Let $\omega^{2003}=1$, with $\omega \not= 1$.  Evaluate $\sum_{n=1}^{2003}\frac{1}{1+\omega^{n}}$.

I've already solved this problem by [hide]pairing terms[/hide] to get an answer of [hide]2003/2[/hide], but am not satisfied.

In pursuit of a complex number solution, I hit a dead end. (got some major work to do) :P

Here's what I have so far, so if someone could provide a tip or solution, that'd be nice.  :)
[hide]Let $z_{k}=1+w^{k}.$ Hence,
\[\frac{1}{z_{k}}=\frac{1}{z_{k}}\cdot \frac{\overline{z_{k}}}{\overline{z_{k}}}=\frac{\overline{z_{k}}}{|z|^{2}}. \]
Now, $\overline{z_{k}}=\overline{1+w^{k}}=1+\overline{w^{k}}$ so we split the sum:
\[\sum_{n=1}^{2003}\frac{1}{1+w^{n}}= \sum_{n=1}^{2003}\frac{1}{|1+w^{n}|^{2}}+\sum_{n=1}^{2003}\frac{\overline{w^{n}}}{|1+w^{n}|^{2}}. \]
I think the second sum equals 0, and the first is a mess. I have no idea what I'm doing.  :P

 [/hide]

How about this:

Since $2003$ is prime, we know the set $\{\omega^{n}\}_{n=1}^{2003}$ contains all the roots of the equation $x^{2003}=1$.  Hence the set $\left \{\frac{1}{\omega^{n}+1}\right \}_{n=1}^{2003}$ contains the roots of the equation $\left (\frac{1}{x}-1\right )^{2003}=1$.

But this equation is equivalent to $2x^{2003}-{2003\choose 1}x^{2002}+\cdots=0$ which combined with the use of Viete's Theorem yields

$\sum_{n=1}^{2003}\frac{1}{\omega^{n}+1}=\frac{2003}{2}$.

Looks good to me. Thanks.  :)

Here's the pairing solution in case anyone wants to see it.

[hide="Solution"]Pairing seems like a good way to exploit $\omega^{2003}=1,$ so let's try it. Note that\begin{eqnarray*}\frac{1}{1+\omega^{n}}+\frac{1}{1+\omega^{2003-n}}&=& \frac{(1+\omega^{2003-n})+(1+\omega^{n})}{(1+\omega^{n})(1+\omega^{2003-n})}\\ &=& \frac{2+\omega^{n}+\omega^{2003-n}}{1+\omega^{n}+\omega^{2003-n}+\omega^{2003}}=1. \end{eqnarray*}Hence, all terms but the last term pair to equal $1.$ The last term equals $\frac{1}{1+\omega^{2003}}=1/2,$ so the total sum is $2003/2.$[/hide]

Does the fact that 2003 is prime have anything to do with it, zgorkster? I think the sum is done the same way regardless if $n$ is composite or prime.

[quote="Elemennop"]Does the fact that 2003 is prime have anything to do with it, zgorkster?[/quote]

Yes, otherwise $\omega$ is not guaranteed to be a primitive root.

[quote="Elemennop"]So it matters that it cycles through all $n$ roots of 1 and doesn't repeat any of them?

I'm not 100% following this primitive root business

edit> even then foreverwise's solution seems to avoid any requirement that $2003$ be prime[/quote]

If $\omega^{p}= 1$, we have

$\sum_{k=1}^{mp}\frac{1}{1+\omega^{k}}= m \sum_{k=1}^{p}\frac{1}{1+\omega^{k}}= \frac{mp}{2}$.

So if the proof only worked for primes, here's an obvious extension to composites.

Its a technical point, but the reason that $2003$ needs to be prime is because all we know is that $\omega^{2003}=1$ and $\omega\neq 1$.  If $\omega^{k_{1}k_{2}}=1$ where $k_{1}$ and $k_{2}$ are integers greater than $1$, then $\{\omega^{n}\}_{n=1}^{k_{1}k_{2}}$ is not guaranteed to have all the $k_{1}k_{2}$th roots of unity.

In that case, what paladin mentions would apply.

the solution is not unique if we are not dealing with primes because we don't necessarily cycle through all the roots...for example

x^{10}-1=0

say we choose $\omega=\exp\frac{2\pi i*2}{10}$

$\omega^{6}=\omega$...we are missing some of the roots

Alright, so you wanted your complex solution - here it is.

[hide]

Let the solutions of  lie on the unit circle. Let this be circle . Now, the numbers  for  lie on the unit circle centered at . Let this circle be called circle . Now, the numbers  lie on the inversion of circle  about circle , and in fact this is the line . Due to the symmetry of these roots, each root  has it's conjugate on the line, whose imaginary part is the opposite of . The sum of all of these has imaginary part zero, clearly, and since each real part is , the sum is equal to .

[/hide]

[quote="Peter VDD"][quote="paladin8"]$\sum_{k=1}^{mp}\frac{1}{1+\omega^{k}}= m \sum_{k=1}^{p}\frac{1}{1+\omega^{k}}$[/quote] and just why is this?  :maybe: if it were true, what prevents us to take p=1, giving an immediate $\frac{m}{2}$ for all $m$?[/quote]Peter, you missed the line which directly preceded that formula, which said
[quote="paladin8"]If $\omega^{p}= 1$, we have $\ldots$[/quote]
and which certainly does make that true.

The theorem as it stands is in any case false: take $p=2,\omega=-1$ as a counterexample.


Find the bounds of
$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$ given that $abc = 1$ and $a,b,c>0$

$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1} < 2$ is easy to see by just reducing to same denominator

This is sharp : $a=b, c \rightarrow \infty$ then $\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1} \rightarrow 2$

Setting $a = x/y, b = y/z, c = z/x$ let's prove $\sum x/(x+2z) \geq 1$
By rearrangement  we have $S_1 = \sum x/(x+2z) \geq S_2 = \sum z/(x+2z)$
$S_1 + 2*S_2 = 3$ gives $S_1 \geq 1$

Hope it's correct


Let $ABC$ be a triangle and $D, E$ two points in exterior such that $\angle ABD+\angle ACE=180^{0}$ and $\angle DAB=\angle CAE$.
If $CD\cap BE=\{P\}$ and $AP\cap BC=\{Q\}$, prove that $\sin \angle AQB=\sin\angle ABD$.


I need an opening I can quickly memorize, so I can play it tomorrow morning, like the Colle's System, except I think I know that opening and don't need it that much. If you all have an opening that's effective and not full of variations, don't hesitate!! =D

Wow a chess tournament tommorow morning; I don't think it's wise learning a new opening. Just stick to whatever you normally play. If you can't play a certain opening just go standard and respond normally to all openings/counters. (i.e. play e4 as white and e5 as black).

[quote="sailingdog"]I had MC chapters this morning.  :)[/quote]

Which has everything to do with the purpose of this thread, of course.


Coulomb's law prescribes an infinitely large force to make two point-like charges of identical sign touch. What is the situation if the charges are uniformly distributed on two circular plates? What is the force required to make two circular plates of charge $Q$ and radius $r$ touch?

[quote="perviz_aze"][quote="PTynan89"]Well, since the plates are much larger than the distance between them as they get close, the electric field by one at the other is $\frac{Q}{\epsilon A}= \frac{Q}{\epsilon \pi r^{2}}$.

Thus, since this is independent of distance, the required force then becomes $E*Q = \frac{Q^{2}}{\epsilon \pi r^{2}}$.[/quote]NO ,no :!:
I think that you musn't write $F=\multiply{Q}{E}$
E is the sum of the electric field of two plate.Only electric field of the one plate effects to the an other one . And that is why we must write  $F=\multiply{\frac{E}{2}}{Q}$[/quote] :maybe: I agree with Perviz.It must be like that


Can you write the question clearly i did not understand!!

Math is my lifw!!!

[quote="mim"]Let $n \in N$. Prove that for $k \in N$ number $k\sqrt{2}$ is in a distant to the nearest integer $\le \frac{1}{n}$.[/quote]

I can't quite understand what your are asking for.  What do you mean by "in a distant".

inom, what does lifw mean?

Number $k\sqrt{2}$ is in distance from the nearest integer less or equal than $\frac{1}{n}$

sorry for my poor english, i hope that it's clear now :>

I think he is trying to say this:

Show that for any natural $n$ we can find natural $k$ such that the number $k \sqrt{2}$ differs by at most $\frac{1}{n}$ from the nearest integer.

By Weyl's equidistribution theorem, the sequence $(n\sqrt{2})_n$ is equidistributed $\bmod 1$ (note that we really only need density here, which is elementary in this case).  The result follows immediately.

Am I using circular reasoning?  No, but I did quote an advanced theorem.  I'll post an elementary proof when I get home, or in a day or two.

[quote="mathgeniuse^ln(x)"]But aren't we trying to prove that theorem.  Aren't you using circular reasoning right now?  Just wondering.[/quote]

He invoked a well-known theorem to prove this special case.  Of course, a problem like this wouldn't be given on an Olympiad or anything of the sort, since it is well-known enough to be a citable result.  :)


In the sides AB and BC of the triangle ABC are choosen points M and N respectively.The segments AN and CM intercept in point O. It is knwon that AO=CO. Can we say that triangle ABC is an isosceles triangle?
a) if AM=CN
b) if BM=BN
so you should prove once for case a and then for case b. The answer are:
a) the triangle can or can't be isoscles
b) the triangle is isosceles

a) clearly the triangle can also not to be isosceles and an exemple is this: in each triangle if O is the symmetric wrt CA of the midpoint of the arch CA of the circumcircle of ABC (not containing A) we have M and N that satisfy AM=CN and AO=CO but not necessary AB=BC.

b) suppose that ABC can be a non isosceles triangle on B and so wlog AB>BC, so let BO meet AC on L, then clearly (cause O is internally to ABC) AL<LC. Also we have MA>NC. But for ceva on O we have $ \frac{AM}{NC}\equal{} \frac{AL}{LC}$ absurd (LHS>1 and RHS<1).
Clearly when the triangle is isosceles b) is ok, then the triangle can be only isosceles.


Let n be an odd integer greater than one.  Prove that the sequence
$ {{n}\choose{1}}, {{n}\choose{2}}, {{n}\choose{3}}, ... {{n}\choose{\frac {n - 1}{2}}}$
has an odd number of odd numbers

[hide]
$ 2( {{n}\choose{1}} \plus{} {{n}\choose{2}} \plus{} {{n}\choose{3}} ...{{n}\choose{\frac {n \minus{} 1}{2}}} ) \plus{} 2 \equal{} 2 ^{n}$ (to see why this is true, let a=b=1 int the binomial theorem)
$ {{n}\choose{1}} \plus{} {{n}\choose{2}} \plus{} {{n}\choose{3}} ...{{n}\choose{\frac {n \minus{} 1}{2}}} \equal{} 2 ^{n \minus{} 1} \minus{} 1$
even numbers always add up to even numbers. odd number of odd numbers add up to an odd number while even number of odd numbers add up to an even number.
Since R.H.S is odd, L.h.S has odd number of odd numbers.
[/hide]


Let a,b,c,d be real number such that  [tex] a;d\ge 0;b,c>0[/tex] and [tex] b-d\ge |a-c|[/tex]
Find the minium value of [tex] P=\frac{b}{a+d}+\frac{c}{a+b} [/tex]

hello, do you mean
Let $a,b,c,d$ be real number such that $ a;d\ge 0;b,c>0$ and $ b-d\ge |a-c|$
Find the minium value of $P=\frac{b}{a+d}+\frac{c}{a+b} $?
Sonnhard.


[quote="11235813"]Too Messy!![/quote]
Not too messy :mad: ,mybe a little :blush: .

[quote="Albanian Eagle"][quote="Tiks"]$a,b,c>0$ and $3(b+c)<13a,3(a+c)<13b,3(a+b)<13c$
Prove that
$\frac{6}{25}(3\frac{\sqrt{3a}+\sqrt{3b}+\sqrt{3c}}{\sqrt{a+b+c}}-4)\geq \frac{a(b+c)}{a^2+(b+c)^2}+\frac{b(a+c)}{b^2+(a+c)^2}+\frac{c(a+b)}{c^2+(a+b)^2}$[/quote]
put $a+b+c=1$ then $f(x)=\frac{18}{25} \sqrt{3x} - \frac{8}{25} +\frac{x^2-x}{2x^2-2x+1}$
we can easily see that $f(x)\geq 0$ when $x\geq \frac{3}{16}$ which is the case for $a,b,c$
so $f(a)+f(b)+f(c)\geq 0$[/quote]
Same with my solution! ;)


In general, when people are having these problems, we recommend switching to Opera (http://www.opera.com).

sometimes i log out of the whole thing and log in again.

[quote="Nerd_of_the_Ages"]Just today, several people have been having trouble with the classrooms; the for me, the text sometimes just stops displaying, but the transcript still gets updated. It continues displaying some text after I press clear, but that it happens again. Just 2 days ago, this did not happen, and it was working just fine.

OP: Windows XP
Internet Browser: FireFox 3.0

I'm guessing that the issue is because of FireFox 3.0, since I didn't have it 2 days ago, and the classroom was working fine.[/quote]
This happened to me a lot yesterday when I was using Firefox 3 with Ubuntu.  Seems to be a bug with Firefox 3 and the Java version I have (Version 6 Update 6)  It has never done this to me before I started using Firefox 3.
I am trying Opera next time and hopefully that will work.

Thx, I used Opera during class, and it worked fine.


Heres the question. I already know the answer but its best not to mention to see if anyone can assist me in finding the right equations to this problem.

How old am  i?

