geometry
trapezoid
symmetry
geometry proposed

Suppose $ABCD$ is a quadrilateral such that a semicircle with the centre at the midpoint of the segment $AB$ and bounding diameter lying on $AB$ touches the other three sides $BC$, $CD$ and $DA$. Show that $AB^2=4\cdot BC\cdot AD$.

First notice that the ABCD must be a trapezoid: AD parallel to BC.  Now let O be the center of the semicircle; in other words, O is the midpoint of AB. Then the equality to be proved becomes

$OB^2 = BC \cdot AD$

Let T be the point where the semicircle touches DC. By symmetry, DO bisects <AOT and CO bisects <BOT. Since <AOT + <BOT = 180 degrees, <DOC = 90 degrees. Now extend DO to meet CB at D'. Since O is the midpoint of AB and DA and CB are parallel, the result is that BD' = AD, so we need to prove that

$OB^2 = BC \cdot BD'$

Now, this is immediate from the fact that D'OC is a right triangle and in a right triangle, *the height to the hypotenuse is the mean proportional of the segments into which it divides the hypotenuse*.

-oh goodness


A certain iodide A on heating with a $ KOH$ solution produces a toxic gas B and a solution containing C. On ignition, B produces D which is a dehydrating agent. Identify A, B, C, D.

Just by curiosity, is A a phosphorus iodide?

Could $ A$ possibly be $ \ce{PH4I}$? Since we have the reaction:

$ \ce{PH4I} \plus{} \ce{KOH} \rightarrow \ce{PH3} \plus{} \ce{KI} \plus{} \ce{H2O}$

We know that phosphine is a toxic gas (so it's a likely candidate for $ B$). Then the dehydrating agent referred to might be $ \ce{P4O10}$. Just speculations.


[quote="darkdevil"]$a,b,c \in \mathbb{R_+}$ , $a+b+c=1 \Rightarrow \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \sqrt{\frac{27}{1+a^2+b^2+c^2}}$[/quote]
By Jensen Inequality for the function $f(x)=\frac{1}{x}$ (where x>0) (and devideing by 3)
$\frac{1}{3}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) \geq \frac{1}{\frac{1}{3}2(a+b+c)}$
It remains to prove then, that
$\frac{3}{2(a+b+c)}\geq \sqrt{\frac{3}{1+a^2+b^2+c^2}}$ (Then squaring and using 1=(a+b+c)^2)
$\frac{3}{4(a+b+c)^2} \geq \frac{1}{(a+b+c)^2+a^2+b^2+c^2}$
Which reduces to $a^2+b^2+c^2 \geq ab+bc+ca$ , which is true.

i found the solution lately  :blush:
we have $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq\frac{9}{2(a+b+c)}$ so ,$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq\frac{9}{2}$
and it's easy to prove $\frac{9}{2}\geq \sqrt{\frac{27}{1+a^2+b^2+c^2}}$ :roll:

[quote="eryaman"][quote="akhrif"]we have $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq\frac{9}{2(a+b+c)}$[/quote]Could you explain to me how you know this?[/quote]

Nesbitt's Inequality:

$\displaystyle \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}$

$\frac{a+b+c}{b+c} + \frac{a+b+c}{a+c} + \frac{a+b+c}{a+b} - 3 \geq \frac{3}{2}$

$\displaystyle \left(a+b+c \right) \displaystyle \left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right) \geq \displaystyle \frac{9}{2}$

$\displaystyle \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \geq \displaystyle \frac{9}{2(a+b+c)}$

By AM-HM inequality:
$[(a+b)+(b+c)+(c+a)](\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a})\geq 9$.

I think AM-HM is more basic.  :)


Find all the natural numbers $\ x,y$ so that the number $\frac{x^3+y^3-x^2y^2}{(x+y)^2}$ is natural.

Sorry, but the solution must be $\ x=y=2$ :)

Yes, I found when X integer $X=km^2+2k^2m+3k-(k^2+3km+k^2m^2)$. X is natural if and only if m=1.


logarithms
function
calculus
derivative
calculus computations

Ehm, so okay. I need to make a graph for y= x*ln(6) - x*ln(x)

I figured out the graph begins at an infinitely small point above 0 and cuts the x-axis at 6. What I need now is at which x value does the maximum value for the function exist. (Hope you can understand what I need, English ain't my first language, hehe - P.S. - I have no idea what Calculus is, so I hope you'll excuse me if it is the aforementioned question). I used a graph plotter to calculate the value, but I would like to know the theory behind it, e.g. how I could get the answer without resorting to an advanced computer.

By my logic, 0=[x(ln(6) - ln(x))]', which would make 1*(1/6 - 1/x)=0, making x=6 and this answer nonsense. Would be great if anyone could point out the faulty logic here, too.
[color=red]
[i]
mod edit: moved from Intermediate[/i][/color]

this board is not for calculus, and you made a mistake in the differentiation, think product rule... you have half, but the other half is missing, and $d_x(\ln{6}) \ne \frac{1}{6}$
and to make your posts easier to read, use dollar symbols around your math

Your attempt at taking a derivative is nonsense; there are several rules of differentiation ignored.
1. The derivative of a constant is zero. If $c$ is a constant, $\frac{d}{dx}cf(x)=cf'(x)$.
2. The product rule: $\frac{d}{dx}f(x)g(x)=f(x)g'(x)+f'(x)g(x)$
Calculate a correct derivative, and then come back. You're better off not factoring the expression for this.

Hmm. So $(a-b)' = a'-b'$
Since a is actually a product, then I follow the rule: $(a*b)'= a' * b + a * b'$
First side, e.g. x*ln(6):
$1 * ln(6) + x * 0 = ln(6)$
Second product:
$1 * ln(x) + x * 1/x = ln(x) + 1$

$ln(x)= ln(6) - 1$, or
$x=e^{\ln(6)-1}$ - correct?
(edit) should be x=e^ln(6) - 1, can't get it to format it correctly :(

[color=green]Moderator edit: that would be x=e^{\ln(6)-1}. If your exponent (or subscript, or denominator, or whatever) is more than one character, enclose it in brackets {}.[/color]

Also known as $x=\frac6e$.

You've got it right now.


The number +1 or -1 is written in each cell of a 2000 x 2000 array.  The sum of all the numbers in the array is nonnegative. Show that there are 1000 rows and 1000 columns such that the sum of the numbers at their intersections is at least 1000.

you may find the solution as following :[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=68971[/url]


[quote="math92"]Solve these questions:

1.  What is the least common multiple of 21 and 6?

2.  Solve this equation for x:

     $\sqrt{x}=\sqrt{3x+22}$

3.  Albatross has 4 blue shirts, 5 green shirts, and 5 black shirts.  What is the probablity that he will not pick a green shirt?[/quote]

[hide="Question 1"]
I just got it using a simple and quick guess and check. So the answer is 42.
[/hide]

[hide="Question 2"]
$\sqrt{x}=\sqrt{3x+22}$
$x=3x+22$
$2x=-22$
$x=-11$
[/hide]

[hide="Question 3"]
The probabillity that he [i]will[/i] pick a green shirt is $\frac{5}{5+5+4}$ or $\frac{5}{14}$

So the probablilty that he won't pick a green shirt is obviously$\frac{9}{14}$
[/hide]

Something's messed up with my cmputer, i can't see your $\LaTeX$
EDIT: i can't even see mine that i just wrote on the line above this


Let $ x,y,z \ge 0$ and $ x \le y\plus{}z$ , then $ \frac{x}{1\plus{}x}  \le  \frac{y}{1\plus{}y} \plus{} \frac{z}{1\plus{}z}$.

$ \frac{x}{1\plus{}x} \equal{} 1 \minus{} \frac{1}{1 \plus{} x} \le 1 \minus{} \frac{1}{1\plus{}y\plus{}z} \equal{} \frac {y \plus{} z}{1 \plus{} y \plus{} z} \equal{}\frac{y}{1\plus{}y\plus{}z}\plus{} \frac{z}{1\plus{}y\plus{}z} \le \frac{y}{1\plus{}y}\plus{}\frac{z}{1\plus{}z}$

Nice, my solution is ham-fisted in comparison(so is my typesetting):

$ x \le y\plus{}z \Rightarrow x\plus{}xy\plus{}xz \le (y\plus{}xy) \plus{} (z\plus{}xz) \le (y\plus{}xy\plus{}zy)\plus{}(z\plus{}xz\plus{}yz) \Rightarrow x\plus{}xy\plus{}xz\plus{}xyz \le (y\plus{}zy\plus{}xy\plus{}xyz)\plus{}(z\plus{}yz\plus{}xz\plus{}xyz) \Rightarrow x(1\plus{}y)(1\plus{}z) \le y(1\plus{}z)(1\plus{}x)\plus{}z(1\plus{}y)(1\plus{}x) \Rightarrow \frac{x}{1\plus{}x} \le \frac{y}{1\plus{}y} \plus{} \frac{z}{1\plus{}z}$

its equal to $ y \plus{} z \minus{} x \plus{} 2yz\plus{}xyz\geq 0$..

[quote="Athinaios"]its equal to $ y \plus{} z \minus{} x \plus{} 2yz\geq 0$..[/quote]
Well, I got $ y\plus{}z\minus{}x\plus{}2yz\plus{}xyz\geq 0$, but it's all good anyway...  :P


The sum of the reciprocals of two consecutive integers is $ \frac{5}{6}$. What is the sum of the two integers?

hello, from the problem we get
$ \frac{1}{n}\plus{}\frac{1}{n\plus{}1}\equal{}\frac{5}{6}$
solving this equation for $ n$ we get $ n\equal{}2$, the second solution is not an integer number, so we get
$ 2\plus{}3\equal{}5$
Sonnhard.

Alternatively, note that $ 2 \plus{} 3 \equal{} 5$ and $ 2 \cdot 3 \equal{} 6$, so the two numbers are 2 and 3, and the sum is 5.

ill answer this in a series of steps

[hide="this is my explanation"]1.when finding something like this we know the following facts

 I.the denominator of the 2 fractions when multiplied must equal to 6

 II.the numerator of the 2 fractions is additionally added from whatever you multiplied from the denominator

2.so the only 2#'s when multiplied and added to the numerator is 2 and 3

3.so we get the equation 2+3=5

4.[hide="ANSWER"]5[/hide][/hide]

[hide="."][size=200][color=#BF4000]PLEASE RATE MY POST THANK YOU SO MUCH!![/color][/size][/hide]

[size=50][b][i][u]please be fair when you're rating my post and please PM me if there is a better way to explain this thank you so much!![/u][/i][/b][/size]


Let the sum of a set of numbers be the sum of its elements. Let $S$ be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of $S$ have the same sum. What is the largest sum a set $S$ with these properties can have?

[hide]Let's start from the large numbers and go down.  We want 15.  We can also have 14, but that removes the possibility of 1.  We can have 13 as well, but then 12 and 2 are removed.  Since 12 has been removed we want to try 11 which works by the fact that the parity does not match for pairs and since these numbers are so close, 3 of them obviously can't equal one of them.  Allowing 11 removes 10 and 9 as well as 3, 4, 5.  However, since 5 is the brink where 3 numbers equals 2, six, seven, and eight should work.  But, picking any one of them invalidates the other two since 13, 14, 15 makes it impossible to have numbers that close together.  We want 8.  This gives 8, 11, 13, 14, 15 as our numbers for a total of 61.  Notice that these numbers are spaced apart by Fibonacci sequence distances.  Is this the answer or is my greedy approach incorrect?[/hide]

Kalva does the same and then proves that it is impossible to find a greater sum.

Bump. I found the approach in post #2 through the [hide="subset (spoiler)"]$\{8, 11, 13, 14, 15 \}$[/hide] but can someone point out why the [hide="concept (spoiler)"]Fibonacci relationship between the differences of these numbers[/hide] is true, perhaps along with a more rigorous solution? Thanks.

Bump. Can anybody answer my question? Thanks.

Bump. I would appreciate any help at all, I'm just curious to know a rigorous solution that proves that the [hide="spoiler"]greedy algorithm[/hide] works, as well as why the [hide="spoiler"]differences between the numbers resemble the fibonacci sequence[/hide].

I may just be bumping at the wrong times so people are not answering, but I just really want to know if there is a rigorous solution, and the reason for the [hide="spoiler"]fibonacci pattern[/hide], which I think is probably due to a [hide="spoiler"]recursion solution[/hide], but I'm not sure.

Bumping this again, this is an interesting problem and I want to know why the Fibonnaci pattern is there.

[quote=ArtyA6]*claps real slow* incredible bump insertionsort. Incredible bump.[/quote]

They had a legitimate question, and that is allowed as this is in contest collections and can be accessed very easily.

[hide=@#12][url=aops.com/community/user/555662][b]insertionsort[/b][/url] · 5 minutes ago [url=aops.com/community/p21199821](view)[/url][color=transparent]helo[/color]
[quote=HamstPan38825]I believe the Fibonacci pattern isn't consistent - for example, why isn't 3 the next element? It might have just have been a coincidence, but I can't be sure.

The greedy algorithm construction is sort of "rigorous" in the context that you know there won't be a greater sum, thus confirming your answer.[/quote]

Thanks, but greedy algorithm doesn't always work optimally, so you would still need to show that greedy does give the maximum sum (unless I'm misunderstanding what you are saying). ;) That is why I wanted a rigorous proof. However, I found out by asking somewhere else (because I didn't want to keep bumping the post) that there doesn't need to be rigor for this problem anyway; all that is required is confidence in intuition because it feels like greedy "should work" (the meaning of which is not really clear) based on coding experience.

In fact, oftentimes greedy algorithm doesn't work optimally compared to more efficient algorithms (consider DP/Knapsack problems). In this context, though, it's quite clear that it is the way to go, but in less obvious contexts choosing to use the greedy algorithm might be deceiving.

-----------
[color=#5b7083][aops]x[/aops] 12[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
Ah, so do you mean that hypothetically skipping the 14 and doing $\{15, 13, 12, 11, \text{more smaller elements}\}$ might yield a larger sum?

