I denote $\sqrt{a}=x$ ,$\sqrt{b}=y$ ,$\sqrt{c}=z$ $\rightarrow$ $x^{2}+y^{2}+z^{2}=2$ and we have to prove that
\[6\sum x^{5}\geq 8xyz+3 \sum x^{7}\]
$\Leftrightarrow$
\[3 \sum x^{5}(x^{2}+y^{2}+z^{2})\geq 2(x^{2}+y^{2}+z^{2})^{2}xyz+3\sum x^{7}\]
which reduces to
$3\sum_{symetric}x^{5}y^{2}\geq \sum_{symetric}x^{5}yz+2\sum_{symetric}x^{3}y^{3}z$ which is Muirhead.

edited.

[quote="bilot4"][quote="e.lopes"][quote="bilot4"]If $a+b+c=2$ with $a,b,c>0$ prove that:

$\sqrt{a^{5}}+\sqrt{b^{5}}+\sqrt{c^{5}}\geq\frac{8\sqrt{abc}+3(\sqrt{a^{7}}+\sqrt{b^{7}}+\sqrt{c^{7}})}{6}$

I have "constructed" this inequality, but I would like to see if there is a shorter solution than the one I have in mind :wink:[/quote]

This is the same that:

$3\sum_{sym}a^{\frac{5}{2}}b \geq 2\sum_{sym}a^{\frac{3}{2}}b^{\frac{3}{2}}c^{\frac{1}{2}}+\sum_{sym}a^{\frac{5}{2}}b^{\frac{1}{2}}c^{\frac{1}{2}}$

True by Muirhead![/quote]
Could you plz explain how this inequality is "the same as mine"

[u]Svejk[/u]:
Very good solution :wink: You have just forgotten a "$3$" in front of $\sum_{sym}x^{7}$ in the second line :!:[/quote]

Homogenize, my friend!  :)

We can consider $6 = 3(a+b+c)$ and $8 = 2(a+b+c)^{2}$. When you open the inequality, you find exactly the expression of my last post!


inequalities
function
inequalities unsolved

Let $f$ be a convex function on $[0,a]$ and also let  $x_i\in[0,a]$ such that $\sum_{i=1}^n x_i\in[0,a]$. Then

\[ f\left(\sum_{i=1}^n x_i\right)\leq \sum_{i=1}^n f(x_i)+(n-1)f(0)  \]

I know a solution for this ineq and its expansion (in a Vietnamese book). If you need , I'll post it .

I guess the inequ barra post may be( or it is wrong):
$\sum_{i=1}^n f(x_i)\leq f\left(\sum_{i=1}^n x_i\right)+(n-1)f(0)$
...
and this is easy

[quote="barra"][quote="zhaobin"]I guess the inequ barra post may be:
$\sum_{i=1}^n f(x_i)\leq f\left(\sum_{i=1}^n x_i\right)+(n-1)f(0)$
...
and this is easy[/quote]

I think i write the inequality correctly, unless you can show that the inequality that I wrote is wrong[/quote]
Assume:$x_1 \le x_2 \le x_3 \cdots \le x_n$
$f(x)=x-x_n,x \in \left[0,a\right]$
it is certainly convex.
but
$LHS>0$ $RHS<0$


real analysis
function
trigonometry
real analysis unsolved

I posted it here, because it's not really a computation....

For which $\alpha \in \mathbb{R}$ is the function

$f: [0,2\pi ] \to [0,+\infty ]:$
$\theta \mapsto |cos \theta|^\alpha$ for $|cos \theta| \neq 0$
and
$\theta \mapsto+\infty$ for $|cos \theta| = 0$

Lebesgue-integrable

Near the zeros of $\cos\theta,\ |f|$ behaves locally like $C|\theta-\theta_{0}|^{\alpha}.$ If you're asking about absolute integrability (which you are). that's the only behavior that matters.

$f$ is integrable if and only if $\alpha>-1.$


calculus
integration
function
absolute value
calculus computations

It definitely exists because $|5x^{2}+x^{30}|\le 6$ whenever $|x|\le 1$.
Don't ask me to evaluate it, though.

Note that for $|x|\le1,\ 7-5x^{2}-x^{30}\ge 7-5-1=1.$ Therefore the denominator never approaches zero in this interval. This is the integral of a continuous function on a closed bounded interval. The integral exists, beyond a doubt. We can further show that the absolute value of the function being integrated never exceeds $1,$ so the value of the integral must lie within the set $[-2,2].$

That was the analyst's answer. Closed-form expressions (or the lack thereof) don't interest me as much.

If you are interested in the practical calculation of the integral, then just use the formula

$\int \frac{1-x^{2}-x^{30}}{\left(7-5 x^{2}-x^{30}\right)^{4/5}}\, dx=\frac{1}{7}x \sqrt[5]{7-5 x^{2}-x^{30}}+C$


[quote="giannis"][color=red]Hint[/color][hide]
$2f(3)+3f(35)=280$
$2f(35)+f(3)=3580$
Hence, $f(3)=10180$
[/hide][/quote]

You call that a hint?  :D

A hint would be:[hide]Let $g(x) = \frac{2x+29}{x-2}$. Then $g(g(x)) = x$.[/hide]

i got the whole solution
one way of doing this is to find the function
$2f(x)+3f(\frac{2x+29}{x-2}) = 100x+80$
now multiple the above equation by 3/2 and substitute $x = \frac{2x+29}{x-2}$
then we will get
$3f(\frac{2x+29}{x-2})+\frac{9}{2}f(x) = \frac{180x+4110}{x-2}$
now subtract 1st from 2nd
$\frac{5}{2}= \frac{-100x^{2}-100x+4190}{x-2}$
now substitute the value of x = 3
$f(3) = 1196$
Another way of doing this is that substitute x= 3 and then x=35
$2f(3)+3f(35) = 380$
$2f(35)+3f(3) = 3580$
solve it


DDR is like... THE BEST GAME EVER.  :P  I play a whole lot like everyday and i'm pretty good at it. (8ft and 9ft.) Its a really great game and i was seeing if anyone else liked it.

[quote="sky9073"]This is the dance-step thing, right?
Yeah, DDR is cool.. but I like PUMP better (more variations, etc)[/quote]

Yeah the Dance-Step thing. What's PUMP?

[quote="sky9073"]DDR has the steps in the shape of a + sign (up, down, left, right).
PUMP has steps in the shape of an x (four corners and center).[/quote]

Oh... I see. I think i've seen that before.

got it 5 years(give or take a year) ago. thought it was pretty cool then. lost it after a week
didnt see it around until recently though

Technically I own it.  My sister got it a year or two ago and used to play it a bunch on the X-Box but now that she is in college it is never player.  In fact, Halo 2 is the only game that has been in my X-Box since the game came out (excluding when my dad plays and arcade demo game  :? ).

And you are guys playing dance dance revolution....  hm... lol well i thought the game was kinda stupid...  :lol: *runs away before darkknight comes and kills me*

elvenchamp777

Aw  its sooo much fun though!!

I suck at it, but me and my friends still have a lot of fun

I love it!!

Let's see, I've AA'ed two or three 9-footers and passed Sakura (a 10).

I'm okay at DDR in general, but don't have a pad at home. I can probably complete most 6-7 steppers. I have gotten ridiculously good at playing it on either controller or keyboard.

Here's my best Sakura

[img]http://www.geocities.com/fredsala/screen00143.jpg[/img]

and my LoM

[img]http://www.geocities.com/fredsala/screen00145.jpg[/img]

I've gotten As on every song that I have (including most 10 steppers) except Paranoia Survivor Max (B on challenge.) AAAed about 6-7 songs (hardest End of Century, 9 steps). AAed almost every 9 stepper (except So Deep.) A on Max 300 and Unlimited. Favorite song - tie between Healing Vision, Witch Doctor, and Classic 10.

[quote="solafidefarms"]I thought that DDR was a type of RAM that doesn't work in my computer? :)[/quote]That's exactly what I thought too :P :)

[quote="metsfan001"][quote="tetrahedr0n"]I'm okay at DDR in general, but don't have a pad at home. I can probably complete most 6-7 steppers. I have gotten ridiculously good at playing it on either controller or keyboard.

Here's my best Sakura

http://www.geocities.com/fredsala/screen00143.jpg

and my LoM

http://www.geocities.com/fredsala/screen00145.jpg

I've gotten As on every song that I have (including most 10 steppers) except Paranoia Survivor Max (B on challenge.) AAAed about 6-7 songs (hardest End of Century, 9 steps). AAed almost every 9 stepper (except So Deep.) A on Max 300 and Unlimited. Favorite song - tie between Healing Vision, Witch Doctor, and Classic 10.[/quote]

My best LoM:
[url=http://www.imageshack.us][img]http://img152.exs.cx/img152/6258/screen000026ae.jpg[/img][/url]

BeMaNiRuLeR's V:
[url=http://www.imageshack.us][img]http://img152.exs.cx/img152/632/screen000035un.jpg[/img][/url][/quote]

Wow. Is this with feet or with ur fingers? I noticed you were using Stepmania.

[quote="metsfan001"]Yeah, hands. I'm still waiting for an adapter for my metal Octane. :)[/quote]

OH. If this was with feet... that would be crazy. With hands i'm a little better but still not that good.

[quote="solafidefarms"]I thought that DDR was a type of RAM that doesn't work in my computer? :)[/quote]

DDR = Deutsche Demokratische Republik (German Democratic Republic, GDR)

I think PUMP was made in Korea...

He He...

I'm a korean.!!!

and I owned both PUMP and DDR

I like PUMP better than DDR...

PUMP is much easier than DDR

hmm... stepmania has marvellous now? Never did my version. I am mediocre in DDR. On feet, I am only 5-6 difficulty? I am a 3rd/4th Mix person. I guess I'm ancient. Kick The Can Maniac is the best I can do, although I can do afronova basic. Almost Hero Maniac, but not anymore.

On hands, I can only pass songs like paranoia evil at best. I've never gotten AAA... only AAs for some songs. Big Blue X. amongst others... haven't played for a while now I can't remember.

[quote="jli"][quote="DarkKnight"][quote="jli"]what's DDR?[/quote]
Hmm... Apparently i should have put what it stood for in the title.  :blush:

DDR stand for Dance Dance Revolution. It's a really fun game where you have four arrows and on the screen arrows are moving and you have to step on the correct arrow when the arrow gets to the top of the screen.[/quote]

why are there two dance's

Dance [b]Dance [/b]Revolution

for some reason, it makes me feel weird[/quote]

Uhh... Good question but i really have no idea.

[quote="Sunny"][quote="DarkKnight"]

Uhh... Good question but i really have no idea.[/quote]

Sounds better to me.  I mean...DR, or DDR.[/quote]

Lol... I guess you do have a point. If you say them, it seems easier to say DDR.


calculus
integration
trigonometry
limit
calculus computations

The sequence $\{c_{n}\}$ is determined by the following equation. \[c_{n}=(n+1)\int_{0}^{1}x^{n}\cos \pi x\ dx\ (n=1,\ 2,\ \cdots).\] Let $\lambda$ be the limit value $\lim_{n\to\infty}c_{n}.$  Find $\lim_{n\to\infty}\frac{c_{n+1}-\lambda}{c_{n}-\lambda}.$

search(sourceScopeID, TagList, destinationScopeID):

$Scope1 \rightarrow F_1 = (T_1 \vee T_2 \vee T_3) \wedge \cdots  \wedge  (T_k \vee T_r \vee T_s) $
$Scope2 \rightarrow F_2 = (T'_1 \vee T'_2 \vee T'_3) \wedge \cdots  \wedge  (T'_{k'} \vee T'_{r'} \vee T'_{s'}) $

$search(Scope1, [T''_1, \cdots , T''_m], Scope2) = (Scope1 \wedge_{i = 1}^{m} T''_i) \vee Scope2 = $
$ (F_1 \wedge_{i = 1}^{m} T''_i) \vee F_2$

After cook Reduction

$ = C''_1 \wedge \cdots \wedge C''_n$  for $C''_i = (T'''_i \vee T'''_{i+1} \vee T'''_{i+2} )$  in $3$-cnf form.


analytic geometry
vector
abstract algebra
category theory
function
group theory
real analysis

Many sources have told me that differential forms (and tensors in general) are a better way of doing things because it allows you to become "coordinate-free." I've looked through numerous sources, and no one has really explained what this precisely means.

I've seen tensors developed as sets of coordinates which change according to certain rules. But this makes no sense, for if you are trying to be coordinate-free, why on Earth are you [i]talking about coordinates[/i]?

I like abstract definitions (for example, a vector space to me means simply an Abelian group with an action of a field on it). I've seen the tensor product as a functor with a certain universal property, and that makes sense to me. I also know what a multilinear function is. But what has never made sense is what it exactly means to be "coordinate-free" and why people talk about coordinates as a way of being coordinate-free.

For example, in Bachman's Geometric Approach to Differential Forms, he writes:

[quote]
Evaluating a 1-form on a vector is the same as pro-
jecting onto some line and then multiplying by some
constant.

In fact, we can even interpret the act of multiplying by a constant geometrically.
Suppose $ \omega$ is given by $ a dx \plus{} b dy$. Then the value of $ \omega(V_1)$ is the length of the
projection of $ V_1$ onto the line, l.
This interpretation has a huge advantage... it’s coordinate free. Recall from the
previous section that we can think of the plane, P, as existing independent of our
choice of coordinates. We only pick coordinates so that we can communicate to
someone else the location of a point. Forms are similar. They are objects that exist independent of our choice of coordinates. This is one of the keys as to why they are so useful outside of mathematics.[/quote]

But it isn't clear what's so "coordinate-free" about this.

So does anyone have any explanations to any of my questions/plaintes/etc above? Do you have any specific examples of what it means to be coordinate-free? If you do, I will thank you very much.