I am twice as old as you were when I was your age. When you are as old as I am, both our ages will add up to 27.

answer in full steps.

I am twice as old as you were when I was your age. When you are as old as I am, both our ages will add up to 27.

[hide]Okay, let's say:
You = x
Me = y

(Me being the older one). At any age, the difference between Me and You will be (y-x). So when You = a, Me = a + (y-x).

When I was your age, Me = x, and You = x- (y-x) = 2x - y.
And my age now is twice yours when I am your age. So,

y = 2(2x-y) = 4x-2y
3y = 4x.

When you are as old as I am, You = y, and Me = y + (y-x) = 2y-x.  Our ages will also add to 27.
Therefore, y + 2y-x = 27
3y-x = 27.

Solving the system of equations, you get y = 12, and x = 9. Tah dah!

[/hide]

I like the humor of this guy. to say [b]"How old am I?" this one will blow your mind.[/b]. In this excercise, as in many others, the problem recides in understanding the problem and be able to translate it correctly to an equation (what [b]pkothari13[/b] did very well! :D ).

Well, he wants his homework done, so he came up with a reasonable excuse.

I was expecting "my age" to cancel out so it could apply to anyone, but that isn't the case and I got the same as PK (except your age is the greater one and my age is the lesser one).

[quote="denizuzun"]didn't blow my mind :)  :)[/quote]

Neither did mine  :D


trigonometry
geometry
algebra
polynomial
advanced fields
advanced fields unsolved

Say we want to show whether or not a set, $S\subset\mathbb{A}_{k}^{n}$, is a variety. If it is, then the proof is quite straight forward. You come up with an ideal $I \subset k[X]$ such that $V(I)=S$, or however you'd think of that notationally (I know it varies quite a bit). You just have to construct the subset in which $S$ is the set of zero loci of your construction.

Here is where I feel it is more complicated (possibly it isn't). How do you show something isn't a variety? Now you have to show that no matter what subset you take of $k[X]$, there is a point in $S$ which is not a zero.

For example, I came across is $\{(t, \sin t) : t\in\mathbb{R}\}$ a variety? I feel like it isn't, but cannot show it. If this is, could someone please provide an example of a set that isn't, and how you show that it isn't. Thanks.

First, a word of warning: "classical" algebraic geometry in the sense of subsets of $k^{n}$ works really well only for algebraically closed $k$.

Concerning $(t,\sin t)$: Suppose $f(X,Y)$ is a polynomial vanishing on this set. For fixed $-1\leq y_{0}\leq1$, the polynomial $f(X,y_{0})$ has infinitely many zeros (as a polynomial in $X$), hence is the zero polynomial, therefore $Y-y_{0}\mid f(X,Y)$. Obviously, this cannot be true for infinitely many values of $y_{0}$ unless $f(X,Y)=0$.

I don't know much algebraic geometry, but I believe the practical way for your general question should be using localization.  But it just reduces global problems to local ones... so it doesn't help any here...


conics
parabola
analytic geometry
graphing lines
slope
calculus
calculus computations

Hi, I've got two questions.

1. Let:
$ f(x)\equal{}\begin{cases}x^{2}&x\le2\\mx\plus{}b&x>2\end{cases}.$

Find the values of $ m$ and $ b$ that make $ f$ differentiable everywhere.

I got an answer of I think...$ m\equal{}4$ and $ b\equal{}\minus{}4$. Are there any more?

2. Where do two perpendicular lines, that are both tangent to the parabola $ y\equal{}x^{2}$, and intersect on the y-axis...intersect?

Thanks.

That has to be the only answer to #1. The line must be the tangent line to $ y\equal{}x^{2}$ at $ x\equal{}2,$ and there's only one of those.

On #2, introduce as a new variable the point of tangency. That is, consider the lines tangent to the parabola at $ (a,a^{2})$ and $ (\minus{}a,a^{2}).$ What would the condition that these be perpendicular look like?

Well, there slopes would be opposite inverses, so...if the point where they intersect is $ (0,b)$, then the slope of the first is:
$ \frac{a^{2}\minus{}b}{a}$
and the second is:
$ \frac{a^{2}\minus{}b}{\minus{}a}$
So...
$ \frac{a^{2}\minus{}b}{a}\equal{}\minus{}(\frac{a^{2}\minus{}b}{\minus{}a})^{\minus{}1}$
...
$ \frac{a^{2}\minus{}b}{a}\equal{}\frac{a}{a^{2}\minus{}b}$
$ (a^{2}\minus{}b)^{2}\equal{}a^{2}$
$ a^{4}\minus{}2a^{2}b\minus{}b^{2}\equal{}a^{2}$

Umm, I have to go sleep now, but...I don't seem to be getting a crucial point here.

[quote]I don't seem to be getting a crucial point here.[/quote]
Tangency. That's what you're missing.

Solution

Let the tangents be $ y_{A}$ and $ y_{B}$.

Let $ y_{A}$ tangent $ y \equal{} x^{2}$ at $ x \equal{} a$. Hence co-ordinates of the point of tangency are $ (a, a^{2})$.

$ \frac{dy}{dx}\equal{} 2x$

$ \Rightarrow m_{A}\equal{} 2a$

Now we have a point and a gradient so the equation of $ y_{A}$ is:

$ y_{A}\minus{}a^{2}\equal{} 2a(x\minus{}a)$

$ \Rightarrow y_{A}\equal{} 2ax\minus{}a^{2}$

Let the point at which $ y_{A}$ intersects the $ y$ axis be $ (0, b)$

$ \Rightarrow b \equal{}\minus{}a^{2}$

$ m_{B}\equal{}\minus{}\frac{1}{m_{A}}\equal{}\minus{}\frac{1}{2a}$

But $ m \equal{} 2x$ so the $ x$ co-ordinate at which the gradient of the tangent is $ \frac{1}{2a}$ must be $ x \equal{}\minus{}\frac{1}{4a}$

So $ y_{B}$ tangents $ y \equal{} x^{2}$ at the point $ \left(\minus{}\frac{1}{4a},\frac{1}{16a^{2}}\right)$

Hence the equation of $ y_{B}$ is:

$ y_{B}\minus{}\frac{1}{16a^{2}}\equal{}\minus{}\frac{1}{2a}\left(x\plus{}\frac{1}{4a}\right)$

$ \Rightarrow y_{B}\minus{}\frac{1}{16a^{2}}\equal{}\minus{}\frac{1}{2a}x\minus{}\frac{1}{8a^{2}}$

$ \Rightarrow y_{B}\equal{}\minus{}\frac{1}{2a}x\minus{}\frac{1}{16a^{2}}$

$ \Rightarrow b \equal{}\minus{}\frac{1}{16a^{2}}$

Substituting $ b \equal{}\minus{}a^{2}$:

$ \Rightarrow b^{2}\equal{}\frac{1}{16}$

$ \Rightarrow b \equal{}\minus{}\frac{1}{4}$ since $ b\leq 0$

Tangents intersect at $ \left( 0,\minus{}\frac{1}{4}\right)$

Where does the $ b\equal{}\minus{}a^{2}$ come from?

$ y_{A}\equal{} 2ax\minus{}a^{2}$, hence $ y_{A}(0) \equal{} b \equal{}\minus{}a^{2}$.

Haha, I feel dumb for asking that question.


This was the first geometry problem in IMO 09 :

Problem 2. Let ABC be a triangle with circumcentre O. The points P and Q are interior points
of the sides CA and AB, respectively. Let K, L and M be the midpoints of the segments BP, CQ
and PQ, respectively, and let [i]c[/i] be the circle passing through K, L and M. Suppose that the line
PQ is tangent to the circle [i]c[/i]. Prove that OP = OQ.


My trig teacher posts an extra credit problem on her whiteboard every week and I don't really get this one: What is the area of a square whose diagonal is one unit longer that the length of its side? Thanks.

Denote length of its side is $x$ , then length of its diagonal is $x+1$. We have $(x+1)^{2}=x^{2}+x^{2}$, or $x+1=x\sqrt{2}$, or $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$. Final, area of that square is equal to $(\sqrt{2}+1)^{2}=3+2\sqrt{2}$. :D


Can you find three integers x, y, and z, such that $(x+y+z)^{3}=xyz?$

[hide="This may help:"]
By AM-GM they all aren't positive.  :)
And there's always the trivial solution $(0,0,0)$.  :P
[/hide]

(I tried to hide the above ^, but it didn't work. Any ideas as to why not?)


How to demonstrate that the sum of the reciprocals of the primes converges?
$ \frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+...$

Thank you for helping. :P

[quote="sos440"]It is a very famous fact that the sum of the reciprocals of of the primes diverges.

[url=http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges]Wikipedia - Proof that the sum of the reciprocals of the primes diverges[/url][/quote]

Thank you sos440. :P


group theory
Galois Theory
superior algebra
superior algebra unsolved

a)  Give examples of Galois extensions E/K and L/E such that L/K is not Galois.
b)  Give examples of finite groups such that N is normal in G and H is normal in N, but H is not normal in G.

I don't know how to start this--how do you find random examples of Galois groups?  Could it be something along the lines of a symmetric group, the group of automorphisms?

We can realize any symmetric group as a Galois group. What theorem connects symmetric groups to arbitrary groups?

[hide="So how can we find a Galois extension with Galois group $ G$ for any finite group $ G$?"]If $ L/K$ is Galois with Galois group $ S_n$, and $ G \le S_n$, then $ G$ is just the Galois group of $ L/L^G$.[/hide]

I think the Fundamental Theorem of Galois Theory will help you get from b) to a).


geometry
MATHCOUNTS
trigonometry
angle bisector
trig identities
Law of Sines

In triangle ABC, the bisector of <B meets AC at K. AB = 18, BC = 24, and AC = 28. Find AK.

[hide="my answer"]   This one is pretty easy, just use the angle bisector formula. Let x be AK, then 18/x=24/(28-x).
Solve for x, that is AK=12. A little bit harder problem is to find the
length of BK. [/hide]

[quote="mathmom"]This must be above Mathcounts level, but I'd be interested to see a proof of the formula cj used.  Trig?[/quote]

Definitely not.

Let $\angle {KBC}=x$

Draw $l$ parallel to $\overline{BK}$ passing through $A$ and extend $\overline{BC}$ to meet $l$.  Let $D$ be $l\cap l_{BC}$.  We claim that $\triangle{ABD}$ is isosceles.

This is easy to show - supplementary angles give $\angle {ABD}=180-2x$; parallel lines give $\angle KBA=\angle BAD=x$.  Thus $\angle BDA=x$, so that we have the desired result.  Clearly, we have $\triangle {DCA}\sim\triangle {BCK}$.  From that, we have $\overline{BC}/\overline{CK}=\overline{DC}/\overline{AC}$.  Use the previously demonstrated fact that $\overline{DB}=\overline{AB}$, and we get:

$\frac {AB+BC}{AK+KC}=\frac {BC}{CK}$

Simplifying, $\frac {AB}{BC}=\frac {AK}{KC}$, the desired result.

If we want to use trig, this is what we might do.

Let $\angle ABK=\angle CBK=x$ and $\angle BKA = y = 180^{\circ}-\angle BKC$

Now use the law of sines twice.

I have a proof without using the laws of sine...
[hide]
First we need to know that in $\triangle{ABC}$ and $\triangle{XYZ}$, if $\angle{BAC}=\angle{YXZ}$ or $\angle{BAC}+\angle{YXZ}=180$, then \[\displaystyle\frac{\triangle{ABC}}{\triangle{XYZ}}=\frac{{AB}\bullet{AC}}{{XY}\bullet{XZ}}\]

So in this case, since $BK$ is the bisector, so we can get that \[\frac{\triangle{ABK}}{\triangle{CBK}}=\frac{{AB}\bullet{BK}}{{BC}\bullet{BK}}=\frac{AB}{CB}=\frac{AK}{CK}\]

Therefore we get
\[\frac{AB}{CB}=\frac{AK}{CK}\][/hide]

I really shouldn't post this last thing here, but although the purely geometrical proof of this theorem is nice, it's so useful to be proficient with trig so you can just bash stuff out.  One identity and the proof here falls out immediately.


If $ x$ satisfies $ \arccos x \plus{} \arccos 2x \plus{} \arccos 3x \equal{} \pi$, then for what triple $ \left( a,b,c \right)$ does $ x$ satisfy $ ax^3\plus{}bx^2\plus{}cx\minus{}1\equal{}0$.
[i][size=75]From High School Math Contests.[/size][/i]

Umm... I don't know the answer, but I do have a question:

What's the difference between $ \arccos$ and $ \cos^{\minus{}1}$ ?

(My trig is getting a little rusty)

:wink: $ \arccos x \equal{} \cos^{\minus{}1} x$.

No one yet?  :maybe: Does anyone want me to post the solution from the book?

