analytic geometry
quadratics
algebra
geometry proposed
geometry

At first sorry for my bad english. I will try to explain my problem.
As you can see in attachment the circles with center B and C are fixed. Circle with center A1 can move on axulary circle(circle with dotted line). In attachment circle moved to position A2. I try to calculate angle x for different radiuses of curve D. Curve D is tangent to circle with center B, circle with center C and circle with center A2.

For example what must the angle x be if the radius of curve D is 400.

Known values

radius of circle B
radius of circle C
radius of circle A
radius of dottwd circle
center point of circle B
and other values shown in attachment

You already have an xy coordinate system with the origin $ C$. $ a$ is radius of the circles $ (A_1), (A_2)$, $ b$ is radius of the circle $ (B)$ (in your case, $ b \equal{} a$), $ c, d$ are radii of the circles $ (C), (D)$ and $ r$ is radius of the auxiliary circle containing $ A_1, A_2.$ The circle $ (D)$ internally tangent to the circle $ (C)$ is centered on a circle concentric with $ (C)$ and radius $ d \minus{} c.$ Equation of this circle containing $ D$ is

$ x^2 \plus{} y^2 \equal{} (d \minus{} c)^2$

Circle $ (D)$ is also externally tangent to $ (B),$ it is centered on a circle concentric with $ (B)$ and radius $ d \plus{} b.$ Equation of this second circle containing $ D$ is

$ (x \minus{} x_B)^2 \plus{} (y \minus{} y_B)^2 \equal{} (d \plus{} b)^2$

To solve these 2 equations, subtract the 1st equation from the 2nd one, yielding a linear relation between x, y, substiute from it to the 1st equation to get 2 quadratic equation for $ x_D, y_D.$ Pick the appropriate (smaller) roots $ x_D, y_D < 0.$ Similarly, the circle $ (A_2)$ is centered on the auxiliary circle concentric with $ (C).$ Equation of this circle containing $ A_2$ is

$ x^2 \plus{} y^2 \equal{} r^2$

Circle $ (A_2)$ is also externally tangent to $ (D),$ it is centered on a circle concentric with $ (D)$ and radius $ d \plus{} a.$ Equation of this second circle containing $ A_2$ is

$ (x \minus{} x_D)^2 \plus{} (y \minus{} y_D)^2 \equal{} (d \plus{} a)^2$

The second pair of equations yields 2 quadratic equation for $ x_{A2}, y_{A2}.$ Pick the appropriate (larger) roots $ x_{A2}, y_{A2} > 0.$ Once you know the coordinates of $ A_2,$ the angle $ \theta_2$ of the ray $ CA_2$ from the y-axis is $ \theta_2 \equal{} \arctan \frac {x_{A2}}{y_{A2}}$ and the desired angle from the ray $ CA_1$, forming the known angle $ \theta_1$ with the y-axis, is $ \xi \equal{} \theta_2 \minus{} \theta_1.$


Find all functions $ f: Q\to Q$:
$ f(f(x)\plus{}x\plus{}y)\equal{}x\plus{}f(x)\plus{}f(y)$

[quote="duongptnkvn"]Find all functions $ f: Q\to Q$:
$ f(f(x) \plus{} x \plus{} y) \equal{} x \plus{} f(x) \plus{} f(y)$[/quote]

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=271589&search_id=2104635645[/url]


Find the maximum value of n and it makes:
With a permutation of 1 to 100 randomly a1,a2,a3,...,a100, and it will exist a group of ten consecutive numbers aj,aj+,aj+2,aj+3,...,aj+9, the sum of these ten numbers won't be smaller than n. find the maximum value of n.

I programmed my computer and found that after 50,000 times, the largest value of [i]n[/i] was 429.  After another 50,000 times the largest value was 435.  (I suspect the maximum value of [i]n[/i] is about 440, but I could be wrong.)

Note that [i]n[/i] can never be greater than 500.

By the way, the title of this forum dosen't match its objective.

i am so sorry for it.
i gonna to alter it now.


inequalities
function
inequalities solved

I don't have a soln, and this is probably obvious to everybody  :D , but it's enough to prove it for n=2. Then we just use induction.

There are mistake in inequality (you can easy find contrary instance: bk=1, ak=epsilon \to 0).
Right inequality:
((a1+....+an)/(b1+....+bn))^(b1+....+bn) \geq (a1/b1)^b1*....*(an/bn)^bn (product)

Let f(x)=be^x-ax. f have exactly one minimum in point x=ln(a/b) and f_min=-b*ln(a/b)

Define fk(x)=bk*e^x-ak*x. It is clearly (for any set of function) that
f1_min+...+fn_min \leq (f1+..+fn)_min
in other words
b1*ln(a1/b1)+...+bn*ln(an/bn)\leq (b1+...+bn)*ln((a1+....+an)/(b1+....+bn))  - which required to be proved

Excuse me. There are small misprints in my proof.
Function f must be ae^x-bx-b and x_min=ln(b/a), f_min=b*ln(a/b)


[quote="pkerichang"]how do you make quotation marks right in $\LaTeX$??? when I type quotation marks, they both points at the same direction instead of pointing to the word inside.  How can I solve this problem???[/quote]

Include the package "quotes" in your document to get quotation marks that face the proper way.

One of them has to be ` and the other has to be '. $`\text {This is how you use quotation marks,'} \text { said chess64.}$

And if you want double quotation things, you have to use `` and ''. Like this: $``\text {This is how you use quotation marks,''} \text { said chess64.}$

edit: Or use zanttrang's way...(I've never heard about that before! :))


if $ A,B,C$ are the angles of a triangle then prove that
$ (1\minus{}cosA)(1\minus{}cosB)(1\minus{}cosC)>\equal{}cosAcosBcosC$

[quote="ishfaq420haque"]if $ A,B,C$ are the angles of a triangle then prove that
$ (1 \minus{} cosA)(1 \minus{} cosB)(1 \minus{} cosC) > \equal{} cosAcosBcosC$[/quote]
The ineq<=> $ (\frac{1\minus{}cosA}{cosA})(\frac{1\minus{}cosB}{cosB})(\frac{1\minus{}cosC}{cosC})$≥1
Replace cosA=$ \frac{1\minus{}x^2}{1\plus{}x^2}$ where x=tan(A/2) and cosB,cos C=....
The ineq<=>$ (\frac{2x}{1\minus{}x^2})(\frac{2y}{1\minus{}y^2})(\frac{2z}{1\minus{}z^2})$≥$ \frac{1}{xyz}$
<=>tanAtanBtanC≥cot(A/2)cot(B/2)cot(C/2)
<=>tanA+tanB+tanC≥cot(A/2)+cot(B/2)+cot(C/2)(easy to prove)


The probability for individual flips of coins is independent from the previous flips.  Simply put, the probability is still $ \boxed{\frac {1}{2}}$

[quote="GameBot"]Suppose you toss a coin and get heads five times in a row. What is the probability, expressed as a common fraction, that you will get heads on the sixth toss?[/quote]

the amount of times you toss the coin does not matter what so ever. So this question is asking what is the probability that you will get heads?

There are only two sides of the coin and one side is heads so the probability is

1/2


inequalities
geometry
3D geometry
inequalities proposed

Let $x,y,z,w\geq0$ with $xyz+yzw+zwx+wxy=4$. Prove that
\[
x^2+y^2+z^2+w^2\geq4.
\]

It follows from Muirhead.

Just square the condition and multiply by 4, and cube the inequality to be proven.  Then the result follows.

If you really want to see the work, the inequality is equivalent to:

$\sum_{sym} w^6+3\sum_{sym} w^4x^2+6\sum_{sym} w^2x^2y^2\geq 64$

and the condition is equivalent to:

$4\sum_{sym} w^2x^2y^2+6\sum_{sym} w^2x^2yz=64$

Since $(6,0,0,0)\succ (4,2,0,0)\succ (2,2,2,0)\succ (2,2,1,1)$, we're done.

Other solution,maybe simple:

Exist $x,y,z \ge 0$:
$\sum{abc} =4xyz$
$\sum{ab}=2\sum{xy}$
$3\sum{a}=4\sum{x}$
Then:
$xyz=1$
We need to prove:
$(a+b+c+d)^2-2(ab+bc+cd+da+ac+bd) \ge 4$
$<->(\frac{4}{3}(x+y+z))^2-4(xy+yz+zx) \ge 4$
$<->4(x+y+z)^2-9(xy+yz+zx) \ge 9$
$<->3(x^2+y^2+z^2)+(x^2+y^2+z^2-xy-yz-zx) \ge 9$
But $x^2+y^2+z^2 \ge 3$ (Because $xyz=1$)
Then we have done!


In a certain year, January had exactly four Tuesday and four Saturdays. On what day did January 1 fall that year?

A. Monday
B. Tuesday
C. Wednesday
D. Friday
E. Saturday

I guessed the answer to be C., and I got it right, but I don't know how to do it...

If there are four Tuesdays and four Saturdays, then the fourth Saturday comes after the fourth Tuesday (think about it.) Now, the latest possible date for the Saturday is January 31st, meaning the fourth Tuesday would be the 27th. That means the first Tuesday is the 6th, so January 1st is $ \colorbox{red}{\text{Wednesday}}$.

[quote="isabella2296"]Now, the latest possible date for the Saturday is January 31st. [/quote]
How do you know that?
EDIT: Heh, stupidity moment.

Because there's no date after January 31st.  :)

Hmm, so this is basically a "working backwards" problem?
And is it possible to eliminate Tuesday from the beginning?

[quote="isabella2296"]the latest possible date for the Saturday is January 31st, meaning the fourth Tuesday would be the 27th. That means the first Tuesday is the 6th, so January 1st is $ \colorbox{red}{\text{Wednesday}}$.[/quote]

Your answer is correct. But your argument is wrong.

Yes, the 1st is on Wednesday.
The 1st Tuesday/Saturday is a Saturday on Jan 4.
The 4th Saturday is Jan 25. The 4th Tuesday is after that on Jan 28.

Jan 31 is a Friday, not a Saturday.


inequalities
logarithms
inequalities unsolved

Let $ a_1,a_2,\cdots,a_n$ are positive real numbers. If $ n$ non-negative real numbers $ x_1,x_2,\cdots x_n$ have sum $ 1$. Prove the inequality:

\[ a_1x_1\plus{}a_2x_2\plus{}\cdots\plus{}a_nx_n \ge a_1^{x_1} a_2^{x_2} \cdots a_n^{x_n}\]

[quote="Obel1x"]Let $ a_1,a_2,\cdots,a_n$ are positive real numbers. If $ n$ non-negative real numbers $ x_1,x_2,\cdots x_n$ have sum $ 1$. Prove the inequality:
\[ a_1x_1 \plus{} a_2x_2 \plus{} \cdots \plus{} a_nx_n \ge a_1^{x_1} a_2^{x_2} \cdots a_n^{x_n}\]
[/quote]

Take a natural logarithm of both sides and you have this inequality:

$ ln(x_1a_1 \plus{} x_2a_2 \plus{} ... \plus{} x_na_n) \geq x_1*ln(a_1) \plus{} x_2*ln(a_2) \plus{} ... \plus{} x_n*ln(a_n)$

Since $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$ follows $ ln(x_1 \plus{} ... \plus{} x_n) \equal{} 0$ So we can subtract this from a LHS and divide RHS with $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n$. After applying rule $ ln(p) \minus{} ln(q) \equal{} ln(p/q)$ we get:

$ ln(\frac {x_1a_1 \plus{} x_2a_2 \plus{} ... \plus{} x_na_n}{x_1 \plus{} ... \plus{} x_n}) \geq \frac {x_1*ln(a_1) \plus{} x_2*ln(a_2) \plus{} ... \plus{} x_n*ln(a_n)}{x_1 \plus{} x_2 \plus{} ... \plus{} x_n}$
$ f(x) \equal{} ln(x)$ is concave so this holds by Jensen inequality.

Since $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} 1$ ,we can make the substitution $ x_i \equal{} \frac {t_i}{t_1 \plus{} t_2 \plus{} ... \plus{} t_n}$ .Let $ t_1 \plus{} t_2 \plus{} ... \plus{} t_n \equal{} T$ ,then Inequality becomes :
\[ a_1 t_1 \plus{} a_2 t_2 \plus{} ... \plus{} a_n t_n \geq T \sqrt [T]{a_1 ^{t_1} a_2 ^{t_2} ...a_n ^{t_n}}\]
which is obvious by AM-GM


[quote="risinglegend"]1st question: 2(1-3a)/3=3-2a/5 wat is the answer?[/quote]

[hide=" The Answer"]
Lets first simplify the top.

$2(1-3a) = 2-6a$

$3-2a = 3-2a$

Bottom could stay.

What you end up with is:

$\frac {2-6a} {3} = \frac {3-2a} {5}$

Then, you solve for a:

Get rid of the denominators first by multiplying both sides of the equal sign by the number of the denominator you are trying to get rid of:

$\frac {2-6a} {3} \times 3$ = $2-6a$

$\frac {3-2a} {5} \times 3$ = $\frac {9-6a} {5}$

Then, multiply by 5:

$10-30a = 9-6a$

Now, it's easy to solve for a: just isolate the variable and solve.

$-30a + 6a = 9 - 10$

$-24 a = -1$

$a = \frac {1} {24}$

[/hide]

Wait... we solved it differently.
Learn latex! WAs it $3-\frac{2a}{5}$ or $\frac{3-2a}{5}$?

[quote="risinglegend"]1st question: 2(1-3a)/3=3-2a/5 wat is the answer?[/quote]

What did you mean?

$3-\frac{2a}{5}$

or

$\frac{3-2a}{5}$?

In your first solution, look at the second and third lines, it's a little flawed I think.

In your second solution, you copied the problem wrong  :lol:


conics
hyperbola
symmetry
rotation
analytic geometry
Asymptote
AoPS Books

Yes, it is, because you can multiply by x on both sides to get xy=1, and so b :^2: -4ac is positive, thus indicating a hyperbola.

