Please try to show all work:
\[\frac{x^{5}+5x^{4}+0x^{3}+0x^{2}+3x+21}{x^{3}+0x^{2}+3x-1}\]
Edit: by the way, the problem originally read:
\[\frac{x^{5}+5x^{4}+3x+21}{x^{3}+3x-1}\]
***Please use division not factoring****

with a easy google search

http://www.sosmath.com/algebra/factor/fac01/fac01.html

the denominator is quasi-equal to yours.

The link is a quadratic, while this is not.

[quote="AstroPhys"]Division of polynomials is just like that of regular numbers, you multiply the divisor to get as close to the dividend as possible, without going over, and the multiple is part of the answer; then you subract like in regular numerical division.
You have the right idea with the $0x^{n}$ terms, it helps keep the orders of $x$ in place.[/quote]

Same in both threads.

[quote="kohjhsd"][quote="DaReaper"][quote="skimnc"]

Thus whats leftover at the end is the remainder.  So the solution is $x^{2}+5x-3-\frac{14x^{2}-17x-18}{x^{3}+3x-1}$[/quote]

is your solution the same as:
\[x^{2}+5x-3\frac{-14x^{2}+17x+18}{x^{3}+3x-1}\]
, or is the way that i wrote it incorrect as a whole?[/quote]
It's same if you intended to write
\[x^{2}+5x-3+\frac{-14x^{2}+17x+18}{x^{3}+3x-1}\]
[/quote]

but if you were going to write \[x^{2}+5x-3+\frac{-14x^{2}+17x+18}{x^{3}+3x-1}\]
would you not have to have put the $-14x^{2}+17x+18$ part in parentheses because you cannot have two signs next to each other.

[quote="DaReaper"]
would you not have to have put the $-14x^{2}+17x+18$ part in parentheses because you cannot have two signs next to each other.[/quote]
As long as the sign is above the division bar its all good

in  $x^{2}+5x-3-\frac{14x^{2}-17x-18}{x^{3}+3x-1}$ doesnt the minus sign only get multiplied by the $14x^{2}$ ??

The minus is in front of the fraction thus it applies to the entire thing.

For example:

$-\frac{x-3}{7}= \frac{3-x}{7}$


Prove that
if
$sin(a)+sin(b)=1$ and $cos(a)+cos(b)=1$
then
$tan(\frac{a}{2})+tan(\frac{b}{2}) =1$

[hide]
Let $x=\cos a+i\sin a$ and $y=\cos b+i\sin b$, then $x+y=1+i$. Also, $x$ and $y$ have absolute value 1, so $x\overline{x}=1,\ y\overline{y}=1$. Conjugating both sides of $x+y=1+i$ gives
\[\frac{1}{x}+\frac{1}{y}=1-i\]
which is equivalent to $xy=i$ after simplifying and substituting for $x+y$.

The equations for $x+y$ and $xy$ give $(x-1)(y-1)=0$. Therefore one of $x$ or $y$ is 1, the other must then be $i$ for their sum to be $1+i$. WLOG let $x=1$ (for the case $y=1$, just switch $x$ with $y$ in the remaining part of the proof). Then $\cos a=1,\sin a=0,\cos b=0,\sin b=1$. This means $a\equiv 0^\circ$ and $b\equiv 90^\circ$ (they can only differ by multiples of $360^\circ$, so
\[\tan \frac{a}{2}+\tan \frac{b}{2}=\tan 0^\circ+\tan 45^\circ =1\]
as desired.
[/hide]

it is not that hard to just solve it normally...just square and add...

2(sina sinb+cosa cosb)+2=2

cos(a-b)=0...

[hide="solution"]$\sin{a}+\sin{b}=1$

$\sin^{2}{a}+\sin^{2}{b}+2\sin{a}\sin{b}=1$

$2\sin{a}\sin{b}=0$

So one of them is 0 and the other is 1.

So one of the angles is 0 and the other is 90(degrees).

So we just need to compute

$\frac{0}{1}+\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$[/hide]

[quote="hien"][quote="1=2"][hide=solution]$\sin{a}+\sin{b}=1$

$\sin^{2}{a}+\sin^{2}{b}+2\sin{a}\sin{b}=1$

$2\sin{a}\sin{b}=0$

[/quote]

I don't think it's true![/quote]
It isn't, he's confusing $\sin^{2}x+\cos^{2}x=1$ with it.


Hi,

The question is: In how many ways can 7 taffy and 8 licorice be distributed to 5 kids, such that each get exactly 3 pieced of candy?

Solution:
   We can rephrase the problem as counting the number of ways to distribute 7 taffy to 5 kids, such that none gets more than 3. Then, we give each kid enough licorice so that each kid has exactly 3 pieces.
   If we didn't have the "no more than 3 pieces" restriction, then the number of ways to distribute 7 pieces of taffy to 5 kids is $ \binom{11}{4} \equal{} 330.$ But we need to exclude the cases where some kid gets [b]4 or more pieces[/b]. Note that only one kid could receive more that 3 pieces, and we have 5 choices for which kid that is. Once we allocate 4 pieces to that kid, we have to distribute the remaining 3 pieces among 5 kids which we can do in $ \binom{7}{4} \equal{} 35$ ways.
   Thus, there are 330-5(35) = 155 ways to distribute the taffy...

My question is, why are they only considering giving one of the kids 4 pieces? For example, here they say: [i]Once we allocate 4 pieces to that kid, we have to distribute the remaining 3 pieces among 5 kids... [/i]

Why can't they allocate 5, 6, or 7 candies to that kid also, why only 4?

Thank you for the help.

They're leaving open the possibility that some of the remaining pieces will also go to that kid.  Note that it says 5 kids at the end of that sentence, not 4.

While I was trying this out, I thought of a follow-up question that I couldn't solve. Does anyone have a solution?

In how many ways can 7 taffy and 8 licorice be distributed to 5 kids, such that each get no more than 3 pieces of candy (not all of the candy must be distributed).

Interesting problem, veezbo. The only thing I can think of right now is casework by the number of undistributed candy. Ex. Case 1: 0 undistributed, Case 2: 1 undistributed...and so on.  That would organize the information but in a rather nasty manner.


Proof the following:

Let $ A_1, A_2, ... , A_n$ be a cyclic polygon, if the sum of inradii does not depend on the triangulation of the polygon, then the polygon is cyclic.


What is the number of inches in
the positive difference between
the perimeters of the two regular
polygons?

[asy]size(100); picture p; path outpath; for(int i = 0; i<8; ++i) outpath = outpath -- dir(360/8*i - 360/16); draw(p,outpath--cycle); label(p,"15''",(1.5,0),fontsize(10pt));

path outpath;
for(int i = 0; i<3; ++i) outpath = outpath -- 1.3*dir(360/3*i-360/12); draw(outpath--cycle); label("33''",(-1.2,.4),fontsize(10pt));
add(shift(3*right+.22*up)*p);[/asy]

Perimeter Triangle: $ 33 \times 3 \equal{} 99$

Perimeter Octagon: $ 15 \times 8 \equal{} 120$

Positive Difference: $ 120 \minus{} 99 \equal{} \boxed{21}$


Hello!
   My friend bets me to find a function that can describe the black hole. I found the picture of black hole in http://www.scienceteecher.com/black_hole.htm . Can anyone attempt to use Computer system Algebra to draw it?
  And anyway, can you show me some orther cool shape please? :D

Looks like a catenoid.  Shouldn't be too hard to plot.

Thats a picture of half of a black hole. And its not very presice at that.


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166848

^^

I don't think any punishment is necessary, just a PM warning.

EDIT: Now that I think about it, this topic could just be renamed and moved, not deleted.

[quote="i_like_pie"]Am I missing something here? :huh:[/quote]

See, the thing is that
1) There's already a MathCounts Marathon, this will confuse users
2) This doesn't belong in the Mathcounts forum.

So, yeah...

Ah, I didn't see the title. I think the topic should simply be locked.


Here is a nice problem of our national olympiad
Hope you enjoy it.
Let $\mathcal O$ be a point in a our plan.and $\mathcal F$ a polygon in our plan.(It can be convex and concave ).
Let $\mathcal P$ be the premeter of $\mathcal O$.
And $\mathcal D$ be sum of the distances between the $\mathcal O$ and vertices of$\mathcal F$.
And $\mathcal H$ the sum of the distances between the sides of the $\mathcal F$ and $\mathcal O$

Prove :

$\mathcal D^2 -\mathcal H^2 \ge \frac{\mathcal P^2}{4}$

This is from a shortlist. I posted it before, but it got no answers.

Really?from a shorlist,
So i think Our instructors have copied it from the shortlist in to one of the TST exams :D .
If any one is interested in the solution ,then i can give it?

Wow,Its challenging indeed.
Inthe solution that i have :
You sould first prove it for the Polygon with two sides.

And another hint is the equality happends if and only if $\mathcal F$ be a convex polygon with equal angles and equal side(what is the english name of this kind of polygon?)and $\mathcal O$ be the center of the circle with goes though the vertices of $\mathcal F$

PS.I know this kind of polygon has a name and the circle going thorugh the vertices has a name too , but i didnt find them in my dictionary.


[color=brown]Let be given a circle  with center $O$ and two points $M,\,N$ inside its so that $OM=ON.$ Let point $S$ lies on the circle and let $E,\,F$ be the points of intersection of $SM,\,SN$ and $(O)$ respectively. The tangents at $E,\,F$ to $(O)$ intersect at $I.$ Prove that the orthogonal bisector of the segment $MN$ passed through the midpoint of segment $SI$[/color]

[color=indigo]To prove this, I use a [b]Lemma.[/b][/color]

[quote][color=darkred][b]Lemma.[/b] Let be given a circle with center $O$ and two points $M,\,N$ inside its so that $O$ be the midpoint of $MN.$ Let point $S$ lies on the circle and let $E,\,F$ be the points of intersection of $SM,\,SN$ and $(O)$ respectively. Assume that $MN$ intersect $EF$ at $P,$ $SO$ intersect $(O)$ at second times $Q.$ Then $PQ$ is a tangent to circle $(O)$[/color][/quote]

[color=red]And do you have another idea?[/color]

are you sure the problem is correct?
[img]http://me.yogurttree.com/junggi/hello.PNG[/img]

[color=red]I'm so sorry, my problem is not correct  :blush: . It must be: [/color]
[quote][color=darkred][b]Problem.[/b] Let be given a circle with center $O$ and two points $M,\,N$ inside its so that $O$ be the midpoint of $MN.$ Let point $S$ lies on the circle and let $E,\,F$ be the points of intersection of $SM,\,SN$ and $(O)$ respectively. The tangents at $E,\,F$ to $(O)$ intersect at $I.$ Prove that: the orthogonal bisector of the segment $MN$ bisect $SI$[/color][/quote]


I was following the instructions on the website <http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php>.
I downloaded the MiKTeX file, and installed it.
Near to the end of the installation, the warning sign comes up and says,
"The operation could not  be completed for the following reasons:Windows API error 1400".
What have i dont wrong??

That's an odd error. Which version of Windows are you using? Anything special on the machine or low RAM or disk space? Is this a new installation of MiKTeX or an upgrade?

Have you tried reinstalling or else removing MiKTeX then reinstalling? How about reinstalling to a different directory?

I am using the Windows XP,, 2002 version.

As stevem suggested,, i reinstalled the MiKTeX to a different directory, but the same error is happening.

Has anyone seen or experienced this error? or anyone knows what the problem is?

I need help here!!

It's very difficult to diagnose the problem. Did you uninstall first before re-installing? If not, try that.

Otherwise you try installing version 2.4 (see http://www.miktex.org/2.4/ ) - you won't lose out by doing so.
You could report it on the support forums but 2 people have already done so in August  http://sourceforge.net/forum/message.php?msg_id=3849824 but without a solution.

I installed MiKTeX on Vista RC1 and it didn't appear to complete with Vista saying there was a problem (but no details). However, MiKTeX works fine so I wonder if you have actually tried to use MiKTeX - does it work?


Who will win the NL Wild Card?  I will vote for the Nationals, they can do it with a little luck and good playing :)I will not do any more baseball posts today, i will finish tomorrow, as to not overflow the forum with baseball :)

i'm tired of all these baseball topics!!!

i guess i shouldn't post any more baseball topics until all these are done and off the first page.  I might not even post baseball again until the playoffs start!


A number of players form seven teams, with each pair of teams having one common member and each player on two teams. How many players are there?

Never heard of the competition.

[hide]Consider a heptagon with each vertex being a team and each diagonal a team member...[/hide]

[quote="paladin8"][quote="aahmadi"][quote][hide]Consider a heptagon with each vertex being a team and each diagonal a team member...[/hide][/quote]
[hide]That makes sense to me, but then wouldn't the answer be 14? disagreeing w/the other amir's answer..[/hide][/quote]

He also meant to include the sides as team members.[/quote]

Yeah, I'm sorry for the poorly stated analogy.

[quote="amirhtlusa"]i think we used this as tiebraker.

[hide="ans"]
21[/hide][/quote]

Isn't that just $\binom{7}{2}$, then?

Is it always that?

[hide="easy to prove its $\binom72$"]
Name the players by the teams they are on; i.e. a player on teams 1 and 2 is called (1,2) which is equivalent to being called (2,1). There are clearly 7 choose 2 ways to name players, and we conclude this from the first condition.[/hide]


function
quadratics
conics
parabola
algebra unsolved
algebra

An orange grower has 400 crates of fruit ready for market and will have 20 more for each day the grower waits. The present price is 60 dollars per crate and will drop an estimated 2  dollars per day for each day waited. In how many days should the grower ship the crop to maximize his income?

Okay. So I know that the answer is 5 days. I cannot solve the problem though.

I tried....

(20x+400)(-2x+60)  and got 30. So obviously that method is wrong. Can someone show me how to solve this or atleast start me off please?