Most of the time, you need to pick a set of coordinates to do explicit calculations. On the other hand, the "coordinate-free" notions let us prove theorems without ever choosing explicit theorems, and apply those theorems in many useful ways without picking the coordinates.
A "coordinate-free" object must make sense no matter what coordinates we choose, and interact with coordinate changes in a specific natural way. The names change; if $ (x,y)$ and $ (x',y')$ are two different coordinate systems, $ dx\plus{}dy$ and $ dx'\plus{}dy'$ are different forms. On the other hand, if we know how these systems are related, we can find $ dx\plus{}dy$ in the $ (x',y')$ coordinates as well.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=228574]Two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=162528]previous[/url] threads with some relation.


find all possible values for $ x$ :

$ \sqrt[7]{x} \minus{} \sqrt[5]{x} \equal{} \sqrt[3]{x} \minus{} \sqrt[2]{x}$

Obviosly $ x=0$ and $ x=1$ are solutions. Let $ x=t^{210},t\ge 0$, then ${ t^{105}-t^{70}-t^{42}+t^30}=0$ had solution $ t=0$ and roots $ t^{75}-t^{40}-t^{12}+1=0$ had only 2 positive roots, one of them $ t=1$, another $ 0<t<1$.


Prove that if $ (g_n)$ is a sequence in $ L^2[0, 1]$ with $ \parallel{} g_n \parallel{}_2 \leq 1$ for all $ n \geq 1$ then $ (g_n/n)$ converges to zero a.e.

This should follow from this:

If $ (f_n)_n$ is a sequence in $ L^1[0, 1]$ is such that $ \sum_{n\equal{}1}^{\infty}  \parallel{} f_n \parallel{}_1 < \infty$ then $ \sum_{n\equal{}1}^{\infty} | f_n(s) | < \infty$ for almost every $ s \in [0, 1]$.

However, I do not see how this would follow.  How do I prove this?

Let $ f_n  \equal{} (|g_n|/n)^2$ and recall if a series converges, then the terms go to zero.


function
geometry
rectangle
real analysis
real analysis unsolved

Prove that every $ f(x,y) \rightarrow \mathbb{R}$ when  $ f(\cdot,y)$ and $ f(x,\cdot)$ are monotonic and continuous f is continuous.

Take any point $ P$ in the plane and consider a horizontal line segment $ S$ with $ P$ in its interior. If $ S$ is small, the values of $ f$ along $ S$ will be close to $ f(P)$. Now think of line segments parallel to $ S$, moving up from below. On each of these segments $ f$ is continuous, and we approach $ f|S$ monotonically from below, pointwise (because $ f$ is continuous vertically). Same idea from above $ S$. Now use Dini's theorem to see we have uniform convergence to $ f|S$ as these segments approach $ S$. This will give you an open rectangle with $ P$ in its interior such that the values of $ f$ are close to $ f(P)$.

It appears we only need monotonicity in one variable.


logarithms
limit
integration
calculus
calculus computations

Root test. The $ n$th roots of the terms go to zero. If we made the terms larger by replacing that $ \ln n$ root with an $ n$th root, the $ n$th roots of the terms would go to $ \frac1e$, and it would still converge.

$ n! \leq \frac{n^{n}}{2^{n-1}}$ so $ (n!)^{n}\leq \frac{n^{n^{2}}}{2^{n^{2}-n}}$, so $ \left(\frac{(n!)^{n}}{n^{n^{2}}}\right)^{1/\ln n}\leq 2^{-\frac{n^{2}-n}{\ln n}}< 2^{1-n}$.

We could use better bounds (e.g. Stirling in place of the elementary AM-GM estimate for $ n!$ I used) to get a better approximation of the actual value of the sum.

[quote="misan"]does this series converge or diverge ?  :maybe:

$ \sum_{n=3}^\infty\;(\;\sqrt[{\ln n}]{\;\;\frac{(n!)^{n}}{n^{n^{2}}}\;\;}\;)$[/quote]
Converge
[hide="hint"]
For example:
Stirling's formula and Cauchy Criterion for Series[/hide]
[hide="Solution"]
Let $ \xi_{n}=\sqrt[\ln n]{\frac{(n!)^{n}}{n^{n^{2}}}}$
By Stirling's formula we have:
$ \xi_{n}=\sqrt[\ln n]{\frac{\frac{n^{n^{2}}}{e^{n^{2}}}(2\pi n)^{\frac{n}{2}}e^{n\lambda_{n}}}{n^{n^{2}}}}=\frac{(2\pi n)^{\frac{n}{2\ln n}}}{e^{\frac{n^{2}}{\ln n}}}e^{\frac{n\lambda_{n}}{\ln n}}$ for $ \frac{1}{12n+1}<\lambda_{n}<\frac{1}{12}$, , so we'll compute [by Cauchy criterion]:

$ \lim_{n\to\infty}\sqrt[n]{\xi_{n}}$, hence:

$ \sqrt[n]{\xi_{n}}=(\frac{2\pi n}{e^{n}})^{\frac{1}{\ln n}}e^{\frac{\lambda_{n}}{\ln n}}$
Ofc $ 0<\sqrt[n]{\xi_{n}}$ and becouse there $ \exists_{m}\ \forall_{n>m}\ \ 0<\frac{2\pi n}{e^{n}}<1$, so: $ \sqrt[n]{\xi_{n}}<(\frac{2\pi n}{e^{n}})^{\frac{1}{n}}e^{\frac{\lambda_{n}}{\ln n}}=\frac{\sqrt[n]{2\pi n}}{e}e^{\frac{\lambda_{n}}{\ln n}}\to \frac{1}{e}$, so we have:
$ 0<\lim_{n\to\infty}\sqrt[n]{\xi_{n}}<\frac{1}{e}<1$, so series $ \sum_{n=3}^{\infty}\xi_{n}$ converges.[/hide]

Or the other way:
[hide="2nd Solution"]
Let $ \xi_{n}=\sqrt[\ln n]{\frac{(n!)^{n}}{n^{n^{2}}}}$.
We have:
$ \frac{(n!)^{n}}{n^{n^{2}}}\leq\frac{(n^{n})^{n}}{n^{n^{2}}}=1$, so $ 0<\frac{(n!)^{n}}{n^{n^{2}}}<1$, hence:
$ \sum \xi_{n}\leq\sum\sqrt[n]{\frac{(n!)^{n}}{n^{n^{2}}}}=\sum\frac{n!}{n^{n}}$, and becouse $ \sqrt[n]{\frac{n!}{n^{n}}}\to\frac{1}{e}$, then $ \sum \xi_{n}$ converges.[/hide]


Iff A,B,C given any real numbers prove that An^2+Bn+C<n! for all sufficiently large intergers n.

It is obvious.

We just have to note that $n! = \sqrt{2 \pi n} \left ( \frac{n}{e} \right ) ^n \left (1 + O \left (\frac{1}{n} \right ) \right )$, by Stirling formula.

n! > n^3 for sufficiently large intergers n.
n^3 > An^2+Bn+C for sufficiently large intergers n. (just take n = A+B+C)


group theory
abstract algebra
linear algebra
linear algebra unsolved

Prove that any finite type torsion subgroup of $GL_n(C)$ is finite.

[quote] Prove that any finite type torsion subgroup of GL_n(C) is finite. [/quote]

What does it mean "  finite type torsion subgroup " ?

It is a finitely generated group in which every element if of finite order.


The location for 2008 hasn't been confirmed yet (I actually heard that it may be on the west coast, so I think it's pretty undecided at the moment :P ). When they find a location, it'll probably be put on the website first.

Btw, I'm not applying, but I'm a '06 alum.

[quote="LadyKn1ght"]Yeah, UCSC was awesome... (even with though the cafeteria was so far away :P )[/quote]

It wasn't really that far away, we just had to make a big loop around the direct path because there was construction

We could have just climbed all the stairs and gone like directly to the front door

i am attending. they just said my handwirting was a little messy and they want me to retake the qualifying quiz

maybe he took it this year and since he wasn't in 8th grade and had messy handwriting they told him to retake it?

waiting is the hardest part...

but if anyone thinks they know, or wants to guess, please post here!

[quote="13375P34K43V312"]The quiz isn't even up yet, how do you know you're going?[/quote]

i qualified last year but i chose not to go

I'm still wondering if I should go again or not...

But I have no clue where MathPath 08 is.

[quote="guyinPA"]don't you have to qualify new each year??[/quote]

Nope, no sense in doing another qualifying quiz if you have shown the aptitude necessary for the camp already, and it's not like it's a timed quiz. However, returning students who fill out their applications (not the quiz, just like name, address, etc) get like a 10 percent discount.

[quote="guyinPA"]don't you have to qualify new each year??[/quote]

thats not what the guy said. atleast... thats not what my mom said the guy said. he just wants me to retake it because last year my handwriting was "messy"

is the qualifying quiz hard. could you give me an example of a problem?

first problem on last years qualifying quiz
$ a \plus{} b \equal{} 2$ $ a^2 \plus{} b^2 \equal{} 3$ find $ a^3 \plus{} b^3$

I think they said something about swarthmore.... i don't know, I was just hanging around the "executive" people during lunch and heard something like that....

[url]http://mathpath.org/camp.htm[/url]

[quote]MathPath 2008 will be held from June 29th to July 27th in Vermont....[/quote]

The qualifying quiz came out

[url]http://www.mathpath.org/Quiz/MP08test.pdf[/url]

Maybe I'll actually submit it this time :wink:

[quote="guyinPA"]Hi - anyone know where Mathpath 2008 will be held? I heard somewhere on the East Coast - but that covers a big area. If anyone thinks they know, please post. Thanks! Also, anyone on here applying?[/quote]

It is in Vermont

Hence the numerous links to the website  :wink:

I don't have to do the quiz, but I'm taking it anyways for fun (I don't even know if I'm going again or not)

Argh, I forgot about the quiz.
Actually, I might not go, since I practicing other stuff by myself...


[u][b]The author of this posting is : Armo[/b][/u]
____________________________________________________________________

Let ABCD is a square and P is a point in it's plane. Find the locus of the points P such that

$\max\left(PA;PC\right) = \frac{1}{\sqrt2} \cdot \left(PB+PD\right)$.

This is not difficult problem, but it has a nice solution.
You can also solve it using coordinates.

[u][b]The author of this posting is : grobber[/b][/u]
____________________________________________________________________

[quote="Armo"]Let ABCD is a square and P is a point in it's plane. Find the locus of the points P such that

$\max\left(PA;PC\right) = \frac{1}{\sqrt2} \cdot \left(PB+PD\right)$.[/quote]

Umm... The circumcircle of the square?  :)

:D
My little students though about half hour on this problem, but with no result :(


The sides of a right triangle are $ \sqrt{3},\sqrt{5}$ and $ x$. Find the sum of the possible values for $ x$, expressed in simple radical form.

$ x$ could be the length of the hypotenuse or $ \sqrt{5}$ could be the length of the hypotenuse

in the first case, we have $ x^2\equal{}\left(\sqrt{3}\right)^2\plus{}\left(\sqrt{5}\right)^2\Rightarrow x\equal{}2\sqrt{2}$
in the second case, we get $ x^2\equal{}\left(\sqrt{5}\right)^2\minus{}\left(\sqrt{3}\right)^2 \Rightarrow x\equal{}\sqrt{2}$

their sum is $ \boxed{3\sqrt{2}}$


In how many different ways can you remove a chip consisting 20 balls if you are allowed to remove 1,2 or 3 balls in one turn? Can you generalize it for any number of balls by the use of recurrence relations? (If the question was asked for 4 balls, the answer would be 7 : 3-1, 1-3, 2-2, 2-1-1, 1-2-1,
1-1-2, 1-1-1-1)

There is a recurrence, $f_{n}=f_{n-1}+f_{n-2}+f_{n-3}$.

Isn't this just partitions? Interesting formula...can anyone derive it?


complex analysis
complex analysis unsolved

$U(x,y) \in C^{1}: R^{2}\to \mathbb{C}$,

$V(x) \in C^{\infty}: R\to R$.

$r>0$ some constant.

If $f(x,y)=U(x,y)+iV(x)$ satisfy $\frac{\partial f}{\partial x}+i \frac{\partial f}{\partial y}=0$, for $|x|<r,0<y<r$.

How to show:
(1) There is a $g(z)$ analytic in $|z|<r$ s.t. $f(x,y)=g(x+iy)$ for $|x+iy|<r$ and $y\geq 0$.

(2) If $V(x)$ is not real-analytic in some nbh of $0$, then there is a contradiction.

[quote="jmerry"]There's something wrong with the statement- as written, we can choose $V$ arbitrarily and adjust $U$ to make the sum analytic.[/quote]
sorry, missed something: $U(x,y)\to 0$ as $y\to 0$, I guess then use Swchartz reflection.


Just to let people know, Wikipedia's Yellow Pigs Day article has been put up for deletion:
http://en.wikipedia.org/wiki/Wikipedia:Articles_for_deletion/Yellow_Pigs_Day

I would encourage you not to post, however, unless you can produce reputable sources - this guy Rklawton is very insulting and stubborn.

Hmm... yeah, a really annoying guy, but it doesn't seem like there's that strong a case for keeping the article.  Here's my view of things.

Reasons to keep it:

It's clear to anyone affiliated with HCSSiM that YP day has been celebrated by everybody in the program (way back to the beginning, I believe, or at least almost all the way back.  My mom celebrated it in '77, for instance.).  Thus it exists to a whole lot of people.  It's not just an inside joke (sorry Michael) but an actual tangible event.

Most of the newspaper articles about HCSSiM would mention YP day, I think.  I don't have any of these articles on me, and most of them are probably old, but they certainly exist.