[quote="joml88"]Is y=1/x an hyperbola?  I would think not but I am not absolutely sure.  My reason for saying that it isn't is because it isn't in the x :^2: -y :^2: =1 form.  I ask this question because my teacher indirectly referred to it during an explanation of rational functions.[/quote]

Yes it is a hyperbola, but it isn't in that form. If you look at all the hyperbolas that form gives you, it only works when the symmetry line is vertical, if you want to say it like that. If you look at the graph of xy=1, its rotated around a bit.

In other words, x :^2: -y :^2: =1 gives some hyperbolas, but not all.

Do you have AoPS books?  A chapter called "That Pesky xy Term" will answer all your questions.  Its called a "rectuangular hyperbola" and it is rotated 90 degrees from the +x axis.

It's somewhat interesting to find the foci of the hyperbola xy=1.  At least, I thought it was...

Yes it is a hyperbola, but it isn't in that form. If you look at all the hyperbolas that form gives you, it only works when the symmetry line is vertical, if you want to say it like that. If you look at the graph of xy=1, its rotated around a bit.

In other words, x :^2: -y :^2: =1 gives some hyperbolas, but not all[/quote]

Thanks for the explanation.  In school they make it seem like x :^2: -y :^2: or y :^2: -x :^2: are the only ways to express a hyperbola.

you can rotate the axes to change it into a form like the one you mentioned. ill just give you the way to do it briefly. if you have

[tex]ax^2 + bxy + cy^2 + dx + ey + f=0.[/tex] take the p= arcot ((A-C)/B) - this is the angle you rotate by. then let x = x'cosp - y'sinp and y=x'sinp + y'cosp. plug into original and the xy term will be eliminated.[/tex]

Is the answer[hide] (2rad2, 2rad2)  and (-2rad2, -2rad2) [/hide]for the foci?

For the radical symbol, you can use ": s q r t :" but without the spaces.  Also, ": r t 2 :" will give you :rt2:.  I don't get quite the same answer as you do for the foci, though.  Perhaps you made an arithmetic mistake?

Okay, im not sure but is it[hide] (2,2) and (-2,-2) [/hide]for the foci?

I still get something not quite the same, although we're definitely in the same neighborhood -- you know, you can always check it by picking two random points on the hyperbola and seeing if they obey the definining law for hyperbolae.  Why not post your solution (not just your answer)?

[hide]OKay, what I did was transform it into a hyperbola w/ a horizontal transverse axis -- a hyperbola that I could apply the regular formulas to:

y^2/1 - x^2/1 = 1

c^2 = a^2 + b^2

c =  :rt2:

No matter how its translated, the distance will be the same.  So, now when you move it into its original setup (y=1/x) you make a right triangle to find the (x,y) of the foci ( :rt2: is the hypoteneuse).  So the coordinates are (1,1) and (-1,-1).

However, it also seems like this is wrong, b/c the vertices of y=1/x are located at its pts of intersection w/the line y=x.

xy =1

x^2 = 1

x = +/- 1

y = +/- 1

Now obviously the foci cant be at the same place as the vertices?!

[/hide]

So I have been wavering around a bunch of diff answers...

[quote]you know, you can always check it by picking two random points on the hyperbola and seeing if they obey the definining law for hyperbolae[/quote]

Yea but the foci are not on a hyperbola, so you cant use the discriminant thing b^2 - 4ac for the foci.  Am I misinterpreting you?

On another note, I am not at the "poincare conjecture" level! =)

The way a hyperbola is defined is that the difference between the lengths to each foci is constant. So you could pick any point on the hyperbola, calculate that difference, and repeat..

Question:  why do you think that when rotated, the equation it will have will be x^2 - y^2 = 1?

Yes, you were misinterpreting what I said.  The definition of a hyperbola is "the set of points for which the difference between the distances between the point and two fixed points is some constant."  In other words, if you have foci F1 and F2, a hyperbola is the set of points P such that | length[PF1] - length[PF2] | = c, where c is some number.  So, you can just check that some points on the hyperbola obey that relationship with respect to your suggested foci.

Okay well I picked a hyperbola that would be the same (ie have the same a, b, and c [u]distance[/u] values) y=1/x and y^2 - x^2 = 1 have these same distance values.  So I used the regular hyperbola values for y^2 - x^2 = 1 (ie I found c), then I knew that the c value for y^2 - x^2 = 1 is the same as the c value for y=1/x.  So then since I knew c, that made a right triangle.

[quote="JBL"]why do you think that when rotated, the equation it will have will be x^2 - y^2 = 1?
[/quote]

I said y^2-x^2=1 not x^2-y^2=1.

Well that is what it is, right?  There is a 90 degree angle between the two asymptotes for both.  Wait, am I digging myself into a hole?  Yikes, okay, the distance from the center (h,k) to the vertex is not the same ( :rt2: for y=1/x and 1 for y^2 - x^2 =1).  Okay, brb w/the equation.

EDIT: Okay, by cheating w/my precalc book I think it is:

y^2 / ( :rt2: )^2 - x^2/( :rt2: )^2

THerefore c= sqrt(a^2+b^2) = sqrt(4) = 2

Since the c is 2, then the coordinates are 2/ :rt2:  or  :rt2: .

Foci are ( :rt2: , :rt2: ) and (- :rt2: ,- :rt2: )

Hehe, be patient -- I'm not actually on this site [i]all[/i] the time ;)

Yes, that's right.  You forgot to check to make sure that your hyperbolas were the same size (not just the same shape), so you needed to expand everything by a factor of :rt2:, which gave you exactly what you needed :)

[quote]Hehe, be patient -- I'm not actually on this site all the time [/quote]

=)  Patience is a virtue...that I don't have! LOL =)  No seariously though I you are very helpful and I would never try to hassle you!

Thanks for the respone.  Yes, I know that a= :rt2:  because that is the distance from (0,0) to (1,1).  Now its obvious a=b, so thats where I got the equation.

Thanks!


What's the general concept behind Part A to this problem?  Is it just the heaviest molecule?

Equal volums of gases under the same conditions of temperature and pressure have the same number of particles (Avogadro's hypothesis). Therefore the chemical quantity is the same in each balloon. Since n = m/M, the balloon having the greatest mass of gas will be the containing the gas with the greatest molar mass.


search
limit
function
real analysis
real analysis unsolved

Let $ f: \mathbb{R}\to\mathbb{R}$ be continuous and satisfy $ |f(y)\minus{}f(x)|\ge k|y\minus{}x|$ for all $ x,y\in\mathbb{R}$ and some fixed $ k>0$. Prove that $ f$ is surjective.

The above link talks about more complicated stuff than the knowledge necessary to solve this problem.

Actually $ f$ is bijective not only surjective. If $ f(x) \equal{} f(y)$ then $ 0\geq k|x \minus{} y|$ so $ |x \minus{} y| \equal{} 0$ which means $ x \equal{} y$ (so $ f$ is injective). Since $ f$ is continous and injective it is obviously monotone. This means that there exists $ \lim_{y\to\infty}f(y) \equal{} l \in \overline{\mathbb R}$. By fixing $ x$ and letting $ y\to\infty$ in the relation we get that $ |l| \equal{} \plus{} \infty$, so $ l \equal{} \pm \infty$. Similarly $ \lim_{y\to \minus{} \infty} f(y) \equal{} \mp \infty$ so the function is also surjective.


Source :  2002 MOP Homework, Part I, Problem 2

In a certain language there are $n$ letters.  A sequence of letters is called a [i]word[/i] if and only if between any pair of identical letters, there is no other pair of equal letters.  Prove that there exists a word of maximum possible length, and find the number of words which have that length.

(I know this is from MOP homework, but it was in the "easy" section.  However, I am currently a rather poor judge of difficulty, and if this belongs in a different forum, please move it there.)

Seems a bit too hard for Intermediate. Off it goes to Pre-Olympiad.

[quote="seamusoboyle"][quote="paladin8"]I think the maximum length is $3n$. Suppose the length is $>3n$. Then there exist $4$ of some letter, but the inner two obviously form a pair between the outer two, contradiction. And we can easily construct one of length $3n$ by just taking blocks of $3$ of the same letter (e.g. $aaabbbccc \ldots$). I'm not quite sure how to count all the possible strings of that length though.[/quote]

im not sure about the inside bit. What about something like $0202$?[/quote]
0202 is allowed.  2002 is not allowed; nor is 210345678902.

[quote="seamusoboyle"][quote="paladin8"]I think the maximum length is $3n$. Suppose the length is $>3n$. Then there exist $4$ of some letter, but the inner two obviously form a pair between the outer two, contradiction. And we can easily construct one of length $3n$ by just taking blocks of $3$ of the same letter (e.g. $aaabbbccc \ldots$). I'm not quite sure how to count all the possible strings of that length though.[/quote]

im not sure about the inside bit. What about something like $0202$?[/quote]

Hmm? I don't quite understand your question... I'm saying that there exist FOUR of a single letter (for instance $a$). Then if you take out all the letters except $a$ it's easy to see that $aaaa$ does not work because the inner two are between the outer two.

[quote="seamusoboyle"]Has it not been solved? I agree with paladins solution[/quote]

We have to find how many strings have that many letters.

I don't have much time to think about this atm, being at school, but here's a start, I suppose.
The three $a$'s divide the word into 4 sections.  For any other letter, the three of that letter must be distributed separately into these sections.  So so it must apply for every letter.  Thus, each letter will not appear in one of the four sections.  And so on, doing this for all of the letters.  I'm not quite sure how to do the and so on part though...

Thinking about the three letters creating 4 sections is a good idea -- try building it up, first one letter, then two, and so on.  Once you've figured out the possible relations between three different letters, you're more or less home.

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=80089 ;)


For my science research, I need to fit a curve to a function that it fits best with...and although I could test out various non-linear regressions, I would prefer to use any software to speed up the process, since this is only one of several parts of what I have to do. So, does anyone know of any good software I could obtain?

Well... I think you could use [url=http://www.pacifict.com/]NuCalc 3.0[/url]

[quote="OVB"]For my science research, I need to fit a curve to a function that it fits best with...[/quote]
CurvFit, [url]http://www.digitalCalculus.com/demo/curvfit.html[/url], has several functions to play with; e.g. Lorentzian, sine, damped sine, etc. series. The Lorentzian series seems to work the best for most real industrial problems.

[quote=OptimalDesigns][quote="OVB"]For my science research, I need to fit a curve to a function that it fits best with...[/quote]
CurvFit, [url]http://fortranCalculus.info/apps/curvfit.html[/url], has several functions to play with; e.g. Lorentzian, sine, damped sine, etc. series. The Lorentzian series seems to work the best for most real industrial problems.[/quote]


calculus
integration
probability
calculus computations

[quote="kunny"]Let $p$ be a constant number such that $0<p<1$.
Evaluate

\[\sum_{k=0}^{2004} \frac{p^k (1-p)^{2004-k}}{\displaystyle \int_0^1 x^k (1-x)^{2004-k} dx}\][/quote]

To solve this I use information regarding the Beta distribution and the binomial distribution.

Firstly, the beta distribution is defined by:

\[f(x) = \frac{\Gamma (\alpha + \beta) x^{\alpha - 1} (1-x)^{\beta - 1}}{\Gamma(\alpha) \Gamma(\beta)}\]
over $x \in (0,1) $.

Note that if$ \alpha , \beta$ are integers then $ \Gamma(\alpha) = (\alpha - 1)! $, for example.

hence the integral in the denominator is \[ \frac{\Gamma (k+1) \Gamma(2004-k+1)}{\Gamma(2006)} \].

At this point we see that the sum is just

\[ 2005 \sum_{k=0}^{2004} \frac{(2004)!}{k! (2004-k)!} p^{2004} (1-p)^{2004 - k} \]

The final step uses the property of a probability distribution for the binomial distribution whereby the sum is just 1 and the final answer is  $2005$.

Yes, your answer is correct,skanderbeg.

kunny


algebra
polynomial
Vieta
Putnam
Diophantine equation
number theory proposed
number theory

[b][color=red]Prove that there are infinitely many pentagonal numbers that can be written as a sum of two perfect squares.[/color][/b]

[i]Remark.[/i] The [i]pentagonal numbers[/i] mean the numbers $p_n$ defined by $p_n = \frac{n\left(3n-1\right)}{2}$ for all $n\in\mathbb N$. See http://mathworld.wolfram.com/PentagonalNumber.html for their properties.

Hope you all like this proposal of mine...

I look forward to reading your replies.  :D

there are infinitely many pentagonal numbers that can be a squre (use Pell equation) so it can be represent as $ 0^2\plus{}x^2$
also easy to prove there are infinitely many pentagonal numbers that can be written as $ 1\plus{}x^2$

[quote="hxy09"] use Pell equation... [/quote]

There's a more elementary solution, esteemed [b]hxy09[/b].  :wink: Would you like me to post it?

Regards,

[quote="coquitao"][quote="hxy09"] use Pell equation... [/quote]

There's a more elementary approach, esteemed [b]hxy09[/b].  :wink: Would you like me to post my [i]official[/i] solution?

Regards,[/quote]
Of course :)

Every pentagonal number can be written as sum of a perfect square and a triangular number:

$ \frac {n(3n \minus{} 1)}{2} \equal{} n^{2} \plus{} \frac {n(n \minus{} 1)}{2}.$

The result follows now from the well-fact that there are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1426331#p1426331][b]infinitely[/b][/url] many square triangular numbers.

Regards,

[quote="coquitao"]Every pentagonal number can be written as sum of a perfect square and a triangular number:

$ \frac {n(3n \minus{} 1)}{2} \equal{} n^{2} \plus{} \frac {n(n \minus{} 1)}{2}.$

The result follows now from the well-fact that there are [url=http://www.mathlinks.ro/viewtopic.php?t=262500][b]infinitely[/b][/url] many square triangular numbers.

