[quote="Forte"]The radiator of a car is full and contains 10 quarts of a mixture that is 30%antifreeze. how many quarts of this mixture must be drained out and replaced with pure (100%) antifreeze in order to bring the mixture up to 50% antifreeze?[/quote]
[hide]
I EDITED into here the top two lines:

Let x = the amount that is drained out and also replaced

(30%)(10) - (30%)(x) + (100%)(x) = (50%)(10)

0.30(10) - 0.30(x) + 1(x) = 0.50(10)

3 - 0.3x + x = 5

0.7x = 2

x = 20/7

x = $ 2\frac67$

$ 2\frac67$ quarts
[/hide]


For any $ x,y,z>0$ and $ \sum x  \plus{} 2 \equal{} xyz$
Prove that $ xyz(x\minus{}1)(y\minus{}1)(z\minus{}1) \le 8$.

[hide]Hint:
Since  $ \sum x  \plus{} 2 \equal{} xyz$
We substitute $ x\equal{}\dfrac{b\plus{}c}{a}$ and analogously for $ y$ and $ z$ with the existence of some positive reals $ a,b,c$.[/hide]

Inequality is equivalent with

 \[ (a\plus{}b)(b\plus{}c)(c\plus{}a)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)\le 8a^2b^2c^2\]
or
\[ (a\plus{}b)(b\plus{}c)(c\plus{}a)(p\minus{}a)(p\minus{}b)(p\minus{}c)\le a^2b^2c^2\]

or

\[ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{p}\cdot S^2\le a^2b^2c^2\]
or
\[ (a\plus{}b)(b\plus{}c)(c\plus{}a)\le 16pR^2\]

 But, because $ p\le \frac{3R\sqrt{3}}{2}$(Mitrinovic) we have

  \[ (a\plus{}b)(b\plus{}c)(c\plus{}a)\le\left(\frac{2(a\plus{}b\plus{}c)}{3}\right)^3\le 16pR^2\]


Let $ a,b,c\geq 0$,no two of which are zero.Prove that:
$ \sqrt{\frac{a(b\plus{}c)}{a^2\plus{}bc}}\plus{}\sqrt{\frac{b(c\plus{}a)}{b^2\plus{}ca}}\plus{}\sqrt{\frac{c(a\plus{}b)}{c^2\plus{}ab}}\geq 2$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$,no two of which are zero.Prove that:
$ \sqrt {\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b(c \plus{} a)}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c(a \plus{} b)}{c^2 \plus{} ab}}\geq 2$.[/quote]
. If exist 1 number equal 0 . To suppose $ c \equal{} 0$ . We have :
Ineq $ \Leftrightarrow \sqrt {\frac {a}{b}} \plus{} \sqrt {\frac {b}{a}} \ge\ 2 \rightarrow$ right
 . If $ a;b;c \ne \ 0$
We have $ 2\sqrt {\frac {a^2 \plus{} bc}{a(b \plus{} c)}} \le\ \frac {a^2 \plus{} bc}{a(b \plus{} c)} \plus{} 1 \equal{} \frac {(a \plus{} b)(a \plus{} c)}{a(b \plus{} c)}$

$ \rightarrow \sqrt {\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \ge\ \frac {2a(b \plus{} c)}{(a \plus{} b)(a \plus{} c)} \equal{} \frac {2a(b \plus{} c)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

Similar $ \rightarrow \sqrt {\frac {a(b \plus{} c)}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {b(c \plus{} a)}{b^2 \plus{} ca}} \plus{} \sqrt {\frac {c(a \plus{} b)}{c^2 \plus{} ab}} \ge\ 2 \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} > 2$

 "=" occur $ \Leftrightarrow a \equal{} b;c \equal{} 0$


It takes Jesse 4 days longer to prepare an order of pies that it takes James. Together they baked pies for 2 days and then James left. Jesse took 7 additional days to complete the order of pies. Working alone, how long would it have taken James to prepare the pies?

Please help me :(

Kindly show the representation, working equation and the solution.

THANKS :D.

Over and out.

Let $ x$ denote the number of days it takes James to complete an order of pies.
We know that it takes Jesse $ x + 4$ days to complete an order of pies.
If James can complete $ 1$ order of pies in $ x$ days, than he can complete $ \frac {1}{x}$ orders of pies in $ 1$ day.
Jesse can complete $ \frac {1}{x + 4}$ orders of pies in $ 1$ day.
Together, they can complete $ \frac {1}{x} + \frac {1}{x + 4} = \frac {x + 4}{x(x + 4)} + \frac {x}{x(x + 4)} = \frac {2x + 4}{x(x + 4)}$ orders of pies in $ 1$ day.
They work together for $ 2$ days, and then Jesse works for an additional $ 7$ to complete $ 1$ order of pies.
We know that $ 2 \times \frac {2x + 4}{x(x + 4)} + 7 \times \frac {1}{x + 4} = 1$.
We can solve this equation. We first clear the denominators by multiplying throughout by $ x(x + 4)$.
\begin{align*} 2(2x + 4) + 7(x) & = x(x + 4) \end{align*}
Expanding, we have:
\begin{align*} 4x + 8 + 7x & = x^2 + 4x \end{align*}
Simplifying, we have:
\begin{align*} 11x + 8 & = x^2 + 4x \end{align*}
This looks like a quadratic equation. Rearranging the terms, we have:
\begin{align*} 0 & = x^2 - 7x - 8 \end{align*}
Factoring, we have:
\begin{align*} 0 & = (x - 8)(x + 1) \end{align*}
So $ x = \{8, - 1\}$. A negative time is meaningless for this problem, so we can conclude that $ x = 8$. Therefore, it takes James $ \boxed{8}$ hours to prepare an order of pies alone.


Hey everyone, I'm new here and I'm in math 436 and we just started learning radicals. I'm kind of confused. I know what to do with √a^2 I know what to do with √ab I know what to do with √a + √a but I don't understand how 18 becomes 3√2.

We got homework, and here's one question "3√8 / 2√18" as well as "√6c^6d^10" Can someone explain how I would go about doing those?

Thanks.

PS. "^" = to the power, "/" = Over or Divided.

This is NOT the right forum for your question, but I'm not a moderator so I can't move it.

Anyway, $18=3^{2}*2$, so $\sqrt{18}=\sqrt{3^{2}*2}=\sqrt{3^{2}}\sqrt{2}=3\sqrt{2}$

In general, prime factorize all numbers inside radicals, and bring squared primes outside the radical.

With variables, just do something like $\sqrt{a^{8}*b^{7}}=\sqrt{a^{8}}\sqrt{b^{6}}\sqrt{b}=a^{4}*b^{3}\sqrt{b}$

I'm not going to do your homework for you.


Prove that $ (3\plus{}\sqrt{5})^{2n}\plus{}(3\minus{}\sqrt{5})^{2n}$ is divisible by$ 2^{n\plus{}1}$.

Note: Actually the original problem is: Prove that the smallest integer greater than $ (3\plus{}\sqrt{5})^{2n}$ is divisible by $ 2^{n\plus{}1}$. But I'm pretty sure these are equivalent. I'm stuck after that, please help me.

Let $ r,r'\equal{}(3\pm\sqrt{5})^2\equal{}14\pm6\sqrt{5}$; then $ (x\minus{}r)(x\minus{}r')\equal{}x^2\minus{}28x\plus{}16$. Therefore $ x_{n\plus{}2}\equal{}28x_{n\plus{}1}\minus{}16x_n$ where $ x_n\equal{}(3\plus{}\sqrt{5})^{2n}\plus{}(3\minus{}\sqrt{5})^{2n}$. Furthermore $ x_0\equal{}2,x_1\equal{}28$. Since both coefficients are divisibile by four, we can prove by induction that it is divisible by $ 4^n$.

[quote="johan.gunardi"]Prove that $ (3 \plus{} \sqrt {5})^{2n} \plus{} (3 \minus{} \sqrt {5})^{2n}$ is divisible by$ 2^{n \plus{} 1}$.

Note: Actually the original problem is: Prove that the smallest integer greater than $ (3 \plus{} \sqrt {5})^{2n}$ is divisible by $ 2^{n \plus{} 1}$. But I'm pretty sure these are equivalent. I'm stuck after that, please help me.[/quote]

*****************************************************************
hi
we know that
3+v5=1+v5(2) and 3-v5=1-v5(2)
notice that (1+v5)/2 and (1-v5)/2 they are golden number that way we have to use "Maillet formule"

Ln __> N so //////        (1+v5)^n=-(1-v5)^n(2^n)
(1+v5)^2n=(1-v5)^2n (2^2n)
your turn to complete solution  (that will help you http://fr.wikipedia.org/wiki/Suite_de_Fibonacci)
ATT     = ==> modulo
*****************************************

[hide="Equivalently..."] Since $ (3 \pm \sqrt{5})^2 \equal{} 14 \pm 6 \sqrt{5}$, an equivalent problem is showing that

$ \frac{1}{2} \left( (7 \plus{} 3 \sqrt{5})^n \plus{} (7 \minus{} 3 \sqrt{5})^n \right)$

is always an integer, which we can prove in the same way as me@home (the recursion) or directly by binomial expansion.  :) [/hide]

[quote="t0rajir0u"][hide="Equivalently..."] Since $ (3 \pm \sqrt {5})^2 \equal{} 14 \pm 6 \sqrt {5}$, an equivalent problem is showing that

$ \frac {1}{2} \left( (7 \plus{} 3 \sqrt {5})^n \plus{} (7 \minus{} 3 \sqrt {5})^n \right)$

is always an integer, which we can prove in the same way as me@home (the recursion) or directly by binomial expansion.  :)
[/hide][/quote]
ACTUALLY USING Maillet formule is much easier "HELLO"

What is the Maillet formula?  It sounds like you're referring to [url=http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html]Binet's formula[/url], and the techniques me@home and I gave are the natural extensions of that.  In any case, the binomial expansion proof is straightforward, if tedious, and the recursion proof is very simple.

http://fr.wikipedia.org/wiki/Suite_de_Fibonacci  there is big different between  maillet and binet you can see in wikipedia ..................

Sorry, I don't speak French. Any English resources?

From a quick skim of the link, it looks like Maillet deals with the Lucas numbers while Binet deals with the Fibonacci numbers.  No clue about any mathematician named Maillet.  I'm guessing he was French.  And also

Ibn Rochde: You may want to use Latex in future posts.  That one was extremely hard to understand, especially the use of "v" to represent the square root symbol.

[quote="Ibn Rochde"]johan.gunardi =you can choose engilsh
ZzZzZzZzZzZz=idont know how to use latex can you help me ?[/quote]

You can write the equation [url=http://www.sitmo.com/latex/]here[/url]. In the forum, put two dollars in front and behind the equation.

[quote="ZzZzZzZzZzZz"][quote="Ibn Rochde"]
ZzZzZzZzZzZz=idont know how to use latex can you help me ?[/quote]

Click on the Latex link on the left sidebar, and download the necessary files for your computer.  After that, it's pretty much self explanatory, but you can always post in the Latex forum.[/quote]

You don't need to download anything to use LaTeX on the forums, just put the latex code between dollar signs ($ \$\text{ stuff }\$$).  Those files are only necessary for creating documents typeset in LaTeX.


20) In a two-person game, one person says an integer, and the other person must say an integer greater than or equal to 1/3 of the previous persons integer, but less than the previous persons integer. They go back and forth following this rule until one person says 1, and that person loses. If Johnny says 2004, what number should Bernie counter him with to guarantee a win for himself?
a) 1820 b) 1821 c) 1822 d) 1823 e) 1824

I was wondering, what is the general approach to this problem?

Thanks

Working backward
If Johnny said 1 at last, then Bernie must have said 2. Then Johnny said a number<6 but greater than 2, Bernie must have said a number k>6 but less than 18, and that k must be 2(3)+1=7. Assume k>7, then Johnny could have said 7, force Bernie to say a number>2, then Johnny says 2 and Bernie loses. Using the same logic we get that Bernie must have said 7(3)+1=22 before he said 7, 22(3)+1=67 before he said 22, 67(3)+1=202 before he said 67, 202(3)+1=607 before he said 202,  607(3)+1=1822 before he said 607. So our answer is 1822

Hmmm...

This problem seems much harder than a standard AMC question.  I remember last year, there was a problem like this on round 1 of the USAMTS, whose problems are considerably harder than AMC level.


Hi again! ;)

So, this time, you have a round table, and there are n people to place (of course, you have n chairs). All the chairs are the same. In how many ways can you arrange the people?
Now, there are n couples (=n men/women, so 2n people ;)). In how many ways can you arrange them if you want to alternate men and women?

Have fun!  :D

Yes, for the first one it's (n-1)! ;)

And now.. the 2nd one! :)

[quote="mathmanman"]Now, there are n couples (=n men/women, so 2n people ;)). In how many ways can you arrange them if you want to alternate men and women?[/quote]
I would say : $n \cdot {(n-1)!}^2$

And you would be right! :)
But, could you show me your method please?
Because, with mine, I find : $n!(n-1)!$, which is, of course, the same thing, but, I was wondering how you could directly get your formula. :)

There are $(n-1)!$ way to put the women, $(n-1)!$ way to put the men, and $n$ way to merge the result.

Ohh! How come I didn't think about it before?!

Thanks a lot ;)

mathmanman, you solved the problem correctly. :first:
tµtµ, you solved it wrongly. (As I think) :oops_sign:

First of all select n alternate seats on the round table. n men can be placed in $(n-1)!$ ways. Then the remaining n places can be filled in $n!$ ways by the women.
Hence the answer muist be $n!(n-1)!$.

So finally I am the winer :thumbup:

[quote="mathmanman"]:rotfl: Yes, you are!

But, why do you say tµtµ solved it wrongly?  :?[/quote]

Oh, I'm sorry [b]tµtµ[/b]. You are correct. I didn't read that you are merging them in n ways.
So, I take back my words.


calculus
derivative
calculus computations

I'm looking at Question #4 over [url=http://www.davidson.edu/math/chartier/CU/Chain/chainans.html#q4]here[/url] and don't understand the step from $ w_s \equal{} w_x \plus{} (3s^2\minus{}r)w_y$ to $ w_{sr}  \equal{} w_{xr} \minus{} w_{y} \plus{} (3s^2\minus{}r)w_{yr}.$  Could someone please walk me through term by term?

No chain rule yet, just the product rule.  When you partial $ w_y$ by $ r$ you get $ w_{yr}$, so when you partial $ (3s^2 \minus{} r)w_y$ by $ r$ you get by the product rule $ ( \minus{} 1) w_y \plus{} (3s^2 \minus{} r)w_{yr}$.


Could someone help me with the 'going backwards' thing with GCD. I'm afraid i don't understand it at all. The question is this:

Let d = gcd(10001, 11899). Find d, and find integers x and y such that
10001x + 11899y = d.

I know how to find the gcd, which is:

11899 = 1 x 10001 + 1898
10001 = 5 x 1898 + 511
1898 = 3 x 511 + 365
511 = 1 x 365 + 146
365 = 2 x 146 + 73
146 = 2 x 73 + 0
So d = 73.

