So is everyone going to the vhhs tourney next week. At least vhhs not competing after their domination last year made everyone think they cheated or it was rigged. Gives us non-vestavia people a much better chance now.

I am coming (from Georgia).
I was just working a copy of last year's written test and it seems very computation heavy. :(
Since the vestavia students aren't competing, are they helping to write it? That would probably make it better.

for the record, they definitely don't cheat. it helps to actually know how to do the problems. like i figure they just practice alot. which is the case. also, seva is a beast at hardcore computation.

Would it be possible to register for this tournament at the last minute (like now)? I can't find any info about it on the internet.

Yea too bad randolph got put in the large school division, being the smallest school there. I guess vestavia knew that we would dominate and get first in Division II. Cant believe Grissom got 1st in comprehensive, they havent done that in a while. Was the pre-algebra test simpler than the rest, cuz the winning score was a 95. For the other levels it was like < 70. O well I SHOULD HAVE AND COULD HAVE PLACED IN THE TOP 10 if it wasnt for stupid mistakes.... Now i dont have a chance anymore since vestavias gonna be competing.

Wow this got longer than i thought

Btw Alan Chou from Grissom is destined to be better than u elemennop. Mwahaha

[quote="lifesapain"]Yea too bad randolph got put in the large school division, being the smallest school there.[/quote]Actually, I believe solafidefarms' school was the smallest there since he's homeschooled (and he was still put in large schools) :lol:

[quote="Elemennop"]what does in addendum mean[/quote]There were some mistakes on the written test that were discovered before the test was administered, so instead of reprinting copies of the test, they printed a sheet of paper that accounted for and corrected the mistakes.

I very much liked the ciphering machine that told us how much time we had left. :)

[quote="solafidefarms"]So vestavia was today. I should say, I'd rather have college students writing the test than high schoolers if it means we don't have 6 problems being changed in addendum. Otherwise, they were good problems. The grading on comprehensive was less than perfect, so Mr. Taylor said that if your results changed he'd send out a new thingy for your plaque. Thus, if you got a comprehensive trophy, it would be rather helpful to him if you sent him your address so it's easy for him to mail a new one. :)

The last results I saw for Comprehensive, DivI:
1: Shawn Pan
2: Eric Morphis
3: Jonathan ...
4: Billy Dorminy
.
.
.

But it might change more if the answer key that we were being graded against had more than 1 mistake.

The ARML team test by Mr. Crawford was very good.[/quote]
Jonathan Mei got 2nd place. I was there too :D

[quote="lifesapain"]Look i know they didnt cheat. But some people who dont know how freakishly good vestavia is at math thought they cheated. Anyway, I heard some florida team thats better than vestavia is coming. Is this true? Is this even possible?[/quote]
That florida team is rickards. And probably they're talking about the algebra2 team. We dominated in algebra2 :D

O well i think i was referring to buchholz, who i somehow thought was really good, probably cuz someone told me they beat vestavia in mu alpha theta national convention. But they didnt do so well here. Are the only smart kids from rickards in ure level/ grade? Like are u guys the same fairview team that did good last year?

[quote="Mathalete"][quote="solafidefarms"][b]The grading on comprehensive was less than perfect[/b], so Mr. Taylor said that if your results changed he'd send out a new thingy for your plaque. Thus, if you got a comprehensive trophy, it would be rather helpful to him if you sent him your address so it's easy for him to mail a new one. :)

The last results I saw for Comprehensive, DivI:
1: Shawn Pan
2: Eric Morphis
3: Jonathan ...
4: Billy Dorminy
.
.
.

But it might change more if the answer key that we were being graded against had more than 1 mistake.

The ARML team test by Mr. Crawford was very good.[/quote]
Jonathan Mei got 2nd place. I was there too :D[/quote]

Observe the bolded section of my post. After fixing the answer key that was in the computer to agree with the other answer keys after the awards were given, [b]Jonathan and Eric swapped ranks[/b]. If you aren't Jonathan, you might want to bug Mr. Taylor about your new and hopefully improved rank :).

i thought he preferred being referred to as t-dawg, if i'm not mistaken

Actually, I'm pretty sure all the smart people from Rickards weren't at the Vestavia competition.  :D

lol wtf.. they used a lot of my questions on comprehensive.

Does any one has Vestavia Hills 2014 Algebra 1 test paper. I appreciate very much if some one could upload it

Does this tournament still happen? If so where is registration? It'd be nice if there was a site like GCTM (which lists all the math competitions in Georgia and gives their information) for Alabama...

@Below  Thanks!

@eashang1 From the looks of it you're a month early.

Registration and tourney info from 2016:
https://sites.google.com/site/talkincircles234/

Registration packet from 2014:
http://madisoncitymath.weebly.com/uploads/2/2/9/3/22939590/2014_vhhs_invitation_packet.pdf

If you'd like, you can inquire with Susan Gifford (giffordms at vestavia.k12.al.us, T 205-402-5250, F 205-402-5262).


[quote="returnofkiladi"]Two similar polygons have perimeters in a ratio of 3 to 7. If the area of the larger polygon is 294 square units, how many square units are in the area of the smaller polygon.

Answer:
[hide]
54
[/hide]

I got 126, but apparently that's wrong.[/quote]

The ratio of side lengths squared is the ratio of the areas. So $ \frac {9}{49} \equal{} \frac {x}{294} \implies x \equal{} \boxed{54}$.

[quote="returnofkiladi"]Two similar polygons have perimeters in a ratio of 3 to 7. If the area of the larger polygon is 294 square units, how many square units are in the area of the smaller polygon.

Answer:
[hide]
54
[/hide]

I got 126, but apparently that's wrong.[/quote]

Remember that the ratio between perimeters will be squared in corresponding areas.


algebra
polynomial
trigonometry
function
analytic geometry

EDIT

old post: [i]Solve with proof: $8\cos^3\theta+4\cos^2\theta-4\cos\theta+1=0$ for $\theta \in [0,2\pi)$.[/i]

I messed the problem up  :blush:

it should be $8\cos^3\theta+4\cos^2\theta-4\cos\theta-1=0$

[hide="A proof why this function is always positive"]
$f(x) = 8x^3+4x^2-4x+1$

$f'(x) = 4(6x^2 + 2x - 1)$

The roots of $f'$ are $x_1,x_2 = \frac{-1 \pm \sqrt 7}{6}$

$f(x_1)$ is a local maximum and $f(x_2)$ is a local minimum

In $(-\infty, x_1),\ f$ is increasing
In $(x_1, x_2),\ f$ is decreasing
In $(x_2, \infty), \ f$ is increasing

$x_1 \approx -0,607625219$
$x_2 \approx 0,274291885$

$-1 \in (-\infty, x_1)$ and $f(-1) = 1$, so $\forall x \in [1,x_1), \ f(x)>0$

The minimum value of $f$ in $(x_1, \infty)$ is the value $f(x_2) \approx 0,36887>0$

So $\forall x \in (x_1, \infty)$ we have $f(x)>0$

We found that $\forall x \in [-1, 1] , \ f(x)>0$

Finally, $\forall \theta \in [0,2\pi)$ we have $f(cos \theta)>0$[/hide]

(I have $[100 \pi]$ posts)

[quote="pontios"][quote="Altheman"]Solve with proof: $8\cos^3\theta+4\cos^2\theta-4\cos\theta+1=0$ for $\theta \in [0,2\pi)$.[/quote]

I made a graph; this function is always positive.

Maybe you mean something like  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59595]this topic[/url]
[quote]$8\cos ^3 u-4\cos ^2 u -4\cos u +1=0$[/quote]

By the way, I'd like to attach the graph of the function in polar coordinates:
 $r(\theta) = 8\cos^3\theta+4\cos^2\theta-4\cos\theta+1$, because it looks nice!

If we wanted to give a name to this function then we'd certainly call it [color=green][b]Fish-function[/b][/color]

See that it has no roots[/quote]

that is pretty awesome :lol:

also, i remembered that other problem, and there is something interesting associated with that problem too, but this is a different problem

Are you sure that that edited problem is correct? Because that would mean that $4\cos\theta$ and $(2cos^2\theta+\cos \theta-1)$ are reciprocals, which is not necessarily true. I randomly chos $\frac{\pi}{6}$ as a value for $\theta$ and the LHS did not equal $0$.

[quote="Adunakhor"]Are you sure that that edited problem is correct? Because that would mean that $4\cos\theta$ and $(2cos^2\theta+\cos \theta-1)$ are reciprocals, [b]which is not necessarily true.[/b] I randomly chos $\frac{\pi}{6}$ as a value for $\theta$ and the LHS did not equal $0$.[/quote]

the whole point is to find values of $\theta$ such that this is true...and why would you expect to choose a random value and it would satisfy the equation  :huh:

Oh wow, somehow I thought the question was asking to prove that that statement was always true for that interval... Sorry,  :oops:

[quote="pontios"]See http://www.mathlinks.ro/Forum/viewtopic.php?p=365612#p365612

$8\cos^3\theta+4\cos^2\theta-4\cos\theta-1=0 \ \ (1)$

Set $v=2 \cos \theta$

$v$ is a [u]complex[/u] number. The equation $(1)$ becomes $v^3 + v^2-2v-1 = 0 \ \ (2)$

Set $v=z+\frac{1}{z}$ (there is a solution for $z$ and it is $z=\cos \theta \pm i \sin \theta$)

Then we have

$v^2 = z^2+2+\frac{1}{z^2}$

$v^3 = z^3 + 3z + \frac{3}{z} + \frac{1}{z^3}$

$v^3 + v^2-2v-1 = 0 \ \ (2)\Rightarrow$

$(z^3 + 3z + \frac{3}{z} + \frac{1}{z^3}) + (z^2+2+\frac{1}{z^2})-2z -\frac{2}{z} -1 = 0 \Rightarrow$

$z^3 + z + \frac{1}{z} + \frac{1}{z^3} + z^2+\frac{1}{z^2} +1 = 0 \Rightarrow$

$z^3 + z^2 + z +1+ \frac{1}{z} +\frac{1}{z^2} + \frac{1}{z^3} = 0 \Rightarrow$

$z^6 + z^5 + z^4 + z^3 + z^2 + z +1 = 0 \Rightarrow$

$(z-1)(z^6 + z^5 + z^4 + z^3 + z^2 + z +1) = 0 \Rightarrow$

$z^7-1 = 0$

So the solutions (for z) are $z = \cos \frac{2k \pi}{7} + i \sin \frac{2k \pi}{7},\ \ k = 1, 2, ..., 6$

(we don't put $k=7$ because the value $z=1$ is not a solution to $(1)$; if $z=1$ then $v=2 \Rightarrow \cos \theta = 1$, which does not satisfy the initial equation)

set for convenience $\theta_k = \frac{2k \pi}{7} \Rightarrow z = \cos \theta_k + i \sin \theta_k$

$v = z+\frac{1}{z} \Rightarrow$

$v = \cos \theta_k + i \sin \theta_k + (\cos \theta_k + i \sin \theta_k)^{-1} \Rightarrow$
$v = \cos \theta_k + i \sin \theta_k + \cos \theta_k - i \sin \theta_k) \Rightarrow$
$v = 2\cos \theta_k \Rightarrow$
$2\cos \theta = 2\cos \theta_k \Rightarrow$

The solutions of $(1)$ are $\theta_k = \frac{2k \pi}{7} \ \ k \in \{1,2, \cdots, 6 \}$[/quote]

very good, the fact that it was a 3rd degree polynomial would trick many people into thinking that there were 3 solutions, but you caught that!


function
invariant
SFFT
special factorizations

Nobody could answer this when I posted it in High School Basics, so I'll put it here  :D

 $A=\{1,2,3...n\}$ and $f(x,y)=xy+x+y$, where $x$ and $y$ are chosen randomly from $A$. the new number is put into $A$ and $x$ and $y$ are removed. find all possible final values of $A$ once the function has been applied $n-1$ times in terms of $n$

[hide="hint"]simon's favorite factoring trick [/hide]

[hide="another hint"]find an invariant[/hide]

[hide="Solution"] $f(x, y) = (x+1)(y+1)-1$

So we are removing $(x+1)-1, (y+1)-1$ and adding in $(x+1)(y+1)-1$.  This suggests an easy invariant:

$\prod_{a \in A}(a+1)$

Which is, of course, $(n+1)!$.  This gives us our final sole element $\boxed{ (n+1)!-1 }$.  [/hide]

[hide]each operation reduces the number of elements in the set by 1, after $n-1$ operations, the set has one element...

this problem screams invariants...let $f(A)$ denote $\prod_{x\in A}(x+1)$

since $(xy+x+y)+1=(x+1)(y+1)$, the operation does not change $f(A)$.

now we have that $f(A)$ starts as $(n+1)!$, and at the end of the operations we have that $(n+1)!=l+1$, so the last element is $(n+1)!-1$[/hide]


When $ n$ is an odd integer, there is a value of $ m$ for which $ n^2 \plus{} 3 \equal{} 4 \times m$. If $ n \equal{} 11$,
what is the value of $ m$?


geometry
geometric transformation
reflection
complex analysis
complex analysis unsolved

i)Show that for any two hyperbolic triangles $ T_1,T_2 \in\mathbb{D}$ there is an automorphism of $ \mathbb{D}$ such that $ M(T_1) \equal{} T_2$, where $ \mathbb{D}: \equal{} \{z\in \mathbb{C}: |z| < 1\}$ is the unit disc.

