If (x-y)/(x+y)=5/2, what is x/y?

The operation # is defined as: x#y=(x:^2:+y:^2:)/(x-y). If a#7=3, what is a?

Simplify: 2^n+2^n+2^n+2^n

If a is any real number, for what value(s) of b does |x+a|=b have no real solutions for x?

Starting at her doorstep, Chelsea walked down the road at 1.5 ft./sec. for 4 sec. Then she stopped for 4 sec., when she realized she forgot something. She then returned to her doorstep at 1.5ft./sec. Describe the graph of her distance from her doorstep as a function of time (i.e. time will be on the x-axis and her distance from her doorstep will be on the y-axis)

The cost of buying noses is given as follows: C(x)={12x if 0:le:x:le:24, 11x if 25:le:x:le:48, 10x if x:ge:49}, where C represents cost and x represents the number of noses you're buying. For how many values of n is it cheaper to buy more than n noses than to buy exactly n noses?

Okay, that's enough for now.
This isn't a contest, that got stupid since on average, 3 peeps participated. Just have fun, solve, discuss, generalize, whatever.

oh, btw, you don't have to be iof a certain age/grade/whatever to do the problems, just don't do them if they're obviously below your level

...technically that's not true since this is the MATHCOUNTS Forum and we post MATHCOUNTS-leveled questions, so only people who qualify for MATHCOUNTS (Is in 8th grade or under) should answer them, but oh well.

meh! I SAID THEY'RE FUN! THAT MEANS TRY THEM! SHEESH! :\ okay sorry little rant there

Sorry to make you all upset, but you sounded like an angry cat to me at first  </jk>

Anyways, I'll go ahead and start solving them.

[hide]Assuming y isn't 0, (x-y)/(x+y)=5/2  2(x-y)=5(x+y)  2x-2y=5x+5y  3x+7y=0  3x/7y+1=0  3x/7y=-1  x/y=-7/3[/hide]

[hide]So a#7=3 and x#y=(x:^2:+y:^2:)/(x-y). a#7=(a:^2:+7:^2:)/(a-7)=(a:^2:+49)/(a-7)=(a+7)(a-7)/(a-7)=a+7 so a+7=3, a=-4. On a side note x#y is simply x+y unless x=y. In that case it is undefined. ((x:^2:+y:^2:)/(x-y)  (x+y)(x-y)/(x-y)  (x+y)[/hide]

[hide]2^n+2^n+2^n+2^n=4*2^n=2:^2:*2^n=2^(n+2)[/hide]

[hide]That's just wrong...[/hide]

[hide]2^n+2^n+2^n+2^n. Ok.

Since multiplication is repeated addition, 2^n plus itself 4 times=2^n x 4. This equals 2^n x 2^2, or 2^(n+2). [/hide]I hope I read the problem right....

NOSES? WHO BUYS NOSES?

Ok.[hide] If you buy 25 noses, 25x11=275. 275/12= about 23.

23 and 24 are therefore cases in which more is cheaper.

49 noses cost 490. 490/11= about 45. 45, 46, 47, & 48 therefore work. If I did this right there should be 6 of these cases. [/hide]

For the "If a is a real #..." problem, the only thing I can see if if B were negative.

As for Chelsea, what do you mean by "describe the graph"? Its shape, or what?

I completely agree with you.

Who's Chelsea?

ah...never mind, yeah you were supposed to describe the shape, but that might turn out to be too subjective on my part. never mind who's chelsea. i put in that name. like the original name was mark or something. but who's going to look at a problem with a mark in it. chelsea is so much hotter.

solutions: come and get em:

If (x-y)/(x+y)=5/2, what is x/y?

[hide]cross multiplying gives 2x-2y=5x+5y => -7y=3x => x/y=-7/3.[/hide]

The operation # is defined as: x#y=(x:^2:+y:^2:)/(x-y). If a#7=3, what is a?

[hide]a:^2:+49=3a-21 => a:^2:-3a+70=0 => some nasty thing with the quadratic formula.[/hide]

Simplify: 2^n+2^n+2^n+2^n

[hide]4*2^n=2^2*2^n=2^(n+2)[/hide]

If a is any real number, for what value(s) of b does |x+a|=b have no real solutions for x?

[hide]all b<0 is impossible[/hide]

Starting at her doorstep, Chelsea walked down the road at 1.5 ft./sec. for 4 sec. Then she stopped for 4 sec., when she realized she forgot something. She then returned to her doorstep at 1.5ft./sec. Describe the graph of her distance from her doorstep as a function of time (i.e. time will be on the x-axis and her distance from her doorstep will be on the y-axis)

[hide]the graph should start from the origin, rise to a plateau, and then drop back to the x-axis[/hide]

The cost of buying noses is given as follows: C(x)={12x if 0:le:x:le:24, 11x if 25:le:x:le:48, 10x if x:ge:49}, where C represents cost and x represents the number of noses you're buying. For how many values of n is it cheaper to buy more than n noses than to buy exactly n noses?

[hide]23, 24, 45, 46, 47, 48 = 6 values[/hide]

Okay, that's enough for now.

This isn't a contest, that got stupid since on average, 3 peeps participated. Just have fun, solve, discuss, generalize, whatever.[/quote]

...so you're solution is "some nasty thing with the quadratic formula."!? "After all I went through..." :lol:


college
MIT
Stanford
University of Chicago

So, where are you going?

In case I haven't said it enough times, I'm going to the University of Chicago.

Congratulations, Magnara and to all the future replyers to this thread...I think just to make it interesting, also include your major, just to spark conversation.

For me, I'm off to UC Berkeley. My major is Molecular and Cell Biology, with an emphasis in Genetics., but I may change that to an emphasis in Immunology.

Congratulations.  From what I understand, Berkeley is a wonderful place to be doing that.

I'm going to major in (pure) mathematics.

Congratulations! :D

I'm going to the University of Waterloo for pure mathematics and computer science.

congrats to everyone :)
MIT class of '09
major in math and Electrical/Comp Sci not very sure though

I was considering University of Chicago as well..
their pure math is one of the best in the nation.

it was a tough decision(i also got into caltech) but I decided to go to MIT.
I'm going to major in mathematics.

I'm headed to Stanford, thinking about a double-major in Mathematics and Philosophy.

I turned down both MIT and Caltech, which was a very hard thing to do. But I think Stanford is where I belong (I'll go to one of the techs for grad school, perhaps).


geometry
rectangle
3D geometry
icosahedron
IMO Shortlist
IMO Longlist

Given three congruent rectangles in the space. Their centers coincide, but the planes they lie in are mutually perpendicular. For any two of the three rectangles, the line of intersection of the planes of these two rectangles contains one midparallel of one rectangle and one midparallel of the other rectangle, and these two midparallels have different lengths. Consider the convex polyhedron whose vertices are the vertices of the rectangles.

[b]a.)[/b] What is the volume of this polyhedron ?

[b]b.)[/b] Can this polyhedron turn out to be a regular polyhedron ? If yes, what is the condition for this polyhedron to be regular ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

A solution for b):

If it's a regular polyhedron, then it must be an icosahedron (because it has $12$ vertices). This means that all its faces must be equilateral triangles. Let $a<b$ be the sides of the rectangles. The condition is, then (we simply compute the distance between a vertex of a rectangle to the closest among the vertices of another rectangle and we make this equal to $a$), $a^2=(\frac b2)^2+(\frac a2)^2+(\frac{b-a}2)^2$, which is equivalent to $b^2-ab-a^2=0$ or, in other words, the rectangles must be golden rectangles, i.e. $\frac ba=\phi=\frac{1+\sqrt 5}2$.

Pretty nice answer :).


by Wilson's theorem, $ (1011)!\times 1012\equiv \minus{} 1\equiv 1012\mod 1013$

dividing both sides by 1012 yields $ (1011)!\equiv 1 \mod 1013$

Quick comment on the above. "Dividing" in a modulo system is less straightforward than multiplying. You will probably make fewer mistakes if you think of it as multiplication.

In this case, we can multiply by $ \minus{}1$ instead of dividing by $ 1012$, and note that $ \minus{}1012 \equiv 1$.

wilson's theorem can be derived by fermat's theorem, but how do you do this using simply only fermat's theorem and yes normal congruences. i've another one: $ 39^{50!}\pmod{2251}$

Why would you want to do anything other than use Wilson's theorem?  Wilson's theorem can be stated as $ (p \minus{} 2)! \equiv 1 \bmod p$, so what you are doing is proving a special case of Wilson's theorem anyway and there is no reason not to use it in full generality.

[quote="TZF"]Quick comment on the above. "Dividing" in a modulo system is less straightforward than multiplying. You will probably make fewer mistakes if you think of it as multiplication.

In this case, we can multiply by $ \minus{} 1$ instead of dividing by $ 1012$, and note that $ \minus{} 1012 \equiv 1$.[/quote]

Well, an often forgotten fact is that you can handly arbitrary fractions $ \frac ab$ modulo $ n$ as long as $ b$ and $ n$ are coprime. All properties of fractional addition, subtraction, multiplication, division and reduction still hold as long as the denominator remains coprime to the modulus.
A simple example: what is $ 2^{\minus{}1}\plus{}3^{\minus{}1}\plus{}6^{\minus{}1} \mod 1234567$¿ (note that the modulus is coprime to $ 2,3,6$ and that $ x^{\minus{}1} \mod n$ is just the normal multiplicative inverse of $ x \mod n$).
Well, it's simply $ \frac 12 \plus{} \frac 13 \plus{} \frac 16 \equal{} 1 \mod 1234567$.
There are more meaningfull cases out there, but this is just a simple example illuminating why it may be easier to handle fractions $ \mod n$ directly.

[hide="Click for some a bit involved reason why this is true"]There is an isomorphism $ \mathbb Z / (n) \to \mathbb Z_{[n]}/(n)$ where $ \mathbb Z_{[n]}\equal{} \{ \frac ab \in \mathbb Q | \gcd(b,n)\equal{}1 \}$ (the localisation at the set of numbers coprime to $ n$).[/hide]


Let R1,R2,R3,R4 be distinct(don't have common point) circles.Find the centre one of them using only ruler.

First, for "distinct," I think you want "non-intersecting."  I don't understand the question, though -- what do you mean, "find the center one of them?"

It's a Steiner (ruler only) construction: you have to construct the center of one of the circles using only the ruler and three more circles. To be precise, you have to construct two lines which intersect in the center of one of the circles.

Can you give more clear solution,please?

That was not meant to be a solution, but an answer to JBL's question.


symmetry
geometry
geometric transformation
rotation
group theory
abstract algebra
graph theory

I have been looking into some stuff in Algebraic Graph Theory recently and have been wondering if the automorphism group of a regular graphs always form a dihedral group.
I feel this is true because of the few regular graphs I've run into, i can always find a graph isomorphic to it that has dihedral symmetry.

AND, as a [b]bonus question[/b]...
Are such regular graphs isomorphic to a graph with dihedral symmetry and with all the vertices lying on a circle.
This is really what I'm asking for, but the first question in think will be enough to get me going in the right direction!

:huh: $ K_5$ has automorphism group $ S_5$, which is not dihedral  :maybe:

Graph isomorphism preserves the automorphism group, so if the original graph didn't have dihedral symmetry neither can any graph isomorphic to it.

What I meant was that, the regular graphs I've seen have [i]at least[/i] dihedral symmetry. I'm not sure how to state it very well. But consider K5, the standard embedding does have dihedral symmetry, however its automorphism group has more than just dihedral symmetry.

I think what I'm trying to ask might be...
Does every regular graph of n vertices contain in its automorphism group the dihedral group Dn?

If this was true would every regular graph have an embedding where  the vertices lie on an n-gon and the embedding has the geometrical dihedral symmetry (rotations and reflectional symmetry)?

[quote="Dragonflare82"]Does every regular graph of n vertices contain in its automorphism group the dihedral group Dn?[/quote]

No.  Not all regular graphs are particularly symmetric.  For instance see the second example of a 3-regular graph on 10 vertices at http://mathworld.wolfram.com/CubicGraph.html

Its automorphism group is not even transitive on the vertices.  You can apply a flip symmetry, (1,3)(4,6)(7,8) in my ordering of the vertices, as well as exchange the inner and outer "rings", (1,4)(2,5)(3,6).  These two symmetries commute, and generate the entire automorphism group of order 4.  Clearly it does not contain the dihedral group on 10 points of order 20.

