[hide]4 hours. If they go in opposite directions, the distance between them after 1 hour is 325 + 400, or 725 miles. If they keep growing at that same rate every hour, in 4 hours they will be 2900 miles apart. 725 x 4 = 2900. [/hide]

[quote="math92"]Tow planes left an airport atr the same time.  One flew north averaging 325 mi/h, and the other flew sount at an average speed of 400 mi/h.  In how many hours will they be 2900 miles apart.[/quote]

[hide]their combined speed will be 725 mi/h

so 2900/725=4 hours[/hide]

[hide]combined distance would be 725. 2900/725 = 4
4 hours.[/hide]

[quote="math92"]Tow planes left an airport atr the same time.  One flew north averaging 325 mi/h, and the other flew sount at an average speed of 400 mi/h.  In how many hours will they be 2900 miles apart.[/quote]

\[d_1=325 \cdot t\]
\[d_2=400 \cdot t\]
\[d_1+d_2=2900\]
Solving for $t$, we see $t=4$.

[hide="wut i got..."]i got 4 hours[/hide]

[quote="math92"]Tow planes left an airport atr the same time.  One flew north averaging 325 mi/h, and the other flew sount at an average speed of 400 mi/h.  In how many hours will they be 2900 miles apart.[/quote]

$v_{\text{separation}}=400\text{ mph North}-325\text{ mph South}$

$v_{\text{separation}}=400\text{ mph North}-(-325\text{ mph North})$

$v_{\text{separation}}=400\text{ mph North}+325\text{ mph North}$

$v_{\text{separation}}=725\text{ mph North}$

$(725\text{ mph})\times t = 2900\text{ miles}$

$\frac{(725\text{ mph})\times t}{725\text{ mph}} =\frac{2900\text{ miles}}{725\text{ mph}}$

$t=\frac{2900\text{ miles}}{725\text{ mph}}=4\text{ hours}$


The largest interior angle in an isosceles trapezoid is 4 times the
measure of the smallest interior angle. What is the measure, in
degrees, of the smallest interior angle in the trapezoid?

smallest interior angle = $ x^{\circ}$

$ 4x\plus{}4x\plus{}x\plus{}x \equal{} 360$
$ 10x\equal{}360$
$ x\equal{}\boxed{36^{\circ}}$


suppose $AB$ , is a segment, and $P$ is a fixed point on it.

a line which is prependicular to $AB$, intersects it at $H$.

$K$ is  moving point on this line.draw circumcircle of triangle$KHA$,

and let $KP$, intersect it at $S$.

what is the locus of $S$.

[quote="rem"]We have $A,S,K,H$ concyclic, so $\angle PAS=\angle PKH, \angle APS = \angle KPH$. Because $\angle KPH + \angle PKH = 90^o$, then $\angle PAS + \angle PSA = 90^o$ hence $\angle ASP = 90^o = const$. Therefore locus of pts $S$ are all pts on circle with diameter $AP$.[/quote]

good. ;)


Find a 1-1 correspondence between the interval [0,1] and the cube [tex]\{(x,y,z)|~0\le x,y,z\le1\}[/tex].

[hide="I think..."]Can someone prove this?

(xyz+xy+xz+yz+x+y+z)/7 goes through all numbers in [0,1] once.
[/hide]

Just do it Peano-Hilbertish :)

[b]1=2:[/b] $ f(x,y,z)\equal{}f(y,x,z)\equal{}\ldots$ so the function is not injective  :huh: it goes through each number though (since it is continouus).

[hide="Perhaps"]
Let's work in base 8.

Divide the cube into two pieces in each of the three dimensions, creating 8 smaller cubes. Set a consistent numbering scheme to number these 0 through 7. Divide each of these cubes into 8 in the same way, and divide those into 8 in the same way, etc.

Now, let the sub-region label at the nth stage be the digit that goes in the 8^(-n) position.

So for $ 0.4 \equiv 0.4\overline{0}$, you would select sub-region 4 on the first step and sub-region 0 on every subsequent step.

This produces all numbers from $ 0.\overline{0} \equiv 0$ to $ 0.\overline{7} \equiv 1$.

[hide="Or if you insist on base 10"]
Divide the cube into ten layers vertically, then each of these into ten layers left to right, then each of the resulting 100 sticks into ten layers front to back, and repeat this process.
[/hide]

My only gripe about this approach is, what about numbers such as $ 0.4 \equal{} 0.4\overline{0} \equal{} 0.3\overline{7}$, there are two ways of representing terminating strings. We can define one point to correspond to such numbers (such as the one with 0's, perhaps), but there would be an infinite number of point-discontinuities.

There may be a way to remedy this by multiplying the number by, say, 0.7_8 and then mapping the numbers from (0.7,1] to these discontinuities in a similar manner. If that is done in a similar manner, then there will still be discontinuities, in which case, squeezing that region by 0.7, too, and performing this cover-up-the-discontinuities process iteratively should ultimately tackle the problem.
[/hide]

[quote="The Zuton Force"][hide="Perhaps"]
Divide the cube into ten layers vertically, then each of these into ten layers left to right, then each of the resulting 100 sticks into ten layers front to back, and repeat this process. At the nth stage of this process, let the 10^-n place value be filled with the digit 0 if the point is in sub-region 1, digit 1 if it is in sub-region 2, ..., and digit 9 if it is in sub-region 10.

This produces all numbers from $ 0.\overline{0} \equiv 0$ to $ 0.\overline{9} \equiv 1$.
[/hide][/quote]

Let $ (x,y,z)\equal{}(0.x_1x_2x_3...,0.y_1y_2y_3...,0.z_1z_2z_2...)$, then we associate that to $ .x_1y_1z_1x_2y_2z_2...$

this doesn't cover the right endpoints...i don't know if there is an easy way to resolve this

Consider the function $ f: [0,1]\rightarrow [0,1]^3$, where
\[ 0.x_1x_2\hdots \equal{} x\mapsto (0.x_1x_4\hdots,0.x_2x_5\hdots,0.x_3x_6\hdots)
\]
and overcome dual representations (to which I have not given any thought).

The interesting thing about such functions is that there is no continuous bijection between $ [0,1]$ and $ [0,1]^3$ (proof by contradiction).

EDIT: I did not pay attention to Altheman's post.


3 cards are chosen at random from a standard 52-card deck.  What is the probability that they are three consecutive cards, all of the same suit? For this problem, an Ace can either come before a 2 or after a King, but not both (so A23 and QKA both count, but KA2 doesn't).

[quote="lowsf"]So order doesn't matter in this problem?  Meaning we are considering the set {7 Club, 5 Club, 6 Club} as "three consecutive cards?"

If the order we drew the cards mattered, what would the solution be?[/quote]

I believe the order actually does matter in this problem, but crazypianist's solution takes this into account.

[quote=bluehurricane][quote=crazypianist1116][hide]Number of correct over number of possible:
There are $ (12)(4)$ possibles since there are 4 suits and 12 possible lowest cards $ \frac{(12)(4)}{\binom{52}{3}} \equal{} \frac{12}{5525}$[/hide][/quote]
I got something close to this, but I did it with order not mattering. Since it's consecutive, order [i]does[/i] matter, right???[/quote]

Yeah, in this problem, order matters.

The question asks "What is the probability that they can be arranged into a group of three consecutive cards, all of the same suit?" so the order they are picked doesn't matter, since they will be rearranged.

[hide=solution]$\frac{4 \cdot 12 \cdot 3!}{52 \cdot 51 \cdot 50}$[/hide]


calculus derivative function integration logarithms trigonometry Support

Can someone give me an example of a discontinuous function in W^(1,1)(R^2), that is a function in L^1(R^2) whose weak derivative is a function in L^1(R^2) as well. I am aware that W^(1,1)(R^1) functions must be C^1, put apparently this does not hold in R^n for n > 1.

Let $\varphi$ be some nontrivial smooth, compactly supported function.

Define $f(x,y)=\begin{cases}\varphi\left(\frac{x}{y}\right)& 0<y\le 1\\ 0&y\le 0\end{cases}.$

Find a way to ramp that down smoothly so that it is zero for $y>2,|x|>2.$

$f$ is bounded and compactly supported, so it is integrable. However, it is discontinuous at the origin.

$f_{x}=\frac1y\varphi'\left(\frac{x}{y}\right)$ for $0<y<1.$

(This classical derivative is clearly also the weak derivative.)

Then, assuming the integral of this causes no trouble for $y\ge 1,$ we get that

$\|f_{x}\|_{1}\le \int_{0}^{1}\int_{-\infty}^{\infty}\frac1y\left| \varphi'\left(\frac{x}{y}\right)\right|\,dx\,dy+C$

$\le \int_{0}^{1}C\,dy+C<\infty.$

I believe $f_{y}$ is also in $L^{1},$ although I haven't fully checked those details.

[quote="bobbydylan"]Can someone give me an example of a discontinuous function in W^(1,1)(R^2)[/quote]
Example: $f(x)=\min\{\log |x|,0\}$. Or $\sin f(x)$ if you want a bounded function.

[quote="bobbydylan"]I am aware that W^(1,1)(R^1) functions must be C^1[/quote]
Counterexample: $f(x)=|x|$.

|x| isn't in L^1, nor is its weak derivative.

on a vaguely related note, would you happen to know what the weak derivative of the characteristic function of the unit circle on R^2 is? any chance the characteristic function of the unit circle is in W(1,,1)(R^2)?

Cheers.

No, the characteristic function isn't in $W^{1,1}$. (It is in a somewhat wider space $BV$.) Its distributional derivatives are not $L^{1}$ functions but signed measures with support on the boundary of the circle.

You may want to find Ziemer's book "Weakly differentiable functions"


In a $ \triangle ABC$, let $ M,N$ be the midpoints of $ CA,AB$ respectively, $ O$ circumcenter, $ H$ orthocenter. $ \omega$ is the circumcircle of $ \triangle ANM$
Prove that, $ PQ \parallel MN$  where, $ P\equal{}\omega \cap AO$ and $ Q\equal{}\omega \cap AH$

[quote="Nemion"]$ AP$ passes through $ O(w)$. Then $ \angle{AQP} \equal{} 90^o$ $ \Rightarrow$ $ PQ\parallel BC$[/quote]
Ok....Harder version: let $ \triangle AMN$ be an arbitrary triangle. Now, Prove that, $ PQ\parallel BC$.
Happy problem solving!
 :)


calculus
integration
calculus computations


can I put yes because MIT is a private school?

Lol. Yes of course why not... it doesn't say whether the poll means a elementary, middle, high, university school or something else...

[quote="gr8sk8r"]I'm just wondering because i'll be starting a private high next year.[/quote]

Which one? Please tell me it's not deerfield.

[quote="banblink182"]"The Sound of Silence"  :lol:[/quote]

What do you mean?  People sure are talking; I'm not yet sure whether they're speaking.

Anyhow... no I am not going to a private school.  However, I did like certain aspects of a private school I went to some years ago.  Mostly it was the active (though not incredibly good) math program.

[quote="probability1.01"]What do you mean?[/quote]

I was referring to the lack of a response from gr8sk8r to yif man12.  :)

Looking forward (um, really?) to my fourth (and, I hope, last) year at a private "international" school where about 80 percent are [i]not[/i] international and which seems to be adopting what I call the "few children left behind, and no children forward" policy. But my public alternatives are even worse. Sigh....

Private schools come in all sorts of shapes and sizes. Hope you like yours, gr8sk8r.

yeah - been at private school for 7 years now, will for another 5

Next year will be my 4th at a private high school.  I only ended up going there when the guidance couselor at my local public school told my mom to get me out of the public school because they didn't have good (read: any) more advanced programs.

Anyone want to actually post what school they're going to?

All through grade school I've gone to public schools.  In Kindergarten/Pre-School, I went to private though.