Assume we are given a hyperbolic triangle $ T_0\in \mathbb{D}$  with three boundary sides $ f_1,f_2,f_3$ in counterclockwise order. Suppose there exists a biholomorphic map $ g: T_0\rightarrow \{z\in \mathbb{C}: Im z > 0\}$(This is the Riemann mapping theorem).

ii)Prove that $ g$ extends continuously and injectively to all three boundary  sides, and it can be chosen so that it sends the three sides $ f_1,f_2,f_3$ onto $ ( \minus{} \infty; 0)$, $ (0,1)$ and $ (0, \infty)$ respectively.

iii)Prove that there are three hyperbolic triangles $ T_1, T_2$ and $ T_3$, all disjoint from each other and from $ T_0$ and so that $ \overline{T_i} \cap \overline{T_0} \equal{} f_i$for$ i \equal{} 1,2,3$, and so that there is a unique antiholomorphic Mobius transformation $ M: D\rightarrow D, z\mapsto\frac {a\overline{z} \plus{} b}{c\overline{z} \plus{} d}$ that sends $ f_i$ onto itself and that maps $ T_i$ onto $ T_0$.

Ok, I have solved part i)(yes really easy, sorry). But, I need for help for parts ii) and iii).
I think for the part ii) we should apply Shwartz reflection principle, but I don't know exactly way.
Please help me if you can.


I got a negative zero on the usamo, but i'm still satisfied b/c at least i made the usamo. How many people got zero on the usamo? What do u feel?

Of course you did, [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamotop12.shtml]Delong Meng[/url]...

EDIT: To avoid a pointless bump...
In response to the post below, Delong likes to say things like that. People who know me would know that I don't talk like that, and I do not expect to make the IMO in 08. :-/

:)

[quote="Dnas (benvid dajamin)"]I am the 2008 IMO perfect scorer! [/quote]
U should say something relevant to the topic.

[quote="nebula46"][quote="Dnas"]Of course you did, [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamotop12.shtml]Delong Meng[/url]...

EDIT: To avoid a pointless bump...
In response to the post below, Delong likes to say things like that. People who know me would know that I don't talk like that, and I only expect to make the IMO in 08. :-/

:)[/quote]
Who is Delong? Dnas, are you David Benjamin, the really smart guy from Indiana? I bet u definitely are going to make the IMO next year because u had a 300 index last year.[/quote]
Delong is [b]you[/b]


algorithm
limit
logarithms
calculus
derivative
algebra proposed
algebra

Yes , I read it in Burton's book but haven't found a proof!!!

This was discovered by Ramanujan -- nobody knows how ... The proof is not quite elementary.

A yet more astounding formula (again 'given' (but not proven!) by Ramanujan) is the following. Let
\[
A := 1 + \frac{1}{1 \cdot 3} + \frac{1}{1 \cdot 3 \cdot 5} + \cdots = \sum_{k \geq 0} \frac{1}{(2k+1)!!} = \sum_{k \geq 0} \frac{2^kk!}{(2k+1)!}
\]
and
\[
B := \frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{1+\frac{4}{2 + \cdots}}}}}.
\]
Then, neither $A$ nor $B$ is expressible in closed form in terms of the classical constants, but their sum $A+B$ turns out to be [i]precisely[/i] $\sqrt{\pi e/2}$!

--Vesselin

Simply Beautiful,,,, Ramanujan Was indeed a GENIUS!!!!

This is however in the journal as a training problem for high school students... however I agree it looks quite hard to solve this elementary. :?

This approximation for $e$ is very interesting:

\[e\approx\left(1+9^{-4^{7\cdot 6}}\right)^{3^{2^{85}}}\]

It is pandigital, but besides that, it is correct to, believe or not, $18457734525360901453873570$ decimal places!

[quote="Bojan Basic"]It is pandigital, but besides that, it is correct to, believe or not, $18457734525360901453873570$ decimal places![/quote]That is almost the number of digits $3^{2^{85}}$ has...

But what do you mean by $^{7.6}$? I mean it's neither 7.60 not 42 since in both cases bernouilly says us the number (which we claim to be an approximation of e) has zillions of digits.

[quote="perfect_radio"]how was such a good approximation found ???[/quote]weider: how was it verified so closely? I mean even if they found an algorithm that takes only 1 basic operation to check if 2 digits are equal, that takes still longer than the universe exists...

I think he means the product $7 \times 6$ ($42$).  The right-hand side is of the form ${(1 + \frac{1}{n})}^n$, which is known to converge to $e$ quickly.  So the approximation is not that much of a surprise.  But the pandigital property is cute.

[quote="Bojan Basic"]Yes, it is $7\cdot 6=42$. The expression is equal to $\displaystyle\left(1+\frac 1{3^{2^{85}}}\right)^{3^{2^{85}}}$ and hence the approximation.[/quote]
That's what I mean: $9^{-4^{7\cdot 6}} = \dfrac1{3^{2^{84}}} \not= \dfrac1{3^{2^{85}}}$?

[quote="perfect_radio"]in any analysis textbook it says there that $\left( 1+\frac{1}{n} \right)^n$ converges (it is bounded & increases)

can we instantenously conclude that $\lim_{x \to \infty} \left( 1+\frac{1}{x} \right)^x$ converges ?

(hope my question is not too dumb  :blush: )[/quote]No, we can't conclude it instantly. (eg. $\sin(2\pi x)$ converges on integer x but not on real x.

It's enough to show that

$\left(1+\frac 1x\right)^{x}$ is increasing.  Taking $\log$s doesn't affect anything, so we need only prove that

$f(x)=x\log\left(1+\frac 1x\right)$ is increasing.

But $f(x)=x(\log(x+1)-\log x)$, and taking derivatives, we get

$f'(x)=\log(x+1)-\log x-\frac {1}{x+1}$

Taking derivatives again, we get

$f''(x)=\frac {-1}{x(x+1)^2}$

so that $f'$ is decreasing on $x\geq 0$.  Since $f'(x)\to 0$ as $x\to\infty$, this means that $f'(x)\geq 0$, which is what we wanted.

i think in calculating the derivative of $\log$ one uses my limit, so your argument is a bit circular... or am I completely wrong?

I don't know how blahblahblah meant it, but I do know a shorter way with de l'hospital:
\[\lim_{x=\infty} f(x)=x\ln\left(1+\frac 1x\right) =\lim_{x=\infty} \dfrac{\log\left(1+1/x \right)}{1/x} = \lim_{x=\infty} \frac{1}{1+1/x} = 1\]
So since the ln converges to 1, the original converges to e.

It's not circular reasoning if we define $e^x$ by its taylor series (as is usually done) and then calculate the derivative of $\log$ from there.


AMC
AIME
USAMTS
Mandelbrot Competition
AoPS Books

Next year, I'll probably be officer of the Math Honors Society at my school. I think it really needs some reform... we only went to 2 out of the 6 MAO regional competitions and we are not going to states. (We did AMC, AIME and Math League)

Anyway, what other things/contests/competitions could our Math Honor Society do? Should we start holding practices?

We deperately need fundraising (for this why we could not go to states, and we have a small amount of motivated people).

Probably, introducing USAMTS to the students?
That's what i;m going to do, since I was the only one who did it this year.

I would also recommend AHSIMC (even though it was not that good last year, I have confidence that it will improve.)

For fundraising, if your school will allow it, I recommend a candy sale.  In our school, both NHS and Band sold candy, and in both cases, it was a resounding success.

The Mandelbrot Competition (http://www.mandelbrot.org) is a wonderful contest -- very likely the best in the country among those which are administered across the entire year (rather than as a one-time event, like the AMC contests).

NYCIML, Mandlebrot, and regional competitions, etc.

And if you bring food, interest increases. Speaking of which, I should suggest the subject of food to TJ's math team officers.

NO WE'RE TOO COOL FOR FOOD HERE AT VARSITY MATH TEAM @ TJ.

your math club may want to start with something easier, attract some people and get them interested first before giving them stuff like nycimls...

first, you could try to get some sponsors from local businesses/companies. we have gotten hoodies, shirts, aops books, funding for trips, etc.
and look into local/state/regional competitions at colleges and stuff.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a torsion-free group with a subgroup $ H$ of finite index isomorphic to $ \mathbb{Z}$. Then $ G$ is isomorphic to $ \mathbb{Z}$.


function
algebra
polynomial
linear algebra
matrix
geometry
3D geometry

Let $M(n,\mathbb{R})$ denote the set of n x n matrices. Let $O(n)$ denote the set of orthogonal matrices. How can i show that $O(n)$ is compact? I know to prove for n=3.

As a topological space, we can identify $M(n,\mathbb{R})$ with $\mathbb{R}^{n^{2}}.$ Then we can use the Heine-Borel theorem: a subset of $\mathbb{R}^{m}$ is compact if and only if it is closed and bounded.

Consider the function $g: M(n,\mathbb{R})\mapsto M(n,\mathbb{R})$ given by $g(A)=A^{T}A.$ If we can show that $g$ is continuous, then $O(n)=g^{-1}(\{I\}),$ which is the inverse image of a closed set under a continuous function, hence closed. And as for showing that $g$ is continuous, note that every element of $A^{T}A$ is a polynomial in the elements of $A.$

As for boundedness, it shouldn't be too hard to show that if $A\in O(n),$ then $|a_{jk}|\le 1$ for each entry of $A.$

thank you Dr kent  :)

my proof for the n=3 case was as follows. we first note that each 3x3 orthogonal matrix defines an isometry of the unit sphere onto itself. the unit sphere is compact. so the problem translates to , if $(i_{n})$ is a sequence of isometries with $i_{n}: M\rightarrow M$ ( where M is compact)then some subsequence of $(i_{n})$ converges to another isometry. this follows easily from arzela-ascoli theorem(?).whats your take on this, is it correct?

I haven't yet thought through your proof to tell whether or not it is valid, but I do have two thoughts.

1. If it is valid, what restricts it to the $n=3$ case? Why shouldn't the same argument work for any $n?$

2. Using Arzela-Ascoli is overkill. This is a concrete, finite-dimensional problem and there should be no need to bring in a tool like that.

In fact, here's another way to look at my proof. Suppose you have a sequence of elements of $O(n).$ This is a bounded sequence in $R^{n^{2}},$ hence it has a convergent subsequence. Now all we have to do is to prove that the limit of that subsequence is also a member of $O(n).$ (You would have to do that last step as well.)


At the Pizza $ \pi$ Paradise, a pizza with a 10'' diameter costs $ \$10$. Given that pizza price is directly proportional to the number of square inches of pizza a person gets, what is the number of dollars in the cost of a pizza with a 14'' diameter? Express your answer to the nearest cent.

The area of a circle is $ pi(d/2)^2$. Now insert the numbers. $ 3.14(10/2)^2$ $ 3.14(5)^2$ $ 3.14(25)\equal{}78.50$ This is the area in square inches. The ratio is 78.50:10

Now to find the area of the second pizza.
$ /pi(d/2)^2$
$ 3.14(14/2)^2$
$ 3.14(7)^2$
$ 3.14(49)\equal{}153.86$
$ 153.86/78.5\equal{}1.96$
$ (1.96)78.50: 10(1.96)$
$ 153.86\equal{}19.60$
So the answer is \$19.60 is the price for a pizza of 14 inches of diameter.
It's probably wrong... just saying...  :D


Find the mean of the terms of the arithmetic sequence:

$ 13,29,45,\ldots,141,157$.

A shortcut: in an arithmetic sequence, the mean=1st term+last term/2.
(13+157)/2=170/2=85.


Le Chatelier's principle questions are usually quite straightforward.  Here's an exception that I often ask my more advanced students:

The solubility (in mass %) of CaCl$_2$ in water at 329K is 56.96%.  Therefore, if 56.96 g of calcium chloride is added to (100-56.96) g of water at 329K, it will all dissolve and a saturated solution will result.  Observation shows clearly that the above process is very exothermic.

(a) Predict, using using this information and Le Chatelier's principle, whether the solubility of calcium chloride should increase or decrease as the temperature is increased.

(b) The solubility of CaCl$_2$ in water at 369K is 60.10% Explain this observation in light of your prediction in part (a).

i've seen this problem before...
i'll wait some time to give others a chance to solve it, then if no one did it, i'll do it.
it's very simple. :)

[quote="pkothari13"](a) Since the reaction is exothermic, one can treat the energy released as a product. Increasing the temperature will favor the consumption of heat, so the equilibrium will shift to the reactant side and the solubility will therefore decrease.

