[hide]$\boxed{6}$. Let's call them A, B, C, D, E, and F. Each of those people saw the other 5 people and everybody else just saw 5 of them.[/hide]

[quote="i_like_pie"]Correct.

[hide="Another Solution"]Lets say I'm one of the students who got an A. I didn't see my own A, but I saw 5 others. Therefore, there are at least 6 students who scored an A on the test.[/hide][/quote]

But you want the minimum, so "at least" doesn't help.

"At least" and "minimum" can both be represented by $\ge$.

Yes but there are at least 3 students who got As but 3 isn't the minimum and minimum here applies to the lower bound.

[quote="i_like_pie"]You're confusing me. "Minimum" and "at least" mean the same thing - the lower limit. "At least 6 students have A's" means the same as "there is a minimum of 6 students who got A's". Why does the wording matter?[/quote]
It doesn't...

At least and minimum are the same thing, mathematically. As mentioned, both are represented by a $\geq$ sign. Ignite is thinking of a common English definition, just like the mathematical "or" and the common English "or" are very different.

i_like_pie's solution is correct: he(she?) shows that there are a minimum of 6 people, and that this minimum is achievable. It's actually Ignite's solution that is not correct:

[quote="Ignite168"]$\boxed{6}$. Let's call them A, B, C, D, E, and F. Each of those people saw the other 5 people and everybody else just saw 5 of them.[/quote]

This only shows that 6 people is achievable, but it gives no reason why less than 6 people is not achievable.

[quote="i_like_pie"]After a math test, each of the twenty-five students in the class got a peek at the teacher's grade sheet. Each student noticed five A's. No student saw all the grades and no student saw his or her own grade. What is the minimum number of students who scored an A on the test?[/quote]

[hide]
Since you want the least, and no one saw their own scores, I have illustrated below:

for example students a b c d e f are there,
then A sees:
BCDEF grades
B sees:
ACDEF grades
C sees
ABDEF
and so on, so the minimum is [b]6[/b]
[/hide]

[quote="Phelpedo"]Another incorrect solution  :roll:[/quote]
[hide="I hope this is correct"] As shown above, many times, 6 is attainable.

We will now prove 5 is unattainable.

Let's call the 5 people who got A's 1, 2, 3, 4, and 5.
Thus, 1, 2, 3, 4, and 5 saw a maximum of 4 A's.

Obviously, 4 or less is impossible, since all students saw 5 A's. [/hide]

[quote="i_like_pie"]The solution is incorrect, but the answer is correct.[/quote]

Obviously, I could do such a thing in the olympiad section.
After the answer is already given, make up a similar solution and "get" the correct answer.

you would have to prove that there can't be any number of students less than 6 for that solution to be valid like lotrgreengrapes did.

[quote="Phelpedo"][quote="bpms"][quote="anirudh"][hide]
but it makes sense. let me relate

lets say the students that were seen are ABCDEF. then all the students A-F see all the grades except there's. However, all of the other students, see 5 out of the 6. So there are a minimum of 6 students :huh: [/hide][/quote]
That is circular logic because you are assuming there are six ;)...[/quote]

[hide]The answer is 5. Lets say the students that were seen are A,B,C,D,E. Then by hypothesis, there are five students with A's (A,B,C,D,E). [/hide][/quote]

[quote="i_like_pie"]No student saw all the grades and no student saw [b]his or her own grade[/b].[/quote]

[quote="Phelpedo"][hide]The answer is 5. Lets say the students that were seen are A,B,C,D,E. Then by hypothesis, there are five students with A's (A,B,C,D,E). [/hide][/quote]
Incorrect.
[hide="The correct answer"]Let's say Bob got an A. He didn't see his own A, but he saw 5 others. In all, there must be at least 6 A's.[/hide]

its 6 since if a person saw got an a and saw 5 more, there has to be at least 6

Obviously I'm not good at being sarcastic (note the circular logic :wink: .


A key point in solving this problem is circumcenter .

Thank you Kouichi Nakagawa.
this is another way.
[img]http://img245.imageshack.us/img245/9334/t43t4343f4rg5.png[/img]

Thanks to Orhan Develi.
[img]http://img175.imageshack.us/img175/7680/s4f6asbn4.png[/img]

Thanks to Azad Oğur Taşçıer.
[img]http://img219.imageshack.us/img219/3156/62517415kq9.jpg[/img]

snagit 7

http://www.dosya.cc/Snagit7.zip.html


A 2005 by 2005 square has entries that are integers between 1 and 2005, inclusively, such that the square is symmetric about the main diagonal. (So the entry in the ith row and jth column is the same as that in the jth row and ith column.) If no integer appears twice in any row or column, prove that each integer from 1 to 2005 appears exactly once in the main diagonal.

Note: The main diagonal is consisted of entries E(i, j) s.t. i=j

Each integer must appear once on each row and column (as otherwise some other number would repeat). Note that when there's an integer on the square $(i, j)$, it appears on the ith column and jth row. Also, it appears on $(j, i)$, so it appears on the jth row and the ith column.

If every integer appears once on each row and column, it must appear $4010$ times in total. That number is divisible by 2 but not 4; and each time an integer appears on a square $(i, j)$ with $i \neq j$, it implies 4 distinct times it appears on a row or column. So for a number to appear $4010$ times in total, it must appear at least once on the diagonal (as otherwise the total amount of appearances would be a multiple of 4). Thus every number must appear on the main diagonal; since there are 2005 numbers and 2005 squares on the diagonal, each number appears once.

For each number $n\in\overline{1,2005}$, let $\pi_n$ be the permutation of $1,2,\ldots,2005$ defined as follows: $\pi_n(i)=j$ iff the entry $(i,j)$ equals $n$. Now, we have $\pi_n(i)=j\iff\pi_n(j)=i$, so if $n$ did not appear on the main diagonal, the set $\{1,2,\ldots,2005\}$ could be partitioned in pairs $\{i,j\}$ with the property that $\pi_n(i)=j,\pi_n(j)=i$, which is, of course, impossible, since $2005$ is odd.

Edit:

Looks like Severius beat me to it.. :)

well,grobber beat me to it(i had the same idea).
it also explains why if $n$ is even the same is possible($n=2k$,then take 2 different squares $S_1,S_2$ of order $k$ each and then form the square
$\begin{array}{cc}S_1&S_2\\S_2&S_1\end{array}$
 and why it is not possible when $n$ is odd.


find the minimum of $ \sum_{i\equal{}1}^{n}\frac{x_{i}}{1\plus{}x_{i\plus{}1}^{2}}$,where $ x_{i}$s are nonnegative reals with sum $ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\equal{}2$ and $ n\geq 3$.p.s.$ x_{n\plus{}1}\equal{}x_{1}$.

[quote="zaya_yc"]$ 2\minus{}\sum_{i \equal{} 1}^{n}\frac{x_{i\plus{}1}^{2}x_{i}}{2x_{i\plus{}1}}\equal{} 2\minus{}\sum_{i \equal{} 1}^{n}x_{i}x_{i\plus{}1}$[/quote]

Er...where did the fact of $ 2$ go?  From here, we can set an even stronger bound of $ 2\minus{}\frac{2}{n}$.  But even that is not the strongest bound: there is no equality case, because AM-GM was applied with $ x_{i}^{2}$ and $ 1$...


find all integral solutions (x,y) of the following:-
$x\sqrt{y}+y\sqrt{x}=20,x\sqrt{x}+y\sqrt{y}=65$

[hide]
Adding the two equations, we get
$x(\sqrt{x}+\sqrt{y}) +y(\sqrt{x}+\sqrt{y})=85 \Longleftrightarrow (\sqrt{x}+\sqrt{y})(x+y)=85$

Since $x$ and $y$ are both integers, $x+y$ is an integer, and thus $\sqrt{x}+\sqrt{y}$ also has to be an integer.

Using $85=17\cdot 5$ and some trial and error gives $(x,y)=(1,16),(16,1)$.[/hide]

[hide="Solution"]Multiplying the first equation by $3$ and adding it to the second, we get

$(\sqrt{x}+\sqrt{y})^3=125\implies \sqrt{x}+\sqrt{y}=5$

Substituting $\sqrt{y}=5-\sqrt{x}$ into the second and simplifiying it we get

$x-5\sqrt{x}+4=0\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-4)=0$

From there $x_1=1, x_2=16$, which gives $y_1=16, y_2=1$. The solutions are $(1,16),(16,1).$[/hide]


Prove that for any pozitive real numbers a,b,c the following ineq holds

sum(a/(a+2b+c))>=3/4.

bye bye

using Cauchy one gets \sum a/(a+2b+c) \geq (\sum a)2/(\sum a2+3\sum ab) \geq 3/4 which comes to
\sum a2 \geq \sum ab which is true. !


A rectangular of a $ 10*14$ is divided into small 140 unit squares and painted in red and white like chess board as below.
We put $ 0$ or $ 1$ in the square such that each row and column has an odd numbers of $ 1$.
Prove that the number of $ 1$ contained in red-painted square is even.

The pattern arranged by a red and a white square alternatively.

RWRWRWRW.....
WRWRWRWR.....
RWRWRWRW.....
WRWRWRWR.....

[quote="kunny"]
We put $ 0$ or $ 1$ in the square such that each row and column has an odd numbers of $ 1$.
[/quote]
What do you mean $ 0$ or $ 1$? Can you provide an example?

It means that the number of either $ 0$ or $ 1$ will be put on the either red or white unit square.

First it's clear that we have an even number of $ 1$-s. Number the rows $ 1$ to $ 10$, and the columns $ 1$ to $ 14$. It's clear that we have to prove that there are even $ 1$-s with the sum of their $ 2$ coordinates even.
Denote by $ a_{i}$ the number of $ 1$-s in the $ i$-th row, by $ b_{j}$ then number of $ 1$-s in column $ j$. We calculate the sum of all the $ 1$-s coordinates in two ways. $ x_{i}$ and $ y_{i}$ are the $ 2$ coordinates of a $ 1$.
First $ S\equal{}\sum (x_{i}\plus{}y_{i})$

But $ S\equal{}\sum_{i\equal{}1}^{10}ia_{i}\plus{}\sum_{i\equal{}1}^{14}ib_{i}$ from here using that $ a_{i}$ and $ b_{i}$ are odd we get that $ S$ is even, so from $ S\equal{}\sum (x_{i}\plus{}y_{i})$ we have an even number of $ 1$-s with the sum of their $ 2$ coordinates odd, but we have an even number of $ 1$-s in total, so we have an even number of $ 1$-s with the sum of their coordinates even.

Label each square of the board with coordinates as used in the Cartesian plane, with the lower-left square of the board being $(1,1)$. Then for every square labeled $(i,j)$, we color it red if $i+j$ is odd and white if $i+j$ is even. Let $r$ denote the number of $1$'s on a red square, and let $w$ denote the number of $1$'s on a white square. We want to show that $r$ must be even.

The number of $1$'s on the board is even, therefore either both of the quantities $w,r$ are odd or both of them are even. If both are even, then the given statement is true. So we assume that both are odd and try to find a contradiction.

If our assumption holds, then the sum of the coordinates of the $1$'s is odd, as each $1$ on a red square has an odd sum of coordinates, and each $1$ on a white square has an even sum of coordinates. So one of the sum of the $x$-coordinates and the sum of the $y$-coordinates (of the $1$'s) must be odd, and the other must be even.

Since there are an odd number of $1$'s in each row, each number from $1$ to $10$ must occur an odd number of times when we are summing the $1$'s $y$-coordinates. So the sum of the $y$-coordinates of all the $1$'s must be odd. However we can apply similar logic to the $x$-coordinates of the $1$'s; there are an odd number of odd terms in that sum, so it must also be odd. So the sum of the coordinates of the $1$'s is even.

Thus we have a contradiction, and we conclude that we cannot have both of $w,r$ be odd. Therefore they are both even, meaning $r$ is even, and we are done.


Given n letters, prove that there are $\binom{2n-1}{n}$ distinct n-letter combinations (I paraphrased this so it might not be correct)

Example, if n = 2, then we can form the combinations AB, AA, BA, BB.  AB, BB, AA are the 3 distinct combinations and $\binom{2(2)-1}{2}= 3$

[quote="13375P34K43V312"]I don't understand the problem, you said there were three combinations, then gave seven. And not all are different.[/quote]
[quote="Phoenix257"]Given n letters, prove that there are $\binom{2n-1}{n}$ distinct n-letter combinations (I paraphrased this so it might not be correct)

Example, if n = 2, then we can form the combinations AB, AA, BA, BB[b][size=150].[/size] [/b] AB, BB, AA are the 3 distinct combinations and $\binom{2(2)-1}{2}= 3$[/quote]
Actually, he said the only three are AB, BB, AA. I think you missed the period.

sorry, I meant to say that

AB, AA, BB, BA were possible combinations but only

AB, AA, BB

count as distinct 2-letter combinations


Prove that in the set $ \left\{\left[k \plus{} \sqrt {k} \plus{} \frac {1}{2}\right]\ |\ k\in\mathbb{N}^{*}\right\}$ it doesn't exist any perfect square .

I think we can prove a stronger statement: $ (k \plus{} \sqrt k \plus{} \tfrac{1}{2}) \not\in \mathbb{N} \quad \forall k \in \mathbb{N}$.