Hi! You shouldn't post this problems in olympiad forums. Next time you should look at the forum index and decide carefully which is the appropiate forum for your questions.
Your function $(20x+400)(-2x+60)$ is OK. You understood the problem.
Notice that it's a quadratic expression (you could expand it in the form $ax^{2}+bx+c$).
It's already factored, so you can see that it has two zeroes: $x=\frac{-400}{20}=-20$ and $x=\frac{-60}{-2}=30$.
But you are not looking for the zeroes of the income. You are looking for the maximum!
You could use the known fact that the maximum or minimum of a quadratic expression $ax^{2}+bx+c$ is at $x=\frac{-b}{2a}$.
But I think that the simplest way to solve it is this one: the maximum must be exactly in the middle of the two zeroes ($-20$ and $30$) because the graph of the parabola is symmetric.


Determine all the ways in which $ e_{3}\in\mathbb{R}^{3}$ can be written as a linear combination of the three vectors $ e_{2}\plus{}e_{3}$, $ e_{2}\minus{}e_{3}$, and $ e_{2}$.

This is a problem that came from an old linear algebra final.  I was told that this problem is actually deceptively difficult so I'd be interested in seeing a solution.

Equivalently, find a solution to Mx = (0,0,1)^T, and compute the nullspace N(M) of M, where

$ \displaystyle M \equal{} \left( \begin{array}{rrr} 0 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & \minus{}1 & 0 \end{array} \right)$.

Then your answer is the set of vectors x + N(M). Do you see why?

What does "Mx = (0,0,1)^T" mean?  Is that supposed to be the transpose?  I understand this part and I certainly see where the matrix comes from.  Solving for $ x$ yields $ x\equal{}(1\plus{}b,b,\minus{}2b\minus{}1)$.  The matrix appears to have null space with dimension 1.  But $ x\plus{}N(M)$ is a vector plus a scalar, which doesn't really make sense, unless you mean the actual vectors in the null space.  In that case, it would be any vector of the form $ (d,d,\minus{}2d)$.  So does that mean that it is any vector of the form $ (1\plus{}b\plus{}d,b\plus{}d,\minus{}2b\minus{}2d\minus{}1)$, $ c,d$ arbitrary numbers?  Why is it necessary to affix the vectors from the null space?

Yes, that's a standard notation for the transpose. How else would you write a column vector in plain text?

Also, [i]of course[/i] we mean the vectors in the null space. Why is it necessary? They're solutions, and we want to find all the solutions.

When I solved this originally, I neglected to consider the null space solutions but that's all.  If this is in fact all there is to it, then this is not a very difficult problem, as I suspected when I first saw it.


http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamowinners.html

Scroll down the page...it says something about a documentary being made about the usamo/imo. Anyone have any idea what it's going to be like or when it's coming out?

I'm deeply involved, but I do not know what it will be like.  It's up to the film-maker.

Plans are for it to be finished in November 2007.  Yes, 2007, that is not a mistake.

Steve Dunbar
AMC Director

woo I just got interviewed

they were all like AHA we are not british

Will it only be about the 2006 USAMO/IMO or will it include the 2005 or 2007 ones?


counting
distinguishability
algorithm
combinatorics open
combinatorics

How many ways can the following be done?

$ n$ indistinguishable balls into $ k$ distinguishable urns where urn $ i$ gets $ n_i$ balls
$ n$ distinguishable balls into $ k$ indistinguishable urns where urn $ i$ gets $ n_i$ balls
$ n$ distinguishable balls into $ k$ indistinguishable urns where no urn gets more than 1 ball

I'm not looking for answers, but how to go about counting these.

Also, are closed solutions to the following problems known?

$ n$ indistinguishable balls into $ k$ indistinguishable urns
$ n$ indistinguishable balls into $ k$ indistinguishable urns where no urn gets more than 1 ball
$ n$ indistinguishable balls into $ k$ indistinguishable urns where urn $ i$ gets $ n_i$ balls
$ n$ indistinguishable balls into $ k$ indistinguishable urns where every urn gets at least 1 ball

[quote]$ n$ indistinguishable balls into $ k$ distinguishable urns where urn $ i$ gets $ n_i$ balls [/quote]

This can be done in either 0 or 1 way. It can be done in one way iff  $ \sum_{i} n_i\equal{}n$.

[quote]$ n$ distinguishable balls into $ k$ indistinguishable urns where urn $ i$ gets $ n_i$ balls[/quote]

This does not make sense.

[quote]$ n$ distinguishable balls into $ k$ indistinguishable urns where no urn gets more than 1 ball[/quote]

This can be done in either 0 or 1 way. It can be done in one way iff  $ n\leq k$.

[quote]$ n$ indistinguishable balls into $ k$ indistinguishable urns[/quote]

This is the integer partition problem. I am not aware of a closed form but we have efficient algorithms for it.
http://en.wikipedia.org/wiki/Integer_partition

[quote]$ n$ indistinguishable balls into $ k$ indistinguishable urns where no urn gets more than 1 ball[/quote]

This can be done in either 0 or 1 way. It can be done in one way iff  $ n\leq k$.

[quote]
$ n$ indistinguishable balls into $ k$ indistinguishable urns where urn $ i$ gets $ n_i$ balls[/quote]

This does not make sense.

[quote]l$ n$ indistinguishable balls into $ k$ indistinguishable urns where every urn gets at least 1 ball[/quote]

Let $ P(n,k)$ be the partition of $ n$ into $ k$ parts. We have efficient algorithms to evaluate $ P(n,k)$. I am not sure of the closed form. If there exists a closed form for this , then we have closed form for $ P(n)$ too.

[b]A list of related posts[/b]
1.http://www.mathlinks.ro/viewtopic.php?t=192171
2.http://www.mathlinks.ro/viewtopic.php?t=185504
3.http://www.mathlinks.ro/viewtopic.php?p=1012575


Who here was at the 2005 GCTM State Math tournament today? I was there, but only as an individual trying out for ARML.

Hey, I was there! I didn't see you, though. Then again, I don't know what you look like so I couldn't look for you. heh. Weren't the tables weird? It's the first time I've seen a school with one sided cafeteria tables. So how'd you do?

Hi, I saw you; great job! My school won, but I wasn't on the team. I didn't do very well - about 165 total, but I made the ARML team because of my AIME score.

Nice! So I'll see you next Sunday, then? What's your last name, by the way?

I'm Sam Brotherton, a freshman at Rockdale Magnet School.

On a side note, do you know how to do #26?

"Let $x_n$ be the recursive relation $x_n=x_{n-1}+6x_{n-2}$, where $x_1=-10$ and $x_2=5$. One of the choices below is equivalent to $x_{20}$. Which is it?"
A. $ \frac{1}{2}(-75(20^2)+255(20)-200)$
B. $275$
C. $18\frac{1}{2}$
D. $\displaystyle \frac{1}{\sqrt5}[(\frac{1+\sqrt5}{2})^2-6(\frac{1-\sqrt5}{2})^{20}]$
E. $7(2^{19})-3^{20}$

(And how do you make the brackets for D bigger and the fractions smaller? I'm still new at LaTeX.  :P )

I just wrote a calculator program to do that one; I don't know how you would do it properly, but it looks like difference equations would be useful. I'm not sure how to use them, though.

As for the Latex, just use \left( and \right) or \left[ and \left]

There are threads in various forums (hopefully I'm not referring to private class posts) that discuss the full derivation for the formulas for linear recursions.  It might be helpful to read up a little on generating functions to learn how to solve them.

Thanks. They look like something really worth knowing, especially for proof-type contests. I'll definitely look that up.


function
integration
calculus
calculus computations

Hello everyone; so Im trying to find the function F(x); that produces a infinite series with the fllowing property:

d/dx = 1 + x*F(x);

using this information is it possible to solve and figure out what function F(x) it must be?

I'm not sure about your notation and what series you're looking for, but if you want to solve the differential equation
\[ \frac {dF(x)}{dx} = 1 + xF(x)
\]
then there are infinitely many solutions.  They are
\[ F(x) = \int_c^x{e^{\frac {x^2 - t^2}{2}dt}
}\]
where $ c\in \mathbb{R}$ is an arbitrary constant.

That is correct; but Im wondering how exacly you would attain that answer;

I can easily figure out the constant, but how did you solve the equation and put it into that form?

First try to solve the homogeneous problem \[ \frac{dF(x)}{dx}\equal{}xF(x)\]
You can solve it simply by dividing both sides by $ F(x)$, integrating, and simplifying to find $ F(x)\equal{}ce^{\frac{x^2}{2}}$.  Now having solved the homogeneous problem, how can you tweak it to add that extra $ 1$?

$ F'(x) \minus{} xF(x) \equal{} 1\Longleftrightarrow e^{ \minus{} \frac {x^2}{2}}F'(x) \minus{} xe^{ \minus{} \frac {x^2}{2}}F(x) \equal{} e^{ \minus{} \frac {x^2}{2}}$

$ \Longleftrightarrow \frac {d}{dx} F(x)e^{ \minus{} \frac {x^2}{2}} \equal{} e^{ \minus{} \frac {x^2}{2}}$

$ \therefore F(x)e^{ \minus{} \frac {x^2}{2}} \equal{} \int e^{ \minus{} \frac {x^2}{2}}\ dx \plus{} C$.

$ \therefore F(x) \equal{} e^{\frac {x^2}{2}}\left(\int e^{ \minus{} \frac {x^2}{2}}\ dx \plus{} C\right)$.


In a pile you have 100 stones. A partition of the pile in $ k$ piles is [i]good[/i] if:

1) the small piles have different numbers of stones;

2) for any partition of one of the small piles in 2 smaller piles, among the $ k \plus{} 1$ piles you get 2 with the same number of stones (any pile has at least 1 stone).

Find the maximum and minimal values of $ k$ for which this is possible.

If $k \geq 14$, then there are at least $1+2+\ldots+14=105>100$ stones, contradiction. So $k \leq 13$. We can take $\{1, 2, \ldots, 11, 12, 22 \}$ for $k=13$; it's easy to check that this works. So the maximum $k$ is $13$.

Note that $k=10$ works, taking $\{1, 3, 5, \ldots, 19 \}$: there are $100$ stones, and if we partition a pile, we get an odd number and an even number; the odd number is among the smaller piles. Now we want to show that $k<10$ doesn't work.

Suppose that there's no 1-stone pile. Then, since $n=(n-1)+1$, if there's a pile with $n$ stones there must be one with $n-1$. So, if $n$ is number of stones in the biggest pile, there are $n+(n-1)+\ldots+2=(n+1)n/2-1$ stones, which can't be $100$. Then there's a pile with just $1$ stone.

Suppose that the biggest pile has $2m+1$ stones. Then at least one of $1, 2m$, one of $2, 2m-1$, ..., one of $m, m+1$ must be present (*), so at least $m+1$ piles in total. If the biggest pile has $2m$ stones, there are at least $m$ piles. Note that (*) can be applied to any pile, in fact.

Now, suppose $k \leq 8$. Then the biggest pile has at most $2k=16$ stones. So there are at most $1+16+15+\ldots+10=92<100$ stones, contradiction. Then $k \geq 9$. If $k=9$, the biggest pile has at most $18$ stones. Take the smallest pile $>10$; by (*), there are at least 5 piles $\leq 10$. If the biggest pile has $17$ or fewer stones, there are at most $1+7+8+9+10+14+15+16+17=97<100$ stones, so it has $18$. However, the smallest pile $>1$ can be at most $4$ by (*), so in any case there are at most $1+4+8+9+10+15+16+17+18=98$ stones. Then $k>9$.

During the Olympiad I could prove that $k_m = 13$ and $k \geq9$ and I got 2 points out of 7 :(

consider the smallest pile with l stones
if $l>2$,then partition it into 1 and $l-1$,contradiction
so $l=1 or 2$
if $l=2$,then the second small pile (for the same reason) contains 3 stones,then 4 stones,5 stones,...
but 2+3+4+...+n cannot equal to 100,contradiction
hence $l=1$
let the largest pile has m stones
if $k\le 9$then
$m+(m-1)+...+(m-7)+1\ge 100$
hence $m\ge 19$
then m has at least 9 methods to be partitioned,with 8 other piles remaining,so there exists a method which contradicts the condition given.
so $k\ge 10$ and 1,3,5,...,19 give an example for 10.

If $k=1$ then it is clearly impossible. So $k\ge2$.

If the smallest pile has at least $3$ stones, we can split it into two piles with $1$ and $2$ stones, so the smallest pile has at most $2$ stones. If it has two stones, then, if there are $x$ stones in another pile, there must be $x-1$ stones in another pile by splitting it into piles with $1$ and $x-1$ stones. Then there are $\sum_{i=2}^{k+1}i=\frac{(k+1)(k+2)}2-1$ stones, and solving for $k$ gives $k=\frac{-3+\sqrt{805}}2$ which isn't an integer.
So the smallest pile has exactly one stone.

Then if the second-smallest pile is at least $4$ in size, we can split it into two piles with at least two stones, so the second-smallest pile can be at most $3$ in size.

If the largest pile has $2k+1$ stones, then there are $k$ partitions $(1,2k),(2,2k-1),\ldots,(k,k+1)$, which each correspond to at least one pile of stones. But the largest pile isn't any one of these piles, so there are at least $k+1$ piles, contradiction.
So the largest pile has at most $2k$ stones.

We can claim now that the minimum and maximum of $k$ are $10$ and $13$, respectively.
Suppose $k\ge14$. Then we have at least $\sum_{i=1}^{14}i=105>100$ stones, so there cannot be a good partition.
On the other hand, suppose $k\le9$. Then there are at most $1+3+\sum_{i=10}^{16}i=95<100$ stones, contradiction.
For $k=10$ we can take piles with sizes $\{1,3,5,\ldots,19\}$ whereas for $k=13$ we can take piles with sizes $\{1,2,3,\ldots,12,22\}$.

Remark: $k=11$ and $k=12$ work too.

[quote=jasperE3]......
Then if the second-smallest pile is at least $4$ in size, we can split it into two piles with at least two stones, so the second-smallest pile can be at most $3$ in size.

.......
On the other hand, suppose $k\le9$. Then there are at most $=1+3+\sum_{i=10}^{16}i=95<100$ stones, contradiction.
......

[/quote]

How does $k\le9$ imply $\sum_{i=1}^{k}a_i \le 1+3+\sum_{i=10}^{16}i=95<100$?
What if $a_k \ge 17$, for example?