EDIT: I noticed another even less symmetric graph that I guess is sort of famous for being regular but completely asymmetric: The [url=http://en.wikipedia.org/wiki/Frucht_graph]Frucht graph[/url] has an automorphism group of order 1, that is, it has no symmetries.

Ok, well that answer one question (thanks JackSchmidt). Now the here's one more important question...

If the automorphism group of a graph of n vertices contains the dihedral group D_n, then does there exist an embedding of the graph that has the geometrical dihedral symmetry?

I think the answer is yes, but have no idea how to prove this.

[quote="t0rajir0u"]That depends at least in part on what you mean by "contains the dihedral group."  Do you mean that it contains the particular copy of the dihedral group you get from the symmetries of an $ n$-gon with vertices labeled $ 1, 2, ... n$, or do you mean that it contains any group isomorphic to the dihedral group?[/quote]

Well, wouldn't it not make a difference, because if it contains a subgroup isomorphic to Dn couldn't we just relabel the indices so that whatever is true for one is true for one automorphism group is true for the other?

No: I haven't worked out and checked the details, but something roughly like the following should work.  Consider a graph $ G$ with two connected components, one of which is $ K_{3 \plus{} 5 \plus{} 7} \equal{} K_{15}$ and one of which is some $ 15 \minus{} 1 \equal{} 14$-regular graph on $ 3\cdot 5 \cdot 7 \minus{} (3 \plus{} 5 \plus{} 7) \equal{} 90$ vertices.  The automorphism group of $ G$ is $ S_{15} \times A$ for some group $ A$.  If I'm not mistaken, $ S_{15}$ contains a copy of $ D_{2 \times 105}$ (the rotation is the product of a disjoint 3-, 5- and 7-cycle).  Thus, the automorphism group of $ G$ contains this dihedral group, but it's the "wrong" copy of the dihedral group -- it's very not transitive on the vertices, for example, and this cannot be fixed by any amount of relabelling.  And, it's easy to see that this graph doesn't have full dihedral symmetry (because, for example, its automorphism group isn't vertex-transitive).

Even if you require that the order of the dihedral group is correct, it's not clear that it will be conjugate to the usual copy of the dihedral group in $ S_n$.  Even if this were true (which I doubt) you would have to prove it first.

[quote="t0rajir0u"]Even if you require that the order of the dihedral group is correct, it's not clear that it will be conjugate to the usual copy of the dihedral group in $ S_n$.[/quote]

Sym(6) contains two conjugacy classes of subgroups isomorphic to the dihedral group of order 12.  One is the "correct copy", operating transitively on six vertices, < (123456), (16)(25)(34) >; the other is a direct product contained already in Sym(5), <(12),(123),(45)>.

I suspect a transitive subgroup of Sym(n) isomorphic to the dihedral group of order 2n is permutation isomorphic to the dihedral group on n vertices.  Without assuming transitivity, examples like dihedral groups of order 8k+4 give an infinite family of counterexamples.


how to take the Ap US test successfully, which book should I use?
By the way, can you guys help me with this DBQ:
Jacksonian democrats viewed them selves as the guardians of the US Constituion, political democracy, individual liberty, and equality od economic opportunity
In light of the followig documents & your knowledges of the 1820s & 18302, to wat extent do u agree w/ the Jacksonians' view of themselves
I'll scan the docs later :oops:
This is the hardest Ap class ever (since I'm not a native speaker  :ninja: , gosh)

Here is some possible background info. you could use in order to enhance your essay beyond just the documents:
-Jackson's humble beginnings
-attitudes toward democracy and individualism in Jackson's West
-economic mobility in the frontier
-Jackson's veto of the Bank of the United States on the basis that the latter was a plutocratic and undemocratic organization
-Bank veto message "Equality of talents, of education, or of wealth can not be produced by human institutions...But when the laws undertake to add to these natural and just advantages artificial distinctions..the humble members of society...who have neither the time nor the means of securing like favors to themselves, have a right to complain of the injustice of their government."
-1828 campaign slogan "Shall the people rule?"
-Jackson's belief in states' rights and local government over federal government, as the former would be more intimately connected with the regular people
-Jackson's success in solidifying the party system and making politics appeal to ordinary people, evidenced by the massive increase in turnout during his time


[quote="tranthanhnam"]let a,b,c be three positive real
prove that :
 $a^3 +b^3+c^3 +3abc \geq ab\sqrt{2(a^2+b^2)}+bc\sqrt{2(c^2+b^2)}+ac\sqrt{2(a^2+c^2)}$[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=schur&t=6532

  It is no interesting when you posted old mathlinks problem over and over gain. It is not a good habit


If P(x) denotes a 2008-degree polynomial such that P(k)=k/(k+1) for k =0,1,2,...,2008. Find P(2009)

For some strange reason, even though the polynomial is uniquely defined, I'm getting a generic answer, so this probably isn't right.

[hide="Solution"]From the definition, we know that $ (k\plus{}1)P(k) \minus{} k \equal{} 0$ for all integers $ 0 \le k \le 2008$. Therefore, $ (k\plus{}1)P(k) \minus{} k \equal{} k(k\minus{}1)(k\minus{}2)...(k\minus{}2008)$. Therefore, when $ k \equal{} 2009$, $ P(k) \equal{} \displaystyle\frac{2009!\plus{}2009}{2010}$.

The generic nature of the solution comes because I assumed the polynomial was monic, although I can't necessarily assume that; there should be a constant $ c$ preceding the polynomial of degree $ 2009$.[/hide]

Imagine doing the same question for small values, for $ k \equal{} 0,1$, find $ P(2)$. Now, is your linear polynomial going to have a constant? But, 2 points determine a line uniquely.  :ninja:  I'll let you find out.

EDIT: Try the same with $ k\equal{}0,1,2$ to find $ P(3)$. Why is there no constant?

Your solution is almost right.  You have an equation in 2 "variables": P(k) and c.  We want to find c, so we need to get P(k) to cancel out.  We can't use k=0, 1, 2, 3, ... 2008, so how do we get P(k) to cancel out?

Remember, the equation relating these is $ (k \plus{} 1)P(k) \minus{} k \equal{} ck(k \minus{} 1)(k \minus{} 2)\ldots(k \minus{} 2008)$.

Edit: See [url=http://www.artofproblemsolving.com/Wiki/index.php/1975_USAMO_Problems/Problem_3]here[/url], too.

i have solved such a problem some time back.i used tellescopic cancellation but i dont remember how i did that


The base of a triangle is 3 inches long. Two lines are drawn parallel to the base, terminating at the other two sides & dividing the triangle into three equal areas. Find the lengths of the two parallel line segments.

Well, the top one will have $ \sqrt {3}$ length since it is a third of the total area, and similar to the original triangle, so has a scale factor of $ 1/ \sqrt {3}$

The second line underneath cuts off a triangle 2 times bigger than the top triangle, and is similar, so has a scale of $ \sqrt {2}$ compared to the top smaller triangle. Therefore, the second line has length $ \sqrt {6}$.


search
number theory proposed
number theory

$a, b, c \in N$.

Prove that

$(ab+1)(bc+1)(ca+1)$ is a perfect square

if and only if $ab+1, bc+1, ca+1$ are all perfect squares.

If I'm not taken wrong, this was already postet three times before, so use the search function.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=kedlaya&t=32104

you will find a link to Kedlaya's proof.

Pierre.


Is
\[ \sum_{n\equal{}1}^{\infty}\frac{1}{n!\plus{}(2n)!}\approx 0,373197\]
rational or not ?

This is the question ...

[quote="Allnames"][quote="knoppix"]Is
\[ \sum_{n \equal{} 1}^{\infty}\frac {1}{n! \plus{} (2n)!}\approx 0,373197
\]
rational or not ?

This is the question ...[/quote]
$ \sum_{n \equal{} 1}^{\infty}\frac {1}{(n!)^2}$ is also irrational
The answer is NO  :lol:[/quote]
I don't see how the irrationality of your number imply the irrationality of my one  :(


[quote="zhaobin"][quote="Nameless"][quote="zhaobin"]just let $abc=1$
I didn't see the solution have some different with [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=68158]www.mathlinks.ro/Forum/viewtopic.php?t=68158[/url] :)[/quote]
No, I think my ineq can't be solved by Mincopxki or AM-GM, so they different  :)[/quote]
sorry,It maybe my first mistake in the new year.
and solution again:
$\sqrt[4]{3a^4+3b^4+10ab^3} \ge \frac{21}{16}a+ \frac{11}{16}b$
is it right now,but it is also  AM-GM ;)
btw the way,nice inequality :lol:、
[/quote]
sorry,It seems I am wrong again :blush:

[quote="zhaobin"][/quote][quote="zhaobin"][quote="Nameless"][quote="zhaobin"]just let $abc=1$
I didn't see the solution have some different with [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=68158]www.mathlinks.ro/Forum/viewtopic.php?t=68158[/url] :)[/quote]
No, I think my ineq can't be solved by Mincopxki or AM-GM, so they different  :)[/quote]
sorry,It maybe my first mistake in the new year.
and solution again:
$\sqrt[4]{3a^4+3b^4+10ab^3} \ge \frac{21}{16}a+ \frac{11}{16}b$
is it right now,but it is also  AM-GM ;)
btw the way,nice inequality :lol:、
[/quote][quote="zhaobin"]
sorry,It seems I am wrong again :blush:[/quote]
I was being a one that just only follow others :(
now I think what I post is right:(but a little mistake with 11 and 21)
$\sqrt[4]{3a^4+3b^4+10ab^3} \ge \frac{11}{16}a+ \frac{21}{16}b$
$\sqrt[4]{3a^4+3b^4+10ab^3} \ge \frac{11}{16}a+ \frac{21}{16}b$
$\iff 196608a^4+196608b^4+655360ab^3 \ge 194481b^4+407484ab^3+320166a^2b^2+111804a^3b+14641a^4$
$\iff 2127b^4+181967a^4+247876ab^3 \ge 111804a^3b+320166a^2b^2$
but we have:
$2127b^4+6381a^4 \ge 8508a^3b$
and
$68864a^4+34432 ab^3 \ge 103296a^3b$
and
$106722a^4+213444ab^3 \ge 320166a^2b^2$

so we proved the last inequality.
(in fact,in post 3 it was my feeling,and haven't check it,sorry.)
now we prove the Nameless's inequality.
arqady,there must be something wrong in your calaclution ;)

[quote="arqady"][/quote][quote="Nameless"][quote="arqady"][quote="Nameless"]$a,b,c>0.$Prove:
$\sum\sqrt[4]{3a^4+3b^4+10ab^3}\ge2(a+b+c)$
It isn't hard. Have fun :D[/quote]
Try $a=2,b=1,c=1$  ;)[/quote]
When $a=2,b=1,c=1\Rightarrow\sum\sqrt[4]{3a^4+3b^4+10ab^3}>8,28\ge8=2(a+b+c)$
So this ineq is still true ;)[/quote][quote="arqady"]
If $a=2,b=c=1$ then $\sum\sqrt[4]{3a^4+3b^4+10ab^3}=$
$=2\cdot\sqrt[4]{3\cdot2^4+3\cdot1^4+10\cdot2\cdot1^3}+\sqrt[4]{3\cdot1^4+3\cdot1^4+10\cdot1\cdot1^3}$[/quote]
I think it is wrong here.

[quote="zhaobin"][quote="arqady"]
If $a=2,b=c=1$ then $\sum\sqrt[4]{3a^4+3b^4+10ab^3}=$
$=2\cdot\sqrt[4]{3\cdot2^4+3\cdot1^4+10\cdot2\cdot1^3}+\sqrt[4]{3\cdot1^4+3\cdot1^4+10\cdot1\cdot1^3}$[/quote]
I think it is wrong here.[/quote]
Why? :?