[quote="123456789"][hide="Hint"]Make pairs of arccos terms on both sides, and use the cosine multiple angle formula.[/hide][/quote]

but did you get (12,14,0) doing that way?  Because the problem with that way is that once you square the square-root term, there would be extraneous values, meaing there probalby could be more than just one (a,b,c)

[quote="AstroPhys"]If $ x$ satisfies $ \arccos x \plus{} \arccos 2x \plus{} \arccos 3x \equal{} \pi$, then for what triple $ \left( a,b,c \right)$ does $ x$ satisfy $ ax^3 \plus{} bx^2 \plus{} cx \minus{} 1 \equal{} 0$.
[i][size=75]From High School Math Contests.[/size][/i][/quote]
[hide="Solution"]
\[ \arccos x \plus{} \arccos 2x \equal{} \pi \minus{} \arccos 3x\implies \cos (\arccos x \plus{} \arccos 2x) \equal{} \cos (\pi \minus{} \arccos 3x)\implies x\cdot 2x \minus{} \sin (\arccos x)\sin (\arccos 2x) \equal{} \cos \pi\cdot 3x \plus{} \sin \pi\cdot \sin ( \minus{} \arccos 3x) \equal{} \minus{} 3x
\]
Notice that $ \sin (\arccos k) \equal{} \sqrt {1 \minus{} (\cos \arccos k)^2} \equal{} \sqrt {1 \minus{} k^2}$.  Hence, we get $ 2x^2 \plus{} 3x \minus{} \sqrt {(1 \minus{} x^2)(1 \minus{} 4x^2)} \equal{} 0\implies (2x^2 \plus{} 3x)^2 \equal{} 1 \minus{} 5x^2 \plus{} 4x^4\implies 4x^4 \plus{} 12x^3 \plus{} 9x^2 \equal{} 4x^4 \minus{} 5x^2 \plus{} 1\implies 12x^3 \plus{}14x^2 \minus{} 1 \equal{} 0\implies a \equal{} 12, b \equal{} 14, c \equal{} 0$
[/hide]

The QuattoMaster 6000 has the solution in the book.  :)
the future , you might be interpreting the question wrong, there [i]is[/i] only one $ x$ value.

Is there a different approach to this problem?

[quote="the future"][quote="AstroPhys"]If $ x$ satisfies $ \arccos x \plus{} \arccos 2x \plus{} \arccos 3x \equal{} \pi$, then for what triple $ \left( a,b,c \right)$ does $ x$ satisfy $ ax^3 \plus{} bx^2 \plus{} cx \minus{} 1 \equal{} 0$.
[i][size=75]From High School Math Contests.[/size][/i][/quote]

umm, this question is kinda weird because solving I get (12,14,0), but technically, I believe only one x would work so technically (beucase there would be three roots but 2 would not work), it could just be (0,0,y) for some y to make it a linear funciton.[/quote]

$ a, b, c$ are rational.  Otherwise the problem is trivial.


So here's a question I've been wondering about.  I have some ideas as to the answer, but I haven't been able to find any resources to verify my intuition.  Why does silver only form +1 ions while copper forms +1 and +2 ions?

Silver also form compounds with oxidation states +2 and +3, but you are right: the oxidation state +1 is much more common and important. Why don't you share with us your ideas about the reason that happens?

Well, I didn't post it in order not to influence people's reasoning.  I honestly have no idea if the following has any merit, but here goes:

[hide]
As we get further and further from the nucleus, the energies of the various orbitals become closer and closer together.  Thus, when they are unoccupied, the difference in energy between the $ 4d$ and $ 5s$ orbitals is less than that between the $ 3d$ and $ 4s$ orbitals.

However, in both silver and copper the relevant $ d$ orbitals are full, which lowers their energy.  It is reasonable to expect that this lowering of energy is about equal.  This means that now the $ 3d$ and $ 4s$ orbitals are closer in energy to one another than are the $ 4d$ and $ 5s$.  Hence, for copper it is not very hard to promote an electron from the $ 3d$ orbital to its valence shell (in particular, to the $ 4s$ orbital).  Once there, the electron can be easily stripped away, resulting in the $ \plus{}2$ cation.

For silver, the distance between the $ 4d$ and $ 5s$ orbitals is large enough that this promotion is impractical, and hence only one electron is available to be removed.
[/hide]

Any thoughts?

Ok I looked up ionization data for silver and copper today.  I think that these data (shown below) reinforces my explanation.  Of course, it's still possible (likely?) that everything I've said is bogus.  Any thoughts?

According to Wikipedia,
Silver
1st: 731.0 kJ/mol
2nd: 2070 kJ/mol
3rd: 3361 kJ/mol

Copper
1st: 745.5 kJ·mol−1
2nd: 1957.9 kJ·mol−1
3rd: 3666 kJ·mol−1

What do you have then to say about gold, whose 1st and 2nd ionization energies are, respectively (and according to Wikipedia) equal to 890.1 and 1980 kJ/mol?

Say in terms of what?  And does my explanation make any sense, or does the effect likely depend on more complicated factors?

It depend from the number of electrons in the extern  electronic layer of each element.

Generally yes.  But copper and silver have the same valence electron configuration.

[quote="Gregx007"]And does my explanation make any sense,[/quote]

Yes it does, but then according to it Gold would be predicted to have even more difficulty to have an oxidation state beyond +1, and however the most stable oxidation state of Gold is the +3.

Hmm.  Well, there is a difference between ion formation and oxidation state in a compound.  For oxidation state, you have to start worrying about MOs, which aren't really a problem for an ion--in the gas phase, anyway.  But what is the most stable charge on a gas phase gold ion?  Probably +3.  Ok, I think this problem is a lost cause.

Ah not quite: I forgot about our good friend, relativistic contraction.  This lowers the energy of the gold $ 6s$ orbital.  (See http://www.madsci.org/posts/archives/1997-05/862179191.Ch.r.html).  Thus, I reinstate my previous explanation.

Can you explain the notion of Relativistic Contraction in this context?

Thanks.

My thought--since relativistic contraction lowers the energy of the gold's valence $ s$ orbital, this orbital is closer in energy to the $ d$ orbitals, and hence all electrons can equally easily be removed.  Since such an effect is absent in silver, there is too much of a gap between $ s$ and $ d$ to remove electrons from both.

Again, not sure how valid this is.

I really want to return to this question. Can someone explain relativistic contraction, in this context, to me. Can this in anyway explain Gold's resistance to corrosion (this is probably miles off track).

Thanks.

Edit: Is this related at al to the Inert Pair Effect.


integration
calculus
calculus computations

Having $ \sum\limits_{k = 1}^n {\binom nk\frac {{( - 1)^k }}{{k + 1}}} .$

This is equivalent to compute $ \sum\limits_{k = 0}^n {\binom nk\frac {{( - 1)^k }}{{k + 1}}} - 1.$

Now $ \sum\limits_{k = 0}^n {\binom nk( - 1)^k \int_0^1 {u^k \,du} } - 1 = \int_0^1 {\left[\sum\limits_{k = 0}^n {\binom nk( - 1)^k u^k \right]} \,du} - 1.$

The last sum should be $ (1 - u)^n,$ but how do I prove that?

[quote="Field"]$ \sum\limits_{k = 0}^n {\binom nk( - 1)^k u^k}$[/quote]

[url=http://en.wikipedia.org/wiki/Binomial_theorem]Binomial theorem[/url]


What is coefficient of $ x^2$ when

\[ \left ( \cdots \left (\left (\left ( x\minus{}2 \right )^2\minus{}2  \right )^2\minus{}2  \right )^2\minus{}\cdots \minus{}2 \right )^2\]

is expanded and like terms are combined?

Sorry I'm not that good but isn't you must state how many $ \cdots$ :?:

[quote="JBL"]Of, you could try to solve in terms of the number of squarings.  For example, with one squaring we have $ 4 \minus{} 4x \plus{} x^2$, with two we have $ 4 \minus{} 16x \plus{} 20x^2 \plus{} \ldots$, with three we have $ 4 \minus{} 64x \plus{} 336x^2 \plus{} \ldots$, and so on.  It may be profitable to first compute the $ x^1$ coefficient.

For those who just want to know the answer, try to compute another term [hide="e.g."]http://www.wolframalpha.com/input/?i=Expand%28%28%28%28x-2%29^2-2%29^2-2%29^2-2%29^2[/hide] and [url=http://www.research.att.com/~njas/sequences/?q=1+20+336+5440&sort=0&fmt=0&language=english&go=Search]look for a sequence[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].[/quote]

Thanks, but this  is not a good solution. This problem was raised on [url=http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv17n1/body/HTML/2?template=CRUX]CRUX Mathematicorum[/url].

JBL is not giving a solution. He is giving a hint. His hint is a good approach to finding a solution. He is saying look at some small cases and you will find an inductive solution.

[hide="Solution:"]
Let the last three terms after squaring n-times be $ b_nx^2 \plus{} a_nx \plus{} 4$.
Then $ ((b_nx^2 \plus{} a_nx \plus{} 4) \minus{} 2)^2 \equal{} (a_n^2 \plus{} 4b_n)x^2 \plus{} 4a_n \plus{} 4 \equal{} b_{n \plus{} 1}x^2 \plus{} a_{n \plus{} 1} \plus{} 4$.
So you get the recurrence $ a_{n \plus{} 1} \equal{} 4a_n$ and $ b_{n \plus{} 1} \equal{} a_n^2 \plus{} 4b_n$, with $ a_1 \equal{} \minus{} 4$ and $ b_1 \equal{} 1$.

So $ a_n \equal{} \minus{} 4^n$, meaning $ b_{n \plus{} 1} \equal{} 16^n \plus{} 4b_n$. Now you just have to solve this recurrence for $ b_{n \plus{} 1}$.
[hide]I believe the solution should be $ b_n \equal{} \frac{1}{3}4^{n\minus{}1}(4^n\minus{}1)$[/hide][/hide]

Thanks gauss202
This is a very good hint and solution!!


I'm looking but can't find the solutions to

(p-1)(p^n+1)=4m(m+1)

where p is an odd prime. I want all possible solution sets (m,n). I don't know what to do at all..

the only solution is $n=1$  $m=\frac{p-1}{2}$
and Rust n>0 :wink:

[quote="Rust"]It is unsolved problem. For example n=0 mean
find all solutions $p=2m^{2}+2m+1$ (unsolved problem).[/quote]

Sorry I should have said n and m are positive integers more than 1.

Thanks for the solutions but could we put proof there are no solutions for n>1?

It equavalent to $p^{n+1}-p^{n}+p=k^{2}, \ k=2m+1.$
If $n>1$ we get $p|k$ (because p - prime), but $p^{2}\not |k^{2}$ - contradition.


Let $a,b,c\in(1,\infty)$. Proof the following inequality: $\frac{\log^2_{ab}{c}}{a+b}+\frac{\log^2_{bc}{a}}{b+c}+\frac{\log^2_{ca}{b}}{c+a}\geq \frac{9}{8(a+b+c)}$

[quote="honglei"]It is not very hard.Just use Cauchy inequality.Then use "a/(b+c)+b/(a+c)+c/(a+b)≥3/2.It can be solved very fast.[/quote]

uhm i believe there are many ways these problems can be solved.
But how exactly did you do it ? its not totally clear to me  :blush:

greetings

By Cauchy, we have

$\frac{\log^2_{ab}{c}}{a+b}+\frac{\log^2_{bc}{a}}{b+c}+\frac{\log^2_{ca}{b}}{c+a} \ge \frac{(\log_{ab}{c}+\log_{bc}{a}+\log_{ca}{b})^2}{2(a+b+c)}$

All that is left to show is $\log_{ab}{c}+\log_{bc}{a}+\log_{ca}{b} \ge \frac{3}{2}$

Notice that $\log_{ab}{c} = \frac{\log{c}}{\log{a}+\log{b}}$. So by Nesbitt's we have

$\log_{ab}{c}+\log_{bc}{a}+\log_{ca}{b} = \sum{\frac{\log{c}}{\log{a}+\log{b}}} \ge \frac{3}{2}$

[quote="paladin8"]By Cauchy, we have

$\frac{\log^2_{ab}{c}}{a+b}+\frac{\log^2_{bc}{a}}{b+c}+\frac{\log^2_{ca}{b}}{c+a} \ge \frac{(\log_{ab}{c}+\log_{bc}{a}+\log_{ca}{b})^2}{2(a+b+c)}$
[/quote]

err, sorry, but i dont really understand how you used to cauchy to derive that  :oops:  would you mind explaining?

an alternate form of cauchy is $\frac{x_1^2}{y_1}+\ldots\frac{x_n^2}{y_n}\ge\frac{(x_1+...x_n)^2}{y_1+...+y_n}$, which is what he used

cauchy gives us $(a_1^2+...+a_n^2)(b_1^2+...+b_n^2)\ge(a_1b_1+...+a_nb_n)^2$.  Let $a_k=\frac{x_k}{\sqrt{y_k}}$, and let $b_k=\sqrt{y_k}$, then the alternate form follows.


ceiling function
geometry
circumcircle
geometric transformation
national olympiad

EDIT: Now we have all the problems!

[b]First day[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=336773#p336773]Problem 1[/url]
Determine all triples of real numbers $(a,b,c)$ such that
$xyz = 8$,
$x^2y + y^2z + z^2x = 73$ and
$x(y-z)^2 + y(z-x)^2 + z(x-y)^2 = 98$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53806]Problem 2[/url]
A flea jumps in a straight numbered line. It jumps first from point $0$ to point $1$. Afterwards, if its last jump was from $A$ to $B$, then the next jump is from $B$ to one of the points $B + (B - A) - 1$, $B + (B - A)$, $B + (B-A) + 1$.

Prove that if the flea arrived twice at the point $n$, $n$ positive integer, then it performed at least $\lceil 2\sqrt n\rceil$ jumps.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=336784#336784]Problem 3[/url]
Let $p > 3$ be a prime. Prove that if
\[ \sum\limits_{i=1}^{p-1}{1\over i^p} = {n\over m},  \]
with $\gdc(n,m) = 1$, then $p^3$ divides $n$.

[b]Second day[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53874]Problem 4[/url]
Denote by $a \bmod b$ the remainder of the euclidean division of $a$ by $b$. Determine all pairs of positive integers $(a,p)$ such that $p$ is prime and
\[ a \bmod p + a\bmod 2p + a\bmod 3p + a\bmod 4p = a + p  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=337092#337092]Problem 5[/url]
Let $O$ be the circumcenter of acutangle triangle $ABC$ and let $A_1$ be some point in the smallest arc $BC$ of the circumcircle of $ABC$. Let $A_2$ and $A_3$ points on sides $AB$ and $AC$, respectively, such that $\angle BA_1A_2 = \angle OAC$ and $\angle CA_1A_3 = \angle OAB$.

