perfect628 tried this before. It didn't work out so well. I'm going to make sure I get some dedicated people this time.

I'm looking for approximately 14 people who will be willing to compete. Description of game rules can be found here: [url]http://en.wikipedia.org/wiki/Big_Brother_%28U.S.%29#Format[/url]

It probably won't be that complicated, with nothing like Food Competitions and the like.

pm me with the following information if you are interested:

1. Username (well, obviously I'll know it).
2. Location (this won't be revealed, don't worry, if you're really uncomfortable telling me, you don't have to).
3. Age (ditto from above)
4. How often you would be willing to check this thread
5. If you do not get selected, would you consider being a potential replacement?
6. How likely is it that you will need to be replaced?

Note: if someone needs to leave the game, he may or may not be replaced.

Just a few other tidbits about the game.

There will be a diary room: you will be expected to keep me updated on what you are thinking (tell the truth) at the time, who you trust, what is happening, etc. none of this will be revealed, other than random stuff that has nothing to do with the game, in selected challenges.

If I get enough applications, you will be receiving PM's about your status in the game (in, out, waiting list, etc.). Once you get this, [b]do not[/b] reveal your status until I announce, publically, the list of players.

Communicating about the game is allowed, encouraged, and expected if you care at all about the game. However, communicating during the course of a challenge is [b]not allowed[/b].

Feel free to post any questions in this thread.

[quote="CatalystOfNostalgia"]
2. Location (this won't be revealed, don't worry, if you're really uncomfortable telling me, you don't have to).
[/quote]

What sort of location? (city, street, full address?) And should we erase that from our profile if we apply?

No, for you, just put Texas. That would be fine.

4 submissions so far...please keep them coming. Also bump.

Don't expect 14+Replacements...IMO start with 12.

[quote="1=2"]So which version of the rules will we be using?[/quote]

The most recent one, used in today's (American) version of Big Brother. That is, HOH puts up two people, 6 people compete for veto (HOH, Nominees, 3 people chosen at random), unless there are less than 6 people, in which all compete. You can take yourself off. Evicted houseguest voted on by everyone except HOH and nominees (unless there are 3 people left, in which HOH decides). 3rd-9th place is the jury, who decides the winner from the final 2.

BUT ALSO THERE ARE TWISTS(I hope...)

For example, someone may get voted back in, double eviction, etc...

[quote="1=2"]OK. I have no clue how to play this.[/quote]

It will become more apparent what is going on once you start playing (if you decide to do so); I remember the first time I played mafia (not here), I didn't know what was going on, but I just went with the flow and understood it after like 2 cycles.

Also since this is the first run and twists take some work to pull off, there probably won't be anything huge. We'll see.

Also, still only 6 submissions...

OK. Here are the basics(should make it a bit more clear):

Everyone enters the house. Each week, we begin with a Head of Household competition. The winner is then responsible for placing(nominating) two other houseguests on "The Block." When you're on the block, you're at risk of being evicted, or out of the game. After initial nominations for the block, there is a Veto competition. The competitiors are the HOH, the two nominees, and three others chosen at random. The winner of this competition has the right to change, or veto, one of the original nominations. They do not have to however. Now, if a nom is replaced, the HOH then has to replace them. Now, after the Veto, all HGs(houseguests) besides the HOH and the nominees vote on whom to evict. The one with the most votes is out. If there is a tie, the HOH breaks it. After this, a new week starts and it all happens again.

Now, once we get to final three, the HOH decides whom to evict by himself. Thus, he chooses his final two partner. For the final two, the last seven evicted all vote on the winner, which decides who wins the game.

[quote="1=2"]I wanna play, then.[/quote]

Send me your thing (see first post)

about how long will it take?

like will it push into end of June??

It depends on how quickly houseguests can submit whatever they need to submit, but mid-June would be a good guess.

I currently have 8 applications, all of which look good, so I think we'll start with 9. Maybe more, depending on how soon I get more.

I have chosen ten houseguests to begin the game. Look out for the offical thread soon.

I currently have 7/11 of the confirmation responses...


Draw the graphic of:
$[x]+[y]=c$
where $[t]$ denotes the closest integer number to $t$ and $c$ is constant.
I know that $(x-a)^2+(y-b)^2=z^2$ is a circle with radius $z$ and centre
$O(a,b)$ (a,b,c - constants) and that $|x-a|+|y-b|=c$ is a square with centre $P(a,b)$ and diagonal $c$, but what about $[x-a]+[y-b]=c$

Just sketch it out and you will notice it is a decending stair-like shape . Each stair is a square of length 1 unit  :D

Hmm, yes it`s truely a stair-like shape but the the top and the right lines of the stairs are not in the graphic right?
 Then what would you say about this graphic:
$[x]^2+[y]^2=4$

[hide]See problem number 3 from http://www.zaba.ru/cgi-bin/tasks.cgi?tour=national.cmo.1975[/hide]

[quote="Beat"]$[x]^2+[y]^2=4$[/quote]
For this one , we can only have $([x],[y])=(2,0),(-2,0),(0,2),(0,-2)$

So plotting out each case we should be getting 4 squares with length 1 unit surrouding the origin
[quote]
[hide]See problem number 3 from http://www.zaba.ru/cgi-bin/tasks.cgi?tour=national.cmo.1975[/hide][/quote]

Well , I dont know the language anyway  :P


If $ 8\cdot2^x \equal{} 5^{y \plus{} 8}$, then when $ y \equal{} \minus{} 8,x \equal{}$

$ \textbf{(A)}\ \minus{} 4 \qquad\textbf{(B)}\ \minus{} 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8$

[hide]

$ 8\equal{}2^3$ so this equation can be rewritten as:

$ 2^{x\plus{}3}\equal{}5^{y\plus{}8}$

When $ y\equal{}\minus{}8$, the equation becomes:

$ 2^{x\plus{}3}\equal{}1$

Clearly, $ x$ has to be $ \minus{}3$, $ \boxed{B}$

[/hide]


geometry
circumcircle
parallelogram
incenter
geometry unsolved

Given four points $B_{1},B_{2},C_{1},C_{2}$ and $A=B_{1}B_{2}\cap C_{1}C_{2},G_{ij}$ are four points satified one of three following cases
$1)$  $G{ij}$ are centroid of triangles $AB_{i}C_{j}$
$2)$  $G{ij}$ are orthocenter of triangles $AB_{i}C_{j}$
$3)$  $G{ij}$ are circumcenter of triangles $AB_{i}C_{j}$
Prove that $G_{ij}$ are four vertex of a parallelogram $(i,j)=\overline{1,2}$
And an open question with incenter !

1), 2) and 3) are easy exercises, while the "open question" with the incenter is easily seen to be false. Are you really sure you wanted to post this in Unsolved Problems?

  Darij

[quote="gemath"]I'm not sure, but it is in Crux with a star therefore I think it is hard, but anyway I haven't tried it ![/quote]

Well, I have solved some asterisked Crux problems, but they were not [i]that[/i] trivial...

  darij

ok Here is the problem that I posted, maybe I don't understand it so much!


superior algebra
superior algebra unsolved

Let $B,N$ be groups, $G=BNB$ a group containing them, such that $(B\cap N)\unlhd N$ and their quotient $W: =N/(B\cap N)\cong S_{3}$.

Let now $s_{1},s_{2}\in W$ be different involutions (so together they generate $W$). Assume also that $Bs_{i}BwB\subseteq Bs_{i}wB\cup BwB$ for all $w\in W$; and assume also that $s_{i}Bs_{i}\not=B$. Given $g_{1},g_{2},h_{1},h_{2}\in G\setminus (B\cup Bs_{1}B\cup Bs_{2}B)$, show that there is a $b\in B$ such that simultaneously $g_{1}bg_{2}^{-1}\in Bs_{1}B\cup B$ and $h_{1}bh_{2}^{-1}\in Bs_{2}B\cup B$. :)

Any help in solving it, or any comments/remarks on how it could possibly be done are very much welcomed.

[hide="Note 1"]One can prove that $G$ is the *disjoint* union of $B\cup Bs_{1}B \cup Bs_{2}B \cup Bs_{1}s_{2}B \cup Bs_{2}s_{1}B \cup Bs_{1}s_{2}s_{1}B$, where of course $S_{3}=\{e,s_{1},s_{2},s_{1}s_{2},s_{2}s_{1},s_{1}s_{2}s_{1}\}$. Assume this as a given if it helps you.[/hide][hide="Note 2"]One can prove that $B\cup Bs_{i}B = Bs_{i}Bs_{i}B$, and that this is a group. Assume as a given if it helps you.[/hide]

Has [i]any[/i]one got [i]any[/i] idea how to tackle this? It's been keeping me awake lately.  :huh:


A single letter will suffice.

Steve Dunbar
MAA Director, American Mathematics Competitions

Congratulations on the award - not too many people get them!

Remember why it's important to send these letters.   Obviously, it's important to show appreciation to those who make donations.  More importantly, those who donate money and time to these efforts want to know if the mission is being accomplished.  The mission varies a bit between organizations but there are usually a few common elements:

1.  improved problem solving skills for all participants, not just those at the top
2.  encourage local groups/clubs/teams to form that enhance the study of serious problem solving
3.  national/international contests often want to help create a community of serious problem solvers by bringing them together at state and national competitions.

Finally, remember that letters and comments can be extremely powerful tools for soliciting future contributions of time, talent, and money - especially letters that directly support the organizational mission.  The math contest community is extraordinarily dependent on the efforts of volunteers to create problems, administer contests, build communities, and solicit contributions used for awards and programs like MOP.  Donors would like to believe that the world is a bit better off because of their efforts.  The "thank-yous" are probably the most convincing evidence that can be presented.

While it's fairly common for award winners to be asked for letters, some of the most effective communications for purposes of future solicitations come from the average people on the math team who never win contests but attest to improved problem solving skills and the value of the community of problem solvers.

[quote="kyyuanmathcount"]I got a CAMC mathematics book for getting some regional AMC12 award, but I had a question about the thank you letters.

"I am pleased to inform you... blablabla"
"It would be most fitting for the student receiving this award to send a letter of appreciation to the donor of their award.  We would appreciate receiving a copy of the letter."

The donor of my award is Mr. Dunbar, so would I still have to send two letters?  In other words, does the "we would appreciate receiving a copy of the letter" refer to some other person(s) in the AMC?

Thanks.[/quote]

Hey

What was the title of the book that you got?

It seems like a good book.

And yeah, I sent a letter of appreciation to Mr. Dunbar too for the plaque coming.


Suppose you have $ n$ hemispheres (boundary included) on a sphere such that every point on the sphere is covered. Prove that there exists a subcollection of at most $ 4$ of the $ n$ hemispheres such that those $ 4$ also cover the sphere.

A hemisphere is half a sphere...

[b]Open Hint:[/b] Suppose the sphere is origin-centered. Then any unit vector defines a unique hemisphere (and vice versa) as follows:

- Construct the plane for which the chosen unit vector is the normal unit vector
- All points in the intersection of the plane and the sphere are part of the hemisphere
- All points on the sphere on the appropriate side of the plane (as indicated by the direction of the unit vector) are part of the hemisphere

I just thought there would be more "restrictions" on the hemispheres.  :)
[hide="Method?"]
Obviously $ n > 1$.
Any completely overlapping hemispheres are considered to be the same hemisphere.

For $ n \equal{} 2$, there is only one case where this works; if the hemispheres are "moved" so they begin to overlap, a "hole" is also created, requiring $ n \equal{} 3$.  When the two hemispheres continue to move until they completely overlap, it brings that case of $ n \equal{} 3$ to $ n \equal{} 2$.  "Moving" with the $ n \equal{} 3$ cases, the biggest "hole" that can be created is the size of a hemisphere - in which case it would collapse to $ n \equal{} 2$ - and as such, without collapsing, the sphere requires $ n \equal{} 4$.  From there, it seems like every other case, $ n > 3$, seem to collapse to $ n \equal{} 3,4$ since there are at least two such overlapping hemispheres on a sphere covered by $ n > 2$ hemispheres.  Each $ n > 3$ is a "sum" of certain cases of $ n \equal{} 2$ and $ n \equal{} 3$ that don't completely cover the sphere.  That should finish it.
[/hide]

Well, I'm confused. The original problem actually said $ n > 4$, since $ n \leq 4$ is trivial; just select all of the hemispheres. I'm not sure how a solution that deals almost exclusively with $ n \leq 4$ can finish the problem. (unless that was intended to be a partial solution case bashing everything except the hard part)

[hide="If the first hint wasnt enough..."] There is a more straightforward construction of the unique hemisphere corresponding to a unit vector $ u$:  it is the set of all unit vectors with non-negative dot product with $ u$.