[quote=Hawk Tiger]A shorter and nicer(maybe not) solution for the minium part:
Let $a_{1}<a_{2}<a_{3}\dots<a_{n}$$a_{1}+a_{2}+\dots+a_{k}=100$
we can prove that $a_{k}\leq 2k$
if $a_{k}>2k$ then consider the pairs of number $(1,a_{k}-1),(2,a_{k}-2)\dots(k,a_{k}-k)$ .there must exist $j$ such that $j\not =a_{i},a_{k}-j\not =a_{i}$for any $i$.this is a contradution.
then we have $a_{1}+\dots+a_{9}\leq 90$
so $k\geq 10$[/quote]

We have only $a_k \le 2k$, where $k$ is the number of stones in the pile with the largest amount of stones. For the rest $i$'s we do not know if $a_i \le 2i$, do we?!


Does there exist a nonempty perfect subset of $ R$ that doesn't contain any rational number ?

Let $ \{q_k\}$ be an enumeration of the rationals and let $ S \equal{} \bigcup_{k \equal{} 1}^\infty \left(q_k \minus{} 2^{ \minus{} k},q_k \plus{} 2^{ \minus{} k}\right)$.  Its complement $ S'$ is closed, contains no rationals, and is uncountable because $ \mu (S) < 2\Rightarrow \mu (S') \equal{} \infty$.  It follows from the [url=http://en.wikipedia.org/wiki/Perfect_set_property]Cantor-Bendixson Theorem[/url] that $ S'$ has a nonempty perfect subset.


Let be $ x\in [\minus{}1,1]$ , and $ a,b,c\in \mathbb{R}$ such that $ |ax^2\plus{}bx\plus{}c|\le 1\ ,\ (\forall)x\in [\minus{}1,1]$ . Without second degree function prove that $ |a|\plus{}|b|\plus{}|c|\le 4$ .

WLOG, assume $ a>0, b\geq 0$.  For $ x\equal{}0,1$, we have $ |ax^2\plus{}bx\plus{}c|\leq 1$, or $ |a\plus{}b\plus{}c|\leq 1, |c| \leq 1$. Taking the first inequality gives $ a\plus{}b\plus{}c\leq 1 \iff \minus{}1\minus{}c\leq a\plus{}b \leq 1\minus{}c$. $ c$ can be anywhere from $ \minus{}1$ to $ 1$, so $ a\plus{}b\leq 2$. Thus, $ |a|\plus{}|b|\plus{}|c|\equal{}a\plus{}b\plus{}|c|\leq 2\plus{}1\equal{}3<4$. Equality for $ 2x^2\minus{}1$.

Try this one now :

If $ |ax^2\plus{}bx\plus{}c|\le 1\ ,\ (\forall)x\in [0;1]$ prove that $ |b|\le 8$


geometry
3D geometry
prism
Pythagorean Theorem

An ant is on an edge of a right rectangular prism one centimeter from a vertex. If the prism has edges measuring 9, 10, and 12 cm, what is the smallest number of centimeters the ant can crawl to reach a point one centimeter from the opposite (furthest) corner?

Assuming that the ant can't go into the cube, we simply 'unfold' the cube along one of the edges of length $ 12$ so that the distance is the hypotenuse of a right triangle of legs $ 12,19\implies\sqrt{12^2\plus{}19^2}\equal{}\boxed{\sqrt{505}}$.

[quote="mr. math"]An ant is on an edge of a right rectangular prism one centimeter from a vertex. If the prism has edges measuring 9, 10, and 12 cm, what is the smallest number of centimeters the ant can crawl to reach a point [u][b]one centimeter[/b][/u] from the opposite (furthest) corner?[/quote]

$ \sqrt{505} \minus{} 1??$

Yeah, that's right--I didn't read the question carefully. :P

No. You take the 1 cm off before using the pythag theorem. So its: $ 12,19\implies\sqrt {(12\minus{}1)^2 \plus{} 19^2} \equal{} \boxed{\sqrt {482}}$. But the answer changes depending wich corner its closest to.

What are you talking about?

The shortest path without restrictions (that goes along the surface) is $ \sqrt{505}$.

Draw a circle around the ending vertex to see that to be one away, we can simply subtract $ 1$ from $ \sqrt{505}$.

Well rt505 is the distance across the surface from corner to corner when you spread out the prism, as you said. But is rt505-1 equal to the segment connecting the corner from 1 cm below the corner...no, if you subtract 1cm then the line leaving the far corner, it will hit the segment that ant is standing on a bit below the ant.
The legs of the right triangle whose hypotenuse you're trying to find are 11 and 19, because 11 is 1 cm less than 12, and the ant is 1 cm from the corner.
I'm not sure if I explain well...draw the prism. Mark a spot 1 cm below a corner as the ant. Then find distance from that point to far corner.

I agree that your way is a possible way to get to $ 1$ away from the opposite corner.

However, $ \sqrt {505} \minus{} 1 < \sqrt {482}$, so my path is shorter (and the shortest).

 [geogebra]65e62c794f90df35bd21ed6fea4504a34d930390[/geogebra]

The red path is the shortest, which is clearly achievable.

Yes, $ \sqrt{505}\minus{}1$ is less than $ \sqrt{482}$, but there is no possible way for $ \sqrt{505}\minus{}1$ to stretch from the point the ant is standing to the corner. You see, the meaning of subtracting 1 is to make up for the part that the ant is 1 cm from the corner ON THE EDGE OF THE CUBE. But the red path does not pass through the edge of the cube, so subtracting 1 has no meaning. Do you see what I mean?

OFF TOPIC: Hey, do you live in St. Louis or something? I do!

My apologies, I thought that the ant started on a corner.

The answer is $ \sqrt {482} \minus{} 1$ if the ant starts on an edge of length $ 12$, then (my argument is similar). If the ant starts on an edge of length not $ 12$, then the answer is $ 6\sqrt {13} \minus{} 1$.

Do you agree?

Uhhh not yet but I think you'll see it this time. There is no need to subtract 1 AFTER the square root. So as you said, you lay out the prism as 2-d and you'd draw this:
 [geogebra]60c0c1238bc9730bc719127a5071931c817bdae4[/geogebra]

So if the ant starts on the 12 segment you do $ \sqrt{((12\plus{}9)\minus{}1)^2\plus{}10^2}\equal{}\sqrt{500}\equal{}10\sqrt{5}.$ I hope I explained well. You see, there is no need to subtract 1 AFTER the pythagorean theorem, because you are calculating the distance directly from the ant's position to the opposite corner.

[quote="mr. math"]An ant is on an edge of a right rectangular prism one centimeter from a vertex. If the prism has edges measuring 9, 10, and 12 cm, what is the smallest number of centimeters the ant can crawl to [b]reach a point one centimeter from the opposite (furthest) corner[/b]?[/quote]

Care to rethink it??? :P And no matter what, you ALWAYS cross a side of length $ 12$ to get the shortest path, so I still believe that the answers are $ \sqrt {482} \minus{} 1,6\sqrt {13} \minus{} 1$.

The two cases are illustrated below.

Oh...OK! So we both helped each other! Thanks! :lol:

Yeah, a whole page on debating the semantics of the problem, because we were too lazy to read it. :P


Hello Sunset
[quote="sunset"]find all  functions like f from R to R that:
f(x+y)=f(x)f(y)f(xy)[/quote]

Let assertion $P(x,y)$ be $f(x+y)=f(x)f(y)f(xy)$

If $\exists$ a such that $f(a)=0$, then $P(x-a,a)$ $\Rightarrow $ $f(x)=0$ $\forall x$ $\Rightarrow $ I'll consider now solutions always different from 0.
$P(x-1,1)$ $\Rightarrow $ $f(x)$ and $f(1)$ have the same sign. So f have a constant signe, say positive (since -f is solution if f is solution). $\Rightarrow $ I'll consider now solutions always > 0.
$P(0,0)$ $\Rightarrow $ $f(0)=0$, $P(1,1)$ $\Rightarrow $ $f(2) = f(1)^{2}$ but $P(2,2)$ $\Rightarrow $ $f(2)=1$  $\Rightarrow $ $f(2)=f(1)=f(0)=1$
$P(\frac{x}{2}+2,1)$ $\Rightarrow $ $f(\frac{x}{2}+2+1)=f(\frac{x}{2}+2)^{2}=f(\frac{x}{2})^{2}f(x)^{2}$
$P(\frac{x}{2}+1,2)$ $\Rightarrow $ $f(\frac{x}{2}+2+1)=f(\frac{x}{2}+1)f(x+2)=f(\frac{x}{2})^{2}f(x)f(2x)$
And so $f(2x)=f(x)$ $\forall$ real $x$
Then $P(x,2)$ $\Rightarrow $ $f(x+2)=f(x)^{2}$ but $P(x+1,1)$ $\Rightarrow $ $f(x+2)=f(x+1)^{2}=f(x)^{4}$ $\Rightarrow $ $f(x)^{2}=f(x)^{4}$ $\Rightarrow $ $f(x)=1$

So the three only solutions are $f(x)=0$ $\forall$ $x$, $f(x)=1$ $\forall$ $x$ and $f(x)=-1$ $\forall$ $x$

--
Patrick

u can refer this

http://www.mathlinks.ro/Forum/viewtopic.php?p=344412#p344412 :wink:


Let A be a real 2x2 matrix with complex eigenvalues. Prove that exist a real invertible matrix P such that $ P^{\minus{}1}AP\equal{}J$ where J=(a,-b;b,a)

Hint: $ a\plus{}ib$ and $ a\minus{}ib$ are the eigenvalues of $ A$. When are two matrices similar?


inequalities
trigonometry
inequalities unsolved

$ Solution$ 1.
[hide]
Using the inequality $ Cauchy-Shwarts$ it's easy to notice, that:
${ (\sum{\sqrt{\frac{ab}{(c+a)(c+b)}}}})^{2}\leq\sum{ab}\sum{\frac{1}{(c+a)(c+b)}},$
so it's enough to prove:
$ \sum{ab}\sum{\frac{1}{(c+a)(c+b)}}\leq 9/4$
<=> $ 8\sum{ab}\sum{a}\leq 9\prod{(a+b)}$
<=> $ 6abc\leq\sum{a(b^{2}+c^{2})}$ <=> $ 0\leq\sum{a(b-c)^{2}}$, which completes our proof.
[/hide]

However, the trigonometrical way is better...

$ Solution$ 2.
[hide]
Let $ cyc(a+b\equiv C)$ $ \Rightarrow $ $ A,B,C$ are side length of a triangle, let it's angles be $ \alpha,\beta,\gamma$ $ \Rightarrow $
<=> $ \sum{\sqrt{\frac{(p-A)(p-B)}{BC}}}\leq 3/2,$ where $ p=\sum{A}/2$

and using the well-known fact:  $ cyc(\sqrt{\frac{(p-A)(p-B)}{BC}}=\sin{\frac{\alpha}{2}})$

<=> $ \sum{\sin{\frac{\alpha}{2}}}\leq 3/2$, which is well-known and can be proved (for instance) by Jensen:
[hide]
Since $ \sin{x}''\leq 0$ for $ x\in [0,\pi]$ $ \Rightarrow$ by Jensen's inequality $ \sum{\sin{\frac{\alpha}{2}}}\leq 3\sin{(\sum{\alpha/6})}=\frac{3}{2}$, which completes our proof.
[/hide]
[/hide]

[quote="nayel"][quote="Twins"]$ \sqrt{ab(a\plus{}b)}\plus{}\sqrt{bc(b\plus{}c)}\plus{}\sqrt{ca(c\plus{}a)}\leq\frac{3}{2}\sqrt{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$[/quote]

By Cauchy-Schwarz inequality, $ \sum\sqrt{ab(a\plus{}b)}\le\sqrt{\left(\sum a\plus{}b\right)\left(\sum ab\right)}$

So we need only show that $ 2\left(\sum a\right)\left(\sum ab\right)\le\frac{9}{4}(a\plus{}b)(b\plus{}c)(c\plus{}a)$

$ \iff 8\left(abc\plus{}(a\plus{}b)(b\plus{}c)(c\plus{}a)\right)\le 9(a\plus{}b)(b\plus{}c)(c\plus{}a)$

$ \iff 8abc\le (a\plus{}b)(b\plus{}c)(c\plus{}a)$

Which is obvious by AM-GM.[/quote]

What's the difference between $ Solution$ 1, and your solution ?  :huh:
Why do you post the same solution again?  :maybe:

[quote="Twins"]$ \sqrt{ab(a\plus{}b)}\plus{}\sqrt{bc(b\plus{}c)}\plus{}\sqrt{ca(c\plus{}a)}\leq\frac{3}{2}\sqrt{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$[/quote]
This one is equivalent to $ \sum\sqrt{\frac{ab}{(c\plus{}a)(c\plus{}b)}}\leq 3/2$. Note that $ \sqrt{\frac{ab}{(c\plus{}a)(c\plus{}b)}}\leq\frac{1}{2}\left(\frac{a}{a\plus{}c}\plus{}\frac{b}{b\plus{}c}\right),...$ we're done!. But it isn't mine. :P


okay this is kinda easy but i want you to try and find more than one solution if you can, or at least one but elegant :D
if $ \left|abc\right|\neq 1$, and $ a \plus{} \frac{1}{b}\equal{} b \plus{} \frac{1}{c}\equal{} c \plus{} \frac{1}{a}$
Show :  $ a\equal{}b\equal{}c$

Clearly $ a\minus{}b \equal{} \frac{b\minus{}c}{bc}$ and $ b\minus{}c \equal{} \frac{c\minus{}a}{ca}$ and $ c\minus{}a \equal{} \frac{a\minus{}b}{ab}$.
Multiplying we get $ (a\minus{}b)(b\minus{}c)(c\minus{}a)(abc)^2 \equal{} (a\minus{}b)(b\minus{}c)(c\minus{}a)$.
Since $ (abc)^2 \neq 1$ it follows that $ (a\minus{}b)(b\minus{}c)(c\minus{}a)\equal{}0$ so that, say, $ a\equal{}b$. It easily follows from the initial equations that $ b\equal{}c$, hence result.

Pierre.

Let $ a \plus{} \frac {1}{b} \equal{} b \plus{} \frac {1}{c} \equal{} c \plus{} \frac {1}{a} \equal{} k\Longrightarrow a \equal{} \frac {bk \minus{} 1}{b},\ c \equal{} \frac {1}{k \minus{} b}$.

