Fie tetraedrul $ABCD$ iar punctele $M,N,P,Q$ pe muchiile lui a.i. $\frac{BP}{PC}=\frac{AQ}{QD}=\frac{CN}{ND}=\frac{BM}{MA}=k$
Determinati punctele $T\in (AC)$ si $S\in (BD)$ a.i. $MN$,$PQ$ si $ST$ sa fie concurente.

Am ajuns la o [hide="relatie"]$M,N,P,Q$  coplanare si formeaza paralelogram [/hide]  dar nu reusesc sa finalizez

[hide="hint"]aplica menelaos in spatiu.[/hide]

Pai am aplicat ca sa dem ca MNPQ sunt coplanare ,iar apoi am aratat ca formeaza paralelogram (dar am mentionat mai sus) , si mai departe nu reusesc sa gasesc pozitia punctelor S,T


function
limit
number theory
number theory unsolved

In a number theory book, I find the following exercise:

prove $ lim sup\frac{pi(x)}{x/logx}>\equal{}1$, where pi(x) denote the number of prime not exceeding x and of course, the direction of x is towards positive infinite.

I don't quite understand what is the meaning of "sup".

$ \limsup$ means "limit supremum":  you can think of it as the [url=http://en.wikipedia.org/wiki/Supremum]supremum[/url] of the limits of every convergent subsequence (or $ \infty$ if a subsequence diverges to $ \infty$).

I still don't understand why "sup" is needed in this context. So is that if we take the "sup" away this problem will still mean the same?

Anybody can give a solution?

[quote="Saradush"]I still don't understand why "sup" is needed in this context. So is that if we take the "sup" away this problem will still mean the same?[/quote]
No; this is true only for monotonic functions.  A function such as $ d(n)$, for example, has a very erratic growth rate, and it's much easier to talk about its limit supremums than about its actual limits (in comparison to other functions).  As an illustration, the sequence $ a_n \equal{} (\minus{}1)^n$ has $ \limsup_{n \to \infty} a_n \equal{} 1$ but $ \lim_{n \to \infty} a_n$ doesn't exist.  Hence this problem is weaker than the full PNT.

If you need help understanding the limit supremum, I'll try to clear it up:

Let's say we have some sequence $ (x_{n})\equal{}\{x_{1},x_{2},\ldots\}$.
Now define
$ T_{n}\equal{}\sup\{x_{n},x_{n\plus{}1},\ldots\}$.
Then the [i]limit superior[/i] is the number $ \inf\{T_{n}\}$.  It's easy to see that the sequence $ T_{n}$ is non-increasing.

The limit superior can also be defined with $ \varepsilon$'s but I think this definition is easier to use the first time around.


function
trigonometry
calculus
derivative
calculus computations

find the primitive of this function:

$ f(x)\equal{}(\sin(x))^{4}(\cos(x))^{5}$

$ 1\minus{}\sin ^2 x \equal{} \cos^2 x$

so $ f(x)\equal{}\sin^4 x(1\minus{}\sin^2 x)^2 \cos x\equal{}[\sin^4 x \minus{}2\sin^6 x \plus{}\sin^8 x ]\cos x$

which is easily the derivative of $ F(x)\equal{}\frac{1}{5}\sin^5 x \minus{} \frac{2}{7} \sin^7 x \plus{} \frac{1}{9}\sin^9 x\plus{}C$


search
function
number theory
complex analysis
advanced fields
advanced fields unsolved

I noticed an interesting limit on site but could not find the proof of it.  Can someone give me the proof or point me in the right direction for:

lim((phi(1)+phi(2)+phi(3)+........+phi(n))/n^2)=3/pi^2, n=1 to infinity

This is an interesting limit. Someone told me to look in "Fundamentals of Number Theory" but I do not have that book handy.

Thank you

I think, we discussed this problem on forum. Try to search!

I can write a sketch.
We just need equality $\varphi(m)=m\cdot \sum_{d\mid m}\frac{\mu(d)}{d}$. Using that we can rewrite initial expression as
\[\sum_{d=1}^{n}\frac{\mu(d)}{d}\sum_{d\mid m,m\leq n}m
=\sum_{d=1}^{n}\frac{\mu(d)}{d}\frac{d\cdot [n/d]\cdot([n/d]+1)}{2}\approx\sum_{d=1}^{n}\frac{\mu(d)\cdot n^2}{2d^2}.\]
We used that $[n/d]+1\approx [n/d]\approx n/d$ (it is only thing, which we must justify in rigorous proof).
So we can write
\[\frac{\varphi(1)+\varphi(2)+...+\varphi(n)}{n^2}\approx \sum_{d=1}^{n}\frac{\mu(d)}{2d^2}\stackrel{n\to\infty}{\longrightarrow}\frac{1}{2\zeta(2)}=\frac{3}{\pi^2}.\]
Here we use another well known equality $\sum_{d} \frac{\mu(d)}{d^s}=\frac{1}{1^s+2^s+3^s+...}=\frac{1}{\zeta(s)}$.

Thank you very much. That is interesting. I am just getting started in complex analysis (i.e. Riemann zeta function) and thought this would be fun.

Thanks again,
Cody

Actually, mentioned equalities don't need complex analysis for their proof.

Hello, me again.  I am sorry to say, I can not follow the sketch you gave me. I thought perhaps this was a typical infinite series summation type proof. I appreciate your help but I do not know what d, mu, mu(d) or any of that is.

Thanks,
Cody

And what did you expected in such number theory problem.
$d\mid m$ means that $m$ is divisible by $d$
$\mu(d)$ is a $\mu$-function. You can find it in any Number Theory Book.
Definition: $\mu(1)=1$, $\mu(p_1^{a_1}...p_k^{a_k})=(-1)^k$ iff $a_1=...=a_k=1$ and $0$ otherwise (here $p_i$ are distinct prime numbers).

I reckon I didn't know what to expect.  Thanks for your last post. I can take it from there.

Cody

That mu function Myth mentions is known as the Mobius function.  It is sometimes used in conjunction with number theoretic functions.

Yes, thank you. I found out what a mobius function is. I believe I understand now.

Cody


number theory
relatively prime
number theory unsolved

For a given integer $n$ , let $S$ be a set of all odd natural numbers less than or equal to $n$ and relatively prime to $n$. For $x\in S$ define $f(x)$ to be the greatest odd divisor of $n-x$. Prove that :

a) for each $x\in S$ there exist $m\leq [\frac{n+1}{4}]$ such thst $f(f(...f(x)...))=x$ (where $[x]$ denote the greatest integer less that or equal to $x$).

b) if $n$ is prime and doesn't divide $2^k-1$ for every $k=1,2,...,n-2$, then the least $m$ from the part a) is equal to $[\frac {n+1}{4}]$.

haaayyy ;)  :lol: it's very beautiful and not hard .... :D

er. *confused* I'm probably making an arithmetic error here but I can't find it...

Take n = 13, x = 7. Then f(x) is the greatest odd divisor of 6, i.e. 3. But then f(f(7)) = f(3) = 5, and then f(f(f(7))) = f(f(3)) = f(5) = 1, but f(1) = 3, so it'll just go 3 - 5 - 1 - 3 - 5 - 1 and never come back to 7?


Prove the formula for the sum of the measures of the interior angles of an n-gon is (n-2)180º. Prove that this formula is correct. Have fun! :lol:

Maybe a little cheap, but...

The sum of the exterior angles of a polygon is 360.

The sum of the interior and exterior angles of an n-gon is 180n.

So the sum of the interior angles is 180n-360 = 180(n-2)

[quote="Carbon57"]The sum of the exterior angles of a polygon is 360.[/quote]

This is the main concept, and can be proven as follows:  as we travel along the sides of an $ n$-gon, we rotate by an exterior angle every time we switch to a new side.  Once we've traveled around the entire $ n$-gon, we've rotated by $ 360^{\circ}$, but we've also rotated by the sum of the exterior angles.  Hence the two are equal.

consider an arbitrary point M inside the n-gon . For M and every 2 consecutive vertexs form a triangle , consequently , we have n triangles . The sum of all the angles of all the triangle is 180 n . However , the 360-degree angle at M is also counted so the sum of all the angle of the n-gon is 180n-360=180(n-2)

u can also do this by proving that a n-gon has (n-2)triangles that has all it's angles as a part of some angles of the n-gon. this can be done by induction.it's true for a triangle.if true for a n-gon then turn it into a (n+1)gon by deleting a side and adding 2 sides originating from the vertices which will create  one more triangle. and n-2 triangles' angles sum to (n-2)180.


calculus
derivative
calculus computations

A higrade company sells two brands X and Y of a commercial soap, in thousand-pound units. If x units of brand X and y units of brand Y are sold, the unit price for brand X is

p(x) = 4000 - 500x

and that of brand Y is

q(y) = 3000 - 450y

Suppose brand X sells for $\$$500 per unit and brand Y sells for $\$$750 per unit. Estimate the change in  total revenue if the unit prices are increased by $\$$20 for brand X and $\$$18 for brand Y.


analytic geometry
function
conics
parabola

Find the product of the non-zero coordinates of all intercepts of the graph of $2|x|+|5y| = 20$

[hide]Let's start with the x-intercepts. For these, $y = 0$, so the equation becomes:

$2|x|+|5*0| = 20$

$|x| = 10$

$x = \pm 10$

For the y-intercepts, $x = 0$, so we get:

$2|0|+|5y| = 20$

$5y = \pm 20$

$y = \pm 4$

Multiplying, $(10)(-10)(4)(-4) = 1600$.[/hide]

Because of simmetry we may consider inf and sup of $xy$ when $2x+5y=20$ where $x\in [0,10]$. It reduces to finding the inf and sup of the function $x(4-\frac{2}{5}x$ on $[0,10]$. The inf is at ends and equals $0$ whereas the sup=$10$ at the vertex $x=5$ of parabola $y= x(4-\frac{2}{5}x)$.
Answer: $[-10,10]$
P.S. excluding $0$ because the author wishes (why?) to find "product of the [b]non-zero[/b] coordinates" ...  I haven't cought what does mean "intercepts" in this context thought.   :(


Find all prime numbers $p\leq q \leq \ r $ such that all the numbers:
$pq+r;pq+r^2;qr+p;qr+p^2,rp+q,rp+q^2$ are prime.

clearly p=2.
at least one of qr+4 and qr+2 is divisible by 3. hence unless q=3.
but the only way that none of the remaining numbers not be divisible by 5 is that r=5
hence p,q,r= 2,3,5.


Euler
algebra
polynomial
search
number theory
prime numbers

as far as I know you pretty much have to memorize them..

If a number is not a multiple of any number less than or equal to its square root, it is prime.

[quote="Spira Mirabilis"]It has been proven that there cannot be a polynomial with rational coefficients that only generates primes.[/quote]

Oh really? I didn't know that  :maybe:  I thought they were still searching for them. Do you have a link?

[quote="7h3.D3m0n.117"][quote="ajai"]Is there any particular formula that determines a prime number, or do the prime numbers just have to be memorized?

All advice is welcome.[/quote]

ahh well it worked until Euler disproved it

$2^{2^{n}}+1$ is a prime number...but he disproved it when he reached $2^{32}+1$

these are called the Mersenne numbers...(although it was interesting to see how Euler would remember up to $2^{32}$ without paper and pencil  :D )

if you can find another version of the Mersenne numbers you couuld get $\$$1 million from a certain company that wants to find out a general prime number formula[/quote]
Please search and note the difference between [b]Fermat[/b] and [b]Mersenne[/b] primes

[quote="Spira Mirabilis"]It has been proven that there cannot be a polynomial with rational coefficients that only generates primes.[/quote]
That is not entirely correct. There are no such polynomials if it is a single variable polynomial. Matiyasevich proved a rather subtle result and there are polynomials with multiple variables that can generate primes.

To miyomiyo: for practical purposes, that is not an efficient method. But, for competitions and the like, that is a very useful way to test primality for small numbers.

[quote="10000th User"]Please search and note the difference between [b]Fermat[/b] and [b]Mersenne[/b] primes [/quote]

 :maybe:  gasp! then AoPS vol 1 made a mistake!!!!


conics
ellipse
geometry
geometric transformation
reflection
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14550[/img]

 :(

You already have 3 ellipse points $ A, B, C.$ The ellipse center is the triangle centroid $ G$ and and reflections $ A', B', C'$ of $ A, B, C$ in $ G$ give 3 more points.

Sorry, I flipped the replies. Amswer to this problem is:

The ellipse center is the triangle centrid $ G$ and it is tangent to $ BC, CA, AB$ at their midpoints $ A', B', C'.$ Reflections $ A'', B'', C''$ of $ A', B', C'$ in $ G$ give 3 more points.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14564[/img]

 :thumbup:


In the equation $ 858 \equal{} a  \cdot b$, $ a$ and $ b$ are both two-digit numbers. What is the greatest possible value of $ a\plus{}b$?


algebra
polynomial
quadratics
complex numbers

An arbitrary fifth degree polynomial cannot be factored (the famous 'insolvability of the quintic' problem).  Fourth degree and lower polynomials can always be factored over C, and their roots are related to the third and fourth roots of unity.

[quote="ComplexZeta"]That doesn't mean you can't factor this one though.[/quote]

Yes, but he said 'something like'.

It does not look like you can factor it into polynomials w/real coefficents.

However, according to the Fundamental Theorem of Algebra, it can be factored into 5 linear factors.  These factors are the same as the solutions to [tex]z^5+z^2+1=0[/tex].

The fifth roots of unity, ie the solutions to [tex]x^5=1[/tex], are more related to factoring [tex]x^5 - 1[/tex].