Given this construction, what can you say about a collection of unit vectors with the property that every unit vector has non-negative dot product with at least one member of the collection? [/hide]


Let  $ p(n) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_1x \plus{} a_0$ and $ n$ is an odd number.
Prove, that $ p(x) \equal{} 0$ has at least one real solution.

Maybe I'm asking nonsense, but what does $ p(n)\equal{}0$ exactly mean? If it was meant to be $ p(x)\equal{}0$, then solution is trivial. If not, please enlighten me.

[hide="p(x)=0"]It has at least one [b]real[/b] solution. And $ n$ complex solutions in total. Why does it have at least one real solution? Non-real complex solutions go in conjugate pairs... And $ n$ is odd, hence one has no pair, and it ought to be real.[/hide]

There is another very simple reason why $ p(x)$ has a real root. Indeed, $ p(x)$ is continuous and has different signs on $ \plus{}\infty$ and $ \minus{}\infty$.

Good ol' Bolzano's theorem  :lol:

BTW, welcome back Myth, haven't seen you around for a long time...

Hm....I was thinking along a different line... if a_n is real, and if x has complex, then it has its conjugate. Therefore, the polynomial has at least one real root.


geometry
rectangle
trigonometry
inequalities
angle bisector

[hide="1"]The bisector AD, the median BM and the height CH of the acute-angled triangle ABC intersect in one point. Prove that the angle BAC is greater than 45 degrees. D is on BC, M is on AC, and H is on AB.[/hide]
[hide="2"]Two equal rectangles are intersecting in 8 points. Prove that the common part area is greater than the half of the rectangle's area.[/hide]

I should have added an s at the end of the title.

In problem (2), are the rectangles congruent to each other or do they just have equal areas?

[quote="socrates"][quote="shobber"]Problem 1 is from Poland, 1961. From the condition we can get that $\tan{A}=\frac{\sin{C}}{\cos{B}}$[/quote]
how to go on ?
 :maybe:[/quote]
i'm not sure. Shobber just left it at that.

What, what is this $\tan$ stuff? Just bust out with Ceva...

Using Ceva's Theorem, we have that:

$BD \cdot CM \cdot AH=DC \cdot AM \cdot HB$, so we have:

$(\frac{DB}{DC})(\frac{CM}{MA})(\frac{AH}{HB})=1$.

The median makes $CM=MA$, so we are left with:

$(\frac{BD}{DC})(\frac{AH}{HB})=1$.

Now using angle bisector theorem, we know that $BD: CD=AB: AC$, so substituting in gives us:

$(\frac{AB}{AC})(\frac{AH}{BH})=1$. From this, we have that $\frac{AH}{AC}=\frac{HB}{AB}$. But $\frac{HB}{AB}<1$ since $H \in AB$. Thus, it follows that $\angle{BAC}>\angle{ACH}$. But since $\triangle{ACH}$ is a right triangle, and $\angle{A}+\angle{ACH}=90^\circ$, it follows that $\angle{BAC}>45^\circ$.

$\boxed{\mathbb{Q}\mathfrak{ED}}$

[quote="Iversonfan2007"] From this, we have that $\frac{AH}{AC}=\frac{HB}{AB}$. But $\frac{HB}{AB}<1$ since $H \in AB$. Thus, it follows that $\angle{BAC}>\angle{ACH}$. But since $\triangle{ACH}$ is a right triangle, and $\angle{A}+\angle{ACH}=90^\circ$, it follows that $\angle{BAC}>45^\circ$.
[/quote]

Ι think this is not true...... :maybe:

arghhhh, no, I know what he means. I made the assumption that my inequalities show that $\angle{A}>\angle{ACH}$ which does not have to necessarily be true. I'll fix it, but I"m sure it's just using the same inequalities and proving that $AH<HC$:wink:


induction
rotation
geometry proposed
geometry

Let A and B be 2 sets of points on a circle, A having 2n points and B having 2m points, and all 2n+2m points are distinct. Show that there is a chord of the circle which separates n points of A and M points of B from the other N points of A and M points of B.

sorry to drag everyone back 2 years in time...but grobber, do you have a solution if the 2n+2m points are not necessarily ON the circle (in which case induction easily takes care of it), but can be INSIDE the circle as well? (perhaps we should add that no three points are collinear.) :)

I've deleted my previous response :).

I think that given any $2m+2n$ points in the plane, no three collinear, we can find a line separating $n$ points from the first set and $m$ from the second set. All you need to do is continuously rotate a line which separates $m+n$ points. Sorry if there's a catch and this doesn't work, but I don't have any more time to spare for this problem right now :).


geometry
combinatorics unsolved
combinatorics

Can we use n sticks to make n triangles which have each side is one of those sticks?

I have post this one in geometry unsolved problems ("equil triangle"), but I think it may be a combinatorics one.
I think we can use many regular tetrahedron.

i think you cant, maybe using the formula for polyedra $f+v=e+2$ and analizing lots of things...


inequalities
probability
function
calculus
integration
real analysis
expected value

$ \textbf{Theorem 1. }\text{(Markov's Inequality)}$
Let $ X$ be a nonnegative random variable. Let $ \lambda$ be any positive number. Then
\[ P( X\geq \lambda) \leq \frac {E(X)}{\lambda}
\]
[hide="Proof"]
Let $ D(x)$ be the cumulative probability function of $ X$, i.e. $ D(x) = P(X\leq x)$, and $ D'(x) = P(X = x)$.
Expected value is defined as an integral over the entire probability space of $ X$:
\[ E(X) = \int_{0}^{\infty} x \ P(X = x)\ dx
\]
The key is to break this up into two intervals:
\[ E(X) = \int_{0}^{\lambda} x \ P(X = x)\ dx \ + \int_{\lambda}^{\infty} x \ P(X = x)\ dx
\]
To bound the expected value, notice that in the first integral, the values of $ x$ summed over are "small" (less than $ \lambda$) but each summand is positive - we can remove this chunk.
\[ E(X)\geq \int_{\lambda}^{\infty} x \ P(X = x)\ dx
\]
Now, each $ x$ summed over is large enough to replace with its lower bound, $ \lambda$.
\[ E(X)\geq \lambda\int_{\lambda}^{\infty}P(X = x)\ dx = \lambda\int_{\lambda}^{\infty}\ \frac {d(\ D(x)\ )}{dx}\ dx
\]

\[ = \lambda\cdot[D(\infty) - D(\lambda)] = \lambda\cdot P(X\geq \lambda)
\]
[/hide]
Yes, I treated this theorem as basically a preolympiad proof problem... what other proofs will you guys come up with?

The classic proof is this:

Let $ W$ be the random variable that is equal to $ \lambda$ on those parts of the sample space for which $ X\ge \lambda$ and equal to zero otherwise. Then $ P(W\equal{}\lambda)\equal{}P(X\ge\lambda)$ and
\[ E(W)\equal{}0\cdot P(W\equal{}0)\plus{}\lambda\cdot P(W\equal{}\lambda)\equal{}\lambda P(X\ge\lambda)\]
But $ W\le X$ everywhere in the sample space, and $ W\le X$ implies that $ E(W)\le E(X).$ Therefore

$ \lambda P(X\ge\lambda)\le E(X),$ so that $ P(X\ge\lambda)\le \frac{E(X)}{\lambda}.$

Now: this absolutely does not belong in the Pre-Olympiad section. And me@home's alleged "proof" isn't a "pre-olympiad style" proof - instead it's a mess of ill-defined and incompatible notation.

To start with, me@home never fairly faces the question of whether $ X$ is a discrete or continuous random variable. And in any case, it is never true that $ D'(x)\equal{}P(X\equal{}x).$ If $ X$ has a density - in other words, $ D'(x)$ makes sense - then $ P(X\equal{}x)\equal{}0.$ If $ P(X\equal{}x)>0$ for any $ x,$ then $ D$ is discontinuous at that point a doesn't have a derivative. It is fair to write this:
\[ E(X)\equal{}\int_0^{\infty}x\,dD(x)\]
But what is that notation? That's actually a Stieltjes integral on the right. It's already true that integration doesn't belong in this forum, but a Stieltjes integral?

My own "classic" argument is really one that uses the outlook and thought patterns that go with the Lebesgue integral, although the fact that such integrals may be used is well hidden. It also makes no difference for that proof whether the distribution of $ X$ is continuous, discrete, or something in between.


http://www.onemorelevel.com/games.php?game=408

It's a fun game where you try to blow up three people before they blow up you.

It's bomberman! With strange homicial pyromaniac (sometimes suicidal) robots!


Salah  wish to you a beautiful and enjoyable times
Bye

Hi Salahcool. For the first one, an obvious solution is $ x \equal{} 5$ and for the second $ x \equal{} 3$.

To see if there are others square both sides.

For the first one we get $ 7x^{2} \minus{}43x \plus{}40\equal{}0$

$ (7x\minus{}8)(x\minus{}5) \equal{}0$

$ x\equal{}\frac{8}{7}$ or $ x\equal{}5$.

For the second, $ 17x \minus{} \sqrt{x^{2}\minus{}5} \equal{} 49$

$ \sqrt{x^{2}\minus{}5} \equal{} 17x\minus{}49$

$ x^{2}\minus{}5 \equal{} 289x^{2}\minus{}1666x \plus{} 2401$

$ 288x^{2} \minus{} 1666x \plus{}2406 \equal{}0$

$ 144x^{2} \minus{} 833x \plus{}1203 \equal{} 0$

$ (x\minus{}3)(144x\minus{}401)$

$ x\equal{}3$ or $ x\equal{}\frac{401}{144}$.

Hope this is helpful.

hello, the only real solution of the equation
$ \sqrt{x\plus{}4}\plus{}\sqrt{2x\minus{}1}\equal{}3\sqrt{x\minus{}1}$
is $ x\equal{}5$.
Sonnhard.

You're right Sonnhard.  I should have checked both possibilities. Thanks.

hello, for the second equation $ \sqrt{17x\minus{}\sqrt{x^2\minus{}5}}\equal{}7$ i get only $ x\equal{}3$ as a real solution and i have checked it.
Sonnhard.

thanks to
             Dr Sonnhard Graubner
               AndrewTom


Find the solutions of $ y \equal{} x ( \frac{dy}{dx} \plus{} ( \frac{dy}{dx} )^{3} )$ .

hello, maybe it is an equation of D'Alembert's type.
Sonnhard.


induction
function
logarithms
real analysis
real analysis theorems

For non-integer powers, take logarithms.

[quote="max_tm"]It seems that I can't find a rigorous proof of the power rule; even in Spivak he uses induction which only proves it for a limited class of functions (unless he goes on to use a better method somewhere else and I've yet to read it)... Is there no all-encompassing proof?[/quote]

Use JBL's suggestion.

Say we defined the function $f(x) = x^{r}$ for $x>0$ and $r$ is an arbitrary real number. Then $f(x)=x^{r}= e^{\ln x r}$ [b]by definition[/b] of exponents. So, $f'(x) = (\ln x r)' e^{\ln x r}= rx^{-1}\cdot x^{r}= rx^{r-1}$


game algorithm circle combinatorics Russia trick

I think this is a Beautiful answer for this question
$O$ IS THE CENTER OF CIRCLE .
 CONSIDER  THAT THE POINTS ARE  $A_{1},A_{2},...,A_{2007}$
1)Amayak consider the out come of vectors $\vec{A_{1}O} ,...,\vec{A_{2007}O}$ and choose nearest point in right side of circle (the nearest point it means that Length of arc from this point to the intersection of out come of vectors and circle is Minimum)
WLOG Suppose that  Amayak removes $A_{1}$
2)Arutyun consider the out come of vectors $\vec{A_{2}O} ,...,\vec{A_{2007}O}$ It is almost obvious this out come is not equal 0
so arutyun consider the intersection of this out come to the circle and he choose the semicircle of  right side of the Diagonal that Passes trough this two point

it is easy to prove that this Algorithm can solve the problem

[quote=pco][quote="N.T.TUAN"][i]A. Akopyan, A. Akopyan, A. Akopyan, I. Bogdanov[/i]

A conjurer Arutyun and his assistant Amayak are going to show following super-trick. A circle is drawn on the board in the room. Spectators mark $2007$ points on this circle, after that Amayak
removes one of them. Then Arutyun comes to the room and shows a semicircle, to which the removed point belonged. Explain, how Arutyun and Amayak may show this super-trick.[/quote]

Consider a virtual rotating oriented (an arrow) diameter. Amayak put (in mind  :) ) the arrow in a position such that the number of points at the left of the arrow minus the number of points at the right is maximum (in case of many such positions, Amayak takes one of them). Two cases then :
a) there is at least on point at the right : Amayak remove one of them (no matter what precise point is removed).
b) there is no point at the right of the arrow : Amayak remove the nearest (distance is considered along the circle) point at the left of the top of the arrow.