Regards,[/quote]
But that's not more elementary, since the proof that there are inifinitly many triangular numbers uses pell's equations..  :roll:

Take a look at the link I set [i]under[/i] the word [b][color=blue]infinitely[/color][/b], man.  :wink:

Showing the infinitude of  a special given Pell equation is very elementary as you don't need to show that a first starting solution exists (it exists by giving one). The Pell way show a more convenient and helpful (!) way. Additionally, coquitao's proof for triangular numbers is at no point simpler than one by Pell.

To the problem itself, I don't like it: much too easy and no teaching for the solver.

[quote="ZetaX"] To the problem itself, I don't like it: [b]much too easy and no teaching for the solver[/b].[/quote]

[b]Let us test this asseveration of yours, esteemed [color=red]ZetaX[/color]:[/b]  :wink:

An integer is of ZetaX-type if each of its prime factors occurs to the second power (at least). Prove that there are infinitely many pairs of consecutive integers of ZetaX-type.

[b]Wow![/b]  Isn't that another application of the ideas used in the [b]elementary[/b] proof of the infinitude of square triangular numbers?

Regards,

Proofs that are elementary at the expense of losing their generality aren't proofs that help you figure out what's really going on.  Does that technique generalize obviously when "second power" is replaced by "third power" or when "two consecutive integers" is replaced by "three consecutive integers"?  Or does it relate to a proof that the problem statement is false in either of these cases?

Certainly the trick of generating new solutions from old solutions can be found in several problems of this type (for example, any problem that involves Vieta jumping), but in the Pell case the full explanation is so much deeper it must be pursued.

Well, I have less objection for the triangular number version. The "problem" here is that considering the "more elementary" (I would be interested to see a definition of that) doesn't help: it just uses a special appliction of that Pell-idea, and as a result gives a much weaker result.

By the way, a related one: http://www.mathlinks.ro/viewtopic.php?t=150428

Thanks a lot for leaving your comments, guys.

To [b]ZetaX[/b]:

-> Do you know of any other Putnam problem where this technique is applied?

To [b]t0rajir0u[/b]:

-> Can you point me to specific examples of similar problems whose solution involves the use of Vieta jumping?

Best wishes to you all...

P.S.  I have another problem proposal of this kind for you guys... Hope you all like it:

Prove that there infinitely many triads $ (n,n \plus{} 1,n \plus{} 2)$ of consecutive natural numbers that can be written as a sum of two squares.

So long!  :wink:

Searching for Vieta Jumping on the forum should reveal enough problems.

[quote="coquitao"]Prove that there infinitely many triads $ (n,n \plus{} 1,n \plus{} 2)$ of consecutive natural numbers that can be written as a sum of two squares.[/quote]
Just the same operation as before.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=116256]Putnam 2000 A2[/url]. Several different constructions are included.

Was that [i]precious[/i] an actual [b]Putnam[/b] problem then? :o

I have concocted my own solution:

If $ (n \minus{} 1, n, n \plus{} 1)$ is a [i]good[/i] triad then $ (n^{2} \minus{} 1 \equal{} (n \minus{} 1)(n \plus{} 1), n^{2}, n^{2} \plus{} 1)$ is also a [i]good[/i] triad. We need to remember here that the product of two numbers that can be written as a sum of two perfect squares is equal to another such number. Voilà!

By the way, our friend [b]ZetaX[/b] has actually proved something more:

If there are infinitely many pairs of consecutive integers that can be written as a sum of two perfect squares then there are infinitely many triads of integers that satisfy the same conditions.

Thanks a lot for leaving your comment, esteemed [b]jmerry[/b].


Let $ a,b,c$ be nonnegative real numbers with sum $ 3$. Prove that
$ \frac{a}{b\plus{}\frac{8}{9}c^3}\plus{}\frac{b}{c\plus{}\frac{8}{9}a^3}\plus{}\frac{c}{a\plus{}\frac{8}{9}b^3} \ge \frac{3}{2}.$
The equality holds if and only if $ a\equal{}b\equal{}\frac{3}{2},c\equal{}0$ and its cyclic permutations. :)

Let $ a \equal{} \max{a,b,c}\geq 1$, if $ a\geq\frac {3}{2}$ then $ f(a,b,c)\geq f(a,b \plus{} c,0)\geq \frac {3}{2}$
The other case is more difficult I'll think it later.


advanced fields
advanced fields unsolved

Is the boundary of the Mandelbrot set connected? Is it measurable?

Wikipedia gives

[quote="Wikipedia"]The Hausdorff dimension of the boundary of the Mandelbrot set equals 2 by a result of Mitsuhiro Shishikura.[4] It is not known whether the boundary of the Mandelbrot set has positive planar Lebesgue measure.[/quote]

And MathWorld concurs with

[quote="MathWorld"]Shishikura (1994) proved that the boundary of the Mandelbrot set is a fractal with Hausdorff dimension 2, refuting the conclusion of Elenbogen and Kaeding (1989) that it is not. However, it is not yet known if the Mandelbrot set is pathwise-connected.[/quote]

But it should be measurable. As a rule of thumb, anything defined by a limit of reasonable things should be measurable. The open question is whether the measure is zero or positive.


Hi.

Does anyone know about how to automatically forward mail from my hotmail account to my new address (for free of course), and preferably not using outlook?

yes, I know it's impossible :)

They'd have to be stupid, taking extra traffic over their servers for someone who has his account on a concurrent server :D

perhaps they have a payfunction for that, but certainly not free!

This thread continues to be erroneously marked as unread as I browse the forum. There is still a cookie management problem on the site.

I don't know what your question is! Do you mean, "How can you place all the messages from your old e-mail account to your new one?" I guess, just copy and paste every single e-mail and send it to yourself. :D Or you could also forward it but when you do then the text gets messed up and symbols interfere in the text.


calculus
integration
derivative
trigonometry
calculus computations

Suppose that $ f(x),\ g(x)$ are differential fuctions and their derivatives are continuous.
Find $ f(x),\ g(x)$ such that $ f(x)\equal{}\frac 12\minus{}\int_0^x \{f'(t)\plus{}g(t)\}\ dt\ \ g(x)\equal{}\sin x\minus{}\int_0^{\pi} \{f(t)\minus{}g'(t)\}\ dt$.

Elementary....

$ f(x)\equal{}\frac{1}{2}\cos x$

$ g(x)\equal{}\sin x$ :rotfl:

$ g(x) \equal{} \sin x \plus{} \int_{t \equal{} 0}^\pi f(t) \minus{} g'(t) \, dt \equal{} \sin x \plus{} F(\pi) \plus{} g(\pi) \minus{} F(0) \minus{} g(0)$, where $ F' \equal{} f$.

Letting $ x \equal{} \pi$ we obtain

$ g(\pi) \equal{} \sin \pi \plus{} F(\pi) \plus{} g(\pi) \minus{} F(0) \minus{} g(0)$,

and hence

$ F(\pi) \minus{} F(0) \minus{} g(0) \equal{} 0$.

Therefore the second condition is satisfied by

$ g(x) \equal{} \sin x \plus{} g(\pi)$,

where $ g(\pi)$ is any constant we choose.  Now, taking the derivative of the first condition with respect to x gives

$ f'(x) \equal{} \minus{} f'(x) \minus{} g(x)$,

or equivalently, $ g(x) \equal{} \minus{} 2f'(x)$.  So we must have

$ f(x) \equal{} \frac {1}{2} (\cos x \minus{} g(\pi) x \plus{} C)$

for some undetermined constant of integration C.  Since $ f(0) \equal{} \frac{1}{2}$, we conclude that $ C \equal{} 0$.  Now substituting this back into the second condition,

$ g(x) \equal{} \sin x \minus{} \int_{t \equal{} 0}^\pi \frac {\cos x \minus{} g(\pi) x}{2} \, dt \plus{} g(\pi) \minus{} g(0)$

$ \equal{} \sin x \minus{} \frac {1}{2} \left[\sin x \minus{} g(\pi) \frac {x^2}{2}\right]_{t \equal{} 0}^\pi \plus{} g(\pi) \minus{} (\sin 0 \plus{} g(\pi))$

$ \equal{} \sin x \plus{} \frac {\pi^2 g(\pi)}{4}$.

But since we require this to also equal $ \sin x \plus{} g(\pi)$, we conclude $ g(\pi) \equal{} \frac {\pi^2 g(\pi)}{4}$, or $ g(\pi) \equal{} 0$.  Therefore the unique solution is

$ f(x) \equal{} \frac {\cos x}{2}$, and $ g(x) \equal{} \sin x$.


LaTeX
number theory solved
number theory

Let {m_1, m_2,...} be a (finite or infinite) set of positive integers. Consider the system of congruencies (1) x = 2*m^2_i (mod 2*m_i - 1) for all i = 1,2,... Give a necessary and sufficient condition for the system (1) to be solvable.

[i]Mikls Schweitzer competition 1951 problem 9[/i]

I read (m_i) 2 , am I right?

In that case, the system is solvable if and only if the set E = {m_1, m_2,...} is finite.
It is easy to see that, since 2m_i -1 is odd and that 4(m_i) 2  = (2m_i + 1)(2m_i - 1) + 1, the equation x = 2(m_i) 2  mod[2m_i - 1] is equivalent to 2x = 1 mod[2m_i - 1].

Now suppose that x is a solution. Then, 2x-1 is odd and divisible by each of the element of E, thus E has to be finite.
Conversely, if E is finite, say E = {m_1,...,m_n}. Let y = (2m_1 - 1)(2m_2 - 1)...(2m_n - 1).
Then y is odd, thus x = (y+1)/2 is an integer which is a solution of the system.

Pierre.

Well done, Pierre. About the syntax actually I tried to keep to the LaTex syntax save the fact that I omitted the {}-parenthesis.


geometry
3D geometry
tetrahedron
perimeter

The four triangular faces of a tetrahedron all have the same perimeter.  Prove that they are congruent.

[hide="Solution"]Suppose the sides of the tetrahedron are $a,b,c,d,e,f$.

Draw the tetrahedron such that faces have sides of length:
$a,b,c$
$b,d,f$
$c,e,d$
$a,e,f$

Since all faces have same perimeter, we have:
$a+b+c=p$ $(1)$
$b+d+f=p$ $(2)$
$c+e+d=p$ $(3)$
$a+e+f=p$ $(4)$

Adding all the equations up and halving both sides, we have:
$a+b+c+d+e+f=2p$  $(5)$

Now subtracting each of $(1)$,$(2)$,$(3)$,$(4)$ from $(5)$, we have:

$d+e+f=p$ $(6)$
$a+c+e=p$ $(7)$
$a+b+f=p$ $(8)$
$b+c+d=p$ $(9)$

$(6)-(4) \Rightarrow a=d$
$(7)-(1) \Rightarrow b=e$
$(9)-(2) \Rightarrow c=f$

So our faces have sides of length:
$a,b,c$
$b,d,f$ or $b,a,c$
$c,e,d$ or $c,b,a$
$a,e,f$ or $a,b,c$

Thus, all triangles are congruent. $\blacksquare$[/hide]

darn i was just about to post my solution to this one..it was similar to it..but i think urs is nicer..well done..


find
$\lim_{x \to 0} \frac{sinx-x}{x^2}$

...this is obvious with l'hopital...but calculus does not really belong here


real analysis
function
group theory
abstract algebra
Divisor Functions

The following proof may be folklore, however I thought it is nice to be mentioned here just in case there are users who don't have knowledge of this idea.
Let's consider the set $ S \equal{} \left\{ \frac{k}{n} : k \in \overline{1, n} \right\}$. Simplifying all the fractions we conclude that $ S$ coincides with the union of the disjoint sets $ S_d \equal{} \left\{ \frac{k}{d} : k \in \overline{1, d}, (k, d) \equal{} 1 \right\}$, defined for all $ d | n$. Expressing the number of elements of $ S$ in two ways, by its definition and as the sum of the number of elements of all the $ S_d$'s yields exactly the desired equality, since $ |S_d| \equal{} \varphi(d)$.

Another classic proof is by considering the degrees of the terms in Dedekind's formula:
\[ x^{n} \minus{} 1 \equal{} \prod_{d| n}{\phi_{d}(x)},
\]
where $ \phi_{k}(x) \equal{} \prod_{\zeta \in P_{k}}(x \minus{} \zeta)$ and $ P_{k}$ is the set of the primitive roots of $ z^{k} \equal{} 1$.

[quote="pohoatza"]Another classic proof is done by considering the degrees of the terms in Dedekind's formula:
$ x^n \minus{} 1 \equal{} \prod_{d|n} \Phi_{d}(x)$,
where $ \Phi_{k}(x) \equal{} \prod_{\zeta \in P_k}(x \minus{} \zeta)$ is the $ k$-th cyclotomic polynomial (with $ P_k$ representing the set of the $ k$-th primitive roots of unity, $ |P_k| \equal{} \varphi(k)$).[/quote]
That's true, however the idea for proving your formula is actually the same - we can write
$ x^n \minus{} 1 \equal{} \prod_{k\equal{}1}^n (x \minus{} e^{2\pi ik/n}) \equal{} \prod_{d|n} \ \prod_{k \in \overline{1, d}, (k, d) \equal{} 1} (x \minus{} e^{2\pi ik/d}) \equal{} \prod_{d|n} \Phi_d (x)$,
because each fraction $ \frac{2\pi ik}{n}$ is simplified uniquely as $ \frac{2\pi ik'}{d}$, where $ k', d$ are coprime and $ d$ is a divisor of $ n$.

The ideas in Tiks' proof can be generalized using the concept of [url=http://en.wikipedia.org/wiki/Dirichlet_convolution]Dirichlet convolution[/url].  In particular, we are proving that $ n \equal{} \varphi(n) * 1$.

It has to be noted that there is a quick proof of this identity that relies solely on results related to properties of the finite cyclic groups. I suppose that proof has already been posted in the forum, but I can't find an explicit reference now. Can any of you guys throw me a bone here?

I wonder why nobody mentions the following proof, based on Mobius' inversion formula.