I believe if worst comes to worst, you can just try those cases yourself.

Edit: The wiki solution gives a bit of insight to why 61 is the maximum.


Audited will not turn in assignments.  You will be able to check it against textbook solutions if the book has them, or discuss them at Office Hours or on the message board if you like.

We'll send an email to the Independent Study participants in August asking which classes they'd like to enroll in; by then, the Complex Numbers/Trig class will be listed.  You don't have to pick your two classes when you sign up for Independent Study.

Wow, me and Rep are taking like the same exact classes this fall.  I am taking Intermediate Ind. study and my two classes will be the interm. counting and trig classes.  I think trig will kill me but I need to learn it.  To answer your question you choose the classes closer to the beginning of the class.  So you don't have to worry about that now.


function
limit
calculus
calculus computations

A function $f: (0,\infty) \to \mathbb{R}$ is defined as $f(x) = 1/x$. For a sequence $\{x_i\} \in (0,\infty)$ converging to $0$, $\lim_{n \to \infty} f(x_n) = \infty$. Why does it take infinity even though we know $x_i \neq 0$ for all $i = \{1,2,3, \ldots\}$?

To my understanding, $1/v \in \mathbb{R}$ for all $v > 0$. Isn't that right?

A sequence can have a limit without ever being that limit.

Are you willing to accept that $\lim_{n\to\infty}\frac1n=0?$ For all $n\in\mathbb{N},\,\frac1n\ne0.$ but yet the limit is zero.

To say that $\lim_{n\to\infty}x_n=\infty$ is to say that $x_n$ is [b]eventually[/b] [i]arbitrarily large[/i].

Here's the formal definition. We say this about the limit if and only if

for every $M$,  no matter how large [color=green][that's the "arbitrary" in "arbitrarily large"][/color]

there exists $N$ such that for every $n>N,$ [color=green][that's "eventually"][/color]

$x_n>M$ [color=green][that's the "large" in "arbitrarily large"][/color]

A sequence of finite numbers can tend to infinity in the limit; in fact, that definition was written under the assumption that each of the $x_n$ is finite. And a sequence of positive numbers can tend to zero in the limit, and so on.

I have another question: Every Cauchy sequence is bounded. Let  $\{a_n\}_{n=1}^\infty$ be a sequence where $a_n = \sum_{i = 1}^n \frac{1}{i}$. Since for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $s,t \geq \mathbb{N}$, $\|a_s - a_t\| < \epsilon$, so $\{a_n\}_{n=1}^\infty$ is a Cauchy sequence. But $a_n$ is a harmonic series thus $\lim_{n \to \infty} \|a_n\| = \infty$, and therefore unbounded. What is wrong?

$a_n$ is most certainly [b]not[/b] a Cauchy sequence!

Take any $n$. Look at $a_{2n}-a_n.$

$a_{2n}-a_n=\sum_{k=n+1}^{2n}\frac1k>\sum_{k=n+1}^{2n}\frac1{2n} =\frac{n}{2n}=\frac12.$

For an $\epsilon>\frac12,$ that's not less than $\epsilon.$


Are you taking it?
Goals?
Comments?
(if you're just going to discuss calculators, [u]please[/u] make a new topic.  :mad:)

Yes.  I'm taking the 12.  Goal is to be top in school again.  TI-89 Titanium, for the win. (That's not discussion of calculators, it's discussion of how I'm going to qualify for AIME)

I'm taking the 10A/B. My goal is to get $140\pm 5$ (it'll probably happen as long as I don't miss more than 1 problem).
Yes, I'll be prepared with my compass, ruler, graph paper, 5 pencils, spare lead, 2 erasers, and TI-89T. I'm more worried about AIME.

I took the AMC 10B and got 126, should've got 144 due to retarded mistakes :ninja:


inequalities
integration
logarithms
Cauchy Inequality
real analysis
real analysis unsolved

Holder's inequality states that if $ p,q$ are nonnegative extended real numbers with $ 1/p\plus{}1/q\equal{}1$ and if $ f \in L^{P}, g \in L^{q}$, then $ fg \in L^{1}$ and $ \int |fg| \leq \parallel{}f\parallel{}_{p}\parallel{}g\parallel{}_{q}$.

It can be proved without using Young's inequality as follows (some details omitted):
Consider the case where $ p,q>1$ and assume $ f,g \geq 0$. Let $ h\equal{}g^{q\minus{}1}\equal{}g^{q/p}$ so $ g\equal{}h^{p\minus{}1}\equal{}h^{p/q}$. Then $ ptf(x)g(x)\equal{}ptf(x)h(x)^{p\minus{}1} \leq (h(x)\plus{}tf(x))^{p}\minus{}h(x)^{p}$. Integrate both sides to see that $ pt \int fg \leq (\parallel{}h\parallel{}\plus{}t\parallel{}f\parallel{})^{p}\minus{}\parallel{}h\parallel{}^{p}$. Then differentiate (w.r.t. t) to see that $ p \int fg \leq p\parallel{}f\parallel{} \parallel{}h\parallel{}^{p\minus{}1}\equal{}p\parallel{}f\parallel{} \parallel{}g\parallel{}$.

For the case $ 0<p<1$, how can the proof be modified to get an analogous version of the inequality?

Perhaps an additional comment before that...
Probably Young's inequality will be an easier way to get the version of Holder's inequality for $ 0<p<1$, where the inequality sign is reversed (I suppose). But in this case, since $ 1/p\plus{}1/q\equal{}1$, $ q$ will be negative so does it still make sense to have $ \parallel{}g\parallel{}_{q}$?

Alright thanks.
I have figured out how to get the case $ 0<p<1$ from the case $ p \geq 1$ by using a few substitutions.

[quote="bookworm_vn"]Holder's inequality is a corollary of Cauchy inequality as following
\[ \alpha x \plus{} \beta y \geqslant x^\alpha y^\beta ,\alpha \plus{} \beta \equal{} 1,x > 0,y > 0.
\]
[/quote]

and if you take logarithms you get young's inequality ...


Hey everybody, I see that Vietnamese people got many troubles when discussing in Vietnamse here. So I send to all of you the link of Unikey programme.
 [color=red]Please choose TELEX and "Unicode" then click "Đóng" after downloading.[/color]
http://www.echip.com.vn/echiproot/Softwares/unikey/UniKey36.zip
Good luck. :D
It's very fast!!!!

If you don't have WinRAR, please try this link:

http://www.hanoisoftware.com/webplus/attachments/2bdec8422bded4201c622eb6a6a82eb6-UniKeyNT.exe

Thanks for your helping
now,I can type by vietnamese


geometry
inequalities unsolved
inequalities

Let 0 ≤ b ≤ 4.Show that the area included by the two curves y = 1 - | 1 – x | and
 y = | 2x – b |
can not exceed (1/3)

An easy one.

The first curve is completely symmetric, attaining its maximal value for $ x \equal{} 1$.
The second curve is placed above the $ x$ axis. Changing $ b$ merely translates the curve.

Take $ b\equal{}2$ then the area is equal to $ \frac{1}{3}$. Changing the value of $ b$ reduces the area which is easy to prove.


This #79 in Chapter 5 of AoPS 2.

A circle has the same ellipse and passes through the foci F_1 and F_2 of the ellipse. The two curves intersect at two points. Let P be any point of intersection. If the major axis of the ellipse has length 15 and the area of triangle PF_1F_2 is 26, compute the distance between foci.

Initially, I thought that the circle intersected the ellipse at x=0, and so, letting the center of the circle be O, triangle POF_2 was a 45-45-90 triangle. Since we knew that PF_2 was 15/2, we could easily compute OF_2, and, consequently, F_1F_2. However, my answer was wrong and, after reading the solution in the book, I saw how to do the problem a different way. But I still do not understand: how can the circle intersect the ellipse at 2 points if they do not intersect at x=0?

I don't think it's possible for a circle to pass through the two foci of an ellipse, intersect it in exactly 2 points, and have those points not be the ends of the minor axis.

Then how was my initial reasoning wrong?

Can you go back and check the wording of the question and make sure it's exact?  As written, it seems to me that the stated construction is impossible.

I think the construction works as stated... imagine a roughly circular ellipse with foci very close to one another.  Any circle with a sufficiently large radius that passes through those two points will intersect the ellipse twice.

I'm not sure what "a circle has the same ellipse" means, however.

The 'two points' should be 'four points'.  Sorry for the typo.  Duly noted & should get corrected in the next reprint...

Whoops, should have been "A circle has the same center as an ellipse." And yeah, makes sense now. Thanks Mr. Rusczyk and everyone else.


[quote="silouan"]If $a,b,c\geq 0$ then prove that
$a^{3}+b^{3}+c^{3}-3abc\geq 2 (\frac{b+c}{2}-a)^{3}$[/quote]
Nice inequality, silouan. :lol:
$a^{3}+b^{3}+c^{3}-3abc\geq 2 (\frac{b+c}{2}-a)^{3}\Leftrightarrow$
$\Leftrightarrow4a^{3}-4(b+c)a^{2}+2(b^{2}+c^{2})a+(b-c)^{2}(b+c)\geq0\Leftrightarrow$
$\Leftrightarrow(2a-b-c)^{2}a+(a+b+c)(b-c)^{2}\geq0.$ :)

this problem is the second problem 1st chapter in Vasc's book and it is very easy.
my solution is:
putting b=c with constant sum makes LHS smaller and RHS bigger
so we may assume that b=c
it is then equivalent to:
$a(a-b)^{2}\geq 0$ so we have equality at (k,k,k) and (0,k,k) :)
Vasc shows that $a^{3}+b^{3}+c^{3}-3abc-2(\frac{b+c}{2}-a)^{3}\geq \frac{3}{2}(b-c)^{2}(\frac{b+c}{2}-a)$ and divides in two cases
$\frac{b+c}{2}-a<>0$ both obvious.


AoPS Books
number theory
modular arithmetic

Hello,

I was wondering if anyone knows how to convert different numbers to different bases (in general). I have tried looking on Google and in the AoPs books, but they are not there. For example, how would you convert 555 in base 7 or 873 in base 5?

Or, if there is another explanation in a book or a site or something, please post them here. In addition, information/explanations and such on modular arithmetic would also be helpful, evn though I have already read the info included in the AoPs books, I still dont understand it completely..

Hello, splintercell990!

[quote]I was wondering if anyone knows how to convert different numbers to different bases (in general).
For example, how would you convert 555 in base 7 or 873 in base 5?[/quote]
Going [b]to[/b] another base requires a repeated-division algorithm.

To change 555 to base-7:

(1) Divide by 7; note the remainder:
$.\qquad\qquad 555 \div 7 = 79\;\;\;R = 2$

(2) Divide the [i]quotient[/i] by 7, note remainder:
$.\qquad\qquad\; 79 \div 7 = 11\;\;\;R = 2$

Repeat step (2) until a zero quotient is attained.
$.\qquad\qquad 11 \div 7 = 1\;\;\;R = 4$
$.\qquad\qquad\; 1 \div 7 = 0\;\;\;\;R = 1$

Now read [b]up[/b] the remainders: $1-4-2-2$

Therefore: $1422_7 = 555_{10}$

I think he was asking how to convert 5557 to base 10.  The idea is the same, although it would probably be best to think of 5557 as 5*7^2 + 5*7 + 5 = 28510.

If he's converting, say,  $3426_7$ to base 10, there's another trick available.

Write this out (but don't take it seriously!): $3 \times 7 + 4\times 7 + 2 \times 7 + 6$

You can run this on your calculator, from left to right  ... with one important warning:
$.\qquad Complete\;each\;addition\text{ before the next multiplication.}$

The baby steps look like this: $3\times 7 = 21$

Add 4: $21 + 4 = 25$

Multiply by 7: $25 \times 7 = 175$

Add 2: $175 + 2 = 177$

Multiply by 7: $177 \times 7 = 1239$

Add 6: $1239 + 6 = 1245\;\;\hdots\;There!\;\hdots\;3426_7 = 1245_{10}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

"Why does this method work?" you ask . . .

The usual evaluation is: $3\cdot7^3 + 4\cdot7^2 + 2\cdot7 + 6 $

Factor a 7 out of the first three terms: $(3\cdot7^2 + 4\cdot7 + 2)\cdot7 + 6 $

Factor a 7 out of the first two terms: $([3\cdot 7 + 4]\cdot7 + 2)\cdot 7 + 6 $

Reading from the inside, look at the steps that are performed:
. . . "3 times 7, add 4, times 7, add 2, times 7, add 6."
They are [b]identical[/b] to the steps in the shortcut method described earlier.

If you consider the last expression above and [i]drop the parentheses,[/i]
we have: $3\times7 + 4\times7 + 2\times7 + 6$ ... which neatly "outlines" the steps to take.

Just remember to complete each addition before continuing to the next mutlplication.
(On your calculator, press "=" after each addition.)

Hmm..I think I get it somewhat...also, after reading some material on modular arithmetic, this is what I understand:

Mod arithmetic is really "remainder arithmetic", or  that we are looking for the integer that occurs as a remainder when one integer is divided by another integer. For example:

1.) When 12 is divided by 7, there is a remainder of 5. Therefore,

[b]12 :equiv: 7 (mod 5)[/b]

2.) When 15 is divided by 4, there is a remainder of 3. Therefore,

[b]15 :equiv: 4 (mod 3)[/b]

3.) When 10 is divided by 4, there is a remainder of 2. Therefore,

[b]10 :equiv: 4 (mod 2)[/b]

Now, to perform addition of mods, you basically add the two numbers first, and then divide the sum by the modulus to compute the remainder.  To perform subtraction of mods, you first subtract, and secondly, compute the remainder.

