function
number theory
least common multiple
prime factorization
prime numbers

Determine, with proof, if there are any natural numbers smaller than 240 that have more than 20 positive divisors.

I'm stuck....

So it's not really easy for me....  :(

[hide="hint"]Assume that there is such a number, $ n$ with more than 20 factors.  Then, $ n$ must have more than 10 factors that are less than $ \sqrt{n}<16$.  In other words, at least 11 of the numbers from 1 to 15 divide $ n$.[/hide]

This seems a bit too difficult for HSB; moved to Intermediate.

I sorta got it, but it isn't rigorous, and it doesn't use your hint...so i'm not sure if its right

[hide]
We want to test numbers that have a lot of prime factors.  We can use this because $ 240<256\equal{}2^8$, so any number with a prime factorization with an exponent larger than $ 7$ doesn't need to be tested.  Clearly, we don't want to go over $ 30$ factors.  Composite numbers from $ 20$ to $ 30$ that don't require exponents greater than $ 7$ are $ 21$, $ 24$, $ 25$, $ 27$, and $ 30$.

$ 21$: The minimum value is $ 2^6 \cdot 3^2\equal{}576$, which is greater than $ 240$.
$ 24$: The minimum value is $ 2^3 \cdot 3^2 \cdot 5\equal{}360$, which is greater than $ 240$.
$ 25$: The minimum value is $ 2^4 \cdot 3^4\equal{}1296$, which is greater than $ 240$.
$ 27$: The minimum value is $ 2^2 \cdot 3^2 \cdot 5^2\equal{}900$, which is greater than $ 240$.
$ 30$: The minimum value is $ 2^4 \cdot 3^2 \cdot 5\equal{}480$, which is greater than $ 240$.

If the number of factors is greater than $ 30$, the minimum value will keep going up.  Thus, the answer is $ \boxed{\text{no}}$.
[/hide]

can you please explain your solution clearly i am lost after 'we can use this'. :(

i read somewhere that:
$ n \equal{} p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots \cdots p_n^{a_n}$ ,then
$ d(n) \equal{} (a_1 \plus{} 1)(a_2 \plus{} 1)\cdots \cdots (a_n \plus{} 1)$

@analytic: there are $ k$ prime factors, not $ n$ prime factors.
[hide="proof of analytics theorem"]Now the $ p_i$ are distinct prime numbers, and analytic listed the prime factorization of $ n$. Thus if a number $ d$ is a divisor of $ n$, it can either have no factors of  $ p_1$, one factor of $ p_1$, etc. to $ a_1$ factors of $ p_1$, and so on until $ d$ can have $ a_k$ factors of $ p_k$. There are $ a_1\plus{}1$ ways to choose the number of $ p_1$s in $ d$, $ a_2\plus{}1$ ways to choose the number of $ p_2$s in $ d$, etc. When we multiply, we get that there are $ (a_1\plus{}1)(a_2\plus{}1)\cdots (a_k\plus{}1)$ numbers that $ d$ can be, and thus there are that many factors of $ n$.[/hide]

wut i thought was to replace p's in above expression by primes in increasing order. if they r not in factorisation of the number power can be 0. taking primes as 2 or 3 (takin least) helps in squeezing as many powers as possible inside 240 which is wut is required in expression for d(n).taking higher primes wud reduce powers.

and an easy check suggests it is not possible for 2 and 3. so, solved.

sorry, i m not able to prove it rigorously.
and tx for proof.

[hide="solution"]From the proof mentioned before, the $ a_i\plus{}1$ are either 1 and 20, 2 and 10, 4 and 5, or 2, 2, and 5. The first three can be easily be proved false, but the fourth is very close: $ 2^4*3*5\equal{}15*16\equal{}240$ is the least one.[/hide]

[quote="AIME15"]If the number of factors is greater than $ 30$, the minimum value will keep going up.[/quote]
You can't conclude this from what you've written (this function is far from monotonic as you have already seen), although you can rule out this case using b-flat's original hint; a number with more than $ 30$ factors less than $ 240$ has to be divisible by every number from $ 1$ to $ 15$ (why?).  What is $ \text{lcm}(1, 2, 3, ... 15)$?


inequalities
triangle inequality
complex analysis
complex analysis theorems

Hey guys. I apologise if this isn't the right section for this...just seemed "most relevant".

I'm going through some homework stuff, and I have come across this question which I just can not do. At ll.

[img]http://i104.photobucket.com/albums/m170/mc_meaney/untitled-3.jpg[/img]

Is anyone out there able to show me how to go about this? I'm sure it's probably quite simple but yes...I've got no idea where to go with this.

Oh, btw, the question SHOULD have brackets in the first part ie [b]Lim (f(z)+g(z))[/b]

Anything would be hugely appreciated (and again, very sorry if this is the wrong place!)[/b]

The arguments look exactly the same as those you saw on $ \mathbb{R}$, except that the definition of "$ z\to\infty$" is slightly different. All of those arguments used nothing more than the triangle inequality, and the triangle inequality still applies.


geometry
3D geometry
induction
algorithm

[quote="PenguinIntegral"]Can the cube of a natural number end in 1985 1's?[/quote]

[hide]No, look mod 4...[/hide]

Oh :blush: I was looking at higher mods like 9 and such......So was the rest of the MC fourm.

[quote="Arne"]Maybe I am really missing something here but why is that so obvious when you look modulo 4? I mean, the number in the question is obviously equal to 3 (mod 4) but so are some of the cubes ($3^3$, $7^3$, $11^3$ etc).[/quote]

Cube.... ooops, thought of square...

The number must $288471$ mod $10^6$...
Each time you switch a power of 10 higher, you get exactly one possibility, so I suspect the question will be answered positively. But the hidden structure is left to discover.

$u_n = 111...1$ ($n$ times $1$)
Prove by induction on $n$. We prove for every $n$, there exists an $a_n$ for which $(a_n)^3 = u_n$ (mod $10^n$).

* $n = 1$. Trivial: $a_n = 1$.

* Suppose $(a_n)^3 = c\cdot 10^{n+1}+u_n$ (mod $10^{n+1}$) with $c \in \{0,1,2,...,9\}$.
We will show there exists always a $b \in \{0,1,2,...,9\}$ so that $a_{n+1} = b\cdot 10^{n+1} + a_n$.

$(b\cdot 10^{n+1} + a_n)^3$
$ = 3a^2b\cdot 10^{n+1} + c\cdot 10^{n+1}+a^3$ (mod $10^{n+1}$) -> by Newtons binomium
$= (3b+c)\cdot 10^{n+1} + u_n$ (mod $10^{n+1}$) -> $a^2 = 1$ (mod 10) and some rearrangement
We can choose $b$ mod $10$ so we can choose $3b$ mod $10$ so we can choose $3b+c$ to be $1$ mod $10$, which completes the induction.

[quote="Stijn"]$u_n = 111...1$ ($n$ times $1$)
Prove by induction on $n$. We prove for every $n$, there exists an $a_n$ for which $(a_n)^3 = u_n$ (mod $10^n$).

* $n = 1$. Trivial: $a_n = 1$.

* Suppose $(a_n)^3 = c\cdot 10^{n+1}+u_n$ (mod $10^{n+1}$) with $c \in \{0,1,2,...,9\}$.
We will show there exists always a $b \in \{0,1,2,...,9\}$ so that $a_{n+1} = b\cdot 10^{n+1} + a_n$.

$(a_{n+1})^3$
$ = 3a^2b\cdot 10^{n+1} + c\cdot 10^{n+1}+a^3$ (mod $10^{n+1}$) -> by Newtons binomium
$= (3b+c)\cdot 10^{n+1} + u_n$ (mod $10^{n+1}$) -> $a^2 = 1$ (mod 10) and some rearrangement
We can choose $b$ mod $10$ so we can choose $3b$ mod $10$ so we can choose $3b+c$ to be $1$ mod $10$, which completes the induction.[/quote]
 :| I wish I could do stuff like that.

[quote="Hexaditidom"]I would really enjoy seeing an $n$th power of a number that ends in a string of $x$ $y$'s where n is a pos. int. >2, x is a pos. int. >= 10, and y is a digit.  There seem to be alot of these kind of problems, but then someone has to disprove it with mods. :roll:[/quote]
Try: $7368288471^3$ ;). My proof actually gives a quite good algorithm for calculation such a number digit after digit.

And indeed, at first I also tried to disprove it. But the fact the answer is actually positive, makes this a surprising problem.


$ a > 1$
$ x_{1}\equal{} a$
$ x_{n\plus{}1}\equal{} 1\plus{}ln\frac{(x_{n}^{2}\plus{}3).x_{n}}{3x_{n}^{2}\plus{}1}$
Prove that the sequence converge and find it limit.

See here http://www.mathlinks.ro/Forum/viewtopic.php?p=758718#758718 .


calculus
integration
trigonometry
real analysis
real analysis unsolved

[color=darkred]Study the following sequencies (recurrent relation, monotonicy, boundness, limit a.s.o.) :[/color]

$ 1\blacktriangleright A_n \equal{} \int_0^{\frac {\pi}{2}}(\sin x \plus{} \cos x)^n\ \mathrm {dx}$, $ n\in N\ .$

$ 2\blacktriangleright A_n \equal{} \int_0^{\frac {\pi}{2}}(1\plus{}\sin x \plus{} \cos x)^n\ \mathrm {dx}$, $ n\in N\ .$

1, Because $ \sin x \plus{} \cos x \equal{} \sqrt {2}\sin (x \plus{} \pi/4)$, we need see $ \int_{\alpha}^{\beta}\sin^n x\,dx$, this is well-known.

[quote="Virgil Nicula"]$ 1\blacktriangleright A_n \equal{} \int_0^{\frac {\pi}{2}}(\sin x \plus{} \cos x)^n\ \mathrm {dx}$, $ n\in N\ .$[/quote]
$ A_{n\plus{}2}\equal{}\int_0^{\frac {\pi}{2}} (\sin x\plus{}\cos x)^n(1\plus{}\sin 2x)\ \mathrm {dx}$ $ \implies$

$ A_{n\plus{}2}\equal{}A_n\plus{}\int_0^{\frac {\pi}{2}}(\sin x\plus{}\cos x)^n\left[1\minus{}(\sin x\minus{}\cos x)^2\right]\ \mathrm {dx}$ $ \implies$

$ A_{n\plus{}2}\equal{}2A_n\minus{}B$, where $ B\equal{}\int_0^{\frac {\pi}{2}}(\sin x\plus{}\cos x)^n(\cos x\minus{}\sin x)^2\ \mathrm {dx}$ $ \implies$

$ \boxed {\begin{array}{ccc}
u(x)\equal{}\cos x\minus{}\sin x & \implies & u'(x)\equal{}\minus{}(\sin x\plus{}\cos x)\\\\
v'(x)\equal{}(\sin x\plus{}\cos x)^n(\cos x\minus{}\sin x) & \implies & v(x)\equal{}\frac {1}{n\plus{}1}\cdot (\sin x\plus{}\cos x)^{n\plus{}1}\end{array}}$

$ (n\plus{}1)\cdot B\equal{}A_{n\plus{}2}\minus{}2$ $ \implies$ $ \boxed {\begin{array}{c}
A_0\equal{}\frac {\pi}{2}\ \ ,\ \ A_1\equal{}2\\\\
(n\plus{}2)\cdot A_{n\plus{}2}\equal{}2(n\plus{}1)\cdot A_n\plus{}2\ ,\ n\in N\end{array}}$

[color=darkred][b]I"ll return soon here ![/b][/color]


abstract algebra
vector
linear algebra
linear algebra unsolved

Is there an example of a free $ A$-module $ E$ with finite bases $ I,J$ such that $ \operatorname{Card}(I)>\operatorname{Card}(J)$?

If $ A$ is a commutative ring, then a standard theorem states that any two bases for $ E$ have the same cardinality. But one can even give an example of a noncommutative ring $ A$ and an $ A$-module $ M$, such that $ M$ possesses a basis for every positive integer $ n$.

For example let $ V$ be a vector space with a countably infinite basis, then consider $ L(V)$ as an $ L(V)$-module. (This example is discussed in Roman's Advanced Linear Algebra.)


geometry
trigonometry
cyclic quadrilateral
trig identities
Law of Sines

A convex pentagon is inscribed in a circle. If it's sides are 16, 16, 16, 16, and 6, it's area will be k:sqrt:7. What is the number k?

By 'number' do you mean 'integer' ?

Thanks

It happens to be an integer, but I don't think there's anything significant about that.

From the way the question is worded it sounds like knowing it is of the form integer root 7 gives you an extra piece of information. But maybe not. Does anyone know if given the sides of an inscribed pentagon (or any shape) there is only a unique one? Or can there be more.. if theres more then I doubt they have the same area.

In general, just given the sides of a polygon it's area is not uniquely determined (except for triangles of course). But if it is inscribable in a circle (i.e. cyclic) then it's area is uniquely determined by the lengths of it sides. A formula for the area of cyclic polygons of 3 to 6 sides is known (though very complicated for 5 and 6), and I believe a general formula for more than 6 sides is conjectured but not proven. I may be wrong about that.