Let $ n, k \in \mathbb{N}$. Solving

$ k \plus{} \sqrt k \plus{} \tfrac{1}{2} \equal{} n$

for $ \sqrt k$ and than squaring we obtain

$ k \equal{} \bigg( \frac {\sqrt {4n \minus{} 1} \minus{} 1}{2} \bigg)^2 \equal{} n \minus{} \frac {\sqrt {4n \minus{} 1}}{2}$

Thus

$ 2(n \minus{} k) \equal{} \sqrt {4n \minus{} 1}$

that clearly doesn't have integer solutions.

We proved there is not a $ k \in \mathbb{N}$ such that $ (k \plus{} \sqrt k \plus{} \tfrac{1}{2}) \in \mathbb{N}$

[quote="alex2008"]I meant the set $ \left\{\left\lfloor k \plus{} \sqrt {k} \plus{} \frac {1}{2}\right\rfloor \ |\ k\in\mathbb{N}^{*}\right\}$ , where $ \lfloor x \rfloor$ denotes the floor function .[/quote]

Oh...

Well, $ \left[ x \right]$ and $ \lfloor x \rfloor$ look quite different  :wink:

AFAIK $ [expression]$ is also the greatest integer function.

Anyway, set $ f(k) \equal{} k \plus{} \sqrt {k} \plus{} \frac {1}{2}$. Obviously this function is increasing. This function defines the set we are looking at for $ k \in \mathbb{N}$.

Assume we have a square $ \lfloor k \plus{} \sqrt {k} \plus{} \frac {1}{2} \rfloor \equal{} x^2$. Then look at $ k \equal{} x^2 \minus{} x$.
$ x^2 \minus{} x \plus{} \sqrt {x^2 \minus{} x} \plus{} \frac {1}{2} < x^2 \minus{} x \plus{} \frac {1}{2} \plus{} \sqrt {x^2 \minus{} x \plus{} \frac {1}{4}} \equal{} x^2$.

Then set $ k \equal{} x^2 \minus{} x \plus{} 1$.

$ x^2 \minus{} x \plus{} 1 \plus{} \frac {1}{2} \plus{} \sqrt {x^2 \minus{} x \plus{} 1} > x^2 \plus{} 1 \minus{} x \plus{} \frac {1}{2} \plus{} \sqrt {x^2 \minus{} x \plus{} \frac {1}{4}} \equal{} x^2 \plus{} 1$.

So $ f(x^2 \minus{} x)$ is too small and $ f(x^2 \minus{} x \plus{} 1)$ is too big.

For all $ k < x^2 \minus{} x$, we have $ f(x^2 \minus{} x \minus{} a) < f(x^2 \minus{} x)$, where a is a positive integer.
For all $ k > x^2 \minus{} x \plus{} 1$, we have $ f(x^2 \minus{} x \plus{} 1 \plus{} b) > f(x^2 \minus{} x \plus{} 1)$, where b is a positive integer.
$ k$ cannot be between these two values as there are no integers between them. Hence, for any perfect square, we have no $ k$ satisfying the condition.


[quote="Varun Nakra"]If there are p intermediate stations between two terminals A and B then in how many ways can a train stop at any three of these stations such that no two of them are consecutive...?[/quote]

[hide="Hint"]Complementary counting[/hide]

[hide="Solution"]The total number of ways to choose three intermediate stations is $ \binom{p}{3}$.

Now, we must find how many ways the train can stop such that two or three are consecutive.

CASE 1: Only two stations are consecutive.
Let S be the superstation block, denoting the pair s-s, where the s's are stations. We have one more s to place, and this can't be next to the superstation block.

If they are 1 station apart, then the number of ways is $ 2(p\minus{}3)$.

If they are 2 stations apart, then the number of ways is $ 2(p\minus{}4)$.

etc...

If they are $ p\minus{}3$ stations apart, then the number of ways is $ 2\cdot 1$.

So the total number of ways for this case is:

$ 2(p\minus{}3)\plus{}2(p\minus{}4)\plus{}\ldots\plus{}2\cdot 1 \equal{} (p\minus{}3)(p\minus{}2)$.

CASE 2: They are 3 stations apart.

This case is pretty obvious, there are $ p\minus{}2$ ways.

So, the total in both cases is $ (p\minus{}2)^{2}$.

Subtracting this from the total, we get:

$ \boxed{\binom{p}{3}\minus{}(p\minus{}2)^{2}}$.[/hide]

Yes thats right. It is C(p,3) - (p-4)(p-3) + (p-3)2 + (p-2) ways

which makes it C(p,3) - (p-2)^2 ways..

But can there be any other method than this ? [b]I am looking for an aliter.[/b]

Any one having a good theory on Generating Functions explaining the concept of
the number of ways in which r groups can be formed out of n different things when one can not take anything less than p times and more than q times. is [b]given as the coeff of x^n  in (x^p + x^p+1 + ... + x^q)^r
Why is this so ?[/b]


MATHCOUNTS
geometry
analytic geometry
rhombus

What is the area enclosed on the
coordinate plane by the graph of the
equation |x | + |4y | = 20 ? http://i44.tinypic.com/27xl1jt.jpg

Since this equation makes a quadrilateral, we want to find the x and y intercepts.

Substituting 0 for x and then 0 for y we have intercepts of (0,5), (20,0), (0,-5), and (-20,0).

The points are all either on the x-axis or the y-axis. Since the equations makes 4 congruent right triangles, we can find one of them and multiply by 4.

The bases of the right triangles are 5 and 20 (distance from origin). So the area of the triangle is 50. So the total area of the area enclosed by the equation is 4*50=200

[hide="solution"] These problems can easily be solved by substituting x=0, finding the points for y, and then substituting y=0 and finding the points for x.

We find that y can be either -5 or 5 when x=0 and x can be either 20 or -20 when y=0. Now, you can either divide into triangles, like Angie did, or note that this is a rhombus. One of the nice things about a rhombus is that the area is half the product of its diagonals. Thus, the area is $ \frac{1}{2}\cdot40\cdot10\equal{}200$.[/hide]


modular arithmetic
number theory
relatively prime
greatest common divisor

I must be so stupid I can't figure this out.  :(

1) Show that the product of five consecutive positive integers cannot be a perfect square.

2) Find all integers n such that 7 divides $ 3n^2 \plus{} 6n \plus{} 5$.

The usual mod arguments don't seem to work.

[quote]The usual mod arguments don't seem to work.[/quote]

Testing cases with mod helps for the second one.

It shows us that $ n \equiv 1, 4, \pmod{7}$

[quote="leoxnlin"]I must be so stupid I can't figure this out.  :( [/quote]
That's good for self-esteem :D

Hint for 1.[hide]  Try using integers $ n \minus{} 2, n \minus{} 1, n, n \plus{} 1$, and $ n \plus{} 2$[/hide]

A long written explanation for 2.[hide]We can rewrite the equation as $ 3n^2 \plus{} 6n \plus{} 3 \plus{} 2 \equiv 0  mod  7$, or $ 3(n \plus{} 1)^2 \equiv \minus{} 2 mod 7$.  Solving for n yields $ n \equal{} \minus{} 1 \plus{} \sqrt(( \minus{} 2 \plus{} 7k)/3)$ where k is a positive integer.  As we examine $ (7k \minus{} 2)/3$, where k is 2, 5, 8...3b-1 to make the expression divisible by 3, each value differs by 7 and the first term is 4.  In other words, we want $ \minus{} 3 \plus{} 7a$, where $ a$ is a positive integer, to be a perfect square.  This occurs when $ a \equal{} 1, 4, 12, 21, 37, 52...$.  Notice how $ \minus{} 3 \plus{} 7a$ represents $ 2^2, 5^2, 9^2, 12^2, 16^2...$  The bases differ in a pattern by 3 or 4, and these determine unique n values.  Solving for n yields 1, 4, 8, 11, 15..., and these show that $ n \equiv 1, 4 mod 7$.[/hide]

Thanks for number 2, azn and SimonM.

azn6021023, that's the exact same hint my friend gave me :| .

Quicker way for #2
[hide]
We want to find $ n$ such that
$ 3n^2 \plus{} 6n \plus{} 5\equiv 0 \pmod 7$
$ \iff 3n^2 \plus{} 6n \minus{} 9 \equiv 0 \pmod 7$
$ \iff n^2 \plus{} 2n \minus{} 3 \equiv 0 \pmod 7$
$ \iff (n \plus{} 3)(n \minus{} 1) \equiv 0 \pmod 7$
$ \iff n\equiv 1, \minus{} 3 \pmod 7$.
[/hide]

For #2 I completed the square and got
$ 3(n\plus{}1)^2\equiv 5\pmod{7}\equiv 12\pmod{7}$
$ (n\plus{}1)^2\equiv 4\pmod{7}$
$ n\plus{}1\equiv 2,\minus{}2\pmod{7}$
$ n\equiv 1,\minus{}3\pmod{7}$

So I just completed the square differently and had a longer solution.   :wink:

Now for number 1

[hide]If we write the product of the five integers as $ n(n^2 \minus{} 4)(n^2 \minus{} 1)$, where n is the middle integer, we can see that there are several cases for the factors to represent a perfect square.

Case 1:  All three represent perfect squares
But this seems impossible since no perfect squares differ by only 4 or only 1.

Case 2:  All three contain separate factors that collectively form a perfect square.

If n is even, then $ n^2 \minus{} 1$ is odd and thus must represent a perfect square on its own (which cannot happen because of case 1) or it must contain factors shared with $ n^2 \minus{} 4$.  Since the two numbers differ by 3, it also appears that the greatest common factor should also be 3.  So now we must see if any of the factors are divisible by 3.

If n is divisible by 3, then $ n^2 \minus{} 4$ and $ n^2 \minus{} 1$ are not divisible by 3.  Furthermore, since one is odd and one is even, they are actually relatively prime to each other and must represent perfect squares, which cannot happen because of case 1.

If n is not divisible by 3, it is congruent to $ 1, \minus{} 1\equiv \pmod{3}$.  $ n^2$ now becomes $ 1 \pmod {3}$, but this implies that $ n^2 \minus{} 4$ and $ n^2 \minus{} 1$ are divisible by 3, which cannot happen.

If there are any assumptions, please tell me.[/hide]

Very confuzzling, azn6021023. I get Case 1, but not Case 2.

I do not see how the different subcases within Case 2 prove anything, and I don't see how they cover all possible subcases of Case 2. Does the case where $ n$ is odd need to be covered? I know $ n$ and $ n^2\minus{}1$ are always coprime, but not for the reasons you said. And why can't $ n^2\minus{}4$ and $ n^2\minus{}1$ be divisible by 3?

I understand each paragraph alone, but I don't see how they fit together and prove the result. Your fatal assumption was that I was smart enough to get it.

Thanks. I'm a noob. :D

If $ n^2 \minus{} 4$ and $ n^2 \minus{} 1$ are both divisible by 3, then we can rewrite $ n^2 \minus{} 4$ as $ 3a$ and $ n^2 \minus{} 1$ as $ 3(a \plus{} 1)$.  Factoring 3 out of the expression gives us $ a$ and $ a \plus{} 1$.  Since they are relatively prime, then $ n$ must be divisible by $ a$ and $ a \plus{} 1$ to make the expression a perfect square.  However, this suggests that n will be greater than $ n^2 \minus{} 4$ or $ n^2 \minus{} 1$, and since we defined n as the middle integer, this cannot occur.

Oh, if $ n$ is coprime with $ n\plus{}1$ and $ n\minus{}1$, then $ n^2\minus{}1$ will also be coprime because each of its factors are coprime with $ n$.


There are 3 solutions to this problem, I believe. Here is a solution to the general case. provided no two points have the same $ x$-coordinate.

Begin by picking any of two of the three points, and draw the line through those. Let the equation of this line be $ f(x)$ (this is a function since no two points have same $ x$-coordinate). Suppose the third unused point has coordinate $ (x_{1},y_{1})$. Now if we shift our line by $ \frac12(y_{1}-f(x_{1}))$, we have that each point is a distance $ \frac12(y_{1}-f(x_{1}))$ from the line.

This can be thought of visually as drawing a line through two points, drawing the line parallel to the first through the third point, and then drawing a final line midway in between the first two. Such a line will always be a solution.

Let $ y=mx+n$ be the equation of a straight line $ d$ and $ M(x_{0},y_{0})$ a point such as $ M\not\in d$.
The distance from $ M$ to $ d$ is $  \frac{|mx_{0}-y_{0}+n|}{\sqrt{m^{2}+1}}$.
In this case we have
$  |2m-4+n|=|5m-8+n|=|7m-9+n|$
so, we have to solve this system.

[b]Generalization:[/b] [i]sans coordinates[/i]

Consider points $ A$, $ B$, and $ C$, and for any points $ P$, $ Q$, let $ M_{P,Q}$ be the midpoint of $ P$ and $ Q$.

Note that any line that passes through $ M_{A,B}$ is equidistant from $ A$ and $ B$.
Similarly, any line that passes through $ M_{B,C}$ is equidistant from $ B$ and $ C$.

Since 2 distinct points define a line, the line passing through $ M_{A,B}$ and $ M_{B,C}$ is a solution.

Symmetrically, the lines $ \overline{M_{A,B}M_{B,C}}$, $ \overline{M_{A,B}M_{A,C}}$, and $ \overline{M_{A,C}M_{B,C}}$ are all solutions.

[quote="rateesquad"]Hey guys, I have no iea how to start this.

