[hide="Solution"]

$\displaystyle \text{The pythagorean theorem:}
a^2+b^2=c^2$

Let's call the side that is 9 b.

Sides:

a=unknown

b=9

c=41

Solution:

$a^2=c^2-b^2$

$a^2=41^2-9^2$

$a^2=1681-81$

$a^2=1600$

$a=40$

$Area = \frac {b \times h} {2}$

$Area = \frac {40 \times 9} {2}$

$Area = \frac {360} {2}$

$Area = 180 \displaystyle \text{ Square Units}$

[/hide]

[quote="236factorial"]You made an eye catching mistake in you area formula for a [b]triangle[/b][/quote]

Oops... :blush:

I knew that formula, it's changed now. :D

[quote="236factorial"]the length of a leg of a right triangle is 9, and its hypotenuse is 41. What is the area of the triangle?[/quote]
[hide]
$\text{The length of the missing leg is}$

$\sqrt{41^2-9^2}$
$\sqrt{(41-9)(41+9)}$
$\sqrt{32*50}$
$\sqrt{1600}$
$40$

$\text{So the area of the triangle is}$

$\frac{9*40}{2}$
$9*20$
$180$[/hide]


calculus integration inequalities trigonometry geometry trapezoid function

Prove that \[\int^\frac{\pi}{4}_0 \frac{\tan x}{x}dx\leqq\frac{4}{\pi}\ln2\]

I can show this numerically by approximating the integral with a composite trapezoid rule with two intervals. (Since the function is convex, the integral is less than any trapezoid rule estimate for it.)

You're looking for a very specific estimation technique, which produces the highly surprising $\frac4{\pi}$ on the right hand side. So far, I have not found that specific technique.

I agree. I also easy obtained estimate using trapezoid rule with two intervals.


trigonometry
logarithms
calculus
derivative
calculus computations

[1]: logarithmic differentation.
$(x^{\sin x})(\frac{\sin x}{x}+\cos x \ln x)$

[4]:
$\sin 2x$

Thank you for your reply,skylarth.

[quote="skylarth"][1]: logarithmic differentation.
$(x^{\sin x})(\frac{\sin x}{x}+\cos x \ln x)$ :)

[4]:
$\sin 2x$ :) [/quote]

kunny

[quote="adidasty"]these are pretty easy.....i wonder how hard derivatives can get. maybe they can get really hairy, i suppose, if you have to do partial derivatives multiple times and in terms of different variables.....[/quote]
See the topics on transforming partial differential equations for this. It's still basically a mechanical process, but it can get very complicated, and you need to be good with algebra.


function
vector
integration
complex numbers
real analysis
real analysis solved

Is there an additive real function f such that for any real interval J which is not a point, f(J) is the set of real numbers?

At first I thought there were no such functions, but now it looks to me like the following idea works:

Let $A=(a_i)_{i\in I}$ be a Hamel basis (basis of $\mathbb R$ over $\mathbb Q$). We may assume $1\in A$. $f$ is, of course, defined by the values it takes for $f(a_i)$. Take $f(1)=0$, so $f(x)=0,\ \forall x\in \mathbb Q$, and take $g$ to be a bijection from $A\setminus\{1\}$ to $A$. We take $f(x)=g(x),\ \forall x\in A\setminus\{1\}$.

$f:\mathbb R\to \mathbb R$ is surjective, and $f(x+r)=f(x),\ \forall x\in \mathbb R,\ \forall r\in\mathbb Q$. It's now clear, I think, that $f$ takes all real values in any non-degenerate interval.

I don't know what a Hamel basis is, so I can't follow the solution. Can someone enlighten me please? (Otherwise I'll look it up.)

I think we can come up with a concrete example by, for example, taking the real part of some function with lots of essential singularities or by looking at a function that is nowhere of bounded variation. I'll think about it.

It's a basis of $\mathbb R$ onsidered as a vector space over $\mathbb Q$. There's an article on Mathworld, of course, but it says the same thing.

[quote="grobber"] and take $ g$ to be a bijection from $ A\setminus\{1\}$ to $ A$..[/quote]
Should it be
"take $ g$ to be a bijection from $ A\setminus\{1\}$ to $ \mathbb{R}$."?

maybe I don't understand something..
if $ f(J) \equal{} \int_{J}...$?
and after take composition with some measure.
measure is aditive and f is aditive hence... e.t.c.  :maybe:

"additive" means here $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$, not $ f(A\cup B)\equal{}f(A)\plus{}f(B)$.

can you explan please there exist a bijection from A\{1} to A??

The Hamel basis $A = \{a_i \mid i \in I\}$ he is using is clearly infinite (in fact the cardinality of $I$ is $\mathfrak{c}$, that of the continuum). Furthermore, he assumes $1 \in A$, say it is $a_{i_0} = 1$. We can isolate a countable subset $J$ of $I$, given by $J = \{a_{i_n} \mid n = 0,1,\ldots \}$.

Now define $g \colon A \setminus \{1\} \to A$ by $g(a_i) = a_i$ if $i \not \in J$, and $g(a_{i_n}) = a_{i_{n-1}}$ for all $n=1,2,\ldots$. Clearly $g$ is a bijection.


We knowed the inequalities $ m_a \plus{} m_b \plus{} m_c\geq\sqrt {3}p$ is not right. But I finded the inequalities
$ \sqrt {m_a^2 \plus{} \frac {3(b \minus{} c)^2}{4}} \plus{} \sqrt {m_b^2 \plus{} \frac {3(c \minus{} a)^2}{4}} \plus{} \sqrt {m_c^2 \plus{} \frac {3(a \minus{} b)^2}{4}}\geq\sqrt {3}p$
is right. Can you prove it? Good luck!:D
........
tk17 nkht

[quote="thegod277"]We knowed the inequalities $ m_a \plus{} m_b \plus{} m_c\geq\sqrt {3}p$ is not right. But I finded the inequalities
$ \sqrt {m_a^2 \plus{} \frac {3(b \minus{} c)^2}{4}} \plus{} \sqrt {m_b^2 \plus{} \frac {3(c \minus{} a)^2}{4}} \plus{} \sqrt {m_c^2 \plus{} \frac {3(a \minus{} b)^2}{4}}\geq\sqrt {3}p$
is right. Can you prove it? [/quote]
Let $ a\equal{}y\plus{}z,$ $ b\equal{}x\plus{}z$ and $ c\equal{}x\plus{}y.$ Then $ \sum_{cyc}\sqrt {m_c^2 \plus{} \frac {3(a \minus{} b)^2}{4}}\geq\sqrt3p\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{5a^2\minus{}6ab\plus{}5b^2\minus{}c^2}\geq\sqrt3(a\plus{}b\plus{}c)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{x^2\plus{}y^2\plus{}z^2\plus{}xz\plus{}yz\minus{}2xy}\geq\sqrt{3}(x\plus{}y\plus{}z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{(z(x\plus{}y\plus{}z)\plus{}(x\minus{}y)^2)(y(x\plus{}y\plus{}z)\plus{}(x\minus{}z)^2)}\geq3(xy\plus{}xz\plus{}yz).$
But $ \sum_{cyc}\sqrt{(z(x\plus{}y\plus{}z)\plus{}(x\minus{}y)^2)(y(x\plus{}y\plus{}z)\plus{}(x\minus{}z)^2)}\geq$
$ \geq\sum_{cyc}\left((x\plus{}y\plus{}z)\sqrt{yz}\plus{}|(x\minus{}y)(x\minus{}z)|\right)\geq\sum_{cyc}\left((x\plus{}y\plus{}z)\sqrt{xy}\plus{}x^2\minus{}xy\right).$
Let $ x\equal{}u^2,$ $ y\equal{}v^2$ and $ z\equal{}w^2,$ where $ u,$ $ v$ and $ w$ are non-negative numbers.
Hence, it remains to prove that
 $ \sum_{cyc}(u^4\plus{}u^3v\plus{}u^3w\plus{}u^2vw\minus{}4u^2v^2)\geq0,$ which obviously true.


let $ a\equal{}e^{(i2\pi)/n}$
evaluate:
$ \sum_{k\equal{}0,n\minus{}1}(1\plus{}a^{k})^{n}$

[quote="drapt@"]let $ a \equal{} e^{(i2\pi)/n}$
evaluate:
$ \sum_{k \equal{} 0,n\minus{}1}(1\plus{}a^{k})^{n}$[/quote]

You already posted it here : [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166827[/url]

And I gave you a solution.


Okay, I have a problem that includes a exponent, and right now I am not too sure how to make the exponent /look/ like an exponent, yet I have seen other users do what I cannot. Any ideas? I think you put a '^' somehwere, but is there other coding that goes along with that?

Do you mean $ \text{\LaTeX}$? If you want to post in the forum, enclose dollar signs around your math.

[code]$5^{3}$[/code]

yields

$ 5^{3}$.

See [[LaTeX]] for more information.

Someone, please move this to the "LaTeX and Asymptote" forum.

Okay, thanks! Let me try it...

 Sorry if this is in the wrong place. I can hate being new. XP

 $ 7^{8}$

 -Looks-


I would have qualified every year in high school had I just gotten AIME #1s right. I have never, in 4 years, gotten a number 1 right on an AIME. I was short by less than one AIME problem in 9th and 10th, and just barely made it this (junior) year despite not getting #1.

And there's no 12th grade option because?

i dunno.. i thought I put it ><
if you're a 12th grader, call it 11th :P

Nah, I'm still a freshman. I just found it strange how you put in a 4th grade but not a 12th grade option.

[quote="abacadaea"][quote="mathemonster"]No way 3 fourth graders made it.[/quote]

um...only one made it.

1 4th grader
1 5th
0 6th
13 7th
27 8th[/quote]

um... read the topic title.

[quote="not_trig"][quote="123456789"]There have been two 4th grade qualifiers in the past 3 years. When's the other time a 4th grade qualified?[/quote]

not everyone tells the truth on polls?[/quote]

That's right, I said 4th grade. LOL JKING!!! tricked ya, ha ha ha!!! :rotfl:

er... That isn't exactly funny. Xeorns is trying to get some information, but you are ruining it because you aren't truthful in the poll.

AoPS polls are notoriously unreliable for precisely this reason.

[quote="Darmani"]Wow....took me about half the thread to realize the question was "When did you first qualify for USAMO?" rather than "When did you first become highly interested in USAMO?"

/me goes off cursing the ambiguities of the English language[/quote]

Oh lol, I just realized that there can be some ambiguity there. Me being a US born, I just assumed whatever popped into my head first (so basically, whatever was the literal sense)

As for the 1st time I got into USAMO, that was in 2010, when I was a 10th grader.

LOLLL.
now there's "10" genii 4th graders that made it to usamo :P

I speak from truth - I got into USAMO in 6th grade (last year).

I am pretty certain that only 2 have made it since 4th grade, 1 from 5th grade, and 3 from 6th grade.

I sense some "wtf" here at the (currently) 13 4th graders that made USAMO.
Yeah, it's probably not true.
To all you liars out there: STOP

I think the reason 4th-graders make it to USAMO is because they accidentally fill in the wrong grade. Perhaps it's actually a 10th grader like me.

This is my first year doing USAMO, and I'm in grade 10, which is the second most popular option, it seems.

[quote="bokagadha"]If there was a 3rd grade option, then we'd have a bunch of 3rd graders making USAMO...[/quote]
And less 4th-graders... XD

Why not discount the 4th-grader option altogether.

NOTE: besides the grade-4 outlier, the votes follow a somewhat normal distribution. Does this strike any as weird?

[quote="worthawholebean"][quote="sdkudrgn88"]Does this strike any as weird?[/quote]No. Those show up a lot. :D[/quote]

Indeed.. Somebody should moderate poll results. Or would that count as censorship? XD

[quote="worthawholebean"][quote="sdkudrgn88"]Does this strike any as weird?[/quote]No. Those show up a lot. :D[/quote]
Yeah, but in this kind of thing? I'd expect there is a spike at 9 or 10, which there isn't. It just kinda gradually goes up.

I have a database going back to 1987 of all USAMO qualifiers. In that time, there have been exactly two 4th graders, two fifth graders, and three sixth graders. All of those have come since 2007. I see no reason to suspect mistakes; one of those fourth graders is listed at "_ Elementary" and the other also qualified as a fifth grader. The other fourth grader apparently took the tests at his local high school, which has had no other qualifiers. The sixth graders are on Mathcounts state teams and qualified again for the USAMO this year, and I've met two of them myself.

It gets a little more crowded at 7th grade, with 46 total. More than half of those are in the last three years.

[quote="Sly Si"]Man, when I was competing, only one or two 8th graders made the USAMO each year. And a 7th grader getting in was eye-popping.[/quote]You're overstating it. 2002-5 averaged two 7th graders and nine eighth-graders per year.
My time (1997-2000), at 1.5 seventh graders and 8 eighth-graders per year, was pretty similar. The emptiness is before that. From 1987-1995, only 8 eighth-graders and one seventh-grader made it. Total. Of those eight people, seven eventually made the IMO team at least once.

then there is brian lawrence who made MOP in 7th grade XD (but it was megamop. oh well, still mop right?)

Or Daniel Kane, who made MOP as a 7th grader in 1999, when it was only 24 people. He skipped to 9th grade the next year.

Umm jmerry, I'm pretty sure Reid Barton made USAMO in the sixth grade, and that was definitely before 2007.

Not according to the lists. He did make MOP as a seventh-grader in 1996, however.