I don't know how you can solve [tex]z^5+z^2+1=0[/tex] though.

Ok, well its easy to check that there are no rational roots, so we won't be able to factor out (x-a) for any integer a. So, if it factors over the integers, it must factor into a quadratic and a cubic. There are 2 possibilities:

[tex](x^2+ax+1)(x^3+bx^2+cx+1)[/tex]

or

[tex](x^2+ax-1)(x^3+bx^2+cx-1)[/tex].

In either case, comparing the coefficient of x and x^4 implies that b=c=-a. Then sub this back in to see that there is no integer a that will work. So this does not factor over the integers. (and consequently, over rationals either)

You can use this technique to get the roots, though, even if it isn't a factorization over the rationals.  If we use the case where we have -1's (you can show now factoring exists for the +1's), we can solve a quadratic equation for a, and then subsequently plug that in and solve a quadratic to find two of the roots for z.  They won't be pretty -- radicals inside radicals, I'm afraid, unless it simplifies and I didn't notice -- but it's very doable.


"What is the area of a semicircular region tangent to two sides of a unit square, with endpoints of its diameter on the other two sides?"
I need help with the problem with a [b][i][u]step by step[/u][/i][/b] solution, because I do not get where I should start. :huh:

Start where you always start in a geometry problem - with a diagram!  :D

Unfortunately, I do not have the tools to provide one for you.  Some tips:

- Draw the radii of the semicircle that end at the tangent points.  (They are parallel to the sides of the square!)
- Draw the semicircle symmetric about the diagonal of the square.
- Assign variables to lengths, then try to find equations for them.  (Try to express as many lengths as you can in terms of the radius.)

Diagram!

[img]http://img.photobucket.com/albums/v194/mukmasterxps/Copyofpic.gif[/img]

[hide="This diagram is a give-away"]
Might take a moment to load...
[img]http://img.photobucket.com/albums/v194/mukmasterxps/pic.gif[/img]
[/hide]

[img]http://i96.photobucket.com/albums/l167/adasarathy/picture_square.png[/img]

Start with a diagram. Then, what you do is that you can see that I have expanded some of the lines.

Then, you can actually do some stuff.  :)

What you now have is an inscribed square. Each side of the square is 1. To find the area of the circle, you need to just do $\pi r^{2}$. So, the area of the circle is $\frac{\pi}{4}$. However, you want the area of semicircle.

[img]http://i96.photobucket.com/albums/l167/adasarathy/picture_square-1.png[/img]

So, you divide by 2.

So, I think the final answer would be $\frac{\pi}{8}$.

It must not be this easy though.  :huh: I probably misunderstood the problem somewhere.  :(

[quote="anirudh"]It must not be this easy though.  :huh: I probably misunderstood the problem somewhere.  :([/quote]

The endpoints of your diameter are not on the square  :(

[quote="bigboy82892"]I still don't get it. How does that particular $45^{0}-45^{0}-90^{0}$ triangle help?[/quote]
You can determine the side length of the square (or vice versa).

How? I need a step-by-step solution. :?: I still don't get it.

Consider the diagram below:

[img]http://img.photobucket.com/albums/v194/mukmasterxps/pic.gif[/img]

We know the side length of the square to be 1.
We see from the diagram that this means $r+\frac{r}{\sqrt{2}}=1$

We can use this to solve for the radius, after which its just $\frac{\pi r^{2}}{2}$ and rationalizing.

OHHHHHHHHHHHH! So the answer should have been $\frac{\pi}{3+2\sqrt{2}}$! OH NO! THAT MEANS I GOT THE ANSWER WRONG!!!!! :wallbash_red: Oh well... at least I got $\frac{5}{6}$ right! :)  :D

Yep, or you could have $(3-2\sqrt{2})\pi$

[quote="bigboy82892"]how could that happen?[/quote]

rationalize the denominator.. multiply the denominator and numerator by the conjugate of the denominator..

this comes out nicely that the denominator is 1

oh snap.. this is the math league contest #2 question


AMC
AIME
USA(J)MO
USAMO
AMC 12
Putnam
inequalities

How well would __*________ do on the AIME, USAMO, IMO?

*
 A. college undergraduate majoring in math
 B. college graduate majoring in math
 C. mathematics professor

The reason that I am curious is that if I had a knowledge of high school mathematics I could probably do fairly well on most middle school competitions like MATHCOUNTS. Therefore, is it the case that, if I knew college-level mathematics or beyond, then would I be able to ace these exams?

If we were to consider middle school mathematics competitions, then most of the problem-solving is reduced to an application of a well-known technique in high school mathematics like basic algebra or trigonometry.

Is this the same with high school competitions?

If you answer "yes" to the question, then this should mean that professors should outperform graduate students on average, and graduate students should outperform undergrad students.

If you answer "no," then each group should perform, on average, about the same because these competitions then are more about actual problem solving and creativity that mere knowledge of more mathematics and experience.

Any comments?

My two cents: What one learns in college and grad school is not at all helpful for such exams except possibly with the exception of introductory courses that teach proof-writing.

taking college math will not directly affect your scores (because many of the things you learn are not applicable for contests), but I think that having more experience with math and spending more time thinking about problems can help.

I agree. Its more the experience you have of having had more time to do math(s) problems that would help you, rather than the fact that you have learnt higher level math(s).

I don't know about that. I don't think most of the graduate students here at UCSB did math competitions when they were in high school, but I have talked to some of them quite a bit, and I am impressed by the amount of ``contest math'' that some of them know, even though I doubt they have ever taken classes in it.

I have a friend who is a sophomore math/physics double-major who didn't do math contests seriously in high school (although he did get 132.5 on AMC12 his senior year, but only 4 on AIME), and I think he would be able to do very well on high school contests if he were to take them now. But he was also on our Putnam team, so training for the Putnam may have helped him. I don't think most math majors would do especially well on contests.

Professors should do pretty well I suppose, but I have found that some professors are only slightly better than I am at problem solving. But then there are some who are far better.

One thing that I have found is that in undergraduate classes, one learns a bit of stuff that is ``standard repertoire'' for the USAMO, and in graduate classes, one is expected to know it already.

The reason I ask is that I remember contest problems from early middle school that were essentially algebra problems. Had I been familiar with algebra, most of those contest problems would have been trivial.

However, were do you learn combinatorics, number theory, inequalities, geometry (I mean theorems like Menalaus's Theorem or Ceva's Theorem), and other topics like that? Are there courses on these in college? Or are these topics limited to math competitions? I have never seen a lesson-by-lesson textbook that covers such things as Ceva's Theorem. The only resources I can find are short books like Geometry Revisited.

Those things are not in the standard curriculm anywhere.  One might find a class on them, but it would probably just be something offered by a professor who was interested in it, not as a part of any kind of systematic learning.

We have undergraduate classes at UCSB in combinatorics and number theory. The combinatorics stuff is exactly the same as the USAMO ``curriculum,'' but the number theory goes far beyond what is required to do well on the USAMO. You'll probably pick up most of the inequalities in analysis classes. In the first quarter of my measure theory class, we covered Hlder's Inequality and Minkowski's Inequality, although they were in integral form (which is a generalization). Cauchy-Schwarz comes up in nearly ever class.

Hmm, what field of math do inequalities really fit into?


geometry
rectangle
calculus
calculus computations

A rectangle is to be inscribed in a semicircle of radius 8, with one side lying on the diamter of the circle. What is the maximum possible area of the rectangle?

4*sqrt(2)
8*sqrt(2)
32
32*sqrt(2)
64

Thanks

Semi-circle can be written as $ f(x) \equal{} \sqrt{64 \minus{} x^{2}}$. Thus, a rectangle inscribed within such a circle has an area of $ 2x\sqrt{64\minus{}x^{2}}$ The rest is just an easy excercise of derivatives.


[quote="taha.hojjati"]about positive numbers a,b,c,x,z,y we know that
1)a<b<c,a<x<y<z<c.
2)abc=xyz,a+b+c=x+y+z.
prove that a=x,b=y,c=z.
help me[/quote]

I think 1) is a<b<c,x<y<z. And it's not true: a=1, b=2, c=3, and x,y,z are the roots of s^2-6s^2+11.1s-6.

[quote="MJ GEO"]hi.could you please use latex.and dont post a problem two times?
i think this problem is in Leningrad olympiad.[/quote]

Hello MJGEO! could you please give us the answer you promised us some days ago ([url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1703347#1703347[/url]) ?
Thanks a lot.


geometry
circumcircle
trigonometry
perpendicular bisector
geometry proposed

In an acute-angled triangle $ABC,\, M$ and $N$ are points on the sides $AC$ and $BC$ respectively, and $K$ the midpoint of $MN.$ The circumcircles of triangles $ACN$ and $BCM$ meet again at a point $D.$ Prove that the line $CD$ contains the circumcenter $O$ of $\triangle ABC$ if and only if $K$ is on the perpendicular bisector of $AB.$

OK Here is a solution

Denote by O circumcenter of triangle $ \triangle ABC$ and let $ CO$ meet $ AB$ at $ C_{1}$. Then we have $ \frac{AC_{1}}{BC_{1}}=\frac{\sin2B}{\sin2A}$. Let $ P$ be a point of intersection of lines $ CD$ and $ AB$. We will prove that bisector of segment$ AB$ goes through $ K$ if and onlyif $ \frac{AP}{BP}=\frac{\sin 2B}{\sin 2A}$ so problem will be solved.

Let $ M_{1}$, $ K_{1}$ and $ N_{1}$ be orthogonal projections of points $ M$, $ K$ and $ N$ onto line $ AB$, respectively.

Then bisector of segment $ AB$ goes trough $ K$ if and only if $ AM_{1}$=$ BN_{1}$, which is equivalent to $ \frac{AM}{BN}=\frac{\cos B}{\cos A}$.

Notice that triangles $ \triangle AMD$ and $ \triangle BND$ are similar since $ \angle MAD=\angle BND$ and $ \angle AMD=\angle NBD$.

From this similarity we have $ {\frac{AM}{BN}=\frac{MD}{BD}}$. From sin theorem we have $ \frac{MD}{BD}=\frac{\sin\angle ACD}{\sin\angle BCD}$. On the other hand, applying sin theorem to triangle $ \triangle APC$ we get $ \frac{AP}{\sin\angle ACP}=\frac{CP}{\sin\angle A}$ and applying sin theorem to triangle $ \triangle BPC$ we get $ \frac{BP}{\sin\angle BCP}=\frac{CP}{\sin\angle B}$.

Finally we have:
$ \frac{AP}{BP}=\frac{\sin \angle ACP \sin\angle B}{\sin\angle BCP \sin \angle A}=\frac{MD\sin \angle B}{BD \sin\angle A}=\frac{AM \sin\angle B}{BN \sin\angle A}=\frac{\cos \angle B \sin \angle B}{\cos \angle A \sin \angle A}=\frac{\sin \angle 2B}{\sin \angle 2A}$

so we have finished our proof.


geometry
geometric transformation
reflection
incenter
circumcircle
geometry proposed

Point $ I_1$ is the reflection of incentre $ I$ of triangle $ ABC$ across the side $ BC$. The circumcircle of $ BCI_1$ intersects the line $ II_1$ again at point $ P$. It is known that $ P$ lies outside the incircle of the triangle $ ABC$. Two tangents drawn from $ P$ to the latter circle touch it at points $ X$ and $ Y$. Prove that the line $ XY$ contains a medial line of the triangle $ ABC$.

[i]Author: L. Emelyanov[/i]

Let $ D$, $ E$, $ F$ be the tangency points of the incircle $ \rho(I)$ with the sides $ BC$, $ CA$ and $ AB$, respectively and let $ M$ be the midpoint of small arc $ BC$, not containing the vertex $ A$. Since $ MB \equal{} MI \equal{} MC$, the reflection $ M_{1}$ of $ M$ into the sideline $ BC$ is the circumcenter of the $ BI_{1}CP$. See that $ PD$ is the $ P$-altitude of triangle $ BPC$ and the reflection $ I_{1}$ of $ I$ into $ D$ lies on the circumcircle of $ BPC$. Thus, $ I$ is the orthocenter of $ BPC$ and since $ XY$ is the polar of $ P$ wrt. $ \rho$ and $ DE$ is the polar of $ B$, it follows that the lines $ XY$, $ DF$ and $ CI$ are concurrent. On the other hand, since $ XY$ is the polar of $ P$, we have that $ PI \perp XY$ and so $ XY \| BC$. In this case, the conclusion follows from [url]http://www.mathlinks.ro/viewtopic.php?t=132338[/url].

It is pretty easy problem and in my opinion last year's third problem was much harder than this one.
[b]Here is a sketch of a solution[/b]:
Suppose that $ AB\equal{}c,AC\equal{}b,BC\equal{}a$,$ p\equal{}\frac{a\plus{}b\plus{}c}{2}$ and $ r$ is incircle's radius.
Since $ PBIC$ is cyclic and $ BD\equal{}p\minus{}b,DC\equal{}p\minus{}c$,it follows that
\[ \boxed{PD\equal{}\frac{(p\minus{}b)(p\minus{}c)}{r}}
\]
Suppose that $ PD$ intersects $ XY$ at $ Z$.
Then $ PZ\cdot PI\equal{}PX^2\equal{}PI^2\minus{}r^2$ and $ PI\equal{}PD\minus{}r\equal{}\frac{(p\minus{}b)(p\minus{}c)}{r}\minus{}r$.
\[ \boxed{PZ\equal{}\frac{\frac{(p\minus{}b)(p\minus{}c)}{r}\left(\frac{(p\minus{}b)(p\minus{}c)}{r}\minus{}2r\right)}{\frac{(p\minus{}b)(p\minus{}c)}{r}\minus{}r}}
\]
So finally $ ZD\equal{}PD\minus{}PZ$ and it is enough to show that $ ZD\equal{}\frac{h_a}{2}$,but this is not really hard...