Prove that the line $A_2A_3$ passes through the orthocenter of $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=337101#337101]Problem 6[/url]
Let $n$ be a fixed positive integer. The points $A_1$, $A_2$, $\ldots$, $A_{2n}$ are on a straight line. Color each point blue or red according to the following procedure: draw $n$ pairwise disjoint circumferences, each with diameter $A_iA_j$ for some $i \neq j$ and such that every point $A_k$ belongs to exactly one circumference. Points in the same circumference must be of the same color.

Determine the number of ways of coloring these $2n$ points when we vary the $n$ circumferences and the distribution of the colors.

Stickyfied, so keep postin'em :)
The contest took place today? (on the 28th?) ...

No, this test was held yesterday, Sep 27;

The second day test is taking (or took) place today.

nooo, uno sólo de geometría, this is terrible, jeje.

ejeje eso fue lo mejor (para mi).. ojala hubiese estado alli

se pueden saber todos los medallistas? solo me enteré que pascual, gabriel tavares, y un mexicano de apellido guevara sacaron oro perfecto y que los cortes estuvieron en 10, 22, 33! ah, leonardo saco plata... se puede saber algo mas?
felicidades a todos los medallistas... especialmente a leo y pascual!

[quote="campos"]i'm sorry  :blush:
i was asking for someone to tell me about all the medallists if possible[/quote]Don't worry. Sometimes we make mistakes :) Just be a bit more carefully next time.

If you want to discuss in spanish there is the hispanic community: [url=http://www.mathlinks.ro/Forum/index.php?f=255][img]http://www.mathlinks.ro/Forum/images/flags/spain.gif[/img][/url]

Of course you can posts results here, that would be great, but please do so in English.

Thanks, Shine!
I am searching to an English version of its paper.
Well, someone has the pontuations an medals? I have only the Brazil results:

                           1     2      3    4     5    6   All
BRA 1 (Thiago)     6     0      6    7     7    7   33
BRA 2 (Thomaz)   7     6      6    7     7    0   33
BRA 3 (Rafael)     7     7      7    7     7    0   35
BRA 4 (Gabriel)    7     7      7    7     7    7   42

I think the cutoffs are 10 for bronze, 22 for silver, 33 for gold... [If I understood the post campos made (in Spanish) correctly...] And three guys got full marks: Pascual (go Pascual!), Gabriel Tavares Bujokas from Brazil, and a Mexican guy.

Am I right?

yes arne, you're right, you have a great translation power...  :D
so that makes all the brazil guys gold medallists... i just have that costa rica won 3 bronzes and 1 honorable mention... i don't have any more details

Hey Thanks Arne and Campos.

All i can give is some of the scores:

Col1  42 points Gold Medal
Col2   3  points
Col3   7  Points
Col4  9 Points HM

This year our team was not very experienced and that is reflected in the results, anyway there is a long way for my teammates.

And for other countries Peru got 4 silver Medals, Mexico 2 Golds, a silver and a bronze. And i dont remember anymore

[quote="Arne"]Congrats Pascual! You did a fantastic job!  :first:[/quote]I second this Pascual!  :winner_first:

More scores:

Severius: 777770=35, Gold medal. :)

Finally I got a gold medal :P

Argentina results:

ARG1  7 7 7 7 7 0   35 Gold (Severius!)
ARG2  6 7 0 4 7 6   30 Silver
ARG3  5 2 2 7 7 0   23 Silver
ARG4  6 0 0 7 7 7   27 Silver

A great job!

[quote="SOL"]I'm Soledad from uruguay, I went to IMO this year and also to Ibero.[/quote]

What were your scores in those olympiads???


There is a 22-digit number that, when multiplied by 7, is the same number but with the 7 at the end transferred to the first digit place. What is tha number?

Hint:
[hide]$7(10x+7)=7\cdot10^{21}+x$
Solve for x, then add a 7 to the end.[/hide]

or you can do it manually, which is obviously a less elgant solution but if using the equation about might lead you to a mistake(if you haven't mastered it) just do it manually. don't be shy, moems doesn't need solutions naman eh, just your final answers... :lol:

didn't you post this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44568]here?[/url]

This type of numbers are known as parasitic numbers.


function
inequalities
algebra unsolved
algebra

Find all $ f: \mathbb{R}^ \plus{} \rightarrow\mathbb{R}$ such that for all $ x,y\in \mathbb{R}^ \plus{}$,

$ f(1) \equal{} 2008$, $ |f(x)|\leq x^2 \plus{} 1004^2$, and $ f(x \plus{} y \plus{} \frac {1}{x} \plus{} \frac {1}{y}) \equal{} f(x \plus{} \frac {1}{y}) \plus{} f(y \plus{} \frac {1}{x})$

Set $ x\plus{}\frac{1}{y}\equal{}a$ and $ y\plus{}\frac{1}{x}\equal{}b$. We get $ f(a\plus{}b)\equal{}f(a)\plus{}f(b)$ => f(x)=ax => a=2008, nd the inequality
$ 2008x\leq x^2\plus{}1004^2$ is obvious.

You need to prove that any pair $ {a, b}$ can be gotten with this replacement - which is false. If I am not wrong, the system is not difficult to solve, and it has solutions if and only if $ ab \ge 4$. Then, we really can obtain $ f(a \plus{} b) \equal{} f(a) \plus{} f(b)$... I concluded this by taking a large $ x$ and observing that

$ f(a \plus{} b \plus{} x) \equal{} f(a \plus{} b) \plus{} f(x)$
$ f(a \plus{} b \plus{} x) \equal{} f(a) \plus{} f(b \plus{} x) \equal{} f(a) \plus{} f(b) \plus{} f(x)$.

Besides that, the equation $ f(a \plus{} b) \equal{} f(a) \plus{} f(b)$ has other solutions. But we can prove that its graphic is dense on the plane if the solution is nontrivial... And this shows that $ f(x) \equal{} 2008x$.


MATHCOUNTS
geometry
perimeter
3D geometry
probability
floor function
calculus

Hello! I would like to start a new marathon, which is a combination of the MOEMS marathon and the MATHCOUNTS marathons that are currently going on. In this thread, you can post problems ranging from very easy MOEMS problems all the way up to hard national MATHCOUNTS problems. Ill first start off with a problem:

NP: An isosceles triangle has a height of 12 cm and an area of 24 square cm. How many cm's are in the perimeter of the triangle? Express your answer in simplest radical form.

P.S.: Whoever posts the solution to a problem, please try and post a new problem.

Thanks.

We have (b=base) $ \frac 12 \cdot 12 \cdot b\equal{}24 \implies b\equal{}4$. Now, using half the base and the height, we find the other two sides to be each $ \sqrt{2^2\plus{}12^2}\equal{}\sqrt{148}\equal{}2\sqrt{37}$. So the final answer is:

$ \boxed{4\plus{}4\sqrt{37}}$

New Problem

When the positive integers up to 2007 are alternately subtracted and added (1 – 2 + 3 – 4 + … + 2001 – 2002 + 2003 – 2004 + 2005 – 2006 + 2007), what is the value of the expression?

Source: 2007 State Countdown

[hide="Solution"]When you pair the first two terms up, and then the next two, you see that you get a sum like this: -1-1-1-1...+2007. In total, there are 1003 negative 1's and so the final sum is -1003+2007=1004[/hide]

[hide="NP"]A 4 by 4 by 4 cube is painted and then cut into sixty-four unit cubes. A unit cube is then randomly selected and rooled. What is the probability that the top face of the rolled cube is painted?[/hide]

It is just the sum of [probability that a cube with [variable] sides is chosen] $ \times$ [the probability that it comes up with a painted side]:

$ \frac{24}{64} \cdot \frac 36\plus{} \frac {24}{64} \cdot \frac26\plus{} \frac {8}{64} \cdot \frac16\equal{}\frac38 \cdot \frac12 \plus{} \frac38 \cdot \frac13\plus{} \frac 18 \cdot \frac16\equal{}\frac{9}{48} \plus{} \frac {6}{48}\plus{}\frac{1}{48}\equal{}\frac{16}{48}\equal{}\boxed{\frac{1}{2}}$

New Problem

1 out of 6 Yummy Cereal boxes has a bobblehead of Zac Efron inside. If Katie buys 3 Yummy Cereal boxes, what is the probability that at least one will have a Zac Efron bobblehead inside? Express your answer as a common fraction.

Ok, i will find the probablity that it doesn't occur.
$ \ (\frac {5}{6})^{3}\equal{} \frac {125}{216}$
Then we subtract from total.
$ \ 1\minus{}\frac {125}{216}\equal{}\frac {91}{216}$
NP
Define a blarghish number as a number when added and subtracted by $ \ 1$, results in a perfect square find the smallest blargish number.

[quote="PowerOfPi"]It is just the sum of [probability that a cube with [variable] sides is chosen] $ \times$ [the probability that it comes up with a painted side]:

$ \frac {24}{64} \cdot \frac 36 \plus{} \frac {24}{64} \cdot \frac26 \plus{} \frac {8}{64} \cdot \frac16 \equal{} \frac38 \cdot \frac12 \plus{} \frac38 \cdot \frac13 \plus{} \frac 18 \cdot \frac16 \equal{} \frac {9}{48} \plus{} \frac {6}{48} \plus{} \frac {1}{48} \equal{} \frac {16}{48} \equal{} \boxed{\frac {1}{2}}$

[/quote]

How come for the first product you get $ \frac {24}{64} \cdot \frac 36$? There are only 8 cubes with 3 sides painted.

[quote="RoFlLoLcOpT"]Clarify please...What do you mean by "added and subtracted by $ 1$"?[/quote]

I think he means when 1 is added and when 1 is subtracted, both those are perfect squares.

Yeah, I mixed it up. It should be:

$ \frac {24}{64} \cdot \frac 16 \plus{} \frac {24}{64} \cdot \frac26 \plus{} \frac {8}{64} \cdot \frac36 \equal{} \frac38 \cdot \frac16 \plus{} \frac38 \cdot \frac13 \plus{} \frac 18 \cdot \frac12 \equal{} \frac {3}{48} \plus{} \frac {6}{48} \plus{} \frac {3}{48} \equal{} \frac {12}{48} \equal{} \boxed{\frac {1}{4}}$

Well, since the other guy's problem doens't make sense,

[hide="NP"]What is the smallest value of x such that absolute value of $ 5x \minus{} 1$ equals the absolute value of $ 3x \plus{} 2$? Express your answer as a common fraction.[/hide]

[hide="Solution?"]
$ 5x\minus{}1\equal{}3x\plus{}2$
$ (5x\minus{}1)\minus{}3x\equal{}(3x\plus{}2)\minus{}3x$
$ 2x\minus{}1\equal{}2$
$ (2x\minus{}1)\plus{}1\equal{}(2)\plus{}1$
$ 2x\equal{}3$
$ \frac{2x}{2}\equal{}\frac{3}{2}$
$ x\equal{}\frac{3}{2}$[/hide]
I only got one solution though. Is it right? If it is, I'll post a new problem.

Nope  :lol: . The other solution is smaller.

[hide="Solution"]
We have:

\begin{align*}
|5x-1|&=|3x+2|
\\ 25x^2-10x+1&=9x^2+12x+4
\\ 16x^2-22x-3&=0
\\ (8x+1)(2x-3)&=0
\\ x&=-\frac{1}{8},\frac{3}{2}
\end{align*}

so the smaller solution is $ \boxed{-\frac{1}{8}}$.
[/hide]

[b]New Problem:[/b]

How many positive integers less than $ 1000$ are not divisible by $ 5$ nor $ 7$?

[hide="Solution"]We can use complementary counting here, so we first count how many positive integers less than $ 1000$ are divisible by $ 5$ and/or $ 7$.  We have $ \lfloor\frac {999}{5}\rfloor \equal{} 199$, $ \lfloor\frac {999}{7}\rfloor \equal{} 142$, and $ \lfloor\frac {999}{35}\rfloor \equal{} 28$.  So there are a total of $ 199 \plus{} 142 \minus{} 28 \equal{} 313$ positive integers less than $ 1000$ that are divisible by $ 5$ and/or $ 7$.  This means there are a total of $ 999 \minus{} 313 \equal{} \boxed{686}$ positive integers less than $ 1000$ that are not divisible by $ 5$ nor $ 7$.[/hide]

[hide="New Problem"]
An $ n\times n\times n$ cube is dipped in paint, and then cut into $ 1\times1\times1$ unit cubes.  If exactly $ 64$ cubes do [b]not[/b] have paint on any of its faces, then what is the volume of the original $ n\times n\times n$ cube?[/hide]

[quote="jxl28"]

An $ n\times n\times n$ cube is dipped in paint, and then cut into $ 1\times1\times1$ unit cubes.  If exactly $ 64$ cubes do [b]not[/b] have paint on any of its faces, then what is the volume of the original $ n\times n\times n$ cube?[/quote]

Since it has $ 64$ cubes..it is a $ 4\times4\times4$ unit cube. You just add 2 to each side to get a $ 6\times6\times6$ unit cube. So $ N\equal{}6$ Volume=$ 216$

NP---
What is the volume of a cube if the surface area of the cube is numerically equall to the volume?

Also: Why so many Marathons? I would prefer 1 or 2 max. :lol:

Well, suface area is $ \ n^{2}*6$
Where 6 is the number of faces, n^2 is surface area of 1 face.
Area is $ \ n^{3}$
so $ \ n^{3}\div\ n^{2}\equal{}n\equal{}6$
$ \ 6^{3}\equal{}216$

Find $ \ P(n,n)$ for any number n? P means choose size r for a quantity n.