Wow, there are now 14 fourth graders who made USAMO, this is getting weird.....   :huh:

One of our math circle members, Michael Ma, is a certified 4th grade qualifier:

[url]http://metroplexmathcircle.wordpress.com/2009/04/18/usamo-qualifiers/[/url]

wow. a perfect score on the AMC10. and at least a 7 on the aime..... :O impressive.

[quote="bokagadha"]If there was a 3rd grade option, then we'd have a bunch of 3rd graders making USAMO...[/quote]
heck, people would be making USAMO (or IMO for that matter) before they were born given that there was an option like that. :ewpu:

Soon we'll will see headlines like

"Rise in women over the age of 30 at International Math Olympiad. Youngest Olympian is -.5 years old.
- Associated Press"

I'm sure I'll be violating a lot of inference conditions, and plus 16 4th-graders is *obviously* not possible, but here goes ...

$ H_0: p \equal{} 0.002$
$ H_a: p > 0.002$
$ n \equal{} 130$
$ \hat{p} \equal{} \dfrac{16}{130} \approx 0.1231$

We get
$ z \approx 30.90$
$ p \approx 0$

We reject $ H_0$ and conclude the proportion of self-purported 4th-graders is significantly larger than 0.002.

QED


In physics, what exactly is a tension?  Please define it and provide a few examples involving tension (like in strings and ropes) because I am quite confused especially when two or more objects are connected and find the tension.

Thank you.

It's just another force that is found in ropes, strings, bands, pulleys...  It's  a force that holds two things together.  Like in a rope, if the pulling force exceeds the tension in the rope, then the rope will snap and the two ends will go flying.  Any problem that involves a pulley involves tension in the rope.  And the tension is always in the opposite direction as the pulling force.


In class we considered the design of a two lane suspension bridge with a spam of 160 meters and a drape which was 10% of the span. The combined weight of the roadway,the traffic and the hanger cables totals about 100kN/m. The weight of the main cables, in this case, can be estimated as being less than 3 kN/m. To support a total weight of 103 kN/m the maximum tension in the cables will need to be 22190kN. If we use 5 mm diameter steel wires with an ultimate strength of 1500MPa and we use a factor of safety of 2.25, a total of 1696 wires will be required. As steel weighs 77kN/m^3, these wires in this example will weight 2.56 kN/m and hence our estimate of 3 kN/m is appropriately conserative. Prepare a plot showing how the number of wires required to safelt support the bridge load of 100kN/m increases as the span of the bridge increases. What is the longest span that could be built?
i tried reading this question a couple of times and find it really confusing, if someone knows how to do it please give me some hints on how to start .. thx

This should be moved to physics (I think). :D


[quote="GameBot"]Find the arithmetic mean of the first thousand positive odd numbers.[/quote]

The sum of the first $ n$ numbers is $ n^2$, so the sum of the first 1000 positive odd numbers is $ 1000^2$. There are 1000 numbers in total, so our answer is $ 1000^2 \div 1000 \equal{} 1000$.

ACTUALLY, the sum of the first $n$ odd numbers can be found by using the arithmetic sum formula. The thousandth odd number is $1999$ so $\frac{2000(1000)}{2}=1000000$. We want mean, so $\frac{1000000}{1000}=1000$

[quote=Mathloser4]The sum of the first n numbers is actually $\frac{n(n+1)}{2}$. So, the sum of the first 1000 positive integers is 500,500. Then you multiply that by $\frac {3}{7}$ to get 214,500. 214,500 is the sum of the first 1000 positive odd numbers. Now to get the arithmetic mean, you just divide that by 500 to get 429. Therefore, the arithmetic mean of the first 1000 positive odd numbers is 429.[/quote]

Why should one multiply by $\frac37$ to get the sum of odd numbers?

Clarification of post from 10 years ago:

The sum of the first n ODD numbers is n^2.


I justed wanted to know if everyone who took the RMO will get $ 15$ points for problem number 1 because of the typo.
[hide]
I feel very bad for the problem setter as he/she will come under a lot of scrutiny. :( [/hide]

well giving 15 pts will just upp the cutoff

i think there shud b marking on basis of what the student has assumed

[hide]i do feel the typo is just a trick :maybe:  :oops: [/hide]

but i called my regional coordinator and he said that perpendicular should have been drawn and perpendicular(produced)will intersect and all that.

so they are not going to give marks.

nothing,problem was eay
but i sam a big fool(or i wasn't in my best touch)
i calculated $ 4\cdot5\cdot4\cdot3$ as $ 60$  :(  :(

this is going to cost me a lot

for solutions please check other topic

i too fooled up in the calculations

any marks ????????????


Prove that in any triangle $ ABC$ there is the inequality

$ \boxed {\ \sqrt[3]{(b\plus{}c\minus{}a)\cdot(c\plus{}a\minus{}b)\cdot (a\plus{}b\minus{}c)}\ \le\ \frac {ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c}\ }\ \le\ \frac {a\plus{}b\plus{}c}{3}\ .$

[quote="Virgil Nicula"]Prove that in any triangle $ ABC$ there is the inequality

$ \boxed {\ \sqrt [3]{(b \plus{} c \minus{} a)\cdot(c \plus{} a \minus{} b)\cdot (a \plus{} b \minus{} c)}\ \le\ \frac {ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c}\ }\ \le\ \frac {a \plus{} b \plus{} c}{3}\ .$[/quote]
See the stronger here:
http://www.mathlinks.ro/forum/viewtopic.php?t=256282

Here is one stronger than the first : $ \frac {(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca}\ \le\ \sqrt 3\cdot\sqrt [3]{\frac pr}\ .$ Try it.

[color=purple] Dear Mr Virgil Nicula,
 It is a nice inequality but it is not hard.
  Here is my solution :
  By $ Rrs$ method we can rewrite your inequality in the following form :
\[ s\plus{}\frac{r(4R\plus{}r)}{s}\ge\frac{4\sqrt{3}}{3}\sqrt[3]{s^2r}
\]
 According to the famous one : $ 4R\plus{}r\ge\sqrt{3}s$ It is suffices to show that
\[ s\plus{}\sqrt{3}r\ge\frac{4\sqrt{3}}{3}.\sqrt[3]{s^2r}
\]
By [b]Cauchy[/b] inequality we have :
\[ s\plus{}\sqrt{3}r\equal{}\frac{s}{3}\plus{}(\frac{s}{3}\plus{}\frac{s}{3}\plus{}\sqrt{3}r)\ge\frac{\sqrt{3}}{3}\sqrt[3]{s^2r}\plus{}3\sqrt[3]{\frac{s^2r}{9}.\sqrt{3}}\equal{}\frac{4\sqrt{3}}{3}\sqrt[3]{s^2r}
\]
which is the desired.
Equality holds iff the triangle is an equilateral triangle.

[/color]


The old countdown system (pre-2004) worked like this:

10th place written vs. 9th place written
winner vs. 8th
winner vs. 7th
etc...

How many different possible combinations of rankings were possible in this system?

[hide] There are two possible choices for 10th, 2 for 9th, 2 for 8th, etc. Then the winner is forced, so there are $2^9$ combinations, or $512$.[/hide]


1. Let $ r_a,r_b,r_c$, $ h_a,h_b,h_c$, and $ r$ be the ex-radii, altitudes, and the in-radius of triangle $ ABC$, respectively.

Show that $ \left(\dfrac{r_a\plus{}r_b\plus{}r_c}{9}\right)\left(\dfrac{h_a\plus{}h_b\plus{}h_c}{9}\right)\ge r^2$

2. Let $ I$ be the incenter, and $ E_a$ be the center of the excircle with respect to vertex $ A$ in triangle $ ABC$. If $ M$ is the midpoint of $ IE_a$, show that:

a)
\[ \angle ACB \cong \angle IMB\]

b)
\[ CM\cong BM\cong E_aM\cong IM\]

1) Use $ r_a\equal{}\frac{2P}{b\plus{}c\minus{}a},h_a\equal{}\frac{2P}{a},r\equal{}\frac{2P}{a\plus{}b\plus{}c}$

We need to prove:

$ \left(\sum \frac{1}{b\plus{}c\minus{}a}\right)\left(\sum \frac{1}{a}\right)\ge \frac{81}{(a\plus{}b\plus{}c)^2}$ This turns into an easy algebraic inequality with $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$

2) It's a well known fact that if X is the midpoint of arc BC on the other side of A, then $ XI\equal{}XB\equal{}XC\equal{}XE_a$, so the midpoint is $ X$, or $ M$ is the midpoint of arc BC.
 [geogebra]1084f7879a210470ed2ce9585a9ad21759ee7c82[/geogebra]

Now use basic properties of angles in a circle.

2b) Follows from that above.


How many ordered pairs of real numbers $(a,b)$ satisfy $(a+bi)^{2002}=a-bi$?

[hide="Solution"]If we put $a+bi=re^{i\phi}$, then

$r^{2002}e^{i2002\phi}=re^{i(-\phi+2k\pi)}$

giving $r^{2002}=r\iff r=0\lor r=1$ and $2003\phi=2k\pi\iff\phi={2k\pi\over 2003}$

Hence the solutions are $z_{k}=\mathrm{cis}\,{2k\pi\over 2003},k=\overline{1,2003}$ and $z_{2004}=0$, so there are $2004$ such ordered pairs.[/hide]

Well, to summarize, [hide] we clearly know it has magnitude 1.  Then $(a+bi)^{2003}=1$ or $a+bi=0$[/hide]


Two of them:

a)  Find the general term of the sequence which begins: 1, 2, 3, 8, 30, 144, ...

b) Find the term $a_{37}$ of the sequence which begins: $a_1 = 3, a_2 = 3, a_3 = 5, a_4 = 4, a_5 = 4, a_6 = 3, a_7 = 5, \ldots$ and satisfies $a_{-n} = 8 + a_n$ for $n > 0$.

For b, what does $a_{-n} = 8+a_n$ have anything to do with it?

[quote="paladin8"][quote="bdpiang"]b) 11 - no of alphabets in the name of n (thirty seven)
 oops! forgot to clarify that for negative nos. add 8 (negative=8) to the nos.
that's what a{-n}=8+a{n} means[/quote]

??? I don't understand.[/quote]

basically, the value of $a_n$ is equal to the number of letters in spelling of $n$.  Thus $a_{37}=11$ because "thirty seven" contains 11 letters.

oh i see now... wonder why we had to add 8 for every "[b]negative[/b]" n's.


Determine all pairs of rational numbers (x,y) such that $ x^{3}\plus{}y^{3}\equal{}x^{2}\plus{}y^{2}$ :P

[quote="tuanct"]Determine all pairs of rational numbers (x,y) such that $ x^{3}\plus{}y^{3}\equal{} x^{2}\plus{}y^{2}$ :P[/quote]
It mean, that equation $ z^{3}\minus{}z^{2}\plus{}c\equal{}0$ had 2 rational roots x,y. Because third root is 1-x-y it is rational too.
We have $ 0\equal{}xy\plus{}(x\plus{}y)(1\minus{}x\minus{}y)\equal{}0$, therefore (x,y) is rational point on the curve $ x^{2}\plus{}xy\plus{}y^{2}\equal{}x\plus{}y$.
One solution (0,0).Let $ y\equal{}tx, t\in Q$, then $ x\equal{}\frac{1\plus{}t}{t^{2}\plus{}t\plus{}1},y\equal{}\frac{t^{2}\plus{}t}{t^{2}\plus{}t\plus{}1},t\in Q$ give all solutions.

Suppose that $ x\equal{}\frac{a}{c}$;$ y\equal{}\frac{b}{c}$  where $ a,b,c\in Z$
My equation is equivalent to $ a^{3}\plus{}b^{3}\equal{}c(a^{2}\plus{}b^{2})$
We only need to find all pairs $ (a,b)$ such that $ (a^{2}\plus{}b^{2})|(a^{3}\plus{}b^{3})$
We have $ a^{3}\plus{}b^{3}\equal{}(a^{2}\plus{}b^{2})(a\plus{}b)\minus{}ab(a\plus{}b)$
hence $ (a^{2}\plus{}b^{2})|(ab(a\plus{}b))$ .Denote $ \gcd(a,b)\equal{}d$ and $ a\equal{}dk$ ,$ b\equal{}dh$,$ \gcd(k,h)\equal{}1$
then we get $ (k^{2}\plus{}h^{2})|(dkh(k\plus{}h))$ We easily see that $ \gcd(kh,k^{2}\plus{}h^{2})\equal{}1$ then $ (k^{2}\plus{}h^{2})|(d(k\plus{}h))$
Other hand, $ \gcd(k\plus{}h,k^{2}\plus{}h^{2})\equal{}1$ or $ 2$ ,where $ k,h$ are the same(or not) parity.
If $ k,h$ are not the same parity then $ d\equal{}(k^{2}\plus{}h^{2})l$ where $ l\in Z$
If $ k,h$ are the same parity then $ d\equal{}\frac{l(k^{2}\plus{}h^{2})}{2}$ where $ l\in Z$


can you find the sum of the series whose terms are the reciprocal of
all positive integers whose [b]only [/b]prime factors are $ 2$ and $ 3$... i.e.  :D

$ \mathbb S\;\equal{}\;\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{6}\plus{}\frac{1}{8}\plus{}\frac{1}{9}\plus{}\frac{1}{12}\plus{}\frac{1}{16}\plus{}\frac{1}{18}\plus{}\frac{1}{24}\plus{}\frac{1}{27}\plus{}\frac{1}{32}\plus{}\ldots\;\;\plus{}\;\;\ldots$


geometry
3D geometry
sphere
inequalities
IMO

Hi, I am a long way removed from high school but sometimes I look at IMO problems for fun when bored at work. Here are my thoughts on this problem.