[quote="discredit"][quote="pohoatza"]it follows that the lines $ XY$, $ DF$ and $ CI$ are concurrent. [/quote]

Why is this?[/quote]

By La Hire's theorem, the point $XY \cap DF = J$ is the pole of $PB$, so $IJ \perp PB$.

Let $r$ be the inradius and $r_a$ be the $A$-exradius and $J$ be the $A$-excenter. Obviously, $BPCJ$ is a parallelogram, so $IP=DP-DI=r_a-r$. Let $P'$ be the inverse of $P$ in the incircle. We have $$IP'=\frac{r^2}{IP}=\frac{r^2}{r_a-r} \Longrightarrow DP'=\frac{rr_a}{r_a-r}=\frac{\frac{\Delta^2}{s(s-a)}}{\frac{a\cdot \Delta}{s(s-a)}}=\frac{\Delta}{a}=\frac{1}{2}\cdot h_a.$$ Clearly, the polar of $P$ wrt the incircle is the line through $P'$ parallel to $BC$ which is the $A$-midline.


Find all real k satisfy that:
$\frac{a^k}{a+b}+\frac{b^k}{b+c}+\frac{c^k}{c+a} \ge \frac{a^{k-1}+b^{k-1}+c^{k-1}}{2}$
for all a,b,c>0

An other problem,maybe simpler:
find all k st this ineq satisfy
$\frac{a^k}{(b+c)^k}+\frac{b^k}{(a+c)^k}+\frac{c^k}{(a+b)^k} \ge \frac{3}{2^k} $

The latter has been discussed in detail at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=88797&highlight=#p88797

The first one is asymmetric. I don't like cyclic inequalities.  :(
Besides, if you make it symmetric, it follows from Muirhead's!  :D

I don't think any of these inequalities has been completely solved. And something tells me that long time will pass till someone will completely solve them.

For the first one:

By Cauchy-Bunhiacopsky inequality

     $ \sum {a^k/(a+b). \sum {a^{k-2}(a+b) \ge (\sum a^{k-1})^2 }}$

But

     $ \sum {a^{k-2}(a+b) \le 2\sum a^{k-1} }$

and the result follows.

Namdung proved first inequality for $k\geq2$.
But the inequality is also valid for some $k<2$.
For example, there is a very nice proof for $k\geq 3/2$. :)

The second inequality is true for $k\geq \frac{ln3}{ln2}-1\approx 0.585$. ;)

[quote="Vasc"]Namdung proved first inequality for $k\geq2$.
But the inequality is also valid for some $k<2$.
For example, there is a very nice proof for $k\geq 3/2$. :)
[/quote]

What is the complete solution for this one?

This inequality deserves a new topic. :)

Can you post your solution if $k \ge \frac{3}{2}$,Vasc?
Thank a lot! :P

Vasc,I have a solution when $k\ge \frac{3}{2}$
Use this ineq which is unexpected:
$\frac{a^{k-1}(a-b)}{a+b} \ge m(a^{k-1}-b^{k-1})$
Where $m=\frac{1}{2k-2}$
Because it $\leftrightarrow (2k-3)a^k+b^k+ab^{k-1} \ge (2k-1)a^{k-1}b$
It's true by Cauchy.
then $\sum{\frac{a^{k-1}(a-b)}{a+b} \ge 0}$
$\leftrightarrow \sum{a^k}{a+b}\ge \sum{a^{k-1}}$
What is your sol?

Dear hungkhtn,

I don't understand your last step.

I have a reversed sign as you can write it as

$\sum \frac {a^k}{a+b} \geq \sum \frac {a^{k-1}b}{a+b}$

and also as

$\sum \frac{a^{k-1}}{2} \geq \sum \frac{a^{k-1}b}{a+b}$

Am I wrong?

Can you explain better your last step, please?

Thank you very much.

Ohp sorry!!! :blush:  :blush:  :blush:  :blush:  :blush:

Fantastic solution :)

What about this ineq:

$\frac{a^k}{b+c}+\frac{b^k}{c+a}+\frac{c^k}{a+b}\ge \frac{a^{k-1}+b^{k-1}+c^{k-1}}{2}$
For all [size=134]Real $k\ge 1$[/size]
Maybe it's easier.

If I am not idiot, then this is not easier, but trivial.Chebyshev+Nesbitt=probable solution.

You are right,It's really trivial.
But I don't like using Trebysep... :(
Express:
$\sum{\frac{a^{k-1}(2a-b-c)}{b+c}}\ge 0$
$\leftrightarrow \sum\frac{(a^k-b^k)(a-b)}{(a+c)(b+c)}+\sum\frac{c(a^{k-1}-b^{k-1})(a-b)}{(c+a)(c+b)} \ge 0$
This is true.

[quote="hungkhtn"]Find all real $p$ satisfy that
$\frac{x^{1+p}}{x+y}+\frac{y^{1+p}}{y+z}+\frac{z^{1+p}}{z+x}\ge \frac{x^{p}+y^{p}+z^{p}}{2}$
for all $x,y,z>0$.[/quote]
$x=1,y=8,z=45\Longrightarrow p\geq 0.1584\cdots$;
[hide="Digits:=11:"]$x=1840,y=14310,z=80431\Longrightarrow p\geq 0.15844121320\cdots$.[/hide]


Let $p$ a prime and $ d_n$ a sequence of integers and $x_n=\sum_{i=0}^{n}{d_i {n\choose i}p^i}$ Prove that if an infinite number of $x_j$ are 0, then they are all 0. Deduce from here Skolem-Mahler theorem for integer sequences.

I see your TIPE is making progress... :D

Pierre.

Yes, it is making progress. But this one is not in the articles you sent me (thanks a lot again), it is a completely elementary lemma that can be even an olympiad problem. A hard one.  ;)


Let $ABC$ be a triangle, $AB=AB=BC$inscribed in circle $C(O,1)$. Prove that for any point M in the interior of the circle the inequality holds:
   $MA+MB+MC$<=$4$.

There is some mistake in the problem, this sum is definitely not less than $4$


This is a question from the AMATYC problem.

  In the expression (AM)(AT)(YC), each different letter is replaced by a different digit 0 to 9 to form three two digit-number. If the products is to be as large as possible, what are the last two digits of the product?
A. 20  B.40   C.50  D.60  E.90

My answer is 40. You? :lol:

[quote="knexpert"][hide]
I got 20. What I did was I let the first number be 97, the second be 96, and the third be 85.[/hide][/quote]

That's not quite right; $95 \cdot 96 \cdot 87$ is bigger. I get $40$ as well.

I got exactly the same answer as you, Paladin8. My solution is
$(AM)(AT)(YC) \leq \left (\frac{AM+AT+YC}{3}\right)^{3}$
so we have to maximize $F=AM+AT+YC$ but $F=11A+YC+M+T$
Hence chose $A=9,YC=87, M=6,T=5$. And got the same answer as you.

  But I have a feel that my answer didn't right. :D

Making the RHS larger doesn't guarantee that you will make the LHS larger.

Ok,ok

$(AM)(AT)(YC)= 100YCA^{2}+10YCA(M+T)+YCMT$
we should maximize $100YCA^{2}$ b/c it is a biggest term in the sum.
Since $98*7^{2}< 9^{2}*87$
So chose $A=9, YC=87$ and so on.
Is it correct?


linear algebra
matrix
linear algebra unsolved

Let $A,B \in {\mathbb R}^{3 \times 3}$ such that $det(AB-BA)=1$. Show that $det \left [ e^{(AB-BA)^3} \right ] = e^3$.

Since $dete^{(AB-BA)}^3=e^{tr(AB-BA)^3}$ we have to find $tr(AB-BA)^3$. If we denote by $\lambda_1,\lambda_2,\lambda_3$ the eigenvalues of the matrix $AB-BA$ we get $\lambda_1\lambda_2\lambda_3=1$, $\lambda_1+\lambda_2+\lambda_3=0$. From where it is not that hard to get $\lambda_1^3+\lambda_1^3+\lambda_1^3=3$, which meant that $tr(AB-BA)^3=3.$


linear algebra
matrix
geometry
3D geometry
pyramid

What is the missing number?
7
84
951
?(missing number)
3

Thanks in advance.

The 3 just throws me off.

i think you are right Frt , that was quite nice , but hey they arent in order like the diagonals , first is 7 , ok, next is  84 , 2nd diagonal , ok third is 951 ,   , then suddenly two jumps to the 5 th  , 3??, 62 came before 3 ,so shouldnt the series be better like this

7,84,951,62,?  where ?=3

That's pretty wise, frt, I guess the pyramid shape of the numbers sort of helped (one-digit, two-digit, three-digit, something, one-digit), but I would've never thought of telephone numbers.  :D

frt, your solution was very creative. :D

I think you are right, that is what most people in my class got. Thanks for the help. I posted this here though because this was a homework assignment.

actually it is not the numbers on a phone, but on the number pad on the keyboard.

[quote="frt"][hide="I guess"]
$62$

[hide="hint"]
Think about the keys of a telephone...
They're arranged as some kind of a 3 by 3 matrix:

1 2 3
4 5 6
7 8 9
(Ignore the last row with * 0 #)

Hope you get it...
[/hide]
[/hide]

By the way, this should be moved to "Other problem solving topics" forum or "Games and Fun Factory" forum.[/quote]
where did you get that from???? COOOL! :|

Ineedhelpbadly has requested that I close this topic.

Frt, that's a very creative solution.  :)


Hey, how come here @ AOPS we don't have sections dedicated to Info Olympiad.

Anyways, can you recommend some good books with challenges in C++/C. And logical problems...

Actually, there's a logic round that you need to clear, so AoPS helps a lot on that side  :D Besides, I guess you're pretty new here, so the next time you come to the AoPS forum, go to other topics. There's a section just for "Computer science and Informatics" with 2000 odd posts as of now  :P As far as books go, I guess the best thing for informatics is to do programs of the last years. It's much much better than referring to any particular book :)


inequalities
induction
inequalities unsolved

a, b   are complex numbers.
|a+b| ≥2   and |a+b|≥1+|ab|
Prove that:  |a^(n+1)  +b^(n+1) |≥|a^n  +b^n|.   for every n from N

[quote="tarta"]Could someone give me an idea?[/quote]
sorry its late here but u tried induction on n?
bye! :blush:


On the horizontal line $ y\equal{}5$, there are two points exactly 5 units from the point $ (7,1)$. What is the sum of the $ x$-coordinates of these two points?

The two points' x-coordinates average to 7, so their sum is 14.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="azotlichid"]1. Prove that if $ H < G$ and $ P$ is a sylow subgroup of $ G$, then $ \exists P'$ - a conjugate of  $ P$ such that $ P' \cap H$ is a Sylow subgroup of $ H$.

2. If $ H$ is normal and $ ([G: H], p) \equal{} 1$ then $ H$ contains every p-Sylow subgroup of $ G$.[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=191327 - same homework sheet?

  darij


calculus
integration
function
derivative
real analysis
geometry
rectangle

When you are evaluating an improper integral like

integral from - :inf: to 2 of (1/(x-1) - 1/(x+1))dx

which has both an infinite lower limit AND several infinite discontinuities, are you supposed to break it up into four or five integrals. I started to do this exercise and it seemed ridiculously complicated because of this.

Is there any easier way to do this?

------------------------------------------------

My other question is about proper notation when evaluating an integral: when you are evaluating an definite improper integral where you use a u-substitution like

integral from 0 to 1 of (x + 1)/( :sqrt: x :^2: +2x) =

limit c --> 0+ of integral from c to 1 of (x + 1)/( :sqrt: x :^2: +2x) =

u = x :^2: +2x
du = 2(x + 1)dx
(1/2)du = (x + 1)dx

u(c) = c :^2: +2c
u(1) = 3

thus you have,

limit c --> 0+ of integral from c :^2: +2c to 3 of (x + 1)/( :sqrt: x :^2: +2x)

is this proper notation, i.e. can I have c :^2: +2c as a lower limit of integration?

---------------------------------

I can't find a way to show that

integral from 0 to ln 2 of x^(-2)e^(1/x)         diverges

I can't find a function that would suit either the Direct Comparison Test nor the Limit Comparison Test. Can someone help me?

For your second question, that's perfectly all right, but instead of
limit c --> 0+ of integral from c  +2c to 3 of (x + 1)/ :sqrt:(x2 + 2x),
you should have the integrand in terms of u.  That is, you didn't make the u-substitution when you re-wrote it, so what you wrote is wrong, but the answer to your question (can you have something complicated-like as a lower limit of the integral) is yes, had you made the substitution correctly.  (I'm sure you knew what you were doing, I'm just being careful to say things as clearly as possible.)

For your third question, why not just compare it to the integral of e^(1/x)?

For 2, yeah, I forgot to rewrite the integrand in terms of u.

For 3, e^(1/x) is just as difficult to integrate as the original function. You can only integrate-- or I only know how to integrate things of the form e^ax. Or, do I consult a table of integrals?

Actually, for number 3, you can ignore my previous advice.  Instead, just try to find the indefinite integral of e^(1/x)/x^2.  The [hide]chain rule[/hide] is a beautiful thing ...

The chain rule applies to taking derivatives. Do you mean integration by parts?

chain rule applies to taking derivatives, and its integral counterpart is u-sub... product rule corresponds to integration by parts...

Anyone have any idea how to do #1?

I'm stumped.

I didn't think you [i]could[/i] Riemann-Integrate things that behaved that badly, even if the "bad parts" should cancel out ...