Here are some links from MathWorld about cyclic polygons:

http://mathworld.wolfram.com/CyclicPolygon.html

http://mathworld.wolfram.com/CyclicPentagon.html

In the case of this problem though, the pentagon is uniquely determined because all of the sides are the same except for one. And the information about the sqrt 7 only helps as far as letting you know what form the answer is in. So it helps to let you know if you got the answer right.

Here is a hint for those who want to look:

[hide]I know two ways to do this one.

For one, think about how to use Ptolemy's Theorem - the sum of the products of opposite sides of a cyclic quadrilateral equals the product of it's diagonals.

Or for another method, try labeling the angles from the orthocenter to the vertices and use the law of sines and the triangle area formula.

[/hide]

It's actually quite simple to prove that given the lengths of the sides of any polygon inscribed in a circle, the order of the sides doesn't affect the area: just imagine drawing in radii from the center to each of the vertices.  Then, if you move the sides around, each of the triangles is moved around as well.  No matter what, at the end all the triangles have the same area, so the polygon does as well.

[quote]ComplexZeta asked:
"But is it necessarily true that if you change the order of the sides around that the resulting polygon can be inscribed in a circle?"[/quote]

[color=cyan]That question doesn't make sense to me the way it is written.  If you give me a polygon inscribed in a circle, I can rearrange its sides however I want to create a polygon of the same area inscribed in the same circle.  Any ordering for the sides will work.  However, for any polygon other than a triangle, having the side lengths doesn't tell you anything about inscribability: you just squish an inscribed quadrilateral a little bit and it can't be inscribed any more.  I think I just don't understand what your question means.[/color][/quote]

Sorry. What I mean is if you are given a bunch of lengths of sides, can you necessarily form a polygon that can be inscribed in a circle with those sides in that order (assuming that they will form a polygon, that is; 1, 1, 1, 7 won't for example)?

[color=cyan]Why wouldn't you be able to?  Having more sides gives you more freedom to mess around with the angles.  I can't imagine that you can't do it for a quadrilateral, and after that it should all follow fairly easily, I think.[/color]

By the case, in case any of u want to kno, k is 123.


What is the greatest integer $ n$ such that $ n^2 \minus{} 11n \plus{}24 \leq 0$?

$ n^2 \minus{} 11n \plus{}24 \leq 0$
$ (n\minus{}3)(n\minus{}8)\leq 0$
$ 3\leq n \leq8$

Hence $ n \equal{} 8$

 :lol:


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Let Z[1/2] be the smallest subring of Q containing Z and 1/2. Let (2x-1)

be the ideal of Z[x] generated by the polynomial 2x-1. Show that

Z[x]/(2x-1) is isomorphic to Z[1/2]. Is the ideal (2x-1) maximal in Z[x]?

If not, find an ideal I which is a subset of Z[x] such that

(2x-1) is a proper subset of I which is a proper subset of Z[x].

To show the first, see that all $a \in \mathbb{Z}$ and $\frac{1}{2}$ are containded in the quotient ring by $a$ and $x$, seen $\mod 2x-1$.
Now note that the map $f$ (from $\mathbb{Z}/ (2x-1)$ to $\mathbb{Z}\left[\frac{1}{2}\right]$) given by $a \mod (2x-1) \mapsto a \in \mathbb{Z}\left[\frac{1}{2}\right]$ and $x \mod (2x-1) \mapsto \frac{1}{2} \in \mathbb{Z}\left[\frac{1}{2}\right]$ is well defined and a homomorphism. But its not to hard to show that it is bijective, too.

For the second: it can't be maximal because $\mathbb{Z}\left[\frac{1}{2}\right]$ isn't a field.
Indeed, the ideal generated by $2x-1$ and $2$ lies in between.

Can you please clarify the second part of the question?  What do you mean when you say the ideal generated by (2x-1) and 2?

Sorry, the one given above is nonsense, a better one is that generated by $(2x-1)$ and $3$.

In general:
Let  $R$ be a (commutative) ring and $a_1,a_2,...,a_n \in R$.
Then the ideal generated by these is $<a_1,a_2,...,a_n> = \{ a_1 r_1 + a_2 r_2 + ... + a_n r_n | r_1 , r_2 , ... , r_n \in R \}$ (the sum of multiples of the so-calles "generators" $a_1, a_2 , ... , a_n$).
The prove that this is indeed an ideal is very similar as the one for principle ideal.
Also for $n=1$, so only one "generator", this is simply the principle ideal generated by $a_1$.

In the above case, it's the ideal of polynomials of type $(2x-1) p(x) + 3 q(x)$ with $p,q$ arbitrary polynomials (with integer coefficient).
This clearly contains all multiples of $(2x-1)$ by definition, but doesn't contain $1$ as can be seen when considering $\mod 3$:
We need that $1 \equiv (2x-1) p(x) \mod 3$, which is impossibly (look at the degree of the polynomial):

There is a more general definition, the "sum of ideals":
Let be any (not necessary finite) set $A = \{ A_i\}$ of ideals given. Then the sum of all these ideals is $\sum_i A_i = \{ \sum_i a_i | a_i \in A_i \text{ and all but finetely many } a_i =0 \}$.
For $A$ finite and $A_i$ the ideal generated by $a_i$, this is the same as defined above.


On a standard die one of the dots is removed at random with each dot
equally likely to be chosen. The die is then rolled. What is the
probability that the top face has an odd number of dots?

Great problem... surprised I haven't seen it anywhere before, is it original?

[hide="Very short"]The even and odd cases are symmetric. Since the top face must either have an odd number of dots or an even number of dots, by symmetry, the answer is $ \frac {1}{2}$.[/hide]

This problem was on the USA 2005 AMC12.

EDIT: I failed. I misread the problem as "a dot is removed from a random face."

No. You fail.

If removed from 6:
4 odds

5: 2

4: 4

3:2

2:4

1: 2

thus we have 2/3x1/2+1/3x1/2=1/2. So you didnt actually fail.  :wink:

Yeah, I did fail.

The probability that the dot is picked from an even face is $ \frac{2\plus{}4\plus{}6}{21}$ because there are $ 21$ dots total and $ 2\plus{}4\plus{}6$ of them belong to an even face. In this case, the probability that the number is odd is $ \frac{2}{3}$.

The probability that the dot is picked from an even face is $ \frac{1\plus{}3\plus{}5}{21}$ because there are $ 21$ dots total and $ 1\plus{}3\plus{}5$ of them belong to an odd face. In this case, the probability that the number is odd is $ \frac{1}{3}$.

Yeah... so the answer is $ \frac{2\plus{}4\plus{}6}{21}\frac{2}{3}\plus{}\frac{1\plus{}3\plus{}5}{21}\frac{1}{3}$ which has a value of $ \frac{11}{21}$.

Oh then i failed too! I think i got this one correct when i did it on alcumus though  :rotfl:

[hide=Solution]There are $1 + 2 + 3 + 4 + 5 + 6 = 21$ dots total. Casework:

The dot is removed from an even face. There is a $\frac{2+4+6}{21} = \frac{4}{7}$ chance of this happening. Then there are 4 odd faces, giving us a probability of $\frac 47 \cdot \frac 46 = \frac{8}{21}$.
The dot is removed from an odd face. There is a $\frac{1+3+5}{21} = \frac{3}{7}$ chance of this happening. Then there are 2 odd faces, giving us a probability of $\frac 37 \cdot \frac 26 = \frac{1}{7}$.
Thus the answer is $\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}$.[/hide]
[hide=*]Source: AoPS [b]not me[/b].  :blush: [/hide]

[hide]you can use casework when we have a odd number and even to get 11/21[/hide]


Find the number of values of $ x$ satisfying the equation
\[ \sqrt {\sin x}  \minus{} \frac{1}
{{\sqrt {\sin x} }} \equal{} \cos x
\]

[hide]Infinitely many: $ x \equal{} {\pi\over 2} \plus{} 2k\pi$[/hide]


[quote="i_like_pie"]In the expression $\frac{a}{b}+\frac{c}{d}+\frac{e}{f}$, each letter is replaced with a different digit from 1, 2, 3, 4, 5, and 6. What is the greatest possible value of this expression?[/quote]
The expression will be the highest with the greatest numbers as numerators.
[hide]
Maximize the value of the 6: $\frac{6}{1}$
Then maximize the 5: $\frac{5}{2}$
And the 4: $\frac{4}{3}$
Add them up: $\frac{6}{1}+\frac{5}{2}+\frac{4}{3}=\frac{36+15+8}{6}=\frac{59}{6}$[/hide]

[quote="star99"]i got that wrong because i wrote it as a mixed number...otherwise i would have placed 1 or 2[/quote]

Really?? The person who was explaining said that they'd would accept it both ways because the problem didn't specify. Thats kinda weird.

yeah, i asked him that...but the chairman's decision is final...

[quote="Ubemaya"]I've never even heard of this competition what was the format/difficulty of the contest and which schools were there?

If I was there I probably would have been killed...[/quote]

The test was like AMC 8 (the junior one). It was pretty easy, didn't really test your mathematical abilities of solving problems, but rather your ability of reading problems correctly and not making stupid mistakes. The senior test scores were much lower than the junior division so I am guessing their test was harder for them.

[quote="mathgeniuse^ln(x)"]Any links of teh questions online.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129655

[quote="Ubemaya"]I've never even heard of this competition what was the format/difficulty of the contest and which schools were there?

If I was there I probably would have been killed...[/quote]

What are you talking about? You're the one who would have killed everone...


[quote="fourierseries"]16. Given reals a,b,c,x,y,z such that

$(a+b+c)(x+y+z)=3$, $(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})=4,$

prove that $ax+by+cz \ge 0$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70625


Can someone please help ?

I saw somewhere that for any number system with base n. If the sum of digits of that number in decimal system is a multiple of n-1 then that number is divisible by n-1. Also if in decimal system the sum of digits in odd places = sum of digits in even places then that number is divisible by n+1.
For example for decimal system, number is divisible by 9 if sum of digits is 9 and divisible by 11 if sums of digits in odd and even places are equal.
Can anyone prove it plz.  :?

If this is easier kindly put it in getting started section.

[hide="hint"]u only need to prove for the decimal system, and then replace the 10 with variable[/hide]


Here's the thing:
1 person says a word.
another person says a word related to the previous word.
and so on

ex. If i say hot dog
someone can say processed meat
someone else can say MickeyD's

The word is:

MickeyD's

cake, pie, oh I don't know, upsidedown pineapple cake?

[quote="randomdragoon"]TEETH!!!!!!!!!!!!!!!!!![/quote]
from what I know, treething doesn't have teeth

[quote="i_like_pie"][quote="randomdragoon"]TEETH!!!!!!!!!!!!!!!!!![/quote]
mouth :D[/quote]
obviously he doesn't have that either.
strangely, he has a tongue  :o

day

I will NOT make the message longer before submitting. Oh wait, I just did.


calculus
derivative
function
trigonometry
calculus computations

How to find derivastive of the function x/y=arctg(y/x)?

First time at my life I see such a thing...

:rotfl:

$ \frac {x}{y} = \tan^{ - 1}\;\frac {y}{x}$

take derivatives of both sides

$ \frac {y\;dx - x\;dy}{y^2} = \left(\frac {x\;dy - y\;dx}{x^2}\right)\;\left[\frac {1}{1 + \left(\frac {y}{x}\right)^2}\right]$

divide by $ dx$ to get

$ \frac {1}{y} - \frac {x}{y^2}\;\frac {dy}{dx} = \left(\frac {1}{x}\;\frac {dy}{dx} - \frac {y}{x^2}\right)\;\left[\frac {x^2}{x^2 + y^2}\right]$

now solve for $ \frac {dy}{dx}$

${ \frac {dy}{dx} = \frac {\frac {1}{y} + \frac {y}{x^2}\left[\frac {x^2}{x^2 + y^2}\right]}{\frac {1}{x}\left[\frac {x^2}{x^2 + y^2}\right] + \frac {x}{y^2}} = \frac {x^2y + 2y^3}{x^3 + 2xy^2}}$


[b][i]hi!
prouve that for evry n£IN and n>=4:
(3+n)^n>3^n+4^n+...+(n+2)^n[/i][/b]


calculus integration logarithms trigonometry derivative algebra

I'm in the process of reading "Surely You're Joking, Mr. Feynman," and have now, twice, heard the process of differentiating under the integral (or using the Leibniz Integral Rule) extolled.

But in all my searching, I haven't been able to come up with much about it.  Wikipedia doesn't even haven an article on it, and [url=http://mathworld.wolfram.com/LeibnizIntegralRule.html]MathWorld[/url]'s is pretty short and not entirely helpful to me.

What's the deal with this thing?  Could someone step me through evaluating an integral by this process, or otherwise shed some light on it?

Curious,
-Starwiz

If $I(t)=\int^b_a f(x,t)dt$ and $f(x,t)$ is 'well-behaved', then

$I'(t)=\int^b_a \frac{\partial}{\partial t}f(x,t)dx$

That's known as Leibnitz' rule for integration. For further explaination about "well-behave" see my post in
http://www.mathlinks.ro/Forum/viewtopic.php?t=62613&start=40

Okay...I think I understand so far.