Based on this info, we can do the following with congruences, if a:equiv: b (mod m):

[b]1.) a + c :equiv: b+c (mod m)
2.) a - c :equiv: b - c (mod m)
3.) ac :equiv: bc (mod m)
4.)a c :equiv: b c (mod m)
5.)(a + b) (mod m) :equiv: a(mod m) + b(mod m)
6.) ab (mod m) :equiv: [a(mod m) b(mod m)][/b]

Now, a couple of questions:

1.) Given that 12 :equiv: 7 (mod 5), what is 12+ 3 congruent to?

In the AoPS book it says the answer is 7+ 3 (mod 5). Can this be written as 10 (mod 5), or does the answer have to be 7+ 3 (mod 5)?

2.) Given that 12 :equiv: 7 (mod 5), what is 122 congruent to?

Can you say that the answer is 49 (mod 5), or do you have to say that the answer is 72 (mod 5)?

Can anyone please go through this post, and comment upon it, and tell me where I am wrong, and why? If you have any other suggestions, or links to other resources about mods, can you also please post them here. Once I have the basics covered, I would like to move on to more advanced material...

[quote]
1.) When 12 is divided by 7, there is a remainder of 5. Therefore,

12 equivalent 7 (mod 5)

2.) When 15 is divided by 4, there is a remainder of 3. Therefore,

15 equivalent 4 (mod 3)

3.) When 10 is divided by 4, there is a remainder of 2. Therefore,

10 equivalent 4 (mod 2)
[/quote]
Shouldn't it be 5 (mod 7) for the first one, 3 (mod 4) for the second one, 2 (mod 4) for the third one, and 2 (mod 4) for the last one?

I don't think so, because if you take a look on page 57 in AoPS Volume 1...under Exercise 5-14, that is where I got the first example...and im assuming the rest are okay, since they are in the same format. I may be wrong, but I am not sure :?

I think that if it was 5 (mod 7), then you would be saying that if you divide 12 and 7, the remainder would be 5, which is not the case...and the same with the rest... :?

It should be the other way around. a = b (mod c) means that a and b have the same remainder when dividing by c.

[quote]I think that if it was 5 (mod 7), then you would be saying that if you divide 12 and 7, the remainder would be 5, which is not the case...and the same with the rest...
[/quote]
It is.

7 * 1 + 5 = 12

Whoops :blush: ...thanks for pointing that out..

check out changing bases in Getting Started it might be helpful


Two cars on Crete left the towns of Knossos and Phaistos at the same time to drive to the other town, passing each other at Gortyns and both travelling at different constant speeds. The car from Knossos travels 20km/h faster than the car from Phaistos. THe distance from KNossos to Gortyns is 10km more than the distance from Phaistos to Gortyns. The car from Phaistos completed the journey from Gortyns in 35 minutes. Find the distance from Knossos to Phaistos.

Could someone verify to me that it is 130km, or if it is not could someone please give me a detailed expanation on how you arrived at you answer, thanks alot!

It is Physic's problem I think. you should use the theorems from Physic.

I'm not sure if it's right, but:

[hide]I'm going to call Phaistos P, Gortyns G, and Knossos K. Also, the car from K to G is going to be car 1, the car from P to G is going to be car 2.

We know
$d = rt$
$\Rightarrow t = \frac{d}{r}$

Let the distance from P to G be x. Thus, the distance from K to G is 10 + x. Let car 2's rate be r. Car 1's rate is then r + 20. We have
$\displaystyle \frac{x + 10}{r + 20} = \frac{x}{r}$
$\Rightarrow xr + 10r = xr + 20x$
$\Rightarrow r = 2x$

We also have
$\displaystyle \frac{x + 10}{r} = 35$
$\displaystyle \Rightarrow \frac{x + 10}{2x} = 35$
$\Rightarrow x + 10 = 70x$
$\displaystyle \Rightarrow x = \frac{10}{69}$

The distance from K to P is 2x + 10, so
$\displaystyle 2x + 10 = 2*\frac{10}{69} + \frac{690}{69} = \frac{710}{69}$[/hide]


geometry
3D geometry
number theory unsolved
number theory

$ a,b,c \in  \mathbb{N^*}$

$ \boxed{\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \in \mathbb{N}}$ then prove that $ abc$ is a perfect cube.

[hide]
$ \frac ab\plus{}\frac bc\plus{}\frac ca\equal{}\frac{ac^2 \plus{} ba^2 \plus{} cb^2}{abc}$ so

$ abc|a^2 c \plus{} b^2 a \plus{} c^2 b$.

Let $ p$ be an arbitrary prime. Suppose $ a\equal{}rp^{x}$, $ b\equal{}sp^{y}$, and $ c\equal{}tp^{z}$. With $ r,s,t$ not divisible by $ p$. Then

$ rstp^{x\plus{}y\plus{}z} | rtp^{2x\plus{}z} \plus{} srp^{2y\plus{}x} \plus{} tsp^{2z\plus{}y}$.

Now I will show that at least two of $ 2x\plus{}z$, $ 2y\plus{}x$, and $ 2z\plus{}y$ are equal. Suppose they are not; then one of them (assume without loss of generality, it is $ 2x\plus{}z$) is strictly less than the other two and strictly less than $ x\plus{}y\plus{}z$, since $ (2x\plus{}z)\plus{}(2y\plus{}z)\plus{}(2z\plus{}y)\equal{}3(x\plus{}y\plus{}z)$. Then

$ rtp^{2x\plus{}z} \plus{} srp^{2y\plus{}x} \plus{} tsp^{2z\plus{}y} \equal{} p^{2x\plus{}z}(rt \plus{} srp^{2y\plus{}x \minus{} (2x\plus{}z)} \plus{} tsp^{2z \plus{} y \minus{} (2x\plus{}z)})$

The right factor is then not divisible by $ p$, so the largest power of $ p$ dividing $ rtp^{2x\plus{}z} \plus{} srp^{2y\plus{}x} \plus{} tsp^{2z\plus{}y}$ is $ p^{2x\plus{}z}$, but $ 2x\plus{}z<x\plus{}y\plus{}z$, so $ rstp^{x\plus{}y\plus{}z}$ cannot divide $ rtp^{2x\plus{}z} \plus{} srp^{2y\plus{}x} \plus{} tsp^{2z\plus{}y}$.

Thus at least two of $ 2x\plus{}z$, $ 2y\plus{}x$, and $ 2z\plus{}y$ are equal, say $ 2x\plus{}z$ and $ 2y\plus{}x$. So $ 2x\plus{}z\equal{}2y\plus{}x\implies x\plus{}z\equal{}2y$ so $ x\plus{}y\plus{}z\equal{}3y$ is divisible by $ 3$. So any prime dividing $ abc$ divides it a number of times divisible by $ 3$ so $ abc$ is a cube.
[/hide]


Given a $ (2m\plus{}1) \times (2n\plus{}1)$ checkerboard where the four corner squares are black, show that if one removes any one red and two black squares, the remaining board can be covered with dominoes (1 $ \times$ 2 rectangles).

Do I just have to show that the number of remaining red and black squares are equal, because that seems way too easy. Or do I have to also show that even with the even number of squares left, dominoes can be placed because we can still have a configuration that doesn't work with an even number of squares?

How do I go about?

[quote="1/(ln x)"]Do I just have to show that the number of remaining red and black squares are equal[/quote]

This is a necessary, but not sufficient, condition.  To show that a tiling is possible, you'll usually need to actually describe it and then prove it always exists.


calculus
integration
function
real analysis
real analysis unsolved

f(x) is a function finite on (0, +inf), and monotone decreasing.
Then for any p>0,
Prove: Int{0->+inf}(x^(p-1)*f(x))dx <+inf
iff        Sum{k=-inf to +inf}(2^(kp)*f(2^k)) <+inf.

So, you want us to prove that

$\int_{0}^{+\infty}x^{p-1}f(x) \, dx <+\infty \Leftrightarrow \sum_{k =-\infty}^{+\infty}2^{kp}f(2^{k}) <+\infty$

What I think you mean is

$\int_{0}^{\infty}x^{p-1}f(x) \, dx \neq+\infty \Leftrightarrow \sum_{k=0}^{\infty}\left( 2^{kp}f(2^{k})+2^{-kp}f(2^{-k}) \right) \neq+\infty$

Makes sense either way.
Break the integral into infinitely many pieces:
\[\int_{0}^{\infty}x^{p-1}f(x)dx=\sum_{k\in\mathbb Z}\int_{2^{k}}^{2^{k+1}}x^{p-1}f(x)dx\]
Since $f$ is decreasing, the integrand is between $2^{k(p-1)}f(2^{k+1})$ and $2^{(k+1)(p-1)}f(2^{k})$. Thus
\[2^{k(p-1)}f(2^{k+1})2^{k}\le \int_{2^{k}}^{2^{k+1}}x^{p-1}f(x)dx\le 2^{(k+1)(p-1)}f(2^{k})2^{k}\]
where $2^{k}$ is the length of the interval of integration. Sum over $k$.


Mr.John was killed; there are 2 suspects involved,X and Y. Speaking of which, there must exactly be a culprit among those two.

At the beginning,exactly the same, each of them had a fair chance of not being the culprit. Later, a blood-covered knife was found; AB was found to be the type of that blood.  If the blood type of X is also AB, and if there are only 10% of people throughout the world to have AB for their blood type.
1.What is the probability of X being the culprit?
2.What is the probability of Y having the same blood type as X?

...I was wondering if the answers must be $ \frac {19}{20}$ and $ \frac {1}{10}$, but well... I don't think that it would be that easy...Did I do something wrong? I'm not quite sure with myself. Help me please...Thank in advance.
 :(

1. In 90% of cases, Y won't have AB, so X is the culprit
 In 10% of cases, it's equal share, so X is the culprit in 5%
95%

2. 10% obviously


I simple had to press that **** cable into the calcutator. But your'e right, Billy. It is very unreliable!

:huh:  :huh:  :huh:  :huh:  no math question????

ummm, didnt you post the same exact thing in the TI-84 help (or something)

locked


calculus
real analysis
real analysis theorems

I was reviewing some material from last semester and picked a random theorem out of my notes to prove just to stay fresh. I picked the monotone convergence theorem and this is how the theorem was stated and how I proceeded.

Monotone Convergence Theorem:
A monotonic sequence converges if and only if it is bounded.

I went ahead and proved the first half for non-decreasing sequences:
NOTE: this was a first semester calculus course, so our sequence is from $ N\to R$ and the set in question is always the reals.

Assume that our sequence $ {a_n}$ is bounded above. Let $ sup(a_n)\equal{}L$ (we are guaranteed this by the least upper bound property of the reals). Since $ L$ is our least upper bound, $ L\minus{}\epsilon$ for some $ \epsilon > 0$ is not an upper bound on $ a_n$. Thus there exists some $ N_0$ in the naturals s.t. $ L\minus{}\epsilon < a_{N_0} \le L$. Since our sequence is non-decreasing, for $ n \ge N_0$ $ \Rightarrow$ $ L\minus{}\epsilon < a_{N_0} \le a_n \le L$. Then in my notes I write that this implies $ |a_n \minus{} L|<\epsilon$ for all $ n \ge N_0$. At the time it must have seemed self evident but I can't seem to get the basic algebra going to show this  :P  any help appreciated!

Uhh, if $ L \minus{} \varepsilon \leq a_n \leq L$ then surely the distance between $ a_n$ and $ L$ is less than $ \varepsilon$?

If you want, it trivially implies $ L \minus{} \varepsilon \leq a_n \leq L \plus{} \varepsilon$ which is equivalent to $ |a_n \minus{} L| \leq \varepsilon$

Yes thank you, I don't know why I felt that I needed to show that but it did follow trivially, so all is well  :) .


function
algebra
polynomial
real analysis
real analysis unsolved

Find an example of a differentiable function $ f\colon {\mathbb R}^2\to {\mathbb R}$ having exactly two critical points both of them maxima.

Impossible. If a differentiable function on $ \mathbb{R}^2$ has two maxima, it has a saddle point between them.

Sorry but i cannot agree with you. A six degree polynomial is known.

[color=green]Redundant quote removed. There is never any reason to quote the entire post immediately preceding yours.[/color]

An outline of the proof: Consider the graph of the function as a mountain range with two peaks. If we look at everything above a certain height, the two peaks start out separate, grow, and eventually merge when we lower that level enough. The point (or multiple points) at which the mountains join is another critical point.

Show me your alleged example, and I'll show you its third critical point.

$ f(x,y) \equal{} (x^2 \minus{} 1)^2 \plus{} (x^2y \minus{} x \minus{} 1)^2$ (the example goes back to 1988 and is not mine).

It would be interesting to have an example such that $ f(x,y)\equal{}f(\minus{}x,y).$ Maybe meanwhile somebody has found it.

Well, technically those are minima.

All right, the "mountain range" argument? They don't merge cleanly, because of the constant values on the entire $ x$ axis. That saddle point gets shunted off to $ \infty$, specifically $ (0,\infty)$.


[size=200][color=red][b][u]Steaua[/u][/b][/color] [color=darkblue][b][u]Bucuresti[/u][/b][/color] ,

[color=orange][b][u]www.mathlinks.ro[/u][/b][/color] and ......

[color=green][b][u]Roger Federer[/u][/b][/color] are the best ![/size]

I'm going with Darij. :D BTW, this is a pretty cool geometry theorem/formula.

all of them are good in geometry... ;)

virgil nicula is more kind...

hi dears...

wirjil nicula is the best but...

its not good to say who is good or who is bad.

this forum is for geometry problems.

is this topic a geometry problem?

I am agree with Ashegh they are all good.

Davron Latipov


probability
geometry
quadratics
algebra
quadratic formula

A disc of diameter D is thrown along a floor covered with 9 inch square tiles. (9 inch by 9 inch) The disc is a winner if its edges are completely within a square and don't intersect the edges of a square. If the probability of throwing a winner is 1/3, calculate D.