You'll get a meaningless statement like [tex]\infty-\infty[/tex] if you try to (Riemann)-integrate it.

What is riemann integration? Is that the normal integrating that I'm used to? Are there other forms of integration?

Riemann integration is the normal one. Lebesgue integration is much better and conceptually not really more complicated. Instead of starting from the x-axis and making rectangles going up, you start from the y-axis and make rectangles going across. You can integrate way more functions with Lebesgue integration than with Riemann integration.

That sounds pretty neat... I'll have to read up on it. Thanks.

[quote="rlee"]Uh, for #1, I think you're pretty much stuck. Break it in two first, then break each of those into two more.

I hate improper integrals =/.[/quote]

Ugh-- that's so messy and tedious. Improper integrals with an infinite limit is ok, improper integrals with an infinite discontinuity in the integrand is ok, but having both in one means that I'll have to plow through a ton of algera and stuff. Ick.


T-shirt design team lacking the most needed characteristic: imagination.
   1,2,3 ...  What the hell is it?

   How about a set of seven sleeping robes  with a complete
   geo problem solution  by Yetti (it should fit on 7 robes with long sleeves).

   As an extra bonus  it will make no difference where you start to read it, one entry
   point is as good as another,  you will be sleeping in no time.

  Maj.  Pestich

[quote="pestich"]T-shirt design team lacking the most needed characteristic: imagination.
   1,2,3 ...  What the - is it?

   How about a set of seven sleeping robes  with a complete
   geo problem solution  by Yetti (it should fit on 7 robes with long sleeves).

   As an extra bonus  it will make no difference where you start to read it, one entry
   point is as good as another,  you will be sleeping in no time.

  Maj.  Pestich[/quote]How about you stop bothering us? ;)

PS Those are team numbers, 7 lucky points and 42 perfect scores. In plus this way they can be used in soccer :P


What is the radius of a circle inscribed in a rhombus with diagonals of length $ 10$ and $ 24$?

$ \textbf{(A)}\ 4 \qquad
\textbf{(B)}\ 58/13 \qquad
\textbf{(C)}\ 60/13 \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ 6$

[hide="Click for solution"]
Let $ E$ be the intersection of the diagonals of rhombus $ ABCD$ satisfying the conditions given.  Because the diagonals are perpendicular bisectors of each other, $ \triangle ABE$ is right with side $ 5$, $ 12$, and $ 13$ with area $ 30$.  Therefore, the altitude drawn to side $ AB$ is $ \frac{60}{13}$, which is the radius of the inscribed circle centered at $ E$, and this is answer choice $ \boxed{\textbf{(C)}}$.
[/hide]

Side is 13, as side and half-diagonals form a right angled triangle.
Radius of the circle is height in that right-angled triangle, s it's $ 60/13$


a) Prove that for all positive reals $u,v,x,y$ the following inequality takes place:
\[ \frac ux + \frac vy \geq \frac {4(uy+vx)}{(x+y)^2} . \]
b) Let $a,b,c,d>0$. Prove that
\[ \frac a{b+2c+d} + \frac b{c+2d+a} + \frac c{d+2a+b} + \frac d{a+2b+c} \geq 1.\]

[i]Traian Tămâian[/i]

(a) The inequality is equivalent to $(x + y)^2 \ge 4xy$, which follows from AM-GM

(b) Take the inequality in (a).  Substitute $u = a$, $v = c$, $x = b + 2c + d$, $y = d + 2a + b$, and $u = b$, $v = d$, $x = c + 2d + a$, $y = a + 2b + c$, and adding, we get
\[LHS \ge \frac{2a^2 + 2b^2 + 2c^2 + 2d^2 + 2ab + 2ad + 2bc + 2cd}{(a + b + c + d)^2}.\]
But $2a^2 + 2b^2 + 2c^2 + 2d^2 + 2ab + 2ad + 2bc + 2cd \ge (a + b + c + d)^2 \Leftrightarrow (a - c)^2 + (b - d)^2 \ge 0$, so
\[\frac{2a^2 + 2b^2 + 2c^2 + 2d^2 + 2ab + 2ad + 2bc + 2cd}{(a + b + c + d)^2} \ge 1.\]

[hide="1"]
$\displaystyle \frac{uy+vx}{xy} \ge \frac{4(uy+vx)}{(x+y)^2}$
$\displaystyle \frac{1}{xy} \ge \frac{4}{(x+y)^2}$
$\displaystyle (x+y)^2 \ge 4xy$
$\displaystyle x^2+2xy+y^2 \ge 4xy$
$\displaystyle x^2-2xy+y^2 \ge 0$

the last part of trivial inequality which does make sense. QED.
[/hide]

Solution for $b)$

Beautiful problem!

Using the result of $a)$,

$\frac{a}{b+2c+d}+\frac{c}{d+2a+b}\geqq \frac{4\{a(d+2a+b)+c(b+2c+d)\}}{\{(b+2c+d)+(d+2a+b)\}^2}$

$=\frac{(a+c)(b+d)+2(a^2+c^2)}{(a+b+c+d)^2}$

$\geqq \frac{(a+c)(b+d)+(a+c)^2}{(a+b+c+d)^2}$

$=\frac{(a+c)(a+b+c+d)}{(a+b+c+d)^2}$,yielding

$\frac{a}{b+2c+d}+\frac{c}{d+2a+b}\geqq \frac{a+c}{a+b+c+d}$.

Similraly, we obtain $\frac{b}{c+2d+a}+\frac{d}{a+2b+c}\geqq \frac{b+d}{a+b+c+d}$

Summing up these inequality, we are done.

[hide="Another solution for b."]
Let $a=x^2,b=y^2,c=z^2,d=w^2$. Using the lemma $\frac{m^2}{p}+\frac{n^2}{q} \ge \frac{(m+n)^2}{p+q}$, which can be reduced to the trivial inequality, and extending it to four fractions, we get:

$\frac{a}{b+2c+d} + \frac{b}{c+2d+a} + \frac {c}{d+2a+b} + \frac{d}{a+2b+c} \ge \frac{(x+y+z+w)^2}{4(x^2+y^2+z^2+w^2)}$

which then follows easily by RMS-HM (or AM-GM).[/hide]

[quote="jhaussmann5"]paladin8,
there's a problem with your solution.  In general,
\[\frac{(x + y + z + w)^2}{4(x^2 + y^2 + z^2 + w^2)} \le 1,\]
so you don't get the bound you need.  I don't think this can be patched.[/quote]

Dang. In my world, the inequality signs are reversed :P

[hide="Number 1 was really easy..."]
Rewrite as
\[
\frac{uy + vx}{xy} \geq \frac{4(uy + vx)}{(x+y)^2}
\]
\[
\frac{1}{xy} \geq \frac{4}{(x+y)^2}
\]
\[
(x+y)^2 \geq 4xy
\]
\[
(x-y)^2 \geq 0
\]
which is just the trivial inequality.[/hide]

Use $\sum a_i/b_i \times \sum a_i*b_i \geq (\sum a_i)^2$

[quote="paladin8"][hide="Another solution for b."]
Let $a=x^2,b=y^2,c=z^2,d=w^2$. Using the lemma $\frac{m^2}{p}+\frac{n^2}{q} \ge \frac{(m+n)^2}{p+q}$, which can be reduced to the trivial inequality, and extending it to four fractions, we get:

$\frac{a}{b+2c+d} + \frac{b}{c+2d+a} + \frac{c}{d+2a+b} + \frac{d}{a+2b+c} \ge \frac{(x+y+z+w)^2}{4(x^2+y^2+z^2+w^2)}$

which then follows easily by RMS-HM (or AM-GM).[/hide][/quote]
    Only on a generalisation. :)  ;)

for b: substituting by a=x^2 (i had doubt between x^3, x^2 V x^2+1) and similar, we get, using the identity \frac{a}{b}+\frac{c}{d}\geq 2\frac{a+c}{b+d} we can sum the denominators, multiply by four and get 2((a+b)^2-(a-b)^2) for each numerator; after normalizing, symply and it's done


Given that $ \frac23$ of the members of a committee use $ \frac34$ of
the chairs in a room, what is the least possible number of members on the committee?

[b]6.[/b]

2x/3 must be divisible by 4 -> 2x must be divisible by 12 -> x must be divisible by 6.

Try again. If you plug in 6, then 4 people take up 3/4 chairs, and you can't have fractional people. 2/3 the people must div 3, and also must be divisible by 2, so the smallest number for 2/3 of the people is 6. Thus, the least number of people is 9. Was that too confusing?

Use a ratio:
2/3 people=3/4 chairs
All people=3/4+3/8=9/8 chairs
9/8 *x =integer
Least possible value is 8.

there are nine pplz in the commitee
2/3 of the pplz (6 pplz) sit in 3/4 of the chairs (6 chairs)


Why didn't you just put this in one of the other .99999...=1 discussions?  I mean do we really need more than 1 discussion on questions that are almost the same?  I m just asking if further discussion of this can be confined to one thread in one forum.

Thanks

[quote="JP"]Here are two quesions I am having,

If x can be any number less than 1, can x be 0.9999999999999999999(to the infinity)?
How about if x can be any real number less than 1, is there a difinite answer for x?[/quote]To put it simple, since 0,99.. to the infinity as you say is actually 1, the answer to the first question is no.

I cannot understand the second question. You should also probably read & post in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19502]this thread[/url] from now on about this.


Hi,
Method of exhaustion from Archimedes ! the same like $\pi r^2$ for the area of the circle ...

[quote="RohanSingh"]How to convince a middle school student that the volume of a sphere is 4/3*pi*r^3?[/quote]
Stick an AoPS book open to page 212 in their face. :)

[quote="mr. math"]i don't really know any actual way. You just have to make sure that they know the formula for tests and competitions

 :starwars:  :weightlift:[/quote]

I don't think so... I think that children have to know why is that formula...And anyone have the proof?????

[url=http://mathforum.org/library/drmath/view/55135.html]mathforum.org/library/drmath/view/55135.html[/url]

It might be too complicated to teach a middle school student, you basically break it up into 4 cones...

Tip[hide]You could prove it yourself[/hide] or [hide]make up a way for the student to memorize it.[/hide]


Probably the freakiest optical illusion I've ever seen:
[url=http://www.ianrowland.com/MiscPages/Mrangryandmrscalm.html]www.ianrowland.com/MiscPages/Mrangryandmrscalm.html[/url]

Makes you wonder if what we see is really what is there.

Very cool. I believe that since the image is not very focused, our eyes play tricks on us and we see a prediction of what we want to see. The effect wouldn't work if it was adjusted to a sharp focus.

This has nothing to do with psychology.  It's all physical.  Essentially, they both have 2 images with varying opacities, but one has a low-pass filter on one level and a high pass filter on the other, and the 2nd is reversed.  So when you look up close, one part of the first image comes out more than the other, and vice versa when you're far away... I wonder how hard it would be to do that in photoshop... maybe I'll try.


Here's a llama there's a llama
and another little llama
fuzzy llama
funny llama
llama llama duck
...

It's probably just a South Afircan-Maths thing... but was just wondering
I know its on albinoblacksheep.com.... but does anyone actually know it off by heart?

I know it... obviously  :)

Llama llama cheesecake llama
tablet brick potato llama
llama llama mushroom llama
llama llama duck!

I was once a treehouse
i lived in a cake
but i never saw he way
the orange slayed the rake

I was only three years dead
but i told a tale
and now listen little child
to the safety rail

 :D

The completeness:

here's a llama
there's a llama
and another little llama
fuzzy llama
funny llama
llama llama
duck

llama llama
cheesecake
llama
tablet
brick
potato
llama
llama llama
mushroom
llama
llama llama
duck

i was once a treehouse
i lived in a cake
but i never saw the way
the orange slayed the rake
i was only three years dead
but it told a tale
and now listen, little child
to the safety rail

did you ever see a llama
kiss a llama
on the llama
llama's llama
tastes of llama
llama llama
duck

half a llama
twice the llama
not a llama
farmer
llama
llama in a car
alarm a llama
llama
duck

is THIS how it's told now?
is it all so old?
is it made of lemon juice?
doorknob
ankle
cold
now my song is getting thin
i've run out of luck
time for me to retire now
and become a duck


Find out all of $ k \in \Re ^ \plus{}$,

$ \forall x,y,z \in \Re ^ \plus{}$,

$ xyz \equal{} k$,

$ \frac {1}{{\sqrt {1 \plus{} x} }} \plus{} \frac {1}{{\sqrt {1 \plus{} y} }} \plus{} \frac {1}{{\sqrt {1 \plus{} z} }} \le \frac {3}{{\sqrt {1 \plus{} k} }}$.

Take $ x\equal{}y\equal{}z\equal{}\sqrt[3]{k}$ we get

$ \frac{3}{\sqrt{1\plus{}\sqrt[3]{k}}}\le \frac{3}{\sqrt{1\plus{}k}}$

thus $ k\le 1$.

Now we prove that for $ 0<k\le 1$ we obtain

$ \frac{1}{\sqrt{1\plus{}x}}\plus{} \frac{1}{\sqrt{1\plus{}y}}\plus{} \frac{1}{\sqrt{1\plus{}z}}\le \frac{3}{\sqrt{1\plus{}k}}$

In fact, assume that $ x\le y\le z$, we have

$ t\equal{}\sqrt{xy}\le \sqrt[3]{k}$.