First put the formula into the following form, M = f( a, b, c ). Now we just need to find f's maximum. One thing to point out, for a constant s, f( sa, sb, sc ) = f( a, b, c ). If we find the maximum value of f on a sphere a^2 + b^2 + c^2 = r^2 for a constant r we basically solved the problem since the maximum on co-centric spheres are the same. Note that the denominator of f on a sphere is simply r^4 so all we need to do is find the maximum of the numerator.

I can think of a way to do this using gradients. However if I accomplish this how would IMO markers mark this problem since I used something technically outside of high school math. Would they mark it as 0 or does that garner partial credit? I do this out of interest so obviously I'm not satisfied with what I have so far for this problem. In seeking the more elegant solution I'm just wondering if I am totally going down the wrong path. I don't really want any more hints besides a 'Yes' or 'No'.

If you do it correctly, you get a 7. If you make even a minor error, however, you get very low marks (I got a 2). No, this is not the way for the elegant solution.

So you are saying even if I use gradients if I arrive at the correct answer I'll get a 7? Good to know.

Also in searching for the more elegant solution I should abandon my current line of thinking and start from scratch?

[quote="Daren"]So you are saying even if I use gradients if I arrive at the correct answer I'll get a 7?[/quote]

Not if you arrive at the correct answer, but if your proof is completely correct.

[quote="Daren"]Also in searching for the more elegant solution I should abandon my current line of thinking and start from scratch?[/quote]

More or less. Well, some things you can keep, e. g. the equality case - when proving an inequality, it is always important to know when it becomes an equality.

  darij

The exact method I wanted to use is called lagrange multiplier. It assums the gradient of the sphere and the gradient of the numerator of f() are parallel. This reduces to 4 equations and 4 unknowns but is not easily solved analytically. Oh well.


Hola

Aqui es un conversacion entre yo y mi amiga para mi espanol clase. Necesito saber si todas cosas son buenas.  :D Estoy en espanol III pero todavia necesita ayudame. Muchas gracias.

Conversacion es de arte. Yo es C y mi amiga es L.

C: Hola
L: Hola
C: ¿Has leído algo de la pintura?
L: Sí, la encuentro formidable.
C: La artista es muy creativa.
L: Cambiando de tema, ¿qué me dices de la danza contemporánea?
C: Para ser sincero, me cae gordo.
L: Eso me recuerda el músico incomprensible.
C: A propósito, ¿qué has oído de la orquesta?
L: La orquesta es hermosa. Quiero patrocinarla.
C: Me parece que el cantante es muy entretenido.
L: Para decir la verdad, no lo soporto porque él es pésimo.
C: Adiós.
L: Chao.

[quote="Silverfalcon"]C: ¿Has leído algo de la pintura?[/quote]
[color=blue]Quedaría mejor decir: [b][i]¿Has leído algo acerca de la pintura?[/i][/b] o [i][b]¿Has leído algo sobre la pintura?[/b][/i][/color]
[quote="Silverfalcon"]C: Para ser sincero, me cae gordo.[/quote]
[color=blue]No comprendo qué quieres decir con [i]…me cae gordo[/i].[/color]
[quote="Silverfalcon"]L: Eso me recuerda el músico incomprensible.[/quote]
[color=blue][i][b]Éso me recuerda al músico incomprensible.[/b][/i] O bien: [b][i]Éso me recuerda al músico, a quien nadie comprendía.[/i][/b][/color]
[quote="Silverfalcon"]C: A propósito, ¿qué has oído de la orquesta?[/quote]
[color=blue]Al parecer, solicita una opinión acerca de la orquesta, más que saber si la oyó. En ese caso, quedaría de este modo: [b][i]¿Qué opinas acerca de la orquesta?[/i][/b] En caso contrario, puedes aplicar lo mismo que en la referencia a la pintura.[/color]
[quote="Silverfalcon"]L: Para decir la verdad, no lo soporto porque él es pésimo.[/quote]
[color=blue]Está bien escrita, mas propongo la siguiente: [b][i]Sinceramente, no lo soporto. Es realmente pésimo.[/i][/b][/color]
[quote="Silverfalcon"]Chao.[/quote]
[color=blue]También se puede decir [b][i]Chau.[/i][/b], aunque no tiene mucha importancia.

Recuerda que en el saludo también debes utilizar los signos de puntuación.

Bonito esfuerzo el tuyo. Éxitos en tu aprendizaje de tan bello idioma.[/color]


I am interested in what people think of the working atmosphere of most quants. In particular...

1) Do you think there is a lot of bureaucratic nonsense that quants need to deal with in many financial firms?

2) Is there a lot of politics at the financial firms that quants work at?

3) I have heard that the financial industry is a very cutthroat environment to work in. What do you think about this?

I appreciate any input you may have.

[quote="ffdbzathf"]i don't know much about the details of being a quant, but in response to 2), I've heard that politics is kept to a minimum.  At DE Shaw, they didn't even let the employees know who the partners were.[/quote]

Things aren't quite that 'idyllic'.  Unless things have changed, you know who the partners are if you want to.

As for the general issue of politics, they exist, of course, but much less so in the profit-producing areas.  There, performers get rewarded.  Politics tend to dominate areas (not just in finance) where performance is less easy to measure or monitor (witness academia).  In my group, there were occasional 'sibling rivalries', but nothing serious.  And politicking simply wasn't part of career advancement.

As for red tape, that depends on the type of firm.  Most hedge funds will have very little.  Large financial firms will have plenty.  I rarely, if ever, had to deal with that sort of stuff.  I had some grimy detail work to do to get my trades done, but it wasn't useless work.  (Legal work, helping the accounting groups deal with the trades, etc - not exciting work, but very important work.)

The environment is very competitive/cooperative, since you both want to stand out among your peers, but you also do better when they do better.  It is competitive, but often much more of 'our firm' against 'everyone else' instead of you against the person sitting next to you.  If you're in the latter situation, find somewhere else to work.  It's not as cutthroat as pre-meds in undergraduate school tend to be, for example.

I worked on the same trading desk as Richard at DE Shaw for about a year.  An educated guess as to my identity might be made from my avatar and nickname :) .  I've also worked as a quant at a couple of other less well known trading firms in the years since then.  So I thought I would offer my thoughts on this as well.

My belief is that a lot depends on how you approach the quant role--to be honest, I think Richard may have been smarter in how he approached it than I was.  You need to take the approach that being a quant is something that you are going to do during a certain phase of your life to attain a certain level of financial security.  You need to set the goal that it is something that you are doing to make a substantial amount of money within a relatively short amount of time--and, although it involves risk, it is probably less risky than other paths (like a tech startup) that technically oriented people could follow to achieve financial security.

So what you are going to find is that you meet a lot of other quantitative people whose goal is to make a lot of money.  What you end up getting is pretty much what you would expect:  a combination of the culture of mathematics and the culture of Wall Street greed.  If you are a mathematician whose goal is to make a lot of money in a short amount of time, it is a far more conducive environment to do so in than, say, trading on the floor of the NYSE.  But at the same time it isn't going to be like attending a number theory colloqium in your college math department.  Quants on Wall Street are interested in making money--but they have the chance to network with other people of a similar background who are also interested in making money.  That background doesn't change, and it can be very supportive working with others whose background is similar, but it must be remembered that the goal is now different--the goal isn't to do great research or achieve tenure but rather to make money.

It is important to, early on in any quant job, get a sense as to how they approach their mathematics.  You aren't (in my experience) going to find complete mathematical rigor in any quant job.  Within any quant group, there are certain things that seem to be accepted more or less on faith--and they vary from one group to another.  It is key, early on, to get a sense as to what the group simply accepts on faith.  Within that basic structure, they then apply rigorous mathematical techniques to maximize their profits.  If you use your mathematical ability to maximize their success within their basic structure, you will be successful.  If you try to use your mathematical aptitude to challenge their basic assumptions--even in cases where those basic assumptions might not be 100 percent true all the time--you will be less successful.


inequalities
limit
algebra
partial fractions

Evaluate $\sum_{n=1}^\infty \frac{n}{(n+1)(n+2)}$.

I tried to do this with partial fraction decomposition. Using $n=A(n+2)+B(n+1)$, I put in -1 for n and found A = -1, and put in -2 for n and found B = 2, so that $\frac{n}{(n+1)(n+2)} = \frac{2}{n+2}-\frac{1}{n+1}$. Then, I wrote out a few terms and found that it seemed to be the same as $\sum_{n=1}^\infty \frac{2}{2n+1}$, but as far as I can tell, that sum diverges (which leads me to believe I am wrong, as this came out of AoPS Vol. 2). Does anyone know where I went wrong?

I think $ \sum_{n=1}^\infty \frac{n}{(n+1)(n+2)} = \sum_{n=1}^\infty \frac{1}{n+2}$
We have a beautiful inequality $ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}>\frac{1}{2}$ then $ \sum_{n=1}^\infty \frac{1}{n+2}=+\infty$

[quote="darkmaster"]I think $ \sum_{n=1}^\infty \frac{n}{(n+1)(n+2)} =-\frac{1}{2}+ \sum_{n=1}^\infty \frac{1}{n+2}$
We have a beautiful inequality $ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}>\frac{1}{2}$ then $ \sum_{n=1}^\infty \frac{1}{n+2}=+\infty$[/quote]

how did you get
$\displaystyle \sum_{n=1}^\infty \frac{n}{(n+1)(n+2)} =-\frac{1}{2}+ \sum_{n=1}^\infty \frac{1}{n+2}$

I wrote it out as $\displaystyle \frac{n}{(n+1)(n+2)} = \frac{2}{n+2}-\frac{1}{n+1} $

And by writing the first 10-12 terms, I found that everything would eventually cancel except the first 2/3. Therefore, I thought that the series would converge and 2/3 would be the answer. Can someone enlighten me?

I wrote out about the first fourteen of the terms (as the series seemed odder than most), and it seemed that all the even-denominator fractions canceled, but for every fraction with an odd denominator had one appearance with a numerator of 2, one with a numerator of -1, and one with a numerator of 1, making it $\frac{2}{2n+1}$.

If you give this problem to Maple, it will answer $ +\infty$
We can write the sum as: $ \frac{2}{3}-\frac{1}{2}+\frac{2}{4}-\frac{1}{3}+\frac{2}{5}-\frac{1}{4}+....$ and rellize that if we have $ +\frac{2}{n+2}$ then we have $-\frac{1}{n+2}$(except $-\frac{1}{2}$). So the infinite sum reduces to $ -\frac{1}{2}+\sum_{n=1}^\infty \frac{1}{n+2}$.

All this stuff you're throwing at this problem - partial fractions, even/odd terms, clever cancelation tricks - is wasted energy for a problem that you should decide within 15 seconds of seeing it. The first question should be, "Does it converge?" For a postive term series (which this is), that's purely a matter of size. So we ask: what do the terms of this series look like for large $n.$ There's a clear answer to that question: the terms look like $\frac1n.$ But we know that the harmonic series $\sum\frac1n$ diverges, so this series must diverge.

Technically, I have just invoked the "limit comparison test": if $a_n$ and $b_n$ are positive sequences and if $\lim_{n\to\infty}\frac{a_n}{b_n}=L$ where $L$ is neither $0$ nor $\infty,$ then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

The limit comparison test lets you "de-clutter" the problem.

^ Exactly what he said.

Modulo a few convergence considerations, we have that:

$\displaystyle \sum_{n=1}^\infty \frac{n}{(n+1)(n+2)} \geq \sum_{n=1}^\infty \frac {n}{2n\cdot 3n}=\frac {1}{6}\sum_{n=1}^\infty \frac {1}{n}$

I get so frustrated and sick whenever I see the symbols sigma and the infinity symbol

[quote="Big Bolzano"]$\sum_{n=1}^\infty \frac{n}{(n+1)(n+2)} = \sum_{n=1}^\infty  (\frac{2}{n+2} - \frac{1}{n+1}) = \frac{1}{2}$[/quote]
What is it? Is it joke?  :D
Did you read previous posts? Series diverges!  :lol:  :lol:  :lol:

Here is another way, that people seem to have been getting towards but nobody explicitly said it:
$\sum_{n=1}^\infty \frac{n}{(n+1)(n+2)} \n = \sum_{n=1}^\infty  \frac{2}{n+2} - \frac{1}{n+1} = \sum_{n=1}^\infty  \frac{1}{n+2} - \frac{1}{n+1} + \frac{1}{n+2}$
The first two parts make a telescoping sum, so
$= -\frac{1}{2} + \sum_{n=1}^\infty \frac{1}{n+2}$
Harmonic series diverges, so this diverges as well.


A student who is 5' 6'' casts a 4' shadow. How many feet tall is a nearby tree which casts a 24' shadow?

We see that $ 24 \div 4\equal{}6$, so our scale factor is $ 6$.  We do $ 5.5 \times 6$, which is $ \boxed {150}$

um im not sure but the answer says its 396 but you got 150 will you explain to me why ???

 :play_ball:  :play_ball:  :play_ball:  :play_ball: I LIKE BALLS : )

[hide=Answer]33 ft[/hide]

Not sure why though. Could someone explain?

Also @above, the question is asking in feet. Even though you probably won't see this.

