My class is working to plot points and trying to form best fit line. This is actually classwork (we're working on the conclusion part on Tuesday) but I don't think I'm doing this right.

Here is the data table (the first row can't be fit in Latex so I just wrote here); M is the molarity and my teacher said that $0.50 M$ is the molarity.

Test Tube \\ Volume of 0.50 M $Pb(NO_3)_2$ (ml) \\ Volume of 0.50 M $Na_2 CO_3$ (ml) \\ Height of Precipitate (mm) \\ Moles $Pb(NO_3)_2$ \\ Moles $Na_2 CO_3$

\begin{tabular}{|c|c|c|c|c|c|} \hline 1 & 0.50 & 7.50 & 5.0 mm & 2.5 \times 10^{-4} & 3.8 \times 10^{-3} & \hline 2 & 1.00 & 7.00 & 8.0 mm & 5.0 \times 10^{-4} & 3.5 \times 10^{-3} & \hline 3 & 2.00 & 6.00 & 22.0 mm & 1.0 \times 10^{-3} & 3.0 \times 10^{-3} & \hline 4 & 3.00 & 5.00 & 23.0 mm & 1.5 \times 10^{-3} & 2.5 \times 10^{-3} & \hline 5 & 4.00 & 4.00 & 19.0 mm & 2.0 \times 10^{-3} & 2.0 \times 10^{-3} & \hline 6 & 5.00 & 3.00 & 21.0 mm & 2.5 \times 10^{-3} & 1.5 \times 10^{-3} & \hline 7 & 6.00 & 2.00 & 12.0 mm & 3.0 \times 10^{-3} & 1.0 \times 10^{-3} & \hline 8 & 7.00 & 1.00 & 18.0 mm & 3.5 \times 10^{-3} & 5.0 \times 10^{-4} & \hline 9 & 7.50 & 0.50 & 11.0 mm & 3.8 \times 10^{-3} & 2.5 \times 10^{-4} & \hline \end{tabular}

Okay.. Here is the problem, from the scale (this can be found in left top corner of the scanned image) and the table above, I can graph the points. BUT, after the first points, the rest of the points seem to be overlapping with each other. For instance, I don't think I can fit two points $(0.25,5.0)$ from the x-axis of Moles of $Pb_(NO_3)_2 \times 10^{-3}$ and $(3.8,5.0)$  from the x-axis of Moles of $Na_2 CO_3 \times 10^{-3}$ in same graph because they seem to be the same point! (or almost really really close). Am I doing something wrong? I'm suppose to draw the best fit line from the points I plot but since the points are overlapping, I don't know what I should do.  :?:  :?

I'm sorry, even with changing the display parameters on my screen, I can't really see your scanned graph.  So my response doen't have the benefit of seeing what you went to the trouble to scan.

I don't know how far you have gone in graphing, but typically real results are graphed with the error bars showing the uncertainity of each measurement.  This is expressed as a capital "I" structure, with the height and width of the top and bottom bars describing a square who's sides show, typically the 95th percentile confidence limits.  If there is another reason or other default or convention, that is followed (1 SD is sometimes used in academic cricles as it is easily calculated).

That being said, even if you are graphing single points there is no reason to have a single point be your only graphic symbol.  I have used visable symbles when graphing points that are too close to resolve.  For example a triangle can be placed over a point, small circle, cross or x with all points visable even thought they overlap.  If you add squares and the listed other symbles in this paragraph, you can have a quite a few very close points graphed.

Alternatively, if all your points are too close, perhaps changing your scale can make reading the graph easier.  (After all, easily grasping all the data is the reason for the graph in the first place.)  I suggest a broken scale where you have a point, say at 1,1 then, with appropriate marking (a gap in the axis and clear jump in the numbers showing the value on the axis) have x-value 100 follow x-value 2.  You can break both axis, sometimes twice, but then you are reaching the limit of an observer being able to make sense of the graph.

A third commomly used solution, in some cases is to use a long scale for one or both axes.  This can be done, even if the range of data doesn't reach across many log cycles.

I hope this is useful.  Note if you use the log axis, your line will be curved. (You can figure our the shape of the curve easily.  I'd suggest using a numerical method to find least square line equation and generate a sufficient number of points to graph to determine the line if you need to graph the curved, best fit "line.")

Have you heard about [url=http://www.chembuddy.com/?left=concentration&right=molar-fraction]molar fraction[/url]?

Try to make your x axis molar fraction of Pb(NO3)2, then your plot will probably make more sense.

I think I found why my graph was like that. I think it's second reason you suggested because the two points were really close on my graph except for the leftmost one, rightmost one, and the center one. So, I just put whatever I can draw and drew the line. It looks OK. I'll ask my chem teacher on Tuesday.

Thank you for all your help!


This was in AoPS 2.

Quadrilateral ABCD is inscribed in a circle with diameter ad=4.  If sides ab and bc each have length 1, find cd.  I used Prolemy's theorum even though it was in a trig chapter :P and got $ 3sqrt5\minus{}1/4$.  However, the book got 7/2.  there is a point in the solution I don't get:

Since AB= CD, arc AB=arc BC, so angle BDA=angle angle BDC.

Huh?

Hopefully I'm not making a really stupid mistake.

It should be ... since AB=BC...
And, then use inscribed angles.

Oops, typo by me there.

But it still doesn't seem to work.  It would work only if d was the center, right?
And it's obviously not, it's on the circumfrence.

let O be the centre
join bo and co
ao=do=bo=co=2
ab=bc=1
tri(AOB) IS CONGRUENT TO TRI(BOC)
angleBOA=angleBOC=x
angleCOD is 180-2x
by cosine rule, cosx=(4+4-1)/(2*2*2)=7/8
cos2x=17/32
cos(180-2x)=-17/32=(ODsq+OCsq-CDsq)/2*OD*OC
solve for CD which comes out to be7/2

note:ODsq means(OD)^2

@abcak
Incorrect. Inscribed angles versus equal arcs are equal, special case is when we're talking about the same arc.

AB=BC, so $ \angle ADB\equal{}\angle BDC$ [geogebra]59aae0afb8ea32ce924eccdd459ada93dce42d29[/geogebra] OK, the picture is not so good :)

Draw diameter BT. Now BCDT is isosceleles trapesoid.
$ BD \equal{} CT \equal{} \sqrt {AD^2 \minus{} AB^2} \equal{} \sqrt {15}$.
Let $ CD \equal{} x$, by Ptolemy for BCDT:  $ BT\cdot CD \plus{} BC \cdot DT \equal{} BD \cdot CT \Rightarrow 4x \plus{} 1 \equal{} 15 \Rightarrow CD \equal{} x \equal{} \frac {7}{2}.$


Sorry for writing some goofy things. It is easy than I thought:
$a^n^+^1 +n =a^n^+^1+1+....+1$ (n times of 1)$\geq (n+1) \sqrt[n+1]{a^n^+^1}=an+a$


Let S={(x,y): x^2-y^3>0}, show that S is open.


How many 6-tuples $ (a_1,a_2,a_3,a_4,a_5,a_6)$ are there such that each of $ a_1,a_2,a_3,a_4,a_5,a_6$ is from the set $ \{1,2,3,4\}$ and the six expressions
\[ a_j^2 \minus{} a_ja_{j \plus{} 1} \plus{} a_{j \plus{} 1}^2\]
for $ j \equal{} 1,2,3,4,5,6$ (where $ a_7$ is to be taken as $ a_1$) are all equal to one another?

Lets consider the values of $ f(a,b) \equal{} a^2 \minus{} ab \plus{} b^2$ for consecutive numbers:

Case a)
$ f(1,1) \equal{} 1$, $ f(2,2) \equal{} 4$, $ f(3,3) \equal{} 9$, $ f(4,4) \equal{} 16$
With solutions $ (1,1,1,1,1,1)$ , $ (2,2,2,2,2,2)$, $ \cdots 4$ cases

Case b)
$ f(1,2) \equal{} f(2,1) \equal{} 3$, there are two possibilities: $ (1,2,1,2,1,2)$ or $ (2,1,2,1,2,1)$
$ f(2,4) \equal{} f(4,2) \equal{} 12$ idem
for $ 4$ more cases

Case c)
$ f(1,3) \equal{} f(3,2) \equal{} f(2,3) \equal{} f(3,1) \equal{} 7$ notice that:
$ (a,3,b,3,c,3)$ with $ a,b,c \in \{1,2\}$ then $ 8$ cases
or $ (3,b,3,c,3,a)$ idem  for another $ 8$ cases
$ f(1,4) \equal{} f(4,3) \equal{} f(3,4) \equal{} f(4,1) \equal{} 13$ idem
Subtotal $ 32$ cases.

Grand total $ 40$ cases

Another interesting problem is to find How many 6-tuples are there such that each of  the six expressions
 are all $ \bold{different}$ to one another?

By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?


Let be given two natural numbers a,b
Prove that for each prime number $p$ exists $\le 1$ pair (x,y) (x,y are two natural numbers) such that:$p=ax^2+by^2$

If $p=ax^2+by^2=az^2+bt^2(x,y,z,t>0)$
$(axz+byt)(axz-byt)=a^2x^2z^2-b^2y^2t^2=(ax^2+by^2)az^2-by^2(az^2+bt^2)=p(az^2-by^2)$
Since $p$ is a prime,$p|axz+byt$ or $p|axz-byt$
And we also have
$p^2=(ax^2+by^2)(az^2+bt^2)=(axz+byt)^2+ab(yz-xt)^2=(axz-byt)^2+ab(yz+xt)^2$
Easy to see $(a,p)=(b,p)=1$
So
(1)$p|axz+byt,p|yz-xt$ and $axz+byt>0,axz+byt\geq p$
So $yz-xt=0$ easy to see $x=z,y=t$
(2)$p|axz-byt,p|yz+xt$
Similar $axz-byt=0,$ also $x=z.y=t$


function
limit
calculus
derivative
logarithms
real analysis
real analysis unsolved

Does this converge to anything? I know for x = 1 it's zero, but what about for numbers above that?
I couldn't get a simple curve to fit it very well.

$ \lim_{n \rightarrow \infty} \left(\sum_{k \equal{} 1}^{n} x^{1 \over k}\right) \minus{} n$

EDIT: I just realized that its derivative diverges at x = 1. Hmm.

It diverges to $ \infty$ for $ x > 1$ and $ \minus{} \infty$ for $ x < 1$, albeit very slowly.  This can be seen by noting, as you did, that $ \lim_{n\to \infty}f_n'(x)\to \infty$ for all $ x > 1$.

$ x^{1/k}\equal{}e^{\frac{\ln x}{k}}\equal{}1\plus{}\frac{\ln x}{k}\plus{}O(k^{\minus{}2})$

$ \sum_{k\equal{}1}^nx^{1/k}\minus{}n\equal{}\sum_{k\equal{}1}^n\left(x^{1/k}\minus{}1\right)\equal{} \sum_{k\equal{}1}^n\left(\frac{\ln x}{k}\plus{}O(k^{\minus{}2})\right)$

$ \equal{}\ln x\sum_{k\equal{}1}^n\frac1k\plus{}O(1)$

As $ n\to\infty,$ this behaves like $ \ln x\cdot\ln n$ and tends to $ \plus{}\infty$ for $ x>1$ and $ \minus{}\infty$ for $ 0<x<1.$

How did you get that $ \displaystyle\sum_{k\equal{}1}^n O(k^{\minus{}2})\equal{}O(1)$?

The sum $ \sum_{k\equal{}1}^{\infty}\frac1{k^2}$ converges to some number $ M$. The technical meaning of that $ O(k^{\minus{}2})$ is that those terms are less than $ Ck^{\minus{}2}$ for some constant $ C$. The sum of those terms is thus less than $ C\cdot M$, so it's bounded. That's the definition of $ O(1)$.

Of course we know what $ M$ is, and we could find an explicit value of $ C$ depending on $ x$. The point of $ O$ language is that we don't really care what those constants are.


arithmetic sequence
number theory
Ramsey Theory

All the natural numbers (1, 2, 3, ...) are colored using a total of n colors.  Prove that there must exist three natural numbers a, b, c of the same color such that a + b = c.

could you clarify what u mean by "coloring" a number?

he's just using coloring as a way to describe partitioning, which is just breaking up the natural numbers into n disjoint subsets whose union is the natural numbers.

So it just asks to prove that there exist a,b,c in one subset s.t. a+b=c

This reminds me of a famous theorem (that is actually very tough to prove, although it seems completely obvious:

Van der Waerden's Theorem: Given any partitioning of the nonnegative integers into k groups, there exists an arithmetic progression in one of the groups of length (at least) n.

Van der Waerden's proof of this theorem is really amazing, but his bound on the smallest number that is the largest of an n-term arithmetic progression is extremely ridiculous. The proof can be found in Khinichin's [i]Three Pearls of Number Theory[/i].


A certain organism begins as two cells. Each cell splits and becomes two cells at the end of three days. At the end of another three days, every cell of the organism splits and becomes two cells. This process lasts for a total of 15 days, and no cells die during this time. How many cells are there at the end of the $ 15^\text{th}$ day?

The organism begins as 2 cells.
After 3 days, it is 4 cells.
6 days=8 cells
9 days=16 cells
12 days=32 cells
[b]15 days=64 cells[/b]

Note that we could also have done $ 2^{\frac{15}{3}\plus{}1} \equal{} \boxed{64}$.

What I would give to reproduce like that  :rotfl:

Anyway this is exponential growth. Thus

$ E \equal{} SR^{\frac {T_T}{T_G}}$

Where E is End pop.
S is Starting pop.
R is rate of growth
$ T_T$ is time (in total)
$ T_G$ is time to grow

Plugging in $ E \equal{} 2\cdot2^{\frac {15}{3}} \equal{} 2\cdot2^5 \equal{} 2\cdot32 \equal{} 64$

Hm I had an uber formula. Pwnt.