Reasons not to keep it:

No reliable source that I know of actually explains thoroughly what YP day is.  Those newspaper articles would probably devote only about a sentence to it.

None of us have the newspaper articles.  And I'm not sure if old local articles count much for verifiability, because how is anyone (besides people with access to Kelly's office) supposed to check them?

The only people who would want to look up what YP day is are soon-to-be students or their parents.  Everyone else who cares what it is has participated and thus doesn't need to look it up.  If future students needed to read about what it was, a description would appear in their informational materials; no such description appears, because you don't need to know what YP day is before you're in it.

Other comments:

I wish people wouldn't conflate (did I use that word correctly, s-m?) YP day with YPMD.  Even some students '06 do it.  Any easy mistake, I guess, because of the similarity in name and because this year YP day was a subset of YPMD.

It's interesting that Rklawton's statement "If it's well known, then you'll have sources that demonstrate this fact. You don't, so it isn't." doesn't seem to hold here.  The statement should hold in general, but something about the nature of HCSSiM stops it from working in this case.

I also am not sure that the YPD article should be kept.  Maybe YPD should simply be a section of the HCSSiM article?

Definitely -- just pick up all the information and insert it into the HCCSiM page.

Also, someone should pursue deletion of Rklawton's page for commercial solicitation because, well, it's commercial solicitation and he seems like a real jerk.

If anyone wants to write a YPD article for the AoPSWiki, we won't delete it ;).  We might even recognize Spivak and Kelly as 'notable'.

Speaking of the AoPS Wiki, I note that HCSSiM is listed in the Summer Programs entry, but that the page is empty---meaning that some alumn or other ought to create/edit it.

Just to let you all know, the page is now deleted.  :( The reason is this: "Rklawton's point that a joke holiday celebrated by 1000 academics is not notable, has not been disputed by any evidence brought up here."


If $0<x<1$, then $0<y<x<1\to 0<z<y<x<1\to x<y<z<x$ - contradition.
Therefore if these system had other solutions, then $x,y,z>1$.
I think it had other solutions $1<x,y,z<2$.

This is my solution.
Obviously $x=y=z=0$ is a root of equation.
From assumtion of problem, we infer $x,y,z>0$
Hence $x^{2}+1\ge 2x\Rightarrow\frac{2x}{x^{2}+1}\le 1\Rightarrow\frac{2x^{2}}{x^{2}+1}\le x$(because of $x>0$), so $y\le x$ (1)
Similarly $y^{4}+y^{2}+1\ge 3y^{2}\Rightarrow\frac{3y^{2}}{y^{4}+y^{2}+1}\le 1\Rightarrow\frac{3y^{3}}{y^{4}+y^{2}+1}\le y$, so $z\le y$
Alternatively $\Rightarrow\frac{4z^{4}}{z^{6}+z^{4}+z^{2}+1}\le z\Rightarrow x\le z$
Therefore $x=y=z=1$
The system has tow roots $x=y=z=0$ or $x=y=z=1$


Several ﬁgures can be made by attaching two equilateral triangles to the regular pentagon $ ABCDE$ in two of the ﬁve positions shown. How many non-congruent ﬁgures can be constructed in this way?
[asy]unitsize(2cm);

pair A=dir(306);
pair B=dir(234);
pair C=dir(162);
pair D=dir(90);
pair E=dir(18);

draw(A--B--C--D--E--cycle,linewidth(.8pt));
draw(E--rotate(60,D)*E--D--rotate(60,C)*D--C--rotate(60,B)*C--B--rotate(60,A)*B--A--rotate(60,E)*A--cycle,linetype("4 4"));

label("$A$",A,SE);
label("$B$",B,SW);
label("$C$",C,WNW);
label("$D$",D,N);
label("$E$",E,ENE);[/asy]$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

We have $ \binom{5}{2}$ ways to choose where the triangles go and we divide by 5 since you can rotate it. Thus the answer is $ \frac{5!}{(5)(3!)(2!)} \equal{} 2 \boxed{\textbf{B}}$

This problem is WAY to brute-force-able (if you even call it brute force... :P). Just stare at the diagram for 2 seconds.

They can either be next to each other, or 2 apart.

We may assume that one of the triangles is attached to side AB . The second triangle can be attached to BC or CD to obtain two non-congruent figures. If the second triangle is attached to AE or DE to , the figure can be reflected about the vertical axis of symmetry of the pentagon to obtain one of the two already counted. Thus the total is 2.

Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\boxed{\boxed{\boxed{\boxed{\boxed{2}}}}}$

[quote=DSL13]Wait... why wouldn't it be 1? You would only have 1 non--congruent figure since all the other figures are congruent to each other.[/quote]

These are different:

You have to include the other figure. therefore there are $2$


:( A man makes a will and dies, leaving his wife pregnant. The man leaves  $ \$$100 so that if his wife has a daughter, the mother sould receive  2/3 and the child receives 1/3, but if his wife have a son, he shall receive 2/3 and the the mother receives 1/3. The mother gives birth to twins, a son and a daughter. How is the inheritance should be divided, with the intent of the father? :(

This is a tricky question.. I'm pretty sure of my answer, but I'm not as beastly as the other people out there: The son gets 1/2 which is 50 dollars. The mother and daughter each get 25 dollars. You could go into further detail, but this is just a basic answer that will have to work for now..    :ninja:

Possibly...

Son - Mother - Daughter ratio, 4:2:1, so divide accoordingly. (400/7, 200/7, 100/7)


When given a development figure of an octahedron die, how do you know that which face is parallel to which?

i know how to do cubic die, but i dont know how to do octahedron...

An 8-sided die is like a two square pyramids.  The bases of both of those are attached together so there are 6 vertices, 8 faces, and 12 edges.

I believe opposite sides should parallel with each other, like when you roll the die, the face on the bottom (covered) will be parallel to the top face which is the number you use.

[quote="FantasyLover"][quote="FantasyLover"]given a development figure of an octahedron die[/quote]

[/quote]

Hm...would that be like a net?

I kinda don't know what an development figure is.... :blush:

Well..the way I do those net problems is to imagine the whole shape in my head.  Or you could draw it out on the paper...I guess.

If you ignore the words...

[img]http://www.emeraldinsight.com/fig/3020010202002.png[/img]

[quote="ernie"]For some reason, I'm not exactly seeing the diagram properly... Can you just tell us the answer? :lol:

Taking what 007math just said, is it $ 3$?[/quote]

I'm not sure, but I think there are $ 4$ parallel faces....I could be wrong...

I think that in this picture the faces with the same color are opposite.


Lines $ OA,OC,OB,OD$ from point $ O$ is in counter-clockwise order $ \angle AOB\equal{}\angle COD$ . Circle $ S$ is tangent with $ OA,OB$, and circle $ T$ is tangent with $ OC,OD$. Circle $ S$ intersect $ T$ at points $ E,F$. Prove that : $ \angle AOE\equal{}\angle DOF$

See the figure below.
Since $ \angle TON\equal{}\angle SOK$ we get $ \Delta TNO\sim \Delta SKO \Rightarrow \frac{TO}{SO}\equal{}\frac{TN}{SK}\equal{}\frac{TF}{SF}\equal{}\frac{TE}{SE}$
So $ O,E,F$ lie on Apollonius circle $ (V)$ of line segment $ TS$ wrt ratio $ R_s/R_t$.
Construct bisector $ OX$ of $ \angle TOS \Rightarrow X\in (V)$
Let $ U, Z$ be the second intersections of $ OT, OS$ and $ (V) \Rightarrow \angle UOX\equal{}\angle ZOX \Rightarrow arc XU\equal{}arc XZ$
But $ arc XE\equal{}arc XF \Rightarrow arc FU\equal{}arc EZ$ or $ \angle EOS\equal{}\angle FOT$. We are done.

[quote="livetolove212"]See the figure below.
Since $ \angle TON \equal{} \angle SOK$ we get $ \Delta TNO\sim \Delta SKO \Rightarrow \frac {TO}{SO} \equal{} \frac {TN}{SK} \equal{} \frac {TF}{SF} \equal{} \frac {TE}{SE}$
So $ O,E,F$ lie on Apollonius circle $ (V)$ of line segment $ TS$ wrt ratio $ R_s/R_t$.
Construct bisector $ OX$ of $ \angle TOS \Rightarrow X\in (V)$
Let $ U, Z$ be the second intersections of $ OT, OS$ and $ (V) \Rightarrow \angle UOX \equal{} \angle ZOX \Rightarrow arc XU \equal{} arc XZ$
But $ arc XE \equal{} arc XF \Rightarrow arc FU \equal{} arc EZ$ or $ \angle EOS \equal{} \angle FOT$. We are done.[/quote]
Why $ \angle UOX \equal{} \angle ZOX$?

Because $ OX$ is the bisector of $ \angle TOS$.

[quote="livetolove212"]See the figure below.
Since $ \angle TON \equal{} \angle SOK$ we get $ \Delta TNO\sim \Delta SKO \Rightarrow \frac {TO}{SO} \equal{} \frac {TN}{SK} \equal{} \frac {TF}{SF} \equal{} \frac {TE}{SE}$
So $ O,E,F$ lie on Apollonius circle $ (V)$ of line segment $ TS$ wrt ratio $ R_s/R_t$.
Construct bisector $ OX$ of $ \angle TOS \Rightarrow X\in (V)$
Let $ U, Z$ be the second intersections of $ OT, OS$ and $ (V) \Rightarrow \angle UOX \equal{} \angle ZOX \Rightarrow arc XU \equal{} arc XZ$
But $ arc XE \equal{} arc XF \Rightarrow arc FU \equal{} arc EZ$ or $ \angle EOS \equal{} \angle FOT$. We are done.[/quote]
nice,thank you very much.  :D


function
superior algebra
superior algebra unsolved

We have $M_{k} = \{nk+1 | n \in N \}$ with $k \in N^{*}$. $(M_{k}, \cdot)$ is a monoid. We consider the monoids: $M_{5}$, $M_{8}$ and $M_{12}$. Prove or disprove that

a) $M_{5}$ and $M_{8}$ are isomorfic.
b) $M_{5}$ and $M_{12}$ are isomorfic.
c) $M_{8}$ and $M_{12}$ are isomorfic.

Thank you very much.

I did not find any bijective functions so far, so I can prove that some of them are isomorfic.... so please.. some help would be great.

Thank you very much! :)

[Moderator edit: This seems to be a particular case of http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=56679 .]


In a circular ring with radii $11r$ and $9r$, we put circles of radius $r$ which are tangent to the boundary circles and do not overlap. Determine the maximum number of circles that can be put this way. (You may use that $9.94<\sqrt{99}<9.95$)


Hi... i cant find solve for this peg game... can any1 help me pls?

[img]http://img301.imageshack.us/img301/7356/peggame7tu.th.jpg[/img]

thx to all.
bye

For those who don't know, this is probably the game where you can jump a peg over another peg into an empty spot and remove the peg which was jumped over, only in straight lines (not diagonal), and only one peg can be jumped at once time. The goal is to have only one peg remaining.

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b]

...and also to get that last peg in the spot that was empty at the start.


I'm having trouble understanding why this is true.  Can someone point me in the right direction please.

Let g(x), f1(x),...,fn(x) be functions in D^n.  Prove that:
w(g(x)f1(x),...,g(x)fn(x))=[g(x)]^n * w(f1(x),...,fn(x))

Let me know if this doesn't make sense.

isn't this just a property of determinants?  multiplying one row by $ \alpha$ multiplies the determinant by $ \alpha$, so multiplying all $ n$ rows by $ \alpha$ multiplies the whole determinant by $ \alpha^n$.

For an illustration of what's going on, let's look at the $ n\equal{}3$ case. We do have to use the product rule in taking derivatives.

$ W(gf_1,gf_2,gf_3)\equal{}$

$ \equal{}\begin{vmatrix}gf_1&gf_2&gf_3\\ gf_1'\plus{}g'f_1& gf_2'\plus{}g'f_2& gf_3'\plus{}g'f_3\\ gf_1''\plus{}2g'f_1'\plus{}g''f_1& gf_2''\plus{}2g'f_2'\plus{}g''f_2& gf_3''\plus{}2g'f_3'\plus{}g''f_3\end{vmatrix}$

Now do one set of row operations: subtract $ \frac{g'}{g}$ times the first row from the second row and subtract $ \frac{g''}{g}$ times the first row from the third row:

$ \equal{}\begin{vmatrix}gf_1&gf_2&gf_3\\ gf_1'& gf_2'& gf_3'\\ gf_1''\plus{}2g'f_1'& gf_2''\plus{}2g'f_2'& gf_3''\plus{}2g'f_3'\end{vmatrix}$

Then subtract $ \frac{2g'}{g}$ times the second row from the third row:

$ \equal{}\begin{vmatrix}gf_1&gf_2&gf_3\\ gf_1'& gf_2'& gf_3'\\ gf_1''& gf_2''& gf_3''\end{vmatrix}\equal{}g^3W(f_1,f_2,f_3)$

(I was tacitly assuming $ g\ne0$ in describing the row operations. However, at any point at which $ g(x)\equal{}0,$ this Wronskian is zero anyway.)

The general case follows the same outlines.


\[
2 < 3 < 5 \Rightarrow 2^{2^p} < 3^{3^p} < 5^{5^p} \Rightarrow \frac{1}{2^{2^p}} > \frac{1}{3^{3^p}} > \frac{1}{5^{5^p}}
\]

And....oops I guess the second inequality isn't true.  Stupid mistakes :oops:

[quote="MCrawford"][quote="nr1337"]
\[
\frac{1}{2^{2^p}} > \frac{1}{3^{3^p}} > \frac{1}{5^{5^p}}
\]
\[
2^{\frac{1}{2^{2^p}}} > 3^{\frac{1}{3^{3^p}} } > 5^{\frac{1}{5^{5^p}}
}
\]
[/quote]

How do you justify this step?[/quote]
shouldnt the signs in the second be reversed?


algebra
polynomial
quadratics
inequalities
quadratic formula

For real numbers $ a,\ b$, let $ P \equal{} a^4 \minus{} 4a^2b \plus{} b^2 \plus{} 6b$.