Could someone guide me through (as i'm completely lost) how to 'go backwards' to find x and y please? Apologies if i'm posting this in the wrong area.

from the first equation you have 1898 = 11899 - 10001. now make that substitution into the second equation:

511 = 10001 - 5 x 1898 = 10001 - 5 x (11899 - 10001) = -5 x 11899 + 6 x 10001.

now substitute this expression (and the one you got previously for 1898) into the next equation to get 365 in terms of 11899 and 10001:

365 = 1898 - 3 x 511 = (11899 - 10001) - 3 x (-5 x 11899 + 6 x 10001) = 11899 x blah + 10001 x blah.

and so on, each time using previous expressions to substitute... if you keep doing this, each remainder will be written in terms of 10001 and 11899, until finally the remainder 73 will be written this way, which is what you want.

I'm still a little stumped with this, could someone please help me out.


Given: parallelogram WXYZ
prove in 2 column form that ---------- triangle ATB is SIMILAR to A1TB1

(show that AT/A1T and BT/B1T both equal to TW/TY)

please help me..NOW!
[img]http://www.geocities.com/hazelmilano23/untitled.bmp[/img]

[quote="nutz_for2.718281828"]Proof: If it couldn't be proved, then you wouldn't have posted it in this forum, and it can't be proved, because you haven't defined points A, T, B, A1, and B1.[/quote]
Hence the picture.
Why are you double posting your homework, once will suffice. I won't give you the answer, but here is a hint:
[hide]Try proving $AB \| A1B1$[/hide]

[quote="bpms"][quote="anirudh"][quote="learnmath2"]Given: parallelogram WXYZ
prove in 2 column form that ---------- triangle ATB is SIMILAR to A1TB1

(show that AT/A1T and BT/B1T both equal to TW/TY) [/quote]

please help me..NOW!

Its been a really long time since I've done 2 column proofs, so here we go:

But one thing, do you know if $AB\parallel A_{1}B_{1}$?
I'll assume yes, since I'm too lazy to prove otherwise

[img]http://www.geocities.com/hazelmilano23/untitled.bmp[/img]
$\\ \text{Figure}\qquad\qquad\qquad\qquad\qquad\because\text{Given}\\ \angle ATB=\angle A_{1}TB_{1}\qquad \text{(A)}\quad\because\text{Vertical Angles are congruent}\\ AB\parallel A_{1}B_{1}\qquad\qquad\qquad\qquad\because\text{Given}\\ \angle A=\angle A_{1}\qquad\qquad\qquad\qquad\because\text{Alternate Interior Angle Theorem}\\ \angle B=\angle B_{1}\qquad\qquad\qquad\qquad\because\text{Alternate Interior Angle Theorem}\\ \therefore\triangle ABT\sim\triangle TA_{1}B_{1}\qquad\qquad\quad\because\text{AAA Similarity}$[/quote]
You don't assume they are parallel. Notice that the parallelogram is there for a reason, shocking I know.[/quote]

I was giving the job of proving them to him, since he would have to put some effort :roll:


Find the sum of the positive values of $ x$, for which
\[ \sqrt[3]{2 \plus{} x} \plus{} \sqrt[3]{2 \minus{} x}
\] is an integer.

[hide="Solution"]

Let $ f(x) = \sqrt [3]{2 + x} + \sqrt [3]{2 - x}$

Clearly there are only finitely many integers which $ f(x)$ can be. Investigation shows that these are $ 1$ and $ 2$. (I used calculus to show it was a decreasing function on $ x > 0$ and then showed that $ f(0) = 2\sqrt [3](2) < 3$

Also note

\begin{align*} [f(x)]^3 & = (2 + x) + 3\times(2 + x)^{(2/3)}\times(2 - x)^{(1/3)} + 3\times(2 + x)^{(1/3)}\times(2 - x)^{(2/3)} + 2 - x \\
& = 4 + 3(2 + x)^{(1/3)}(2 - x)^{(1/3)}(f(x)) \end{align*}

Solving for $ f(x) = 2$

\begin{align*} 8 & = 4 + 3(4 - x^2)^{1/3}2 \\
\left (\frac {2}{3} \right )^3 & = (4 - x^2) \\
x & = \sqrt {4 - \left ( \frac {2}{3} \right )^{3}} \end{align*}

Solving for $ f(x) = 1$

\begin{align*} 1 & = 4 + 3 ( 4 - x^2 ) ^{1/3} 1 \\
- 1 & = (4 - x^2) \\
x & = \sqrt {5} \end{align*}

Therefore the sum of values is

${ \boxed {\sqrt {5} + \sqrt {4 - \left ( \frac {2}{3} \right )^{3}} } \approx 4.16057}$

[/hide]

Nice, SimonM.  :lol:

I think it can be shown that the function is decreasing without Calculus, though.

We can write the function as $ \sqrt[3]{x\plus{}2} \minus{} \sqrt[3]{x\minus{}2}$, from where we can use the fact that $ x^p$ is strictly concave if $ 0<p<1$

I meant to add that I'd shown it without calculus as well, but forgot. Anyway, for that calculus was much better and therefore should be an acceptable method

I agree the Calculus is better. But, eh, variety.   :)


The median of the list

n, n+3, n+4, n+5, n+6, n+8, n+10, n+12, n+15

is 10. What is the mean?

[hide]The median of the list is $n+6=10\Rightarrow n=4$.  That makes all the terms add to $9n+63=9\cdot4+63=99$.  There are 9 terms, so the mean is $\frac{99}{9}=\boxed{11}$.[/hide]

[hide]
You get that $n=4$
Plugging in 4 for n, you get that the sum of all of teh numbers is 99.
$\frac{99}{9}=\boxed{11}$[/hide]


Hi

A real example is the game called spot the ball.

I have an area with only positive values that represents x and y. An example is my x=200 and y=250.
Other people without knowing my x and y are trying to find out the closest values by choosing their x and y.
user 1 x=190 y=200
user 2 x=190 y=100
user 3 x=150 y=400
user 4 x=130 y=260
unlimited users...

Know I need to create a formula that will find out who has found the closest value to my x and y?

Thanks in advance.

You use the distance formula (a.k.a. the Pythagorean theorem) to compute the distance between any two points, then just compare all of the distances.

The distance between $ (a, b)$ and $ (c, d)$ is $ \sqrt{(a \minus{} c)^2 \plus{} (b \minus{} d)^2}$.  In fact, though, you might as well just compare the values $ (a \minus{} c)^2 \plus{} (b \minus{} d)^2$ (since taking the square root doesn't change the relative order of the values).


calculus
integration
function
calculus computations

if a planet of radius $R$ has an atmosphere whose density varies with respect to distance from the planet's surface and is proportional to $\frac{1}{(r+1)^{2}}$, what is the mass of the atmosphere in terms of the proportionality constant $k$ and the planet's radius $R$

In the three-dimensional world the answer is infinity: the corresponding integral diverges. Are you sure that you quoted the problem right?

actually I wrote this problem myself, it seemed interesting, I somewhat arbitrarily chose this function because i knew that 1/n^2 converges regularly, can u show me how this diverges and solve for the smallest exponent it converges?

The problem is that in $n$-dimensional space, the spherical shell contributes a volume element that is $dV=Cr^{n-1}\,dr$ for some constant $C.$ (Note: I know the exact values of that constant, I'm just trying not to distract from the argument.)

So, what would it take for $\int_{r>R}\frac1{r^{m}}\,dV$ to converge? Integrate with respect to $r$ and you get

$C\int_{R}^{\infty}\frac1{r^{m-n+1}}\,dr.$

For this to converge, we would need $m-n+1>1$ or $m>n.$

For example, in three-dimensional space if you use a density of $\frac1{(r+1)^{4}},$ then you would get convergence. Any exponent greater than $3$ would work.

That comparison with $\sum\frac1{n^{2}}$ that you were thinking of? That's a one-dimensional thought.

thanks for clearing that up Kent, now back to the original problem lets "pretend" that i originally had $\frac{1}{(r+1)^{4}}$ and solve the problem using this new data


A pair of positive integers $ (x, y)$ satisfies the equation $ 31x \plus{} 29 y \equal{} 1125$. What is $ x \plus{} y$ ?

First, what I did was I wrote the x's as $ 29x \plus{} 2x \plus{} 29y \equal{} 1125$, which I simplified to get $ 29x \plus{} 29y \equal{} 1125 \minus{} 2x$. I divided 1125 by 29 and the closest multiple of 29 to 1125 is 1131, which makes $ x \equal{} \minus{} 3$. Substituting this  into the original equation, I got $ y \equal{} 42$. Therefore, I got the sum $ x \plus{} y \equal{} 39$.

I tried this another way and I did $ 31x \plus{} 31y \minus{} 2y \equal{} 1125$, which is nothing but $ 31x \plus{} 31y \equal{} 1125 \plus{} 2y$. I tried finding the closest multiple of 31 to 1125 and I got 1147, which means that $ y \equal{} 11$. Substituting this into the original equation, I got $ x \equal{} 26$. Therefore, I got the sum $ x \plus{} y \equal{} 37$.

I know that the second solution is what is required (positive integers), but in general, if we didn't need to find only positive integers, would the first solution be correct?

If you didn't need just positive integers, then there would be an infinite number of solutions.

hello, the only positive integer solutions are $ x\equal{}26$ and $ y\equal{}11$ so we have $ x\plus{}y\equal{}37$.
Sonnhard.

First of all, Dr.Sonnhard, that was a rude comment.
He needed some help.

Here we go: The second solution is mathematically correct. There's nothing wrong about it.
The problem specifically said that the numbers are positive, or else there would be an infite number of pairs.
Since we need positive integers only, (x,y) can only be (26,11). Not (-3,42). There's no way that the two solutions are equal.

Is that what your question was?

Actually I believe that canton is asking that if the question did not ask for positive integers as answers and just asked for answers, then would his first solution be correct?

The answer to that would be yes, and as everyone else has said, there would be infinitely many solutions.

Take it mod 29.

You get 2x=23 mod 29 so 2x=52mod29 so x=26mod29

When x=26, y=11 so 37 and we win.


vector
calculus
derivative
linear algebra
matrix
quadratics

Trying to solve the problem in the attached jpeg.
As I understand a linear system must be made up and the unknowns are $ \alpha_i$s
The only problem is that when I try to "unfold" the term in 3rd.jpeg(attached image) terms like $ \alpha_i\alpha_j$ appear
and this is not a linear system any more.

This is a [i]least squares[/i] problem.

Here's the general structure: We have a set $ \{e_0,e_1,\dots,e_n\}$ of vectors in an inner product space. (Make this a real inner product space for now.) Assume this is an independent set, but (to our numerical detriment) it is not an orthogonal set. We have another element $ f$ in this inner product space, and we would like to approximate $ f$ as closely as possible by a linear combination of the given set. That is, we would like to find $ \alpha_i$ so as to minimize
\[ \left\|f \minus{} \sum_{i \equal{} 0}^n\alpha_ie_i\right\|^2 \equal{} \left\langle f \minus{} \sum_{i \equal{} 0}^n\alpha_ie_i,f \minus{} \sum_{j \equal{} 0}^n\alpha_je_j\right\rangle
\]
Call that quantity $ r(\alpha_0,\dots,\alpha_n).$ Now take the partial derivative of $ r$ with respect to $ \alpha_k.$ We get that this partial derivative is

$ \left\langle \minus{} e_k,f \minus{} \sum_{j \equal{} 0}^n\alpha_je_j\right\rangle \plus{} \left\langle f \minus{} \sum_{i \equal{} 0}^n\alpha_ie_i, \minus{} e_k\right\rangle$

$ \equal{} 2\left\langle f \minus{} \sum_{i \equal{} 0}^n\alpha_ie_i, \minus{} e_k\right\rangle$

$ \equal{} \minus{} 2\langle f,e_k\rangle \plus{} 2\sum_{i \equal{} 0}^n\alpha_i\langle e_i,e_k\rangle$

If we set this equal to zero, we get the equation

$ \sum_{i \equal{} 0}^n\alpha_i\langle e_i,e_k\rangle \equal{} \langle f,e_k\rangle$

Do this for each $ k$ and we have a matrix equation for the critical point of $ g.$ It's clear that if $ g$ only has one critical point, then it is a minimum.

Expressed in matrix form: Let $ H$ be the square matrix whose entries are $ \langle e_i,e_j\rangle.$ This is called the [i]gramian[/i] of this particular set. A gramian is always symmetric and positive semidefinite; if the underlying set is independent, then the gramian is positive definite and hence invertible. (If the underlying set is orthogonal, the gramian is diagonal; we don't have that here, rather badly.) Let $ a$ be the column vector of the $ \alpha_i$ that we want to solve for. Let $ b$ be the column vector of $ \langle f,e_i\rangle.$ Given that, the equation becomes $ Ha \equal{} b.$

This is the so-called [i]normal equations[/i] for the least squares problem. The problem with the normal equations approach can be numerical instability, and we have that here in a major way (although not that badly with the $ 4\times 4$ matrix). The gramian here - the Hilbert matrix - is notoriously [i]ill-conditioned[/i]. The cure for the numerical instability would be to replace the set $ \{1,x,x^2,\dots x^n\}$ with an [i]orthogonal[/i] set of the same span. (In this case, that would be a version of the Legendre polynomials.)

This theory could have been developed algebraically without using the partial derivatives; but the partial derivatives are a fairly simple approach.

In answer to your question: Yes, the expression $ r$ is quadratic, not linear. But the minimization of a quadratic expression does lead to a linear problem.

Thanks Kent Merryfield , I didn't know this was a least squares problem (altough I did read something very similar in a statistics course).
The ill-condition which you talk of , is related to angles between the lines/planes described by the linear equation(for 2 and 3 equation systems).
But how can we generalize this to n-dimensional systems and how can we transport this property there ?
It is known that ill-conditioned systems appear when the angle between the lines/planes described by the system is very small so that it causes very big changes in the exact solutions for very small changes in the coefficients of the system(this is the defintion of ill-conditioned systems).

P.S. after some pages of calculations I was a able to reach the same system described in the jpegs above

The [i]condition number[/i] of a square matrix $ A$ is defined as follows: Choose a norm for our vector space, and compute the induced operator norms of our matrices with respect to this norm. Then $ \kappa(A)\equal{}\|A\|\cdot\|A^{\minus{}1}\|.$

The smallest possible condition number is 1. An ill-conditioned matrix is one whose condition number is large.

What's the issue? Suppose you're trying to solve $ Ax\equal{}b,$ but what you actually find is $ \tilde{x}$ such that $ A\tilde{x}\equal{}\tilde{b},$ where the relative error of $ \tilde{b}$ compared to what it's supposed to approximate, namely $ b,$ is small. How big could the relative error in $ \tilde{x}$ as an approximation of $ x$ be? The answer is that it could be as big as the condition number times the relative error in $ \tilde{b}.$ That is, with an ill-conditioned matrix, we can get something which comes very close to satisfying the equation but still fails to be close to the true solution.