[quote="zhaobin"][/quote][quote="arqady"][/quote][quote="Nameless"][quote="arqady"][quote="Nameless"]$a,b,c>0.$Prove:
$\sum\sqrt[4]{3a^4+3b^4+10ab^3}\ge2(a+b+c)$
It isn't hard. Have fun :D[/quote]
Try $a=2,b=1,c=1$  ;)[/quote]
When $a=2,b=1,c=1\Rightarrow\sum\sqrt[4]{3a^4+3b^4+10ab^3}>8,28\ge8=2(a+b+c)$
So this ineq is still true ;)[/quote][quote="arqady"]
If $a=2,b=c=1$ then $\sum\sqrt[4]{3a^4+3b^4+10ab^3}=$
$=2\cdot\sqrt[4]{3\cdot2^4+3\cdot1^4+10\cdot2\cdot1^3}+\sqrt[4]{3\cdot1^4+3\cdot1^4+10\cdot1\cdot1^3}$[/quote][quote="zhaobin"]
I think it is wrong here.[/quote]
in fact:
$\sum\sqrt[4]{3a^4+3b^4+10ab^3}=$
$=\cdot\sqrt[4]{3\cdot2^4+3\cdot1^4+10\cdot2^3 \cdot1}+\cdot\sqrt[4]{3\cdot2^4+3\cdot1^4+10\cdot2\cdot1^3}+\sqrt[4]{3\cdot1^4+3\cdot1^4+10\cdot1\cdot1^3}$ ;)

[quote="arqady"]Wow! You are right. Sorry.[/quote]
and it is very fun and exicting to discuss with you :)

Zhaobin's solution is similar to mine. But I have only one solution.
Can anyone have another solution?


$u_n=\sqrt[4]{n+\sqrt[4]{(n-1)+\sqrt[4]{(n-2)+...+\sqrt[4]{1}}}}$

Find an asymptotic expansion with two terms in $u_n$  when n tends to +oo

Isn't it clear that $\sqrt[4]{n}$ is an equivalent? It suffices to observe that our sequence is always dominated by $n$ and then to apply this observation to $\sqrt[4]{n-1+...}$.  :?

Not much more difficult. Since $(x_n-\sqrt[4]{n})=\frac{x_{n-1}}{x_n^3+...+\sqrt[4]{n^3}}$, you find the development.


algebra
polynomial
LaTeX
number theory theorems
number theory

I wrote a paper on a restricted sum sets using polynomial methods that I need to send together with my application for the undergrad studies. As I am not quite sure that everything is alright in the paper can someone help me by going through it and point out mistakes or give me suggestions? I hope you will find it interesting and useful.

thx

A quick look reveals plenty of stupid mistakes in $ \text{\LaTeX}$ that make some formulae next to unreadable. Despite one can guess what was meant in all cases, I suggest you take care of those before asking somebody to proofread the text. :). Your English is also quite clumsy in a few places but this is not so important now. We can discuss it later after the formulae are cleaned. Mathematics, on the other hand, seems OK (though I haven't taken a really close look yet).


conics
parabola
analytic geometry
graphing lines
slope
trigonometry
rotation

Find the length of a edge of the regular triangle whose three vertices  $P,Q,R$  lie in parabola $y=x^2  $ in x-y plane, assuming that the slope of the line which pass through two points $P$ and  $Q$ is $\sqrt2$.

sqrt(6)?
It is quite strange..
what is regular triangle?

You are wrong,siuhochung.

Regular triangle is the triangle which have three equal sides.

Since $P, Q, R$ lie on the parabola, their coordinates are related as $y_P = x_P^2$, $y_Q = x_Q^2$, $y_R = x_R^2$. Equations of the lines $PQ, QR, RP$ are

$PQ:\ \ y - y_P = \frac{y_Q - y_P}{x_Q - x_P}(x - x_P) = (x_Q + x_P)(x - x_P)$

$QR:\ \ y - y_Q = (x_Q + x_R)(x - x_Q)$

$RP:\ \ y - y_R = (x_R + x_P)(x - x_R)$

Slope of the line $PQ$ is $\tan \theta = x_Q + x_P = \sqrt 2$ - the 1st equation for the 3 unknowns $x_P, x_Q, x_R$. Since the lines $QR, RP$ and $PQ$ form an angle of $60^\circ$, slope of the lines $QR, RP$ (the 2nd and 3rd equations) are

$x_R + x_Q = \tan {(\alpha \pm 60^\circ)} = \frac{\tan \alpha \pm \tan {60^\circ}}{1 \mp \tan \alpha \ \tan {60^\circ}} = \frac{\sqrt 2 \pm \sqrt 3}{1 \mp \sqrt 6}$

$x_P + x_R = \tan {(\alpha \mp 60^\circ)} = \frac{\tan \alpha \mp \tan{60^\circ}}{1 \pm \tan \alpha \ \tan{60^\circ}} = \frac{\sqrt 2 \mp \sqrt 3}{1 \pm \sqrt 6}$

Taking the upper sign case and subtracting the last 2 equations,

$x_Q - x_P = \frac{\sqrt 2 + \sqrt 3}{1 - \sqrt 6} - \frac{\sqrt 2 - \sqrt 3}{1 + \sqrt 6} = -\frac{6\sqrt 3}{5}$

$PQ = \sqrt{(x_Q - x_P)^2 + (x_Q^2 - x_P^2)^2} = |x_Q - x_P| \sqrt {1 + (x_Q + x_P)^2} = \frac{18}{5}$

The lower sign case is treated in exactly the same way. Both solutions correspond to the same equilateral triangle with the side $PQ = \frac{18}{5}$. The upper resp. lower sign cases correspond to the point $Q$ being the upper resp. lower intersection of the line $PQ$ with the parabola.

Thank you for your detail solution as usual,yetti. :)

The answer is right.

kunny

[quote="ma_go"]how about changing point of view?
let P = (a,a²) and Q = (b,b²). we could take P as origin, and rotate the parabola round this new origin by an angle of 60°...
then we check if Q' lies again on the same parabola... is it as long as your solution?
(sorry, i'm too lazy to try  :lol:  :blush:  :P )[/quote]

Yes.Your idea is super solution. My solution may be same as yours.

kunny

Actually, we can simplify the working quite a lot:

$x_P + x_Q = \sqrt{2}, x_Q + x_R = \frac{\sqrt{2} + \sqrt{3}}{1 - \sqrt{6}}, x_R + x_P = \frac{\sqrt{2} - \sqrt{3}}{1 + \sqrt{6}}$

or

$x_P + x_Q = \sqrt{2}, x_Q + x_R = \frac{\sqrt{2} - \sqrt{3}}{1 + \sqrt{6}}, x_R + x_P = \frac{\sqrt{2} + \sqrt{3}}{1 - \sqrt{6}}$

(since the gradients of the 3 sides are clearly unique).  From here, we can solve for $x_P, x_Q, x_R$ easily and thus find $PQ$.


This is my first year in Physics Olympiad, and I got the Tipler physics book recently. Should I also purchase Halliday/Resnick?

I've heard many stories of people simply reading Halliday or Tipler books and making team. Should I really focus on doing all of the end-of-chapter problems?

Also, I was wondering if anyone has the 2-3 tests that used to be at this link
http://www.compadre.org/psrc/items/detail.cfm?ID=390
those lucrative AAPT people took away those sample exams and replaced it with a single 1997 exam
By the way this is a simultaneous effort to self-study AP Physics C, so any study tips are welcome. Thanks.


symmetry
blogs
integration
real analysis
real analysis unsolved

A tough one: let $K$ be a convex compact in $R^2$. Prove that its gravity center is the midpoint of at least 3 segments
whose endpoints are on the border of $K$. Can we improve it to 4 such segments?

An easy answer to the second part: In a triangle, there are exactly three such segments, which are parallel to the sides.

Let me try.

Let $BB$ be the boundary of $K$, $O$ the center  of gravity. Let $K'$ be the symmetric of $K$ vrt $O$ and $BB'$ it boundary. The points of intersection of $BB$ and $BB'$ group in pairs of symmetric ones, and we have to prove there are at least 6 such ppoints.

Suppose there are just two points, $A$ and $A'$. The compacts $K\backslash K'$ and $K'\backslash K$ are symmetric wrt $O$ and are contained in the compact sets $C$ (respectively $C'$) bounded by $AA'$ and a part of boundary $BB$ (respectively $BB'$). However, as $O$ is the center of gravity of $K\bigcap K'$ (actually it's center of symmetry), $O$ must also be the center of gravity of $K\backslash K'$, so must lie in the interior of $C$ contradicting the fact that it is actually on its boundary.

Now suppose that there are four points $A,B,A',B'$ that form a parallelogram. $K \bigcap K'$ is a convex symmetric body containing $A,B,A',B'$ and so the interior of $ABA'B'$. The symmetric sets $K \backslash K'$ and $K'\backslash K$ consist of two compacts each, which have as boundary a part of $BB$ and a part of $BB'$ (for example the small part of $BB$ between $A,B$ and the small part of $BB'$ between $A,B$). Wlog the compact $C_1$ having $A,B'$ on boundary and $C_2$ having $A,B$ on the boundary belong to $K\backslash K'$ . The compacts $C_1',C_2'$ symmetric to them make $K'\backslash K$. clearly $C_1$  belogs to $D_1$: the conver set containing $C_1,A,B'$ and analogously we build $D_2,D_1',D_2'$ containing $C_2,C_1',C_2'$.

Again $O$ must be the center of gravity of $K\backslash K'=C_1\bigcup C_2$ thus it lies on $(GH)$ where $G\in D_1$ is the center of gravity  of $C_1$ and $H\in D_2$ is the center of gravity of $C_2$. Analogously if lies on $[G'H']$ (the symmetrics of $G,H$). However $O$ belongs to $[GG'], [HH']$ implying that $G,H,G',H',O$ are collinear. We will actually disprove this statement, showing that no two points from $C_1,C_2$ or $C_1,C_2'$ can be collinear with $O$.

We note that $OB'$ can't intersect $C_1$ or $C_2$ in the interior as $B'$ is on the boundary of convex $K,K'$, and analogously for $OA',OB,OA$. Therefore the lline $OH'$ passes through the interior of angle $\angle {B'OA'}$ and $OG$ through the interior of angle $\angle AOB'$. But this interiors and their symmetrics are disjoint, so the lines $OG$ and $OH$ can't coincide, finishing the proof.

Here's what I did. I'm not sure it's Ok, since I don't really "know my physics", but I looked up what I needed here on Google :).

Assume there are only two such segments, $AC,BD$, with $A,B,C,D$ in this order, if we go in a counterclockwise direction (it's easy to prove that there are at least two such segments). Let $G$ be the centroid. The assumption that $G$ is the midpoint only of $AC,BD$ says that we may assume WLOG to have $GX>GY$ for all $X\ne A,C,D$ on the portion of the border of $\mathcal K$ bounded by $A,C$ which contains $D$, and $Y$ being the intersection between $(XG$ with the border of $\mathcal K$.

Now take $AC$ to lie along the $x$-axis, and assume $G$ to be the origin of the plane. Also take the portion of $\mathcal K$ lying above $AC$ to be that containing $D$. Denote it by $\mathcal K_1$, and denote the other part of $\mathcal K$ by $\mathcal K_2$. Now, the magnitude of the force moment exerted by $\mathcal K_1$on the axis $AC$ is $\int_{\mathcal K_1}y\ \mbox ds\ (*)$, while the magnitude of the force moment exerted by $\mathcal K_2$ around the axis is $\int_{\mathcal K_2}|y|\ \mbox ds\ (**)$. By the remark at the end of the previous paragraph, $-\mathcal K_2$ is contained in $\mathcal K_1$ strictly, so $(*)$ is equal to $\int_{-\mathcal K_2}y\ \mbox ds$ plus something positive, i.e. clearly greater than $(*)$. This is impossible however, since the lines through $G$ are supposed to divide $\mathcal K$ into two bodies with the same force moment around the axes represented by those lines.

Sorry if it's total nonsense :).


In the 6*4 grid shown, 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column.  Let N be the number of shadings with this property.  Find the remainder when N is divided by 1000.
I want to find some problems similar to these for practice :lol:

Try the AoPS Counting & Probability book :D

or AoPS books in general.

I solved this on with lots of casework, if I recall correctly...

Can you explain what kinda of casework did you use?


One must learn huge amounts of mathematics these days. To master a subject one has to make a lot of exercises. It happens very often I see the solution right away. When it would have been a test, I would wrote down a complete solution, but when studying, I just don't have the time for this. Where does one draw the line? When is it useful to write up a full proof (of a solution in your head), and when is it just a waste of time?

Or does one not have to worry about this anymore if one becomes 'more mature in mathematics'. I believe it was very useful in the early days (just one example: the classical $ \epsilon$-$ \delta$ techniques in real analysis), but now it just slows you down instead of improving your ability to do mathematics? What are your thoughts?

With enough experience, you should know when the "proof in your head" (or even proof in your head plus written scratch work) translates easily to a rigorous, written proof, and when it doesn't.