(1) Find the range of value of $ a$ such that $ P\geq 0$ for any real number $ b$.

(2) Find the range of value of $ a$ such that $ P\geq 0$ for any real number $ a$.

Ah thank you Zuton Force, I see why my solution was incorrect now.

[b]@azn:[/b] I think your answer to the first one is incomplete... consider $ (a,b)\equal{}(0,\minus{}1)$

Note that for any $ b$, the polynomial is positive for sufficiently large $ a$, and that for any $ a$, the polynomial is positive for sufficiently large $ b$. Thus, to prove positivity, we must show that roots without multiplicity are impossible.

[b](1) a such that P>=0 for all b[/b]

[hide]$ P \equal{} b^2 \plus{} (6 \minus{} 4a^2)b \plus{} a^4$, so by the quadratic formula we must have $ (6 \minus{} 4a^2)^2 \minus{} 4a^4 \leq 0 \iff a^4 \minus{} 4a^2 \plus{} 3 \leq 0 \iff 1 \leq a^2 \leq 3$

$ \therefore a \in [ \minus{} \sqrt 3, \; \minus{} 1] \cup [1, \; \sqrt {3}]$[/hide]

[b](2) b such that P>=0 for all a[/b]

[hide]$ P \equal{} a^4 \minus{} (4b)a^2 \plus{} (b^2 \plus{} 6b)$, and we don't want any positive solutions in $ a^2$, unless it is a double-root.

Go in two parts. First, the case of no roots/roots with multiplicity:

$ 16b^2 \minus{} 4(b^2 \plus{} 6b) \leq 0$
$ 12b^2 \leq 24b$
$ 0 \leq b \leq 2$

Next, two non-positive roots. Again from the quadratic formula:

$ 4b \leq \pm \sqrt {12b^2 \minus{} 24b}$ (we are now assuming the discriminant is positive)

Clearly, we see that $ b$ is negative, so we can just show that:

$ 4b \leq \minus{} \sqrt {12b^2 \minus{} 24b}$

Since both sides are negative, we can use $ a < b \implies a^2 > b^2$ for negative numbers:

$ 16b^2 \geq 12b^2 \minus{} 24b$
$ 4b^2 \geq \minus{} 24b$
$ b^2 \geq \minus{} 6b$
$ b \leq \minus{} 6$

The total solution set is the union of both solutions:
$ \boxed{b \in ( \minus{} \infty, \minus{} 6] \cup [0,2]}$[/hide]

2). I assume you mean find range of b s/t p>= 0 for any real a.

b must be less than or equal to -6 (if it is negative), otherwise the expression b^2 + 6b would be negative, causing the inequality to fail (you can just choose a = 0).
b=0 always holds.

If b is positive, we look at the case where a^4 -4a^2b is negative. If a^4 < 4a^2b, -2sqrtb < a < 2sqrtb. The minimum of the graph a^4 -4a^2b occurs when 4a^3 - 8ab = 0, 4a(a^2 - 2b) = 0, a = sqrt2b. This falls into the range of a since sqrt2b < 2sqrt b.

If a = 2sqrt2b, a^4 - 4a^2b = -4b^2. The whole expression becomes 6b - 3b^2.

For 6b - 3b^2 >= 0, b =< 2.

b=< -6 or  0=< b =< 2.

While you've shaved off many (in fact, all) invalid intervals, you still have the burden of proof: just because the square-term and constant term of a general quadratic are both positive does not mean that the quadratic is always positive; to complete the solution, you will need to prove that  :)

Here is my solution.

(1) $ P\equal{}a^4\minus{}4a^2b\plus{}b^2\plus{}6b\equal{}b^2\minus{}2(2a^2\minus{}3)b\plus{}a^4$

$ \equal{}\{b\minus{}(2a^2\minus{}3)\}^2\plus{}a^4\minus{}(2a^2\minus{}3)^2$.

$ P\geq 0$ for $ \forall{b}\in\mathbb{R}\Longleftrightarrow P_{min}\geq 0$

$ P_{min}\equal{}P_{b\equal{}2a^2\minus{}3}\equal{}a^4\minus{}(2a^2\minus{}3)^2\geq 0$, yielding $ \boxed{1\leq |a|\leq \sqrt{3}}$.

(2) Let $ x\equal{}a^2$, $ P\equal{}x^2\minus{}4bx\plus{}b^2\plus{}6b\equal{}(x\minus{}2b)^2\minus{}3b^2\plus{}6b$. We are to find the condition for which $ P_{min}\geq 0$ for $ x\geq 0$.

Calssifying the position of the axis $ x\equal{}2b$,

Case 1 : $ 2b<0\Longleftrightarrow b<0$

$ \therefore P_{min}\equal{}P_{x\equal{}0}\equal{}b^2\plus{}6b\geq 0\Longleftrightarrow b\leq \minus{}6,\ 0\leq b$. Since $ b<0$, we have $ b\leq \minus{}6$.

Case 2 : $ 2b\geq 0\Longleftrightarrow b\geq 0$

$ P_{min}\equal{}P_{x\equal{}2b}\equal{}\minus{}3b^2\plus{}6b\geq 0\Longleftrightarrow 0\leq b\leq 2$, which satisfies $ b\geq 0$.

Cansequently the desired range of $ b$ is $ \boxed{b\leq \minus{}6,\ 0\leq b\leq 2}$.


why does the rate of this reaction[i] increase [/i]with [i]increase[/i] in the polarity of the solvent when carried out under conditions which strongly favour $S_{N}2$ mechanism

$RX$ $+$ $NH_{3}$ $\rightarrow$ $RN^{+}H_{3}$ $+$ $X^{-}$

First, describe the [EDIT: $S_{N}1$] mechanism of this reaction to yourself.  First, there is the dissociation step of $RX$ into $R^{+}$ and $X^{-}$, followed by nucleophilic attack of $NH_{3}$ on the $R^{+}$.

So far so good.  Now ask yourself, what is the rate determining step?  And how would a polar solvent increase its rate?

[hide]
Rate determining step (RDS): dissociation of $RX$.

A polar solvent has a high dielectric constant, meaning it can support separated opposite charges very well.  This lends stability to the products of the RDS.  Now, by Hammond's Postulate, we see that the [i]transition state[/i] of the RDS must also be stabilized.  Therefore, the activation energy of this reaction is lowered, and its rate increases.
[/hide]

the mechanism u have described is for $S_{N}1$  :rotfl:
so please change it since the rest that follows is obviously wrong

Wow.  Good point (laughs at self).  Anyway, the same reasoning works: the product of the RDS (in this case, the only step) are stabilized by the solvent; we can again apply Hammond's postulate.

well could you elaborate yor answer

[hide="please note"]normally the rate of $S_{N}2$ varies inversely with the polarity of the solvent[/hide]

You are correct in your statement.  Thus, the question becomes, how is this reaction different from a normal $S_{N}2$ reaction?  Notice that we start out with two uncharged species and end with two charged species.  You "typical" $S_{N}2$ will start and end with one charged species.

My explanation is therefore that in your run of the mill reaction, a polar solvent stabilizes the reactants and products approximately equally.  Here, only the products are stabilized by increasing polarity.

[hide="My answer"]
although this proceeds via $S_{N}2$ mechanism it ideally behaves like that of an $S_{N}1$ here the $NH_{3}$ is hardly solvated no matter what the solvent is so9 the thing important here is how 'easily' the $X^{-}$ leaves the system which clerly increases with increase in polarity due to increase in coulombic force of attraction between them so the activation enegy which is to be provided to the attackin reagent is reduced.[/hide]

You state that "the $NH_{3}$ is hardly solvated no matter what the solvent is."  Why is this?

that's because here the solvent is polar(normally for substitution reactions we use this) since $NH_{3}$ is  not so polar it can't form polar bonds though it contains lone pair of electrons but those are not so useful here because polar protic solvent ssolvate excess electron species via $H-Bond$ but here it is not possible as $N$ has little  charge and hydrogen itself is well kept by it;'s adjoining atom. as far as aprotic polar solvents goes the mainly solvate the cations no matter what the raectants be.

NH3 is a polar compound. This reaction is an SN2 reaction (if R is primary or secondary) and SN1 if R is tertiary. Assuming the reaction Pardesi is referring to in his original post is SN2, then obviously the conditions that favour SN2 reactions will favour this one: and those conditions are increasing the polarity of the solvent by using polar aprotic solvents, since these are able to stabilize cations (in this case RNH3+), and do not surrond the incoming nucleophile (NH3) by a layer of molecules, which would difficultate the SN2 attack.


analytic geometry
geometry unsolved
geometry

Let $ \omega_1$ and $ \omega_2$ be circles with respective centers $ O_1$ and $ O_2$ and respective radii $ r_1$ and
$ r_2$, and let $ k$ be a real number not equal to 1. Prove that the set of points $ P$ such that
                                                           $ {PO_1}^{2} \minus{} {r_1}^{2} \equal{} k({PO_2}^{2} \minus{} {r_2}^{2})$

is a circle.

Thanks!

$ {PO_{1}}^{2}\minus{}{r_{1}}^{2}\equal{} k({PO_{2}}^{2}\minus{}{r_{2}}^{2})$. From thus $ PO_1^2\minus{}kPO_2^2\equal{}const$. If $ a_1,a_2$ and $ b_1,b_2$  are coordinats of the $ \omega_{1}$ and $ \omega_{2}$ centers, then
$ (x\minus{}a_1)^2\plus{}(y\minus{}a_2)^2\plus{}k(x\minus{}b_1)^2\plus{}k(y\minus{}b_2)^2\equal{}c$ This is a conic.The сoefficients by $ x^2$ and $ y^2$ are equal. Also this coefficient is not $ 0$ because $ k\neq 1$. So this is a circle.

P.S. Oh the set of points $ P$ such that \[ \alpha_1PO_1^2\plus{}\alpha_2PO_2^2\plus{}..\plus{}\alpha_nPO_n^2\equal{}const\] could be a line, whole plane, empty set or a circle. Using this you can also solve it.

[quote="KrokoZyabr"]$ {PO_{1}}^{2} \minus{} {r_{1}}^{2} \equal{} k({PO_{2}}^{2} \minus{} {r_{2}}^{2})$. From thus $ PO_1^2 \minus{} kPO_2^2 \equal{} const$. If $ a_1,a_2$ and $ b_1,b_2$  are coordinats of the $ \omega_{1}$ and $ \omega_{2}$ centers, then
$ (x \minus{} a_1)^2 \plus{} (y \minus{} a_2)^2 \plus{} k(x \minus{} b_1)^2 \plus{} k(y \minus{} b_2)^2 \equal{} c$ This is a conic.The сoefficients by $ x^2$ and $ y^2$ are equal. Also this coefficient is not $ 0$ because $ k\neq 1$. So this is a circle.

P.S. Oh the set of points $ P$ such that
\[ \alpha_1PO_1^2 \plus{} \alpha_2PO_2^2 \plus{} .. \plus{} \alpha_nPO_n^2 \equal{} const\]
could be a line, whole plane, empty set or a circle. Using this you can also solve it.[/quote]
cant' we WLOG assume $ a_1\equal{}a_2\equal{}0\equal{}b_2 ; b_1 \ge 0$?
that will make the computations become more simpler and we get:
$ PO_1^2\minus{}kPO_2^2\equal{}(x\minus{}0)^2\plus{}(y\minus{}0)^2\minus{}k(x\minus{}b_1)^2\minus{}k(y\minus{}0)^2\equal{}const$

$ \implies (1\minus{}k)(x^2\plus{}y^2) \minus{} 2x(kb_1) \minus{}kb_1^2\equal{}const$

$ \implies x^2\plus{}y^2\minus{}2x(\frac{kb_1}{1\minus{}k})\plus{}\frac{\minus{}kb_1^2}{1\minus{}k}\equal{} \frac{const}{1\minus{}k}$

$ \implies x^2\plus{}y^2\minus{}2x(\frac{kb_1}{1\minus{}k})\plus{}(\frac{kb_1}{1\minus{}k})^2\equal{}\frac{const}{1\minus{}k}\minus{}\frac{\minus{}kb_1^2}{1\minus{}k}\plus{}(\frac{kb_1}{1\minus{}k})^2\equal{}const'$

$ \implies (x\minus{}(\frac{kb_1}{1\minus{}k}))^2\plus{}(y\minus{}0)^2\equal{}const'$

in this way we can find both the coordinates of the center of the circle passing through $ P$ ,which is $ (\frac{kb_1}{1\minus{}k},0)$ , and also its radius expressed in the terms of $ r_1,r_2,b_1$.
now we can find the center of this circle by doing this:
we have two cases:
1-$ k<1$
then consider the point $ S$ on the semiline $ O_1O_2$ for which we have $ \frac{O_1O_2}{O_1S}\equal{}\frac{1\minus{}k}{k}$
then we can easily prove that $ PS$ is constant (just a matter of computations).

2-$ k>1$ then consider the point $ S$ on the semiline $ O_2O_1$ for which the similar condition in the case $ 1$ holds.
i wonder if there exists a solution without using $ (x,y)$ coordinates...


How many fifteen-minute appointments can a doctor fit into her schedule if she works from 8:30 am to noon and from 1 pm to 4:30 pm?