Blah. I go to Shanghai American School. Private, as you might assume.

Im goin to Suff. Academy next year! Woohoo :lol:


algorithm
algebra
polynomial
geometry
linear algebra
linear algebra unsolved

I'm looking for [b]any[/b] hints.  If you can think of any subject (whether in abstract algebra, numerical optimization, etc.) that might reveal some insight into the following problem, please respond!  This problem only seems complicated because of formalities; it is in fact quite simple.

We are given a set of equations, such as
\[ a_0 \plus{} a_1 \plus{} a_3 \plus{} a_1a_3 \plus{} a_2a_3 \equal{} 1 \\
a_0 \plus{} a_1 \plus{} a_2 \plus{} a_0a_1 \plus{} a_1a_3 \equal{} 1 \\
a_2 \plus{} a_3 \plus{} a_0a_2 \plus{} a_0a_3 \plus{} a_1a_3 \equal{} 0
\]
where $ a_i\in \{0,1\}, \forall i$ and the following is true of the algebraic structure,
\[ 0 \plus{} 0 \equal{} 0 \\
0 \plus{} 1 \equal{} 1 \\
1 \plus{} 1 \equal{} 0 \\
0\cdot 0 \equal{} 0 \\
0\cdot 1 \equal{} 0 \\
1\cdot 1 \equal{} 1
\]
along with additive and multiplicative commutativity and associativity, but [i]no inverse defined[/i].

[b]I need an algorithm that finds ONE solution to the problem above (and other similar, larger problems).[/b]  Note that all terms are either a single variable or the product of two variables.  The system will always be under-constrained, implying that multiple solutions exist.

Trial and error works fine for small problems such as the above, but I need something efficient.  I will need to solve problems of over 100 variables (over 5000 possible terms).

Thanks in advance!  Any help is appreciated.

- Andrew Cheong

[size=75]Edit: I've thought about generalizing this problem such that I can simply apply a nonlinear optimization algorithm (Steepest Descent, Ellipsoid), but I don't know how it works with the algebraic structure given above.  Speaking of which, what is the formal name of the structure to which I am referring?[/size]

The algebraic structure to which you refer is usually denoted $ \mathbb{F}_2$, the [url=http://en.wikipedia.org/wiki/Finite_field]field with two elements[/url].  It is isomorphic to $ \mathbb{Z}/2\mathbb{Z}$, the integers $ \bmod 2$, or to the [url=http://en.wikipedia.org/wiki/Boolean_ring]Boolean ring[/url] of order $ 2$.

Inverses are, in fact, defined for all nonzero elements (as in any field); the multiplicative inverse of $ 1$ is $ 1$, as is plain from your definitions.  I don't know anything about nonlinear optimization, but any technique that uses field operations alone should work with no modifications, and general techniques for understanding the solutions to systems of polynomial equations fall under the category of [url=http://en.wikipedia.org/wiki/Algebraic_geometry]algebraic geometry[/url].  I believe the keywords you want are "algebraic geometry over fields of positive characteristic," although the case of $ \mathbb{F}_2$ might be sufficiently degenerate that much simpler techniques are available.

Edit:  Since it appears that want you want is an algorithmic solution, you might want to look at [url=http://www.amazon.com/Ideals-Varieties-Algorithms-Computational-Undergraduate/dp/0387946802]Cox, Little, and O'Shea[/url].  It's a very readable introduction to the subject from an algorithmic perspective.  For now, the notion of a [url=http://en.wikipedia.org/wiki/Gr%C3%B6bner_basis]Groebner basis[/url] is something that might help, although certain denominators might be zero more often than desired.  In any case, picking a basis carefully enough could potentially make the brute-force approach feasible.

Thank you!  It's gonna take me a few days to read and comprehend all of the things you have pointed out, but I think you are dead on to what I'm doing.  I'm actually reversing a cryptographic algorithm, and I was able to reduce everything to XOR's and AND's, which translate nicely to addition and multiplication in $ \mathbb{F}_2$.  If the sample space of solutions is small enough, perhaps you're right that a brute force approach would be best.  I was hoping for some sort of integer programming optimization problem, which I see is related by the notion of a Groebner basis.  Thanks again.

Let $ S$ a system of $ n$ equations with $ n$ unknowns in $ F_2$; moreover we assume that $ S$ is of degree $ 2$ (like yours) and is a RANDOM system (coefficients are randomly chosen).
    Then solving $ S$ using theory of Grobner basis is slower than solving with brute force. In particular nobody can practically solve such a system if $ n\equal{}80$.


These are Blundon inequalities. I'll post a solution later for the second one. :)  ;)


I was wondering how you become a moderator? I'd really like to be one somewhere. :huh:

To get a moderator, you need to be a good member of the community.
With 2 posts, this is very unlikely to be the case for you.
And you could at least have taken a search for what you want. Then you would have seen a rather new thread like http://www.mathlinks.ro/Forum/viewtopic.php?t=185384 instead of asking again...

Evidence tells me that those asking to get some moderator somewhere without any reasoning never get moderators  :wink:

is there anything specific that you have to do?

See again the link. It will not get more specific.


The diagonals of quadrilateral ABCD intersect in point E. Given that |AB|=|CE|, |BE|=|AD|, and ∠AED = ∠BAD, determine the ratio |BC| / |AD|. (|XY| means the length of the segment from point X to point Y.)

Indeed $ \angle AED \equal{} \angle BEC$ by vertical angles so $ \angle BAD \equal{} \angle BEC$. It follows that $ \triangle ABD\cong \triangle ECB$. So $ \angle ADB \equal{} \angle EBC$ so $ AB$ is parallel to $ BC$. Furthermore, $ BE \equal{} BC$. Let $ AD \equal{} BE \equal{} y$, $ BD \equal{} BC \equal{} z$. $ \triangle AED\sim\triangle CEB$, so $ DE \equal{} \frac {y^2}{z}$. Indeed, we can see that $ \triangle AED\sim\triangle BAD$ by AA similarity, so
\[ \frac {ED}{AD} \equal{} \frac {DA}{DB}\implies\frac {\frac {y^2}{z} }{y} \equal{} \frac {y}{y \plus{} \frac {y^2}{z}}
\]
We may normalize to $ AD \equal{} y \equal{} 1$, and get $ z \equal{} 1 \plus{} \frac {1}{z}\implies z^2 \minus{} z \minus{} 1 \equal{} 0\implies z \equal{} \frac {1 \plus{} \sqrt {5}}{2}$.

Nice problem  :)

couldn't you also use stewart's theorem multiple times?

Wait whaat? I had this one right and then when writing up couldn't figure out how I got phi, I ended up putting 1 as the ratio....

Hmm... I did this problem correctly too.... >_____< arg....


This the problem and they say the answer is E why?

A right triangle has the sides 8 and 15. What must the length of the hypotenuse be?

A. 7
B. 15
C. 21
D. 17
E. NOTA

[color=red][Moderator edit: This is definitely not Olympiad Section, so I move it.][/color]

[quote="white_horse_king88"]The answer isn't because you have $8^2 + 15^2 = 17^2$. So, it should be 17.

Just a fun problem based on this:

On a coordinate system, if a circle of radius 17 is placed such that it's center is at the origin of the coordinate system, how many coordinate pairs with integral coordinates does the circumference pass through?

Badly worded, but the idea is there.[/quote]

[hide="Quick Guess"]12[/hide]

Folks, if you have a right triangle with sides 8 and 15, the hypotenuse need not be any number -- it could be either 15 or 17.

Oh, good point. The sides measured 8 and 15 are not necessarily the legs.

[quote="white_horse_king88"]Good. Any explanation?[/quote]

[hide="Solution"]The circle will intersect two times with the x-axis and two times with the y-axis. The intersections will have integer coordinates because one coordinate will be 0 and the other +/-17. The points (+/-8,+/-15) will be on the circle because 8^2+15^2=17^2. Same for the points (+/-15,+/-8). So we have 4+4+4=12 points.

However, I can't prove that these are the only points.[/hide]

Since the coordinates are integers, and the radius is 17, you have all integral solutions to the pythagorean equation, $a^2 + b^2 = 17^2$. There are only 12 integral pairs $(a, b)$, which satisfy that. Namely, (0,  :pm: 17), (:pm: 17, 0), (:pm: 15, :pm: 8), (:pm: 8, :pm: 15), which totals to 12.


Duke
college
MATHCOUNTS
geometry
3D geometry
trigonometry
sphere

Who all's going (see title)?  I just found out that I'll be able to go on a South Carolina team.  A big competition that in SC that was planned for the same day was cancelled recently so one of the all state coaches has been acting quickly to organize a team.  I look forward to the competition :)

I am going, and some other Georgian AoPSers are going; I don't know which ones though.

I'm going to the Duke Math meet.  I'll be the one with the really really good looks.  How will I know who is who?

I'll be wearing my AoPS shirt with my username embroidered upon it. :) :lol: It's my competition uniform.

edit: now i'm not going.  :(
good luck to the SC team though.

To finish out those results the Atlanta Math Club got 4th and South Carolina got 5th (  :lol:  :lol: ).

Individual: 4 way tie with a score of 7 b/w Arnav Tripathy, Veronica Pillar, Menyoung Lee, and Noah Blach.  In the lighthning tiebreaker round ARNAV TOTALLY PWNED EVERYONE.  It was awesome...Arnav hardly gave the others a chance he was so fast and he even showed his sense of humor along the way  :D .

How fast did he do the questions? Was the tibreaker thingie like Countdown round in Mathcounts?

Let's see his first problem had something to do with a cone shaped popsicle...I didn't start reading the problem immediately and he got it before I could even read the problem...the guy proctoring the questions was reading it out loud and didn't even come close to finishing it before Arnav rang in. Probably about 5 to 10 seconds elapsed overall. Another question he had was $\sqrt[3]{\sqrt{56+\sqrt{56+\sqrt{56+\cdots}}}}.$  He got that in about 3 seconds (I'm guestimating the time).  Another one was $\cos(\sin^{-1}{1/3}).$  Took him no more than 5 seconds.  Then another was $\frac{10!\cdot5!}{12!+13!}$ which he got in about 3 seconds.

Yes, very Mathcountish questions.....My Devil Round team, containing MysticTerminator, was 2nd, though I don't know how much I contributed...

[quote="FMako"]The math meet was pretty excellent.  Although, the duke union should post their solutions or should have handed them out after the contest.  Did anyone hear how to solve that brainteaser?  I met solafidefarms, he stood out with his shirt and all.
Note: TJ A = 1st, TJ B = 2nd, Chapel Hill = 3rd, ... don't remember the rest.[/quote]

East Chapel Hill = 3rd thank you very much :lol:

although I'm not sure it has anything to do with how good we are as a team; each year it's pretty much just the sum of our individual abilities. Like I think we were 2nd last year b/c we had a couple more ppl? eh w/e

joe, i think the (ice cream) cone question was something like finding the radius of a half-sphere that has half the area of another half sphere...

i didnt have time to read it either but i asked one of the guys running the competition to let me see a copy of the question since i was sitting near him..

maybe they call them semispheres, i dont really know, lol

-asif

The ice cream cone problem, if I recall correctly, had a fictional character eating a cone with radius $8$ and height $8$ (these numbers may be incorrect).  The character eats the cone such that the radius decreases at the same rate as the height.  What is the height of the cone when the volume of the cone is half of what it was originally.  Arnav came up with $\frac{8}{\sqrt[3]{2}}$ in about 5 seconds.