(b) At 369K, some of the water will have evaporated, so the mass of $CaCl_2$, per $H_2O$ will decrease.

Are these correct explanations?[/quote]

Your reasoning in (a) is "correct" in the sense that you arrived at the expected answer for the question as worded.  However, the solubility actually does increase with temperature.

The objective in part (b) is therefore to explain why the simple prediction in part (a) is incorrect.  It has nothing to do with evaporation.  A method for measuring the concentration at saturation is to extract some solution while carefully excluding solid, then measuring the amounts of water and salt gravimetrically (i.e., weigh the solution immediately, then again after the water has all evaporatated).   Slow evaporation before the extraction doesn't affect the concentration as long as the solid/solution equilibrium is established at a reasonable rate.

I'm not sure why Le Chatlier's principle isn't obeyed, but to make a guess... does it have something to do with the concentration of water?  Help?

Here's my answer:[hide] In Le Chatelier's principle, what matters is the behavior of the system [i]at equilibrium[/i].  Dissolving calcium chloride in water is initially very exothermic, but as the solution concentration increases, the reaction eventually becomes somewhat endothermic.   Formation of the saturated solution from salt and pure water is exothermic overall since the process is exothermic over most of the dissolution process.  Nevertheless, it is endothermic [i]at saturation (equilibrium) [/i] and this leads to the correct Le Chatelier's principle prediction that solubility should increase with increasing temperature.  [/hide]

Wow, that's pretty cool. Btw, are all solution processes at equilibrium endothermic?


geometry
perimeter
trigonometry
symmetry
calculus
integration
geometric transformation

In scalene, but non-right triangle ABC, tan A = 5/12.  If the length of each of the sides of triangle ABC is an integer, find the smallest possible perimeter.

The answer key says 80, but I don't see it.  Any ideas?

the fact that $ \tan A = \frac{5}{12} $ implies that $ \cos A = \frac {12}{13} $.  so, by the law of cosines, $a^2 = b^2+c^2-2bc \cdot \frac {12}{13} $, so for $ a, b, c $ to all be integers, at least one of $ b, c $ must be divisible by 13.  by symmetry you may assume it's $ b $: $b = 13k$.  now, drop the altitude $ CH $ from $ C $ to $ AB $ (with H on the line AB).  then $ CH = 5k $ and $ AH = 12k$ (because $ tan A = 5/13 $) and $ CHB $ is a right triangle with integral sides (because $CB$ is an integer and $AH \pm HB = AB$).  now, note that the smallest perimeter occurs when angle $ B $ is obtuse.  (if $ B $ is acute, then if we reflect $ B $ in line $ CH $ to get $ B' $, triangle $CAB'$ also has integral sides, but smaller perimeter.)  so $A$ and $B$ should be on the same side as $H$.

now, we need $CHB$ to be a right triangle with integral sides and with a side divisible by 5 (because $ CH = 5k $).  the smallest we can take $CHB$ is the $15-20-25$ triangle.  (we can't have it be the $5-12-13$ because then $ABC$ wouldn't be scalene.)  the smallest perimeter of $ABC$ occurs when $BH$ is taken to be the 20 side (so that $AB$ is minimized).  in that case, $CH = 15$, $k = 3$, $CA = 39$, $ AB = 36 - 20 = 16$, and $CB = 25$, and the perimeter of $ABC$ is $39+16+25=80$

Thanks.   For some reason I completely overlooked the possibility that triangle ABC could be obtuse.


Let $a_1,a_2,...,a_n$ nonegative reals. Prove that
$\frac{m}{2}\leq\frac{a_1+a_2+...+a_n}{n}-\sqrt[n]{a_1a_2...a_n}\leq\frac{(n-1)M}{2}$
where $m=\min{(\sqrt {a_i}-\sqrt {a_j})^ 2}$ and $M=\max{(\sqrt {a_i}-\sqrt {a_j})^ 2}$ where $1\leq i,j\leq n$

[What is your opinion for this ineq Harazi? I am curious to hear your opinion.

My opinion is that it follows from another problem I posted on the forum. If I'm not wrong, this one is even weaker. That one was solved by Myth and Ravi B.


Prove or disprove (this is (was) my conjecture):

Between the numbers $p_n-p_{n-1}+1$ and $p_n-p_{n-1}-1$ at least one of them is a prime. ($p_n$ & $p_{n-1}$ are consecutive primes)

I think you mean: Prove that it's prime infinitely many times.
Since as good as all primes are odd this is equivalent to the existence of infinetly many prime twins.

He might mean to prove (or disprove) that at least one of $p_n-p_{n-1}+1$ and $p_n-p_{n-1}-1$ is always prime.  If so, the pair 1327, 1361 is a counterexample.

If that is the conjecture, then we can disprove it.  Note that 2477 and 2503 are consecutive primes.  Their difference is 26.  But neither 25 nor 27 is prime.

This is wrong for sure, but the first counterexamples are bigger, so it's just some looking around...

Edit: ok, Ravi B is more specific

My edit came too late!  Sorry Ravi B -- you can claim priority.

Okay, but then you can claim credit for the smallest counterexample.  :)

:lol: Well, we haven't proved that yet.  Small[i]er[/i].

According to [url=http://mathworld.wolfram.com/PrimeGaps.html]MathWorld[/url], for primes less than your prime 1327, the largest prime gap is 22.  So your counterexample is indeed smallest.


There's person A and person B.  They each have a 1/6 chance of winning a game.  What is the probability that person A wins 5 games before person B wins one?

Now they each have a 1/24 chance of winning a game.  Now what is the probability that person A wins 5 games before person B wins one?

are the events of A winning and B winning mutually exclusive? like is it russian roulette with reloaded guns, except person B is iron man? or can both occur in the same game, like as if it were a carnival sideshow game or something like that?

Okay, put it this way, instead.  Persons A and B take turns playing a game.  On each turn, the player has a 1/6 probability of earning a point.  Person B goes first.  What is the probability that Person A earns 5 points before person B earns one point?

If they each instead have a 1/24 probability of earning a point on each turn, then what is the probability that person A earns 5 points before person B earns one point? (Person B still goes first.)

Solution?

[hide]For Person A to win the first 5 times, the probability is $ \dfrac{1}{6^5}\equal{}\dfrac{1}{7776}$.  Then, we see that turns in which neither player wins can also occur.  If (here I'm not sure) the order of Person A's points and neither players' points don't matter, then we have an infinite geometric series where $ r\equal{}\dfrac{2}{3}$ and the final probability would be $ \dfrac{1}{2592}$.  If order does matter, it is now $ \displaystyle\sum_{n\equal{}0}^{\infty}\dbinom{n\plus{}5}{n}$, which I have no clue how to sum.[/hide]

The order does matter, as each different ordering is a different case.  But the sum you came up with, azn, is definitely incorrect (the series diverges), but you were on the right track.  Lets try again.

Lets assume that the game lasts $ 2x$ turns before the score is A:5 to B:0.  Person A has a $ \dbinom{x}{5}(1/6)^5(5/6)^{x \minus{} 5}$ chance of scoring like this and person B has a $ (5/6)^x$ chance of scoring like this, for a total probability of $ \sum_{x \equal{} 5}^{\infty} \dbinom{x}{5}(1/6)^5(5/6)^{2x \minus{} 5}$ which I am approximating as $ .0635033171$.  I do not know any way to find the exact value.

In the other case, with a probability of $ 1/24$, we have a total probability of $ \sum_{x \equal{} 5}^{\infty} \dbinom{x}{5}(1/24)^5(23/24)^{2x \minus{} 5}$ which I am approximating to be $ .343933529$

facis, I went the same way you did, but I don't think your combinations are disjoint.  That is, if we're allowed to choose 5 turns for A to win from his x turns, then he may win all five before the 2x turns happen.

To fix that, I have that Person A has a $ \dbinom{x \minus{} 1}{4}(1/6)^5(5/6)^{x \minus{} 5}$ chance of scoring his fifth win on turn 2x.  (that is, we choose the first 4 from the first x-1 turns, but the last win is fixed at turn x)

yeah... good point.

Adjusting: in the $ p\equal{}1/6$ case, the total probability is $ \sum_{x \equal{} 5}^{\infty} \dbinom{x\minus{}1}{4}(1/6)^5(5/6)^{2x \minus{} 5} \approx .0194037913$.

in the $ p\equal{}1/24$ case, the total probability is $ \sum_{x \equal{} 5}^{\infty} \dbinom{x\minus{}1}{4}(1/24)^5(23/24)^{2x \minus{} 5} \approx .0280640206$.

Wow.  Huge difference.  Thanks for catching my mistake.

The version of the game described in the second post by the O.P. is much more complicated than the original version.  In the way the game was originally described, the game is played and the possible outcomes are that A wins (1/6 of the time), that B wins (1/6 of the time) or neither (2/3 of the time).  For the purposes of the question asked, this is exactly the same as a game that A wins half the time and B wins half the time, and so the answer is 1/32.


The following game is played on a rectangular chessboard $ 5 \times 9$ (with five rows and nine columns). Initially, a number of discs are randomly placed on some of the squares of the chessboard, with at most one disc on each square. A complete move consists of the moving every disc subject to the following rules:
$ (1)$ Each disc may be moved one square up, down, left or right;
$ (2)$ If a particular disc is moved up or down as part of a complete move, then it must be moved left or right in the next complete move;
$ (3)$ If a particular disc is moved left or right as part of a complete move, then it must be moved up or down in the next complete move;
$ (4)$ At the end of a complete move, no two discs can be on the same square.
The game stops if it becomes impossible to perform a complete move. Prove that if initially $ 33$ discs are placed on the board then the game must eventually stop. Prove also that it is possible to place $ 32$ discs on the boards in such a way that the game could go on forever.

Color the squares of the odd rows 1, 2, 1, 2... alternatively and the squares of the even rows 3, 4, 3, 4... in the same way. We see that we have 15 squares with a 1 inside of them, 12 with a 2, 10 with a 3 and 8 with a 4. After two moves, a disc that was on a square with a 1 or a 4 respectively will go on a square with a 4 or a 1 respectively. Likewise a discs on a square with a 2 or a 3 will go on a square with a 3 or a 2. If we have 33 discs on the board, then there must be at least 17 discs on the 1 and 4 squares (in total), or on the 2 and 3 squares (in total). Considering the first case. Since there are at most 8 discs sitting on squares with a 4, then there must be at least 17-8=9 discs on the squares with a 1. That implies that after two moves all of the discs that were on squares with a 1 are now on squares with a 4. But this will stop the game since we have 8 squares with a 4 and at least 9 discs which must sit on such squares. If instead we had at least 17 discs on the 2 and 3 squares, then after one move all of these discs will go on squares labeled with a 1 or a 4, again stopping the game. For an example with 32 discs, consider 8 (disjoint) 2x2 squares and put 4 discs in each on them. Then one can preform a cycle, moving the disc in the top-left corner to the right, the top-right disc down, the bottom-right to the left, and the bottom-left disc up. This movement will keep the game going forever.


anyone know the exact fraction of pi. A question i solved gave me the answer 2199/700. it says it was right. i checked the calculator and it doesnt equal pi. anyone know the exact fraction of pi?

Regrettably $ \pi$ can't be described as fraction, because it is not rational number.

As kunny said, $ \pi$ is not a rational number. There is, however a fairly famous fraction that approximates $ \pi$ more closely than it has any right to. That fraction is $ \frac{355}{113},$ and to get a rational number that approximates $ \pi$ any better than that, the numerator and denominator would have to have at least 5 digits each.

[quote="shaun ong"]MY FAVOURITE HERO ANTIMAGE!!!!!!!!!!!!!!!!!!!!!!!!!!!! pi= 22/7 or 3.14159265358979323846 to be exact . Wait a minute , what do u mean by its not a rational number? I jz now that pi = ratio of circumference to the radius . You can find more about it here http://en.wikipedia.org/wiki/Pi[/quote]
It's not rational though...rational means it can be expressed as an irreducible fraction with integer numerator and denominator. Pi has been proven to be irrational, and in fact has been proven to be transcendental.

[quote="shaun ong"]pi= 22/7[/quote]

$ \pi$ is [i]approximately[/i] $ \frac {22}{7}$.  The approximation deviates starting from the thousandths place.

[quote="shaun ong"] or 3.14159265358979323846 to be exact .[/quote]

This is also an approximation of $ \pi$.  The decimal expansion of $ \pi$ does not terminate at the number of digits your calculator can show.  It does not, in fact, terminate at all.  The next few digits are $ ...26433832795028841971...$

[quote="shaun ong"] Wait a minute , what do u mean by its not a rational number? I jz now that pi = ratio of circumference to the radius . You can find more about it here http://en.wikipedia.org/wiki/Pi[/quote]

First of all, $ \pi$ is the ratio of the circumference (of a circle) to its [i]diameter[/i].  Second of all, the simplest way of putting it is that $ \pi$ cannot be written as a fraction.