Hi,

Some of this site's new members might not know me quite well but I was once, very active member in this site, posting over 100 posts on one month and averaging two to three posts per day. But then I quit it and it's been since I really did anything with math.

But math is still topic that interests me and now, I'm thinking of burning a new contest out from my head. It won't be so easy but to keep myself from some math, I'll do it. Here's the basic rules:

1) Problems will be posted on my website (which I'll soon make adjustment so I can have a forum under topic contest.
2) To get into the contest, you must register to my forum and the rule for the registering is that you must be middle school student or high school student.
3) Once you registered, post to let me know that you have been registered.
4) When all people for contest has been registered, I'll post out 5 problems.
5) If you get ALL right, you pass. If you got at least one wrong, you're OUT.
6) No prize for winning and no shame for loss.

It would be a good chance for me to look over extra math plus, I hope this will interest many people who will prepare for AMC.

NOt that those problems will be like that way.

Soon, I'll build problem and add the link to my website to it.

RULE: Please respect each other. And you can't post any where else than in the certain forum I made for this contest. It is because others are for my school and my buddies  :P

If interested, please post to let me know. I'll remove this post if no one's interested.

Sincerely,

I'm interested.

Can elementary schoolers participate?

Sure. The reason I said they can't is because some stuffs might be little bit hard for them. But sure. Why not? :D

The link for the my forum where question will be posted is here:

http://www.dhost.info/jwec05/Forum/phpBB%20Program/phpBB2/index.php

If interested, you must go to this site and register and post on the that forum I made for this and show me that you're interested!

THanks.

I'm in 7th grade, but I'll give it a shot.

By the way, I average about 10 posts a day on the iDB...

Sure, why not. It won't be a waste of time.

But, isn't it a bit too strict?

Strict.. hmm.. Well, I guess it kinda is because you're out of the run if you get at least one wrong. On the other hand, it's easy to pick out the best winner who will be careful on the work and getting the right answer.

And if you're interested, pleaes go to the link above and regsiter, then post to let me know.

Thanks.

Soon and hopefully, I'll try to get more people in and we can start this contest.

Hello Silverfalcon, I remember you!
I registered.

Yes, I remember you.

Anyway, so far I have three people signed and registered. But. I only can identify one.

drspyder - I don't know who that is... I'm assuming it's Treething but please identify it.

themonster - I have no clue.

eryaman - eryaman

Keep registering! We need more people than three.

Lol.. That was bit obvious. :blush:

Hey, can you send me some sample problems, to see if I should sticky the topic on the Mathcounts forum?

I don't want to sticky it and find that it's way too easy or way too hard.

Just making sure. Thanks.

PS- Welcome back man. There's been some change...

I can try but I don't have any. Only thing I can tell you is that it would be in minimum 8th grade to maximum 10th grade. Some maybe harder but not too much.

I'll now post two practice problems for people to look at.

The link to the site where problem is available is:

http://www.dhost.info/jwec05/Forum/phpBB%20Program/phpBB2/viewtopic.php?p=172#172

Good luck! You may discuss it in here or there but try to do there with identifying yourself. I still need to know who I have now.

So far contestants:

6 People have joined.

Remember, you can still JOIN NOW.

I'm trying to get Latex available for my website.

If this works out, the contestants will able to read the questions in Latex and maybe perhaps write them with Latex.

Not enough joined, still.

How long do we have to work on the problems?

Since we're still getting people to join and stuffs, in the first round, I'm thinking of giving the time till next Friday, 11/12. After that, I'll take the answes I have and grade it and hopefully in the week that comes after that, I can see who has been passed.

Don't think about when to turn the answers now yet. Just do it and do your best, and check it and then turn it.

Okay?

Please remember that the answers must be turned to me by this FRIDAY!!

I have only got one answer from the contestant.

Please turn them as soon as possible.

Silverfalcon

Are the first two problems the first round?

No. I hope you checked the sticky in this forum.

There should be a sticky saying, "Round 1 Problems Available" with five questions.

That's the round one.

Those two questions are just practices except no one really did it.

Two contestants have survived.

Those two are: theone853 and RC-7th

Since it's two people left, I'll post ten questions on the last round.

Anyone who gets most will get win.

If tie occurs, we repeat one more round.

Questions will be, I'm going to try, within next week. I'm going to make this pretty challenging because I know both users are very smart contestants and if the questions are too easy, I'm pretty sure they will have tie on perfect score.

Silverfalcon

can I join the contest NOW?
the links for the problems didn't work.

No, sorry. The contest that's going on now will end in next week though.

I might start one right after this one and when that happens, I'll let people to know.

Silverfalcon


MATHCOUNTS
percent
AMC
AIME
calculus
AMC 10

I will be hosting a MathCounts Math Jam on December 17th.

I have decided to run this Math Jam as a game.  Points will be awarded to the students who get the correct answers most quickly.  Points will also be awarded to excellent explanations of solutions.

Let's see who is most prepared for the countdown round!

ha, this will be sweetness. and its the day before school is let out, meaning more fun. m00t

Arg notsoconfidentnotsoconfidentnotsoconfident...I always get my butt kicked in both Math Jams and Online Classes, not to mention competitions so :cry:

I think it's a great idea. It doesn't hurt anyone if they don't do well, and the winners can be happy. If you don't want others to know how you did, you can always get another account for the Math Jam so as to be more anonymous.

The second might not be such a good idea for that is double-registration which is usually considered not good.

Is double-registration considered not good? I hadn't heard that, but maybe you are right.

Mmm, come to think of it it may be OK here but last time I thought "smurfing" or whatever was OK when a person registered twice or something with a name just like one other and Mr. Rusczyk deleted his/her account so it's case-by-case I guess...

urg, THE BOB's just going to sweep the whole thing

that doesnt stop him from entering the mathjam though, maybe Mr. C would use his explanations afterwards if no one does it, or just wont post them or something. i dunno. it dunt matter.

its also good for the >8th graders cause it may bring back good memories... and i agree on posting his answers if we don't get it

Oh wait I didn't all of you guys (Arnav, Neal, and Chinaboy) were above 8th...am I the only one here that qualifies for MATHCOUNTS?

too true... and fortunately, its true. i just said it would be good for them, but it didn't really say that i was one also (fortunately, im not one of the hs people yet... one more shot at mc.)

urg...
i can't type fast enough.
in the other online classes while i'm still typing the answer is already up.
:(

Same here, especially in countdown round explanation, the one time I actually get it they leave the problem before I'm finished explaining...*sigh* wonder why AMC10's a breaze for me and MATHCOUNTS is not...:?:

Well I've never been in MATHCOUNTS (and probably never will) but I stink and constantly embarass myself in Math Jams, Online Classes, and competitions like toppers

As for AMC I scored a 100 (without the percent sign :oops:) but when I look back at it now it's easy stuff.

Yeah I know, I was in 7th geez, laugh at me, hahaha, wootwoofwoot, I did 2.8 times better on the [b]English[/b] MCAS *bragbrag*

Don't be so hard on yourself Tare, you do really good in my competiiton.

This will be fun.

[quote="Syntax Error"]um, dont you need a bit over a 120 to qualify for aime

for amc 10 that is[/quote]
for last year's i saw many people who got 119.5 and qualified for aime.

:lol: !!!!!!!TOMORROW IS THE MATH JAM!!!!!!! :lol:

I seriously am scared after seeing my butt kicked on UNL Math Day and Getting Started Contest...(and probably on the AMC Mock Contest also)
*sigh* I have no idea why I do better with harder more challenging problems around AMC 10 to easy AMC 12 than Getting Started or MATHCOUNTS level question...or so I think? :?:

So list of people who'll probably in the Math Jam:
Leeyuhang
Topper
Tare (if I don't forget :))
MysticTerminator (although >8th)
Syntax Error
Chinaboy
theone853?

just listing my competitors...hmm with Bob and Neal this is going to be tough...

well yeah they'll be pretty good. martin went to nats last year...i don't know the others

hehehe... thanks for reminding me about it!

[quote="Tare"]Yeah but after seeing certain things recently I'm not seeing you much of a threat anymore :P </jkjkjk :wink:>[/quote]
Tare, if that's directed against me... well, you'll just pay for it in the math jam, whether it was or not ;)

Well this isn't really my level but I've never been to a math jam before but I finally worked out the time difference to america so you may see me in there.. or u may not.

I agree. I am a slow, slow, sloooooow thinker :?:

[quote="Chinaboy"]LoL, I was never much of a threat in countdown... last year, i had to defend my 1st written test in countdown...

in chapter ... i lost 2-3
in state... hehehe... i lost 0-3....

BUT im determined to change that.[/quote]

and i will change that change you are making

i need some easy mac, and i went to return of the king midnight showing. so ill either be later, or just not go, but i need easy mac

ok... w/e...

there is a new element in this year's countdown that we found while doing ABLE... buzz before you get the answer... hehehe... its risky though

Yeah, to quote Mr. Rr.... "I'll tell you how you are with time, but I'm not going to let you rough up the middle schoolers"

Basically, he didn't post the things I said. That is all.

Oh...why? I thought you were 8th since you were planning to participate...:oops:

Wha? No wonder he's so confident he can kick my butt...

Still if he is then whoa...

hehe that was fun!!!!!! :lol:
though i did better than what i had expected
how many of you posted answer but you were too late?

ya, when i faltered on the keyboard, i didn't even try again... since i lost lots of time in messing up on the keyboard... oh well...

i would post excuses, but thats pansyish and anywho, its not the real thing, and yeah, im starting to make excuses, so ill just stop here

i just posted one... but that's not really okay... but overall, it was fun! and a good warm up!

FUNFUNFUN! i'll stop... before i copy and paste that all over...

Tare, I'm a senior taking AP Calculus BC and AP Statistics (not that the second one is much of a math class)


geometry circumcircle inradius incenter trigonometry

This is a [b]really[/b] bad solution and I am not sure if it even works for all cases.  Furthermore, it is explained in a really bad manner.  Well, here it is:
[hide] Well, this is kind of guessing.  Assume that x+y+z is fixed (I have no idea if it is fixed or not).  Create a right triangle with with circumradius 12 and inradius 5, say that Angle C is 90.  Immediately we can realize that AB is 24 since the circumcenter of a right triangle is on the midpoint of the hypotenuse.  Next we can take the incircle.  Say the incenter is I and the points of tangency is X on AB, Y on AC, and Z on BC.  We know that IX=IY=IZ=5=CZ=CY.  Now, assume that BC=a.  This means that BZ=a-5=BX.  Since AB=24, we have that AX=29-a=AY.  Now, since CY=5, we have that AC=34-a.  So, we have a right triangle ABC with AC=34-a, BC=a, and AB=24.  The perpindicular from O to AB is 0, and the perpindicular to BC+perpindicular to AC=(BC+AC)/2=$17$, which is our answer.  Notice that it is the circumradius+the inradius[/hide]

I assume that the lengths of the segments from O to sides $\overline{AB}$, $\overline{AC}$ and $\overline{BC}$ are perpendicular, aren't?

Then we can aply Carnot's theorem, which says:

\[x+y+z=r+R\]

$\therefore \ x+y+z=17 \ \blacksquare$

[quote="MathTeX"]I assume that the lengths of the segments from O to sides $\overline{AB}$, $\overline{AC}$ and $\overline{BC}$ are perpendicular, aren't?

Then we can aply Carnot's theorem, which says:
\[x+y+z=r+R \]
$\therefore \ x+y+z=17 \ \blacksquare$[/quote]

I think the point is to derive Carnot's theorem.

[quote="mysmartmouth"]Let triangle ABC be inscribed in circle with center O.  Let x, y, and z, be the lengths of the segments from O to sides, AB, AC, and BC, respectively.