Problem ( In exact words): It is [u]not[/u] possible to draw a straight line through three of the points (2,4), (5,8) and (7,9). However, it [i]is[/i] possible for a straight line to [i]miss[/i] all three points by the same amount: that is, therere is a line that makes the vertecal distance [i]d1[/i], [i]d2[/i], and [i]d3[/i] equal. Find the equation of this line. [Hint: Let the equation of the line be y=mx+b. Then the point P has coordinates (2, 2m+b).[/quote]

As the points are not collinear, they form a triangle.  The required lines are the lines joining the mid points of the sides.

Hence there will be three answers for this question.


I have created a document, and the left spacing is pefect, but it seems that the text wraps to the next line too early, meaning  that the right margin is way too large... is there any way to fix this?

The geometry package makes this very easy. MiKTeX will isnatll it for you if it's not already there and the manual is [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]here[/url].
So if you just want to make the right-margin 2cm then just say
\usepackage[right=2.5cm]{geometry}
It's very flexible so
\usepackage[margin=1.5in]{geometry} gives a margin of 1.5in all round and
\usepackage[left=2.5cm, right=2cm]{geometry} gives a left margin of 2.5cm and a right margin of 2cm.


inequalities
logarithms
inequalities unsolved

Isn't it Rado[b]n[/b]'s inequality ?  :? Nevertheless I'm quite sure I've seen proof of it by Jensen - I'll try to find later....

Weighted AM-GM $\Rightarrow$ Jensen's applied to $\ln x$

[quote="billzhao"]I don't know about deducing AM-GM with Rado, but proving Rado with AM-GM is surely a piece of cake.

1) is equivalent to $a_n^n \geq x_na_{n-1}^{n-1}$, which comes from weighted AM-GM:
\[
(x_na_{n-1}^{n-1})^{1/n} \leq \frac{x_n + (n-1)a_{n-1}}{n} = a_{n}
\]

2) is equivalent to $x_n \geq ng_n - (n-1)g_{n-1}$, which we derive again using weighted AM-GM:
\[
\frac{x_n + (n-1)g_{n-1}}{n} \geq (x_ng_{n-1}^{n-1})^{1/n} = g_n
\][/quote]

This is the [b]circular resoning[/b].Don't you think so,billzhao?

kunny

I don't think so. Because Rado isn't the only way to prove AM-GM. In fact, it often isn't.

The point of problem solving is to demostrate a desired result using other known/accepted results. That is exactly what we're doing here. We know that AM-GM is true, so we can use it like a blackbox (not caring how it's proved originally), because we know that it's [b]true[/b].

[quote="pengshi"]Let $x_1$, $x_2, \ldots$ be a sequence of positive real numbers. if $a_k$ and $g_k$ are the arithmetic and geometric means, respectively, of $x_1, \ldots ,x_k$, prove that
a) ${a_n^n \over g_n^n} \geq {a_{n-1}^{n-1} \over g_{n-1}^{n-1}}$
b) $n(a_n - g_n) \geq (n-1)(a_{n-1} - g_{n-1})$

Note that if you put these inequalities from 1 to n together, you get AM-GM.[/quote]
b) $n(a_n - g_n) \geq (n-1)(a_{n-1} - g_{n-1})$
Rado's inequality?In which year was created?

I think the inequality Jacobstahl's inequality.


calculus
integration
inequalities
function
trigonometry
real analysis
real analysis solved

Prove:
If the function $ f(x) \in C^1 \  ;\  x\in(a,b) $ and is zero in a or in b,
we get:
\[
\displaystyle \int^b_a (f(x)^2 \, dx \leq 4\frac {(b-a)^2} {\pi^2}
\int^b_a (f'(x))^2 \, dx
\]
We obtain the equality iff:
\[
f(x) \equiv k \sin \frac {\pi(x-a)} {2(b-a)}
\]
or
\[
f(x) \equiv k \sin \frac {\pi(x-b)} {2(b-a)}
\]

This one was also Oral Examination Polytehnique P' around 1995
P' today is PC*

Suppose $a=0 ,\; b=1$, $f(0)=f(1)=0$

$J_1=\int_{0}^{1}f'(x)^2dx, \; J_2=\int_{0}^{1}f(x)^2dx, \; J=\int_{0}^{1}\frac{f(x)^2}{\tan^2(\pi x)}dx, \;I_1=\int_{0}^{1}\frac{f(x)f'(x)}{\tan(\pi x)}dx, \;
I_2=\int_{0}^{1}\frac{f^2(x)(1+\tan^2(\pi x))}{\tan^2(\pi x)}dx$

$f$ is $C^1$ and conditions on $f(0)=f(1)=0$ imply the integrals exist.

Cauchy-Schwarz to $I_1$ gives $I_1^2\leq J J_1$ (1)

$I_2=J+J_2$

Integration by part gives $I_1=\frac{\pi}{2}I_2$

$I_1^2=\frac{\pi^2}{4}I_2^2=\frac{\pi^2}{4} \left ( (J-J_2)^2+4JJ_2 \right )\geq \pi^2JJ_2$ (2)

(1) & (2) gives $J_1\geq \pi^2J_2$

Isn't this direct application of Parseval formula?


Find $ f:R \to R $ such that :
 $ f(x)f(yf(x)-1)=x^2f(y)-f(x) $

[quote="Chang Woo-JIn"]I have a solution of this problem.

Let the funtional equation above $T$.

put $x=y=1$. we get $f(1)f(f(1)-1)=0$.

so we can know there is some real number $p$ s.t $f(p)=0$

put $x=p$. then $p^2 f(y)=0$. If $p$ is not $0$. $f(x) \equiv 0$ for all $x \in R$.

Now, we need only to consider $a=0 \Leftrightarrow f(a)=0$.

by $f(1)f(f(1)-1)=0$ (mentioned above) we get $f(1)=1$.

pu $x=1$ in $T$. $f(y-1)=f(y)-1$. or $f(x+1)=f(x)+1$. --------$(1)$

so we can change $T$ as $f(x)f(yf(x))- f(x)=x^2 f(y)-f(x)$ or $f(x)f(yf(x))= x^2 f(y)$. ----------$(2)$.

put $y=x$ in $(2)$. we get $f(xf(x))=x^2 $.

and put $x=xf(x)$ in $(2)$. we get $f(x^2 y )= f(x)^2 f(y)$. ----$(3)$.

we can know $f(x)^2 = f(x^2 ) $ by putting $y=1$ in $(3)$.

so $(3)$ is as same as $f(xy)=f(x)f(y)$. $(x\geq 0)$

How ever, cuz we know $f(-1)=-1$ from $(1)$. we get $f(xy)=f(x)f(y)$ $(\forall x, y \in R)$.

Let's product $f(y)$ each side of $(1)$. then we get $f(x+y)=f(x)+f(y)$. ---------$(S)$

we already know $f(n)=n$ where $n$ is integer.

and from $(S)$, we get $f(nx)=nf(x)$ $\Rightarrow f(q)=q$ where $q \in Q$.

cuz $f(x^2 )= f(x)^2$. we get $f(x)\geq 0 $ for all $x\geq 0$.

so $f(x)$ is increasing function.

Let $u_i, v_i$ be the two rational sequence, s.t $u_i < x < v_i$ and $u_i , v_i \rightarrow x$.

then we get $f(u_i)=u_i\leq f(x) \leq f(v_i)=v_i$.

so $f(x)=x$ for all $x\in R$.

$Q.E.D.$[/quote]

    Nice solution although you forgot $ f(x) \equiv 0 $

:D  I wrote it above. (line 5)

thank you for nice problem!


Έστω τρίγωνο $ABC$  και ο περιγεγραμμένος του κύκλος που έχει κέντρο το $O$ .Έστω ευθεία που περνά από το $O$ και τέμνει τις $AB$ , $AC$  στα  $K,L$  αντίστοιχα .Έστω $M,N$  τα μέσα των $BL$  και  $CK$  αντίστοιχα .Να δείξετε ότι
$\widehat{MON}=\widehat{BAC}$

Νομίζω ότι υπάρχει λάθος. Την κοιτάς ξανά ??
 :)

έχεις δίκιο απόλυτο .Είχα γράψει ΒC αντί ΑC .
Κοίτα το πρώυο ποστ διορθωμένο πλεον


group theory
abstract algebra
superior algebra
superior algebra solved

Let G be a finite group and let f be an isomorphism of G. M={x in G | f(x)=x<sup>-1</sup>}. Show that if |M|>(3/4)|G| then G is commutative.

Some ideas:
Let g(x)=f(f(x)) for all $x\in G$.
If $x\in M$ then $f(x)=x^{-1}\Rightarrow f(f(x))=f(x^{-1})=f(x)^{-1}=(x^{-1})^{-1}=x$.

Let $A=\{x\in G|g(x)=x\}$. It is easy to see that A is a subgroup of G and $M\subset A$. This means $|A|\geq \frac{3}4|G|$. From this we conclude that A=G.
So f(f(x))=g(x)=x for all $x\in G$.

Denote by $C(x)$ the centralizer subgroup of $x$ in $G$, that is $\{y \in G | xy = yx\}$. Further, I use $f^2 = id$ as proved by amfulger.

1. $C(m) = G$ for any $m \in M$.
Proof: Let $n \in M, nm \in M$. By the given cardinalities, the set $N(m)$ of such $n$ has cardinality $|N(m)| >\frac 12 |G|$. Then $mn = f(m)^{-1}f(n)^{-1} = (f(n)f(m))^{-1} = f(nm)^{-1} = nm$. So $C(m) \supseteq N(m)$. But since $C(m)$ is a subgroup of $G$ and $|C(m)| \geq |N(m)| > \frac 12 G$, we have $C(m) = G$.

2. $xy = yx$ for any $x,y \in G$
Write $x,y$ as $x = m_1m_2, y = n_1n_2$ with $m_1,m_2,n_1,n_2 \in M$, which again is possible by the given cardinalities. Elements of $M$ commute freely as proved above, so $xy = m_1m_2n_1n_2 = n_1n_2m_1m_2 = yx$.

yuk, i hope this is correct.

You could have finished the problem easier.

Let $Z(G)=\{x\in G| C(x)=G\}$.

Part (1) in luqash's proof shows that $M\subset Z(G)$. But Z(G) is a subgroup of G with more than (3/4)|G| elements, hence Z(G)=G, hence G is commutative.

you are right, your solution is more elegant. I wonder whether this can be sharpened to $|M| > \frac 12 G$...


function
real analysis
topology
advanced fields
advanced fields unsolved

Hello,

Can anyone solve the following problems in Rudin Chapter 11?

3. If $ \{f_n\}$ is a sequence of measurable functions, prove that the set of points $ x$ at which $ \{f_n(x)\}$ converges is measurable.

14. Prove that a complex function $ f$ is measurable if and only if $ f^{ \minus{} 1}(V)$ is measurable for every open set $ V$ in the plane.

3. Using the Cauchy criterion, $ \{f_n(x)\}$ converges iff $ \forall \epsilon > 0$ $ \exists N$ such that $ |f_n(x) \minus{} f_m(x)| < \epsilon$ for all $ m,n\ge N$. In fact, it suffices to check this condition for countably many values of $ \epsilon$ such as $ \epsilon \equal{} \frac {1}{k}$. Thus, the set of convergence is
$ \bigcap_{k \equal{} 1}^{\infty} \bigcup_{N \equal{} 1}^{\infty}\bigcap_{m \equal{} N}^{\infty}\bigcap_{n \equal{} N}^{\infty} \{x\colon |f_m(x) \minus{} f_n(x)| < 1/k\}$.

I don't know what can happen if $ f_n$ take values in a noncomplete metric space, or in a topological space.


How many integers are there between $ \frac{13}{5}$ and $ \frac{15}{2}$?

If you look at the fractions, the first one is a little over two and the second one is a little over 7. Between 7 and 2, there are 5 integers so the answer will be 5.

Just calculate the fractions: 2.6 and 7.5. {3~7} all work, for a total of 5


[hide="The problem"]The teacher wrote the whole numbers from 1 to something,[hide="continue"] but somebody erased one number so that the average of the numbers on the board is 13 and 9/13.[/hide] [hide="continue"]Which number was erased?[/hide][/hide]  [hide="Will You Solve It?"] I don't know [/hide]

What are you saying?

This, [hide] n(n+1)/2n - x = 126/13
[/hide]

or [hide] n(n+1)/2n -x = 13 or n(n+1)/2n -x = 9/13[/hide]

Please avoid posting the same challenge to multiple Forums.  Choose the forum that is most appropriate for the topic.

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=239341#p239341]duplicate[/url] in the MATHCOUNTS Forum has already been solved.

ok, i wasn't sure where it belonged so i put it in both

Ok, I wasn't quite sure if you meant that it was a mixed number or not.

If you want to continue to post on this problem, post it under the topic in the Mathcounts section.


geometry
circumcircle
trigonometry
similar triangles

$ ABCD$ is a cyclic qudrilateral with $ AB\equal{}8$, $ BC\equal{}4$, $ CD\equal{}1$, $ DA\equal{}7$.  Let $ O$ and $ P$ denote the circumcenter and intersection of $ AC$ and $ BD$ respectively.  The value of $ OP^2$ can be expressed as a common fraction $ \frac{m}{n}$.  Find the remainder when $ m\plus{}n$ is divided by $ 1000$.