Dschafer, you got an account!  Oh, right, this is a Duke thread.   Hmmm,  well, there was a distinct and disturbing lack of geometry or number theory (number 7 was too easily guessed that it doesn't count).  It was heavily focused on clever algebra and combinatoric type calculations.  I admit that they were good problems, but it was a little unbalanced.  Overall it was pretty fun though.  The relay was way too hard, did you guys see those number ones?  They purposefully built in like 5 places to mess up.  I was surprised to see the Lightning Round not only follow the exact format of Countdown rounds, but also to follow the difficulty level.  Fun contest though.

The relays [i]were[/i] ridiculous.  I was sitting in the third spot waiting for an answer but on both relays I didn't get any legit answers to plug in.  I was a little ticked...but then I looked at my what the one and two spots had and was like....OH...OK  :o

I've tried to contact the Duke Math Union in order to retrieve solutions to problems from various rounds.  Unfortunately, they haven't returned my email.  Has anyone been able to contact them or obtain a solutions sheet?
I am mainly refering to solutions to power round and team round questions, since at least the answers were given for the individual.  I would also like to see the first and second tier relay round questions, creating a general formula for the third question seemed to take less time than anyone of the beginning two problems.  The only question I saw was "What is the first rightmost non-zero digit in $25!$".

Paraphrased versions of the two first tier problems in the relay round:

What is the rightmost non-zero digit in $25!$?

and

$45$ is the leg of a right triangle with integer sides, with $a$ and $b$ representing the lengths of the other two sides.  How many ordered pairs $(a,b)$ exist?

I have good news.  The Duke Math Union hopes to post the solutions in two weeks.
As I have solved every problem on the individual, team, and relay round, I can answer any questions about those problems, if you are too eager to wait for the official solutions.  I know I was.  As for the devil round and power round, I tore up most of the devil round papers to pass to my teammates, so I don't have most of the questions, and I need work on proofs, so I'm no expert at the power round.

Whoa... I've never heard about this. I might try to get my school (Vestavia) there next year.

Hmm when is this? This looks interesting. Is there a website or something?

Does anyone have an online copy of the most recent Duke Math Meet Contest???  I really want to look it over.


Given a triangle $ABC$, $A_1$ divides the length of the path $CAB$ into two equal parts, and define $B_1$ and $C_1$ analogously. Let $l_A$, $l_B$, $l_C$ be the lines passing through $A_1$, $B_1$ and $C_1$ and being parallel to the bisectors of $\angle A$, $\angle B$, and $\angle C$. Show that $l_A$, $l_B$, $l_C$ are concurrent.

See this problem (post nr. 4):
http://www.mathlinks.ro/Forum/topic-38929.html


The number of geese in a flock increases so that the difference between the populations in year $ n \plus{} 2$ and year $ n$ is directly proportional to the population in year $ n \plus{} 1$. If the populations in the years $ 1994$, $ 1995$, and $ 1997$ were $ 39$, $ 60$, and $ 123$, respectively, then the population in $ 1996$ was

$ \textbf{(A)}\ 81\qquad \textbf{(B)}\ 84\qquad \textbf{(C)}\ 87\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ 102$

My copy of the tests shows this very clearly to be $ n\plus{}1$.  Moderator, please edit.

[hide]
Let the population in year 1996 be $ x$. Equating the proportion for years 1994-1996 and 1995-1997 yields

$ \frac {x \minus{} 39}{60} \equal{} \frac {123 \minus{} 60}{x}$
$ x^2 \minus{} 39x \equal{} 60 \cdot 63$
$ x^2 \minus{} 39x \minus{} 3780 \equal{} 0$
$ x \equal{} 84$ or $ x \equal{} \minus{} 45$

The positive answer is equal to the alternative $ \boxed{\textbf{(B)}}$.
[/hide]

Sorry  :blush:. Thanks for pointing it out.

[quote="AIME15"][hide="Click for solution"]
Let $ x$ be the number of geese in $ 1996$ and let $ k$ be the constant of proportionality.  Then $ x\minus{}39\equal{}60k$ and $ 123\minus{}60\equal{}kx$.  Solving the second equation for $ k$ and using that to solve for $ x$ in the first equation yields $ x\minus{}39\equal{}60 \left (\frac{63}{x} \right ) \implies x^2\minus{}39x\minus{}3780\equal{}0$.  Factoring yields $ (x\minus{}84)(x\plus{}45)\equal{}0$, and since $ x$ is positive, $ x\equal{}84$, or $ \boxed{\textbf{(B)}}$.
[/hide][/quote]

May be slightly OT and don't mean to be a backseat moderator but this is plagiarism as it is word for word the solution in the AHSME Solutions Manual for 1997:

http://grab.by/p3u8

Otherwise, does anyone have some good proportion problems like this AHSME/AMC level that I could practice on? I'm particularly weak in 'directly proportional'.

TIA

Um, you have [b]all[/b] of the AHSMEs... there have to be a few of those problems in there :P I recommend starting from 1980ish.


geometry
rectangle
floor function
trigonometry
national olympiad

Who has 2001 Chinese Western mathematical olympiad? (chinese is ok)
I only have 2002 and 2003. :(

1) The sequence $\{x_n\}$ satisfies $x_1 = \frac{1}{2}, x_{n+1} = x_n + \frac{x_n^2}{n^2}$.  Prove that $x_{2001} < 1001$.

2) $ABCD$ is a rectangle of area 2.  $P$ is a point on side $CD$ and $Q$ is the point where the incircle of $\triangle ABC$ touches the side $AB$.  The product $PA \cdot PB$ varies as $ABCD$ and $P$ vary.  When $PA \cdot PB$ attains its minimum value,
a) Prove that $AB \geq 2BC$,
b) Find the value of $AQ \cdot BQ$.

3) $n, m$ are positive integers of different parity, and $n > m$.  Find all integers $x$ such that $\frac{x^{2^n} - 1}{x^{2^m} - 1}$ is a perfect square.

4) Let $x, y, z$ be real numbers such that $x + y + z \geq xyz$.  Find the smallest possible value of $\frac{x^2+y^2+z^2}{xyz}$.

5) Find all real numbers $x$ such that $\lfloor x^3 \rfloor = 4x + 3$.

6) $P$ is a point on the exterior of a circle centered at $O$.  The tangents to the circle from $P$ touch the circle at $A$ and $B$.  Let $Q$ be the point of intersection of $PO$ and $AB$.  Let $CD$ be any chord of the circle passing through $Q$.  Prove that $\triangle PAB$ and $\triangle PCD$ have the same incentre.

7) Find, with proof, all real numbers $x \in \lbrack 0, \frac{\pi}{2} \rbrack$, such that $(2 - \sin 2x)\sin (x + \frac{\pi}{4}) = 1$.

8) We call $A_1, A_2, \ldots, A_n$ an $n$-division of $A$ if
(i) $A_1 \cap A_2 \cap \cdots \cap A_n = A$,
(ii) $A_i \cap A_j \neq \emptyset$.
Find the smallest positive integer $m$ such that for any $14$-division $A_1, A_2, \ldots, A_{14}$ of $A = \{1, 2, \ldots, m\}$, there exists a set $A_i$ ($1 \leq i \leq 14$) such that there are two elements $a, b$ of $A_i$ such that $b < a \leq \frac{4}{3}b$.

[quote="zhaoli"]Hi, here is the problems of chinese western mathematical olympiad 2001.[/quote]

Why don't you post further past problem sets ?  ;)

[quote="zhaoli"]2001 is the first year to have CWMO![/quote]

Hmm. And what did you guys use before in order to qualify for the Chinese National Olympiad ?  :)

In general you could enlighten us about the competition structure in China. Thanks.

[quote="orl"][quote="zhaoli"]There are competitions in schools, provinces respectivly to select students to take part in Chinese National mathematical Competition.
And the best 3-6 students in every province were qualified for Chinese Mathematical Olympiad(CMO).
Then the best 20 or more students were selected to training camp.
At last, the best 6 students were selected for the IMO.[/quote]

So what is the CWMO ? Is it a province olympiad ? What about the problems of the other province olympiads ?  :)[/quote]
orl, the full name of CWMO is China Western Mathematical Olympiad. As you can see here, this is not a province olympiad. In fact, every western province in China will have a team for it. However, I don't think this is really a olympiad for western......

hello i have a solution for part one of problem 2 here goes:
well let angleAPB=x then we have that since (APB)=(ADP)+(BCP)=1 then we have PA*PB=2/sinx, then PA*PB obtains its minimum value when angleAPB=90 therefore when the circle with diameter AB  intersects CD if it does, then let d be the distance between the midpoint of AB  and the line CD. note that d=BC. also we have that the radius of the circle diameter AB wich is AB/2 is greater or equal than d=BC. So wehave our result. for part i have no solution. am i right??


Here's the first one:
How can u find the height of a building using aneroid barometer? :D

well...there can be $ \infty$ answers...but rem it is "Lateral thinking"
answer watever u think and then we'll find the best. :D

to start with,just [b]Post one solution at a time [/b] :)

[quote="rituraj007"]
aneroid barometer?
[/quote]

 :huh:  :?:

[hide="Roflmao"]will sell barometer to someone at exactly twice the money that i can convince a person to measure the height of the building for me  so that i finally know the height of the building and also have enuf money for my dinner :P :rotfl:  :rotfl:  :rotfl: [/hide]

drop it from the top of the building and calculate the time required(if you have a watch ofc)  :P  :D

[quote="hell_ever"][quote="rituraj007"]
aneroid barometer?
[/quote]

 :huh: [/quote]
well its not necessarily req to kno wat it is...since this is lateral thinking :p:

@par--one of the best answers :wink:
others shud continue to fow in :)

I'll take out [b]meter[/b] from barometer  :rotfl:  :rotfl:  :rotfl:

but how'll that measure height :P  :oops:

i think the [b]meter[/b] in the barometer will measure the height

the meter of barometer is just like that of ammeter--so how do u measure ht with that meter??? :P  :oops:


algebra
function
domain
advanced fields
advanced fields unsolved

For each of the following, state whether the statement is True or False. If True, explain why in English. If false, find a counterexample (that is, find statements P(x) and Q(x) and a domain D such that the hypothesis is true and the conclusion is false).

a. If [ (For all x in D) P(x) v (For all x in D) Q(x) ] then ( For all x in D) (P(x) v Q(x))

b. If (For all x in D) (P(x)) v Q(x)) then [ (For all x in D) P(x) v (For all x in D) Q(x) ]

a) How is the conclusion false? $\neg\forall x\in D(P(x) \vee Q(x))\Rightarrow \exists x\in D\neg (P(x) \vee Q(x))\Rightarrow \exists x\in D(\neg P(x) \wedge \neg Q(x))$, but there cannot be an $x$ in $D$ where both $P(x)$ and $Q(x)$ are false, since by the hypothesis either $P(x)$ is true or $Q(x)$ is true.

b)How is the conclusion false? $\neg[(\forall x\in D) P(x) \vee (\forall x\in D) Q(x)]\Rightarrow (\exists x\in D) \neg P(x) \wedge (\exists x\in D) \neg Q(x)$, but this can't happen since either $P(x)$ is true or $Q(x)$ is true for all $x\in D$.

[hide="Stupid joke"]If this is "discreet" math does that mean we shouldn't tell anyone about it?[/hide]

Actually I am pretty sure that all math jokes are a subset of 'discreet math'.


Let's have r = red and s = something else. How can we draw two reds in three turns? rrs, rsr, and srr, of course. Since you only considered the rrs case, you need to multiply your answer by three.

I had these type of problems when I started mathcounts  :maybe:   Well, the other guy already explained it most of it... This is just a furthur detailed explanation: If you have 3 colors and you choose 2 of them, how many ways can you choose them? It's just 3 C 2... If you learn Combination, this will be ezpz... The combination formula is: n!/(n-r)!*r!  .... I wish I could use the other complicated (and even more organized) text as the other experts, but I'm new, and I want it to be ez.  So in this case, u plug it in and u get...