But how would I solve the example Mathworld gives using this method?

They tell us that $\phi(\alpha) = \int^\pi_0 \ln \left(1 - 2\alpha\cos x + \alpha ^ 2 \right) \, dx$ can be solved to $\phi(\alpha) = 2 \pi \ln |\alpha|$.

My first step was to take the partial derivative of the expression within the integral with respect to alpha:

$\phi'(\alpha) = \int^\pi_0 \frac{2\alpha - 2 \cos x}{1 - 2\alpha \cos x + \alpha ^2} \, dx$

That may or may not be solvable.  But if it is, I don't know how.  My only guess, partial fractions, doesn't seem to help; I get (through my calculator):

$\phi'(\alpha) = \int^\pi_0 \left(\frac{-\alpha}{2\alpha \cos x - \alpha^2 - 1} + \frac{1}{\alpha \left(2\alpha \cos x - \alpha^2 - 1 \right)} + \frac{1}{\alpha} \right)\, dx$

The first two terms of which I don't know what to do with.

Taking the partial derivative with respect to alpha again doesn't help, either.  It becomes really ugly.

Any ideas as to how to make this work?

Thanks again,
-Starwiz

Please take a look. :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17769]www.mathlinks.ro/Forum/viewtopic.php?t=17769[/url]


A right circular cone of volume $ A$, a right circular cylinder of volume $ M$, and a sphere of volume $ C$ all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then

$ \textbf{(A)}\ A \minus{} M \plus{} C \equal{} 0 \qquad \textbf{(B)}\ A \plus{} M \equal{} C \qquad \textbf{(C)}\ 2A \equal{} M \plus{} C$
$ \textbf{(D)}\ A^2 \minus{} M^2 \plus{} C^2 \equal{} 0 \qquad \textbf{(E)}\ 2A \plus{} 2M \equal{} 3C$

[hide="Answer"]All have the same radius, and the height of the cylinder and the cone are both equal to the diameter of the sphere, so the height is $2r$. Therefore, we have the volume of the cone as $A=\frac{1}{3}\pi r^2(2r)=\frac{2\pi r^3}{3}$, the volume of the cylinder is $M=\pi r^2(2r)=2\pi r^3$, and the volume of the sphere is $C=\frac{4}{3}\pi r^3$. Therefore, we have $A-M+C=\frac{2\pi r^3}{3}-2\pi r^3+\frac{4\pi r^3}{3}=0\Rightarrow \boxed{A}$.[/hide]


Given three positive real numbers $ x$, $ y$, $ z$ such that $ x \plus{} y \plus{} z \equal{} 1$, prove that

$ \frac {xy}{\sqrt {xy \plus{} yz}} \plus{} \frac {yz}{\sqrt {yz \plus{} zx}} \plus{} \frac {zx}{\sqrt {zx \plus{} xy}} \le \frac {\sqrt {2}}{2}$.

I've just found out that this problem was at China TST:
http://www.mathlinks.ro/viewtopic.php?p=550608#550608


trigonometry
geometry
trapezoid
trig identities
Law of Cosines

Quadrilateral ABCD is inscribed in a circle with diameter AD = 4, if side AB and BC each have length 1, then find CD?

Probably made some errors...

[hide]Let $ CD \equal{} x$. We can see that $ BD \equal{} \sqrt {15}$ and $ AC \equal{} \sqrt {16 \minus{} x^2}$.

By Ptolemy's theorem:

$ 4 \plus{} x \equal{} \sqrt {240 \minus{} 15x^2}$
$ 16 \plus{} 8x \plus{} x^2 \equal{} 240 \minus{} 15x^2$
$ 16x^2 \plus{} 8x \minus{} 224 \equal{} 0$
$ 2x^2 \plus{} x \minus{} 27 \equal{} 0$
$ x \equal{} \frac {7}{2}$[/hide]

EDIT: Subtraction error  :mad:

[hide="Trig"]
Law of Cosines on $ \triangle AOB$ ($ O$ center) yields $ \theta \equal{} \angle AOB \equal{} \cos^{ \minus{} 1}\left(\frac {1^2 \minus{} 2^2 \minus{} 2^2}{ \minus{} 2 \cdot 2 \cdot 2}\right) \equal{} \cos^{ \minus{} 1}(7/8)$. Thus $ \angle COD \equal{} 180 \minus{} 2\cos^{ \minus{} 1}(\theta)$, and $ CD^2 \equal{} 2 \cdot 2^2 \plus{} \minus{} 2 \cdot 2^2 \cos \angle COD$, and $ \cos \angle COD \equal{} \minus{} (2 \cos^2 \theta \minus{} 1) \equal{} \minus{} \frac {17}{32}$. Hence $ CD \equal{} \sqrt {8 \plus{} \frac {17}{4}} \equal{} \boxed{\frac {7}{2}}$.
[/hide]

[quote="azjps"][hide="Trig"]
Law of Cosines on $ \triangle AOB$ ($ O$ center) yields $ \theta \equal{} \angle AOB \equal{} \cos^{ \minus{} 1}\left(\frac {1^2 \minus{} 2^2 \minus{} 2^2}{ \minus{} 2 \cdot 2 \cdot 2}\right) \equal{} \cos^{ \minus{} 1}(7/8)$. Thus $ \angle COD \equal{} 180 \minus{} 2\cos^{ \minus{} 1}(\theta)$, and $ CD^2 \equal{} 2 \cdot 2^2 \plus{} \minus{} 2 \cdot 2^2 \cos \angle COD$, and $ \cos \angle COD \equal{} \minus{} (2 \cos^2 \theta \minus{} 1) \equal{} \minus{} \frac {17}{32}$. Hence $ CD \equal{} \sqrt {8 \plus{} \frac {17}{4}} \equal{} \frac {7}{2}$.

Probably not right, I'll fix when I get back, just temporarily submitting.
[/hide][/quote]

lol I was just about to post a law of cosines proof. Anyways, you actually were right.

[quote="Carbon57"]Probably made some errors...

[hide]
$ 4 \plus{} x \equal{} \sqrt {240 \minus{} 15x^2}$
$ 16 \plus{} 8x \plus{} x^2 \equal{} 240 \minus{} 15x^2$
$ 16x^2 \plus{} 8x \minus{} 216 \equal{} 0$
$ 2x^2 \plus{} x \minus{} 27 \equal{} 0$
$ x \equal{} \frac { \minus{} 1 \plus{} \sqrt {217}}{4}$[/hide][/quote]

The problem with that: $ 16\minus{}240\neq \minus{}216$ =) if you fix that you'll get the right answer.

Consider quadrilateral AEFD with AE = FD = 1.It is obvious that EF = CD.As AEFD is isosceles trapezoid $ AF \equal{} ED \equal{} \sqrt {15}$.Applying Ptolemy we get $ 15 \equal{} 4EF \plus{} 1\Rightarrow \;EF \equal{} \frac {7}{2 }$


inequalities
Cauchy Inequality
inequalities unsolved

$ (a,b,c)>\equal{}0$

Prove that : \[ \sum_{cyc} \sqrt{a^2\plus{}bc} \leq \frac{3(a\plus{}b\plus{}c)}{2}\]

does the stronger inequality hold? :

\[ \sum_{cyc} \sqrt{a^2\plus{}bc} \leq \sqrt{2}(a\plus{}b\plus{}c)\]

[quote="anas"]$ (a,b,c) > \equal{} 0$

Prove that :
\[ \sum_{cyc} \sqrt {a^2 \plus{} bc} \leq \frac {3(a \plus{} b \plus{} c)}{2}
\]
[/quote]
It's the hungktgtn's very nice inequality. :lol:

maybe we can use jenesen because $ \sqrt{x}$ is concave....

no offense to Vasc , but is there any solution more elementary or involving Cauchy-shwarz???  :blush:

I don't remember who is author of the following proof:
Let $ a\geq b\geq c.$ Then $ 2a\plus{}c\geq2\sqrt{a^2\plus{}bc}.$
Thus, it's remains to prove that $ a\plus{}3b\plus{}2c\geq2\left(\sqrt{b^2\plus{}ac}\plus{}\sqrt{c^2\plus{}ab}\right),$ which equivalent to
$ a^2\plus{}5b^2\plus{}2ab\plus{}12bc\geq8\sqrt{(b^2\plus{}ac)(c^2\plus{}ab)}$
and use $ b^2\plus{}ac\plus{}c^2\plus{}ab\geq2\sqrt{(b^2\plus{}ac)(c^2\plus{}ab)}.$
We give a quadratical inequality of $ a,$ which easy to prove. :wink:

yes thank you again arqady ( but a cauchy solution would be great  :D )

How would you prove the inequality with Cauchy-Schwarz?
Maybe that wouldn't be useful.

[quote="kunny"]How would you prove the inequality with Cauchy-Schwarz?[/quote]Using the Cauchy inequality,

$ \left(\sum{\sqrt {a^2 \plus{} bc}}\right)^2\leq\sum{\left(3a^2 \plus{} 4bc \plus{} ca \plus{} ab\right)}\sum{\frac {a^2 \plus{} bc}{3a^2 \plus{} 4bc \plus{} ca \plus{} ab}}$

$ \equal{} 3(a \plus{} b \plus{} c)^2\sum{\frac {a^2 \plus{} bc}{3a^2 \plus{} 4bc \plus{} ca \plus{} ab}}.$

Hence, it suffices to prove that

$ \frac {a^2 \plus{} bc}{3a^2 \plus{} 4bc \plus{} ca \plus{} ab} \plus{} \frac {b^2 \plus{} ca}{3b^2 \plus{} 4ca \plus{} ab \plus{} bc} \plus{} \frac {c^2 \plus{} ab}{3c^2 \plus{} 4ab \plus{} bc \plus{} ca}\leq \frac {3}{4}.$

But $ \left[3 \minus{} 4\sum\frac {a^2 \plus{} bc}{3a^2 \plus{} 4bc \plus{} ca \plus{} ab}\right]\prod\left(3a^2 \plus{} 4bc \plus{} ca \plus{} ab\right)$

$ \equal{} 8(b \minus{} c)^2(c \minus{} a)^2(a \minus{} b)^2 \plus{} abc\left[5\sum{a^3} \plus{} 27\sum{a^2(b \plus{} c)} \plus{} 66abc\right]\geq0.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=245304


was that all you wrote or did you write more explaining why given those are the only triples, you get 13 and 31?

that was just the end of my solution, i had more before that

hm i proved why the only solutions for (a,b,c) are (2,3,4) and (2,3,5)

I would've proved why those are the only solutions. The proof would however be basically the exact same thing as the contradictions in the official solutions, which was the main point, so you may end up losing a point or so.

first  i proved x
then i proved y
then i proved z
so all in all my solution was a page
hoping for a decent grade

Well I said

Note that the maximum value for (n-1)/a+(n-1)/b+(n-1)/c is (n-1)/2+(n-1)/3+(n-1)/4, since if we have n-1, then (n-1)/b+(n-1)/c=1, which is a contradiction (I said earlier the minimum value was 1)

The next largest is (n-1)/2+(n-1)/3+(n-1)/5.

The next largest is (n-1)/2+(n-1)/3+(n-1)/6=n-1, which can't be equal to n.

Anything else is smaller, so the only solutions are 13,31. QED.

Would this be a 5? It's essentially the same solution, just without all the fancy wording...

[quote="BOGTRO"][quote="Yongyi781"][quote="BOGTRO"]
The next largest is (n-1)/2+(n-1)/3+(n-1)/6=n-1, which can't be equal to n.
[/quote]

You would have to prove that changing the (n-1)/5 to (n-1)/6 would be the next largest. I avoided that altogether by saying "incrementing any of the denominators by 1" would result in a value less than or equal to 1.[/quote]

That isn't a trivial fact??

I found that so simple I didn't have to put it...[/quote]

Well... that's a fine line that might be the difference between a 4 and a 5.

[quote="andersonw"][quote="abacadaea"]in my solution, i said something along the lines of "The only (a,b,c) with a>b>c that satisfy 1/a+1/b+1/c>1 are (2,3,4) and (2,3,5). Thus, the answers are 13 and 31."

How many points will i lose for assuming this? Its not trivial, but its pretty obvious in my opnion.[/quote]

Well, on something completely unrelated, 2<3<4 and you said a>b>c   :wink:
If you proved statements that implied those two were the only two solutions, you should be okay.[/quote]

yeah, i realized that. I didnt do that in my solution though. :)
Wait. I proved 1/a+1/b+1/c>1. is that enough?

Look guys, if you were in a room with someone nitpicky, and they kept asking you "Why," I'm sure at some point you would have to think more than you thought you would have to about your answer. When you're writing your solutions try to do this to yourself, be annoying and keep asking yourself "Why." If there's any doubt as to all in your mind for whether or not you're being detailed enough, add more detail.