8:30 to 12:00       this is 3.5 hours
1:00 to 4:30          this is 3.5 hours
Each time slot is 210 minutes. 210/15= 14 appointments
Since both time slots are equal in length, we double for 14*2 = $ \boxed{28}$


So far qualifiers from Round 1:

pkothari13
Xantos C. Guin
dk2
dangvh
seungsoo
YSJeong
JesusFreak197
IPERFERPI
BornForMath
xdl615x
Lucky707
Zach
eryaman
tarquin

This will be also the same thread where you'll find the qualifiers from B.

So far:

Assassin of Death
Secret Asian
The Original Pi Guy
[b]JesusFreak197[/b]
[b]dangvh[/b]
chess64
Shogia
Shambles
SnowStorm

*A person with his/her name bold means they already qualified from A.*

when does Round 2 start?

BTW, congratulations to everyone who qualified!

If the dates are correct it would be tomorrow, May 22.

The 29th I think he said.  He is really busy with finals or something, I think.

Following are qualifiers to COP Round 2.

pkothari13
Xantos C. Guin
dk2
dangvh
seungsoo
YSJeong
JesusFreak197
IPERFERPI
BornForMath
xdl615x
Lucky707
Zach
eryaman
tarquin
Assassin of Death
Secret Asian
The Original Pi Guy
chess64
Shogia
Shambles
SnowStorm
shinwoo
DarkKinght

This is total 24 people who will be competiting in Round 2.

Sorry for week delay. :)


1. Consider all functions f from the set {1, 2, 3, . . . n} to itself for any positive integer n.  For each value n, some of these functions have the property that f(f(x)) = x for every x in the set {1, 2, . . ., n}.  Let the number of such functions be A(n).

For example:

A(1) = 1:

f(1) = 1.

A(2) = 2

f(1) = 1, f(2) = 2  OR

f(1) = 2, f(2) = 1

A(3) = 4

f(1) = 1, f(2) = 2, f(3) = 3 OR

f(1) = 2, f(2) = 1, f(3) = 3 OR

f(1) = 1, f(2) = 3, f(3) = 2 OR

f(1) = 3, f(2) = 2, f(3) = 1.

Prove that A(n) satisfies a recursion of the form:

A(n) = A(n - 1) + C(n) * A(n - 2),  where C is some function of n.

A hint, telling you what function of n C is, which might make life easier:

[hide]C(n) = n - 1[/hide].

2. Consider all functions f from the set {1, 2, 3, . . . n} to itself for any positive integer n.  For each value n, some of these functions have the property that f(f(x)) = f(x) for every x in the set {1, 2, . . ., n}.  Let the number of such functions be B(n).

For example:

B(1) = 1

f(1) = 1

B(2) = 3

f(1) = 1, f(2) = 2 OR

f(1) = 1, f(2) = 1 OR

f(1) = 2, f(2) = 2.

Consider the number of fixed points of one of the functions f to find a summation expression for B(n).  (That is, write B(n) as a summation of some function of k for k = 0, 1, . . . n.)

Prove the recursive formula B(n) = n*B(n - 1) + 1.

*You need to know much more for this question.*

Compute the limit as n goes to infinity of B(n)/n!

Moderator:  if you think these are too hard for intermediate, feel free to move them to advanced.  Actually, I suspect you probably should do so.

1. [hide]every number can only be in a cycle of length 1 or 2 (basically, a divisor of 2). If the nth number is in a cycle of length 1, we just have A(n-1). If the nth number is in a cycle of length 2, we choose the other number that it cycles with in C(n)=n-1 different ways, and then multiply that by A(n-2) (the number of ways to partition up the rest). Summing these two possibilities, we have A(n)=A(n-1)+(n-1)A(n-2).[/hide]

2. what is a fixed point? possible soln later g2g now.

[hide] a thought has occured to me. if the function x has image I, where I is a subset of the first n integers, then i believe a necessary and sufficient condition is that f(y)=y for all y that are elements of I. [/hide]

I'll leave them here for the time being. If interesting_move chooses to move them, that's fine too.

2. blah [hide]alright, let's try classifying by their image size. if the image size of function f is k, 0:le:k:le:n, we have C(n,k) ways to choose those k elements; then for those k elements, we simply have 1 ways of deciding what they are mapped to (namely, each one to itself). For the rest (n-k) elements, each of them must be mapped to one of the k elements, so we have sum k^(n-k)*C(n,k) of k from 0 to n. That what you were looking for?

for the last part, if we assume the recursion to be true, we have B(n)/n!=B(n-1)/(n-1)!+1/n!=1/1!+1/2!+1/3!+1/4!+... hmmm dunno what that is. [/hide]

Commontary on what MysticTerminator wrote:

[hide]Number (1) is all good.

For number 2, that A(k) bit isn't right -- it shouldn't be involved in the answer, although I see where you got that from.  I think it would be A(k) in there if we had f(f(f(x))) = f(x).  You're pretty close, though.  Try searching online for the value of 1/1! + 1/2! + . . .[/hide]


How to convert equation z^28-z^8-1 in such a form that we can use roots of unity to determine its complex roots?

Please make it into Intermediate Please.

Anyway...

I don't get what you want...

I guess you would want it in the form of:

$\omega+\omega^{2}+\dots+\omega^{2}8$?

[quote="A M Y"]Well you [i]could[/i] try factoring it...
Let $x=z^{4}$

$z^{28}-z^{4}-1=x^{7}-x^{2}-1=(x^{2}-x+1)(x^{5}+x^{4}-x^{2}-x-1)$

See what you can do with that?[/quote]

Wow that is awesome.

Did you figure out a factor by a root of unity check?

Well, we know that the first one has a solution of $\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}$ and $\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}$

But how do oyou do for the second one

There were many AIME problems like this btw.

This was posted in the trig/complex numbers marathon
(though the solution doesn't use roots of unity)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126345&start=20[/url]
My first post on that page.

[quote="mathgeniuse^ln(x)"]Well, we know that the first one has a solution of $\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}$ and $\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}$ [/quote]
huh? Those are the 3rd roots of unity, and don't we want the 3rd roots of negative unity?

Yes, we do as a matter of fact, I was just being stupid.

Multiply them by -1, then.

|z^{28}-z^8}|=1, since $|z|=1$, $|z^{20}-1|=1$,

${z^{20}=\text{cis}\pm \frac{\pi}{3}}$

then we have to check the solutions back in the original equation, etc


I am thinking of two positive integers.  If the first number is equal to the product of the two numbers decreased by twice one more than te second, what are the possibilities for the two numbers?

[hide="Solution"]Let $a,b$ be the two numbers.  Then $a=ab-2(b+1)\implies (a-2)(b-1)=4$.  The rest is easy.[/hide]

[quote="GlenthemanN"][hide]

$a = ab-2b+2$
$a-2 = b(a-2)$
$0 = (a-2)(b-1)$

so $\boxed{a = 2, b = 1}$
[/hide][/quote]
my teacher said there would be three answers...for a and b each, do you just guess and check?

[quote="The QuattoMaster 6000"][quote="GlenthemanN"][hide]

$a = ab-2b+2$
$a-2 = b(a-2)$
$0 = (a-2)(b-1)$

so $\boxed{a = 2, b = 1}$
[/hide][/quote]
[hide] That is not true, the equation should be $a=ab-2(b+1)=ab-ba-2\Rightarrow 2=ab-2a-a\Rightarrow 4=ab-2b-a+2\Rightarrow (a-2)(b-1)=2^{2}$.  From here, there are quite a few possibilities.  We could say that $a-2=2=b-1$ so, $(a,b)=(4,3)$ or we could say that $a-1=4$ and $b-1=1$ so $(a,b)=(5,2)$, or finally, we can say that $a-2=1$ and $b-1=4$ thus $(a,b)=(3,5)$.  So, our solutions are $\boxed{\text{(3,5), (5,2), and (4,3)}}$ [/hide][/quote]

wouldn't the second answer be 6 and 2?


conics
geometry
cyclic quadrilateral
geometry proposed

$A$ is a point outside a circle, $AM$ and $AN$ are tangents to the circle and the lines $APQ$ and $ARS$ meet the circle at $P,Q,R,S$. Prove that the point of interchange of $PS$ and$QR$ lies on the line $MN$

This property can be used to construct a tangent to a circle drawing three lines through $A$ : $APQ,ARS,AR'S'$ that meet the circle at $P,Q,R,S,R',S'$. Let $C$ the point of intersection of $PS$ and$QR$ and $C'$ the point of intersection of$PS$ and$QR'$. Then $CC'$ meets the circle at $M$ and $N$ the two points of contact for the tangents from $A$.

It's easy if we take into account the polar propertirs of the cyclic quadrilateral and the complete quadrilateral in general:

Assume $X=PS\cap QR,\ Y=PR\cap QS$. Then the polar of $Y$ wrt angle $QAS$ is $AX$, but this is also the polar of $Y$ wrt the circle. Then if $Z,T$ are the points where $AX$ cuts the circle, then $(A,X;Z,T)=-1\Rightarrow X$ is on the polar of $A$ wrt the circle, which is $MN$.


[quote="fanicha"][b]Hola, será que alguien me puede facilitar los enunciados de la Olimpíada Bolivariana de Matemática de años pasados, pues sólo he encontrado los de 2000 y 2002. Gracias de antemano y saludos.

Estefanía Ordaz[/b][/quote]
Hola, disculpa podrias poner los enunciados de la  Olimpíada Bolivariana de Matemática  2000 y 2002. Gracias.

[b]Acá estan los enunciados de la I y III OBM. Si consigues otros me avisas. Saludos.

Estefanía Ordaz[/b]

Yo se que ha pasado bastante tiempo, pero los lograste conseguir Estefania?


search
function
complex analysis
algebra
polynomial
algebra proposed

Here you can find Mason's incredible lemma, which allows an immediate solution of the problem.
http://www.mathlinks.ro/viewtopic.php?search_id=141498873&t=15566

[quote="ciprian"]I wait for some solutions :)[/quote]

What are you exactly  waiting for ?

There are infinitly many solutions to this equation, for any value of $n$.
The limitation is only the degree of the solutions.
For $n<4$, we have solutions with any degree
For $n=4$, we have solutions with degrees 0 and 1 (plus trivial solutions $(x,0,x)$)
For $n>4$ we have solutions with degree 0 (plus trivial solutions)

@ harazi: Indeed a very beautiful and extremely useful theorem.... The only solution I know also uses this Theorem:D

But there are other proofs as well. For example, for $n \geq 4$ the nonexistence of nontrivial solutions follows from the fact that the Fermat curve $z_{1}^{n}+z_{2}^{n}= 1$ has genus $\binom{n-1}{2}> 1$ and hence (viewed as a (chart in a) Riemann surface $X$) it is [i] hyperbolic[/i], i.e. covered by the unit disc $\Delta$. The existence of nonconstant meromorphic solutions $p(z), q(z) \in \mathcal{M}(\mathbb{C})$ to $p(z)^{n}+q(z)^{n}= 1$ (in particular, nonconstant rational functions) would then yield a nonconstant holomorphic map from $\mathbb{C}$ to $X$, and hence a nonconstant holomorphic map from $\mathbb{C}$ to the universal cover $\Delta$, which is absurd (recall Liouville's theorem: every bounded entire function is constant). This argument thus shows the more general statement that for $n \geq 4$ there do not exist a nontrivial solution in [i] entire functions[/i] $f,g,h \in \mathcal{O}(\mathbb{C})$ to $f^{n}+g^{n}=h^{n}$.

For $n = 3$, additional work as above shows that there are no [i] polynomial[/i] solutions as well; however, in this case, there ARE solutions in entire functions! Thus I leave my challenge problem for this thread: construct a solution to $f^{3}+g^{3}=h^{3}$ in nonconstant entire functions $f,g,h$.


function
limit
integration
floor function
calculus
irrational number
real analysis

I have read that $ x_{n}$ is equidistributed in $ [0,1]$ iff for all continuous functions $ f$ on $ [0,1]$ we have

$ \lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f(x_{k})=\int_{0}^{1}f(x)\,dx.~~~(1)$

(for example see http://en.wikipedia.org/wiki/Equidistributed_sequence#Equidistribution_modulo_1)
The famous example of Weyl's says if $ \alpha$ is an irrational number and $ x_{n}=n\alpha$, then $ x_{n}$ is equidistributed in $ [0,1]$.

But, what am I doing wrong when I use the function $ f(x)=x$, because the LHS of (1) diverges and the RHS converges?
Or, look at $ x_{n}=\sqrt{n}$ and $ f(x)=x^{2}$ again the same problem as above (this sequence is also supposed to be equidistributed in $ [0,1]$.)

Should (1) really be $ \lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f(\{x_{k}\})=\int_{0}^{1}f(x)\,dx$ where $ \{a\}=a-\lfloor a \rfloor$?  In which case, the theorem with (1) is useless in proving $ x_{n}=\sqrt{n}$ is equidistributed in $ [0,1]$ (I guess just use the definition).

I think you're just a little confused.  $ x_{n}= \sqrt{n}$ can't be equidistributed in $ [0, 1]$ because it only has two elements which fall into that interval.  Your statement (1) is only meant to hold for sequences $ x_{n}$ where $ x_{n}\in [0, 1]$ for all $ n$.  That is why we take $ f$ as a function on $ [0, 1]$.  Your proposed fix (letting $ x_{n}$ take any real value but considering $ f(\{x_{n}\})$) is a perfectly fine alternative statement.

JBL,  thank you...
Is there an analogous result to theorem in (1) for any interval?
Really I am just trying to find ways to see if sequences like $ \sqrt{n}$, $ \cos(n)$, $ \sin(n)$, $ \ln(n)$ (and other calculus functions) are uniformly distributed.  Now, some of these are not too bad just using the definition below while others are messy.