  3!/(3-2)!*2! = 3!!!

So.. does this sound ez to u?

[quote="returnofkiladi"]A bag contains 10 buttons (4 red, 1 blue, and 5 white). If Sally chooses 3 buttons, randomly, without replacement, what is the probability that she chooses exactly 2 red buttons?

It's confusing because I thought that the answer is 1/10 because there are currently 4/10 red marbles, then after I take 1 out, I'll have 3/9 red marbles, and then the next time, I don't want a red marble, so I multiply the result by 6/8 (since there are 2/8 red marbles).

2/5 * 1/3 * 3/4 = 1/10, but the answer is 3/10, why is that?[/quote]
[hide="Hint/answer"]Remember, there are 3 ways you can order that. If R=red button and S=something else, it can be RRS, RSR, or SRR, so it's $ \frac{2}{5}\times\frac{1}{3}\times\frac{3}{4}\times3\equal{}\frac{3}{10}$[/hide]


geometry
circumcircle
trigonometry
inequalities
rectangle
exterior angle
geometry unsolved

i am not sure if this uses circular logic, but it does not seem to

if they are all equal, then this is trivial, otherwise

WLOG AB>AC, let $D$ be on $AB$ so $AC=AD$

let $\angle BAC=2x$, then $\angle ADC=\angle ACD=90-x$, and $\angle CDB=90+x$, by exterior angle theorem, $\angle ABC <90-x$, so
$\angle CDB>90-x>\angle ABC$, and that implies that $BC>DB$, adding $AC$, $AC+BC>AB$

[quote="Ashegh"]ok here is a simple proof :rotfl:

Prove that in each triangle $ABC$, we have these facts:

$AB+BC>AC$

$AB+AC>BC$

$BC+AC>AB$

I like this theorem. maybe this post wont be a basic prove, how ever I like to prove

It by inversion.

suppose $O$ is a point on circumcircle of $ABC$, in which

$OB=OC$, I choose $O$ as the center of inversion.(consider that $O$ can

be each point on circumcircle,but I choose it with this property, to be more

comfortable).

Before going on the inversion, I write a lemma:

Suppose $O$ is center of inversion. And $AB$ is a segment,then the length of

$A'B'$ is :                                    $A'B'=\frac{K.AB}{OA.OB}$

($K$, is the radius of inversion. Here is $OB^2=OC^2$)

……………………………………………………………………..

ok , lets start to invert the shape.i want to know what is the length of$A'B',B'C',A'C'$.

as the lemma said, we have these facts:

$A'B'=\frac{OB^2 .AB}{OA.OB}=\frac{AB.OB}{OA}$

$B'C'=\frac{OC^2 .BC}{OC.OB}=BC$

$A'C'=\frac{OC^2 .AC}{OA.OC}=\frac{AC.OC}{OA}$

it is knows that $A'B'+B'C'=A'C'$.(because $A',B',C'$, are collinear)

lets get instead of them:

$\frac{AB.OB}{OA}+BC =\frac{AC.OC}{OA}$

lets do multiple this equation at $OA$.then:

$AB.OB +BC.OA=OC.AC$

As we know $OB=OC$, then lets divide this equation at $(OB=OC)$.

The we have:      $AB+\frac{BC.OA}{OB}=AC$

Ok …before going on the next part I want to say a lemma:

Suppose $(C)$, is a circle. $O$ is a point on exterior side of $(C)$.$OA$ is tangent to

$(C)$, and an arbitrary line which goes through $O$, intersects $(C)$, at points

$B,C$ respectively.($B$ is nearer to $O$).

Prove that:  $OB<OA$   and $OC>OA$.

Ok .we know that, $OB.OC=OA^2$. Suppose $OB<OA$  and $OC<OA$.

Then we should have: $OB.OC<OA^2$  , which is wrong.

Again suppose :  $OB>OA$   and $OC>OA$, then we should have:

$OB.OC>OA^2$, which is also wrong.

Then one should be less and another should be more than OA.

And because $OB<OC$, we conclude that: $OB<OA$   and $OC>OA$

…………………………………………………………………………….

Ok . according to the inversive shape, we know that: $OA.OA'=OB^2$

Then according to lemma we will have:    $OA<OB$  and $OA'>OB$

$OA<OB$, then $\frac{OA}{OB}<1$. Do multiple at $BC$. Then:

$BC>\frac{BC.OA}{OB}$.

I have been said that:  :      $AB+\frac{BC.OA}{OB}=AC$

Then if we choose $BC$, instead of $\frac{BC.OA}{OB}$, the left side of

Equality will get larger, and we have:

$AB+BC>AC$

with similar way we can conclude that :

$AB+AC>BC$

$BC+AC>AB$

solving easy problems easily is very good, but solving them with very hard ways

have special enjoy.and I like it…. ;)

.[/quote]

wow ...

inversion? :!:   :?:   :idea:   :blush:   :|   :huh:   :ninja:

what a simple solution

what is inversion?????

i hanged....   system failur....

why did u hang?

every thing is ok my darling

ask each part, and then i will explain it to u...

dont worry dear...ok? ;)  :D

dear $Ashegh$, i cant figure out u problem,then what should i do.

to learn the tools u used to prove, it take me too times.

i think i should wait for a nother solution :D

shobber s solution was nice. :P

Ashegh, how many strange solution u have for this problem?

because my theacher appreciate u solution.

 :clap2:

[quote="britney spears"][quote="Davron"]:)

All we know that if $AB,BC,CA$ are the sides of a triangle, then they can be written in this form :
$(*)$
$AB=x+y$
$BC=y+z$
$CA=z+x$.

Now we need to prove that : $AB+BC > CA$ which is $x+y+y+z \geq z+x$ which is obviously true,
since $2y > 0$.

If you need the proof of $(*)$ i can show it.

Davron[/quote]

hmmmmm...

is it one of the facts of incircle?[/quote]

Yes, it is.

[quote="britney spears"][quote="Ashegh"][quote="britney spears"]shobber s solution was nice.
Ashegh, how many strange solution u have for this problem?

because my theacher appreciate u solution.

 :clap2:[/quote]

hmmmm...

really :D .................

hmmm...

with ptolemy inequality it will also be proved.my way to prove this problem by invrsion, was exactly the way of

ptolemy equality by inversion. :!:  :?:  :!:

since know i dont have another one[/quote]

ashegh could u plz post ure solution ?[/quote]

hmmmmmmmmmmmmm..............................

ok ok ok ok ok ok ok ok ................................

suppose $ABC$ is atriangle and $O$ is its circumcircle.

we wrote ptolemy inequality for quadrilaterall $ABCO$.

we have:

$AB.OC+AC.OB>AO.BC$

and because $BO,CO,AO$ are radius and are equal we can devide the inequality at $BO,CO,AO$.

then we have: $AB+AC>BC$...

but as u know this proof may be wrong.

because each of the proofs which i saw about ptolemy inequality ,have used this fact that: $AB+AC>BC$

then it is clear that we cant proof this theorem by ptolemy inequality ;)

[quote="Ashegh"][/quote][quote="britney spears"][quote="Ashegh"][quote="britney spears"]shobber s solution was nice.
Ashegh, how many strange solution u have for this problem?

because my theacher appreciate u solution.

 :clap2:[/quote]

hmmmm...

really :D .................

hmmm...

with ptolemy inequality it will also be proved.my way to prove this problem by invrsion, was exactly the way of

ptolemy equality by inversion. :!:  :?:  :!:

since know i dont have another one[/quote]

ashegh could u plz post ure solution ?[/quote][quote="Ashegh"]

hmmmmmmmmmmmmm..............................

ok ok ok ok ok ok ok ok ................................

suppose $ABC$ is atriangle and $O$ is its circumcircle.

we wrote ptolemy inequality for quadrilaterall $ABCO$.

we have:

$AB.OC+AC.OB>AO.BC$

and because $BO,CO,AO$ are radius and are equal we can devide the inequality at $BO,CO,AO$.

then we have: $AB+AC>BC$...

but as u know this proof may be wrong.

because each of the proofs which i saw about ptolemy inequality ,have used this fact that: $AB+AC>BC$

then it is clear that we cant proof this theorem by ptolemy inequality ;)[/quote]

really? are u sure, all of proofs for pytagorian, have used this fact?

[quote="M4RI0"][quote="britney spears"]can anybody please show me some simple proofs for this theorem?

in each triangle $ABC$ ,it is known that:

$AB+BC>AC$

$AC+BC>AB$

$AC+AB>BC$

i know it may be easy problem.but...pleas help me :blush: .[/quote]

Let $ABC$ be an acute triangle:

$P$ is the feet of the altitude $AP$, then we get two rectangle triangle, we know the hypotenuse is the greatest side(from the Pytaghoras Theorem), then:
$AB>BP$ and $BC>PC$  :arrow:  $AB+BC>BP+PC=BC$

For the cases for an obtuse triangle we proceed the same way.[/quote]

how nice :)

can u prove the other case?plz?

[quote="britney spears"][quote="Ashegh"]i didnt say pytagorian theorem,

i said ptolemy inequality.

if u mean that, im not sure, because i think , i dont say all of its proof.

but it seems that in all of the proofs it is used that:$AB+AC>BC$[/quote]

u think right, my teacher said,all of ptolemy proofs had used that fact ;)[/quote]

maybe ure teacher is right.

but i think she conclude this fact as me.

how many proofs, she saw for this theorem?

did she read all of them?

or no...

she only guess that...

 :roll:   :huh:


Yeah, that's a good method, but I got a different answer.

[hide="my answer"]
63
[/hide]

[quote="Treething"]I believe that the length of that chord is $\sqrt{15}$, meaning that $x = \boxed{15}$

Using similar triangles, the chord is $\frac12$ away from its parallel diameter.[/quote]

[hide]
If the chord is $\frac12$ away, then shouldn't the length be $2*\sqrt{4^{2}-{\left(\frac12\right)}^{2}}= \sqrt{63}\implies x=63$?
[/hide]

Well, beaten to it.

Arghahrawifjadoigjfasodijaslkfja this is like... my sixth careless mistake out of six problems?

I accidentally labeled the diameter as 4... yeah, I got 63.

wow... treething you are reading stuff wrong... it happened to me one day in school while practicing a practice vestavia in which i got an 18 on it. I got a 52 and a 64  two weeks later, proving how bad I was doing on problems because of carelessness and lack of sleep. I read a number that asked for the longest distance and i calculated the shortest distance.  :o  I made silly mistakes on 9 problems lolol oh well


inequalities
geometry
area of a triangle
Heron\u0027s formula
inequalities proposed

If A,B,C are the angles of a triangle prove that

 \sum (sinA) 2 / \sum (cos(A/2)) 2  \geq

sinAsinBsinC/cos(A/2)cos(B/2)cos(C/2)

\sum (sinA) 2 = \sum a 2/4R 2m where  a,b,c are the sides of the triangle and R is the radius of the circumcircle.

cos(A/2)= \sqrt (p(p-a)/bc), where p is the semiperimeter of ABC.

 \pi sinA=area[ABC]/2R 2  and  \pi cos(A/2)= \sqrt (p 2p(p-a)(p-b)(p-c)/(abc) 2).

After computations using Heron's formula we must prove
 \sum (a 2 ) \geq 2 \sum a(p-a). If we replace a,b,c by y+z,x+z,x+y we must prove
 \sum (x+y) 2  \geq 2 \sum x(y+z), or  \sum x 2  \geq  \sum xy which is well known.

I would say "ah ce scarbos" which means "ah so yucky" :D

The ideas I used are natural.
Math is not always nice and playful  :(

This is the type of inequalities I dislike most. They are so simple, but they look like hell. And they have no interesting idea. I do like inequalities that look like a monster, but they should have an intelligent idea and should be hard. This one is simple, stupid and ugly. Ihhhhh!