[quote="bassamali01"]Can we use the geometric sequence which is   a*r^(n-1)

I tried but it went wrong. Can anyone explain it plz[/quote]
Good evening, I believe that isabella2296's solution is similar to what you're talking about. I don't think that a(r^(n-1)) quite works for this problem, although I may be wrong. I think it is just exponential growth, as others have stated.


parameterization
quadratics
algebra unsolved
algebra

Solve
$(x+a)^{4}+(x+b)^{4}=c$.
a,b,c real parameters

let$y=x+\frac{a+b}{2},t=\frac{a-b}{2}$ then:
$(y+t)^{4}+(y-t)^{4}=c\Rightarrow{2y^{4}+2(6y^{2}t^{2})+2t^{4}=c}$
now let $y^{2}=p$ and use the quadratic formula.

[quote="Farenhajt"]That should be $y=x+{a+b\over 2}$[/quote]
  yes, you are right.
thank Farenhajt.


Jena starts on a trip at 8:00 a.m. and travels at a rate of 50 mph. Harvey starts out at 10:30 a.m. the same day and travels the same route at a rate of 75 mph. What is the sum of the total miles traveled by Harvey and Jena when Harvey catches up with Jena?

They start out $ 2.5$ hours apart, and Greg rides $ 25$ mph faster. $ \frac{150}{25}\equal{}6$, so they each rode $ 6$ hours. $ 6(75\plus{}50)\equal{}6(125)\equal{}\boxed{750}$.

(Sigh, sorry if this is sorta hard to understand, I'm just too lazy to write out all the equations  :sleeping:)

[quote="RoFlLoLcOpT"]...so they each rode $ 6$ hours. [/quote]
Actually, Harvey rode $5$ hours and Jena rode $5+2.5=7.5$ hours.  That makes the total mileage to be:
\begin{align*}5\times75+7.5\times50&=5\times75+75\times5\\&=2(5\times75)\\&=(2\times5)75\\&=10\times75\\&=\boxed{750}\end{align*}


This problem has three parts, the first two parts are easy to solve, but the last part I couldn't figure out at all, and I couldn't tell if it needed the MVT or not.

the problem:
Let $ f: [0\plus{}\infty) \rightarrow \mathbb{R}$ be differentiable on $ (0,\plus{}\infty)$

a)If$ f'(x) \rightarrow a \in \mathbb{R}$ as $ x\rightarrow \plus{}\infty$, show that for $ h>0, \lim_{x \rightarrow \plus{}\infty}\frac{f(x\plus{}h)\minus{}f(x)}{h} \equal{} b$.
(I just applied MVT for some $ y \in [x, x\plus{}h]$, then took the limit)

b) If $ f(x) \rightarrow a \in \mathbb{R}$ and $ f'(x) \rightarrow b \in \mathbb{R}$ as $ x \rightarrow \plus{}\infty$ then $ b\equal{}0$
(I took the limit from part a) and show that $ f(x\plus{}h) \rightarrow a$)

c)If $ f'(x) \rightarrow b \in \mathbb{R}$  as $ x\rightarrow \plus{}\infty$, then $ f(x)/x \rightarrow b$ as $ x\rightarrow \plus{}\infty$.
(I can't tell how to get  $ f(x)/x$ part)

For the last one:

Given $ \epsilon > 0,$ find $ N$ such that if $ x > N,\ |f'(x) \minus{} b| < \frac {\epsilon}2.$

Now for $ x > N,$ write

$ \frac {f(x)}{x} \equal{} \frac {f(N)}{x} \plus{} \frac {f(x) \minus{} f(N)}{x}$

$ \frac {f(x)}{x} \minus{} b \equal{} \frac {f(N) \minus{} bN}{x} \plus{} \frac {f(x) \minus{} f(N) \minus{} b(x \minus{} N)}{x}$

$ \left|\frac {f(x)}{x} \minus{} b\right|\le\left|\frac {f(N) \minus{} bN}{x}\right| \plus{} \left|\frac {f(x) \minus{} f(N) \minus{} b(x \minus{} N)}{x}\right|$

But $ f(x) \minus{} f(N) \equal{} f'(c)(x \minus{} N)$ for some $ c\in(N,x)$ by the MVT, so

$ f(x) \minus{} f(N) \minus{} b(x \minus{} N) \equal{} (f'(c) \minus{} b)(x \minus{} N).$ Note that $ |f'(c) \minus{} b| < \frac {\epsilon}2.$

So, $ \left|\frac {f(x)}{x} \minus{} b\right|\le\frac {|f(N) \minus{} bN|}{x} \plus{} \frac {\frac {\epsilon}2(x \minus{} N)}{x}$

$ < \frac {|f(N) \minus{} bN|}{x} \plus{} \frac {\epsilon}2.$

Now, since $ N$ is fixed, we may choose $ x$ large enough that $ \frac {|f(N) \minus{} bN|}{x} < \frac {\epsilon}2.$

Hence, for sufficiently large $ x,$ $ \left|\frac {f(x)}{x} \minus{} b\right| < \epsilon.$

This clearly has some relationship to the Cesaro-Stolz family of theorems.

[quote="Kent Merryfield"]This clearly has some relationship to the Cesaro-Stoltz family of theorems.[/quote]
As a side note, the correct name is [b]Stolz[/b], not [b]Stoltz[/b]. At least, that's how I see it spelled everywhere.

And, yeah, [url=http://planetmath.org/encyclopedia/StolzCesaroTheorem.html]Cesaro-Stolz[/url] is a discrete version of L'Hospital.


linear algebra
matrix
graph theory
combinatorics unsolved
combinatorics

You have an n by m array of 0s and 1s. Let a be the most 1s you can pick so that no
two are in the same row or column. Pick a subrectangle (not necessarily non-empty) of zeros which has dimensions b by c; that is, pick b special rows and c special columns so that any entry which is in a special row and special column is 0. Suppose b+c is maximal. Show that a+b+c=n+m.

Bomb

It's a consequence of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Konig&t=43344]Konig-Egervary Theorem[/url]:

In a bipartite graph, the maximal size of a matching is equal to the minimal size of a minimal vertex cover (check that link for definitions).

Consider the bipartite graph $G$ having the rows of the matrix in one class and the columns as vertices in the other class of the bipartition. Two vertices are connected iff there is a $1$ at the intersection of the row and column corresponding to the two vertices.

The maximal size of a matching equals your $a$, and the minimal size of a vertex cover is equal to the minimal number of rows and columns which contain all the $1$'s, which is, in turn, readily seen to be equal to your $m+n-(b+c)$.

Thx, but can you please elaborate on these maximal vetrex stuff etc. Thx

Bomb


probability
logarithms
trigonometry
AoPSwiki
function
complex numbers
algebra

What is the nature of $ \sum \frac {\sin(n^2)}{n}$

thank's

I suspect it converges, but I do not see the tools for showing that.

What about this? Suppose $ (X_n)$ is a sequence of i.i.d. random variables with finite variance and mean zero. What is the probability that $ \sum_{n\equal{}1}^{\infty}\frac{X_n}{n}$ converges?

Since the problems of this type have been discussed a few times already, I'll post a sketch of the solution. To show that the series converges, it is enough to show that $ S_n \equal{} \sum_{k \equal{} 1}^n e^{2\pi i k^2 x}$ is $ O(n^{\delta})$ for some $ \delta < 1$ provided that $ x$ is not Liouvillian. Write
\[ |S_n|^2 \equal{} \sum_{k,m}e^{2\pi i(k^2 \minus{} m^2)x} \equal{} (k \minus{} m \equal{} h) \equal{} \sum_{ \minus{} n\le h\le n}e^{2\pi ih^2 x}\sum_{m\in I_h}e^{4\pi i mhx}
\]
where $ I_h \equal{} \{m: 1\le m\le n,1\le m \plus{} h\le n\}$ is a "subinterval" of $ \mathbb Z$ of length at most $ n$. Now, the sum over $ I_h$ is bounded (possibly, with some constant) by $ \min\left(n,\frac 1{\|2hx\|}\right)$ where $ \|y\|$ is the distance from $ y$ to the nearest integer. Now choose an irreducible fraction $ p/q$ with $ q\le n$ such that $ \left|2x \minus{} \frac pq\right| < \frac 1{nq}$. Since $ x$ is not Liouvillian, we must have $ q > n^\beta$ with $ \beta > 0$. Now, when $ h$ goes from $ \minus{} n$ to $ n$, $ 2hx\mod 1$ goes essentially over all possible fractions with denominator $ q$ hitting each fraction about $ n/q$ times (it would be literally true if we had the exact equality $ 2x \equal{} \frac pq$ but the error never exceeds $ \frac 1{nq}n \equal{} \frac 1q$, so it doesn't change the picture much). Therefore, we get the estimate
\[ \sum_h \min\left(n,\frac 1{\|2hx\|}\right)\approx \frac nq (n \plus{} q \plus{} \frac q2 \plus{} \frac q3 \plus{} \dots \plus{} \frac q{q/2})\approx \frac {n^2}q \plus{} n\log q
\]
Hence, $ |S_n|^2\le n^{2 \minus{} \beta} \plus{} n\log n$ and we are done.

I was thinking in Dirichlet's Test, with the argument that the partial sums of sin(n^2) are limited (since those of sin(n) also are) and 1/n is monotonic with zero limit - and so the series converges. However,

[quote="Kent Merryfield"]I suspect it converges, but I do not see the tools for showing that.[/quote]

So, aren't the partial sums of sin(n^2) limited?

No, they aren't, and this has also been discussed a couple of times (though not recently :)). Assume that they are bounded by $ M$. Then, the expressions $ \text{Im\,}\left[e^{ih^2}\sum_{n\equal{}1}^N e^{in^2}e^{2inh}\right]$ are also bounded by $ 2M$. Now fix $ N>4M^2$ and let $ h$ run over positive integers. Since $ (2h,h^2)\mod 2\pi$ is dense in $ [0,2\pi]^2$ (actually, it is equidistributed there), we see that $ \left|\sum_{n\equal{}1}^N e^{in^2}e^{iny}\right|\le 2M$ for all $ y$. It remains to square this estimate and to integrate it with respect to $ y$ over $ [0,2\pi]$ to get $ N\le 4M^2$, which contradicts the choice of $ N$. I suspect that $ \limsup_{n\to\infty} \frac{|S_n|}{\sqrt n}>0$ but it would be harder to prove.

On the other hand, you were right thinking of Dirichlet test. The proof I gave is based precisely on this idea: we rewrite the series as $ \sum_n \frac {S_n}{n(n\plus{}1)}$ and try to compare it to a power series. But since the expressions $ 1/{n(n\plus{}1)}$ tend to zero quite fast, we do not need boundedness of $ S_n$; some moderate growth is also OK.

Thanks Fedja for the detailed answer. However, I apologize but I have to ask this:

1) What is the meaning of "$ \mod 2 \pi$"?
2) What is the meaning of "is dense in $ [0, 2 \pi]^2$"?

Also, how about if we had $ \sin(n^{3/2})$ instead of $ \sin(n^2)$?

$ \mod 2\pi$ means what is sounds like: to compute $ x \mod 2\pi$, you subtract multiples of $ 2\pi$ from $ x$ until you get a number in the interval $ [0, 2\pi)$.

"Dense" means that if you take any point and any neighborhood that point, the set in question has nonempty neighborhood with that neighborhood.  E.g., the set of rationals is dense in $ [0, 1]$ but the set $ \{\frac {1}{n} \mid n \in {\bf Z}_{ > 0}\}$ is not, since there exists a ball around $ \frac {2}{3}$ of radius $ \frac {1}{7}$ which does not intersect our set.  "$ [0, 2\pi]^2$ is the set of ordered pairs of real numbers, each appearing in the interval $ [0, 2\pi]$.

:o

[quote="fedja"]

$ |S_n|^2 \equal{} \sum_{k,m}e^{2\pi i(k^2 \minus{} m^2)x}$[/quote]
I dont understand this moment  :( . as I know
$ |S_n|^2 \in\ R$ because $ |S_n| \in R$
but
$ \sum_{k,m}e^{2\pi i(k^2 \minus{} m^2)x} \in C$ because
$ e^{2\pi i(k^2 \minus{} m^2)x} \in C$
how can you write this equality  :huh: ?
maybe this is true
$ |S_n|^2 \equal{} |\sum_{k,m}e^{2\pi i(k^2 \minus{} m^2)x} |$
because really :
$ |a \plus{} bi|^{2} \equal{} a^{2} \plus{} b^{2} \equal{} (a^{2} \minus{} b^{2})^{2} \plus{} 4a^{2}b^{2} \equal{} |a^{2} \minus{} b^{2} \plus{} 2abi| \equal{} |(a \plus{} bi)^{2}|$
I note that the fedja's proof is really great.
I also can to prove simple that this series will diverge $ \sum \frac {|\sin n^{2}|}{n}$
I have a question:
as fedja wrote ".... $ x$ is not Liouvillian". then if I understand right to take this case $ \sum \frac {\sin n^{2}}{n}$ in your sum
$ \sum e^{2\pi k^{2}xi}$ you must put instead $ x$  value $ x \equal{} \frac {1}{2\pi}$ but why $ \frac {1}{2\pi}$ is not Liovillian ?? , I know that not every transcendental number is a Liouville number, but it's not obviusly for me that $ \frac {1}{2\pi}$ is not liovillian number  :huh:
I think it's will be good example that if ($ \frac {1}{2\pi}$  is not liovillian) then we know number which is not liovillian but is transcedental number  :)
I think the next lemma is true

if $ a$ is (edit) not  liovillian number  then $ \frac {1}{a}b$ is not liovillian number for $ b \in Q$  rational

As to the first question, does it bother you that $ |e^{ia}\plus{}e^{ib}|^2\equal{}(e^{ia}\plus{}e^{ib})(e^{\minus{}ia}\plus{}e^{\minus{}ib}) \equal{}e^{i(a\minus{}a)}\plus{}e^{i(a\minus{}b)}\plus{}e^{i(b\minus{}a)}\plus{}e^{i(b\minus{}b)}$? :? You, indeed, have a real number on the left and a sum of complex numbers on the right but, since each complex number appears together with its conjugate, the entire sum is real. :)

As to the second question, you may now find the proof of the fact that $ \pi$ is not Liouvillian on AoPSWiki: [url]http://www.artofproblemsolving.com/Wiki/index.php/Pi/Liouvillian[/url] and yes, you are right saying that if $ a$ is [b]not[/b] Liouvillian, then any rational multiple of $ 1/a$ is not Liouvillian (I hope that's what you meant to say).