I suppose in the case of these real symmetric matrices in a least-squares context, we should be using the $ 2$-norm. As a matrix operator norm (for real symmetric matrices, anyway), that's the same as the spectral radius, which is the largest eigenvalue. For the $ 4\times 4$ matrix $ H$ given above, I get that $ \|H\|\approxeq 1.5$ while $ \|H^{\minus{}1}\|\approxeq 1.03\times 10^4,$ giving a condition number of $ \approxeq 1.55\times 10^4.$

By contrast, the condition number of an orthogonal matrix (one whose columns form an orthonormal set) is exactly $ 1,$ which is as small as possible.


Given are $\triangle ABC, L$ is Lemoine point,$L_{a},L_{b},L_{c}$ are three Lemoine point of triangles $LBC,LCA,LAB$ prove that $AL_{a},BL_{b},CL_{a}$ are concurrent!
A question:Locus $P$ such that  $AL_{a},BL_{b},CL_{a}$ are concurrent with $L_{a},L_{b},L_{c}$ are three Lemoine point of triangles $PBC,PCA,PAB$ :)

[quote="gemath"]Given are $\triangle ABC, L$ is Lemoine point,$L_{a},L_{b},L_{c}$ are three Lemoine point of triangles $LBC,LCA,LAB$ prove that $AL_{a},BL_{b},CL_{a}$ are concurrent![/quote]

You have a nice proof of that??

  darij

[quote="gemath"]I have a solution by vector menthod, but I'm not sure it is nice, you have a solution too Darij ? :)[/quote]

I use barycentric coordinates in quite an explicit way. I don't plan to post my solution here since it is just computation. I wanted to know whether you have something better.

  darij

[quote=gemath]

Given are $\triangle ABC, L$ is Lemoine point,$L_{a},L_{b},L_{c}$ are three Lemoine point of triangles $LBC,LCA,LAB$ prove that $AL_{a},BL_{b},CL_{a}$ are concurrent!

[/quote]

[b]Lemma[/b] [b]:[/b] Given a $ \triangle ABC $ and a point $ P. $ Let $ \triangle \widetilde{A}\widetilde{B}\widetilde{C} $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ and let $ K, \widetilde{K} $ be the symmedian point of $ \triangle ABC, $ $ \triangle \widetilde{A}\widetilde{B}\widetilde{C}, $ respectively. Then $ K, P, \widetilde{K} $ are collinear.

[b]Proof[/b] [b]:[/b] Let $ O $ be the circumcenter of $ \triangle ABC, \triangle \widetilde{A}\widetilde{B}\widetilde{C} $ and let $ T,\widetilde{T} $ be the intersection of $ \odot (O) $ with $ AK, $ $ \widetilde{A}\widetilde{K}, $ respectively. Let $ \overline{DEF}, \overline{\widetilde{D}\widetilde{E}\widetilde{F}} $ be the Lemoine axis of $ \triangle ABC, $ $ \triangle \widetilde{A}\widetilde{B}\widetilde{C}, $ resp. and let $ \overline{XYZ} $ be their perspectrix. Obviously, $ X, Y, Z $ lie on the polar $ \tau $ of $ P $ WRT $ \odot (O). $ Since $ ABTC, \widetilde{A}\widetilde{B}\widetilde{T}\widetilde{C} $ are harmonic quadrilateral, so $ T, $ $ P, $ $ \widetilde{T} $ are collinear $ \Longrightarrow $ $ AK $ $ \cap $ $ \widetilde{A}\widetilde{K} $ lies on $ \tau. $ Similarly, $ BK $ $ \cap $ $ \widetilde{B}\widetilde{K} $ and $ CK $ $ \cap $ $ \widetilde{C}\widetilde{K} $ lie on $ \tau. $

Since the tangent of $ \odot (O) $ at $ B, $ $ C, $ resp. and $ AK $ are concurrent, so $ D $ lies on the polar of $ AK $ $ \cap $ $ \widetilde{A}\widetilde{K} $ WRT $ \odot (O). $ Similarly, $ \widetilde{D} $ lies on the polar of $ AK $ $ \cap $ $ \widetilde{A}\widetilde{K} $ WRT $ \odot (O), $ so $ D, $ $ P, $ $ \widetilde{D} $ are collinear. Analogously, $ P $ lies on $ E\widetilde{E}, $ $ F\widetilde{F}, $ so from Desargues' theorem for $ \triangle YE\widetilde{E}, \triangle ZF\widetilde{F} $ we get $ \tau,  EF,  \widetilde{E}\widetilde{F} $ are concurrent, hence their pole $ P, K, \widetilde{K} $ WRT $ \odot (O) $ are collinear.
____________________________________________________________
[b]Back to main problem[/b] [b]:[/b]

Let $ \triangle DEF, \triangle XYZ $ be the pedal triangle of $ L $ WRT $ \triangle ABC, $ circumcevian triangle of $ A $ WRT $ \triangle LBC, $ resp. and let $ K, K_a $ be the Symmedian point of $ \triangle DEF, $ $ \triangle XYZ, $ respectively. Clearly, $ \triangle DEF \cup K $ and $ \triangle XYZ \cup K_a $ are directly similar, so notice $ L $ is the centroid of $ \triangle DEF $ we get $ \triangle LBC $ $ \stackrel{-}{\sim} $ $ \triangle K_aYZ. $ $ (\ddagger) $

Let $ L_A $ be the reflection of $ L $ in the midpoint of $ BC, $ then from $ (\ddagger) $ we get $ \triangle ABC \cup L_A $ $ \stackrel{-}{\sim} $ $ \triangle AZY \cup K_a $ $ \Longrightarrow $ $ AK_a, AL_A $ are isogonal conjugate WRT $ \angle A, $ hence $ AK_a $ passes through the isogonal conjugate $ T $ (WRT $ \triangle ABC $) of the complement of $ L $ WRT $ \triangle ABC. $ On the other hand, by [b]Lemma[/b] we know $ A, L_a, K_a $ are collinear, so we conclude that $ T $ lies on $ AL_a. $ $ \qquad \blacksquare $
____________________________________________________________
[quote=gemath]

A question:Locus $P$ such that  $AL_{a},BL_{b},CL_{a}$ are concurrent with $L_{a},L_{b},L_{c}$ are three Lemoine point of triangles $PBC,PCA,PAB$

[/quote]

 $ AL_a, BL_b, CL_c $ are concurrent $ \Longleftrightarrow P $ lies on [url=http://bernard.gibert.pagesperso-orange.fr/curves/q016.html]Lemoine quintic[/url]

[b]Lemma:[/b] Let $\triangle ABC$ be a scalene triangle with symmedian point $K.$ $AK$ cuts $BC$ at $D$ and circles $\odot(KAB)$ and $\odot(DAC)$ meet again at $X.$ Then $AX$ is the A-symmedian of $\triangle AKB.$

[b]Proof[/b]: Let $U \in AD$ be the intersections of the tangents of $\odot(ABC)$ at $B,C.$ Invert with center $A,$ denoting inverse points with primes. $U'$ is on the A-median of $\triangle AB'C',$ such that $\odot(AU'B'),\odot(AU'C')$ are tangent to $B'C'.$ $D'$ is the 2nd intersection of $AU'$ with $\odot(AB'C')$ and since $(K,A,D,U)=-1$ $\Longrightarrow$ $K'$ is the reflection of $U'$ on $D'.$ Since $\angle U'BC'=\angle U'AB'$ and $\angle U'C'B'=\angle U'AC'$ $\Longrightarrow$ $\angle B'UC'=\pi-\angle B'AC'=\angle B'D'C'$ $\Longrightarrow$ $B'U'C'D'$ is a parallelogram. Thus if $M'$ is the midpoint of $B'C',$ then $\overline{D'M'}:\overline{D'K'}=-1:2$ $\Longrightarrow$ $D'$ is centroid of $\triangle K'B'C'$ $\Longrightarrow$ $D'C'$ cuts $K'B'$ at its midpoint $X',$ i.e. $AX'$ is A-median of $\triangle AK'B'$ $\Longrightarrow$ $AX$ is the A-symmedian of $\triangle AKB. \ \blacksquare$

Back to the proposed problem. Let $BL,CL$ cut $CA,AB$ at $E,F$ and let $BL_a,CL_a$ cut $\odot(LBC)$ again at $Y,Z.$ From the previous lemma, it follows that $Y\in \odot(BAE)$ and $Z \in \odot(CAF)$ $\Longrightarrow$ $L_a$ is the radical center of $\odot(LBC),$ $\odot(BAE),$ $\odot(CAF)$ $\Longrightarrow$ $AL_a$ is the radical axis of $\odot(BAE),\odot(CAF).$ Now according to [url=https://artofproblemsolving.com/community/c6h282732]Three concurrent radical axes[/url], $AL_a$ passes through the isogonal conjugate of the complement of $L$ and same for $BL_b$ and $CL_c.$


function
trigonometry
calculus
derivative
limit
real analysis
real analysis unsolved

Let f(x) = x when x irrational, p(sin(1/q)) when x= p/q in lowest terms. Where is this function continuous.

I've managed to prove that f is discontinuous when is in Q/{0}, by picking epsilon= |p/q-psin(1/q)|/2, for every delta < epsilon there will obviously be some irrational y in that interval, and it will give us |f(x)-f(y)|= |psin(1/q)-y|>= ||psin(1/q)-x| - |x-y||> eps. This obviously doesn't work at x=0, since f(x)=0 there.

For x in the irrationals, here's what I tried: pick E>0, then there is some delta <E/2 s/t for all rational y in (x-delta, x+delta) |1/q-sin(1/q)|<E/2. Then |f(x)-f(y)|<= |x-p/q| + |p/q-psin(1/q)| <= E/2 + p(E/2). Of course, the problem here is that p can arbitrarily large, so we run into problems. Anyone can help?

Thanks

Rationals with small denominator $q$ are "rare": they are at distance $1/q$ one from another. So, if $x$ is irrational, it has a neighborhood in which all rationals have large $q$. Since $\sin$ has derivative $1$ at $0$, it follows that
$\lim_{q\to \infty}q(\sin(1/q)-1/q)=0$ -- this is just the definition of a derivative. You can pick $\delta$ so that $q|\sin(1/q)-1/q|\le \epsilon/2$ - this is better than the estimate you were using.


[b]Problem1[/b] :Let $ ABC$ be a triangle and let $ D,E,F$ be the points of tangency of the incircle $ (I)$ of $ ABC$ with the sides $ BC,CA,AB$ , respectively.Let $ A',B',C'$ be three points on the rays $ ID,IE,IF$ respectively such that $ IA' \equal{} IB' \equal{} IC'$.Denote by $ M,N,P$ the midpoints of $ EF,DF,DE$.Prove that $ AA',BB',CC'$ concur at the one point $ X$   ; $ A'M,B'N,C'P$ concur at the one point $ Y$ and the line$ XY$ passes through the [b]Gergonne [/b]poin of triangle$ ABC$.

[i]Reference [/i]
A topic of Linh :[url]http://www.mathlinks.ro/viewtopic.php?t=315748[/url]

 :)

Denote $\frac{_{\overline{ID}}}{^{\overline{IA'}}}=\frac{_{\overline{IE}}}{^{\overline{IB'}}}=\frac{_{\overline{IF}}}{^{\overline{IC'}}}=k.$ Then lines $AA',BB',CC'$ concur at $X$ due to Kariya's theorem. Nevertheless, we'll dispense with this theorem to show that $XY$ passes through the Gergonne point $G_e$ of $\triangle ABC.$ Since $\triangle DEF$ and $\triangle A'B'C'$ are centrally similar with similarity coefficient $k,$ then it follows that $\triangle MNP$ and $\triangle A'B'C'$ are centrally similar with similarity coefficient $-\frac{_{k}}{^{2}}$ $\Longrightarrow$ $Y \equiv MA' \cap NB' \cap PC'$ is the similarity center of $\triangle A'B'C'$ and $\triangle MNP.$ Let $AA'$ cut the line connecting $Y$ and the Gergonne point $G_e$ at $X,$ we'll show that $BB',CC'$ pass through the same point $X.$ We have then

$\frac{\overline{XY}}{\overline{XG_e}}=-\frac{AM}{MA'} \cdot \frac{YA'}{AG_e} \cdot \frac{ \sin \widehat{IAA'}}{\sin \widehat{DAA'}}=-\frac{IA'}{A'D} \cdot \frac{AD}{AI} \cdot \frac{AM}{MA'} \cdot \frac{YA'}{AG_e}$

$\Longrightarrow \frac{\overline{XY}}{\overline{XG_e}}=-\frac{2}{(k-1)(k+2)} \cdot \frac{AM}{AI} \cdot \frac{AD}{AG_e}$

Then, substituting $\ : \ \frac{AM}{AI}=\frac{s(s-a)}{bc} \ , \ \frac{AD}{AG_e}=\frac{(s-b)(s-c)+a(s-a)}{a(s-a)}$

$\frac{\overline{XY}}{\overline{XG_e}}=\frac{2}{(1-k)(k+2)}\cdot \frac{s[(s-b)(s-c)+a(s-a)]}{abc}$

$\Longrightarrow \frac{\overline{XY}}{\overline{XG_e}}=\frac{1}{(1-k)(k+2)} \cdot \left ( 2+ \frac{r}{2R} \right)$

Since this latter expressions is obviously independent of the chosen vertex, it follows that $BB',CC'$ intersect $YG_e$ at the same point $X,$ and the proof is completed.

There's a remarkable corollary coming from this result. If $A',B',C'$ coincide with the points at infinity of $ID,IE,IF,$ then $X,Y$ coincide with the orthocenter $H$ of $\triangle ABC$ and 9-point center $U$ of the intouch triangle $\triangle DEF,$ moreover $k=0.$ Therefore

In any scalene triangle $\triangle ABC,$ its orthocenter $H,$ Gergonne point $G_e$ and the nine point center $U$ of its intouch triangle $\triangle DEF$  are collinear such that $\frac{\overline{HU}}{\overline{HG_e}}=1+\frac{r}{4R}.$


In a long corridor, there are one thousand windows along one wall. Coincidentally, there are exactly one thousand people in the corridor. The first person is ordered to open the blinds on every window. Then, the second person is ordered to close the blinds on every second window. Then the third person is told to go to every third window, and close the blinds if they are open, and open the blinds if they are closed. The fourth person does this for every fourth window, and so on. After all 1000 people complete the process, how many blinds are open?

All integers except perfect squares have an even number of factors and will be closed.

Perfect squares have an odd number of factors and will be open.