Heyyy...
I"m new here and I just accidently stumbled on this site, and I thought that this site is cool.
Unlike everybody else, I've never been in a math competition, but there's one I want to try to compete next month. I've been studying math for quite a while, and i got this trig question from my pre-calc class:

My teacher talked about Principal Inverse Function, and I didn't get it much except every trig functions have some inverse function such as arccos, arcsine, etc. Also, she said there has to be ONE value, and what one value?

Would you guys explain this concept to me in a language that would be easy for me to understand, and no techinal terms, unless you define it. I'm dumb, and needs some time to sink it in.

Hey welcome to this place, im seamus.

Say your looking at $\sin{x}$

Notice that $\sin{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$

Also notice that if we change the number in the $\sin{}$ part by $2\pi$, we get the same thing. So for example:

$\sin{(\frac{\pi}{4}+2\pi)}=\sin{\frac{9\pi}{4}}=\frac{1}{\sqrt{2}}$

So what is $\arcsin \frac{1}{\sqrt{2}}$?

For above, it can be either $\frac{\pi}{4}$  or $\frac{9\pi}{4}$

Since we dont like this, it can get in the way of otherwise nice theorems, we mathematicians decided that we'd only let it have one value. In this case $\frac{\pi}{4}$

All clear?  :P

Thanks for clearing this up!! Thanks!!!!!


Prove that the set M of all five digits integers where each of the digits 1,2,3,4,5 occurs exactly once can be separated into two subsets M_1 and M_2 such that:

\sum_(x \in M_1) x^2 = \sum_(x \in M_2) x^2 .

For each five-digit number abcde we consider another five-digit number edcba. It is clear that if xyzuv is a cyclic shift of abcde and iklmn is a cyclic shift of edcba then xyzuv<>iklmn. So for each abcde we sum up squares of all cyclic shifts. It is easy to show this sum equals the analogous sum for edcba.

That's all.


This is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck onThis is a question I am stuck on

[quote="mstoenescu"]hi !

$DC$ is not equal to $1.7$ ! $DC=\sqrt{3}$ !

Use trigonometric functions in triangle $BDC$ ![/quote]

 lol you're happy =)

umm on another note, learningfish, please do not spam saying you need help over and over...we get the message =)

Or... without trig...

[b]Hint[/b]

[hide]Use your knowledge of special right triangles[/hide]

[b]Answer[/b]

[hide]You know that X is 45, so BD=AD = 1. Then you have right triangle BDC weith BD = 1 and DC = rt3. Use your knowledge of 30-60-90 right triangles to figure out z is clearly 30![/hide]


Let $\mathbb{P}$ be the set of primes and $\mathbb{N}$ be the set of natural numbers
then $|\mathbb{P}|=|\mathbb{N}|$ where the symbol $|\mathbb{A}|$ denotes the cardinality of $\mathbb{A}$

note that $|\mathbb{N}| = \aleph_0$
let $\mathbb{S}=2^\mathbb{P}$ (The power set of $\mathbb{P}$ or the set of all subsets of $\mathbb{P}$). Thus we know that $|\mathbb{S}|=|\mathbb{R}|=\aleph_1$

However, define mapping $u: \mathbb{S} \mapsto \mathbb{N}$ such that each element of $\mathbb{S}$ maps to to the product of all its elements.
eg. if $\mathbb{A} = \{2, 7, 11\} \in \mathbb{S}$, then $u(\mathbb{A}) = (2.7.11) = 154$

Note that this mapping is injective.
However note that $\mathbb{S}$ is a proper subset of $\mathbb{N}$ since only the numbers which are not product of prime powers are in $\mathbb{S}$.
Thus $|\mathbb{S}| < |\mathbb{N}|$ and no bijection exists from $\mathbb{N} \mapsto \mathbb{S}$ (we cannot map to numbers like $2.2.3=12$ using the mapping $u$)

Isn't this a contradiction?

[Note: I edited this post to make it latex compatible]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49289

darij was right, y'know; no need to look for second opinions

The mapping $u$ isn'tdefined on all of $S$ because you can't take the product of an infinite set of primes.  Your argument shows that the set of [i]finite[/i] subsets of $P$ is countable.  Incidentally, $\aleph_1$ is the next largest cardinal after $\aleph_0$, not necessarily $2^{\aleph_0}$ --- that's a form of the [url=http://en.wikipedia.org/wiki/Continuum_Hypothesis]Continuum Hypothesis[/url].

I read the link...
Just have a question, hopefully it doesn't sound too stupid...
Is it true that if we define $\mid \mathbb{R} \mid = 2^{\aleph_0}$ as having cardinality $\aleph_k$ for any $k \in \mathbb{N}$, $k$ not necessarily $1$, it is still consistent with the Zermelo-Fränkel axiomatic system and of the axiom of choice? What are the implications if $k>1$, or does it really matter?

Even I was surprised when I first found out.

$\aleph_1 = \aleph_i$ for all integers $i > 1$.

The first cardinal greater than $\aleph_1$ is $\aleph_\omega$ where $\omega = \aleph_0$


In the figure shown, segments AB and CD are parallel.
What is the sum of the measures of angles BAE, AEC
and ECD? http://i42.tinypic.com/azgt3b.jpg

Draw BD.  Because the sum of angle measures in a pentagon is 540, and angle ABD+angle BDC=180 by same side interior angles, the other three add to 360.

If you draw a line segment from A to C, you get triangle AEC, the sum of whose interior angles equals 180 degrees. Then since AB and CD are parallel, the same side interior angles add up to 180 also. So the answer is 360 degrees.


How many of you have thought about what you are actually saying when you pledge allegiance to the flag at school in the morning?  My US Government teacher talked about this topic in class recently and I found it interesting so I thought I would start a conversation on the board.

Just think about each part and what it means.  Have you ever thought about the pledge or do you just say it?

Personally, I have thought about this before my teacher brought it up.  However, I am pretty sure that many, probably most, don't even give the pledge a second thought.  You could extend this to many other instances, not just pledging allegiance to the flag.  For example, although this is a touchy issue, religion.  Many people go to church just because their parents did it and it is engrained in their brain that it is the 'right' thing to do.  Or what about just when someone sneezes.  It is customary to say "bless you".  I wonder how many people have thought about that, or just say it because they don't want to be rude.

I'm not saying I am for or even against any of this.  I am merely trying to point out the growing trend of indifference that has become the norm of society.  More often than not, people only care enough to get what is good for them.  This is human nature, and it can't be changed really.

Ok, so hopefully we can get a discussion started about this.  I would like to hear what everyone else thinks since I've pretty much only heard my teachers opinion and the opinion I have.  Also, he lent out this book called [i]The Children's Story[/i] which is nice little story about how children are usually just taught the pledge without being given it's meaning.

Here is a short history of the Pledge of Allegiance - http://history.vineyard.net/pledge.htm

And another account of its history - http://www.homeofheroes.com/hallofheroes/1st_floor/flag/1bfc_pledge.html

I'm not a fan of the "daily dose of loyalty". but its not like i got a choice.

here are some interesting lyrics that can be related to this topic.
"Ya standin' in line
Believin' the lies
Ya bowin' down to the flag
Ya gotta bullet in ya head"

I have noticed both the allegiance thing and the nose thing. I believe that everywhere where the world God in a government document is mentioned should be deleted because it is not proper for the US government to state they believe in God. As far as sneezing, if you say "bless you" instead of "God bless you" that shows that you r not talkin about God so that should do even though u might start arguing about the bless but that doesnt count.

[quote="nonie"]hey, I just realized that we don't say the pledge in my middle school. We used to in elementary though... I've never really thought about it means or what it impliew. You don't HAVE to say it you can just mouth it or just stand up o r whatever.[/quote]

Students in many schools are required by law to say it.

I don't say the pledge; I haven't recited it since I was forced to in 8th grade.

The pledge of allegiance is absolutely useless. You can prrove you're a good citizen by being active in civil society: participate in political discussions, vote, do something. All those things are much better than saying something which, I think, pretty clearly violates the lemon test(a test to decide whether something breaches separation of church and state)

The Court's decision in this case established the "Lemon test", which details the requirements for United States legislation concerning religion. It consists of three prongs:

   1. The government's action must have a legitimate secular purpose;
   2. The government's action must not have the primary effect of either advancing or inhibiting religion; and
   3. The government's action must not result in an "excessive entanglement" of the government and religion.
-curtesy of wikipedia

Never gave it a second thought, except for maybe in second grade, where we had to do this sickeningly patriotic project about the Pledge.

agolsme, I'm not sure I understand the link between the pledge and the lemon test.  Can you explain?

[quote="MCrawford"][quote="furious"]I'm not a fan of the "daily dose of loyalty". but its not like i got a choice.[/quote]

It's ironic that you have no choice in saying the pledge in a "free country".  It's more than ironic, it's sad.

A pledge is not a pledge if it's not made voluntarily.  When it's made involuntarily, it's propaganda.[/quote]

I'm pretty sure schools aren't allowed to force children to say the pledge...I don't, at least, and I don't get in trouble.

When I just moved to America from Canada, I didn't know the pledge, so I sorta "faked" it.

Now that I know it, I sorta just say it without a second thought. We don't have to say it in middle school.

[quote="MCrawford"][quote="furious"]I read the book that Joe talked about in the first entry : [i]A Children's Story[/i] by James Clavell. I really liked it as it shows how children can be brainwashed and particularly how most children don't even know what they are saying when they are pledging.[/quote]

It's easy to brainwash people.  Within a large group of people, most of them will believe anything their culture tells them during their formative years -- unless they're taught how to think to high levels.[/quote]
People with fanatical religious and political beliefs, see above.

[quote="MCrawford"]Also, something that few people seem to be responding to is the fact that there are many states that have laws on the books now that do require the pledge regardless of religious beliefs.  Several more have proposed such laws and they may be enacted soon.  I feel that such a movement is worthy of some discourse.  What will its effects be and how much of a chance does it have of succeeding?[/quote]

I find that disgusting and highly pointless. After all, a student has the option of whispering it or mouthing it without actually "saying" it. Such a law would be impossible to enforce.

All of my classes are messed up. Nobody does the pledge...

Another interesting point is that the question remains: when you "pledge allegiance to the flag", to what are you ~really~ pledging allegiance? Am I pledging allegiance to my country? I love my country. I love the unprecedented ideals upon which my country was founded. Or am I pledging allegiance to my government? I believe the current government to be a disgrace (and no, I'm not inviting a political debate, I'm just stating my opinion). I would never pledge allegiance to the government, at least not unless some serious change happens. I say the pledge when invited to (which is rarely), because I love my country, despite its many problems. But I always make a mental point to remind myself that just because I love my country doesn't mean I love the Bush administration and the conservatism that runs the place.

Just thought that was an important question to bring up.

Just a note: It's lawful to stand in respect, but one does not have to recite the pledge. That's the ruling before the current issues.

Most people don't consider why the phrase "under God" is used. God isn't used as some representation of ecclesiastical power over the government, it's used more in terms of one nation, spiritually united. It's not propaganda. Sure, such a statement could affect the learning process of children growing in America, but it's not the same as a statement, "one nation, who does not believe in God" because it refers to a specific God as compared to the idealism which the phrase "God" represents in the original. That's my take on it. We should not abolish it. There's no basis except for a few atheists who are too bored to mind their own matters and decide to create a fuss. I mean, if Christians who go to atheists and tell them, "you are going to [that place]" are immoral, so are atheists who just like to bother a Christian community (which is what the United States more or less is - choose whether or not to believe this but fifty percent of Americans are Christians) by telling them, "the thing you believe in should be abolished from the following object because I don't like it." Sure, there's separation of church and state, but only for the purpose of thwarting a theocracy. The religious ideals govern the layer behind the actual politics. Don't you think that other factors (race, sex, economic status, AND religion) all play a subliminal role in one's decisions, especially those of our leaders? Is it right to just say, the traces of the largest religion should be abolished from any secular matter when it inheres in every matter along with other minority religions and the "pseudo-religion " of atheism?