For $ f$ a multiplicative arithmetic function, one has $ \sum_{d \mid n} \mu(d)f(d) \equal{} \prod_{\textrm{prime } p \mid n} (1\minus{}f(p))$. For $ f(x) \equal{} \frac {1} {x}$ therefore $ \sum_{d \mid n} \frac {\mu(d)} {d} \equal{} \prod_{\textrm{prime } p \mid n} (1\minus{}\frac {1} {p}) \equal{} \frac {\phi(n)} {n}$.
So $ \sum_{d \mid n} \mu(d) \frac {n} {d} \equal{} \phi(n)$, and now the inversion formula yields $ n \equal{} \sum_{d \mid n} \phi(d)$.

Let $G=\mathbb{Z}/ n\mathbb{Z}$ be the cyclic group of order $n$.
Since all of its subgroups are of the form $k \mathbb{Z} /n \mathbb{Z}$ with $k |n$ we know that $\forall$ $d|n$ there is a unique subgroup $H \leq G$ with $|H|=d$;
If $A_d=\{x \in G : \text{ord}(x)=d \}$, then from the above observation and the fact that $ord(x^s)=\frac{ord(x)}{(ord(x),s)}$, we deduce that $|A_d|=\varphi(d)$.Since $G$ is the disjoint union of all the sets $A_d$, the conclusion follows.

[quote="mavropnevma"]
For $ f$ a multiplicative arithmetic function, one has $ \sum_{d \mid n} \mu(d)f(d) \equal{} \prod_{\textrm{prime } p \mid n} (1\minus{}f(p))$. [/quote]
Nice proof!
For clarity, I would like to point out that the above holds only for [url=http://mathworld.wolfram.com/CompletelyMultiplicativeFunction.html]completely multiplicative functions[/url], not all [url=http://mathworld.wolfram.com/MultiplicativeFunction.html]multiplicative functions[/url].

[hide=J2]Let us define a subset of $S=${$1,2,3,...n$} as $T(d)$. $T(d)$ contains all elements $k \in S$, such that $gcd(k,n)=d$. If $k$ is such a number then $k=dm$ for some $m$ relatively prime to $\frac{n}{d}$. There are $\phi(\frac{n}{d})$ such m in the range $[1,n]$ by the definition of $\phi$. Thus the number of elements of $T(d)$ is also $\phi(\frac{n}{d})$ because this is a one to one relation. Since each subset of the form $T(s)$ is disjoint for different values we can add them all up to get the total set $S$. Thus the cardinality of each of the disjoint sets add up to n. We know that the cardinality of each of the disjoint sets is $\sum_{d|n}\phi(\frac{n}{d})=\sum_{d|n}\phi(d)$ which is supposed to equal to total cardinality $n$. Thus we are done.[/hide]

[quote=Tiks][quote="Peter"]Show that for all $n \in \mathbb{N}$, \[n = \sum^{}_{d \vert n}\phi(d).\][/quote]

The problem is obvious in case when $n=1$.

Let's suppose that we have proved the desired identity for all $s<n$. We have that $n=p^{t}m$ for a positive integer $m$ and a prime $p$ so that $gcd(m;p)=1$.
We know that $\phi(mn)=\phi(m)\phi(n)$ when $gcd(m;n)=1$.
=>
\[m=\sum^{}_{d|m}\phi(m)\]
=>
\[n = \sum^{}_{d \vert n}\phi(d)=\sum^{}_{d|m}\phi(m)+\sum^{t}_{i=1}\sum^{}_{d|m}{\phi(dp^{i})}=\sum^{}_{d|m}\phi(m)+\sum^{}_{d|m}\phi(m)(\sum^{t}_{i=1}\phi(p^{t}))=\sum^{}_{d|m}\phi(m)(1+(p-1)(\sum^{t-1}_{i=1}p^{i}))=\sum^{}_{d|m}\phi(m)(p^{t})=p^{t}m\]

We are done! :razz:[/quote]

Very nice soloution


Παιδιά  ισχύει ότι για  α, β, γ >0  είναι 1/(α+β)^2 + 1/(α+γ)^2  >= 1/( α^2 + βγ)  ???

Μόνο αυτό.  Αν ισχύει μη δώσετε την απόδειξη φανερά,   αφήστε με να το προσπαθήσω λίγο...

Ισχύει
 :D

[hide]Αποδεικνύεται με Cauchy -Swartz ή με ισοδυναμίες . Είναι ισοδύναμη με

$ a^4\minus{}2a^2bc\plus{}bc^3\plus{}cb^3\minus{}b^2c^2\geq 0$ που ισχύει αφού $ a^4\plus{}b^2c^2\geq 2a^2bc$ και $ bc^3\plus{}cb^3\geq 2b^2c^2$

Προσθέτοντας ... :) [/hide]

Σιλουανέ είσαι απίστευτος !!!  :)   Πότε πρόλαβες να απαντήσεις ....  Τέλος πάντων. Εσύ ξέρεις  :rotfl:

Θα την προσπαθήσω πριν δώ τη λύση.  Thanks

Όλοι με κατηγορούν γιατί απαντάω γρήγορα τι θα γίνει πλέον με αυτό  :P  :mad:

[hide]Πάντως η λύση με το Cauchy Swartz αξίζει πραγματικά ... Μπορείτε να τη βρείτε ?  :wink: [/hide]

Σιλουανέ θα μας πεις plz τη λύση με Cauchy-Swartz???

Με μια ματιά, λογικά είναι:

$ \left[\left(\frac {1}{a \plus{} b}\right)^2 \plus{} \left(\frac {1}{a \plus{} c}\right)^2\right]\cdot\left[\left(\frac {1}{2}\right)^2 \plus{} \left(\frac {1}{2}\right)^2\right]\geq \frac {1}{2(a^2 \plus{} bc)}$

Ε, μετά λογικά με πράξεις βγαίνει  :P

μαλλον δεν βγαινει  :huh:

Εγώ το προσπάθησα με τις βασικές ανισότητες 2(α^2+β^2) >= (α+β)^2  και  2(α^2+γ^2)>= (α+γ)^2  αντιστρέφοντάς τες και προσθέτοντας κατά μέλη αλλά δεν μου βγήκε τίποτα ...

Ούτε από ΑΜ - GM έβγαλα κάτι ...  Δεν ξέρω. Ενώ φαίνεται απλή με έχει δυσκολέψει πολύ... Κάτι μου διαφεύγει σίγουρα . Θα το ξανα κοιτάξω ... :read:  και αυτό που προτείνεις εσύ mostel  :(

Δεν είναι και ο πιο εύκολος Cauchy Swartz ομολογουμένως αλλά έχει τη λογική του  :)

Here it goes ... Είναι $ (b\plus{}c)(a^2\plus{}bc)\geq (a\sqrt{b}\plus{}c\sqrt{b})^2\equal{}b(a\plus{}c)^2$  άρα

$ \frac{1}{(a\plus{}c)^2}\geq \frac{b}{b\plus{}c}\cdot \frac{1}{a^2\plus{}bc}$  (1)

Με εντελώς ανάλογο τρόπο βγάζουμε ότι $ \frac{1}{(a\plus{}b)^2}\geq \frac{c}{b\plus{}c}\cdot \frac{1}{a^2\plus{}bc}$ (2)

(1)+(2) => το ζητούμενο  :D

Hope you liked it  :P

Τρομερή λύση. Η πιο δυνατή που έχω δει σε άσκηση εδώ και καιρό.

Οντως ωραια λυση!  :omighty:


geometry
circumcircle
trigonometry
trapezoid
analytic geometry
angle bisector

Let $ABC$ be an acute triangle. Circle $\omega_{BC}$ has segment $BC$ as its diameter. Circle $\omega_{A}$ is tangent to lines $AB$ and $AC$ and is tangent externally to $\omega_{BC}$ at $A_{1}.$ Points $B_{1}$ and $C_{1}$ are defined analogously.
Prove that lines $AA_{1},\,BB_{1},$ and $CC_{1}$ are concurrent.

[quote="April"][color=darkred]Let $ABC$ be an acute triangle. The circle $\Omega_{a}$ has the segment $[BC]$ as its diameter. The circle $\omega_{a}$ is tangent to lines $AB$ and $AC$ and is tangent externally to the circle $\Omega_{a}$ at the point $T_{a}.$ The points $T_{b}$ and $T_{c}$ are defined analogously. Prove that lines $AT_{a},\,BT_{b}$ and $CT_{c}$ are concurrently.[/color][/quote]
[color=darkred][b][u]Lemma I.[/u][/b] Let $ABC$ be an acute triangle with the circumradius $R$. The circle $w$ is tangent to the lines $[AB]$ , $[AC]$ and is externally tangent

to the circle with the diameter $[BC]$ in the point $T$. Suppose $b>c$ and denote the intersection $X\in BC\cap AT$. Prove that $\boxed{\frac{XB}{XC}=\frac{c(b+2R)}{b(c+2R)}}$.[/color]

[color=darkblue][b]Proof.[/b] Denote the area $S=[ABC]$, the circle $w=C(U)$ and the point $D\in AU\cap BC$. Apply the Menelaus' theorem to the transversal $\overline{ATX}$ for $\triangle UDM$ :

$\frac{AU}{AD}\cdot\frac{XD}{XM}\cdot\frac{TM}{TU}=1$ $\implies$ $\frac{XD}{XM}=\frac{AD}{TM}\cdot\frac{UT}{UA}=$ $\frac{2\cdot AD}{a}\cdot \sin\frac{A}{2}=$ $\frac{4bc\cos\frac{A}{2}}{a(b+c)}\cdot\sin\frac{A}{2}=$ $\frac{2bc\sin A}{a(b+c)}=$ $\frac{bc}{R(b+c)}=$ $\frac{4S}{a(b+c)}$ $\implies$ $\boxed{\frac{XD}{XM}=\frac{4S}{a(b+c)}}$ $\implies$

$\frac{XD}{4S}=\frac{XM}{a(b+c)}=\frac{a(b-c)}{2[4S+a(b+c)](b+c)}$ $\implies$ $XM=\frac{a^{2}(b-c)}{2[4S+a(b+c)]}$ $\implies$ $\frac{XB}{XC}=\frac{\frac{a}{2}-XM}{\frac{a}{2}+XM}$ $\implies$ $\frac{XB}{XC}=\frac{2S+ac}{2S+ab}$ $\implies$ $\boxed{\frac{XB}{XC}=\frac{c(b+2R)}{b(c+2R)}}$.

[b]Remark.[/b] $\boxed{\frac{XD}{XM}<1}$. Indeed, $XD<XM$ $\Longleftrightarrow$ $4S<a(b+c)$ $\Longleftrightarrow$ $ab\sin C+ac\sin B<ab+ac$ $\Longleftrightarrow$ $ab(1-\sin C)+ac(1-\sin B)>0$, what is truly.[/color]

April: Nice problem. What group was this problem from?

Virgil: Nice proof. Just so it is clear: $U$ is the center of $\omega_{a}$. Also $\frac{UT_{A}}{UA}=\frac{r_{\omega_{a}}}{UA}=\sin\frac{A}{2}$ since $U$ lies on the angle bisector of $\angle A$. [It took me a while to see this ratio.]

[quote="April"][color=indigo][b]It's a problem for Black Group, Altheman! [/b][/color][/quote]

where can i get the problems of other group....and what's about the solution? :?:
and i have some problems of previous MOSPs, can i get all?if then where?

I don't know, Moonmathpi496. I get these problems at the site: http://www.unl.edu/amc/
And can you give me (or post) some problems of previous MOSP? Thank you.

[quote="nsato"]I have another approach.  As a hint, here's the diagram.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/5dae2eb29df0bde91389bdeaf4ec9513.png[/img][/quote]

Can you express it?Thank you!

Denote $(K) \equiv \omega_{A}.$ $X,Y$ are the orthogonal projections of the midpoint $M$ of $BC$ on $AB,AC,$ $P,Q$ are the tangency points of $(K)$ with $AB,AC$  and $B',C'$ are the orthogonal projections of $A_1$ on $AB,AC,$ respectively. In the trapezoid $XMKP,$ we have

$A_1B' \equal{} \frac {XM \cdot KA_1 \plus{} KP \cdot MA_1}{MK} \equal{} \frac {KA_1 \cdot(XM \plus{} MA_1)}{MK} \equal{} \frac {KA_1 \cdot (a \plus{} h_c)}{2MK}.$

Analogously, in the trapezoid $YMKQ,$ we have $A_1C' \equal{} \frac {KA_1 \cdot(a \plus{} h_b)}{2MK}$

$\Longrightarrow \frac {A_1B'}{A_1C'} \equal{} \frac {a \plus{} h_c}{a \plus{} h_b} \equal{} \frac {1 \plus{} \sin B}{1 \plus{} \sin C } \equal{} \frac {\sin \widehat{ BAA_1}}{\sin \widehat{CAA_1}}.$

By cyclic exchange we get the relations

$\frac {\sin\widehat{CBB_1}}{\sin \widehat{ ABB_1}} \equal{} \frac {1 \plus{} \sin C}{1 \plus{} \sin A} \ , \ \frac {\sin \widehat{ ACC_1}}{\sin \widehat{BCC_1}} \equal{} \frac {1 \plus{} \sin A}{1 \plus{} \sin B}$

$\frac {\sin \widehat{ BAA_1}}{\sin \widehat{ CAA_1}} \cdot \frac {\sin\widehat{CBB_1}}{\sin\widehat{ ABB_1}} \cdot \frac {\sin\widehat{ ACC_1}}{\sin\widehat{ BCC_1}} \equal{} \frac {(1 \plus{} \sin B)}{(1 \plus{} \sin C)} \cdot \frac {(1 \plus{} \sin C )}{(1 \plus{} \sin A)} \cdot \frac {(1 \plus{} \sin A)}{(1 \plus{} \sin B)} \equal{} 1.$

Therefore, by Trig Ceva's theorem, the lines $AA_1,BB_1,CC_1$  concur.