But the way to go is this. We want to find positive integral 1<a<b<c such that 1/a+1/b+1/c>1. Note that since we are dealing with integers the condition is equivalent to $ 2\le a\le b\minus{}1\le c\minus{}1$. We suspect that (a,b,c)=(2,3,4),(2,3,5) are the only solutions. First we show that a=2. Assume for contradiction that $ a\ge3$. Then $ b\ge4$ and $ c\ge b\plus{}1\ge5$. So $ 1/a\plus{}1/b\plus{}1/c\le 1/3\plus{}1/4\plus{}1/5<1$. So $ a\equal{}2$.

Then we now want to show that b=3. Clearly $ b\ge a\plus{}1\equal{}3$. So assume $ b\ge4$. Then $ c\ge5$. So $ 1/a\plus{}1/b\plus{}1/c\le 1/2\plus{}1/4\plus{}1/5<1$. So b=3. The rest is easy. $ 1/a\plus{}1/b\plus{}1/c\equal{}1/2\plus{}1/3\plus{}1/c>1$, so $ 1/c>1/6$, or $ c<6$. This gives $ c\equal{}4,5$.

This is the easiest way to rigorously establish that those are our only solutions. If I were a betting man, I'd put down some cash that the graders will deem it necessary to have something like the above.


Ola!
Verdaderamente este es un buen problema, me gusto!!!
La solucion es algo larga de escribir, y en este momento no tengo tiempo, pongo un adelanto.

Las soluciones son:
$(x,y)= (1,0) (-1,0) (4,3) (-4,3) (4,5) (-4,5)$

Ojala pongan mas problemas interesantes como este!!

 :lol:

si esta bien seria bueno que lo postees por q ami me salio solo viendo las cotas de x e y se llega a que $x+1=y$ ó $-(x+1)=y$ ó $y=x+2$ ó $y=x+1$ luego reemplazando tenemos esos valores   :D

Pues si, mi solucion fue muy diferente a la tuya!!
La posteare luego, ojala que tu tambien pongas la tuya!!
 :lol:

[quote="mhuarancca"]bueno un problema mas:

hallar todas las soluciones enteras de la ecuacion :

     $(x^{2}-y^{2})^{2}=1+16y$[/quote]

Bueno, espero que la mia sea diferente a la de los dos :)

$(x^{2}-y^{2})^{2}= 1+16y \Rightarrow (x^{2}-y^{2})^{2}+(8y)^{2}= (1+8y)^{2}$

Esta es una ecuación $a^{2}+b^{2}= c^{2}$ (Fermat para $n=2$). Todas las soluciones con $a$, $b$ y $c$ primos relativos están parametrizadas por $m$ y $n$ en $(m^{2}-n^{2})^{2}+(2mn)^{2}= (m^{2}+n^{2})^{2}$. Como $8y$ y $8y+1$ son primos relativos, las soluciones enteras tienen que estar dadas por

$(8y)^{2}= (2xy)^{2}$

De aquí resultan tres posibilidades $y=0$ o $x = \pm 4$. Analizando cada caso se obtiene la solucion de conejita arriba.

Daniel


function
algebra
polynomial
floor function
calculus
derivative
algorithm

Hi everybody!

I have some problem with Hölder (Lipschitz) exponent!
From what I know, Lipschitz refer to integer values whereas Hölder to non-integer ones...

The usual definition roughly states that:

A function $ f$  is pointwise Lipschitz (local Hölder) $ \alpha \geq 0$ at $ \nu$ if
there exist K>0 and a polynomial $ P_{\nu}$ of degree $ m \equal{} \lfloor \alpha \rfloor$, where $ \lfloor \alpha
\rfloor$ is defined as the largest integer less than or equal to $ \alpha$, such that
$ \forall  x \in  \mathbb{R} , |f(x)\minus{}P_{\nu}(x)|\leq
K|x\minus{}\nu|^{\alpha} .$

The above definition refers only to functions which have a positive alpha. I read somewhere , but I can't find no more the reference :( ,  that "if we want to consider negative values
too, as in this case the concept of function is not longer valid, we have to shift to the class of tempered distribution"

What does it mean? Why if such exponent is negative the concept of function is no longer valid? Can you give me some intuition for that? :(

Many many thanks in advance!!
:)

Vale

There's no point in looking at exponents greater than 1; that forces the derivative to be zero everywhere, giving only (locally) constant functions. Lipschitz is an exponent of exactly 1, and no other cases.

If you apply the definition unchanged using a negative exponent, you get that faraway points must have similar values, but being nearer doesn't help. This gives strange results. On $ \mathbb{R}$, the only such functions are constants. On a finite interval $ [a,b]$, we get the class of all bounded functions, with no further restrictions. If you want something interesting, you need a different definition.

Hi Jmerry!
Thank you for your answer! Maybe I haven't been so precise in the statement of my question...
You are right, and the definition I've provided is not valid for negative holder exponent.

I'm currently dealing with singularity detection algorithm and try to understand well the theory behind...

If you want to generalize the definition of holder exponent to negative values too, I've found the following definition:
 [i] Let f(x) be a tempered distribution on an interval (a, b). The distribution
f(x) is said to have a global Holder exponent alpha on (a, b) if and only if its
primitive has a global Holder exponent alpha+1 on (a, b).[/i]

My point is try to well understand conceptually why to contemplate negative exponent, we have to shift to the distributions domain. What does happen topologically speaking in that point if I detect a negative holder? how is that functional in that point? I can't get it very well.

I used to read the following legend about holder exponent:
$ \alpha \geq 1$ denotes a continuous and differentiable function; $ 0 < \alpha < 1$ represents the continuous but non-differentiable case; $ \minus{}1 < \alpha \leq 0$  is valid for
discontinuous and non-differentiable functions and finally for $ \alpha \leq \minus{}1$   we are dealing with functions no longer locally integrable, like the Dirac distribution and we have to
shift to the tempered distributions domain.

Can you help me somehow why negative value make sense only for distributions?
Many many thanks to everybody!!!

Vale

Between $ 0$ and $ \minus{}1$, you can make some sense of it with ordinary functions. The antiderivative of $ |x|^{\minus{}\frac12}$ has Holder exponent $ \frac12$, for example. Beyond $ \minus{}1$, the singularities are so bad that you can't integrate across them, and don't get anything new unless you introduce distributions.


Aca un lindo problema!

[b]100a+10b+c[/b]=[b]n(a+b)[/b] donde [b]n[/b] es múltiplo de 7 más 4. Hallar el valor de n si [b](a+b)[/b] es mínimo.

Tienes que especificar qué son a, b y c: enteros, enteros positivos, dígitos?  :mad:
Por la forma del problema pareciera que son dígitos, en tal caso la escritura debe ser $\overline{abc}=n(a+b)$.

$Tipe$

:oops: upss! se me escapo... :huh:
y si...es de esa manera
Gracias por la observacion


superior algebra
superior algebra unsolved

Please help me to prove and explain the geometric meaning of the primary decomposition.Thanks alots.
a/ $ J\equal{}<xy> \equal{} <x>\cap<y>$ when $ n\equal{}2$.
b/$ J\equal{}<xy,xz> \equal{} <x>\cap<y,z>$ when $ n\equal{}3$


The average height of ten players on one of the local basketball teams is $ 6^{\prime}1^{\prime\prime}$. Two more players, $ 6^{\prime}5^{\prime\prime}$ and $ 6^{\prime}9^{\prime\prime}$ tall, join the team. by how many inches did the average height of the team increase?

Let's only use the inches part of each player. The average height was 1 inch, so the total height of 10 players was 10 inches. A 5 inch and 9 inch player join. This makes the total height 10+5+9=24, with 10+2=12 total players. The average height now is 24/12=2 inches. So, the total height increased by 2-1= $ \boxed{1}$ inch


calculus
integration
calculus computations

Suppose that F= <x, y, z >.  Calculate the flux of F through the outwardly oriented ellipsoid E given below,  (x^2/a^2) + (y^2/b^2)+ (z^2/c^2)=1

Are you allowed to use the [url=http://en.wikipedia.org/wiki/Divergence_theorem]divergence theorem[/url]?

Note that $\nabla \cdot \vec{F}= 3.$ Using the divergence theorem sets up the integral of that constant over the interior of the ellipsoid, which makes it $3$ times the volume of the ellipsoid.

The most efficient way to compute the triple integral, assuming you don't already know the answer, would be to make the changes of variable $u=\frac{x}{a},v=\frac{y}{b},w=\frac{z}{c}.$ This would convert the ellipsoid into a sphere.

How do you calculate the volume of the ellipsoid.   How to use the changes of variable u=x/a, v=y/b, w=z/c?  Can you be a little more specific?  Thanks a lot.  you help is really appreciated.

any one knows how to do it?  please reply.  thanks.

how to use the change of variable to find the volume of the elipsoid?  Can anyone help me out?  I don't know how to do it.

Acting completely helpless, and then begging, are not traits that make people want to answer you. The change of variables is straightforward. Just write down the triple integral and do it.


conics
ellipse
calculus
calculus computations

In the $ xy$ plane, Let $ L$ be the part of an ellipse $ \frac {x^2}{4} \plus{} \frac {y^2}{3} \equal{} 1\ (y\geq 0)$ and denote  the parts of 2 circles $ (x \minus{} 1)^2 \plus{} y^2 \equal{} 1\ (y\leq 0),\ (x \plus{} 1)^2 \plus{} y^2 \equal{} 1\ (y\leq 0)$ by $ M,\ N$ respectively. For moving points $ P\in{L},\ Q\in{M},\ R\in{N}$, find the maximum value of $ \overline{PQ} \plus{} \overline{PR}.$

This question is almost trivial.

[hide]
There are two key observations that must be made.  (1) The foci of the ellipse are also the centers of the circles.  (2) In a plane, the maximum distance from any given fixed point to a circle is measured along the line joining the given point and the circle's center.

Once these facts are noted, then it is obvious that for any given point $ P$ on the ellipse, the maximum value of $ \overline{PQ} \plus{} \overline{PR}$ is attained when $ \overline{PQ}$ and $ \overline{PR}$ intersect (-1,0) and (1,0), respectively.  But this is equivalent to the sum of the radii of the two circles, plus the sum of the distances from $ P$ to the foci.  All of these distances are constant for any choice of $ P$, and the sum is $ 4 \plus{} 1 \plus{} 1 \equal{} 6$.
[/hide]

That's right. This problem can be solved by using the definition of an ellipse; $ \overline{PF_1} \plus{} \overline{PF_2} \equal{} Const.$


Let a triangle $ABC$ inscribed in circle $\Gamma$ be given. Circle $\Theta$ lies in angle $Â$ of triangle and touches sides $AB, AC$ at $M_1, N_1$ and touches internally $\Gamma$ at $P_1$. The points $M_2, N_2, P_2$ and $M_3, N_3, P_3$ are defined similarly to angles $B$ and $C$ respectively. Show that $M_1N_1, M_2N_2$ and $M_3N_3$ intersect each other at their midpoints.

Orl, it would be nice if you, at least, gave the links to the places were the problems were posted before by yourself. This one was discussed in Hyacinthos messages #9537, #9570, #10148, #10150:

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/9537?expand=1]Hyacinthos message #9537[/url]"][b]From:[/b] "orl_ml" (Orlando Doehring)
[b]Subject:[/b] triangle ABC inscribed in circle

Let a triangle ABC inscribed in circle (O) be given . Circle (K1)
lies in angle Â of triangle and touches sides AB, AC at M1, N1 and
touches internally (O) at P1. The points M2, N2, P2 and M3, N3, P3
are defined similarly to angles B and C respectively. Show that M1N1,
M2N2 and M3N3 intersect each other at their midpoints.[/quote]

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/9570?expand=1]Hyacinthos message #9570[/url]"][b]From:[/b] Juan Carlos Salazar
[b]Subject:[/b] Re: triangle ABC inscribed in circle

Dear Orlando:
You wrote:
> Let a triangle ABC inscribed in circle (O) be given . Circle (K1)
> lies in angle Â of triangle and touches sides AB, AC at M1, N1 and
> touches internally (O) at P1. The points M2, N2, P2 and M3, N3, P3
> are defined similarly to angles B and C respectively. Show that M1N1,
> M2N2 and M3N3 intersect each other at their midpoints.