I don't think the pledge is a good thing. Probably a little unhealthy. But people get way too upset about "under god". Most religions have a god. And if yours doesnt, and it upsets you, just don't say the pledge.

This case is reasonable.

If you don't want to say: "Under God.", then don't say it. That's true. People around you are saying it, though. It may irritate you. Schools shouldn't force students to say the Pledge of Allegiance. The original one didn't even have "Under God".

I want to stop reciting the Pledge. It is meaningless. Lots say it for the heck of saying it. Very few mean it.

[quote="The Pledge of Allegiance"]I Pledge Allegiance,
to the flag,
of the United States of America,
and to the republic,
for which it stands,
one nation, " "
Indivisible,
and justice for all.[/quote]

I don't even know what is means.

People from my school recite it every day. No offense, but I think it is a waste of time.

Well, keep in mind that law prohibits public schools to force you to say the pledge.

I get to say it in Spanish every day this year. I don't really mind saying it; it's just a routine thing to do. I doubt much would be lost by not saying it just as not much is gained by saying it. Well after you're out of school, it's unlikely you will continue to recite it every day anyway, so I'd say just put up with it for a few more years if you don't like it.

I simply refuse to say the pledge of allegiance every morning.  I stand up, but I don't even try to pretend that I say anything or put my hand on my heart.  I believe that putting "under god" in an important document is unconstitutional and that it is also unconstitutional to require anybody to recite it, whether it says "under god" or not.

Exactly samath.

I just stand there while everybody else chants it like mindless zombies. I don't see the need to pledge to a flag belonging to a nation of God. I might as well go to church on Sunday and pray there. Besides the fact that political doctrines highly influenced by religion are not necessarily the most fair and objective things...

I think that the people that are saying "Oh look, they're saying it, they must believe it!" are just seeing a bunch of people who don't know or care what they are saying.

[quote="mathclass"]in all honesty i never think about what the pledge says. not until whatshisname stood up and said he didn't want his daughter saying "under god." i AM loyal to the united states, but as for a lot of people, our loyalty doesn't really show until something drastic happens.

really, the pledge is pointless. no one cares. in our class every monday we mumble the pledge with our minds on something else. maybe we could do something that really has meaning (like community service) instead of an empty piece of fluff.[/quote]

so true.  the pledge is just something we memorized in kindergarten.  who cares?  and for anyone who has actually thought about it, liberty and justice for all?  where did that come from?  indivisible?  politics defies this.  we have to have laws for "liberty and justice"  but politics divides us into sides.  contradictory.  We really should be doing.

simple math:  the pledge is about 10 seconds long.  We say it 5 times a week for about 40 weeks a year.  thats 2000 seconds, which is about 33 minutes.  just in one school alone, there are usually 500 to 1200 kids.  just think about how much the world would improve if every single person who said the pledge just helped on a community project for 33 minutes!!

I've been saying it ever since I learned how to say it in sixth grade.

I'm not religious but saying the pledge does not bother me one bit.

[quote="MCrawford"][quote="furious"]I'm not a fan of the "daily dose of loyalty". but its not like i got a choice.[/quote]

It's ironic that you have no choice in saying the pledge in a "free country".  It's more than ironic, it's sad.

A pledge is not a pledge if it's not made voluntarily.  When it's made involuntarily, it's propaganda.[/quote]
No one can force you to pledge allegiance to the flag. Just keep your mouth closed if you don't want to say it! What you said about the propoganda vs. pledge is true though. I wonder how many of us would actually do that on our own if we weren't "required" to in certain situations.


solve:
$ sin^2x\minus{}\frac{5}{2}sinx\minus{}\frac{3\sqrt{3}}{2}sin(x\plus{}\frac{\pi}{4}) \equal{} 0$

hello, using the addition formulas we get
$ \sin^2(x)\minus{}\frac{5}{2}\sin(x)\minus{}\frac{3}{4}\sqrt{6}\sin(x)\minus{}\frac{3}{4}\sqrt{6}\cos(x)\equal{}0$.
By substituting $ t\equal{}\tan(\frac{x}{2})$ we have to solve
$ 9t^4\minus{}(10\sqrt{6}\plus{}18)t^3\plus{}8\sqrt{6}t^2\minus{}(18\plus{}10\sqrt{6})t\minus{}9\equal{}0$
Sonnhard.


Was flipping through my notes once again and saw something jotted down in the corner: "Does there exist a sequence $ { a_n }$ such that $ LIM(a_n)=\bb{R}$?" (Note that in this context $ LIM(a_n)$ denotes the set of limit points of $ a_n$, that is, the set of all points to which some subsequence of $ a_n$ converges to. Thinking about this from an intuitive perspective, I think the answer is no, here is my reasoning:

1. $ a_n$ must diverge, since if it converged, $ LIM(a_n)$ would equal whatever $ a_n$ converges to.

2. $ a_n$ must be bounded, since if it was unbounded then no subsequence of $ a_n$ would be bounded and hence none would converge. (Is it true that there would be no bounded subsequence? I think I can find an argument through contradiction but I haven't thought about this too hard, this may be where my reasoning is flawed).

3. If $ a_n$ must be bounded, then this intuitively tells me that there exists no such sequence which satisfies our property (I have no solid proof of this either... maybe I should think about this harder too  :P ).

Any help appreciated!

[quote="max_tm"]$ a_n$ must be bounded, since if it was unbounded then no subsequence of $ a_n$ would be bounded[/quote]
False. Consider $ 1,\frac{1}{1},2,\frac{1}{2},3,\frac{1}{3},4\dots$

The usual way to obtain such a sequence is to enumerate all rational numbers. Such an enumeration is possible because the set of rationals is countable (this is really a tautology). If you want an explicit enumeration, you can list all rational numbers of the form $ \frac{m}{n}$ with $ |m|\plus{}|n|\equal{}1$ (well, there aren't many of those), then those with $ |m|\plus{}|n|\equal{}2$, then those with $ |m|\plus{}|n|\equal{}3$...

If you need an intuitive example, just look at the sequence

$ a_n \equal{} \{ \boxed{\minus{}1, \minus{}0.9, \minus{}0.8, \cdots, 0.8, 0.9, 1}, \boxed{\minus{}2, \minus{}1.99, \minus{}1.98, \cdots, 1.98, 1.99, 2}, \boxed{\minus{}3, \minus{}2.999, \minus{}2.998, \cdots, 2.998, 2.999, 3}, \cdots \}.$

Let $ r$ be any real number and $ N \plus{} \sum_{k\equal{}1}^{\infty} \frac{c_k}{10^k}$ be the decimal expansion of $ r$. (Of course, $ N \in \mathbb{Z}$ and $ 0 \leq c_k < 10$.) Then $ b_n \equal{} N \plus{} \sum_{k\equal{}1}^{|N|\plus{}n} \frac{c_k}{10^k}$ is a subsequence of $ a_n$ convergent to $ r$.

[quote="mlok"]The usual way to obtain such a sequence is to enumerate all rational numbers.[/quote]

This is because $ \mathbb{Q}$ is dense in $ \mathbb{R}$ right?


geometry
analytic geometry
geometry proposed

IN triangle ABC, K,L and M are points on BC,CA and AB respectively

AK,BL and CM meet at P

Triangles BKP,CLP and AMP have equal areas

Prove P is the centroid of triangle ABC

This means that $P$ has barycentric coordinates $x:y:z$ with $x,y,z\in\{\pm 1\}$, and it's easy to see that if two of $x,y,z$ have opposite signs, then two of the points $K,L,M$ will be improper points (infinity points), so this only leaves $1:1:1$ as barycentrics of $P$, and this defines the centroid.


What is the range of the parametric equation

$x(t)=t^{2}-1$
$y(t)=t^{2}+1$
$z(t)=1-t$

If you mean how look like this curve in spase consider projection of your curve to the plane z=0 and those to y-x=2.
They are evidently a part of a line and a two peaces of parabola (for t>0 fnd for t<0).

The same problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=117003]here[/url].


geometry
parallelogram
probability
calculus
integration
analytic geometry

$ABCD$ is a parallelogram on a $2$dimensional plane. The co-ordinates of $A$, $B$, $C$ and $D$ are $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, $(x_{3}, y_{3})$ and $(x_{4}, y_{4})$ respectively. The fact of the matter is that all those co-ordinates are integers.

What is the probability that a parallelogram, whose corners lie at points that have integral coordinates, is having an area that also takes a integral value?

[hide="Pretty good"] WLOG every such parallelogram is congruent to a parallelogram of the form $(0, 0), (a, b), (c, d), (a+c, b+d)$, which has area $|ad-bc|$, an integer. [/hide]


function
induction
number theory proposed
number theory
functional equation

Find all functions $ f: \mathbb Q\longrightarrow\mathbb Q$ such that:
$ f(x)+f(\frac1x)=1$
$ 2f(f(x))=f(2x)$

We can think of a graph of $ \frac{n}{m} $ and $ m+n $ is costant. This is very crucial. ( Or, You can use $ Ord2 $ )

[quote=Rust]By induction $ f(x)=\frac{x}{x+1}.$[/quote]
[quote=Omid Hatami]Yes, your answer is correct. But the main problem is how to use induction, and it is not easy though.  :)[/quote]
No, his answer is wrong (not to mention useless "solution").

The answer is
$$ f(\frac{n}{m})=\frac{n}{|m|+n}\text{ } (n>0)$$
and $f(0)\in\{0,1\}$
This is a very nice problem :)

[quote=MNJ2357]The answer is
$$ f(\frac{n}{m})=\frac{n}{|m|+n}\text{ } (n>0)$$
and $f(0)\in\{0,1\}$[/quote]
Well, after some fix of the problem statement, you're right. But if the proposer [url=https://artofproblemsolving.com/community/c6h1167559]forgot to say that the domain[/url] of the first equation is $\mathbb Q \setminus \{0\}$, then [i]pco[/i] or any other member has every right to say that the problem has no solution, like [url=https://artofproblemsolving.com/community/c6h1508308]here[/url].

[quote=MNJ2357]This is a very nice problem :)[/quote]
I agree. So why no one posted a solution yet? There are at least 5 copies of this problem on the site, this particular topic is 11 years old. I guess some people think that knowing a buzzword like [i]induction[/i] would won them maximal number of points at a real Olympiad automatically. :D

[quote=MNJ2357]The answer is
$$ f(\frac{n}{m})=\frac{n}{|m|+n}\text{ } (n>0)$$
and $f(0)\in\{0,1\}$[/quote]
I thought the same thing for a while. But now it appeared that even this is wrong :oops_sign:
This answer is right for the case $f \colon \mathbb Q^+ \to \mathbb Q^+$ or at least $f \colon \mathbb Q_{\ge 0} \to \mathbb Q$. But for this particular variant of the problem, when $f \colon \mathbb Q \to \mathbb Q$, this answer is wrong! See the details at [url=https://artofproblemsolving.com/community/c6h1704861]the revival attempt[/url].

The original statement presented at the 1991 Irish Olympiad contained $\mathbb Q^+$ as both domain and codomain.
[hide=For a solution]follow the white rabbit [aops][url=http://artofproblemsolving.com/community/c6h1521559p10979015]>[/url][/aops][/hide]


set theory
combinatorics proposed
combinatorics

Is there an ordinal greater than which all ordinals are indescribable? Prove your result.

I'm not sure exactly where to post set theory problems, but combinatorics should work.

Hmm... maybe you could explain when "ordinals" are considered "indescribable".