thanks.
As fedja wrote this sum $ \sum \sin n^{2}$ diverges it's already mean that this series $ \sum \frac {|\sin n^{2}|}{n}$ will diverge.
here is other proof.
suppose it's convrege, then this serie will be converge too :
$ \sum \frac {|\sin (n \minus{} 1)^{2}| \plus{} |\sin^{2}n| \plus{} |\sin (n \plus{} 1)^{2}|}{n}$
this mean that it's enough to show that this serie will diverge
$ \sum \frac {|\sin (n \minus{} 1)^{2}| \plus{} |\sin^{2}n| \plus{} |\sin (n \plus{} 1)^{2}|}{n}$
if it's not true then
for big $ n$ we have :
$ (n \minus{} 1)^{2} \equal{} \pi k \plus{} a$
$ n^{2} \equal{} \pi k' \plus{} a'$
$ (n \plus{} 1)^{2} \equal{} \pi k'' \plus{} a''$
where
$ k,k',k''$ is integer and
$ |a|,|a'|,|a''| < \frac {1}{4}$
therefore
$ 2 \equal{} (n \minus{} 1)^{2} \minus{} 2n^{2} \plus{} (n \plus{} 1)^{2} \equal{} \pi (k'' \plus{} k \minus{} 2k') \plus{} a \plus{} a'' \minus{} 2a'$
which can't be.
we are done

[quote="Extremal"]for big $ n$ we have :
$ (n \minus{} 1)^{2} \equal{} \pi k \plus{} a$
$ n^{2} \equal{} \pi k' \plus{} a'$
$ (n \plus{} 1)^{2} \equal{} \pi k'' \plus{} a''$
where
$ k,k',k''$ is integer and
$ |a|,|a'|,|a''| < \frac {1}{4}$
therefore
$ 2 \equal{} (n \minus{} 1)^{2} \minus{} 2n^{2} \plus{} (n \plus{} 1)^{2} \equal{} \pi (k'' \plus{} k \minus{} 2k') \plus{} a \plus{} a'' \minus{} 2a'$
which can't be.
we are done[/quote]

Could you give some precisions Extremal

Firstly:

i think k,k',a,a',a'' depends with n

$ (n \minus{} 1)^{2} \equal{} \pi k_{n} \plus{} a_{n}$
$ n^{2} \equal{} \pi k_{n}' \plus{} a_{n}'$
$ (n \plus{} 1)^{2} \equal{} \pi k_{n}'' \plus{} a_{n}''$
where
$ k_{n},k_{n}',k_{n}''$ sequence of  integer and

Secondly:

How do you prove that
$ |a_{n}|,|a_{n}'|,|a_{n}''| < \frac {1}{4}$

Thirdly:

$ \sum \frac {|\sin (n \minus{} 1)^{2}| \plus{} |\sin^{2}n| \plus{} |\sin (n \plus{} 1)^{2}|}{n}$

will be

$ \sum \frac {|\sin a_{n}| | \plus{} |\sin a'_{n}|\plus{} |\sin a''_{n}|}{n}$

what is the nature of this serie

[quote="fedja"]
  if $ a$ is [b]not[/b] Liouvillian, then any rational multiple of $ 1/a$ is not Liouvillian [/quote]

How do you prove that ?
I did not find anything on google

do you have a reference for this result ?

thank's for your replies

[quote="Moubinool"]

Could you give some precisions Extremal

Firstly:

i think k,k',a,a',a'' depends with n

[/quote]
yes this is true.for each $ n$ we have $ a_{n}$ and $ k_{n}$

[quote="Moubinool"]

Secondly:

How do you prove that
$ |a_{n}|,|a_{n}'|,|a_{n}''| < \frac {1}{4}$

[/quote]
if someone is big ten $ 1/4$ for all $ n$ then I can already say that this series will diverge.
$ \sum \frac{|\sin a_{n\minus{}1}|\plus{}|\sin a_{n}|\plus{}|\sin a_{n\plus{}1}|}{n}$
but this is what I need  :)

[quote="Moubinool"][quote="fedja"]
  if $ a$ is [b]not[/b] Liouvillian, then any rational multiple of $ 1/a$ is not Liouvillian [/quote]
How do you prove that ?
I did not find anything on google
[/quote]
Isn't it sort of "obvious"? :? If $ a\ne 0$ can be approximated by fractions $ \frac pq$ with precision $ 1/q^M$, and $ r\equal{}\frac uv$ is rational, then  $ \frac ra$ is approximated by $ \frac{uq}{vp}$ with error $ Cq^{\minus{}M}$, which, up to a constant factor, is the same as $ (vp)^{\minus{}M}$ (I think here of $ M,u,v$ as fixed and $ p,q\to\infty$). This tells you that any rational multiple of the inverse of a Liouvillian number is Liouvillian, which is equivalent to the statement you asked about. Similarly, if you have a Liouvillian number $ a$, then $ R(a)$ is also Liouvillian for every rational function $ R$ with rational coefficients.


The inradius of $\triangle ABC$ with side lengths $a$, $b$, and $c$ is $a-\frac{1}{c}$.

If $b = 1$, find the area of $\triangle ABC$.

I think we might need more information given in the problem.

Unless you want it in terms of a, b, and c, and in this case, we just use the formula $A=rs$

It is solvable in its current form. I'm quite sure no more information is needed.

Any ideas?  I ran into some pretty horrible algebra when I tried to do this with Heron's, and it's too late at night for me to think of anything smarter.

If we use $A = rs = \text{herons}$ we could get $a$ in terms of $c$, wouldn't we need one more equation?

Think about the [i]properties[/i] of a triangle with an inradius of $a-\frac{1}{c}$. What type of triangle will it be?

Are there any ideas?


abstract algebra
Ring Theory
superior algebra
superior algebra solved

assume that $R$ is Noetherian ring such that all maximal ideal are principle . show that all ideal are principle.

Assume that there exist non-principal ideals. The ring $R$ is noetherian, so there exists an ideal $I$ that is maximal among this non-principal ideals.
Because $I$ is not principal, it is not the unit ideal $(1)$ (which is principal).
Let $I=(y_{1},\ldots,y_{n})$. This is possible because the ring is noetherian, so any ideal is finitely generated.

Let $m=(m)$ (a little abuse of notation) be a maximal ideal containing $I$. The since $I\subset m$ we have $y_{i}=mx_{i}$ for some $x_{i}\in R$.
Let $J=(x_{1},\ldots,x_{n})$. Obviously $I\subseteq J$ and $I=mJ$. If $J$ properly contains $I$, then by the maximality assumption on $I$, $J=(j)$ for some $j\in R$. But $I=mJ=(mj)$ would mean that $I$ is a principal ideal which is not what we assumed of it. Therefore $J=I$, and since $I=mJ$, $I=mI$ also. Moreover, we can assume that $I=mI$ for any maximal ideal $m$ containing $I$ (we have seen how we contradict the non-principalness of $I$ otherwise). Actually we have not used that $m$ is a maximal idea, so we have that $I=Iu$ for any $u\in R$ such that $I\subset (u)$.

By a result on finitely generated modules over noetherian rings in Atiyah and Macdonald's "An Introduction to Commutative Algebra", there exists $x\in R$, $x\equiv 1(mod\ m)$ such that $xI=0$.
$x\equiv 1(mod\ m)\Rightarrow x\not\in m\supset I\Rightarrow x\not\in I$. Then the ideal $(I,x)$ generated by $I$ and $x$ strictly contains $I$, so it must be a principal ideal i.e. $(I,x)=(y)$ for some $y\in R$.
Since $I\subset (I,x)$, we get $I\subset (y)$, so by what we first proved $I=yI$. But $yI=(x,I)I=xI+I^{2}$. Since $xI=0$, $yI=I^{2}$.
$I=yI=I^{2}\Rightarrow I=I^{2}$. By the same result about finitely generated modules over noetherian rings, there exists $a\in I$ such that $(1-a)I=0\Rightarrow (1-a)y_{i}=0\ \forall i=\overline{1,n}\Rightarrow y_{i}=ay_{i}\Rightarrow I\subset (a)$. But $a\in I$, so $(a)\subset I$. We have that $I=(a)$ which again contradicts the assumption made on $I$.

Therefore all ideals in $R$ are principal.


According to early results, here are the scores (subject to change)

Note, everyone received #1 (you could use the points on the test):

 :first: alkjash 120

2nd: Brut3Forc3 112.5

3rd: andersonw 109.5

4th: zhou_qiang 106.5

5th: Just_Beginner: 105

6th: Bacteria 102

Yes, only 6 ppl scored above 100 on this test.  :o The index of passing is still being determined.

[quote="Bacteria"]In your PM to me you said I got 108...[/quote]

As I said, subject to change. gfour84 might have not graded properly  :o

Ah, I see.
 Is an official answer set going to be released?

Mock AMC 9/09 V. 3.1 Answer Key:
EBBEA BCACB EDEBA AEDDC CABDD

Mock AMC 9/09 V. 3.2 Answer Key:
DEEAC BBCEB CDAEB EDDAA CCABD

Bacteria, I think you missed 8 problems, so I miscounted the wrong time  :oops:  :P

I just compared the answers to the message in my sentbox and I got 18 right  :maybe:

EB[b]E[/b]EA BC[b]B[/b]CB EDEBA [b]C[/b]E[b]C[/b]D[b]A[/b] C[b]CEA[/b]D  vs
EBBEA BCACB EDEBA AEDDC CABDD

You sure?

Oh wow, sorry I was seeing the B as an E in my sleepy state  :oops:
I looked it over like 5 times too.

And congratulations to everybody, especially to Caelestor and gfour84 and whoever else helped make such a hard test.

We spent time on these, we hope you like them.

That was V. 3.1 solutions. I shall fix up a V. 3.2 solution tomorrow.

The qualifying index was set to 87 for this test. The following people were successful:

Bacteria
mlbmlb37
bokagadha
Just_Beginner
SorcerorofDM
andersonw
Brut3Forc3
alkjash
zhou_qiang
zapi2007

I'm giving this thread a bump just to let everybody know about this mock, which I thought was good quality and very difficult. The link to the mock is [url=http://agmath.com/media//DIR_37352/MockAMC12hardCAELESTOR.pdf]here[/url], and the solutions are provided on this thread. (I got the link from agmath.com)

[quote=SohumUttamchandani][quote=Caelestor]Separate solution threads have been formed throughout the forum. Access below:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1635289#1635289]Easy problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1635291#1635291]Still simple problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1635297#1635297]Moderate problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1635302#1635302]Harder problems[/url]

The results will come later. For now, we know that the highest score was 120 on V. 3.1. How many points did you improve on V. 3.2?

EDIT: Argh, poll was deleted. There were 3 on 2001, and 4 on above average last time I checked.

EDIT 2: There appears to be a bug with the poll. Oh well, just vote for whatever you feel.[/quote]

none of your problem links work :/[/quote]

The threads were probably deleted.


In the following $ R_j ,j\in \{1,2,...,26\}, $ denotes perfect one ohm resistors.  Consider that between two points $ A , B $  we have following network
\[A\bullet-R_1-R_2-R_3-
\begin{array}{|lcr|}
 --------&R_4&------   \\
  --------&R_5&------   \\
   --------&R_6&------   \\
    --------&R_7&------   \\

 --------&R_8&------   \\
 --------&R_9&------   \\
 --------&R_{10}&------   \\
  --R_{11}--R_{12}-&\cdots& ---R_{26}-   \\
\end{array}-\bullet B
\]
 Let [b]r[/b] be  the resistance, measured in ohms, between $ A $ and $ B $ .[b] Find  r  with five decimals.  [/b]

Key words: Continued fractions

Some references:
[1] Amer.Mat.Monthly (1974), Problem E 2459, proposed by A.A. Mullin.
[2] Matematicki  Vesnik , Problem 393,13(28)1976, solution in 15(1978) pag.298.

This looks like a problem requiring as much math as every 10-year old should know. If they were not requiring 5 decimal places, the answer could be calculated as 3.1 ohms without a calculator.
$ R_1$, $ R_2$, and $ R_3$ are in series and add up to an equivalent resistance of $ 3$ ohms.
In the set of resistance between the two vertical lines, there is the series group of $ R_{11}$ through $ R_{26}$ that adds up to $ 16$ ohms, and is in parallel with the other seven resistors. The total resistance between the two vertical lines, R, is about 1/7 ohm and can be calculated with:
$ 1/R \equal{} 1/1 \plus{} 1/1 \plus{} 1/1 \plus{} 1/1 \plus{} 1/1 \plus{} 1/1 \plus{} 1/1 \plus{} 1/16 \equal{} 7 \plus{} 1/16 \equal{} 7.0625$
$ R \equal{} 1/7.0625 \equal{} 0.14159$ ohm
Adding that $ R$ to the $ 3$ ohms for the $ R_1$ to $ R_3$ set you have
$ r \equal{} 3 \plus{} 0.14159 \equal{} 3.14159$ ohm

[quote="KMST"]This looks like a problem requiring as much math as every 10-year old should know. [/quote]
Then you've missed the point entirely.  The circuit encodes a convergent of the continued fraction of $ \pi$ ;)  It's the structure of the answer, not the answer itself, that is of interest.


Is she your HERO too?

She should be on HEROES.

SHE IS BETTER THAN EVAN O'DORNEY.

I may have to replace

"Evan O'Dorney = JESUS" in my dig with
"Azia Kim = JESUS"


Which do you like better?

GO TELETUBBIES :D

Shoulda put neither and both seperate.  I love both of them so much!  I just couldn't choose!

waaaaaaaaaaaaaaaaaah!

i didnt wanna choose neither 'cause i always feel guilty about being a party pooper, so i chose teletubbies (just 'cause i absolutely HATE barney. sorry, barney fans  :( )

GO TELETUBBIES! yay dreambold!

do u wanna be teletubby number 5?

hey fans, who wants to be? check in my blog for more details! 5th teletubby contest COMING SOON!