$ a\equal{}AB, b\equal{}BC, c\equal{}CD, d\equal{}DA$ are given sides, $ p\equal{}AC,q\equal{}BD$ are diagonals, $ s \equal{} \frac{_{a\plus{}b\plus{}c\plus{}d}}{^2}\equal{}10$ is semiperimeter, $ R$ the circumradius. Brahmagupta's formula gives the area as $ S \equal{} \sqrt{(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)} \equal{} 18.$ Areas of the $ \triangle DAB, \triangle ABC, \triangle BCD, \triangle CDA$ are $ S_A, S_B, S_C, S_D.$

$ \frac{daq \plus{} bcq}{4R} \equal{} S_A \plus{} S_C \equal{} S \equal{} S_B \plus{} S_D \equal{} \frac{abp \plus{} cdp}{4R}$

Multiplying the 2 formulas, $ S^2 \equal{} \frac{(da \plus{} bc)(ab \plus{} cd)pq}{16R^2},$ where $ pq \equal{} ac \plus{} bd \equal{} 36$ by Ptolemy theorem $ \Longrightarrow$

$ R^2 \equal{} \frac{(da \plus{} bc)(ab \plus{} cd)pq}{16S^2} \equal{} \frac{60 \cdot 39 \cdot 36}{16 \cdot 18^2} \equal{} \frac{65}{4}.$

Since $ S \equal{} \frac{pq}{2} \sin \widehat {APB},$ where $ S \equal{} 18, pq\equal{}36,$ it follows that $ \sin \widehat{APB} \equal{} 1,$ $ AC \perp BD.$ Then $ abcd \equal{} 4R^2 (R^2 \minus{} OP^2)$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=191162[/url]),

$ OP^2 \equal{} R^2 \minus{} \frac{abcd}{4R^2} \equal{} \frac{65}{4} \minus{} \frac{224}{65} \equal{} \frac{3329}{260}.$

$ AB^2\plus{}CD^2\equal{}BC^2\plus{}DA^2$ so $ AC\perp BD$.
let $ M$ and $ N$ be the projections of $ O$ onto $ AC$ and $ BD$, respectively.
if we let $ PC\equal{}x$, by similar triangles we have $ PD\equal{}\frac{7}{4}x$, $ PA\equal{}14x$ and $ PB\equal{}8x$.
Ptolemy's theorem: $ \left(14x\plus{}x\right)\left(\frac{7}{4}x\plus{}8x\right)\equal{}8\plus{}28\implies x^2\equal{}\frac{16}{65}$.
$ MP\equal{}\frac{14x\minus{}x}{2}\equal{}\frac{13}{2}x$ and $ NP\equal{}\frac{8x\minus{}\frac{7}{4}x}{2}\equal{}\frac{25}{8}x$
$ OP^2\equal{}MP^2\plus{}NP^2\equal{}\frac{3329}{64}x^2\equal{}\frac{3329}{260}$


[quote="liyi"]$R$ is a ring. $L$ is the intersections of all non-zero left ideals. $L^2\neq 0$. Show that $R$ has (multiplicative) identity.[/quote]

What is your precise definition of a ring ? For me it always have a multiplicative
unit ....
:-(

Fabrice

A ring is a set R together with two binary operators $+,\cdot$ and satisfying the following conditions:
1) $(R,+)$ is an additive group; (Here it is additive implies that it is commutative)
2) $(R,\cdot)$ is a semigroup;
3) Left and right distributivity

I'll start. Maybe someone else will pick it up.

1. $L=Ry$ for some $y\in L$ and $y^2\neq 0$.
Proof: [b]$L$ is obviously a left ideal of $R$.[/b]
$L^2\neq 0$ so there exists $x,y\in L$ such that [b]$xy\neq 0$[/b]. This means that $xy\in Ry\neq 0$.
Now because $L$ is a left ideal, $Ry\subset L$. By $L$'s definition, $L\subset Ry$. [b]This leads to $L=Ry$. Also, $y\in L$  because it was chosen as such.[/b]

$x\in L$ means there exists $r\in R$ such as $x=ry$. So $ry^2\neq 0$, so $y^2\neq 0$.

2. $Rz=0\Rightarrow z=0$.
Proof: $Rz=0$ implies that the additive subgroup, $<z>$, generated by $z$ is a left ideal.
If $z\neq 0$ then $<z>\supset L=Ry\ni y$.
$z\in R$ and $Rz=0$ lead to $z^2=0$. So $<z>^2=0$, leading to the contradiction $y^2=0$.
So $z=0$.

That's all for now.

$L=Ry\ni y\Rightarrow (\exists ) e\in R,\ ey=y$

(3) $e$ is a right identity.
Proof: Let $a\in R$ and let $u=ae-a$. If $u\neq 0$ then using (2), $Ru\neq 0$, so $Ru\supset L=Ry\ni y$ so there exists $l\in R$ such as $l(ae-a)=lu=y$.
Multiply to the right by $y$ to get $y^2=l(ae-a)y=l(aey-ay)=l(ay-ay)=0$. This contradicts $y^2\neq 0$.

(4)$e$ is the multiplicative identity.
Proof: $r(ea-a)=(re)a-ra=ra-ra=0$. So $R(ea-a)=0$. Using (2) again the last yields $ea=a$. So $e$ is a multiplicative identity.

Nice. Thank you.

Now go on to prove this one: (based on the problem above)
If $R$ is simple then it is a division ring.


algorithm
number theory
Euclidean algorithm
relatively prime
greatest common divisor

prove the chicken mcnuggets theorem will only work for numbers which are relativley prime

[hide="solution"]
Assume the two numbers, say $ a$ and $ b$ are not relatively prime. Then we can write them as $ a=kx$ and $ b=ky$ where $ k,x,y$ are positive integers and $ k>1$. By the Chicken McNugget theorem, the maximum integer unachievable in the form $ am+bn$ for positive integers $ m$ and $ n$ should be $ ab-a-b=k^{2}xy-kx-ky=(k)(kxy-x-y)$. Assume that $ ab-a-b$ is in fact the greatest integer unacheivable in the form $ am+bn$ Notice that, by the Euclidean algorithm, all numbers in the form $ am+bn$ are multiples of $ k$. Let us consider $ z=(k)(kxy-x-y)+1$. Obviously, $ z$ is not a multiple of $ k$ and thus cannot be expressed in the form $ am+bn$. Also, $ z>ab-a-b$. This contradicts the assumption that $ ab-a-b$ is the greatest integer unacheivable in the form $ am+bn$. Thus we have proven that the Chicken McNugget theorem only applies to relatively prime integers.[/hide]

here's a bit simpler solution (might be a condensed version of the other one)

if the two numbers are not relatively prime, we have a=kx and b = ky, as FightTheTide pointed out.

Then any combination of the two will be a multiple of k because knx+kmy = k(nx+my).

So only numbers congruent to 0 mod k can be attained, meaning there are infinitely many unattainable numbers.

Yes, I just felt like typing the whole thing out. The basic point was to say that all different combinations of sums of multiples of the number would be a multiple of their greatest common divisor, so any number that is not a multiple of that greatest common divisor is unattainable.


Using every digit only once, write three numbers that fit this sequence:

1:3:5

(not found in MATHCOUNTS)

And the ratio is:
[hide="look if you dare"]269 : 807 : 1345[/hide]
 :)  :)  :)  :lol:

[quote="hwenterprise"]So...how did you get 801...next time, may you please show your WORK?![/quote]

huh?

I dont see an 801 anywhere!!! I see an 807...and how do you show your work on this type of problem??

[hide]I kept guessing until I got it. All I knew for sure is that:

 the first 2 numbers had to be 3 digits each, and the third one had to be 4 digits.

the first number had to be greater than or equal to  213 (so that the 3rd number would have 4 digits), and that it had to be less than or equal to 329, since any number greater than that would make the second # have 4 digits.

the last number had to end in a 5 or a 0

the first number couldn't end in a 5 or a 0, because any number ending in 0 would have the other numbers ending in 0, and any number ending in 5 would have the last number ending in 5.

Umm.... that's all I can think of, but if I think of more things that you can't have, I'll post them.
 :)  :)  :)  [/hide]

I believe this same problem was posted in the intermediate forum, with the condition that all three numbers be three digits.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33297

I and some other members of this forum posted our solutions there.

I think this was posted somewhere in The Basics middle school forum too :D

Sorry, GoBraves. I didn't mean to write 801. I was probably off task. I meant 269. elementsofmathkid, thanks for telling me.

Sure. :) Sometimes I have all of the work in my head and on paper and don't remember to show it. So... thenother people don't have a clue as to what I'm talking about.  :(

 :)  :)  :)


Let $m_{1},m_{2},...,m_{n}$ be mutually distinct integers. Prove that there exists a $f(x)\in\mathbb{Z}[x]$ of degree $n$ satisfying the following two conditions:
a)$f(m_{i})=-1\forall i=1,2,...,n$.
b)$f(x)$ is irreducible.

It was proved on the forum that $(X-a_{1})(X-a_{2})...(X-a_{n})-1$ is always irreducible if $a_{i}$ are pairwise distinct integers.

Suppose that $f(x)=(x-m_1)(x-m_2)\dots (x-m_n)-1=g(x)h(x)$ for nonconstant monic polynomials $g(x), h(x)\in \mathbb{Z}[x]$. Then for all $1\leq i\leq n$ we have $g(m_i)h(m_i)=-1$.

Because $g(m_i), h(m_i)$ are integers, we have $g(m_i)=\pm 1, h(m_i)=\mp 1$ so $g(m_i)+h(m_i)=0$.

But the polynomial $p(x)=g(x)+h(x)$ has degree less than $n$ and has $n$ roots (namely $m_1, m_2\dots, m_n$) so $p(x)=0\implies g(x)=-h(x)$.

Then $f(x)=(x-m_1)(x-m_2)\dots(x-m_n)-1=-g(x)^2$. But $(x-m_1)(x-m_2)\dots(x-m_n)-1$ has leading coefficient $1$, and $-g(x)^2$ has leading coefficeint $-1$, a contradiction.

From the above we can always take $f(x)=(x-m_1)(x-m_2)\dots(x-m_n)-1$ satisfying the two conditions of the problem.


calculus integration function logarithms derivative algebra domain

Integrate (x/(x-r))^1/2 with respect to x, where r is a constant.

Btw, how could I type in those mathematical symbols?

[quote="huweizhi"]So, what do you do next??? Obviously, you can't integrate with respect to u at this stage.[/quote]

$\int f(x) \cdot dx \, = \, \int f(x(u) \cdot d(x(u)) \, = \, \int g(u) \cdot du \, = \, G(u)+C \, = \, G(u(x))+C \, = \, F(x)+C$

g(u) is a rational function easy to integrate

[quote="greengrass"][quote="benorin"]Assuming that $x>r,$ and that $x>0$ then we have

$\int \sqrt{\frac{x}{x-r}}\, dx = \sqrt{x(x-r)}+r\log \left( \sqrt{x}+\sqrt{x-r}\right)+C$

This I know because I went to [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic]www.quickmath.com[/url] and put sqrt(x/(x-r)) in the box marked "Integrate".[/quote]

You don´t need assume anything writing solution as

$\sqrt{x(x-r)}+r \ln \left( \sqrt{x}+\sqrt{x-r}\right)+C \, = \, \sqrt{x(x-r)}+\frac{r}{2}\ln \left( \sqrt{x}+\sqrt{x-r}\right)^{2}+C \, = \sqrt{x(x-r)}+\frac{r}{2}\ln \left| 2x-r+2 \sqrt{x(x-r)}\right|+C$[/quote]

You may well be correct! I made little effort to inspect the integration process or check by differentiation, yet the basis of my assumptions stand thus: 1) I figured that the reader was only interested in real-valued functions [so I limited the domain of x to produce such]; and 2) I made simplifications to the answer given to me by the webmathematica engine (here http://www.quickmath.com) that require such assumptions to be valid and respect (1).

[quote="benorin"]You may well be correct! I made little effort to inspect the integration process or check by differentiation, yet the basis of my assumptions stand thus: 1) I figured that the reader was only interested in real-valued functions [so I limited the domain of x to produce such]; and 2) I made simplifications to the answer given to me by the webmathematica engine (here http://www.quickmath.com) that require such assumptions to be valid and respect (1).[/quote]

OK  :)

[quote="huweizhi"]Is your webmachine really reliable? See what you get if you put in lnx![/quote]

$\int \ln x \cdot dx$ must be expresed Log[x] in these webmachine

Otherwise ln is a constant for quickmath

Was this the problem?


AoPSwiki
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Unfortunately, the AoPSWiki hasn't been very popular recently. It's a terrific resource that I feel will become as valuable as Google and Mathworld are to you already. So to get more people to contribute to the AoPSWiki we're going to hold a little contest!

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$ \bullet$ No prior experience with wikis is required. You can learn how to edit and create new articles here.
$ \bullet$ DO NOT COPY WORK FROM OTHER SOURCES. Everything must be in your own words. You will be disqualified immediately if you are caught doing this. This is a violation of the copyright policies of the AoPSWiki.

Now for the prize! The user whose contributions are judged to be best will receive bragging rights and will be able to talk about how they won the AoPSWiki PageDay #1. Sorry, no monetary prizes :). But bragging rights are almost as good.  It's likely that we will hold future PageDays which will have prizes.  So this can be viewed as practice for those.

I split the meta-discussion about the wiki and the PageDay - it's in the AoPSWiki forum.  Please keep discussion in this thread about contributions to the Wiki, such as what's needed and what you can add.  If you'd like to talk about why you do or don't want to participate in the wiki, you can continue that discussion in the thread in the AoPSWiki forum.