There are $\boxed{31}$ perfect squares under 1000.


function
logarithms
inequalities
inequalities unsolved

Let x \geq y \geq z>0
Prove that :
x y y z z x  \leq y x z y x z

Lets see:

We apply the function $\ln$ to $x^{y}y^{z}z^{x} \leq y^{x}z^{y}x^{z}$ and we need to prove that:
$y \ln(x)+z \ln(y)+x \ln(z) \leq x \ln(y)+y \ln(z)+z \ln(x)$. Lets consider the function: $f(t)=y \ln(t)+z \ln(y)+t \ln(z)-t \ln(y)-y \ln(z)-z \ln(t)$ with $t\leq y$ and $0<z \leq y$. We need to prove that: $f(z) \leq 0$. We compute the first derivate of f and get: $f'(t)=\frac{y}{t}+ \ln(z)- \ln(y)- \frac{z}{t}$ and now the second derivate: $f''(t)=-\frac{y}{t^{2}}+\frac{z}{t^{2}}=\frac{z-y}{t^{2}} \leq 0$ so $f'$ is a decreasing function thus $f'(t) \leq f'(y)=1+\ln(z)-\ln(y)-\frac{z}{y}$. Let $g(t)=-\ln(t)-\frac{z}{t}+1+\ln(z)$ with $t \leq z$. The derivate of g is: $g'(t)=-\frac{1}{t}+\frac{z}{t^{2}}=\frac{z-t}{t^{2}}\leq 0$, so g is a decreasing function thus: $g(y) \leq g(z)=0$. And because $f'(t) \leq f'(y)=g(y) \leq g(z)=0$ we get that $f(t) \leq f(y)=0$. And with this we ended the solution.

I hope it is okay!
cheers! :D :D

This can be easily generalized to:

Prove that for any $0<x_{1} \leq \ldots \leq x_{n}$ we have:
$x_{1}^{x_{2}} \cdot x_{2}^{x_{3}} \cdots x_{n-1}^{x_{n}} \cdot x_{n}^{x_{1}} \geq x_{2}^{x_{1}} \cdot x_{3}^{x_{2}} \cdots x_{n}^{x_{n-1}} \cdot x_{1}^{x_{n}}$.

Right?cheers! :D :D

Can't we just use the ordering inequality?

Can you please post your solution? Thank you very much!

[quote="Lagrangia"]We apply the function $\ln$ to $x^{y}y^{z}z^{x} \leq y^{x}z^{y}x^{z}$ and we need to prove that:

$y \ln(x)+z \ln(y)+x \ln(z) \leq x \ln(y)+y \ln(z)+z \ln(x)$. [/quote]

We may assume $x\not=z$, otherwise $x=y=z$ and the inequality is clearly true.

Since $\ln$ function is concave, we have
\[\frac{y-z}{x-z}\cdot\ln{x} + \frac{x-y}{x-z}\cdot\ln{z} \leq \ln\left(\frac{y-z}{x-z}\cdot{x} + \frac{x-y}{x-z}\cdot{z} \right) =\ln{y} \]

Hence  $(y-z)\ln{x} + (x-y)\ln{z} \leq (x-z)\ln{y}$.

That is, $y \ln(x)+z \ln(y)+x \ln(z) \leq x \ln(y)+y \ln(z)+z \ln(x)$.

Q.E.D.

Very nice solution fuzzylogic .
Here my solution :
If $ x=y $ or $ y=z $ it's clear .
We assume : $ x>y>z $
So the inequality becomes :
\[ \frac {lnx-lny}{x-y} \leq \frac {lny-lnz}{y-z} \] ( It's obvious )


The year 2006 is the most recent calendar year with digits that
are each even.  What was the most recent year before 1999
with digits that were each even?

Obviously 1 isn't even, so the year can't begin with 1 - hence, the year we're looking for must have three digits. 9 is also odd, so it has to have all 8s, to be as large as possible. Therefore, the year is 888.

 :welcomeani:
 :arrow: since you want to find the most recent year before 1999 that has even numbers for all digits, you should start finding the largest even digit.

  :arrow:  The largest even digit is eight, so you would probably think about doing 1888,   1 is a odd digit though, so the largest year before 1999, all even digits is  :idea: [u]888[/u]
 :winner_first:   :jump:

I've seen people get 1888 on this exact problem


geometry
circumcircle
rectangle
ratio
trigonometry
inradius
similar triangles

It is important to know how to find the area of a triangle.  There are many different ways to do it (i.e. there are a ton of formulas out there).  Knowing how to find the area of a triangle can have many important uses and I'll include a few here.  NOTE: In general, we'll be working with $\triangle ABC$ with $BC=a, CA=b,AB=c, R$ the radius of the circumcircle, $r$ the radius of the incircle, and $s$ the semiperimeter (1/2 the perimeter).  Bracket notation will also be used for area, i.e. $[ABC]$ denotes the area of triangle $ABC$.

\[ \text{Basics} \]

The basic notion of area is fairly simple.  It's just how much space something occupies.  One way to find the area of a triangle is to draw a rectangle around it as such ($AD$ is an altitude): [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=616[/img]

We can see that $\triangle ADB\cong \triangle BEA$ and $\triangle ADC\cong \triangle CFA$ thus implying that $[ABC]$ is one half the area of the rectangle.  The rectangle's area is just $bh$ ($b$ in this case is the base...not the length of the side opposite from vertex $B$).  So $[ABC]=\frac 12 bh$.

[b]Exercise 1[/b]: Find the smallest altitude in a 3-4-5 right triangle.
[b]Exercise 2[/b]: Find the area of an equilateral triangle with side length 2.
[b]Exercise 3[/b]: Find the area of an equilateral triangle with side length $x$ (we're looking for a general formula here).

\[ \text{Similar Triangles} \]

When going between two similar triangles we know that there is a common ratio (call it $r$) between the lengths of one triangle and the lengths of the other.  But when going between the [i]areas[/i] of two similar triangles we instead have to take into account that area is two dimensional.  In our formula for area, $[ABC]=\frac 12 bh$, we will inrease [i]both[/i] $b$ and $h$ by a factor of $r$ thus resulting in a ratio of $r^2$ between the areas of the two triangles.

Example: Triangles $ABC$ and $DEF$ are similar.  If $AB=2$, $DE=4$, and $[DEF]=64$ find $[ABC]$.
[i]Solution[/i]: The ratio between the two triangles is $\frac 24=\frac 12$.  So $\frac{[ABC]}{[DEF]}=\left(\frac 12\right)^2.$  Simplifying we find that $[ABC]=16$.

[b]Exercise 4[/b]: The area of triangle $ABC$ is 12.  We then connect the midpoints of the three sides of $ABC$ to form three sides.  Find the area of the triangle bound by these three sides.

\[ \text{Sharing a Base/Altitude} \]

Always be on the look out for triangles that share a common base or height.

Example: The areas of the triangles below are in the ratio $\frac{h_1}{h_2}$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=617[/img]

\[ \text{Bringing Trig Into the Picture} \]

We again go back to the formula $[ABC]=\frac 12 \text{base}\times\text{height}$.  Now consider the following picture:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=618[/img]

We know that $\sin C=\frac h{AC}=\frac hb\Leftrightarrow h=b\sin C$.  So thus $[ABC]=\frac 12 \text{base}\times\text{height}=\frac 12 ab\sin C$.

Additionally, we could then use the Extended Law of Sines which states that \[ \frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=2R. \]

Plugging in $\sin C=\frac c{2R}$ we get $[ABC]=\frac{abc}{4R}.$

\[ \text{Using the Inradius} \]

Let's draw in the incircle of our triangle and see what we can find.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=619[/img]

The radii that are drawn are to the points of tangency.  We know that these radii make right angles with the sides.  Therefore $[ABC]=\frac 12 ar+\frac 12 br+\frac 12cr=\frac{a+b+c}2\cdot r=rs$.

\[ \text{Formulas} \]

Here are some common formulas for the area of a triangle:

\begin{eqnarray*} [ABC] &=& \frac 12 bh\\ &=& \frac 12 ab\sin C\\ &=& \sqrt{s(s-a)(s-b)(s-c)} \qquad \text{(Heron's Formula)}\\ &=& \frac{abc}{4R}\\ &=& rs \end{eqnarray*}

Additionally, here's [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=4071]some stuff on exradii.[/url]

I'll get some problems together and post them later.

Nice job :) !, Although the four or five people that come to this forum know all this stuff already  :P , but I think it's nice to have all these lessons just in case more people just trying to learn come to this forum.

I am going to make one too on circles, and probably more advanced theorems of geometry, (Ceva, Stewarts, Angle Bisector, etc) Hopefully we'll be able to make these lessons on most of the concepts of math useful in contests....


function
calculus
integration
LaTeX
real analysis
real analysis unsolved

A function $ f$ is continuous on $ [a,b]$ and $ M \equal{} sup${$ f(x),x \in [a,b]$}, prove that $ ({\int(f(x))^n dx)}^{\frac {1}{n}}$ converges to $ M$ where the lower limit of the integral is $ a$ and the upper limit of the integral is $ b$. Thank you for your attention.

Notation: \int_{a}^{b} gives $ \int_a^b$. Also, \left( and \right) (or other delimiters) are useful: $ \left(\frac12\right)$. For { and } in $ \text{\LaTeX}$, use \{ and \}.

The limit you're looking at is as $ n\to\infty$, of course- and you need to add the assumption that $ f$ is nonnegative.

Hint: $ f$ is very close to $ M$ on some small subinterval. Use this to get lower bounds for the integral of high powers.

Without loss of generality I will assume $ b \minus{} a > 1$ the proof of the opposite case is similar.

Clearly $ \int_a^b (f(x))^n \leq \int_a^b M^n$ so that $ (\int_a^b (f(x))^n )^{1/n} \leq (\int_a^b M^n)^{1/n} \equal{} M(b \minus{} a)^{1/n}$ and given $ \epsilon > 0$  $ \exists N$ such that for $ n \geq N$ we have $ M(b \minus{} a)^{1/n} \leq M(1 \plus{} \epsilon)$ if $ b \minus{} a > 1$

Now there exists $ s \in [a,b]$ such that $ f(x) > M \minus{} \epsilon$ in a neighborhood about $ s$
So that  $ \int_a^b (f(x))^n dx \geq \int_{s \minus{} \delta}^{s \plus{} \delta} f(x)^n dx \geq \int_{s \minus{} \delta}^{s \plus{} \delta} (M \minus{} \epsilon)^n dx$ for some $ \delta > 0$ by the continuity of $ f$

Hence $ (\int_a^b (f(x))^n dx)^{1/n} \geq (2\delta)^{1/n}(M \minus{} \epsilon) \geq (M \minus{} \epsilon)(1 \minus{} \epsilon)$ for large enough $ n$

Putting this all together we that for any $ \epsilon > 0$

$ (M \minus{} \epsilon)(1 \minus{} \epsilon) \leq (\int_a^b (f(x))^n dx)^{1/n} \leq M(1 \plus{} \epsilon)$ for large n.  This proves the result.


prove that there are infinitely many primes $p$ s.t. $p\equiv 3 \pmod 4$

[hide]

Simply modify the standard Euclidean argument:

Say there was a last prime $p \equiv 3 \pmod 4$.  Multiply together all of the primes $\equiv 3 \pmod 4$.  If this number is $\equiv 1 \pmod 4$. add 2.  If this number is $\equiv 3 \pmod 4$. add 4.  Either way, none of the previous primes divide the new number, and it cannot be composed of only primes $\equiv 1 \pmod 4$ since any number of 1's multiplied together is still 1.  So there must be more primes $\equiv 3 \pmod 4$.  Contradiction.

[/hide]

Nice exersize:
Show that there exists $\infty$ many primes $p$ with:

a) $p \equiv 2 \mod 3$.
b) $p \equiv 1 \mod 4$.
c) $p \equiv 1 \mod 3$.
d) $p \equiv 1 \mod 5$.
e) $p \equiv 4 \mod 5$.
f) $p \equiv 1 \mod m$, $m$ any prime.
g) $p \equiv 1 \mod m$, $m$ any integer.

(I think it's ordered by difficulty)
(and only use things you can proof ;) )

g) Suppose there are only finitely many primes of the form $p\equiv1\pmod{m}$, namely $p_{1},p_{2},\ldots p_{r}$. Let $M=mp_{1}\cdots p_{r}$ and let $N\in\mathbb{Z}$. Then $\Phi_{m}(NM)\equiv \Phi_{m}(0)=\pm 1\pmod{M}$, therefore mod $p_{i}$ and mod $m$. In particular $\Phi_{m}(NM)$ is not divisible by $p_{i}$ and none of its prime factors divides $n$. $\Phi_{m}(NM)\to\infty$ as $N\to\infty$, so for large $N$ $\Phi_{m}(NM)\ne \pm 1$. Therefore there exists a prime $p$ dividing $\Phi_{m}(NM)$. Now if $p\not |m$, then $p$ divides $\Phi_{m}(a)$ for some $a\in\mathbb{Z}$ if and only if $p\equiv 1\pmod{m}$. Thus $p\ne p_{i}$ as $1\leq i\leq r$, and we have found a new prime.

Yes, that's the "standard" way to do it. Note that there is a nicer way for $m=p$ prime, e.g. using the BalticWay problem with the divisors of $(n+1)^{p}-n^{p}$ from 2004.

More:
h) $p \equiv 3 \mod 8$.
i) $p \equiv 5 \mod 8$.
j) $p \equiv 7 \mod 8$.
k) $p \equiv 6 \mod 7$.
l) $p \equiv 8 \mod 9$.
m) $p \equiv 104 \mod 105$.
n) $p \equiv 71 \mod 105$.


Οι συναρτησιακές εξισώσεις , είναι συνιθισμένο θέμα σε διαγωνισμούς. Έχουν ελάχιστη θεωρία και πολλές ωραίες τεχνικές.
  Δε θα βάλουμε ασκήσεις με συνέχεια ή διαφορικό λογισμό, παρά μόνο ασκήσεις που χρησιμοποιούν την έννοια της συνάρτησης , το 1-1 , τη μονοτονία και  την έννοια του φραγμένου ή του ακρότατου.
  Με 20 ασκήσεις θα είμαστε καλά.

              [color=red]Καλή συγκομιδή !!! [/color]

ΑΣΚΗΣΗ 1-(Κίνα 1993)
Έστω η συνάρτηση $f: (0,+\infty)\rightarrow(0,+\infty)$ η οποία  για κάθε $x,y>0$ ικανοποιεί τη σχέση $f(xy)\leq f(x)f(y).$ Για κάθε $x>0$ και για κάθε $n\in\mathbb{N^{*}},$ να αποδείξετε ότι:
$f(x^{n})\leq f(x)\cdot(f(x^{2}))^{\frac{1}{2}}\cdots(f(x^{n}))^{\frac{1}{n}}.$

ΛΥΣΗ-(Γιάννης Παναγέας)
Θα δουλέψω επαγωγικά χρησιμοποιώντας τη δεύτερη μορφή της επαγωγής ( δηλ θεωρώ ότι ισχύει το p(1),...,p(n) και δείχνω ότι ισχύει το p(n+1) και όχι θεωρώ απλά ότι ισχύει το p(n)).

  Για n=1 φαίνεται προφανώς ότι ισχύει.

 Έστω ότι ισχύει μέχρι και για $n=k$ ,δηλαδή

  $f(x)\leq f(x)$
  $f(x^{2})\leq f(x)(f(x^{2})^{\frac{1}{2}})$
  .
  .
  .
  .
  $f(x^{k})\leq f(x)( f(x^{2})^{\frac{1}{2}}).....( f(x^{k})^{\frac{1}{k}})$

  Θα δείξω ότι ισχύει για  $n=k+1$.