$ \therefore c \plus{} \frac {1}{a} \equal{} k\Longleftrightarrow \frac {1}{k \minus{} b} \plus{} \frac {b}{bk \minus{} 1} \equal{} k$

$ \therefore (1 \minus{} k^2)b^2 \minus{} k(1 \minus{} k^2)b \plus{} 1 \minus{} k^2 \equal{} 0$

$ \Longleftrightarrow (1 \minus{} k^2)(b^2 \minus{} kb \plus{} 1) \equal{} 0$.

If $ k \equal{} 1$, then $ a \equal{} \frac {b \minus{} 1}{b},\ c \equal{} \frac {1}{1 \minus{} b}\Longrightarrow abc \equal{} \minus{} 1$.

If $ k \equal{} \minus{} 1$, then $ a \equal{} \minus{} \frac {b \plus{} 1}{b},\ c \equal{} \minus{} \frac {1}{1 \plus{} b}\Longrightarrow abc \equal{} 1$.

Thus $ k\neq \minus{} 1$, we have $ b^2 \minus{} kb \plus{} 1 \equal{} 0\ \therefore b \plus{} \frac {1}{b} \equal{} k$, yielding $ a \equal{} b \equal{} c$.


floor function
algebra
polynomial
number theory proposed
number theory

hi,
let $ (u_{n})_{n \in \mathbb{N}}$ and $ (v_{n})_{n \in \mathbb{N}}$ s.t.
\[ u_{n\plus{}1} \equal{} 3u_{n} \plus{} 4v_{n},\]
\[ v_{n\plus{}2} \equal{} 2u_{n} \plus{} 3v_{n}, u_{1}\equal{}3 , v_{1} \equal{} 2.\]
We define the sequences $ (x_{n})_{n \in \mathbb{N}}$ and $ (y_{n})_{n \in \mathbb{N}}$ with
\[ x_{n} \equal{} u_{n} \plus{} v_{n},\]
\[ y_{n} \equal{} u_{n} \plus{} 2v_{n}.\]
Prove
\[ y_{n} \equal{} \lfloor x_{n}\sqrt{2} \rfloor.\]

Let $ T=\left(\begin{array}{cc}3 & 2 \\ 4 & 3\end{array}\right)$ and note, that $ \left({x_{n+1}\atop y_{n+1}}\right)=T\left({x_n\atop y_n}\right)$ and $ T^2=6T-1$ hold. From this we conclude that the characteristic polynomial of $ T$ has the roots $ r_{\pm}=3\pm\sqrt{8}=3\pm 2\sqrt{2}$ with corresponding eigenvectors $ \left({1\atop\pm\sqrt{2}\right)}$, i.e.

$ T^n=\left(\begin{array}{cc}1 & 1 \\ -\sqrt{2} &\sqrt{2}\end{array}\right)\left(\begin{array}{cc}r_-& 0 \\ 0 & r_+\end{array}\right)^n\frac{1}{\sqrt{8}}\left(\begin{array}{cc}\sqrt{2} & -1 \\ \sqrt{2} & 1\end{array}\right)=\left(\begin{array}{cc}\frac{r_+^n+r_-^n}{2} & \frac{r_+^n-r_-^n}{\sqrt{8}}  \\ \frac{r_+^n-r_-^n}{\sqrt{2}} & \frac{r_+^n+r_-^n}{2}\end{array}\right)$ . Define $ a_n,b_n\in\mathbb{N}\cup\{0\}$ via $ a_n+\sqrt{2}b_n=r_+^n$. This yields $ r_-^n=a_n-\sqrt{2}{b_n}$. Using $ x_1=5,y_1=7$ we conclude $ x_{n+1}=5a_n+7b_n$ and $ y_{n+1}=7a_n+10b_n$ .

Now, direct calculation using $ 0<r_-^n\leq 1$ gshows that $ \forall n\in\mathbb{N}: \,\sqrt{2}x_n-y_n=(5\sqrt{2}-7)a_{n-1}+(7\sqrt{2}-10)b_{n-1}=(5\sqrt{2}-7)r_-^{n-1}$ is a number greater than zero, but smaller than one, hence $ \lfloor \sqrt{2}x_n\rfloor =y_n$.


modular arithmetic
quadratics
number theory proposed
number theory

Show for prime $ p$, any odd solution to $ 18^{x} \equal{} 49 \pmod{p}$ requires p = 1 or 7 (mod 8)

If $ x$ is odd, $ 18$ must be a quadratic residue $ mod p$. Obviously $ p\neq3$, so $ \left(\frac{18}{q}\right)\equal{}1\implies \left(\frac{2}{q}\right)\equal{}1$, and it happens when $ p\equiv\pm 1 (mod 8)$, and our proof is complete.


logarithms
algebra
function
domain
calculus
calculus computations

$ f(x,y)\equal{}x(y\log y\minus{}y)\minus{}y\log x$,

Find  $ \max_{x\in\left[\frac{1}{2},2\right]}\min_{y\in\left[\frac{1}{3},1\right]}f(x,y)$.

What does $ \max_{x\in\left[\frac{1}{2},2\right]}\min_{y\in\left[\frac{1}{3},1\right]}f(x,y)$ mean? Do you mean 'find the maximum and minimum values of $ f$' in the domain?

I take the problem to mean the following:

Define $ f : [\frac12, 2] \times [\frac13, 1] \to \mathbb{R}$ by what he said.

Determine the maximum (over $ x$) of the minimum (over $ y$) of said $ f$.

Is this correct?

Well, first we compute $ f_y \equal{} x \log y \minus{} \log x$.  Then we solve $ f_y \equal{} 0$ to get $ y_0 \equal{} y_0(x)$.  Then we ask, "for which values of $ x$ is $ y_0(x)$ contained in the interval $ \left[\frac{1}{3}, 1\right]$?"  Then for those values of $ x$ we compare $ f(x, 1/3)$, $ f(x, y_0(x))$ and $ f(x, 1)$ and choose the minimum, while for other values of $ x$ we just compare $ f(x, 1/3)$ and $ f(x, 1)$ and choose the minimum.  This gives you a new piecewise function of $ x$, which you then have to maximize.  None of this is very much fun.


This one should be classical..

{x_n} is a sequence of numbers.
x_n > 0 for all n
lim   n (x_n / x_{n+1} - 1) > 0

prove that \sum_{n=1}^\infty (-1)^{n+1} x_n coverges.

Obviously, we try to use Leibniz. First of all there is eps>0 and k>0 such that for all n>=k we have n(x_n/x_(n+1)-1)>eps. It follows immediately that the sequence (x_n)n>=k is decreasing. If we prove that it tends to 0, we are done. But it is easy to see that in fact x_n<x_k*1/[(1+eps/k)...(1+eps/(n-1))]. But (1+eps/k)...(1+eps/(n-1))>1+eps(1/k+...+1/(n-1)) which tends to infinity. Thus, x_n tends to 0.

hmmm.  A very classical problem...
my proof is exactly the same to yours. :)


function
induction
algebra proposed
algebra

Denote $X=[1,+\infty)$. Find all functions $f: X\to X$ such that
a)$f(x)\leq 2(x+1)\forall x\in X$, and
b)$f(x+1)=\frac{f^{2}(x)-1}{x}\forall x\in X$.

Very fast, olorin !  :wink:
This is my solution.
Suppose $f$ is a function satisfy condition of problem.
We have
$f(x)\leq 2(x+1) \Rightarrow f(x+1)\leq 2(x+2) \Rightarrow f^{2}(x)\leq 1+2x(x+2)<2+2x(x+1)=2(x+1)^{2}$

$\Rightarrow f(x)<2^{\frac{1}{2}}(x+1)\Rightarrow f^{2}(x)<1+2^{\frac{1}{2}}x(x+2)<2^{\frac{1}{2}}(x+1)^{2}\Rightarrow f(x)<2^{\frac{1}{2^{2}}}(x+1)$, etc.

final, we obtain $f(x)<2^{\frac{1}{2^{n}}}(x+1)\forall n\forall x\in X$. Therefore $f(x)\leq x+1\forall x\in X$(*).

Now use (*) we have $xf(x+1)=(f(x)-1)(f(x)+1)\leq (x+2)(f(x)-1)$(**)

$\Rightarrow x\leq xf(x+1)\leq (x+2)(f(x)-1)\Rightarrow f(x)\geq\frac{2x+2}{x+2}$. From (**) we so have $(x+2)(f(x)-1)\geq xf(x+1)\geq \frac{x(2x+4)}{x+3}\Rightarrow f(x)\geq \frac{3x+3}{x+3}$, etc.

final, we have obtain $f(x)\geq\frac{nx+n}{x+n}\forall n\forall x\in X$, therefore $f(x)\geq x+1\forall x\in X$(***).

By (*) and (***) we have $f(x)=x+1\forall x\in X$.
... :D


Speaking of countdown, there was this problem in the 04-05 chapter countdown that was really hard for countdown:

$ \text{For how many values of}x\text{is the expression}\frac{x-5}{x^{2}-9}\text{not defined}$

[quote="archimedes1"]At the risk of being yelled at for continuing the off-topic problem,
I'd just like to say that if you know what "undefined" means, you could probably solve it before you finished reading it.  When you divide by $ 0$, an expression becomes "undefined". So there are [i]two[/i] values ($ \pm 3$) that cause the denominator to become zero and thus make the expression undefined.[/quote]

yea thats the point i was trying to make...it just seemed weird he said it was a super hard countdown problem because it can be solved in 1 sec mentally..

actually whenever the denominator is in the form $ x^{2}+k$ where $ k$ is a negative real number, there are [b]always two solutions[/b]
if it was $ x^{2}-\pi$,its still two..

sorry if i went off topic, it just seemed weird

But, it was chapter... you can normally solve chapter countdowns in like 2 secs.


There are two colors Jessica can use to color this 2 by 2 grid. If non-overlapping regions that share a side cannot be the same color, how many possible ways are there for Jessica to color the grid?

Call the colors 1 and 2. These are the two possible colorings:

1 2
2 1

2 1
1 2

So there are just $ \boxed{2}$ possible ways for Jessica to color the grid.


Has anyone had any experience with this? What's the best way to prepare for it?

I'm doing old contests right now and seeing which areas I need to review and learn about, but it makes me mad that there aren't explanations to the answers.

Yeah, I won it last year... Basically, you should review stuff from your chemistry textbook, and learn basic quantum theory, and learn to be fast with calculations. This contest is one that doesn't require too much advanced knowledge - just logic and calculation skill...  :)

Good luck!

Im using barron's SAT subject test chem book to prepare + classroom notes.

hey... where do you get the past/old contests from?
I can't find them anywhere. =(

I did it last year too and placed 75th. (Hint: flourine is the most electronegative atom xP).  I was fully prepared for anything that was on it, and i became so by studying some past contests and using the rest for practice tests, and referencing wikipedia/an old textbook for more detail in some areas such as VSEPR and I forget what else.  There isn't much quantum theory on it...

I'm doing it this year with only my grade 11 course. So basically I'm not very prepared.

Not that I'm some mod or something, but isn't this a math forum, not a science forum?

But good luck to all who are taking it!

Can anyone provide a list of the specific concepts one should know?


Let ABC be a triangle. Prove that
     $\frac{m_a}{l_a} \geq \sqrt{\frac{b^2+c^2}{2bc}}$
   In the book "He thuc luong trong tam giac va tu giac " of Mr Phan Huy Khai and Mr Nguyen Dao Phuong, there is one inequality which is much weaker and much more mundane : $\frac{m_a}{l_a} \geq \sqrt{\frac{(b+c)^2}{4bc}}$

Using  Ma2 = (2(b2+c2)-a2)/4
          La2 = bc ((b+c)2-a2)/(b+c)2
Ma2/La2 = (2(b2+c2)-a2)(b+c)2 / 4bc((b+c)2-a2) ≥ (b2+c2) / 2bc

or directly after opening parathesys :

2a2(b2+c2)  ≥ a2(b+c)2   q.e.d.

If you are interesting in inequality similary to this look:
http://www.mathlinks.ro/Forum/viewtopic.php?t=18137&highlight=


ARML
LaTeX
modular arithmetic
pigeonhole principle
rotation
number theory
SFFT

[size=150][color=darkblue]Day 1[/color][/size]

[quote="scorpius119"]The total surface area of both boxes is given by $2(4\cdot 6+4\cdot 8+6\cdot 8)+2(2\cdot 3+2\cdot 5+3\cdot 5)$ (because each box contains 2 faces of each combination of two sides), which is equal to $2(104)+2(31)=270$.

The exposed area will be given by subtracting off the area of the face of the larger box touching the floor and then subtracting [i]twice[/i] the area of the face of the smaller box touching the larger box, since the face of the smaller box covers that much area of the larger box. The exposed area will be minimized by maximizing these two areas.

The largest area of a face of the larger box is $6\cdot 8=48$ and the largest area of a face of the smaller box is $3\cdot 5=15$, making the minimum exposed area equal to $270-48-2(15)=\boxed{192}$.[/quote]

You are not required to submit a solution.  However, you can get bonus points if your solution is chosen to be posted in this thread.  Grammar, spelling, $\LaTeX$, etc. DO count.  I generally tend to dislike * being used for multiplication.  I wont always choose just one solution.  I'll try to have days where I choose maybe 2 or 3.

[size=150][color=red]Day 2[/color][/size]

[quote="matt276eagles"]Problem #2.  Answer: $\boxed{4000}$.

Note that $|f(x)|=|x-1||x-2|^{2}|x-3|^{3}\cdots|x-1999|^{1999}|x-2000|^{2000}$.

If $x<0$ or $x>2001$, then all of the factors are greater than or equal to $1$.  Also, it is easy to verify that if $x=0,2001$, then $f(x)>1$.  Therefore, $0<x<2001$.