Here's the start of a list of things that people can work on:

$\bullet$ Adding competitions to the wiki (e.g. [url=http://www.artofproblemsolving.com/Wiki/index.php/Arkansas_mathematics_competitions]Arkansas math competitions[/url] doesn't have any competitions listed)
$\bullet$ Adding contest problems (use [url=http://mathideas.org/public/scripts/AoPSWiki/]chess64's webpage[/url] to do most of the work for you)
$\bullet$ In general, the non-math subjects in the wiki are very sparse.  There's a lot to be added in those areas.  Please restrict what you put in the wiki to topics that could get their own forum on AoPS, e.g. an article on Shakespeare isn't needed in the wiki.
$\bullet$ Add historical information about mathematicians or historical math events.
$\bullet$ Expanding stubs.
$\bullet$ Improving already existing articles.  Many tutorial articles are in need of example problems, or excercise problems.

Others are welcome to add to this list.

Many contest problems need to be added, and even more need solutions.

Other things to work on:

$\bullet$ Making new pages on wanted topics.
$\bullet$ Adding images to pages that need them (Many AIME and AMC problems have no images).
$\bullet$ Adding to new pages.

You mean math journalism? Or for that matter non-math? Whoa! I am good at this! I'll give you as much style, formatting, grammar, spelling, readability and clarity as you want, not to mention content, as long as you do not ask for mathematial knowledge ...

I'm in!

[color=blue]May I make a contribution in another language?[/color]

[quote="solafidefarms"]$\bullet$ DO NOT COPY WORK FROM OTHER SOURCES. Everything must be in your own words. You will be disqualified immediately if you are caught doing this.[/quote] And if you've written the thing you copied yourself? (e.g. if you're the author of a wikipedia entry)

And how will you coordinate this contest? The system isn't provided on people trying to add the same content at the same moment (which will inevitably happen if you just let everything run).

I don't think we'll have to worry about the issues you're raising about coordinating the contest.  It will be very rare that two people will be editing the same thing.  Don't worry too much about these details - if you do, nothing will ever get done.

[quote="Peter VDD"]And if you've written the thing you copied yourself? (e.g. if you're the author of a wikipedia entry)

And how will you coordinate this contest? The system isn't provided on people trying to add the same content at the same moment (which will inevitably happen if you just let everything run).[/quote]

If it's in a wikipedia article, then it is subject to their copyleft policy.  Anyone can reprint it iff they give the right for others to reprint it.  AoPS doesn't accept the copyleft.  I'm not really sure about the specifics of this.  You'd have to get AoPS to change their policy (i.e. through a private discussion with one of the admins and actually, talking to MCrawford would probably be the best since he dealt with that though he's not an admin anymore).

If you plan on making a huge edit to a page and you don't plan on actually updating it for a while (i.e. you want to do all the editing first and then submit the page) such that it very plausible that someone else will edit the page then you should quickly edit the page to say that it's Under Construction before you start working on your edit so that others know what you're doing.  (I can't remember if we have a template for this or not...probably not.)

Please use English for all of the articles.

[quote]You mean math journalism? Or for that matter non-math? Whoa! I am good at this! I'll give you as much style, formatting, grammar, spelling, readability and clarity as you want, not to mention content, as long as you do not ask for mathematial knowledge ...
[/quote]

We're asking for good, solid, [i]understandable[/i] articles in the wiki.  Obviously the quality of an article depends on the math (mathematical knowledge) in it.  If you need me to explain further what was meant by that statement, I will.

[quote="joml88"]If you plan on making a huge edit to a page and you don't plan on actually updating it for a while (i.e. you want to do all the editing first and then submit the page) such that it very plausible that someone else will edit the page then you should quickly edit the page to say that it's Under Construction before you start working on your edit so that others know what you're doing.[/quote] That was indeed the sort of things I'd add as guidelines for the contest. It's not the type of worrying that will reduce the amount of submissions, but it's just a good habit to do such things, also for later contributions. It's not necessary that everyone finds out these things by losing work from not doing it. ;)

[quote="funcia"]So you might have another (wiki day)too??? I'm probably flying home from NJ the 4-5(not sure of which but packing and all...)[/quote]

Yes, it's likely we will.  Not sure when though.  It also depends on how this one goes :)

I may host an unofficial wikiday some time in future.
By the way, I need to know the kind of topics allowed. Can I write about Edward Lear and Lewis Carroll and Douglas Adams, and maybe Age of Empires Conquerors and fifa 98 and fifa 07, and perhaps linked lists and binary search trees - you know, you get the general idea.

I just took a really good look at the wiki, and couldn't help being reminded of some yellow pages. Why, I myself can double it over a weekend  :P

Sadly my ms internet explorer at home is not working, so I must use either the bitcomet explorer or the mozilla firefox, and neither seemed to eager to open the Aops page till last evening. And I am leaving college right now ...

Let's see ...

It's not volume you'll be judged on, it's quality. If you do a lot of bad pages, you won't win if somebody does a couple of really good pages.

I won't do a single bad page - that's a promise. All I need to know is whether I can post on anything (well, anything G-rated).

And yes, my home mozilla is working fine! I am all rearing to go at it! I believe it starts in 12 and a half hours right?

Bubka, I think it's pretty clear that the AoPSwiki is about math, not about Age of Empires... Anything that relates somehow to the highschool math done here.

[quote="The Front Page"]The AoPSWiki is a project for supporting educational content useful to students of math, science, computer science, technology, and other problem solving topics.[/quote]

Does that clarify what you may and main't write?

Only serious matters?
I was hoping to put in something on Brother Bear :(

problem solving....like if i can't make spaghettios?  :maybe:
OK well does that mean we can put thinga about books conerning mathematicians?  :) I will be able to write tomorrow after all!!!!

Nice idea! I have a book called Men of Mathematics!

Did quite a bit on physics, but a lot more to go. How are you guys doing? Keep reporting here.

I am going to start putting in a bunch of AMC problems (and solutions).

Just curious, when writing solutions to the AMC problems, do you want us to write our own solutions, or write one based on the official solution?

EDIT:  I started 2005 AMC 10A Problems, but I have to go for a little while, but I will finish later.

can we have an update as to the current top 5?

(i think i'm in it)
-jorian

I'm just wondering, is it ok to copy and paste from pages already in the AOPS wikie?

Like what I'd like to have is a list of computer scince compeitions, like this list for math - http://www.artofproblemsolving.com/Wiki/index.php/List_of_mathematics_competitions#United_States

So would it be ok to copy and paste from that for the computer science page?

I shouldn't be surprised if it is between me and i_like_pie at the top, and if I am in the lead. Delving into vector spaces now ...

Going to bed now. Will return tomorrow with renewed vigour!

[quote="Xantos C. Guin"]I believe your own solutions.

BTW: Most of the 2005 AMC 10A solutions have already been written. Also, if you want to post problems and/or solutions, be sure to follow the correct format.[/quote]

Yes, I know that, I am going to finish the few that aren't, make the page of all the problems, and then work on other years.

I'm also posting other problems.

I can't believe that I forgot to make the problems page for 2005 AMC 10A. Thanks for doing that.

So who won? HAs the winner been decided?

It will take time to go through the edits and choose a winner.  Expect a decision by Friday, hopefully sooner.


First, the cases n>=1. In this case, let us prove that it is at least 3/4. This becomes  \sum a^n+ \sum a^(n+1)>=3/4(2+ \sum a+ \sum ab). But this is easy, since  \sum a^n \geq  \sum a and  \sum a^(n+1) \geq  \sum a 2  \geq  \sum ab and  \sum a,  \sum ab>=3. Thus, the minimum is 3/4. Maximum doesn't exist. Take a=x, b=1 and c=1/x, with x very large.
  For n=0 the infimum is 0. For this take a=x^2, b=x and c=1/x^3 with x large. The supremum is 1. It can be easily shown.

yes, the min is 3/4.
For n=k with k is a real number (k>1) , I have a long sol for this pro. I'll send it to you by mail, master.
    Darkmaster

And I have a very short solution! Of course, it's enough to prove that  \sum a^k \geq  \sum a and  \sum a^(k+1) \geq  \sum a 2  and then we repeat all computations above written. For this we have from Cebasev that   \sum a^k \geq ( \sum a)( \sum a^(k-1))/3 \geq  \sum a from AM-GM. And the same for the other one.


a+b+c = abc , positive reals

$\sum_{cyc} \frac{1}{\sqrt{a^2+1}} \le \frac{3}{2}$

It can be solved by triangle function. :)

We use a trigonometric substitution.
To avoid confusion  use variables X,Y,Z  instead of a.b.c
Then X+Y+Z =XYZ.
Since X,Y and Z are positive real numbers there exist acute angles A,B and C such that X=TanA Y=TanB, Z=TanC and the given condition on X,Y and Z becomes TanA +TanB +TanC =TanATanBTanC which implies that A,B and C are the angles of a triangle.

The inequality to be proved now becomes : CosA +CosB +CosC <=3/2

3-2(CosA +CosB+CosC) =(SinA-SinB)^2  + (CosA+CosB-1)^2  >=0

There are many other possible proofs

This is a classical trig sub ineq. But here is a solution without trig.

Let $x=1/a$, etc. Then the constraint becomes $xy+yz+zx=1$. And the inequality becomes $\sum_\mathrm{cyc}\sqrt\frac{x^2}{x^2+1} \leq \frac{3}{2}$.

Proof is as follows, by AM-GM
\[
\sum_\mathrm{cyc}\sqrt\frac{x^2}{x^2+1}
= \sum_\mathrm{cyc}\sqrt\frac{x^2}{x^2+xy+yz+zx}
= \sum_\mathrm{cyc}\sqrt{\left(\frac{x}{x+y}\right)\left( \frac{x}{x+z} \right)}
\leq \frac{1}{2}\sum_\mathrm{cyc}\left( \frac{x}{x+y} + \frac{x}{x+z} \right)
=\frac{3}{2}
\]


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Can anyone help me prove this:
Prove the inverse of triangular matrix is triangular matrix?

Here's another way: if the given matrix is an $ n\times n$ matrix over a field $ F$ (such as $ \mathbb{R}$ or $ \mathbb{C}$), let $ V_k$ denote the subspace of $ F^n$ where all parts beyond the $ k$th are zero. (So in $ F^3$, $ V_1$ is all elements of the form $ (a,0,0)$, $ V_2$ is all elements of the form $ (a,b,0)$, $ V_3$ is the whole space.)

Then any matrix $ M$ is upper triangular if and only if $ M(V_k)$ is contained in $ V_k$ for all $ k$. In particular, this is true of $ A$, and we will show true of $ A^{\minus{}1}$ (provided the inverse exists, of course: if $ A\equal{}O$, we would have problems).

So $ A(V_k)\leq V_k$ for every $ k$. If for some $ j$, we had $ A(V_j)<V_j$, then multiplication by $ A$ would map the space $ V_j$ to a space of strictly smaller dimension. So there would exist nonzero $ {\bf u}$ in $ V_j$ such that $ A({\bf u})\equal{}{\bf 0}$, but this contradicts $ A$ being invertible.

So $ A(V_k)\equal{}V_k$ for all $ k$. Applying $ A^{\minus{}1}$, we get $ V_k\equal{}A^{\minus{}1}(V_k)$ for all $ k$, so $ A^{\minus{}1}$ is upper triangular.

more concretely, say that $ A$ is $ n\times n$ upper triangular with nonzero entries on the diagonal, and form the augmented matrix $ (A|I_n)$.  Then we perform the same row operations on both sides of $ |$ to get $ (I_n|B)$ where $ B \equal{} A^{ \minus{} 1}$.  But if we do this in the 'obvious' way, all of the row operations which transform $ I_n\to B$ preserve upper triangularity, and $ I_n$ is upper triangular, so ...

[quote="The Captain"]It follows immediately from Cayley-Hamilton's theorem.[/quote]

this is really cute and I'll elaborate for the OP.  By cayley hamilton, there is a polynomial (the characteristic polynomial) $ P$ with $ \text{deg }(P) \equal{} n$ such that $ P(A) \equal{} 0$.

So $ \alpha_n A^n \plus{} \alpha_{n \minus{} 1} A^{n \minus{} 1} \plus{} \ldots \plus{} \alpha_0I_n \equal{} 0_n$.  Moreover $ \alpha_0\not\equal{}0$ (why?).  Multiplying both sides by $ A^{ \minus{} 1}$ gives
\[ \minus{}\frac{1}{\alpha_0}(\alpha_n A^{n \minus{} 1} \plus{} \ldots \plus{} \alpha_1I_n) \equal{} A^{ \minus{} 1}
\]
But all of the summands on the LHS are upper triangular.


1. A 30 kg mirror is hung at the midpoint of a 68 cm long wire. The ends of the wire are attached to the mirror at point sthat are 40 cm apart. If the wire will withstand a tension of 200 N before it breaks, is the mirror safe?

2. A 20 kg bundle of newspapers is lifted up 0.72 m by a 65 N force and placed in a wagon. A child exerts a constant force of 42 N at an angle of 25 degrees to the horiontal. The wagon is pulled 10 m over level ground and then 8 m down a hill with a grade of 5 degrees. Calculate the total work done by the child.