The following sequence of rational numbers converges to $ \pi.$ Each term in the sequence is "best possible" in the sense that it approximates $ \pi$ more closely than any fraction $ \frac pq$ where $ q$ is smaller than the [i]next[/i] denominator in the sequence.

$ 3,\;\frac{22}{7},\;\frac{333}{106},\;\frac{355}{113},\;\frac{33102}{103993},\;
\frac{33215}{104348},\;\frac{66317}{208341},\;\frac{99532}{312689},\;
\frac{265381}{833719}\,\dots$

There is no repeating pattern that generates this sequence; if there were, $ \pi$ would satisfy an integer-coefficient quadratic equation, and it doesn't.

What was the problem you solved?

Pi is irrational, meaning that it can't be written as a fraction, and, most importantly, is the ratio of a circle's circumference to its diameter. This has been known since ancient times.


After typing a solution in TeX, you could get a numerical answer, such as 52 or x+1. What code do you use to make a box around the answer? Thanks.


Why is this logic wrong?

Find the probability of at least one coin landing heads when two coins are flipped.

The probability of the first coin landing heads is 1/2, and the probability of the second coin landing heads is 1/2. The probabilities are independent of one another, so p = 1/2 + 1/2 = 1.

[You used "first coin" twice, but I think you meant "second coin" in the second instance.]

You are confusing the words "independent" and "disjoint".  The two events you described are independent, but are not disjoint.

Let A and B be two events.  If A and B are disjoint, then
Pr(A or B) = Pr(A) + Pr(B).  If A and B are independent, then
Pr(A or B) = 1 - Pr(not A and not B) = 1 - Pr(not A) Pr(not B) = 1 - (1 - Pr(A)) (1 - Pr(B)).

In your case, the two events are independent, so the second formula applies.  Namely, Pr(A or B) = 1 - (1/2)(1/2) = 3/4.

Ah, I see, thanks for the clarification.


function
real analysis
calculus
integration
real analysis unsolved

Let us define $f_{n}(x)=\frac{1}{4^{n-1}}\cdot f_{n-1}(4^{n-1}x)$
  and $f(x)=\sum_{i\geq 1}f_{i}(x)$. It is well-known that $f$ is continous and not differentiable at all. Prove that $f$ is not monotonous on any interval.

We need a little more structure on the $f_{n}.$ We could have $f_{n}(x)=x,$ although then the series for $f$ would diverge. You need to specify a little more about the behavior to keep me from doing things like that.

A cheap argument using an advanced theorem: a function that is monotone on an interval (more generally, that is of bounded variation on an interval) is differentialble almost everywhere. It follows from this that any nowhere differentiable function must also be nowhere monotone.

While true, that's a big gun of a theorem, and uses Lebesgue tools. I'm sure Cezar wanted a direct proof using tools from a previous course (the one in which uniform convergence and the Riemann integral were developed.)

So, can you prove this directly without strong tools? :P

Well, the statement is false without additional assumptions on $f$, as Kent Merryfield pointed out. If $f_{0}(x)=\arctan(x)$, then for each $n$ the function $f_{n}$ is increasing and bounded by $(\pi/2)/4^{n-1}$. Therefore, $f$ is a bounded strictly increasing function.


complex analysis
complex analysis unsolved

Hi would you please help me would these questions  :(

(i) Let [i]D[/i] be the closed unit disk, $n$ $\in$ $\mathbb{N}$ and
define $f_{n}$: [i]D[/i] $\to$ $\mathbb{C}$ by $f_{n}$([i]z[/i]) = $z^{n}$/$n$.
Prove that {$f_{n}$} converges pointwise.

(ii) Let $n$ $\in$ $\mathbb{N}$ and define $f_{n}$ : [0,$\pi$] $\to$
$\mathbb{R}$ by $f_{n}$($x$) = sin $nx$. Prove that {$f_{n}$} does not
converge pointwise.

Thanks

[quote="ueamath"]Hi would you please help me would these questions  :(

(i) Let [i]D[/i] be the closed unit disk, $n$ $\in$ $\mathbb{N}$ and
define $f_{n}$: [i]D[/i] $\to$ $\mathbb{C}$ by $f_{n}$([i]z[/i]) = $z^{n}$/$n$.
Prove that {$f_{n}$} converges pointwise.

(ii) Let $n$ $\in$ $\mathbb{N}$ and define $f_{n}$ : [0,$\pi$] $\to$
$\mathbb{R}$ by $f_{n}$($x$) = sin $nx$. Prove that {$f_{n}$} does not
converge pointwise.

ii) if $x=\frac{\pi}{2}$ the limit does not exists.


logarithms
quadratics
algebra
system of equations

The solutions to the system of equations

\begin{eqnarray*} \log_{225}{x}+\log_{64}{y} &=& 4\\ \log_x{225}-\log_y{64} &=& 1 \end{eqnarray*}

are $(x_1,y_1)$ and $(x_2, y_2).$ Find $\log_{30}{(x_1y_1x_2y_2)}.$

[hide="Solution"]
Let $a=\log_{225}{x}$ and $b=\log_{64}{y}.$  We know that $\log_x{225}=\frac 1{\log_{225}{x}}=\frac 1a$ and $\log_y{64}=\frac 1{\log_{64}{y}}=\frac 1b.$  The system is thus

\begin{eqnarray*} a+b &=& 4\\ \frac 1a-\frac 1b &=& 1. \end{eqnarray*}

We can solve for $a$ in the first equation, substitute in into the second equation, and get $b$ to be $3\pm\sqrt{5}$.  Using the same procedure again, we find that $a$ is $1\pm\sqrt{5}.$  Therefore $x_1y_1x_2y_2=\left(225^{3+\sqrt{5}}\right)\left(64^{1+\sqrt{5}}\right)\left(225^{3-\sqrt{5}}\right)\left(64^{1-\sqrt{5}}\right)=30^{12}.$  Therefore our answer is $\log_{30}{30^{12}}=\boxed{12}.$[/hide]

You accidentally wrote $\log_30 30^{12}$ instead of $\log_{30} 30^{12}$.

There must be a more elegant way... when I first glanced at the problem, your way came to mind but it seemed too complex...

Use someting along these lines to find roots quickly:

$a^2 - 4a - 1 = 0$
$r_1*r_2 = -1$

its a stab at a better solution...

Uhhh...joml88's solution is much simpler, elegant, nice, cool, awesome, better-looking, shorter, clearer, better, and better than your feeble attempts at solutions. :D Seriously though, his solution is very simple. What else would you want? :?

Joml's is a fairly straightforward solution; there's no reason to shy away from a quadratic just because it gives you irrational exponents.

Just label the terms $a$ and $b$ for the first, and $\frac{1}{a}$ and $\frac{1}{b}$ for the bottom one, it becomes trivial from there, the answer simplifies to $\boxed{12}$

Edit: Looking at Joml88s solution now, I see that's exactly what he did :P

[quote="Iversonfan2005"][quote="arkmastermind"]lol, no reason to even complain about the quadratic in joe's solution - you dont have to solve the quadratic, all you need is the sum of the roots  :)[/quote]

oh yeah, i missed that too when I did it..oh well, i always liked to see everything that's going on with the roots, less careless mistakes that way[/quote]

really? seems like theres more room for mistake that way..  :P

Nice log practice. Denote $\log_{225}{x} = a, \log_{64}{y} = b$. Then $a+b=4, \frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}=1$, so $b-a = ab$, or $a(b+1) = b \implies a = \frac{b}{b+1}$. We can now solve to get $b = 1 \pm \sqrt5, a = 3 \pm \sqrt5$, for an answer of $\left(225^{3+\sqrt{5}}\right)\left(64^{1+\sqrt{5}}\right)\left(225^{3-\sqrt{5}}\right)\left(64^{1-\sqrt{5}}\right)=30^{12} \implies \boxed{12}$.


function
algebra
polynomial
algebra unsolved

Find all the functions $f: R \rightarrow R$ such that:

$f(x.f(x)+f(y))=[f(x)]^{2}+y$

[hide]
Suppose $x=0$.  Then we get \[f(f(y)) = [f(0)]^{2}+y\] This tells us that $f(f(x))$ is linear; thus there is some $x$ such that $f(f(x))=0$.  Let that number be $r$.  So letting $x=f(r)$, we get \[f(f(y)) = y.\] Replacing $x$ with $f(x)$ in the given equation, we get \[f(f(x)\cdot f(f(x))+f(y)) = f(x\cdot f(x)+f(y)) = [f(f(x))]^{2}+y = x^{2}+y\] However, the original equation says $f(x\cdot f(x)+f(y)) = [f(x)]^{2}+y$ and so we have $[f(x)]^{2}+y = x^{2}+y \implies f(x) = \pm x$.[/hide]

[quote="lordWings"]About the last solution, just one of my usual comments.
[hide="Comment"]$[f(x)]^{2}=x^{2}$ $\to$ $f(x)=\pm x$.
It's $f(x)=\pm x$ for each separate $x$.
That doesn't prove that $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$.

There could still be solutions in this form:
$f(x)=x$, if $x\in S$,
$f(x)=-x$, if $x\not\in S$,
for a certain set $S\subset\mathbb{R}$.

There's still some work to be done to decide if there are any sets $S$, other than the trivial $S\equiv\mathbb{R}$ or $S\equiv\emptyset$, that may give solutions to the functional equation.
[/hide][/quote]

Ah, I [i]knew[/i] my proof was missing something! :)  But that's quite easy to prove:

[hide]
Using your definition of $S$:

We have that $f(0) = 0$ and $f(0) =-0$, so we will disregard that case.  Suppose $x \in S$ and $y \not\in S$ and neither is equal to $0$.    We have \[f(x^{2}-y) = x^{2}+y\] Suppose that $x^{2}-y$ is in $S$.  Then we have \[x^{2}-y = x^{2}+y \implies y=0\] which is a contradiction.  Similarly, if $x^{2}-y$ is not in $S$, we get \[y-x^{2}= y+x^{2}\implies x=0\] which is a contradiction.  So we either have $x=0$ or $y=0$.  However, since the equality $f(x^{2}-y) = x^{2}+y$ must work for ALL $x \in S$ and $y \not\in S$, that means that either $S = \{ 0 \}$ or $\mathbb{R}\backslash S = \{ 0 \}$.  However, we don't need to make an exception for $S = \{ 0 \}$ or $\mathbb{R}\backslash S = \{ 0 \}$ since $0=-0$.  This means that either $f(x) = x$ for all x or $f(x) =-x$ for all x and we're done.

... I hope.... :)
[/hide][/hide]

Yes, it's usually easy, but you must remember that the proof isn't complete until you've checked that.
I've seen this situation several times in these forums.

[quote="nr1337"]
$f(f(y)) = [f(0)]^{2}+y$This tells us that $f(f(x))$ is linear[/quote]

[url=http://en.wikipedia.org/wiki/Linear]Wikipedia[/url] admits the two definitions of linear function:
- function with additivity $f(x+y)=f(x)+f(y)$ and homegeneity $f(kx)=kf(x)$,
- polynomial of degree $1$, because the graph of this function is a line; if you prefer, you can call it an "affine" function instead of linear.


calculus
integration
geometry
geometric transformation
rotation
calculus computations

For the positive constant number $ a$, let $ D$ be the part surrounded by the curve $ \sqrt{x}\plus{}\sqrt{y}\equal{}\sqrt{a}$ and  the line $ x\plus{}y\equal{}a$.

(1) Draw the outline of $ D$ and find the area.

(2) Find the volume of the solid by rotating $ D$ about the line $ x\plus{}y\equal{}a$ as the rotation axis.

[quote="atomicwedgie"]1.
[hide]
We employ a rotation of $ \pi/4$ radians counterclockwise about the origin:

$ (u,v) \equal{} \left(\frac {x \minus{} y}{\sqrt {2}}, \frac {x \plus{} y}{\sqrt {2}}\right)$.

Note that

$ a \equal{} (\sqrt {x} \plus{} \sqrt {y})^2 \equal{} x \plus{} 2\sqrt {xy} \plus{} y$

$ 4xy \equal{} (x \plus{} y \minus{} a)^2$

$ 2a(x \plus{} y) \equal{} (x \minus{} y)^2 \plus{} a^2$,

for $ 0 \le x, y \le a$.  Under the transformation, this becomes

$ 2av\sqrt {2} \equal{} 2u^2 \plus{} a^2$,

or

$ v \equal{} \frac {u^2}{a\sqrt {2}} \plus{} \frac {a}{2\sqrt {2}}$,

which describes an arc of a parabola defined over the interval $ \minus{} a/\sqrt {2} \le u \le a/\sqrt {2}$.  Since the line is transformed to the segment $ v \equal{} a/\sqrt {2}$, the area of the enclosed region is simply

$ |D| \equal{} 2 \int_{u \equal{} 0}^{a/\sqrt {2}} \frac {a}{\sqrt {2}} \minus{} v(u) \, du \equal{} 2 \int_{u \equal{} 0}^{a/\sqrt {2}} \frac {a}{2\sqrt {2}} \minus{} \frac {u^2}{a\sqrt {2}} \, du \equal{} \frac {a^2}{3}$.