Find the sum $x+y+z$, if the circumradius of the triangle is 12, and the inradius is 5.[/quote][color=darkred]Must the condition "the acute triangle $ABC$ ..."
In this case, if denote the middlepoints $M$, $N$, $P$ of the sides $[BC]$, $[CA]$, $[AB]$ respectively,then exists the [b][u]Euler's relation[/u][/b] : $OM+ON+OP=R+r$.[/color]

[color=darkblue][b][u]Proof 1.[/u][/b] Observe that $a\cdot OM+b\cdot ON+c\cdot OP=$ $2\cdot [ABC]=$

$2pr=(a+b+c)r$. Apply the [u]Ptolemeu's relation[/u] in

the cyclic convex quadrilaterals $APON$, $BMOP$, $CNOM$ :

$\left\{\begin{array}{c}b\cdot OP+c\cdot ON=a\cdot R\\\ c\cdot OM+a\cdot OP=b\cdot R\\\ a\cdot ON+b\cdot OM=c\cdot R\\\ a\cdot OM+b\cdot ON+c\cdot OP=(a+b+c)r\end{array}\right\|\ \bigoplus$ $\Longrightarrow$
\[(a+b+c)(OM+ON+OP)=(a+b+c)(R+r)\Longrightarrow OM+ON+OP=R+r \]
. [b][u]Proof 2.[/u][/b] $\left\{\begin{array}{c}OM=R\cos A\\\ ON=R\cos B\\\ OP=R\cos C\\\ \cos A+\cos B+\cos C=1+\frac{r}{R}\end{array}\right\|\ \bigoplus$ $\Longrightarrow$
\[OM+ON+OP=R\sum\cos A=R\left(1+\frac{r}{R}\right)=R=r \]
.[/color]

[quote="Virgil Nicula"]Must the condition "the acute triangle $ABC$ ..."
[/quote]

Perhaps, I just made the problem up.  However, I think Carnot's theorem works regardless of whether or not the triangle is acute.  Cut-the-knot proves the acute case, then "leaves the other case to the reader."


Can anybody tell the meaning of [x] and {x} in maths?

Generally:

$ \lfloor x\rfloor$ means the greatest integer smaller than or equal to $ x$.

$ \{x\}$ means the not integer part.

So $ x \equal{} \lfloor x\rfloor \plus{} \{x\}$

Or if $ a\le x < a \plus{} 1\implies \lfloor x\rfloor \equal{} a$  and $ \{x\} \equal{} x \minus{} a$ where $ a$ is an integer and $ x\in \mathbb{R}$.
Examples:

$ \lfloor 3.4\rfloor \equal{} 3$

$ \{3.4\} \equal{} .4$

$ \lfloor \pi\rfloor \equal{} 3$

Edit: more stuff, and no problem Nidhi

Thanks for such instant reply tony....

[quote="tonypr"]Generally:

$ [x]$ means the greatest integer smaller than or equal to $ x$.

$ \{x\}$ means the not integer part.

So $ x \equal{} [x] \plus{} \{x\}$
Examples:

$ [3.4] \equal{} 3$

$ \{3.4\} \equal{} .4$

$ [\pi] \equal{} 3$[/quote]


function MATHCOUNTS probability AMC USA(J)MO USAMO

[quote="espark52"][hide="#1"]if $2^{x}*9^{y}$ is equal to the four- digit number $2x9y$ where $x$ is the hundreds digit and y is the units digit, what is the product of $x$ and $y$?[/hide]
[hide="#2"]It's possible to earn 0,1,3,7 or 10 points with each shot in the game of Blippy. How many positive scores less than 30 cannot be made in 3 shots?[/hide]
[hide="#3"]Joe bought a pumpkin that cost 10 cents more per pound than his sister's. together, the two pumpkins weighted 20 poundes, but Joe's pumpkin was heavier. Joe paid 3.52, and his sis paid 0.48. How many pounds did Joe's pumpkin weigh?[/hide][/quote]
[hide="1"] y=1 doesn't work, but y=2 gives 2592, so our answer is 10. 10 I tell you.[/hide]
[hide="2"]Quickly going through the numbers we see that 19, 22, 24, 25, 26, 28, and 29. [/hide]
[hide="3"]Factoring 3.52 and .48 we get 22*16 and 12*4. So Joes pumpkin was 16 pounds. THis can probably be solved with systems too but I was too lazy to set them up.[/hide]

[hide]#1... y has to be even. However, 4, 6, and 8 are too big and y can't be zero because 2^11 = 2048. Y is 2. Then 2X92 is divisible by 9, so X = 5. (2+5+9+2 = 18) 2*5 = [b]10.[/b][/hide]

[quote="humblemastermind"]For your number 2, you forgot 15, so the final answer would be 8.[/quote]15=7+7+1 :wink:

For some reason when I plug make equations for #3 it doesn't work...

[hide="#3 with equations"]To find the number of pounds, you can divide the cost by the cost each pound. Let x be the cost per pound for sister.

$\frac{3.52}{x+.1}+\frac{.48}{x}=20$
Solve...and $x=.12$ And $\frac{3.52}{.22}=16$[/hide]

Hmm... that's a good solution. I had a system in two variables, where x was Joe's pumpkins weight in pounds and y was Joe's pumpkins cost.

@davidli: Good Solution, I always did Guess and Check with that, noticing that 352=16*22.

[quote="jhredsox"]that is correct. i already knew the answer so i didn't post my solution before hand as a common courtesy so other people could try. (yes, i did check the answers just to make sure :P )

-jorian[/quote]

Well, good for you.

The answer is 6, for #2, not whatever the people had been posting, the possibilities.

[quote="Altheman"]here is a tricky way to do #2...

consider the degree of x, i will define this as the sum of the scores in blipo (w/e)

look at (1+x+x^3+x^7+x^10)^3

now consider the coefficient of the x^2 term, when we multiply it out, we do 1*x*x, x*1*x, x*x*1, x^3*1*1, 1*x^3*1, x^3*1*1

expanding that mess, there are no coefficients for x^19, x^22, x^25, x^26, x^28, x^29, hence they cannot be made from scores in the game, and that means that the answer is 6[/quote]

Yeah, I like that.  It's always nice to know more advanced methods (like generating functions) for mathcounts, but they aren't tested much, there is always another method.   :D

Hmm.  I don't really see how to figure out the coefficient of any given term.

Is there a way to do this?

I think Altheman is intending it to be expanded out.

I think he's just trying to show us there's a mathematical basis beyond the simple "guess and check" problem.

Personally, I was in awe of how that worked.

[quote="mathgeniuse^ln(x)"]Who would do that though during a competition.[/quote]

Well of course I wouldn't, but if you really were a hater of guessing and checking and you wanted to be absolutely sure, here's a method.
But like kyyuanmathcount and Fanatic said, it's good to know something beyond mathcounts that's possible to use to your advantage on mathcounts.

would this be something like mods?

-jorian

Generating Functions isn't really relating to mods.

But mods are good for MathCounts.

[quote="jhredsox"]i meant for the fact that they are above mcounts level, but can help w/ mcounts.

-jorian[/quote]
mods are DEFINITELY not beyond mathcounts. If you don't know them good luck with about a fifth of the NT problems on mathcounts.

AKA your brother correct?  :wink:

Yeah, I think you can't have a list of exactly what you need and what you don't need.  it's whatever you want, the more the merrier, but you can survive without lots of things.

[quote="Fanatic"]I just asked a former nats champ if he knew mods during his mathcounts years.

Response: "kinda but not really".

so  :wink:  to your post.[/quote]
You may not know that they are called mods, but he probably understood the basic concepts that were underlying basic modular math. Then again, maybe I have no clue what I'm talking about (quite possible).

Considering he's now 5 years older, I'm pretty sure he knows what mods are and what information could be deemed as modular and what couldn't.

Probably correct on the basic concepts though, hence the "kinda" in "kinda but not really".

[quote="jhredsox"]wait, fanatic's brother was a national champ?

-jorian[/quote]
Yes, and a USAMO winner 3 years into High School...

did he used to live in illinois? (i won't go any farther, because i don't wanna sound suspicious)

-jorian

[quote="kyyuanmathcount"]Might I suggest AoPS's Intermediate Counting and Probability class.  I think they focus two lessons on methods using generating functions.[/quote]
But isn't it only offered in the summer?


geometry
circumcircle
induction
ratio
geometric transformation
reflection
power of a point

In the plane are given a circle with center $ O$ and radius $ r$ and a point $ A$ outside the circle. For any point $ M$ on the circle, let $ N$ be the diametrically opposite point. Find the locus of the circumcenter of triangle $ AMN$ when $ M$ describes the circle.

[quote="April"][color=darkred]In the plane are given a circle $w=C(O,r)$ and a fixed point $A$ outside of this circle. For a mobile point $M\in w$ let $[MN]$ be a diameter of the circle $w$. Denote the circumcircle $w_{m}=C(L)$ of the triangle $AMN$. Ascertain the geometrical locus of the point $L$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the power $p_{w}(X)$ of the point $X$ w.r.t. the circle $w$. Observe that $O\in MN$- the radical axis of the circles $w_{m}$ and $w$ $\implies$ $p_{w_{m}}(O)=p_{w}(O)$ $\implies$
$LO^{2}-LA^{2}=-r^{2}$ $\implies$ $LA^{2}-LO^{2}=r^{2}$ (constant) In conclusion, the geometrical locus of the point $L$ is [b]a perpendicular line on the fixed line $OA$.[/b][/color]

Nice!

This line will intersect $OA$ at a point $K$. Let's find $K$

The circle $w_{m}$ intersects the line $OA$ at $A$ and $T$, where $T$ is a fixed point because $OA\cdot OT = r^{2}$
$O$ is between $A,T$.

$LA=LT\Rightarrow L$ is on the perpendicular bisector of $AT$

Namely, if $A(a,0)$ then $T\left(-\frac{r^{2}}{a},0\right)$

Then $x_{K}= \frac{a^{2}-r^{2}}{2a}$

[u]Note[/u]
We have the same solution if we choose the point $A$ in the interior of $w$

this is a very beautiful problem...locus questions are quite difficult to come up with, in my opinion, because there are so few options for the result

[b]Approach by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=34189]tdl:[/url][/b]

Call $ I$ is circumcenter of $ (AMN)$, we have:
$ AI^2 \minus{} OI^2 \equal{} MI^2 \minus{} OI^2 \equal{} OM^2 \equal{} R^2$
Then locus of $ I$ is a line perpendicular with $ AO$!

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=26]grobber:[/url][/b]

Let $ P$ be that circumcenter. We have $ R^2 \equal{} PM^2 \minus{} PO^2 \equal{} PA^2 \minus{} PO^2$, so $ PA^2 \minus{} PO^2$ is constant, meaning that the locus is a line perpendicular to $ AO$

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=651]Pascual2005:[/url][/b]

Let A' be the inverse of A over an induction with center O and ratio R, and let A'' be the image of A' under the reflection across the center O. Then: A,O,A'' are collinear and M,O,N are collinear, but OA*OA''=OA*OA'=R 2 and also OM*ON=R 2  wich implies that AMA''N are conciclic and then the circuncircle Is in the perpendicular bisector of AA''. (R=r)

Inversion with center $ O$ and negative power $ \minus{} r^2$ (ordinary inversion with positive power $ r^2$ followed by reflection in the inversion center) takes $ A$ into $ A' \in AO,$ such that $ \overline{OA'} \cdot \overline{OA} \equal{} \minus{} r^2.$ The cirumcircle $ (P)$ of $ \triangle AMN$ cuts the inversion circle $ (O)$ at diametrically opposite points $ \Longrightarrow$ this inversion takes the circle $ (P)$ into itself. Since $ A, A' \in (P),$ locus of circumcenter $ P$  is the perpendicular bisector of $ AA'.$


group theory
superior algebra
superior algebra unsolved

Find all homomorphisms from $ \mathbb{Z}/4\mathbb{Z}$ to $ \mathbb{Z}/8\mathbb{Z}$.
In my book it states $ \mathbb{Z}_n$ is a group under addition. But in class we've used this as a group under both addition and multiplication.
Which should I assume it is under? Or just state which operation I am assuming both to be under (no reason to believe one is mulitiplication and one is addition)

Without qualification, you're talking about the additive group.  The multiplicative group is usually denoted $ (\mathbb{Z}/n\mathbb{Z})^{*}$.

Thanks =)
I did it assuming it was under addition and arrived at a total of 4 homomorphisms. This is their $ \gcd$. Is that a coincidence?

The answer should be $ 8$.  A homomorphism from a cyclic group to any other group is determined by its value at a generator.

Umm what do you mean specifically as "its value at a generator" ?
Also I found in my text just now, pg 207 (chapter 10 - Group Homomorphisms exercises) in Contemporary Abstract Algebra (Gallian's)
that actually the number of homomorphisms from $ \mathbb{Z}_n$ to $ \mathbb{Z}_k$ is $ \sum \phi{(d)}$ and $ d$ is a common divisor of $ n$ and $ k$. And from NT we know it is $ \gcd(n,k)$. :S

Whoops.  :oops:  Let me give a full solution.