  Πολλαπλασιάζω και έχω:
  $f(x)f(x^{2})...f(x^{k}) \leq ( f(x)^{k})( f(x^{2})^{\frac{k-1}{2}}) ... ( f(x^{k})^{\frac{1}{k}}) ( f(x)f(x^{2})...f(x^{k}) )^{2}\leq ( f(x)^{k+1})( f(x^{2})^{\frac{k+1}{2}}) ... ( f(x^{k})^{\frac{k+1}{k}})$

  όμως  $f(x^{m})f(x^{k+1-m})\geq f(x^{k+1})$ και άρα

 $( f(x^{k+1}))^{k}\leq ( f(x)f(x^{2})...f(x^{k}) )^{2}\leq ( f(x) ( f(x^{2})^{\frac{1}{2}})... ( f(x^{k})^{\frac{1}{k}})^{ k+1})$

  $( f(x^{k+1}) )^{k+1}\leq ( f(x) ( f(x^{2})^{\frac{1}{2}}) ... ( f(x^{k})^\frac{1}{k}) ( f(x^{k+1})^{\frac{1}{k+1}) )^{ k+1}}$

  και άρα
$f(x^{k+1}) \leq f(x) ( f(x^{2})^{\frac{1}{2}}) ... ( f(x^{k})^{\frac{1}{k}}) ( f(x^{k+1})^{\frac{1}{k+1}}) )$
  και έτσι τελειώνει η επαγωγή

[quote="Protonios"]ΑΣΚΗΣΗ 2: (έχουμε εγκαταλείψει το τόπικ αυτό)(ας ξεκινήσουμε με κάτι απλό)
Eotvos Kurchak Mathematics Competition 1979

Ορίζουμε για όλους τους πραγματικούς αριθμούς τη συνάρτηση f τέτοια ώστε f(x)<=x και f(x+ψ)<=f(x)+f(ψ) για κάθε χ,ψ πραγματικούς. Να δειχθεί ότι η f(x)=x.

Λύση: (by me)

f(x)<=x (1)
f(x+ψ)<=f(x)+f(ψ) (2)

Στη σχέση (1) θέτουμε χ=0 και παίρνουμε:
f(0)<=0

Στη σχέση 2 θέτουμε χ=ψ=0 και παίρνουμε f(0)>=0.

Άρα f(0)=0.

Είναι επίσης σύμφωνα με τη (2) f(x)+f(-x)>=f(0)=0.

Η (1) δίνει ακόμη f(x)<=x και f(-x)<=-x από τις οποίες με πρόσθεση κατά μέλη λαμβάνουμε f(x)+f(-x)<=0.

Συνεπώς f(x)+f(-x)=0 ή ισοδύναμα f(x)=-f(-x).

Από την (1) προκύπτει f(-x)<=-x ή ισοδύναμα x<=-f(-x) ή τελικά f(x)>=x, από την (1) και την οποία ολοκληρώθηκε η απόδειξη.[/quote]

 [color=blue]Για εισαγωγική άσκηση είναι καλή! Είναι βέβαια πια τώρα σχολική ,  αλλά αυτό δεν πειράζει καθόλου. Μπαίνει !!![/color]

Απλά ήθελα να επαναφέρω το τόπικ. Θα βάλλω και δυσκολότερες. :D  :wink:

[size=150][color=red]ΑΣΚΗΣΗ 3 [/color][/size]

 Να βρείτε όλες τις συναρτήσεις  $f : \mathbb{Z}\rightarrow \mathbb{Z}$ , για τις οποίες ισχύει ότι :

$f(m-n+f(n)) = f(m)+f(n)$      για κάθε  $m,n \in \mathbb{Z}$

   [b]ΛΥΣΗ[/b]

 Για $m=n$ παίρνουμε   $f(f(n)) = 2f(n)$ για κάθε  $n\in\mathbb{Z}$. Θέτουμε  $n: =f(n)$ στην εξίσωση

$f(m-n+f(n)) = f(m)+f(n)$

και παίρνουμε :

$f(m+f(n)) = f(m)+2f(n)$  για κάθε  $m, n \in \mathbb{Z}$.

Από την άλλη μεριά , αν ξαναθέσουμε  $m: =m+n$   στην αρχική εξίσωση , παίρνουμε:

$f(m+f(n)) = f(m+n)+f(n)$ $\forall$l $m, n \in \mathbb{Z}$.

Οι τελευταίες δύο εξίσώσεις δίνουν :

 $f(m+n) = f(m)+f(n)$ για κάθε  $m, n \in \mathbb{Z}$. Επομένως  $f(n) = kn$  για κάποια σταθερά  $k\in\mathbb{Z}$ ($k = f(1)$).

Από δω  εύκολα προκύπτει τελικά ότι  $k=0$, ή $k=2$.

[quote="Protonios"]Δεν ξέρω αν συμφωνείτε αλλά προτείνω να βάλλει κάποιος την άσκηση συναρτησιακής που έπεσε στη Βαλκανιάδα με μια καλή λύση. Όποιος μπορεί ας το κάνει. :wink:[/quote]

 Προτείνω να το κάνει ο Αλέξανδρος , αφού στη Ρόδο είχε την ευκαιρία να δει όλες τις λύσεις ή τις προσπάθειες. Μπορεί βέβαια να μας τη στείλλει και ο Σιλουανός που την έλυσε καλά.

 Μπάμπης

Θα παρουσιάσω τη λύση που βρήκα κατά τη διάρκεια του διαγωνισμού .
[size=150][color=red]ΑΣΚΗΣΗ 4[/color][/size]

Να βρεθούν όλες οι συναρτήσεις $f: \mathbb{R}\to\mathbb{R}$ έτσι ώστε να ισχύει
\[f\left(f(x)+y\right)=f\left(f(x)-y\right)+4f(x)y \]
για κάθε ${x,y \in \mathbb{R}}$.

[color=red]
ΛΥΣΗ (by me)[/color]
Mia προφανής λύση είναι η $f(x)=0$ . Έστω ότι δεν είναι $f(x)=0$ για κάθε $x$ στο R
Έστω $D$ το σύνολο τιμών της $f$ (η εικόνα της f)
Αρχικά σταθεροποιούμε το $x$ και έχουμε
$f(f(x)+y)-f(f(x)-y)=4f(x)y$ και αφού το $y$ σαρώνει όλο το $R$ θα είναι
$D-D=\mathbb{R}$ (*)

Τώρα θέτουμε $g(x)=f(x)-x^{2}$ και λόγω της αρχικής λαμβάνουμε :
$g(g(x)+x^{2}+y)=g(g(x)+x^{2}-y)$ για κάθε $x,y\in\mathbb{R}$
Θέτουμε στην τελευταία όπου $y$ το  $g(x)+x^{2}$ και παίρμουμε
$g(2g(x)+2x^{2})=g(0)=f(0)=c$ . Άρα $g(2f(x))=c$ για κάθε $x$
Άρα $f(x)=x^{2}+c$ για κάθε $x\in\ 2D$  (1)
Τώρα παρατηρούμε ότι αν $y_{1},\ y_{2}\in D$ τότε $y_{1}+y_{2}\in\ 2D$
Θέτουμε τώρα $f(x_{1})=y_{1}$ όπου $x_{1}\in R$  και προφανώς $y_{1}\in D$
Θέτουμε τώρα όπου $x$ to $x_{1}$ kai opou $y$ to $y_{1}$ στην αρχική και παίρνουμε .
$f(y_{1}+y_{2})=f(y_{1}-y_{2})+4y_{1}y_{2}$ . Όμως $y_{1}+y_{2}\in 2D$ άρα λόγω της (1) έχουμε
$(y_{1}+y_{2})^{2}+c=f(y_{1}-y_{2})+4y_{1}y_{2}$  ή $f(y_{1}-y_{2})=(y_{1}-y_{2})^{2}+c$ .

Όμως σύμφωνα με την σχέση (*) το $y_{1}-y_{2}$ σαρώνει όλο το R
Άρα $f(x)=x^{2}+c$ όπου $c=f(0)$ για κάθε $x$ στο $\mathbb{R}$

[b]Σιλουανέ εύγε ! [/b]

 Η λύση έχει ένα σημείο για κουβέντα , όμως αυτό παρακάμπτεται , αν η $f$ έχει σύνολο τιμών το $\mathbb{R}$.ΑΥτόόμως  δεν το γνωρίζουμε αλλά  είναι και  κάτι που δεν πρέπει να ισχύει για τη συνάρτηση που αναμένουμε. Θα έχουμε πρόβλημα!

[quote="silouan"]
$\frac{f(f(x)+y)-f(f(x)-y)}{2y}=2f(x)$  , $(1)$

[/quote]

[b]ΣΧΟΛΙΟ[/b]

 [color=red]Δε βλέπω να  προσέξατε το προηγούμενο μήνυμά μου ! [/color]

Η παραπάνω λύση από μόνη της δε λύνει το πρόβλημα.  Η παραπάνω πορεία (σχέση $(1)$)  αποδεικνύει ότι
 $f'(f(x) = 2f(x)$ . Αν τώρα η $f$ έχει σύνολο τιμών το $\mathbb{R}$ ,  τότε θα είναι και $f'(u) = 2u , \forall u \in \mathbb{R}$ και η απόδειξη τελειώνει, αρκεί η συνάρτηση που προκύπτει να είναι δεκτή.

[u]Μπάμπης[/u]

[quote="stergiu"][quote="silouan"]
$\frac{f(f(x)+y)-f(f(x)-y)}{2y}=2f(x)$  , $(1)$

[/quote]

[b]ΣΧΟΛΙΟ[/b]

 [color=red]Δε βλέπω να  προσέξατε το προηγούμενο μήνυμά μου ! [/color]

Η παραπάνω λύση από μόνη της δε λύνει το πρόβλημα.  Η παραπάνω πορεία (σχέση $(1)$)  αποδεικνύει ότι
 $f'(f(x) = 2f(x)$ . Αν τώρα η $f$ έχει σύνολο τιμών το $\mathbb{R}$ ,  τότε θα είναι και $f'(u) = 2u , \forall u \in \mathbb{R}$ και η απόδειξη τελειώνει.

[u]Μπάμπης[/u][/quote]

Μισό λεπτό να πάρουμε τα πράγματα με τη σειρά! Θέτουμε $u=f(x)+y$. To $u$ επομένως δεν διατρέχει όλο το $R$?Συνεπώς $f'(u)=2\cdot f(u) \forall u \in R$ $q.e.d.$

[quote="Nick Rapanos"] Πρέπει όμως να αναφερθεί και ο cretanman που μου έδειξε τη λύση αυτή (που αν δεν κάνω λάθος και του cretanman του την είχε δείξει η Ε. Μήτσιου)!![/quote]

Η λύση (η ιδέα για τη λύση αυτή ) ανήκει στον Ρουμάνο αρχηγό.  :P

[quote="Nick Rapanos"][quote="Nick Rapanos"][quote="silouan"]Να και μια άλλη λύση στην παραπάνω(με βοήθεια παραγώγου  :wink: ) .
[color=red]ΑΣΚΗΣΗ 4- ΛΥΣΗ 2 [/color]

Προφανώς η $f(x)=0$ για κάθε x είναι λύση .
Η αρχική σχέση για $y\neq 0$ γράφεται
$\frac{f(f(x)+y)-f(f(x)-y)}{2y}=2f(x)$

Θέτουμε $f(x)+y=u$ (το x και το y είναι προφανώς ανεξάρτητα )
Άρα $\frac{f(u)-f(u-2y)}{2y}=2(u-y)$ ή
$\frac{f(u-2y)-f(u)}{-2y}=2(u-y)$

Θέτουμε $-2y=h$ και η παραπάνω δίνει
$\frac{f(u+h)-f(u)}{h}=2u+h$  . Άρα

$\lim_{h\rightarrow 0}\frac{f(u+h)-f(u)}{h}=\lim_{h\rightarrow 0}2u+h$
άρα
$f'(u)=2u$ άρα $f(u)=u^{2}+c$ όπου βρίσκουμε $c=f(0)$ . QED .

Ελπίζω να την απολαύσετε.. :wink:[/quote]

Είδες οτι τελικά γίνεται άσκηση Τρίτης Λυκείου με αυτό τον τρόπο?Πρέπει όμως να αναφερθεί και ο cretanman που μου έδειξε τη λύση αυτή (που αν δεν κάνω λάθος και του cretanman του την είχε δείξει η Ε. Μήτσιου)!![/quote]

Lathos mou!Loipon to u saronei olo to R alla oxi otan to h teinai sto 0, opote sosta epesimane o kyrios stergiu oti yparxei provlima!!!! :lol:[/quote]

Νίκο , χαίρομαι που βρήκες μόνος σου το  αιχμηρό σημείο. Δεν θέλησα να το γράψω στο προηγούμενο μήνυμα , για το ψάξετε λίγο μόνοι σας αλλά και για να δω  μήπως μου ξεφεύγει κάτι.

 Ας το κοιτάξουν και άλλοι!

[u]Μπάμπης [/u]

Ta parapano (opos kai to parakato) den einai sosta. I lisi opos exei dothei einai lathos kai nomizo oti o litis (autis me tin paragogo) tin ekane mia alli proseggisi...

[quote="SorrowMan"]I lisi pou edosa egw stin askisi einai i eksis :

An thesw f(x)=y , vgenei f(2y)=f(0)+4y^2 , gia 2y=u vgenei f(u)=f(0)+u^2 , u<--->x , f(x)=x^2+f(0)[/quote]

 Αυτή είναι η συνάρτηση για τις τιμές της μεταβλητής που βρίσκονται [b]στο σύνολο τιμών της [/b]και όχι στο $\mathbb{R}$
  Αυτή είναι η μόνη δυσκολία του προβλήματος.Πρέπει να αποδείξουμε ότι η συνάρτηση έχει σύνολο τιμών το $\mathbb{R}$κάτι όμως που θα δημιουργήσει άλλο σοβαρό πρόβλημα και θα δημιουργήσει αντίφαση με αυτό που ψάχνουμε  , . Η απλή εναλλαγή των $y,u$ δε λύνει δυστυχώς την άσκηση.

 Μπάμπης

[quote="stergiu"][quote="SorrowMan"]I lisi pou edosa egw stin askisi einai i eksis :

An thesw f(x)=y , vgenei f(2y)=f(0)+4y^2 , gia 2y=u vgenei f(u)=f(0)+u^2 , u<--->x , f(x)=x^2+f(0)[/quote]

 Αυτή είναι η συνάρτηση για τις τιμές της μεταβλητής που βρίσκονται [b]στο σύνολο τιμών της [/b]και όχι στο $\mathbb{R}$
  Αυτή είναι η μόνη δυσκολία του προβλήματος.Πρέπει να αποδείξουμε ότι η συνάρτηση έχει σύνολο τιμών το $\mathbb{R}$,κάτι όμως που θα δημιουργήσει άλλο σοβαρό πρόβλημα και θα δημιουργήσει αντίφαση με αυτό που ψάχνουμε Η απλή εναλλαγή των $y,u$ δε λύνει δυστυχώς την άσκηση.