[quote=dragonfly][quote="isabella2296"]No problem. :)

If it's divisible by 6, then it must also be divisible by 2 and 3. Similarly, if it's divisible by 8, it's divisible by 4. So we really just need to worry about 8 and 9.

Now, let's take the case of 9 first. $ 4 \plus{} A \plus{} B \plus{} 8$ must be a multiple of 9, so our possibilities for A and B are:

(1,5)
(2,4)
(3,3)
(4,2)
(5,1)
(0,6)
(6,0)
(6,9)
(7,8)
(8,7)
(9,6)

Now, for 8, the last three digits have to be a multiple of 8. So, we're down to:

2,4
6,0
9,6

So our answer is $ 4248 \plus{} 4608 \plus{} 4968 \equal{} 13824$.

NP: Find the least common multiple of 5, 6, 7, 8, and 12.[/quote]
I don't think you got all of the numbers such as 4428, 4068, 4788.[/quote]

For the np. 1?
A positive multiple of 45 less than 1000 is randomly selected. What is the probability that it is a two-digit integer? Express your answer as a common fraction.

[hide=solution]there are 2 2 digit multiples of 45 and 22 multiples of 45 below 100, so it is simply $2/22=1/11                        [/hide] sorry have to go eat someone else can post a problem

[hide=NP]Find the next smallest positive integers such that when it is divided by 3 the remainder is 2, when divided by 5 the remainder is 4, and when divided by 7 the remainder is 6.[/hide]

[quote]Find the next smallest positive integers such that when it is divided by 3 the remainder is 2, when divided by 5 the remainder is 4, and when divided by 7 the remainder is 6.[/quote]

[hide=answer]104[/hide]

[hide=Next problem]Find (x,y)

−5y+8x=−18
5y+2x=58
[/hide]

[quote=Panagiotis11][quote]Find the next smallest positive integers such that when it is divided by 3 the remainder is 2, when divided by 5 the remainder is 4, and when divided by 7 the remainder is 6.[/quote]

[hide=answer]104[/hide][/quote]

Answer was 209 but I'll accept yours anyways (I know you know how to solve these types of problems).

Oh ok. (BTW I've already answerd it and I wasn't trying to cheat. Mods or admins and confirm that.)

New question:
[hide]Many calculators have a reciprocal key $\boxed{\frac{1}{x}}$ that replaces the current number displayed with its reciprocal. For example, if the display is $\boxed{00004}$ and the $\boxed{\frac{1}{x}}$ key is depressed, then the display becomes $\boxed{000.25}$. If $\boxed{00032}$ is currently displayed, what is the fewest number of times you must depress the $\boxed{\frac{1}{x}}$ key so the display again reads $\boxed{00032}$? [/hide]

[quote=shantang]
New question:
[hide]Many calculators have a reciprocal key $\boxed{\frac{1}{x}}$ that replaces the current number displayed with its reciprocal. For example, if the display is $\boxed{00004}$ and the $\boxed{\frac{1}{x}}$ key is depressed, then the display becomes $\boxed{000.25}$. If $\boxed{00032}$ is currently displayed, what is the fewest number of times you must [b]depress[/b] the $\boxed{\frac{1}{x}}$ key so the display again reads $\boxed{00032}$? [/hide][/quote]

???

[hide="Solution if I'm interpreting this right"]
Well, you can press it once and it will read $\frac{1}{32}$. Press it again and it will read $32$, so we see that the answer is $\boxed{2}$.
[/hide]

[hide="Next Problem"]
What is the value of $1^2-2^2+3^2-4^2+\dots+99^2-100^2$?
Source: Mathcounts i think (i don't remember what year)
[/hide]

Were you supposed to solve your own problem? I don't think so...

[quote=Iamawesome1]
[hide="Next Problem"]
What is the value of $1^2-2^2+3^2-4^2+\dots+99^2-100^2$?
Source: Mathcounts i think (i don't remember what year)
[/hide][/quote]

....

[hide=P100000]how many ways are there to fill a $6*6*6$ cube using $3*3*1$ squares? [/hide]

What do you mean "ways"? There should only be one.

He's probably talking about the different ways to orient the $3*3$'s within the $6*6*6$...

[hide=S100000]By combinatorics the answer is $24+1+1=\boxed{26}$[/hide]

[hide=P100001]$9x+9=i$. Find the value of x:[/hide]

[hide=S100001]$x=-1+\frac{i}{9}$[/hide]
[hide=P100001]rectangular prism. sum of edges is $144$. volume is $63$. find surface area.[/hide]


http://web.archive.org/web/20040531004745/www.kalva.demon.co.uk/canada/can69.html

On problem 6, they are trying to get a factorial of a decimal! I understand that this is possible via the gamma function but isn't the Canada Olympiad non-calculus?

Using . for multiplication.  Or, you could have tried a search!  "Canada 1969" worked nicely:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=385582481&t=87628
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=385582481&t=157404


Provide a general method for making the synthesis of a terminal allene from a terminal alkene:

$ RCH \equal{} CH_2 \longrightarrow RCH \equal{} C \equal{} CH_2$.

Yes probably simmons smith reaction to get R- substituted cyclopropane.

Good question Chemrock: then how would you proceed?

dunno if this will work, but will give ti try neways,
bromination follwed by double dehydrohalogenation to give alkyne, foll by addition of carbene +heat

how abt this then alkyne + nanh2 +ch3i follwoed by treatment with nanh2 to get terminal allene? :|

Then treat it with $ \ce{Cl2}$ in presence of light to get 1,3-dichloro 1-alkyl propane. Then hehydrohalogenation with alc.KOH.

[quote="chemrock"]Then treat it with $ \ce{Br2}$ in presence of light to get 1,3-dichloro 1-alkyl propane. Then hehydrohalogenation with alc.KOH.[/quote]
treatment with bromine to get chloro product :rotfl:  also ring opening can take place at any of the carbons

even my method was a guess. but how do you canvert alkyne to allene? will treatment with a strong base b enuf? :huh:
oh sorry i think a weak base like naoh is req

Ya that is acetylene allene rearrangement :)

That's not very good, because it's hard to stop at the allene stage. Much better methods are available.

Here’s a method that I think will work:
Alkene + ozonolysis to give acid + reaction with SOcl2 + Ch2n2 foll by wolff rearrangement to get a ketene.
Ketene + wittigs reagent to give desired terminal allene

And what solvent would you use for the Wolff rearrangement?

how about this i know this works if those hydrogens didn't exist....
not sure though
1)$ \ce{t\minus{}BuO}^{\minus{}}\ce{K} / \ce{CHCl3}$
2)$ \ce{CH3Li}$

Very good Pardesi, that was the answer I was looking for.

[quote="Pardesi"] if those hydrogens didn't exist....[/quote]

What hydrogens?

the vinylic hydrogens...i thought capturing them by $ \ce{CH3}^{\minus{}}$could be a unwanted side reaction during the second step :)

i dont thin kthe vinylic hydrogens are that acidic.
btw, pardesi can you give the products in each step. i know the first part, carbene is formed which adds to the double bond, what does ch3 li do? :huh:

[quote="Pardesi"]the vinylic hydrogens...i thought capturing them by $ \ce{CH3\minus{}}$ could be a unwanted side reaction during the second step[/quote]

What a strange thing to say: the pka of vinylic protons is arround 44, while acetylenic protons have a pka arround 25.

[quote="VARUNARASU"]what does ch3 li do?[/quote]

Methyl lithium substitutes one chlorine by a lithium atom, which then undergoes alpha elimination to form a carbenic cyclopropane; this reactive species then immediatly rearranges to the allene.

[quote="VARUNARASU"]i dont thin kthe vinylic hydrogens are that acidic.[/quote]
yes they aren't now but sfter the dichlorocarben attacks they might be susceptible to attack by $ \ce{CH3}^{\minus{}}$ as they are the $ \beta$ hydrogens.

the mechanism is that given by Carcul :)

[quote="Carcul"][quote="Pardesi"]the vinylic hydrogens...i thought capturing them by $ \ce{CH3 \minus{} }$ could be a unwanted side reaction during the second step[/quote]

What a strange thing to say: the pka of vinylic protons is arround 44, while acetylenic protons have a pka arround 25.

[quote="VARUNARASU"]what does ch3 li do?[/quote]

Methyl lithium substitutes one chlorine by a lithium atom, which then undergoes alpha elimination to form a carbenic cyclopropane; this reactive species then immediatly rearranges to the allene.[/quote]

how do you know that you get a terminal allene?

But you don't have $ \ce{CH3\minus{}}$ in the reaction medium: what you have is methyl lithium.

i mean is there any other choice of allene...well the explaination i can think of is It happens :)

[quote="Carcul"]But you don't have $ \ce{CH3 \minus{} }$ in the reaction medium: what you have is methyl lithium.[/quote]
yes that's what i satisfied myself with but was unsure of the fact because Lots of Indian Books refer to Corey-house  type reactions and other reactions of gilman reagents as attack by $ \ce{CH3}^{\minus{}}$ :(

isnt CH3- almost as good as methyl lithim, considering that the bond is ionic :huh:

[quote="VARUNARASU"]how do you know that you get a terminal allene?[/quote]

Because the one of the carbons of the intermediate cyclopropane is unsubstituted. Here's a picture:

[quote="VARUNARASU"]isnt CH3- almost as good as methyl lithim, considering that the bond is ionic[/quote]

The carbon-lithium bond has a considerable covalent character. On the other hand, $ \ce{CH3Na}$ is pratically ionic.

[quote="pardesi"]isn't it almost like explaining the more nucleophillic character of grignard reagents ver their basicity even there the bond is ionic yet it behaves as a uch better nucleophille :)[/quote]

i still have one doubt abt that. the alpha hydrogens of aldehydes and ketones( pka around 20), are more acidic than acetylenic hydrogens(pka 25). and when aldehydes or ketones, if not sterically hindered, are treated with an organometallic, you get a nuclephilic addition product which is bocz they are better nucleophiles. but when the two groups (alkyne and carbonyl) are present together, why does the acid base reaction take place first and i dont think the caidity of either of the hydrogens is increased that much.? :roll:

can u give an example of that i don't get u ...
also i think the grignard reagents prefer nucleophillic attack to michael addition again because of theit nucleophillicity :)

no, im not referring to michael addition. what i meant was, suppose there is a compound with a keto group say, and a terminal alkynyl group.
on adding one eq. of the grig reagent, acid base reaction takes place btw the acetylenic hydrogen and the grignard reagent and not nuclephilic addition i.e you dont get an alcohol.
even if acid base reaction takes place, i dont get why the alpha hydrogen(to keto) shouldnt be involved, seeing that it is more acidic.

[quote="Carcul"][quote="VARUNARASU"]and when aldehydes or ketones, if not sterically hindered, are treated with an organometallic, you get a nuclephilic addition product which is bocz they are better nucleophiles.[/quote]

That's not the reason. And I have already explained this at least two times on this forum.[/quote]

it was actually in one of [url=http://www.mathlinks.ro/viewtopic.php?search_id=1457769441&t=146980]my old threads[/url] :D

yeah ,i sort of remember posting this question :oops:. but can you give the reason again, and waht abt the second part, why isnt the alpha hydrogen involved? and also i have another stupid doubt  :fool: regarding the mechanism you prposed earlier, how is cl replaced by li. i mean isnt a nucleophile being replaced by and electrophile over here?

[quote="VARUNARASU"]how is cl replaced by li.[/quote]

A radical mechanism is probably involved.

what radical, here you have li+ not li rite. can ch3- donate electrons to the electron deficient carbons and form a free radical? :huh:

1) We do not have $ \ce{Li\plus{}}$, but a compound with a [i]covalent[/i] bond $ C\minus{}Li$.

2) The mechanism is not completely understood, but for some cases it is known to occur through radicals. Therefore, we may assume that, in the general case, radicals are involved.

how do you prepare Ch3Na as it cannot be prepared by reaction of na with alkyl halides?

$ 2 \ce{Na} \plus{} \ce{CH3}\minus{}\ce{Hg}\minus{}\ce{CH3} \longrightarrow 2\, \ce{CH3Na} \plus{} \ce{Hg}$

Apparently this is called the Skattebøl rearrangement:

http://en.wikipedia.org/wiki/Skatteb%C3%B8l_rearrangement


geometry
perimeter
Miscellaneous Problems

Let $p$ be an odd prime. Show that there is at most one non-degenerate integer triangle with perimeter $4p$ and integer area. Characterize those primes for which such triangle exist.

Let $ \Delta ABC$  be  a triangle with sidelengths $ AB=c$, $ BC=a$, $ CA=b$ , semiperimeter   $ s=\frac{1}{2}( a+b+c  )$ and area  $ A$.

From the Heron formula $ A =\sqrt{s(s-a) (s-b ) ( s-c ) }$.
We know that   $ s=2p$ and the area  $ A$ is an integer number  . Thus
$ 1\leq a,b,c\leq 2p-1$,   $ {s(s-a) (s-b ) ( s-c ) }$  needs to be a perfect square.

 Consequence : there exists such a side $ m$ ($ m= a,b,c$ )  of $ \Delta ABC$  that $ p\mid  (s-m  )$.
We can choose $ m=a$ with out lost of generality. Then  $ s-a= p$, since $ 1\leq (s- a) \leq 2p-1$, then $ a=p$.