[hide = Solution]$5$ feet $6$ inches is $66$ inches. So someone $66$ inches casts a $4$ inch shadow. A $24$ inch shadow means $66 \cdot 6 = 396$ inches, or $\boxed{33}$ feet.[/hide]


The price of an article was increased $ p\%$. Later the new price was decreased $ p\%$. If the last price was one dollar, the original price was:

$ \textbf{(A)}\ \frac {1 \minus{} p^2}{200} \qquad\textbf{(B)}\ \frac {\sqrt {1 \minus{} p^2}}{100} \qquad\textbf{(C)}\ \text{one dollar} \qquad\textbf{(D)}\ 1 \minus{} \frac {p^2}{10000 \minus{} p^2}$
$ \textbf{(E)}\ \frac {10000}{10000 \minus{} p^2}$

[hide]

Let the original price be $ x$. We have $ (1\plus{}\frac{p}{100})(1\minus{}\frac{p}{100})x\equal{}1$. Hence, $ x\equal{}\frac{1}{1\minus{}(\frac{p}{100})^2}\equal{}\frac {10000}{10000 \minus{} p^2}$, or $ \textbf{(E)}$

[/hide]


quadratics
function
analytic geometry
algebra
polynomial
sum of roots

A quadratic function, in the form $f(x) = ax^2+bx+c$, has vertex $(h,f(h))$. If $h=\frac{578}{2}$, what is the sum of the function's zeroes?

[hide="A hint"]
x-coord of vertex = $\frac{-b}{2a}$
[/hide]

[hide="A giveaway"]
sum of roots = $\frac{-b}{a}$
[/hide]

The zeroes are always the same distance away from the vertex. It stands to reason that the answer would be double the x-coordinate of the x-axis, [b]578[/b].

The x-coordinate of the vertex is always $-\frac{b}{2a}$, while the sum of the zeros is $-\frac{b}{a}$, since you know that the first part is $\frac{578}{2}$, then the sum of the zeros is $\frac{578}{2}\cdot2=578$


Hi All,
My goal is to create a set of values with a predetermined average.  The problem is that each value has a condition such that they can not be set to a % that would create this average value.    I will explain the issue verbally and then mathematically.

Verbally,

I have a set of values.
Each value has another value associated with it and the total cannot be greater than 100%
I wish to make the average of all values say 50%
Right now some values are 50%, others are less (because of the conditional values)
If I add  (1 - (1 - <the current value>) * <a constant> to any values that can be increased,
I can make the final average 50%.
I can determine the constant value through trial and error, but would like to use an equation to figure out the constant.

Mathematically,

i = each value
C = The current i
T = The highest number that this i could be based on the condition
x = The constant <= 1
N = The total of all values
Y = The current average value
Z = The target average value
min = The lower of the values

Find x such that N=N*Z
In order to do so;
N = ∑(min(T[i], 1-1-C[i]*x)

This is a GIS problem, so it is kind of hard to explain.  Any help I can get
would be truely appreciated!!
Thanks
Alexis

Dr. Math,
I hope I can explain myself well enough for you to change your mind.
First let me describe the problem.

C = The current value of each square
T = The highest number that this square could be based on the condition
x = The constant <= 1
M = The total area of all squares
N = The desired total
Z = The target average value
Let me better explain the issue.

I have a raster dataset (think of it like a piece of graph paper where each square represents a value).  Each square represent a piece of land.  I want say, 50% of the entire page to be covered with trees or N=M*Z.  Easy enough, but there is a condition for each square.  Some squares have too much other stuff to fill them half full with trees.  So, some of the squares will have to be more than 50% covered and some with < 50% tree cover (due to the condition), such that the total tree cover is 50%.

I have so far set up my dataset so that each square is set as close to 50% as possible based on the condition.  I have figure out that in order to determine how much to add to these cells I can run the following equation;
1-1-C*x and when this value is less than T (the portion of the square available to be tree cover) I add this value to the current value for this square.
By trial and error, I can get a constant (x) to make the total coverage = 50%.
I would like to somehow calculate this value on the fly???

If you could help that would be great!
Thanks
Alexis


analytic geometry
trigonometry
integration
logarithms
calculus
calculus computations

A man starts walking from the origin of a rectangular coordinate system. His first step is of unit length along the positive $ x$-axis. He then spins on his heels through the counterclockwise angle of $ \theta$ and walks a second step of distance one-half unit. Then he spins on his heels through another positive angle of $ \theta$ and walks a third step of distance one-third unit. He continues this process endlessly.

For $ \theta$ other than $ 0$ and $ 2\pi$, it should be clear that this walk converges to a definite point in the plane.

Find the exact value of the distance from the origin of the point of convergence when $ \theta \equal{} \pi/2.$

maybe the approach should be

$ p(n)= x(n)+ iy(n)$ for n=1,2,3...

$ x(n)=\sum_{k=0}^n \frac{\cos (k \theta)}{k+1}$

$ y(n)=\sum_{k=0}^n \frac{ \sin (k \theta)}{k+1}$

so  $ p(n)=  \sum_{k=0}^n \frac{\cos (k \theta)+ i\sin ( k \theta)}{k+1}  =  \sum_{k=0}^n \frac{ e^{i k\theta}}{k+1} = ie^{-i\theta}\sum_{k=0}^n \frac{ e^{i (k+1)\theta}}{i(k+1)}$

so $ p( \infty ) =  ie^{-i\theta}\sum_{k=0}^\infty \frac{ e^{i (k+1)\theta}}{i(k+1)}=  ie^{-i\theta} \sum_{k=0}^\infty \int e^{i (k+1)\theta} d\theta= ie^{-i\theta} \int \sum_{k=0}^\infty e^{i (k+1)\theta} d\theta$

$ =  ie^{-i\theta}\int  \frac{e^{i \theta}}{1-e^{i \theta}} d\theta=  -e^{-i\theta} \ln (1-e^{i \theta})$

so if $ \theta = \frac{ \pi}{2}$ then we get

${ p(\infty)=i \ln (1-i)= i \ln (\sqrt 2 e^{-i \frac{\pi}{4}})= \frac{i}{2} \ln 2 +i \ln(e^{-i \frac{\pi}{4}})=\frac{i}{2} \ln 2 + \frac{\pi}{4}}$

so the distance is $ \sqrt { \left( \frac{\ln 2}{2}\right)^2+ \frac{\pi^2}{16}}$

hopefully.


[quote="jmerry"]It's obviously false.[/quote]
sorry,the question was suppose to be prove or disprove.

The best kind of disproof is a counterexample. Try to construct one.

[quote="guile"]Let $ A$ and $ B$ be square matrices, prove that $ {(AB)}^2 \equal{} A^2B^2$.[/quote]
This is false due to the obvious fact that matrix multiplicaton is not necessarily commutative.


hi
I have  a strange problem.

I am trying to build macro that will separate the code for
html and pdf version of my book.
The undesirable effect of this macro that for some reasons
the number the figure given in this format \thechapter.\thesection.thefigur
when used in the \ref macro.
That it for chapter 1 section 1 and figure 3 it will be 1.1.3
Yet the label of the figure is correct one.
While the desirable numbering should be the standard as
\thechapter.\thefigure that is for chapter 1 and figure 3 -> 1.3
and the same as the label of the figure and \ref command.
not 1.3 in figure and 1.1.3 in \ref.

The snip of the program are below any help will be appreciated.

\documentclass[10pt,dvipdfm]{book}

\usepackage[usenames,dvipsnames]{color}
\usepackage{html}
\begin{htmlonly}
\usepackage[dvips]{graphics}
\end{htmlonly}
%begin{latexonly}
\usepackage{wrapfig}
\usepackage[dvips]{graphicx}
\usepackage[hang,sl,small]{caption}
\renewcommand{\figurename}{Fig.}

%
\newcommand{\wrapImg}[5]{
\begin{makeimage}
\end{makeimage}
%\begin{wrapfigure}[15]{r}[+4pt]{#3\textwidth} %}[h,t,b]
\begin{figure}[h,t,b]
        \label{#4}
\begin{center}
        \latex{\includegraphics[width=#3\textwidth, clip]{#1.eps}}
        \html{%\htmlimage{scale = 1.8, thumbnail=1.5}
                \htmladdimg[ALT="#2"]{#1.jpg}}
\end{center}
        \caption{#5}
\end{figure}
%\end{wrapfigure}
}
%

\begin{document}

\chapter{Review of Mechanics}
\label{chap:mech}

\section{Center of Mass}

\subsection{Moment of Inertia for Mass}

\wrapImg{intro/mech/parallelTheorem}
        {Summation on Moment of Inertial  }
        {0.4}
        {mech:fig:sumIxx}
        {The schematic that explain the  summation of moment of inertia.}

\subsubsection{The Parallel Axis Theorem}

this figure \ref{mech:fig:sumIxx} is not the right number
\end{document}

The differing formats of caption and labels has nothing to do with your macro, it is the default style of the book class. You have loaded the caption package so use it to give Fig. 1.1.3. For example:
[code]\DeclareCaptionLabelFormat{addchapter}{#1\nobreakspace\thechapter.#2}
\captionsetup{labelformat=addchapter} [/code]

Yes, thank you.
Now the label becames the same for the figure and \ref macro.
for example in aux file
\newlabel{mech:fig:sumIxx}{{1.1.1}{1}}
However, the desire is to have the figure numer only 1.1

that is only \thechapr.\thefigure when \thefigure is the continuous and running number of the figures in the chapter.
what else I need to change.

Put \refstepcounter{figure} inside the figure environment and then the label will be the same format as the figure counter.

I insert the command and now the macro looks like this
\setcounter{figure}{0}
\newcommand{\wrapImg}[5]{
   \refstepcounter{figure}
\begin{makeimage}
\end{makeimage}
%\begin{wrapfigure}[15]{r}[+4pt]{#3\textwidth} %}[h,t,b]
\begin{figure}[h,t,b]
   \label{#4}
\begin{center}
   \latex{\includegraphics[width=#3\textwidth, clip]{#1.eps}}
   \html{%\htmlimage{scale = 1.8, thumbnail=1.5}
      \htmladdimg[ALT="#2"]{#1.jpg}}
\end{center}
   \caption{ #5}
\end{figure}

But I get \thefigure is jumping by 2 that it
start as 1.2 and go to 1.4 while the
latel is
\newlabel{mech:fig:sumIxx}{{1.1}{1}}

what the next step

thanks for the help so far


trigonometry
number theory
least common multiple
trig identities
Law of Cosines

A)  Prove that, if the sides of a triangle are all integers, then the cosines of the angles of the triangle are all rational numbers.
B)  Prove that, if cos A, cos B and cos C are rational numbers, and A + B + C = 180°, then there is a triangle with integer sides and angles A, B and C.

[hide="A)"]from the law of cosines you know that
$a^{2}=b^{2}+c^{2}-2*b*c*\cos(\alpha)$
It follows that
$\cos(\alpha)=\frac{a^{2}-b^{2}-c^{2}}{-2*b*c}$
a rational number is defined as a number that can be written as a broken number of two integers. as you can easily see, that is the case.[/hide]

Yes. Generally, proofs are inappropriate for Getting Started. ;)

[quote="Melissa"]OK, is nr B too hard for this forum? Can somebody please solve it?[/quote]
Don't know if you want help or its just a chalange...
So don't  peek .. unless you want to see a solution  .. or a  hint or two :)
[hide="Hint 1 "]All you have to  prove  that you can find a triangle where each side is rational.
[hide="Because"] ...Because  if you multiply by lcm of the denominators .. you get an  similar  triangle with integer sides.[/hide] [/hide]

[hide="Click if more hint is needed"]For that if  can prove $\frac {sin \ A }{sin \ B} $ is rational you are done. [/hide]

[hide="Ok this gives away everything"]Or for that matter if $sin \ A \cdot sin \ B $ is rational  .. Not that  $sin^2 \ B $ has to be rational because it is $1-cos^2B$

But $cos \  A cos \  B - sin \  A sin \  B $  is ...  $-cos \ C$ [/hide]

Thanx, Gyan, I really needed help. But I understood it after the first hint :)

Dunno if law of cosines could be used...but yeah, was looking for a way to use the law of sines...


let a,b,c>0. Prove that
$ \frac{1}{{{a^2} \plus{} 2bc}} \plus{} \frac{1}{{{b^2} \plus{} 2ca}} \plus{} \frac{1}{{{c^2} \plus{} 2ab}} \ge \frac{{ab \plus{} bc \plus{} ca}}{{{a^2}{b^2} \plus{} {b^2}{c^2} \plus{} {c^2}{a^2}}} \plus{} \frac{{27}}{2}.\frac{{{{\left[ {\left( {a \minus{} b} \right)\left( {b \minus{} c} \right)\left( {c \minus{} a} \right)} \right]}^2}}}{{{{\left( {a \plus{} b \plus{} c} \right)}^6}\left( {ab \plus{} bc \plus{} ca} \right)}}$

[quote="VIMF"]let a,b,c>0. Prove that
$ \frac {1}{{{a^2} \plus{} 2bc}} \plus{} \frac {1}{{{b^2} \plus{} 2ca}} \plus{} \frac {1}{{{c^2} \plus{} 2ab}} \ge \frac {{ab \plus{} bc \plus{} ca}}{{{a^2}{b^2} \plus{} {b^2}{c^2} \plus{} {c^2}{a^2}}} \plus{} \frac {{27}}{2}.\frac {{{{\left[ {\left( {a \minus{} b} \right)\left( {b \minus{} c} \right)\left( {c \minus{} a} \right)} \right]}^2}}}{{{{\left( {a \plus{} b \plus{} c} \right)}^6}\left( {ab \plus{} bc \plus{} ca} \right)}}$[/quote]

stronger we have

$ \frac {1}{{{a^2} \plus{} 2bc}} \plus{} \frac {1}{{{b^2} \plus{} 2ca}} \plus{} \frac {1}{{{c^2} \plus{} 2ab}} \ge \frac {{ab \plus{} bc \plus{} ca}}{{{a^2}{b^2} \plus{} {b^2}{c^2} \plus{} {c^2}{a^2}}} \plus{} 310\frac {{{{\left[ {\left( {a \minus{} b} \right)\left( {b \minus{} c} \right)\left( {c \minus{} a} \right)} \right]}^2}}}{{{{\left( {a \plus{} b \plus{} c} \right)}^6}\left( {ab \plus{} bc \plus{} ca} \right)}}$


Celebrating the creation of the "graphical design" forum...
(Although I'm not 100% sure if this qualifies as "graphical design")
Here's something I made some time ago. Whee~ planes!