[size=200]Go Barney [/size]:clap:

" I lov u
  u luv me
  wer a happy family
  w/ a great big hug & a kiss from me 2 u
  wont u say u luv me 2"

I think its been like 12 years since ive watched either (stopped watching when I was 2 because my mom said it was a bad influence) But I said teletubbies cus there not as obese.

[quote="splatyango"]i like teletubbies!!!! (sorry i cant be the fifth teletubby, b/c i already have a place.)[/quote]
u could actually have multiple spots in aops family tree if you prefer, just no more than two please...thx

yeahh you shud've put sesame street

and how are teletubbies or barney a bad influence??

:huh: This is sooooo weird...  And it's disturbing that RussianRocket knows the song...

It's pretty obvious that you teletubby fanatics just want to attract attention to yourselves by pretending to like watching...well...a form of television program in which the protagonists communicate through some kind of grunting.  If you aren't trying to attract attention, well, I shudder to try to find an end to this sentence...

i haven't watched either show in a looooooooooooooonnggggggggggggg time. i HAVE seen veggie tales :D  recently though....don't ask.

I'm not sure if Gauss actually even knew what Teletubbies are  :dry: . I [b]absolutely hate[/b] Barney, so I just put Teletubbies, though I'm not a huge fan.

good job man.

i have a problem calming down :D  :dry:

Math Geek
[quote]This is sooooo weird... And it's disturbing that RussianRocket knows the song...
[/quote]

Ofcourse it is thats why i picked it  :rotfl:

Anyone watch dragon tails or george shrinks??? GO SHRINKS!!! :D

*Faints*...  good thing Dora isn't up there...

why?

stargirl u don't like teletubbies or barney?

[quote="sea-wolf"]i dont know... teletubbies?? barneys a bad influence on young kids...  :P[/quote]
yup i agree. barney is the only preschool show with constant criticism by teens...

See Anti-Barney Humor under Wikipedia.

[quote="mz94"][quote="dynamo729"]wow i have just lost 10% of my faith in the human race.....[/quote]

I lost faith in wayyy over 50% of all AoPSers right now...[/quote]

OH, my faith of AoPSers is just about gone. But I was talking about the whole human race.


[color=darkblue]Suppose we choose a permutation on $ n$ elements uniformly and at random. What is its average number of fixed points? [/color]

I think it's 1 - there are $ n!$ permutations and $ n!$ fixed points (recall IMO 1987 problem 1), so the average is 1. Kind of surprising actually...


geometry
3D geometry
probability
number theory
least common multiple

1.  A monument made of a certain number of rows consisting of cube shaped bricks starts with a row of 34 bricks. The row above has 31 bricks, the row above that has 28 bricks, and so on such that each row has 3 fewer bricks than the row below it. Later, a student notices that the total number of bricks used in the monument is just enough to create the brick floor of a rectangular patio that is just one layer of bricks, seven times long as it is wide. (no bricks are broken to make the floor) How many rows of bricks does the [u]monument[/u] have

2.while staying at a 15 story hotel, Polya plays the following game. She enters an elevator in the 6th floor. Each time she flips "heads", she goes up 1 floor. each time she flips "tails" she goes down 1 floor. What is the probability that each of her next 5 stops is on the 7th floor or higher?

3. Two circles, one with radius 5 inches and the other of radius 2 inches, are tangent at point P. Two bugs start crawling at the same time from point P, one crawling along the larger circle at 3pi inches per minute, the other crawling along the smaller circle at 2.5pi inches per minute. How many minutes is it before their next meeting at point P?

[hide="a bit complicated solution"]let the length of the side of the patio be x. the area is $ 7x^2$. that means the number of bricks on the monument must be congruent to 0 mod 7. we can say the number of bricks is $ \sum_{y\equal{}1}^{n}{37\minus{}y}$ where n is the number of rows.this must be congruent to 0 mod 7 so we try multiples of 7 for n.[/hide]

[hide="1"]Let $ x$ be the width of the rectangle.  Then the area is $ 7x^2$.
We wish to find some number $ 34\minus{}3k$ such that $ \frac {68\minus{}3k}{2}*(k\plus{}1)\equal{}7x^2$
$ 65k\plus{}68\minus{}3k^2\equal{}14x^2$
$ 2k\minus{}2\plus{}4k^2\equiv 0 \mod 7$
$ 2k^2\plus{}k\minus{}1 \equiv 0 \mod 7$
$ k\equal{}4$
errr... ya $ 5$ rows.[/hide]
[hide="2"]
The first flip must be heads and so must the second.  From there, we can have heads or tails, and we split it into cases.

Case 1: The third flip is tails.
The probability of this is $ \frac {1} {2}^3$
Then, we have to have a heads.  $ \frac {1}{2}^4$.
Then it does not matter what we have.
Total probability: $ \frac {1} {16}$

Case 2: The third flip is heads.
If this is true, we end up on the ninth floor, and it does not matter what we flip anymore.
Total probability: $ \frac {1} {8}$

Add these up, $ \frac {3} {16}$[/hide]

[hide="3"]
The circumference of the larger circle is $ 10\pi$ and the smaller circle has circumference $ 4\pi$

It takes one bug $ \frac {10} {3}$ minutes to go around the larger circle, and it takes the other bug $ \frac {8} {5}$ minutes to go around the smaller circle. Now, we wish to find the least common multiple of these two fractions.
$ k*\frac {10}{3}\equal{}x*\frac{8}{5}$
$ \frac {k}{x}\equal{}\frac{12}{25}$ Where $ k,x\in\mathbb {Z}$ and $ >0$.
Thus, the two smallest values will be $ k\equal{}12$ and $ x\equal{}25$

$ \frac {10}{3}*12\equal{}40$ minutes before their next meeting.[/hide]

1 (a somewhat different approach - you would have to do this fast though)
[hide]
well you need 0 mod 7 to make a 1*7 thingamajobober

34 mod 7 + 31 mod 7 + 28 mod 7

6+3+0+4+1

14, 0 (mod 7)

so 5 rows [/hide]

(ma+math/mamath)+bug

[hide="A different approach for 3"]The first bug gets to P every $ \dfrac{10}{3}$ minutes, and the second bug takes $ \dfrac{8}{5}$ minutes. That's $ 200$ and $ 96$ seconds, respectively. Now we just need to find the least common multiple of those two numbers.

$ LCM(200,96)\equal{}2^5*3*5^2\equal{}2400$

It takes $ 2400$ seconds $ \equal{}\dfrac{2400}{60}\equal{}40$ minutes for the bugs to meet again.[/hide]

[quote="1=2"][hide="A different approach for 3"]The first bug gets to P every $ \dfrac{10}{3}$ minutes, and the second bug takes $ \dfrac{8}{5}$ minutes. That's $ 200$ and $ 96$ seconds, respectively. Now we just need to find the least common multiple of those two numbers.

$ LCM(200,96) \equal{} 2^5*3*5^2 \equal{} 2400$

It takes $ 2400$ seconds $ \equal{} \dfrac{2400}{60} \equal{} 40$ minutes for the bugs to meet again.[/hide][/quote]
nice- I like it better finding LCMs with whole numbers, not fractions
=)


4 pairs of couple are lining in a straight line such that no two couples are standing side by side . How many ways are there for such permutation.

[hide]
The total number of all arrangements is equal $8!$. To count the number of desired arrangements, we can count the complement: the number of arrangements with at least one couple sitting next to each other.

The number of arrangements with one couple sitting next to each other is equal to \[\binom{4}{1}7!\] This double counts the arrangements with two couples sitting next to each other so we subtract that \[\binom{4}{2}6!\] Using PIE to correctly count everything, the number of arrangements with at least one couple sitting next to each other is equal to \[\binom{4}{1}7!-\binom{4}{2}6!+\binom{4}{3}5!+\binom{4}{4}4!\] The number of desired arrangements is then equal to \[8!-\binom{4}{1}7!-\binom{4}{2}6!+\binom{4}{3}5!+\binom{4}{4}4! = 24241\] [/hide]

I think you forget that each couple can interchange their place also .


limit
trigonometry
real analysis
real analysis unsolved

Given $ f'(0)= \lim_{n\ \rightarrow \ \infty}\ nf (\frac{1}{n}) \ , \ \frac{-\pi}{2}\ < \cos^{-1}\frac{1}{n}\ <\ \frac{\pi}{2}\ and \ f(0)=0;$  find $ \lim_{n\ \rightarrow \ \infty}\left( \ (n+1)\frac{2}{\pi}\cos^{-1}\frac{1}{n}-n \right )$

i am getting this as 1 but the answer isnt so. Plz help

Answer
[hide]$ \frac{\pi-2}{\pi}$[/hide]

$ \lim_{n\to\infty}\left[(n+1)\frac{2}{\pi}\arccos(1/n)-n\right] = \lim_{n\to\infty}\left[n\left(\frac{2}{\pi}\arccos(1/n)-1\right)+\frac{2}{\pi}\arccos(1/n)\right]$

$ = \lim_{n\to\infty}\frac{\frac{2}{\pi}\arccos(1/n)-1}{\frac{1}{n}}+\frac{2}{\pi}\cdot \frac{\pi}{2}$

$ = \lim_{n\to\infty}-\frac{2}{\pi \sqrt{1+1/n^{2}}}+1 = 1-\frac{2}{\pi}= \frac{\pi-2}{\pi}$.

Thank You for the solution, but in the steps below, I feel you have used L'hopital's rule. But I cant understand whats the use of the other GIVEN data. Could you plz explain.
Thx
[quote="Carcul"]
$ = \lim_{n\to\infty}\frac{\frac{2}{\pi}\arccos(1/n)-1}{\frac{1}{n}}+\frac{2}{\pi}\cdot \frac{\pi}{2}$

$ = \lim_{n\to\infty}-\frac{2}{\pi \sqrt{1+1/n^{2}}}+1 = 1-\frac{2}{\pi}= \frac{\pi-2}{\pi}$.[/quote]

You can also solve the limit by putting n=secx, so as n approaches infinity ,x would approach pi/2.

Isn't this an IIT-JEE'2004(mains)problem

Since there is no $ f$, something's wrong with the statement.

Ignoring that and using a bit of power series, $ \arccos(\frac1n)=\frac{\pi}2-\frac1n+O(n^{-3})$. Using this, the quantity we're taking the limit of is $ n+1-\frac2{\pi}-\frac2{\pi n}+O(n^{-2})-n=1-\frac2{\pi}+O(n^{-1})$, which tends to $ 1-\frac2{\pi}$.


trigonometry
algorithm
search
geometry
angle bisector
exterior angle
geometry solved

Let $M, N$ be the midpoints of the sides $BC$ and $AC$ of $\triangle ABC$, and $BH$ be its altitude. The line through $M$, perpendicular to the bisector of $\angle HMN$, intersects the line $AC$ at point $P$ such that $HP = \frac{1}{2}(AB+BC)$ and $\angle HMN = 45$. Prove that $ABC$ is isosceles.

This is a cute problem, but the condition $\angle HMN = 45^\circ$ is irrelevant - it only allows to calculate the actual angles of the triangle $\triangle ABC$ after it is known to be isosceles.

[color=blue][b]Lemma:[/b][/color] Let M, N be the midpoints of the legs AB, CA of a triangle $\triangle ABC$, respectively, and $H \in CA$ the foot of the B-altitude BH. Let $O, P \in CA$ be the feet of the internal and external bisectors of the angle $\angle HMN$. Then the segment $OP = BC$ and the segment $HP = \frac{AB + BC}{2}$ if and only if the triangle $\triangle ABC$ is isosceles with the base a = BC and the legs b = AB = CA.

[color=blue][b](1) Case a < b[/b][/color] Assume that the triangle $\triangle ABC$ is isosceles and denote $\alpha = \angle A$. Then $a = 2b \sin \frac \alpha 2$. First assume that a < b, which means that the points C, H, O, N, P follow on the line CA in this order. The sides of the triangle $\triangle HMN$ are then equal to

$n = HM = \frac b 2$ (right angle median of the triangle $\triangle ABH$ with hypotenuse b = AB)

$h = MN = \frac a 2$ (midline of the triangle $\triangle ABC$)

$m = NH = AH - AN = b \cos \alpha - \frac b 2$

The distances of the vertex N from the feet O, P of the internal and external bisectors MO, MP of the angle $\angle HMN$ and the distance between them are

$NO = \frac{mh}{n + h},\ \ NP = \frac{mh}{n - h}$

$OP = NO + NP = \frac{mh}{n + h} + \frac{mh}{n - h} = \frac{2hmn}{n^2 - h^2} =$

$= \frac{ab(2b\cos \alpha - 1)}{b^2 - a^2} = \frac{ab^2 (2 \cos \alpha - 1)}{b^2\left(1 - 4 \sin^2 \frac \alpha 2\right)} = \frac{a(2\cos \alpha - 1)}{1 - 2(1 - \cos \alpha)} = a$

$HP = HN + NP = m + \frac{mh}{n - h} = \frac{mn}{n - h} =$

$= \frac{b^2(2 \cos \alpha - 1)}{2(b - a)} = \frac{b^2[1 - 2(1 - \cos \alpha)]}{2b(1 - 2 \sin \frac \alpha 2)} =$

$= \frac{b(1 - 4\sin^2 \frac \alpha 2)}{2\left(1 - 2 \sin \frac \alpha 2\right)} = \frac{b\left(1 + 2 \sin \frac \alpha 2\right)}{2} = \frac{b + a}{2}$

Thus we found $OP = a$ and $HP = \frac{a + b}{2}$. Conversely, assume that $HP = \frac{a + b}{2}$ and denote m = HN as before. Since $MH = \frac{AB}{2} = \frac a 2$ and $MN = \frac{BC}{2}$, we have

$NP = \frac{mb}{a - b}$

$HP = HN + NP = m + \frac{bm}{a - b} = \frac{am}{a - b} = \frac{a + b}{2}$

$HN = m = \frac{a^2 - b^2}{2a}$

$OP = ON + NP = \frac{bm}{a + b} + \frac{bm}{a - b} = \frac{2abm}{a^2 - b^2} = a$

Let Q be the midpoint of OP = a. Since ON < NP (see the above formulas), the points C, N, Q follow on the line CA in this order. The median MQ of the right angle triangle $\triangle OMP$ is equal to $MQ = \frac{OP}{2} = \frac a 2$. MN is a midline of the triangle $\triangle ABC$, hence, $MN = \frac{BC}{2} = \frac a 2$ as well. Thus the triangle $\triangle NMQ$ is isosceles. The segment $HM = \frac{AB}{2} = \frac b 2$ is the median of the right angle triangle $\triangle AHB$ with the hypotenuse AB and the segment $HQ = HP - QP = \frac{a + b}{2} - \frac a 2 = \frac b 2$.  Thus the triangle $\triangle MHQ$ is also isosceles. These 2 isosceles triangles $\triangle MNQ \sim \triangle MHQ$ are similar, having the base angle at the vertex Q common. The triangle $\triangle AMH$ is isosceles with $MA = MH = \frac{AB}{2}$, hence, the angles $\angle BAC \equiv \angle MAH = \angle MHA$ are equal. Since $MN \parallel BC$ and the points C, N, Q follow on the line CA in this order (see above), the angles $\angle BCA = \angle MNQ$ are also equal. As a result, the triangle $\triangle ABC$ is similar to the isosceles triangles $\triangle HMQ \sim \triangle MNQ$, i.e., it is isosceles.