Aruyun put (in mind) the arrow in a position such that the number of points at the left of the arrow minus the number of points at the right is maximum (in case of many such positions, Aruyun takes one of them).  Two cases then :
a) there is at least on point at the right : Aruyun shows the semicircle at the right of the arrow
b) there is no point at the right of the arrow : Aruyun rotates the arrow (top towards the left side) till the top of the arrow is on the nearest point. Aruyun then shows the semicircle at the right of the arrow.

--
Patrick[/quote]

Maybe I am being dumb but doesn't that fail in the picture below?

When we talk about $MN$, we refer to the arc segment starting from $M$ and travelling clockwise to $N$.

Scale so that our circle is the unit circle. Let $A_1$ be the point in the first quadrant closest to $(1,0)$ and number the 2006 remaining points $A_2,\dots, A_{2007}$ clockwise from $A_1$ around the circle. (Note: $A_1$ may be the point $(1,0)$). Consider the $2016$ arcs formed by all the points, and select the largest clockwise arc (if there are multiple, arbitrarily select one). Suppose $A_iA_{i+1}$ was the largest one. Amayak now removes $A_{i+1}$.

Now, when Arutyun enters the room, he knows that the removed point lied on the largest remaining arc, i.e. $A_iA_{i+2}$. Clearly, this is the unique largest arc.

Let $B$ be the midpoint of  $A_iA_{i+2}$. If the removed point $A_{i+1}$ was on $A_{i}B$ not on $B$, then $|A_{i+1}A_{i+2}|>|A_iA_{i+1}|$ i.e. Amayak would've removed a different point. Hence the removed point lied on $BA_{i+2}$, from which Arutyun can select a suitable semicircle.

Solved with [b]L567[/b] and [b]mueller.25[/b]

The assistant only looks at the two points immediately clockwise of $(0,1)$, $A, B$, with $B$ clockwise of $A$. If there is a semicircle worth clockwise distance between $A$ and $B$, the assistant deletes $A$, otherwise deleting $B$. The magician looks at point $C$ (which was either $A$ or $B$) immediately clockwise of $(0,1)$, and picks the semicircle starting at $C$ and going clockwise. This works... $\blacksquare$


For the first one should it be $5x \equiv 1(\bmod3)$ instead of $5x \equiv 11(\bmod3)$?

Hmm..... I know you should use the CRT here because I remember someone saying it is used with a system of congruences, but I never learned how to use it. Any help?

[quote="felipesa"]Find all solutions of the systems :

[1]
$\begin{gathered} 3x \equiv 1(\bmod 5) \hfill \\ 4x \equiv 6(\bmod 14) \hfill \\ 5x \equiv 11(\bmod 3) \hfill \\ \end{gathered}$

[2]
$\begin{gathered} x \equiv 1(\bmod 3) \hfill \\ x \equiv 2(\bmod 4) \hfill \\ x \equiv 3(\bmod 7) \hfill \\ x \equiv 4(\bmod 11) \hfill \\ \end{gathered}$[/quote]

$3x \equiv 1\equiv 6 \mod 5$
$x \equiv 2 \mod 5$ (1)

$4x \equiv 6\mod 14$
$2x \equiv 3\equiv 10 \mod 7$
$x \equiv 5 \mod 7$ (2)

$5x \equiv 11\mod 3$
$2x \equiv 2\mod 3$
$2(x-1) \equiv 0\mod 3 \implies 3|(x-1)$
$x-1 \equiv 0 \mod 3$
$x \equiv 1 \mod 3$ (3)

we need all $x$ that satisfy (1), (2), and (3)
so using CRT to combine them, in any order...it does not matter

$x \equiv 2 \mod 5$ (1)
$x \equiv 5 \mod 7$ (2)
CRT states there will be a solution $\mod 5*7$
searching a bit, we get
$x \equiv 12 \mod 35$
combine with:
$x \equiv 1 \mod 3$ (3)
we get
$\boxed{x\equiv 82 \mod 105}$

Hey, Al (or someone else), would you mind posting the CRT and an easy sample problem? My fundamentals are lacking. :blush:

Use the search function, you'll find what you want ;)


In the super mario galaxy boxart the letters with asterisks(*) are (in order) UR MR GAY.
i find it hilarious
[url]http://www.nintendolife.com/articles/2007/08/30/super_mario_galaxy_final_boxart[/url]

haha nice lawrence...
you got this from william, right?

ill go update my profile now..

XD

That's hilarious!  I never noticed that!


algorithm
number theory proposed
number theory

find the least $ N$ from the set: $ S \equal{} \{1,11,111,1111,11111,...\}$ wich is a multiple of $ 49$

where it was proposed, please?

thanks

we have $ 49|10^{\phi(49)}\minus{}1\equal{}10^{42}\minus{}1$.and if $ r$ is an integer such that $ 49|10^r\minus{}1$ we must have $ 42|r$.

[quote="tchebytchev"]and if $ r$ is an integer such that $ 49|10^r \minus{} 1$ we must have $ 42|r$.[/quote]

why is this implication true??

let $ r$ be the primitive root of $ 49$ ,then $ r|42$ so $ r \in{1,2,3,6,7,14,21,42}$the only possible case is $ 42$ , and by  Division algorithm we see the results.


function
complex analysis
complex analysis unsolved

Prove that any entire function that satifies the condition $f(z+i)=f(z)$ and $f(z+1)=f(z)$ is equal to a constant

The values that $f$ takes in the whole plane are precisely the values that it takes in the square $\mathcal S$ with vertices $0,1,i,1+i$. Since $\mathcal S$ is compact, the restriction of $f$ to $\mathcal S$ has a maximum, so $f$ would be an entire bounded function. By Liouville's Theorem, it must be constant.

Thank Grobber alost! The solution is very short and very easu to understand.


Let $ a,b,c$ be real numbers and $ n,k$ be  positive real numbers. Suppose that we have

$ a+b+c=n$
$ ab+bc+ca=k$

Prove that there are no real $ a,b,c$ that satisfy the equations if $ k\geq\frac{n^{2}}{2}$.

i assume you meant $ k>\frac{n^{2}}{2}$.
[hide="solution"] Assume the contrary, we have such numbers. then $ 2(ab+bc+ca)> (a+b+c)^{2}\iff 0> a^{2}+b^{2}+c^{2}$, impossible.[/hide]

That's right, good job!

But it also holds for $ k=\frac{n^{2}}{2}$ since that gives us the condition $ a^{2}+b^{2}+c^{2}=0$ which means that $ a=b=c=0$. By this we'd have $ n=0$ but we have $ n>0$ so it's also true for equality!


When the mean, median, and mode of the list

10, 2, 5, 2, 4, 2, x.
are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ?

I understand how to do it but confused about "non-constant arithmetic progression"; what does that mean. thanks for helping.

I believe that that just means that all the terms are not the same. In other words, the mean, median, and mode are not all the same number.

I believe this problem has been posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1930810171&t=195141]here.[/url]

Yea, it just means that the arithmetic difference is not 0, so the mean, median, and mode are different numbers.

thanks for all responses, that really helps


function
integration
inequalities
limit
induction
calculus
calculus computations

For $n=0,1,2,\cdots,$ the function $f(x)$ is defined by $f_0(x)=e^x,\ f_{n+1}(x)=\int_0^x f_n(t) dt.$

(1) For $n\geq 1,$ Show that  $e^x=1+x+\frac{x^2}{2!}+\cdots +\frac{x^{n-1}}{(n-1)!}+f_n(x).$

(2) Let $n\geq 1.$ For $x\geq 0,$ show that the inequality  $0\leq f_n(x)\leq \frac{x^n}{n!} e^x.$

(3) Find $\lim_{n\to\infty} f_n(1).$

Note:You are not allowed to use Taylor expansion.

(1) is not exactly the Taylor expansion with integral rest? :?


Let $ P(x)$ be a polynomial with integer coefficients such that \[ P(m_1) \equal{} P(m_2) \equal{} P(m_3) \equal{} P(m_4) \equal{} 7\] for given distinct integers $ m_1,m_2,m_3,$ and $ m_4.$ Show that there is no integer m such that $ P(m) \equal{} 14.$

[quote="orl"]Let $ P(x)$ be a polynomial with integer coefficients such that
\[ P(m_1) \equal{} P(m_2) \equal{} P(m_3) \equal{} P(m_4) \equal{} 7
\]
for given distinct integers $ m_1,m_2,m_3,$ and $ m_4.$ Show that there is no integer m such that $ P(m) \equal{} 14.$[/quote]
Suppose P(m)=14 so $ G(x)\equal{}P(x)\minus{}7$ so $ G(m)\equal{}(m\minus{}m_1)(m\minus{}m_2)(m\minus{}m_3)(m\minus{}m_4)L(m)\equal{}7$ and $ m\minus{}m_1,m\minus{}m_2,m\minus{}m_3,m\minus{}m_4$ distinct integers and 7 is prime so there is no such m


calculus
integration
logarithms
calculus computations

Hello!
Please help me solve these integrals
1.
$ \int {\frac{{dt}}
{{e^{\frac{t}
{2}}  \plus{} e^t }}}$
2.
$ \int {\frac{{dt}}
{{t\left( {2 \plus{} 3\ln ^2 t} \right)}}}$
3.
$ \int {\frac{{\arcsin t}}
{{\sqrt {2 \minus{} t} }}} dt$
4.
$ \int {\frac{{e^{\arctan x} dx}}
{{\sqrt {\left( {1 \plus{} x^2 } \right)^3 } }}}$
5.
$ \int {\frac{{dt}}
{{\sqrt {e^{2t}  \plus{} 1} }}}$

Hello
[hide="5"]
just u need to put $ {e^{2t} \plus{} 1} \equal{} x^2$
then $ dt \equal{} \frac {x}{x^2 \minus{} 1}dx$

[/hide]
Thank u.

Hello
[hide="1"]
just take $ e^{\frac{t}{2}}$ common from denominator.
Then put $ e^{\frac{t}{2}}\plus{}1$=$ x$
then $ dt \equal{}\frac{2dx}{x\minus{}1}$
 [/hide]
Thank u.

hello, to 2) let $ z\equal{}\ln(t)$, so we have $ dz\equal{}\frac{1}{t}\,dt$ and our integral is $ \int \frac{1}{2(1\plus{}(\sqrt{\frac{3}{2}}z)^2}$, further we put $ u\equal{}\sqrt{\frac{3}{2}}z$ so we have finally $ \sqrt{\frac{1}{6}}\int\frac{1}{1\plus{}u^2}\,du$ which elementary.
Sonnhard.

Hello
[hide="2"]
just put $ ln t \equal{} x$
now ur equation is
$ \int\frac {dx}{2 \plus{} 3x^2}$
=$ \int\frac {1}{3}\frac {dx}{(\sqrt\frac {2}{3})^{2} \plus{} x^2)}$
=$ \frac {1}{\sqrt6}tan^{ \minus{} 1}(x{\sqrt {\frac {3}{2}}})$[/hide]
Thank u.

hello, to 3) do you mean $ \int\frac{\arcsin(t)}{\sqrt{2\minus{}t}}\,dt$ or $ \int\frac{\arcsin(t)}{\sqrt{1\minus{}t}}\,dt$?
Sonnhard.