As you may have guessed, my other sn is snow.

Pretty good, it looks quite professional.  :)

Thanks, EFuzzy.. :)

I use 7.0 because I haven't been able to, er, get a CS. (sadly)

[quote="sky9073"]...
[/quote]No warez mentioning please ;)

Hmm...is CS2 worth the price tag? (I have CS...)

[quote="Valentin Vornicu"][quote="xxreddevilzxx"]is it a bad thing to mention warez?? :?[/quote]Yes. Our general recommendation for all the mods is to delete on site and warn the user which has posted links to or about warez activities.[/quote]

Sorry about that. Something I learned today! not to mention warez... I will keep that to myself also.

[quote="Valentin Vornicu"][quote="xxreddevilzxx"]is it a bad thing to mention warez?? :?[/quote]Yes. Our general recommendation for all the mods is to delete on site and warn the user which has posted links to or about warez activities.[/quote]I know that as a student, so many times the price of the software we want (or we need) is somehow inaccesible for our pockets, and therefore warez are quite attractive. Nevertheless, their use is, you wanted or not, a crime. Anyway, most of the nice copyrighted software, there is a freeware opensource version, that, sometimes is not as good as the original (you would get surprised to know that some times it is even better) and its use is always legal.

[size=200][color=red]Open Source FOREVER!!![/color][/size]

I doubt that photoshop has a THAT good opensource version, but you may try to find something on the web, or if you have the time and the expertice, to start such project  :P .

i was playing with CS2 in a Apple store, it was awesome but too

[i]Moderator: Let's try to not make large white spaces that take two seconds for everyone to scroll through. :P[/i]

.  the upgrade is about 150$
i had elements :(  :blush: but it's good enough for me

I was really bored yesterday.  I mean [i]really[/i] bored.  So I was playing around with Photoshop:

http://img17.echo.cx/img17/7537/blahisbored0sa.png

Well, this is basically my first picture with photoshop so it's not that great (its a background).

[url]http://img292.echo.cx/img292/7860/firstbgcopy1wm.jpg[/url]

[quote="nr1337"]What version of Photoshop do you have?  I have CS1.[/quote]

I have Photoshop Elements...it came with the computer.

Is that good :?:

[quote="hwenterprise"]
Is that good :?:[/quote]

It's adequate for most purposes. I happen to have CS2 now.

[quote="pkothari13"]question mithsapprentice: what does cs2 have that is better than cs? is it worth the upgrade?[/quote]

From what I would think, no.

I'm using photoshop 5, and i can't say there's that much difference from ps 7, or CS, so it's probably not.

Most likelym, it just adds more bloat to the software.

To tell the truth, there is no difference between 7.0,CS, or CS2 other than that the custom intallaton(which takes up much more space) is more user-friendly. No one uses the custom installation. Everyone just does typical installation. Other than that, CS/CS2 comes with Illustrator, which is very helpful when creating 3D images. Also, CS and CS2 come with a few more brushes. So what? Download or make your own brushes. It's not very hard. I work with CS right now, and in the future, I might upgrade to CS2...WITH Custom installation, otherwise there's not much point.

Eh actually I think CS was better than CS2, but CS2 is prettier ("friendlier"...w/e). CS2 takes like 5 second to load but it eats up a mountain of RAM for no good reason. I usually have to shut off most of my other programs so that CS2 doesn't start stalling. Quite annoying.

Yeah CS2 isn't so much of an upgrade, i use it anyway though - along with Painter IX (corel) - Painter is REALLY good and REALLY easy to use i think its worthwhiel to look into it. It's a 'natural' art program. Macromedia Fireworks does a decent job w/ web graphics i guess. That's about all ican think of, and heres my 'fun with photoshop/painter' : (ToBe Rendition of 'ICE')
Edit: i had to convert to jpg for now. i might upload a png versoin that looks better...

You need to improve on your brushing, and actually use the tools. Try getting online tutorials. It's the best way to become a better designer. Also, your background should've been completely brushed, then layer it, go to layer>blending options, and test different backgrounds. There you go, that's the basic background.

I think that pic just looke bad cuz of the conversion. Look at this one and tell me if it still needs improvement.


1) Find $ f:R--->R $ satisfying :
    $ f(xf(x) + f(y)) = f^2(x) + y $
 2) Find $ f:R--->R $ satisfying:
    $ (f(x) + f(y)) . (f(u) + f(v)) = f(xu-yv) + f(xv-yu) $

I think in the first problem you mean
$f(xf(x)+f(y))=f(x)^{2}+y$

making $ f(x)=0$
we get $ f(f(y))=y$

making $x=y=0$
$f(f(0))=f(0)^{2} \equiv f(0)^{2}=0 \equiv f(0)=0$

making $y=0$
$f(xf(x))=f(x)^{2}$

making $x=f(z), y=0$
$f(f(z)f(f(z)))=f(f(z))^{2} \equiv f(zf(z))=z^{2}$
so $ f(z)^{2}=z^{2} \equiv f(z)= \pm z$

I hope it is everything correct,  :)

On Bajoukx's solution:
You can't just assume that there is some $x$ with $f(x)=0$; we need to prove that first.
On the other hand, by setting $x=0$, we have $f(f(y))=y+C$, where $C=f(0)^2$. Then, if $a=f(-C)$, $f(a)=0$. Then we have your first identity, so $y+C=f(f(y))=y$ and $C=0$, so $f(0)=0$. The remainder of your proof works, although I would put equations on separate lines rather than use $\equiv$.

Is #2 a misprint? As written, we can take $x=y=u=v$; then $4f(x)^2=2f(0)$, so either $f(x)=0$ or $f(x)=\frac12$.

Sorry , I wrote wrongly problem 2 .
   $ (f(x) + f(y)) . (f(u) + f(v)) = f(xu-yv) + f(xv + yu) $


Construct a triangle ABC knowing the midpoints of AB, AC and the midpoint of the smaller arc BC of the circumcircle of ABC.

[quote="matheus"]why not?. i think the triangle is unique[/quote]

If you think it is, prove it. I am pretty sure it is not.. and I think there are up to two solutions for every given triple of points.

But at least you know how to build those traingles???

A friend of mine told me that it would b a good idea to look at the 9 points circumference (circunfrencia de los 9 puntos -in spanish-). It's just an idea.


Given four positive numbers $ a$, $ b$ and $ c$, and $ d>1$
1]Find the number of solutions to the equation:

                      $ a\sin(bx\plus{}c)\equal{}\log_{d}x$

2] Find the number of solutions to the equation

                      $ a\sin(bx\plus{}c)\equal{}\frac{x}{d}$

 :)

Hint:

Think about the intesections of the graphs.

 :)


number theory
least common multiple
greatest common divisor

Without using the descomposition of a number in their prime factorization. How to prove that

$lcm(a, b)=\frac{a*b}{gcd(a,b)}$

GCF=Greatest Common Factor, GCD=Greatest Common Divisor, it's the same.

[hide="Proof"]If $d=\gcd(a,b)$, then we can write $a=dp, b=dq$, where $p,q$ are coprime. Therefore $\mathrm{lcm}(a,b)=\mathrm{lcm}(dp,dq)=dpq$ and the result follows.[/hide]

[hide]$gcd(a,b)=d,lcm(a,b)=k:$
$I)d\mid a \wedge d\mid b\Longleftrightarrow db\mid ab \wedge da\mid ab \Longleftrightarrow dk\mid ab$
$II)a\mid k \wedge b\mid k \Longleftrightarrow ab\mid ka \wedge ab\mid kb \Longleftrightarrow ab\mid kd$
$\Longrightarrow (I),(II) ab=kd$ or$lcm(a,b)=\frac{a.b}{gcd(a,b)}$[/hide]

[quote="Kurt Gödel"]Let \[a = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\alpha_{i}}\ \wedge\ b = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\beta_{i}}\] Because $\max(m,n)+\min(m,n) = m+n,\ \forall m,n\in\mathbb{N}$ we know that \[\begin{aligned}\gcd(a,b)\cdot\text{lcm}(a,b) & = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\min(\alpha_{i},\beta_{i})}\cdot \prod_{p_{i}\in\mathbb{P}}p_{i}^{\max(\alpha_{i},\beta_{i})}= \prod_{p_{i}\in\mathbb{P}}p_{i}^{\min(\alpha_{i},\beta_{i})+\max(\alpha_{i},\beta_{i})}\\ & = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\alpha_{i}+\beta_{i}}= \prod_{p_{i}\in\mathbb{P}}p_{i}^{\alpha_{i}}\cdot \prod_{p_{i}\in\mathbb{P}}p_{i}^{\beta_{i}}= a\cdot b\end{aligned}\] [/quote]
Also my solution (I think it is the best way to solve this problem), but
[quote="Andy"]Without using the descomposition of a number in their prime factorization[/quote]
@$\text{Andy}$: Why without prime factorisation?

[quote="Rzeszut"]@$\text{Andy}$: Why without prime factorisation?[/quote]
For example, because one wants to see if it holds outside (unique) factorisation domains.
But I'm also not a friend of unreasoned restrictions on the way of proving...


calculus
derivative
limit
function
calculus computations

Suppose that $ f$ is differentiable at $ x$ prove that:

$ f'(x) \equal{} \displaystyle\lim_{h\to\ 0}\frac {f(x \plus{} h) \minus{} f(x \minus{} h)}{2h}$

If $ f(x) \equal{} x^3$ then find: $ f'(x^2)$ and $ f'(a^2)$

1)$ f'(x) \equal{} \lim_{h\to\ 0}\frac {f(x \plus{} h) \minus{} f(x)}{h} \equal{} \lim_{h\to\ 0} \minus{} \frac {f(x \minus{} h) \minus{} f(x)}{h}$
$ \lim_{h\to 0}\frac {f(x \plus{} h) \minus{} f(x \minus{} h)}{2h} \equal{} \frac {1}{2}(\lim_{h\to 0}\frac {f(x \plus{} h) \minus{} f(x)}{h} \plus{} \lim_{h\to 0} \minus{} \frac {f(x \minus{} h) \minus{} f(x)}{h}) \equal{} \frac {1}{2}(2f'(x)) \equal{} f'(x).$

2) By the chain rule:

$ f'(x^2) \equal{} (x^2)^3(2x) \equal{} 2x^7$
$ f'(a^2) \equal{} (2a^7)(a')$ (not sure about the last one; is $ a$ a constant or a function of $ x$?)

Is there a way to find $ f'(x^2)$ given $ f(x) \equal{} x^3$ without the chain rule?  This was an exercise before the chain rule was introduced.

[quote="gauss202"]Elixir, your answer to the second problem is incorrect. $ f'(x^2)$ literally means the derivative of $ f$ evaluated at the input $ x^2$. What you gave is $ D_x[f(x^2)]$, which is something different.

So the answer should be $ f'(x) \equal{} 3x^2$, so $ f'(x^2) \equal{} 3(x^2)^2 \equal{} 3x^4$[/quote]I just realized it was incorrect, but for a totally different reason. I misread $ f(x)$ as $ f'(x)$. Maybe I need my eyes checked.  :maybe:

But if I were to go by what my teacher taught about derivatives, then $ f'(x^2) \equal{} 3(x^2)^2 * 2x \equal{} 6x^5$. He said that $ f'(x^2)$ is still with respect to $ x$. (Though, now that I think about it in terms of the chain rule, you are most probably right.)  :)

[quote="Stokes93"]Oh, I see you're just evaluating the derivative at $ x^2$.  So for the third one you would have $ 3a^4$ right?

What is the difference between $ D_{x}[f(x^{2})]$ and $ f'(x^2)$?[/quote]That would be correct.

The difference, I believe, is that $ D_{x}[f(x^{2})]$ is the derivative with respect to $ x$ while the latter is with respect to $ x^2$. $ D_{x}[f(x^{2})] \equal{} \lim_{h\to\ 0} \frac {f((x \plus{} h)^2) \minus{} f(x^2)}{h}$ while $ f'(x^2) \equal{} \lim_{h\to\ 0} \frac {f((x^2 \plus{} h) \minus{} f(x^2)}{h}.$ You'll get two different limits.