Either you mean some other word, or it's higher university stuff in which case it should be in college playground...


The periodic functions f(x):  R -> R and g(x): R -> R satisfy the condition:
lim (f(x)-g(x))=0. Prove that f(x)=g(x) for every real number x.

Here?s a solution:
Suppose T1 and T2 are the periods of f and g respectively. For a fixed x lim(f(x+nT1)-g(x+nT1)) =0 as n->infinity, so lim g(x+nT1)=f(x) as n->infinity. By the same way lim f(x+nT2)=g(x) as n->infinity. We also have lim(f(x+nT1+nT2)-g(x+nT1+nT2)) =0 as n-> infinity, so lim(f(x+nT2)- g(x+nT1))=0 as n->infinity, but this limit is g(x)-f(x) (this is for a fixed x), so g(x)-f(x)=0 for any x. It?s not the first time I?m proving this, but this is the only proof I know. I haven?t struggled to find another one though. Any other proofs?

Here is something i've noticed:
Let T(x)=f(x)-g(x).
As x tends to infinity T'(x) tends to 0.
This means that as x tends to infinity, T(x) tends to be a constant. The periodicity thing wud give us that it is indeed a constant, and that T(x)=0, any x \varepsilon R


limit
real analysis
real analysis unsolved

Let $ (a_n)_{n \geq 0}$ be a sequence such that: $ a_0 \in (0,1)$ and $ a_{n \plus{} 1} \equal{} a_n^3 \minus{} a_n^2 \plus{} 1$. Evaluate the limit $ \lim_{n \to \infty}{a_0 \cdot a_1 \cdot \dots \cdot a_n}$.

It is obvious that $ a_n \in (0,1)$ , $ \forall n\geq 0$.From the initial relation we get that $ a_{n\plus{}1}\minus{}1\equal{}a_n^2(a_n\minus{}1)$.So $ a_n^2\equal{}\frac{a_{n\plus{}1}\minus{}1}{a_n\minus{}1}$.Therefore we have that

$ (a_0a_1 \cdots a_n)^2 \equal{}\frac{a_{n\plus{}1}\minus{}1}{a_0\minus{}1}$.

Because $ \lim_{n\to \infty}a_n \equal{}1$ (it's easy to prove this one) we get that the limit we seek is $ 0$.

How about finding the limit :

$ \lim_{n \to \infty}{ \left(a_n \right)^n}$ ?

Let $ a_n \equal{} 1 \minus{} \frac {b_n}{n}$; you should be able to show $ b_n\to \frac{1}{2}$ and hence $ \lim_{n\to \infty}a_n^n\equal{}e^{\minus{}\frac{1}{2}}$.


geometry
combinatorics unsolved
combinatorics

A polygon-shaped city is divided into areas by straight streets. At each vertex of a polygon there is a city square. Each street connects two squares and passes through no other square. Each street is one-way and it is possible to (i) enter every square, (ii) leave every square, (iii) go round the city along its boundary. Show that it is possible to go round at least one of the city areas.


Find the positive difference between the median and smallest number in this set: $ \{21.06, 21.5, 21.51, 21.06, 21.5\}$.

Order these numbers:

$ {21.06, 21.06, 21.5, 21.5, 21.51}$

Thus, the median is $ 21.5$.  Obviously, the smallest number is $ 21.06$, thus, the answer is $ \boxed{0.44}$.

Perhaps, considering the timing of the asking this question, and the thread shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241758]here[/url] BUrnorth simply got this and the other question mixed up. Or the game did.


algebra
polynomial
function
algebra unsolved

Solve the equation:$ x^7\plus{}2x^6\minus{}5x^5\minus{}13x^4\minus{}13x^3\minus{}5x^2\plus{}2x\plus{}1\equal{}0$

It's a symmetric equation! We can divide LHS to $ x^3\sqrt{x}$. Okie?

BY keen observation, $ {(x \plus{} 1)}$ is a root of the given equation....

Then divide the given equation by $ {(x \plus{} 1)}$ and then u get a polynomial function of degree 6...
which can be solved becuse the resulting function is al;so a symmetric equation ...

Now divide by $ {x^3}$ and solve....


geometry
3D geometry
combinatorics proposed
combinatorics

A cube of side length $n$ is divided into unit cubes by partitions (each partition separates a pair of adjacent unit cubes). What is the smallest number of partitions that can be removed so that from each cube, one can reach the surface of teh cube wihout passing through a partition?

draw a graph (i know because manuel likes graphs) with vertex at the center of the squares (not outter layer sqares, only interior ones) and an edge between two squares if they are adjacent and the partition between them was remove.  Now draw an edge from the squares in the outter layer of interrior squares to a new vertex outside if they are an exit to the last layer. It is clear that this graph is conected and therefore it has at least $(n-2)^3$ edges. Since every aprtition we remove gives at most $(n-2)^2$ new edges then we must remove at least $n-2$ partitions.  Now we show it is enough. in any direction there are $n-1$ partitions which are parallel planes, so remove the $n-2$ leftmost of them in some direction, thats enough.


Before Arne reads your message and roasts you up like he did to BEE(NO offence Arne,I sort of enjoyed that :D ),i'll tell you that no one is going to post problems "specially" for you,at least i wouldn't...I think you should browse through all the questions and read new posts and find what you want.As for goals and propositions,i don't get what you mean...I'm sure there will be some people ready to help you if you be a little more specific.

This isn't a homework site as Arne said(I think)  but help will be provided if necessary.

I am really sorry all the mods and administerators if the above things I said were for you to say and I had no right to say them...You might as well delete my post if you feel so :)

Ah... well can't you at least give me a link where I can find some problems, I would truly appreciate it. And what's with this "specially"? I thought people love to help each other.

Well, the homework questions that i see posted here are all obvious, (if you looked at the section in the book, you should be able to solve them). Also, those problems give no satisfaction when you solve them, it is not that we don't want to help, but you have to be challenging yourself to get that help, most people won't waste their time with homework problems, but you can always bother your teacher if you need help, in fact, you should if you are having difficulty in math.

But here are a few things that were similar to the problems on my test (maybe a little easier).

Solve for $x$: $7x^4+8x^2-25=0$

Write a quadratic with the solutions $x=5+9i$ and $x=5-9i$

In the quadratic $5x^2-7x+85=0$, What is the sum of the roots? What is the product of the roots? (Roots are the same thing as solutions)

Solve for $x$ in: $x -3\sqrt{x}-24=0$

[quote="falling_cross"]Hi, I was wondering if you guys could help me out with something. I need some interesting tasks(goals, propositions, problems?) with quadratic equations. And I need them fast. Thanks.[/quote]

Google: quadratic equations. There will be tons.  :D

[quote="236factorial"]No homework questions?  :D

[quote="rrusczyk"]Actually, I disagree - this is a great place to get homework help.  What's the point of a community if people can't get help from it?[/quote]

Meditate on that  ;)

Who's BEE?[/quote]

Yeah i read what Mr.rrusczyk wrote befor i gave my post but i also mentioned that help will be providede if necessary and if he was a bit specific(Read my post again ;) )Also BEE didnot even mention any questions he needed help on...first he should have done all the questions he copuld and ask them here if he ahd any doubts.....and yeah ...A+Math is some site with homework help.


Is there a committe forming argument for the combinatorial identity

\[ \dbinom {n \minus{} m}{k \minus{} m} \dbinom {n}{k \plus{} m} \dbinom {n\plus{}m}{k} \equal{} \dbinom {n \minus{} m}{k} \dbinom {n}{k \minus{} m} \dbinom{n \plus{} m}{k \plus{} m}?\]

Thanks in advance.

If no one has a solution, I would still appreciate any thoughts on the problem. I was thinking that maybe it was like choosing a total of $ 3k$ people out of a total of $ 3n$ people, where there are 3 different types of people, or something like that, but I could not get it to work.


They had a question about partitions on a past mathcounts test (2001 state sprint, number 24). is there a general formula to find the number of partitions (p(n)) for a given integer n? I know that when n is negative, p(n)=0, and i know a few values.

Any help is appreciated. :)

There is a general formula, but it isn't closed form, using generating functions. But for the MC problem, writing them out is the best way.

Closed form...so it's not terminating? Or does that mean something else?

So here's how the thing goes. Think about $(x^{a})(x^{b}) = x^{a+b}$. Hey look at that, addition became multiplication.

Now consider the function

$(1+x+x^{2}+\cdots)(1+x^{2}+x^{4}+x^{6}+\cdots)(1+x^{3}+x^{6}+\cdots)(\cdots)$

The first term corresponds to any number if 1s, the second to any number of 2s, etc. If we expand this as a normal product of polynomials, we get the series

$\sum_{n=1}^{\infty}p(n)x^{n}$ where $p(n)$ is the partition function.

However, we rewrite this product as

$\prod_{n\in\mathbb{N}}\frac{1}{1-x^{n}}$, since we know that $\frac{1}{1-x^{n}}= 1+x^{n}+x^{2n}+x^{3n}+\cdots$.

When we expand the product, the coefficient of $x^{n}$ will be the number of partitions of $n$.


How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

$ \textbf{(A)}\ 96 \qquad \textbf{(B)}\ 104 \qquad \textbf{(C)}\ 112 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 256$

[hide]
The numbers must be of the same parity; otherwise their average would not be an integer. There are $\binom{5}{2}=10$ ways to pick the two digits out of the even digits, as well as 10 combinations for the odd digits, so 20 combinations total. The average of them is then determined by the other two. There are $3!=6$ ways to order the digits. $20(6)=120$

Now we subtract the ones that start with 0. There are 4 choices for the highest digit, and $2!=2$ ways to position that digit and the mean of it and 0. That's 8 numbers that start with 0.

$120-8=\boxed{112}$
[/hide]

Elegant solution.
I just brute forced  :maybe:
That looks nice to me.

smart solution
i just wrote them all out


Σγχαρτήρια σε όλου τους επιτυχόντες στο Θαλή και καλή συνέχεια με τον Ευκλειδη.

Δημήτρη, τα αποτελέσματα βγήκαν για όλες τις τάξεις; Εμένα μου εμφανίζει μόνο της Β Γυμνασίου.
Καλή συνέχεια σε όσους πέρασαν.

Ναι βγήκαν αλλά δεν τα έχουν ανεβάσει ακόμα στο site της Μαθηματικής .


Let $ ABC$ be acute triangle with altitude $ AH$ and centroid $ G$ let $ A',B',C'$ be midpoint of $ BC,CA,AB$ let $ HC'\cap AC\equal{} D,HB'\cap AB\equal{} E$ let $ GD \cap BC\equal{}M,GE\cap BC\equal{}N$ prove that $ MC\cdot NB\equal{}4MA'\cdot NA'$.

Just using Menelaus Theorem, we can get the result easily :P
1. Show that $ \frac{AD}{CD}\equal{}\frac{BE}{AE}\equal{}\frac{BH}{CH}$
2. $ \frac{A'M}{CM}\equal{}\frac{AD}{2CD}$ and $ \frac{A'N}{BN}\equal{}\frac{AE}{2BE}$
$ \Rightarrow MC.NB\equal{}4NA'.MA'$

Hi thanks, but I don't know you used Menelaus Theorem for what triangle and what transversal  :blush: ?


what's your favorite book genre?
I liek mystery.

Science Fiction and will always be!!!  :D
(And my favourite Subgenre: Utopia)

[quote="lotrgreengrapes7926"]I like fantasy and realistic fiction. :D[/quote]

same here :)

I like humor. my favorites are dilbert books.

humor, mystery, adventure,...

that kind of stuff.

calculus books, mysteries, humor, info, etc.