Now, I use the definition that a sequence of real numbers $ (u_{n})$ is uniformly distributed modulo 1 if for all $ a,b$ such that $ 0\le a<b\le 1$,
\[ \lim_{n\rightarrow \infty}\frac{\#\{r \le n : \{u_{r}\}\in [a,b] \}}{n}=b-a \]
where $ \{u\}=u-\lfloor u \rfloor$ denotes the fractional part of $ u$.

I see my error...statement in (1) only holds for sequences in $ [0,1]$ i.e. for the sequence $ (\{u_{r}\})$ via my definition.


Let $M=\{A_{1},A_{2},\ldots,A_{5}\}$ be a set of five points in the plane such that the area of each triangle $A_{i}A_{j}A_{k}$, is greater than 3. Prove that there exists a triangle with vertices in $M$ and having the area greater than 4.

[i]Laurentiu Panaitopol[/i]

I think that $4$ can be improved to something like $3\cdot\frac{1+\sqrt 5}2$, but it's so much easier to draw than to explain.. :) And besides, I'm not even entirely sure it's true, so I won't post anything else right now :).

I want to revive this post. Been working on this problem for 2 hrs now, and neither head nor tail of it. Somebody pls post a solution.


geometry
circumcircle
inradius
geometric transformation
reflection
incenter
Euler

Let $ABC$, $A'B'C'$ be two triangles which have the same circumcircle and the same incircle. Let $K,L,M$ be the tangency points of the incircle with $\triangle ABC$, and $K', L', M'$ be the tangency points of the incircle with $\triangle A'B'C'$. Prove that $\triangle KLM$ and $\triangle K'L'M'$ have the same orthocenter.

The nine-point circles of $KLM, K'L'M'$ are one and the same, i.e. the image of the common circumcircle through the inversion of pole $I$ and power $r^2$ ($r$ is the inradius of $ABC$). This means that $KLM,K'L'M'$ have the same nine-point center, and, since they have the same circumcenter, they must have the same orthocenter, since the orthocenter is the reflection of the circumcenter in the nine-point center.

If this problem form Russia, I want to see a human solution.

[quote="Kuba"]wow.. that's something interesting if I understood correctly. The inversion of pole I and power $r^2$ of the circumcenter of $ABC$ goes into the nine points circle of $KLM$?[/quote]

Yup :). Think about it:

Let $T$ be the midpoint of $LM$ (I took $L\in CA,M\in AB$). In the right triangle $ILA$, we have $r^2=IL^2=IA\cdot IT$. I'm sure you can see it immediately if you draw it.

The fact that Grobber used is that, if the incircle of a triangle ABC touches the sides BC, CA, AB at the points K, L, M, then the nine-point circle of triangle KLM is the image of the circumcircle of triangle ABC in the inversion with respect to the incircle of triangle ABC. This is very easy to prove:

Let X, Y, Z be the midpoints of the sides LM, MK and KL of the triangle KLM. Then, the circumcircle of triangle XYZ is the nine-point circle of triangle KLM. Now, we will prove that the points X, Y, Z are the images of the points A, B, C in the inversion with respect to the incircle of triangle ABC. Then, it will follow that the circumcircle of triangle XYZ is the image of the circumcircle of triangle ABC in the inversion with respect to the incircle of triangle ABC; in other words, the nine-point circle of triangle KLM is the image of the circumcircle of triangle ABC in the inversion with respect to the incircle of triangle ABC, and the problem will be solved.

So it remains to show that the points X, Y, Z are the images of the points A, B, C in the inversion with respect to the incircle of triangle ABC.

Let I be the incenter of triangle ABC. The points L and M, being the points of tangency of the incircle of triangle ABC with the sides CA and AB, must be symmetric to each other with respect to the angle bisector of the angle CAB, i. e. with respect to the line AI. Thus, $LM\perp AI$, and the midpoint X of the segment LM lies on the line AI. Consequently, < IXL = 90°. On the other hand, since I is the center of the incircle of triangle ABC, while L is the point where this incircle touches the line CA, we have $IL\perp CA$, so that < ILA = 90°. Thus, < IXL = < ILA. Also, < XIL = < LIA, since the point X lies on the line AI. Thus, the triangles IXL and ILA are similar, and it follows that IX : IL = IL : IA. Equivalently, $IA\cdot IX = IL^2$. Since we also know that the point X lies on the line AI, and since the point I is the center and the distance IL is the radius of the incircle of triangle ABC, this equation yields that the point X is the image of the point A in the inversion with respect to the incircle of triangle ABC. Similarly, the points Y and Z are the images of the points B and C in this inversion. And we are done.

  darij

This problem can also be solved without inversion, mainly using this 2 results:
(1) Let the incircle touch the sides of a triangle ABC at D,E,F. Then the circumcenter of triangle ABC lies on the Euler line of triangle DEF. (I suppose there is a name for triangle DEF :? just like orthic triangle and medial triangle)
(2) The inradius of the orthic triangle equals $2R\cos A\cos B\cos C$.
The second one is pretty easy to prove. For the first one, I remember it appeared in Iran MO, but I forgot which year :blush:

And another nice solution of this one, is by considering the ortic triangle of $KLM$. Then it's easy to see that this one has sides parallel to those of $ABC$ so they are homothetic. Thus the incenter of $ABC$,which is $I$, transforms in the incenter of the ortic triangle of $KLM$, which is $H$. Moreover the circumcenter of $ABC$ transforms in the circumcenter of the orthic triangle which is the center w of the 9-point circle of $KLM$. But $H,O,w$ are collinear on the Euler line of $KLM$ so $H$ lies on $OI$, and the homothety has the ratio fixed, $\frac {2R}r$, so it transforms $OI$ in $Hw$ so $Hw=wO$ and has a fixed length, so $H$ is fixed.


Let $p_{1}, p_{2}, \cdots, p_{n}$ be distinct primes greater than $3$. Show that \[2^{p_{1}p_{2}\cdots p_{n}}+1\] has at least $4^{n}$ divisors.

It can be proven that number of divisors of this number is not less than $ 2^{2^{n\minus{}1}}$.

[quote="baysa"]It can be proven that number of divisors of this number is not less than $ 2^{2^{n \minus{} 1}}$.[/quote]Really? Can you prove that? If that is true, it's very interesting.

See here, page 7
http://reflections.awesomemath.org/2008_2/cyclotomic.pdf

Hello ZetaX , isn't it possible for $ s_i$ to be a prime. Forgive me if im wrong

[quote="ZetaX"]Set $m=p_1 p_2 ... p_n$. It's well knwon that $2^{p_i}+1 | 2^m+1$.

If $p,q$ are different odd primes, we have that $\gcd(2^p+1,2^q+1) | \gcd(2^{2p}-1,2^{2q}-1) = 2^2-1=3$.
For all primes $p,q>3$ we get that $\frac{2^p+1}3, \frac{2^q+1}3$ are coprime (and clearly $>0$).
We set $s_i=\frac{2^{p_i}+1}3$.
For each $p_i$ we choose a prime $q_i$ with $q_i|s_i$. Then we set $r_i = \frac{s_i}{q_i}$. Now $r_i \neq q_i$ because otherwise $2^{p_i}+1=2q_i^2$, impossible $\mod 4$.

Consider the set $S_i := \{1,q_i,r_i,s_i\}$:
As we've seen, it really has four different elements, and elements from different sets are coprime.
Thus we get a total number of $4^n$ different products $\prod x_i$ with $x_i \in S_i$. And every such product divides $2^m+1$.[/quote]
si is prime number isnt it

I think i have found another solution to this exercise based only on Zgimondy theorem

my solution is a little bit long but the main idea that for each $2^{p_i}+1$ can be divided by two distincts primes wich are either distinct with other primes wich divide $2^{p_k}+1$ such that k>i

Maybe I'm making a mistake, but can't Zsigmondy give us an even stronger estimate than baysa's?

If we let $a_1, a_2, \dots, a_{2^n}$ be, in order, the $2^n$ divisors of $p_1 p_2 \dots p_{n}$ (there are exactly $2^n$ since the $p_i$ are distinct) by Zsigmondy's Theorem (since $p_i > 3$ we don't have to worry about the exceptional case) \[2^{a_i} + 1 | 2^{p_1p_2\dots p_n} + 1\] has a primitive prime divisor $q_i$; that is \[q_i | 2^{a_i} + 1\] and \[q_i \nmid 2^{a_j} + 1\] for any $j < i$. Then \[q_1q_2 \dots q_{2^n} | 2^{p_1p_2\dots p_n} + 1\] where the $q_i$ are distinct, so that this number has at least $2^{2^n}$ divisors.

I was referring to the user baysa's previous post in this thread.


geometry perimeter probability conics parabola quadratics algebra

Yif man12 and I have collaborated to produce a large set of getting started challenge problems which we feel you all will enjoy. We hope theres something for everyone! [b]We request, however, that you select only one divisionnamely the one that you feel will provide the greatest challenge for you without being too hardand complete some or all of the problems only in that division.[/b] We ask this because we do not want solvers that are more advanced solving problems that are too easy for them, so these problems can be left for those who will be challenged by them. The divisions are in increasing order of difficulty.

Solve away!!!

[b]Division I[/b]

1) A circle has circumference [tex]32 \pi[/tex]. What is its area?

2) Simplify: [tex]\displaystyle \frac{ (x+1) + (x-1) }{ (1+x) + (1-x) }[/tex]

3) A given square has its perimeter equal to its area. Find its side length.

4) 2/5 of all the carts are tarts, 1/3 of all the tarts are darts and 3/7 of all the darts are zarts. If there are 420 carts, then how many zarts are there?

5) If [tex]\displaystyle \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{x} = 2[/tex], find x.

[b]Division II[/b]

1) In triangle ABC, angle A is twice angle B and half of angle C. Find its measure.

2) Yasha has a special robo-dog that moves distances that follow a geometric sequence. After one second, he jumps forward one foot. A second later, he jumps back 3 feet. At the end of 6 seconds, how far is the dog from his starting point?

3) Dan receives an average of 4.5 phone calls per hour and picks up his calls after 3 rings. Dina receives an average of 2 phone calls per hour, but he picks up his calls after 5 rings. Today, Dina has already answered 2 phone calls and Dan has answered 4. If the time right now is 9:00 AM, after what hour will Dan have more rings than Dina?

4) Let the sum of the first 231 positive integer multiples of 4 be Q, and let the sum of the first 231 positive integer multiples of 3 be D. Find D - Q.

5) Ogg has a bag with 8 marbles in it, 4 red and 4 green. Suddenly a hole opens up in the bottom of the bag and the marbles begin to fall out. What is the probability that the first two marbles that fall out are different colors?

[b]Division III[/b]

1) A segment connecting the midpoints of two adjacent sides of a square has length 7. What is the area of the square?

2) Find the value of: 1  3 + 5  7 + 9  11 + ...  735 + 737.

3) An isosceles right triangle is inscribed in a circle. The circle has area 8pi. Find the sum of the lengths of the sides of the triangle.

4) If  [tex] \displaystyle \frac{x}{y} = \frac{y}{x}[/tex], find the value of [tex](x+y)^2 + (x-y)^2 - (2y)^2[/tex].

5) Let A equal the number of positive integers less than or equal to 100 with exactly 3 positive integer divisors. Let J equal the number of positive integers less than or equal to 100 with exactly 5 positive integer divisors. Find the product AJ.

[b]Division IV[/b]

1) Find all the points where the parabola [tex]y=x^2[/tex] intersects the circle [tex]x^2 + y^2 = 1[/tex].

2) (yif man12 made this one, so any complaints about the cruelty of this problem should be directed to him, not me  :D ) David is playing golf on the par-4 18th hole. He has a 3/5 chance of hitting his first shot onto the fairway, and a 2/5 chance of hitting his first shot into the rough. If his first shot goes on the fairway, he has a 2/3 chance of hitting it onto the green and a 1/3 chance of missing into a bunker. If his first shot goes into the rough, he has a 1/5 chance of getting it on the green and a 4/5 chance of missing into a bunker. If he's in the bunker, he has a 1/10 chance of getting it into the hole and a 9/10 chance of getting it onto the green. When he's on the green, he has a 4/7 chance of putting the ball into the hole. lf he misses his first putt, he will always make his second. What is the probability that he will get a par (hit exactly 4 shots) on the hole?

3) Find the value of [tex]\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}}+ \dots + \frac{1}{\sqrt{2003} + \sqrt{2004}}[/tex].

4) Prove that for positive integers a and b, [tex]\sqrt{a} + \sqrt{b}[/tex] is an integer if and only if a and b are both perfect squares.

5) Find all right triangles such that their side lengths form an arithmetic sequence.

I'm glad people are still making questions -- I bet Scrambled has some he would donate if you do this again.  I would suggest, however, that you make the three different divisions three different posts in order to make it clearer who is answering what.

Well, the reason I posted them all together was so it would be clear that people should only answer one set of problems. If the post were split into four smaller ones, it would be harder for me to made the instructions clear. When people solve these problems, perhaps they can state at the beginning of their post which division of problems they are solving.

Plus, if the post were split up then it wouldn't be a lollapalooza anymore  :D . I like lollapaloozas.

Lollapalooza got canceled  :(  seems like this is a bad summer for concerts in general...

1) In triangle ABC, angle A is twice angle B and half of angle C. Find its measure.

[hide]

A=2x

B=x

C=4x

7x=180

x=25 5/7

A=51 3/7[/hide]

2) Yasha has a special robo-dog that moves distances that follow a geometric sequence. After one second, he jumps forward one foot. A second later, he jumps back 3 feet. At the end of 6 seconds, how far is the dog from his starting point?

[hide]

1 second :  1

2 seconds: -2

3 seconds: -1

4 seconds: -4

5 seconds: -3

6 seconds: -6

6 units[/hide]

4) Let the sum of the first 231 positive integer multiples of 4 be Q, and let the sum of the first 231 positive integer multiples of 3 be D. Find D - Q.