For the second problem, $ W\equal{}\mathbf{F}\cdot\mathbf{d}\equal{}Fd\cos\theta$.


probability
analytic geometry
graphing lines
slope

i dont get the rules

all i see is a ashtray (for smokin'  with no smoking in it!)

wat am i supposed 2 do?

i think these are called [i]oxymorons[/i]  :wink:

Actually, I think oxymorons are just phrases that consist of two words that are antonyms of each other, e.g. "jumbo shrimp". These are probably just contradictions. (But Post-A-Paradox sounds better than Post-A-Contradiction.  :P )

EDIT: Gah, the image's too huge. I'll just post the URL: http://lh3.ggpht.com/_Z6aJpvdS97M/SU8u9VkO2aI/AAAAAAAADkk/UD2x6 Zb8C4/St+George+Snow+and+House+Pictures+002.JPG

[quote="saszs"]someone clear up the rules.

what kind of ashtray has a no smoking sign on it.

many won't stop smoking.[/quote]

You post something that contradicts itself. For example, in the first one, the object is an ashtray, but it has a no smoking sign. However, ashtrays are where people place their cigarettes or whatever (jeez, i dont smoke! How should I know?)

You have to copy the entire thing, not just the blue part. It got messed up.

delete the space between the 6 and the Z

is it showing up for u?
cuz it's still not showing up

instead of whining about why it doesn't make sense :roll:
follow the rules and stop spamming

solution: it is not actually an ashtray.
the fallacy of hasty generalization
next paradox:
[img]http://abyss.uoregon.edu/~js/images/jourdain_cards.gif[/img]

They are not the same card.

I don't have a paradox so anyone else is free to post  :lol:

post an image (if possible) of your solution
if you don't have a new paradox then please don't post
next paradox
say you are in a game show
at the end, the host shows you three doors that have 2 goats and a car
you pick #1
he opens #3; it has a goat
is it advantageous to switch to #2?
If so, why?

yes, its simple probability, 2 has 2/3 chance and 1 has 1/3 chance

[quote="Eragon"][b]post an image (if possible) of your solution
if you don't have a new paradox then please don't post[/b]
[/quote]
 :spam:  :furious:
[i]read the rules[/i]
[i][b][color=red][size=200]PLEASE,READ THE RULES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/size][/color][/b][/i]

the slopes of the triangles are not equal, so technically they are not triangles. for more see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275916[/url]

[img]http://www.guy-sports.com/fun_pictures/no_entry.jpg[/img]

gee i wonder...


Hello it's just me again basically the same question as my other post but i decided that that one was a little long and thought I would prolly get more replies if I downsized. Unfortunately it would no longer let me edit so I decided to just write a new one instead.

I try not to really let this worry me much but sometimes the thought does creep up on me and it's rather depressing. I mean I really love math and want to like be a mathematician (I mean seriously what could possibly be cooler than that) but sometimes the thought does creep up on me that based on my current performance on the AMC etc. it would seem I am not very good at math. Especially when I consider that many of the great mathematicians that I read about and admire apparently were quite good at this sort of thing. It's a problem cause I like math a lot it really is a very big part of who I am and how I see the world around me. A very very small part of me very rarely but sometimes wonders whether I would be better off not liking math then I wouldn't care or need to worry, but then again I can't imagine myself actually disliking something so beautiful.I just don't know how bad of a sign this is. Like I said it's quite depressing so I make it a point not to think about it but every once in awhile it does make me wonder just a little bit whether or not I can actually be successful in what I want to do. So what I am wondering is how much concern do you guys think this warrants?

Chill out.  You're still young.  Do what comes naturally for now, and later you'll know if math is right for you.  A lot of high school math enthusiasts find out that they really like other things in college.  (High school versions of all subjects are pathetic versions of the real thing.  It's just that a pathetic version of math still allows for some glimmers of excitement.)

As for your AMC concern, don't worry about it.  It's not a big deal.  AMC says *something* about your math aptitude, but not as much as some people think.  To take an extreme example, if you decide in 9th grade that your goal is to qualify for AIME before graduating high school, and you *actively work toward that goal* and fail it, then *maybe* you should give up on becoming a mathematician.


To a prime number m, written in base 10 as

m = \sum(1\leq k\leqn)ak10k

where a1, a2, ..., an are digits, we associate  the polynomial

pm(x) = \sum(1\leq k\leqn)akxk.

Prove that any such polynomial is irreducible in Z[x].


Cho 20 diem tren 1 duong tron.Co 2 nguoi choi voi luat choi nhu sau:nguoi thi nhat ke 1 duong thang qua 2 diem bat ki.Nguoi thu 2 cung ve 1 duong thang bat ki nnhung ko duoc cat duong thang cua nguoi 1.Nguoi nao ko ve duoc duong thang nua se thua.Hoi ai se la nguoi chien thang?Vi sao? :)

Nobody helps me?Please giev me a solution!!!! :(

ghjk nen post o muc nao do tren forum thao luan thi hon  :)

Tùy vào sự phân bố của 20 điểm trên đường tròn nữa.
Nhưng nếu là người vẽ đầu thì cơ hội thắng cao hơn rất nhiều.

chắc chắn người thứ 2 sẽ thắng
ở bất kì trường hợp nào người thứ 2 đều thắng

Ko nguoi thu 1 luon thang! Cach giai la nguoi thu nhat ke 1 duong thang chia duong tron thanh 2 phan cung chua 10 diem(tinh luon 2 diem la 2 dau mut cua doan thang ay). Bay gio choi y nhu tro choi song Nim, nguoi thu 1 chi can bat chuoc cach di cua nguoi thu 2 la nguoi thu 1 lun thang. :)

Hãy thử mở rộng bài toán. Nếu mỗi lần phải kẻ hai đường thì sao!!

[quote="o0oTheBestOffo0o"]hinh nhu em cung gap 1 bai tuong tu the nay
co 2 nguoi va 1 cai ban hinh chu nhat moi nguoi lan luot dat 1 chai bia vao 1 vi tri bat ky nguoi thua cuoc la nguoi ko dat duoc nua hoi xem ai la nguoi thang cuoc va lam the nao?:):)[/quote]
sao bác ko post vào forum trên thế giới :)


See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]

Find all positive integers $a, b, c$ such that $3abc+11(a+b+c)=6(ab+bc+ac)+18.$

$3abc+11(a+b+c)=6(ab+bc+ac)+18\Leftrightarrow$
$\Leftrightarrow3xyz=x+y+z,$ where $x=a-2,$ $y=b-2,$ $z=c-2.$
And number of cases...
not big, since $3xyz-x-y-z=x(yz-1)+y(xz-1)+z(xy-1).$ ;)

W.L.O.G $1\leq a\leq b\leq c.\ 3abc=6(ab+bc+ca)+18-11(a+b+c)<6(ab+bc+ca)+18.$

$\therefore abc<2(ab+bc+ca)+6<\Longleftrightarrow a<2\left(\frac{a}{c}+1+\frac{a}{b}\right)+\frac{6}{bc}$

$\leq 2(1+1+1+)+6=12,$ yielding $a\leq 12.$

Comment:

Since $a+b+c$ shoud be devided by 3, we can set $a+b+c=3k$ for some positive integers $k.$

Thus we can rewrite the given equation as $abc=2(ab+bc+ca)+6-11k.$

Remark the symmetry for $a,b,c,$ set $A=a-1,\ B=b-1, C=c-1,.....$


ratio
geometry
rectangle
geometric transformation
homothety
circumcircle
symmetry

[b]i.)[/b] Consider a circle $K$ with diameter $AB;$ with circle $L$ tangent to $AB$ and to $K$ and with a circle $M$ tangent to circle $K,$ circle $L$ and $AB.$ Calculate the ration of the area of circle $K$ to the area of circle $M.$

[b]ii.)[/b] In triangle $ABC, AB = AC$ and $\angle CAB = 80^{\circ}.$ If points $D,E$ and $F$ lie on sides $BC, AC$ and $AB,$ respectively and $CE = CD$ and $BF = BD,$ then find the size of $\angle EDF.$

[b](ii)[/b] The triangles $\triangle ABC, \triangle CDE, \triangle BDF$ are all isosceles with AB = AC, CD = CE, BD = BF. Hence, $\angle ABC = \angle ACB = \frac{180^\circ - \angle CAB}{2} = 50^\circ$, $\angle CDE = \angle BDF = \frac{180^\circ - \angle ACB}{2} = \frac{180^\circ - \angle ABC}{2} = 65^\circ$, $\angle EDF = 180^\circ - (\angle CDE + \angle BDF) = 50^\circ$.

[b](i)[/b] Let $r_K, r_L, r_M$ be radii of the circles (K), (L), (M). Let T, U, V be the tangency points of the circle pairs (L), (M); (M, K), (K, L). Since the circle (L) is tangent to the diameter AB of the circle (K), their center line KL is perpendicular to AB, their tangency point V is on this center line. Hence, $r_L = \frac{r_K}{2} = \frac{AB}{4}$. Invert the circles (L), (M) in the circle (K). The line AB passing through the inversion center is carried into itself, the circle (L) tangent to the inversion circle and passing through the inversion center K is carried into the single common tangent v of the circles (K), (L) parallel to the line AB and the circle (M) is carried into a circle (M') tangent to the inversin circle at the point U and to the parallel lines $v \parallel AB$. Hence, radius of the circle (M') is $r_M' = r_L = \frac{r_K}{2}$. The tangency point T of (L), (M) goes to the intersection T' of the ray KT with the line v (the inverted circle (L)). If W' is the tangency point of the circle (M') with the line AB, then KW'T'V is a rectangle. From here, the radius $r_M$ can be calculated in various ways.

For example, since the circles (L), (M') are congruent, it is clear that $TU \parallel AB$, just by symmetry. The center M of the circle (M) lies on the perpendicular bisector of TU, which is a midline of the rectangle KW'M'L, and on the ray $KM \equiv KM'$, which is its diagonal. So M is the diagonal intersection of this rectangle. Since the circle (M) is tangent to the line AB, $r_M = \frac{r_L}{2} = \frac{r_K}{4},\ \frac{\pi r_M^2}{\pi r_K^2} = \frac{1}{16}$.

Or, the circles (M), (M') are centrally similar with the similarity center at the inversion center K and coefficient of similarity $\frac{r_M}{r_M'} = \frac{r_K^2}{t_M'^2}$, where $t_M' = KW' = VT'$ is the tangent length from the inversion center K to the inverted circle (M'). This tangent length can be calculated by Pythagorean theorem from the right angle triangle $\triangle KW'M'$ with one leg $M'W' = r_M' = \frac{r_K}{2}$ and the hypotenuse $KM' = r_K + r_M' = \frac{3r_K}{2}$:

$t_M'^2 = KM'^2 - M'W'^2 = \frac{9r_K^2}{4} - \frac{r_K^2}{4} = 2r_K^2$

Substituting $r_M' = \frac{r_K}{2}$ and $t_M'^2 = 2r_K^2$ into the formula for the homothety coefficient $\frac{r_M}{r_M'}$ gives again $\frac{2r_M}{r_K} = \frac 1 2,\ \frac{r_M}{r_K} = \frac 1 4,\ \frac{\pi r_M^2}{\pi r_K^2} = \frac{1}{16}$

It would also be possible to use the concurrency point O of the common tangents t, u, v at the tangency points T, U, V of the circles (K), (L), (M). This is the circumcenter of the triangle $\triangle TUV$, which is perpendicular to the inversion circle (K), hence, it is carried into itself. But I have to leave some room for armpist to re-hash this.

[quote="yetti"][b](ii)[/b]

          But I have to leave some room for armpist to re-hash this.[/quote]

   How so very considerate of you, Yetti!  Believe me when I say that settling your

   hash is not my intention at all, quite the opposite.

   Beautiful drawing, top draw, as always, I must say.

   Here is my solution, not too much of it, but that's all I came up with:

   Draw a circle X congruent to L and concentric to K.

   Circle M has to be tangent to it simply out of symmetry considerations.

   It follows immediately that  radius of M is  1 / 4 of radius K.

   I like those Indian problems.

   Thank you.

   M.T.

[quote="armpist"]
    Here is my solution, not too much of it, but that's all I came up with:

   Draw a circle X congruent to L and concentric to K.

   Circle M has to be tangent to it simply out of symmetry considerations.

   It follows immediately that  radius of M is  1 / 4 of radius K.

   I like those Indian problems.

   Thank you.

   M.T.[/quote]
Oh yes? What symmetry considerations do you have in mind, with respect to a hedgehog ?