[hide]
My answer $ 2.25$.
[/hide]

[hide]we look at the position of the disc relative to the center. the center can be located in an area of $ (9\minus{}D)^2$. we have $ \frac{d^2\plus{}81\minus{}18D} {81} \equal{} \frac{1} {3}$
$ d^2\minus{}18d\plus{}81\equal{}27$
$ d^2\minus{}18d\plus{}54\equal{}0$
quadratic formula gives us $ 9\pm3\sqrt{3}$
we cant do $ 9\plus{}3\sqrt{3}$ because the probability of winnning is 0.
therefore we get $ 9\minus{}3\sqrt{3}$[/hide]

[quote="abacadaea"][hide]we look at the position of the disc relative to the center. the center can be located in an area of $ (9 \minus{} D)^2$. we have $ \frac {d^2 \plus{} 81 \minus{} 18D} {81} \equal{} \frac {1} {3}$
$ d^2 \minus{} 18d \plus{} 81 \equal{} 27$
$ d^2 \minus{} 18d \plus{} 54 \equal{} 0$
quadratic formula gives us $ 9\pm3\sqrt {3}$
we cant do $ 9 \plus{} 3\sqrt {3}$ because the probability of winnning is 0.
therefore we get $ 9 \minus{} 3\sqrt {3}$[/hide][/quote]

I think that is correct. Although you could have skipped use of the quadratic.
$ (9 \minus{} D)^2\equal{}27$
$ 9 \minus{} 3\sqrt {3}\equal{}D$

im stupid i cant answer it :blush:  :mad:

Please don't revive threads unless you have something useful, like a different approach to the problem.  If you need help with a problem, post a topic and we'll try to do our best.


calculus
integration
real analysis
real analysis unsolved

If the exact answer cannot be work out , how to do the approximation , I need the answer .

$\int_{0}^{1}\left(\int x^{x}\ dx\right)\ dx$

Use Fubini's theorem: $\int_{0}^{1}\left(\int_{0}^{x}t^{t}\,dt\right)dx=\int_{0}^{1}\left(\int_{t}^{1}dx\right)t^{t}\,dt =\int_{0}^{1}(1-t)t^{t}\,dt$. Then use some CAS, with a unsurprising result. [quote]Mathematica could not find a formula for your integral. Most likely this means that no formula exists.[/quote] Finally, use some rule for approximate integration (discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141704]here[/url]).


The sequence a_n is defined by a_1 = 20, a_2 = 30, a_(n+2) = 3a_(n+1) - a_n.
Find all n for which 1 + 5a_(n+1)a_n is a square.

it is not hard to show that $1+5a_na_{n+1}=(a_n+a_{n+1})^2+Something$ so from here the analisis is easy ;)


inequalities
geometry
circumcircle
IMO Shortlist
triangle inequality

Let $ABCD$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that \[\angle{PAB}+\angle{PDC}\leq  90^\circ\qquad\text{and}\qquad\angle{PBA}+\angle{PCD}\leq  90^\circ.\] Prove that $AB+CD \geq  BC+AD$.

[i]Proposed by Waldemar Pompe, Poland[/i]

Just consider the intersection point $ Q$($ \not=P$) wrt a circumcircle of $ APB$ and a circumcircle of $ CPD$.
Then a circle passing through the points $ A,Q,D$ and a circle passing through the points $ B,Q,C$ are also externally tangent at $ Q$,
and given condition reduced to $ AQD, BQC \ge 90$.
Now, the problem is easy.

Could anyone please elaborate on the above proof? I don't quite get it.

I can see why the circles through AQD and BQC are tangent only through a quick inversion through P, but can someone give a synthetic proof?

[hide][b]Lemma[/b]: Let $WXYZ$ be an arbitrary quadrilateral, and let $M$ and $N$ be the midpoints of $WZ$ and $XY$, respectively. Then $2MN \leq WX + YZ$.
[i]Proof:[/i] Consider the points as vectors. We have
\[ 2|M-N| = 2 \left |\frac{W+Z}{2} - \frac{X+Y}{2} \right | = |(W-X) + (Z-Y)| \leq |W-X| + |Z-Y|, \]
that is, $2MN \leq WX + YZ$, as desired. $\blacksquare$

Let $\omega_1$ be the circumcircle of $\triangle APD$ and let $\omega_2$ be the circumcircle of $\triangle BCP$. Let $E$ be the intersection of lines $AB$ and $CD$ (possibly at infinity), let $\theta = \angle AED$ (and set $\theta = 0$ if $E = \infty$), and suppose without loss of generality that $A$ lies between $E$ and $B$. Then we have
\[ \angle DPA = 180^{\circ} - (\angle PAD + \angle ADP) = (\angle BAD + \angle ADC - \theta) - (\angle PAD + \angle ADP) = \angle BAP + \angle PDC - \theta \leq 90 - \theta, \]
and
\[ \angle CPB = 180^{\circ} - (\angle PBC + \angle BCP) = (\angle ABC + \angle BCD + \theta) - (\angle PBC + \angle BCP) = \angle ABP + \angle PBC + \theta \leq 90 + \theta. \]
Now consider the circles $\Gamma_1$ and $\Gamma_2$ such that the directed angle $\angle DXA$ equals $90^{\circ} - \theta$ for all $X \in \Gamma_1$, and the directed angle $\angle CXA$ equals $90^{\circ} + \theta$ for all $X \in \Gamma_2$. We claim that $\Gamma_1$ and $\Gamma_2$ cannot intersect twice. Suppose for the sake of contradiction there exists some point $P'$ lying on both $\Gamma_1$ and $\Gamma_2$. Since $\angle DP'A \geq \angle DPA$, $P'$must lie within $\omega_1$, and since $\angle CP'B \geq \angle CPB$, $P'$ lies within $\omega_2$. But since $\omega_1$ and $\omega_2$ are tangent at $P$, we must have $P' = P$, so our claim is proven.

We now have two cases:
[list][*][b]Case 1[/b]: $E = \infty$. We have $\angle DPA = \angle CPB =90^{\circ}$. Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively. Because $\Gamma_1$ and $\Gamma_2$ have diameters of $AD$ and $BC$, they are centered at $M$ and $N$, respectively, and have radii of $\frac{AD}{2}$ and $\frac{BC}{2}$, respectively. Since $\Gamma_1$ and $\Gamma_2$ do not intersect intersect, we have $MN \geq \frac{AD+BC}{2}$, (since $\frac{AD+BC}{2}$ is the sum of the radii of $\Gamma_1$ and the $\Gamma_2$.) But since $AB || CD$, we have $MN = \frac{AB+CD}{2}$, so $AB + CD \geq AD + CB$, as desired.
[*][b]Case 2[/b]: $E \neq \infty$. Consider an inversion about $E$ with radius 1, and suppose it maps $A,B,C,D,\Gamma_1, \Gamma_2$ to $A', B', C', D', \Gamma_1', \Gamma_2'$, respectively. We have that $B'$ is between $E$ and $A'$, $C'$ is between $D'$ and $E'$, the interior of $ABCD$ is mapped to the interior of $A'B'C'D'$, and that $\Gamma_1'$ and $\Gamma_2'$ intersect in at most one point. Furthermore, for any point $X \in \Gamma_1$ in the interior of $ABCD$, we have
\[ 90^{\circ} - \theta = \angle DXA = \angle DXE + \angle EXA = \angle ED'X' + \angle EA'X' = 360^{\circ} - \theta - \angle D'XA' = \angle D'XA' - \theta, \]
so $\angle D'XA' = 90^{\circ}$, whence $A'D'$ is a diameter of $\Gamma_1'$. Similarly, we can find that $B'C'$ is a diameter of $\Gamma_2'$.

Let $M$ and $N$ be the midpoints of $A'D'$ and $B'C'$, respectively. Because $\Gamma_1'$ and $\Gamma_2'$ have $A'D'$ and $B'C'$ as their diameters, respectively, they are centered at $M$ and $N$, respectively, and have radii of $\frac{A'D'}{2}$ and $\frac{B'C'}{2}$, respectively. Because they do not intersect, $MN \geq \frac{A'D' + B'C'}{2}$. By the lemma, $2MN \leq A'B' + C'D'$, so we have $A'D' + B'C' \leq A'B' + C'D'$.

We wished to show that $AB + CD \geq AD + BC$. This is equivalent to $EB- EA + EC - ED - AD - BC \geq 0$ which, after our inversion, is equivalent to
\[ \frac{1}{EB'} - \frac{1}{EA'} + \frac{1}{EC'} - \frac{1}{ED'} - \frac{A'D'}{EA' \cdot ED'} - \frac{B'C'}{EB' \cdot EC'} \geq 0, \]
or
\begin{align*} \frac{EB' + EC' - B'C'}{EB' EC'} \geq \frac{EA' + ED' + A'D'}{EA' \cdot ED'} \tag{2}. \end{align*}
Now, by the law of cosines, we have
\begin{align*}
1 + \cos \angle E &= \frac{EB'^2 + EC'^2 - B'C'^2}{2EB' \cdot EC'} + 1 = \frac{(EB' + EC')^2 - B'C'^2}{2EB' \cdot EC'} \\
& = \frac{(EB' + EC' - B'C')(EB' + EC' + B'C')}{2EB' EC'}. \end{align*}
Similarly,
\[ 1 + \cos \angle E = \frac{(EA' + ED' + A'D')(EA' + ED' - A'D')}{2EA' \cdot E'D'}. \]
Thus, if we divide both sides of (2) by $2(1 + \cos \angle E)$, we see that (2) is equivalent to
\[ \frac{1}{EB' + EC' + B'C'} \geq \frac{1}{EA' + ED' - AD'} \iff EA' + ED' \geq EB' + EC' + BC' + AD' . \]
Recalling that $EA' = EB' + B'A'$ and $ED' = EC' + C'D'$, we see that this last inequality is equivalent to $A'B' + C'D' \geq A'D' + B'C'$, which we have proven.[/list]
[/hide]
EDIT: Fixed reversed inequality signs.

[hide="combining ideas from the above posts"]
Conisder the second intersection $Q$ of circumcircles $APB$ and $CPD$. First, we show that the circumcircles $AQD$ and $BQC$ are tangent as well. It suffices to show that $\angle DAQ + \angle CBQ = \angle CQD$, since then the tangents at $Q$ to the two circumcircles $AQD$ and $BQC$ must coincide. But since

$\angle DAQ + \angle CBQ = \angle DAP - \angle PAQ + \angle CBP + \angle PBQ$

and $\angle PAQ = \angle PBQ$, we have

$\angle DAQ + \angle CBQ = \angle DAP + \angle CBP = \angle CPD = \angle CQD$

as desired. Thus the circumcircles $AQD$ and $BQC$ are tangent. Next, note that

$\angle BQC = \angle BQP + \angle CQP = \angle BAP + \angle CDP \leq 90^{\circ}$

and similarly $\angle AQD \leq 90^{\circ}$. It follows that there exists at most one point $R$ with $\angle ARD = \angle BRC =  90^{\circ}$. Therefore, the distance from the centers of the two circles with diameters $AD$ and $BC$ must be at least as great as the sum of their radii; i.e. $MN \geq \frac{1}{2} (AD + BC)$ if $M$ and $N$ are the midpoints of $AD$ and $BC$, respectively. Finally, since

$AB + CD \geq 2MN$

is equivalent to the triangle inequality in complex numbers, it follows that $AB + CD \geq AD + BC$.

[/hide]

Why $Q$ is inside the convex quadrilateral?

Let $Q$ be the second intersection of $(ABP)$ and $(CDQ)$. We claim that $(ADQ)$ and $(BCQ)$ are tangent. We will make use of the following lemma:
\\$~$\\
If $ABCD$ is a convex quadrilateral and $X$ is an interior point then $(ABX)$ and $(CDX)$ are tangent if and only if $\angle AXD=\angle ABX+\angle XCD$. The proof of this lemma is omitted.
\\$~$\\
Note that $\angle ADP+\angle BCP=\angle APB$ since $(ADP)$ and $(BCP)$ are tangent. However, $\angle AQB=\angle APB$ and $\angle ADP+\angle BCP = \angle ADQ+\angle BCQ$ by angle chasing, so we have
\[\angle AQB=\angle ADQ+\angle BCQ\] and so the claim is true by the lemma.
\\$~$\\
Now, WLOG let $P$ be closer to $AD$ than $Q$ then
\begin{align*}
\angle BQC &=360^\circ-\angle PQC-\angle PQB\\
&=\angle BAP+\angle CDP \\
&\le 90^\circ
\end{align*}
and similarly $\angle CDQ$ is not obtuse.
\\$~$\\
Now, we can discard the diagram except $A,B,C,D,Q$. Let $A'$ and $D'$ be diametrically opposite on $(ADQ)$ such that $A'D'\parallel AD$, where $A,A',D',D$ are in that order. Define $B'$ and $C'$ similarly. We have because that $\angle AQD$ and $\angle BQC$ are acute, that $AD\le A'D'$, $BC\le B'C'$, and $AB\ge A'B'$ and $CD\ge C'D'$.
\\$~$\\
Let $O_1$ and $O_2$ be the midpoints of $A'D'$ and $B'C'$, respectively, which happen to be the centers of $(ADQ)$ and $(BCQ)$. Let $M$ be the midpoint of $AC$. We have
\begin{align*}
AB+CD &\ge A'B'+C'D' \\
&= 2O_1M+2O_2M \\
&\ge 2O_1O_2 \\
=\ge 2O_1Q + 2O_2Q \\
=\ge A'D'+B'C' \\
\ge AD+BC
\end{align*}
as desired.


function
calculus
derivative
trigonometry
calculus computations

Now I have a point x such that f(x)=x
I know that if abs(f'(x))>1, the point is attractive
and if it is smaller then 1 it is pushing points away.
But my problem is, that f'(x)=1, How do I know now what kind of point this is
Is there a theorem I can use?

Too many $ x$s in the notation for my tastes. Let's say the function $ x\mapsto f(x)$ has a fixed point at $ x\equal{}x_0.$

The best thing to do is to draw graphs; what I'm going to say will become clear from those graphs.

As you said, if $ |f'(x_0)|<1,$ the fixed point is attracting and if $ |f'(x_0)|>1$ it is unstable (repelling).