$ f'(x)$ is a function; $ f'(x^2)$ is what you get when you evaluate that function at $ x^2$.  $ f(x^2)$ is a function;  $ \frac {d}{dx} \left(f(x^2)\right) \equal{} D_x(f(x^2))$ is what you get when you take the derivative of that function.  The two objects $ f'(x^2)$ and $ D_x(f(x^2))$ are the result of starting with $ f(x)$ and applying a derivative and a substitution in both possible orders.  Substitutions and derivatives don't commute with each other, and the extent of the not-commuting is precisely captured by the chain rule.  If your teacher told you that these two things are the same, s/he was very, very wrong (but I think (hope!) it more likely that you're misremembering or confused about what your teacher said).


How would you do this problem:

What is the largest prime factor of 95!?

[hide]Basically the greatest prime from 2..95 = 89[/hide]

[hide]find primes. then divide 95! by it. eventually you will see that the greatest prime bet 1-95 is the highest. 93 isn't a prime, neither is 91. so it is 89[/hide]

[quote="Boridi"]The prime factorization of 95 is  19 * 5

19 is its largest prime factor[/quote]

The problem is very clumsily written. It's not "95 + exclamation mark with a question mark", rather "95 factorial + a question mark".


geometry
circumcircle
incenter
power of a point
radical axis
geometry proposed

For a triangle ABC with circumcircle C1, incenter I, and excenter E relative to BC. IE intersect BC at F; IE intersect C1 at G. H is the midpoint of arc AG. HI intersect C1 at M and HE intersect C1 at N. Proved that M, F, and N are collinear.

$ \angle HNM = \frac {1}{2}(arc AH + \overarc{AM}) = \frac {1}{2}(arc HG + arc AM) = \angle HIG$
Therefore $ NIME$ is cyclic quadrilateral.
We know that $ IBEC$ is cyclic quadrilateral.
$ IE, BC, MN$ is radical axis of $ (IBEC)$ and $ (INEM), (IBEC)$ and $ (ABC), (INEM)$ and $ (ABC)$ so they concur at radical center of three circles.


Show that $\exp \bigl( \mathcal{M}_n(\mathbb{R})\bigr) = \{M^2 ~|~M\in \mathrm{GL}_n(\mathbb{R}) \}$

It was discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=exponential+matrix&t=28711]here[/url] but not solved. There's something there about another topic.

here it is http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=43500

Very nice proof alekk, much shorter than what I had :blush:

what was your proof for this pb ? It is always very interesting to see many approaches !


geometry
function
3D geometry
topology
advanced fields
advanced fields unsolved

Find all maps $ f$ from the euclidean plane into itself which send each triangle in a triangle whose area  is at most $ k$ times the area of the initial triangle, $ k$ being a constant non depending on the triangle chosen.

I am not sure what you mean: should the image of a triangle be a triangle, or should it just be contained in a triangle? Your subject line suggests the former, but the latter sounds much more interesting. owk

not just "every three points are sent in three points such that bla bla"?

I also think it's rather vague :?. When speaking about triangles, it can't hurt to mention whether one means (a) three vertices of a triangle, or (b) three sides of a triangle, or (c) a triangular surface.

wouldn't any other request be too "rigid"?
maybe it's just an idea of mine, but.. anyway, it's too phylosophical.. unless until anyone proves anything (me included :p)

Maybe it should also be clarified whether degenerate triangles are allowed, or whether we need to prove continuity first. owk

Oh, sorry for being so unclear. Degenerate triangles are accepted and by triangle I understand only the set of its vertices.

I think I can show that such a function either sends everything on a single line or is affine, but the proof gets kind of boring (and I'm sure it can be simplified quite a bit), so I'm not going to post it all.

Suppose $ f$ takes three non-collinear points $ x,y,z$ into non-collinear points. The aim is to show that it must be affine. First notice that $ f$ sends collinear points into collinear points, and hence the image of every line is contained in a line. In order to prove the continuity of $ f$, suppose the sequence $ p_{n}$ converges to a point $ p$. Since $ f(p)$ lies off at least one of the three lines determined by the images of $ x,y,z$, we can assume WLOG that $ f(p)f(x)$ and $ f(p)f(y)$ are distinct lines. The areas of $ p_{n}px$ and $ p_{n}py$ both go to zero, so the areas of $ f(p_{n})f(p)f(x),\ f(p_{n})f(p)f(x)$ must likewise tend to zero. In other words, $ f(p_{n})$ approaches both lines $ f(p)f(x),\ f(p)f(y)$, which is possible only if $ f(p_{n})\to f(p)$. This proves the continuity of $ f$.

Assume now that $ f$ sends a line $ \ell$ into a point $ p$. If $ x$ is a point such that $ f(x)\ne p$, then $ f$ sends every point off the parallel through $ x$ to $ \ell$ on the line $ f(x)p$ (because every such point belong to a line passing through $ x$ and a point on $ \ell$). By continuity, $ f$ sends the whole plane on $ f(x)p$, contradicting the initial hypothesis. Assume further that there are two different lines $ \ell_{1},\ell_{2}$ such that $ f(\ell_{i})$ are both contained in the same line $ \ell$. Now let $ D$ be an open set mapped by $ f$ off $ \ell$ (such a set exists by continuity and the initial assumption that $ f$ doesn't map everything into a single line). For every two points $ x_{i}\in\ell_{i},\ i=1,2$ with $ x_{1}x_{2}$ passing through a point $ p$ in $ D$, we must have $ f(x_{1})=f(x_{2})$. Otherwise $ f(p)$ would lie on the line $ f(x_{1})f(x_{2})$, which coincides with $ \ell$. It follows now that $ f$ must be locally constant (and hence constant) on $ \ell_{1}$, say: every $ x_{1}\in\ell_{1}$ can be connected to a point $ x_{2}\in\ell_{2}$ through a line which cuts $ D$. By the previous observation, $ f$ must take the value $ f(x_{2})$ on the intersection between $ \ell_{1}$ and the "cone" connecting $ x_{2}$ to $ D$; this intersection contains a neighborhood of $ x_{1}$ on $ \ell_{1}$, proving the claim that $ f$ is locally constant on $ \ell_{1}$. This, however, contradicts the fact proven before that $ f$ cannot be constant on a line. To sum it all up, we've shown that the images through $ f$ of different lines lie on different lines; in particular, it follows that $ f$ is one-to-one.

After this, I showed that $ f$ is actually onto (this isn't difficult at all, but it would occupy another paragraph at least as long as the previous one; this is harder to write down than I thought :)). This, together with the injectivity, implies that $ f$ must preserve parallelism. In order to show that $ f$ is affine, it suffices to prove that it sends midpoints to midpoints, for we can then use continuity and the conclusion follows easily. To this end, construct the midpoint of a segment $ xy$ as follows: choose an arbitrary point $ p$ off the line $ xy$, let $ x',y'$ be the intersections of a line parallel to $ xy$ but different from it and which doesn't contain $ p$ with $ px,py$ respectively, and let $ q$ be the intersection of $ xy',yx'$. The midpoint $ m$ of $ xy$ is precisely the intersection between $ pq$ and $ xy$. Since $ f$ preserves all operations and relations used in the construction (it is one-to-one, so different points are mapped into different points, it sends intersections of lines into the intersections of their images, it preserves parallelism, etc.), it must send $ m$ into the midpoint of $ f(x)f(y)$, as desired.


AMC
AIME
geometry
3D geometry
sphere
pyramid
percent

Well lets do it w/o a picture then. The original problem didn't have one.

[hide]
Is it [tex]\sqrt{2} + \frac{3}{2}[/tex] ?
[/hide]

Wow, I'm an idiot.  I used 1 and .5 and did a plane ....

[quote="sonar"][hide]
Is it [tex]\sqrt{2} + \frac{3}{2}[/tex] ?
[/hide][/quote]

I'm pretty sure that's wrong.

your answer get's us approx 2.914, while the answer is definately at least 4 (double the largest radius).[/hide]

Sorry about the picture, I thought I attached it.  It is attached to this one.  And no, your answer is wrong.
This was on last year's AMC.  I didn't answer it on the test, and it took me a couple of hours to solve.  Once you get the right idea, however, it is fairly routine.

Here is a hint: WARNING it will spoil the challenge of the problem.
[hide]Take the center points of the spheres and draw a triangular pyramid[/hide]

Um, this shouldn't be that difficult. I saw it on one of the AMC 12's. You draw up a triangular pyramid connecting all 4 origins, and find the height, which be Pythagorean isn't all that difficult. Add 2 and 1, to get your final answer.

Which is:

[hide]$\frac{\sqrt{69}}{3} + 3$[/hide]

Problem: Easy One:

The fraction: $\frac{2(\sqrt{2}+\sqrt{6})}{3(\sqrt{2 + \sqrt{3}})}$ is equal to?

You're right, but now you have to post a question of your own.

sorry... I didn't see that you had posted a problem.

Okay here is the answer:
[hide]
First, square the whole expression.
We get:     (32+16 :sqrt: 3)/(18+9:sqrt:3)

obviously, this is 16/9.

Since we squared it at the beginning, we take the square root and find that the answer is 4/3.

(It might have been -4/3, but we check and find that it is clearly positive.)[/hide]

THE NEW QUESTION: (simple concept problem)
Factor completely:  A2B + AB2+ A2 + B2 - (A+B)

sorry :(
it should read
  A2B + AB2+ A2 - B2 - (A+B)

[quote]Um, like alc said, can you continue this by posting a problem?[/quote]

Sure. I just needed a good problem:

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a -bi$

LOL. I just did that two days ago. On one of the sets - I think one of the new ones.

wowww, those were all very clever solutions, and the answer was indeed 144. i suppose anyone who got that answer can post the next one.

on another note, did anyone post a solution to this problem:

$\displaystyle \frac{1}{sin45sin46}+\frac{1}{sin47sin48}+...+\frac{1}{sin133sin134} = \frac {1}{sinn}$

[quote="BornForMath"]How did you percieve that 154 is too large :?:[/quote]

Well, that's a little more complicated than showing that 134 is too small, but I'll attempt it (it was partly intuitive)...

$110^5=16105100000$. Now, take the possible values of $x$: $134$, $144$, $154$. We'll express them as $n-10$, $n$, and $n+10$. $(n+10)^5=n^5+50n^4+1000n^3+10000n^2+50000n+100000$. $133^5$ is only slightly lower than $(n-10)^5=n^5-50n^4+1000n^3-10000n^2+50000n-100000$, so for convenience we'll just use that to represent $133^5$. If $x=154$, then we have approximately $n^5-50n^4+10^3n^3-10^4n^2+5(10^4)n-10^6+16105100000+84^5+27^5=n^5+50n^4+10^3n^3+10^4n^2+5(10^4)n+10^6$ (Sorry for the split in the equation.) Therefore, $16104900000+84^5+27^5=100(144)^4+20000(144)^2$, which is essentially equal to $110^5+84^5+27^5=100(144)^4+20000(144)^2$, and then I guess I intuitively saw that the right side was bigger than the left, meaning that $154$ is too large.

Does that basically make sense? :P

I am trying to come up with a nice solution to PK's problem but I can not seem to find anything. Does anyone else have any ideas that may simplfiy the problem?

Well you know that the number is 0 mod 3 and ends in 4.  So it's simply a question of showing that i's in the interval from 125 to 163.  You know that it's greater than 125 because 133>125.  You can show that it's less that 163 because
$163>\frac{6}{5}*133$
$\frac{6}{5}^5$ is approximately 2.5.  You can then show that $\frac{133}{110}$ is more than 6/5, $\frac{110}{84}$ is more than 6/5, and then 27^5 is less than $\frac{1}{3^5}$ of 81^5.  Thus it's in that range and so the number is 144.

That one was answered before; he's referring to the $\frac{1}{\sin 45\cdot \sin 46}+\frac{1}{\sin 47\cdot \sin 48}+...+\frac{1}{\sin 133\cdot \sin 134}=\frac{1}{\sin n}$ problem.

[b]EDIT:[/b] $\sin 134=\sin 46$, $\sin 133=\sin 47$, etc., so we have $\frac{1}{\sin 45\cdot \sin 46}+\frac{1}{\sin 46\cdot \sin 47}+...+\frac{1}{\sin 88\cdot \sin 89}+\frac{1}{\sin 89}=\frac{1}{\sin n}$. Does that help at all?

Hey everyone --

I found this down on the eighth page and figured I'd move it back up so we could get it going again.

Anyway:

A convex polyhedron has faces of 12 squres, 8 hexagons, and 6 octagons (all faces are regular).  How many diagonals does the polyhedron have which are NOT along a face of the polyhedron (AIME 1988, #10)

JB

[hide]The number of edges is \[ \frac{1}{2}(12(4)+8(6)+6(8))=72 \]  Thus, because $V=E+2-F$, where V, E and F are the numbers of vertices, edges, and faces, respectively, (Euler proved this) we have $V=48$.  Thus, the number of space diagonals in question is the number of ways to choose two vertices, minus the number of edges, minus the number of face diagonals
\[ =\binom{48}{2}-72-12(\binom{4}{2}-4)-8(\binom{6}{2}-6)-6(\binom{8}{2}-8)=\boxed{840} \]
[/hide]

I remember I saw this one when I was in eight grade (about a year before I came to AoPS) and couldn't do it...but I just found it now and it was easy...so yeah...

Source: AIME 2001, #9

We have $\triangle{ABC}$ such that $AB=13$, $BC=15$, $AC=17$.  $\triangle{ABC}$ has points $D$, $E$ and $F$ on sides $AB$, $BC$ and $CA$, respectively.  We have \[ \frac{AD}{AB}=\alpha, \frac{BE}{BC}=\beta, \frac{CF}{CA}=\gamma \] where $\alpha+\beta+\gamma=\frac{2}{3}$ and $\alpha^2+\beta^2+\gamma^2=\frac{2}{5}$.  Find $\frac{area(DEF)}{area(ABC)}$.