Notice that

$ \quad \frac{1}{\sqrt{1\plus{}x}}\plus{} \frac{1}{\sqrt{1\plus{}y}}\\
\le \sqrt{2}\cdot \sqrt{ \frac{1}{1\plus{}x}\plus{} \frac{1}{1\plus{}y} }\\
\equal{} \sqrt{2}\cdot \sqrt{\frac{1\minus{}xy}{1\plus{}x\plus{}y\plus{}xy}\plus{}1}\ (1\minus{}xy\ge 0)\\
\le \sqrt{2}\cdot \sqrt{\frac{1\minus{}xy}{\left( 1\plus{}\sqrt{xy} \right)^2}\plus{}1}\\
\equal{}\sqrt{2}\cdot \sqrt{\frac{1\minus{}\sqrt{xy}}{1\plus{}\sqrt{xy}}\plus{}1}\\
\equal{}\frac{2}{\sqrt{1\plus{}t}}$

It suffices to show

$ f(t)\equal{}\frac{2}{\sqrt{1\plus{}t}}\plus{} \frac{1}{\sqrt{1\plus{}\frac{k}{t^2}}}\le \frac{3}{\sqrt{1\plus{}k}}$

Now we prove that for $ 0\le t\le \sqrt[3]{k}$ we have $ f'(t)\ge 0$, which is equivalent to

$ \frac{k}{\left(1\plus{}\frac{k}{t^2}\right)^{\frac{3}{2}}t^3}\minus{}\frac{1}{(1\plus{}t)^{\frac{3}{2}}}\ge 0$

i.e.

$ k^2(1\plus{}t)^3\minus{}\left(k\plus{}t^2\right)^3\ge 0$

Notice that $ k\ge t^3$ and $ k\ge k^2$ we get

$ k^2(1\plus{}t)^3\minus{}\left(k\plus{}t^2\right)^3\equal{}\left(k\minus{}t^3\right)\left(k\minus{}k^2\plus{}3kt\plus{}t^3\right)\ge 0$

So we obtain $ f'(t)\ge 0$, and it yields

$ f(t)\le f\left(\sqrt[3]{k}\right)\equal{}\frac{3}{1\plus{}\sqrt[3]{k}}\le \frac{3}{\sqrt{1\plus{}k}}$

According to all the above, the answer is $ 0<k\le 1$.

Sorry,there's a mistake on my handwriting.

Edit $ xyz \equal{} k$ into $ xyz \equal{} k^3$.

I think it can be solved in a similar way, but I've forgotten the concrete.


AMC AIME algebra polynomial probability logarithms geometry

[size=167] Mock AIME 3 for 2007[/size]

Test written by: Alex Anderson (Altheman)

[b]Answers should be sent from 10/15/06 to 10/22/06[/b] (11:59 CST at latest for you procrastinators  :P )

[b]Answers along with a short description of what you did should be sent to me, Altheman, through a private message.[/b] I want to be able to give some feedback about your results beyond just you are wrong. Please do not discuss the problems before 10/22/06. After the test is over (pending any extensions), make a single topic for each problem you want to discuss.

The rules are standard:

1. Only the following items may be used: pens, pencils, paper, protractors, compasses, and graph paper. Absolutely NO calculators.
2. 3 hour time limit.
3. All answers are integers between $000$ and $999$ inclusive.
4. Answers such as $54$ or $3$ should be written as $054$ and $003$ respectively.

A few notation notes:

1. $\cap$ means intersection.
2. $n\in\mathbb{N}$ means $n$ is a natural number (1,2,3,...)

Here is the .pdf version if it is not at the bottom:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=5426[/url]

[hide="Also the test is here"]1. $x$ and $y$ are real numbers such that $x^{2}+3y=9$ and $2y^{2}-6x=18$. If the $4$ solutions to the system are $(x_{1},y_{1})$, $(x_{2},y_{2})$, $(x_{3},y_{3})$, $(x_{4},y_{4})$, then find $(y_{1}-x_{1})+(y_{2}-x_{2})+(y_{3}-x_{3})+(y_{4}-x_{4})$.

2. If $abc-ab-bc-ca+a+b+c=2008$ for positive integers $a$, $b$, and $c$, then determine the minimum value of $a+b+c$.

3. Let $P(x)=2x^{59}-x^{2}-x-6$. If $Q(x)$ is a polynomial whose roots are the $59$th powers of the roots of $P(x)$, then find the sum of the roots of $Q(x)$.

4. Let $x=(i-11)(i-10)....(i+8)(i+9)$ (where $i=\sqrt{-1}$). If $a$ and $b$ are positive integers such that $x\cdot (a+bi)$ is an integer, then find the minimum value of $a-b$.

5. You have a fair 12 sided die with sides 0, 1, 2,..., 10, and 11. After rolling that die $4$ times, you construct a base $12$ number where the $n$th roll corresponds to the $n$th place value starting at the units digit, then going left. So if you roll $1$, $10$, $0$, and $11$ (in that order), the resulting number is $B0A1$ (where A is the base $12$ representation of $10$, and B is $11$). If $\frac{p}{q}$ is the probability that $11\cdot 9\cdot 8$ (base 10) divides the resulting number, then find $p+q$ (where $p$ and $q$ are relatively prime positive integers).

6. Denote $S(n,k)=\log_{k}\frac{2}{1}+\log_{k}\frac{3}{2}+...+\log_{k}\frac{n-1}{n-2}+\log_{k}\frac{n}{n-1}$. If $n$ and $k$ are positive integers greater than $1$ and less than $500$, find the number of ordered pairs $(n,k)$ such that $S(n,k)$ is an integer.

7. Consider a regular hexagon $ABCDEF$. A point, $P$, is randomly chosen in $\triangle AEF$. If $\frac{p}{q}$ is the probability that $BP^{2}+EP^{2}>2\cdot AP^{2}$, then find $p+q$ (where $p$ and $q$ are relatively prime positive integers).

8. A cube has vertices $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, $(0,1,1)$, $(1,0,0)$, $(1,0,1)$, $(1,1,0)$, and $(1,1,1)$. At $t=0$, a particle is at $(\frac{1}{2},\frac{1}{2},0)$. After one second, the particle travels to $(\frac{9}{16},\frac{5}{8},\frac{1}{13})$. The particle bounces until it hits an edge or a corner at which point it stops. If the particle travels in straight lines with uniform speed, find the time (in seconds) when the particle stops motion.

9. $a_{n}$ is a sequence such that $4\cdot a_{n}=a_{2n}$ and $a_{2n}=2\cdot a_{2n-1}+\frac{1}{4}$ for all $n\in\mathbb{N}$. Find the sum $S=a_{1}+a_{2}+...+a_{31}$.

10. Let $S_{n}$ be the set of integers between $10^{n-1}$ and $10^{n}-1$ lacking zero digits (so $111$ counts, but $101$ does not). Let $f(S_{n})$ denote the sum of the elements in $S_{n}$. For all $n\in \mathbb{N}$, $\frac{f(S_{2n})}{f(S_{n})}$ can be written explicitly as $\alpha\cdot(a^{n})+\beta\cdot(b^{n})$ for real constants $\alpha$, $\beta$, $a$ and $b$. Find $a\cdot b$.

11. If $x$ and $y$ are real numbers such that $2xy+2x^{2}=6+x^{2}+y^{2}$ find the minimum value of $(x^{2}+y^{2})^{2}$.

12. $\triangle ABC$ has $F$ on $AB$ and $D$ as the midpoint of $BC$. Suppose that $CF \cap AD=X$, $BX\cap AC=E$ and $AFXE$ is a cyclic quadrilateral. If $AE=3$, $CD=4$, $CE=5$ and $DX^{2}=\frac{p}{q}$, then find $p+q$ (where $p$ and $q$ are relatively prime positive integers).

13. Let $a_{n}$ be a sequence such that:

i)  $a_{k}\in{1,2,3,4}$ for all $k\in\mathbb{N}$
ii) $a_{1}>a_{2}$, $a_{2}<a_{3}$, $a_{3}>a_{4}$, $a_{4}<a_{5}$,....

Find the number of different septuples: $(a_{1}, a_{2}, a_{3},...,a_{7})$.

14. A piece of paper is black on one side and white on the other side. Cut out a regular dodecagon $ABCDEFGHIJKL$ with $AG=8$. Look at the white side. Color $AE$ red, $CF$ green, and $DH$ blue. Look at the white side through the entire folding process. Fold along the red line (fold in the direction so you do not see any black after the fold). Then fold along what you can see of the green line, after that, fold along what you can see of the blue line (folding in the direction so you do not see any of the back side). Now flip the paper around. If the area of the visible black region is $a+\sqrt{b}$ where $a$ and $b$ are positive integers, then find $a+b$.

15. If $a_{n}+i\cdot b_{n}=(1+17i)^{n}$ where $a_{n}$ and $b_{n}$ are real sequences and $c_{n}=\gcd (a_{n}, b_{n})$, determine the last 3 digits of $c_{2007}$.[/hide]

A few words for the wise: even if you cannot get some of the problems, you should work at them later and check hints (that I will provide later). All problems have reasonable solutions; the questions require minimal computation if you find the most elegant solution.

Also if there are any mistakes (which I highly doubt since I checked over so many times), send me your problem through a PM with your answers. If something like an answer does not fit the required model, just send me an unsimplified solution.

I hope you think some of the problems are interesting! And finally, good luck  :) .

Thanks...this week has been pretty hectic, and I've yet to do the WOOT AIME and Problem Set  :(

[quote="Altheman"]i will be accepting answers for another week...(due to low participation )[/quote]

Thanks, I didn't have any time to work on it yet.

How long do you think it will be before we get the results back?

[quote="Altheman"]1. i sent out grades (with comments, sparce if you just gave me answers), sorry it took so long, the quarter is ending for me in school, so i am really busy

2. [b]date is extended to 10/30/06[/b] (by 11:59 PM CST)

(it won't let me edit my intial post  :mad: )[/quote]
We are now only allowed to edit posts within 24 hours...

[quote="13375P34K43V312"]Then what if you have something to say, then have something else to say in the same thread 30 minutes later, before someone else posts? You can't double post within 1 hour...[/quote]
You can edit your post. You just can't edit it after 24 hours have passed.

Well I think the first appearance was #6 on the 2004 AIME1.

[quote="bpms"][quote="Altheman"][quote="darktreb"]Hmm, I proved a generalized version of #13 and wrote it as a problem be used in some official contests. Quite a coincidence to see it here![/quote]

this is really just a rehashing of an old problem, i messed it up because it was too easy to brute force, but my intended solution is basically considering a one to one correspondence with a light bouncing between 4 parallel panels of glass, (where the number of the sequence is the panel you bounce off of), the problem i saw before was 3 panes of glass, so i thought to add a panel, then to solve, you can use two recursive statements and find an explicit fomula from there[/quote]
It looks very similar to USAMTS problem 2 Round 1.[/quote]

Maybe at first, but the solution is considerably more complicated, I think, because values can be repeated.

Did anyone get the answer for 15? Can someone post the answer (not necessarily the solution) for that? Thank you.

Here are the answers, I will put together a more complete summary of the test when I have more time, school is killing me now.

[hide]1. $\boxed{000}$
2. $\boxed{232}$
3. $\boxed{177}$
4. $\boxed{130}$
5. $\boxed{769}$
6. $\boxed{533}$
7. $\boxed{011}$
8. $\boxed{052}$
9. $\boxed{602}$
10.$\boxed{810}$
11.$\boxed{018}$
12.$\boxed{091}$
13.$\boxed{353}$
14.$\boxed{196}$
15.$\boxed{008}$[/hide]

a few comments:

i particularly liked problems 4,5,7,8,9,12,14,15, because i felt they were completely original (as far as my inspiration goes), yet they were within problem solving ideas

i thought some problems like 1,2,3,6 were just rehashing common problem solving ideas

too bad 13 was so easily solved, i was expecting people to find a recursion to solve it, i should have reframed the question something to the effect of the number of 20, 21,22,23,24 number tuples would be given, and you would have to fidn the number of 25-tuples, but the question was very wordy as it was, so i ono

unsure about 11, i thought that it was pretty conspicuous that i did not collect terms, so people would look at the x^2+y^2 in a more geometric sense,

i am considering making another mock aime, maybe around winter break, i need some time to come up with problems, i am pretty busy now

for #4, i got a=109 and b=21. 109-21=88?
Maybe on the PDF, it said a+b instead of a-b?
Because the one on top said a-b.

Maybe someone could update the wiki with solutions?

Could someone post a complete set of solutions? I'm still confused by the wordings of some of the problems.

I will start with a few ideas, I will see if I have time to write up full solutions right now. Maybe on the weekend, if there are specific questions that you want a solution for, then ask. If it is a geometry problem, you may have to wait for a diagram, those take a while.