 Μπάμπης[/quote]

Den katalava kala(mallon) ti thelete na peite kai pou einai to provlima :blush: ....

[quote="Nick Rapanos"][quote="staretak"]Apo pote i $f(x)=x^{2}+c$ (i kai i f(x)=0) exei sinolo timon to $R$????[/quote]

Ontos auto pou leei o stef to exo dei ki ego kai gi auto pisteuo oti i askisi de lynetai me auti ti methodo (ektos eab kapoios exei na proteinei kanena tromero trick....) :P[/quote]

Den grafei pouthena stin diatiposi tis askisis oti i f prepei na kinite se olo to R aparetita .. Mi berdevete sto f:R->R , to deftero R den einai aparetita to sinolo(pedio) timwn tis sinartisis,genika leme pws einai to sinolo afikseos tis f , diladi ti times tha mporei na parei i f , den tis pernei kai aparetita oles , stin prokimeni periptosi pernei apo to c ews kai to +00 , diladi pali pernei times mesa sto R...An twra kapoios apodiksei oti i sinartisi etsi opws orizete prepei na einai opwsdipote EPI(den nomizw vevaia sto likio na iparxei afti i enia,ean thimamai kala) tote tha eprepe i sinartisi na kinite se olo to R aparetita , alla de nomizw pws einai kai olas.....

Pantws akoma den exw katalavei giati einai lathos i dikia mou apodiksi :oops:  :blush:

Opos ta les einai, alla basika mallon auta pou les ta apeuthineis se sena...dioti esi perneis f(2y) opou y stin eikona tis f, ara den exeis orisei tin f se olo to R, alla mono stin 2*Imf (i gia na to poume pio austira stin antistrofi eikona tis eikonas tis f meso tis tautotikis...exei simasia auti i diaforopoiisi sti Theoria Katigorion). Omos i f fisika den einai epi (oute itan oute bgenei, opos eipothike parapano) opote i idea sou odigei basika se kati pou einai tetrimmeno.

Basika, gia na min pedeueste, kalo tha itan na asxoleithite ligo me tin lisi tou korifaiou Antonakoudi, dioti auti captures the real point!

[quote="staretak"]Apo pote i $f(x)=x^{2}+c$ (i kai i f(x)=0) exei sinolo timon to $R$????[/quote]

Δεν λέμε ότι η $f$ έχει σώνει και καλά σύνολο τιμών το $\mathbb{R}$ , αλλά ότι για να συμπεράνουμε από τη σχέση
 $f(f(x)) = f^{2}(x)+c$  ότι $f(x) = x^{2}+c , \forall x\in \mathbb{R}$ θα έπρεπε η $f$ να έχει σύνολο τιμών το $\mathbb{R}$.
 Φυσικά , η συνάρτηση που αναμένουμε δεν έχει σύνολο τιμών το $\mathbb{R}$ - συγνώμη που σε προηγούμενα μηνύματά μου δεν εκφράστηκα καλά , διότι είχα εστιάσει σε αυτό ακριβώς σημείο το πρόβλημα(εξέλαβα και μερικά μηνύματα άλλων ως δεδομένα και δεν εμβάθυνα όσο έπρεπε για να κάνω σαφείς διατυπώσεις) - οπότε αυτή η προσέγγιση με την παράγωγο φαίνεται να έχει πρόβλημα και δε λύνει την άσκηση.

[u]Μπάμπης [/u]

[quote="vag5858"]ΣΙΛΟΥΑΝΕ
ΘΕΣΕ f(x)+y=u ΚΑΙ f(x)-y=v
OΠΟΤΕ ΠΑΡΑΚΑΜΠΤΟΝΤΑΙ ΤΑ ΠΕΡΙ ΣΥΝΟΛΟΥ ΤΙΜΏΝ
(ΠΟΥ ΣΩΣΤΑ ΕΠΙΣΗΜΑΙΝΕΙ Ο ΦΙΛΟΣ ΜΠΑΜΠΗΣ)
ΤΗ ΛΥΣΗ ΑΥΤΗ ΜΟΥ ΤΗΝ ΕΙΠΕ Ο ΣΩΤΗΡΗΣ ΛΟΥΡΙΔΑΣ[/quote]

Ayto  akougetai eksairetika endiaferon...Tin idid idea eixa ki ego sto diagonismo mono pou tin egrapsa alla den ti synexisa...(logo m*l*ki*s...)

P.s. Anypomono na do ti lysi tou lourida-tha mas apodeiksei gia akomi mia fora me ti lysi tou oti einai [u]koryfaios??[/u]

[quote="vag5858"]ΣΙΛΟΥΑΝΕ
ΘΕΣΕ f(x)+y=u ΚΑΙ f(x)-y=v
OΠΟΤΕ ΠΑΡΑΚΑΜΠΤΟΝΤΑΙ ΤΑ ΠΕΡΙ ΣΥΝΟΛΟΥ ΤΙΜΏΝ
(ΠΟΥ ΣΩΣΤΑ ΕΠΙΣΗΜΑΙΝΕΙ Ο ΦΙΛΟΣ ΜΠΑΜΠΗΣ)
ΤΗ ΛΥΣΗ ΑΥΤΗ ΜΟΥ ΤΗΝ ΕΙΠΕ Ο ΣΩΤΗΡΗΣ ΛΟΥΡΙΔΑΣ[/quote]

To ίδιο προσπάθησα να κάνω και εγώ στον διαγωνισμό αλλά προκύπτουν μετά άλλα προβλήματα,όπως  ότι to u και το v δεν είναι ανεξάρτητα μεταξύ τους(π.χ. δεν μπορεί να είναι και τα 2 αρνητικά) :wink:
Θα χαρώ να δω μια ολοκληρωμένη λύση που χρησιμοποιεί το παραπάνω   :)

[quote="cretanman"]Xwris na kserw ton tropo me ton opoio doulepse o kos Louridas tha ithela na anaferw oti pragmati thetontas $f(x)-y=u$ kai $f(x)+y=v$, vgazeis an thimamai kala sto telos oti $f(v)-v^{2}=f(u)-u^{2}$. Omws apo edw den mporeis na isxiristeis (opws swsta episimainei kai o bilot4) oti gia tin sinartisi $g(x)=f(x)-x^{2}$ isxiei $g(v)=g(u) \forall v,u \in \mathbb{R}$ dioti ta $v,u$ den einai aneksartita metaksi tous alla ta sindeei mia sxesi.

Gia tin istoria kai mono na anaferw oti o Alvanos arxigos edwse mia lisi stirizomenos se auto kai i opoia dothike kanonika se olous tous vathmologites kai stous arxigous kai h opoia telika aporrifthike (swsta) ws lanthasmeni to apogeuma tis imeras tou diagwnismou. Einai ontws poli lepto to simeio autis tis askisis i opoia den einai katholou tetrimmeni. Den kserw ton tropo me ton opoio mporoume na apallagoume apo to provlima auto me tin parapanw lisi. Isws na exei kati diaforetiko o kos Louridas sto mialo tou (to opoio pisteuw oti tha kataligei se isodinami lisi me tin episimi).

Alexandros[/quote]

[b]Είναι όντως περίεργη άσκηση ! [/b]

 Διότι από πρώτη άποψη , αν σταθεροποιήσεις το $x$ και αφήσεις το $y$ να διατρέχει το IR, τότε τα $u,v$  διατρέχουν το IR. Ωστόσο αυτό δεν αρκεί , διότι μετξύ τους υπάρχει σύνδεση. Ίσως ένα αντιπαράδειγμα θα φώτιζε την κατάσταση!

 Μπάμπης

Παιδιά έχετε απόλυτο δίκαιο .Τώρα άνοιξα το <<ελληνικά>> και είδα όλες αυτές τις συζητήσεις περί της δεύτερης λύσης . Για την πρώτη λύση όμως δεν σχολίασε κανείς ....Πείτε μουτη γνώμη σας .(όπου D ουσιαστικά εννοώ Ιm f)

Exw kai mia akoma aporia epi tis askisis....Wraia vrikame oti i f(x)=x^2+c kai i f(x)=0  ikanopoioun tin sxesi...Giati omws prepei na stamatisoume edw?Den prepei na diksoume pws einai OLES oi sinartiseis pou tim ikanopoioun?Pou kseroume oti den iparxei kai alli f(x) pou ikanopoiei tin arxiki mas sxesi?

an apodeikniame mono auto tha kaname mono mia apli epalitheusi...gi auton akrivos to logog ginetai oli auti i istoria mexri na kataliksoume se lysi.s auto pou les dld exeis apolyto dikio alla to exoume kanei (p.x. sti lysi tou silouanou gt stin alli den to kataferame)

Pou apodiktike oti i f(x)=x^2+c kai i f(x)=0 einai kai oi monadikes liseis?Toulaxiston egw vlepw pws apodikniete pws einai liseis....
 Siga siga oso psaxnete i askisi olo kai kati vriskoume....Mia iposimiosi akomi...Apodikniete,an thelei kaneis tha to grapsw se epomeno quote, oti i f(0) = 0 diladi mia sinartisi pou ikanopoiei tin arxiki mas sxesi einai i f(x)=x^2

[quote="Nick Rapanos"][quote="SorrowMan"][quote="mostel"]nai, h $x^{2}$ einai profanhs... me to "mati" opws leei kai o RAKAR

 :D[/quote]

Mafto pou eipa , ennow pws apo tis f(x)=x^2+c pou grapsame , mono i f(x)=x^2 einai lisi kai kamia alli :)[/quote]

 :blush:  $f(x)=x^{2}+1$ ena paradeigma[/quote]

An valeis afti tin sinartisi mesa stin arxiki sou sinartisi kataligis sto oti prepei na isxiei  y^2 =4*y*x^2 ...Giati na isxiei afto?

[quote="silouan"]Ρε παιδιά καθήστε καλά .....
Η λύσεις είναι $f(x)=0$ και $f(x)=x^{2}+c$ όπου $c=f(0)$ .
Η πρώτη προφανώς επαληθεύει .
Για τη δεύτερη έχουμε και λέμε :
$(f(x)+y)^{2}+c=(f(x)-y)^{2}+c+4f(x)y$ ή $4f(x)y=4f(x)y$  που ισχύει........... :)[/quote]

Exete dikio :blush:  ekana 4 sinexomenes fores akrivos to idio arithmitiko lathos kai mou evgene oti prepei to f(0) = 0 !!!!!Signwmi :maybe:


$5!$, $7!$, $9!$, and $11!$ all end in zero, so they are out of this sum. $1!$ ends in 1, and $3!$ ends in 6, so 1+6=7.

[hide]
1! ends in 1
3! ends in 6
5! ends in 0
7! ends in 0
9! ends in 0
11! ends in 0
so add them all up and it ends in a 7 [/hide]

[quote="joejia"]What is the units digits of $1!+3!+5!+7!+9!+11!$ ?[/quote]

[hide]Since 5! ends in a zero, 7!, 9!, and 11! all end in zero. Therefore, the only thing we care about for the units digit is the units digits of 1! and 3!, which are 1 and 6, respectively. Therefore the answer is 7.[/hide]

[quote="nemesis"]can anyone post other problems[/quote]

Be patient, the forum is sort of dead right now...  :D


Given $ n\plus{}2$ integers, show that there exist two of them whoe sum, or else whose difference, is divisible by $ 2n$.

[hide="I think this works"] We find the maximum value for the number of terms in such a set.

Consider the set modulo 2n. Note that if we use the same residue twice, then the difference of the two terms is divisible by 2n, so every residue must be distinct. If some term is congruent to b mod 2n, where b is not 0 or n, then there is a corresponding residue that we cannot put in the set.

Thus, the maximum number of terms in the set is 2+n-1=n+1<n+2, so the proof is complete.[/hide]

[quote="BanishedTraitor"]Given $ n \plus{} 2$ integers, show that there exist two of them whoe sum, or else whose difference, is divisible by $ 2n$.[/quote]
[hide="Hint"]
Let the holes be remainders of $ n$ so that they a minimum in absolute value (meaning $ 0,\pm 1,\pm 2,\pm 3,...,\pm (n\minus{}1)$).
[/hide]

I don’t think there is always a difference; for example, the set {1,2,3,4,5} for $ n \equal{} 3$ doesn’t have any pair of numbers whose difference is divisible by 6. (I presume we want [b]distinct[/b] pairs of numbers, otherwise the whole problem is just trivial.) However, there is always a sum which is divisible by $ 2n$.

[hide="This is my proof."]A sequence of $ n \plus{} 2$ integers always contains at least one which is divisible by $ n$. Say, this number is $ kn$. If $ kn$ is not the first or last number in the sequence, then $ kn \minus{} 1$ and $ kn \plus{} 1$ are in the sequence and their sum is divisible by $ 2n$. If $ kn$ is the first number in the sequence, then both $ (k \plus{} 1)n \minus{} 1$ and $ (k \plus{} 1)n \plus{} 1$ are in the sequence (since the sequence is long enough); the sum of these two, then, is divisible by $ 2n$. Similarly, if $ kn$ is the last number in the sequence, then both $ (k \minus{} 1)n \minus{} 1$ and $ (k \minus{} 1)n \plus{} 1$ are in the sequence and we can take their sum again.[/hide]

How does the Pigeon-Hole Principle work when solving combinatorics problems?

gauss: Look at Catalyst's proof. Basically, you show that there are only $ n\plus{}1$ "holes" available that allow you to prevent a sum/difference divisible by $ 2n$.

Jane: the $ n\plus{}2$ integers can be anything, not necessarily consecutive, so there does not have to be a number divisible by $ n$.


Putnam
AMC
USA(J)MO
USAMO
AIME
geometry
projective geometry

I'm just curious. One guy who got silver medal in IPhO said that IPhO can be easily solved if I'm a 'A' physics major student in junior of college. How about IMO(-maybe after undergrad?) or Putnam(-maybe after grad school?)?

i don't think IMO and Putnam work exactly the same way.  they deal with very challenging problems using not more than the high school level (IMO) or undergraduate college level (Putnam) of mathematics, but they require a very high level of problem solving.  i imagine that most people with degrees in math could not solve the majority of IMO and Putnam problems.  IPhO is similar, but it is true that having been through a significant amount of college physics classes will give you insights into the problems that most people at the high school level do not have.

The way somebody has explained to me, they told me you only need to really know your stuff to get to Physics Camp while in Math you need to really know your stuff, study a whole ton, and then also have tons of natural ability to get into MOSP.  Here is some of my speculation, I hope it doesn't offend anybody.