The problem does not say that the circle (L) touches the diameter AB of the circle (K) at its midpoint K. Let S, W be the tangency points of the circles (L), (M) with the line AB. Let T, U, V be the tangency points of the circle pairs (L), (M); (M), (K); (K), (L). Using Pythagorean theorem for the right angle triangle $\triangle KLS$ with the hypotenuse $KL = KV - LV = r_K - r_L$ and one leg $LS = r_L$, the other leg KS is equal to

$SK = \sqrt{KL^2 - LS^2} = \sqrt{(r_K - r_L)^2 - r_L^2} = \sqrt{r_K(r_K - 2r_L)}$

Similarly,

$KW = \sqrt{r_K(r_K - 2r_M)}$

The tangent distance SW of 2 tangent circles (L), (M) is equal to $SW = 2\sqrt{r_Lr_M}$. WLOG, assume that the points A, S, K, B follow on the line AB in this order. The circle (M) can be on either side of the circle (L), closer to the point A or closer to the point B. Assume that the circle (M) is closer to the point B and the circle (L) sufficiently large, $r_L > \frac{r_K}{4}$, so that the points A, S, K W, B follow on the line in this order. Then

$2 \sqrt{r_L r_M} = SW = SK + KW = \sqrt{r_K(r_K - 2r_L)} + \sqrt{r_K(r_K - 2r_M)}$

$2 \sqrt{r_L r_M} - \sqrt{r_K(r_K - 2r_M)} = \sqrt{r_K(r_K - 2r_L)}$

Rearranging and squaring this:

$4 r_L r_M - 4 \sqrt{r_K r_L r_M(r_K - 2r_M)} + r_K^2 - 2r_K r_M = r_K^2 - 2r_K r_L$

$2 r_L r_M + r_K(r_L - r_M) = 2 \sqrt{r_K r_L r_M(r_K - 2r_M)}$

Squaring again,

$4r_L^2 r_M^2 + 4 r_K r_L r_M (r_L - r_M) + r_K^2(r_L^2 - 2r_L r_M + r_M^2) = 4 r_K r_L r_M (r_K - 2r_M)$

$4r_L^2 r_M^2 + 4 r_K r_L r_M (r_L + r_M) + r_K^2(r_L^2 + 2r_L r_M + r_M^2) = 8r_K^2 r_L r_M$

$2r_L r_M + r_K (r_L + r_M) = 2r_K \sqrt{2r_L r_M}$

which is the simplest equation for the radius $r_M$ given the radii $r_K, r_L$. As a check, if we put $\frac{r_L}{r_K} = \frac 1 2$, we get $\frac{r_M}{r_K} = \frac 1 4$. Assume now that the circle (M) is closer to the point B and the circle (L) sufficiently small, $r_L < \frac{r_K}{2}$, so that the points A, W, S, K, B follow on the line in this order. Then starting with

$2 \sqrt{r_L r_M} = WS = WK - SK = \sqrt{r_K(r_K - 2r_M)} - \sqrt{r_K(r_K - 2r_L)}$

we arrive to the same equation for $r_M$ in terms of $r_K, r_L$ as before. No other cases have to be considered. Denoting $m = \frac{r_M}{r_K},\ l = \frac{r_L}{r_K}$, this quadratic equation for m can be written as

$2lm + l + m = 2 \sqrt{2lm}$

$4l^2m^2 + 4 l^2m + 4lm^2 + l^2 + 2lm + m^2 = 8lm$

$(2l + 1)^2 m^2 + 2l(2l - 3)m + l^2 = 0$

$m = \frac{l}{(2l + 1)^2}\ \left(3 - 2l \pm 2\sqrt{2 - 4l}\right)$

Since the problem asks for the area ratio equal to $\frac{S_{(M)}}{S_{(K)}} = \frac{r_M^2}{r_K^2} = m^2$ in terms of $r_K, r_L$ or $\frac{r_L}{r_K} = l$, the result is

$m^2 = \frac{l^2}{(2l + 1)^4}\ \left(4l^2 - 28l + 17 \pm 4(3 - 2l)\sqrt{2 - 4l}\right)$

It checks with the actual Sketchpad construction.

[quote="yetti"]

Oh yes? What symmetry considerations do you have in mind, with respect to a hedgehog ?

The problem does not say that the circle (L) touches the diameter AB of the circle (K) at its midpoint K.[/quote]

 Nice observation. I was surprised when you opted the center point tangency in

 that wonderful GIF file. Based on the obvious solution it simply could not have

 been an Olympiad caliber problem.  It is for the better. Let's use this generous,

 sophisticated interpretation (it does not ask for the answer in terms of radii ratio

 either, so let's have some reasonable freedom here).

 Here is my solution:

 Construct a normal to AB through it's midpoint K. It will intersect circle K at X (on

 the other side of circle L wrt  AB).

 Draw tangents to circle K from X.  These tangents make angles ALPHA and BETA

 with AB ( the ones in where circles M' and M" are inscribed).

  Then  the area ratios simply are:

    Ratio' = tan ( ALPHA / 2 )^4 * m ^ 2

    Ratio'' = tan ( BETA / 2 )^4 * m ^ 2         where m = RofL / RofK

[quote="yetti"]
It checks with the actual Sketchpad construction.[/quote]

   How does this answer endures  the Sketchpad test?

  Thank you.

   M.T.

  P.S.   As you know,  hedgehogs are on the list of endangered species,

  so watch out for Audubon Society when you mention these creatures in vain.

  On second thought, Audubon should be protecting you, because you are more

  endangered than hedgehogs, you are unique.

[quote="armpist"]
 Here is my solution:

 Construct a normal to AB through it's midpoint K. It will intersect circle K at X (on

 the other side of circle L wrt  AB).

 Draw tangents to circle K from X.  These tangents make angles ALPHA and BETA

 with AB ( the ones in where circles M' and M" are inscribed).

  Then  the area ratios simply are:

    Ratio' = tan ( ALPHA / 2 )^4 * m ^ 2

    Ratio'' = tan ( BETA / 2 )^4 * m ^ 2         where m = RofL / RofK

   How does this answer endures  the Sketchpad test?

  Thank you.

   M.T.
[/quote]

There is only one tangent to from the point X to the circle K. This tangent makes the angles ALPHA = 0 and BETA = 0 with AB.  This is easy to see, even though the angles ALPHA and BETA are not given. Then the area ratios are simply:

    Ratio' = 0

    Ratio'' = 0

Wow! Now, you do not need the Sketchpad to check it out, do you  :? ?

Yetti

Congratulations, Yetti ! It is a fine and very detailed solution. In no case I will not find a other solution for this problem.

The tangents certainly are to the circle L, sorry for typo.

  Thank you.

  M.T.


How to prove that Det(A)<>0  (not equal zero)
implies the column (or row) vectors are independent?

thanx


Put $a=2^x\ge0$ etc. CS in Engel form:
$\sum_{cyc}\frac{ab}{a/b+1}=\sum_{cyc}\frac{ab^2}{a+b}\ge\frac{(\sum b\sqrt{a})^2}{2(a+b+c)}\ge\frac{9abc}{2(a+b+c)}$ by AM-GM :)


geometry
inequalities
calculus
function
perpendicular bisector
geometry solved

We are given n>=3 points in the plane. Prove we can find three of the points, say A,B,C satisfying 1 <= AB/AC <= (n+1)/(n-1)

This is strictly just an idea, I don't know if it leads anywhere - I'll think about it overnight, but I'm posting it anyway to get things rolling.

Proof by contradiction.  Let the points be [tex]P_{1}, \dots, P_{n}[/tex].  There are [tex]\frac{n^2 - n}{2}[/tex] lengths, no two of which differ by a factor less than [tex]\frac{n+1}{n-1}[/tex].  WLOG, suppose the smallest length, [tex]P_{1}P_{2}[/tex], is one.  Show that there must exist [tex]P_{i}[/tex] such that [tex]P_{1}P_{2}P_{i}[/tex] violates the triangle inequality.

Since [tex]2(n - 2)[/tex] lengths emanate from [tex]P_{1}[/tex] and [tex]P_{2}[/tex], at least one length equals or exceeds [tex]\left(\frac{n+1}{n-1}\right)^{2n-5}[/tex].  Can we show that

[tex]
\displaymath{
\left(\frac{n+1}{n-1}\right)^{2n-5} \ge \left(\frac{n+1}{n-1}\right)^{2n-6} + 1
}[/tex]

or something similar?

Mildorf: I believe that an approach like ours works (my approach, similar to yours, led me to the exact same inequality), but unfortunately, it fails for large $n$ because for $f(n)=(\frac{n+1}{n-1})^{2n-5} - (\frac{n+1}{n-1})^{2n-6}, lim_{n \rightarrow \inf} f(n) = 0$.

This is true, the above inequality becomes false for sufficiently large [tex]n[/tex].  I don't know if computing the smallest [tex]n[/tex] who which it is false is of any use.  However, we can substantially improve this inequality.  The largest distance, which we assume to be [tex]P_{3}P_{4}[/tex], is at least [tex]\left(\frac{n+1}{n-1}\right)^{\frac{n^2 - n}{2} - 1}[/tex].  At least one of [tex]\{P_{3}, P_{4}\}[/tex] is half of this distance from [tex]P_{1}[/tex] and [tex]P_{2}[/tex]. Hence, it would suffice to show that

[tex]
\displaymath{
\frac{1}{2} \cdot \left( \frac{n+1}{n-1} \right)^{\frac{n^2 - n}{2}} \ge
\frac{1}{2} \cdot \left( \frac{n+1}{n-1} \right)^{\frac{n^2 - n}{2} - 1} + 1
}[/tex]

or, equivalently

[tex]
\displaymath{
\left( n+1 \right)^{\frac{n^2 - n}{2} - 1} \ge \left( n-1 \right)^{\frac{n^2 - n}{2}}
}[/tex] [tex]\equiv[/tex] [tex]\displaymath{
\left( n+1 \right)^{n^2 - n - 2} = \left( n+1 \right)^{(n+1)(n-2)} \ge \left( n-1 \right)^{(n)(n-1)} = \left( n-1 \right)^{n^2 - n}
}[/tex]

Potentially useful fact: With a little bit of calculus, we can show that [tex]f(x) = x^{\frac{k}{x}}[/tex] is a differentiable function over the positive reals with absolute maximum at [tex]x = e[/tex], where [tex]f[/tex] is strictly increasing over [tex](0,e)[/tex] and strictly decreasing over [tex](e, \infty)[/tex].

To Andrew:

Sorry but I have difficulties following your proof. Consider the following:

Let A_i (i=1,2,...,n) be n points in the plane with |A_1A_n| being the longest line segment. Now place A_i (i=2,3,...,n-1) on the perpendicular bisector of A_1A_n. I cannot find a point, say A_k, such that |A_1A_k|<|A_1A_n|/2 or |A_nA_k|<|A_1A_n|/2. And certainly, I cannot find (n-[n+1/2]) many points satisfying the relations you claimed.

Please correct me if I am wrong.

yeah...okay...i see it, thanks.
silly me...  :blush:


calculus
derivative
calculus computations

g(v)=f(v)/v
after differentiation i got (vf'(V)-f(v))/v^2, but I don't know what to do next, please help.

Is f(v) = v by any chance?
because that's what i'm gettin after solving g'(v) = 0  :maybe:

[quote="colin_java"]I guess you want to find the minimum value of g by differentiating it like you have.

Then you would equate the derivate to 0.
Solve this equation, to give you the stationary points, and assuming g is continuous, it will give you the local maxima/minima.

Test the second derivative g''(v), if it is >0 you have a minimum, if its <0, a maximum.[/quote]

Your response has left open several questions which may be important:

What if g is not continuous at some points?
What if g' does not exist at some point(s)?
What if g'' does not exist at some point(s)?
What if g''(v) = 0?


limit
integration
probability
calculus
real analysis
real analysis unsolved

Show that $\lim_{n\rightarrow+\infty}\frac{1}{\Gamma(n+1)}\int_{0}^{n}x^{n}e^{-x}dx=\frac{1}{2}$. :D

We could quote the Central Limit Theorem.

Let $X_{1},X_{2},\dots$ be a sequence of independent identically distributed distributions each of which is has an exponential distribution with mean $1.$

Let $Y_{n}=\sum_{k=1}^{n}X_{k}.$

It can be shown that $Y_{n}$ has a so-called Gamma distribution. That is, its density is

$f_{n}(y)=\frac1{\Gamma(n)}\,y^{n-1}e^{-y}.$

Hence, the quantity we're trying to find the limit of is the probability that $Y_{n+1}$ is less than $n,$ which is one less than its own mean.

Let $Z_{n}=\frac{Y_{n}-n}{\sqrt{n}}.$ By the Central Limit theorem, $Z_{n}$ tends in distribution to a standard normal random variable $Z.$

$\lim_{n\to\infty}P(Y_{n+1}<n)=\lim_{n\to\infty}P\left(Z_{n+1}<\frac{n-(n+1)}{\sqrt{n}}\right)$

$=\lim_{n\to\infty}P\left(Z_{n}<-\frac{1}{\sqrt{n}}\right)=P(Z<0)=\frac12.$

Of course, a more direct analytic argument would also be welcome.

Computing the integral directly, using integration by parts, and making certain algebraic simplifications, arrives at the following formula:

$\frac1{n!}\int_{0}^{n}x^{n}e^{-x}\,dx= 1-e^{-n}\left(1+n+\frac{n^{2}}2+\frac{n^{3}}{3!}+\cdots+\frac{n^{n}}{n!}\right).$

(Just in case anyone else wants to start with this.)