[hide] For simplicity's sake, I will use [XYZ] to define the area of triangle XYZ.

The first thing we notice is that

[ABC] = [DEF] + [FAD] + [DBA] + [ECF]

From this we derive

[DEF]/[ABC] = 1 - [FAD]/[ABC] + [DBA]/[ABC] + [ECF]/[ABC]

We can find the ratios on the right using the identity Area = a*b*sin(theta)

Examining triangle FAD, we see that it shares an angle with the triangle ABC.  Therefore, [FAD]/[BAC] = (AD)(AF)/(AC)(BC).  We know AD/AB = a, and since CF/CA = c, then AF/AC = 1-c.  Therefore, our ratio is equal to a(1-c).  Similarly, we find the ratios [DBE]/[ABC] and [ECF]/[BCA] to be b(1-a) and c(1-b) respectively.

Therefore, [DEF]/[ABC] = 1 - (a(1-c) + b(1-a) + c(1-b))
=1 - (a + b + c - ab - ac - bc)
=1 - (a + b + c -
[(a + b + c)^2 - (a^2 + b-2 + c^2) / 2)
=1 - (2/3 - [(2/3)^2 - 2/5]/2)
=1 - (2/3 - 1/45)
[b]=16/45[/b]
[/hide]

Al walks down to the bottom of the escalater that is moving up and he counts 150 steps.  His friend Bob walks up to the top of that escalator and counts 75 steps.  If Al's walking speed (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time (assume this value is constant).

JB

[hide="My Solution"]Let the number of steps on the escalator be $x$.  Let the rate of Bob's speed be $y$.  Then, we find that:
$x=(y+x)*\frac{75}{x}$ and $x=(3y-x)*\frac{150}{x}$.  We can easily manipulate the expression so that we get $y=\frac{3x}{5}$ and using this we find that $x=120$[/hide]

[hide="My Problem"]In right triangle ABC AB is the hypotenuse and the altitude from C meets AB at X.  All of triangle ABC's sides are integers.  $BX=31^3$ and $cos(B)=\frac{m}{n}$ where m and n are integers.  Find $m+n$.[/hide]

[hide="solution"]
Let $\frac{q}{m}$ be $sin(B).$  Then $\frac{n}{m} 31^3, (\frac{q}{m})^2 31^3,$ and $\frac{qn}{m^2} 31^3$ are all integers (they are CB, AX, AC respectively).  Looking at the three values we can conclude that $m=31$.  The only pythagorean triple with 31 is (31,480,481), hence our answer is 31+481 = 512.
[/hide]

Problem:

The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$. Find the unique number that is not in the range of $f$.

[quote="bubala"]Since you're not posting the next problem, I will...again, it's an AIME problem between 1950 and 2005

Find the least positive integer $n$ such that $\frac{1}{sin45sin46}+\frac{1}{sin47sin48}+...+\frac{1}{sin133sin134} = \frac {1}{sinn}$.

In my opinion, this is slightly contrived, so here's a

[hide="helpful hint"]Multiply both sides of the equation by $sin1$.  [/hide][/quote]

What exact year is this from?


Can you help me to differentiate this function :
1) $f(x)$ $=$ $\vert \lfloor x \rfloor \vert$ $*$ $\lfloor \vert x \vert \rfloor$
2) $f(x) = sign(1- \sqrt[3]{x}) * max(x,1- \sqrt[3]{x})$
3) $f(x) = x^2 * f(x-2)$
with respect to $x$

[quote="MCrawford"]Hi.  I see that you are new.  Try to keep posts in appropriate forums.  Middle school students aren't all that likely to help you with differential calculus.[/quote]

Sorry, ............... :P
thank's for your attention............ :oops:


trigonometry
geometry
rectangle
Pythagorean Theorem

Post any pythagoren theorem proofs - include diagrams and everything.  The person with the highest number of unique proofs wins.

Look [url=http://www.cut-the-knot.org/pythagoras/index.shtml]here[/url] for a ton of proofs.

James Garfield (20th President) had a neat proof of the Pythagorean Theorem using the Trapezoid.

[quote="inom"]Try to proof by area square[/quote]

Consider a square with sides $a+b$, divided each side in two segments, one with length $a$ and the other with length $b$. Now, join the four points that divide each side of the square in two segments. Doing this, you have a small square, inside the original one, with side $c$. Now, the area of the original square is $(a+b)^2$. This square is divided in $4$ equal triangles, each with area $\frac{ab}{2}$, and the square with area $c^2$. Now, we have:
$(a+b)^2=c^2+4 \frac{ab}{2}$
$a^2+2ab+b^2=c^2+2ab$
$a^2+b^2=c^2$, which is what we wanted to prove.

i saw a nice one, i think its in Roger Penrose's Road to Reality.  He uses a grid of two sized squares and then connects some points giving one side along the bigger square and another along the smaller.  Then by shifting the grid it cuts part of a bigger square along the hypotenuse which divides the triangle into a part which is outside the hyp square but that triangle cut off simply fits back into the hyp square along with the big square.  The little square is made up of two of those triangles which were cut off.  Its hard to describe but the picture in the book really helps.

[quote="pkerichang"]kind of off topic, but: is it possible to prove that $\sin^2{\theta} + \cos^2{\theta} = 1$ WITHOUT pythagorean, so we can prove puthagorean with it?? Sounds impossible, tho...[/quote]

The proof would be [b]equivalent[/b] to a proof of the Pythagorean identity, given the triangle definitions of $\sin, \cos$.

I find the Pythagorean theorem rather boring.  It's a lot more interesting to see what happens in spaces where it doesn't hold.  :)

I have seen a 100% visual proof of this with no words at all, just a picture.  I will try to find it...

Books have been published with over four hundred proofs of countless different kinds.  This really is beating a dead horse.

In fact, the little animation on the bottom has a proof.

I know a proof using Ptolemy's Theorem:
We know that a rectangle is cyclic.
Let $a=AB$, so $a=CD$.  Let $b=BC$, so $b=AD$.  Let $c=AC$, so $c=BD$.  See the diagram below, which has rectangle $ABCD$.
By Ptolemy's Theorem, $(AB)(CD)+(AD)(BC)=(AC)(BD)$.  Using $a$, $b$, and $c$, we get $a^2+b^2=c^2$.

But still, the one that I know the best is the one with a square with side length $a+b$, which José posted.


function
algebra
floor function
geometric sequence
arithmetic sequence
number theory proposed
number theory

consider the sequence a(n) = |n*sqrt 2003| where |a| denotes floor function of a.
prove we can find an arbitrarily long geometric progression within the sequence a(m).

actually solve wifout szemederi, bcoz wif szemederi it is obvious (consider
b(n) = In a(n). b(n) contains an arbitrarily long arithmetic progression so we are done  :P )


Let $ I$ be an interval in $ \mathbf R$, $ \gamma \in C^1(I,\mathbf R^n)$ a $ C^1$ curve in $ \mathbf R^n$. Prove that image $ \gamma(I)$ is nowhere dense in $ \mathbf R^n$.

Since $ \gamma(I)$ is a compact set, it's enough to prove that it does not contain any ball. But $ C^1$ implies Lipschitz, hence $ \gamma(I)$ has measure zero.


Le $p$ be a given odd prime. Find all functions $f: Z\to Z$ satisfying the following conditions:
$i)$ If $m\equiv n$ ( mod p) for $m, n\in Z$ then $f(m)=f(n)$
$ii)$ $f(mn)=f(m).f(n)$ for all $m, n\in Z$

think this is from korea 2000.you can put it $f: Z\to R$.hope it's good.
[hide]
from condition $i)$ we have that $f$ attains at most $p$ different values.if for some $a$ $0<|f(a)|<1$ then the set $f(a^{n})$ attains infinetly many values, the same argument goes if $|f(a)|>1$. so $f(a)=-1,0,1$. now note that $f(0)=0,1$ because $f(0)=f(0)^{2}$. now consider $r$, the primitive root of $p$(not sure how to prove it's existance,but somebody else might know).for it we know that $r,r^{2},r^{3},...,r^{p-1}$ all give different residues modulo p.
if $f(r)=0$ then $f(x)=0$ for all $x$ such that $x\not|p$ and from that we have $f(x)=0$ for all $x$.
if $f(r)=1$ then $f(x)=1$ for all $x$ such that $x\not|p$ and $f(np)=0$ for all $n$ or $f(np)=1$ for all $n$.
if $f(r)=-1$ then $f(x)=-1$ iff there exists natural $k$ such that $x\equiv r^{2k+1}$ $mod$ $p$ and $f(x)=1$ iff there exists natural $k$ such that $x\equiv r^{2k}$ $mod$ $p$ and $f(np)=0$.
to check that these functions work it's enough to take the residue of every number modulo $p$ and write it as $r^{k}$ for some natural $k$(since the definition of the primitive root allows us to do so).[/hide]


inequalities
vector
inequalities proposed

Can you prove or disaprove this theorem:

$\sum_{cyc}(\prod^{k}x_i ^{\alpha _i}) \geq \sum_{cyc}(\prod^{l}x_i ^{\beta _i})  \Longleftrightarrow \sum_{cyc}(\prod^{k}x_i ^{\pi _{(\alpha _i)}}) \geq \sum_{cyc}(\prod^{l}x_i ^{\pi_{(\beta _i)}})$

where $\pi _{(\alpha _i)}$ and $\pi_{(\beta _i)}$ are permutations of $\alpha _i$ and $\beta _i$

Hmm... $\sum_{cyc}(\prod^{k}x_i ^{\alpha _i})$ and $\sum_{cyc}(\prod^{k}x_i ^{\pi(\alpha _i)})$ are the same, aren't they?

No, maybe I write it not enough precisely but here's an example:

$x^2y+y^2z+z^2x$ and permutation $x^2z+z^2y+y^2x$

When I think about it now I think it should be reformulated a bit so it's not true for arbitrary permutation but there [b]exist[/b] a permutation that it is true. Nevertheless I'm lookin for some prove or disaprove for general one.

And I don't know if it shouldnt rather be somethin like:

${\sum_{cyc}(\prod^{k}x_i ^{\alpha _i}) \geq \sum_{cyc}(\prod^{l}x_i ^{\beta _i})  \Longleftrightarrow \sum_{cyc}(\prod^{k}x_i ^{\pi _{(\alpha _i)}}) \geq \sum_{cyc}(\prod^{l}x_i ^{\beta _i}})$

(right side doesnt change)

so many doubts  :lol:

Megus, unfortunately, my answer to your question will be bound to the disclaimer "if I understand you correctly", since the way you posed the question is rather unsatisfactory, what is actually due to the fact that questions like this are generally very hard to put into words. Anyway, in my understanding, your conjecture would yield that the inequality

$a^4b^4+b^4c^4+c^4d^4+d^4a^4\geq a^3b^4c+b^3c^4d+c^3d^4a+d^3a^4b$ for nonnegative reals a, b, c

is equivalent to the inequality

$a^4c^4+b^4d^4+c^4a^4+d^4b^4\geq a^3b^4c+b^3c^4d+c^3d^4a+d^3a^4b$ for nonnegative reals a, b, c.

(Here, we permuted the exponents on the left hand side: $\left(4;\;4;\;0;\;0\right)\to\left(4;\;0;\;4;\;0\right)$, leaving the right hand side invariant.) Unfortunately, this is not the case, since the first inequality always holds (I'm leaving the proof as a little exercise), while the second one doesn't hold e. g. for a = 1, b = 2, c = 3, d = 2. So, your conjecture, at least if I got it right, is not true. :(

  darij

Thanks Darij for reviving this topic. You understand me correctly - it is wrong what i realized soon [one week :)] after i post it, though some other conjectures came to my head. But first:

[b]Notation:[/b]
$R_{\alpha}=(\pi _{(\alpha _1)},...,\pi _{(\alpha _k)})$, $R_{\beta}=(\pi _{(\beta _1)},...,\pi _{(\beta _k)})$
$\sum _{i=1}^k \alpha _i = \sum _{i=1}^k \beta _i \Longleftrightarrow \sum _{i=1}^k \pi _{(\alpha _i)} = \sum _{i=1}^k \pi _{(\beta _i)}$
$\forall _i  \ x_i, \alpha _i, \beta _i \geq 0$
$\alpha _1 \geq \alpha _2 \geq \alpha _3 \geq ... \geq \alpha _k$
$\beta _1 \geq \beta _2 \geq \beta _3 \geq ... \geq \beta _k$

[b]Conjecture 1[/b] [I cannot prove it, nor disaprove it]:

$(\sum_{symm}(\prod^{k}x_i ^{\alpha _i}) \geq \sum_{symm}(\prod^{k}x_i ^{\beta _i})) \wedge$
$(\sum_{cyc}(\prod^{k}x_i ^{\alpha _i}) \geq \sum_{cyc}(\prod^{k}x_i ^{\beta _i})) \Longrightarrow \exists _{R_\alpha} \exists _{R_\beta}(\sum_{cyc}(\prod^{k}x_i ^{\pi _{(\alpha _i)}}) \geq \sum_{cyc}(\prod^{k}x_i ^{\pi_{(\beta _i)}}))$

[b]Conjecture 2[/b] [this is obvious, just reangarrement to prove -> and by adding all cyclic inequalities we get <-]:

$(\sum_{symm}(\prod^{k}x_i ^{\alpha _i}) \geq \sum_{symm}(\prod^{k}x_i ^{\beta _i})) \wedge$
$(x_1 \geq x_2 \geq x_3 \geq ... \geq x_k) \Longleftrightarrow$
$\forall _{R_\alpha} \exists _{R_\beta}(\sum_{cyc}(\prod^{k}x_i ^{\pi _{(\alpha _i)}}) \geq \sum_{cyc}(\prod^{k}x_i ^{\pi_{(\beta _i)}}))$

plus something more of my "deep"  ;) thinking [some variations of first conjecture; use of vectors in inequalities...] but I dont want to tire you.


trigonometry
function
algebra
domain
inequalities unsolved
inequalities

Find the minimum value of

\[ |cos\alpha| \plus{} |cos\beta| \plus{} |cos\gamma| \plus{} |cos(\alpha \minus{} \beta)| \plus{} |cos(\beta \minus{} \gamma)| \plus{} |cos(\gamma \minus{} \alpha)|,\]

where $ \alpha,\beta,\gamma\in\mathbb{R}$.