I'm a student from China.
Recently we took a exam but quite a few of us thought the test paper was NOT written in idiomatic English.
As we can not find any native English speaker, we ask for your help.
If you are a native English speaker, could you spend some time helping us judge whether it is idiomatic?
And we want you to post your answers to the questions.

P.S. Perhaps my English is not idiomatic but I hope you can understand...  :)

Problems can be found here.
[url=http://picasaweb.google.com/wasltone/HxfMqJ#]http://picasaweb.google.com/wasltone/HxfMqJ#[/url]

Thank you!

I really don't understand what you mean by "idiomatic sentences", all that was present in the picassa images that you had 'linked' to the post were  "Fill in the blanks with the appropriate word" type questions.

I can give you an instance of an Idiom, it is as follows-- Jack has lost his marbles; This idiom conveys to us that Jack has become crazy (lose your mind and behave in an eccentric manner) , and NOT that Jack has really lost his set of beloved marbles.  :lol:

An interpretation of an idiom is that -- "A phrase whose meaning is different from the individual words present in it"; I don't know for sure whether the sentences present in the "images" you had "linked" are idiomatic or not, they sound like simple/plain English. Most of your answers to the "blanks" were correct.

I think you should encapsulate what you actually want us to help you with, or what you intend to convey; I know my explanation/ reply is discursive but I feel an aspect of your query is slightly ambiguous.

I am not a NATIVE ENGLISH speaker myself. :lol:  so I am confronted with this ambiguity (at least for me) that seem to have cropped with your question.

I mean,
Do you think the article in the test paper is real, fluent and idiomatic English?
I guess the article was not written by native English speaker, do you think so?

What i mean by idiomatic is
  Mother-tongue competence, whether or not it is the speaker's first language.
  pertaining or conforming to the mode of expression characteristic of a language
NOT
  resembling or characteristic of an idiom

The paper doesn't seem as if it was written by a native speaker. There are a few mistakes along with some things that a native speaker normally wouldn't say. I only read the first paper but I'll give some examples:

There are no quotations around "The sky is (blank) all over - I'll wear my straw hat today..."
The sentence "There was (blank) advantage for Mrs. Cockle in living at the very (blank) of the house" doesn't seem correct. "In living" seems misplaced.
The blank in the phrase "and the biggest of the three was larger than umbrellas that are (blank) made nowadays" seems like it shouldn't belong. Yes, one could put  an adverb but I don't think there's one that would fit.

I don't know maybe I'm just being too critical. My English teachers have made me think that way. I'd seek someone else's advice though, before mine.

Thank you.

P.S.
The answer to
and the biggest of the three was larger than umbrellas that are (blank) made nowadays
is "ever".


Let,
$ A: ( \minus{} r, 0)$
$ B: (r, 0)$
$ C: (, \sqrt {3}r)$
$ P: (a, b)$.

So,
$ BC: y \equal{} \minus{} \sqrt {3}x \plus{} \sqrt {3}r$
$ CA: y \equal{} \sqrt {3}x \plus{} \sqrt {3}r$
$ AB: y \equal{} 0$

Therefore,
$ D: \left(\frac {a \minus{} \sqrt {3}b \plus{} 3r}{4}, \frac { \minus{} \sqrt {3}a \plus{} 3b \plus{} \sqrt {3}r}{4}\right)$
$ E: \left(\frac {a \plus{} \sqrt {3}b \minus{} 3r}{4}, \frac {\sqrt {3}a \plus{} 3b \plus{} \sqrt {3}r}{4}\right)$
$ F: \left(a, 0\right)$

Hence,
$ PD \equal{} \sqrt {\left(\frac {a \minus{} \sqrt {3}b \plus{} 3r}{4} \minus{} a\right)^2 \plus{} \left(\frac { \minus{} \sqrt {3}a \plus{} 3b \plus{} \sqrt {3}r}{4} \minus{} b\right)^2} \equal{} \frac {\sqrt {3}(r \minus{} a) \minus{} b}{2}$
$ PE \equal{} \sqrt {\left(\frac {a \plus{} \sqrt {3}b \minus{} 3r}{4} \minus{} a\right)^2 \plus{} \left(\frac {\sqrt {3}a \plus{} 3b \plus{} \sqrt {3}r}{4} \minus{} b\right)^2} \equal{} \frac {\sqrt {3}(r \plus{} a) \minus{} b}{2}$
$ PF \equal{} \sqrt {\left(a \minus{} a\right)^2 \plus{} \left(0 \minus{} b\right)^2} \equal{} b$

Thus, $ PD \plus{} PE \plus{} PF \equal{} \sqrt {3}r \equal{} const.$

[quote="murgi"]Solution is quite nice......But I would like to see a pure geometry approach(i.e. without coordinate)[/quote]

I see.

This solution is "proof without words".
[geogebra]fb00a0eab46b2574d699e8356eda2afe59ad8412[/geogebra]

I can't view Kouichi Nakagawa's graph. What do I have to install on my computer to see it?

Anyway, I have a solution. I hope it's not the same as Kouichi's.

First let

$ BD \equal{} a$
$ CE \equal{} b$
$ FA \equal{} c$
$ PD \equal{} x$
$ PE \equal{} y$
$ PF \equal{} z$

and $ AB \equal{} BC \equal{} CA \equal{} 1$
We have,
$ x^2 \plus{} a^2 \equal{} x^2 \plus{} (1 \minus{} c)^2$
$ y^2 \plus{} b^2 \equal{} z^2 \plus{} (1 \minus{} a)^2$
$ z^2 \plus{} c^2 \equal{} y^2 \plus{} (1 \minus{} b)^2$

Adding them all up,
$ a \plus{} b \plus{} c \equal{} \frac {3}{2}$

$ a \plus{} b \plus{} c \equal{}$ half the perimeter of the triangle.

[quote="chrischris"]I can't view Kouichi Nakagawa's graph. What do I have to install on my computer to see it?[/quote]

Is the JAVA installed in your PC?
If not installed please install it in the following site.
http://java.com

[quote="murgi"]Inside an equilateral Triangle An Arbitary pt.   [b] P[/b] is taken...Perpendiculars  [b]PD,PE  [/b]and [b]PF[/b] are dropped onto the sides[b] BC[/b],[b]CA[/b] and [b]AB[/b] respectively.Show that [b]BD+CE+AF[/b] doesnot depend on the choice of[b] P[/b].[/quote]
By Pythagoras,
   BD - DC + CE - EA + AF - FB
= PB -  PC + PC - PA + PA - PB
= 0
implies
(BD + DC)(BD - DC) + (CE + EA) + (AF + FB)(AF - FB) = 0
a(BD – DC) + a(CE – EA) + a(AF – FB) = 0
where a is the length of each side
BD + CE + AF = DC + EA + FB
Each = (BD + DC + CE + EA + AF + FB) / 2
       =3a/2 which is constant

The area of the triangle is constant, but it equals [PAB]+[PBC]+[PCA]= the sum of the three perpendiculars x a side, so the sum of the three perpendiculars is constant.

Wait, aren't we supposed to prove the sum BD+CE+AF instead of the sum of the perpendiculars PD+PE+PF?

[quote="gfour84"]Isn't there a name for this theorem, Its quite famous.[/quote]For the sum of perpendiculars = h, yes it's called [url=http://mathworld.wolfram.com/VivianisTheorem.html]Viviani's Theorem[/url].

[quote="gfour84"]I thought it started with a V . . .[/quote]The name starts with a V as I have already mentioned in my last post... So, what's your point in the reply?


algebra polynomial function calculus integration

Prove that there exists no polynomial $ P(x)$ such that $ P(x)$ represent  the number  of positive divisors of $ x$ for each $ x \in \mathbb{N}$

May i take a chance?  :D
There are many numbers (Prime for instance) which have no. of divisors = 2. And no. of divisors = integer for all n. $ \implies$ P(x) shd be a constant function (Accord to a similar prob given by you) but it is not (as there are different values and a polynomial is conts).
$ \implies$ there is no such polynomial  :)   :P

Plz verify and tell if its right

be clear...i can't get u
u speak and let the smilies be silent for some time :P

Firstly: A polynomial is conts
Secondly: There are many primes which have no of +ve divisors = 2
Thirdly: There are many composite numbers which have different no of divisors
Fourthly: For the fn to be conts, it has to be constant accord to my lemma earlier
Fifthly: But it cant be accord to 3

So $ \implies$ such a fn cannot exist  :)

Ok now? The previous one was typed in haze  :D

no that's not correct my problem was $ f$ was constant for all $ x$ may it be natural number or real number...here u don't know how the polynomial will behave for the real numbers :)

just a doubt  :D  How can you have a non conts polynomial?  :huh:

who said if a polynomial tales a certain value many times itt will be non continious :P

How is that polynomial defined for reals?  :huh:  It isnt right? Then is it conts?  :maybe:

that's what i said...
it's just known that at natural numbers it gives u their divisors but it may give jsut anything for real numbers u don't know how it behaves :)

so is it conts at all?  :huh:  If it isnt defined how can it be conts?  :huh:

Can you define a polynomial?? Or is it a function?

[quote="madness"]Ok is this right? For a polynomial to exist x-m|P(x) - P(m)  But 3 doesnt divide P(6) - P(3) = 2  :)[/quote]

Is this a rule for all polynomials?? I don't think pardesi will post such standard theory questions.

That is surely a rule by remainder theorem da  :D  :D  And i dont think the ques is nicely framed even if its right  :maybe:

[quote="madness"]Ok is this right? For a polynomial to exist x-m|P(x) - P(m)  But 3 doesnt divide P(6) - P(3) = 2  :)[/quote]
dei loose that's for polynomials with integral coeficients :P

oh yes  :blush:  But anyways is ur ques right?  :huh:

I'm talking abt the polynomial part  :maybe:

Just one doubt. Can you explain to me how its conts? Plz do that first  :)

dei loose such a polynomial doesn't exist da then hw can it be continiou da :D

No accord to your hypothetical assumption, you claim that it has a value for reals  :D  Is that what u claim  :huh:  If so it shouldnt have any value for reals shd it?  :maybe:

dei the problem is plain da u don't now what values it assumes for reals all u know is for natural numbers it is the number of divisors

Ok assume its a nth order polynomial . Then P(x)-2 will have infinite solns right so n shd tend to infinity so there cannot exist such a polynomial  :)
Right?  :maybe:

yep rohit i think you are right ..let p(x) be the required ploynomial ...let g(x) = p(x)-2.. then g(x) has infinte roots but a polynomial cannot have infinite distinct roots

finally i think someone agrees  :D  :D  :D

Rohith you beat me to it da. I was thinking of calculus approach

Since f(x) = 2 for all primes, f'(x) = 0 has 1 root between each prime and since there are infinite primes, it has infinite roots but f'(x) is a polynomial cause f(x) is one.

I love calculus!!!!

yes that's right ...actually this  a pretty standard problem mostly given in the 2nd or 3rd year of RMO coachong :)

so u can clearly see that i am not a nv  :P  :P  :P

Mine is non-standard soln cause they don't teach calculus in RMO!! :D

Here's a simple extension to the problem

Prove that there exists no polynomial $ P(x)$ which represents the sum of divisors of $ x$ and $ x$ being natural.

Let F(x) = P(x) - (1 + x)

Consider X = Prime!! Problem is killed!


I think i keep messing up.

When the numerator of p/q, not necessarily in lowest terms, is decreased by 2 and its denominator is increased by 6, the value of the fraction is reduced by two-thirds. However, if the numerator of p/q is increased by 6 while its denominator is decreased by 2, its value is tripled. Determine the sum of the reciprocals of p and q.

Note that 1/p+1/q= p+q/pq
Setup your equation as (p-2)/(q+6)=p/3q and cross multiply
We have 3pq-6q=pq+6p
So pq=3p+3q=3(p+q)

So the answer is 1/3