Yes, this point is the incenter of ABC triangle.
Next a simple proof:
Take only circle (K1), circle (O) and ABC triangle with the tangency
points M1,N1 and P1,denote in ABC triangle:ang(B)=2beta,ang(C)
=2gamma,also it is well know that P1M1 is bisector of ang(AP1B)in
AP1B triangle and P1N1 is bisector of ang(AP1C)in AP1C triangle,
hence: ang(BP1M1)=ang(M1P1A)=ang(C)/2=gamma and ang(AP1N1)=ang(N1P1C)
=ang(B)/2=beta, also produced P1N1 cut (O) at D, then BD is bisector
of ang(B), also BD cut M1N1 at E, now:
ang(AM1N1)=ang(AN1M1)=beta+gamma ,due to (AM1=AN1).
ang(M1BE)=ang(M1BD)=ang(EBC)=ang(DBC)=beta.
ang(M1EB)=ang(AM1E)-ang(M1BE)=ang(AM1N1)-ang(M1BE)=(beta+gamma)-
beta=gamma.
Therefore:M1BP1E is cyclic, due to:ang(BP1M1)=ang(M1EB)=gamma,then:
ang(M1P1E)=ang(M1BE)=beta, so ang(EP1D)=gamma.Also EP1CN1 is cyclic
due to ang(AN1M1)=ang(EP1C)=beta+gamma, now: ang(ECN1)=ang(EP1N1)
=gamma, so CE is bisector of ang(C).Hence E is incenter of ABC
triangle and midpoint of M1N1, similarly for the anothers circles.QED.
Sincerely
Juan Carlos[/quote]

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/10148?expand=1]Hyacinthos message #10148[/url]"][b]From:[/b] Juan Carlos Salazar
[b]Subject:[/b] Re: triangle ABC inscribed in circle

Dear friends:
--- In "Hyacinthos", "orl_ml" (Orlando Döhring)
wrote:
> Let a triangle ABC inscribed in circle (O) be given . Circle (K1)
> lies in angle Â of triangle and touches sides AB, AC at M1, N1 and
> touches internally (O) at P1. The points M2, N2, P2 and M3, N3, P3
> are defined similarly to angles B and C respectively. Show that M1N1,
> M2N2 and M3N3 intersect each other at their midpoints.
If we consider the triangle XYZ bounded by the tangent lines at
tangency points of the circles (K1),(K2),(K3)with circle(O).
Now ABC and XYZ are perspective with perspector Ko.
Can someone to find other properties of Ko?
Thanks
Sincerely
Juan Carlos[/quote]

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/10150?expand=1]Hyacinthos message #10150[/url]"][b]From:[/b] Jean-Pierre Ehrmann
[b]Subject:[/b] Re: triangle ABC inscribed in circle

Dear Juan Carlos
> --- In "Hyacinthos", "orl_ml" (Orlando Döhring) wrote:
> > Let a triangle ABC inscribed in circle (O) be given . Circle (K1)
> > lies in angle Â of triangle and touches sides AB, AC at M1, N1 and
> > touches internally (O) at P1. The points M2, N2, P2 and M3, N3, P3
> > are defined similarly to angles B and C respectively. Show that M1N1,
> > M2N2 and M3N3 intersect each other at their midpoints.
> If we consider the triangle XYZ bounded by the tangent lines at
> tangency points of the circles (K1),(K2),(K3)with circle(O).
> Now ABC and XYZ are perspective with perspector Ko.
> Can someone to find other properties of Ko?

Some remarks :
1) the line M1N1 is the perpendicular at I to AI
2) P1P2P3 is the circumcevian triangle of X(56) = external center of
similitude of the circumcircle and the incircle
3) If A1B1C1 is the circumcevian triangle of a point P and A2B2C2
the triangle bounded by the tangents to the circumcircle at A1, B1,
C1, then A2B2C2 and ABC are perspective at the isogonal conjugate of
the anticomplement of the isogonal conjugate of P.

As X(56) is the isogonal conjugate of the Nagel point, it follows
that Ko is the isogonal conjugate of the anticomplement of the Nagel
point or, equivalently, Ko is the isogonal conjugate of the
reflection of the Nagel point in the incenter.
Friendly. Jean-Pierre[/quote]

With Inversion Prove that Midpoint of $M_iN_i$isincenter of $ABC$

Can we express the radius of the incircle as a function of the radii of the three circles in the problem


Let $ O$ be the circumcenter of $ \Delta ABC$.Points $ A',B',C'$ are taken on $ AO,BO,CO$ such that $ A'O \equal{} B'O \equal{} C'O$. Denote by $ A_1,B_1,C_1$ the intersection of perpendicular lines from $ A',B',C'$ on $ AO,BO,CO$ with $ BC,AC,AB$.Prove the collinearity of $ A_1,B_1,C_1$.
Added problem: if $ A',B',C'$ are the midpoints of $ AO,BO,CO$ then the line $ A_1B_1C_1$ is perpendicular to $ ON'$
($ N$ is the center of the nine-points circle of $ \Delta ABC$ and $ N'$ is chosen such that $ \angle NAB\equal{}\angle N'AC$ and $ \angle NBC\equal{}\angle N'BA$ and ...)

An old problem.

[b]Happy the Independence Day of Vietnam! [/b] :thumbup:


I have to appear for NTSE 2nd round

While solving previous years papers (prev 2 years) i got an average of around 70-80 in both papers(MAT &SAT)marks in MAT were higher by 3-4 marks

1)whether i'll clear it or not?
2)do cut offs are for individual papers?

PS-To mod,I think this topic can be discussed here
           If not Sorry and please let me know

75 in SAT is good enough
1 of my friends said he expected 95 in MAT bit couldn't clear
i feel cut-off is on a combined basis :rotfl:

i really doubt if der's any cut off system in ntse
coz wen i had been to our state camp dey had clearly informed us of absence of cut offs
thing is if u can be in top 1000 dat's it u r thru
and to provide dat rank dey take combined marks off sat mat and interview
had dey used cut offs den i dont think evryone in 2nd level would be interviewed
even den from my own experience i can assure of no cut offs for individual paper as my score in sat was pathetically bad yet i could make it to d top 1000

cutoffs r there similar to jee 4 2nd round
but they do give preference to overall score
suppose 999th rank got 150(total)-u score 100 in SAT & 50 in MAT,i don't think u will b the 100th
cutoffs r used during ties

and ofc  there r no cutoffs at national level i.e.interview

How do you get the results of the first round?
Here, in Tamil Nadu, the 1st level was on November 18th 2007 and still there is no news about the results.
Though I know I will surely qualify, I don't know when to start preparing for 2nd level as it is not confirmed.
Almost just a month more for the 2nd level, they still haven't given the results.

[quote="Aravind Srinivas L"]
Though I know I will surely qualify, I don't know when to start preparing for 2nd level as it is not confirmed.
Almost just a month more for the 2nd level, they still haven't given the results.[/quote]

1. He's in class X
2. Results are released by the individual government?
3. How do you know that you will 'SURELY QUALIFY'? Thats called overconfidence.

[quote="Cyber-Shreyas"]
3. How do you know that you will 'SURELY QUALIFY'? Thats called overconfidence.[/quote]
Now thats not necessary bcos he might have worked d a m n hard for it :) and/or must have done the exam exceptionally well :)

but still there cud b uncertainties :D

i worked d a m n nothing 4 it but still qualified(mayb there was some mixing of  papers :rotfl: )

here the results r displayed in newspapers(last yr it was in HT)

no cutoffs as far as I know......
Rank is declared on the basis of the combined marks of written exam and interview....and i think they give more importance to interview because I had done the written pretty decently but couldn't qualify for the scholarship bcoz i had done very badly in the interview........ :ninja:

i think that chap is just confident, not overconfident!

however my advice to him will be, don't carry this attitude to the interview, or you will ruin your chances!!

as for the overall results, they do consider interview marks, but generally, interview marks are less than 50 percent as there is a small scope for bias in it (some questions may be easy, while others tough, as opposed to written round where everyone gets the same questions).
i could say that i had a decently good interview (they asked me something about e-governance, and i blabbered something about internet and governance, and they asked me if i knew exactly what it was, and i was honest with them- (i dont know anything abt it, and i think they appreciated it!!!) and hence got the scholarship.

P.S. i think 75 is a high score for SAT (i believe i was getting abt 50-60). however the MAT was very easy and i got more than 95 so i think they consider a minimum cutoff, above which they select the top 1000 students on the basis of total marks.

[quote="anandghegde"] i think they give more importance to interview because I had done the written pretty decently but couldn't qualify for the scholarship bcoz i had done very badly in the interview........ :ninja:[/quote]
200 marks 4 written & just 25 4 interview--and u guys r taking abt 50% weightage 2 interview!!!!

interview is just a FORMALITY!!

thanks to all 4 your advice
@ Aravind Srinivas L
we got results firstly through newspaper and then through letter from SCERT
but i've not received any admit card till now
in letter they told to wait till 1st week of April and gave 2 telephone nos to contact
Best of luck .May you get selected
Anyways nice to see someone of same class

Again thanks to all

But I'll really thank you after final results :wink:

Can anybody suggest some good books for NTSE as i have only books of class +1 this time and time is running out

Thanks in advance

NTSE guide by TMH publications is nice....And the NTSE series by Macmillan publications is also good.  :)

TMH sux.........--

i wud prefer Upkar guide--it really guided me through

Well
I have qualified in level 1 with 1st position in the state in 8th std.
I am going to start preparing after 13th April as my Annual exams get over only by that time.
Could any one suggest a good book for preparing for the second level?

I believe only Arihant has a book meant for class VIII. As for class X, Dhanpat Rai is excellent.

i just cleared the 1st round of ntse . but i dont know how to prepare for the 2nd round. please guide me and tell me the books i would require

what is your class,nobody from my class   :(  :(

:P finally we have few eleventhies!!!!!!!!! :D  :D

yes

i referred JOI and this site to some of my friends
they all left after some time
they were more interested in orkut,etc

but now i have got one from my class :D  :D  :rotfl:

hey i am not orkut coz i really am not interestred in ll those stufd!

but most of my friends i referred to these sites were not interested
i use orkut also :D

but not interested much in orkut

one can't hav n opinion about orkut as it is really i BBIIGG site
if u want 2 see all "THOSE" stuff ,u check out my id and the communities i hav joined
[hide]my ID is "shiny pokes :D "[/hide]
i m there just 2 ask my teachers not 2 punish me if i don't do my homework :rotfl:  :rotfl:
......... :P

well i don't dislike it ,but don't like it much

same here rituraj even i think there are some good things and some bad things in orkut .IT is just on you as to how u tackle it .some people take the good of it while somee just creep on!
well i am rather a little awaaaaay from those things as my mind gets quite often but actually i dont hate orkut .i just dont think i am for it!
hey leaving besides those things  i just wanted to know the detailed syllabus of kvpy plz plz plz plz rituraj or someone who has cleared plz tell me wat to study for kvpy
and also increase my RATING
thanx in advance
bye !
see ya!
plz plz plz plz reply
its quite urgent!

hey man
the biggest syllabus about kvpy is  don't worry -- remain calm and composed

what was the syllabus of NTSE?? it is the same 4 kvpy science

thanx a lot and y is my rating 1.2
can u or anyone help?

no i am much more active than rituraj on orkut coz it is place where many of my family members and friends are present

btw,how do we know our ratings????

i dont know but mine is 1.2
plz help choronz or rituraj or pasrdesi plz mods

my rate is just 1.3 but y to worry about that
ur rating not even visible to others usually.............


vector
linear algebra
matrix
linear algebra unsolved

hello,

here is a rather hard pb that my teacher gave me:

let E a linear space of dimension n. F=(F0,F1,..,Fn) ,where Fi is a subspace of E of dimension i and Fi is included in F(i+1) , is called a flag. Let F and F' be 2 flag.Prove that there exist a base B=(e1,e2,..,en) of E so that:
1) Fi=Vect(e1,e2,...,ei)
2) F'i=vect(e_p(1), e_p(2), ..,e_p(i)) where p is a permutation of [1,n]

My teacher gave me a quite artificial solution but i don't know if there exists a more natural solution .. any idea ?

[quote="Maja"]I'm not absolutely sure about this solution, but here it is.

Consider standard base of E i.e. (1,0,0,...,0), (0,1,0,...,0),..., (0,0,0,...,1) and arbitrary flag F. Then we have that one of these vectors is base for F1, and denote it with e1. Since F1 is included in F2, then we can supplement e1 with another vector (from standard base of E) which we call e2, so that e1 and e2 are base for F2. Then again, we find e3 such that e1, e2, e3 are base for F3,... and so on. Finally, for Fn, we will obtain that its base is standard base (only these vectors are probably in different order).
Then, repeat all this for F', but vectors you obtain denote with ei1, ei2,..., ein, respectively. Doing this, we also get and permutation p (i1, i2,..., in) which satisfies 1) and 2).
So, is this ok?

Bye[/quote]

Hi Maja,

in fact i don't think it's so simple .. Why "one of these vectors is base of F1" and why "Since F1 is included in F2, then we can supplement e1 with another vector (from standard base of E)" these statements are not correct i think ..

bye

Yes, alekk, you're right.

I misunderstood the problem , so I made stupid mistakes  :blush: ;  and definitely its not so easy.
I'll try to find corret solution after my exams, maybe I'll be more concetrated then  :)

... or maybe I can try to correct it.
I shouldn't start from the standard base (I have no idea why I did that), but the rest is very similar.

Well, F1 has its own base e1. Because, F1 is included in F2, we have that e1 is in F2 and it is linearly independat sequence of vectors, so it can be suplemented to the base of F2 with one vector, say e2 because dimension of F2 is 2 (explanation: there is theorem which claims that every sequence of linearly independant vectors in one linear space is base or it can be extended to the base of that linear space). Then, from the facy that F2 is included in F2, we get that e1, e2 are in F3 and they form linearly independant sequence of vectors, so there exsists vector e3 such that e1, e2, e3 is base for F3,...