I wonder we would solve the problem in using the triangle inequality.

I think minimum is $ 2$ when $ \alpha \equal{} \beta \equal{} \frac{\pi}{2}$ and $ \gamma \equal{} \pi$. Left to prove this...

You were right, Yuriy! (By the way, nice Russian Blue cat!)
Reduce modulo $ \pi$ to have wlog $ \minus{} \pi/2 \leq \alpha \leq \beta \leq \gamma \leq \pi/2$. The function $ f : [ \minus{} \pi/2 \minus{} \alpha, \pi/2 \minus{} \gamma] \to [0, \infty)$ given by $ f(x) \equal{} \cos(x \plus{} \alpha) \plus{} \cos(x \plus{} \beta) \plus{} \cos(x \plus{} \gamma)$ is non-negative and concave on the domain, hence it reaches its minimum at one end.
Denote the given expression by $ E(\alpha, \beta, \gamma)$; it is obvious that $ E(x \plus{} \alpha, x \plus{} \beta, x \plus{} \gamma) \equal{} E(\alpha, \beta, \gamma) \plus{} f(x) \minus{} f(0)$.
It means the expression reaches its minimum at one end, so it is enough to look at triplets where one of the angles is $ \pm \pi/2$.
Then the expression becomes $ \cos \varphi \plus{} \cos \theta \plus{} | \sin \varphi | \plus{} | \sin \theta | \plus{} | \cos(\varphi \minus{} \theta) |$. But, in general, for $ x \in [ \minus{} \pi/2, \pi/2]$, one has $ \cos x \plus{} | \sin x | \geq 1$.
Denote $ \cos x \equal{} t$, so $ \cos x \plus{} | \sin x | \equal{} t \plus{} \sqrt {1 \minus{} t^2}$, easily seen to reach its minimum value $ 1$ at $ t \equal{} 0$ and $ t \equal{} 1$, i.e. $ x \in \{ \minus{} \pi/2, 0, \pi/2 \}$.
It follows that the minimum for the expression is $ \framebox{2}$, e.g. for $ \varphi \equal{} 0$, $ \theta \equal{} \pi/2$.


We have the Fibonacci series defined like so:
f_1=1,f_2=1 and f_(n+1)=f_n+f_(n-1), for any natural number n>=2. Prove that f_n<3^(n/2) and compute the limit when n-->oo of (f_1/2+f_2/2^2+..+f_n/2^n.

cheers!

for the part with  f_n<3^(n/2) is simple... it's pure induction.. but how about the 2nd part?

this was given for the national tests in the Republic of Moldavia 2003!

cheers! :D

my intuition says that perhaps the ineq they gave to prove helps us to solve the limit! can't find no link! :?

cheers!

Here's how it should be done (I'm not going to actually calculate the limit). Just write the general expression of   f(n). f(n)=(1/5^1/2)*(((1+5^1/2)/2)^n-((1-5^1/2)/2)^n). We replace every f(k) with this expression and then we calculate the infinite sums of 2 geometric progressions and we are done. I don't know if the first part was meant to help us, maybe the limit is the same it would be if we replaced f(n) by 3^n/2, but I don't know (like I said, I didn't calculate it).


modular arithmetic
LaTeX
number theory unsolved
number theory

for which prime do we have $ \left( \frac {p\minus{}1}{4} \right)^{ \frac {p\minus{}1}2 } \equiv 1 \pmod p$

I don't understand the problem...
And please use $ LaTeX$.

what don't u understand ?
i ask to find any odd prime for which
((p-1)/4)^((p-1)/2) is congruent to 1 mod p
or in another way there exist 1<x<p such that
(p-1/4) is congruent to x^2 mod p
(i don't have LATEX )

$ (\frac{p\minus{}1}{4})*(\minus{}4)\equal{}1\mod p$. Therefore your question equavalent to $ (\minus{}4)^{(p\minus{}1)/2}\equal{}1\mod p$ or $ (\minus{}1)^{p\minus{}1)/2}\equal{}1\mod p$ or $ p\equal{}1\mod 4$/


$ a,b,c > 0$ . Prove that:

$ \sum_{cyc}\frac{a}{\sqrt{a\plus{}b}} \ge \frac{1}{\sqrt{2}}.\sum_{cyc} \sqrt{a}$

[quote="Duy_VN"]$ a,b,c > 0$ . Prove that:

$ \sum_{cyc}\frac {a}{\sqrt {a \plus{} b}} \ge \frac {1}{\sqrt {2}}.\sum_{cyc} \sqrt {a}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265475


Ok, i know this is simple, but i cant seem to figure out the logic. Here is my problem
I understand for example,
convert liters to kiloliters
1.0L = 1.0 x 10^-3 kL
I get that by
1.0L/1 x 10^-3/1L

what i dont understand,
convert liters to dekaliters, when deka is 10^1
i do it like this,
1.0L x10^1/1 which is 1.0 x 10^1.
the aswer is 1.0 x 10^-1
I assume you are supposed to take the recipical, which would be 1/10^-1, but why?

Thanks!

Your answer of 1.0 x 10^-1 is correct.  Keep in mind that a dekaliter is larger than a liter.  So one liter isn't enough to fill up a liter, and you should have an answer that is smaller than 1.

So its just a matter of if your conversion is smaller then then the initial quantity, then you through a negative infront of it? I can do that, but i thought that was cheating. Good to know. Next thing i totally dont understand. Another simple problem

convert 2.0 lbs to grams.
the way i do it,
2.0 lb x 454g/1lb =908 grams.

the book gives me the answer of 9.1 x 10^2g
why would you break out scientific notation when the answer is so simple?

Probably because of significant figures. In multiplication, the result is rounded down to the number of significant figures in the factor with the lowest number of significant figures, which in this case is $2$. So the book expresses the number in scientific notation as 9.1 x 10^2 instead of 908 since 908 has 3 significant figures.

[quote]So its just a matter of if your conversion is smaller then then the initial quantity, then you through a negative infront of it? I can do that, but i thought that was cheating.[/quote]

Woops, my bad.  I misread your original question (although you do have a misleading typo in there).  Your original calculation is incorrect.  You should use the following:

$1.0 \mbox{ L x }\frac{10. \mbox{ L }}{1.0 \mbox{ dekaliters}}= 1.0 \mbox{ x }10^{-1}\mbox{ dekaliters}$


integration
logarithms
calculus
geometry
function
limit
calculus computations

Show that lim (1+1/2+1/3+....+1/n)/n =0
                  n->infinity

my method:
$ 0\leq(1\plus{}1/2\plus{}1/3\plus{}....\plus{}1/n)/n <(  \int_{1}^{n}1/x dx) /n$
=$ (1\plus{}1/2\plus{}1/3\plus{}....\plus{}1/n)/n <Log(n) /n$

take limit

lim Log(n) /n  ->0   as n->infinity

so

lim (1+1/2+1/3+....+1/n)/n =0
                  n->infinity
 is it correct? thanks~

The sequence is monotonic decreasing and the lower limit of $ (a_n)_n$ is $ 0$

You can do this without knowing anything about the logarithm or the integral of $ \frac1x.$ In fact, the only property of the sequence $ \frac1n$ that you really need is that it tends to zero. That's because this is an instance of Cesaro summability of a series, or the Cesaro-Stolz theorem. We could get the answer here either by quoting that theorem, or by an argument (which isn't too difficult) that reconstructs most of the key parts of the proof of that theorem.

Let me try something else, which will echo Lomisedu's proof.

I do have one criticism of that proof: the statement that
\[ 1\plus{}\frac12\plus{}\frac13\plus{}\cdots\plus{}\frac1n<\int_1^n\frac1x\,dx\]
is false. The sum is bigger, the integral smaller. I would suggest drawing a picture, representing the sum as the areas of boxes, and the integral as the area of the curve. The problem here is that the tops of the boxes lie above the curve.

That's fixable. Just slide the boxes one unit to the left. The left-most box now has $ x$ between $ 0$ and $ 1$ and isn't related to the integral any more - so we just include that one box separately. The true statement is:
\[ 1\plus{}\frac12\plus{}\frac13\plus{}\cdots\plus{}\frac1n<1\plus{}\int_1^n\frac1x\,dx\equal{}1\plus{}\ln n\]
From there, the rest of your proof works.

Now let's do that in general - or at least in semi-general. I'm not going to go for the most general case. Suppose $ f(x)$ is a decreasing positive continuous function on $ [0,\infty)$ such that $ \lim_{x\to\infty}f(x)\equal{}0.$ Let $ a_n\equal{}f(n)$ for all $ n\to\infty.$ Then:
\[ \lim_{n\to\infty}\frac1n\sum_{k\equal{}1}^na_k\equal{}0.\]
Let's follow Lomisedu's argument. By the same "boxes under the graph" argument:
\[ \sum_{k\equal{}1}^na_k<1\plus{}\int_1^nf(x)\,dx\]
\[ 0<\frac1n\sum_{k\equal{}1}^na_k<\frac{1\plus{}\int_1^nf(x)\,dx}n\]
Is $ \lim_{n\to\infty}\int_1^nf(x)\,dx$ finite or infinite?

Case 1: $ \lim_{n\to\infty}\int_1^nf(x)\,dx$ is finite. Then there is a bound $ M$ such that we can say that $ \frac{1\plus{}\int_1^nf(x)\,dx}n\le\frac{M}n.$ That clearly tends to zero as $ n\to\infty.$

Case 2: $ \lim_{n\to\infty}\int_1^nf(x)\,dx$ is infinite. Then L'Hôpital's rule applies.
\[ \lim_{y\to\infty}\frac{1\plus{}\int_1^yf(x)\,dx}y\equal{}\lim_{y\to\infty}\frac{f(y)}1\equal{}0.\]
(That's by the hypothesis that $ f(x)\to0$ as $ x\to\infty.$) It follows that
\[ \lim_{n\to\infty}\frac{1\plus{}\int_1^nf(x)\,dx}n\equal{}0.\]
From either case, the limit we seek must be $ 0$ by the squeeze theorem.

OK, that's not anywhere close to full generality. We don't really need $ f$ to be monotone, or positive, or even continuous. But I used all of those things in order to re-capture the heart of Lomisedu's proof. And I probably shouldn't really quote L'Hôpital here - the proof should be more direct than that. In fact, Cesaro-Stolz is often referred to as "L'Hôpital for sequences."


geometry
area of a triangle
Heron\u0027s formula

given a triangle with sides 33, 52, 69. find the average of the longest and shortest altitudes.

can i check my answer with you guys?

[hide="answer?"]$\frac{680\sqrt{14}}{69}$

is there a better way than just heron's formula?
[/hide]

i cant think of any better method (my sister says to use the sickler theorem...no idea what it means and im pretty sure it doesnt exist... but she sucks at math and has no idea what shestalking about)

anyways:

herons fomula gives

330 rt 14 as the area

the argest alt. is when the base is the smallest

so are people getting the same answer?  :blush:

Yeah, I got the exact same thing using the same method.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and let $ p$ be the smallest prime which is a factor of $ |G|$. Prove that a subgroup of $ G$ which has index $ p$ must be a normal subgroup of $ G$.

I'm thinking of using the Sylow theorems, but I can't seem to get anywhere since the Sylow theorems deal with some $ p^{m}$.

Let $ H$ be a subgroup of $ G$ with index $ p$.
$ H$ induces a permutation representation
$ \pi _H : G\minus{}\minus{}\minus{}\minus{}>S_p$.
Let $ K\equal{}ker(\pi _H)$.
Then $ K$ is a subgroup of $ H$ and a normal subgroup of $ G$.
Since $ G/K$ is a subgroup of $ S_p$ (1st isomorphism theorem), $ [G: K]\equal{}[G: H][H: K]\equal{}p[H: K]$ is a factor of $ |S_p|\equal{}p!$. So $ [H: K] | (p\minus{}1)!$.
Since $ [H: K]$ is a factor of $ |G|(\equal{}[G: H][H: K]|K|)$,  $ [H: K]\equal{}1$.
In all, $ H\equal{}K$ is a normal subgroup of $ G$

Could you explain the reasoning behind $ [H: K] \equal{} 1$? I still don't seem to get that point.

$ [H: K]$ divides $ |G|$ and $ (p\minus{}1)!$, which are relatively prime. (by assumption!)