN/BN\equal{}AM/BM$. prove that E,F and N are collinear.

[b]Generalization:[/b] Given a convex quadrilateral $ ABCD. E\equal{}AD\cap BC, F\equal{}AC\cap BD$. $ EF\cap DC\equal{}\{N\}. (ABN)\cap DC\equal{}\{J,M\}$. $ (ABN)\cap EF\equal{}\{N,J\}$. Prove that $ \frac{AN}{BN}\equal{}\frac{AM}{BM}$.
[b]Solution:[/b]
Denote $ I\equal{}AB\cap CD, K\equal{}AB\cap EF$. We have $ (IJDC)\equal{}\minus{}1$ then $ (KIAB)\equal{}\minus{}1 \Rightarrow J(KIAB)\equal{}\minus{}1$ or $ J(NMAB)\equal{}\minus{}1$.
So $ AMBN$ is a harmonic quadrilateral $ \Rightarrow AN.BM\equal{}BN.AM$ and we are done.

can you tell me more about your solution, please  :blush:
I still don't know. if $ EF \cap DC \equal{} {N}$, so I think $ (ABN) \cap DC \equal{} (N, another point)$