$ \begin{gathered}
  2.{\text{ }}\int {\frac{{dt}}
{{t\left( {2 \plus{} 3\ln ^2 t} \right)}}}  \equal{} \frac{1}
{2}\int {\frac{{{1 \mathord{\left/
 {\vphantom {1 t}} \right.
 \kern\minus{}\nulldelimiterspace} t}}}
{{1 \plus{} \left( {\frac{{\sqrt 6 }}
{2}\ln t} \right)^2 }}} dt \equal{} \frac{{\sqrt 6 }}
{6}\int {\frac{{d\left( {\frac{{\sqrt 6 }}
{2}\ln t} \right)}}
{{1 \plus{} \left( {\frac{{\sqrt 6 }}
{2}\ln t} \right)^2 }}}  \equal{} \frac{{\sqrt 6 }}
{6}\arctan \left( {\frac{{\sqrt 6 }}
{2}\ln t} \right) \plus{} C. \hfill \\
  4.{\text{ }}\int {\frac{{e^{\arctan x} dx}}
{{\sqrt {\left( {1 \plus{} x^2 } \right)^3 } }}}  \equal{} \int {\frac{{e^{\arctan x} dx}}
{{\left( {1 \plus{} x^2 } \right)\sqrt {1 \plus{} x^2 } }}}  \equal{} \int {\frac{{d\left( {e^{\arctan x} } \right)}}
{{\sqrt {1 \plus{} x^2 } }}}  \equal{}  \hfill \\
   \equal{} \frac{1}
{{\sqrt {1 \plus{} x^2 } }}\int {d\left( {e^{\arctan x} } \right)}  \minus{} \int {e^{\arctan x} d\left( {\frac{1}
{{\sqrt {1 \plus{} x^2 } }}} \right)}  \equal{}  \hfill \\
   \equal{} \frac{{e^{\arctan x} }}
{{\sqrt {1 \plus{} x^2 } }} \plus{} \int {\frac{{xe^{\arctan x} }}
{{\sqrt {\left( {1 \plus{} x^2 } \right)^3 } }}} dx \equal{} \frac{{e^{\arctan x} }}
{{\sqrt {1 \plus{} x^2 } }} \plus{} \int {\frac{{xd\left( {e^{\arctan x} } \right)}}
{{\sqrt {1 \plus{} x^2 } }}}  \equal{}  \hfill \\
   \equal{} \frac{{e^{\arctan x} }}
{{\sqrt {1 \plus{} x^2 } }} \plus{} \frac{x}
{{\sqrt {1 \plus{} x^2 } }}\int {d\left( {e^{\arctan x} } \right)}  \minus{} \int {e^{\arctan x} d\left( {\frac{x}
{{\sqrt {1 \plus{} x^2 } }}} \right)}  \equal{}  \hfill \\
   \equal{} \frac{{e^{\arctan x} }}
{{\sqrt {1 \plus{} x^2 } }}\left( {1 \plus{} x} \right) \minus{} \int {\frac{{e^{\arctan x} dx}}
{{\sqrt {\left( {1 \plus{} x^2 } \right)^3 } }}} {\text{ }} \Leftrightarrow {\text{ }}\int {\frac{{e^{\arctan x} dx}}
{{\sqrt {\left( {1 \plus{} x^2 } \right)^3 } }}}  \equal{} \frac{{e^{\arctan x} }}
{{2\sqrt {1 \plus{} x^2 } }}\left( {1 \plus{} x} \right) \plus{} C \hfill \\
  5.{\text{ }}\int {\frac{{dt}}
{{\sqrt {e^{2t}  \plus{} 1} }}}  \equal{} \int {\frac{{\frac{1}
{{e^t }}}}
{{\sqrt {1 \plus{} \left( {\frac{1}
{{e^t }}} \right)^2 } }}} dt \equal{}  \minus{} \int {\frac{{d\left( {\frac{1}
{{e^t }}} \right)}}
{{\sqrt {1 \plus{} \left( {\frac{1}
{{e^t }}} \right)^2 } }}}  \equal{}  \minus{} arcsinh\frac{1}
{{e^t }} \plus{} C. \hfill \\
\end{gathered}$

4/ put $ x \equal{} tan(t)$ so $ dx \equal{} (1 \plus{} tan^2(t))dt$

$ \int \frac {e^{arctan(x)}}{\sqrt {(1 \plus{} x^2)^3}}dx \equal{} \int \frac {e^t}{\sqrt {1 \plus{} tan^2(t)}}dt$

$ \equal{} \int e^t.cos(t) \equal{} \frac {1}{2}(cos(t) \plus{} sin(t)) \equal{} \frac {1}{2} e^{arctan x}(\frac {x \plus{} 1}{\sqrt {x^2 \plus{} 1}})$

[quote="Dr Sonnhard Graubner"]hello, to 3) do you mean $ \int\frac {\arcsin(t)}{\sqrt {2 \minus{} t}}\,dt$ or $ \int\frac {\arcsin(t)}{\sqrt {1 \minus{} t}}\,dt$?
Sonnhard.[/quote]

Yes, you're right.

Thank you all for help!

hello, if your integral is $ \int \frac{\arcsin(t)}{\sqrt{1\minus{}t}}\,dt$ so we get with integration by parts
$ \displaystyle{\int \arcsin(t)\frac{1}{\sqrt{1\minus{}t}}\,dt\equal{}\minus{}2\arcsin(t)\sqrt{1\minus{}t}\plus{}2\int\frac{1}{\sqrt{1\plus{}t}}\,dt\equal{}\minus{}2\arcsin(t)\sqrt{1\minus{}t}\plus{}\sqrt{1\plus{}t}\plus{}C}$.
Sonnhard.


For positive real numbers a,b,c with a+b+c =abc, show that

\[ \frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac32  \]

and determine when equality occurs.

(use trig substitution)

$\alpha = arctan a$
$\beta = arctan b$
$\gamma = arctan c$

this inequality is from Korea 1998.
http://www.mathlinks.ro/Forum/viewtopic.php?p=314932#p314932
the equality occurs when $a=b=c= \sqrt{3}$  ;)


limit
real analysis
real analysis unsolved

Let [b](a_n)n>=1[/b] where [b]a_1=1[/b], and [b]a_(n+1)=a_n+1/a_n[/b]  and [b](b_n)n>=1[/b] where [b]0<a_1<1[/b] and [b]b_(n+1)=b_n-(b_n)^3[/b].
Calculate [b]lim(a_n*b_n)[/b].

I find that [b]lim(a_n)=+00[/b] and [b]limit(b_n)=0[/b] but now I don't know how to use ,"maybe" Cesaro Stolz to find [b]lim(a_n*b_n)[/b]

More precisely, $ \lim_{n\to\infty}\frac{a_n}{\sqrt{n}}\equal{}\sqrt{2}$, which follows from $ a_{n\plus{}1}^2\equal{}a_n^2\plus{}2\plus{}1/a_n^2$. I also think that $ \lim_{n\to\infty}b_n\sqrt{n}\equal{}\sqrt{2}$, but don't see an algebraic trick for this.

The "trick" (used uncountably many times in Romanian olympiads and contests :() is to make the power of $ n$ equal to $ 1$. That is, let's compute
\[ \lim_{n \to \infty} b_n^2 n \equal{} \lim_{n \to \infty} \frac {n}{\frac1{b_n^2}} \stackrel{CS}{ \equal{} } \lim_{n \to \infty} \frac1{\frac1{b_{n \plus{} 1}^2} \minus{} \frac1{b_n^2}} .
\]
Cesaro-Stolz works because $ b_n$ decreases to $ 0$. Also, to finish the problem, use the recurrence relation and then treat the limit in $ x : \equal{} b_n \to 0$.


[color=white]I noticed that the number of users on AoPS is almost 300!  Yay!  Will we have a celebration or something when it reaches 300?  :wink: [/color]

I don't think that 300 needs a celebration (1000 is another story...) since it's simply 3*100 as with 2*100 and 1*100.  However, 500 is a great number to celebrate because it's (1/2)*1000 although you could also write 300 as (3/10)*1000.  Ah, 1000....

-interesting_move

Let's celebrate when we get to 496. Do others like that idea?

Here's an idea: let's have a race.  Which will get to 496 first?  ComplexZeta (# of posts) or the user number?

LOL, I think ComplexZeta will win because he posts about 4.7 posts per day, and there are only about 2 new users per day.  It will be a close race though, especially if you take into account that ComplexZeta's posting # will be more linear, and the number of people will be more exponential.

-interesting_move

ComplexZeta will also be on vacation for some time in the summer, during which time he will not be posting at all. Furthermore, he is disappointed by the small number of active posters from the bay area and will thus inform other people from his high school as to the existance of this forum, so it seems likely that the members will reach 496 before ComplexZeta reaches 496 posts (but not guaranteed).

Why are you speaking in third person?

-interesting_move

I just feel like doing that at random times.

jojobird, we now have 300 members. Thank you Lord Venom.

*patpat* Poor Simon. SFBA pple just aren't interested in posting...maybe I should start bugging people too...

HEY! *patpat* is a Lynbrook(e) math geek thing. (The (e) at the end is due to Z9R4C3 at an ARML practice, and I insisted that she kept writing it like that. But then again, you probably already knew that.) Don't get me involved in a Lynbrook(e) math activity. I'm not one of you!

Haha - *imagines Lynbrook math habits pervading AoPS forum* Very interesting. *patpatpatpatpatpatpatpat*

I'll bet you anything that there are more people on artofproblemsolving.com from SFBA than from where I come from.

I go to a pretty big high school (~3600 students) so I'll try to help out with the members.  What should I do though?  Any ideas?

-interesting_move

Tell your math club buddies about it.   Send e-mails to everyone you know about it...hijack their computers and set http://www.artofproblemsolving.com to their home page!   :P

Hang on!  Yes, we have 300+ users, but how many of them have made more than 5 posts?  How many have made [i]any[/i] posts?  It seems like some users may be people looking for help on a single problem (who, after their question is answered, go away but leave their username behind to take up space.)

[color=white]WOW...I left this post for a day and already it's on 2 pages...[/color]

[color=white]That seems to be true, mathfanatic.  I wish we could have some way to display only registered users, but then again we do - the member list.  Just sort it by ascending/descending and draw a line where the posts you decide as active reach.  Oh well.

I'm betting ComplexZeta will reach 496 posts.  It would be almost certain if we were only counting active useres (regulars).  I mean, I'm fairly new here and I'm already number 15 or so.[/color]

Some people can not help that they do not visit this site for weeks.  I went away to a camp for four weeks, where they would not let the campers use computers.  During the summer many people are going on vacations, so they will not post for months at a time. Others don't want to post, merely to read the posts of others.

I'm from the Bay Area, I go to the worst school, Sequoia.  We had 2 people qualify for AIME, and I was not one of them  :( , I was in the upper 90's unfourtenetly.  I look foward to meeting some of you from the Bay Area at ARML, other contests and I am thinking about doing BMC...yea, just felt like ranting...

[color=white]Hmm...that's true, bookworm.  Some people are gone at camp for weeks at a time during the summer and don't get to post very often.  Maybe I'm just being a little too scared at the prospect of losing my ranking  :lol:  :oops: [/color]

[color=white]WE HAVE 12 USERS ONLINE!!!!  MORE!!! MORE!!!!!!!

I get a bit excited when there's so many people on...[/color]

[quote]WE HAVE 12 USERS ONLINE!!!! [/quote]

Yeah, we reached 13, but then I think it began to drop.

Fierytycoon

i think theres 4 people on aops from okie-homa, but one of them has never posted

i've only posted like 10 times, and I'm still in the top 50.  But then there are like 200 people who have posted 0 times.  Still, they could be like me.  Although I don't post a lot, I consider myself an active member, because I visit the site about 2 times a week, I read through a lot of posts, although normally I dont have anything to say.

Same with me... I visit the forum almost every day, but I just don't feel like posting most of the time.

hmmmmmmmmmmmm...i visit the forum prbly 10 times a day.

Hehe. I visit the forum 80 times a day, or maybe even more. I really should find something else to do, but I don't want to spend more time reading about how to guess solutions to differential equations, which seems really pointless to me.

im wondering how much youre gonna spend on aops once your in college.

Yes, I am too. It depends how many classes I take and how much work I have, I guess. In college, unlike high school, I shall try to remember that I'm there to learn, and all else is secondary. But at least I'll be here every day.

I wonder, How come you guys start univ one year earlier than us in Sweden? We start school at the same age. :O It's not fair.