From the other side
$ (s-b ) + (s-c )= a$, and since
$ (s-b) (s-c ) =  \frac {A^2}{s  (s-a ) }$ then
$ (s-b )(s-c ) = 2 k^2$, where k is natural number.
It is evident that $ ((s-b ),(s-c )) =1$ . Because otherwise there would be a number $ h$,  such   that $ h\mid (s-b)$ and $ h\mid  (s-c )$ and so $ h\mid  (s-b) + (s-c )$ that is $ h\mid a$, therefore
$ h\mid p$ and that implies that $ h=1$ or $ h=p$ .If $ h=p$ then $ b=c=p$ then $ 2s= a+b+c =3a=3p$ , hence there is a contradiction.
We know (Sierpinski) that there exists a unique representation of prime p in the form $ p=tx^2 + y^2$ with $ t,w,x,y$
natural numbers and $ (x,y)=1$. So we take $ p=2x^2 +y^2$. That means
Suppose
 $ (s-b)= 2x^2$
and
$ (s-c)= y^2$ with
 $ ( x,y)=1$, ($ k= x^2y^2$), then $ (s-b )+(s-c) =a= 2x^2 +y^2=p$.
 Hence, $ \Delta ABC$ has the sides
 $ a=p$, $ b= 2p-2x^2$ and  $ c= 2p-y^2$ .Thus the perimeter
 $ 2s= p +(2p-2x^2) + (2p-y^2)= 4p,$
and the area
$ A^2= s (s-p )(s-(2p-2x^2))(s-y^2 )= 4p^2 x^2y^2$

Next, it is possible if and only if $ p = 1 (mod8)$ or $ p=3 (mod8)$.
That is when
1° case $ 2x^2 +y^2 = 1 (mod8)$
or
2° case  $ 2x^2 +y^2 =3 (mod8)$.

Example
case 1°
Taking $ x=4$ and $ y=3$
$ x^2=0(mod8)$ and $ y^2=1(mod8)$
we have $ p=41$ ,$ {s(s-a) (s-b ) ( s-c )=2*41*2*4*9}$, which is a perfect square.

 case 2°
For example let  $ x^2=1(mod8)$ and $ y^2=1(mod8)$.
Taking $ x=3$ , $ y=5$, we have $ p=43$, then $ a=43$, $ b=68$,$ c=61$,
$ {s(s-a) (s-b ) ( s-c )= 1664100}$ which is a perfect square.

Sorry, small mistake :blush: :
[quote] s(s-a) (s-b ) ( s-c )=2*41*2*4*9[/quote]
I meant of course: $ s(s\minus{}a) (s\minus{}b ) ( s\minus{}c )\equal{} 4\cdot 41^2 \cdot 16 \cdot 9$
and then the triangle $ \Delta ABC$ has $ a\equal{}41$ ,  $ b\equal{} 50$ ,  $ c\equal{}73$ .


articles AMC AIME email USA(J)MO USAMO probability \/closed

WOOT is a 7-month Olympiad preparation and testing program that brings together many of the best students to learn Olympiad problem solving skills.  WOOT includes 8 problem solving articles and sets of practice problems, 16 classes, 2 AIME-style tests, 3 practice Olympiads, 3 problem sets, and lots of interaction with many of the very best problem solvers.

July 15 is the last day to receive the discounted enrollment price of $\$595$.  Thereafter, the enrollment fee is $\$695$.  WOOT runs from mid-September until mid-April.  Due to sponsorship from AoPS and D. E. Shaw group, all 2006 MOP participants are invited to participate in WOOT for free.

[url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]Click here[/url] to read more about WOOT, and to see the WOOT schedule.

[quote="jb05"]Us MOSPers filled out some info for you last week. Have we already been signed up, or do we need to sign up again?[/quote]

You'll be signed up when I get the forms from MOP - the AMC will send them to me.  You'll get an email sometime this month with more details.

[quote="bubka"]Funny how so called maths lovers sell their maths knowledge to make money.[/quote]

Unlike what some people believe, Administrators are not invincible, so they do have human needs (though they do have other super powers).  :o They have to meet their needs somehow, so charging for courses is a good way to do that.

I think that for what you get out of the courses they offer, they are worth a lot more than what they charge.

[quote="PenguinIntegral"][quote="bubka"]Funny how so called maths lovers sell their maths knowledge to make money.[/quote]
I have an intense urge to stab you.
[/quote]

Same here :mad: .

Thanks for the WOOT reminder, I almost forgot :D

I moved the very long discussion that didn't belong in this thread into the Round Table where it belongs.

[quote="rrusczyk"][quote="pkerichang"]
One question I have: How will I know that I can understand and utilize the stuffs taught in WOOT, instead of being puzzled and wasted the money because of the difficulty??? This is the one thing keeping me back.  Also, how much time on average per week should we expect to spend on homework and studying and stuff??? (I have some other stuffs going on so my time is kind of limited).[/quote]

If you're getting 5+ on the AIME now, you'll be able to understand a fair amount of WOOT, though you'll have to spend some time thinking about some concepts.  You'll also be able to save transcripts/papers for later consumption, and you'll have a big community and a message board to discuss concepts in more detail.

As for time, if you only have a few hours available a month, then WOOT is probably not for you.  Expect to spend 2-5 hours a week if you want to get the most out of it, where you'll have a fair amount of flexibility (i.e. if you're busy this week and next, but have a lot of time the week after that, you'll be able to not do much for a couple weeks, then do a bunch when you have time).[/quote]
What level of difficulty are the assignments (such as problem sets, etc.), and how much time do they usually take?

The level of the assignments will vary - you'll be able to choose the problems that are right for you.  As for the time, this will depend on how much time you want to spend on WOOT.  We suggest expecting to spend 3-4 hours per week on WOOT, but if you want to spend 8 hours a week on it, you'll be able to once the program is underway.  The program is quite flexible in this regard.

[quote="pkerichang"][quote="mysmartmouth"]I got a 4 on the AIME in 8th grade.  I think I have improved my math skills over a year, but on practice AIMEs I usually get around 4 or 5.  I don't expect to make USAMO this year (my freshman year).  When would be the opportune time to take WOOT?  Money in not really an issue, but I wouldn't want to waste a large amount.  I was thinking about taking WOOT my sophmore year, when I'll probably make USAMO, or else come very close.  What do you think?[/quote]

Wow,from your post count I thought you're older and more experienced...
Anyways, if I'm in your situation, I'll sign up for it (if the cost is not too much a deal) and study hard this summer.  In that case, I believe you can improve to get at least 5 on AIMEs (the first 5 are not too hard anyways, although I suck at counting :( )[/quote]

Well, I've improved a lot.  I spend a lot of time on AoPS doing problems, and I think I will take some classes too.  But I talked to other people, and they said try it out.  How does the refund work?  I heard something about getting your money back within a certain number of days if it's too hard...

I would recommend that you do WOOT, and try the lowest level. WOOT will prove to have been very valuable in the long run.

The drop policy is that we'll permit full refunds before the second class (i.e. after one article and one class).

We'd recommend that a rising 9th grader with a 4 on the AIME in 8th grade and a lot of enthusiasm for mathematics try WOOT.  Such a student would also find our intermediate classes appropriate.

I am in 7th grade, and considering taking the WOOT next year as an 8th grader.  I am taking Intermediate Counting and Probability and Intermediate Number Theory this summer, and understand almost all of the material in those classes so far.  I am a little over half-way done with Algebra II.  On the AIMEs, last school year, I got 4-6 on most of the practice ones without any Algebra II knowledge, but got a 1 on the real thing (5 stupid mistakes).  I would say that I have improved a lot since then.  The biggest problem I had was getting started (in the right direction) on a lot of the problems, but the Intermediate classes have helped a lot with that.

I don't know if I could do many of the higher level things (USAMO), so I am wondering if I should take WOOT.

If you can handle our intermediate level classes, then you should learn a lot in WOOT.

[quote="pkerichang"]Hmm...from that, how does the Olympiad subject classes compare with WOOT??[/quote]

Some of WOOT is easier, some of it is the same, and some of it is more advanced than the average Olympiad subject class lesson.

[quote="pkerichang"]one more question, will we go over trig/complex numbers in WOOT (it doesn't say in the schedule)???[/quote]

A little in the algebra classes, but not exhaustively.  (We did cover complex numbers, particularly in geometry, in WOOT 2005-06; the Intermediate Complex/Trig class is your best bet for AIME-level trig/complex numbers, and for an introduction to complex number geometry.)

Just curious- in terms of difficulty, how would you compare the WOOT material to the AoPS books' content?

The AoPS Volume 2 covers some material that will be covered in WOOT, but there will be a lot of material on other subjects that are just as difficult as well as plenty of material that will be more difficult by varying degrees.

[quote="rzsolt"]Is WOOT good putnam prep?[/quote]

Yes in the sense of focusing on proof writing and problem solving.  But not entirely in that the Putnam usually has questions on different areas of math like calculus and linear algebra.  However, the Putnam students who get past the first few problems to the point of working on those are already pretty much top Putnam students.

Sorry but I can't resist to comment on MCrawford's signature :P :

if I'm rotating around the ball back and forth, then I'm in an accelerating frame (a.k.a. non-inertial) because of the presence of the centripetal force.  Then I should be able to know if I'm in an accelerating frame or not, so I think the ball's spinning...

Just to make this not a spam, exactly what subjects do you cover in algebra (factoring, polynomials, etc..)??

[quote="ch1n353ch3s54a1l"]I was looking at the WOOT thing and I didnt really figure out what WOOT meant by having two classes a week. Are they the same classes and they let you choose 1 class just in case you can't go to one, or are they different classes on the same subject?[/quote]
What they do is that they do every class twice a week (Monday and Thursday) so that if someone can't do the class one of the days (like me, I have orchestra rehearsal Monday evenings) he/she can do it the other day.
I've been wondering, if you're available both days, can you take both classes in case you want to reinforce the material you learned, etc.?

There isn't anything stopping you from attending both.  But if you do go to both, it's courteous to just observe the second session instead of answering all the questions (which you will have already seen).

I think joml88's answer is good.  We want to focus on each student as much as class time allows and taking a passive role once (if you attend both sessions) helps that process.

[quote="ch1n353ch3s54a1l"]Ok.
If you couldn't make both classes for one week will there be transcripts or something for you to look at after the class is done?[/quote]

Yes, classes are transcripted (both of them in fact).

[quote="Ichverstehenicht"]Unfortunately, I did really bad on AIME (first time doing this contest) and got maybe a 2 or a 3. I am a rising junior; should I sign up or just do some practice and come back next year?[/quote]

That's up to you.  If you can get nearly all the problems posted in the Getting Started forum and some of the problems in the Intermediate forum, then it may be that you scored lower than you usually would on the AIME.  You might want to take a look at the two problems I walked through in tonight's Math Jam (transcript is posted in the Math Jam area) and see if you can follow them well.  If you can, you'll learn a lot from WOOT.

I couldn't make it, but the math jam answered more than a few questions.  Here is the link:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=165

[quote="computer_nuke"]I was really thinking about taking WOOT, but I have a few questions. Is the class more focused on proof writing (as in how to write a good and complete proof) or is it more focused on thinking problems through? Or is it doing both?[/quote]

Definitely both.  And the way we teach it, thinking things through is conducive to learning proof-writing more quickly.  Be prepared to participate heavily on the message board if you want your proof-writing skills to improve most rapidly.  We can point student by student to those who improved the most by those who have participated in WOOT and other class message boards heavily (zabelman and MysticTerminator for instance -- two IMOers).

[quote="quantum leap"]So, if the problem sets are of varying degrees of difficulty, can you take more than one problem set?[/quote]

You'll get to pick and choose which problems to do.  For example, you might be given 15 problems and told to submit solutions to 8 of the problems (I'm making the numbers up for an example).  Therefore, you can pick which of your solutions to receive feedback on.  For subjects where you are more comfortable, you might choose many of the hardest problems, and for other subjects, stay in shallower water.

[quote="Tecnogram888"]My parent's credit card limit is only $\$$500....which isnt large enough to cover the cost of WOOT.  Is there anyway I can mail a check to you guys or something before july 15th and still get the early bird discount?
or do credit card companies allow u to go over..as long as u pay the balance immediately?[/quote]

I don't know about ways to fudge credit card limits, but you can certainly mail us a check and we'll honor the discount according to the postmark date.


algebra
function
domain
quadratics
Ring Theory
group theory
abstract algebra

Let R = {x+yi : x,y 2 Z}. R is a Euclidean domain, and 2+i is irreducible in R. Hence the ideal I = <2+i> is maximal;
hence R/I is a field. Which familiar field is it?

I have approached this problem in a variety of ways, in my quest for the answer have come across quadratic fields and various other facts about Gaussain integers, but still have not arrived at the answer for this question.
Will be grateful if someone could help.

[quote="ZetaX"]Some checking gives that the field has $5$ elements (e.g.: this field is a subring of the ring $\mathbb{Z}[i] \mod 5$ having $25$ elements, so by Lagrange it has $5$ elements ($1$ or $25$ is also possible but simply excluded to be the case)), and there is just one such field (the same as $\mathbb{Z} \mod 5$).[/quote]

Thank you.. but how do you go about checking there are 5 elements.
I tried to use the approach to find the cosets of R/I but got stuck..
I am just starting  on this branch...and hence need some help... :(

A bit better view of this: the field is the quotient of this ring $\mod (2+i)$.

When looking $\mod 5$ in $\mathbb{Z}[i]$, every element can be represented as $a+bi$ with $a,b \in \{0,1,2,3,4\}$ since we can change $a,b$ to their remainder by division through $5$ (same as in the integer case).
But this representation is unique since when $5 | a + bi$, then $5| a,b$.
Thus there are $25$ elements.