[hide="1"]eliminate one of the variables, then by vieta's the sum of the solutions is 0, likewise for the other variable, for a total sum of 0[/hide]

[hide="2"]factorize into $(a-1)(b-1)(c-1)=2007=3*3*223$[/hide]

[hide="3"]for each root r, we can make that statement, so if $r_{i}$ are the roots, then we add them up to get: $\sum (r_{i})^{59}-\sum (r_{i})^{2}-\sum r_{i}=59*6$
by vieta's sums, $\sum (r_{i})^{2}=0=\sum r_{i}$[/hide]

[hide="4"]the conjugates multiply into reals, so you are left with $i(i-11)(i-10)$, then the conjugate of that gives you the answer[/hide]

[hide="5"]you get every base $12$ number from $0$ to $12^{4}-1$, the total numbers divisible by $72*11$ is just $1+\lfloor \frac{12^{4}-1}{72*11}\rfloor+1$ (adding for 0)[/hide]

[hide="6"]the expression becomes $\log_{k}n$, so $n$ is a power of $k$, just some case work now, $n=k$, $n=k^{2}$, etc[/hide]

[hide="7"]by the british flag theorem, $BP^{2}+EP^{2}=AP^{2}+DP^{2}$, so the inequality becomes $DP>AP$ so draw the perpendicular bisector, easy triangle areas now[/hide]

[hide="8"]this is equivalent to the ball passing through a lattice point, find the first time at which the happens, the position is described by $t*(\frac{1}{16},\frac{1}{8},\frac{1}{13})+(\frac{1}{2},\frac{1}{2},0)$, then find t such that all positions are integers, easy diophantine[/hide]

[hide="9"]
first: let $S_{2^{n}}=\sum_{1}{2^{n}}a_{k}$
$S_{2^{n}}=\sum a_{2k-1}+\sum a_{2k}$
$=\sum \frac{a_{2n}-\frac{1}{4}}{2}+4S_{2^{n-1}}$
$=6S_{2^{n-1}}-2^{n-4}$

$a_{1}=4a_{2}$ and $a_{2}=2a_{1}+\frac{1}{4}$
$a_{1}=\frac{1}{8}=S_{2^{0}}$

$n=1$ $\frac{3}{4}-\frac{1}{8}=\frac{5}{8}$
$n=2$ $\frac{15}{4}-\frac{1}{4}=\frac{7}{2}$
$n=3$ $21-\frac{1}{2}=\frac{41}{2}$
$n=4$ $123-1=122$
$n=5$ $732-2=730$
$a_{32}=4a_{16}=...=4^{5}a_{1}=128$
$730-128=602$
[/hide]

[hide="10"]consider a weighted average of any digit, it is obviously $\frac{1+2+...+9}{9}=5$, then for each place value, multiply by $111....1=\frac{10^{n}-1}{9}$, then there are $9^{n}$ numbers of this kind, giving the sum to be $\frac{5}{9}*(90^{n}-9^{n})$ the desired expression comes from the difference of squares,[/hide]

[hide="11"]let $x=r\cos\theta$, $y=r\sin\theta$, then $r^{2}=\frac{6}{\sin 2\theta+\cos 2\theta}=\frac{6}{ \sqrt{2}*\sin(2\theta+45)}$[/hide]

[hide="15"]
this problem leads to a generalization,
$a_{n}+i*b_{n}=(1+[2k+1]i)^{n}$ where $\{a_{n},b_{n}\}\subset\mathbb{R}$
$c_{n}=\gcd(a_{n},b_{n})=2^{\lfloor\frac{n}{2}\rfloor }$

first, $a_{n+1}=a_{n}-(2k+1)b_{n}$ and $b_{n+1}=b_{n}+(2k+1)a_{n}$
$c_{n+1}=\gcd(a_{n}-(2k+1)b_{n},b_{n}+(2k+1)a_{n})$

[hide="so"]
note $b_{n}$ is never divisible by $2k+1$, (obvious through induction with the recursion, so
$c_{n+1}=\gcd([a_{n}-(2k+1)b_{n}][2k+1],b_{n}+(2k+1)a_{n})$
[/hide]
$c_{n+1}=\gcd( [4k^{2}+4k+2]b_{n},b_{n}+(2k+1)a_{n})$,
note that $c_{n}=\gcd(a_{n},b_{n})=\gcd(b_{n},(2k+1)a_{n}+b_{n})$ so
$c_{n}|c_{n+1}$, and any other factors of $c_{n+1}$ are those of $[4k^{2}+4k+2]$, it follows inductively that

all factors of $c_{n}$ are from $2(2k^{2}+2k+1)$ $\star\star$

factoring $(1+[2k+1]i)^{n}=(k+1+ki)^{n}\cdot (1+i)^{n}$, note the factors of 2 produced by $(1+i)^{n}$

take the sequence $(k+1+ki)^{n}=a_{n}'+ib_{n}'$

--
lemma 1: ignoring the 2's from $(1+i)^{n}$, $c_{n}=c_{n}'$

[hide="because"] when we have an even power, we take out all the (1+i)^{2k} since it just contributes $2^{k}$, otherwise, for odd powers, a $(1+i)$ is still left, but this does not change the gcd (because)
let $t$ be an odd positive integer,
first, $(1+i)(a_{t}'+ib_{t}')=(a_{t}'-b_{t}')+i(a_{t}'+b_{t}')$

a quick check shows that $a_{t}'$ and $b_{t}'$ have different partity mod 2, so we can say that

$\gcd( a_{t}'-b_{t}',a_{t}'+b_{t}')=\gcd( 2a_{t}',a_{t}'+b_{t}')=\gcd( a_{t}',a_{t}'+b_{t}')=\gcd( a_{t}',b_{t}')=c_{t}'$[/hide]
--

let $p$ be a prime such that $p|[2k^{2}+2k+1]$, by $\star\star$, those $p_{i}$ are the only other primes that can divide $c_{n+1}$ other than $2$,

--
lemma 2 $p_{i}\not|a_{n}'$

easily obtain the recursion (for $n\in\mathbb{N}_{0}$)
${a_{n+1}'-2(k+1}a_{n+1}+(2k^{2}+2k+1)a_{n}'=0$ so
${a_{n+1}'-2(k+1}a_{n+1}\mod p_{i}$

note $\gcd(k+1,2k^{2}+2k+1)=1$, so $p\not|(k+1)$, so $a_{n+1}'\equiv a_{n}'\mod p_{i}$, it follows inductively that $p_{i}\not|a_{n}'$
--

so $c_{n}'$ cannot have any prime factors other than $p_{i}$ by $\star\star$ and lemma 1

furthermore, $a_{n}'$ has none of $p_{i}$ as factors, thus $c_{n}'$ has no factors, therefore $c_{n}'=1$,

by lemma 1, $c_{n}$ cannot have any prime factors other than those from $(1+i)^{n}$ and $c_{n}$, it can easily be seen that $c^{n}=2^{\lfloor\frac{n}{2}\rfloor}$ for $n\in\mathbb{N}$

back to the problem, some trivial computation,

$x\equiv 0\mod 8$,
$x\equiv 2^{1003}\equiv2^{903}\equiv...\equiv2^{3}\mod 125$ since $\phi(125)=100$, then $x\equiv 8\mod 1000$
phew
[/hide]

[quote="PenguinIntegral"]What is the British Flag Theorem?[/quote]
An obscure theorem that is proven with pythagorean that says that if a point P is chosen inside rectangle ABCD than $AP^{2}+PC^{2}=BP^{2}+DP^{2}$. I haven't done this AIME yet, but I have only seen one application of it in a math problem so far and that was in Mathcounts.

Actually, the point need not be inside or in the plane of the rectangle.

the theorem is not necessary in the problem, it is just the fastest, rigirous way to do it, you can always just find a few points on the locus, and connect

[quote="PenguinIntegral"]Could you give more solutions? The ones you ommited are the coolers ones...

For 15, I don't think it's AIME level at all...  :maybe:

I'm not understanding your manipulations in 9 either, could you be more verbose?[/quote]

first, somebody got it correct, second, i did the general form, the specific case is much easier, third, the only step that you really need is the factoring out the (1+i), the rest you could conjecture during the test,

for problem 9, i don't know what else to say other than i was using the recursive statements to go from step to step

i am in 8th grade. Should i start preparing for AMC 10 and AIME already? My dream is to make USAMO Someday.  :blush:

[quote="mustafa"]I would take what you study for one step at a time. First work on your middle school competitions until can completly and utterly pwn them, then do the same for the AMC 10, then the AMC 12, then AIME......[/quote]
Um, no.  Work on harder competitions carries over to lower competitions, but work on lower competitions does not really carry over to harder competitions in a meaningful way.  Therefore this proposal is inefficient.  If you are getting to the point to where you can easily answer almost every question on a certain level of competition, it's been time to move on to harder stuff for quite some time.  In general, you should study problems that are a little beyond you.  now a ranger, I would advise you to work on AMCs, and if you can get over 100 on one of those, work on AIMEs, and a little before you feel ready, start work on easy olympiad problems.  It is quite easy to make USMO after a year to a year and a half of good work, even if you couldn't even qualify for MATHCOUNTS nationals before. :D   To do well on it takes some more time, though...

Anyhow, read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10806]this[/url].

thanks boy soprano u give good advice, just like that thing when i asked about how to improve on vestavia when i had a 26, i now can get two times higher.

and i never tried out for mathcounts nationals because i didnt even know what mathcounts was lol until this year when i joined the math team.

and our school doesn't do it either

OMG don't make this another advice thread, there are plenty of threads that say the same thing

[hide="Jumping off point for 12"]
By Power of a point, $ CX \cdot CF\equal{}40$. By Ceva, AF=3n and BF=5n. By Menelaus, $ \frac{CX}{XF}\equal{}\frac{8}{3}$, so $ \frac{CX}{CF}\equal{}\frac{8}{11}$
[/hide]

Can someone give the "easy" solution for 13? Mine is a terrible brute force where I got 409.

[quote="Altheman"][hide="8"]this is equivalent to the ball passing through a lattice point, find the first time at which the happens, the position is described by $ t*(\frac {1}{16},\frac {1}{8},\frac {1}{13}) \plus{} (\frac {1}{2},\frac {1}{2},0)$, then find t such that all positions are integers, easy diophantine[/hide][/quote]

In fact, it is impossible to have all $ x,y,z$ integral with $ (t/16 \plus{} 8/16,t/8 \plus{} 4/8,t/13)$ since $ t \plus{} 8\neq t \plus{} 4\mod 8$. If I'm not mistaken, the particle will NEVER hit a corner even if it does not stop when hitting an edge.

By the way could you please post solutions to the rest of the problems? :)

sorry to revive this; I just want to make sure... For the problem the poster above me talked about, because edges can work, you don't need all three to be integers, rite? Only 2 of the three? (since, as mathwizarddude said, it IS impossible)

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3751056_2007_mock_aime_3_by_alex_anderson]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS. Answer key is [url=https://artofproblemsolving.com/community/c5h114749p675524]here[/url]


Find all $n$ such that the sum of the inverses of all divisors of n! is an integer.

Note that one of the divisors of n! is n! itself, and all other divisors will divide this, so we take as our denominator n! and our expression can be rewritten as $\frac{\sigma(n!)}{n!}$, since for any divisor d there exists another e such that de=n!.

Since n!/n!=1, we can ignore the n! divisor in the sum, and concentrate on the others, denoting this $\tilde\sigma(n!)$. If our expression is to be an integer, $\tilde\sigma(n!)=kn!$ where k is a positive integer.

So, if k=1, n! must be a perfect number (of which 6 is one, and I can't be bothered checking the rest).

If k=2, I have a feeling we have no solutions, but am feeling apathetic now, sorry.

That should help though.

We will show that for any $n>5$, $\sigma(n!)$ is not divisible by $n!$. The following lemma (a slightly stronger form of Zsigmondy's theorem) provides the key: For any integer $k>6$, either $k+1$ is prime and $(k+1)^2|2^k-1$, or $2^k-1$ has a prime divisor $\geq 2k+1$.

Now assume $n$ is large. Let $v$ be the power index of 2 in $n!$, then $v>\frac{n}{2}$. By lemma, $2^{v+1}-1$ does not divide $n!$, and the part that doesn't divide is at least $v+1>\frac{n}{2}$. Since $2^{v+1)}-1$ is a factor of $\sigma(n!)$, we know there is a part in $\sigma(n!)$ that is at least $\frac{n}{2}$ and that does not divide $n!$. But note that $\frac{\sigma(n!)}{n!} < \frac{n}{2}$, so this ratio could not be an integer.


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

This is related to the problem posted by -oo- [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111780]here[/url].

Let $X$ be an infinite set, and let $S(X)$ be the group of bijections from $X$ onto itself. Prove that $S(X)$ has no proper subgroups of finite index.

With X finite we have that the derived group (commutator subgroup) of S(X) is A(X). A(X) is simple when X has more than 4 elements.

When X is infinite we have the same result, that the derived group of S(X) is simple. It was shown in the other thread that this derived group is S(X) itself.

Now we have a theorem in group theory saying that: Let G be a simple group with a subgroup of finite index n. Then G is isomorphic to a subgroup of Sn (the symmetric group on n letters).

Since S(X) is infinite that is impossible.

The theorem involving the simple group is clear (G acts faithfully on the left cosets), but what about the simplicity of S(X)?

i don't think it's simple: take the set $H \subset S(X)$ of functions with a finite number of non-trivial finite orbits (and the rest of point fixed). it seems it's a normal subgroup of $S(X)$, isn't it?

Yeah ma_go, that's right. $H$ is called the restricted symmetric group, and consists of those permutations with only a finite number of points not fixed.

Yes you are right, mago. H is a normal subgroup. I was too hasty there.

Any embedding of {1,...,n} in X induces a permutation of X that only moves finitely many points, so we have an embedding of Sn in S(X). H is the union of all the images.

Notice, however, that H has infinite index in S(X): Let N be a countably infinite subset of X. Consider the permutations:

s2 : (1 2)(3 4)...
s3 : (1 2 3)(4 5 6)...
etc.

They differ pairwise at infinitely many points, so are not congruent mod H.

Let me try to correct my assertion as follows: S(X) has only "small" normal subgroups.

I haven't proved it, but if it is true then my proof goes through as before: Take a maximal normal subgroup and consider the simple quotient.


vector
number theory unsolved
number theory

Does someone know how I could to solve the recurrence relation $ f(x) \equal{} 2f(x\minus{}1) \minus{} f(x\minus{}4)$ ?
The initial conditions are: $ f(1) \equal{} c1, f(2) \equal{} c2, f(3) \equal{} c3, f(4) \equal{} c4$ and $ c1$, $ c2$, $ c3$ and $ c4$ are integers and have different values.

Thanks in advance!