[quote="ComplexZeta"]Hehe. I visit the forum 80 times a day, or maybe even more.[/quote]

Well, I don't visit it 80 times a day, usually only 2 or three times.  But then I stay on for about 4 hours each "time"!

Okay, that was an exaggeration.  Still, I spend plenty of time here that I could be using for more academic purposes.

And Mindspa - maybe having 1 more year makes you guys better prepared for university than here in the USA.  I don't know for sure, though.


game
USA(J)MO
USAMO
algorithm
combinatorics
game strategy

The Y2K Game is played on a $1 \times 2000$ grid as follows. Two players in turn write either an S or an O in an empty square. The first player who produces three consecutive boxes that spell SOS wins. If all boxes are filled without producing SOS then the game is a draw. Prove that the second player has a winning strategy.

This was the last USAMO that I took as a high school student.  I remember being very frustrated with this problem - I wasn't very well-educated about inequalities, so I was able to do #4 only after a couple of hours of effort.  That would have left only an hour or so for this problem.  Also, I think I misinterpreted the problem as saying that the first player is only allowed to write the letter S and the second player can only write the letter O.

I recently worked on this problem some more (ahh, nostalgia!), this time using the correct interpretation of the problem.  Here is the solution I came up with.  I don't know if the out-of-high-school-for-well-nigh-a-decade crowd is allowed to post here.  If not, feel free to remove this post.  But I felt a deep sense of joy when I found this solution and I hope I can share that with the interested reader.

[redacted - I will repost this tomorrow]

Let's call the first player A, the second player B.

B's strategy is as follows:
On his first turn, place an S in a square far away from both ends of the board and A's first counter.
On his second turn (unless he can finish an SOS somewhere), place an S three squares to one side of his first S, and if A has put his second counter 'near' B's first, make sure it is on the other side from this one.

After this, B follows the following algorithm:

If he can finish an SOS somewhere, do so.
Otherwise, if there is an empty square with a square each side empty, put an O in it.
Otherwise, where there is an empty square with neither adjacent square empty, put an O in it.

[Whenever B plays, there are an odd number of empty squares, so at least one of the above is always possible].

If I have got it right, it should be fairly apparent that A cannot win with B following this. To ensure that B actually wins-
At the beginning, B created a sequence of four squares that went S-empty-empty-S. At some point, A must place an O or S in one of these squares: then B just has to put the other one in the other empty square and he wins.

Is there also a winning stategy for the first player, player A?
gr, C

By the definition of winning strategy (a strategy that one player can follow such that she can win regardless of the actions of all other players), the number of players who can have winning strategies is at most the number of possible simultaneous winners.  Since at most one player wins this game, at most one player can have a winning strategy.

The second player wins. The strategy is to place an S on his first turn far enough away from the ends of the board and the first player's first move, and play an S three boxes down on the next move. Note that if anyone is ever forced to play in one in a pair of boxes between two S's (S/_/_/S), let this be a domino of death, the other player will automatically win on the next turn, and furthermore it is impossible that the original player can win.

We will prove that the only way for anybody to play such that the other player wins on the next turn is if he is forced to play in a domino of death. We claim that if there is an empty box that is not in a domino of death, the person whose turn it is can either (a) win or (b) play in such an empty box such that the other player can't win on the next turn.

If the player P has the chance to win, then he can just win. Thus, if he can't win and furthermore has to play such that the other player Q wins on the next turn, the Q's SOS must include the letter that the P played. We have 3!=6 cases. For example,

P plays the first S, Q plays the O. We seek all situations such that if P played O instead, Q still could have won. Basically, the board is going to look like _/_/S before P plays, and O/_/S in the alternative strategy. If this still allows Q to win, then it must be because there's an S in the middle and an S to the left of an O; we have a domino of death.

The other five cases work the same way, we find that we need Q to have played in a domino of death. Note that the second player in our game can always avoid playing in a domino of death because he always plays with an odd number of boxes available (as 2000 is even), and dominoes of death contain two boxes each, with one box only being a member of at most one domino of death. Thus, there's at least one un-paired empty box and thus at least one box not part of a domino of death for the second player to play in.

Nothing new here, but I love this problem :)

[hide=Solution]
For brevity, let the first and second players be named A and B, respectively. Also, we call the three configurations of boxes (S, empty, S), or (S, O, empty), or (empty, O, S) [i]winning configurations[/i] (since if any of these configurations are ever achieved, the next player to play can win).

Define a group of four consecutive boxes to be [i]deadly[/i] if the first and last contain the letter S, and the middle two boxes are empty. Notice that if any move is made in the two empty boxes of a deadly group, the next player to play can win: if an S is written, the (S, empty, S) configuration is achieved, and if an O is written, the (S, O, empty) or (empty, O, S) configuration is achieved.

[b]Lemma:[/b] B can always force the existence of a deadly group.

[b]Proof:[/b] If A writes an S on his first turn, then B can write an S three boxes away, producing a [i]deadly[/i] group. If A writes an O on his first turn, then B can write an S somewhere at least $100$ tiles away from the O and from each end of the grid (the exact number is not important, only that it is far enough from the O that no winning configurations can be achieved next). Then, on B's second turn, he will either win immediately if it is possible, or he can place a second S three boxes away, opposite the direction of A's second move.

After B produces a deadly group, he can proceed as follows until the only empty boxes remaining lie in deadly group. After each of A's moves, if it is not possible for B to win immediately, then B chooses an empty box $X$ that is not in a deadly group. If the neighboring boxes are both empty, B writes an O in box $X$. If a neighboring box contains an O, B writes an O in box $X$. If a neighboring box contains an S but neither contain O, B writes an S in box $X.$ These moves will not produce any winning configurations. (Specifically, in the third case, notice that writing an S cannot produce a (S, empty, S) configuration opposite the neighboring S, since that would mean that box $X$ is in a deadly group.)

Eventually, the only squares left will lie in deadly groups. Since deadly groups have two empty boxes and the game began with an even number (2000) of empty boxes, this means that it will be A's turn when this occurs. Then A is forced to make a move in a deadly group, so B wins. $\square$

(Looking at the previous solution, I see the theme of "this S, empty, empty, S configuration is deadly" is common here ;))
[/hide]

View the board position as a string of the characters S,O,-. Call the players Alice and Bob.

We prove a series of lemmas to prove that Bob will win if both players play optimally.

[u]Lemma 1:[/u] Bob can make sure that there is some turn of his where the board has a substring S--S.

[u]Proof of Lemma 1:[/u] If Alice opens with placing an S somewhere, then Bob simply places another S to get the substring S--S somewhere. We're not done yet since it is now Alice's turn. But if she plays in one of the two blanks in the S--S, it is easy to see that Bob wins, so Alice must play outside the --. Therefore, it is Bob's turn, and the substring S--S exists.

Now, assume she instead opens with playing an O somewhere. Bob should then place an S far from the O and the edges of the board (fifty squares from the O and the edges should suffice). Alice won't play so that Bob can complete an SOS, since she is playing optimally, so she plays something else. Now, on Bob's second move, he has two ways to make the S--S, but at most one of them could lead to a winning position for Alice because of her second move. So he just plays the one that won't lead to a winning position for Alice, so Alice's third turn is not a win, so Bob's third turn arrives with the substring S--S existing somewhere on the board. $\blacksquare$

An important consequence of this lemma is that the game will now not draw - eventually someone must play in the blanks in S--S, and then they lose.

[u]Lemma 2:[/u] Once Bob has S--S somewhere, Bob can have the ends of the board to be filled on one of his turns.

[u]Proof of Lemma 2:[/u] Suppose Bob has S--S and it is his turn, and WLOG Bob can't immedeatly win (else game over). If an edge square is empty, by putting an O on the edge, Bob adds no winning positions, so Alice makes a move that doesn't finish, so its Bob's turn again. Thus, Bob can fill up the edges. $\blacksquare$

[u]Lemma 3:[/u] If Bob can't win on a turn, then he can play so that he will be around to play another turn.

[u]Proof of Lemma 3:[/u] Suppose there is a substring of the form -O or O-. Certainly if Bob adds an O to make them OO, then he could not have added any new winning substring (a winning substring is SO-, S-S, or -OS). Thus, WLOG assume that there are no substrings of the form -O or O-.

Thus, the board looks like
\[X_1(-)^{a_1}X_2(-)^{a_2}\cdots (-)^{a_n}X_{n+1}\]
where $X_i$ is a string of S and Os that starts and ends with S (except maybe $X_1$ and $X_{n+1}$ that could start and end with O respectively), and has no substring of the form SOS, and $a_i\ge 2$ for all $i$. If any of the $a_i\ge 3$, then Bob should simply put an S in the first - in $(-)^{a_i}$. It is easy to see that this adds no winning substrings, so it suffices to examine the case $a_1=\cdots=a_n=2$. But this means that the number of squares placed so far is $2000-2n$, which is even, so it can't even be Bob's turn! Therefore, we are done. $\blacksquare$

Lemma 1 implies that there is no draw, and lemma 3 implies that Bob won't lose, so Bob must win.

QED

If we generalize $2000$ to $n$, the first player wins  for all odd $n(\geq7)$ and the second player wins for all even $n(\geq14)$
The rest becomes a tie, checking the remaining cases.

My second combinatorics solution here :D

Let the players be X and Y. In Y's first move, place an S far from the boards corners and X's move. Remember, whenever SOS possible, do it. In second move create a S[][]S next to the first move, in case X played something near the first move, play the second S in the opposite side. Now do either: 1) If x[]x (x=S or O) then put O. 2) If [][][] then put O in middle. This is possible because there are odd squares after X plays. Continuing like this X must once put a S or O in the S[][]S, then Y wins.

OH YEAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH  I GET IT NOW NOOOOOWWWWWWWW IT MAKES SENSSEEEEEEEEEEEEEEEEE

whatever A plays, B plays >3 squares away from all sides and >5 squares away from A a letter S. Obviously if A threatens to win then B wins, so henceforth assume A is never one move from winning when it's B's turn. Next move B can guarantee to place another S three spots away from the first S. Now he can just fill up the rest of the board accordingly by spamming Os wherever there are three empty spots (he places it in the middle), or wherever both spots are occupied. Note that if he runs out of moves in this algorithm, he can just place an O next to another O, and if he runs out of these moves, the config is made up of some blocks of [block]xx[block=SOO...OOS]... etc., so everything is filled except for some areas of SxxS. However, in each one of these, there are an even number of unfilled, so actually, it must be 1's move. Now wherever 1 moves in one of these blocks player 2 just moves the opposite type in the other x, and wins.


lines
geometry
point set
combinatorial geometry
IMO Shortlist

Let $ k$ and $ n$ be integers with $ 0\le k\le n \minus{} 2$. Consider a set $ L$ of $ n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $ I$ the set of intersections of lines in $ L$. Let $ O$ be a point in the plane not lying on any line of $ L$. A point $ X\in I$ is colored red if the open line segment $ OX$ intersects at most $ k$ lines in $ L$. Prove that $ I$ contains at least $ \dfrac{1}{2}(k \plus{} 1)(k \plus{} 2)$ red points.

[i]Proposed by Gerhard Woeginger, Netherlands[/i]

[hide]Let $ l$ be the line with minimal distance to $ O$ and $ P$ be the point of$ I$ on $ l$ that minimizes $ OP$. A standard argument shows that $ OP$ cannot intersect any line of $ L$. This solves $ k=0$. The rest can be done by induction over $ k$: Sorting the points $ P_1,P_2,\dots,P_{n-2}$ of $ L%Error. "capI" is a bad command.
\setminus\{P\}$, it is obvious that the segment $ OP_i$ can cut at most $ i$ lines (Induction over $ i$). Then, remove $ l$; the induction hypothesis asserts that there are at least $ \frac{1}{2}k(k+1)$ points $ P$ so that $ OP$ intersects at most $ k-1$ lines of $ L$. Adding $ l$, $ P,P_1,\dots,P_k$ back, we see that there must be $ \frac{1}{2}k(k+1)+k+1 = \frac{1}{2}(k+1)(k+2)$ red points.[/hide]

Overall not that difficult, especially for a G[u]5[/u]...

This one can be done using the dual principle (i.e., changing the words "point" and "line"). We transform each point in its polar line wrt a circle centered in $ O$. Using the fact that $ a \in B \Leftrightarrow b \in A$ and an approximation of $ O$ and $ X$ makes $ x$ moves away of $ O$, one can prove the following:

$ OX$ intersects at most $ k$ lines in $ L \text{ } \Leftrightarrow$ the half plane determined by $ x$ which does not contain $ O$ has at most $ k$ points in $ L'$, the set of the poles of the lines in $ L$.