In addition, the concurrency point is a known Kimberling center.

Let $D \equiv AA_1 \cap  BC.$ Then $\frac {BD}{CD} \equal{} \frac {c \cdot \sin \widehat{BAA_1}}{b \cdot \sin \widehat{CAA_1}} \equal{} \frac {c(1 \plus{} \sin B)}{b(1 \plus{} \sin C)}.$

Now, using barycentric coordinates with respect to $\triangle ABC,$ we obtain

$D \equiv \left (0: \frac {b}{1 \plus{} \sin B}: \frac {c}{1 \plus{} \sin C} \right ) \equiv  \left (0: \frac {1}{ac \plus{} S}: \frac {1}{ab \plus{} S} \right)$

By cyclic exchange, the lines $AA_1,BB_1,CC_1$ concur at

$X_{1123} \left (\frac {1}{bc \plus{} S}: \frac {1}{ac \plus{} S}: \frac {1}{ab \plus{} S} \right) ,$ the Paasche point of $ \triangle ABC.$


How many different positive 2-digit numbers are divisible by 5 or divisible by 3, but not divisible by
both (3 and 5)?

a.42 b. 36 c. 30 d. 33 e. 39

[hide=" click to reveal hidden content"]

$ 19$ positive two-digit integers, $ 5, 10, 15, ..., 95$, are divisible by $ 5$.

$ 33$ positive two-digit integers, $ 3, 6, 9, ..., 99$, are divisible by $ 3$.

Of these, $ 6$ positive two-digit numbers, $ 15, 30, 45, 60, 75, 90$, are divisible by $ 15$.

So, I make the answer $ 19\plus{}33\minus{}6 \equal{} 46$, which is not one of the options.

[/hide]

To get rid of numbers divisible by 3 and 5, you just need to get rid of the numbers divisible by 15 (and there are 6 of those which are two-digit).  There are 18 two-digit numbers divisible by 5 and 30 divisible by 3.

@ AndrewTom Since when are $ 3$, $ 5$, $ 6$ and $ 9$ postive two-digit integers? :wink: Subtracting these out will get you the answer of $ 42$ or A.


induction
number theory
strong induction

Having some issues with a couple number theory problems:

1) Prove that no integer in the sequence 11, 111, 1111, 11111,... is a perfect square.

2) Assuming that for every n>1 there is a prime number between n and 2n, prove that every positive integer can be written as the sum of distinct primes. (For this problem 1 is treated as a prime)

[hide="Solution #1"]
It is known that a square mod 4 always equals 0 or 1. It is also known that n*100 for all positive integers n is divisible by 4. Thus, we need only consider the 11, which is 3 mod 4. It follows that all other numbers with only 1s are also 3 mod 4, which does not satisfy the requirement for a perfect square.
[/hide]

[hide="Question for #2"]
What do you mean every positive integer? (There's an infinite amount of them) Are you talking about those between n and 2n?
[/hide]

[hide="Solution for #2"]
Strong induction argument. Base cases:

$ n\equal{}1$: obvious (it is just 1).
$ n\equal{}2$: obvious (it is just 2).
$ n\equal{}3$: obvious (it is just 3).

From now on we can assume $ n>3$. If $ n$ is even, then by the assumption in the problem, there exists a prime $ p$ between $ \frac n2$ and $ n$, therefore, $ n\minus{}p<\frac n2$ and by the inductive hypothesis we can write $ n\minus{}p$ as the sum of distinct primes $ p_1\plus{}p_2\plus{}\dots\plus{}p_k$. Then attach $ p$, which is clearly distinct from $ p_1,p_2,\dots,p_k$ because $ n\minus{}p<\frac n2$ while $ p>\frac n2$.

If $ n$ is odd, then $ n\plus{}1$ is even, and there exists a prime $ p$ between $ \frac{n\plus{}1}2$ and $ n\plus{}1$. If this number is $ n$ we are done, otherwise, $ n\minus{}p<\frac{n\plus{}1}2$ and an identical argument as above shows that we can write $ n$ as the sum of distinct primes as well.[/hide]


LaTeX
inequalities unsolved
inequalities

Let $ a,b,c > 0$ and $ abc \plus{} a \plus{} c \equal{} b$.

Find maximum of $ A \equal{} \frac {2}{a^2 \plus{} 1} \minus{} \frac {2}{b^2 \plus{} 1} \plus{} \frac {3}{c^2 \plus{} 1}$

[quote="kabi"]Here is the solution:[/quote]

Thank you! Your picture is not emergent. But I see your way. It's nice. Answer is $ \max P \equal{} \frac {10}{3}$

Can you find $ a,b,c$ when  $ P \equal{} \frac {10}{3}$[size=150] ?[/size]

[quote="thanhnam2902"]

Thank you! Your picture is not emergent. But I see your way. It's nice. Answer is $ \max P \equal{} \frac {10}{3}$

Can you find $ a,b,c$ when  $ P \equal{} \frac {10}{3}$[size=150] ?[/size][/quote]

Hello I m really sorry for the pics.I was in hurray and I don't have the Latex too so that I decided to post the pics.
[hide="hint to find the value of a,b,c"]
as sin B=1/3
find c=tan B for a and b as I assumed a=tan A and B=tan(A+B)  use the other eq.
[/hide]
Thank u.

[quote="kunny"]Why $ ca\neq 1$?[/quote]
Assume $ ac \equal{} 1$.
$ abc \plus{} a \plus{} c \equal{} b$, then $ b \plus{} a \plus{} c \equal{} b$ or $ a \plus{} c \equal{} 0$, And then $ (a \plus{} c)^2 \equal{} a^2 \plus{} c^2 \plus{} 2ac \equal{} a^2 \plus{} c^2 \equal{} 0$, so $ a \equal{} c \equal{} 0$, but $ ac \equal{} 1$. Contradiction. So $ ac \neq 1$.


geometry
trigonometry
angle bisector
trig identities
Law of Sines

In $\triangle{ABC}$, $AC=1$, $AB=2$, the length of the angle bisector of $A$ is $1$.
Find the area of the triangle.

[hide]
this problem doesn't seem to be correct.[/hide]

Draw a triangle with BC as the base and A above.  Then draw in the angle bisector, AD.  ACD is an isosceles triangle with legs 1 and base a.  Then ABD is a triangle with sides 1, 2 and 2a.  Let angle ADC be $\theta$.  Then apply the law of cosines.
$a^2-2a\cos{\theta}=0$
$4a^2+4a\cos{\theta}=3$
We see that cos theta is half of a.  Thus,
$a^2-a^2=0$
$4a^2+2a^2=3$
$a=\frac{\sqrt{2}}{2}$
From this, you find the height of the triangle to be $\sqrt{\frac{7}{8}}$.  Multiply that by half the base, $\frac{3\sqrt{2}}{4}$ to get the area, which i think it  $\frac{3\sqrt{7}}{8}$.
If I made any mistakes here, correct me.

i dont think this problem is flawed. and i even found a solution. to know what i'm talking about look at the image (let $H$ be the intersection of $\overline{BC}$ and the angle bisector through $A$).

[hide]first we know that $\overline{AC}=\overline{AH}$, so we know the following about the angles:
$\measuredangle ACH = 90^{\circ} -\frac12 \measuredangle HAC$, or
$\measuredangle HAC = 180 ^{\circ}-2 \measuredangle ACH$

also we know from $\triangle ABC$ and the angle bisector that
$\measuredangle ACH + 2\cdot \measuredangle HAC= 180 ^{\circ}- \measuredangle CBA$, or
$360 ^{\circ}-3\cdot \measuredangle ACH = 180 ^{\circ} - \measuredangle CBA$, or
$\measuredangle CBA =3\cdot\measuredangle ACH -180 ^{\circ}$

also we see from the law of sine:
$\sin \measuredangle ACH = 2\sin \measuredangle CBA$

putting both equations together, we get:
$\sin \measuredangle ACH = 2\sin 3\cdot \measuredangle ACH$

putting $\measuredangle ACH = x$ we transform:
$\sin x - 2\sin 3x$
$=\sin x - 2\left(3\sin x-4\sin^3x\right)$
$=7\sin x - 8\sin^3 x = 0$

since we know $0<x<\pi$:
$\sin^2 x =\frac78$
so $ \measuredangle ACH \approx 69.295^{\circ}$
and $\measuredangle HAC = 180 ^{\circ} - 2\cdot\measuredangle ACH $
so the area is $A= \frac12 \overline{AB}\cdot\overline{AC}\cdot \sin 2\cdot \measuredangle HAC = \sin 2\cdot 41.410 ^{\circ}=\boxed{0.9077}$[/hide]

[hide]Let $H$ be the intersection of the bisector of angle $A$ and $BC$. Since $AH$ is an angle bisector, we know the $BH:HC = AB:AC = 2:1$, so let $BH=2x$ and $HC=x$. Let angle $HAC=\theta$ and $ACH=AHC=\phi$, since triangle $AHC$ is isosceles.

By Law of Sines in triangle $AHC$, $\sin{\theta}=x\sin{\phi}$. In triangle $ABC$, we find $\frac{\sin{2\theta}}{\sin{\phi}}=\frac{3x}{2}$. Using the double angle formula for $\sin{2\theta}$ and substitution from the first equation, we find $\cos{\theta}=\frac{3}{4}$.

By double angle formula, we find $\cos{2\theta} = 2\cos^2{\theta}-1=\frac{1}{8}$. Then, by Pythagorean identity, we find $\sin{\theta}=\frac{3\sqrt{7}}{8}$. The area of triangle $ABC$ is then $\frac{1}{2}*2*1*\frac{3\sqrt{7}}{8}=\frac{3\sqrt{7}}{8}$.[/hide]

Let the angle bisector cut the side BC at D.  Let $BD=2x$ and $CD=x$

considering the triangles ACD and ABD,  using cosine rule

Cos(DAC)=$\frac{{1 + 1 - x^2 }}{2} = \frac{{1 + 4 - 4x^2 }}{4}$ = Cos(DAB)

solving for x, gives $2x^2=1$

rest is simple.


If every divisor of $ n$ can be expressed as $ a^r\plus{}1$ $ a \ge 0,r \ge 2$ $ a,r \in Z$
Prove:either $ n$ is a prime or $ 1$

Sorry,it can be $ 2p$ where $ p$ is a prime :blush:


geometry
circumcircle
geometric transformation
homothety
incenter
geometry solved

Let a triangle $ABC$ , the bisector $AD$  , the median $AM$ and a point $P$ on $AD$. Let $E,F$ the projections of $P$ on sides $AB , AC$. If $EF$ meets $AM$ at $K$ , prove that $KP \perp BC$.

[u] Babis[/u]

 [i]( Untill this moment I have not a solution to the problem.Probably it is harder than what I think , but probably easier.It is proposed in the preolympiad section too, but no solution was sent.)[/i]

Draw the circumcircle of ABC. The bisector of angle A meets it in P'. Let E',F' be the projections of P on AB,AC. Since P' is the midpoint of arc BC, M is the projection of P' on BC. Therefore, from Simson's theorem, E',F',M are collinear.

Now it's easier: the homotety with center A that moves P' to P, moves E' to E, F' to F, and then M (intersection of the median and E'F') to K (intersection of the median and EF), and then the line P'M to PK, which are both perpendicular to BC.  :wink:

[color=red] Congratulations ! [/color]
 This fantastic solution deserves a [color=red]gold[/color]  medal ! It's '' from the book !  '' , as Erdoes would say !

 [u] Babis[/u]

[quote="stergiu"]Let a triangle $ABC$ , the bisector $AD$  , the median $AM$ and a point $P$ on $AD$. Let $E,F$ the projections of $P$ on sides $AB , AC$. If $EF$ meets $AM$ at $K$ , prove that $KP \perp BC$.[/quote]

In other words, you want to show that the following three lines concur at one point:
- the perpendicular to the line BC through the point P;
- the line EF;
- the line AM, i. e. the line joining the point A to the midpoint of the segment BC.
This is Theorem 4 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124774#124774]http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4[/url], with different notations.

  Darij

[quote="stergiu"][color=darkred]Let a triangle $ABC$ , the bisector $AD$  , the median $AM$ and a point $P$ on $AD$. Let $E,F$ the projections of $P$ on sides $AB , AC$ respectively. If $EF$ meets $AM$ at $K$ , prove that $KP \perp BC$.[/color][/quote]
[color=darkblue][b]Very nice problem and very short and syntetical Edriv's proof[/b]

through the choosing of the point $P: =A'$- the intersection

between the ray $(AI$ with the circumcircle of the triangle $ABC$ !

[b][u]A metrical proof.[/u][/b] For $b\ne c$ and, through homotety, suppose w.l.o.g. that $P: =D$ !

$L\in EF\cap BC$ and $S\in BC\ ,\ AS\perp BC$ $\Longrightarrow$ $\boxed{\frac{MD}{MS}=\left(\frac{a}{b+c}\right)^{2}}\ \ (1)$.

The [u]Menelaus' theorem[/u] for $\overline{LEF}$ and $\triangle ABC$ $\Longrightarrow$ $\frac{LC}{LM}=\frac{a^{2}+b^{2}-c^{2}}{a^{2}}$.

The [u]Menelaus' theorem[/u] for $\overline{LKF}$ and $\triangle AMC$ $\Longrightarrow$ $\boxed{\frac{MK}{MA}=\left(\frac{a}{b+c}\right)^{2}}\ \ (2)$.

The relations $(1)$ and $(2)$ $\Longrightarrow$ $\frac{MK}{MA}=\frac{MD}{MS}$, i.e. $KD\parallel AS$ $\Longleftrightarrow$ $KD\perp BC$.

[b]Remark.[/b] Can choose $P: =I$- the incenter of $\triangle ABC$. It is more easily.

Indeed, for $b\ne c$ denote $T\in BC$, $IT\perp BC$ and $K_{1}\in AM\cap IT$.