Then, we repeat all this with flag F' and get vectors e1', e2', e3',..., en'. Permutation p is obvious...

Now, if this doesn't work, I'll definitely wait the end of my exams  :)

[quote="Maja"]...

Then, we repeat all this with flag F' and get vectors e1', e2', e3',..., en'. Permutation p is obvious...
[/quote]

Hi maja,

I'm sorry but i don't understand why e1', e2', e3',..., en' is a permutation of e1,e2,..,en ?? To have this statement you have to choose very carefully the vectors e1,e2,..,en. In your proof they are randomly chosen, so it can't work ..

bye.

Hi alekk,

well, again I've made a mistake. I'm not sure whether it can be removed or not (I guess not), so I'll quit for now and will try to solve this problem later.

Bye.

Hi!
I guess I just can not stop trying to solve this.  :)

Anyway, the point is that Fn=Fn'=E (because Fn is included in E and dim(Fn)=dim(E) so we have that Fn=E; and same for Fn').
Then, F1' is include in E, so those vectors e1, e2,...en, described in previous letters, are generators for F1', so some of them are base for F1' (explanation: there is a theorem2 which says that every sequence of generators of linear space V includes a subsequence which is base for V) ; and because dim(F1')=1 we know that only one of them is base for F1'. And denote it with e1'.
Now, we continue with the procedure of getting e2', e3',...,en' and from the proof of theorem1 (every sequence of linearly independant vectors in one linear space is base or it can be extended to the base of that linear space) we can see that {e1, e2,..., en}={e1', e2',..., en'} so the permutation p is defined with e1->e1',...,en->en'

I hope that we are closer to the solution now.

Maja

So, what was your solution Alekk?

Pierre.

Take the columns of $ A,B$ to be vectors in some bases of the flags $ F, F^\prime$. Then any other bases should be of the form $ AU, BV$ with $ U,V$ invertible upper-triangular matrices. Thus, we have to prove that we can find two such matrices $ U,V$ such that $ AU \equal{} BVP$, where $ P$ is a permutation matrix. Let $ C \equal{} B^{ \minus{} 1} A$, so that $ V^{ \minus{} 1} C U \equal{} P$. Using modified Gaussian elimination, we can transform any invertible matrix $ C$ into a permutation matrix $ P$, via left- and right-multiplying with upper-triangular elementary matrices, and this implies the claim of the problem.

why upper triangular elementary matrix ,what is known is just elementary matrix


algebra
polynomial
Vieta
number theory unsolved
number theory

Find all positive integers pairs $ (a,b)$ such that $ \frac {(ab\plus{}1)}{(ab\minus{}a^2\plus{}b\plus{}1)}$ is an integer.

notice that
$ ab\minus{}a^2\plus{}b\plus{}1\equal{}(a\plus{}1)(b\minus{}a\plus{}1)$  :wink:

Further more we can let $ b\minus{}1\equal{}k(a\plus{}1)$, but i dun know how to prove there is juz a solution $ (a,a^2)$ when $ k>1$

you can use x = a+1 and y = b-a+1 -> a=x-1 and b = x+y-2....it's easy to solve!

can bel.jad5 explain yr solution and can anyone provide a full solution?

when i repalce a and b by x-1 and x+y-2. I found that xy/(x²+xy-3x-y+3)
so x/(y-3)
now we can just put y = tx+3 to find all solutions.
I found 2 solutions: (a,a²) and (3a-1,3a+1)
is there another solution without using the factorization ab-a²+b+1 = (a+1)(b-a+1) ?( for exemple by vieta-jumping or another method!)


function
trigonometry
calculus
derivative
analytic geometry
graphing lines
slope

How do I find the all maximums, minimums and inflection points of $ y \equal{} x \plus{} \sin x$ on the interval $ \minus{} \frac {\pi}{2}\le x\le\frac {3\pi}{2}$?

[hide="My progress so far"]
$ y \equal{} x \plus{} \sin x$
$ y^{\prime} \equal{} 1 \plus{} \cos x$
$ 0 \equal{} 1 \plus{} \cos x$
$ \minus{} 1 \equal{} \cos x$
$ x \equal{} \pi$
[/hide]

[quote="Kent Merryfield"]That's a critical point, but you haven't yet characterized it, as a max, min, or something else.

Is $ f'(x)$ positive or negative on $ \left( \minus{} \frac {\pi}2,\pi\right)\,?$

Is $ f'(x)$ positive or negative on $ \left(\pi,\frac {3\pi}2\right)\,?$

And for this bounded closed interval, don't forget that you must also consider endpoints when thinking about max and min.

Also: the word "inflection point" requires you to look at concavity, and that is second derivative information. So go ahead and take the second derivative, and find where that is positive, negative, or zero.[/quote]

The second derivative is $ \minus{} \sin x$. And when I plug in $ \pi$, i get 0, meaning that that point could be anything. Is that right?

The original function is increasing in $ [ \minus{} \frac {\pi}2\,0)$ and $ (\pi\,\frac {3\pi}2]$.
It is decreasing in $ (0\,\pi)$.
After plugging in the right and left endpoints, they happen to be the global maximum and minimum  repectively.

As for the concavity and inflection points part, according to the increasing/decreasing part, there is an inflection point at x=0 and $ \pi$.

I'm not sure how you would find the local maximum and minimums, if there are even any.
Is that right?

hello, the point $ x=\pi$ which fulfilled the condition
$ f^'(\pi)=f^{''}(\pi)=0$ and $ f^{'''}(\pi) \neq 0$ is a so called horizontal inflexion point.
Sonnhard.

[quote="Kent Merryfield"][quote="iamhe"]It is decreasing in $ (0\,\pi)$.[/quote]
Why did you say this?[/quote]

Because the derivative was negative in that interval

It's not.

You seem to have some fundamental problems with this. You consistently confuse first derivatives with second derivatives, and you don't seem to know what an inflection point is.

The first derivative is the slope of the tangent line. First derivative information gives you the intervals of increase and decrease. A point at which the first derivative is zero is a critical point - a place with a horizontal tangent that might (or might not) be a maximum or a minimum. (Endpoints have the same status as critical points.) It is far, far more important for curve sketching that you get the first derivative information correct that anything about the second derivative.

The second derivative tells you about concavity - which way it's curving. An inflection point is a point at which the sense of concavity changes. A point at which the second derivative is zero is a possible inflection point, but you don't know for sure until you determine the concavity on each side of that point.

$ f'(x)\equal{}1\plus{}\cos x,$ and as you established in your very first post, the only point at which this is zero is $ x\equal{}\pi.$ Nothing happens at $ x\equal{}0,$ and you should never have thought of $ x\equal{}0$ as having anything to do with increasing and decreasing. The only intervals to consider are the ones from $ \minus{}\frac{\pi}2$ to $ \pi$ and from $ \pi$ to $ \frac{3\pi}2.$ But $ f'(x)\equal{}1\plus{}\cos x$ is positive on both of those intervals. Hence $ f(x)$ is increasing on its entire domain - which makes the left endpoint the minimum, the right endpoint the maximum, and the point at $ x\equal{}\pi$ a point at which there is a horizontal tangent but which is neither a maximum nor a minimum.

The only context in which $ x\equal{}0$ appears is as one of the two places at which $ f''(x)\equal{}0.$ Hence we only consider it in thinking about convexity. The second derivative is positive for $ \minus{}\frac{\pi}2<x<0,$ negative for $ 0<x<\pi,$ and positive again for $ \pi<x<\frac{3\pi}2.$ Hence both $ 0$ and $ \pi$ are inflection points, and the concavity is upwards on the outer two intervals and downwards in the middle. Note that a point with a horizontal tangent that isn't a max or a min must turn out to be an inflection point - for a smooth function, that's the only way that can happen.


quadratics algebra polynomial function domain search

find a general formula for the sequence.
$ 2,3,5,8,12,17,\cdots$

Does: $ a_{n} \equal{} a_{n\minus{}1} \plus{} n$ where $ n \geq 1$ and $ a_{0} \equal{} 2$ work?

Hypothesis: it's a quadratic sequence, so the general term takes the form of $ ax^2 \plus{} bx \plus{} c$
Plugging the first three values in, it turns out to be:
$ a_n \equal{} \frac {1}{2}n^2 \minus{} \frac {1}{2}n \plus{} 2$

By the way, I have a question related to this:
For $ n^{th}$ degeree polynomial generated sequence $ f(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_0$
Can we prove that (which is my hypothesis) the "$ n$th difference" of them is constant, and for a special case, $ f(x)\equal{}x^n$, its value is $ n!$?
"nth difference": e.g. $ 2^n \minus{} 1^n$ is 1st difference, $ (3^n \minus{} 2^n) \minus{} (2^n \minus{} 1^n)$ is second difference. (I don't know what the exact term is, so sorry for this rough description)
This is also posted in my previous topic, but my level is not high enough to understand their explanations...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170097

[quote="i_like_pie"]$ a_n \equal{} 2a_{n \minus{} 1} \minus{} 2^{n \minus{} 3}$[/quote]

That's wrong:

$ a_{6} \equal{} 17$
$ a_{5} \equal{} 12$
$ 2 a_{5} \minus{} 2^{3} \equal{} 24 \minus{} 8 \equal{} 16$.

And:

[quote="i_like_pie"]Yes, that is the [much more] simple way of expressing my answer.[/quote]

Those two are not equivalent.

[quote="timwu"]For $ n^{th}$ degeree polynomial generated sequence $ f(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_0$
Can we prove that (which is my hypothesis) the "$ n$th difference" of them is constant[/quote]

Yes.  Generally, if $ f(x)$ is a polynomial of degree $ n$ with leading coefficient $ a_n$, you can prove that

$ \Delta f(x) \equal{} f(x \plus{} 1) \minus{} f(x)$

is a polynomial of degree $ n \minus{} 1$ with leading coefficient $ na_n$.  This is known as the [i]difference operator[/i], and is a discrete analogue of differentiation.  The "$ n^{th}$ difference" to which you refer is $ \Delta^n f(x)$, where

$ \Delta^{n \plus{} 1} f(x) \equal{} \Delta ( \Delta^n f(x) )$

This is a polynomial of degree $ 0$ (hence a constant) with leading coefficient $ n! a_n$.

thank you very much every one, but i dont get this part
[quote]Just sum that all up to get $ a_n \equal{} 2 \plus{} {n(n \minus{} 1)\over 2} \equal{} {n^2 \minus{} n \plus{} 4\over 2}$[/quote]

sum up what ? sorry can you explain in abit more detail (abit more specific) please of what u mean , i am not as clever as you.i dont get things stright away i dont know why.

Sum up the equations. On the LHS, by telescoping, you get just $ a_n$, and on the RHS you get $ 2 \plus{} (1 \plus{} 2 \plus{} \cdots \plus{} (n \minus{} 1))$, and the sum of consecutive naturals is a well-known result.

[quote="archimedes1"][quote="kunny"]
By the way, what do you call the sequence $ a_{n \plus{} 1} \minus{} a_n$ in English?[/quote]
I think it is called the sequence if "first differences", but I'm not sure.  If the first differences are constant, then it is a linear function.  If the second differences are constant (ie, the first differences of the first difference sequence), then it is a quadratic function, etc.[/quote]

Thank you for your reply, may be it is called ''the progression of differences''. The equation I have showed in the last my post is called the formula of  ''the progression of differences'' in Japan. Isn't this a formula in your country?

[quote="kunny"]$ \{a_n\}: 2,3,5,8,12,17,\cdots$

$ \{a_{n \plus{} 1} \minus{} a_n\}: 1,\ 2,\ 3,\ 4,\ 5,\ \cdots$, yielding $ a_{n \plus{} 1} \minus{} a_n \equal{} n\ (n\geq 1)$.

$ \therefore \boxed{a_n \equal{} a_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}\ (a_{k \plus{} 1} \minus{} a_k)\ (n\geq 2)}$, which is same as Farenhajt's result.

$ \equal{} 2 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} k$

$ \equal{} 2 \plus{} \frac {(n \minus{} 1)n}{2}$

$ \equal{} \frac {1}{2}(n^2 \minus{} n \plus{} 4)\ (n\geq 1)$.

By the way, what do you call the sequence $ a_{n \plus{} 1} \minus{} a_n$ in English?[/quote]
thnx Farenhajt
Kunny really nice explanation, thankx  i get it now.
Is this formula true for any sequence no matter arithemtic or geometric? for different number of differences the general formula stays the same ?

If by "this formula" you mean

$ a_n \equal{} a_1 \plus{} \sum_{k\equal{}1}^{n\minus{}1} (a_{k\plus{}1} \minus{} a_k)$

then yes.  This is just a rearrangement of terms.  The problem with general sequences is that if you have a weird enough one, $ a_{k\plus{}1} \minus{} a_k$ won't be any simpler than $ a_k$.  This "formula" is only really useful for polynomial sequences, as I have already explained.

That "formula" is not an "universal formula" (and no formula is), it's just specifically used for polynomial function, which is already mentioned by other people who posted before.
For the first one, I don't see how it's generated, so I've no idea, either.
For the second one, it's obviously a geometric sequence.