[color=blue][b](2) Case a > b[/b][/color] Again, assume that the triangle $\triangle ABC$ is isosceles, but now a > b. Then the points C, N, O, H, P follow on the line CA in this order and we can similarly calculate

$n = HM = \frac b 2,\ \ h = MN = \frac a 2$ (as before)

$m = NH = AN - AH = \frac b 2 - b \cos \alpha$

$HO = \frac{mn}{h + n},\ \ HP = \frac{mn}{h - n}$

$OP = HO + HP = \frac{mn}{h + n} + HP = \frac{mn}{h - n} = \frac{2hmn}{h^2 - n^2} = a$

$HP = \frac{mn}{h - n} = \frac{b^2(2 \cos \alpha - 1)}{2(b - a)} = \frac{b^2[1 - 2(1 - \cos \alpha)]}{2b(1 - 2 \sin \frac \alpha 2)} =$

$= \frac{b(1 - 4\sin^2 \frac \alpha 2)}{2\left(1 - 2 \sin \frac \alpha 2\right)} = \frac{b\left(1 + 2 \sin \frac \alpha 2\right)}{2} = \frac{b + a}{2}$

and again, $OP = a$ and $HP = \frac{a + b}{2}$. Conversely, assuming $HP = \frac{a + b}{2}$, we can show in an entirely similar way as in the case (1) that the triangle $\triangle ABC$ is similar to the similar isosceles triangles $\triangle HMQ \sim \triangle MNQ$, i.e., it is isosceles.

[color=red][b]Case <HMN = 45o[/b][/color] If we know that the angle $\angle HMN = 45^\circ$, it is easy to determine the angles of the triangle $\triangle ABC$.  Denote $\alpha = \angle MHQ = \angle NMQ = \angle BAC$. First, assume a < b, i.e., the points C, H, O, N, Q, P follow on the line CA in this order. Then

$\angle HMQ = \angle HMN + \angle NMQ = 45^\circ + \alpha$

$\angle HMQ = \frac{180^\circ - \angle MHQ}{2} = 90^\circ - \frac \alpha 2$

$45^\circ + \alpha = 90^\circ - \frac \alpha 2,\ \ \alpha = 30^\circ$

$\angle ABC = \angle ACB = \frac{180^\circ - \alpha}{2} = 75^\circ$

Second, assume that the points C, N, O, H, P follow on the line CA in this order. Then we can similarly calculate

$\angle HMQ = \angle NMQ - \angle HMN = \alpha - 45^\circ$

$\angle HMQ = \frac{180^\circ - \angle MHQ}{2} = 90^\circ - \frac \alpha 2$

$\alpha - 45^\circ = 90^\circ - \frac \alpha 2,\ \ \alpha = 90^\circ$

$\angle ABC = \angle ACB = \frac{180^\circ - \alpha}{2} = 45^\circ$

Whew!!!  Fixed. :yup:

nice solution yetti but very very long.is there any more solution for this problem?

A different solution:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle, and let M and N be the midpoints of its sides BC and CA. Let H be the foot of its altitude from the vertex B. Let the external angle bisector of the angle HMN intersect the line CA at a point P. Then,

[b](a)[/b] We have < HMN = |C - A|.
[b](b)[/b] We have $HP=\frac{c+a}{2}$ if and only if a = b.[/color]

Note that, while part [b](b)[/b] of this theorem proves that, given the conditions of the problem, the triangle ABC is isosceles, part [b](a)[/b] allows to compute its angles from < HMN = 45°.

Here is a [i]proof of Theorem 1[/i]:

We assume that $A\geq C$ throughout the following proof. In the case $A\leq C$, everything is analogous.

See the figure below.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=390[/img]

Since < BHC = 90°, the point H lies on the circle with diameter BC. The center of this circle is obviously the midpoint M of the segment BC. Thus, MH = MC. Hence, the triangle CMH is isosceles; thus, < MHC = < MCH, or, equivalently, < MHC = C. On the other hand, since M and N are the midpoints of the sides BC and CA of triangle ABC, we have MN || AB, thus < MNC = < BAC, what rewrites as < MNC = A. By the exterior angle theorem in triangle HMN, we have < MNC = < HMN + < MHN. Thus,

< HMN = < MNC - < MHN = < MNC - < MHC = A - C = |C - A|,

and Theorem 1 [b](a)[/b] is proven.

Note that, since M is the midpoint of the segment BC, we have $MC=\frac{BC}{2}=\frac{a}{2}$, so that $MH=MC=\frac{a}{2}$.

Now, since the line MP is the external angle bisector of the angle HMN, we have $\measuredangle NMP=90^{\circ}-\frac{\measuredangle HMN}{2}$, so that

$\measuredangle HMP=\measuredangle HMN+\measuredangle NMP=\measuredangle HMN+\left(90^{\circ}-\frac{\measuredangle HMN}{2}\right)$
$=90^{\circ}+\frac{\measuredangle HMN}{2}=90^{\circ}+\frac{A-C}{2}$.

Thus, $\sin\measuredangle HMP=\cos\frac{A-C}{2}=\cos\frac{C-A}{2}$. Further, by the sum of angles in triangle HMP, we have

$\measuredangle MPH=180^{\circ}-\measuredangle HMP-\measuredangle MHP=180^{\circ}-\measuredangle HMP-\measuredangle MHC$
$=180^{\circ}-\left(90^{\circ}+\frac{A-C}{2}\right)-C=90^{\circ}-\frac{C+A}{2}=90^{\circ}-\frac{180^{\circ}-B}{2}$            (by the sum of angles in triangle ABC)
$=\frac{B}{2}$.

Thus, by the Sine Law in triangle HMP, we have

$\frac{HP}{MH}=\frac{\sin\measuredangle HMP}{\sin\measuredangle MPH}=\frac{\cos\frac{C-A}{2}}{\sin\frac{B}{2}}$.

But by the Mollweide formulas, $\frac{\cos\frac{C-A}{2}}{\sin\frac{B}{2}}=\frac{c+a}{b}$. Thus, $\frac{HP}{MH}=\frac{c+a}{b}$. Hence,

$HP=MH\cdot\frac{c+a}{b}=\frac{a}{2}\cdot\frac{c+a}{b}$.

Thus, $HP=\frac{c+a}{2}$ holds if and only if $\frac{c+a}{2}=\frac{a}{2}\cdot\frac{c+a}{b}$, what is obviously equivalent to a = b. Hence, Theorem 1 [b](b)[/b] is proven as well.

  Darij

I will use the Darij's notations and will prove the point $(b)$.

I suppose w.l.o.g. $a\ne c$. Thus, $2\cdot NA=b,\ 2\cdot MH=a,\ 2\cdot MN=c$.

$\frac{PH}{MH}=\frac{PN}{MN}\Longrightarrow \frac{PH}{a}=\frac{PN}{c}=\frac{NH}{|a-c|}$.
$NH=NA-HA=\frac b2 -\frac{b^2+c^2-a^2}{2b}=\frac{|a^2-c^2|}{2b}\Longrightarrow NH=\frac{|a^2-c^2|}{2b}$.
$PH=\frac{a}{|a-c|}\cdot NH=\frac{a}{|a-c|}\cdot \frac{a(a^2-c^2)}{2b}=\frac{a(a+c)}{2b}\Longrightarrow HP=\frac{a(a+c)}{2b}$.
Therefore, $HP=\frac{a+c}{2}\Longleftrightarrow \frac{a(a+c)}{2b}=\frac{a+c}{2}\Longleftrightarrow a=b$.

[quote="yetti"][quote="armpist"]In my sketch the property HP = 1/2 * (AB +BC) happens for isosceles triangles

   independently of the angle HMN.

   Please verify.

   Thank you.

   M.T.[/quote]

You can try the following version of greedy algorithm:

[b]Step 1:[/b] Read the reply one. If this is sufficient, stop. Otherwise, go to step two.
[b]Step 2:[/b] Read the reply two. If this is sufficient, stop. Otherwise, go to step one.

Hope it helps.

Yetti.[/quote]

I don't get it.

[quote="mysmartmouth"]I don't get it.[/quote]

Feel free to ask for details.

is it like [i]if and only if[/i], yetti ?

Davron Latipov

Yes, it is "if and only if", but that's not the point. The converse of the posted problem is actually easier to prove, which is what I did first. But the 1st reply begins

[quote="yetti"]... the condition $\angle HMN = 45^\circ$  is irrelevant - it only allows to calculate the actual angles of the triangle after it is known to be isosceles.[/quote]
The reply to armpist was a part of self-defence against his permanent attacks, which you can find with the search button, if you are really curious.

Yetti


function
analytic geometry
algebra
domain
calculus

Use the vertical line test to determine whether the graph is a function or not. Basically, if you slide a pencil which is positioned vertically from left to right across the graph, if the pencil ever touches the graph at more than one point, then it is not a graph.
Functions are one-to-one and many-to-one (i.e. each x-coordinate maps onto at most 1 y coordinate).

In your case, both are functions.

[quote="BlackSwordsman"]A function only gives one output for each input.
So the equations you gave are indeed functions.

But something like $ \sqrt {x}$ isn't a function because for each input there are 2 outputs (negative and positive)[/quote]

One might define $ \sqrt{x}$ as the positive root of x; then it is of course a function. However, the equation $ y^2\equal{}x$ can't be represented by a function $ f: x\mapsto y$ since there are 2 y's for every x. :)

[quote="BlackSwordsman"]A function only gives one output for each input.
So the equations you gave are indeed functions.

But something like $ \sqrt {x}$ isn't a function because for each input there are 2 outputs (negative and positive)[/quote]

I thank you for your simple explanation.

[quote="sharkman"]Determine if each equation given is a function. If it is or is not a function, explain the reason why. How do I know if an equation is a function or not?

(1) y = |x| + 4

(2) y = (1/x)[/quote]
Both of these equations are functions, although they aren't one-to-one (meaning that no two x-values can generate the same y-value). You can tell if an equation is a function or not by graphing it and using the vertical line test. In other words, an equation is not a function if it has two possible y-values that can be generated from one x-value.

[quote="gauss1181"]Both of these equations are functions, although [b]they aren't one-to-one [/b](meaning that no two x-values can generate the same y-value). You can tell if an equation is a function or not by graphing it and using the vertical line test. In other words, an equation is not a function if it has two possible y-values that can be generated from one x-value.[/quote]
$ f(x) \equal{} \frac {1}{x}$ is one-to-one if you use the understood domain and range.  That is, $ \iff f: \mathbb{R} \minus{} \{0\} \to \mathbb{R} \minus{} \{0\}$.  So technically, it is one-to-one from general observation.

[quote="JRav"]You people and your simple definitions.  In multivariable calculus, nothing is that simple.  How I long for those days. :lol:[/quote]
Really haha?  I thought multi is so much easier than my basic proofs class.

Well I'm sure MV Calc is easier than analysis, but it's still harder than HS algebra and HS trig.


algebra
polynomial
modular arithmetic
geometry
3D geometry
induction
bir tiyinga qimmat masala

Prove that there is no polynomials with integer coefficients,$ Q_1(x),Q_2(x),\dots,Q_n(x)\in\mathbb{Z}[x]$,such that
$ 3x\equal{}\sum_{i\equal{}1}^{n}(Q_i(x))^3$

The $ 3$ of the [i]RHS[/i] has made me spend some time on trying to work $ \pmod 3$. Then I remembered the identity $ 18k \equal{} (2k \plus{} 14)^3 \plus{} (3k \plus{} 30)^3 \plus{} ( \minus{} 2k \minus{} 23)^3 \plus{} ( \minus{} 3k \minus{} 26)^3 \plus{} ( \minus{} 1)^3$ and tried to work $ \pmod 9$. This fails. But from this identity we deduce that if $ x$ is representable as sum of cubes of polynomials, then $ 3x \equal{} 18x \minus{} 15x$ would be representable in that form too. The same holds for $ 5x$, since $ 3x \equal{} 18x \minus{} 3\cdot5x$.
Having these in mind, it is natural to try to prove that
[i]If $ k$ is an odd integer, then there are no polynomials $ Q_1,\ldots,Q_n\in \mathbb{Z}[X]$, so that $ kx \equal{} \sum_{i \equal{} 1}^n[Q_i(X)]^3$.[/i]
Alternatively, we have to prove that there are no polynomials $ Q_1,\ldots,Q_n\in\mathbb{F}_2[X]$, so that $ x \equal{} \sum_{i \equal{} 1}^n[Q_i(X)]^3$.