In a formal point of view, the difficulty of the problem is the same after the transformation. However, I find points more intuitive than lines, which allowed me to get the solution easily.

[hide="Is this right?"]We will keep $k$ fixed and induct on $n$. For the base case $n=k+2$, we note that there are $\binom{n}{2} = \binom{k+2}{2}$ points here, so it is sufficient to show that for all points of intersection $X$, the open segment $OX$ intersects no more than $n-2$ lines. But this is obvious, as there are only $n$ lines, $X$ lies on two of them, and $X$ does not lie on the open segment $OX$.

Suppose that for some $n$ and $k$, the result is true. Consider any $n+1$ lines $\ell_1, \ell_2, \ldots, \ell_{n+1}$. Applying the inductive hypothesis to $\ell_1, \ell_2, \ldots, \ell_n$, we can find points $\mathcal{X} = X_1, X_2, \ldots, X_{\binom{k+2}{2}}$ such that the open segments $OX_1, OX_2, \ldots, OX_{\binom{k+2}{2}}$ each intersect at most $k$ lines.

Let $\mathcal{Y}$ be the set of all $X_i$ such that $OX_i$ intersects exactly $k$ lines before $\ell_{n+1}$ is added and $\ell_{n+1}$ intersects $OX_i$. It is clear that for $X \notin \mathcal{Y}$, $X \in \mathcal{X}'$.  For each $Y \in \mathcal{Y}$, let $f(Y)$ be the intersection of $OY$ and $\ell$. Because each ray emanating from $O$ can contain at most one element of $\mathcal{Y}$, $f$ is injective, so $f(\mathcal{Y})$ has at least $|Y|$ elements. No three lines are concurrent and $Of(Y)$ intersects at most $k$ lines after $\ell_{n+1}$ is added, so $f(\mathcal{Y}) \subseteq \mathcal{X}'$. Hence, \[ |\mathcal{X}'| = |\mathcal{X} \backslash \mathcal{Y}| + |\mathcal{Y}| \geq |\mathcal{X}| - |\mathcal{Y}| + |\mathcal{Y}| \geq |\mathcal{X}| \geq \binom{k+2}{2}\]
by the inductive hypothesis, so we are done.[/hide]

I use induction on $k$, the base case is easy by considering a corner of the region of $O$. Now assume the statement is true for $k-1$, I prove it for $k$. Consider a point $Q$ that is a corner of the region in which $O$ is in. In a line through $Q$ there are at least $k+1$ good intersection points of this line with other lines. If we forget about this line, by induction there are $1+...+k$ good points $P$ (such that $OP$ cuts at most $k-1$ lines), and adding this line back adds at most one intersection point to the segment $OP$ to them, so they are still good. All in all we have $1+...+(k+1)$ good points so we are done.

fix $n$ and apply induction on $k$. the base case for $k=0$ is obvious consider the intersection of the nearest lines to $O$.

consider that the statement is true for $k$ we will prove for $k+1$
now note that if we consider a point $X$ on a line $l$ such that no line intersects $OX$ then the nearest $k+1$ points near to $X$ on the line $l$ are red.
if you don't consider so after removing $l$ there are $\binom{k+2}{2}$ and after restoring $l$ there are $k+2$ new red points including $X$. so in general we have $\binom{k+3}{2}$.$\blacksquare$

Let $\ell$ be the nearest to $O$ line from $L$ and $\ell \cap I=\left\{ X_1,X_2,...,X_{n-1}\right\}$ with $|OX_i|\leq |OX_{i+1}|$ for every $i.$

[b]Claim.[/b] Open segment $OX_i$ intersects at most $i-1$ line from $L.$
[b]Proof.[/b] Suppose the opposite, so $OX_i$ intersects $i$ lines and in particular it intersects line $\ell '$ such that $Y= \ell \cap \ell '\notin \bigcup_{t=1}^{i-1}X_t.$
If $Z=OX_i \cap \ell '$ we get the contradiction with $$|OY|\geq |OX_i|\implies \angle OYX<90^\circ \implies d(O,\ell ')=|OY|\cdot \sin \angle OYZ<|OY|\cdot \sin \angle OYX_i=d(O,\ell )\text{ } \Box$$

Now with fixed $n$ we apply induction by $k.$ By the claim $X_1$ is always red, so the base case $k=0$ follows. Now consider problem statement for $L\backslash \ell$ and $k=p,$ so there exist at least $C_{p+2}^2$ red points. After backing $\ell$ to $L$ and increasing $k$ to $p+1$ all red points are preserved and by claim all points $X_1,X_2,...,X_{p+2}$ are red, so totally there are at least $C_{p+3}^2$ red points $\blacksquare$


Do you know the Peking University and the National Taiwan University? My friend will go to university and he want to be a Physics major. Now he has 2 choices: Peking University and National Taiwan University. The Peking University and the National Taiwan University are 2 of the top universities in China, and they are also strong in Science. Besides, he want to go to graduate school overseas after getting the BSc. How much do you know these 2 universities? What do you think about them? Could you please give him some advice?

I have been an international student at National Taiwan University (in the Chinese department, many years ago). N.T.U. ("Tai Da" in Chinese) is a great university with great students, and would be a great learning opportunity. I know many alumni of Beijing University ("Bei Da") and have every reason to believe it has a very strong math department (and physics department, etc.) but have never attended there. Living in Taiwan is sufficiently different from living in Beijing in terms of weather, politics, interaction with local people, etc. that that might be a good reason to choose one school over the other. I like countries with a free press a LOT better than countries without one, and on that basis I like living in Taiwan better than living in China, but maybe someone living in China in the next several years will see China undergo some changes in the direction of democratization something like those that Taiwan underwent during the years that I have lived and visited there.

It's hard to go wrong with two great choices like that. Best wishes in making a suitable decision. What was the application process like? I think other participants here would like to know how to apply to each of those universities from overseas.

[quote="tokenadult"]What was the application process like? I think other participants here would like to know how to apply to each of those universities from overseas.[/quote]
There are two exams for students in Macao or Hong Kong who want to apply the universities in Mainland China and Taiwan. But I don't know the application process of international students. If you have interest in applying these two universities, you can browse the following website:
[url]http://www.pku.edu.cn[/url] (Peking University)
[url]http://www.ntu.edu.tw[/url] (National Taiwan University)
[url]http://www.overseas.ncnu.edu.tw[/url] (Overseas application of Universities in Taiwan)

I can't see much difference between them.I voted for PKU.But if I had an opptunity,I would have gone to NTU.

[quote="yisun88"]Maybe I'm misinformed, but how does your friend have the opportunity to choose between PKU and NTU.... I didn't know students in China went to college in Taiwan or vice versa?[/quote]
Although Macao and Hong Kong are parts of China (PRC), students in Macao and Hong Kong can choose the universities in mainland of China and also Taiwan.

Ah, I didn't see your location.  That's indeed a very interesting choice.  This is perhaps a bit extraneous, but the simplified/traditional transition might be tough  :wink:


Euler
geometry
circumcircle
cyclic quadrilateral
geometry unsolved

In a non-equilateral acute-angled triangle $ ABC$ with $ \angle C\equal{}60^{\circ}$, $ U$ is the circumcenter, $ H$ the orthocenter and $ D$ the intersection of $ AH$ and $ BC$. Prove that the Euler line $ HU$ bisects the angle $ BHD$.

[hide="Suggest"]A well-known result: $ AHUB$ is cyclic quadrilateral when $ \angle C\equal{}60^o$.
Then $ \angle UHB\equal{}\angle UAB\equal{}30^o$...[/hide]


Suppose $a,b,c$ are positive real numbers. Find the greatest $\alpha$ such that \[\alpha\left(\sum_{\mathrm{cyclic}}\frac{2\sqrt{ab}}{a+b}\right)+2\left(\sum_{\mathrm{cyclic}}\frac{a}{\sqrt{bc}}\right)\geq 3\alpha+6.\]

The answer is $\min_{x>0}\frac{(x+2)(x^2+1)}{x}$. :maybe:


14.  When written as a decimal, the fraction m/n (with m < n) contains the consecutive digits 2, 5, 1 (in that order). Find the smallest possible n.

15.  AB = 360, BC = 507, CA = 780. M is the midpoint of AC, D is the point on AC such that BD bisects angle ABC. F is the point on BC such that BD and DF are perpendicular. The lines FD and BM meet at E. Find DE/EF.

Number fourteen is interesting.

[hide]Suppose we have the decimal $0.A251 ...$.

We only consider $\frac{m}{n} = 0.251 \ldots$, since for some $k$, $10^k \frac{m}{n} -  A  = 0.251 ... $.

So for certain, $\frac{251}{1000} \le \frac{m}{n} < \frac{252}{1000}$. (1)

From there, we see that $251n \le 1000m < 252n$, so the difference a multiple of 1000 and a multiple of 251 is less than $n$.

So we have for our possible $n$, $251(4m - 1) \le 1000m$ - is a good range (up to a certain value of $m$, which is the value we want). We don't want multiple of 251 greater multiple 1000 for obvious reasons!

From our statement about the difference less than $n$ inferred from (1), it's clear $1000m - 251(4m-1) < n \Rightarrow 251-4m < 4m - 1$. And so for smallest $m$, $\Rightarrow m > 31$, $m = 32$.

And so $n = 127$.[/hide]

[quote="white_horse_king88"]Number fourteen is interesting.

[hide]
So we have for our possible $n$, $251(4m - 1) \le 1000m$ - is a good range (up to a certain value of $m$, which is the value we want). We don't want multiple of 251 greater multiple 1000 for obvious reasons!

From our statement about the difference less than $n$ inferred from (1), it's clear $1000m - 251(4m-1) < n \Rightarrow 251-4m < 4m - 1$. And so for smallest $m$, $\Rightarrow m > 31$, $m = 32$.

And so $n = 127$.[/hide][/quote]

I understand everything up to here.  I don't see how you suddenly determined that $(4m-1) /le n$.

It's sort of obvious,
since 251n \le 1000m \le 252n so that
4m = 1000m/250 > 1000m/251 > 251n/251 = n
since both m, n are nature numbers, so that 4m is no less than n+1
so that 251n is no less than 251(4m-1)
so 251(4m-1) \le 251n \le 1000m

frank the goblin tinker


There was once a king who wanted to determine who the smartest person in the land was ). After a bunch of tests, it was down to two men (P and S), who claimed to be "perfect" logicians (meaning there was never any flaw in their logic).

The king then told them their final test:

"I am thinking of two different numbers greater than 1, whose sum is less than 100. I will tell P the product of my two numbers, and I will tell S the sum of the two numbers," the kind stated.

The king whispers in P's ear the product of his two numbers, and whispers to S the sum of the two numbers...The following is the conversation that the logicians had:

P: I don't know the two numbers...
S: I know you don't
P: Ah...now I know them...
S: So do I

Together, both men then stated the two numbers to the king, and he stared in disbelief...

What were the two numbers?

Shouldn't this be in Puzzles and Brainteasers?


search
function
number theory unsolved
number theory

find all positive integers such that :

$ n / 2007^n \minus{} 2006 ^n$

/ means divides.

sorry if it is already posted, I used search function and didn't find it (my country's team selection test is in less than 1 week)

Posted before . Only $ n\equal{}1$ satisfy condition .
$ n|(a\plus{}1)^n\minus{}a^n$
Call $ p$ is the smallest prime divisor of n then
$ p|\gcd ((a\plus{}1)^n\minus{}a^n, (a\plus{}1)^{p\minus{}1}\minus{}a^{p\minus{}1})\equal{}(a\plus{}1)^{\gcd (n,p\minus{}1)}\minus{}a^{\gcd (n,p\minus{}1)}\equal{}1$
So $ p|1$ this give contradition .
So only $ n\equal{}1$ satisfy the condition.


Suppose that a certain binary security code allows n-digit sequences such that each digit in a sequence is a 0 or a 1, but no two successive digits can be a 0.  How many codes of length 10 exist?

Solution.
[hide]
We wish to choose the $0$'s out of a $10$ slots. Basically this the case:
[b]Case 1[/b]
1_1_1_1_1_<-This diagram shows that $5$ $0$'s can be put into those spots, with the first digit starting as $1$

[b]Case 2[/b]:Same as case 1 except by symmetry we have $0$ as the first digit.