A homomorphism $ \phi : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/k\mathbb{Z}$ (not to be confused with the totient) is determined by the value of $ \phi(1)$ (since $ \phi(a) \equiv a \phi(1) \bmod k$) subject to the condition that $ \phi(n) \equal{} 0$; that is, $ \phi(1)$ is an element of order dividing $ n$ in $ \mathbb{Z}/k\mathbb{Z}$.  Since an element in $ \mathbb{Z}/k\mathbb{Z}$ has order dividing $ k$, it follows that $ \phi(1)$ has order dividing $ \gcd(n, k)$.  A cyclic group of order $ k$ has exactly $ d$ elements of order $ d, d | k$, and the result follows.


Of the following sets, the one that includes all values of $ x$ which will satisfy $ 2x \minus{} 3 > 7 \minus{} x$ is:

$ \textbf{(A)}\ x > 4 \qquad \textbf{(B)}\ x < \frac {10}{3} \qquad \textbf{(C)}\ x \equal{} \frac {10}{3} \qquad \textbf{(D)}\ x > \frac {10}{3} \qquad \textbf{(E)}\ x < 0$

[hide]
Solve for x.
$ 2x\minus{}3>7\minus{}x$
$ 3x > 10 \Longrightarrow x > \dfrac{10}{3}$
$ \boxed{D}$ $ x > \frac{10}{3}$
[/hide]


vector
superior algebra
superior algebra solved

We know that a transformation from V to W is linear if the following hold:
1.) For every x, y in V, T(x+y) = T(x) + T(y)
2.) For every x in V and for every a in R (real numbers), T(ax) = aT(x)

I need two nonlinear transformations from R to R.  One must satisfy #1 above and violate #2.  The other must violate #1 and satisfy #2.

For the first: consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. It has a basis. Permute that basis.

For the second: You need more dimensions. In one dimension, you get $T(a)=aT(1)$.


Find the simplest form for $ R=$$ \sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}$

[quote="Ponnamperuma"]$ R=\sqrt{1+3i}+\sqrt{1-3i}$
$ R^{2}=(1+3i)+(1-3i)+2\sqrt{(1+3i)(1-3i)}$
$ R^{2}=2+2\sqrt{1-9i^{2}}$
$ R^{2}=2+2\sqrt{10}$
$ R=\sqrt{2(1+\sqrt{10})}$.

Is this what you mean? I've got some doubts...[/quote]
[hide]You made a mistake in your first step. It should be,
\begin{eqnarray*}R & = & \sqrt{1+\sqrt{3}i}+\sqrt{1-\sqrt{3}i}\\ R^{2}& = & 1+\sqrt{3}i+4+1-\sqrt{3}i\\ R & = & \sqrt{6}\end{eqnarray*}[/hide]


1.)  Im using Moodle but  Im wondering why the could not be shown.

Here's the link:
http://upnismed.moodle4free.com/blog/index.php?userid=2

2.) The Latex shown there is big and not smooth. Is there anyway that I could change like the one on this forum?

Unfortunately the link gives an error but I can probably guess what is happening. The default for Moodle is to assume there is no LaTeX system available and use [url=http://www.forkosh.com/mimetex.html]mimeTeX[/url]instead. This is a LaTeX emulator and doesn't give such high quality images.

The solution is to install LaTeX, ImageMagick and Ghostscript and then use the LatexRender syste. Or failing that, replace mimetex by [url=http://www.forkosh.dreamhost.com/source_mathtex.html]mathTeX[/url].

This is what the Moodle instructions say:
[quote]TeX notation is primarily used to generate inline mathematical formulae within Moodle resources. For example $ \$$ $ \$$\sqrt{a+b}$ \$$ $ \$$ (the $ \$$ $ \$$  is the 'tag' for the start and end of this format).
Moodle uses external binaries to render the images which may need setting up for your system. The filter will first check for a LaTeX renderer (Ghostscript is also required for this method) and failing that will use the simpler mimetex binary included in the Moodle distribution. LaTex permits the whole range of syntax, whereas mimetex only supports a mathematical subset.

Ghostscript is available for Windows and there is a LaTeX implementation called MikTeX available also.

The settings page allows the LaTeX and Ghostscript binary paths to be configured, along with the LaTeX preamble.[/quote]


search
function
linear algebra
matrix
linear algebra unsolved

I'm almost sure that I've seen this problem on these forums before, but most threads are named so atrociously that I've been unable to find it.  So feel free to link to the original thread, but no snarky responses like "search function" please.

The problem is to show that for any real square matrix $ A$, that $ A A^t$ has the same image as $ A$.  I suppose that this is generalizable to complex matrices with $ A^*$ instead of $ A^t$.

This seems to be immediate from the [url=http://en.wikipedia.org/wiki/Singular_value_decomposition]singular value decomposition[/url] of $ A$, but I am somewhat unsatisfied with this proof, half because I have some homework to do to understand the proof of singular value decomposition, and half because I seem to remember seeing an elementary proof of this at some point.

Any ideas?

Well, it would help if you used the proper words- you're talking about rank, not range. The "range" is the column space, and the column space of $ A^T$ is not the same as that of $ A$. The rank is the dimension of that space, which is the same.

In generalizing to $ \mathbb{C}$, that transpose must become a Hermitian adjoint. This absolutely requires a positive definite inner product.

If $ A^TAx\equal{}0$, $ x^TA^TAx\equal{}\|Ax\|^2\equal{}0$, and the null space of $ A^TA$ is the same as the null space of $ A$. We don't need $ A$ to be square- any real matrix will do.

I meant image, actually, and yes, I used about five different similar words when searching.  I've made the change.

Thanks for the simple proof. (though it took me a minute to figure out why it implies the result)

Oops, I was wrong here- $ A^TA$ has the same null space as $ A$, while $ AA^T$ has the same image.

But this is really a problem about rank, so your proof is basically correct, as $ rank(A)\equal{}rank(A^t)$.


AMC
USA(J)MO
USAMO
induction
trigonometry
analytic geometry
search

How was USAMO for the MA people?
Personally, I thought it was a little easier than previous years, at least #1 and #4, which were the only ones I solved.

For #1, I solved it completely but my solution was kind of messy.
I showed that each term of the sequence $ a_n\equal{}2^{2^{n\plus{}1}}\plus{}2^{2^{n}}\plus{}1$ could be factored as n+1 different pairwise coprime factors.
So maybe 5 or 6?

For #2, I was able to find the answer and prove it worked (not very rigorously though).
However, I couldn't find a way to prove that the rest of the regular polygons couldn't be triangulated into isosceles triangles.
I'm guessing I only got a 1 or 2 on that.  :(

I think it was about the same as last year in that there weren't any free points like last year on #4 and #5, but the problems were slightly easier.

I got 1 and 4 completely, and hopefully what I did on #2 merits a point, putting me in good shape for MOP, or so it looks. Our freshmen seem to have done pretty well, David/Me/Elizabeth/SamL all appear to have fallen in the 13-18 range. David was the only one of us that solved #5, but he said he had 'minor errors' in his proofs for 4 and 5.

Good job!
Can anyone tell me how to prove that all the other regular polygons can't be triangulated?

Oh dang I just realized I have a minor hole in my solution to #4: the original one I wrote up proved that the isosceles triangle with the longest legs contains the center, because if it's obtuse, you get a contradiction, but I never said anything about it being right. All parts of my proof still work, but I never explicitly stated the isosceles right triangle case.

Anyway it turns out that proving what I proved is completely useless. Anyway consider the isosceles triangle that contains the center (either within or on). It's not too hard to show that given a minor (or semicircular) arc, with the endpoints connected there have to be 2^k-1 points of the polygon on it. This follows from the fact that the chord has to be the longest segment on there, so it has to be the base, and you do an induction-type thing. My written up proof of that wasn't so great, so maybe I'll get docked another point for that.

Then you be like two of those arcs are the same, so add up all the points.

1,4,5. I had the same errors as David: I didn't take care of the case when one of the $ a_i$ is $ 0$, which is pretty easy but eh.

I solved #1, 4 completely. I made a lot of progress on #2(I got close to official solution #1 except I didn't think of the law of sines argument during the test), I made a litte bit of progess on #5.

#4 took me a RIDICULOUS amount of time to write up

Expecting somewhere in 15-17 range

[quote="CatalystOfNostalgia"]David/Me/Elizabeth/[b]SamL[/b][/quote]
Um since when. I failed utterly. I got 1 totally. I may have scraped out a point on 2. On 4 I left out a whole case, and may have scraped out a point on 5. So how is that 13-18?

I got 1 and 4 pretty well, but my proof for 1 may be a bit flawed and/or convoluted.

I just wrote a bunch of trivial stuff on 2 and 5. My solution to 3 was a failed version of induction.

And for 6, I just wrote stuff about how there are a power of two ways to put a person's friends (or remaining unplaced friends) in the rooms. I don't think I'll get full credit, because I missed a bunch of cases (Example: If A has friends B, C, and D, B has friends A, C and E, from E we place B in 2 and from D we place A in 1, then where does C go if C has no other friends?). In retrospect it probably would have taken only like 20 minutes to prove that but I gave up and was in a state of "whaaa???" for the last few minutes

Throughout the thing I did not use pictures at all, and the lengths of my proofs never exceeded 2 pages.

I'll probably get like 5/1/0/7/1/1 for 15?

[quote="Phelpedo"]You might be too modest, but I'll just point out that even 1's are not that easy to get. "Trivial stuff" on #2 and #5 will probably not even earn 1 point. Your #6 looks pretty shoddy because it doesn't appear to guarantee that the friends have an even number of friends in the room. Finally, note that 5's are rarely ever given. If your #1 has a minor flaw, it will get a 6, no matter how convoluted. If it has a major flaw, you'll probably get no more than 2.[/quote]

 :) lol...This is my first USAMO, so I guess I'm just happy to be here. Hopefully I can work at it so my next two years will be a bit better.

#1 I got it, but with a huge flaw though (I feel really dumb about the mistake)

#2 I made some progress working with coordinates. I got the coordinates of the 4 points in question. I don't know what it merits.

#3 I drew a diagram and said that the answer follows trivially.

#4 I got this one with two lemmas.

Lemma 1: If an n-gon can be isoceles triangulated, then a $ n*2^a$-gon can also be isoceles triangulated.

Lemma 2: Call a 'slash' a diagonal of longest possible length. If an (2n+1)-gon can be isoceles triangulated, then it has exactly 2 slashes.

Lemma 1 reduces the problem to a search for odd numbers that work. From lemma 2, it is easy to show that these odd numbers must be in the form $ (2^b \plus{} 1)$.

That left $ 2^a(2^b \plus{} 1)$ with the exception of (a,b)=(0,0)

#5 I got it under the assumption that there existed a constant such that $ cr_1, cr_2, cr_3 \in \mathbb{Z}$. Whether or not this constant exists... I'm not sure. Probably won't get any points.

#6 Due to laws forbidding dismemberment, no mathematician can be split, therefore there is only $ 1$ way to split them. Since 1 is indeed a power of 2  (k=0), the proof is complete.

[quote="chickendude"]
#6 Due to laws forbidding dismemberment, no mathematician can be split, therefore there is only $ 1$ way to split them. Since 1 is indeed a power of 2  (k=0), the proof is complete.[/quote]

Haha, I did a something similar.
I said the host doesn't care about what the mathematicians insist by Euler's Theorem that all humans are lazy.
Therefore he just splits them randomly with $ 2^n$ ways.

@chickendude: For #4, I think you needed to prove the converse of your Lemma 1 to make the claim that you only need to look at odd $ n$, which is why I didn't use that Lemma.

For #5, try $ 1,\sqrt{2},3\minus{}2\sqrt{2}$, and $ (3,\minus{}2,\minus{}1)$. Your lemma isn't true.


Twelve students are to be divided among Mr. Mirus's and Ms.Batty's classes. No teacher is to have more than 8 students. How many different groups of students could be in Mr. Mirus's class?

[quote="vishalarul"]Simple.
[hide="Here is the solution"]
12C1 = 12 (# of ways if 1 child is in Mr.M's class)
12C2 = 66 (# of ways if 2 children are in Mr.M's class)
12C3 = 220(# of ways if 3 children are in Mr.M's class)
12C4 = 495(# of ways if 4 children are in Mr.M's class)
12C5 = 792(# of ways if 5 children are in Mr.M's class)
12C6 = 924(# of ways if 6 children are in Mr.M's class)
12C7 = 792(# of ways if 7 children are in Mr.M's class)
12C8 = 495(# of ways if 8 children are in Mr.M's class)

Adding all the numbers gives us 3796 ways.
[/hide][/quote]
[hide]
but you have to subtract the ways of 1 child, 2 children, and 3 children, since each of these would put more than 8 kids in the other class.
So the answer would be 3498[/hide]

[quote="daermon"]I thought of klebians method but im too lazy to add it all up :jump:  :roll:  :P[/quote]

...
That also is spam.
Sorry, but if you're not going to contribute anything, then why even post? To increase your post count?