[quote="binomial_4eva"]i tried using that formula for these 2, but i couldnt get one that works, may be i have gone wrong somewhere
[color=green][size=150]1[/size][/color]
$ 1,5,10,15,31,51,\cdots$

[color=blue][size=150]2[/size][/color]
$ 1,2,4,8,16,32,\cdots$

can someone help please.[/quote]

I don't get #1... the differences are 4, 5, 5, 16, 20? I am confused.

But #2 is just $ a_n \equal{} 2^{n\minus{}1}$.

[quote="binomial_4eva"]the first one is a polynomial one isnt it ? a cubic .
by the way isnt this true that we can turn any formula to a polynomial ? i think we can. if we can then that formula should work for all of sequences because (''it's just specifically used for polynomial function'').[/quote]

First, we can't. An example is your second question, which has a general term $ f(x) \equal{} 2^{x \minus{} 1}$: it can't be turn into a polynomial function.

Secondly, if we assume it's a polynomial function, we actually can do that, since for any $ n$ points, we can always find a $ n \minus{} 1^{th}$ degree polynomial to fit them. Howeve it will not be fixed, we can't claim it's the general term; that can be safely stated only when you got lower than $ n \minus{} 1^{th}$ degeree.

For example, if we continue on the difference operating refered by @t0rajir0u, you'll actually get a 5th difference, but it's just [b]one[/b] difference. What if we add one more? The general term still exists, but changes, which implies it's not the exact formula.

What? This is all horrible. I could have a sequence that starts like this:

$ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024$, and ask you what the next number is. You'll (most likely) say $ 2048$. I say you're wrong. My sequence was ($ a_0 = 1$)

$ a_n = \begin{cases}{2^n\text{ if } n\leq 10\\0\text{ otherwise}\end{cases}}$.

Nothing stops me from doing such a thing. This is a perfectly valid sequence.

Also, we can always find a polynomial of [b]at most[/b] degree $ n-1$ for any $ n$ points. I challenge you to find a $ 2$nd degree polynomial such that $ f(0)=f(1)=f(2)=0$. $ f(x) = 0$ is not a $ 2$nd degree polynomial ;). The difference method works for finding these polynomials.

Really, if you give me the first $ n$ terms of the sequence, I know absolutely nothing about the rest of it (see my example above). Any next point can be fit to a function easily - it can even be polynomial!

There are plenty of polynomials that start out $ 1,2,4,8,16,32$. For all we know, one of those could have been the actual sequence. Who knows?

For a more natural example of the problem Hamster is pointing to, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151093]this thread[/url].

@Hamster1800: Sorry, I make a mistake at that part.

Correction:

For any $ n$ points, if we assume it's a polynomial of degree at most $ n \minus{} 1$, we'll find one and only one solution within this restriction. However, if we extend the domain of degree to any positive integer, there will be infinite such functions for each degree. (because the system of coefficients becomes dependent)

For example, given sequence $ \minus{} 1,0,1,...$

With the assumption that it's at most a quadratic sequence, we can easily confirm it's $ f(x) \equal{} x \minus{} 2$.

But then try this one $ f_1(x) \equal{} (x \minus{} 2)^3$, and $ f_2(x) \equal{} (x \minus{} 2)^5$,...

$ f(x)$ is not the only polynomial be able to fit this three points, neither is any other polynomial for its corresponding points.

A similar example:  http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1938136038&t=166812

The lesson to draw from this discussion is that, as far as analyzing an actual infinite sequence of numbers derived from a problem, initial patterns can break.  Another great example:  http://www.research.att.com/~njas/sequences/A066178 starts with several powers of two and then stops.  In a sequence derived from a problem, the smart thing to do is to look at the problem (although the numbers themselves may be useful).

On the other hand, given a finite number of terms of a sequence (not derived from a problem), many experienced problem-solvers on this site feel that there is a point (which this discussion has not yet reached) after which such problems become arbitrary because the rules that can generate them are extremely diverse and can come from literally any field of mathematics.  Consider, for example, the sequence

$ 1, \infty, 5, 6, 3, 3, ...$

The "rule" that generates this sequence is absurd to rediscover.

[hide="It is this."] The $ n^{th}$ term is the number of $ n$-dimensional "regular" polytopes.  For example, there are an infinite number of regular polygons and five platonic solids.  In dimensions higher than $ 4$, the only regular polytopes are the hypercube, the simplex, and the dual (edited: thanks, Hamster1800!) hypercube. [/hide]

[quote="t0rajir0u"]
$ 1, \infty, 5, 6, 3, 3, ...$

The "rule" that generates this sequence is absurd to rediscover.

[hide="It is this."] The $ n^{th}$ term is the number of $ n$-dimensional "regular" polytopes.  For example, there are an infinite number of regular polygons and five platonic solids.  In dimensions higher than $ 4$, the only regular polytopes are the hypercube, the simplex, and the dual simplex. [/hide][/quote]

[hide="correction"]
The polytopes are actually the hypercube, simplex, and dual hypercube. The dual of a simplex is a simplex.
[/hide]


Suggesting to move the post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27335]Geometry problem[/url] by fisnik in Olympiad section, Geometry, Unsolved problems, to High school, Intermediate topics.

yetti

Done. Next time, please PM one of the moderators of the section to do it :) Thanks.


geometry
circumcircle
incenter
geometric transformation
homothety
Euler
geometry proposed

Let $ABC$ be a triangle. $P$, $Q$ and $R$ are the points of contact of the incircle with sides $AB$, $BC$ and $CA$, respectively. Let $L$, $M$ and $N$ be the feet of the altitudes of the triangle $PQR$ from $R$, $P$ and $Q$, respectively.

a) Show that the lines $AN$, $BL$ and $CM$ meet at a point.

b) Prove that this points belongs to the line joining the orthocenter and the circumcenter of triangle $PQR$.

[i]Aarón Ramírez, El Salvador[/i]

I noticed that this problem appeared in the Bulgarian Mathematics Olympiad (2002) as well.

Are you serious??? :(  Well, this problem was supposed to be original from Aarón (he is my instructor :)) but anyway...

arent BL,CM AND BN  thE ALTITUDES OF ABC?

[quote="Jutaro"]Let $ABC$ be a triangle. $P$, $Q$ and $R$ are the points of contact of the incircle with sides $AB$, $BC$ and $CA$, respectively. Let $L$, $M$ and $N$ be the feet of the altitudes of the triangle $PQR$ from $R$, $P$ and $Q$, respectively.

a) Show that the lines $AN$, $BL$ and $CM$ meet at a point.

b) Prove that this points belongs to the line joining the orthocenter and the circumcenter of triangle $PQR$.

This problem is due to Aarón Ramírez (El Salvador)  ;)[/quote]

The problem is hyper-well-known; it was given to us at the German IMO training seminar in Oberwolfach 2004, just to mention one of its occurences. Of course, it's possible that Aarón Ramírez independently discovered this fact, as one often discovers old things when doing geometry.

In http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 post #8, I proved that the triangles NLM and ABC are homothetic; this immediately yields the concurrence of the lines AN, BL and CM, so that part [b]a)[/b] is proven. I denoted the center of homothety - i. e., the point of concurrence of the lines AN, BL and CM - by S, and then showed that this point S lies on the Euler line of triangle PQR, i. e. on the line joining the orthocenter and the circumcenter of triangle PQR. So we have also proven part [b]b)[/b].

@ Marko: No, the lines AN, BL and CM are not the altitudes of triangle ABC (else, the orthocenter of triangle ABC would lie on the Euler line of triangle PQR, what is wrong ;) ). The lines AN, BL and CM concur at the so-called isogonal Mitten point of triangle ABC; this is $X_{57}$ in Kimberling's ETC. If this says something to you.

  Darij

[quote="Arne"]Nice problem.

I will reformulate it as follows.

Let $A_1A_2A_3$ be a non-equilateral triangle with circumcentre $O$ and incentre $I$, and  let $T_1$, $T_2$, $T_3$ be the tangency points of the incircle of $A_1A_2A_3$ with $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively. Let $S_1S_2S_3$ be the orthic triangle of $T_1T_2T_3$ (that is, the triangle formed by the feet of the altitudes of $T_1T_2T_3$). Let $H$ be the orthocentre of triangle $T_1T_2T_3$.

a) Prove that the lines $A_1S_1$, $A_2S_2$ and $A_3S_3$ are concurrent.

b) Prove that $O$, $I$ and $H$ are collinear.

[b]Solution.[/b]

a) By an easy angle chase we deduce that $A_1A_2 \parallel S_1S_2$, $A_2A_3 \parallel S_2S_3$ and $A_3A_1 \parallel S_3S_1$. Hence the triangles $A_1A_2A_3$ and $S_1S_2S_3$ are homothetic. This means that the lines $A_1S_1$, $A_2S_2$ and $A_3S_3$ are concurrent at some point $P$ (which is the centre of the homothety mapping $A_1A_2A_3$ to $S_1S_2S_3$.)

b) Notice that $H$ is the incentre of $S_1S_2S_3$ (it is wel known that the orthocentre of a triangle is the incentre of the orthic triangle). The homothety with centre $P$ which maps $S_1S_2S_3$ to $A_1A_2A_3$ maps the incentre of $S_1S_2S_3$ to the incentre of $A_1A_2A_3$, so $H$ is mapped to $I$. Hence, $P$, $H$ and $I$ are collinear.
Now, since $I$ and $H$ are the circumcentre and orthocentre of $T_1T_2T_3$ respectively, the points $P$, $H$ and $I$ lie on the Euler line of $T_1T_2T_3$. It is well-known that the circumcentre $W$ of $S_1S_2S_3$, which is the centre of the nine-point circle of $S_1S_2S_3$, lies on this Euler line as well (in fact, $W$ is the midpoint of $IH$). It follows that the points $P$, $H$, $I$ and $W$ are all collinear.
The homothety mentioned above, mapping $S_1S_2S_3$ to $A_1A_2A_3$, maps the circumcentre of $S_1S_2S_3$ to the circumcentre of $A_1A_2A_3$, so $W$ is mapped to $O$. Hence $P$, $W$ and $O$ are collinear.
Since $P$, $H$, $I$ and $W$ are collinear, we can conclude that $O$, $I$, $H$, $W$ and $P$ lie on the Euler line of $S_1S_2S_3$. Hence result.[/quote]

I don't understand, isn't the idea to prove that $P$, $H$ and $I$ are collinear ?

[quote="marko avila"]thanks darij its just that inmy figure it looks like it was true. anyway the centroamerican olympiad is a contest for people less than 15 years and though dariij´s solution is okay there has to be a more geometrical way because the arguments darij´s uses are most of the time not known bye students who participate in this competition. so anyone??[/quote]

It's basically the same, let $ H $ the orthocenter of the triangle  $PQR $. Let $ I $ the incenter of the triangle $ ABC $. Notice that $ I $ is the circumcenter of the triangle $ PQR $.
Part a)
Since $LMN$ is the triangle orthic of the triangle $PQR$, then $ \angle PLN = \angle PRQ $. Then, as $ I $ is the circumcenter of the triangle $PQR$, then $\angle PIQ = 2 \angle PRQ $, but $ IB $ is bisecting the angle $ \angle PIQ $, then $ \angle BIQ = \angle PRQ $. But see that $BQIP$ is cyclic ($ \angle BPI + \angle IQB = 90 +90 = 180 $) therefore $ \angle BPQ = \angle BIQ $. From the above, $ \angle BPL = \angle BPQ = \angle $ PLN and thus $ BA \parallel  LN$. Similarly one has $ NM \parallel AC $ and $ RQ \parallel  AB$.  Then $LMN$ is similar to the triangle $ ABC $.

Let $ X $ the intersection of the lines $ AN $ y $ CM $. We will show that $ A, L, X $ are collinear, and thus lines must be satisfied in one spot. To do this, see that being $ MN \parallel AC $, then  triangles $XAC$ and $XNM $ are similar. In turn, as the triangle $ LNM $ is similar to the triangle $ ABC $, then $ \displaystyle k = \frac{LN}{AB} = \frac{NM}{AC} = \frac{XN}{AX}$. In turn, $ \angle LNX = \angle BAX$ because $LN \parallel  AB$. This approach implies that  $\triangle LNX \ sim \triangle  BAX$. From the above, $ \ angle AXB = \angle NXL = \angle AXL $ where $ L $ is over $ BX $ and conclude that $ A, L, X $ are collinear.

Part b) It is well known that the orthocenter of a triangle is the incenter of its orthic triangle. Also that the circuncenter of the triangle $ PQR $ is the incenter of the triangle $ ABC $. Thus $ H $ is the incenter of the triangle $ LMN$. See triangles $ ABC $ and $ LMN $, for his likeness, $ H $ in $ \triangle  LMN$ would be $ I $ in $ \triangle ABC $, but $ X $ is the same in both. Of this, $ X, H, I $ are collinear and this follows the result.