Suppose $ f$ is smooth enough in a neigbhorhood of $ x_0$ that we can ask about its higher order derivatives - that is, assume $ f\in C^2$ when I talk about second derivatives, $ C^3$ when I talk about third derivatives, and so on.

If $ f'(x_0)\equal{}1$ and $ f''(x_0)<0$ then the fixed point $ x_0$ is attracting (slowly) from the right and repelling from the left.

If $ f'(x_0)\equal{}1$ and $ f''(x_0)>0$ then the fixed point $ x_0$ is attracting (slowly) from the left and repelling from the right.

If $ f'(x_0)\equal{}\minus{}1,$ we know we can't split it by sides since the sequence will alternate sides. I'll have to consider that a little further.

If $ f'(x_0)\equal{}1$ and $ f''(x_0)\equal{}0,$ then we have a (very slowly) attracting fixed point if $ f'''(x_0)<0$ and a repelling fixed point if $ f(x_0)>0.$

Use a calculator or spreadsheet to try this out for either $ f(x)\equal{}\sin x$ or $ f(x)\equal{}\tan x$ near zero.


Solve the differential equation
\[ 2y^2 e^{2x} \textup{ d}x \plus{} (4ye^{2x} \minus{} y^{\minus{}2}e^{\minus{}y}) \textup{ d}y \equal{} 0.\]Writing the equation as $ M \textup{ d}x \plus{} N \textup{ d}y \equal{} 0$, this equation is not exact, since $ \frac{\partial M}{\partial y} \equal{} 4ye^{2x}$ and $ \frac{\partial N}{\partial x} \equal{} 8ye^{2x}$. How I am supposed to convert this to an exact equation like the book instructs?

hello, the solution is
$ y(x)^4e^{2x}\plus{}e^{\minus{}y(x)}\plus{}C\equal{}0$.
Sonnhard.


inequalities
number theory unsolved
number theory

Find, with proof, all $n \in \mathbb{N}$ such that $2^n \mid (3^n - 1)$.

There are some trivial solutions - $n = 1$, $n = 2$.

Now, if $n$ is odd and $n \geq 3$ then $3^n - 1 \equiv 2$ (mod 8) will give a contradiction.

So, $n = 2k$ must be even. Hence $2^n = 4^k$ divides $3^n - 1 = (3^k - 1)(3^k + 1)$. Since $3^k + 1$ and $3^k - 1$ differ by 2, one of them contains at most one factor 2 and hence the other one must contain at least $2k - 1$ factors 2. Hence we have (in either case!) that $3^k + 1 \geq 2^{2k - 1}$. Hence $k = 1$ or $k = 2$ (the inequality is false for the other possible values of $k$). The only additional solution we get is $n = 4$ and this works as well. Hence all solutions are givenby $n = 1,2,4$.


inequalities
probability
induction
inequalities solved

Let $n$ be a positive integer, and let $x_{1}$, $x_{2}$, ..., $x_{n}$, $y_{1}$, $y_{2}$, ..., $y_{n}$ be $2n$ nonnegative real numbers such that $x_{i}+y_{i}=1$ for every $i\in\left\{1,2,...,n\right\}$.
Prove that
\[(1-x_{1}x_{2}\cdots x_{n})^{m}+(1-y_{1}^{m})(1-y_{2}^{m})\cdots (1-y_{n}^{m}) \geq 1. \]
for all positive integers $m$.

Consider the following probabilistic experience :
Let c(1), c(2), ... , c(n) be coins such that, for each i the probability for c(i) to give head is x_i.

Throw these coins independantely, m times each. Then :
(1 - x_1*...x_n) m  is the probability p(A) of the event A : " for each of the m throws, at least one of the coins has given head".
Note that A = B  \cup C, where B is " there is a coin which has given head at each of the m throws", and C is "at least one coins has given head at each throws, but not the same for each of the throws".
Moreover B \cap C =  \emptyset , thus p(A) = p(B) + p(C).

In another hand, (1 - (y_1)m)(1 - (y_2) m )...(1 - (y_n) m ) is the probability that each of the coins does not gives head at least once on the m throws = p(B - ), where B -  is the opposite event of B.
Then the LHS of the desired inequality is p(A) + p(B - ) = p(B) + p(B - ) + p(C) = 1 + p(C)  \geq  1.

Note that equality occurs iff n = 1.

Pierre.

A very ingenious proof, thank you very much.

Yes, it is an incredible solution and I was amazed when I have read it from an excellent training book written by Titu Andreescu and Svetoslav Safchev: Mathematical Miniatures. Isn't there a more natural solution?

It is very nice solution. And it is natural for those who know probability. It can be done by induction. If I'm not mistaken, it also used in Poland MO.

Namdung

I have another idea:
I will prove by induction on n :
to prove this ineq is true with n we must use lemma :
 (a+b-ab)^m \geq a^m+b^m-(ab)^m ,for all a,b belong to [0,1],m is natural number.

Well, I think that Pierre should mention the reference where he took this solution (otherwise it is unfair!).

Other (analytic) solutions may me found in Crux Mathematicorum or in the book "from Erdos to Kiev".

[quote="Kantor"]Other (analytic) solutions may me found in Crux Mathematicorum or in the book "from Erdos to Kiev".[/quote]

I am interested in an analytic solution too.
Can you please give me the exact reference ?
I was not able in find it in the book "from Erdos to Kiev".


Simplify the following expressions.

1. $ \frac{2\minus{}\frac{x\minus{}2}{x}}{x\plus{}2}$

2. $ \frac{a\minus{}1\plus{}\frac{2}{a\plus{}2}}{a\plus{}1\minus{}\frac{2}{a\plus{}2}}$

3. $ \frac{\frac{a\plus{}x}{a\minus{}x}\minus{}\frac{a\minus{}x}{a\plus{}x}}{\frac{a\plus{}x}{a\minus{}x}\plus{}\frac{a\minus{}x}{a\plus{}x}}$

4. $ \frac{1\plus{}x}{1\minus{}\frac{1}{1\minus{}\frac{1}{1\plus{}x}}}$


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reflection

Do you stand behind Obama or McCain and why?

BTW, take a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216480

If I have to choose between these two, I'd stand behind McCain because McCain isn't like Obama. I think the 2 biggest problems I have with Obama are 1-that he's Pro-Choice (or pro-abortion or anti-life, whichever way you want to refer to it), and 2-his small-town comment:
[quote="Obama"]You go into these small towns in Pennsylvania and, like a lot of small towns in the Midwest, the jobs have been gone now for 25 years and nothing's replaced them...And they fell through the Clinton administration, and the Bush administration, and each successive administration has said that somehow these communities are gonna regenerate and they have not.

And it's not surprising then they get bitter, they cling to guns or religion or antipathy to people who aren't like them or anti-immigrant sentiment or anti-trade sentiment as a way to explain their frustrations.[/quote]

Note: If you ask someone else, those 2 reasons might be their reasons to [b]support[/b] Obama rather than McCain. So it depends on how people view them. I see those 2 reasons as very important issues that I feel are important enough to outweigh all the faults of McCain and cause me to stand behind McCain just so that Obama won't be president.

There are numerous other reasons I don't like Obama, but those 2 are the biggest.

[quote="ProtestantT"]I'd stand behind McCain because McCain isn't like Obama[/quote]
That isn't really a logical response.  Then I stand by Obama because he's not like McCain :what?:

I stand by Obama for many reasons, two of which are the exact opposite of what you cited.  I agree with his pro-choice views, and I agree with his small-town comment.  Frankly, it is true that many small-town folk do "cling" to guns and religion, viewing the war in Iraq as a mandate from God, and the like.  I am ardently anti-religion and although Obama is a devout Christian (HE'S NOT A MUSLIM!!!), it would be foolish to go against a candidate simply because they aren't atheist-there aren't too many atheist politicians out there because religious values are important to constituents.  I also disagree with McCain's views on tax cuts for the wealthy, a 100-year Iraq war (obviously he doesn't want that to happen, but he would allow for it), health care, and government regulation.

[quote="JRav"][quote="ProtestantT"]I'd stand behind McCain because McCain isn't like Obama[/quote]
That isn't really a logical response.[/quote]
That is a logical response. Since McCain is the only candidate from the OP that is not Obama (The OP only gave two choices, so wanting one choice to not win is as swaying as wanting the other choice to win), if I want Obama to not win, I should stand behind McCain. Since only one of the two candidates in the OP can win, then McCain winning means Obama doesn't win.

[quote="ProtestanT"]...
if I want Obama to not win, I should stand behind McCain. Since only one of the two candidates in the OP can win, then McCain winning means Obama doesn't win.[/quote]
Yes, but voting in a "lesser of two evils" standpoint isn't a great reason to vote for someone...

[quote="JRav"]I am ardently anti-religion and although Obama is a devout Christian (HE'S NOT A MUSLIM!!!), it would be foolish to go against a candidate simply because they aren't atheist-there aren't too many atheist politicians out there because religious values are important to constituents.[/quote]
Refer to [url=http://isbarackobamamuslin.com/]here[/url] :P

[quote="ProtestanT"]BTW, take a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216480

If I have to choose between these two, I'd stand behind McCain because McCain isn't like Obama. I think the 2 biggest problems I have with Obama are 1-that he's Pro-Choice (or pro-abortion or anti-life, whichever way you want to refer to it), and 2-his small-town comment:
...

Note: If you ask someone else, those 2 reasons might be their reasons to [b]support[/b] Obama rather than McCain. So it depends on how people view them. I see those 2 reasons as very important issues that I feel are important enough to outweigh all the faults of McCain and cause me to stand behind McCain just so that Obama won't be president.

There are numerous other reasons I don't like Obama, but those 2 are the biggest.[/quote]
I personally am pro-choice, and although I respect your view, I take great issue with your view that Obama being pro-choice is a [b]very important[/b] issue to determine that you don't want him to take office. There are many other vastly important issues to consider, like the recent major economic problems, healthcare, and foreign policy. I anticipate that, given your stance on abortion, you may believe human life to be the issue of most importance. But if you take this stance, foreign policy is just as, if not more, important: people are dying because of the conflicts in Iraq, Afghanistan, Darfur, and Israel/Palestine, and the different courses either candidate will take regarding these conflicts will have a great impact on lives around the world.

That being said, I support Obama. His proposed ideas to raise taxes for the wealthy and close tax code loopholes from large corporations and his exit strategy for Iraq are things I agree with. I also am a proponent of more government intervention in the economy like Obama. (The subprime mortgage crisis shows what happens when regulations go away.) That doesn't mean I see Obama as the perfect candidate: both he and McCain seemed pretty sketchy about what spending priorities should be taken away due to the recent financial crisis. Also, McCain does raise a good point with his reduction of pork-barrel spending (although corporate windfalls (to my knowledge) suck up much more money from our Treasury).

Keep in mind that although Palin touts herself as a maverick, Alaska has the highest pork-per-capita ratio in the nation.  Not to mention voting for the bridge to nowhere, and demanding that a state official be fired for not firing her brother-in-law.

Here is my view. It has been made so completely abundantly clear that Palin is completely incompetent and unqualified for the job that no rational person would vote for McCain. [i]Especially[/i] single issue voters like pro-lifers. You can have your view, but come on. Vote in alignment with it next time. Voting to put Palin into office is such a destructive decision that I have made a personal vow to never speak to anyone that votes for McCain ever again.

Think about it. America is being so selfish at this point. They think that they can vote for who they want without regard for the rest of the world. This is not the case, though. You have to base your decision with the rest of the world in mind. We are electing a world leader. You don't want people to have abortions. Then don't have an abortion. Palin can't name a single major newspaper in the world. She has been sheltered her whole life making her completely ignorant of the current state of the world. Her world view is completely skewed by her religion. This is not someone that can be put into a position of power in the world. I could go on for paragraphs upon paragraphs, but you get the point.

McCain is completely respectable and would be able to handle the position (despite my disagreeing with almost everything he says), but his choosing Palin as VP has moved beyond party lines, opinions, issues, etc. If we care about the future of the US and the world we must move beyond these petty selfish issues and vote against McCain.

Very nice post.  I only hope when you mean vote against McCain, you mean vote for Obama, not Bob Barr or Ralph Nader.

[quote="HilbertThm90"]You don't want people to have abortions. Then don't have an abortion.[/quote]You're begging the question here. If you think abortion is murder, then it is a VERY IMPORTANT issue.

People who vote on abortion are voting based on their own opinion. At this stage, that is just plain selfish. You have to put your opinions behind you, and keep Palin out of office. The first step is to realize that any opinion you have that you are voting on has an opposite opinion. If your "belief" was as clear-cut as you think it is, then everyone would agree on it and it wouldn't be an issue you needed to vote on.

Suppose for the next four years people continue to have abortions. The world continues. Suppose McCain is elected and then dies a few months or even years in. I'm not saying the world will end, but it will be drastically changed and not in a good way. How can Palin have reasonable conversations with other world leaders when at this point they probably don't respect her. She won't be taken seriously and her presence at these meetings will be purely for show.

Outside of America, abortion isn't really even considered an issue. To ignore the rest of the world in this decision and vote for a world leader based on something the rest of the world is not concerned about is to vote irresponsibly.

P.S. Changing abortion laws will be so difficult that even voting McCain into office isn't likely to change them, so the vote is basically wasted if that is your single issue anyway.

I would like to elaborate on Hilbert's last comment.

Suppose Roe v. Wade was overturned. Then, it would just go back to the states. Then people might have to travel to get an abortion. A recent article, [url=http://ncronline3.org/drupal/?q=node/2058]found here[/url], written by an anti-abortion Christian, explores this further.