People have been doing math for all of their lives.  I would suspect that a far greater number of people study intensely for math contests than for physics contests.  People have so much more math knowledge than physics knowledge considering that the first year of physics you get could be junior year in high school.  Thus, not nearly as many people are seriously comfortable with the basics in physics.  Knowing this, contest writers aren't going to make physics problems that require knowing every trick in the book to solve.  Thus, people who have strong math backgrounds and know their physics pretty well can go very far.  In math, people have already seen so many problems.  Old IMOs, USAMOs, AoPS, etc.  The competition is much larger and has existed for so long.  Thus, problem writers are forced to make problems that continue to challenge those of us who have seen so many old problems and basically know the precalculus subjects really darn well.  That is why after getting an undergraduate degree, I don't expect IMO to get much easier.  They really aren't testing if you know the material, they are testing if you are smart enough to apply that material and find that particular cleverness or trickiness.  Maybe by then you've seen more standard tricks, but the effect is not nearly as big and is not the primary goal of learning undergraduate math.

This year taking the semifinals in physics, I had a much more solid understanding of physics (I hadn't really trained directly for contests), and the problems on the test all seemed doable.  It was just an issue of time, not of not being able to see how to do the problems.

I'm editting this to mention what the point of that last bit was.  I feel that AMC, and AIME problems probably are similar in that they aren't too bad given that you know the topics fairly well and had unlimited time.  USAMO and IMO problems can easily stump me for months even if every step of a correct solution is understandable.  It seems to be less a game of recognizing the best technique and relevant theorems and more of a creative endeavor.

I've met many good mathematicians who can't solve IMO problems

[quote="filletwho"]hmmm...personally, I've always felt math competitions had more depth than competitions in other subjects. This is just probably an elitist attitude, and people of other subjects probably feel the same way about their own subject. However, I can't help feeling that in math, it's more of a journey while other subjects involve cramming and memorizing information a month before combined with one or two practice exams. Any thoughts?[/quote]
This is why I enjoy math, and hate geography, language arts, and other subjects. Math has its unique characteristic of challenging a person to find new things with the tools they have. Geography and other subjects is simple memorization.

In my own case, after having gotten a Ph.D. in mathematics and teaching in college for more then 25 years, I'm far, far better at solving Putnam problems than I was when I was an undergraduate and taking the Putnam, and I'm far, far better at solving AMC/AIME type problems than I was when I took the AHSME myself. In neither case am I anywhere close to being as good (or as quick) at solving those problems than was my own son when he was in high school and when he was an undergraduate.

None of which contradicts anything that pdiao, manuel, or filletwho just said.

Actually, maybe what I posted before is all junk.  I have certainly seen problems that have been like, "oh!  p-adic analysis" or "glad I recognized this is just finite projective geometry" or something like that.  Problems often have inspiration from more advanced topics.  As with all topics, how deeply you understand something can vary quite a bit.   Maybe math isn't so different, the playbook is maybe just bigger.  Number 6 was just a spiral similarity, use isolated fudging, etc.  Aren't they just other standard expert tricks that are memorized and learned?  Maybe IMO is a creative endeavor for me since I just haven't mastered it yet.  I hope (and assume) math research at least wouldn't just eventually become standard but would always remain creative.  Whoa, this is getting way off topic.


Let be $ a,b,c\in \mathbb{R}_\plus{}$ such that $ a>b\minus{}1\ ,\ b>c\minus{}1\ ,\ c>a\minus{}1$ . Prove that :

$ \sum_{cyc}\frac{9a^6}{(1\plus{}b\minus{}c)^4}\ge (a^2\plus{}b^2\plus{}c^2)^3$

It's the known barra's inequality. He posted it 4 years ago. :wink:

Can you give me a link , arqady ? I'd like to see a solution .  :)

Here is a link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63663


function
analytic geometry
graphing lines
slope
calculus
derivative
calculus computations

oooo, man do i feel stupid. thanks for your help.

[quote="mr.yellow"]hi im stuck on this one question so im wondering if anyone can help
heres the question

Consider the function f(x) = (5x^3) - 3x on the interval [ -2 , 2 ]. Find the average or mean slope of the function on this interval.
By the Mean Value Theorem, we know there exists at least one c in the open interval
( -2 , 2 ) such that f'( c) is equal to this mean slope.
For this problem, there are two values of c that work. find the smaller one and the larger one.

ok so im stuck on the last part trying to get the c values. for the mean slope i came up with 17. when trying to find the c's i used the equation (15c^2)-3 and solved for c which gave me sqrt(4/5) and -sqrt(4/5) which i think is incorrect.

so if anyone can help me it would be great. thanks[/quote]

mr. yellow,

You were correct in the first place.

$ \text {For the function } f(x) = 5x^3-3x\text{ the mean slope is } 17.\\
f'(x)=15x^2-3\\
\\
\therefore f'(c)=15c^2-3=17\\
\\
15c^2-20=0\\
\\
c^2=\frac{4}{3}\\
\\
c=\pm\sqrt{\frac{4}{3}}\\
\\
$
The point of the problem is that the derivative is quadratric (has 2 roots), the mean-value theorem guarantees $\textbf{at least one}$ point $c$ in the interval, and not all solutions to all problems are integers. If the answer is $c=\pm\sqrt{\frac{4}{3}}$, then sometimes you just have to leave it that way.


You don't have to give all your work. Just the basic structure and thought process.

[quote="Inspired By Nature"]I came across a problem today and I was wondering how many different kinds of proofs there are for:

RMS+HM ≥ GM+AM[/quote]

Sorry, but I don't get your point... what are you trying to prove?

oh right  :oops: . For two positive integers a and b, prove that their
RMS+HM ≥ GM+AM

[b]if[/b] this is too hard, could one of the mods please move this to pre-olympiad?

Hmm... I've actually never seen that inequality before.  Are you sure it's true?

Yes, it is true.  It was on a Mandelbrot team round once.  I think they also have this problem in AoPS V2.

[quote="Inspired By Nature"]I came across a problem today and I was wondering how many different kinds of proofs there are for:

RMS+HM ≥ GM+AM[/quote]

What's $RMS$ :?:

[quote="M4RI0"]What's $RMS$ :?:[/quote]

$RMS$ stands for $\text{Root-Mean-Square}$:

$\sqrt{ \frac{a_1^2 + a_2^2 + ... + a_n^2}{n} }$

It has many real-world applications because it is analogous to finding the "representative" velocity of an ideal gas based on a summation of the gas's kinetic energy.  :)

Essentially the inequality to be proved is this:

\[ \displaystyle \sqrt{\frac{a_1^2 + \ldots + a_n^2}{n}} + \frac{n}{\frac{1}{a_1} + \ldots + \frac{1}{a_n}} \geq \frac{a_1 + \ldots + a_n}{n} + \sqrt[n]{a_1 \ldots a_n}  \]

It seems he would be satisfied with the case $n = 2$, though.

For the two variable case: using very tedious algebra (and I cheated by using a TI-89 to expand some large binomial powers)

We wish to show $\sqrt{\frac{x^2+y^2}{2}} + \frac{2}{\frac1x+\frac1y} \geq \sqrt{xy} + \frac{x+y}{2}$ for $x,y>0$

Because both $x$ and $y$ are greater than zero and all the means are greater than zero, squaring is a reversable step.

$\sqrt{\frac{x^2+y^2}{2}} - \sqrt{xy} \geq \frac{x+y}{2} - \frac{2}{\frac1x + \frac1y}$

$\sqrt{\frac{x^2+y^2}{2}} - \sqrt{xy} \geq \frac{x+y}{2} - \frac{2xy}{x+y}$

$\sqrt{\frac{x^2+y^2}{2}} - \sqrt{xy} \geq \frac{(x+y)^2 - 4xy}{2(x+y)}$

$\sqrt{\frac{x^2+y^2}{2}} - \sqrt{xy} \geq \frac{x^2+2xy-4xy+y^2}{2(x+y)}$

$\sqrt{\frac{x^2+y^2}{2}} - \sqrt{xy} \geq \frac{x^2-2xy+y^2}{2(x+y)}$

$\sqrt{\frac{x^2+y^2}{2}} - \sqrt{xy} \geq \frac{(x-y)^2}{2(x+y)}$

$\frac{x^2+y^2}{2} + \frac{xy} - \sqrt{2x^3y+2xy^3} \geq \frac{(x-y)^4}{4(x+y)^2}$

$\sqrt{2x^3y + 2xy^3} \leq \frac{x^2 + y^2}{2} + \frac{2xy}{2} - \frac{(x-y)^4}{4(x+y)^2}$

$\sqrt{2x^3y+2xy^3} \leq \frac{(x+y)^2}{2} - \frac{(x-y)^4}{4(x+y)^2}$

$\sqrt{2x^3y+2xy^3} \leq \frac{2(x+y)^4-(x-y)^4}{4(x+y)^2}$

$2x^3y + 2xy^3 \leq \frac{4(x+y)^8 - 4(x^2-y^2)^4 + (x-y)^8}{16(x+y)^4}$

$(2x^3y+2xy^3)(16(x+y)^4) \leq 4(x+y)^8 - 4(x^2-y^2)^4 + (x-y)^8$

Expanding.. (I used a calculator here but it's perfectly possible by hand)

$32x^7y + 128x^6y^2+224x^5y^3+256x^4y^4+224x^3y^5+128x^2y^6+32xy^7 \leq x^8 + 24x^7y + 156x^6y^7 + 168x^5y^3 + 326x^4y^4 + 168x^3y^5 + 156x^2y^6 + 24xy^7+ y^8$

$0 \leq x^8 - 8x^7y + 28x^6y^2 - 56x^5y^3 + 70x^4y^4 - 56x^3y^5+28x^2y^6-8xy^7+y^8$

$0 \leq (x-y)^8$ which is true by the trivial inequality.

[quote="Inspired By Nature"]thanks for the input guys, appreciate it.   :)
Hamster1800, using the TI-89 wasn't necessary, it was all possible by hand. But it would take up a page depending on your handwriting, so no worries :P.[/quote]

I realise that I didn't need to but I didn't feel like expanding it manually and the TI-89 was so much faster.

Now, how do we prove the general case?

This is true for $n$ variables, isn't it? More generally $f_j+f_k\ge f_r+f_s$ with $j>r\ge s>k$ and $j+k=r+s$?  ($f_j$ being the $j$-th power mean)

I think it is a consequence of my generalisation of a problem from the Dutch university competition. (haven't checked in detail though)

[quote="Peter VDD"]More generally $f_j+f_k\ge f_r+f_s$ with $j>r\ge s>k$ and $j+k=r+s$?  ($f_j$ being the $j$-th power mean)[/quote]

Karamata, perhaps?


(This is #7 of the alternate AIME from 2000)

Calculate the value of 1/(2!*17!) + 1/(3!*16!) + ... + 1/(9!*10!).

[Note: Answer need not be a positive integer from 1 to 999.  This makes it a little harder.]

Okay, so, first of all you have typos.  But other than that, you're correct.  Those numbers aren't that hard to deal with at all.


Find the indicated value of the function.

http://img370.imageshack.us/img370/9883/aaaxp4.jpg

the answer is a number ????

[hide]
$ f(x) = \sqrt{x+8}-x+2$

Solve $ f(\sqrt{2}-2)$

To solve this, just subtitute $ \sqrt{2}-2$ in for $ x$ in $ f(x) = \sqrt{x+8}-x+2$.

$ f(x) = \sqrt{x+8}-x+2$

$ f(\sqrt{2}-2) = \sqrt{(\sqrt{2}-2)+8}-(\sqrt{2}-2)+2$

and simplify  :wink: [/hide]


abstract algebra
Ring Theory
advanced fields
advanced fields unsolved

I have some questions on the page pp51:

1. In the proof of Prop7.4, I cannot understand the last sentence:

"This shows that $ M_{\text swab{p}}$ is an $ S_{\text swab{p}}$-module of finite length equal to the number of times $ \text swab{p}$ occurs in the set $ \{ \text swab{p}_1, ... , \text swab{p}_r \}$";

2. In the proof of Thm7.5, how to prove that

" $ Z(AnnM'') = Z( \text goth{p} ) \cap H$ , where $ H$ is the hyperplane $ x_i = 0$ ";

3. How to understand the concept of "the presheaf kernel of $ \phi$", where $ \phi : \mathcal{F} \to \mathcal{G}$ is a morphism

of presheaves.

I am not having the book at hand at the moment I am afraid, so just a little hint to your third question:
[quote="foggy"]Why the presheaf kernel of $ \phi$ is a sheaf, where  is a morphism $ \phi : \mathcal{F}\to\mathcal{G}$?[/quote]Use the left exact sequence coding the sheaf axiom (cf., e.g., [url]http://en.wikipedia.org/wiki/Sheaf_%28mathematics%29#Sheaves[/url]). Note that taking kernels behaves nicely in the resulting diagram to get your claim.

I have solved the sheaf problem, but the questions of Hartshorne's still remain!

And something more on 1. and 2. thanks to google-books:
[quote="foggy"]I have some questions on the page pp51:

1. In the proof of Prop7.4, I cannot understand the last sentence:

"This shows that $ M_{\text swab{p}}$ is an $ S_{\text swab{p}}$-module of finite length equal to the number of times $ \text swab{p}$ occurs in the set $ \{\text swab{p}_1, ... , \text swab{p}_r \}$";
[/quote]
First a roadmap:
- Each $ \mathfrak{p}_i$ is a minimal homogeneous prime ideal containing $ Ann(M)$.
- If $ \mathfrak{p}$ and $ \mathfrak{q}$ are minimal prime ideals containing $ Ann(M)$, then $ (S/\mathfrak{p})_\mathfrak{q}\neq 0$ (forget about the grading here) iff $ \mathfrak{p} = \mathfrak{q}$.
- If $ \mathfrak{p}$ is a homogeneous prime ideal of $ S$, then $ (S/\mathfrak{p})_\mathfrak{p} = Quot(S/\mathfrak{p})$ is a simple $ S$-module. Again, forget about any grading here.

Use the facts above and localize the filtration given in the proposition at $ \mathfrak{p}$ to see that at every step the quotients have length 1 when $ \mathfrak{p} = \mathfrak{p_i}$ and 0 otherwise to calculate the length of $ M_\mathfrak{p}$ over $ S_\mathfrak{p}$.

[quote]
2. In the proof of Thm7.5, how to prove that

" $ Z(AnnM'') = Z( \text goth{p} ) \cap H$ , where $ H$ is the hyperplane $ x_i = 0$ ";[/quote]
Show that $ Ann(M)$ and $ (\mathfrak{p},X_i)$ have the same radical.


Let $p$ and $q$ be two positive integers that have no common divisor. The set of integers shall be partioned into three subsets $A$, $B$, $C$ such that for each integer $z$ in each of the sets $A$, $B$, $C$ there is exactly one of the numbers $z$, $z+p$ and $z+q$.

a) Prove that such a decomposition is possible if and only if $p+q$ is divisible by $3$.

b) In the case we omit the restriction that $p$, $q$ may not have a common divisor, prove that for $p \neq q$ the number $\frac{p+q}{\gcd(p,q)}$ is divisible by 3.

Easy part of a:
3|p+q => Z can be partitioned.
If 3|p then 3|q, so p,q are not coprime. So 3 divides neither p nor q.
Take A=3Z, B=3Z+1, C=3Z+2. It is easy to see that these sets are good for the problem.