Now there are clearly two different elements in the field, so it has more than $1$ element. Also the elements represented by $2+i$ and $0$ are not the same $\mod 5$, but $\mod (2+i)$, thus there are less than $25$ elements. By Lagrange, only $5$ is left as possibility.

I do not understand why you applied Lagrange since that applies to subgroups.  :?:
Also my ring R is the ring of Gaussain integeres i.e R-={ x+iy: x,y are elements of Z}

I use Lagrange for the additive subgroup (neglecting totally the multiplicative property). Since that ring of $\mathbb{Z}[i] \mod 5$ has $25$ elements, meaning especially the additive group of that ring has $25$ elements, so any additive subgroup (any subring must imply such a subgroup!) and any quotiont group (dito!) must have $1,5$ or $25$ elements.


[quote="linzk3"]im having trouble finding out what y equals in terms of x in the following equation....
x/5+y/4=1[/quote]
[hide]
multiply both sides by 4
$\frac{4x}{5}+y=4$
subtract $\frac{4x}{5}$ from both sides
$y=4-\frac{4x}{5}$
put over common denominator
$y=\frac{-4x+20}{5}$
[/hide]


function
trigonometry
complex analysis
complex analysis unsolved

Let $f : \mathbb{C}\to \mathbb{C}$ - analytic function such that $f(z) = z^{2}$ for all $z \in \mathbb{Z}$. Does it follow that $f(z) = z^{2}$ for all $z \in \mathbb{C}$ ?

Yes, $f(z) = z^{2}+\sin{2\pi z}$ also works.

I think such a phenomenon is always possible if we have given countably many values on a discrete set by Weierstrass product


Can anyone tell me what a good (free) program is for working with large integers 25+ digits? I've been told that Python is good. Anything else?

Python isn't a program, it's a language.  Any language worth its salt has functions in its library for dealing with arbitrary-precision arithmetic; for example, in Java you use the BigInteger class.


inequalities
geometry
geometric transformation

Let a,b,c are greater than zero. Prove that

$ \frac{a^2}{ 6\sqrt{abc}\minus{}3\sqrt[3]{abc} \minus{}a^2}\plus{}\frac{b^2}{ 6\sqrt{abc}\minus{}3\sqrt[3]{abc} \minus{}b^2}\plus{}\frac{c^2}{ 6\sqrt{abc}\minus{}3\sqrt[3]{abc} \minus{}c^2} \ge \frac{3}{2}$

[quote="4865550150"]Let a,b,c are greater than zero. Prove that

$ \frac {a^2}{ 6\sqrt {abc} \minus{} 3\sqrt [3]{abc} \minus{} a^2} \plus{} \frac {b^2}{ 6\sqrt {abc} \minus{} 3\sqrt [3]{abc} \minus{} b^2} \plus{} \frac {c^2}{ 6\sqrt {abc} \minus{} 3\sqrt [3]{abc} \minus{} c^2} \ge \frac {3}{2}$[/quote]

Are you sure about it? Because it's not homogeneus at all.

[quote="colorfuldreams"]柯西不等式
$ LFS \equal{} \frac {(a \plus{} b \plus{} c)^2}{18\sqrt {abc} \minus{} 9\sqrt [3]{abc} \minus{} a^2 \minus{} b^2 \minus{} c^2} \geq \frac {3}{2}$
$ 2(a \plus{} b \plus{} c)^2 \plus{} 3(a^2 \plus{} b^2 \plus{} c^2) \plus{} 27\sqrt [3]{abc} \geq 54\sqrt {abc}$
然後很容易就知道
$ a \plus{} b \plus{} c \geq 3\sqrt [3]{abc}$
$ 2(a \plus{} b \plus{} c)^2 \geq 18(abc)^{\frac {2}{3}}$
$ 3(a^2 \plus{} b^2 \plus{} c^2) \geq 9(abc)^{\frac {2}{3}}$
然後
$ LHS \geq 27(abc)^{\frac {2}{3}} \plus{} 27{abc}^{\frac {1}{3}} \geq 54\sqrt {abc}$
 :lol:[/quote]

漂亮！ :roll:  :lol:

不知道你是哪間學校 :lol:
我在東區

[quote="Dimitris X"]Hey guys,will you stop spaming????[/quote]
 :maybe:  :maybe:
My English is poor

[quote="colorfuldreams"][quote="4865550150"][quote="colorfuldreams"]不知道你是哪間學校
我在東區[/quote]

我係將軍澳個度架 :blush: ，依家學習學習不等式 :lol:
唔知你讀緊中幾呢？我讀中七.......[/quote]

 :maybe: 中5

不過你學這些是不是為了競賽?

有好幾本是為不等式學習的書籍
Pham Kim Hung - Secrets in Inequalities (volume 1)
old and new inequality
New book: Inequalities with Beautiful Solutions - V. Cirtoaje, V. Q. B. Can, T. Q. Anh
不過我想香港應該買不到.有幾本是越南人寫的[/quote]

純粹為興趣 :D
請問你話個幾本書, 你係邊度找到?我好想學多d不等式 :help:

我們會不會大壩道 :lol:
在國外論壇打中文
外國人全部看不懂 :D

[quote="colorfuldreams"]我們會不會大壩道 在國外論壇打中文
外國人全部看不懂[/quote]

i agree with you. We must stop....... :P

this is highly irritating... IF U GUYS R'NT FLUENT WITH ENGLISH STICK TO MATHEMATICAL THINGS.. WE WILl try to understand it somehow...(afterall we arent fools) if u r uncomfortable even to do that then go to chinese forum or sthng where u prob have many speaking the same lang and converse there... and i do think u 2 know enough english to communicate... :mad:

[quote="colorfuldreams"]
外國人全部看不懂 :D[/quote]
This translated to  "All foreigners can not read"

But just for you
所有外國人可以使用谷歌翻譯  :wink:
which means "All foreigners can use Google Translation"

[quote="dgreenb801"][quote="colorfuldreams"]柯西不等式
$ LHS \equal{} \frac {(a \plus{} b \plus{} c)^2}{18\sqrt {abc} \minus{} 9\sqrt [3]{abc} \minus{} a^2 \minus{} b^2 \minus{} c^2} \geq \frac {3}{2}$
[/quote]
You can't use Cauchy like this because the denominator of the original fractions might be negative.[/quote]

I think the question is wrong
because
we let $ a \equal{} 300 , b \equal{} 1 , c \equal{} 1$
$ LHS \equal{} \minus{} 1.00093 \plus{} 0.012 \plus{} 0.012 < \frac {3}{2}$
 :o

[color=red]Don't use Chinese here![/color] :wink:

Japanese Communities Moderator

kunny

[quote="colorfuldreams"]不知道你是哪間學校 :lol:
我在東區[/quote]

[quote="4865550150"][quote="colorfuldreams"]不知道你是哪間學校
我在東區[/quote]

我係將軍澳個度架 :blush: ，依家學習學習不等式 :lol:
唔知你讀緊中幾呢？我讀中七.......[/quote]

[quote="colorfuldreams"][quote="4865550150"][quote="colorfuldreams"]不知道你是哪間學校
我在東區[/quote]

我係將軍澳個度架 :blush: ，依家學習學習不等式 :lol:
唔知你讀緊中幾呢？我讀中七.......[/quote]

 :maybe: 中5

不過你學這些是不是為了競賽?

有好幾本是為不等式學習的書籍
Pham Kim Hung - Secrets in Inequalities (volume 1)
old and new inequality
New book: Inequalities with Beautiful Solutions - V. Cirtoaje, V. Q. B. Can, T. Q. Anh
不過我想香港應該買不到.有幾本是越南人寫的[/quote]

[quote="4865550150"][/quote][quote="colorfuldreams"][quote="4865550150"][quote="colorfuldreams"]不知道你是哪間學校
我在東區[/quote]

我係將軍澳個度架 :blush: ，依家學習學習不等式 :lol:
唔知你讀緊中幾呢？我讀中七.......[/quote]

 :maybe: 中5

不過你學這些是不是為了競賽?

有好幾本是為不等式學習的書籍
Pham Kim Hung - Secrets in Inequalities (volume 1)
old and new inequality
New book: Inequalities with Beautiful Solutions - V. Cirtoaje, V. Q. B. Can, T. Q. Anh
不過我想香港應該買不到.有幾本是越南人寫的[/quote][quote="4865550150"]

純粹為興趣 :D
請問你話個幾本書, 你係邊度找到?我好想學多d不等式 :help:[/quote]

[quote="colorfuldreams"]我們會不會大壩道 :lol:
在國外論壇打中文
外國人全部看不懂 :D[/quote]


Points $ A$ and $ B$ are fixed on a circle $ C_{1}$. Circle $ C_{2}$, whose center lies on $ C_{1}$, touches line $ AB$ at $ B$. Another line through $ A$ intersects $ C_{2}$ at points $ D$ and $ E$, where $ D$ lies between $ A$ and $ E$. Line $ BD$ intersects $ C_{1}$ again at $ F$. Prove that line $ EB$ is tangent to $ C_{1}$ if and only if $ D$ is the midpoint of the segment $ BF$.

Hello!

The points $ O_{1},O_{2}$ are the centers of the circles $ C_{1},C_{2}$ and $ C_{1}\cap C_{2} \equal{} \left\{B,C\right\}$.
$ AB\bot BO_{2}\Rightarrow O_{1}$ is the midpoint of $ AO_{2}$.
$ \angle O_{2}FB \equal{} \angle O_{2}AB \equal{} \angle O_{2}BC \equal{} \angle O_{2}AC \equal{} \angle O_{2}FC \equal{} \alpha$.
$ \angle O_{2}FC \equal{} \angle O_{2}FD\Rightarrow CF \equal{} DF$   (1)
The lines $ AB,AC$ are tangents to $ C_{2}\Rightarrow AB \equal{} AC$   (2)
The quadrilateral $ ABCF$ is cyclic $ \Rightarrow\angle ABD \equal{} \angle ACF$   (3)

$ BD \equal{} DF$ and (1) $ \Rightarrow BD \equal{} CF$   (4)
(2),(3),(4)$ \Rightarrow\Delta ADB\equiv\Delta AFC\Rightarrow\angle ADB \equal{} \angle AFC\Rightarrow\angle BDE \equal{} \angle ABC\Rightarrow$
$ \Rightarrow\angle BCE \equal{} \angle BEC\Rightarrow BC \equal{} BE\Rightarrow\angle O_{2}BC \equal{} \angle O_{2}BE \equal{} \alpha\Rightarrow$
$ \Rightarrow\angle O_{2}AB \equal{} \angle O_{2}BE \Rightarrow EB$ is tangent to $ C_{1}$

Reciprocal
$ EB$ is tangent to $ C_{1}\Rightarrow\angle O_{2}BE \equal{} \angle O_{2}AB \equal{} \alpha\Rightarrow\angle O_{2}BE \equal{} \angle O_{2}BC\Rightarrow BE \equal{} BC\Rightarrow$
$ \Rightarrow\angle BEC \equal{} \angle BCE\Rightarrow\angle ABC \equal{} \angle BDE\Rightarrow\angle AFC \equal{} \angle ADB$   (5)
(2), (3), (5) $ \Rightarrow\Delta ADB\equiv\Delta AFC\Rightarrow BD \equal{} CF$   (6)
(1), (6) $ \Rightarrow BD \equal{} DF$

Petrisor Neagoe :)


khahesh mikonam tamame azizany ke jadid ozv shodand.dar inja khodeshoono moarefi konand.mer30.

ghabl az inke ye topic e jadid baaz konid, lotfan had e aghal be safhaye aval ye negahi bendaz baad.

azizan mitoonan inja khodeshoon o moarefi konan:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55375

topic Locked.


:arrow: Guys , help out of this freaking S**t:

A right circular tank(pointy end down) is being filled with liquid at a constant rate of 2m^3/s. The tank has a height of 12m and its circular face(at the top) has a radius of 5m. At what rate is the radius of the upper surface of the liquid increasing when the tank is half full?

given formula: V=3/Pi*r^2*h

There's a copy in the calculus forum where it belongs, so this thread can be safely deleted.


OK i created a little childish Problem( Dont hit me too much)

Problem: Let $a, L \in \mathcal{R}-\{0\}\text{and}f,g: \mathcal{R}\longrightarrow \mathcal{R}$.

Let \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=L\] Can we say \[\lim_{x \rightarrow a}[f(x)-Lg(x)]=0 :\] :D

[quote="Kent Merryfield"]Well - no.

Let $a=0,$ just to keep things simple, and let $f(x)=\frac1{x^{2}}+\frac1x$ while $g(x)=\frac1{x^{2}}.$

Then $\frac{f(x)}{g(x)}\to1$ but $f(x)-g(x)\to\pm\infty.$

Question: can you save the result with some additional hypothesis? Maybe $f,g$ bounded?[/quote]
I think the problem first assumes that $a, L \not= 0.$ He is interested in knowing if the result holds precisely when those conditions are met.

$a\ne0$ is unimportant; you could just translate. Change my example by replacing all my $x$'s by $x-1,$ then take the limit as $x\to 1$ instead of $x\to 0.$ We have $L=1$ there. Then it's still a counterexample to the original statement.

Kent i think you are right , If they are bounded then the result holds.
If \[\frac{f(a+h)}{g(a+h)}= L+\epsilon\] then \[f(a+h)-Lg(a+h) = \epsilon*g(a+h)\] So when $h \longrightarrow 0$ and by hypthesis $\epsilon \longrightarrow 0$ then if $g$ is bounded \[g(a+h)*\epsilon \longrightarrow 0\]


$n$ is called [i]diamond 2005[/i] if $n=\overline{...ab999...99999cd...}$, e.g. $2005 \times 9$. Let $\{a_n\}:a_n< C\cdot n,\{a_n\}$ is increasing. Prove that $\{a_n\}$ contain infinite [i]diamond 2005[/i].