My solution:By Stirling formula:$\lim_{n}\frac{\sqrt{2\pi n}n^{n}e^{-n}}{\Gamma(n+1)}=1$ we know that it is sufficient to show that
 $\lim_{n}\frac{1}{\sqrt{2\pi n}n^{n}e^{-n}}\int_{0}^{n}x^{n}e^{-x}dx=\frac{1}{2}$.
In fact $\frac{1}{\sqrt{2\pi n}n^{n}e^{-n}}\int_{0}^{n}x^{n}e^{-x}dx= \frac{1}{\sqrt{2\pi }}\int_{0}^{1}\sqrt{n}x^{n}e^{n-nx}dx=$
$\frac{1}{\sqrt{2\pi }}\int_{0}^{+\infty}\sqrt{n}e^{-(n+1)x}e^{n-ne^{-x}}dx= \frac{1}{\sqrt{2\pi }}\int_{0}^{+\infty}e^{-\frac{(n+1)x}{\sqrt{n}}}e^{n-ne^{-\frac{x}{\sqrt{n}}}}dx$.
Clearly $e^{-\frac{(n+1)x}{\sqrt{n}}}e^{n-ne^{-\frac{x}{\sqrt{n}}}}= e^{-\frac{x^{2}}{2}+O(\frac{1}{\sqrt{n}})}\rightarrow e^{-\frac{x^{2}}{2}}$.
Thus $\lim_{n}\frac{1}{\sqrt{2\pi }}\int_{0}^{+\infty}e^{-\frac{(n+1)x}{\sqrt{n}}}e^{n-ne^{-\frac{x}{\sqrt{n}}}}dx= \lim_{n}\frac{1}{\sqrt{2\pi }}\int_{0}^{+\infty}e^{-\frac{x^{2}}{2}+O(\frac{1}{\sqrt{n}})}dx=$
$\frac{1}{\sqrt{2\pi }}\int_{0}^{+\infty}e^{-\frac{x^{2}}{2}}dx$(which is just the standard Gaussian distribution)$=\frac{1}{2}$.


Important!! I'm asking how the story is told in your school, not what do you think of that.
Please vote [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=392544#392544]here[/url] too.

I think the question is hard, perhaps it would be interesting if you would explain what the different intervals mean, because putting a number on contribution is weird

Sorry for bringing something else up here, but I do not understand military history of USA

in 1776 they started fighting Britain for indepence , with as i was taught french help\

then in 1812 they sided with france again, howver as they had no navy, while the british had, their only option was to attack the still loyalist canada, resulting in the burning down of toronto and washington

but in between, there was this weird conflict called quasi war in which USA sided with britain AGAINST the french

Have i read this wrong?  Could someone clarify any of these things?  Againm I am not expressing my opinion about all this, I just like to know my history( well it is not my history :roll:   )

It's obviously impossible to make a quantitative description of such things, so the numbers are just some feelings by your intuition, for example, more than 75% means if there were no France, American can hardly win, maybe will even fail, 50%~75% means France dominates, but American could still win without France, although harder.

3 words for you, over and over:

Omlette du fromage!!!
Omlette du fromage!!!
Omlette du fromage!!!

Omlette du fromage!!!
Omlette du fromage!!!
Omlette du fromage!!!

Omlette du fromage!!!
Omlette du fromage!!!
Omlette du fromage!!!

It is taught as France being a minor, but colorful player.  The reality is that (hold on for a few facts) the defeat of the British General Charles Cornwalis at Yorktown, widely considered the military turning point in the war, was a trap Cornwalis planned for General George Washington.  His plan was to retreat down the peninsula, get picked up at the end by the British fleet and quickly be deposited again at the top of the peninsula, thereby trapping the rebels and finally being able to engage them.  However, when poor Charles reached the end he saw the French fleet, which, having just beat off the British fleet, now occupied the critical area.

This is widely taught, what is not taught is that the French also made up more than half of the army opposing Charles.  This was so bad that when the French required that Charles surrender to a rebel, he refused and sent a representative.  (Washington also had a representative accept the surrender.)

In my opinion, this war for independence, was won by a then superpower, who left the newly independent state with its freedom, sort of like our plan with Iraq.

I was taught, fairly recently, that the french weren't with us from the beginning,and didn't want to commit to helping us fight, until certain battles like the Battle of Trenton, or something. This is when the french may have realized we had a chance and it would be to thier profit to try and help us out. This is just what I learned in social studies last month.

and can anyone elaborate about that quasi war where usa seemed to join with the british?

I was taught that the French began helping after some battle (forget the name - U.S. history was 2 years ago for me in 8th grade).  However, the French help was very useful, for example when the British general surrendered most of the troops were actually French.  France also provided a lot of supplies, a navy (which the colonies, or United States did not have).  However, I was also taught that the fact that America was able to last until the French decided to help, though, plus the stupidity of the British war methods with rows, bright red uniforms, etc. did give the Americans a chance to win even without other help.

Then I read an article in Time magazine (I think it was in an issue about Ben Franklin, specifically an article about his role as a diplomat to the French) about the importance of French help.

I vote 75%.

Edit: okay, since that isn't a choice, 50-75%.

France, Spain and England were the most important. So, I´ll choose 25%-50%.

I was taught that the US would lose without France but would have come close to winning...

I was taught that the French were pretty much critical to the success of the war. Basically, it was their training that gave the Americans any chance at all and that Americans had not actually mastered a kind of ragtag warfare as is popularly imagined. Also, as was mentioned above, the U.S. didn't really have any resources to resist Britain if it weren't for the French.

After the war, France and Britain were the two superpowers by 1800 or so. At this time, nobody really liked either of them due to their impressment of foreign sailors, among other things.

From AP US History this year,  The French did not join the United States until after the battle of Saratoga where the Americans won a substansial victory and it became clear that it was possible for the colonists to beat Britian. (The reason they were not defeated earlier was due in large part to tactical blunders by the british)  After the Battle of Saratoga many nations including Frnace, Spain and Prussia aided the colonies.  France was very important because of the navy they provided and the money.  The other countries were also very important and a prussian officer (whos name i forget) was much more important for training American troops then the French.  The French and Spanish also helped the colonies by declaring war on England and fighting elsewhere in the wrold so that the English could not use all of their forces in the Americas.  I would say France was not that critical but that European support was (of which france was the most).  However had France not joined it is likely that the colonies might stil have won because of the victory at saratoga.

The reason for the quasiwar with France is that during the American revolution, France and the United States made a permanent treaty of friendship but after the French revolution during the napoleonic wars, both france and england violated American neutrality.  In addition, France made some diplomatic blunders including the xyz affair nad the citizen genet affair that angered Americans.  This resulted in the Americans and French signing new treaties that ended the permanent treaty.  during the war of 1812, the Americans and French were both simultaneously fighting the english but there was little communication between the two countries.  Thats what I learned in ap history hope that helps.


algebra
polynomial
induction
algebra proposed

let a>=3 be a real number, and p(x) a polynomial of degree n with real coefficients. prove that:

max[|a^j-p(j)|]>=1 for 0<=j<=n+1.

I solved it using Lagrange's interpolation polynom plus small computation using calculus.
Do you have elementary solution?

I dont know if it is elementary, but i solved it using the identity:

 :Sigma: (-1)^i.[(n+1)C(i)].P(x-i)=0 for all x and P a polynomial of degree at most n.

*.[(n+1)C(i)]. is the binomial coefficient (n+1)n......n+1-i+1)/i(i-1).....1.

I think these are equaivalent approaches.

Another solution
We will prove the stronger result: $ \max_{0\leq i \leq n+1}|a^i - P(i)|\geq (\frac{a-1}{2})^{n+1}$ by induction
- For $ i=0 (x)=c$.Clearly,
$\max \{|1-c|,|a-c|\}\geq \frac{a-1}{2}$
-Assume that it holds for some polynomial $ P:\deg P\leq n-1$
Consider $Q(x) =\frac {P(x+1)-P(x)}{a-1}\Rightarrow \deg Q=n-1$.Then use induction hypethosis for $Q(x)$ we have $\exists i \leq n$:
$|\frac{P(i+1)-P(i)}{a-1} - a^i| \geq(\frac{a-1}{2})^{n}$,
$\Rightarrow |(P(i+1)-a^{i+1})-(P(i)-a^i)| \geq (\frac{a-1}{2})^{n}(a-1)$,
${\Rightarrow  \max \{|a^{i+1}-P(i+1)}|;|a^i-P(i)|\}\geq (\frac{a-1}{2})^{n+1}$
So we conclude that $\max_{0\leq i\leq n+1}|a^i-P(i)| \geq ( \frac {a-1}{2})^{n+1}\geq1$,and we are done


if u want please check this link
http://www.sparetimegroup.net
it's about informatics

[quote="neelnal"]thx. it helped a lot. :rotfl: who knew it would influence someone 3 years later.[/quote]
This is G&FF, but your post is still :spam:.

Nice 10 year revive. That was so long ago.

If you scroll down long enough, you start seeing serious topics about math.


How many numbers can be expressed as the distinct sum of two numbers from the following set:

{1,2,3,5,7,9,11}

I got 21, because 7*6/2 = 21, but the answer is 11.

How do you get 11?

well, 10 are the same, since some choices end up equal

ex. 9+3=5+7

But how can you quickly determine the number of sums that are the same?

You just have to list them out.

Primes less then n/2.

There aren't that many primes less then 20, so do it.


algebra
polynomial
number theory open
number theory

It is well known that if $4^n + 2^n + 1$ is a prime then $n = 3^k$ (and of course $2^n + 1$), is this a part of any generalization?

do you suggest:

sum 2^in from i=0 to k is a prime only if n=(k+1)^a for some a?

the sum is: 2^kn+2^(k-1)n.....+2^n+1

The one suggested by Pascual2005 has one problem: it can even only get prime when $k+1$ is prime.

Let $\Phi_{n}(x)$ denote the $n$-th cyclotomic polynomial, defined on any way you like, e.g. by $x^{n}-1= \prod_{d|n}\Phi_{d}(x)$ for all $n$ or by $\Phi_{n}(x) = \prod_{m \in (\mathbb{Z}/n \mathbb{Z})^{*}}(x-\zeta_{n}^{m})$, with $\zeta_{n}= e^{\frac{2 \pi i}{n}}$ a $n$-th primitive root of unity.

[b]Generalisation:[/b]
[b]Theorem 1[/b]
Let $a \geq 2$ be an integer and $p$ be prime. Then $a^{(p-1)m}+a^{(p-2)m}+...+a^{2m}+a^{m}+1$ can only be prime if $m=p^{k}$ for some $k$.
[u]Proof:[/u]
This is the case $n$ prime of theorem 2.

[b]Generalisation of generalisation:[/b]
[b]Theorem 2[/b]
Let $a \geq 2$ and $n \geq 1$ be integers. Then for $\Phi_{n}(a^{m})$ to be prime, it is necessary that $m$ has only prime divisors also dividing $n$, except of the case $a=2, n=1$ where it is only necessary for $m$ to be prime.
[u]Proof:[/u]
The case $n=1$ is well know stuff from elementary results on Mersenne numbers. So let $n \geq 2$.
When $m$ is not of the desired type, see theorem 3 for a factorisation of $\Phi_{n}(a^{m})$.
It's enough to show that $\Phi_{mn}(a)>1$ and $\Phi_{kn}(a)>1$ now, since then $\Phi_{n}(a^{m})$ will surely not be prime.
For this it suffices to show that $\Phi_{w}(b)$ is greater than one for all $w,b \geq 2$. This is easy when looking at the roots of $\Phi_{w}(x)$: they all are on the unit circle and $\neq 1$ (the latter by $n \neq 1$), so their distance to $b$ is $>1$, so also the product of those distances, giving what we wanted.

[b]Generalisation of generalisation of generalisation:[/b]
[b]Theorem 3[/b]
Let $a \geq 2$ and $n \geq 1$ be integers. Then the polynomial $\Phi_{n}(x^{m})$ is irreducible iff $m$ is only divisible by prime divisors also dividing $n$; when $q|m=q^{v}k$ with $q\nmid k,n$, two nontrivial factors are explicitely given as $\Phi_{mn}(x)$ and $\Phi_{kn}(x)$.

[u]Proof:[/u]
If $q$ is some prime dividing $m=q^{v}k$ ($q \nmid k$) but not $n$, we see that the two factors proposed have only roots $\Phi_{n}(x^{m})$ has, and they have no common roots (all this may be hard to see at first). Thus these are indeed different factors. Another way to get these factors is by using the following Lemma sometimes.

You can stop reading all that generalisations, since only the already proven part is needed for the proofs of theorem 1 and 2.
But here goes the other direction:

We need a important property of cyclotomic polynomials (only the first part is needed but the other is good to know):
[b]Lemma:[/b]
Let $n$ be any positive integer and $q$ be prime. Then:
- $\Phi_{n}(x^{q}) = \Phi_{nq}(x)$ iff $q|n$
- $\Phi_{n}(x^{q}) = \Phi_{nq}(x) \cdot \Phi_{n}(x)$ iff $q \nmid n$
This is essentially done by comparing the roots of both sides using that $\Phi_{w}(x) = \prod_{a \in (\mathbb{Z}/w \mathbb{Z})^{*}}(x-\zeta_{w}^{a})$.

Back to the problem:
If $m$ has only such prime factors, by repeated use of the Lemma, part 1, we get that $\Phi_{n}(x^{m})=\Phi_{mn}(x)$. And the cyclotomic polynomials are irreducible as was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=15750 .

[mod.: quoting the (long) post before for writting a one-line comment is not a very good idea]

 Very interesting :wink:

Bulgarian Olympiad 1981 3.4
(among others)

sum 2^in from i=0 to k is a prime only if n=(k+1)^a for some a?

the sum is: 2^kn+2^(k-1)n.....+2^n+1.This sum be prime only when n=(k+1)^a, where (k+1) is prime! I can give solution if it's interests anyone.


HELLO????????????

HOw do i make something happen

You just use keyboard to type numbers and words to make Math Problems. :lol: Nothing else..and Welcome to Oklahoma Forum! I hope to you often in here!

dont wory eric it's easy once you get the hang of it  :D

Ps this is arnav :rotfl:  :rotfl:

[quote="hunter34"]Hi Arnav, is this your third account or something?[/quote]
.... how do you know him..? :?: (he is not from Texas)