Theres $2\binom{5}{k}$ ways to pick $k$ 0's. For $0\le k \le 5$.

So summing up all the ways:
$A=2\binom{5}{0}+2\binom{5}{1}+2\binom{5}{2}+2\binom{5}{3}+2\binom{5}{4}+2\binom{5}{5}=2(2^6)=\boxed{144}$.

And it is the same answer as the previous.
[/hide]

[quote="gotztahbeazn"]hmm... where did u get the 2 from? in front of ur $2 \binom{5}{k}$?[/quote]

2 represents the 2 cases that it it can be picked $\binom{5}{k}$ ways for example the first way is if 0 weas picked from 1_1_1_1_1_ and the other is from _1_1_1_1_1. I hoep that made it more clear..

Jimhu, that sums up to 64, not 144.  But the correct answer is 144.

[quote="epi14159"][hide]A=10C0+10C1+9C2+8C3+7C4+6C5=144[/hide][/quote]
[hide="I did the same thing"]
If we have no $0$'s, there is simply $1$ possible code.
If we have one $0$, there are $\binom{10}{1}=10$ possible codes.
If we have two $0$'s, there are $\binom{9}{2}=36$ possible codes, and so on...
[hide="Reasoning"]
We basically place all the $1$'s first, and then place the $0$'s between (or on the very left/right) of the $1$'s.
For example, if we have two $0$'s, we have eight $1$'s, which means we have $9$ spaces to put two $0$'s.
(This is similar to jimhu's idea.)[/hide]

So there are $\binom{10}{0}+\binom{10}{1}+\binom{9}{2}+\binom{8}{3}+\binom{7}{4}+\binom{6}{5}=144$ possible codes.
[/hide]


Do you have a solution to this problem? Is my solution correct?

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=151375  :wink:


Halo is awesome.  But the series ended... :(


The student lockers at Olympic High are numbered consecutively beginning with locker number $ 1$. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number 9 and four cents to label locker number 10. If it costs $ \$137.94$ to label all the lockers, how many lockers are there at the school?

$ \textbf{(A)}\  2001 \qquad \textbf{(B)}\  2010\qquad \textbf{(C)}\  2100 \qquad \textbf{(D)}\  2726\qquad \textbf{(E)}\ 6897$

[quote="4everwise"]The student lockers at Olympic High are numbered consecutively beginning with locker number $1$. The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number 9 and four cents to label locker number 10. If it costs $\$137.94$ to label all the lockers, how many lockers are there at the school?

$\text{(A)} 2001 \qquad \text{(B)} 2010\qquad \text{(C)} 2100 \qquad \text{(D)} 2726\qquad \text{(E)}6897$[/quote]

[hide]18 cents for first 9 lockers
next two digit lockers cost total of 360 cents
next three digit lockers cost total of 5400 cents
1000-1999 lockers cost 8000 cents, keep doing that until you get 18+360+5400+8000, each number then costs 8 cents, so 18+360+5400+8000=13778, there are two more numbers (13778+8+8=13794) so total there are 9+90+900+1000+1+1=(A) 2001

i think... :? [/hide]

[hide]Lockers 1-9 cost $\$0.18$
Lockers 10-99 cost $\$3.60$
Lockers 100-999 cost $\$54.00$
There is $\$80.16$ left to pay for the 4-digit locker labels, which cost $\$0.08$ each. So $80.16=0.08x$, which gives $x=1002$. So the total number of lockers is $1002+999=2001\implies\boxed{A}$[/hide]

[quote]Happiness - Eat, work, play and DO MATHS

18 cents for first 9 lockers
next two digit lockers cost total of 360 cents
next three digit lockers cost total of 5400 cents

$137.94-$0.18 - $3.60-$54= $80.16 80.16 divide by$0.08 = 1002

1002 + 900 +90 +9 = 2001.

Answer is  2001 or A.


$ a$ is chosen from the set $ \{1,2,3,4\}$, and $ b$ is chosen from the set $ \{5,6,7,8,9\}$. What is the probability that the product $ ab$ is a prime number?

The product of any two positive integers can only be prime if one of the numbers is $ 1$ and the other is prime.

This simplifies the problem immensely: $ (a,b)$ can only be $ (1,5)$ or $ (1, 7)$. That's $ 2$ possibilities out of a possible $ 4\cdot5$ which is $ 20$. The final answer is thus $ \frac{2}{20}$ which is more simply $ \frac{1}{10}$.

[hide=Solution]
For $a$ we have $4$ numbers to choose from. For $b$ we have $5$ numbers to choose from. This means we have $4\cdot5=20$ total.

For the successes, we have only $1\cdot5$ and $1\cdot7$ which work because they come out to be prime.

This leaves us with the probability $\frac{2}{20}$ which further simplifies to $\frac{1}{10}.$

[/hide]


For what value of $ x$ does $ 10(10^x)^2\equal{}100\;000$?


Find the largest positive integer $N$ so that the number of integers in the set $\{1,2,\dots,N\}$ which are divisible by 3 is equal to the number of integers which are divisible by 5 or 7 (or both).

It is equavalent to $[\frac N3 ]+[\frac{N}{35}]=[\frac N5]+[\frac N7].$ Let $a=N(mod 3),b=N(mod35),c=b(mod 5),d=b(mod 7)$.
We have $\frac{2N}{105}=\frac a3 +\frac{b}{35} -\frac c5 -\frac d7$.
Maximum expression $\frac{b}{35}-\frac c5 -\frac d7$ is $\frac{14}{35}$ (b=21). Maximum $\frac a3=\frac 23$.
Therefore $N=105(2/3+14/35)/2=56$.

Hey Rust, I think you are wrong because $[\frac{65}{3}] =[\frac{65}{5}]+[\frac{65}{7}]-[\frac{65}{35}]$

We can set really bad bounds on the right hand side and left hand side of $[\frac{N}{3}] = [\frac{N}{5}]+[\frac{N}{7}]-[\frac{N}{35}]$ and find the largest value from there.

We have that $[\frac{N}{3}] \geq \frac{N}{3} - \frac{2}{3}$ and $[\frac{N}{5}]+[\frac{N}{7}]-[\frac{N}{35}] < \frac{N}{5}+\frac{N}{7}-\frac{N}{35}+1 $.

From this, we get that the equality is broken for all N that satisfy $\frac{N}{3} - \frac{2}{3} > \frac{N}{5}+\frac{N}{7}-\frac{N}{35}+1$ which is equivalent to $n>67.5$ From here, it is easy to see that 65 is the largest number that works and is our answer.

Let f(n) = [n/3] - [n/5] - [n/7] + [n/35].

We are looking for the largest n with f(n) = 0.

Now [n/5] + [n/7} <= [n/5 + n/7] = [12n/35] = [n/3 + n/105].

So for [n/5] + [n/7] to exceed [n/3] we certainly require n/105 ≥ 1/3 or n ≥ 35.

Hence f(n) ≥ 0 for n ≤ 35.

But f(n+35) = [n/3 + 11 + 2/3] - [n/5 + 7] - [n/7 + 5] + [n/35 + 1] = [n/3 + 2/3] - [n/5] - [n/7] + [n/35] ≥ f(n) (*).

Hence f(n) ≥ 0 for all n.

But f(n+105) = [n/3 + 35] - [n/5 + 21] - [n/7 + 15] + [n/35 + 3] = f(n) + 2.

Hence f(n) ≥ 2 for all n ≥ 105.

Referring back to (*) we see that f(n+35) > f(n), and hence f(n+35) > 0, unless n is a multiple of 3.

But if n is a multiple of 3, then n + 35 is not and hence f(n+70) > f(n+35) > 0. So f(n) > 0 for all n ≥ 70.

f(70) = 1. So f(69) = 2 (we have lost 70, a multiple of 7). So f(68) = f(67) = f(66) = 1 (we have lost 69, a multiple of 3).

Hence f(65) = 0 (we have lost 66, a multiple of 3).

Answer: 65.

This issue is equivalent
floor(N/3)=floor(N/5)+floor(N/7)-floor(N/35)   *
N/3-2/3<=floor(N/3)<=N/5+N/7-N/35+34/35  so N<=86
we can note N=35 work so N=35+k where 0<=k<=51 now Equation * equivalent
floor((k+2)/3)=floor(k/5)+floor(k/7)-
floor(k/35)
From this we can see that floor(k/35)<=k/105<1 so k=<34 and
floor((k+2)/3)=floor(k/5)+floor(k/7)
And it's easy to check that k=30   is the biggest solution and N=65  is the solution


inequalities
geometry
circumcircle
inequalities proposed

[quote="Virgil Nicula"][color=darkred]Prove that in any triangle $ ABC$ there is the inequality $ \boxed{\ \sum\left(\frac{b\minus{}c}{a}\right)^{2}\ge 2\cdot\left(1\minus{}\frac{2r}{R}\right)\ }\ge 0\ ,$

where $ R$ is the length of circumradius and $ r$ is the length of inradius.[/color][/quote]
It's equivalent to $ (a\minus{}b)^{2}(a\minus{}c)^{2}(b\minus{}c)^{2}\geq0.$ :wink:


geometry
circumcircle
angle bisector
power of a point
geometry solved

In an acute triangle $ABC$ with $AB \neq AC$, let $V$ be the intersection of the angle bisector of $A$ with $BC$, and let $D$ be the foot of the perpendicular from $A$ to $BC$. If $E$ and $F$ are the intersections of the circumcircle of $AVD$ with $CA$ and $AB$, respectively, show that the lines $AD, BE, CF$ concur.

Triangles $AEV$ and $AFV$ are right-angled triangles. Because $AV$ is an angle bisector they are similar. Because they have one side in common they are also congruent. Therefore $|AE|=|AF|$.
From the Power of a Point: $|BF|*|AB|=|BV|*|BD|$ and $|CE|*|CA|=|CD|*|CV|$ which implies that
$\frac{|BF|}{|CE|}*\frac{|AB|}{|CA|}=\frac{|BD|}{|CD|}*\frac{|BV|}{|CV|}$
But from the Angle Bisector Theorem: $\frac{|BV|}{|CV|}=\frac{|AB|}{|CA|}$
so
$\frac{|BF|}{|CE|}=\frac{|BD|}{|CD|}$
Because of that: $\frac{|AF|}{|BF|}*\frac{|BD|}{|CD|}*\frac{|CE|}{|AE|}=1$ and the result follows from Ceva.

Since $\measuredangle ADV=90^{\circ}$, the segment AV is a diameter of the circumcircle of triangle AVD. Thus, $\measuredangle AFV=90^{\circ}$ and $\measuredangle AEV=90^{\circ}$, so that the points F and E are the feet of the perpendiculars from the point V to the lines AB and AC, respectively. Let P be the point of intersection of the lines BE and CF. Then, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=5460 (or also http://www.mathlinks.ro/Forum/viewtopic.php?t=41286 ), the line AP is the altitude of triangle ABC from the vertex A if and only if the line AV is the angle bisector of the angle CAB. But we know that the line AV is the angle bisector of the angle CAB. Hence, we obtain that the line AP is the altitude of triangle ABC from the vertex A; in other words, the line AP coincides with the line AD (since the line AD is the altitude of triangle ABC from the vertex A). Thus, the lines AD, BE and CF concur - namely, at the point P.

It is strange in how many different equivalent versions a problem can occur...

  Darij


Find a triple of rationnal numbers $(a,b,c)$ such that :

$\sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}$

I would like to see a solution to this problem please!


geometry
3D geometry
prism
Pythagorean Theorem

What is the length of the longest, straight, thin, rigid, solid, glass, green, striped stick that can fit into a right into a right rectangular prism measuring 3 feet by 4 feet by 12 feet?

This is a variation of the Pythagorean theorem.

Note: The reason is because basically, we're trying to find the length of the [b]space diagonal[/b] of the prism. The formula for that is an extension on the Pythagorean Theorem, $ \sqrt{a^2 \plus{} b^2 \plus{} c^2}$.

EDIT: Beaten to it.

Hrm... "space diagonal of a prism", that's a bit misleading.

It certainly is the [b]longest diagonal of a right rectangular prism[/b] (AKA a cardboard box).