Compare with [url=http://www.mathlinks.ro/Forum/topic-15091.html]this problem.[/url]

The density of positive integers without $2005$ consecutive 9's in their decimal representation goes to 0 (at exponential speed in the number of digits; it is a subset of the integers whose base $d$ representation contains no digit $d-1$, where $d = 10^{2005}$).  So it cannot contain any sequence such as $a_n$, that has positive upper density.

Moreover, $\sum a_n^{-1}$ converges, where $a_n$ is a sequence of positive integers without given digit in $d$-representation. It is classics.

[quote="grobber"]Yes, given a positive integer $b>1$ and $k\in\overline{0,b-1}$, a veru crude estimation shows that the sum of the reciprocals of those numbers which do not have the digit $k$ in base $b$ is less than $\sum_{n\ge 0}\left(\frac{b-1}b\right)^n$, so it's finite. This cannot happen if $\left(\frac{a_n}n\right)_n$ is bounded.[/quote]

If the bound is weakened to $a_n \leq f(n)$ with the sum of $1/f(n)$ finite, such reasoning doesn't work.
However, you can simply observe that there are $(d-1)^n$ numbers of at most $n$ digits in base $d$ with no digit equal to
$d$, and at least $f^{-1}(d^n)$ members of sequence $a_k$.  So the "boundary" between possibility and impossibility lies around $f(n) = n^K$  where $K = \log(d) / \log(d-1)$, and this is achievable by letting $f(n) = a_n$ simply be the sequence of all numbers whose base $d$ representation is of the required type (note that this is a subset of the numbers allowed in the problem, but its density is within a constant factor of the larger set).


$k_1,k_2,k_3, ...$ are disjoint circles. (circles cannot be inside eachother and not intersect)
The radius of the circle $k_i$ is $\frac{1}{i}$ and its centre is $P_i$. Is it possible for the sequence of points $P_i$ to be convergent?

I believe this has been discussed before. It's from the Vojtech Yarnk conpetition (in $2004$, I think), so it should be somewhere in the "College Playground".


LaTeX
inequalities unsolved
inequalities

a+x=b+y=c+z=1                                     a,b,c,x,y,z  > 0
prove :

  (abc+xyz)(   1            1            1        )   > 3
                    ___  +    ___  +    ___
                     ay          bz           cx

when does equality hold?[/list][/b]

[color=red]Edit by Megus: moved to the right forum[/color]

Moreover, try to post in the appropriate section and not twice the same problem  ;)

Pierre;

As for LaTeX, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 .
As for the problem, it was solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=2926 .

  Darij


A six place number is formed by repeating a three place number; for example, 256,256, or 678,678, etc.  Any number of this form is always exactly divisible by:

(A) 7 only
(B) 11 only
(C) 13 only
(D) 101
(E) 1001

[hide]Great problem. Let's say we have a number xyz and we want to make that number to xyzxyz. What do we do? Times it by 1001 !(dont get this? times xyz by 1000 = xyz000 and then add xyz which is equivilent to multiplying xyz by 1001) E. This works for any number xyz.[/hide]

[hide]to repeat a three digit number we multiply it by 1000 and add the number (1000N+N) so we get 1001 N $\boxed{E}$[/hide]

[hide]If we divide any number abcabc by abc, we get 1001, always. [/hide]


inequalities
trigonometry
calculus
calculus computations

For $ 0 < b < a < \frac {\pi}{4}$ prove this inequality:
$ \tan^2{a} > \tan{(a \plus{} b)} \cdot \tan{(a \minus{} b)}$
Can somebody help me prove this by using calculus...  :?:

$ \frac{a\plus{}b}{2}\plus{}\frac{a\minus{}b}{2}\equal{}a$

$ \frac{a\plus{}b}{2}\minus{}\frac{a\minus{}b}{2}\equal{}b$

hello, applying the addition formulas we get $ \minus{}2\sec^2(a)\sec(a\minus{}b)\sec(a\plus{}b)\sin^2(b)\sin(a\minus{}\frac{\pi}{4})\sin(a\plus{}\frac{\pi}{4})>0$ because of $ 0<b<a<\frac{\pi}{4}$.
Sonnhard.

[quote="mptlu6"][quote="kunny"]$ \frac {a \plus{} b}{2} \plus{} \frac {a \minus{} b}{2} \equal{} a$

$ \frac {a \plus{} b}{2} \minus{} \frac {a \minus{} b}{2} \equal{} b$[/quote]

And... ?  :o[/quote]

Let $ x\equal{}\frac{a\plus{}b}{2},\ y\equal{}\frac{a\minus{}b}{2}$, the inequality can be written $ \tan ^ 2(x\plus{}y)>\tan 2x\tan 2y$.


In saal, saal besiyar zibaiiye va rahe hale zibaii ham dare ,khali az lotf nist ke roye an fekr konid  :lol:

een sooale manam deedam ama motasefane ziad natavanestam jelo beravem. rahnamayee??

[quote="Ramtin Madani"]cheraa hal nemishe?!!
soale kojaast :o[/quote]

Saale Romanie :lol:

[quote="ehsan2004"]$n \in\mathbb{N}$ va bozorgtar as $3$ mibashad , va $x\in\mathbb{R}$ s.t. \[ \{x\}=\{x^2\}=\{x^n\}  \] (ke ${x}$ yani joze aashari $x$) .Sabet konid $x$ adad sahihe. :D[/quote]

pas chi shood ide ha toon :P

haaay bacheha kojaiin mesle inke in chand rooze tatil kheyli behetoon khosh gozashte  :lol:


quadratics
number theory proposed
number theory

Find $ x,y$ integer numbers $ x^4\minus{}2y^2\equal{}\minus{}1$

I don't know if it's right but i found
(x,y)=(1,1) or (1,-1) or (-1,1) or (-1,-1)

It is quite easy...
$ x^{4}\plus{}2y^{2}\minus{}1$
First it is obvious that $ x\equal{}2k\plus{}1$, and after expansion it follows that $ y\equal{}2l\plus{}1$
We land in:
$ k(2k^{3}\plus{}2k^{2}\plus{}3k\plus{}1)\equal{}l(l\plus{}1)$
Since $ k|l(l\plus{}1)$ and $ (l,l\plus{}1)\equal{}1$ hence $ k|l$ or $ k|l\plus{}1$

a)$ l\equal{}nk$
$ k(2k^{3}\plus{}2k^{2}\plus{}3k\plus{}1)\equal{}n^{2}k^{2}\plus{}nk$
$ n^{2}\equal{}2k(k\plus{}1)\plus{}3$ which cannot be possible, because $ 4|2k(k\plus{}1)$ for all integers k, thus $ 2k(k\plus{}1)\plus{}3\equiv 3 \ (mod4)$, but 3 is not a quadratic residue modulo 4, hence no solutions

b)$ l\equal{}nk\minus{}1$
$ 2k^{3}\plus{}2k^{2}\plus{}3k\plus{}1\equal{}n^{2}k\minus{}1$
$ 2\equal{}k(n^{2}\minus{}2k^{2}\minus{}2k\minus{}3)$, so $ k|2$
Now we only have to look through $ k\equal{}1,\minus{}1,2,\minus{}2$ and we get the desired result :wink:

[quote="xxp2000"]polskimisiek says
"Since $ k|l(l \plus{} 1)$ and $ (l,l \plus{} 1) \equal{} 1$ hence $ k|l$ or $ k|l \plus{} 1$".
Why is that?
$ 6|45*46$, but $ 6\not|45$, $ 6\not |46$. It seems you need other condition like $ k$ is prime to make this claim right.[/quote]

Aaah, you are totaly right, I overlooked that simple fact. Well, we could proceed a little further similarily, denoting $ (k,l)\equal{}g \ (k,l\plus{}1)\equal{}h$ we have $ (g,h)\leq (l,l\plus{}1)\equal{}1$ thus $ (g,h)\equal{}1$ and if $ k\equal{}k'gh$ then $ k'\equal{}1$ so $ k\equal{}gh$ and we will have $ (g,h)\equal{}(l,l\plus{}1)\equal{}(g,\frac{l}{g})\equal{}(h,\frac{l\plus{}1}{h})\equal{}1$, but from here I am not sure how to obtain anything more...  :oops:


Hello!

$ABC$ is a triangle such that $\angle ABC=80^\circ$. $AD$ is the angle bissector of $\angle ABC$ with $D\in (AC)$. Determine $\angle BAC$ if $AD=DB+BC$.

how can $AD$ be the bisector of $\angle ABC$ with $D\in (AC)$? please correct the problem.

[quote="nayel"]how can $ AD$ be the bisector of $ \angle ABC$ with $ D\in (AC)$? please correct the problem.[/quote]

Sorry!!, the angle bissector is $ BD$ and not $ AD$.

I think the answer is $ \angle BAC \equal{} 20^{o},$ as it is consequent from the configuration of a regular enneagon.

We have $ DB \plus{} BC \equal{} BC \plus{} BE \equal{} BC \plus{} EF \equal{} BC \plus{} CF \equal{} AG \equal{} AD$ and the solution is completed.

Kostas Vittas.

By the way of this problem, I would like to present an approximately construction of a regular enneagon, as a recreative geometric intermission.

[b][size=100]1)[/size][/b] – A circle $ (K)$ centered at $ K,$ with radius $ R \equal{} 1$ is given and let $ MG,$ its diameter. We draw the circle $ (M),$ centered at point $ M,$ with radius the segment $ MK \equal{} R,$ which intersects the circle $ (K),$ at points so be they $ A,$ $ D.$ As well known, the points $ A,$ $ D,$ $ G,$ are the vertices of an equilateral triangle and also of a regular enneagon, inscribed to the given circle $ (K).$

[b][size=100]2)[/size][/b] – We denote the point $ L\equiv AD\cap MG$ and let $ N$ be, the point between $ L,$ $ K,$ such that $ LN / NK \equal{} 6 / 1.$ We draw the circle $ (A),$ centered at point $ A,$ with radius the segment $ AN,$ which intersects the extension of $ AK$ towards $ A,$ at one point so be it $ Q.$

[b][size=100]3)[/size][/b] – Through the point $ Q,$ we draw a line, tangents to the circle $ (M)$ at point $ T,$ which intersects the diameter $ MG$ of $ (K),$ at one point so be it $ S.$

[b][size=100]4)[/size][/b] – Let $ P$ be, the point outwardly to the circle $ (K),$ as the third vertex  of the equilateral triangle with base the segment $ LD.$ The line connecting the points $ P,$ $ S,$ intersects the circle $ (K)$ at one point so be it $ C,$ as another vertex of the regular enneagon inscribed to $ (K).$

[b][size=100]NOTE[/size][/b] – I have checked the approximation of the above construction, only by computer. That is I have not in mind any synthetic or analytical proof.

[b][size=100]REMARK[/size][/b] – The difference $ d$ may be better, if we consider the point $ N$ between $ L,$ $ K,$ such that $ LN / NK \equal{} 6,046 / 1$ $ ($ instead of $ LN / NK \equal{} 6 / 1$ $ ).$ Then, we will have $ d \equal{} 0,68404030 \minus{} 0,68404028 \equal{} 0,00000002.$

It is easy to construct the ratio $ \kappa \equal{} 6,046 / 1$ by computer, or when we draw not on a paper page (radius in [b][size=100]cm[/size][/b]), but on a bigger area (radius in [b][size=100]m[/size][/b]).

The division by computer, of a circumference with radius $ R \equal{} 1$ in to nine equal arcs, give us $ \lambda_{(9)} \equal{} 0,68404029$ where $ \lambda_{(9)}$ is the side length of the inscribed regular enneagon.


modular arithmetic
number theory unsolved
number theory

[quote="Jean-Louis"]How to find all the primes of the form n^n + 1, where n is a natural number, that have no more than 300000 digits ?[/quote]

First, $n^{n}+1<10^{300000}$ $\implies$ $n\log10(n)<300000$ $\implies$ $n<65000$

Second, if $n=k(2p+1)$ have an odd divisor $2p+1$ then $n^{n}+1=(n^{k})^{2p+1}+1=0\pmod{(n^{k}+1)}$ and $n^{n}+1$ is not prime.

So $n=2^{k}$ with $k\leq 15$ (since $n<65000$) and $n^{n}+1=2^{k2^{k}}+1$.

Again, if $k$ has an odd divisor $2^{k2^{k}}+1$ can't be prime (same thing as two lines above). So $k$ must be 0 or a power of two.

So the only values for k may be $0,1,2,4,8$ which gives five possibilities for $n$ : $1,2,4,16,256$

For $n=1$, $n^{n}+1=2$ is prime : OK
For $n=2$, $n^{n}+1=5$ is prime : OK
For $n=4$, $n^{n}+1=257$ is prime : OK
For $n=16$, $n^{n}+1=$ ... to be checked (I can't)
For $n=256$, $n^{n}+1=$ ... to be checked (I can't)

And these values are the only possibilities for primes of the structure $n^{n}+1$ with less than 300000 decimal digits.

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In fact, $k=2^{u}$ implies $n^{n}+1=2^{2^{u+2^{u}}}+1=F_{u+2^{u}}$ where $F_{k}$ is the $k^{th}$ Fermat number.
So, for $u=0,1,2,3$ The five numbers we have are : $2,F_{1},F_{3},F_{6}$ and $F_{11}$

And it is well known (?) that $F_{1}=5$ is prime, $F_{3}=257$ is prime, and $F_{6}$ and $F_{11}$ are not primes.