There is a video out there (not sure what site) of Palin in a church, with a witch doctor, trying to remove her demons (I wouldn't call it an "exorcism" but it's pretty close).  Yet she doesn't consider herself a religious nut?  Unfortunately, too many people in this country let their religious beliefs govern their lives.  People in Alabama go crazy when the 10 Commandments are removed from the courthouse.  Members of the religious right bomb abortion clinics.  We have a separation of church and state, but too many people choose to ignore this.  Even though I disagree with McCain on basically everything, at least he doesn't go out touting his religion or missions from God, like Palin or Bush.  We need to keep Palin as far from the White House as possible (like Alaska  :wink:), so we can avoid 4 more years of religious extremism.

I really think that her skewed view of the world due to her religious affiliation is much scarier on a global scale than a domestic one. When trying to deal with countries that have strong religious affiliations, she will always have her own divine agenda of what is right and wrong (obviously [i]always[/i] the other religion is the work of Satan). When trying to deal with the more secular parts of the world like Europe, they won't take a lot of what she has to say seriously knowing that it comes from a dogmatic rather than scientific or rational point of view. How is she to employ diplomacy on a global level?

Just watch the clip about how we shouldn't question Israel [i]ever[/i] to find out how scary this world view is.

Continuing on your point, how about negotiations with the middle east?  For some reason, I don't think Ahmadinejad would take too kindly to Palin coming to Iran (or at the UN, wherever) and saying "We're going to bring Jesus Christ to your country and reform it so it is better for world peace".  Sort of like having Pat Robertson as your secretary of state.

Honestly, though, she's been very good at separating her religion from her politics as governor. She hasn't pushed anti-abortion stuff, she hasn't pushed creationism, etc.

[quote="jmerry"]The second-biggest close race is the congressional district across the lake. I get ads for it anyway, because Seattle broadcast stations reach at least some of the district.[/quote]  My brother is working for the challenger there -- I just interrupted his canvassing to wish him good luck.  For those who like the 538-type approach, the "Meta-Analysis" at http://election.princeton.edu/ is fun -- it's algorithm is more transparent than 538, for one thing.  His analysis was also spot-on in 2004, modulo some assumptions about splitting the undecided vote that he's not making this year.  The current (last?) median prediction is O 352 (95% confidence interval appears to be 310-377 or so), and his median Senate result is D 58.

[quote="Kent Merryfield"]Nate Silver at 538.com just updated the overall probability of Obama winning the election to 98.1%.[/quote]

what kent is really referring to is http://www.fivethirtyeight.com, which is a site which tracks polling data and makes electoral predictions using said data and simulations.  if you haven't already heard of it it's definitely worth checking out. (538.com does not redirect to the right site, though)

[quote="1=2"]I just want some clarification: What does Obama mean by a "civilian force to meet the security goals that we have set for ourselves"?[/quote]If you see the full context of the quote, he was talking about the Peace Corps, People-to-People, and other goodwill-type activities.

I'm not sure if Silver has updated since yesterday but currently, he has Obama at 346.5 and the Democrats at 57.1 seats in the Senate.

My polling place is about 150 feet away, and I can see it from my window. Just now, a news van pulled up.
I already voted a few hours ago, but that's pretty neat anyway.

It's over. Obama is our next president.  :D

Well, CNN just called it for Obama a few minutes ago.  Senate races look pretty good, too.

58 for senate is looking ok right now also, for democrat.
fivethityeight.com says 15% chance for 60+

Congratulations to Obama!

I just have to say, have to say, HAVE to say, Obama went to my school!!! :D  :D  :D  (I'm so glad... I don't think we'll even have "real" classes tomorrow... Day long discussion  :D )

And so this comment is really biased, but I think McCain made an epic concession speech, and it was one of the best speeches I've ever heard... And then Obama's speech, of course...

Finally, maybe this nation is moving in the right direction.

Go Obama.

Local reaction: the news radio station is warning drivers away from certain spots in the city due to crowds in the street celebrating. Long lines and rain discouraged a few, but not many (no lines at my polling place). The governor's race has been called for the incumbent Democrat Gregoire, who has a lead of about five points even with relatively few Seattle-area ballots counted so far.

And now, that's enough for tonight. I'll check again in the morning.

[quote="jmerry"]The governor's race has been called for the incumbent Democrat Gregoire[/quote]  That struck me as pretty incredible.  Still feeling good this morning, but waiting for the last few Senate & House seats to fall. What's wrong with Alaskans -- they like having the most corrupt congressional delegation around?  I mean, how often does a convicted felon get elected to anything, let alone senator?  Also, I'd like to see Indiana go blue for purely aesthetic reasons :).

So as of now, it still seems that Alaska and Minnesota have yet to call their Senate races.  Either way, the Dems aren't getting 60.  Still a good night, though.  In Maryland, the 1st district Congressional seat has yet to be called between Frank Kratovil and Andy Harris.  If Kratovil wins, the Dems pick up another House seat.

North Carolina and Missouri have still yet to call, even though it doesn't matter.  According to RCP, Obama leads NC (50-49) and McCain leads Missouri (50-49).

Actually, it seems Minnesota was called, according to RCP.  Georgia, Oregon, and Alaska have yet to be called.

[quote="JRav"]Minnesota was called[/quote]  The final margin of victory seems to be between 500 and 600 votes out of 2.4 million -- that leads to an automatic recount under MN law.


Define your "equations".
And moved, it's not Number Theory.

[quote="mathsrox"]Using four 4s and any mathematical equation generatye 224. If you are using exponentiation / log then generate the required bases and exponents. i.e. sqrt(4) is not allowed[/quote]

I don't understand "generate the required bases and exponents". Are the following expressions allowed, and if not, how could there be written ? :
log(4)
exp(4)

And why sqrt(4) is not allowed since "any mathematical equation" is allowed? And what could be the good way of write this ?


Denote by [x] the largest interger not exceeding x and write {x}=[x]-x . find the limit
                 $ lim_{n\minus{}\minus{}>\plus{}\infty}${$ (7\plus{}4\sqrt{3})^{n})$}

Please lock this topic, this is a problem from M&Y, this Issue.

[quote="N.T.TUAN"]from M&Y, this Issue.[/quote]

what is M&Y?

It is //http://www.nxbgd.com.vn/toanhoctuoitre/?p=8   . :lol:


trigonometry
function
calculus
calculus computations

x^2-2asin(cos x)+a^2=0
for what values of (a) this equation has only one solution?
thanks.[/hide]

hello, do you mean the equation
$ x^2\minus{}2a\sin(\cos(x))\plus{}a^2\equal{}0$?
Sonnhard.

Note that the left hand side is an even function. ;)

Next tip: if $ f$ is even, then $ f(x)\equal{}f(\minus{}x)$. So, if the solution of $ f(x)\equal{}0$ exists and is unique, it must be __________ (fill the blank).

x=0.
how does it help for the question?
the equation will be:
0^2-2asin(cos0)+a^2=0
??

Are you thinking at all? :roll: You now know the root and you need to find the parameter. So, what should you do?

Almost. It remains to show that for those values of $ a$ there are no other solutions.

for a=0:
x^2=0
x=0 =>aonly one solution.

for a=1.682:
x^2-3.3658sin(cosx)+2.88322=0
whats next?

Apparently you need to show that $ f(x)>0$ for all $ x\ne 0$. Just think a bit. My advice would be to write $ a$ as $ \sin(\cos 1)$ rather than an approximate decimal; it makes things clearer. Anyway, that's about as far as I wanted to go. Try to finish yourself :)

you mean a= 2sin(cos0) right?
because we said that f(0)=0.

Right, $ a \equal{} 2\sin(\cos 0) \equal{} 2\sin 1$. Sorry for the stupid misprint :blush:

how?
2sin(cos0)=1.682
2cos1=1.0806

and I didnt find a way how to proove that the f(x)>0.
i only found out that the function has a minimum point in x=0.

Well, I tend to make misprint after misprint today, I meant $ \sin 1$, of course (edited) :lol:. As to the proof, $ x^2>0$ for $ x\ne 0$. Now, what can you say about $ a^2\minus{}2a\sin(\cos x)$ when $ a\equal{}2\sin 1$? Note that $ \cos$ is always between $ \minus{}1$ and $ 1$ and $ \sin$ is increasing on $ [\minus{}1,1]$.


modular arithmetic
floor function
algebra
binomial theorem

determine the unite digit in the decimal expansion of
$(20+\sqrt{397})^{2674}$

[hide]
\[\left\lfloor \left( 20+\sqrt{397}\right)^{2674}\right\rfloor =\left( 20+\sqrt{397}\right)^{2674}+\left( 20-\sqrt{397}\right)^{2674}-1\]
\[\equiv 2\cdot 397^{1337}-1\equiv 2\cdot 7^{1337}-1 \pmod{10}\]
because after expanding out by the binomial theorem, all parts with radicals cancel and the other remaining integers are divisible by 10.

Since $7^{2}\equiv-1$, $7^{1337}\equiv 7(-1)^{668}\equiv 7$ so $2\cdot 7^{1337}-1$ has a units digit of $\boxed{3}$.
[/hide]

[quote="white_horse_king88"][quote="scorpius119"]
\[\left\lfloor \left( 20+\sqrt{397}\right)^{2674}\right\rfloor =\left( 20+\sqrt{397}\right)^{2674}+\left( 20-\sqrt{397}\right)^{2674}-1 \]
[/quote]

How do you know that this is true? There's probably something obvious I'm missing...[/quote]

$\left( 20+\sqrt{397}\right)^{2674}+\left( 20-\sqrt{397}\right)^{2674}-1$ is an integer (can be shown by expanding both using the binomial theorem) and $0<\left( 20-\sqrt{397}\right)^{2674}<1$.


Prove that \[ x^2\plus{}xy\plus{}y^2\equal{}7^n\] has a integer solution $ (x,y)$ for every $ n\in\mathbb{N}$

A very easy problem:
I)for $ n \equal{} 2k$ just take $ y \equal{} 0$ and $ x \equal{} 7^{k}$
II)for $ n \equal{} 2k \plus{} 1$ proceed inductively
$ 7^{2*0 \plus{} 1} \equal{} 7 \equal{} 2^{2} \plus{} 2*1 \plus{} 1^{1}$
Suppose we have $ 7^{2k \plus{} 1} \equal{} x^{2} \plus{} xy \plus{} y^{2}$
Then $ 7^{2k \plus{} 3} \equal{} (7x)^{2} \plus{} (7x)(7y) \plus{} (7y)^{2} \equal{} x_{1}^{2} \plus{} x_{1}y_{1} \plus{} y_{1}^{2}$
Q.E.D. :wink:

A note:
An interesting question is how many of these solutions can we obtain in temrs of n.
For example: $ 7^{2} \equal{} 7^{2} \plus{} 7*0 \plus{} 0^{2} \equal{} 5^{2} \plus{} 5*3 \plus{} 3^{2}$ and $ 7^{3} \equal{} 14^{2} \plus{} 14*7 \plus{} 7^{2} \equal{} 18^{2} \plus{} 18*1 \plus{} 1$
With $ n\rightarrow \infty$ do we have that the number of solutions also grows infinitely? :wink:


Let $ a,b,c$ satisfy $ a\plus{}b\plus{}c\equal{}0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}6$.
Determine the least value of the expression:
$ P\equal{}a^2b\plus{}b^2c\plus{}c^2a$

[quote="duongptnkvn"]Let $ a,b,c$ satisfy $ a \plus{} b \plus{} c \equal{} 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 6$.
Determine the least value of the expression:
$ P \equal{} a^2b \plus{} b^2c \plus{} c^2a$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162227


modular arithmetic
search
superior algebra
superior algebra unsolved

Show that if G is a finite group with identity e and with an even number of elements,then there is a b in G,where b not equal to e, such that b*b=e.
(* is the binary operation of G)

i'd tried to prove by contradiction,that is, i assume b*b=b.Then b^n = b^(n-1) for n>0,which contradict the fact that b^n=e.But the problem is,my proof did not ultilize the "even number of elements" which i think should be part of the proof.

Can anyone help?

Suppose that $ \forall x\in G\minus{} \{e\}$ we have $ x^2 \neq e$, so $ x\neq x^{\minus{}1}$ $ \forall x\in G\minus{}\{e\}$.Therefore $ G$ has the following structure : $ G\equal{}\{x_1,x_1^{\minus{}1},x_2,x_2^{\minus{}1},...,x_n,x_n^{\minus{}1},e\}$ , but in this case $ |G|$ is odd, contradiction.

[hide]Define the set $ A \equal{} \{g\in G|g\neq g^{ \minus{} 1}\}$. Since for every $ a\in A$ we also have $ a^{\minus{}1}\in A$ (and they are not equal), $ A$ must be even, hence $ G \minus{} A$ is even and consists of at least 2 elements. Every element that squares to the identity is either the identity or has order 2, therefore $ G \minus{} A \minus{} \{e\}$ has at least one element of order 2.[/hide]

[u]Solution[/u]. Let $ S \equal{} \{(x_1,x_2)\in G^2: x_1x_2 \equal{} 1\}$.  Then $ |S| \equal{} |G|$.  Define an equivalence relation on $ S$ by saying that two elements are equivalent if one can be obtained from the other by a cyclic permutation.  Then each of these equivalence classes have order $ 2$ unless the element has the same componenet elements, i.e., $ (x,x)\in S$ s.t. $ x^2 \equal{} 1$.  Let $ T \equal{} \{(x,x)\in S\}$.  So $ |G| \equal{} |S| \equal{} 2k \plus{} |T|\implies |T| \equal{} |G| \minus{} 2k\equiv 0\pmod 2$.  Because $ (1,1)\in T$, $ T\neq\emptyset$.  Therefore, there is an $ x\in G$ such that $ |x| \equal{} 2$, as required. $ \Box$

[u]Remark[/u]. If $ p$ is a prime such that $ p\parallel{}G|$, this can be generalized to produce the result that $ \exists x\in G$ such that $ |x|\equal{}p$ (Cauchy's Theorem).

Hi there, MaY...

Take a look at the following post:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=875870790&t=225332[/url]

You'll find a neat generalization of this fact there.  :wink:

Best of luck,

José.