The graph of $f(x)$ is continuous and has roots at $x=1,2,3\cdots 1999,2000$.  We claim that between each set of adjacent roots, there exists at least one point on the graph such that $|f(x)|>1$.  To prove this, consider any two consecutive integers $n,n+1$ in the range $(0,2001)$.  We want to prove that between $n$ and $n+1$ there exists a value of $x$ such that $|f(x)|>1$.  Put $x=n+\frac{1}{2}$.  Two of the factors of $\left|f\left(n+\frac{1}{2}\right)\right|$ are $\left(\frac{1}{2}\right)^{n}$ and $\left(\frac{1}{2}\right)^{n+1}$  The product of these is $\left(\frac{1}{2}\right)^{2n+1}$, which is at least $\left(\frac{1}{2}\right)^{3999}$.  However, no other factors are less than $1$, and there are many more than $3999$ factors (after considering the exponents) greater than $2$ to cancel out all of the $\displaystyle\frac{1}{2}$'s.  Therefore, our claim is correct.

Between each pair of adjacent roots there are two points such that $|f(x)|=1$, and there is an additional point such that $|f(x)|=1$ on each end.  Therefore, the number of values for $x$ is $\boxed{4000}$.[/quote]

[size=150][color=orange]Day 3[/color][/size]

[quote="chess64"]
Let $a=2^{10}$. Using the identity $\log x^{k} = k\log x$, we have \[ \left[(2b)(\log_a x)\right]^2 = 4\log_a x,  \] or \[ \log_a x = \frac{1}{b^2}\qquad (\text {if } x\ne 1).  \] Since the LHS is equal to $\frac{\log x}{\log a}$, we can write \[ \log x = \frac{10\log 2}{b^2}\implies x=10^{(10\log 2)/b^2} = 2^{10/b^2}.  \] If $b^2>10$, then the RHS cannot be an integer. Therefore $b=\boxed {\sqrt{10}}$ is the largest value of $b$ such that $x$ is an integer.[/quote]

[size=150][color=darkred]Day 4[/color][/size]

[quote="chess64"][u][b]A.[/b][/u] Let $p$ be the largest prime less than $c$. Assume for the sake of contradiction that $a\equiv 0 \pmod {p}$. Then inspection of the equation modulo $p$ gives $a\equiv b\equiv 0\pmod {p}$. Therefore we can write $a=pj$, $b=pk$ for integers $j,k$. Substituting into the equation, we have $p^3(j^3+k^3)$. However, $p^3$ is clearly greater than $c$, and its only factors are $\{1,p,p^2,p^3\}$. Its only factors that divide $c!$ are 1 and $p$, so $p^3\nmid c!$ and we have reached a contradiction. Hence, $a\not\equiv 0 \pmod p$, as desired. $\Box$

[u][b]B.[/b][/u] We will prove that $a+b$ is a multiple of 2, 3, 5, and 11, so it must be a multiple of $\text{lcm}(2,3,5,11)=330$.

[u]1.[/u] $a+b\equiv 0 \pmod {2}$

Parity analysis shows that $a$ and $b$ must have the same parity (otherwise we will have $0^3+1^3\equiv 1\pmod {2}$ but $2\mid c!$), so $a+b\equiv 0 \pmod {2}$, as claimed.

[u]2.[/u] $a+b\equiv 0 \pmod {3}$

Modulo 3, we have $a^3\equiv(-b)^3\pmod {3}\iff a\equiv -b \pmod {3}$, and the desired conclusion follows.

[u]3.[/u] $a+b\equiv 0 \pmod {5}$

Similarly, we must have $a^3\equiv(-b)^3\pmod {5}$, and $a+b\equiv 0 \pmod {5}$.

[u]4.[/u] $a+b\equiv 0 \pmod {11}$

We still have $c!\equiv 0 \pmod {11}$ so $a^3\equiv (-b)^3\pmod {11}$ and again the conclusion is obvious; $a+b\equiv 0\pmod {11}$.

Hence, $a+b\equiv 0\pmod {330}$, as needed. $\Box$[/quote]

(chess64's original text very slightly modified to fix one typo)

Chess64's proof seems to assume that $a^{3}\equiv (-b)^{3}\pmod{p}$ directly implies $a\equiv -b\pmod{p}$.  Although this does happen to be true for $p=2,3,5,11$, it is not always true.  For example, if $a^{3}\equiv (-b)^{3}\pmod{7}$, then one possible solution is $(a,b)\equiv (1,3)\pmod{7}$, in which case $a\not\equiv -b\pmod{p}$.  Am I missing something?

[quote="matt276eagles"]Although this does happen to be true for $p=2,3,5,11$...[/quote]

So what's the problem? :D

[size=150][color=brown]Day 5[/color][/size]

[quote="Solution 1 - Iversonfan2005"]We can use the angle bisector theorem to see that $AE: EC=7: 3.$  Therefore $AE=\frac{7}{2}$ and $EC=\frac{3}{2}.$  Now notice that $AP$ is also the angle bisector of $\triangle{AEB}$. Using angle bisector theorem, again, we have that $BP: PE=x: y=7: 7/2=2: 1$, or simply $\boxed{2}$.[/quote]

[quote="Solution 2 - jackb1115"]We will use mass points.  Let M(x) be the mass assigned to point x.

Because AD is an angle bisector, BD/CD = c/b = 7/5.  This leaves M(B)/M(C) = 5/7.  Ergo, we will assign a value of 7 to point C and a mass of 5 to point C.  In turn, since AE/CE = c/a = 7/3, we have M(A)/M(C) = 3/7.  Since M(C) = 7, this leaves M(A) = 3.  To balance, we must assign a value of M(A) + M(C) to E, so M(E) = 10.

Now, we find PB/PE, which is equal to M(E)/M(B) = 10/5 = [b]2[/b][/quote]

[size=150][color=orange]Day 6[/color][/size]

Part (1) was pretty much straightforward.  To show that $K$ is an integer, you need to show that $|(a-b)(b-c)(c-a)|$ is divisible.  This can be shown easily by using modular arithmetic and the pigeonhole principle.  You could also break it into cases.

[quote="Shogia"]

1.  To show that $K$ must be an integer it suffices to show that at least one of $(a-b), (b-c), (c-a)$ is divisible by $2$ since it is given that $a, b, c$ are all integers themselves.  Each of $a, b, c$ is congruent to either 0 or 1 mod 2.  Thus by the pigeonhole principle a least one of the three pairs must have the same parity so that their difference is congruent to 0 mod 2.  Thus at least one of the factors is even and $K$ must be an integer.[/quote]

Part (2) was considerably harder.  Still, with organized casework it is manageable.  I'll note that at ARML, you should give the values of $(a,b,c)$ that make $K=3$.  It's not too hard to find an example (a=2, b=1, c=4, for example) but not doing so could get points docked.

[quote="Shogia"]2.  If $K$ is some prime $p$ then $2K$ must either factor as $(\pm1)(\pm2)(\pm{p})$ or as $(\pm1)(\pm1)(\pm2p)$.  Since the factors are $(a-b), (b-c), (c-a)$ they must sum to $0$.  For the second case we have $\pm1\pm1\pm2p=0$ or $\pm2p=\pm1\pm1$  The LHS is either $0$ or $\pm2$, so $p$ is $1$ or $0$, neither of which is prime.  For the first case we have $\pm1\pm2\pm{p}=0$ or $\pm{p}=\pm1\pm2$.  The LHS is either $\pm1$ or $\pm3$, so the only prime value for p is $3$.  Therefore $K=3$ is the only prime value for $K$.
[/quote]

If you got part (2), then part (3) was very similar.

[quote="Shogia"]
3.  If $K$ is the square of some prime, $p$, then $2K$ must factor as one of the following:
Case 1: $(\pm1)(\pm{p})(\pm2p)$--->$\pm2p\pm{p}=\pm1$--->$\pm3p=\pm1$ or $\pm{p}=\pm1$, for which there are no primes p that satisfy either equation.
Case 2: $(\pm2)(\pm{p})(\pm{p})$--->$\pm{p}\pm{p}=\pm2$--->$\pm2p=\pm2$ or $0=\pm2$, for which there are no primes p that satisfy either equation.
Case 3: $(\pm1)(\pm1)(\pm2p^2)$--->$\pm1\pm1=\pm2p^2$--->$\pm2p^2=\pm2$ or $0=\pm2p^2$, for which there are no primes p that satisfy either equation.
Case 4: $(\pm1)(\pm2)(\pm{p^2})$--->$\pm1\pm2=\pm{p^2}$--->$\pm{p^2}=\pm1$ or $\pm{p^2}=\pm3$, for which there are no primes p that satisfy either equation.
All possible factorings have been exhausted so $K$ cannot be the square of a prime.
[/quote]

Part (4) was easier than part (3) and probably should have been worth only 2 points (my fault).  You mostly just had to use what you proved (or were asked to prove) in the earlier parts.  At ARML on later parts of the power round you can assume the validity of earlier parts.

[quote="Shogia"]
4.  In parts 1, 2, and 3 it was shown that $K$ must be an integer that is not the square of a prime or a prime other than 3.  This means that the area of triangle $ABC$ cannot be $2$, $4$, $5$, or $7$ for values less than or equal to $8$.  Nor can the area be $0$ because the 3 points are distinct.  Thus the only possible values for the area are $1$, $3$, $6$, or $8$.  There is nothing special about triangle $ABC$ the same will hold for any three points chosen on the parabola so the area of triangle $BCD$ must also be one of these values.  The sums of the areas of these two triangles is equal to the area of quadrilateral $ABCD$.  However, there is no pair of possible areas that sums to $8$ so the area of quadrilateral $ABCD$ cannot be $8$.[/quote]

[size=150][color=blue]Day 7[/color][/size]

The trick to this problem is noticing that Simon's Favorite Factoring Trick can be implemented.  A couple people misread the question statement and gave all the numbers, $n$, that could be written in the form $a+b+ab$.  I was tempted to give these people credit based on the fact that their solutions clearly showed they had the concept down but changed my mind since at ARML errors such as this one do occur (they occur on all math competitions, of course).

[quote="Iversonfan2005"]We have that $ab+a+b=n$, and since we have an $ab$ term we are tempted to think of Simon's Factoring trick. The equation factors into $(a+1)(b+1)=n+1$. Because $a$ and $b$ are positive integers, we know that neither $(a+1)$ or $(b+1)$ can equal $1$, therefore, $n+1$ is composite. Thus, the only numbers $n$ that cannot exist are those which are one less than a prime number. Because we want to find these kinds of numbers in the interval $[90,100]$, the only two which work are $\boxed{96,100}$.[/quote]

[size=150][color=cyan]Day 8[/color]
[/size]
This one had a really slick solution that once you see it makes you think "duh!".  Seeing it, however, is not at all easy.  You could also have "brute forced" it though...

[quote="Solution 1 - Elemennop"]
Consider the intersection of lines $AF$ and $CD$. Call this intersection $Q$. Rotate the figure $90^{\circ}$ about $Q$ three times, so we have the original figure and three images. Our figure is now a large square with a rotated inner square.

The inner square is composed of the sides $EF+DE$, which we are given is $12$. The outer square is composed of sides $AB + BC$, which is $17$ from our given info. The total area inside the larger square but outside the smaller square is $17^2-12^2=145$.  However, this is the area of the original figure and it's four images; each of the images are congruent to the original figure, so our desired area is a quarter of what we found, and our final answer is $\frac{145}{4}.$[/quote]

[quote="Solution 2 - xxreddevilzxx"]Extend $AF$ and $CD$ and call the intersection $G$. Also extend $DE$ and call the intersection of that and $AB$ point $H$. Then by simple angle chasing, we see that $\quad AHEF$ and $\quad EFGD$ are similar. Setting $AF=x$, $DE=m$, we get

$\frac{12-m}{x} = \frac{7-x}{m}$

$(12-m)m = x(7-x)$ ----------------- 1.

Now denote $GD = 10-x = o$, $GF = 7-x = p$, $EF = 12-m = n$

Then we know that since $GD^2 + GF^2 = DF^2 = DE^2+EF^2$,
$o^2 + p^2 = m^2 + n^2$.
By given condition, $m+n = 12$, $o+p = 17-2x$.
$(o+p)^2-(m+n)^2 = o^2+p^2-m^2-n^2+2op-2mn=2(op-mn) = 145-68x+4x^2$
Since by 1., $mn = x(7-x)$, then

$2[(10-x)(7-x)-x(7-x)] = 4x^2-68x+145$
$2[(10-2x)(7-x)] = 4x^2-68x+145$
$4x^2-48x+140 = 4x^2-68x+145$
$20x = 5$, $x = \frac{1}{4}$

The area of the hexagon then could be found by
Area of ABCG - Area of EFGD, or
$ABCG - EFD - DGF$
$\Rightarrow 70 - \left(\frac{mn}{2}\right) - \left(\frac{op}{2}\right)$
$\Rightarrow \frac{140 - (mn+op)}{2}$
$= \frac{140 - (x(7-x)+(10-x)(7-x))}{2}$
$= \frac{140 - 10(7-x)}{2}$
$= \frac{140 - \frac{270}{4}}{2}$
$= \frac{\frac{290}{4}}{2}$
$= \frac{290}{8} = \frac{145}{4}$[/quote]

[size=150][color=orange]Day 9[/color][/size]

If you are fluent in modular arithmetic and are faced with a number theory problem, it always serves as a good first step to try and interpret the problem in terms of mods.  This will often times make the problem a bit more straightforward.

[quote="matt276eagles"]Problem #9.  Answer: $\boxed{5}$.

Since the repetend has $96$ digits, we have $\frac{1}{97}=\frac{\text{repetend}}{\underbrace{999\cdots 999}_{96\text{ nines}}}$.  Cross-multiplying gives $\underbrace{999\cdots 999}_{96\text{ nines}}=97(\text{repetend})$.  Consider this equation $\pmod{1000}$.  The equation becomes $97(100A+67)\equiv 999\pmod{1000}$, which simplifies to $700A+499\equiv 999\pmod{1000}$.  Therefore, $700A\equiv 500\pmod{1000}$, and $7A\equiv 5\pmod{10}$.  Since $A$ is a digit, the only possibility is $A=\boxed{5}$.[/quote]

Heh I used FLT and multiplicative inverses :D

EDIT: Huh I just checked my sentbox and it wasn't in there, so I probably just imagined myself typing up the solution and PMing to joml88...

[color=brown][size=150]Day 10[/size][/color]

[quote="lingomaniac88"]I will use $\lfloor x \rfloor$ as the greatest integer less than or equal to $x$, and $\{ x\}$ for $x-\lfloor x \rfloor$, or the fractional part of $x$.  So $x=\lfloor x \rfloor+\{ x\}$.