We could have done this more easily in the xy-plane, but the volume calculation would still require a rotation.  We now immediately find

$ |V| \equal{} 2 \int_{u \equal{} 0}^{a/\sqrt {2}} \pi \left(\frac {a}{2\sqrt {2}} \minus{} \frac {u^2}{a\sqrt {2}}\right)^2 \, du \equal{} 2\pi \int_{u \equal{} 0}^{a/\sqrt {2}} \frac {a^2}{8} \minus{} \frac {u^2}{2} \plus{} \frac {u^4}{2a^2} \, du \equal{} \frac {\pi \sqrt {2}}{15} a^3$.
[/hide][/quote]

That's the correct answer. :)


[b]A water pipe can fill a 250 liter tank in  2 hours 30 minutes .In how much time will it fill 9 /10 of the tank ?[/b]

We know $ 2$ hours $ 30$ minutes is equal to $ 150$ minutes. Since we want $ \frac {9}{10}$ of that, we multiply $ 150$ by $ \frac {9}{10}$ to get $ \boxed{135}$ minutes, or $ \boxed{2\text{ hours }15 \text{ minutes}}$.

[size=67][color=red]AIME15: Fixed your TeX.[/color][/size]

Or, you could use ratios, where you set the ratio of time to volume, and put $ x$ for the desired time:

$ \frac{150 \text{ min}}{250 \text{ L}}\equal{}\frac{x \text{ min}}{225 \text{ L}}$

Solving for $ x$, we get the same answer as \$LaTeX\$.


inequalities
calculus
derivative
trigonometry
geometry
circumcircle
geometry proposed

Let $ ABC$ be a triangle. Prove that $ \boxed {\ h_a \plus{} \max\ \{b,c\}\ \le\ 4R\cos^3\frac A3\ }$

where $ h_a$ is the length of the $ A$-altitude and $ R$ is the circumradius of $ \triangle ABC$ .

The trigonometrical form of this inequality is : $ \boxed {\ \cos\ (B\minus{}C)\plus{}2\max\ \{\sin B,\sin C\}\ \le\ 3\cos\frac A3\ }$ .

[b]Partcular cases.[/b]

$ 1\blacktriangleright\ A \equal{} 90^{\circ}\ \implies\ h_a \plus{} \max\ \{b,c\}\ \le\ \frac {3a\sqrt 3}{4}$ .

$ 2\blacktriangleright\ A \equal{} 135^{\circ}\ \implies\ h_a \plus{} \max\ \{b,c\}\ < \ a$ .

[u][b]Sorry, sorry[/b][/u] ! I lost a condition - "$a\ge b\ \wedge\ a\ge c$".

[b][u]Remark[/u].[/b] $\ h_a \plus{} \max\ \{b,c\}=4R\cos^3\frac A3\ \Longleftrightarrow$ $2B=90^{\circ}+C$ and in this case the rays
$[AH$ and $[AO$ trisect angle $\widehat {BAC}$ , where $H$ - orthocenter and $O$ - circumcenter of $\triangle ABC$ .


How many 5 digit integers begin with 3 and end with 5, or have 7 as a digit?

The answer I get is different from the claimed answer:
[hide]
Number of integers beginning with 3 and ending with 5
$10^{3}=1000$

Number of integers containing at least one 7
$9000-8\cdot9^{4}=37512$

Both starting with 3, ending with 5, and containg at least one 7:
$10^{3}-9^{3}=271$

$1000+37512-271=38241$

The book claims the answer is:38212 :oops:
[/hide]

Anyone else?

Are there really 38212 possible integers when there are only 90,000 five digit integers. makes no sense.

my mistake: it says "or" not "and"


let $ X \equal{}${ set of all bijective functions from $ {1,,..,n}$ to $ {1,,..,n}$} let $ f \in x$
we define $ t_{f}(a) \equal{}$ $ 1$ if $ f^{12}(a) \equal{}$  $ a$ else $ \equal{} 0$

find $ \sum_{f \in X}^{} \sum_{i \equal{} 1}^{n}t_{f}(i)$

:D i did it  :!:
can anybody answer me now :?:  :)  :)

[quote="SOUMYASHANT NAYAK"]@gaurav

You seem d.a.m.n serious abt getting into the team... :P
I mean...I can see it from your ardent desire to get hold of the IMO shortlist - 2007...
 :rotfl:

Just chill guy... Dont overwork yourself.... :rotfl:
[quote]can anybody answer me now[/quote]
Yes that question can be answered now... :P[/quote]

Why are you bullying and teasing little kids da? Bad boy... You should be hanged in Andaman.


Let $ a_1\equal{}2, a_2\equal{}5$ and $ a_{n\plus{}2}\equal{}(2\minus{}n^2)a_{n\plus{}1}\plus{} (2\plus{}n^2)a_n$ for $ n\geq 1$. Do there exist $ p,q,r$ so that $ a_pa_q \equal{}a_r$?

[hide="My idea"]Prove that $ a_n\equiv 2(mod 3)$[/hide]


geometry
incenter
circumcircle
trigonometry
geometry proposed

Let $ ABC$ be a non-equilateral triangle. Denote by $ I$ the incenter and by $ O$ the circumcenter of the triangle $ ABC$. Prove that $ \angle AIO\leq\frac{\pi}{2}$ holds if and only if $ 2\cdot BC\leq AB\plus{}AC$.

It is only calculation
\[ \angle AIO\le\frac{\pi}{2}\Leftrightarrow\cos AIO\ge 0\Leftrightarrow IA^{2}+IO^{2}-OA^{2}\le 0\Leftrightarrow \frac{r^{2}}{\sin^{2}\frac{A}{2}}+R^{2}-2Rr-R^{2}\ge 0\Leftrightarrow \sin^{2}\frac{A}{2}\le \frac{r}{2R}\Leftrightarrow \frac{(s-b)(s-c)}{bc}\le \frac{2S^{2}}{sabc}=\frac{2s(s-a)(s-b)(s-c)}{sabc}\Leftrightarrow 2(s-a)\ge a\Leftrightarrow b+c\ge 2a \]
here I used $ OI^{2}=R^{2}-2Rr$ and $ IA=\frac{r}{\sin\frac{A}{2}}$

Some obvious observation: if we have triangle $ ABC$ with $ |AB|=|AC|$ and point $ D$ on side $ BC$ then $ BD \leq CD$ is equivalent to $ \angle BDA \geq 90^\circ$.
Let the $ M$ be the midpoint of arc $ BC$ not containing $ A$. Then $ A, I, M$ are collinear, the triangle $ AOM$ is isosceles and from the Ptolemy Theorem we have: $ |AM\parallel BC|=|MI|(|AB|+|AC|)$ (since it is well known that $ |MI|=|MB|=|MC|$). Now $ \angle AIO \leq 90^\circ$ is equivalent to $ |MI| \leq |AI|$ or $ |AM| \geq 2 |MI|$ or $ |MI|(|AB|+|AC|) = |AM\parallel BC| \geq 2|MI\parallel BC|$ or $ |AB|+|AC| \geq 2 |BC|$.


geometry trapezoid rotation parallelogram

Suppose that $ABCD$ is a trapezoid in which $\overline{AD}\parallel\overline{BC}$. Given $\overline{AC}\perp\overline{CD}$, $\overline{AC}$ bisects angle $\angle BAD$, and $area(ABCD)=42$, then compute $area(ACD)$.

[hide="My solution"]
Rotate $D$ about $A$ by $2\angle CAD$ degrees to get $D'$, which lies on line $AB$. Notice that $\triangle ACD' \cong \triangle ACD$, so $D'C = DC$. Clearly $\triangle BD'C\sim\triangle AD'D$, so $\frac{AB}{AD'}= \frac{D'C}{D'D}= \frac{1}{2}= \frac{AB}{AD}$. Writing $area(ACD)$ and $area(ABC)$ as $\frac{1}{2}\sin(CAD)(AC)(AD)$ and $\frac{1}{2}\sin(CAB)(AB)(AC)$, respectively, we find that $area(CAD) = 2area(BAC)$ so our answer is 28.
[/hide]

ooh i remember doing this question

well in these types of problems, it always works out so that no matter which case you use, you'll get the same answer, so i made a right trapezoid with $\angle ADC$ 45 degrees and $\angle DBC$ 135 degrees

it worked out nicely, and the area of triangle ACD turned out to be yeah, what you got, 28.

[hide]
We first have $\angle BCA=\angle CAD=\angle BAC$, so $AB=BC$.

Let $E$ be the point such that $BCEA$ is a parallelogram (then $E$ lies on $\overline{AD}$). Then $AE=BC$ and $EC=AB$, that makes $\angle CAD=\angle ACE$. Also, $\angle ECD=90^\circ-\angle ACE$ and $\angle EDC=90^\circ-\angle CAD$, so $\angle ECD=\angle EDC$ and $EC=ED$. Finally, we have $AE=ED$. The triangles $\triangle ABC$, $\triangle ACE$, $\triangle ECD$ split the trapezoid into three parts of equal area. Then each part has area 14, so $[ACD]=\boxed{28}$.
[/hide]


Currently I am planning to spend a week in Birmingham in July.  If I do, I am willing to work with students preparing for Mu Alpha Theta.  Actually, my intention would be to run a kind of mini-camp for exploration beyond Mu Alpha Theta types of problems.  Students outside of Vestavia would also be invited.

My goal would be to put together a good group of several students willing to work for several hours a day.  I would bring with me problems designed to help students with harder AIME and easier USAMO problems.

I might also help with a couple of the Vestavia Mu Alpha Theta practices.

Anyone who is interested in forming such a group (or participating), please let me know.

[quote]I'd love to attend if I'm going to be in town at the time.[/quote]

Same here.

One thing that Mrs. Tipton mentions but hasn't acted on is for a group to get together after school to practice Mandelbrot/ARML-style team proof rounds.  So this could be a good chance to get that group organized and maybe do some practice.

That kind of work is exactly what I had in mind.

[quote="Cicatriz"]Do you know what hours in the day you would want to do it?[/quote]

I would probably want to start mid-morning, take a break around lunch, and then work until the middle of the afternoon, but I can be flexible if students come up with their own plan.

What time in July are you planning on coming? I'm only able to come if it's in the last week to week and a half of it (the week before MAO, pretty much).

Approximately July 17 to 24, but I haven't made firm plans yet.

That sounds like an awesome opportunity.  But I am going to be out of town 18-22.

I am starting to plan out my time for the trip.  I would like to hear suggestions for a time and place to hold the work sessions.  Are we going to work at VHHS?  For part of each day I might simply have students meet me at the Bean Coffee House and hand out problems/concepts to work on.  Students can work on these anywhere and then we can meet up later in the day to discuss the work.

Does this sound like a good plan?

Mrs. Tipton said the high school would be fine. Your plan sounds good. If we wanted to do anything over a meal, then a lot of us like Moe's by the Rave theater. Any time is good for me, but it seems like most people aren't getting up until 11 or so.

This thread is over two years old.  At that time I spent some time in Birmingham and worked with some of the Vestavia students.


Depends on whether it trivializes the problem or not..

You don't really need to cite anything obscure to solve any of these problems. So it is likely that your solution will be unnecessarily complicated if you cite things.

You should concentrate on writing good solutions rather than looking for excuses.

[quote="ZzZzZzZzZzZz"]Just wondering, will citing a formula or equation that is relatively difficult to prove affect whether or not a solution receives a commended?[/quote]
Commendations are very subjective. One grader will commend a short, crisp solution that skips unnecessary detail, yet another grader will commend a solution that gives complete detail and generalizes the problem.

As for me, I like solutions that cite classic theorems in detail before applying them. Something like:
[quote][b]Ceva's Theorem.[/b] Let $ ABC$ be a given triangle and $ P$ any point of the plane. If $ X$ is the intersection point of $ AP$ with $ BC$, $ Y$ the intersection point of $ BP$ with $ CA$ and $ Z$ is the intersection point of $ CP$ with $ AB$, then
\[ \frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}\equal{}1.\]
(this wording copied from PlanetMath)[/quote]
Note that the above is only half of Ceva's Theorem, presumably the half being used in the solution.

Erin Schram
USAMTS grader

In the past, I've commended both solutions that were really clever and short and also solutions that were very thorough AND generalized the problem.  However, there are variety of graders and you'll never know who's going to grade yours.  It's possible that some graders wont commend anything and that others might commend more than they should.  There's definitely an element of chance involved with commended solutions.

My advice, in bulleted form:

$ \bullet$ Optimize your proof as much as possible (e.g. if you must use casework, find the most efficient way to do it).
$ \bullet$ Explain your logic thoroughly but not verbosely.
$ \bullet$ Make it look nice (use $ \text{\LaTeX}$, adequate whitespace, pictures, tables, diagrams and whatever else might enhance your solution).
$ \bullet$ (optional) If you can find a generalization, put it down after your solution to the main problem.