[quote="m.r650200"]Does someone know how I could to solve the recurrence relation $ f(x) \equal{} 2f(x \minus{} 1) \minus{} f(x \minus{} 4)$ ?
The initial conditions are: $ f(1) \equal{} c1, f(2) \equal{} c2, f(3) \equal{} c3, f(4) \equal{} c4$ and $ c1$, $ c2$, $ c3$ and $ c4$ are integers and have different values.

Thanks in advance![/quote]

If you are speaking about a sequence $ a_{n\plus{}4}\equal{}2a_{n\plus{}3}\minus{}a_n$, then the general method is :

1) Consider that the space of solutions is a $ \mathbb R$-vector space whose dimension is at most $ 4$

2) Consider that $ a_n\equal{}r^n$ is a solution as soon as $ r^4\equal{}2r^3\minus{}1$

3) So the general solution is $ a_n\equal{}\alpha r_1^n\plus{}\beta r_2^n \plus{} \gamma r_3^n \plus{} \delta r_4^n$ where $ r_i$ are the four roots (real and complex) of $ x^4\minus{}2x^3\plus{}1\equal{}0$


Show that for $ A, B \subset \mathbb{R}$ and $ f: A \times B \rightarrow \mathbb{R}$ bounded, sup$ _{(x,y) \in A \times B} f(x,y)$ = sup$ _{y \in B} ($ sup$ _{x \in A} f(x,y))$.

Still more requests for people to solve your homework for you.  :roll:

You'll find that people will be a lot more receptive to requests for help if you tell us what work you've done so far and what part of the problem you're stuck on.


function
LaTeX
floor function
number theory
relatively prime
algebra

$ (\forall)m,n\in \mathbb{N}$ prove that :

$ \left[\frac{2^n\minus{}1}{2^m\minus{}1}\right]\cdot 2^{m[\frac{n}{m}]}\equal{}\frac{2^{m[\frac{n}{m}]}\minus{}1}{2^m\minus{}1}\cdot 2^n$

Is there no one interested ? This equation is easy .  :)

Let $ n\equal{}mq\plus{}r$ where $ 0\le r<m$ and then it is easy enough.

From $ n \equal{} mq \plus{} r$ it is easy to find that the only solution is $ m \equal{} n \equal{} 1$

note that numbers $ 2^{m} \minus{} 1$ and $ 2^{n} \minus{} 1$ are relatively prime ,$ m,n$ distinct natural numbers.

edit : [b] It seems I am very drunk today  :(   [/b]

[quote="enndb0x"]note that numbers $ 2^{m} \minus{} 1$ and $ 2^{n} \minus{} 1$ are relatively prime ,$ m,n$ distinct natural numbers.[/quote]

$ 2^2 \minus{} 1 \equal{} 3$.  $ 2^4 \minus{} 1 \equal{} 15$.  $ (3, 15) \equal{} 3 \neq 1$.

Do the square brackets imply a certain operation (say, floor function)?

The brackets represent the integer part . I don't know if it's the same as the floor function but see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=268298&search_id=452031732]here[/url] what is the integer part .

I think that floor function=integer part  :lol:

LaTeX has built-in commands \lfloor and \rfloor, yielding $ \lfloor \; \rfloor$, an unambiguous notation for the greatest integer (= floor = integer part) function.

Ok , thanks JBL , but in my country the brackets are the notations for the integer part


Find the minimum value of $xy+yz+xz$. given that $x, y, z$ are real and $x^{2}+y^{2}+z^{2}= 1$. No calculus, please!

that's the maximum value tho...

[hide]
$x^{2}+y^{2}+z^{2}+2xy+2yz+2xz = (x+y+z)^{2}= 1+2(xy+yz+xz) \ge 0$
so I have $xy+yz+xz \ge-\frac{1}{2}$.  Can't find an equality case tho.  But there suppose to be infinitely many solutions to $x^{2}+y^{2}+z^{2}= 1$ and $x+y+z = 0$
[/hide]

Oops, my bad xD

If you let $(x,y,z)=(k,k,-2k)$ and set the sum of squares equal to 1, you get:
$(x,y,z)=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\right)$
which does in fact yield the minimum value.


Find a function $f: \mathbb N \to \mathbb N$ such that for all positive integers $n$, \[ f(f(n))\equal{}n^2.\]

[quote="stvs_f"]find $ F: N \to N$ such that :
$ \forall n \in N$ :
\[ f(f(n)) \equal{} n^2\]
[/quote]

Let $ A\equal{}\{x$ such that $ x$ is a natural number but not the square of a natural number$ \}$

For any $ n>1$, let $ h(n)\in A$ and $ p(n)\in\mathbb N\cup\{0\}$ such that $ n\equal{}h(n)^{2^{p(n)}}$

Obviously $ h(n)$ and $ p(n)$ exist and are unique for any $ n>1$

Let $ U$ and $ V$ a split ($ U\cup V\equal{}A$ and $ U\cap V\equal{}\emptyset$) of $ A$ in two equinumerous subsets and $ b(x)$ a bijection from $ U\to V$ (such subsets exist since $ A$ is an infinite set).

Then the general solution of the required equation is :

$ f(1)\equal{}1$
$ \forall n$ such that $ h(n)\in U$ : $ f(n)\equal{}b(h(n))^{2^{p(n)}}$
$ \forall n$ such that $ h(n)\in V$ : $ f(n)\equal{}b^{\minus{}1}(h(n))^{2^{p(n)\plus{}1}}$


Printing a book costs 2 cents per page plus 2 dollars for the cover. How many pages does a book have if its total cost is twice as much as the total cost of a 100-page book?

[hide="What I Got"]
The cost for a 100-page book would be $ 0.02 \times 100 \plus{} 2$, or 4 dollars. So that cost for this book is 8 dollars. When you subtract the cost of the cover, the price for the pages is 6 dollars. Each dollar buys 50 pages, so there are 300 pages in the book. [/hide]


calculus
function
topology
real analysis
real analysis unsolved

Hi,

cause I´m a principant in this field I've encountered 2 questions in the calculus classes that I can´t response. Could anybody help me?

1st: Let [i]d[/i] be discrete metric and [i]e[/i] Euclidean metric. Show that if the function f: (R,d) ---> (R,e) is continuous then f is necessairly constant.

2nd: Let A be an open subset of R. Find any real continuous function f: R ---> R such that A = f^(-1) [(0,+ infinity)].

Thanks for any help.  :roll:

1. This is wrong. Every function $f\colon (\mathbb{R},d)\to (\mathbb{R},e)$ is continuous (since every subset of $\mathbb{R}$ is open in the dicrete topology, preimage of an open set is open). Do you mean  $f\colon (\mathbb{R},e)\to (\mathbb{R},d)$ ?

2. Consider the function $f(x)$ equal to the distance from $x$ to the complement of $A$: $f(x) = \inf_{y\notin A} |x-y|$.

I have only copied it from the task... maybe there's a mistake. And what would be the solution if we considered your suggestion $f: (R,e) --- > (R,d)$? Thanks for the given function in the 2nd task.

In that case, the only way we can have $|f(x)-f(y)|_d < \epsilon$ for any $\epsilon \leq 1$ is if $x = y$.  So around any point $x_0$ there is an open ball in which $f$ is constant.  But then $f$ is constant everywhere.


calculus
integration
function
derivative
real analysis
real analysis unsolved

Let $f(x)$ be the Dirichlet function, does there exist a differentiable function $g(x)$ on $\mathbb{R}$ so that $g'(x)=f(x)$?

[quote="didilica"]The answer is no. If such a function $g$ would exit then its derivative, in this case $f=g'$, should have the Darboux property. However the Dirichlet function does not verify the Darboux property.[/quote]

I do not understand.

It helps if you say [i]what it is[/i] that you don't understand.  "Darboux property" = "intermediate value property."

Here is an informative article on the [url=http://eom.springer.de/D/d120050.htm]Darboux property[/url].


modular arithmetic
quadratics
number theory
number theory proposed

Let $ p$ be a prime such that $ p \equal{} 4k \plus{} 1$ for some $ k\in \mathbb N$. Prove that $ p\mid (k^k \minus{} 1)$.

$ k^k = (\frac {p - 1}{4})^\frac {p - 1}{4} \equiv (p - 1)^\frac {p - 1}{4}(4^\frac {p - 1}{4})^{ - 1} \equiv ( - 1)^\frac {p - 1}{4}(2^\frac {p - 1}{2})^{ - 1} \equiv ( - 1)^\frac {p - 1}{4}\left(\frac {2}{p}\right)^{ - 1} \equiv (*) ( - 1)^\frac {p - 1}{4} (( - 1)^\frac {p^2 - 1}{8})^{ - 1} \equiv ( - 1)^\frac {p^2 + 2p - 3}{8} \pmod {p}$
So it's equivalent to showing that $ 16 | p^2 + 2p - 3$ when $ p \equiv 1 \pmod {4}$. Just calculate the expression $ p^2 + 2p$:
$ p^2 + 2p = (4k + 1)^2 + 2(4k + 1) = 16k^2 + 16k + 3 \equiv 3 \pmod {16}$.

(*) We used a special elementary case of quadratic reciprocity: $ 2^\frac {p - 1}{2}= \left(\frac {2}{p}\right) = ( - 1)^\frac {p^2 - 1}{8} \pmod {p}$, where the fraction is Legendre symbol.

Using Legendre's Symbol is really well-known, and we could just do a slightily change, transforming it in:
$ p$ divides $ ( \minus{} 1)^k \minus{} ( \minus{} 1)^{2k^2 \plus{} k}$.
I'd like to look any different solution, not using Legendre's symbol.

[quote="mr.danh"]Let $ p$ be a prime such that $ p = 4k + 1$ for some $ k\in \mathbb N$. Prove that $ p\mid (k^k - 1)$.[/quote]

For erudito, a solution without even using fractions modulo $ p$:

Since $ p$ is a prime and $ p=4k+1$ for some $ k\in \mathbb N$, a known theorem yields that $ -1$ is a quadratic residue modulo $ p$. That is, there exists an integer $ j$ such that $ j^2\equiv -1\mod p$. Now, $ j\not\equiv 1\mod p$ (since $ j\equiv 1\mod p$ would yield $ j^2\equiv 1\mod p$, what would contradict $ j^2\equiv -1\mod p$), so that $ p\nmid 1-j$, and thus Fermat's Little Theorem entails $ \left(1-j\right)^{p-1}\equiv 1\mod p$. That is, $ \left(1-j\right)^{4k}\equiv 1\mod p$ (since $ p-1=4k$, as $ p=4k+1$). Now,

$ \left( 1-j\right) ^{2}=1-2j+j^{2}\equiv 1-2j-1$ (since $ j^{2}\equiv -1\mod p$)
$ =-2j\mod p$,

so that

$ \left( 1-j\right) ^{4}=\left( \left( 1-j\right) ^{2}\right) ^{2}\equiv \left( -2j\right) ^{2}=4j^{2}\equiv 4\left( -1\right)$ (since $ j^{2}\equiv -1\mod p$)
$ =-4%Error. "modp" is a bad command.
$,

hence

$ \left( 1-j\right) ^{4}k\equiv -4k=1-\left( 4k+1\right) =1-p\equiv 1\mod p$.

Therefore,

$ \left(\left( 1-j\right) ^{4}k\right)^k\equiv 1^k=1\mod p$.

But

$ \left(\left( 1-j\right) ^{4}k\right)^k=\left(1-j\right)^{4k}k^k=\left(1-j\right)^{p-1}k^k\equiv 1k^k=k^k\mod p$.

Thus, $ k^k\equiv 1\mod p$, qed.

  darij


summer program
MathZoom
MATHCOUNTS
MathPath
geometry
AMC
AIME

Hi. I know people who went to Math Path and there was a lot of theory taught there. But I've hear Mathzoom focuses more on teaching how to suceed at competitions. If true, can someone explain HOW they do this? For ex, how did they train for MATHCOUNTS?   beyond maybe practicing old problems, what special things do they do?  Any special tricks taught? You can pm me your answer if you feel more comfortable.  Thanks!

I will give a brief answer here.  Other "Zoomers" feel free to add your own.

MathPath is for middle school students, so the focus is more on exposure of the variaous topics in math, including advanced theories and math history.  At Math Zoom, we have different levels of curricula for students with different experiences, and have classes for middle school and high school levels.  The focus is on problem solving techniques, so our curriculum is more in depth.  Success in math competitions is one of the goals, and comes naturally if you do well at Math Zoom, but the main purpose of Math Zoom is far beyond that.

In our classes, lectures are given on topics including algebra, geometry, number theory, and combinatorics.  Problem solving strategies are taught along with the concepts, and there are plenty of problems to work on.  Based on experiences of students, the camp consists of three groups (at least): Red Group (Mathcounts/AMC 8 level), Blue Group (AMC 10/12/Beginning AIME level), and Black Group (Advanced AIME/Olympiad level).  Special tricks?  A lot of them.  Old contest problems?  Sure.  But as mentioned above, it is much more than just succeeding in math contests.  A main part of the lectures is on the concepts in depth.  The curriculum is structured to help you build a solid foundation in math, improve analytical thinking skills, and develop abilities in rigorous mathematical reasoning.  In the future, the skills you learned will continue to benefit you in any of your endeacors, because analytical thinking is not just limited to math.

That partially answers your question.  To fully understand what we do exactly, there is one straightforward way to find out :)

You won't be disappointed.
 :D

DrKevin thanks for the reply! (wish I knew some of the "tricks" though - especially for MATHCOUNTS!!) Actually, I'm trying to really figure out not what gets someone to MATHCOUNTS Nationals, but what truly distinguishes between say #12 or 10th place and the number 1 or 2 place competitors (uh, beyond luck, that is)? But I suppose that question is best suited for another area of this site ...?