The Menelaus' theorem for $\overline{TIK_{1}}$ and $\triangle ADM\Longrightarrow \boxed{\frac{K_{1}A}{K_{1}M}=\frac{b+c-a}{a}}\ \ (1)$.

$LC=\frac{a(p-c)}{|b-c|}$, $LM=\frac{a^{2}}{2|b-c|}$ $\Longrightarrow$ $\frac{LC}{LM}=\frac{2(p-c)}{a}$.

The Menelaus' theorem for $\overline{LKF}$ and $\triangle AMC\Longrightarrow \boxed{\frac{KA}{KM}=\frac{b+c-a}{a}}\ \ (2)$.

Therefore, $K_{1}\equiv K$, i.e. $KI\perp BC$.[/color]

Consider the circle with centre P and radius PE=PF. Draw the line l from A which is perpendicular to PK and let G be the intersection point of l and EF.  Since K lies on the polar of A, l is the polar of K. This means that (E,K,F.G)=-1 and since BM=CM we have l//BC  from which we deduce that PK is perpendicular to BC.


vector
trigonometry
geometry
real analysis
real analysis unsolved

"Say you have the Statue of liberty (92 feet high), but it is tilted at an angle $b$.  Assuming that the Statue starts at the water, we need to find out how far away you should be from that point where it touches the water, in order for the angle between the line connecting you to the midpoint of the statue and the line connecting you to the top of the statue is maximized.  Express the answer in terms of the distance you are from the statue $d$, the angle that the statue forms with the water $b$, and the angle that forms between the line from you to the midpoint and you to the top $x$."

There is a picture of this attached.  I tried to write this in terms of vectors, then use the dot product to find the angle, then maximize it, but it seemed like it would be really long.  I was wondering if anyone has a shorter / easier solution.

$\displaystyle \mathbf{v}=-d-46\cos{(b)}\mathbf{i}+46\sin{(b)}\mathbf{j}$
$\displaystyle \mathbf{w}=-d-92\cos{(b)}\mathbf{i}+92\sin{(b)}\mathbf{j}$

Thanks for your help.

It is actually an elementary geometry problem. You can start with the observation that the set of points $X$ such that the angle $AXB$ is fixed and $X$ is on a given side of the line $AB$ is an arc of a circle through the points $A$ and $B$. The final thing to do is to recall one nice theorem about the length of the tangent line to a circle. You can get the answer in almost no time this way.  :)


What is a '[b] Pseudo[/b] Prime[b] Numbers[/b]?[b]

                                                                        Respectfully,
                                                                       Arthur Eddington[/b]

This sort of information is easy to find out with a quick Google search:

http://mathworld.wolfram.com/Pseudoprime.html
http://en.wikipedia.org/wiki/Pseudoprime


x,y,z \geq 0 x+y+z=1. Prove that:
x^3+y^3+z^3+6xyz \geq 1/4

we have from shur that 3\sum x3 + 9xyz \geq 3 \sum xy(x2 + y2) and adding \sum x3 + 6xyz to this inequality we obtain

4  \sum x3 + 15 xyz \geq (\sum x)3 that is

4 \sum x3 + 24 xyz \geq 1 which is the required inequality. We added up 9xyz in the end, but since x,y,z\geq 0, it's ok.

We also see that 1/4 is the best constant, even if x,y,z>0, because taking x -> 0, y -> 1/2, z -> 1/2 we obtain equality.


COP 5th will take place sometime on August. If you don't know what COP is, please read the COP (Contest of Opportunity) thread. It's a good competition that test your knowledge on math. It's not that hard neither (at least the first round  :D ).

If you are interested, please sign-up in here by "I'm in," "I'm interested." Those who sign-up will get private message when Round 1 is posted.

Regards,

Silverfalcon

[quote="sirorange"]It depends on when in August it is.[/quote]

same here, my school starts again in the middle of August :mad:

I'm in, but I will chime in with the other two in saying that if it's after when my school starts I may not be able to get it in- but I will try.

I think most North America school starts in August.

Well, I don't think it's that big deal though. It's only 5 problems in Round 1, 7 problems in Round 2, and 2 problems in Round 3. The only big round is Round 4, Championship Round where you have 15 problems.

5+7+2+15 = 29 problems. Not bad.  :D

2 problems on Round 3 are proof, by the way.

Me too. Hope this time po pass first round :). My school starts in September :P but i have time to solve the problems durring it :)

i am in...hope this time i will have enough time to post the answers.

If the competition starts $\pm$ 14th of august, I'm certainly in.
I'm on holiday until then, and with no internet it could be pretty hard for me to read the problems...  :D

But I'm in, no matter when it starts. Maybe i'm not home yet, so you won't recieve any answers from me, but is that a big problem? If it is you'll have to tell me when the problems will be posted...

Eh, I'll give it a shot.  Hopefully come august I'll remember that I signed up...

[quote="martha"]started yet? may i try?[/quote]

For the first question, no not yet. It will start very soon, however.

For the second question, yes, you can join.

 :lol:

If it hasn't started yet (I don't think it has), I'm in.

[quote="Silverfalcon"]The 5th COP will start in 8/28/05, which is two weeks from today.

I already have problems done but I have some work that needs to be done so...  :)[/quote]
Oh no...Two days before my school starts.....I really hope that I will have time to do the problems this time.

Problems are posted!

This thread will soon be removed from stickied.


[quote="mws21"]Can somebody tell me which sections of the book is the most important in regards to AMC 12 and AIME?

I'm thinking of cyclic quadrilaterals, diophantine equations, probabilities, and sequences but I want to hear other people's opinions.[/quote]

It really depends on wheter you're thinking of the AMC 12 or AIME.
Diophantine equations are important for the end of the AMC 12 and are scattered throughout the AIME, so that's important.

Number theory is always important if you want to ace the AMC 12.
The AIME is also heavy on number theory.

Any other subjects important for AMC12 and AIME besides those?

Any suggestions are welcomed.

[quote="Boy Soprano II"][quote="buzzer11"]The AIME is also heavy on number theory.[/quote]
I disagree.  In my opinion, the AIME topics are relatively evenly distributed.  However, we receive almost no instruction in number theory in schools, so we may perceive there to be much number theory.

And you really have time to read through the whole book and do several problems in all the chapters between now and the next AMCs, so I don't see what the rush is.  Everything in the book is worthwhile.[/quote]

Really? All of the AIMEs I have looked at have had a considerable amount of NT.
Well "heavy" was too strong of a word, I guess. What I mean is that studying NT will definately be worthwhile.


function
real analysis
real analysis unsolved

Suppose $f_n$ is a sequence of real functions with the property that for any real $x$ and any sequence $x_n$ that goes to $x$ the sequence $f_n(x_n)$ is convergent. Does it follow that $f_n$ converges uniformly on any compact of interval of $R$?

You can also look here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Berkeley&t=34597#

The answer is yes.
For every $x$ define a function $g(x)=limf_n(x_n)$,where $x_n=x,\forall n$.So we have a simple convergence to $g$.
Suppose that $f_n$ doesnt converge uniformly to $g$.Then there exists $\epsilon$ and $(n_k,p_k,x_{k})$ such that
$|f_{p_k}(x_{k})-f_{n_k}(x_{k})|\geq \epsilon$.
For 2 sequences $a_n$ and $b_n$ which converge to the same limit x we have $lim_{n\to \infty} f_n(c_n)=lim f_n(a_n)=limf_n(b_n)$,where $c_{2n}=a_{2n}$ and $c_{2n+1}=b_{2n+1}$.So for every sequence $x_n$ which converges to x we have $lim_{n\to \infty} f_n(x_n)=g(x)$.
Since $x_{k} \in$ compact subset we can find a subsequence $x_{m_k}$ such that $lim_{k\to \infty} x_{m_k}=x$.
And we're done,because $f_{p_k}(x_{m_k})$ and $f_{n_k}(x_{m_k})$ have the same limit.
I hope it's ok.

This is a test for uniform convergence over a compact inteval.


If the area of a trapezoid is a^2-b^2 what is the height of the trapezoid?

            a
     __________
   / 1               \
 /   1  h              \
___1____________
            b

The area is $ (a\plus{}b)(a\minus{}b)$ by factoring the difference of squares.  Using our formula, the area is also $ h \left (\frac{a\plus{}b}{2} \right )$, so we have $ h \left ( \frac{a\plus{}b}{2} \right )\equal{}(a\plus{}b)(a\minus{}b)$.  Dividing by $ a\plus{}b$ yields $ \frac{h}{2}\equal{}a\minus{}b \implies h\equal{}\boxed{2(a\minus{}b)}$, which is what we wanted.

Could you simplify that, im still in the middle school level.

What don't you understand? Do you not know how to find the area of a trapezoid, etc.?

. Dividing by  yields \frac{h}{2}=a-b \implies h=\boxed{2(a-b)}, which is what we wanted.

Thats what kinda confused me, i read it to fast.

Perhaps it's easier if we write $ h \left (\frac{a\plus{}b}{2} \right )\equal{}(a\plus{}b)(a\minus{}b)$ as $ \frac{h(a\plus{}b)}{2}\equal{}(a\plus{}b)(a\minus{}b)$.  Then we divide by $ a\plus{}b$ on both sides which leaves us with $ \frac{h}{2}\equal{}a\minus{}b$.

I assume you just mean $ x\equal{}\minus{}5$; plugging in yields $ f(\minus{}5)\equal{}\boxed{\minus{}20}$. :wink:

[quote="uanaka"]that helps, ok, i need to stump anyone that answers my question.

What i the value of the dependent variable when the independent variable is -5 and the function is f(x)=-x^2+2x+15??

I WILL GET YOU[/quote]

Just so you know, i don't think you can stump an AoPSer in AoPS.  These questions are middle school level which typically you know how to solve (unless you are learning, or aren't there yet)

Thats me, Im still learning Algebra 1 and 2 (If i can find something for it)


Let $ x_1,x_2,...,x_{9999}$ be all roots of the equation
$ x^{9999} \plus{} 99x^{9998} \plus{} 1 \equal{} 0$
Define: $ P(x)$ such that $ P(x_i \plus{} \frac {1}{x_i}) \equal{} 0$ for $ i\in[1,2,...,9999]$
if $ \frac {P(1)}{P( \minus{} 1)} \equal{} \frac {m}{n} ;m,n\in N$ and $ (m,n) \equal{} 1$
find the value of  m+n


inequalities
function
integration
calculus
derivative
real analysis
real analysis unsolved

Function $ f: [a,b]\to\mathbb R$ is in the class $ C^1$. Prove the inequality:

$ \int_a^b(f(x)\minus{}f(a))^2dx\le\frac{(b\minus{}a)^2}{2}\int_a^b(f'(x))^2dx.$

Assume $ a\equal{}0$ and $ f(a)\equal{}0$. The inequality is true for $ b\equal{}0$; therefore it suffices to compare the derivatives of both sides with respect to $ b$, i.e. to prove that
\[ f(b)^2\le b\int_0^b (f'(x))^2\,dx\plus{}\frac{b^2}{2}f'(b)^2\]
To this end, use Cauchy-Schwarz:
\[ f(b)^2\le \left(\int_0^b 1\cdot |f'(x)|\,dx\right)^2\le b\int_0^b (f'(x))^2\,dx\]


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let A be a matrix in M_3(R). Show that:

a)If f is a polynomial in R[x] which doesn't have any real roots then f(A)=/= 0_3.
b)There is a positive integer n s.t. (A+A*)^(2n) = A^(2n) + (A*)^(2n) iff det(A)=0.
--------------------------------------------------------------------------------

a) We can see A as a matrix in M_3(C), A is similar to a triangular superior
A~
(a * *)
(0 b *)
(0 0 c)

Suppose that f(A)=0, f(A) is similar to
(f(a) * *)
(0 f(b)*)
(0 0 f(c))
f(a)=f(b)=f(c)=0

The degree of the polynomial caracteristic of A is 3 at least one eigenvalue among a,b,c is real.This imply f has a real root contradiction.

So f(A)=/=0

b) We will use the relation A.A*=detA.I_3 where
A*=transpose Comatrix of A

Suppose detA=0, then A.A*=A*.A=0 these matrices commute
(A+A*)^2 = A^2+A*^2

If there exist n  N and detA=/=0 (A is invertible)
(1) (A+A*)^(2n)=A^(2n)+A*^(2n) we multiply (1) by A^(2n)
(A^2+k)^(2n) = A^(4n)+k^(2n)I
A vanishes the polynomial P(x)= (x^2+k)^(2n)-x^(4n)-k^(2n)
we verify that p(x)=kx^2.w(x) where w(x) has no real roots
kA^(2n).w(A)=0 but A is invertible, then w(A)=0 we get a contradiction with a)

what means that A~(similar) to a diagonal superior matrix??? and when can we say this?? can u give me some references moubi??? thx :D

[quote="Kuba"]what means that A~(similar) to a triangular superior matrix??? and when can we say this?? can u give me some references moubi??? thx :D[/quote]

A similar to a triangular superior means there exist P invertible such that

P^(-1).A.P =
(a * *)
(0 b *)
(0 0 c)  where a,b,c are eingenvalues of A

Any matrice A \in M_n(C) is trigonalisable

I am sure it is possible to solve a) without triangular form.

References: It is probably done in any linear algebra book chapter
reduction diagonalisation, trigonalisation. Maybe your math teacher
has a book about that, he can borrow you.


Given $x_0=0$ and $0<x_1,x_2,...,x_n<1$ and $\sum_{i=1}^n x_i=1$ Prove that :
$1\le \sum_{i=1}^n \frac{x_i}{\sqrt{1+x_0+x_1+...+x_{i-1}}.\sqrt{x_i+...+x_n}}<\frac{\pi}{2}$
The first ineq is very easy but the second one is very nice  :)

Yes,not new.You can see here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=25233


Fie x astfel incat $sinx \ne0$.Demonstrati ca sirul $(x_{n})$ , $x_{n}=sin(nx) $ nu are limita.