Where can I find the answers for 2006 AIME?

I really need them.

I'm practicing them, but I don't have the answers...

lol :rotfl:

http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I

and

http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_II


Hello Guys,

I have found a pretty nice solution for the inequality, but it would be nice to see some other ones (without using brute force-bunching...).

Here is the inequality:

Let a,b,c be non-negative reals such that a+b+c=3. Prove that

$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} \geq \frac{3}{2}$

Just wanted to say that this has been also discussed in

http://www.mathlinks.ro/Forum/viewtopic.php?t=13943&highlight=peter

  Darij


calculus
derivative
induction
real analysis
real analysis theorems

is it true that if $f$ and $g$ are derivable $n$ times then $(fg)^{n}=\sum_{k=0}^{(n)}{n \choose k}g^{(k)}f^{(n-k)}$
by induction on n?

oh, it's beautiful when it's true ,but is it true for example when $f(x)=ax^n$ and $g(x)=bx^n$ , i don't know? :huh:


I've a question.
In my area of Germany it is rumoured that if a student participate at the IMO, the student goes to a math-camp for one year. During this year he doesn attend the school.
I belive it's not true. Maybe anybody can tell me, how the way to IMO is in China.

Well, during the training camp, the student is not necessary to go to school.

[quote="Gauß"]Thank you. :)

Some of my mates thought it will take a year and so on.
For instance in Turky the traing camp takes half a year.[/quote]Those are rumours too. Nobody can stay in camp for half a year!! :)

[quote="Gauß"]My professor ( Gronau ) had told us.
Then he is wrong?![/quote]I would say that his informations may be inaccurate. It's best to ask the Turkish IMO people. There are some from their team on the site.


For $x_i=0$ we have equality for any $C \geq 0$. Thus we are interested in $\sum x_i >0$. Since the inequality is homogeneous, we can normalize $\sum x_i=1$. Using that you can rewrite the inequality and solve similarly to IMO 1999/2.

[hide]
We claim that $C = 5^{15}$. It's easy to see that equality holds at $x_1 = x_2 = x_3 = x_4 = x_5$.

Now we want to prove that it is always true for $C = 5^{15}$.

From Chebyshev, we have

$5^{15} \sum x_i^{2005} \ge 5^{14}\left(\sum x_i^5 \right)\left(\sum x_i^{2000} \right)$.

By AM-GM, $\sum x_i^5 \ge 5x_1x_2x_3x_4x_5$.

By Holder, $5^{15}\sum x_i^{2000} \ge \left(\sum x_i^{125}\right)^{16}$.

Multiplying these together, we have

$5^{14}\left(\sum x_i^5 \right)\left(\sum x_i^{2000} \right) \ge x_1x_2x_3x_4x_5 \left(\sum x_i^{125} \right)$

so

$5^{15} \sum x_i^{2005} \ge x_1x_2x_3x_4x_5 \left(\sum x_i^{125} \right)$

as desired. Hence $C = 5^{15}$ works.[/hide]


function
algebra
polynomial
calculus
calculus computations

Let $\mathcal{F}(f)$ denote the fourier transform of $f$.

What is then (somehow simplify) $\mathcal{F}(\mathcal{F}(f))=\mathcal{F}^{2}(f)$

What is $\mathcal{F}^{4}{f}$?

The latter is identity, but how we we prove that and the earlier question above?

Thanks for reply.

The formula for the inverse Fourier transform differs only from the formula for the Fourier transform in having a sign change in the complex exponential. (Incidentally, this problem implicitly has a unitary version of the Fourier transform in which the inverse transform and the transform have the same constant outside the integral.)

Follow through on that and you get that $\mathcal{F}^{2}(f)(x)=f(-x).$

Once you have that, $\mathcal{F}^{4}(f)=f$ follows immediately.

Question: what is the set of all possible eigenvalues of the Fourier transform operator? And can you find at least one eigenfunction for each of those eigenvalues?

(The answer to that question is not unique - all of the eigenspaces are infinite dimensional.)

So there is no need to split the function in even an odd terms?

What is the set of all possible eigenvalues of the Fourier transform operator?

You get more detail about what's going on by splitting into even and odd parts, so it's worth doing. But strictly speaking, you don't have to for this result.

In an algebra of operators, if an operator satisfies a polynomial equation, then any eigenvalue of that operator must also satisfy that polynomial equation as well. The Fourier transform satisfies the polynomial equation $x^{4}-1-0.$ The roots of that equation are $\pm 1,\pm i.$ That's the complete set of possible eigenvalues. Now, are those all actually eigenvalues?

By the way, your quote of my previous post was completely unnecessary. Please don't use the "quote" button unless you have a specific purpose, and use it selectively - to quote portions of posts.


In $\triangle ABC$, $\angle C$ is a right angle. Point $M$ is the midpoint of $\bar{AB}$, point $N$ is the midpoint of $\bar{AC}$, and point $O$ is the midpoint of $\bar{AM}$. The perimeter of $\triangle ABC$ is 112 cm, and $ON=12.5$. What is the number of square centimeters in the area of quadrilateral MNCB?

[hide="Solution"]
$252$ square centimeters :D
Isn't it some kind of theorem that says the median to the hypotenuse is always half the length of the hypotenuse? Well, I figured that out (somehow). So $NO=AO=12.5$, which is one-fourth of the hypotenuse. So, the $AB$ must be 50. Looking at the Pythagorean triples, we can quickly see that 7-24-25 or 14-48-50 works. The area of $ABC=\frac{1}{2}\times14\times48=336$ and the we know the trapezoid $MNCB$ must be $\frac{3}{4}$ the area of $ABC$, so it is $252$. :D[/hide]

[quote="ch1n353ch3s54a1l"]Why is trapezoid MNCB $3/4$ ths the area of ABC?[/quote]

Was my answer correct? $\triangle{AMN}$ is $\frac{1}{4}$ the area of $\triangle{ABC}$ since it is similar and $AN$ is $\frac{1}{2}$ of $AC$. So $\triangle{ABC}-\triangle{AMN}=MNCB \Longrightarrow 1-\frac{1}{4}=\frac{3}{4}$.

Your answer's right, I just couldnt see where the 3/4 came from...thanks


Guys, is there a solution book for irodov.(i want if possible a link for a ebook of solutions of irodov). And also a link for e book of dc pandey.

@tanmoy

search it on google... or esnips.com or better still http://www.4shared.com
when u search on google add .djvu because most books are in this format.....
waise irodov will require quite a bit of calculus.....

there is a solution available in the market name evergreev irodov's solution

i think irodov's sol wont benefit much

doing the problems on ur own is real fun (coz u kno u wont be do it mostly)
thats 4 me only :oops:  :P

same here and thats i didnt buy the solution!

Guys i got a e book solution of irodov by abhay............... :D
just dont know whether all the solutions in this are correct................... :blush:
P.s. anurag i am not tanmoy :rotfl:

oopss but i read somewhere that u mentioned urself as tanmoy bhunia???
anyways sorry for the mistake.... :oops:

the book is quite famous(so solutions got to be right)

but I am not going to buy it,ootherwise i'll end up looking the solutions and won't attempt the questioms seriously :D

he was probably lakshya

he was "fbaggins"

lakshya is lakshya :P

oks... oops!!!! befuddlers and fbagins ???
f & b played the trick  :rotfl:
anyways whats your real name befuddlers??

[quote="anurags92"]anyways whats your real name befuddlers??[/quote]
That's a lot complicated than you think :wink:  :P . Guess why :rotfl:

Aarghh...
Why so??? No idea!!! Bad in guessing.....
Waise till now i was thinking that he was tanmoy and was bit uneasy since he knew so much... hehehehee
now a bit relieved..... lolz.....
Anyways u preparing for this years rmo??
if so how ;)
and with which books  :maybe:

PS - Who is BEFUDDLERS????

[quote="anurags92"]Anyways u preparing for this years rmo??
if so how ;)
and with which books  :maybe: [/quote]
Preparing=will sit for exam
Not studying much theory...
Only doing some problems... (about 4 a week)
Not much preparation... I have board exams ahead...

[quote="anurags92"]PS - Who is BEFUDDLERS????[/quote]
He told me a bit about the issue in pm. Let's keep the suspense and wait for him to disclose ... :roll:
It's nothing much really, so don't imagine too many things :P

How many people do use Irodov SOlution currently and may be looking for more.

Thanks,

Calculus is my worst weakness in physics. Anyway I could learn how to properly use physics in physics questions?  :blush:


I'm writing a formal essay.  How would I quote (from Love's Labour Lost, 1588)

[code]Good Lord Boyet, my beauty, though but mean,
Needs not the painted flourish of your praise:
Beauty is bought by judgement of the eye,
Not utter'd by base sale of chapmen's tongues[/code]

in the document?  Does it need to be centered/indented/italicized/in quotes/etc?  Also, how would I cite Love's Labour Lost, 1588?  Quotations, italics, underlining?

Thanks in advance!

There are many, many different citation formats, so my best advice would be to talk to the person you are writing the essay for and ask what format they prefer.  It's also hard to advise you about how to format the quote-- depending on the paper and the context of the quote, it might be better to work it into the text or block quote it.

How would I block quote it?  Can you (or anyone else) show me an example?  (I'm not going to have a chance to ask the person who I'm submitting this to what they prefer.)

Thanks!

I'd italicize and put in quotes the four lines and italicize or underline the citation.

MLA format is what my school requires, and is pretty common all over the world (the most used, I think). Google it and you will find templates or guidelines for quote integration

I've always been taught to do this:
"Good Lord Boyet, my beauty, though but mean,/Needs not the painted flourish of your praise:/Beauty is bought by judgement of the eye,/Not utter'd by base sale of chapmen's tongues" ([i]Love's Labour's Lost[/i], 1588).


[quote="SplashD"][quote="buzzer11"]Show that for all integers [i]n[/i] greater than $2$:

$1+2+3+...+n$ is a composite number.[/quote]

[hide]If $n$ is an odd integer greater than or equal to 3, then $n$ can be expressed in the form $2k+1$ where $k$ is a positive integer.

$1+2+3+...+n=\frac{n(n+1)}{2}=(2k+1)(k+1)$ and is therefore composite.

If $n$ is an even integer greater than2, then $n$ can be expressed in the form $2k$ where $k$ is an integer such that $k\ge2$

$1+2+3+...+n=\frac{n(n+1)}{2}=k(2k+1)$ and is therefore also composite.[/hide][/quote]

right. amcavoy, please try to expand your proof next time.

is this solvable through induction?

[quote="computer_nuke"][quote="buzzer11"]

is this solvable through induction?[/quote]

It is definitely solvable through induction. First you assume n = 3 and it holds, then you assume that it holds for n, and use that to prove that it holds for n+1 and you are done.[/quote]

demonstrate, please.  :lol:

[quote="computer_nuke"]For $n=3$, we have $1+2+3=6$ which is a composite number, then we assume that for n it holds that the sum $1+2+3+...n$ is a composite number. Now we have $1+2+3+...n=(n)(n+1)/2$ and we must prove that for n+1, it is also a composite number. So we have assumed $(n)(n+1)/2$ is a composite number and proving that $(n)(n+1)/2+(n+1)$ is a composite number. We simplify to get:

$(n)(n+1)/2+(n+1) = (n+1)(n/2+1)=(n+1)(n+2)/2$, which through the same reasoning buzzer11 gave, is a composite number as well, therefore, for all integers $n>2$, $1+2+3+...n$ is a composite number.[/quote]

 :)


function
complex analysis
complex analysis unsolved

Let $f,g$ not zero entire functions s.t. $f(5)=g(5)$ .
If $\frac{f' \left( \frac{1}{n^{2}+1}\right) }{f \left( \frac{1}{n^{2}+1}\right) }= \frac{g' \left( \frac{1}{n^{2}+1}\right) }{g \left( \frac{1}{n^{2}+1}\right) }$ for $n=1,2,...$ show that $f=g$ .

Complete solution please  :)

[quote="julien_santini"]I would like to add the condition that this equality also holds at point $0$ ... in wich case $(\frac{f}{g})'$ would equal $0$, meaning $f=g$.[/quote]

Thanks  :)  . I finally solved it . It was easier than i thought   :oops:

Well, the equality [b]did[/b] hold at $0$ by continuity, so everything's fine ...


calculus
integration
geometry
ratio
function
probability
expected value

Given a triangle $ABC$, pick points $P$, $Q$, and $R$ randomly on sides $a$, $b$, and $c$ respectively. Find the expected value of $\frac{ [PQR] }{ [ABC] }$.

(The more generalized problem of picking $P$, $Q$, and $R$ from anywhere within $\triangle ABC$ can be found [url=http://mathworld.wolfram.com/TriangleTrianglePicking.html]here[/url].  But I think this may be a different problem altogether.)

this uses the exact same integration technique

This question is way, way easier than that one.