[hide]

D= 4*1+4*2+4*3+...+4*230+4*231

D= 4(1+2+3+...+230+231)

D= 4(26796)

D= 107184

Q= 3*1+3*2+3*3+...+3*230+3*231

Q= 3(1+2+3+...+230+231)

Q= 3(26796)

Q= 80388

107184-80388=26796

26796[/hide]

5) Ogg has a bag with 8 marbles in it, 4 red and 4 green. Suddenly a hole opens up in the bottom of the bag and the marbles begin to fall out. What is the probability that the first two marbles that fall out are different colors?

[hide]

R G

R R

G G

G R

2/4=1/2[/hide]

Division IV:

[hide]

1.  We can rewrite x^2 + y^2 = 1 as y=sqrt[1-x^2].  Now we just need to find x such that x^2 = sqrt[1-x^2].  Squaring both sides and putting all the terms on one side, we get x^4 + x^2 - 1.  Let z = x^2.  By the quadratic formula, the solutions are (-1 :pm: sqrt[5])/2, thus x = sqrt[(-1 :pm: sqrt[5])/2].  But sqrt[(-1 - sqrt[5])/2 is imaginary.  But we still have two points: the positive square root of (-1+sqrt[5])/2, and the negative square root of (-1+sqrt[5])/2.  Thus our two points are (sqrt[(-1 + sqrt[5])/2], (-1 + sqrt[5])/2) and (-sqrt[(-1 + sqrt[5])/2], (-1 + sqrt[5])/2).

[/hide]

I'm stuck on number two, though

[hide]

3.  Rationalizing the denominator in each term, we get sqrt[2]-sqrt[1]+sqrt[3]-sqrt[2]+sqrt[4]-sqrt[3]+sqrt[5]-sqrt[4]...+sqrt[2004]-sqrt[2003] and a lot of terms cancel and leave sqrt[2004]+1.

5.  x^2 + (x+d)^2 = (x+2d)^2

Expanding and simplifying,

2x^2 + 2xd + d^2 = x^2 + 4dx + 4d^2

x^2 = 2xd + 3d^2

x^2 - 2xd - 3d^2 = 0

(x-3d)(x+d) = 0

The first factor gives x=3d.  Plugging this into the first equation gives

(3d)^2 + (4d)^2 = (5d)^2

(9+16)d = 25d

9+16 = 25

This is always true.  But the other factor of (x-3d)(x+d)=0 implies that x is negative, and we can't have a triangle with negative side lengths.  Thus if our arithmetic sequence has common difference d, then our triangle has side lengths 3d, 4d, and 5d.

[/hide]

nr1337 wrote:Division IV:
[hide]
1.  We can rewrite x^2 + y^2 = 1 as y=sqrt[1-x^2].  Now we just need to find x such that x^2 = sqrt[1-x^2].  Squaring both sides and putting all the terms on one side, we get x^4 + x^2 - 1.  Let z = x^2.  By the quadratic formula, the solutions are (1 :pm: sqrt[5])/2, thus x = sqrt[(1 :pm: sqrt[5])/2].  But sqrt[(1 - sqrt[5])/2 is imaginary.  But we still have two points: the positive square root of (1+sqrt[5])/2, and the negative square root of (1+sqrt[5])/2.  Thus our two points are (sqrt[(1 + sqrt[5])/2], (1 + sqrt[5])/2) and (-sqrt[(1 + sqrt[5])/2], (1 + sqrt[5])/2).
[/hide]

Hmmm...[hide]it said all points, which I think includes two imaginary intersections. y=x^2 is intersecting x^2+y^2=1. Substitute and use the quadratic to get y=(-1 :pm:  :rt5: )/2=x^2, so there are four points:

[tex]\left(\sqrt{\frac{-1+\sqrt(5)}{2}},\frac{-1+\sqrt{5}}{2}\right)[/tex]

[tex]\left(-\sqrt{\frac{-1+\sqrt(5)}{2}},\frac{-1+\sqrt{5}}{2}\right)[/tex]

[tex]\left(\sqrt{\frac{-1-\sqrt(5)}{2}},\frac{-1-\sqrt{5}}{2}\right)[/tex]

[tex]\left(-\sqrt{\frac{-1-\sqrt(5)}{2}},\frac{-1-\sqrt{5}}{2}\right)[/tex]

The last two, of course, are imaginary. A minor detail.[/hide]

MathFiend wrote:2) Yasha has a special robo-dog that moves distances that follow a geometric sequence. After one second, he jumps forward one foot. A second later, he jumps back 3 feet. At the end of 6 seconds, how far is the dog from his starting point?
[hide]
1 second :  1
2 seconds: -2
3 seconds: -1
4 seconds: -4
5 seconds: -3
6 seconds: -6
6 units[/hide]

this was partially my fault because i made the problem a little unclear. the robo-dog's movements are the distances in a geometric progression. so, your progress for the dog from the 3 second on is wrong.

but all the other problems you've done are correct. try number 3 and correct number two.

[quote="nr1337"]But how can we have imaginary intersections?[/quote]
An intersection is just where the values are equal.

Hmm, an intersection is where the graphs meet. You don't usually include imaginary values in that.

Yes, confuted, the original problem used the words "parabola" and "circle", which I believe are confined to real coordinates.

Scrambled wrote:nr1337 wrote:Division IV:[hide]
3.  Rationalizing the denominator in each term, we get 2-1+3-2+4-3+5-4...+2004-2003 and a lot of things cancel and leave 2003.

[/hide]
[hide]Um... isnt it (( sqrt 2004)-1)?[/hide]

Oops. I somehow forgot the radicals.

Edit:Fixed A Slight Error in #4

DIViv

2.[hide]A little notation will make this solution easier. Fairway=F, Rough=R, Bunker=B, Green=G, Hole=H, and Missed Putt=M. So, we find the paths that lead to four shots and then add the probabilites of each. They are: (F,G,M,H), (F,B,G,H), (R,G,M,H), and (R,B,G,H). So, the probability is:

(3/5)(2/3)(3/7)(1)+(3/5)(1/3)(9/10)(4/7)+(2/5)(1/5)(3/7)(1)+(2/5)(4/5)(9/10)(4/7)=414/875[/hide]

4.[hide]First, let's assume a is not a perfect square, and no specifications for b. For rt(a) + rt(b) to be an integer, rt(a) has to be in the form N_1 + d where N_1 is an integer and 0<d<1 and rt(b) has to be in the form N_2 - d where N_2 is an integer. So, squaring each we find: a = (N_1):^2: + 2(N_1)(d) + d:^2: and b = (N_2):^2: + 2(N_2)(d) + d:^2:, since N_1 and N_2 are integers, (N_1):^2: and (N_2):^2: are also integers. Since a and b must be integers, b - a must also be an integer, so:

b - a = ((N_2):^2: - (N_1):^2:) + d(2(N_2)  - 2(N_1)). For b - a to be an integer, d(2(N_2)  - 2(N_1)) must be an integer, because we know ((N_2):^2: - (N_1):^2:) is an integer. Since a is not a perfect sqaure, rt(a) is irrational and as a result, d is irrational because we defined N_1 to be an integer. Furthermore, we know 2(N_2) - 2(N_1) is an integer, which is rational, so d(2(N_2)  - 2(N_1)) is only an integer and also only rational when d is rational and since we already said d is always irrational, we have a contradiction, so a has to be a perfect square. So, since a is a perfect square, rt(a) = N_1 + 0 = N_1 and for rt(a) + rt(b) to be an integer, rt(b) must be an integer. So, rt(b) = N_2 -> b = (N_2)^2. So, we know a and b must be squares of integers in order for rt(a) + rt(b) to be an integer.[/hide]

Bassooner wrote:in spoiler [hide]
In saying that the golfer gets exactly par 4, it can be assumed that he will get 2, 3, 5, 6, 7, 8 ... shots.  However, with what we are given, the highest number of shots possible is five; onto the rough/fairway, onto the bunker, onto the green, miss the hole, and into the hole.  The least possible number is three: onto the rough, into the bunker, into the hole; onto the rough, onto the green, into the hole; onto the fairway, onto the green, into the hole; onto the fairway, into the bunker, and into the hole.  There is no possible way to get two shots: You would need to go staight onto the green, or from the fairway/rough into the hole.  There is no possibilities of this, or a hole in one mentioned, so we can be assured that the least possible number of shots is three.  I took the probabilities of all of the pathways for the golfer to get 5 shots, and multiplied each sequence, then added the two sequences together.  These would be [tex]\frac{2}{5}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{3}{7}\cdot\frac{1}{1}=\frac{216}{775}[/tex] and [tex]\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{9}{10}\cdot\frac{3}{7}\cdot\frac{1}{1}=\frac{27}{350}[/tex].  When these two probabilities are added, the result is [tex]\frac{3861}{10850}[/tex].
The next section that needs to be solved is the 3 shot venture.  [tex]\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{1}{10}=\frac{1}{50}[/tex], [tex]\frac{3}{5}\cdot\frac{2}{3}\cdot\frac{4}{7}=\frac{24}{105}[/tex], [tex]\frac{2}{5}\cdot\frac{1}{5}\cdot\frac{4}{7}=\frac{8}{175}[/tex] and, [tex]\frac{2}{5}\cdot\frac{4}{5}\cdot\frac{1}{10}=\frac{8}{250}[/tex].  These are the fairway, bunker, hole; fairway, green, hole; rough, green, hole; and rough, bunker, hole (respectively).  Adding these three probabilities, we get the fraction [tex]\frac{571}{1750}[/tex].
Adding the two probabilities of getting 5 or 3 shots, we get [tex]\frac{18503}{27125}[/tex].
To find the probability of getting a par 4, we subtract the above probability from 1, because the golfer can't have more that 5 shots, or less than 3.
The final answer is [tex]\frac{8622}{27125}[/tex]!:-)
[/hide]

lol, sorry about making you do so much work and having nothing to show for it, but you made a  mistake somewhere in the problem. you may have added the fractions wrong, but what you got isn't the correct answer

Division III

1) A segment connecting the midpoints of two adjacent sides of a square has length 7. What is the area of the square?

[hide] The segment is the hypotenuse of a 45 45 90 right triangle. Then the length of half a side of the square is (7sqrt2)/2. Then the length of a side of the square is 7sqrt2. Then the area of the square is 49(2)=98.[/hide]

2) Find the value of: 1  3 + 5  7 + 9  11 + ...  735 + 737.

[hide] If you group the terms together so that there are two in each group, you see that the sum is -2. (except for the last term, it's solo) To find the group number, you can use positive term=1+4(group number-1). Solving for the last group, which includes 733 and -735, the group number of 733 and -735 is 183. So, we we have the sum of the terms is -2(183) + 737 (which is the last term without a group). Solving, we have 371. [/hide]

3) An isosceles right triangle is inscribed in a circle. The circle has area 8pi. Find the sum of the lengths of the sides of the triangle.

[hide] An isosceles right triangle has a 90 degree angle in it. Therefore, the hypotenuse is the diameter. I'm sorry , but I don't remember the name of the theorem. So, lets call the diameter of the circle x. Then 8pi=(x/2)^2 pi. Solving, x=sqrt32. Then x=4sqrt2. Then the length of hte hypotenuse is 4sqrt2. Then the lengths of the other two sides are 4. So, the sum of the lengths of the triangle is 4+4+4sqrt2 = 8+4sqrt2.[/hide]

4) If \displaystyle \frac{x}{y} = \frac{y}{x}, find the value of (x+y)^2 + (x-y)^2 - (2y)^2.

[hide] Expanding the sum of the terms, and simplifying, you get 2x^2 - 2y^2. Since x^2=y^2 by the equation, the sum of the terms is 0. [/hide]

5) Let A equal the number of positive integers less than or equal to 100 with exactly 3 positive integer divisors. Let J equal the number of positive integers less than or equal to 100 with exactly 5 positive integer divisors. Find the product AJ.

[hide] According to the def of A, there are three divisors-- 1, the number, and a third divisor. Since the third divisor is not part of "a pair" of different divisors  that multiply to achieve the number, its "partner" must be itself. Thus it the number is a square number. There are 9 of those between 1 and 100. (Squares of 2-10) Remember that 1 cannot be used; it would be repeated. According to the def of J, there are five divisors. Using the same thinking, they are 1, the number, the square root of the number, and a pair of different divisors. This means that the number is the product of a square root and two other divisors. Also, this pair of divisors must be low enough to accomodate for the 'less than or equal to 100" rule. Therefore, 25 cannot be the the square number. It is too great. (2*3*25=150) 16 can. (2*3*16=96) So, if we examine all of the possibilities for 16,9,4 we can find J. Remember that 1 cannot be used. Since 16 is a multiple of 4, it suffices that we find the possibilities for 4 and 9. If we look at the possible divisors, they must be 3,5,7, and 11 if 4 is the square number used. Also, 2 and 5 in the case 9 is the square number used. In the case that 4 is the square number used, we have 4C2 = 6 possibilities. In the case that 9 is the square number used, we have 1 possibility. Still, we must eliminate any intersections. There is only one, of course, 90. So this makes 6 possibilities alltogether. J=6 So, AJ = 9(6)=54. AWD![/hide]

mlee06--

[hide]For #5, not all squares have just 3 divisors--what about 64?[/hide]

Ok, now I'll try to salvage the last problem... I really don't know what I was thinking...