After playing around with cubes of simple polynomials (with all coefficients $ 0$ or $ 1$) I discovered a wonderful
[b][i]Lemma[/b][/i]
Let $ f\in\mathbb{Z}[X]$. Let $ f^3 \equal{} \sum_{i\ge0} a_ix^i$. Then $ \boxed{\sum_{i\ge0}a_{3i \plus{} 1} \equiv\sum_{i\ge0}a_{3i \plus{} 2}\pmod2}$. Call this property $ P$.
[b][i]Proof of Lemma[/i][/b]
Firstly note that if polynomials $ f,g$ have property $ P$, then $ f \plus{} g,f \minus{} g$ and $ x^{3t}f$ also have property $ P$, where $ t\in Z$.
We can proceed by induction over the degree of $ f$. We can take $ f\in\mathbb{F}_2[X]$. $ \deg f\in\{0,1,2\}$ can be checked by hand (there are just $ 2^3$ polynomials).
Now assume the statement true for polynomials $ f$ with $ \deg f\le n \minus{} 1$ and consider $ g \equal{} x^n \plus{} f$, where $ \deg f\le n \minus{} 1$. Then $ g^3 \equal{} x^{3n} \plus{} 3x^{2n}f^2 \plus{} 3x^nf^2 \plus{} f^3$. We are not interested in $ x^{3n} \plus{} f^3$ as it has property $ P$. We have to show that $ 3x^{2n} f \plus{} 3x^nf^2$ has propery $ P$ by the induction hypothesis.

[i]Case 1: $ n\equiv0\pmod3$[/i]
Then $ 3x^{2n}f \plus{} 3x^nf^2$ has property $ P$ iff $ 3f \plus{} 3f^2$ has property $ P$, which is true, because $ 3f \plus{} 3f^2 \equal{} (f \plus{} 1)^3 \minus{} f^3 \minus{} 1$, so $ 3f \plus{} 3f^2$ is a linear combination of polynomials having property $ P$ (since $ \deg f \equal{} \deg (f \plus{} 1)\le n \minus{} 1$, by the induction hypothesis, these polynomials have $ P$).

[i]Case 2: $ n\equiv1\pmod3$[/i]
Then $ 3x^{2n}f \plus{} 3x^nf^2$ has property $ P$ iff $ 3x^2f \plus{} 3xf^2$ has property $ P$. But $ 3x^2f \plus{} 3xf^2 \equal{} (f \plus{} x)^3 \minus{} f^3 \minus{} x^3$, and $ f \plus{} x,f,x$ have property $ P$.

[i]Case 3: $ n\equiv2\pmod3$[/i]
Then $ 3x^{2n}f \plus{} 3xf^2$ has $ P$ iff $ 3x^4f \plus{} 3x^2f^2$ has $ P$, which is true, since $ 3x^4f \plus{} 3x^2f \equal{} (f \plus{} x^2)^3 \minus{} f^3 \minus{} x^6$, and $ f \plus{} x^2,f$ have $ P$ by the induction hypothesis.$ \blacksquare$

It is clear that $ Q \equal{} Q_1^3 \plus{} Q_2^3 \plus{} \ldots \plus{} Q_n^3$ has property $ P$ for any $ Q_1,\ldots,Q_n\in \mathbb{Z}[X]$ while $ x$ has not and we're done.
Moreover, the proof shows that no polynomial in $ \mathbb{Z}[X]$ for which the sum of the coefficients near $ X,X^4,X^7,\ldots$ has different parity than the sum of the coefficients near $ X^2,X^5,X^8,\ldots$ is representable as a sum of cubes of integer polynomials.

Prove the lemma as above by substituting in the third root of unity $ \omega$ into $ f^3$. Then compare the coefficients of $ \omega$ and $ \omega ^2$, and find they have the same parity. The coeffiecient of $ \omega$ and $ \omega ^2$ are the sum of coefficients of terms with exponents 1,2 (mod 3), plus the same constant, and the lemma follows.
(We use the fact that if $ a\plus{}b\omega \plus{} c \omega^2 \equal{} 0$ then $ a\equal{}b\equal{}c$.)


Suppose that $p(x)\in \mathbb C[x]$ and all roots of $p$ are in the unit circle. Prove that all roots of $p(z)+z^n\overline{p(\frac1{\overline{z}})}$ are on the unit circle.

The problem is exciting I have a hint.
[hide="Hint"]

You can use this that if $|z-\omega|^2-|z\overline{\omega}-1|^2=(|z|^2-1)(|\omega|^2-1)$
[/hide]

Nice hint!
Here is the proof : we show that neither inside nor outside there are no roots.
Suppose that $p(x)=c(x-a_1)(x-a_2)\ldots (x-a_n)$. Then $q(x): =p(x)+x^n\overline{p(\frac{1}{\overline x})}$. We have
\[ q(x)=c(x-a_1)(x-a_2)\ldots (x-a_n)+\overline c(1-\overline{a_1}x)(1-\overline{a_2}x)\ldots (1-\overline{a_n}x) \]
Suppose that $|x|<1$ then for each $i$ we have $|x-a_i|^2-|1-\overline{a_i}x|^2=(|z|^2-1)(|a_i|^2-1)>0$.
So $|x-a_i|>|1-\overline{a_i}x|$ so we have
\[ |q(x)|\geq |c(x-a_1)(x-a_2)\ldots (x-a_n)|-|\overline c(1-\overline{a_1}x)(1-\overline{a_2}x)\ldots (1-\overline{a_n}x)|>|c||(1-\overline{a_1}x)(1-\overline{a_2}x)\ldots (1-\overline{a_n}x)|-|\overline c||(1-\overline{a_1}x)(1-\overline{a_2}x)\ldots (1-\overline{a_n}x)|=0 \]
And in the case $|x|>1$ we have $|x-a_i|<|1-\overline{a_i}x|$ and so
\[ |q(x)|\geq |\overline c(1-\overline{a_1}x)(1-\overline{a_2}x)\ldots (1-\overline{a_n}x)|-|c(x-a_1)(x-a_2)\ldots (x-a_n)|>|\overline c||(x-a_1)(x-a_2)\ldots (x-a_n)|-|c||(x-a_1)(x{}_a{}_2)\ldots (x-a_n)=0 \]
So inside and outside the unit circle there are no roots.


first place - Aleksandra Kasman - student of Yakov Kasman

second place - Forrest Moody - student of Kevin Chance

third place - Xinke Guo-Xue - student of Delores Howard

Honorable mentions:

Jean Yu, student of Jyesook Jung

Eugene Wu, student of Delores Howard

Botong Ma, student of Tatiana Kasman

why is this in alabama math?

jeez, aleksandra is good. her mom told me that she had like 4 honorable mentions and like 2 winners.

congrats hurdler!

Well, I'm not sure how many AL users check the piano discussions forum on aops, and this is an AL forum, not necessarily an AL math forum I think.

I wonder how much botong and jean practice each day.

well the real question is, how long do YOU practice everyday?

an hour ...

.. so less than first and second place i'm guessing


[quote="Quevvy"]$2005^{1002.5}<2006^{1002.5}<2005^{1003}$
[/quote]

Er.  The first part contributes nothing to your solution, and the second part isn't at all obvious.

Well, there could be some rearrangement issues.


Number 163, chap 10, vol 2.

My problem is in the solution (because i had no idea where to start) I can see that it is the result of a dot product.
But
1. I dont see how the max value of a dot product is the product of the lengths... What about when cos x > 1?  I also dont understand how the maximum is attained when the two vectors are parallel. Woudlnt that mean that cos x = 1? But once again, what about when cos x > 1? And I dont see how the min is just neg the max.

Btw.... the equation was f(x,y) = 2sinxcosy + 3sinxsiny + 6cosx
find f min and fmax

The cosine of an angle can never be greater than 1.

heres how i solved it..i didnt use the dot product, just the sum formula stuff.

2sinxcosy+3sinxsiny+6cosx=sinx(2cosy+3siny)+6cosx

2cosy+3siny can be expressed as a single sine, let this expression = some g(y), and scale it to K, or Kg(y)=2Kcosy+3Ksiny. sum formula for sine is sinzcosy+coszsiny. 2K would equal sinz (see why?) and 3K would be cosz. squaring and adding, we get 4K :^2: +9K :^2:=1, or 13K :^2: =1, so K=1/sqrt(13). Kg(y) = sin(y+Arcsin[1/sqrt(13)]), so g(y)=sqrt(13)sin(blah blah blah). [b]The max and min can be found by checking for the amplitude, which is sqrt(13).[/b]

Then f(x), our original function, becomes sqrt(13)sinx+6cosx, because [b]our choice of y doesnt effect our choice of x, so we could just maximize that expression and be fine[/b].

It doesn't matter whether we chose it to be negative or positive sqrt(13), because the maximums and minimums still occur at the ends of the waves (THE AMPLITUDE).

so we can express our new f(x) as a single sine (see why?), and after some work, we get f(x)=7sin(x+Arcsin[sqrt(13)/7]).

[b]Because the tops of the waves are the amplitude (draw sinx and see), we get the max and min to be the low and high points of the wave, given by the amplitude, or plus or minus 7 is our answer.

btw, the dot product is maximized at when cosx=1 because, dot product being ||x|| ||y|| cosx, if we took cosx :le: 1, then we would just be taking a fraction of ||x|| ||y|| part, and we want the entire thing. tell me if this is confusing... :?

I cant believe I made such a stupid mistake..... Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 Cos X cant be greater than 1 ....

And now it all makes so much sense...when they are going in opposite directions, then it is cos 180 = -1.... ARG... So easy, how could i make such  A STUPIEEED mistake. Arg!!

Time to get back to this art history stuff. GRR.. Art history (eh.. free ap.)


geometry
geometric transformation
reflection
geometry unsolved

Let $ABC$ be an isosceles triangle with $AB=AC$. A circle is tangent to sides $AB,AC$ at $R,S$ and cuts the side $BC$ at $K,L$, where $B,K,L,C$ lie in that order. $AK$ cuts the circle again at $M$. Let $P$ be the reflection of $K$ with respect to $B$. Show that $M,R,P$ are collinear.

Easy one :)   We need to prove that $AM/MK*PK/PB*BR/RA=1$ (@)Let $T$ the intersection of $RS$ and $AK$. Then $A,M,T,K$ is an armonic division, so $AM/AK=TM/TK$. Moreover we have that $BR/RA=KT/TA$. After an easy computation we get that the expression from @ is $1$ and the result follows.


LaTeX
function
limit
calculus
calculus computations

Ok, so i have officially started Calc I in college :D

Anyways, here is a simple problem that i think i got correct, but i just want to make sure my reasoning is correct. It has to do with Einstein's theory of relativity. ONCE AGAIN, i apologize for my lack of Latex D:

lim (as v--> c from the left) of [Mo/(1-(v^2/c^2)^1/2] where Mo is the rest mass (a constant) and c is the speed of light (another constant, obviously). In this problem, mass is a function of velocity.

Well, here is my reasoning. as V--> c from the left, the quantity (v^2/C^2) approaches something infinitely close to 1. Thus, the whole expression in the denominator approaches zero, but it's still a little above zero because (v^2/C^2) is just a little less than 1. Thus the limit is Mo(infinity) which is just infinty.

Your reasoning is correct.

In $\text{\LaTeX},\ \lim_{v\to c^{-}}\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\infty.$


trigonometry
ratio
inequalities
inequalities proposed

Let $ a,b,c\geq 0$ and $ ab\plus{}bc\plus{}ca\equal{}3$
Prove that $ \frac{1}{1\plus{}a^2}\plus{}\frac{1}{1\plus{}b^2}\plus{}\frac{1}{1\plus{}c^2}\geq\frac{3}{2}$

Let $ a\equal{}\tan \alpha ,\ b\equal{}\tan \beta ,\ c\equal{}\tan \gamma$, we have ....

[quote="mufc"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}$[/quote]
The expanding gives:
$ \sum_{cyc}\frac {1}{1 \plus{} a^2}\geq\frac {3}{2}\Leftrightarrow(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 \plus{} 4abc(a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc)\geq0.$

[quote="kunny"]Let $ a \equal{} \tan \alpha ,\ b \equal{} \tan \beta ,\ c \equal{} \tan \gamma$, we have ....[/quote]
Continue,please.

[quote="mufc"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}$[/quote]

Here let us apply Cauchy-Schwartz-Buniakowski inequality...
an Other way is
You know a^2 +b^2 +c^2 >=ab +bc +ac =3 so we have then (a +b +c) >= 3 by collecting them in the ratio one / two ; so you will go.
Another way to do is:
Let us denote by $ f(a,b,c)$ the $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}$. Arithmetical mean is bigger than harmonical mean, so $ {\frac{(a^2 \plus{} 1) \plus{} (b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}{3}\geq \frac{3}{f(a,b,c)}}$ but left hand side is bigger than $ 2$ (because $ ab \plus{} bc \plus{}ca \equal{} 3$). So $ f(a,b,c)\geq\frac{3}{2}$ by dividing.
Thanks for you reading.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164187

$ {(a \minus{} b)^2}\geq{0}$
$ {(b \minus{} c)^2}\geq{0}$
$ {(c \minus{} a)^2}\geq{0}$
adding gives $ {2(a^2 \plus{} b^2 \plus{}c^2)}\geq2{ab \plus{} bc \plus{} ca}$ so $ {a^2 \plus{} b^2 \plus{}c^2}\geq{ab \plus{} bc \plus{} ca} \equal{} 3$ so $ {(a^2 \plus{}1) \plus{}(b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}\geq6$ and $ \frac{(a^2 \plus{}1) \plus{}(b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}{3}\geq2$ so $ {f(a,b,c)}$ will go right side with becoming the $ \frac{3}{2}$ left side

[quote="madshiet"]$ {(a \minus{} b)^2}\geq{0}$
$ {(b \minus{} c)^2}\geq{0}$
$ {(c \minus{} a)^2}\geq{0}$
adding gives $ {2(a^2 \plus{} b^2 \plus{} c^2)}\geq2{ab \plus{} bc \plus{} ca}$ so $ {a^2 \plus{} b^2 \plus{} c^2}\geq{ab \plus{} bc \plus{} ca} \equal{} 3$ so $ {(a^2 \plus{} 1) \plus{} (b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}\geq6$ and $ \frac {(a^2 \plus{} 1) \plus{} (b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}{3}\geq2$ so $ {f(a,b,c)}$ will go right side with becoming the $ \frac {3}{2}$ left side[/quote]
you have
$ \frac{(a^2 \plus{}1) \plus{}(b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}{3}\geq2$
and $ \frac{(a^2 \plus{}1) \plus{}(b^2 \plus{} 1) \plus{} (c^2 \plus{} 1)}{3}\geq \frac{3}{f(a,b,c)}$
but is not implie that$ f(a,b,c) \ge \frac{3}{2}.$