Cazurile $\sin x=\pm1$ sunt simple. Presupunem $\sin x\neq0$ si $\cos x\neq0.$
Daca $\sin nx\rightarrow L\in\mathbb{R},$ atunci $\sin(n+2)x-\sin nx=2\sin x\cos(n+1)x\rightarrow0,$ deci $\cos nx\rightarrow0.$ Dar atunci $\cos(n+2)x-\cos nx=-2\sin x\sin(n+1)x\rightarrow0,$ deci $\sin nx\rightarrow
0.$ Acest lucru e imposibil, deoarece $\sin^{2}nx+\cos^{2}nx=1.$


There are 12 diagonals that can be drawn on the faces of a cube. Two different diagonals are selected at random. What is the probability that the diagonals selected share a common point.
Thanks!

share a common point as in they intersect?

There are 12 diagonals on a cube, 2 on each side (so 2 X 6 diagonals).  They're saying what is the probability that the diagonals share a point-in other words, they intersect.  The only diagonals that intersect are the ones on the same side, so this contributes 6 to the total.  However, it can also mean that they intersect at the same corner (hence, they share the corner point).  Three diagonals intersect at every corner, for a total of 8 X 3=24 points.  Hence, the answer is 30/C(12,2).

[hide]
First, arbitrarily select a diagonal.
Now, you see that there are two diagonals that share each vertex with the chosen one and one that intersects it.
Thus, there are five diagonals that share with it a common point, and six that don't, so the probability is $ \boxed{\frac{5}{11}}$.
[/hide]


bueno soy de colombia y  la cosa es que acabo de comenzar ing.quimica y pues me entere de las olimpiadas de matematica universitaria y me gustaria presentarme para las del proximo año pero no manejo los temas que se requieren para la resolucion de los problemas:Calculo,teoria de numeros,algebra abstracta,algebra lineal,geometria. entonces queria pedirles me ayudaran aconsejandome libros sobre cada uno de los temas que deo estudiar para comenzar a prepararme.hable con un profesor y me recomendaba mas que estudiara librosn de analisis pues es mas riguroso que uno de calculo ahi si pues lo dejo a su opinion.en todo caso les agradezco cualquier ayuda

Hola Lambert, personalmente creo que una de las mejores maneras de mejorar es participando activamente en los foros de olimpiadas de AoPS, donde se han posteado numerosisimos problemas de las IMO. Alli podrás apreciar la gran variadad de soluciones y herramientas disponibles para la resolución de problemas. En cuanto a geometría, mi experiencia me indica que la  esencial para las olimpiadas es la sintética, es decir, aquella que no involucra el analisis puramente algebraico para las resoluciones sino que ésta emplea recursos puramente geométricos como: relaciones angulares, razones simples y dobles, transformaciones en el plano, etc.

Tengo un material muy bueno (en ingles) sobre las diversas tecnicas que se emplean para resolver los problemas geométricos de nivel de olimpiadas. Es pesado para adjuntarlo por aqui. Yo con mucho gusto te lo enviaré a algun correo que me indiques a través de un mensaje privado.


[quote="bubka"]Heard of that joke?

In a room there were reportedly four people: Superman, Spiderman, a hot lady mathematician, and a very ordinary looking lady mathematician. At this time a man entering the room was murdered. Who did it?
[hide="answer"]The very ordinary looking lady mathematician, because fictitious characters are incapable of murder! :rotfl: [/hide][/quote]

LOL :rotfl:  That's funny. But some girls who go to the Olympiad are pretty hot aren't they?

dude yeah IMO has some surprisingly hot girls

I think many of them may have been like guides or other noncontestant people

but there were probably some hot contestants too

I've only met one hot girl that was smart at math.

coughcoughLIZZYcoughcough

oh...
i thought this was gonna be a discussion on why there arent as many girls in math as there should be
darn
i'm not gonna get involved in this
it could be insulting to girls who do like math :ninja:

they don't exist, oh well

although it would be funny watching the socially inept math guys drool

DUDE THE COLLEGE STATION GUYS ARE PRETTY PROUD OF SOMEONE OVER THERE BUT I DON'T REALLY KNOW WHO IT IS

ALSO THERE ARE A COUPLE OTHERS WHOM I'VE MET BUT IT WOULD BE STUPID TO JUST BE LIKE HEY GUYS GUYS HERE ARE THE NAMES OF SOME HOT MATH GIRLS: ET CETERA

[quote="daermon"][quote="trigonometry1.01"]Hey guys,

Hot girls (ooh, sizzle) just aren't good at math.

So like dude girls just aren't impressed by math in general. This might change if you're on the varsity football team, but other than that it is pretty much a fundamental fact of the universe.[/quote]
i dont think its a fundamental fact of the universe :o
i think the reason is that girls, for the most part, are more interested in socializing than boys
and math generally doesn't seem to be a social activity :(
though it can and should be :)
for math to be more popular, it really needs to shed the image of nerdiness, because there are a lot of people interested in math who are not nerds
like me for example :D
if the image of math is changed
there will be a lot more 'social' people involved
and math will get more and more popular[/quote]
same goes for metal
but
it just isnt possible
stereotypes in the society are pretty much impossible to change
(i would know, i tried pretty hard to get people into metal without any success )

hott girls do math

asmallgoat is uber hott

sure she suck(she says so at least)

i do math too
(you didn't know that one)

WAIT WAIT ARE YOU A HOTT GIRL I HAD NOT BEEN PREVIOUSLY APPRISED OF THIS FACT

[quote="bubka"]sorry if we sound offensive, but a hot girl good at maths is simply an experimental error and should be ignored while making statistical predictions (you have heard of outliers I hope?) :roll:[/quote]

all the offense in the world taken by my friend asmallgoat

sure she sucks in her mind...

I think we have to define "good" as well. I mean of course most/all the girls are going to look inferior compared to people like Terence Tao, Reid Barton etc. So I think we should have some sort of bottom line for being "good".

[quote="BanishedTraitor"]I mean of course most/all the girls are going to look inferior compared to people like Terence Tao, Reid Barton etc.[/quote]I assume by this you haven't heard of Ana Caraiani until now ;)

... and she's not the only female Romanian to make Putnam Fellow.

I won't say beauty+brains (under some reasonable statistical quantitavtive measure) is constant, but I may venture to say that it follows the normal distribution or something similar. :rotfl:

guys guys

Reid Barton is pretty awesome.

guys guys yes he is HOW ARE YOU NOT ZHAI

also dude didn't reid own until like 2001 and then tiankai started owning and then lawrence and then zhai THOSE ARE ALL GUYS

also dude I bet of the set {iurie boreico, alexandr magazinov, zhiyu li - he solves problems with his heart}, they are all guys

hotness * intelligence * availability = c

where c is a constant

fortunately it is a pretty damn large constant

y'know, that factor ot 10^8 is always helpful

Well, Russia has tons of hot girls and mathmeticians, so there must be some with both skills... :D  :rotfl:

Not in America, though.

Dang... I bet girls do not solve problems with their hearts...

I LIKE PRETTY GIRLS

THEY ARE PRETTY

UNFORTUNATELY PRETTY GIRLS ARE PRETTY HARD TO FIND SINCE MOST GIRLS THAT PEOPLE THINK ARE PRETTY ARE NOT REALLY PRETTY THEY JUST HAVE 100 TONS OF MAKEUP ON WHICH IS PRETTY LAME

[quote="sonny"]DUDE ITS REALLY HARD TO FIND SOMETHING THAT DOESN'T EXIST. Admit it guys, our math skills aren't going to help us get a beauty, WE ARE DOOMED. DOOMED......................anywayz heres a question do you think all pretty girls hate guys good at math?[/quote]

What... being good at math is A W E S O M E

like if some girl is there

you could just be like

HELLO I AM AWESOME

I bet it would probably work pretty well... you should try it some time

[quote="hunter34"][quote="sonny"]anywayz heres a question do you think all pretty girls hate guys good at math?[/quote]

Not really I got 2 or 3 friends from being good mathematician, but they were there also  when I scored on a bicycle kick. I wonder if that has to do with anything?[/quote]

the question is WERE THEY PRETTY( and you didnt specify they were girls)?

yes all 3 were girls and they were somewhat pretty

[quote="i_like_pie"][quote="SoxFan34"]Well, Russia has tons of hot girls and mathmeticians, so there must be some with both skills... :D  :rotfl:[/quote]
I agree with the first part, but the second is questionable...[/quote]

I guess you are saying that since you are a Russian guy...
The second is probably true, seeing all of the new Russian mathematicians.


Wth go mod n?  :huh:
A remainder can only be less than or equal to its divisor. You want me to prove that  :huh: . Dont infuriate with such problems   :ninja:

@pardesi: WOP is well ordering principle

and plz look at ur pms


logarithms
geometry
geometric transformation
dilation
pigeonhole principle
probability
combinatorial geometry

Speaking of distance sets of points in the Euclidean plane ( http://www.mathlinks.ro/Forum/viewtopic.php?p=143942#143942 ), here is a related (beautiful  :lol: ) classical problem which, although probably very difficult for high school students, can be proven entirely elementarily:

[b]Problem.[/b]  Call a finite set $S$ of points in the Euclidean plane 'well-distributed' if: 1) any two distinct points in $S$ are at least a unit distance apart; and 2) every circle with radius $10$ centered inside the convex hull of $S$ contains at least one point of $S$ in its interior.

Show that there exists a constant $c > 0$ with the following property: if $S$ is any 'well-distributed' set of $n$ points in the plane, then there are at least $cn^{3/4}$ different distances determined by $S$.

Of course, the constant $10$ may be replaced by any other. Perhaps this result holds without the assumption that the set is 'well-distributed,' but this (as far as I am concerned) has not yet been proven. Actually, the exponent $3/4$ may be improved to $6/7$ (Solymosi-Toth), but I don't know an elementary proof of this. (In contrast, the proof for $3/4$ may be understood by any student involved in olympiads.)

Cheers,

--Vesselin

This sounds strange because an old result from Chung and Erds states that any set of $n$ points in the plane determines at least $f_2(n)$ distinct distances whith :
\[c n^{\frac 4 5} < f_2(n) < c' \frac n {\sqrt {\ln (n)}}\]

It is conjectured that the upper bound is assymptotically best possible (i.e. the right order of magnitude).

Pierre.

:blush:  :blush:  :blush:
Apologies --- I didn't know that this lower estimate (with the exponent $4/5$) has been proven. As for the upper bound, it is obvious, as demonstrated by the example of the square grid. Nevertheless, under the assumption that the poinst are 'well-distributed,' the $3/4$ problem is quite straightforward. Here is a rough sketch of the solution I had in mind:

We may assume, via a dilation, that the minimal distance between two points of $S$ is $1$, and let us choose two points $A,B \in S$ with $|AB| = 1$. Draw enough concentric half-annuli of radius, say, $20$, centered at the midpoint of $AB$ to cover a positive fraction (at least $n/2$) of the points of $S$. Suppose we have drawn $K$ half-annuli, none of which is redundant. Because the distance between any two points in $S$ is $\geq 1$ and the half-annuli cover at least $n/2$ of the points in $S$, it must be that $K \gtrsim \sqrt{n}$ (by saying this, I mean that there is a constant $c>0$, independent of $n$, such that $K > c\sqrt{n}$).

By the pigeonhole principle and the assumption on well-distribution, a positive fraction of the strips between two successive annuli contain $\approx n/K \lesssim \sqrt{n}$ points of $S$ (that is, there are constants $c,c'>0$, independent on $n$, such that a positive proportion of the strips contain between $cn/K$ and $c'n/K$ points). Consider closely an arbitrary of these strips. Assume that there are $k$ distances $d_1,\ldots,d_k$ measured from the points from that strip to either $A$ or $B$. Let $A_i$ be the set of points whose distance from $A$ is $d_i$; similarly, denote by $B_i$ the set of points whose distance from $B$ is $d_i$. By the pigeonhole principle, there is an $i_0$ such that either $|A_{i_0}| \geq n/kK$ or $|B_{i_0}| \geq n/kK$; we may assume that the first case holds. On the other hand, because two half-circles intersect in at most one point, we have that $|A_i \cap B_j| \leq 1$, for all $i,j$, and it follows from
\[
A_{i_0} = \bigcup_{j=1}^k A_{i_0} \cap B_j
\]
that $|A_{i_0}| \leq k$. Now it follows from $k \geq |A_{i_0}| \gtrsim n/kK$ that $k \gtrsim \sqrt{n/K}$.

Hence, a positive fraction of the $K$ strips determine $\gtrsim \sqrt{n/K}$ different distances from either $A$ or $B$. Therefore, the total number of distinct distances determined by $S$ to either $A$ or $B$ is $\gtrsim K\sqrt{n/K} = \sqrt{nK} \gtrsim n^{3/4}$, hence the desired result.
Nice argument, isn't it?  :P
Pierre, could you provide a reference for the result of Erdos and Chung that there are always $\gtrsim n^{4/5}$ distinct distances?

--Vesselin

In fact there could be a slight correction about the $n^{\frac 4 5}$ bound. This bound is given in the 'Handbook of Dicrete and computational geometry', p. 836 and the given reference is :
L.A.Szekely, 'Crossing numbers and hard Erds problems in discrete geometry', Combinatorics, Probability and Computing, 7 (1997), 353-358.

But, in a paper from Szekely in the memory of Erds, he said that the best known lower bound is due to Chung, Szemeredi and Trotter who showed that
\[ \frac {n^{\frac 4 5}} { ( \ln (n) )^c} < f_2(n).\]
And he gives the reference :
F.R.K.Chung, E.Szemeredi, W.T.Trotter, 'The number of different distances determined by a set of points in the euclidean plane', Discrete Comp. Geometry, 7 (1992), p. 1-11.

I do not have these papers in hand, so...

Pierre.

Also a great book that covers this topics of repeated distances in space is Combinatorial Geometry by Janos Pach. Great introduction to this beautiful topic.