[hide="Hint 1"]If you consider the complement, you can use the fact that [hide="Hint 2"]expectation is linear.  Thus, you only need to figure out the expected value of [hide="Hint 3"]the area of triangle $AQR$.  In other words, you just need to know what $E(p\cdot q)$ is while $p$ and $q$ vary uniformly from 0 to 1.  Then, since $p$ and $q$ are independent, $E(p\cdot q) = E(p) \cdot E(q)$.[/hide][/hide][/hide]

i guess i was a bit vague; you find the area in terms of BP/PC,[ say BP=x, etc] and the ratio is:
$x(1-y)+y(1-z)+z(1-x)$ where x,y,z range from 0 to 1 [uniform distribution]

by integration technique i just mean to find the average value of this function

that other problem is not that different, you just need to be careful with the bounds on the integral

Except that one can do this problem without actually taking any integrals, which is not the case in the more general version.

yea...once you have that expression, it is pretty trivial...

you can average each variable independently...

x+y+z-xy-yz-zx
x(1-y-z)+y+z-yz

average of that expression is just when x=1/2, etc for all...

so the answer is 3/2-3/4=3/4

Right, except that that's actually the expected size of the complement.

yea...woops...i was thinking that 3/4 was a bit too large


geometry conics parabola calculus integration

Find the area bound by $ y\equal{}x^2$ and $ y\equal{}x\plus{}2$.

This is easily done without calculus or other methods above HSB level.

How do you find the area without calculus? Maybe you will use Archimedes method, won't you?

You are correct, the solution I know of uses Archimedes' method, and I know of no others besides calculus.

All right, i_like_pie. :)

Let $ x\equal{}\alpha ,\ \beta$ are the intersection points of a parabola $ y\equal{}ax^2$ and the line $ y\equal{}mx\plus{}n$,

the area of the region bounded by these graphs is $ \boxed{\frac{|a|}{6}(\beta \minus{}\alpha )^3}$

Using your formula, $ A \equal{} \frac {(2 \minus{} ( \minus{} 1))^3}{6} \equal{} \frac {27}{6} \equal{} \frac {9}{2}$. Using Archimedes' method, I get the same result. :)

Unfortunately, my humble attempts at proving that formula failed miserably, so would you please post a proof of it?

I'm afraid the calculus solution would be quite beyond HSB level, so I will not post it here.

I'm probably not using the fastest and easiest way because I'm not too skilled in calculus, but I integrated the area under the parabola from $ \minus{} 1$ to $ 2$, and subtracted that from the area below the line $ y \equal{} x \plus{} 2$, which I found without calculus.

Now although the process is rather simple in this case, I'm still not sure we're allowed to discuss this in HSB.

I see. You can find the area in using integration, but we can't argue in this thread.
I will post the proof of the formula in another thread later.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=170850]Take a look![/url]


Hi all
i think all of u must be aware of the [url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]AoPS-Wiki[/url].As u can see over there the [url=http://www.artofproblemsolving.com/Wiki/index.php/Physics]Phsyics[/url] deserves lots and lots of contribution from u all .SO it's a sincere request to all to contribute as many articles as u can :)

Thanks
Pardesi
Physics Forums Moderator

I've been considering this for quite some time. My problem is, how do we present a problem-solving based introduction to Physics? In ([b]elementary[/b]) Physics, the relation between 'theory' and 'problem-solving' is not as intimate as it is with mathematics. Further, we are not allowed to follow Wikipedia in either content or style.

Yes that will be good...in fact it is quite nice to analyse problems and ;earn physics than 'mugging up' things...
I think we shud do what some books do i.e. write the definition of something and then post ore or two problems of three stages of difficulty  that use the concept...
i posted something on momentum ...i think i will add three problems of varied difficulty to it...
what's ur take :?:

WHat is AOPSWiki???????? :maybe:  :maybe:

it is a concept similar to wikipedia where u have to contribute articles..u can edit,add anything u want but ofc relevant.
[url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]here[/url] is the link :)

why is the physics wiki so empty.  :(

I edited a few pages, but...there is so much more work to do.   :|


Ok I got this logical puzzle from a friend and I'm stuck!

I got a clue and five dices.
The point is to figure out the correct amount of bears & waterholes.

I got this clue first:
[b]Polarbears are social animals that like to gather around waterholes and drink[/b]

Then I got to hit my five dices, and ended up with these numbers:
[b]2 2 5 1 3[/b] the she looked at my dices and told me that the sollution to it was that there are [b]6 Bears[/b] & [b]2 Waterholes[/b].
This was the example.

Now I made another toss with the dices, and got [b]1 5 1 3 2[/b].
Now I have to figure out how many bears & waterholes there is, and I'm dead stuck.

Can anyone see the logical way of getting the amount of bears & waterholes?
Cause I sure can't...

[quote="laughinghead505"][hide="Im not sure but"] I think waterholes are the dots that would be found in the middle of the die and the bears are the dots around the waterholes. Hence there would again be 6 bears and 2 waterholes. Again, I'm not sure... there could be other soutions that work. You should ask your friend to tell you  one more time so you can test some of them out...[/hide][/quote]
Wouldn't the 1 also be counted as a water hole?

[quote="laughinghead505"][hide="Im not sure but"] I think waterholes are the dots that would be found in the middle of the die and the bears are the dots around the waterholes. Hence there would again be 6 bears and 2 waterholes. Again, I'm not sure... there could be other soutions that work. You should ask your friend to tell you  one more time so you can test some of them out...[/hide][/quote]

And I'm not sure how 2 would work for that. Both watering holes?

This is a toughie.  :lol:

Bears a social, so if u have only two dots on of which is a waterhole, there is one bear around it, but he is social so there must be another bear so two does not increase the number of bears and waterholes.
Also, if you have one dot, and that is a waterhole, there must be at least one bear near it, so one does not increase the number of bears and waterholes.
I agree with laughinghead's answer.

[quote="laughinghead505"][hide="Im not sure but"] I think waterholes are the dots that would be found in the middle of the die and the bears are the dots around the waterholes. Hence there would again be 6 bears and 2 waterholes. Again, I'm not sure... there could be other soutions that work. You should ask your friend to tell you  one more time so you can test some of them out...[/hide][/quote]

Don't know how you became so sharp, but thanks alot :P

This game is similar to [url=http://en.wikipedia.org/wiki/Petals_Around_the_Rose]petals around the rose[/url]

I think that they are the same game, just with different names.

yes but petal around the rose is more... elegant. You do not have to say that you give a clue, it's in the name!!!


complex numbers
algebra proposed
algebra

let $p(x) = x^5 + x$, $q(x) = x^5 + x^2$.
 Find all pairs $(w,z)$ of complex numbers with $w\neq z$ such that $p(w) = p(z)$ and $q(w) = q(z)$.


vector
inequalities proposed
inequalities

prove that $abc(ab+ac+bc)\leq a^3b^2+b^3c^2+c^3a^2$ for all $a,b,c>0$.

a=0,b=c=1
I guess you have reversed the sign...

[quote="dule_00"]prove that $abc(ab+ac+bc)\geq a^3b^2+b^3c^2+c^3a^2$ for all $a,b,c>0$.[/quote]
$a^3b^2+ab^2c^2+a^2b^2c \geq 3 a^2b^2c$

Davidlam, elegant solution indeed. I'm interested in more solutions. I know next one:

Aplaying  Cauchy-Schwarz on vectors $({\sqrt{a} \over c},{\sqrt{b}\over a},{\sqrt{c}\over b)}$ and ${1 \over \sqrt{a}},{1 \over \sqrt{b} },{1 \over \sqrt{c}}$ we get $({a \over c^2}+{b \over a^2}+{c \over b^2})({1 \over a}+{1 \over b}+{1 \over c})\geq ({1 \over c}+{1 \over a}+{1\over b})^2$. After dividng by ${1 \over a}+{1 \over b}+{1 \over c}$ we have ${a \over c^2}+{b \over a^2}+{c \over b^2}\geq {1 \over a}+{1 \over b}+{1 \over c}$ which is desired equality after multiplaying both sides with $a^2b^2c^2$.

sign $=$ ocurs when ${a \over c}={b \over a}={c \over b} \leftrightarrow a=b=c$

[quote="dule_00"] I'm interested in more solutions.[/quote]

$abc(ab+ac+bc)\leq a^3b^2+b^3c^2+c^3a^2\Leftrightarrow\sum_{cyc}7a^3b^2\geq7\cdot\sum_{cyc}a^2b^2c.$
But $\sum_{cyc}7a^3b^2=\sum_{cyc}(4a^3b^2+2b^3c^2+c^3a^2)\geq7\cdot\sum_{cyc}\sqrt[7]{a^{12+2}b^{8+6}c^{4+3}}=7\cdot\sum_{cyc}a^2b^2c.$ :)


geometry
3D geometry
sphere
geometry open

Hi Guys can anyone help me with this:

A sphere which has a diameter of 100mm is to have attached to it 70 spikes with a base diameter of 6mm.  I need to work out how far apart each can be spaced if they are to be evenly spread out accross the sphere in any direction.

Any clues?

I've gone through volume surface area equations but the best figure I have reached is 20.46mm between each spike.

cheers

Hint:

Think of how many traingles can be made putting 70 points equally distanced in a sphere.


algebra
binomial theorem
combinatorics proposed
combinatorics

Consider the set $ E\equal{}${$ 1,2,3,...,2002,2003$}.For each subset $ A\equal{}${$ a_1,a_2,...,a_k$} of $ S$($ k$≥$ 1$),we let $ r(A)\equal{}a_1\minus{}a_2\plus{}a_3\minus{}a_4\plus{}...\plus{}(\minus{}1)^{k\minus{}1}a_k$,where $ a_1>a_2>a_3>...>a_k$.Find the value of $ S\equal{}\sum_{A\subset\ E}r(A)$

[quote="JBL"]Every element $ k < 2003$ occurs exactly as many times with a plus sign as a minus sign ($ \sum_{j \text{ even}} \binom{2003 \minus{} k}{j} \equal{} \sum_{j \text{ odd}}\binom{2003 \minus{} k}{j}$), while 2003 occurs always with a plus sign.  And 2003 occurs in exactly $ 2^{2002}$ subsets.  So the answer is $ 2003 \cdot 2^{2002}$.[/quote]
I don't understand.For example,$ 1$ is always occurs with a plus sign because it is the smallest element.So why do we have ($ \sum_{j \text{ even}} \binom{2003 \minus{} k}{j} \equal{} \sum_{j \text{ odd}}\binom{2003 \minus{} k}{j}$) ?

[quote="JBL"]The way you wrote the question, 1 occurs with a plus sign for the subset $ \{1\}$ but with a minus sign for the subset $ \{1, 2\}$.  For each subset of $ \{2, \ldots, 2003\}$ with an even number of elements, 1 occurs with a plus sign, and for each subset with an odd number of elements, 1 occurs with a minus sign.  The equality I asserted is a standard combinatorial result (you can get it by expanding $ (1 \minus{} 1)^{2003 \minus{} k} \equal{} 0$ with the binomial theorem, for example), and together these show that the total contribution of 1 to the sum is 0.[/quote]
Thanks a lot.May be the way I wrote this problem isn't very clear,so you don't understand the problem.
For example
If $ A\equal{}${$ 3,5,9,65$},then $ r(A)\equal{}65\minus{}9\plus{}5\minus{}3$
If $ A\equal{}${$ 1,3,9,6$},then $ r(A)\equal{}9\minus{}6\plus{}3\minus{}1$
If $ A\equal{}${$ 2,3,...,2003$},then $ r(A)\equal{}2\minus{}3\plus{}4\minus{}5\plus{}6\minus{}7\plus{}...\plus{}2002\minus{}2003$ and $ 1$ is not occur in the expression $ r(A)$ in this case.
So if $ A\equal{}${$ a_1,a_2,...,a_k$},then $ r(A)\equal{}a_1\minus{}a_2\plus{}a_3\minus{}a_4\plus{}...\plus{}(\minus{}1)^{k\minus{}1}a_k$ where $ a_i>a_{i\plus{}1}$(the sequence $ (a_i)$ is arrangeed in descending order)

You keep switching back and forth between ascending and descending order: in your initial post, you have descending, but in your first reply to me you write as if you meant ascending.  In your second reply, you have descending three times and ascending once.

I understand the problem and have solved it, modulo the fact that you can't decide between ascending and descending order.  If you still don't understand my solution (as further explained in my second post), ask some question about it and I'll be happy to explain further.

I am really sorry for my mistake.In fact, I mean the descending order.I really want to see the solution.I won't make anything wrong anymore.Thank you very much.

Small errors happen -- don't worry about it :)

Perhaps you could explain what part of my solution (given in post 2, and elaborated on in post 4) is unclear to you?

Edit: Okay, another forum member asked me via PM to explain further, so here is a duplicate of the same solution but with all the details in one place.

First, it's clear that 2003 appears in $ 2^{2002}$ different subsets (once for every subset of $ \{1, \ldots, 2002\}$), and in each subset it appears with a positive sign, so its total contribution to our answer is $ 2003 \cdot 2^{2002}$.