[hide]  According to the def of A, there are three divisors-- 1, the number, and a third divisor. Since the third divisor is not part of "a pair" of different divisors that multiply to achieve the number, its "partner" must be itself. Thus it the number is a square number. But we must take care that the square root of the number does not have any divsiors in itself; it must be prime. There are 4 of those between 1 and 100. (Squares of 2,3,5, and 7) Remember that 1 cannot be used; it would be repeated. According to the def of J, there are five divisors. Using the same thinking, they are 1, the number, the square root of the number, and a pair of different divisors. The numbers between 1 and 100 that are square and have 5 divisors are 16 and 81. (Notice that they are squares of squares...) So this makes 2 possibilities alltogether. J=2 So, AJ = 4(2)=8. AWD! [/hide]

Division III

1) A segment connecting the midpoints of two adjacent sides of a square has length 7. What is the area of the square?

[hide]The segment of length 7 cuts off a 45-45-90 triangle with legs half the length of the side of the square.  Since it's a 45-45-90, the legs of the square are length 7/:sqrt:2, so the side of the square is 7:sqrt:2 and the area of the square is 98.[/hide]

2) Find the value of: 1  3 + 5  7 + 9  11 + ...  735 + 737.

[hide]...=(1-3)+(5-7)+...+(733-735)+737

=-2-2-...-2+737

(There are (733-1)/2 = 366 odd numbers from 1 to 733 -- that makes 183 pairs.)

...=-2(183)+737

=-366+737

=371[/hide]

3) An isosceles right triangle is inscribed in a circle. The circle has area 8pi. Find the sum of the lengths of the sides of the triangle.

[hide]An isoceles right triangle is a 45-45-90.  An angle inscribed in a circle intersects an arc whose measure is twice as much as the angle, so the right angle intersects an arc of 180 degrees, making the hypotenuse of the triangle the diameter.

The area of the circle is 8:pi:, its radius is 2:rt2: the diameter is 4:rt2:.

Again, the triangle is 45-45-90. So the legs are 4:rt2:/:rt2:=4, the sum of the lengths of the sides is 8+4:rt2:.[/hide]

4) If  [tex] \displaystyle \frac{x}{y} = \frac{y}{x}[/tex], find the value of [tex](x+y)^2 + (x-y)^2 - (2y)^2[/tex].

[hide]Cross-multiply the first eqn to get x^2=y^2.

(x+y)^2 + (x-y)^2 - (2y)^2

=2x^2+2y^2-4y^2

=2x^2-2y^2

Plug x^2=y^2 back in -- the expression = 0[/hide]

5) Let A equal the number of positive integers less than or equal to 100 with exactly 3 positive integer divisors. Let J equal the number of positive integers less than or equal to 100 with exactly 5 positive integer divisors. Find the product AJ.

[hide]An integer with an odd number of divisors is a perfect square, since divisors always come in pairs except when two of them are equal.  The 3 divisors are 1, sqrt of n, and n, so n must be the square of a prime.  Thus, A: {4, 9, 25, 49}, A=4

For 5 divisors, you need squares of squares of primes. =)  The first divisor is 1, the 2nd is :sqrt:p, the 3rd is p, the 4th is p:sqrt:p, and the 5th is p:^2:

J: {16, 81}, J=2

AJ=8[/hide]

whew, I'm glad that I had the same answer for #5 division 3 as fluffybunny123...

I'm going to Division i right now, then probably the others later -

1.[hide]A circle has circumference 32(pi)32. What is its area?

Circumference = 2r(pi)

Area = r2(pi)

2r(pi) = 32(pi)

r= 16

Area = r2(pi) = 162(pi) = 256(pi)[/hide]

2.[hide]([x+1] +[x-1])/([1+x] + [1-x])

(x+1+x-1)/(1+x+1-x) = (2x)/(2) = x[/hide]

3.[hide]A given square has its perimeter equal to its area. Find its side length.

Perimeter = 4s

Area = s2

4s =  s2

4=s

length = 4[/hide]

4.[hide] 2/5 of all the carts are tarts, 1/3 of all the tarts are darts and 3/7 of all the darts are zarts. If there are 420 carts, then how many zarts are there?

2/5 * 1/3 * 3/7 = 6/105

6/105 = x/420

105x = 2520

x=24

24 zarts[/hide]

5.[hide]1/2 + 1/3 + 1/4 + 1/5 + 1/x = 2

1/2 + 1/3 = 5/6

5/6 + 1/4 = 13/12

13/12 + 1/5 = 77/60

77/60 + 1/x = 2

1/x = 43/60

43x = 60

x = 60/43[/hide]

Well, that's it for now.

Now I'll try Division II

1.[hide] In triangle ABC, angle A is twice angle B and half of angle C. Find its measure.

Total measure of Angles in a triangle = 180

so, A + B + C = 180

B = A/2

C = 2A

A + A/2 + 2A = 180

(2A+A+4A)/2 = 180

7A/2 = 180

7A = 360

A = 51 6/14

B = 25 13/14

C = 102 12/14[/hide]

2.[hide]Yasha has a special robo-dog that moves distances that follow a geometric sequence. After one second, he jumps forward one foot. A second later, he jumps back 3 feet. At the end of 6 seconds, how far is the dog from his starting point?

1 second travel = 1-3

1 second travel = (-2)

6 second travel = 6*(1 second travel)

6 second travel = 6*(-2) = (-12)

abs(-12) = 12

12 feet

[/hide]

3.[hide]Dan receives an average of 4.5 phone calls per hour and picks up his calls after 3 rings. Dina receives an average of 2 phone calls per hour, but he picks up his calls after 5 rings. Today, Dina has already answered 2 phone calls and Dan has answered 4. If the time right now is 9:00 AM, after what hour will Dan have more rings than Dina?

Dan's R/H = 4.5*3 = 13.5

Dina's R/H = 2*5 = 10

Dan so far = 4*3 = 12

Dina so far = 2*5 = 10

Dan already has more rings than Dina.[/hide]

Are you sure this one was written correctly?

4.[hide]Let the sum of the first 231 positive integer multiples of 4 be Q, and let the sum of the first 231 positive integer multiples of 3 be D. Find D - Q.

2-1 = 4-3

1+2+3+4 = 10 = the first 4 positive integer multiples of 1

2*(1+2+3+4) = 20 = the first 4 positive integer multiples of 2

10 - 20 = -10 = -(1+2+3+4)

D-Q = -(1+2+3+4...231) = -([231(231+1)]/2) = -([231*232]/2) = -(26796)

D-Q = -26796[/hide]

5.[hide]Ogg has a bag with 8 marbles in it, 4 red and 4 green. Suddenly a hole opens up in the bottom of the bag and the marbles begin to fall out. What is the probability that the first two marbles that fall out are different colors?

Probability that the first marble will be a color = 1

Probability that the second marble will be a different color than the first one = 4/7 (since one marble of the different color fell out)

1*(4/7) = 4/7[/hide]

[quote=Symmetry24]5) Find all right triangles such that their side lengths form an arithmetic sequence.

You set least side be x. x^2+(x+1)^2=(x+2)^2, So x=-1 or 3

So two right triangles work: (-1,0,1) And (3,4,5)[/quote]

Normally, I would complain about ppl necroing long gone posts, but this one has quite a nice set of good problems for others to have fun with.

Anyways, Symmetry24, one of your answers is too silly to be realistic.

[hide=Division IV 5]
Let the right triangle have sides $a,a+d,a+2d$ where $a$ and $d$ are positive.
$2a^2+2ad+d^2=a^2+4ad+4d^2$
$a^2-2ad-3d^2=0$
We see this is a quadratic in $\frac{a}{d}$.
Let $\frac{a}{d}=x$
This leads to $a^2=d^2x^2$ and $a=dx$.
$d^2x^2-2d^2x-3d^2=0$
$d^2(x^2-2x-3)=0$
Since $d^2$ is always positive, $x^2-2x-3=0$
This factors into $(x-3)(x+1)=0$.
Clearly, $\frac{a}{d}$ is positive since both are positive so $x=3$.

We now have that $(3d,4d,5d)$ is the only possible right triangle.


geometry
parallelogram
trigonometry
Pythagorean Theorem
area of a triangle
Heron\u0027s formula

I'll list four geometry problems that are very basic.

I'll list one by one.

#1

Given that RE is an altitude of :tri: RST and RM = TM, find the area of the triangle NME.

Image can be found on my website:

http://dhost.info/jwec05/Silverfalcon/

Click on that page and click the first image on the top. My site doesn't seem to allow hotlinking.. ;)

Next problem:

Given that l || m || n, find the value of x,y, and z.

The link to the problem is same as above except the second picture.

To DePopov,

You can figure it out whether they're parallel or not.

How can you be sure they'ren't parallel?

#3

Given that PQRS is a parallelogram and PA = PB

Prove that AS = BQ

Again, the same link as above except the third image. :lol:

Last one :?

Given that PR || VO; RO || PV; PR :cong: RO

Prove that $\angle ROP$ and $\angle RVO$ are complementary.

Again, this is fourth image.

Solution

[hide]
First, we draw $RE$. Since it is an altitude we can use the Pythogrean theorem and see that $RE=24$. Again using the Pythagorean Theorem on $\triangle RET$ we find that $RT=26$. Thus, $RM=MT=13$. From here we will subtract the areas of $\triangle SNE$, $\triangle RMN$, and $MET$ from $\triangle RST$ to find the area of $\triangle MNE$. First note that $\sin \angle SRT=\frac{28}{26} \cdot \frac{24}{30}=\frac{56}{65}$ from Law of Sines.

\[[RST]=\frac{1}{2} \cdot RS \cdot ST \cdot \sin \angle RST\]
\[[RST]=15 \cdot 28 \cdot \frac{24}{30}=336\]
\[[SNE]=\frac{1}{2} \cdot NS \cdot SE \cdot \sin \cdot \angle RSE\]
\[[SNE]=9 \cdot 15 \cdot \frac{24}{30}=108\]
\[[MET]=\frac{1}{2} \cdot TE \cdot MT \cdot \sin \angle RTE\]
\[[MET]=5 \cdot 13 \cdot \frac{24}{26}=60\]
\[[RMN]=\frac{1}{2} \cdot RN \cdot RM \cdot \sin \angle SRT\]
\[[RMN]=7.5 \cdot 13 \cdot \frac{56}{65}=84\]
\[[MNE]=[RST]-[SNE]-[MET]-[RMN]=336-108-60-84=\fbox{84}\]

[/hide]

Easier way:

[hide="NO_TRIG"]Since RE is an altitude, this forms a right triangle and by the theorem, the segment that contains midpoint of hypotenuse and the opposite vertex is equal to the half as long as the hypotenuse therefore:

SN = RN = NE = 15.

For NM, since it's a triangle and NM is a median, it's (18+10)/2 = 14.

For ME, it's 13, given.

Using a Heron's Formula of a triangle that has side 13,14,15, the area is 84.[/hide]

$MN$ is not a median although it does have length 14.

I'm not sure whether MN is really median or not but MN connects the midpoint of the two segments of the triangle and is parallel to the third side, therefore, it has half as long as the third side.

This is a theorem, if you didn't notice in case. ;)

My answer is different from your guys's for the first one.

[hide]Using the pythagorean theorom for $\triangle ERS$, we can find out that $RE = 24$.  We can use the pythagorean theorom again, for $\triangle ERT$, and get $RT = 26$.  $RM=MT=13$, since $M$ is the midpoint of $RT$.  We know that $\triangle RST \sim \triangle RNM$ by SAS, since $\frac{15}{3} = \frac{13}{26}$, and $\angle R$ is in both triangles.  We know that the sides of $\triangle RNM$ is one half of the sides of $\triangle RST$, making $NM = 14$.  I'm going to call the point of intersection for $NM$ and $RE$ point $P$.  We know that $RP = \sqrt{15^2-x^2}$, and $RP = \sqrt{13^2-{(14-x)^2}}$.  From those equations, we get $x = 9$, making $RP = 12$, and $PE$, an altitude of $\triangle NME$, equal $12$.  So we have a base, $NM=14$, and an altitude, $PE=12$, so the area of $\triangle NME = 84$.[/hide]

No offense but if you see that RE is an altitude (which I forgot to draw in a diagram), you should apply this theorem without using the Pythagorean Theorem.

This theorem is especially nice because in case of triangle where only thing you're given is the side of the hypotenuse and ask to find the segment that contains the opposite vertex and midpoint of the hypotenuse.

Then here's easy challenge:

Prove that above theorem, the segment connecting the midpoint of the hypotenuse and the opposite vertex is half as hypotenuse.

 :)

For the problem, 'last one' posted by Silverfalcon, here's a hint:

[hide]It's fairly clear that $PROV$ is a parallelogram, but can you say more than that?[/hide]

Furthermore,

You have to prove the detailed fact of that parallelogram before doing anything else.

 :)


As a whole, I am both safe and secure.
Behead me, and I become a place of meeting.
Behead me again, and I am the partner of ready.
Restore me, and I become the domain of beasts.

Who or what is the “I” referred to in the problem?

Source: Puzzle Periodical

[quote="daine"][hide]Stable
Table
Able
What exactly does it mean by "restore me?" does it mean to put any letter on the beggining? If so, then perhaps the last word is fable.[/hide][/quote]

[hide]I think it means to add the letters back on that you took off.

i.e. Stable

Not sure, it could mean something else, though[/hide]

a stable can hold animals so that makes sense


factorial
number theory unsolved
number theory

I came across this problem on the internet (I think it is from MOP '00).  I know how to find the zeroes of a factorial, but would like to see how someone would approach this.

How many zeroes does $6^{5^4}+4^{5^6}$ end in?

Thanks.

I think the answer is 625, but I am unsure about the validity of my justification. I'll come back to this when I am feeling more convinced. :weightlift:
It'll probably take me a day, because I have a lecture to prepare for tomorrow  :?

Just wondering:  Why did you choose mod 25?

Thank you.

I chose mod 25 because it worked. If I had to go larger, I would have.