$\lfloor x \rfloor$ and $\lfloor y \rfloor$ must be integers.  We can rewrite our equations as follows:
$\lfloor x \rfloor+\{ x\}+\lfloor y \rfloor = 5.3$
$\lfloor y \rfloor+\{ y\}+\lfloor x \rfloor = 5.7$
Therefore, we can conclude that $\{ x\}=0.3$ and $\{ y\}=0.7$.
So $\lfloor x \rfloor+\lfloor y \rfloor=5$.  Let $a$ be $\lfloor x \rfloor$ and $b$ be $\lfloor y \rfloor$.

So $a+b=5$.  Since they are asking for first quadrant pairs, $x>0$ and $y>0$, or $a>-0.3$ and $b>-0.7$.  The number of ordered pairs $(x,y)$ that satisfy those conditions are $(0.7,5.3)$, $(1.7,4.3)$, $(2.7,4.3)$, $(3.7,2.3)$, $(4.7,1.3)$, and $(5.7,0.3)$.  So the answer is $\boxed6$.[/quote]

[color=blue][size=150]Day 11[/size][/color]

[quote="minsoens"]The total number of points that the whole group of players gets is in variant:
--If there is a winner, then the winner gets 2 points and the loser gets none. The total number of points is increased by two.
--If there is a tie, then each player gets 1 point, and the total number of points is increased by two.

Let $x\ge 10$ represent the number of ninth graders. Then $Nx+20$ is the total number of points gained by all of the players by the end of the $\binom{x+2}{2}$ games. Each game adds 2 points overall, so:
$\frac{Nx+20}{\binom{x+2}{2}}=2$
$Nx+20=(x+2)(x+1)$
$N=\frac{x^2+3x-18}{x}=x+3-\frac{18}{x}$
$N$ is an integer, so $\frac{18}{x}$ is an integer. But $x\ge 10$, so $x=18 \Longrightarrow N=\boxed{20}$[/quote]

[size=150][color=cyan]Day 12[/color][/size]

[quote="Iversonfan2005"]WLOG, let the sidelength of the cube be $1$. Then, the starting surface area of the cube is $6$. After six days, we are given that the surface area reduces in half, and in this case, it becomes $3$. Therefore, $6x^2=3$, and $x=\frac{\sqrt{2}}{2}$, where $x$ is the sidelength of the cube. Thus, the sidelength after six days must be equal to $\frac{\sqrt{2}}{2}$. Therefore, $1-6n=\frac{\sqrt{2}}{2} \Longrightarrow n=\frac{2-\sqrt{2}}{12}$. Now that we have found $n$, we need to find the number of days until the volume of the cube is $\frac{1}{8}$ for this case. Thus, $x^3=\frac{1}{8} \Longrightarrow x=\frac{1}{2}$, where $x$ represents the side length of the cube once again. So, we have that $1-\frac{2-\sqrt{2}}{12}(d)=\frac{1}{2}$, where $d$ represents the number of days. Solving, we have that $d=\boxed{6+3\sqrt{2}}$ days.[/quote]

Huh I sent you day 11... can you check your inbox please? :(

Did you get it right? (I delete all of the PM's from a given day after I grade them...I thought you answered 18 which was a common wrong answers...since 18 was the number of 9th graders or w/e).

Oh darn, you're right. I need to start reading these questions more carefully :mad:

[size=150][color=darkblue]Day 13[/color][/size]

[quote="matt276eagles"]Problem #13.  Answer:  $\boxed{16}$.

We claim that the smallest positive integer that can not be the difference between a square and a prime is $16$.  For $n=1,2,3\cdots 15$, it is sufficient to find some positive integer $m$ and prime $p$ such that $n=m^{2}-p$:

$n=1,(p,m)=(3,2)$
$n=2,(p,m)=(2,2)$
$n=3,(p,m)=(13,4)$
$n=4,(p,m)=(5,3)$
$n=5,(p,m)=(11,4)$
$n=6,(p,m)=(3,3)$
$n=7,(p,m)=(2,3)$
$n=8,(p,m)=(17,5)$
$n=9,(p,m)=(7,4)$
$n=10,(p,m)=(71,9)$
$n=11,(p,m)=(5,4)$
$n=12,(p,m)=(13,5)$
$n=13,(p,m)=(3,4)$
$n=14,(p,m)=(2,4)$
$n=15,(p,m)=(181,14)$

For $n=16$, the equation becomes $p=m^{2}-16=(m-4)(m+4)$.  Note that $m^{2}>16\Rightarrow m>4$.  Therefore, both factors are positive.  If neither factor is $1$, then the expression is composite.  Therefore, $m-4=1$, and $m=5$.  But this implies that $p=9$, which is a contradiction because $p$ is prime.  Therefore, $\boxed{16}$ is the smallest positive integer that cannot be the difference between a square and a prime.[/quote]

[size=150][color=yellow]Day 14[/color][/size]

[quote="minsoens"]Here we have the 2 pole problem, with the height of the "poles" being equal to the length of the radii of the bases. The radii of the bases are $\sqrt{\frac{400}{\pi}}=\frac{20}{\sqrt{\pi}}$ and $\sqrt{\frac{900}{\pi}}=\frac{30}{\sqrt{\pi}}$
The formula for the "height" of the intersection of the lines from the poles is $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$. Let's call $a=\frac{20}{\sqrt{\pi}}$ and $b=\frac{30}{\sqrt{\pi}}$. Then $c=\frac{12}{\sqrt{\pi}}$. $c$ is also the radius of the circle formed by the intersection of the the two reverse cones, so the area of this intersection is $A=\pi r^2=\pi c^2=\boxed{144}$[/quote]

Some formulas you ought to have memorized.  This is one of them.  However, you should know how to derive/prove most/all of the formulas you use since you will often be faced with problems that won't be immediately solved with that formula but can be attacked similarily to the way that you proved the formula.  Anyway, here's a proof of the formula minsoens used:

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=866[/img]

We have a setup as above where the lines containing $x,y,$ and $z$ are parallel.  The theorem states that \[ \boxed{\frac 1x + \frac 1y = \frac 1z.}  \]

Seeing parallel lines we think similar triangles.  Indeed we have from similar triangles that \[ \frac az = \frac{a+b}y \ \ \text{and} \ \ \frac bz = \frac{a+b}x.  \]

Adding these two equations gives us $\frac{a+b}z = \frac{a+b}x + \frac{a+b}y.$ Which, after cancelling the $a+b$'s on both sides, gives us our desired result.

[size=150][color=blue]Day 15[/color][/size]

[quote="wobster109"]Define a "block" to be a section of the sequence consisting of $n$ consecutive integers of the same parity.

Claim: the last number in a block of n integers is $n^2$.

I will show this by induction.  First, the block of 1 integer has the last number 1, which is 1 squared.

Now assume the block of length $n$ integers has last number $n^2$.  the numbers in the block with $n+1$ integers will be $n^2+1, n^2+3, \ldots, n^2+2n+1$.  Note that the last number is equal to $(n+1)^2$ so the claim is proven.

Now find the largest integer where $1+2+3+. . . +n<1999$.  The largest integer satisfying this condition is 62.  The sum of the integers from 1 to 62 is 1953.  So the 1953rd term is $62^2=3844$.  There remain 1999-1953=46 terms until the 1999th term.  The next 46 terms will be the next 46 odd numbers.  They are $3845, 3847,\ldots, 3935.$

So the 1999th term is $\boxed{3935.}$[/quote]

OMG, THE LATEX, IT BURNS MY EYES  :surrender: I HOPE IT GETS FIXED BY TOMMORROW

[size=150][color=violet]Day 16[/color][/size]

[quote="minsoens"]Let the 10-gon be $ABCDEFGHIJ$. We shall group the diagonals by ones that have an even number of sides between its endpoints and those that have an odd number.
For even, $AC$ is parallel to $JD$, $IE$, and $HF$. $\binom{4}{2}=6$ but we can use pairs parallel to $BD$, $CE$, $DF$, and $EG$, so there are $30$ pairs for the even group.
For odd, $AD$ is parallel to $JE$ and $IF$. $\binom{3}{2}=3$ but we can use pairs parallel to $BE$, $CF$, $DG$, and $EH$, so there are $15$ pairs for the even group.
There are a total of $\boxed{45}$ pairs.[/quote]

[size=150][color=green]Day 17[/color][/size]

[quote="alan"]The volume of $ABCD$ is found by $\left(\frac 13\right)(b)(h) = \left(\frac 13\right)\left(\frac{25\sqrt{3}}4\right)\left(\frac{5\sqrt{6}}3\right) = \frac{125\sqrt{2}}{12}$

Let $AMN$ be the new base, and $AP$ be the new height of the pyramid.

The sum of the volumes of the pyramids with height $AP$ and base $AMN$ with $AM, AN, AP \in \{1,2,3,4\}$ is given by $\frac 13$ times the sum of all the possible heights times the sum of all the possible bases.

$AMN = \frac 12 (AM)(AN) \sin 60^\circ$
Let $B$ be the area of all of these bases.
$B = \frac 12 (1+2+3+4)(1+2+3+4) \sin 60^\circ = 25\sqrt{3}$
Let $H$ be the sum of the possible heights.
$H = \frac 15 (1+2+3+4)\left(\frac{5\sqrt{6}}{3}\right) = \frac{10\sqrt{6}}{3}$
Sum of volumes $= \frac 13 (B)(H) = \frac 13 (25\sqrt{3})\left(\frac{10\sqrt{6}}{3}\right) = \boxed{\frac{250\sqrt{2}}{3}}$[/quote]

[quote="Iversonfan2005"]oh, man, the day 15 solution was a lot better than mine. Good job wobster109![/quote]

A lot of people submitted similar solutions.  Wobster's was chosen since it was more rigorous than others.  Unfortunately, nothing was capitalized in the original solution so I had to edit it  :mad:  :P

whoops sorry!  i habitually don't capitalize anything unless i'm writing an essay for school which i will actually turn in.  sorry!!!  i'll capitalize solutions from now on! :blush:

[quote="solafidefarms"]How're the newer days' checking coming?[/quote]

Please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=518539#p518539]this[/url] post.

I actually could start checking them now but I don't want to go back on my word since some people might be putting them off until the weekend.

I need to protest for a bit about the solution for Day 8. Rotating the figure repeatedly by 90 degrees around the intersection of the extensions of CD and AF does not give you a square. There does exist a point that you can rotate the figure around to get a square, yes, but it is not the point Q that Elemennop gave.

Oh yeah, and on your proof for the theorem used for Day 14 there is an error; the second formula should be:
\[ \frac{b}{z} = \frac{a+b}{x} \text{ and } \frac{a}{z} = \frac{a+b}{y} \]

since day 18 is graded, may we have a solution posted?

[size=150][color=orange]Day 19[/color][/size]

[quote="wobster109"]All angles will be in degrees.

Let G be the intersection of CF and the square on side AB.

First, show CF and AD are perpendicular.
<DBA = 90 + <CBA = <CBG.
Also, DB = CB and AB = GB.
So  triangle DBA is congruent to triangle CBG.

So <DBA = < BCG.  this gives DBE is similar to CFE.
From which follows 90 = <DBE = <CFE.

Since CF is perpendicular to AE, CF is an altitude of triangle ACE.
So ACE ~ CFE ~ DBE.  let the length of CE be x.

Using the similar triangles, we have
$\frac{x}{204} = \frac{68-x}{68}$,
yeilding x=51.

So BE is 68-51 = 17.
So $DE^2 = 17^2 + 68^2 = 4913$, while $CE^2 = 51^2 = 2601$.
This gives the ratio of the area of CFE to the area of DBE as $\frac{2601}{4913}=\frac9{17}$.

The area of DBE is $\frac12\cdot17\cdot68 = 578$.
So the area of CFE is $\frac9{17}\cdot578 = 306$.

So the absolute value of the differences in area is 578 - 306 = [b]272[/b].[/quote]

[size=150][color=yellow]Day 20[/color][/size]

[quote="alan"]consider each number $\pmod 3$.
$21\equiv 51 \equiv 81 \equiv 0 \pmod 3$
$31\equiv 61 \equiv 1 \pmod 3$
$41\equiv 71 \equiv 2 \pmod 3$
the problem simplifies into arranging $0,0,0,1,1,2,2$
Let $a,b,c,d$ be the first $4$ terms in the chain.
$a+b+c+d \equiv 0 \pmod 3$ by definition.
Let $e$ be the next number.
$b+c+d+e \equiv 0 \pmod 3$
$a \equiv e \pmod 3$
similar logic will show that the general chain is $a,b,c,d,a,b,c$
because of parity, $1$ or $2$ must be $a,b,$ or $c$.
Therefore $d \equiv 0 \pmod 3$
The remaining $0,1,2$ can be mapped to $a,b,c$ in $3!$ ways.
There are $3!,2!$ and $2!$ ways to distribute the $0, 1$, and $2$, respectively.
Therefore there exist $3!*3!*2!*2! = \boxed{144}$.[/quote]

[color=blue][size=150]Day 21[/size][/color]

Like I said, this one isn't really [i]as[/i] bad as it seems initially.  One thing you could've done is look at the analogous problem from 1D, 2D, and 3D where visualizing the problem is easier (props to solafidefarms for pointing this out).

[quote="matt276eagles"]Problem #21.  Answer:  $\boxed{\frac{1}{6561}}$.