[quote="MaY"]i assume b*b=b.[/quote]
This is not how a correct proof by contradiction goes; those aren't the only two outcomes.  The negative of $ b^2 \equal{} e$ is $ b^2 \equal{} \text{anything except } e$.  (It is never possible for $ b^2$ to be equal to $ b$ in a finite group unless $ b \equal{} e$.)


Stie cnva programu olimpiade(cand e proba)??
stiti link-uri???

Proba o sa fie Luni 17 Aprilie de la ora 09,00 :)
Afisarea rezultatelor probabil Marti la pranz, contestatii Marti dupa amiaza, barajele pentru OIM / OBM / OBMJ probabil miercuri si joi.

Am atasat un fisier text cu [b]schita de program a ONM 2006[/b].

Interesant... totusi sunt niste typo`s unde se precizeaza data........ :roll:  Sa vad daca am inteles bine: Daca mergi la baraje nu poti sa participi la Myller?

Myller a fost in aceasi zi cu primul baraj :)


In the voltaic cell given by $ Co | Co^{2\plus{}}\parallel Fe^{3\plus{}}, Fe^{2\plus{}}| Pt$ the cathode would be?

A)cobalt metal
B)$ Fe^{3\plus{}}$
C)Piece of Pt
D)standard hydrogen electrode

[hide="Answer"]The correct option is c), although it would be more correct to say that the cathode is made up of a piece of platinum emerged in a solution containing Fe3+ and Fe2+ ions.[/hide]

How did you come to that reasoning please sir?

In the diagramatic representation of galvanic cells (in this case Co|Co2+||Fe3+, Fe2+|Pt), the cathode is (by international agreement) always written on the right hand side, and for each electrode, the order is reactant/product. Also, because for a galvanic cell to be spontaneous it must verify Eº(cell) > 0, then the cathode will be the electrode with the highest reduction potential.


Ring Theory
calculus
integration
modular arithmetic
number theory
relatively prime

Let $ \{x,y\}\subset\mathbb Z.$ Solve the following equation:
\[ x^5 \plus{} 31 \equal{} y^2
\]

[quote="arqady"]Let $ \{x,y\}\subset\mathbb Z.$ Solve the following equation:
\[ x^5 + 31 = y^2
\]
[/quote]
[hide]
$ x^{5}=(y+\sqrt{31})(y-\sqrt{31)}$
For some reason I'm fairly certain that rt(31) is a UFD, but I'm not sure how to prove it.
Since the two terms on the RHS are coprime, they must both be fifth powers.
$ (a+b\sqrt{31})^{5}=y+\sqrt{31}$
$ 5a^{4}b+10a^{2}b^{3}+b^{5}=1$
$ b|1$
If b=1, then $ 5a^{4}+10a^{2}=0$
$ a=0$ is the only integral solution.
If b=-1, $ -5a^{4}-10a^{2}=2$
This has no real solutions.
(0,1) doesn't exactly work, so no solutions.[/hide]

[quote="arqady"][quote="diophantient"]
$ (a \plus{} b\sqrt {31})^{5} \equal{} y \plus{} \sqrt {31}$
[/quote]
Why? :wink:[/quote]
$ (y\plus{}\sqrt{31})$ and $ (y\minus{}\sqrt{31})$
are relatively prime, so if their product is a fifth power, then so must each of them.

[quote="ZzZzZzZzZzZz"][quote="diophantient"]If b=-1, $ \minus{} 5a^{4} \minus{} 10a^{2} \equal{} 2$
This has no real solutions.
(0,1) doesn't exactly work, so no solutions.[/quote]
The problem says that the solutions are complex.[/quote]

No, the problem asks for solutions in integers  :wink:

[quote="polskimisiek"]It has been solved before here:
http://www.mathlinks.ro/viewtopic.php?t=184045
 :wink:  :P

P.S. Am I missing sth? How can we say that 2 irrational numbers are relatively prime?[/quote]
We're working in $ \mathbb{Z}\sqrt{31}$
http://mathworld.wolfram.com/UniqueFactorization.html

[hide]
Try mod 31.
Then x^5 is either 0, 1, or -1.
That means y^2 is 0, 1, or -1.
This only happens when y is 0, 1, or -1 mod 31.

EDIT: Whoops, I had x and y switched.

[/hide]

[quote="bertram"]$ 3^5 \equiv \minus{} 5\pmod{31}$  ...[/quote]

That's odd.  Let me go check my calculations.

Well, 31 is a prime, and thus there is a number $ 31\not{|}a$ with $ a^n \not\equiv 1 \pmod{31}$ for $ a<30$. This implies that this number satisfies $ a^5\not\equiv\pm 1\pmod{31}$ since otherwise $ a^{10}\equiv 1\pmod{31}$ ;)

There are, in fact, six different values of $ a^{5}$ in mod 31, since if $ a^{5}\equal{}k$ then $ k^{6} \equiv 1 \mod  31$ and since $ a^{n} \equiv 1 \mod p$ has n solutions for $ n<p$

[quote="bertram"][quote="diophantient"]There are, in fact, six different values of $ a^{5}$ in mod 31, since if $ a^{5} \equal{} k$ then $ k^{6} \equiv 1 \mod 31$ and since $ a^{n} \equiv 1 \mod p$ has n solutions for $ n < p$[/quote]

To be exact, there are $ \operatorname{lcd}(n,\varphi(p))$. ;)[/quote]
Whoops, luckily in this case it's the same.


calculus
derivative
trigonometry
search
real analysis
real analysis unsolved

Hello. :)
I've gotten stuck with an ordianary differential equation while trying to solve an electric flux problem:

I'm trying to find: $ Y(y)$ through:

$ Y''(y)\cdot(1\plus{}x_d\cdot e^{\minus{}\beta y})\plus{}Y'(y)\cdot(\minus{}\beta\cdot x_d\cdot e^{\minus{}\beta y})\minus{}\left(\frac{n\pi}{a}\right)^2\cdot(1\plus{}x_d\cdot e^{\minus{}\beta y})\cdot Y(y)\equal{}0$

Where $ \beta,x_d$ are positive constants.

------------------------------------------------------

I'll also post how I got here, just if somebody would like to check whether my calculations are correct.

[hide][u][b]2-D problem[/b][/u]

Find the electric flux $ \Phi(x,y)$ in $ (x,y)\in(0,a)\times(0,b)$.

Boundary conditions:
[list][*]$ \Phi(0,y)\equal{}\Phi(a,y)\equal{}\Phi(x,0)\equal{}0$
[*]$ \Phi(x,b)\equal{}U_0$[/list]
Permittivity: $ \epsilon(y)\equal{}\epsilon_0[1\plus{}x_d e^{\minus{}\beta y}]$

Start using Gauss's Law: $ \nabla\cdot\left[\epsilon(y)\cdot(\minus{}\nabla\Phi)\right]\equal{}0$

[b][u]Solution[/u][/b]

So, I start out using Gauss's Law:

$ \nabla\cdot\left[\epsilon(y)\cdot(\minus{}\nabla\Phi)\right]\equal{}0\\
\nabla\cdot\left[\epsilon(y)\cdot(\nabla\Phi)\right]\equal{}0\\
(\nabla\Phi)\cdot(\nabla\epsilon(y))\plus{}\epsilon(y)\nabla^2\Phi\equal{}0\\
(\nabla\Phi)\cdot\left(\vec{i_y}\frac{\partial\epsilon(y)}{\partial y}\right)\plus{}\epsilon(y)\nabla^2\Phi\equal{}0\\
\vec{i_y}(\nabla\Phi)\cdot\left(\minus{}\cancel{\epsilon_0}\cdot\beta\cdot x_d\cdot e^{\minus{}\beta y}\right)\plus{}\cancel{\epsilon_0}[1\plus{}x_d\cdot e^{\minus{}\beta y}]\nabla^2\Phi\equal{}0\\
\vec{i_y}(\nabla\Phi)\cdot\left(\minus{}\beta x_d e^{\minus{}\beta y}\right)\plus{}[1\plus{}x_d e^{\minus{}\beta y}]\nabla^2\Phi\equal{}0$

Separation of Variables: $ \Phi(x,y)\equal{}X(x)\cdot Y(y)$

$ [X(x)Y'(y)]\cdot\left(\minus{}\beta x_d e^{\minus{}\beta y}\right)\plus{}[1\plus{}x_d e^{\minus{}\beta y}][X''(x)Y(y)\plus{}X(x)Y''(y)]\equal{}0$

Dividing by $ X(x)Y(y)$:

$ \frac{Y'(y)}{Y(y)}\cdot\left(\minus{}\beta x_d e^{\minus{}\beta y}\right)\plus{}[1\plus{}x_d e^{\minus{}\beta y}]\left[\frac{X''(x)}{X(x)}\plus{}\frac{Y''(y)}{Y(y)}\right]\equal{}0\\
\frac{Y''(y)}{Y(y)}[1\plus{}x_d e^{\minus{}\beta y}]\plus{}\frac{Y'(y)}{Y(y)}\cdot\left(\minus{}\beta x_d e^{\minus{}\beta y}\right)\plus{}\frac{X''(x)}{X(x)}[1\plus{}x_d e^{\minus{}\beta y}]\equal{}0$

Taking the x-derivative:

$ 0\plus{}0\plus{}[1\plus{}x_d e^{\minus{}\beta y}]\frac{d}{dx}\left[\frac{X''(x)}{X(x)}\right]\equal{}0\\
\frac{d}{dx}\left[\frac{X''(x)}{X(x)}\right]\equal{}0\\
\frac{X''(x)}{X(x)}\equal{}K_x\\
X''(x)\equal{}K_xX(x)$

Boundary conditions:
[list]
[*]$ \Phi(0,y)\equal{}0\Rightarrow X(0)\equal{}0$
[*]$ \Phi(a,y)\equal{}0\Rightarrow X(a)\equal{}0$[/list]

In order to satisfy these conditions: $ K_x\equal{}\minus{}k^2<0$

$ X(x)\equal{}\{\sin(kx),\cos(kx)\}\equal{}\sin(kx),\text{ so that } X(0)\equal{}0\\
X(a)\equal{}0\Rightarrow k\cdot a\equal{}n\pi\\
\Rightarrow k\equal{}\frac{n\pi}{a},\: n\equal{}0,1,2,3,\ldots$

So: $ X(x)\equal{}\sin\left(\frac{n\pi}{a}x\right)$

Substituting:

$ \frac{Y''(y)}{Y(y)}[1\plus{}x_d e^{\minus{}\beta y}]\plus{}\frac{Y'(y)}{Y(y)}\cdot\left(\minus{}\beta x_d e^{\minus{}\beta y}\right)\minus{}\left(\frac{n\pi}{a}\right)^2[1\plus{}x_d e^{\minus{}\beta y}]\equal{}0$

And I'm having problems in finding Y(y). :([/hide]

I would really appreciate any help in that. :)

I'm not sure if you can get a closed form solution, but the following simple idea may help (after all, you get a series anyway, so why not a double series?).

Your equation can be rewritten as $ Y''\minus{}K^2Y\equal{}e^{\minus{}\beta y}P(D)Y$ where $ P(D)$ is a second order linear differential operator with constant coefficients. Now, search your solutions in the form $ Y_{\pm}\equal{}\sum_{m\ge 0}c_m e^{(\pm K\minus{}m\beta)y}$. You'll get $ c_{m\plus{}1}((m\plus{}1)^2\beta^2\mp 2K(m\plus{}1)\beta)\equal{}P(\pm K\minus{}m\beta)c_m$ and the series will converge if the leading coefficient of $ P$ is less than $ 1$ , i.e., if $ x_d<1$.

The big problem with this approach is that you'll also need to satisfy the boundary conditions and the values of $ Y_{\pm}$ at $ 0$ and $ b$ are given by rather ugly series, so to find their linear combination that fits the boundary condition explicitly will be a nightmare. Still, I do not see what else can be done :(. Maybe somebody will have a better idea. :? Where does this question come from?

[quote="fedja"]Where does this question come from?[/quote]

Exams at university.

Well, that's what I wanted to see, that indeed we cannot get any further. Thank you for your time, you've been very helpful. :)


calculus
integration
trigonometry
calculus computations

$ \int \frac{1\plus{}\sin(x)}{1\minus{}\sin(x)} dx$

hmm any help?

MUltiply top and bottom by $ 1 - \sin{x}$ to obtain

$ \int{\frac {(1 + \sin{x})^2}{\cos^2{x}}}$

Then use integration by parts with:

$ u = (1 + \sin{x})^2$
and
$ dv = \frac {1}{\cos^2{x}}dx$

I get:

$ \tan{x}(1 + \sin{x})^2 + 2\cos{x} - x + \frac {1}{2}\sin{2x} + C$

Maple agrees with me, though for some odd reason it insists that C=2 without being given any initial conditions ^_^

hello, i have got
$ \minus{}4\, \left( \tan \left( 1/2\,x \right) \minus{}1 \right) ^{\minus{}1}\minus{}x$.
Sonnhard.

A much simpler method follows:

$ \frac{1\plus{}\sin x}{1\minus{}\sin x} \equal{} \frac{(1\plus{}\sin x)^2}{1\minus{}\sin^2 x} \equal{} \left(\frac{1\plus{}\sin x}{\cos x}\right)^{\!\!2}$

$ \equal{} (\sec x \plus{} \tan x)^2 \equal{} \sec^2 x \plus{} 2 \sec x \tan x \plus{} \tan^2 x \equal{} 2 \sec^2 x \plus{} 2 \sec x \tan x \minus{} 1$.

Now integration is trivially simple:

$ \int \frac{1\plus{}\sin x}{1\minus{}\sin x} \, dx \equal{} 2 \int \sec^2 x \, dx \plus{} 2 \int \sec x \tan x \, dx \minus{} \int \, dx$

$ \equal{} 2 \tan x \plus{} 2 \sec x \minus{} x \plus{} C$.