It depends on what state you're in.  If you're in CA it's probably just luck/arithemtic errors.  (Btw sorry AIME,  you can make it next year)

And yeah I think that question's more suited for the mathcounts forum.. :P

In a normal state, usually the best three competitors go to Nationals.  Sometimes one of the top four isn't necessarily in the smartest four.  It depends on luck, usually.

As for CA, there were 8 people within 1 point of the state champion.  It's all misreading, luck, and arithmetic errors that distinguish the top 10 spots.

[quote="AIME15"]As for the "levels", I think I'm definitely past AMC 10/12/Beginning AIME, but not quite at Advanced AIME/Olympiad, will there be something for people like me?[/quote]
I have been thinking about your case.  There are reasons for putting you in Blue Group and reasons for putting you in Black Group.  Right now I am leaning toward the Blue Group.  Don't think "AMC 10/12/Beginning AIME" as non-challenging.  The materials are definitely challenging enough, and you will need to build the solid foundation to be ready for more advanced materials.  And it is also flexible.  If I decide that you will benefit more in the Black Group, then you will be promoted to that group.  That happened a couple times before and worked well.

[quote="helagha"]I think I will be in the Blue Group. This is my first year going (if I do hopefully go)[/quote]
Did you send your application materials yet?

[quote="turak"]Since there are four groups this year, what is the approximate level of each of those four?[/quote]
Approximately:

[color=red]Red: Beginners or close to beginners.  Practice on the AMC8 and Mathcounts level.[/color]
[color=green]Green: Advanced AMC 8, up to AMC 10.[/color]
[color=blue]Blue: Advanced AMC 10, AMC 12, AIME.[/color]
Black: Math Olympiad level.  Some advanced AIME will be included, but mostly on Olympiad topics and techniques.

Levels are adjusted according to the levels of experiences of enrolled students.  For this year, the above is the general guideline.

Eventually, we are going to have 8 groups: complete rainbow colors ([color=red]Red[/color], [color=orange]Orange[/color], [color=yellow]Yellow[/color], [color=green]Green[/color], [color=blue]Blue[/color], [color=indigo]Indigo[/color], [color=violet]Violet[/color]) plus Black.  Then the levels will be adjusted accordingly ... :D


It is now time for the board members to be elected. Ernie and I spent a long time deciding and we decided these would be the best people for the job:

[11:14] mewto55555: romaniangenius
[11:14] mewto55555: xoangieexo
[11:14] mewto55555: isabella2296
[11:14] mewto55555: vallon22
[11:14] mewto55555: violin321
[11:14] mewto55555: dragonlaird7
[11:14] mewto55555: superllamel
[11:14] mewto55555: fantasylover

Now, ernie will be in charge of this election as I will be unable to access AOPS until Monday. Due to the large number of candidates, no votes will be accpeted on FTW! You must pm ernie the ballot or post here.

[hide="For ernie only"]
Ernie when they pm results, please pm those to me so I can double-check.
If anyone else read this SHAME ON YOU!
[/hide]

[b]IMPORTANT:[/b]This election is not the candidates pitted against each other. Instead you simply vote whether each one should be on the board.

All ballots must be in this format

RomanianGenius: Your vote (yes or no)
Xoangieexo: Your vote
Isabella2296: Your vote
Vallon22: Your vote
Violin321: Your vote
Dragonlaird7: Your vote
Superllamel: Your vote
FantasyLover: Your vote

Here are my votes:

RomanianGenius: Yes
Xoangieexo: Yes
Isabella2296: Yes
Vallon22: Yes
Violin321: Yes
Dragonlaird7: Yes
Superllamel: Yes
FantasyLover: Yes

Ernie keep a seperate tally for each candidate.

ok, two questions

so you can vote for more than one person right?

and doesn't there need to be 8 board members? and don't we have 8 there already?

i guess that was 3 questions....

Vallon each person is a seperate issue.

Ernie here is a list of everyone eligible to vote:

Skaldjfhrulez
Violin321
Mewto55555
Kl2836
(^_^)
Mathpassion101
Isabella2296
Xoangieexo
Lotsofmath
Ernie
FantasyLover
Yoh2241
Vallon22
AIME15
Jongao
Dojo
Pacman2812
Budi713
Myyellowducky82
Vamathletes
Guineapiggies
Pinkpiglet
Superllamel
Jain_harshi
Splatyango
Mathblitz
Cherry_pi
RomanianGenius
Weird888
Dragonlaird7
Dragon96
Dannyhamtx
Kookamunga
Nikeballa96
Pi guy
27r
Yclarinet
Chenhsi

i vote yes for all
but i have never seen dragonlaird7 on though....
different times maybe

I wil do what I can until I must leave.

Skaldjfhrulez
Violin321
Kl2836
(^_^)
Mathpassion101
Xoangieexo
Lotsofmath
Ernie
FantasyLover
AIME15
Jongao
Dojo
Pacman2812
Budi713
Myyellowducky82
Vamathletes
Jain_harshi
Splatyango
Mathblitz
Cherry_pi
Weird888
Dragon96
Dannyhamtx
Kookamunga
Nikeballa96
Pi guy
27r
Yclarinet
Chenhsi

RomanianGenius: 8 yes 1 no
Xoangieexo: 8 yes 0 no
Isabella2296:  8 yes 0 no
Vallon22: 8 yes 0 no
Violin321: 8 yes 1 no
Dragonlaird7: 7 yes 2 no
Superllamel:  7 yes 1 no
FantasyLover: 8 yes 0 no

RomanianGenius: Yes
Xoangieexo: Yes
Isabella2296: Yes
Vallon22: Yes
Violin321: Yes
Dragonlaird7: no (no offense i just don't know dragon that well) :|
Superllamel: Yes
FantasyLover: Yes

RomanianGenius: Yes
Xoangieexo: Yes
Isabella2296: Yes
Vallon22: Yes
Violin321: Yes
Dragonlaird7: Yes
Superllamel: Yes
FantasyLover: Yes

If mewto and ernie thought these were the best people for the job, then they probably are.

I need to go now. ERnie, it's up to you!

Left to vote:

Skaldjfhrulez
Violin321
Kl2836
(^_^)
Mathpassion101
Xoangieexo
Lotsofmath
Ernie
FantasyLover
AIME15
Jongao
Dojo
Pacman2812
Budi713
Myyellowducky82
Vamathletes
Jain_harshi
Splatyango
Mathblitz
Cherry_pi
Weird888
Dragon96
Dannyhamtx
Kookamunga
Nikeballa96
Pi guy
27r
Yclarinet
Chenhsi

RomanianGenius: 8 yes 1 no
Xoangieexo: 9 yes 0 no
Isabella2296:  9 yes 0 no
Vallon22: 9 yes 0 no
Violin321: 8 yes 1 no
Dragonlaird7: 7 yes 2 no
Superllamel:  8 yes 1 no
FantasyLover: 9 yes 0 no

[quote="iin77"]RomanianGenius and mewto are the same person, you know.  Just thought it might be important...[/quote]

RG and mewto are bros.  Just thought that might be important...

RomanianGenius: 10 yes 1 no
Xoangieexo: 12 yes 0 no
Isabella2296: 12 yes 0 no
Vallon22: 12 yes 0 no
Violin321: 10 yes 1 no
Dragonlaird7: 9 yes 2 no
Superllamel: 11 yes 1 no
FantasyLover: 12 yes 0 no

[quote="Yoh2241"]RomanianGenius: (Never seen him/her)
Xoangieexo: Yes
Isabella2296: Yes
Vallon22: Yes
Violin321: (Never seen him/her)
Dragonlaird7: (Never seen him)
Superllamel: Yes
FantasyLover: Yes[/quote]

So do you vote no for RG, violin, and dragon?

And also with your votes, if you vote someone no, please supply someone else.

whoa i didnt know this existed

yes for all!

IF  are going to keep track, please also keep track of who is left to vote!

[quote="vallon22"]it is now 15-5 75%

gl kl

edit: @mewto, fine... =P

[hide="people left to vote"]
Skaldjfhrulez - 1200
Violin321
Mathpassion101
Yoh2241
Vallon22
AIME15
Jongao
Vamathletes
Pinkpiglet
Superllamel
Mathblitz
Weird888
Dragon96
Kookamunga
Nikeballa96
Pi guy
Yclarinet
Chenhsi
Textangle
Jjx1
Aime_is_hard
Mind Storm
Bisbmard9
Emilylambkin
[/hide][/quote]
I thought super voted?

Just another question.
Whos the other board member going to be?
I mean RomanianGenius fell out right?

Weren't there other people already elected?

[quote="Dojo"]No I believe that kl is subbing in for Dragonlaird, not RomanianGenius[/quote]
dragon's gone?

Hes not gone, but if you had read the other parts, he fell out of the run because too many people voted against him.

Can i run for being one of the board members? I want to be a candidate.

[quote="myyellowducky82"][quote="ernie"][quote="Dojo"]you don't 'run' for board.
the prez and vp nominate you.[/quote]

Yes, and if kl2836 is a "success" (too lazy to read previous posts), we have enough people.  Thank you for your willingness, however. :D[/quote]
but then why are we voting for or against kl?[/quote]

Because we need to see if the CTC members think that kl2836 is a good candidate (I think so).

[quote="myyellowducky82"]but the other slots are filled right?[/quote]

Read the posts before yours. :wink:


Given a triangle $ABC$. Suppose that $\exists X \in [BC], Y \in [CA] , Z \in [AB]$ such that: $AZ=BX=CY$ and $\widehat{BAX}=\widehat{CBY}=\widehat{ACZ}$.
Prove that: $ABC$ is equilateral.


trigonometry
calculus
algebra unsolved
algebra

[quote="thanhnam2902"]Solve this equation in $ R$:

$ 3\cos^2{(4x)} \plus{} 5\sin^2{(4x)} \equal{} 2\left ( 1 \minus{} \frac {\sqrt {3}}{2}\sin{x} \right )$[/quote]

Where is this problem coming from ? I did not see any hint for this equation. So basic computation :

Let $ \sin x\equal{}y$, then : $ \sin^22x\equal{}4y^2(1\minus{}y^2)$ and $ \sin^24x\equal{}16y^2(1\minus{}y^2)(1\minus{}4y^2(1\minus{}y^2))$

The equation may be written $ 3\plus{}2\sin^24x\equal{}2\minus{}\sqrt 3\sin x$ and so :

$ 1\plus{}32y^2(1\minus{}y^2)(1\minus{}4y^2(1\minus{}y^2))\plus{}\sqrt 3y\equal{}0$

$ 128y^8\minus{}256y^6\plus{}160y^4\minus{}32y^2\minus{}\sqrt 3 y\minus{}1\equal{}0$

I found no intelligent method to solve this and used numeric calculus (Newton) to find only 3 real roots in $ [\minus{}1,\plus{}1]$ :
And so :
$ y\equal{}\minus{}0.987294589...$
$ y\equal{}\minus{}0.779083771...$
$ y\equal{}\minus{}0.658213258...$

Which implies $ 6$ values for $ x$ in $ [0,2\pi]$ and the solutions :
$ x\equal{}3.86003558199229...\plus{}2k\pi$
$ x\equal{}4.03479566332100...\plus{}2k\pi$
$ x\equal{}4.55281199632294...\plus{}2k\pi$
$ x\equal{}4.87196596444644...\plus{}2k\pi$
$ x\equal{}5.38998229744838...\plus{}2k\pi$
$ x\equal{}5.56474237877709...\plus{}2k\pi$

And these values dont seem to be rational multiples of $ \pi$, neither "well known" values :(


find all integer solutions to

$x^3+y^3=9xy$

[quote="Treething"]Ooh, wait, I've seen this equation before...

This is the graph that looks like a loop (I can't think of the exact name >_< ), right?[/quote]

You are thinking of the Folium of Descartes:
[url=http://mathworld.wolfram.com/FoliumofDescartes.html]mathworld.wolfram.com/FoliumofDescartes.html[/url]

It is easy:

Rewrite this as (x+y+3)(x^2+y^2+9-xy -3x -3y) = 27
[hide](Now  note:
x^2+y^2+9 -xy -3x - 3y  is  non negative  ( because it is (1/2) ((x-3)^2+(y-3)^2+(x-y)^2) ) ), hence[b] x+y+3 is non-negative)[/b]

Rewrite tthe above again as:

(k+3) (k^2 - 3m -3k+9 ) = 27 where   k= x+y and m= xy

obviously  k must be a multiple of 3 ( if k is not 0 mod 3, LHS is not 0, but RHS is 0 mod 3)
so let k = 3n
we have  9(n+1)(3n^2 - 3n - m+9) = 27
or  (n+1)(3n^2-3n-m+3) = 3

so we have n+1 = 1 or n+1 = 3     (And then m = 0 or 8)
which gives [/hide]  (0,0) or (4,2)  as (non-negative)   integer values of x and y

here is my solution, in spoiler...it is very similar to Gyan's.

[hide]
let $a = x+y, b = xy$

substituting, we have $a^3-3ab = 9b$ and we see that $3|a$

then let $a = 3c$ and we have $27c^3 - 9cb = 9b$ or $3c^3 - cb = b$

then $b = \frac{3c^3}{1+c} = 3c^2 - 3c + 3 - \frac{3}{c+1}$

and since b is an integer, $c+1|3$ and the only possible values for c are 2, 0, -2, -4 which means the only possible pairs of (a,b) are:

(6,8), (0,0) (-6,24), (-12,64) and trying each we end up with the solutions

(0,0), (2,4), (4,2)[/hide]