Now choose any $ k < 2003$.  We have that $ k$ appears in a subset with positive sign if and only if there are an even number of elements in the subset larger than $ k$.  (It doesn't matter what elements smaller than $ k$ appear in the subset.)  There are $ k \minus{} 1$ elements smaller than $ k$ and $ 2003 \minus{} k$ elements larger than $ k$, so $ k$ appears with a positive sign in exactly $ 2^{k \minus{} 1} \cdot \left(\binom{2003 \minus{} k}{0} \plus{} \binom{2003 \minus{} k}{2} \plus{} \ldots\right) \equal{} 2^{k \minus{} 1} \sum_{j \text{ even}} \binom{2003 \minus{} k}{j}$ subsets.  Similarly, $ k$ appears with a negative sign in exactly $ 2^{k \minus{} 1} \cdot \left(\binom{2003 \minus{} k}{1} \plus{} \binom{2003 \minus{} k}{3} \plus{} \ldots\right) \equal{} 2^{k \minus{} 1} \sum_{j \text{ odd}} \binom{2003 \minus{} k}{j}$ subsets.

Now it's a standard result that for $ n > 0$, $ \sum_{j \text{ odd}} \binom{n}{j} \equal{} \sum_{j \text{ even}} \binom{n}{j} \equal{} 2^{n \minus{} 1}$ (and I suggested that one way of proving this is expanding $ (1 \minus{} 1)^n$ using the binomial theorem and separating the positive and negative terms).  Thus in fact $ k$ appears in subsets with a positive sign exactly the same number of times as it appears in subsets with a negative sign, so its total contribution to our answer is just 0.  This is true for every $ k < 2003$, and so the final answer is the same $ 2003 \cdot 2^{2002}$.  (In general, for any finite nonempty $ S \subset \bf R$, if we perform this process we get exactly $ (\text{largest element of } S) \cdot 2^{|S| \minus{} 1}$ as the total sum.)

By the way, here's a natural "$ q$-analogue" of this question:  suppose that for the subset $ A \equal{} \{a_0, a_1, \ldots, a_k\}$ with $ a_0 > a_1 > \ldots > a_k$, we compute the sum $ r_q(A) \equal{} a_0 \plus{} q a_1 \plus{} q^2 a_2 \plus{} \ldots \plus{} q^k a_k$, and then we sum this over all subsets.  What is the resulting value $ \sum_{A \subseteq S} r_q(A)$?  (Note that we recover our original problem by setting $ q \equal{} \minus{}1$.)  It makes the most sense to give the answer for an arbitrary finite $ S \subset \bf R$.

We can also divide all the subsets of $ E$ into two parts.$ A$ is the union of subsets which contain$ \{2003\}$,$ B$ is the union of subsets don't contain$ \{2003\}$
There is a bijection $ f$ from $ A$ to $ B$ while $ f(S)\equal{}S\cup{2003}$($ S \in A$)
Easily to show $ f(S)\plus{}S\equal{}2003$
So the answer is $ 2003*2^{2002}$
p.s.By convention,we denote $ r(\emptyset)\equal{}0$


quadratics
number theory unsolved
number theory

Hi, friends! I'm posting MMO 2005 (9th grade questions)...

    9th Grade

   [u][b]Problem1:[/b][/u]   The discrimnants of 3-quadratic equations with non-zero coefficient are 1,4 and 9.  Prove that it's possible to take one of the roots of equestion such that the sum of them is equal to the sum of the rest of the roots.

   [u][b]Problem2: [/b][/u]  Are there 2005-distinct natural numbers such that the sum of any-2004 of  them can be divisible by the rest number.

   [u][b]Problem3: [/b][/u]  Circle O_1 passes through the center of the circle O_2. Two tangent lines drawn from a point C on the cicle O_1 to the circle O_2 cut the circle O_1 at the points A and B. Prove that AB is perpendicular to the line passes through the centers of the circles.

   [u][b]Problem4: [/b][/u] Is it possible to make a square from 1000-pieces obtained by breaking into pieces a triangle.(any triangle)

   [u][b]Problem5: [/b][/u] Given a point P (not intersection of altitudes) in an acute triangle ABC. Prove that the circles pass through the midpoints of the triangles PAB, PAC, PBC, ABC and the circle passes through the projections of O to the sides intersect at one point.

As always this problems are far from originality. :(


Who took the "log 1" and "ciphering time trials"?

I took them.  I didn't do nearly as well as I would have like or even as I should have.  I got in the 30's for the log1 and I got exactly 60 on the ciphering.  I made a bunch of mistaked on the ciphering...but that is why it is timed.

i got a 54 on ciphering... many careless mistakes... but as said above, it was a time trial... got 2nd highest score in my school

I got a 27 on the log1 and a 50 on the ciphering...I thought I did horrible on both, I wasn't thinking those days.


(page 30)

I was doing the algebra to derive the Stewart's Theorem for angle bisectors and here's my work.

[hide]
$ dad \plus{} man \equal{} cnc \plus{} bmb$
$ d^2 \equal{} \frac {cnc \plus{} bmb \minus{} man}{a}$
$ d^2 \equal{} \frac {cnc}{a} \plus{} \frac {bmb}{a} \minus{} \frac {man}{a}$
$ d^2 \equal{} \frac {bc^2}{b \plus{} c} \plus{} \frac {b^2 c}{b \plus{} c} \minus{} mn$
$ d^2 \equal{} \frac {bc(b \plus{} c)}{b \plus{} c} \minus{} mn$
$ d^2 \equal{} bc \minus{} mn$
$ d \equal{} /sqrt {bc \minus{} mn}$
[/hide]
The solution manual goes on and gets a completely different form but I'm pretty sure that this form is correct or am I missing something?


[quote="pvthuan"]Prove that if $v,a,s,c$ are positive real numbers, then
\[ (v^2+a^2+s^2+c^2)^2\geq4(v^3a+a^3s+s^3c+c^3v).  \][/quote]
Try $v=3,a=2,s=c=1.$ ;)

[quote="pvthuan"]Prove that if $v,a,s,c$ are positive real numbers, then
\[ (v^2+a^2+s^2+c^2)^2\geq4(v^3a+a^3s+s^3c+c^3v).  \][/quote]

   Nice imagination, Thuan. All of terms spells " Vasc" and have the number "4".
   I spend some hours to SCP this one and go no where. I think you should check carefully your problem before posting. It just waste other people time.

What a pity, I strongly believed that it's true since we have
\[ (a+b+c+d)^2\geq4(ab+bc+cd+da).  \]
I just want to have a stronger result as what Vasc did with
\[ (x+y+z)^2\geq3(xy+yz+zx)  \]
to have
\[ (x^2+y^2+z^2)^2\geq 3(x^3y+y^3z+z^3x).  \]
Unfortunately, it is not true. I am sorry for its incorrectness. Any way, it is nice enough, isn't it?


Find all polynomials $p_n(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ ($n\geq 2$) with real and non-zero coeficients s.t. $p_n(x)-p_1(x)p_2(x)...p_{n-1}(x)$ be a constant polynomial. ;)

Well if all the coefficients are non-zero, then the degree of $p_n(x)$ is $n$, whilst the degree of $\prod_{i=1}^{n-1}p_i(x)$ is $(n-1)!$.  For the difference of these two to be constant, we need $n=(n-1)!$, which holds only for $n=1$.  Therefore there are no solutions for $n\ge2$.

Right?

[quote="Diarmuid"]whilst the degree of $\prod_{i=1}^{n-1}p_i(x)$ is $(n-1)!$[/quote]
No, the degree of $\prod_{i=1}^{n-1}p_i(x)$ is $1+2+\ldots +(n-1)$, which implies $n=3$. A short computation shows that $P(x)=2a^3x^3+2a^2x^2+ax+\frac 12$ for some $a \in \mathbb{R}^*$.

Oops...  :blush:

Sorry, I make a lot of mistakes when I'm tired...


What is the sum of all positive two digit integers with exactly 12 positive factors?

[hide]For a number to have 12 divisors, it must be in one of 3 forms: $p_1^{11}$, $p_1^5\cdot p_2^1$, $p_1^1\cdot p_2^1\cdot p_3^2$, or $p_1^3\cdot p_2^2$.  The first one won't create any 2-digit numbers.  Now, write out all the 2-digit numbers for the other 3 cases.

$p_1^5\cdot p_2^1$
$2^5\cdot3=96$ and that is all.

$p_1^1\cdot p_2^1\cdot p_3^2$
$2^2\cdot3\cdot5=60$
$2^2\cdot3\cdot7=84$
$2\cdot3^2\cdot5=90$ and that is all.

$p_1^3\cdot p_2^2$
$2^3\cdot3^2=72$ and that is all.

The sum of these numbers is $96+60+84+90+72=\boxed{\boxed{402}}$[/hide]

Can any of the 12 factors be repeated, such as $2^2$ or $5^3$ or can all the factors
 be different?

The smallest two-digit number possible is 10 and the largest two-digit number possible is 99.
The smallest prime possible is 2 and the largest prime possible is 47.

That means that there are 15 prime numbers that can be used to create two-digit numbers that are between 10 and 99.

I don't understand what you're saying. It specifies two-digit integers with $12$ distinct positive factors.


Let $ x,y > 0 : x+y=2 $ . Find the maximum value of $ \alpha $ such that :
  $ x^{\alpha}.y^{\alpha}(x^{\alpha}+y^{\alpha}) \leq 2 $

Don't know whether this helps but for $\alpha = 2$ it is true so $\alpha \geq 2$.

To prove that, write $x = 1 + s$ and $y= 1 - s$ where $0 \geq s < 1$, then \[x^2y^2(x^2 + y^2) = (1 + s)^2(1 - s)^2(2s^2 + 2) = 2(1 - s^4)(1 - s^2) \leq 2.\]

I think it doesn't help, since the answer is alpha=3 ;)

Proof:

* For $\alpha=3$, this is clearly true as well: using the same subsitution, we get to prove $(1+s)^3(1-s)^3((1-s)^2-(1-s)(1+s)+(1+s)^2)\leq 1$, which is $-s^4(6-8s^2+3s^4)\leq 0$, thus true for all s in the given interval.

* For $\alpha>3$: I disproved it by showing that f is strictly convex/concave in both left and right environment of s=0, that is: $f''(0)\not=0$, therefor there must exist an s in [-epsilon,epsilon] for which the inequality is false, as we have equality in s=0.

[hide="For those who want the actual calculations for part b"]We want to know lim(f''(s), s=0). Around 0 we can put (1+s)^a=1, (1-s)^a=1, which simplifies the expression a lot, and limiting gives $2a(a-3)$, which is not zero since a>3. This also confirms that for a=3 this counterargument is not valid.[/hide]


Find $ min(1 \minus{} a \minus{} b \minus{} c \plus{} ab \plus{} bc \plus{} ca)$ if $ 0<a,b,c<1$

hello, the $ \min(1\minus{}a\minus{}b\minus{}c\plus{}ab\plus{}ac\plus{}bc)\equal{}\frac{1}{4}$ for $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the $ \min(1 \minus{} a \minus{} b \minus{} c \plus{} ab \plus{} ac \plus{} bc) \equal{} \frac {1}{4}$ for $ a \equal{} b \equal{} c \equal{} \frac {1}{2}$.
Sonnhard.[/quote]

I don't think so. If we treat the expression $ 1 \minus{} a \minus{} b \minus{} c \plus{} ab \plus{} ac \plus{} bc$ as a linear function of $ a$ (we could do this because there is no constraint between $ a,b,c$) then the extreme values always occurs at extreme points of $ a$. Similar argument for $ b,c$. So the expression does not have minimum value.

hello, yes, you have right, the searched minimum is $ 0$, this will reached for example if $ a$ tends to zero, $ b$ tends to one and $ c$ tends to zero.
Thank you, Sonnhard.

I think the question has no answer.right??? :maybe:

I think if we have $ a\plus{}b;b\plus{}c;c\plus{}a\leq 1\Rightarrow Min(1\minus{}\sum a\plus{}\sum ab)\equal{}\frac{1}{4}$

[quote="Flakky"]Find $ min(1 \minus{} a \minus{} b \minus{} c \plus{} ab \plus{} bc \plus{} ca)$ if $ 0 < a,b,c < 1$[/quote]
We have that
\[ 1\minus{}a\minus{}b\minus{}c\plus{}ab\plus{}bc\plus{}ca\equal{}(1\minus{}a)(1\minus{}b)(1\minus{}c)\plus{}abc \ge 0\]
Now, let $ a \rightarrow 1,b\equal{}c\rightarrow 0$, we will have
$ 1\minus{}a\minus{}b\minus{}c\plus{}ab\plus{}bc\plus{}ca\equal{}0$
And the conclusion follows!


When a question asks the distance between a point and a line, does it mean the distance of the perpendicular segment from the point to the line?

Yes. This is the convention also used for the distance between a point and a plane, and I believe it can be generalized.

basically, the shortest distance between the two objects

By definition, distance is always the shortest length.