None of the answers so far are correct. I will wait for some more time before posting the answer.

[quote="espark52"]you could count $12C1+12C2+12C3+12C4=12+66+220+495=793$[/quote]
Remember that no teacher can have more than 8 students. For example, if you allocate 12C1 for one teacher, then it will effectively assign the remaining 11 students to the other teacher. So this is not allowed.

Made a subtraction error in my problem edited it now, please check it again.
[hide="The error was.."]Subtracted 40 too much.[/hide]

[quote="life=tennis+math"][hide]$\binom{12}{4}$+$\binom{12}{5}$+$\binom{12}{6}$+$\binom{12}{7}$+$\binom{12}{8}$ which equals $2(495)+2(792)+924$. Simplify to $990+1584+924$=$3498$.
ANSWER: $3498$.
EDIT: Hey weird this is what Klebian and Ignite got too, are you sure it's wrong?[/hide][/quote]
It is wrong. Your answer is only one step away from the correct answer. There is a subtle issue you all are not considering. I will post the solution at the end of the day.

[quote="SplashD"][quote="mthiyaga"][quote="life=tennis+math"]I think I might know...
[hide]$\binom{12}{4}$+$\binom{12}{5}$+$\binom{12}{6}$+$\binom{12}{7}$+$\binom{12}{8}$ which equals $2(495)+2(792)+924$. However, the group of 6 in Ms. Batty's class could be in Mr. Mitus' class also, so we double $924$ to get $1848$. Simplify to $990+1584+1848$=$4422$.
ANSWER: $4422$.
[/hide][/quote]
You are correct! Yes, we need to double 924 as Ms. Batty's selection can be switched to Mr. Mitus also.[/quote]

I don't understand why you have to double because if you select 6 kids for one teacher's class, the other class is now fixed.[/quote]
Since we are dealing with groups of individuals, each group is distinct. For example, (John, Frank) and (Suzan, Brat) are two distinct groups. So If Ms. Batty can select a group of 6 students in $\binom{12}{6}$ ways, all these groups can also be assigned to Mr. Mitus.

Well say there are 2 teachers and 2 students to put into their classes. Obviously, there are 4 ways to do this.

If you do $\binom{2}{0}+2(\binom{2}{1})+\binom{2}{2}$, you'll get 6.


Let's say I want to submit some math assignment that I have done in LaTeX and I want to have my name at the top of every page, along with the page number in the lower right corner or middle, and maybe some horizontal lines that divide the header and footer from the body of the page... I have yet to find a tutorial that thoroughly addresses these issues, so I need help...

Does anyone know how?

Also, I do not have the fancyhdr package because I cannot figure out how to install it, the instructions I have found have so far made no sense. Help on that issure would be greatly appreciated as well.

I should have a tutorial on this up by the end of the day.

Title: \title{Whatever you want}
Name: \author{Your name}
\maketitle
the last piece is to compile the header.
You could also include \date{\today} to add today's date to the paper. (Note: this changes everytime you recompile the paper.)

is there a way to center the header? the entire thing is off to the left on my document, and it's also short... are there ways to extend the header father across the page and also center it?

Not sure what you mean.  \chead isn't doing what you want?

The \begin{center} \end{center} environment can be used in the header, as can \parbox (we'll have stuff up about \parbox in the next few days), and \includegraphics for images.

[quote="Spoon"]here is what a I mean: http://www.tobinsight.com/math/format.pdf

Notice how the header is to the left, it's also low and the footer is almost off the page... I've tried to correct these but haven't had any luck.[/quote]
Try changing the size of the region LaTeX displays text in.

[quote="Spoon"]so you're saying that the only way for me to get it centered is for me to reduce my textwidth? I was hoping there was a way to expand my header width... but if I must reduce the textwidth, then so be it... do you know what the maximum length would be?[/quote]
I was guessing.

Set

\textwidth  <your width>

When I use this, the header and the text match that.

How are you doing your multiple choice?  It looks like you're doing that in some way that extends out into the margins.

actually i am using your exact \prob command, I really like it so I decided to adapt it... I didn't change the coding in any way at all, it's the exact form you have it on the site... so I dunno.

\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}
\usepackage{fancyhdr}

\pagestyle{fancy}
\headheight 35pt

\rhead{Right top}
\chead{Middle top}
\lhead{Left top}
\rfoot{Right bottom}
\cfoot{\thepage}
\lfoot{Left bottom}

\pdfpagewidth 8.5in
\pdfpageheight 11in
\headsep .25in
\textheight 8.5in
\textwidth 6in
\oddsidemargin 0in
\evensidemargin 0in

that's the header I am using currently... I am not sure if something is wrong...

Try jacking up that textwidth.  Make it 7in, maybe even 7.5.   How does that look?

I tried 7.5 and it just expanded the body... off the page a bit, but the header didn't change a bit... how strange.

Also, the header isn't close enough to the top, is there a command that moves it up?

Ah - put the textwidth before the header junk.

Stick the textwidth before the \pagestyle statement. That'll do it.


[quote="luimichael"]Prove that for any natural number $n$,

$\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + ...+\lfloor \sqrt{n^2} \rfloor = \frac 1 6 n(4n^2-3n+5)$.[/quote]

[hide="Hint"]
$\lfloor {\sqrt{n^2}} \rfloor = \lfloor {\sqrt{n^2+1}} \rfloor = ... = \lfloor {\sqrt{(n+1)^2-1}} \rfloor = n$, $n \in \mathbb N$
[/hide]

[quote="luimichael"]Prove that for any natural number $n$,

$\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + ...+\lfloor \sqrt{n^2} \rfloor = \frac 1 6 n(4n^2-3n+5)$.[/quote]

[hide]
Let $f(x) = \lfloor x \rfloor$, then $f(a^2) = f(a^2 + 1) = \ldots = f(a^2 + 2n)$. So the summation is equivalent to:
\[ \sum\limits_{i=1}^{n}i(2i+1) -2n^2= 2\sum\limits_{i=1}^{n}i^2 + n(n+1)/2 -2n^2  \]
this simplifies into \[ \frac{2(n)(n+1)(2n+1)}{6} + n(n+1)/2 - 2n^2 = \frac{4n^3 -3n^2 + 5n}{6}  \][/hide]

[quote="M4RI0"]
$\lfloor {\sqrt{n^2}} \rfloor = \lfloor {\sqrt{n^2+1}} \rfloor = ... = \lfloor {\sqrt{(n+1)^2-1}} \rfloor = n$, $n \in \mathbb N$[/quote]How do you prove this?  :?


Let $ ABC$ and $ A_{1}B_{1}C_{1}$ be two triangles having the same circumcircle $ \mathcal{O}(O, R)$ and the same incircle $ \mathcal{I}(I,r)$. If the sidelines $ BC$ and $ B_{1}C_{1}$ are parallel, prove that $ A_{1}$ is the tangency point of the $ A$-mixtilinear incircle of triangle $ ABC$ with $ \mathcal{O}$.

Thank pohoatza for your nice problem :)
PROOF：
Let A1C2,A1B2 are the tangent to the incircle. C2,B2 are the intersections
IA1*IG=R^2-OI^2=2Rr   but we have △ILA1 ∽ △GC2H  => C2G/r=C2H/IA1
IA1*C2G=2Rr  =>  C2G=B2G=IG   => B2C2 is a tangent to the incircle  and
OG ⊥ C2B2  but we have OG ⊥ BC  => BC ∥C2B2  => C2=C1,B2=B1
DONE!

My solution:

Let $ D $ be the tangent point of $ (I) $ with $ BC $ and $ A_1D \cap (ABC) =D' $ .

From the [b]Dual theorem of Desargues' Involution[/b]
we get $ (A_1B, A_1C), (A_1A, A_1D'), (A_1B_1, A_1C_1) $ are in the same involution ,
so $ BC, AD', B_1C_1 $ are concurrent at the pole of this involution  .
Combine with the condition we get $ AD'//BC//C_1C_1 $ ,
so $ AA_1 $ is the isogonal line of $ A $ -Nagel line .
ie. $ A_1 $ is the tangent point of $ A $ -Mixtilinear circle with $ (ABC) $

Q.E.D


quadratics
analytic geometry
graphing lines
slope
conics
parabola
algebra

Find the locus of the points $ M$ in the plane $ Oxy$ such that the tangents from $ M$ to $ y\equal{}x^2$ are perpendicular.

 :)

[quote="ElChapin"]Find the locus of the points $ M$ in the plane $ Oxy$ such that the tangents from $ M$ to $ y \equal{} x^2$ are perpendicular.

 :)[/quote]

Let $ y\equal{}mx\plus{}c$ is a tangent to the curve $ y\equal{}x^2$.  So, the equation $ mx\plus{}c\equal{}x^2$ must have equal roots.  This gives us $ m^2\plus{}4c\equal{}0$.  Hence we have $ c\equal{}\minus{}\frac {m^2}{4}$.  So any tangent to the given curve can be written as $ y\equal{}mx\minus{}\frac {m^2}{4}$
Let this tangent passes through $ M(h, k)$ a point in the plane.
Then we have $ k\equal{}mh\minus{}\frac {m^2}{4}$
This is a quadratic equation in m and will have two roots for m.  Let the roots are $ m_1, m_2$.  If the tangents through M are perpendicular, the product of the slopes must be equal to $ \minus{}1$. So we should have $ k\equal{}\minus{}\frac {1}{4}$.  Hence the locus of M is the line $ y\equal{}\minus{}\frac {1}{4}$ which is the directrix of the parabola given.


Shared with Javagal Srinath.

Guess it. [hide=" :D "]31st August.[/hide]

Mine: 5 months after Christmas.

(I don't know anybody else that has that birthday.)

[quote="bubka"]Let us bulid a birthday roster. Mine is 21 July. I share it with Ernest Hemmingway. Give your own birthdays and famous people's birthdays if you know them. Whoever likes it may wish you here or through a PM. I will compile the list once in a while.[/quote]

I also have July 21,

just like C. C. Sabathia,

(the ace of the Cleveland Indians roster)

The first November doomsday. (By doomsday I mean John Conway's, not Hitchhiker's guide's, even though those are not necessarily mutually exclusive sets)

[quote="Iversonfan2005"]December 25<---- :( , only get one set of presents.... :mad:[/quote]

your the first person i met born on that day. It's my half birthday.

October 20.  :lol:  1st person with an october birthday!

how come everyone you meet seems to have a birthday in the summer? shouldn't they be evenly spread?

March 24th, hehehe.   :D And I know a person with my half birthday, very cool.

Nice response but be more explicit.
I have not got Vihang's or randomdragoon's. Elucidate. I have got the rest though.

Can't you see the smiley below "Guess it".(in my first post)

You should click such things ... hehe.
[hide=" :rotfl: "]PWNED[/hide]

16 oct!!!!
tis' the day this 'would be' fuel research scientist was born!!
 :thumbup:

My birthday is on:  Pick between these two smilies, [hide=" :rotfl: "]May 7, 1992[/hide][hide=" :ninja: "]Guess again :P [/hide]

May 16!!!  :roll:
 three famous people born on this day:
 I, ME and MYSELF  :lol:    :lol:  :lol:

Too many smilies in your post: 4.
You can have at most 2 smilies in this message. Please remove some smilies before submitting this message.

Okay I got Vihang's.
But randomdragoon's has no such smilie - none that I can see at least.
And the next person please count yourself once only in the list of famous persons.

January 24th Nothing interesting happened on that day and no one famous was born then either.
It apparently is the [url=http://www.msnbc.msn.com/id/6847012/]saddest[/url] day of the year though.

MY 13TH BIRTHDAY IS TOMORROW
MY 13TH BIRTHDAY IS TOMORROW
MY 13TH BIRTHDAY IS TOMORROW
MY 13TH BIRTHDAY IS TOMORROW
MY 13TH BIRTHDAY IS TOMORROW
I do not know anyone who shares my birthday. (actually I do know one...3 hours older than her!!)
Otherwise, noone famous.