Given $ a,b,c \geq0 : ab \plus{} bc \plus{} ca \equal{} \frac {3}{2}$. Prove that:
 $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2} \leq 2$

Even more similar:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103762

[quote="mufc"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}$[/quote]
This inequality is [b]Problem 2.26 - page 64[/b] in new book (where it is presented with two simple solutions, and one of them is by Cauchy-Schwarz only):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote="mufc"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}$[/quote]

more similar:
 Let $ a,b,c$ be positive reals such that $ a\plus{}b\plus{}c\equal{}3$

Prove that: $ \frac{a}{1\plus{}b^2}\plus{}\frac{b}{1\plus{}c^2}\plus{}\frac{c}{1\plus{}a^2}\ge \frac{3}{2}$

[quote="xyy"][quote="mufc"]Let $ a,b,c\geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that $ \frac {1}{1 \plus{} a^2} \plus{} \frac {1}{1 \plus{} b^2} \plus{} \frac {1}{1 \plus{} c^2}\geq\frac {3}{2}$[/quote]

more similar:
 Let $ a,b,c$ be positive reals such that $ a \plus{} b \plus{} c \equal{} 3$

Prove that: $ \frac {a}{1 \plus{} b^2} \plus{} \frac {b}{1 \plus{} c^2} \plus{} \frac {c}{1 \plus{} a^2}\ge \frac {3}{2}$[/quote]
$ \sum \frac{a}{1\plus{}b^2}\equal{}\sum\left(a\minus{}\frac{ab^2}{1\plus{}b^2}\right)\geqslant \sum\left(a\minus{}\frac{ab}{2}\right)\equal{}3\minus{}\frac{\sum ab}{2}$
$ \sum ab\leqslant \frac{\left(\sum a\right)^2}{3}\equal{}\frac{1}{3}$

$ ...$

[quote="xyy"]
 Let $ a,b,c$ be positive reals such that $ a \plus{} b \plus{} c \equal{} 3$

Prove that: $ \frac {a}{1 \plus{} b^2} \plus{} \frac {b}{1 \plus{} c^2} \plus{} \frac {c}{1 \plus{} a^2}\ge \frac {3}{2}$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422697#422697


probability
algorithm
combinatorics proposed
combinatorics

If we're a given a fair coin and we're allowed to make any finite number of flips, we can measure any probability that has form $ a/2^b$.

But if the coin has probability of head as $ \frac{1}{3}$, then which rational probabilities can we measure with finite number of flips?

Your first statement seems incorrect to me.  Let $ \alpha\in (0,1)$ be an arbitrary real number.  Here's an algorithm that with probability 1 uses a finite number of flips of a fair coin and generates a random variable $ Z$ taking value 1 with probability $ \alpha$ and taking value 0 otherwise:

Let $ (a_1,a_2,a_3,\ldots)$ be any sequence of digits after the dot in the binary expansion of $ \alpha$ (any means that in case of ambiguity like $ 0.1$ or $ 0.0(1)$ we chose whichever form).

Let us denote a random variable $ X_i$ equal 0 if a head occured on the $ i$-th flip and 1 otherwise.
We consecutively flip the coin as long as $ X_i\equal{}a_i$.
Now, if $ X_i<a_i$, we take $ Z\equal{}1$, and if $ X_i>a_i$ we take $ Z\equal{}0$.
The algorithm stops with probability 1 (the probability that you flip more than $ n$ coins is $ 2^{\minus{}n}$).
The algorithm can only stop for $ i$ such that $ a_i\equal{}1$ and then for each such $ i$ there's exactly one trajectory that occurs with probability $ 2^{\minus{}i}$, so that:
$ P(Z)\equal{}\sum_{i: a_i\equal{}1} 2^{\minus{}i}\equal{}\alpha$, which completes the proof.

Wojtek

So that means that you can get any real number whose binary representation in base two is finite. This is equivalent to saying it has form $ \frac{a}{2^b}$.

For example, $ \frac{1}{3}$ in base two is $ 0.0101010101.....$ so there is no way to measure $ \frac{1}{3}$ with finite number of flips.

Moreover, irrational numbers are out of the question because they have infinite binary representations as well (if it were finite, it would be rational).

No, the binary representation needs not be finite.
What I wrote is that you can generate arbitrary probabilities in finite number of steps as long as an upper bound for steps is not given in advance.

Wojtek

Hello,

Thank you. Hmm strictly speaking you are correct, we will terminate with a high probability. But that's not what I'm looking for. Let me reword the problem then.

Suppose you are given a fair coin, and a real number $ r$ between 0 and 1 inclusive. You need to then mention a natural number $ n$ such that, after throwing the coin at most $ n$ times, you can simulate the probability $ r$.
For which $ r$ is this possible?
Answer: $ r$ whose binary representation is finite.

now if we replace the coin with an unfair coin whose probability is $ \frac{1}{3}$, for which $ r$ is this possible?
For example, I know that if $ r \equal{} k/9$ for $ k$ integer, we can do it. I generated the algorithm by hand. But I don't know if it's always possible for $ r \equal{} k/27, k/81, ...$. Also, I don't know how to prove that there are no others.

Even trickier alternative: you have 2 coins. One is fair, and one is unfair like the coin above. For which $ r$ is this possible? My guess is that $ k/(2^a3^b)$ but again, I can't prove it yet.

[quote="hendrata01"]
Suppose you are given a fair coin, and a real number $ r$ between 0 and 1 inclusive. You need to then mention a natural number $ n$ such that, after throwing the coin at most $ n$ times, you can simulate the probability $ r$.
For which $ r$ is this possible?
Answer: $ r$ whose binary representation is finite.
[/quote]

I still have to disagree.  Or, at least, ask for some kind of definition of what you mean by simulating a probability.  For example, I can flip two fair coins and ask "What is the probability that they are both heads given that they are not both tails?"  The answer is $ \frac{1}{3}$, which is not of the form $ \frac{k}{2^n}$.  And I can, in the conventional sense of the word, 'simulate' this just fine.  Are conditional probabilities not allowed?  More generally, are you restricting yourself to just one particular probability measure?


IN CELEBRATION OF INTERNATIONAL CAPS LOCK DAY, ALL POSTS I'M WRITING TODAY WILL BE AT LEAST 95% CAPS.

THAT IS AWESOME! I WISH EVERYONE A MERRY CAPS LOCK!

stop spamming immediatly! <-- has to be seen as big capslockes letters.

@moderators: please really delete this kind of stupid thread, it's in no point better than the action with $3$'s.

[URL=http://www.google.com/search?q=international+caps+lock+day&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official]HTTP://WWW.GOOGLE.COM/SEARCH?Q=INTERNATIONAL+CAPS+LOCK+DAY&START=0&IE=UTF-8&OE=UTF-8&CLIENT=FIREFOX-A&RLS=ORG.MOZILLA:EN-US:OFFICIAL
[/URL]

RESULTS 1 - 10 OF ABOUT 1,970,000 FOR INTERNATIONAL CAPS LOCK DAY. (0.10 SECONDS)

1,970,000 SOURCES SAY YOU'RE WRONG.

NOW HOW ABOUT YOU GET THAT SAND OUT AND STOP BEING SO CRANKY.

[quote="ZetaX"]No one is cranky.
It's the people like you that caused the falldown of G&FF. You have no idea how it was here before all you spammers joined. Period.[/quote]

REMEMBER, ADMITTANCE OF YOUR CRANKINESS IS THE FIRST STEP TO RECOVERY. THE NEXT IS TO STOP CONCEALING A SANDBOX.

THE POINT OF GAMES & FUN FACTORY IS FUN, WHICH EVERYBODY WITHOUT A SANDBOX SEEMS TO BE HAVING.

NOW PLEASE STOP SPAMMING IN THIS TOPIC, YOU HAVE NO IDEA WHAT IT WAS LIKE BEFORE YOU STARTED POSTING IN THIS TOPIC. PERIOD.

AND YOU SHOULD GET AN IDEA BY READING THE PAST POSTS.

[quote="fredbel6"][quote="ZetaX"]No one is cranky.
It's the people like you that caused the falldown of G&FF. You have no idea how it was here before all you spammers joined. Period.[/quote]
I keep stressing that Games and Fun does earn its place, but that does not mean there are no limitations here.  If it gets out hand and it is locked down forever... there will be spam refugees all over our beloved Mathlinks.
That will make Mathlinks look like this :
[img]http://www.daf.uni-mainz.de/landeskunde/2003_1/Musik/Images/Love%20parade.jpg[/img]
So lock this! :mad:[/quote]

STOP SPAMMING PLEASE.

YOU SPAMMERS ARE THREATENING THE WHOLE REMOVAL OF G&FF. THERE ARE LIMITATIONS TO HOW OFF TOPIC YOU CAN GET EVEN IN G&FF.

International caps lock day does not exist.

Proof:

Assume it exists. Then there is a wikipedia article about it.
http://en.wikipedia.org/wiki/Special:Search?search=international+caps+lock+day
Contradiction.


calculus
integration
conics
ellipse
geometry
trigonometry
calculus computations

In the $ x \minus{} y$ plane, let $ l$ be the tangent line at the point $ A\left(\frac {a}{2},\ \frac {\sqrt {3}}{2}b\right)$ on the ellipse $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2}\equal{}1\ (0 < b < 1 < a)$. Let denote $ S$ be the area of the figure bounded by $ l,$ the $ x$ axis and the ellipse.

(1) Find the equation of $ l$.

(2) Express $ S$ in terms of $ a,\ b$.

(3) Find the maximum value of $ S$ with the constraint $ a^2 \plus{} 3b^2 \equal{} 4$.

1.

$ y\equal{}\frac{\minus{}2b}{a\sqrt 3} x \plus{}\frac{5b}{2\sqrt 3}$

2.

$ S\equal{}\frac{3ab\sqrt 3}{16}\minus{} \int_{\frac{a}{2}}^a \frac{b}{a} \sqrt{a^2\minus{}x^2}dx \equal{} \frac{ab 3\sqrt 3}{16}\minus{} \frac{ab}{2}(\frac{\pi}{3}\minus{}\frac{\sqrt 3}{4})\equal{}(\frac{7\sqrt 3}{16}\minus{}\frac{\pi}{6})ab$

$ S(b)\equal{}\frac{\pi}{6} b\sqrt{4\minus{}3b^2}, S'(b)\equal{}0$ when $ b\equal{}\sqrt{\frac{2}{3}}$

so $ S(\sqrt{\frac{2}{3}})\equal{}\frac{7}{8}\minus{}\frac{\pi}{3\sqrt 3}$

You seemed to make a mistake in finding the equation of $ l$.

whoops extra two flying around on my derivative.. sucks. ill let someone else fix it =/

[quote="kunny"]In the $ x \minus{} y$ plane, let $ l$ be the tangent line at the point $ A\left(\frac {a}{2},\ \frac {\sqrt {3}}{2}b\right)$ on the ellipse $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (0 < b < 1 < a)$. Let denote $ S$ be the area of the figure bounded by $ l,$ the $ x$ axis and the ellipse.

(1) Find the equation of $ l$.

(2) Express $ S$ in terms of $ a,\ b$.

(3) Find the maximum value of $ S$ with the constraint $ a^2 \plus{} 3b^2 \equal{} 4$.[/quote]

(1)

$ \frac {x}{2a}\plus{}\frac {\sqrt{3}y}{2b}\equal{}1$

[quote="kunny"]In the $ x \minus{} y$ plane, let $ l$ be the tangent line at the point $ A\left(\frac {a}{2},\ \frac {\sqrt {3}}{2}b\right)$ on the ellipse $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (0 < b < 1 < a)$. Let denote $ S$ be the area of the figure bounded by $ l,$ the $ x$ axis and the ellipse.

(1) Find the equation of $ l$.

(2) Express $ S$ in terms of $ a,\ b$.

(3) Find the maximum value of $ S$ with the constraint $ a^2 \plus{} 3b^2 \equal{} 4$.[/quote]

(2)

$ S\equal{} \frac ab\int_0^{\frac {\sqrt {3}}{2}b}2b\minus{}\sqrt{3}y\minus{}\sqrt{b^2\minus{}y^2}\ dy$

$ \equal{}\frac ab\left(2by\minus{}\frac{\sqrt 3}{2}y^2\minus{}\frac y2\sqrt{b^2\minus{}y^2}\minus{}\frac {b^2}2 \sin^{\minus{}1}\frac yb\right)_0^{\frac {\sqrt {3}}{2}b}\equal{}\boxed{\boxed{\frac{3\sqrt {3} \minus{}\pi}{6}ab}}$

[quote="kunny"]In the $ x \minus{} y$ plane, let $ l$ be the tangent line at the point $ A\left(\frac {a}{2},\ \frac {\sqrt {3}}{2}b\right)$ on the ellipse $ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1\ (0 < b < 1 < a)$. Let denote $ S$ be the area of the figure bounded by $ l,$ the $ x$ axis and the ellipse.

(1) Find the equation of $ l$.

(2) Express $ S$ in terms of $ a,\ b$.

(3) Find the maximum value of $ S$ with the constraint $ a^2 \plus{} 3b^2 \equal{} 4$.[/quote]

(3)

By applying A.M-GM we get

$ \frac {a^2 \plus{} 3b^2}{2}\ge \sqrt {3}ab\implies ab\le\frac {2}{\sqrt 3}$ and equality holds for $ \ a\equal{}\sqrt {3} b\implies a\equal{}\sqrt2;\ b\equal{}\sqrt{\frac23}$

$ \implies\boxed{\boxed{\max S \equal{} 1 \minus{} \frac {\pi}{3\sqrt 3}}}$


Euler
quadratics
calculus
integration
algebra
number theory solved
number theory

This has been solved by Kiran Kedlaya in a paper which appeared in Mathematics Magazine, vol.71, n1 (1998), p.61-63.

Pierre.

Well, in fact the proof is not too long so I can write it here :
A P_1 set is a set of positive integers, the product of any two distinct elements of which is 1 less than a perfect square.