$QWEXT$

From [url=https://artofproblemsolving.com/community/u378952h1773492p11651097]here[/url], we get that $\Delta LMN$ and $\Delta ABC$ are homological, hence, the conclusion follows from Desargues' Theorem

For the second part, Let the concurrency be $X$, then, $\Delta XLM \sim \Delta XBC$ and cyclically, hence, $X$ is the homothety center mapping $\implies$ $X$ maps the incircle of $\Delta LMN$ to $\Delta ABC$ $\implies $ hence done!


MATHCOUNTS
greatest common divisor
geometry
search
number theory
relatively prime
prime numbers

I am still trying to derive it myself... I got to a point...and then I died.  :P By the by...that formula only works for relatively prime numbers  :lol:

It does only work for relatively prime integers

snicker snicker mysmartmouth... Administrator!  :?  ;)  :P

The CMT is very useful I would agree!

Either he very recently became an admin, or he just hasn't been in the forums too much. (Sorry if you aren't a he)

[quote]
The proof to that formula is so long... fell asleep while it was being told to me...[/quote]

Really?  It shouldn't be.

Suppose m,n have no common factor > 1.

Step 1: Write the numbers 0,1,2,... in m columns.  For example, if m=5
0 1 2 3 4
5 6 7 8 9
....

Step 2: Circle the first m multiples of n (the numbers 0,n,2n,...,(m-1)n) in this chart.  Because m,n have no common factor > 1, there will one circled number in each column.

Step 3: A number is representable if and only if it lies on or below the circled number in its column.

Step 4: The largest nonrepresentable number is therefore the number directly above the largest circled number.  The largest circled number is (m-1)n; the number directly above it is (m-1)n -m = mn-n-m.

Yakko

I guess I was listening to a different proof.

Who found this formula? Why would he look for a formula about buying the most chicken McNuggets?

Its just crazy name stuff.
There's probably a better name for it (and a worse one, too)
And I thought commercialism was for sports! (Now I've seen everything)
Wasn't there another thread about this? (I think it was the "what we really need to know" thread)

wil somebody please tell me why its called the CHICKEN MCNUGGET THEOREM?!?!?!!?!?! what does it have ANYTHING to do at all about chicken mcnuggets?!?!?! TELL ME!!!!!!!!!! AHHHHHHHHHHH!!!!!! :mad:  :what?:

[quote="mathlover"]Will somebody please tell me why its called the Chicken McNugget Theorem?[/quote]
At one time, McDonald's sold Chicken McNuggets in 6, 9, and 20-packs.  Read more on [url=http://mathworld.wolfram.com/McNuggetNumber.html]mathworld[/url]

This is one of the few formulas that occasionally pops up in MATHCOUNTS that I strongly believe shouldn't ever be in MATHCOUNTS, primarily because the proof of the the theorem is well beyond MATHCOUNTS level.  A student can't be expected to reason his/her way to the result in any sensible amount of time without knowing this formula well in advance, and probably can't be expected to be able to derive the formula when still in middle school (though there are very rare exceptions to this).  Therefore, it shouldn't be included (and only rarely is - and even then, the numbers are usually at least small enough that trial and error is feasible.)

It's not that hard if you do it the common sense way o_o .

I forgot exactly how it went, but Ian and I derived it to a fellow New Jersey team member on the train while playing Spades, so it couldnt've been too hard...

It went something like this...

If the numbers are relatively prime, you can get ab by using either b as or a bs. So, you can get ab - a, which factors to a(b - 1) OR ab - b, which factors to b(a - 1). You can't get ab - a - b because that can be simplified to a(b - 1) - b, or b(a - 1) - a. Since b and a are relatively prime, you can't get an = b or bn = a (n is an integer), so ab - a - b is the highest number that can't be reached.

thats what I thought, but I couldn't figure out whetherWe could make proof  that it was the GREATEST

[quote="mysmartmouth"]The Chicken McNugget formula states that the largest number that can't be obtain by the sum of two numbers m and n, is [i][b]mn-m-n[/b][/i].  This is occasionally useful, but you can do without it.  Nice to have it in the Sprint, as I've seen Sprint Round questions such as "In a game called Snoof, a basket is worth 7 points, and a goal is worth 3.  What is the largest score that cannot be obtained  in a game of Snoof?"  It would obviously be 11 using this formula, but otherwise, it takes about 30 seconds longer.  [i]What is really interesting, anyway is that this was named because McNuggets came in certain sizes, and someone wanted to know what was the most McNuggets he couldn't buy.[/i]  I saw this in another area of the Forum, but I thought it was more appropriate for MathCounts, so I brought it here[/quote]

Wait, isn't the formula [i][b]mn-(m+n)[/b][/i]? In the game Snoof, if the basket was worth 7 points and the goal was worth 3 points, for the score to be 11, it would have to be [b][i]21-10[/b][/i], not [b][i]21-4[/b][/i] or [b][i]21-(-3)[/b][/i].  If anyone made a post about the change, I'm sorry I didn't catch it but I tried to see if anyone had.

mn-m-n = mn+(-m-n) = mn-(m+n)

It is just easiest to express as mn-m-n. ;)

Ok, I forgot to do order of operations.  Now I understand.  Thanks for helping me.  :D

Boy this is useful and would have saved 35 min. last Mathcounts meeting.

treething is your proof actually valid?  you proved ab-a-b is not obtainable, but you also jumped in steps of a and b down from ab; how do we know that while you were dropping from ab you didn't skip anything greater than ab-a-b that was also not obtainable?  im not 100% convinced  :P

Is there a formula for the number of impossible scores/numbers of Chicken McNuggets? If so, what is it? And how is it derived? This shows up often in Sprint too, and I need help.

I don't know that there is a formula, but you vould use the Chicken McNugget Formula and then just count the number of scores that don't work.  This won't be hard to do, because MathCounts doesn't expect you to know the Chicken McNugget Formula, and therefore keeps the numbers small.

[quote="Icy"]A relatively prime number is just a number that a layman (like me) would call prime, right? What can you do if the numbers aren't prime?[/quote]
to be relatively prime is  when the only common factor(gcd) between two integers is 1

for your second question i am not sure. depending on the numbers i think i'd brute force. someone else must have a better idea...

EDIT: yeah can't go from one number to two... :oops:

There is no such thing as "A" relatively prime number.

TWO numberS are relatively prime if their GCF = 1

Being relatively prime requires two or more numbers.  And they certainly don't have to be prime. 64 and 81 are relatively prime for example.

[quote="Icy"]A relatively prime number is just a number that a layman (like me) would call prime, right? What can you do if the numbers aren't prime?[/quote]

If the two numbers aren't relatively prime then there are an infinite number of integers that aren't attainable. Say the GCD is X then M can be written as AX and N can be written as BX. adding together gets a number of AX's (lets say Y) and a number of BX's (lets say Z). the sum being

YAX+ZBX distribute and you get
X(YA+ZB) therefore all sums are a multiple of X

@Treething Your proof only shows MN-M-N is unattainable, it makes no effort to show that it's the largest unattainable one.

Huh?? What does reviving topics have to do with competing in Mathcounts?

I don't know if this has been answered before, but is there a formula for three relatively prime numbers? How about three numbers where two of them have a common factor d?

[quote="zserf"]Considering people are only allowed to compete in mathcounts for three years maximum I think that this is acceptable.[/quote]

Dude...with the exception of the contest threads from the contests forum, never ever ever ever revive old threads, it's been unacceptable on this forum since its conception (to the best of my knowledge, at least)

[quote="Klebian"]It's funny. People often complain that people revive old topics, but if that user creates a new topic asking a similar question, people complain that he should search the forum and bump that topic.

Kind of a catch-22 situation here.[/quote]

uh, no, people only complain that they're asking the same question. as far as i know at least...


geometry
videos
college
combinatorics proposed
combinatorics

I don't know if this appeared in a forum before, but I think it is a well known.
Prove that any family of $n$ closed convex sets in the plane has at least $n^{1/5}$ members that are either pairwise disjoint or pairwise intersecting.

[i]Edited by Myth[/i]

I found this paper in the internet which deals with this problem:
http://www.math.nyu.edu/~pach/publications/larmanramsey.pdf

The problem was harder than what I thought but the idea seems very beautiful.

[quote="manuel"]

The problem was harder than what I thought[/quote]

But, where did you find it?

Pierre.

I found it in a book as an exercise but without the solution...

May I ask you which book?
With this kind of exercise, it seems to be very interesting.

Anyway, thanks for the links too.

Pierre.

Pierre,
the book is a really great and interesting book. I'm sure you will like to take a look at it.

Combinatorial Geometry, by Janos Pach.

Ok, thanks!

So, if I'm not wrong J.Pach gives as exercises the problems he solved in his research... ;)  I understand why they are so difficult...

Pierre.

yes .. janos is a great mathematician and a great teacher as well.. you can find some lectures in video Real One format he gave in MSRI at Berkeley in the MSRI site..


Since $ AC \equal{} BC$, we have $ m \angle BAC \equal{} m \angle ABC \equal{} 40^{\circ}$. Therefore,

$ x \equal{} 180 \minus{} \angle ABC \equal{} 180 \minus{} \angle BAC \equal{} 180 \minus{} 40 \equal{} \boxed{140^{\circ}}$


Ok... My self-confidence has gone down after I found that I couldn't decipher the vast majority of Wolfram articles. Either because I'm reading too quickly or my math skills are not up to par.

I'm only in the equivalent of AB Calculus right now (anything before that doesn't seem to be covered. I really haven't investigated much formal mathematical reading beyond my textbook since I am still struglling with the hardest problems on my textbook). Is Mathworld designed for college and graduate students? After all, the site was designed as the author took notes on his university-level courses (and these only started during his high school). SO should I just forget about mathworld for now?

Hmm... Ive never really heard anyone say that.
I really only use mathworld as a huge database, something that I can use as a rather advanced math encyclopedia. I dont think many people use it as a place to just go and casually read.

I am pretty comfortable with most of the math that people usually learn as undergraduates, and I still can't understand a lot of MathWorld properly.

I got really motivated to learn a lot more math back when my son was much younger, already noticeably interested in math, and I found out that I hit the wall in the SECOND PARAGRAPH of the Encyclopedia Britannica article "Arithmetic." I got farther in the article "Geometry," which is very good, but I realized then that I needed to learn more about the higher-math perspective on math. One book that has been very helpful for getting me ready for other reading about mathematics is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0486284247/learninfreed]Concepts of Modern Mathematics[/url] by Ian Stewart, available in an inexpensive Dover reprint.

Having read most of some Calculus I and Calculus II books...
Usually, a Mathword page is understandable for the first half of it.  Then it gets hard.

But if you look hard enough, you'll find some easier mathworld articles

http://mathworld.wolfram.com/Addition.html

 :D

Thanks! =) So I shouldn't feel too humiliated! Though definitely I could set a goal of understanding some of it once I'm done with high school (for me, it's also being done with multivariable analysis).


geometry
3D geometry
sphere
analytic geometry
graphing lines
slope
calculus

An hourglass is formed by rotating $y=e^{x}$ around $y=x$.  Find the radius of the largest sphere that can be passed through the neck of the hourglass.

[hide]
aren't $y=e^x$ and $y=x$ closest together when the slope of the tangent line at $x=c$ for $y=e^x$ equals 1?
so it would be
$\frac{dy}{dx}=e^x=1$
$x=0$
perpendicular line from $(0,1)$:
$y=x+1$
distance between $y=x+1$ and $y=x$ is $\sqrt{\frac{1}{2}^2+\frac{1}{2}^2}=\frac{\sqrt{2}}{2}=r$
[/hide]

edit
oops, calculation error... thanks amcavoy

[hide="Solution"]You are right about the slope being $1$.  The x-coordinate of the point on $y=x$ may be different though.  We already have the point $\left(0,1\right)$, so we have

\[ \frac{x-1}{x}=-1\implies x=\frac{1}{2}. \]

This means our points are $\left(0,1\right)$ and $\left(\frac{1}{2},\frac{1}{2}\right)$, so the distance (the radius) is $\sqrt{2\left(\frac{1}{4}\right)}=\boxed{\frac{\sqrt{2}}{2}}$.[/hide]


I'm getting hundreds of emails from this site for some reason. Everytime I post, I get like 3 emails...why? And is there anyway to stop this?

Do you have the "notify me when a topic I posted in is posted in again" thing checked? That's probably why.

Go to Options > Posting a Message.  Change the one that says something about email notification.  That should work.

Thanks, I did that...hopefully it won't do it again :?

That happened to me!  I had 1000 e-mails from AoPS!!!How do you say AoPS anyways.  Are you supposed spell it out or say AoPS.  Or do you say A-ops???

[quote="math92"]Or do you say A-ops???[/quote]

That one, to the extent of my knowledge.

Everyone i know who does it (most everyone) or has the book says A-O-P-S

i say A-O-P-S

But I also say AIM...not AMY

[quote="paladin8"]I say A-ops. It's quicker. I don't like saying all the letters. But most people I know spell it out.[/quote]
how do you pronounce it?

A-ops...I don't know how to explain the pronunciation any further  :?

I say "A-ops" also.

this is getting off topic, and will turn into SPAM if this discussion keeps going, i suggest mods or admins lock this topic...