August 12th.
No one interesting was BORN on that day.
But Cleopatra died on it.
AND I JUST FOUND THAT OUT DX
 :ewpu:

my birthday is November 29, the 333th day in a non-leap year and the birth date of C.S.Lewis (Chronicles of Narnia), William Tubman, the retreat of UN troops from North Korea and of Soviet troops from Afghanistan.
The other deaths and births and events I didn't find interesting

Feb. 14

a lot of people
apparently that i don't know...

Ok, so I'm on Feb. 3rd

I share a birthday with Elizabeth Blackwell (first female doctor).

Here's a whole list of famous people who were born on my  birthday.

http://www.historyorb.com/birthdays/february/3

(there's 257 of them :P )

18th February

[hide="People:"][url=http://en.wikipedia.org/wiki/Alessandro_Volta]1745 - Alessandro Giuseppe Antonio Anastasio Volta, inventor (battery)[/url]

[url=http://en.wikipedia.org/wiki/Sri_Ramakrishna]1836 – Sri Ramakrishna, Indian Bengali guru of Swami Vivekananda[/url]

[url=http://en.wikipedia.org/wiki/Ernst_Mach]1838 - Ernst Mach, Chirlitz-Turas Austria, physicist/philosopher/psychologist[/url]

[url=http://www.ferrari.com/English/about_ferrari/History/Pages/Enzo_Ferrari.aspx]1898 - Enzo Ferrari, racing car manufacturer, Modena[/url]

[url=http://en.wikipedia.org/wiki/John_Travolta]1954 - John Travolta, Englewood NJ, actor[/url]

[url=http://en.wikipedia.org/wiki/Roberto_Baggio]1967 - Roberto Baggio, Italian footballer[/url]

[url=http://en.wikipedia.org/wiki/Claude_Mak%C3%A9l%C3%A9l%C3%A9]1973 - Claude Makélélé, French footballer[/url]

[url=http://en.wikipedia.org/wiki/Gary_Neville]1975 - Gary Neville, English footballer[/url]

[url=http://en.wikipedia.org/wiki/Jermaine_Jenas]1983 - Jermaine Jenas, English footballer[/url]

[url=http://en.wikipedia.org/wiki/Anton_Ferdinand]1985 - Anton Ferdinand, English footballer[/url]
 yeah, Rio Ferdinand's brother.

So, that makes five famous footballers  :) [/hide]

[hide="Events:"]3102 BC – Epoch (origin) of the [url=http://en.wikipedia.org/wiki/Kali_Yuga]Kali Yuga[/url]  :evilgrin:

1884 – Mark Twain's Adventures of Huckleberry Finn is published for the first time

1901 – Winston Churchill makes his maiden speech in the House of Commons

1911 – The first official flight with air mail takes place in Allahabad, British India, when Henri Pequet, a 23-year-old pilot, delivers 6,500 letters to Naini, about 10 km away

1929 – The first Academy Awards are announced

1930 – Elm Farm Ollie becomes the first cow to fly in a fixed-wing aircraft and also the first cow to be milked in an aircraft

1930 - US astronomer Clyde Tombaugh discovers Pluto

1979 – Snow fell in the Sahara Desert in southern Algeria for the only recorded time in history[/hide]

April 19th. I don't know who else was born on that day.

Since we so innocently revived this topic, I share my birthday [hide="with..."]AIME15USAMO!!! WOOHOO!!![/hide]

I share my birthday [hide="with..."] MAYBACH!!! WOOHOO!!! [/hide]

June 25

he message is too small. Please make the message longer before submit

17th June 1994:D
check out this link for more details
 :wink:
http://en.wikipedia.org/wiki/June_17

January 22, 1998.

I share this birthday with Bill Bixby


[quote="pkerichang"]I send in my solutions a while ago (2-3 days), yet I haven't seen a check mark nor receive a confirmation mail.  Should I send it again??[/quote]We have not yet filed any solutions.  Please be patient -- it will take a few days for us to go through our of our received papers.

I saw the check mark, thanks :)

It's just that the last 2 rounds I submit my solutions 2-3 days before the deadline and I got the check immediately, so this time I kind of paniced...

Out of curiosity, what is the standard policy for people who thought they had sent it but did not, after the deadline has passed?

Would they have to prove somehow that they did indeed try to send it (i.e. forward the log of the email from "sent messages")? Or could they just resend it, even if the deadline has passed?

Don't worry; we haven't filed the papers yet.

[quote="perfectnumber628"]yeah because i sent mine on the day it was due and i haven't seen a checkmark yet (i'm getting worried...)[/quote]

What is your USAMTS user id number?

my id number is 6839, and there's still no checkmark.  (is that bad?)

I also sent my solutions the day they were due, and still have not received the green checkmark. My ID number is 6227.

yessss!!! i have the checkmark now.  thank you.

I haven't gotten a check mark, either. I sent in my solutions on Thursday, March 9, and my USAMTS ID is 6219. It's been almost two weeks...


articles
geometry
3D geometry
sphere
function
calculus
probability

I am not an expert in those fields (or any other fields;-), but this is my opinion (influenced by what I've read in some popular-style articles) - there is practically no such thing as a diameter of an electron or a photon, because you cannot just take it and measure it. You would have to have a really small something to measure it with (a high-energy photon?). And also those particles are not really well defined spheres bouncing from each other and floating in space ... modern physics describes them as wave functions (?). You could probably measure something like a "characteristic dimension" or something like that ...

yes the radius probably can't be defined. I haven't read this anywhere but my intuition tells so.

Also, I am talking about any sort of "characteristic dimension" which might be theoretically possible to measure!!

So now what do you think??

hey kubus, my PM got lost so I am telling you here.

I am in class 12 and will be attending college next year.
Also I don't wish to hide from you but rather gain. My resources , as I told you earlier are Irodov's book and "200 Puzzling Physics Problems".The former has  more calculus and is simple but highly algebraic while the latter has zero algebra and rquires greta thought!!!!

Thanks for finally posting!!! I persuade you to continue to the other forums as well!!!!

In case particles you can't measure anything because the particle has a quality to be everywhere by some probability, which is described by function  $\psi $ which I actually don't know.

that probablity refers to the motion of the particle so you cant determine the positionj of the electron properly.

i'm talking about the characteristic dimensions. Is it theoretically possible. has somebody ever devised or thought of such an experiment or dispoved it can't be done. It's sure somebody must have done that!!!

I have read that the proton and neutron have defined radii (defined similarly to the "radii" of atomic orbitals) based on the charge distribution.  According to my understanding of the Standard Model, these charge distributions are the result of the wave functions of the orbiting quarks within the nucleon.  Recently, I have also read about attempts to define the radius of a proton based on the orbital radii of the quarks inside (again, based on their wave functions).

An electron is significantly different, in that it has no (known) internal structure, but is believed to be a fundamental unit of mass.  [url=http://www.physicsforums.com/archive/t-27712_radius_of_an_electron.html]Upper limits[/url] have been put on the radius of an electron, but it is difficult to define such a radius (note that this is the first kind of radius in the discussion of a proton, which may not exist at all if an electron is a fundamental unit).

It would be very difficult to even define a radius of a photon.

First of all, the link you showed mentioned some Casimir force?? Can you explain!!

Also, I think I have found an excellent link--- the experiment I said is being carried out at ( at least) MIT [url=http://ldt.mit.edu/proton/]ldt.mit.edu/proton/[/url]

I think I'll be able to find more links through google. Why didn't I think of that before!!! :fool: I'll post whenever I find something intersting!!!

Well, it's good to hear and know that i was on the right path atleast. Even if I don't know ANYTHING about QED or QCD, it seems that the dimensions of nucleons are measurable due to properties of their constituent quarks and their wave functions(as you have said, inteluser).But everone is saying that probably the dimensions of quarks are NOT measurable. Is this true? What does theory say??

I don't know too much about it, but the casimir effect is this attractive force between two metal plates when they're placed very close together.  In the space between the two plates, there will be virtual particles snapping in and out of existance; however, because of their proximity, there can be a fewer number of wavelengths for virtual particles.  So there's a smaller energy density in between the two plates than outside, and so the pressure from outside the plates pushes them together

[quote="Rushil"]This must surely look foolish!!!!

[b]Has somebody ever devised an experiment or though of one to measure the dimensions of elementary particles???[/b]

i.e. has somebody thought of an experiment to measure e.g. the radius of photon , electron if such a quantity can be defined???[/quote]

The Heisenberg Uncertainty Principle says that for any particle: $\Delta p_r \Delta r \ge \frac{\hbar}{2}$

If we let ${\Delta p_r = \frac{h}{\lambda}}-\frac{h}{\hat{\lambda}}$  then

$\Delta r \ge \frac{1}{4\pi} \frac{\lambda \hat{\lambda}}{\lambda -\hat{\lambda}}$

For a photon $\lambda =\hat{\lambda}$ by definition, and thus $\Delta r = \infty$ .

For any massive particle e.g. electron, proton, neutron there is some uncertainty in the DeBroglie wavelength. Nonetheless, the uncertainty in $r$ is driven by the uncertainty in the projection of momentum along the radial line. If electrons, for example, follow crisply defined elliptical orbits then $\Delta p_r = 0$ ; the consequence of which is that once again $\Delta r = \infty$ .

Conclusion: Heisenberg guarantees that the radii of subatomic particles cannot be measured.


trigonometry
limit
real analysis
real analysis unsolved

Let x_o be a point in (0,1] and define a sequence {x_n}_n>=0 recursively by setting x_n = sin(x_n-1), for all n>=1.

First, prove that {x_n}_n>=0 is a sequence in (0,1] such that lim_n->oo x_n = 0.

Now prove that lim_n->oo sqrt(n)*x_n = sqrt(3).

I think that this sequence, the second part of the problem, has been discussed before in the forum, see a problem proposed by Cezar with a solution given by Harazi, however for the first part you should use this technique:

$-1 \leq x_{n}=\sin x_{n-1}\leq 1$, hence the sequence is bounded.

On the other hand,

$x_{n}=\sin x_{n-1}\leq x_{n-1}$, hence the sequence decreases, therefore it is convergent. Now take the limit as n converges to $\infty$ in the recurrence relation and we get that $l=\sin l$, where $l=\lim x_{n}$, from which it follows that $l=0$.

hmm I could not find  the post by Cezar but here's a simple way to get the second: just note that $(1/a_{n+1}^{2}-1/a_{n}^{2})\to 1/3$ then use Stolz method and the first result to get the second.


calculus
integration
logarithms
calculus computations

These three definite integrals seem beautiful and interesting
But I can't solve it . Need your help
[size=150]1 . $\int_{0}^{1}x^{x}\,dx$
2 . $\int_{0}^{1}\frac{1}{x^{x}}\,dx$
3 . $\int_{0}^{1}\sqrt[x]{x}\,dx$[/size]

May I ask what exactly is beautiful and interesting about them?  :maybe:

As far as I know, you can't express these integrals into normal functions.
I tried to use an expansion:
\[x^{x}= \sum_{n=0}^\infty \frac{(x \log x)^{n}}{n!}\]

\[\int_{0}^{1}\frac{(x \log x)^{n}}{n!}= \frac{(-1)^{n}}{(n+1)^{n+1}}\]

\[\int_{0}^{1}x^{x}= \sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^{n+1}}\]
I guess this sum won't help you any further.

Edit:
I just computed it's numerical value and it only diviates by 0.2% form $\pi/4$

For the first two ones, check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147211]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147210]here[/url].

For the third one we could also write

$\int^{1}_{0}x^{1/x}\,dx = \int^{1}_{0}e^{\ln(x^{1/x})}\,dx = \int^{1}_{0}e^{\frac{\ln x}{x}}= \int^{1}_{0}\sum_{n=0}^{\infty}\frac{(\ln x/x)^{n}}{n!}\,dx = \sum_{n=0}^{\infty}\frac{1}{n!}\int^{1}_{0}\left(\frac{\ln x}{x}\right)^{n}\,dx$.

However, $\int^{1}_{0}\frac{\ln x}{x}\,dx$ diverges to infinity, and so we cannot use that expansion in that way.


Find the area of the shaded area of the unit square.

[hide]I got $ 1\plus{}\frac{\pi}{3}\minus{}\sqrt{3}$[/hide]

Can you tell us what the lines are made from?

Good. I got the same answer as you, maxschindler.

EDIT: What the lines are made from? What do you mean?

I think there are made from quarter circles, right? And.........

The other lines are just for help. I drew them in.