Euler found a general construction of four element P_1 sets using the following lemma :
If {p,q,r} is a P_1 set then so is {p,q,r,s} where
s = p+q+r+2pqr \pm 2 \sqrt [(pq+1)(qr+1)(rp+1)] , as long as s > 0.

Proof of the lemma :
The values of s defined above are the roots of the quadratic equation :
p 2  + q 2  + r 2  + s 2  - 2(pq+pr+qr+ps+qs+rs) - 4pqrs - 4 = 0, (1)
which can be written in the following ways :
(p+q-r-s) 2  = 4(pq+1)(rs+1)
(p+r-q-s) 2  = 4(pr+1)(qs+1)
(r+q-p-s) 2  = 4(rq+1)(ps+1)
Since rs+1 is an integer which is the quotient of two perfect squares, it is also a square, as are ps+1 and qs+1, hence the conclusion.

Now, let's prove the desired result.
Suppose that there exist triples p,q,r of positive integers such that (pq+1)(qr+1)(rp+1) is a square but not all of pq+1, qr+1, rp+1 are squares.
Choose such a triple that minimize p+q+r. Wlog we may assume that p < q < r. Define s as in the lemmausing the negative square root.
We will show that 0 < s < r and that (pq+1)(qs+1)(sp+1) is a square but not all of pq+1, qs+1, sp+1 are squares, contradicting the minimality of p+q+r.

By the equivalent forms of (1), we know that :
16(pq+1)
 2 (pr+1)(qr+1)(ps+1) = (pq+1) 2 (p+r-s-q) 2 (p+s-q-r) 2  is a perfect square.
Since (pq+)(qr+1)(rp+1) is a square so then is is (pq+1)(qs+1)(sp+1).
Moreover pr+1 is a square if and only if qr+1 is a square, and qr+1 is a square iff qs+1 is square, so not all of pq+1,qs+1,sp+1 are squares.
We also have
(rs+1)(p+q-r-s) 2 /[4(pq+1)]  \geq  0 ans so s  \geq  -1/r.
Note that r = 1 implies (since r > q > p) that p=q=r in which case (pq+1)(qr+1)(rp+1) is not a square, a contradiction. Hence r > 1 and so s  \geq  0.
Moreover if s = 0, then, we have 4(pq+1) = (p+q-r) 2
4(qr+1) = (q+r-p) 2
4(rp+1) = (r+p-q) 2
contradicting the assumption that not all of pq+1, qr+1, rp+1 are squares. Therefore s is a positive integer.

If s' is the other root defined in the lemma, then we have
ss' = p 2  + q 2  + r 2  - 2pq - 2pr - 2qr - 4
< r 2  - p(2r-p) - q(2r-q)
< r 2
Since s is the smaller of the two roots, we have s 2   \leq  ss' and so s < r, yielding the desired contradiction.

Pierre.

A related problem may be found here :
http://www.mathlinks.ro/viewtopic.php?t=4516&highlight=

Pierre.

[color=red][moderator edit: the links needs to be corrected][/color]

[quote="K09"]Let x,y,z be positive intergers,Prove that (1+xy)(1+yz)(1+zx) is perfect square iff
1+xy,1+yz,1+xz are perfect square numbers[/quote]
Well, the sufficient condition is VERY HARD  :P  :P , but to the neccesary, well IO only want to say that if $(1+xy)(1+xz)(1+yz)=A^2$ then
\[
(xyz)^2+(xyz+x+y+z)^2+2=x^2+y^2+z^2+2A^2.
\]
But I suppose someone knows a beautiful and simply proof of this one part of the result.

B.R.

RX TCM

I had solution.It very hard. I use Euler and Ferma Problem .
If a,b,c sastify ab+1,bc+1,ca+1 are perfect squares then exist d such 5that ad+1,bd+1,cd++1 are perfect squares (a,b,c,d are positive interger

[b]k09[/b]! could you possibly  explain more clear ?
 I don't really understand your idea but i really fascinate of this strage and nice result.

Ok,i have had solution(i suppose it's true).Here's hint:
Assume the contrary [s true.Choose [b](x,y,z)[/b] positive integers s.t
      [b](xy+1)(yz+1)(zx+1)[/b] is perfect square and at least one of [b]xy+1,yz+1,zx+1[/b]   isnot perfect square.
   With [b](xy+1)(yz+1)(zx+1)=A^2[/b] ,let [b]2A=2xyz+x+y+z-t[/b]
 Also assume that [b]x>=y>=z[/b]
  We have[b]x^2+y^2+z^2+t^2 -2(xy+xz+xt+yz+yt+zt)-4xyzt -4=0[/b]
 this is a quardritic equation with vary t if we fixed x,y,z which have two integral roots.
 Easily we get one of two roots ,called,[b]t[/b] such that [b]0<t<x[/b] .
 Add that [b](x+y-z-t)^2=4(xy+1)(zt+1)[/b] and three other identities we have descent contradiction.

See this page: http://www-math.mit.edu/~kedlaya/papers/

[quote="hominot"]Let $ x$, $ y$, $ z$ be positive integers. Prove that: $ \left(xy \plus{} 1\right)\left(yz \plus{} 1\right)\left(zx \plus{} 1\right)$ is a perfect square if and only if $ xy \plus{} 1$, $ yz \plus{} 1$, $ zx \plus{} 1$ are all perfect squares.[/quote]
I have a document about this problem  :)


symmetry
inequalities
quadratics
inequalities unsolved

[quote="hien"]My friend have just gave me an inequality problem as a gift. But i don't think it is true. The problem is:
Let $a,b,c>0$ and $a+b+c=1$. Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{25}{1+48abc}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119301

[hide]
WLOG let $a=\max \{ a,b,c\}$. The inequality is equivalent to
\[(1+48abc)(a-a^{2}+bc)\geq 25abc \]
which is a "quadratic in $bc$" of the form $P(a)(bc)^{2}+Q(a)(bc)+R(a)\geq 0$, where $0\leq bc\leq \frac{(1-a)^{2}}{4}$. The vertex of the quadratic occurs at $bc=-\frac{Q}{2P}$, which is equal to $\frac{48a^{3}-48a^{2}+25a-1}{96a}$. However, this vertex lies outside the interval for $bc$! This is so because
\[\frac{48a^{3}-48a^{2}+25a-1}{96a}>\frac{(1-a)^{2}}{4}\iff 1-a-24a^{3}<0 \]
and as $a=\max \{ a,b,c\}$, we have $a\geq \frac{1}{3}$, so
\[1-a-24a^{3}\leq 1-\frac{1}{3}-\frac{24}{27}=-\frac{2}{9}<0 \]
Therefore, the quadratic is either increasing on the interval $0\leq bc\leq \frac{(1-a)^{2}}{4}$ or decreasing on that interval. It suffices to check the cases when $bc=0$ and $bc=\frac{(1-a)^{2}}{4}$. The first case simply reduces to $a(1-a)\geq 0$. For the second case, multiply both sides by 4:
\[(1+12a(1-a)^{2})(4a(1-a)+(1-a)^{2})\geq 25a(1-a)^{2}\]

\[\iff (1-a)((1+12a(1-a)^{2})(1+3a)-25a(1-a))\geq 0 \]

\[\iff (1-a)(3a-1)^{2}(2a-1)^{2}\geq 0 \]
so the second case works.
[/hide]


conics
parabola
analytic geometry
quadratics
function
calculus
derivative

The graph of $ y \equal{} 2x^2 \plus{} 4x \plus{} 3$ has its:

$ \textbf{(A)}\  \text{lowest point at } {(\minus{}1,9)}\qquad
\textbf{(B)}\ \text{lowest point at } {(1,1)}\qquad \\
\textbf{(C)}\ \text{lowest point at } {(\minus{}1,1)}\qquad
\textbf{(D)}\ \text{highest point at } {(\minus{}1,9)}\qquad \\
\textbf{(E)}\ \text{highest point at } {(\minus{}1,1)}$

[hide]
Complete the square:
$ x^2 \plus{} 2x \plus{} \frac {3}{2} \equal{} (x \plus{} 1)^2 \plus{} \frac {1}{2}$
I divided by 2 in the beginning, so multiply it back to get
$ 2(x \plus{} 1)^2 \plus{} 1$
The lowest point is when the square part is 0, so x = -1. Plugging in, the answer is (-1, 1)

$ \boxed{\text{C}}$[/hide]

Well we know that there is only a minimum for the parabola because the x^2 coefficient is positive. So we use -b/2a and get that the x coordinate -1. So plug in the value and get y as 1. So lowest point is (-1,1)
C.

[hide]It is quadratic function with $ a \equal{} 2 > 0$ so it has minimum,you can eleminate solutions $ A,B$ and $ C$.Both solutions has number $ 1$ so now check that $ f( \minus{} 1) \equal{} 1$ ,the solution is $ \boxed {D}$ [/hide]

Um...choice $ D$ represents a maximum, not a minimum, for one thing...

The highest point is undefined (infinity, infinity). The parabola opens upward. Therefore, choices (D) and (E) are eliminated.

Sorry I misread the problem carefully

The parabola is concave and have minimum at point $ \left(\frac{\minus{}b}{2a},f\left(\frac{\minus{}b}{2a}\right)\right)\equal{}(\minus{}1,1)$ or $ \boxed {C}$

here's a super beast calculus solution:
find the first derivative of 2x^2+4x+3. 4x+4=0 to find the critical point, at -1. Plug in to get (-1,1). Find the second derivative of 2x^2+4x+3=2. The derivative is positive, so the critical point at x=-1 is negative. Thus, the solution is C: lowest point at (-1,1).
Just a few notes:
1) I was kidding about the "super beast" part. Actually calculus is pretty dumb in this situation, but I like using it because it makes me feel smart.
2) I thought it was interesting to note how the second derivative+concavity test was a proof for if a parabola opens up or down. I've learned that if the leading coefficient is positive, it points up, and if its negative, it points down, but I nevver thought calc would prove it. On another note, is there a way to prove if the parabola opens up or down without calc?

[quote="lukezli"]
2) I thought it was interesting to note how the second derivative+concavity test was a proof for if a parabola opens up or down. I've learned that if the leading coefficient is positive, it points up, and if its negative, it points down, but I nevver thought calc would prove it. On another note, is there a way to prove if the parabola opens up or down without calc?[/quote]
You answered your own question, I think. To figure out which way it opens, just look at the sign of the coefficient of the $ x^2$ term. Positive means up and negative means down.


calculus
derivative
logarithms
calculus computations

differentiate
$ \frac {d}{dx}x^x$

$ \frac {d^2}{dx^2}x^x$

thanks

Consider $ y \equal{} x^x$.

Hence we have $ \ln{y} \equal{} x \ln{x}$. Differentiating both sides gives us $ (\frac {1}{y})(\frac {dy}{dx}) \equal{} \ln{x} \plus{} 1$. We isolate $ \frac {dy}{dx}$ and substitute $ y \equal{} x^x$ to get $ \frac{dy}{dx}\equal{}x^x(\ln{x} \plus{} 1).$

$ \frac{d}{dx} x^{x}\equal{}\frac{d}{dx} e^{x lnx}\equal{}{(x lnx)}'e^{x lnx}\equal{}(ln x\plus{}1)e^{x lnx}$

Hi centry57.

You forgot to end with $ \equal{}(\ln x \plus{} 1)x^{x}$

For the second derivative, we differentiate $ x^{x}(\ln{x}\plus{}1).$ Using the product rule we get $ x^{x}(\frac{1}{x}) \plus{} [x^{x}(\ln{x}\plus{}1)](\ln{x}\plus{}1) \equal{} x^{x\minus{}1} \plus{} x^x(\ln{x}\plus{}1)^2$.


Well, verbalized you can say a negative of a number is the opposite of a number. And say if a number is positive, it is something, and if it is negative it is nothing. And because the opposite of nothing is something, it must be true  :D
Or if $x<0$ then the opposite of $x$ has to be $>0$.


linear algebra
matrix
function
trigonometry
college contests

the problems were in three groups.i'll translate only the advanced section(A).the olympiad was held in one day.13May in Weliko Turnovo.

problem1.given is a square matrix A,nxn,a(i,j)=i*j. f(x)=det(Ax+I).
a)calculate f'(0), for n=4.
b)calculate f'(0), for every n natural.

problem2.the set of natural numbers is written as a sequence alpha={a1,a2,...an,...}.
a)prove that the limit g(alpha)=lim_(n->inf)(1/a1 +1/2a2 +...1/nan) exists.
b)prove that inf_(alpha)g(alpha)=0 and that max_(alpha)g(alpha)=g(w).where w={1,2,...,n,...}.

problem3.f:[0,1]->R is contunuous function,f(0)=f(1).
a)prove that for every natural n, there exists a chord on the sketch of f that is parallel to Ox and its length is 1/n.
b)does there exists a continuous function,f:[0,1]->[0,1],f(0)=f(1) and every chord parallel to 0x has lengh different from 2/3.
c)the same question but 2/5.

Nice problems -- they would be even nicer Latexed  :)

[hide="Number 3"]a. Let $g(x) = f(x + \frac1n)- f(x)$.  Then $g$ is continuous and
$g(0) + g(\frac1n) + \ldots + g(\frac{n-1}n) = f(\frac1n) - f(0) + f(\frac2n) - f(\frac1n) + \ldots + f(1) - f(\frac{n-1}n) = f(1) - f(0) = 0$
So either $g$ is zero at every one of these values, in which case we're done immediately, or $g$ is positive at at least one and negative at at least one, in which case we're done by the intermediate value theorem.
b. Yes.  $f(x) = \frac{1 + \sin 2\pi x}2$ works nicely.
c. Yes.  The piecewise linear curve
$0 \leq x \leq \frac16 \Rightarrow f(x) = x + \frac12, \frac16\leq x\leq\frac5{12} \Rightarrow f(x) = -x + \frac56, \frac5{12}\leq x\leq\frac7{12} \Rightarrow f(x) = x, \frac7{12}\leq x\leq\frac56 \Rightarrow f(x) = -x + \frac76, \frac56\leq x \leq 1 \Rightarrow f(x) = x - \frac12$
should do the job.[/hide]