For each box, let $A$ be the set containing all values of $i$ such that $x_{i}\ge\frac{1}{3}$ for every point in the box.  For each box, let $n$ be the number of elements in $A$.  The volume of a box is then $\left(\frac{1}{3}\right)^{8-n}\left(\frac{2}{3}\right)^{n}=\frac{2^{n}}{3^{8}}$.  For the red box on the origin, $A=\emptyset$.  If we add or take away a single element from $A$, thereby increasing or decreasing $n$ by one, then we get an adjacent box.  Therefore, if $n$ is even, then the box is red, and if $n$ is odd, then the box is blue.  We have

$R-B=\binom{8}{0}\frac{2^{0}}{3^{8}}-\binom{8}{1}\frac{2^{1}}{3^{8}}+\binom{8}{2}\frac{2^{2}}{3^{8}}-\cdots +\binom{8}{8}\frac{2^{8}}{3^{8}}$
$R-B=\frac{1}{6561}-\frac{16}{6561}+\frac{112}{6561}-\frac{448}{6561}+\frac{1120}{6561}-\frac{1792}{6561}+\frac{1792}{6561}-\frac{1024}{6561}+\frac{256}{6561}$
$R-B=\boxed{\frac{1}{6561}}$.[/quote]

Joe, you said that if i do this practice series and win i get candy right? well, bring a pack of jolly ranchers on the bus.


calculus
integration
trigonometry
real analysis
real analysis unsolved

How do I determine this integral
\[\int \cos{\cos{x}}dx\]

Again, the [url=http://integrals.wolfram.com/index.jsp]Integrator[/url] does not give any sort of closed form.


calculus
geometry
trigonometry
number theory

I am a freshmen and I'm puzzled. Does Calculus help you with problem solving? I'm not sure whether to take calculus next year or not. Please help me decide.

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php

Calculus is useful for problem solving when you're solving [i]calculus problems[/i], though.  ;)

Well, if it's a choice between calculus or boring high school math, then go with the calc.

But I'm pretty sure that [i]most[/i] of the mathematical problem solving (in contests, etc.) is based more on a good understanding of methods and topics taught before calculus such as geometry, trig, number theory and the like. That's not to say that calculus isn't useful, as it can be used as a shortcut in many problems; but if you're planning on competing, I would think that a non-calculus class (if it's good, and is in a topic that you are not as strong in,) would be better. But that's just my opinion.  :)


geometry
3D geometry
modular arithmetic
symmetry
quadratics
function
algebra

If $x,y \in \mathbb{Z}$ and

$k=\frac{x^2+y^2+6}{xy} \in \mathbb{Z}$ then $k$ is the cube of an integer.

This problem can be easily solved if we recall the solution to IMO 1988 Problem 6. I'll write out the solution soon.

Firstly, since if $\frac{x^2 + y^2 + 6}{xy}$ is a perfect cube, then so is $\frac{\mid x \mid ^2 + \mid y \mid ^2 + 6}{\mid x \mid \mid y \mid}$, thus we can assume WLOG that $x, y > 0$.

Next, we want to show that if $k \in \mathbb{Z}$, then $\gcd (x, y) = 1$. We note that $y^2 + 6 \equiv 0 \pmod{x}$ and $x^2 + 6 \equiv 0 \pmod{y}$. So, if $\gcd (x,y) \neq 1$, then $\gcd (x, y) = 2, 3$ or $6$.
Let $x = 2^{\alpha_1} 3^{\alpha_2} x_0, y = 2^{\beta_1} 3^{\beta_2} y_0$, where $\gcd (6, x_0) = \gcd (6, y_0) = 1$. We thus have $k = \frac{2^{2\alpha_1} 3^{2\alpha_2} {x_0}^2 + 2^{2\beta_1} 3^{2\beta_2} {y_0}^2 + 6}{2^{\alpha_1 + \beta_1} \cdot 3^{\alpha_2 + \beta_2} \cdot x_0y_0}$.
If any of $\alpha_1, \beta_1 > 0$, WLOG say $\alpha_1 \geq 1$, then taking the numerator of $k \mod 2$, we have $\beta_1 \geq 1$. Thus, we have $4 \mid 2^{\alpha_1 + \beta_1}$. Yet $\beta_1 \geq 1 \Rightarrow$ numerator of $k$ is $\equiv 2 \pmod{4}$ (Contradiction). Thus $2 \nmid x$ and $2 \nmid y$.
By a similar argument, $3 \nmid x$ and $3 \nmid y$. $\therefore \gcd (x,y) = 1$.

Thus, if $x=y$, then since $\gcd (x,y) = 1$, it necessarily follows that $x = y = 1$, which in this case yields the perfect cube $k = 8$. If exactly one of $x,y = 1$, say $y = 1$, then $k = x + \frac{7}{x} \Rightarrow x = 7$ (which yields the perfect cube $k = 8$) From the previous part, it also follows that $x, y \neq 2, 3, 4, 6$. We next claim that if $k \in \mathbb{Z}$, then $x, y \neq 5$.

Suppose $y = 5$, then we have $k = \frac{x^2 + 31}{5x} \Rightarrow x \equiv \pm 2 \pmod{5}$ for $k$ to be an integer. Letting $x = 5x_0 \pm 2$, we have $k = \frac{5{x_0}^2 \pm 4 x_0 + 7}{(5x_0 \pm 2)}$, which reduces to $k = x_0 + \frac{7 \pm 2x_0}{5x_0 \pm 2}$, yet $\frac{7 \pm 2x_0}{5x_0 \pm 2} < \frac{7 + 2x_0}{5x_0 - 2} = \frac{7}{5x_0 -2} + \frac{2}{5} + \frac{4}{5(5x_0 - 2)} \leq \frac{7}{3} + \frac{2}{5} + \frac{4}{15} = 3$, so for $k$ to be an integer, we must have $\frac{7 \pm 2x_0}{5x_0 \pm 2} = 1$ or $2$. Checking the $4$ cases, we find that none of these cases yield an integer for $k$. Thus $k \in \mathbb{Z} \Rightarrow x, y \neq 5$.

We shall now proceed to prove the problem by contradiction, with our motivation coming from the solution of IMO 1998 Problem 6.
Suppose $\exists$ pairs $(x,y)$ such that $k = \frac{x^2 + y^2 + 6}{xy} \in \mathbb{Z}$ but $k$ is not a perfect cube. From the set of possible pairs $(x,y)$, choose a pair $(x_1, y_1)$ such that $\max (x_1, y_1)$ is the smallest. Let $k_0 = \frac{{x_1}^2 + {y_1}^2 + 6}{x_1y_1}$.

From earlier considerations of the possible values of $x$ and $y$, and by the symmetry of $x$ and $y$ in the expression for $k$, we can assume WLOG that $x_1 > y_1 > 6$. Note that $k_0 = \frac{{x_1}^2 + {y_1}^2 + 6}{x_1y_1} \Rightarrow {x_1}^2 + {y_1}^2 + 6 - k_0x_1y_1 = 0$ $\Rightarrow {x_1}^2 - k_0y_1 (x_1) + ({y_1}^2 + 6) = 0$. Consider the quadratic equation $x^2 - k_0y_1 x + ({y_1}^2 + 6) = 0$ in variable $x$, whose roots have sum $k_0y_1$ and product ${y_1}^2 + 6$. Note that $x = x_1$ is a root. Since $(x, y) = (x_1, y_1)$ satisfies $k_0 = \frac{x^2 + y^2 + 6}{xy}$, $\exists$ another pair $(x,y) = (x_2, y_1)$ satisfying $k_0 = \frac{x^2 + y^2 + 6}{xy}$, such that $x_1 + x_2 = k_0y_1$ and $x_1x_2 = {y_1}^2 + 6$.

Since $x_2 = k_0y_1 - x_1$, we have $x_2 \in \mathbb{Z}$. Furthermore, since $x_1 > 0$ and ${y_1}^2 + 6 > 0$, we also have $x_2 \in \mathbb{Z}^+$. Note that $x_2 = \frac{{y_1}^2 + 6}{x_1} < \frac{{y_1}^2 + 6}{y_1} = y_1 + \frac{6}{y_1} < y_1 + 1$. This implies that $x_2 \leq y_1 \Rightarrow \max (x_2, y_1) = y_1 < x_1 = \max (x_1, y_1)$ (Contradiction)

If I didn't remember wrongly, there are quite a few problems on the forum which use similar techniques.

I would like to show another solution, which is not so complicate as
this before. I think that this technique, on English, is called "infinite
descent" but I am not sure. As this guy before me said, we can take
$x,y\in N$ and all this stuff what I am going to write is with $x,y \in N$.

 $A$ is a set of all pairs $(x_0,y_0)$ s.t.
$\frac{x_0^2+y_0^2+6}{x_0y_0}\in N$. Take $(x,y)$ from $A$ with
minimal $x$ (in other words, for any pair $(x_1,y_1) \in A$ it
stands $x_1\geq x$). Without the lost of generality we may assume
$x\geq y$.

Now take \[ k=\frac{x^2+y^2+6}{xy} \] or, after some writing
\[ x^2-ky\cdot x +y^2 +6=0 \]

Write this as a quadratic function of $x$ ($f(x)=x^2-ky\cdot x +y^2 +6=0$). One solution of this quadratic equation is $x$ and other is
(by Viet's formulas) $ky-x$. Now if $ky-x<0$ then $f(-1)\leq 0$ and
that  is not possible, also if $ky-x=0$ then $f(0)=0$ which is not
possible, so $ky-x>0$. Now see that pair $(ky-x,y)$ is also in $A$,
and it has to be $ky-x>x\geq y$ because we said that $x$ is minimal
(and we also said $x\geq y$). From that we have $f(y)\geq 0$ or
$2y^2+6\geq ky^2$ and after a little discussion we get $y=1, k=8$
and $x=1$, and as we can see $k=8=2^3$. End of story :)

bye!

Sorry, I didn't see that you have also solution via infinite descent, I saw only the beginning of your solution and I thought that you didn't solve it like me :)

I agree with you when you say that there aren't many problems solveable with this technique not only on this forum, but in whole problem-solving material. This technique is very important, and problems that can be solved with it are, ussualy easy to notice, as this for example, as I saw it I knew how it's suppose to be solveb, but on the other hand, there are someones that are not so obvious.

Like this: Find all pairs of natural numbers (a,b) s.t. $a|b^2+1$ and $b|a^2+1$. Solution $(F_{2n-1},F_{2n+1})$ where $F_i$ is $i$th Fibonnaci number :)

sorry again..

bye

Well, I was trying to be rigorous in my proof, hence the first part of my proof, which takes care of the finite cases not covered by the infinite descent. :)

Well you see, it's OK to be rigorous when you are solving the problem, but when you are writing solution in places like this (forums, e-mails, etc.) it's better not to stay too long on some trivial stuff, becose it makes your solution look ugly big, and unreadable. It's better just to say main stuff and trivial things let to the reader :)

bye

Ok, point taken.  :)
I guess I can get too carried away with being precise and complete at times.  :P

I enjoyed the nice long precise complete proof more. Without it, I would have been unconvinced (maybe because trivial stuff isn't quite as trivial to me, which is my fault - but hey!).

When you are solving problems (not only mathematics) it's very
important that you have patience, and not to look in a solution
after only 15 minutes or 30 or whatever... If you look to solution
eventually, it would be nice only to see an idea, or some hint, and
then continue solving with that hint, in that case, although you saw
the solution, you steel have some enjoyment :) because you solve the
problem in the end :) This of course is only my opinion, so if you
disagree, that OK :)

Bye


Given convex hexagon ABCDEF prove that at least on of the triangles ABC, BCD, CDE, DEF, EFA and FAB has area that is not greater then $ \frac 16S$ where S is the area of ABCDEF.

BYe

The lines $ AD,BE,CF$ form in the interior of the hexagon 3 quadrilaters and 1 triangle(maybe degenerated).
Hence 1 of the quadrilaters has area$ \leq S/3$, so one of the halves of quadrilater(witch is a triangle) has area$ \leq S/6$.
By the other hand 2 of the triangles from the 6 mentioned have one side comon with the one witch area is smaller then $ S/6$.
But one of the 2 triangles has area smaller(or equal) to our triangle and one has area grater(or equal) to our triangle.
So it exist one triangle from the 6 ones who's area is smaller then $ S/6$.


integration
calculus
function
calculus computations

[quote="rorylando45"]Hey, not sure if this is the right section, but I'm pretty sure it is. How do I go about solving:

$2xy\frac{dy}{dx}+y^{2}=x^{3}$

I think I have to multiply through by an integrating factor $e^{\int Pdx}$ or something where P is the coefficient of y, but I have a $y^{2}$ term not a y term. Argh I'm stuck.

Thanks.[/quote]

Write as,
$M(x,y)+N(x,y) \frac{dy}{dx}= 0$
Where,
$M(x,y) = y^{2}-x^{3}$ and $N(x,y) = 2xy$
Use the cross partials test,
$\frac{\partial M}{\partial y}= 2y = \frac{\partial N}{\partial x}$

This is an exact differential equation.
That means, we need to solve the system:
$\frac{\partial \phi}{\partial x}= y^{2}-x^{3}$ and $\frac{\partial \phi}{\partial y}= 2xy$

The solution to the first one is:
$\phi(x,y) = xy^{2}-\frac{1}{4}x^{4}+g(y)$
But then,
$\frac{\partial \phi}{\partial y}= 2xy+g'(y)$
We see that we can chose $g(y)=0$.
Thus,
$4xy^{2}-x^{4}= C$
Is the solution, you can use the implicit function theorem from here.


Choose the following five randomly once for an ordering (a statement**), and allow concatenation
where possible:

 .
$ \le$
4
5
6

What is the probability that an inequality is formed which is a true statement?

** A statement here consists of the five written in a line in a row.

[hide="Solution"]
Valid orderings take three forms (let $ a,b,c$ denote digits):
1. $ a. \leq bc$   Choose two of the three numbers in $ \binom{3}{2}$ ways and order in $ 2!$ ways. Place the chosen numbers after the inequality sign. The remaining characters can be placed in only one way.
2. $ a \leq bc.$   (Same number of orderings as 1.)
3. $ a \leq b.c$   Valid orderings of this form are $ 4\leq 5.6, 4\leq 6.5, 5\leq 6.4$.

Thus, the probability is $ \frac {2!\binom{3}{2} \plus{} 2!\binom{3}{2} \plus{} 3}{5!} \equal{} \boxed{\frac {1}{8}}$. Note: It is unclear as to whether cases 1 or 2 should be included; if they are excluded, the probability is simply $ \frac {3}{120} \equal{} \frac {1}{40}$.[/hide]

[quote="eganviking1"][quote="mismatchtea"]
Valid orderings take three forms (let $ a,b,c$ denote digits):
1. $ a. \leq bc$   Choose two of the three numbers in $ \binom{3}{2}$ ways and order in $ 2!$ ways. Place the chosen numbers after the inequality sign. The remaining characters can be placed in only one way.
2. $ a \leq bc.$   (Same number of orderings as 1.)
3. $ a \leq b.c$   Valid orderings of this form are $ 4\leq 5.6, 4\leq 6.5, 5\leq 6.4$.