It'd be a simple exercise to make a picture and figure out that formula: $ \sqrt {a^2 \plus{} b^2 \plus{} c^2}$.  Trust me, all you need is the Pythagorean Theorem (twice).

How is one supposed to know that's the longest distance through the box?  Just intuition I guess.[/b]

No, that formula will give you the longest segment possible.


What is the greatest three-digit number ``abc'' such that $ 4,a,b$ forms a geometric sequence and $ b,c,5$ forms an arithmetic sequence?

i think its 697, but i'll let someone confirm before posting a solution.

It's 697 because those digits satisfy the given conditions. $ 4, 6, 9$ is a geometric sequence, and $ 9, 7, 5$ is an arithmetic sequence.


geometry
circumcircle
inequalities
geometry unsolved

Problem:
Given an [b]acute-angled [/b]triangle ABC.Let BC=a,CA=b,AB=c.The radius of the circumcircle of ABC is R.Prove that $ (\frac {ab}{c})^2 \plus{} (\frac {bc}{a})^2 \plus{} (\frac {ca}{b})^2$≥$ 9R^2$

[quote="mihai miculita"]$ \mbox{Assertion is FALSE!} \\
\mbox{Let: } \\
 \begin{array}{c}a = 2 \\
b = c = x > 1 \end{array}\}\Rightarrow \{\begin{array}{c} R = \frac {x^2}{2.\sqrt {x^2 - 1}} \\
{(\frac {bc}{a})^2 + (\frac {ac}{b})^2 + (\frac {ab}{c})^2 = \frac {x^2 + 32}{4}} \end{array}\Rightarrow 9.R^2 - [(\frac {bc}{a})^2 + (\frac {ac}{b})^2 + (\frac {ab}{c})^2] = \frac {8x^4 - 31x^2 + 32}{4.(x^2 - 1)} > 0.$
$ \mbox{Because: } \\
\triangle = 31^2 - 32^2 < 0\Rightarrow 8x^4 - 31x^2 + 32 > 0;\ (\forall) x\in\mathbb{R} \mbox{ and } x^2 - 1 > 0; \ (\forall)x > 1.$[/quote]
Today I find out the way to solve this problem.
First we know the famous ineq(IRAN 96):
If x,y,z>0,then we have:
$ \frac {1}{(x + y)^2} + \frac {1}{(y + z)^2} + \frac {1}{(z + x)^2}$≥$ \frac {9}{4(xy + yz + zx)}$
Because A,B,C are acute angle,cotA,cotB,cotC>0.
Using IRAN 96 INEQUALITY with x=cot A,y=cot B,z=cot C.
Note that cotAcotB+cotBcotC+cotCcotA=1 and:cotA+cotB=$ \frac {sinC}{sinAsinB}$
After that,we have:
$ \sum (\frac {sinAsinB}{sinC})^2$≥$ \frac {9}{4}$
Since sinA=$ \frac {a}{2R}$,sinB=$ \frac {b}{2R}$,sinC=$ \frac {c}{2R}$
Hence
$ (\frac {bc}{a})^2 + (\frac {ac}{b})^2 + (\frac {ab}{c})^2$≥$ 9R^2$ :rotfl:


inequalities
LaTeX
algebra
partial fractions

1. Factorise x3 +4x2 +x − 6 using the factor theorem.
Hence solve the equation
x3 +4x2 +x −6=0 (5)

2. Use the remainder theorem to find the remainder
when 2x3 +x2 −7x −6 is divided by
(a) (x −2) (b) (x +1)
Hence factorise the cubic expression (7)

3. Simplify
6x2 +7x −5
2x −1
by dividing out (4)

4. Solve the following inequalities:
(a) 2 − 5x ≤ 9 + 2x (b) |3 + 2t| ≤ 6
(c)
x − 1
3x + 5
> 0 (d) (3t + 2)2 > 16
(e) 2x2 − x − 3 < 0 (14)

5. Resolve the following into partial fractions
(a)
x − 11
x2 − x − 2
(b)
3 − x
(x2 + 3)(x + 3)
(c)
x3 − 6x + 9
x2 + x − 2
(24)

6. Evaluate, correct to 3 decimal places,
5 e−0.982
3 ln 0.0173
(2)

7. Solve the following equations, each correct to 4
significant figures:
(a) ln x = 2.40 (b) 3x−1 = 5x−2
(c) 5 = 8(1 − e−x2
) (10)

8. The pressure p at height h above ground level
is given by: p=p0e−kh where p0 is the pressure
at ground level and k is a constant. When p0 is
101 kilopascals and the pressure at a height of
1500m is 100 kilopascals, determine the value
of k. Sketch a graph of p against h (p the vertical
axis and h the horizontal axis) for values
of height from zero to 12 000m when p0 is 101
kilopascals
(10)

9. Evaluate correct to 4 significant figures:
(a) sinh 2.47 (b) tanh 0.6439
(c) sech 1.385 (d) cosech 0.874 (6)

10. The increase in resistance of strip conductors due
to eddy currents at power frequencies is given
by:
λ = αt
2
sinh αt + sin αt
cosh αt − cos αt

Calculate λ, correct to 5 significant figures, when
α=1.08 and t =1 (5)

11. If A ch x −B sh x ≡4ex −3e−x determine the
values of A and B. (6)

12. Solve the following equation:
3.52 ch x + 8.42 sh x = 5.32
correct to 4 decimal places (7)

You mean you want us to do your homeworks for you? Nah, that's too boring... :P. You can get help here though: just post one problem a time, show what you've done and where you got stuck and we'll give you a hint about how to proceed :). Also keep in mind that this site uses $ \text{\LaTeX}$ for displaying math. formulae and it will take you less time to learn it than it'll take us to guess what you mean by e-x2 and other constructions like that.


Given real constants A, K, can we explicitly determine
all real X such that:

A = X * sin(KX) ?

Thanks.

No, there is no general formula for the roots. Since such questions arise again and again, I think we should write something about that on Wiki :).

Perhaps its really a good idea. I know that, usually, a general formula for such expressions
can't be found. Just thought that in this particular case, maybe, someone could find a tricky
solution.
Thanks.  :)


It is easy to show the impossibility of covering a unit square with $5$ squares whose sides have lengths $<\frac{1}{2}$ (note that I do not ask for a tiling, just a covering).

For $7$ such squares, a covering exists.

What about $6$ squares ? Surely, the problem is impossible; but I can't find a proof ...

oh...its possible: on unit square:

take on two adjacent edges, 5 squares with side length $1/3$ (so that each edge has 3)

Then the remaining area is a square of side length $2/3$

Have I misunderstood the problem?

But $2/3$ is larger than $1/2$, so that covering does not satisfy the conditions.

julien_santini, what is the easy proof that 5 squares are impossible?  We had previously discussed this problem in this thread:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12615[/url].

There, rcv came up with a complicated proof.

Well, I thought I had an easy answer based upon the same idea as Ravi's (about the maximal length enclosed by a small square) but I fell in the same fallpit !

Okay.  By the way, was this problem in Newman's book (A Problem Seminar)?  If so, have you seen his solution?

Nope. However, I've seen it has been asked at the Iberoamerican olympiads. This means that there must be some reasonably easy solution somewhere ... I've tried the official website for those olympiads but didn't find anything.

since each of the squares side $<1/2$, thus on each side you need at least three squares touching it. Count the number of squares: you need at least 8

Thus how did you come up with an arrangement of 7 squares? Or am I too tired to think  :blush:  .........................


modular arithmetic
Additive Number Theory

Show that there are infinitely many positive integers which cannot be expressed as the sum of squares.

Maybe the correct version is this one

[quote="Peter"]Show that there are infinitely many positive integers which cannot be expressed as the sum of three squares.[/quote]

My logic is that we all know that any positive integer number can be represented as a sum of four squers.The case for two squers is easy.So the left one is for three squers. :roll:

[quote="Peter"]Something's weird here... trying to find the correct version.[/quote]

Definitely something wrong. I don't remember the correct version.

Maybe it is "the sum of any number of different squares"? Would that be true?

No it would not (all integers $>128$ are a sum of distinct squares). I tend to the three squares, too.

But 3 squares is still trivial since every $n\equiv7 \pmod 8$ is a counterexample... I think we'll just wipe this one out then.

Then the set of powers of $2$ does the trick...


From a PM I sent to rrusczyk:

[quote]Check the Missouri forum. Apparently the craze about these stupid rickrolls is just becoming extremely disturbing that it makes people afraid to click on any hyperlinks posted in the forums, blogs, or anywhere.

Shouldn't us AoPSers be doing something about that? For example, there are rickroll links in algebra2000 and nikeballa96's blogs, as well as multiple threads in the MO forum. Seriously, that HAS TO STOP. It's becoming an AoPS addiction that isn't funny, but very annoying. It's like having AoPSers trying to set trojan horses everywhere, here and there, on AoPS that makes it dangerous to click hyperlinks.

It just isn't funny. Can you guys please do something about it? [/quote]

The rickroll craze seriously has to stop. Not only is it disturbing to the general community, but it pretty much makes people afraid to click on any hyperlinks posted on AoPS.

I don't see how trojan horses are like rickrolls.

People are basically posting things that look like something that others want to click on but actually lead to rickrolls.
For example, people post stuff like "Click to reveal hidden content" or "here's a beastly math problem" which actually link to the rickroll, which is just annoying.

[quote="AIME15"]Hover your mouse over the link.  If it goes to a website it's probably a rickroll, unless you recognize the website.[/quote]
Bleh, well but not EVERY person is smart enough to do that...esp. those new to the forums or something...some people think hide tags are always regular text hidden rather than a hyperlink...

Also, not every person has a high degree of gullibility avoidance here.

For opera, even if you hover over the link, it doesn't display anything.

First off, shouldn't this be in Questions, Suggestions, & Announcements?

Second off, I am pretty sure hovering over the link works. Opera 9.64 here if you look. ( the mouse in the picture is missing, but you can see the URL has changed from its usual semi-dark blue to another.)

To kill a rickroll without CTRL+ALT+DELETE
1. attempt to close it while its moving
2. a screen should pop up saying part of the song.
3. hold ENTER until it dissappears

Um, use CTRL-W to finish step 1.  Then, right after you click "OK" on one of the pop-ups, hit CTRL-W again.  One more pop-up will appear, and then they will subside.

Hey, another problem with the rickrolls is that sometimes the links are [url=http://tinyurl.com]tinyurls[/url]. (don't worry, that's perfectly safe)

We can't really help people using tinyurl.com ...

If you go to tinyurl.com, you can enable previews so when you click a tinyurl, it will take you to a page that says where it will go, then you can click continue or just exit it.

[quote="Differ"]Solution:

Post rickroll once: warning
Twice: temp ban (one day)
Three times: temp ban (one week)
Four times: permanent ban[/quote]

I believe this is seriously way too nice. Rickrolls are seriously problems, and many such websites introduce serious viruses into computer.

The solution is to stop rickrolling and start [url=losethegame.com]gamerolling[/url].


[quote="chien than"]Find min :$A=\frac{x}{y}+\sqrt{1+\frac{y}{z}}+\sqrt[3]{1+\frac{z}{x}}$
where $x;y;z$ are positive real numbers[/quote]
Use the fact,

$\frac{x}{y}+\frac{1}{2}\sqrt{1+\frac{y}{z}}+\frac{1}{2}\sqrt{1+\frac{y}{z}}+\frac{1}{3}\sqrt[3]{1+\frac{z}{x}}+\frac{1}{3}\sqrt[3]{1+\frac{z}{x}}+\frac{1}{3}\sqrt[3]{1+\frac{z}{x}}$

Now use AM-GM to get:

$\frac{x}{y}\left(1+\frac{y}{z}\right)\left( 1+\frac{z}{x}\right)$
(Along with a constant in front).


let ABC be a triangle and (C) is its circumcircle. M,N are the mid points of AB,AC.

another circle goes through M and N and also is tangent to (C) in point A'.

find B' and C' ,by this way. and prove that:

                                                            AA',BB',CC' are concurrent.

for the $999...$ times ,You know the solution of this problem ,so post it in the Proposed section ,is it so hard to understand this?

[color=blue]Solution[/color]
By Davode Vakily:

Let $AA`$ cut $BC$ in $A``$ and Let $E$ be the feet of altitude of $A$ to $BC$ .
Invert to $A$ and see that $EB=A``C$ .thus by ceva thoerem lines $AA``$ are concurrence