For a)
Suppose WLOG: 0 \in A, p \in B and q \in C.
Consider p, 2p and p+q. Suppose 2p \in A ==> p+q \in C. So q and p+q \in C - contradiction. Thus 2p \in C and more general conclusion (!): 3kp \in A, 3kp+p \in B and 3kp+2p \in C. By the same method: 3nq \in A, 3nq+q \in C and 3nq+2q \in B.
It is necessary equations 3xp=3yq+q, 3xp=3yq+2q, 3xp+p=3yq+q, 3xp+p=3yq, 3xp+2p=3yq, 3xp+2p=3yq+2q must have no slutions in integers (x,y). It follows that p, q, 2p-2q, p-q, 2p and 2q are not divisible by 3 ==> 3|p+q.

For b)
It easy follows from a)


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I did search for this and couldn't find any; if you happened to find a thread that addresses these problems, just attribute it to my being a newbie  ;) .

How do you calculuate the integral of this: $\int xe^xdx$

I used partial integration where $u=x$ and $dv=e^x$ and got $xe^x-\int e^x$ then $xe^x-xe^x+C$ but this is just C. Where did I make the mistake?

Futhermore, what is the derative of $x{}^a{}^a$?
I did $x{}^(a{}^a)=e{}^(a{}^2)\log x$ and got stuck. Is it $2ax{}^(a{}^2)\log x$? BTW, if the laTex looks weird, I mean to say x^(a^a).

Hello, Sekein.

[quote="Sekein"]I used partial integration where $u=x$ and $dv=e^x$ and got $xe^x-\int e^x$ then $xe^x-xe^x+C$ but this is just C. Where did I make the mistake?[/quote]

$\int e^x dx=e^x+Const.$ :)

[quote="Sekein"]Futhermore, what is the derative of $x{}^a{}^a$?
an to say x^(a^a).[/quote]

$\frac{d}{dx} x{}^{a{}^a}=\frac{d}{dx}e^{\ln x{}^{a{}^a}}=\frac{d}{dx}e{}^{a{}^a\ln x}=e{}^{a{}^a\ln x}\cdot \frac{a^a}{x}=x{}^{a{}^a}\cdot \frac{a^a}{x}=a^a\cdot x{}^{a{}^a-1}$

Consequently this result is the same as plugg $p=a^a$ into the formula of $\frac{d}{dx} x^{p}=px^{p-1}\ (p\in{\mathbb{Q}})$.

kunny

Ah yes I knew the first one, that was a stupid mistake and I realized it before you posted. Thanks for the second one, I did not know that (sorry about using log and not ln, my calculus textbook treats log the same way as ln).

No problem. :)  We also use $log$ as $\ln$ in Japanese highschool class.

kunny

[quote="Sekein"]I did $x^(a^a)=e^(a^2)\log x$ and got stuck. Is it $2ax^(a^2)\log x$? BTW, if the laTex looks weird, I mean to say $x^(a^a)$.[/quote]

Use {}: x^{a^a} becomes $x^{a^a}$. You need the brackets whenever the expression you want to superscript/subscript/whatever is more than one character.


Here's a trigonometry problem I'm struggling on:
(3- (root6)) / 2 =  (sin^2 x)/ (1 sin 2x)[/u]

I will do a favor to everyone and rewrite the problem in LaTeX:

$\frac{3-\sqrt{6}}{2}=\frac{\sin^{2}x}{1+\sin(2x)}$

Is this right?

Edit:
Just to let you know, you really should use LaTeX.  Just put dollar signs around your math.

Oh, I'm reall sorry but I forgot the + sign. It is 1+ sin 2x


a lot like tabular

$\left( \begin{array}{c c c}
a&b&c\\
d&e&f\\
g&g&h\\
\end{array} \right)$

\left(
\begin{array}{c c c}
a&b&c\\
d&e&f\\
g&g&h\\
\end{array}
\right)

That's the way I do it

ledurt

If you want nicer spacing and simpler code, use bmatrix or pmatrix instead of array:
$\begin{bmatrix}a&b\\c&d\end{bmatrix}$
$\begin{pmatrix}a&b\\c&d\end{pmatrix}$


[size=150]QUESTION 1:

If a > 0, show that the solution set of the inequality x^2 < a consists of all numbers x for which -(sqrt{a}) < x < (sqrt{a}).

QUESTION 2:

If a > 0, show that the solution set of the inequality x^2 > a consists of all numbers x for which
x > (sqrt{a}) OR x < -(sqrt{a}).

Why does question 2 use the word OR to separate two inequalities?

Thanks.[/size]

[hide="Question 1"]Setting $x^{2}= a$ to find boundary points, we get that the boundary points are $x = \pm \sqrt{a}$. Putting these points on a number line and testing regions yields the desired solution.[/hide]

[hide="Question 2"]Use the exact same methodology as in question 1. This time, however, the intervals that are valid solutions turn out to be $(-\infty ,-\sqrt{a})$ and $(\sqrt{a}, \infty)$. Since any number in [i]either[/i] of these intervals will satisfy the inequality, we say that the solution consists of all numbers $x$ such that $x$ lies in the interval $(-\infty ,-\sqrt{a})$ OR the interval $(\sqrt{a}, \infty)$.[/hide]

the first one could also be written as $-\sqrt{a}< x$ and $x < \sqrt{a}$.  or is used because its not correct to write $-\sqrt{a}>x > \sqrt{a}$.

btw, whats up with the big font


[hide]
Each term has up to 3 x's and 3 y's, so the total is $4^{2}= 16$[/hide]

can you explain how [hide] 3x's and 3y's become 4^2? [/hide] sorry, but i don't understand. Thank you.

And how do you suppose a = 3 and b = 3 is attainable

[quote="xxxxx"]Find the number of terms in the simplified and expanded version of $(2x-3y+7)^{3}$[/quote]
[hide="I think I got this"]
Let's ignore the 7 for now, it adds only 1 term.
Since each term of the expanded form is a cubic term, the sum of the powers of x and y is 3.
So it is simply the number of solutions to
$x+y=3$ which is 4, plus the one for $7^{3}$ so the answer is 5.[/hide]

[quote="bpms"]
Let's ignore the 7 for now, it adds only 1 term. Since each term of the expanded form is a cubic term...
[/quote]

That's wrong! Try writing the expression this way: $(2x-3y+7)(2x-3y+7)(2x-3y+7)$. You can easily see that if I multiply the 7 with an x- and a y-term (and, sooner or later, I'll have to do this!), then I get a term of degree 2! :wink:

[quote="Ponnamperuma"][quote="bpms"]
Let's ignore the 7 for now, it adds only 1 term. Since each term of the expanded form is a cubic term...
[/quote]

That's wrong! Try writing the expression this way: $(2x-3y+7)(2x-3y+7)(2x-3y+7)$. You can easily see that if I multiply the 7 with an x- and a y-term (and, sooner or later, I'll have to do this!), then I get a term of degree 2!
t0rajir0u's reasoning is definitely correct!  :wink:[/quote]
[hide]Okay, let me rephrase that as
$x+y \leq 3$
has 10 solutions for nonnegative x and y[/hide]

Oops.  Wasn't thinking there.  bpms is correct.

Well, you could say that:
$7$ adds only one term, so you are left with $blahx^{3}+blahx^{2}y+blahxy^{2}+blahy^{3}+blahx^{2}+blahxy+blahy^{2}+blahx+blahy$. Wait, that means there are ten terms. Did I do something wrong? :?:
EDIT: Sorry bpms, skipped the bottom half. :oops:


27. How many three-digit numbers between 100 and 300 are there for which one of the digits equals the sum of the other two?

30. Two integers are randomly selected from the set of integers greater than or equal to -5 and less than or equal to 5. The two numbers need not be different. What is the probability that the sum of the two integers is less than their product?

Can you please clarify? For No. 27, say the number is $ abc$. Does only a = b + c (or any combination like that), or a = b + c, b = a + c, and c = a + b?

For the next one, I will give you a hint: use complimentary counting (if you don't know what it is, check it out on Alcumus).

akhil0422: All of your listed possibilities are valid. The wording generally implies so in these situations where it doesn't specify.

As for the second one, why complementary probability? The conditions are $ \minus{} 5\le x,y\le5$ where $ x,y$ are not necessarily distinct, and $ x \plus{} y < xy\implies(x \minus{} 1)(y \minus{} 1) > 1$. Complementary probability actually makes it more difficult (i.e. counting possibilities to $ (x \minus{} 1)(y \minus{} 1)\le1$).

My hint is to utilize that as long as $ (x,y)$ isn't $ (0,0)$ or $ (2,2)$, $ x,y\ne1$, and both $ x\minus{}1$ and $ y\minus{}1$ are the same sign (positive or negative), it works.

Ohhhhh, I get it now. I forgot to subtract (2,2) from the list.

As for the first problem, I kept on getting 34, because I forgot 211. I'm so careless. Thanks very much!


Over the past year, I've heard that, around this winter, the book [i]Algebra[/i] by Michael Artin would be having a new edition coming out soon, within a matter of months.  Can someone confirm or disaffirm this, or possibly even post a month?  Or if I've missed it somehow?

How would you learn this kind of information (referring to when it will be released) besides through the grapevine?  Who to ask?  Where to look?

Artin teaches at MIT, and for the past couple of semesters he's been using preprints of his new edition to teach his algebra class instead of the edition that's currently in print.  However, I don't actually know when he plans to release it.


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Is it possible to find an ordered and complete field that has $ \mathbb R$ as a proper subfield? If we don't need the archimedean property, I mean.

Take $ ^*\mathbb R$, the hyperreal numbers ( http://en.wikipedia.org/wiki/Hyperreal_number ).

But the hyperreals don't form a complete field, or do they?

Complete and ordered is enough; they imply the archimedean property. You have to drop the completeness property and substitute something weaker to get nonarchimedean ordered fields.

A simple proof:
We can derive the limit laws easily in a topological field of any kind; in particular, $ \lim ax_n\equal{}aL$ if $ \lim x_n\equal{}L$ and $ a$ is a constant, and $ \lim a_{n\plus{}1}\equal{}\lim a_n$ if either exists.
Now, what is the limit of $ a_n\equal{}\left(\frac12 \right)^n$? This sequence is decreasing and bounded below, so it has a limit $ u$.
Now, $ a_{n\plus{}1}\equal{}\frac12 a_n$; taking limits on both sides, $ u\equal{}\frac12u$, so $ u\equal{}0$. This is the archimedean property; there is a sequence of integers $ n$ so that $ \frac1n$ tends to zero, and any positive number must be larger than some $ \frac1n$.

The classic example of a non-archimedean ordered field: the field of rational functions. That is, the elements of this field are functions of the form $ f(x) \equal{} \frac {a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} a_0} {b_mx^m \plus{} b_{m \minus{} 1}x^{m \minus{} 1}\cdots \plus{} b_0},$ where $ a_n$ and $ b_m$ are nonzero (except for the zero element of the field). Identify elements as usual for forming a field of quotients of an integral domain.

Our definition of positive: $ f > 0$ iff $ \frac {a_n}{b_m} > 0.$ It's not hard to show that we have an ordered field. All we really need to to is to show that the sum of two positive is positive, the product of two positives is positive, and any element of the field is positive, negative, or zero.

Then the set $ \{1,2,3,\dots\}$ is bounded above by $ x.$ Or to get at jmerry's example, the sequence $ \frac1n$ is bounded below (and hence bounded away from zero) by $ \frac1x.$

This field is, of course, very far from being complete. Just one example: $ \{1,2,3\dots\}$ is bounded above but does not have a least upper bound. If $ b$ is any upper bound of this set, then so also is $ b \minus{} 1.$

Theorem: any two complete ordered fields are isomorphic, where the isomorphism is order-preserving as well as being a field isomorphism. I've heard this called "Dedekind's Theorem," although the specific language is not something Dedekind would have used. The proof is fairly laborious, and essentially replicates the work needed to construct the real numbers in the first place ([i]e.g.[/i], Dedekind cuts.) The argument jmerry gave points the way to one key element of the proof: the rational numbers must be dense in this field. (And "the rational numbers" makes sense in any field of characteristic zero.)

That is, up to order-preserving isomorphism, there is only one complete ordered field: the real numbers.


The following code (\|u\|_\infty) should produce $ \parallel u\parallel_\infty$ and not $ \|u\|_\infty$, am i right?

Note that (\|\cdot\|_\infty) produces $ \|\cdot\|_\infty$.

This bug has been reported at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1503472#1503472 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277944 along with a workaround.


Let $ x_{i}$ and $ y_{i}$ be real numbers such that

         $ x_{1}\leq x_{2}\leq\cdots\leq x_{n}$  and  $ y_{1}\leq y_{2}\leq\cdots\leq y_{n}$

Let $ z_{i}$ be a permutation of $ y_{i}$

Prove that:

$ \sum{(y_{i}\minus{}z_{i})(y_{i}\plus{}z_{i}\minus{}2x_{i})}\leq0$

 :)

Is it just me or is the LHS equal to zero?

[hide="it is just you  :D "]$ \sum{(y_{i}\minus{}z_{i})(y_{i}\plus{}z_{i}\minus{}2x_{i})}\equal{}\sum (y_{i}\minus{}z_{i})(y_{i}\plus{}z_{i})\plus{}(y_{i}\minus{}z_{i})(\minus{}2x_{i})\equal{}\sum y_{i}^{2}\minus{}\sum z_{i}^{2}\plus{}\sum(y_{i}\minus{}z_{i})(\minus{}2x_{i})\equal{}\sum(y_{i}\minus{}z_{i})(\minus{}2x_{i})\equal{}2\sum x_{i}z_{i}\minus{}2\sum x_{i}y_{i}\le 0$ by rearrangement[/hide]


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The question is the following:

We would like to factor $ N \equal{} 1659629823585814114799165699914477904363$ using pollard rho with the function $ f(x) \equal{} x^2 \plus{} 2$ and the seed $ x_0 \equal{} 13$. One notices that $ N \equal{} 19\cdot 87348938083463900778903457890235679177.$ After how many iterations will the algorithm stop?

Note that [b]no calculators[/b] can be used to answer the question. Is there a nice 'trick' involved?

Thanks!

Los

I read the article in Wikipedia. In step one you can choose a bound freely. Therefore I guess the answer is not unique.

Where the problem comes from? I have never done Pollard rho by hand so I was wondering.

This is wrong forum. Number theory Unsolved problems should be fine. You should read what kind of theorems are discussed in which forum and then choose the suitable.

hi.

one usually studies the algorithm in the context of an applied algebra class, so..

No, i think there is something that one should see here.

Run it mod 19. The sequence:
$ 13,13$
$ 0,2$
$ 2,0$
$ 6,6$
($ x^2\plus{}2$ cycles as $ 13,171\equiv 0,2,6,38\equiv 0,2,6,\cdots$)
Done, at step 3. Whatever we do mod $ N$, it reduces to those numbers mod $ 19$, and we get a gcd of $ 19$ at step 3.