Thus, the probability is $ \frac {2!\binom{3}{2} \plus{} 2!\binom{3}{2} \plus{} 3}{5!} \equal{} \boxed{\frac {1}{8}}$. Note: It is unclear as to whether cases 1 or 2 should be included; if they are excluded, the probability is simply $ \frac {3}{120} \equal{} \frac {1}{40}$.[/quote]

I don't think that's all the cases. What about the statement $ a.b \leq c$?[/quote]

Thanks, I forgot about that case and another one.
4.  $ a.b \leq c$ Valid orderings of this form are $ 4.5 \leq 6, 5.4 \leq 6, 4.6 \leq 5$.
5.  $ .ab \leq c$ (Same number of orderings as 1 and 2).
Thus, the final probability becomes $ \frac{1}{5}$.


This is the funniest thing I've seen in a looong time.
[url=http://winterson.com/2005/06/episode-iii-backstroke-of-west.html]winterson.com/2005/06/episode-iii-backstroke-of-west.html[/url]

It's only that funny if you've seen SW3, though.
"They're in my behind!"  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

Ninja edit: Watch out, the "f word" pops up a few times.

that is hilarious! :rotfl:  :rotfl:  :rotfl:


geometry
circumcircle
geometric transformation
reflection
trapezoid
probability

The incircle of ABC touches BC, CA, AB at K, L, M respectively. The line through A parallel to LK meets MK at P, and the line through A parallel to MK meets LK at Q. Show that the line PQ bisects AB and bisects AC.

ABC is an acute-angled triangle with circumcenter O and orthocenter H (and O ≠ H). Show that one of area AOH, area BOH, area COH is the sum of the other two.

This time I won't tell you what competition it is from shobhit  ;)

First of all, I am assuming that the question asked for the extention of $PQ$ since $Q$ is not necessarily $\in AC$ and $M$ is not necessarily $\in AB$. Let $AX$ and $AY$ be extentions of $AP$ and $AQ$ respectively such that $X,Y \in BC$. Because we are given that the extention of $PQ$ (let us call this $C'B'$) bisects $CA$ and $BA$, then $\triangle{ACB} \sim \triangle{AC'B'}$ and $\frac{[\triangle{AC'B'}]}{[\triangle{ACB}]}=\frac{1}{4}$. It is suffficient to prove that $\triangle{AQP} \sim \triangle{AYX}$ and $\frac{[\triangle{AQP}]}{[\triangle{AYX}]}=\frac{1}{4}$ to prove the previouis statement since the bases of $\triangle{AQP}$ and $\triangle{AYX}$ lie on those of $\triangle{ACB}$ and $\triangle{AC'B]}$ along with sharing the same altitudes. Now, by using parallel segments, we see that $\angle{QYX}=\angle{PKX}$ and that $\angle{PXK}=\angle{QKY}$. From this, we get that $\angle{YQK}=\angle{KPX}$ and $\triangle{YQK} \sim \triangle{KPX}$. Now, also notice that $\angle{A}=\angle{YQK}=\angle{KPX}$ as well, so that $\triangle{AYX} \sim \triangle{YQK} \sim \triangle{KPX}$. Also, using angle chasing through transversals, we have that $\angle{QKP}=\angle{YQK}=\angle{KPX}=\angle{A}$. Since $\triangle{AQP}$ and $\triangle{QKP}$ share the same bases, their angles are all equal and the two triangles are congruent. It follows that $\triangle{AQP}=\triangle{YQK}=\triangle{KPX}=\triangle{QKP}$, and $[\triangle{AQP}]=\frac{[\triangle{AYX}]}{4}$ and $QP$ bisects $AY$ and $AX$. Thus, $C'B'$ bisects $AC$ and $AB$. $\text{QED}$.

EDIT: typos like crazy

yeah, i just realized that danielhstennis08 just gave me a national olympiad problem to start off proofs... thanx a lot ;)  :P

http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=46&year=2000&p=208231

i guess taht's the solution, but it's completely different from mine, so if anyone wants to check mine, be my guest :D

EDIT: BTW, what is inversion. I don't understand grobber's solution... :?

u didn't realize shobhit... I told you haha

[quote="DanielHStennis08"]
This time I won't tell you what competition it is from shobhit  ;)[/quote]

well, i hoped it wasn't another national olympiad problem (but i guess it is once again). Oh well...atleast it's geometry so it shouldn't be too bad :D

well, it is not too bad now that I saw the solution... but i couldn't do it before... then again, it is me. haha

um..yeah, this problem isn't that easy.

Here's what I got so far.

Without loss of generality, draw $\triangle{ABC}$ such that $\triangle{AOH}$ be the largest triangle and thus is the sum of the other two. Since $OH$ is the base for all three triangles, then $\triangle{AOH}$ has the largest altitude and is the sum of the other two altitudes of the other two triangles. Let $CX$ be the altitude of $\triangle{COH}$ and $YB$ be the altitude of $\triangle{OHB}$. Now it comes down to proving that $XC=YB$. Consider the reflection of $YB$ to be $Y'B'$, where $Y'=C$, and $XB'$ a segment which equals the length of $AP$ (the altitude of $\triangle{AOH}$). Now consider a point $Q$ outside of circle $O$ in which $B'Q$ is a straight line segment and $\angle{B'QA} \not=90^\circ$. Now, construct a piont $R$ on circle $O$ such that $AR$ is a chord and $\angle{QB'R}\not=90^\circ$. Of course, $\angle{B'RP} \not= \angle{QAR} \not=90^\circ$ or else $H=O$. Now, by looking at this figure, $AQB'R$ must be an iscoceles trapezoid with $QA=B'R$ in order for $RA$ to be parallel to $B'Q$ (which must be the case since $PX$ intersects both $AR$ and $QB'$ at right angles).

Now it comes down to proving that the ONLY position $PX$ can take is such that $AP=XB'$ inside the trapezoid. Maybe power of point?....

[quote="DanielHStennis08"]I don't think anyone but us is looking at this topic... ask joe to try it... i'm sure he will...[/quote]

I'll look at it tomorrow.

[quote="DanielHStennis08"]looks like joe didnt like the problem[/quote]

well, these problems are very nontrivial :D

New Problem:

Shobhit2005 (3:58:06 PM): A fair coin is tossed repeatedly. Find the probability of obtaining five consecutive heads before two consecutive tails.


probability
symmetry
combinatorics unsolved
combinatorics

Consider a $n$-letter-word consisting of either an $A$ or an $B$.
Let $S$ denote the number of times the strings $AB$ or $BA$ occurs in the word.
[color=green]EDIT: I mean that we have a given number of letters to choose among, namely $n$ letters, where $k$ of them are $A$ and $n-k$ of them thus $B$.[/color]

a)Find the average value of $S$ among all possible $n$-words.

b)Is there any generalizations? Like $ABC\ldots$?

[quote="JBL"]The probability that adding another letter gives you two consecutive different letters is $\frac{1}{2}$.  So it's just $\frac{n-1}{2}$.  This can be doubtless be generalized in lots and lots of ways.[/quote]
Well, sorry but actually I meant that we have a given number of letters to choose among, namely $n$ letters, where $k$ of them are $A$ and $n-k$ of them thus $B$.

So given these number of $A$ and $B$ we cinsider the position $(i, i+1)$ in the word where $1\geq i \leq (n-1)$. Moreover let $P_{i}$ be the probability of having an pair ($AB$ or $BA$) at the position mentioned above. By symmetry all $P_{i}$ are equal, so the average number would be $(n-1)P_{i}$

But here is where the edit apply: If the number of each letter were not bounded $P_{i}=\frac{1}{2}$, so you are correct. :wink:

But if they are bounded as above we must consider the remaining positions apart from $(i, i+1)$. The $n-2$ positions with $k-1$ $A$ and $n-1-k$ $B$ can be combined in $\binom{n-2}{k-1}+\binom{n-2}{n-k-1}=2\binom{n-2}{k-1}$ ways.
In total we have $\binom{n}{k}$ words, thus the answer is
\[\langle S\rangle=\frac{(n-1)2\binom{n-2}{k-1}}{\binom{n}{k}}\]

[color=blue]Am I correct?[/color]
[color=red]How about three letters? anyone?[/color]

Yes, it is correct, although it simplifies to $\frac{2k(n-k)}n$.  We can also see this more directly than in your proof, which allows us also to generalize more easily, I think:  follow one $A$ through all permutations.  $\frac{n-1}{n}$ of the time, it is not the last letter, in which case it is the first letter is a pair $AB$ $\frac{k}{n-1}$ of the time.  The other $\frac{1}{n}$ of the time, it is the last letter, in which case it's the first letter in an $AB$ pair none of the time.  So we get $\frac{k}{n-1}\cdot \frac{n-1}{n}= \frac{k}{n}$ pairs for this letter, and so $\frac{k(n-k)}{n}$ $AB$ pairs on average.  But the number of $BA$ pairs is surely the same, by symmetry.  Thus the answer.

With $n$ letters $A_{1}, A_{2}, \ldots, A_{n}$ of multiplicity $a_{1}, a_{2}, \ldots, a_{n}$, respectively, with $a_{1}+\ldots+a_{n}= N$, a typical $A_{i}$ is the first letter of a pair in which the other letter is $A_{j}$ an expected $\frac{a_{j}}{N}$ times, so we get $\frac{a_{i}a_{j}}N$ pairs $A_{i}A_{j}$ on average, etc.


Make sure you read about both on their websites. Also, look here
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261914[/url]

If you're thinking about applying now, you need to hurry up!


trigonometry
inequalities
geometry proposed
geometry

[quote="Ligouras"]Let $ ABC$ be an triangle. Prove o Disprove that
$ \sin^2\frac {A}{4} \plus{} \sin^2\frac {B}{4} \plus{} \sin^2\frac {C}{4} \ge \ \frac {3(2 \minus{} \sqrt {3})}{8}$.

_____________________
Ligouras[/quote]

Get $ f(x)\equal{}sin^2x$, then $ f''(x)>0$, where $ x$ belong to $ (0;\frac{\pi}{4})$.
So by Jensen ine. we have:
$ \frac{\sin^2\frac {A}{4} \plus{} \sin^2\frac {B}{4} \plus{} \sin^2\frac {C}{4}}{3} \geq sin^2\frac{\pi}{12}\equal{}\frac{1\minus{}cos\frac{\pi}{6}}{2}\equal{} \frac {2 \minus{} \sqrt {3}}{4}$

So we have:
$ \sin^2\frac {A}{4} \plus{} \sin^2\frac {B}{4} \plus{} \sin^2\frac {C}{4} \ge \ \frac {3(2 \minus{} \sqrt {3})}{4}>\frac {3(2 \minus{} \sqrt {3})}{8}$.
and that mean  we can't achieve $ \frac {3(2 \minus{} \sqrt {3})}{8}$


calculus
integration
limit
function
trigonometry
real analysis
real analysis unsolved

[size=75]According to some results in Mathematica, it seems that for each $ p\ge 0$

$ \lim_{\epsilon\rightarrow 0\plus{}0} \left( \int_0^{\plus{}\infty} \frac{x^p}{1\plus{}x^{p\plus{}1\plus{}\epsilon}} \mathrm{d}x\minus{} \frac{1}{\epsilon} \right)\equal{}0$

I don't know how to prove that. Is that related to the Hypergeometric Functions or something else?[/size]

Although there is probably some trick using standard techniques, I would use a complex contour integral.  The integral $ \int_{0}^{\infty}\frac{x^{\alpha}}{x^{\alpha\plus{}1}\plus{}1}\,dx$ can be evaluated using an indented contour around 0.

[quote="JRav"]... The integral $ \int_{0}^{\infty}\frac {x^{\alpha}}{x^{\alpha \plus{} 1} \plus{} 1}\,dx$ can be evaluated using an indented contour around 0.[/quote]
It should  diverge, is not it ?  :)

[quote="JRav"]Although there is probably some trick using standard techniques, I would use a complex contour integral.  The integral $ \int_{0}^{\infty}\frac {x^{\alpha}}{x^{\alpha \plus{} 1} \plus{} 1}\,dx$ can be evaluated using an indented contour around 0.[/quote]

Hi but :) !!

$ \int_0^x \frac {t^a dt}{t^{a \plus{} 1} \plus{} 1} \equal{} \frac {1}{a \plus{} 1} ln(|1 \plus{} x^{a \plus{} 1}|)$

if $ x \rightarrow \plus{} \infty$ the integral diverge....

I think that the integral can be calculated without using any contour integration. Simply use the substitution $ x^{p\plus{}1\plus{}\epsilon}\equal{}y$ and

the integral becomes

$ \int\limits_{0}^{\infty}\frac{x^{p}}{1\plus{}x^{p\plus{}1\plus{}\epsilon}}dx\equal{}\frac{1}{p\plus{}1\plus{}\epsilon}\beta((p\plus{}1)/(1\plus{}p\plus{}\epsilon), 1\minus{}(p\plus{}1)/(p\plus{}1\plus{}\epsilon))$

$ \equal{}\frac{\pi}{(p\plus{}1\plus{}\epsilon)\sin \frac{(p\plus{}1)\pi}{p\plus{}1\plus{}\epsilon}}.$

Now you can calculate your limit.


How can I participate in this competition? Is there a website about this competition?

Also,

I'm interested in trying out as many math competitions I can through the internet, mail, or local schools without paying a lot of money or having to get my entire school involved. I am a rising ninth grader

So far, I think I'll be able to participate in the AMC (through school)and USAMTS(through the internet) for sure. I'm also looking into the iTest, but am unsure of the rules to join and whether I need to pay...-

[[National Online Math League]]

For the official website, check the external links.

that link leads to the wrong place (mathworld)

itest is free. the website seems down, probably for updating for this years competition.

Actually I seriously doubt this competition will happen this year. They haven't run the contest since Cody got busy with school.

Yeah, NOML is in hibernation. We'll announce it when it comes back to life, possibly after Cody gets his PhD.

@aufha: Here's another inexpensive, long distance competition. ;) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163891