$ \int \frac{1\plus{}\sin x}{1\minus{}\sin x}\ dx\equal{}\int \left(\minus{}1\plus{}\frac{2}{1\minus{}\sin x}\right)\ dx$

$ \equal{}\minus{}x\plus{}2\int \frac{1}{1\minus{}\sin x}\ dx$

$ \equal{}\minus{}x\plus{}2\int \frac{1}{2\sin ^ 2 \left(\frac{\pi}{4}\minus{}\frac{x}{2}\right)}\ dx$

$ \equal{}\boxed{\minus{}x\plus{}\frac{2}{\tan \left(\frac{\pi}{4}\minus{}\frac{x}{2}\right)}\plus{}C}$


geometry
cyclic quadrilateral
geometry proposed

[quote="Niichavo"]Let $ ABCD$ be a cyclic quadrilateral,$ M$ and $ N$  be the midpoints of $ AB$ and $ CD$,respectively
If $ O$ be the intersection of $ AC$ and $ BD$,$ P$ and $ Q$ be the projection of $ O$ on $ AD$ and $ BC$,respectively
Find the angle between the lines $ MN$ and $ BD$[/quote]

 I think the angle between the lines $ BD$  and  $ PQ$ is $ \frac{\pi}{2}$

It was noticed on this forum before, it's an old problem.
However, take R and S the midpoints of AO and BO respectively. See that MR is the midline of triangle AOB and PR median in the right angled triangle APO, so the triangles PRM and MSQ are equal, hence MP = MQ. Similarly PN = NQ and PMQN is a kite.

Best regards,
sunken rock


Τι ειναι στην πραξη αυτο? Μπορειτε να το εξηγησετε με πιο απλα λογια?

Καλος ηρθες στην παρεα μας... :wink: Προφανος ειναι ο συντελεστης του διωνυμου.Τι ταξη πας αν επιρεπεται?

Πανεπιστημιο και εχω να αναπτυξω προγραμμα σε γλωσσα C που να υπολογιζει τον διωνυμικο συντελεστη δυο αριθμων...

Σαν διευκρινιση δινουν αυτο:

[quote]Dιευκρίνιση: Wς διωνυμικό συντελεστή C(n, k) δύο ϕυσικών αριθμών 0 · k · n, εννοούμε έναν ϕυσικό
που ορίζεται αναδρομικά απο την σχέση:
C(n, k) = 1, όταν k = 0 ή k = n
ή
C(n, k) = C(n − 1, k − 1) + C(n − 1, k), όταν 0 < k < n[/quote]

Είναι ο συντελεστής μιας δύναμης του x του αναπτύγματος της παράστασης $ (x \plus{} 1)^n$ (ή οποιουδήποτε διωνύμου).
Ορίζεται και ως ένας αριθμός που γράφεται ως συνδυασμός n ανά k ή διαφορετικά διατυπωμένο ως
 $ \frac {n!}{k!(n \minus{} k)!}$ με n, k φυσικούς

Η ιδιότητα που δίδεται ως διευκρίνηση είναι η τριγωνική αναγωγική ισιότητα που ισχύει σε διωνυμικούς συντελεστές

Καλα εκανα που σε ρωτησα.... :) Ειναι απλο....Για λογους συντομιας και απλοτητας εχουμε τον συντελεστη διωνυμου και τον συμβολιζουμε με $ C(n,k)\equal{}n!/k!(n\minus{}k)!$ και επειδη ειναι συντελεστης του πολυωνυμου $ (x\plus{}1)^n$ ειναι παντα ακεραιος.....

Βασικά Billopoulos δε σου χρειάζεται ο αναλυτικός τύπος, αφού εσύ πρέπει να δουλέψεις με τον αναδρομικό

[quote]Dιευκρίνιση: Wς διωνυμικό συντελεστή C(n, k) δύο ϕυσικών αριθμών 0 < k < n, εννοούμε έναν ϕυσικό
που ορίζεται αναδρομικά απο την σχέση:
C(n, k) = 1, όταν k = 0 ή k = n
ή
C(n, k) = C(n − 1, k − 1) + C(n − 1, k), όταν 0 < k < n[/quote]

Χρησιμοποιώντας τον αναδρομικό τύπο ανάγεις (στο πρόγραμμα) τον υπολογισμό του C(n,...) στον υπολογισμό των C(n-1, ...)
τον οποίο ανάγεις στον υπολογισμό των C(n-2, ...) και πάει λέγοντας...

Έχεις ξανα-ασχοληθεί με αναδρομικές συναρτήσεις στη C;

Λουκάς

Οχι δεν εχω ασχοληθει ξανα...

Προφανως ειναι αυτο που λες και προφανως θα υπαρχουν οι δυο περιπτωσεις και στην δευτερη μια επαναληπτικη δομη!

Παράδειγμα αναδρομικής συνάρτησης: [b]Εύρεση του ΜΚΔ δύο αριθμών α,β[/b]

Χρησιμοποιούμε την αναδρομική σχέση

[color=olive]ΜΚΔ(α,β) = ΜΚΔ(α, β-α) αν α<β
ή
ΜΚΔ(α,β) = ΜΚΔ(α-β, β) αν α>β
ή
ΜΚΔ(α,β) = α, αν α=β[/color]

Τότε η συνάρτηση στη C γράφεται:

[code]int GCD(int A, int B) {

   if (A==B) return (A); else
   if (A<B) return (GCD(A,B-A)); else
   return (GCD(A-B,B));

}[/code]

Έτσι η GCD καλεί τον εαυτό της μέχρι να βρει τη συνθήκη τερματισμού που εδώ είναι η A=B

Με παρόμοιο τρόπο γράφεται και η συνάρτηση που ζητάς

Αν και είναι λίγο άσχετο, αυτός ο αναδρομικός τύπος μπορεί να σχεδιαστεί μόνο στις τελευταίες γλώσσες προγραμματισμού (σε παλαιότερες, όπως Pascal δεν υπάρχουν τόσο ισχυρές αναδρομικές δομές).


topology
advanced fields
advanced fields unsolved

Let $X$ be the union of circles of $(x-\frac 1n)^2+y^2=\left(\frac 1n\right)^2$, its topology is the subspace topology of $\mathbb R^2$. $Y$ is the quotient space $\mathbb R/\mathbb Z$. Show that $X$ and $Y$ are not homoeomorphic.

$Y=\mathbb R/\mathbb Z$ is the circle with the usual topology so its fundamental group is $\mathbb Z$. On the other hand, the fundamental group of $X$ is free on countably many generators, so they're not homeomorphic.

Any other observation without using fundamental group?

Looks like exactly the sort of thing I'm doing in one of my classes right now, including that we've used the specific examples.
If you remove the origin, $X$ has infinitely many connected components. If you remove any point of $Y$, it remains connected.

[quote="grobber"]Sure: the removal of any two points of $Y$ disconnects it, while we can easily find two points in $X$ which can be removed without disconnecting $X$.[/quote]
Seems to me that the removal of any point from $Y$ disconnects it?
And I still don't get it. Suppose $h: Y\to X$ is a homoeomorphism. We remove $y\in Y$ makes $Y$ disconnected, but we don't know what $f(y)$ is. Maybe $X-f(y)$ is also disconnected, if $f(y)$ contains two intersections of two circles.

[quote="jmerry"]A homeomorphism is a bijection. If you use what I said, the origin in the "Hawaiian earring" set must map to a point that disconnects the circle, and this is impossible.[/quote]
Hawaiian earring contains the origin? The book says only that it is the union of the circles.

And doesn't it look to you like the locus of the points $(x,y)$ with $\left(x-\frac 1n\right)^2+y^2=\frac 1{n^2}$ contains the origin $(0,0)$ of the plane?

I thought completely wrongly  :oops:  :oops:  :oops:  I was thinking of $(x-\frac 1n)^2+(y-\frac 1n)^2=(\frac 1n)^2$...  Now I see it. Thank you all.


inequalities
induction
inequalities unsolved

Prove that $ \frac{1}{ 2^{2}}\plus{}\frac{1}{ 3^{2}}\plus{}...\plus{}\frac{1}{ n^{2}}<\frac{n\minus{}1}{ n}$, for $ n$ natural numbers greater then 1. (Prove it by induction.)

[quote="ridgers"]Prove that $ \frac {1}{ 2^{2}} \plus{} \frac {1}{ 3^{2}} \plus{} ... \plus{} \frac {1}{ n^{2}} < \frac {n \minus{} 1}{ n}$, for $ n$ natural numbers greater then 1. (Prove it by induction.)[/quote]

when $ n \equal{} 2$,LHS=$ \frac {1}{ 2^{2}} < \frac {1}{ 2}$=RHS

assume that when $ n \equal{} k$,the given inequality holds,
when $ n \equal{} k \plus{} 1$

by induction hypothesis,we only need to prove $ \frac {1}{(n \plus{} 1)^2} \plus{} \frac {n \minus{} 1}{n} < \frac {n}{n \plus{} 1}$

$ \Longleftrightarrow n \plus{} (n \minus{} 1)(n \plus{} 1)^2 < n^2(n \plus{} 1)$

$ \Longleftrightarrow \minus{} 1 < 0$
which is obviously true.

can we solve it without using induction??? :?:

[quote="ridgers"]Prove that $ \frac {1}{ 2^{2}} \plus{} \frac {1}{ 3^{2}} \plus{} ... \plus{} \frac {1}{ n^{2}} < \frac {n \minus{} 1}{ n}$, for $ n$ natural numbers greater then 1. (Prove it by induction.)[/quote]

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

So

$ \sum_{k \equal{} 2}^{n}{\frac {1}{k^2}} < \sum_{k \equal{} 2}^{n}{(\frac {1}{k \minus{} 1} \minus{} \frac {1}{k})} \equal{} \frac {n \minus{} 1}{n}$

By the way, easy to know

$ \sum_{n \equal{} 1}^{ \plus{} \infty}{\frac {1}{n^2}} \equal{} \frac {\pi^2}{6}$.


[quote="gemath"]Given $x,y,z>0$ then
$(x+y+z)^{4}\ge 9(x^{2}+y^{2}+z^{2})(xy+yz+zx)$
is it true ?[/quote]
It's wrong: $x=y=1$ and $z\rightarrow0^{+}.$ :wink:

thank you but it is a pity for this result ! :D


Hello,

$ x_n\equal{}3x_n \plus{} n \minus{}1$

$ x_1\equal{}1$

Find $ x_n$.

[quote="Tomekk"]
$ x_n \equal{} 3x_n \plus{} n \minus{} 1$
[/quote]

$ x_{n\plus{}1} \equal{} 3x_n \plus{} n \minus{} 1$   :?:

I got:

$ x_n\equal{}\frac {1}{4}\left( 5\cdot3^{n\minus{}1}\minus{}2n\plus{}1  \right)$

I assume you meant $ x_n \equal{} 3x_{n \minus{} 1} \plus{} n \minus{} 1$.

Then $ \left (x_n \plus{} \frac {n} {2} \plus{} \frac {1} {4} \right) \equal{} 3\left (x_{n \minus{} 1} \plus{} \frac {n \minus{} 1} {2} \plus{} \frac {1} {4} \right )$, so $ x_n \plus{} \frac {n} {2} \plus{} \frac {1} {4} \equal{} 3^{n \minus{} 1}\left (x_1 \plus{} \frac {1} {2} \plus{} \frac {1} {4} \right )$. Finally, $ x_n \equal{} \frac {7\cdot 3^{n \minus{} 1} \minus{} 2n \minus{} 1} {4}$.

If, like in the post above, $ x_{n\plus{}1} \equal{} 3x_{n} \plus{} n \minus{} 1$, you work it out in a similar way  :) .

Hello, I apologise for typo, I meant $ x_{n\plus{}1}\equal{} 3x_{n}\plus{}n\minus{}1$.

I dont understand why you both did that kind of factorization, is there some kind of alghoritm for solving such problems?

I usually write values of $ x_1, x_2, ... x_5$ and try to "guess" formula, then prove it by induction.

hello, i got
$ x_n\equal{}\frac{5}{12}3^n\minus{}\frac{1}{2}n\plus{}\frac{1}{4}$ for
$ x_{n\plus{}1}\equal{}3x_n\plus{}n\minus{}1$ and $ x_1\equal{}1$.Sonnhard.


Prove that for integers $ n,p\geq 4$, for some positive $ x_1,x_2,...,x_n$ such that the sum is $ n$, we have that $ \sum_{i\equal{}1}^n\frac{1}{x_i^p}<\sum_{i\equal{}1}^nx_i^p$

The question just asks for "for some". Read it carefully!  :)

[quote="indybar"]Prove that for integers $ n,p\geq 4$, for some positive $ x_1,x_2,...,x_n$ such that the sum is $ n$, we have that $ \sum_{i \equal{} 1}^n\frac {1}{x_i^p} < \sum_{i \equal{} 1}^nx_i^p$[/quote]
$ x_1 \to 0$ ?

Choose $ x_1\equal{}x_2\equal{}x_3\equal{}\frac{2}{3},x_4\equal{}2,x_n\equal{}1$ for $ n\geq 5$. Then the left hand side is
\[ (n\minus{}4)\plus{}3\cdot \frac{3^p}{2^p}\plus{}2^{\minus{}p}\]
and the right hand side is
\[ (n\minus{}4)\plus{}3\cdot \frac{2^p}{3^p}\plus{}2^p\]
so after simplifying, we want to show
\[ 3\cdot 3^{2p}\plus{}3^p<3\cdot 4^p\plus{}4^p3^p\]
This is true because
\[ 3\cdot 3^{2p}\plus{}3^p<3^{2p\plus{}1}\plus{}3^{p\plus{}2}\equal{}(3\plus{}3^p)3^{p\plus{}1}<(3\plus{}3^p)4^p\]
as long as $ (4/3)^p>3$. But indeed, $ (4/3)^4\equal{}256/81>3$.

So this works for $ n\equal{}4$ as well, right?

When $ n\equal{}4$, same example works, except none of the variables are equal to 1. (In fact, setting all but 4 variables to 1 reduces the problem to the 4-variable case.)