Although this maybe off topic, I had Endre Szemeredi as a guest lecturer for our undergrad seminar on computational complexity last year and, believe it or not, I could hardly understand his English from the heavy Hungarian accent and sloppy handwriting... Fan Chung is a great teacher along with R. Graham!

Yes. R. Graham is great.. btw where do you study 3X.lich ?

I guess [url=http://www-cse.ucsd.edu/users/rgraham/]UC San Diego.[/url]  :)

[quote="pbornsztein"]I would like having some people like Szemeredi, Chung, Graham as teachers  :?

Pierre.[/quote]

Pierre, you are such an excellent mathematician, especially combinatorics. I think you don't need it.  :)

Are you kidding?
These people are among the world best in combinatorics. I'm not as good as to have not all to learn from them. :(

Pierre.

[quote="manuel"]Btw, who is the best alive graph theorist or combinatorist?
Among the best I know Paul Seymour, Lovasz, Graham  who else is on the world bests?[/quote]

Noga Alon ( http://www.math.tau.ac.il/~nogaa/ ) is also an outstanding combinatorialist. But there should not be such term as 'best mathematician.'

Hmmm...I was sure to have seen recentely something about this problem.
Now, that I have found the paper back, as far as I know, the best current bound is :
\[c \cdot n^{\frac 6 7} < f_2(n)\]

Reference :
J. Solymosi, Cs.D. Toth, 'Distinct distances in the plane', Discrete and computational geometry, 2001, p. 629-631.

Pierre.


I'm going to write my second Mock Mathcounts test. It's a chapter/state level test. Since Mathcounts is almost here, this will be a good oppurtunity to practice. If MORE THAN 10 people is interested, I will do it.

[quote="jclarkemathL314159"]I'm interested

Where was the first?[/quote]

Search for it. It's here somewhere...

EDIT: Here's the website.
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mathcounts+mock&t=78075

Sure as long as people explain how they did it...

[quote="bpms"]You're original was back when biffanddoc posted here... Anyways, I'm in.[/quote]

Why doesn t biffandoc post here anymore?

Probably age...I didn't really know them but I'm making a hypothesis. Time to go collect data! JK

Interested in the test... (not for myself, though)

are the problems from previous mathcounts competitions? If they are, I'll sign up, but please post them over weekends, so I have time to solve them.

howcome splashd seems to be on aim sometimes, but it says "user is not available" when i im him ???

-jorian

[quote="13375P34K43V312"]He blocked you.[/quote]

I did?

i think its working now

leet:did U block me???

-jorian

[quote="jhredsox"]i think its me vs now a....
-jorian

oh, u asked when, i think its whenever[/quote]

You can face him when he's finished eating dinner.

These are, after consulting with goldendomer, a graphical representation of the final bracket.

[hide="Click to unveal the countdown bracket"][img]http://i96.photobucket.com/albums/l167/adasarathy/countdownbracket-1.png[/img]
[img]http://i96.photobucket.com/albums/l167/adasarathy/loser_bracket.png[/img][/hide]

These are, I believe, the final countdown format.

small problem. i'm facing now a ranger.

MY AIM'S BROKE SOMEHOW!!!!!!

-jorian

its working, i just think it needed to cool off

-jorian

The format is perfect, only part that needs to change is jhredsox is playing now a ranger. :)

ok, we can probably do it tonight

-jorian

I can't face bpms tonight, I have a class.


The sum of two numbers is 25 and their difference is 11. What is
the larger of the two numbers?


[hide]Really sorry about the poor use of numbers. Don't know how to use latex.   :)
Adding 9 gets m+13=23m+23/m+7. Multiply m+7 to get rid of the fraction. This results in m(squared)+20m+91=23m+23. Subtract the entire right side to the left to get m(squared)-3m+68. Using quadratic we finally get 3 plus/minus i sqr.263 /2.[/hide]

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]
if you need help with LaTeX programming


1. Find f:N\{0}->R s.t. f(1)=1 and  \sum {d|n} f(d)=0

2. Find f:N\{0}-.R s.t.  \sum {d|n} f(d)=n

3. Find f:N->R s.t. f(m+n)+f(mn)=f(m)f(n)+1 for all m, n in N

I have a solution for the 3rd one! :D :D

if we put m=n=0 we get that f(0)=1
if we put n=1 and from that we get f(m)=f(m)f(1)-f(m+1)+1. if f(1)=1 then we get that f(n)=1 is a solution.
lets now consider f(1)=alpha; alpha<>1.
if we put m=n=1 in the relation we get that f(2)=ALPHA^2-ALPHA+1.
if m=1,n=2 we get that f(3)=alpha^3-2alpha^2+2alpha
if n=3,m=1 we get that f(4)=alpha^4-3alpha^3+4alpha^2-2alpha+1
if n=m=2 we get that alpha=0 or alpha=2
1)if alpha=0==>f(1)=0 we put m=1 and we get that f(n)+f(n+1)=1 and we also know that f(0)=1 and f(1)=0 we get that f(n)=(1+(-1)^n)/2 (use induction to prove it!!!! :D)
and using the fact that f(n)=(1+(-1)^n)/2 we replace it in:  f(m+n)+f(mn)=f(m)f(n)+1 we get that 1+(-1)^m+(-1)^n=2(-1)^(mn)+(-1)^(m+n) verifying a few cases m odd, m even we get that f(n)=(1+(-1)^n)/2 is a solution.
2) if alpha=2 then putting m=1 we get that f(n)=n.

all done!!!!! cheers! :D :D

Thxforthe soln! Try the first 2. They're well-known functions!

[quote="grobber"]
2. Find f:N\{0}-.R s.t.  \sum {d|n} f(d)=n
[/quote]

The Mbius inversion formula to get f
http://mathworld.wolfram.com/MoebiusInversionFormula.html

 \cap  \cap  \emptyset  \cup \beta  \int

Now, I wouldn't call them well known! :)
I'd rather call them famous results! :D :D
cheers! :D :D

Now following the link that Moubinool posted I managed to give a solution! :D :D

If we replace n in the equation with a^n, with a prime and n natural we get that
f(1)+f(a)+..+(a^n)=0 thus we have that f(a)=-1 and f(a^k)=0, for any k>=2.
Lets prove now that: f(xy)=f(x)f(y) for any x,y natural, (x,y)=1. we do this using induction after alpha=mn. for alpha=1 it is obious.we consider it true for any alpha<=betha, where betha is fixed and it is pozitive. lets prove it for betha+1 :)
we have that x,y are natural with (x,y)=1. If x=1 then we have that:
f(xy)=f(1y)=f(y)=f(1)f(y)=f(x)f(y).if x,y>1 we now have that:
0=sum(d|xy) f(d)=sum(d1|x)sum(d2|y) f(d1d2)=sum(d1|x)sum(d2|y) f(d1d2)+f(xy)-f(x)f(y)=
(sum(d1|x)(f(d1)))(sum(d2|y)(f(d2)))+f(xy)-f(x)f(y) and the afirmation is  proved :D.
Then we have that f(1)=0 and f(n)=0, so if there exists a prime with a^2|n then f(a1a2..an)=(-1)^n if all a1,a2,..,an are distinct primes. this means that f(n)=moebius function. cheers! :D :D

I forgot to say that:

For the other one.. you do the same and you get that f(n)=Euler's indicator! :D :D

cheers! :D :D


Given is an integer sequence $\{a_n\}_{n \ge 0}$ such that $a_{0}=2$, $a_{1}=3$ and, for all positive integers $n \ge 1$, $a_{n+1}=2a_{n-1}$ or $a_{n+1}= 3a_{n} - 2a_{n-1}$. Does there exist a positive integer $k$ such that $1600 < a_{k} < 2000$?

By induction $ a_{i \minus{} 1} < a_i < 2 a_{i \minus{} 1}$
$ i \equal{} 1$ $ a_0 < a_1 < 2 a_0$
$ \forall i < j$ $ a_{i \minus{} 1} < a_i < 2 a_{i \minus{} 1}$
Operation $ 1)$
$ a_{j \minus{} 1} < 2 a_{j \minus{} 2}$
$ a_{j \minus{} 2} < a_{j \minus{} 1}$
$ a_{j \minus{} 1} < a_j \equal{} 2a_{j \minus{} 2} < 2a_{j \minus{} 1}$
Operation $ 2)$
$ a_{j \minus{} 2} < a_{j \minus{} 1}$
$ a_{j \minus{} 1} < 2a_{j \minus{} 2}$
$ a_{j \minus{} 1} < a_j \equal{} 3a_{j \minus{} 1} \minus{} 2a_{j \minus{} 2} < 2a_{j \minus{} 1}$

It remains to prove that $ a_i$ is a sum of at most two power of two
then there is no $ k$ such $ 1024 \plus{} 512 \equal{} 1536 < a_k < 2048$

A simple induction shows $ (a_n,a_{n\plus{}1})$ must have the form $ (2^a\plus{}2^b,2^{a\plus{}1}\plus{}2^b)$

Exactly $ a_n \equal{} 2^{b_n} \plus{} 2^n,b_0 \equal{} b_1 \equal{} 0$. If $ a_{n \plus{} 1} \equal{} 2a_n$, then $ b_{n \plus{} 1} \equal{} b(n) \plus{} 1$. If $ a_{n \plus{} 1} \equal{} 3a_n \minus{} 2a_{n \minus{} 1}$, then
$ 2^{b_{n \plus{} 1}} \equal{} 2^{b_n \plus{} 1} \plus{} 2^{b_n} \minus{} 2^{b_{n \minus{} 1} \plus{} 1}$.
If $ b_{n \minus{} 1} \equal{} b_n$,then $ b_{n \plus{} 1} \equal{} b_n$ else $ b_{n \plus{} 1} \equal{} b_n \plus{} 1$.
Therefore $ a_n \equal{} 2^{n \minus{} c} \plus{} 2^n$, were $ c\ge 1$ is first case, when used $ a_{c \plus{} 1} \equal{} 2a_c$ (if it is not used c=n $ a_n \equal{} 1 \plus{} 2^n$).


[b]A[/b](a1,a2), [b]B[/b](b1,b2) and [b]C[/b](c1,c2) are three collinear points. Prove that [b]A[/b], [b]B[/b] and [b]C[/b] are collinear if and only if [b]AB[/b]=k[b]AC[/b] where k [i]E[/i] R.

Yea, I know I'm no good  :oops:

Hm! I thinks it mean:$ \overrightarrow{AB}\equal{}k\overrightarrow{AC}$ and it is clearly!

I know but my teacher want me to prove that AB=kAC (AB and AC are vectors and k is a constant)

[b]let $ A,B,C$ be three distinct points in the plane,and $ O$ be an arbitary point,then $ A,B,C$ are collinear iff there exists a real number $ \lambda$ in the interval $ (0,1)$ s.t:
$ \overrightarrow{OC} = \lambda\overrightarrow{OA} + (1 - \lambda)\overrightarrow{OB}$[/b]

[u]proof:[/u]
let $ \frac {|AC|}{|CB|} = \frac {m}{n}$ then we get that $ |AC| = \frac {m}{m + n},|CB| = \frac {n}{m + n}$ therefore we have:

$ \overrightarrow{AC} = \frac {m}{m + n}.\overrightarrow{AB}$

now note that:

$ \overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OA} + \frac {m}{m + n}.\overrightarrow{AB}$

we have $ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$

so the last equality becomes:

\begin{eqnarray*}\overrightarrow{OC} = \overrightarrow{OA} + \frac {m}{m + n}\left(\overrightarrow{OB} - \overrightarrow{OA}\right) = \frac {\left(m + n\right)\overrightarrow{OA} + m\left(\overrightarrow{OB} - \overrightarrow{OA}\right)}{m + n} = \left(\frac {n}{m + n}\right)\overrightarrow{OA} + \left(\frac {m}{m + n}\right)\overrightarrow{OB}\end{eqnarray*}

now let $ \lambda = \frac {n}{m + n}$

we get the desired result...

now just simplify this result,you'll get what you want...


function
inequalities
inequalities proposed

Let $ a,$ $ b$ and $ c$ are non-negative numbers such, that $ a^4\plus{}b^4\plus{}c^4\equal{}3.$ Prove that:
\[ a^5b\plus{}b^5c\plus{}c^5a\leq3\]

I'm not quite sure whether my solution is correct since I have just started learning analysis.

Let:
$ F(a,b,c)\equal{}a^{4}\plus{}b^{4}\plus{}c^{4}$
and
$ f(a,b,c)\equal{}a^{5}b\plus{}b^{5}c\plus{}c^{5}a$
Then:
$ \bigtriangledown F(a,b,c)\equal{}(4a^{3},4b^{3},4c^{3})$
$ \bigtriangledown f(a,b,c)\equal{}(5a^{4}b,5b^{4}c,5c^{4}a)$
From Lagrange's multiplier theorem the equation given in the problem will follow only when both gadients are linear dependent (looking for maximum value of function), thus there exists x that:
$ 4a^{3}*x\equal{}5a^{4}b \\ 4b^{3}*x\equal{}5b^{4}c \\ 4c^{3}*x\equal{}5c^{4}a$
$ x\equal{}\frac{5}{4}ab\equal{}\frac{5}{4}{bc}\equal{}\frac{5}{4}ca$
Since $ a\equal{}b\equal{}c$ is sufficient for this equation the inequality given follows
 :wink: Of course it'd be nice if sb tried to evaluate my proof.

[quote="polskimisiek"]
Let:

$ f(a,b,c) \equal{} a^{5}b \plus{} b^{5}c \plus{} c^{5}a$
Then:
$ \bigtriangledown f(a,b,c) \equal{} (5a^{4}b,5b^{4}c,5c^{4}a)$
[/quote]
I think, here you are wrong. :wink:

Aaah, yes. You're right.
Well, it turns out that it's not that easy.
So the problem is brought to proving that the equation:
$ \frac {a^{4}b\plus{}c^{5}}{a^{3}} \equal{} \frac {b^{4}c \plus{} a^{5}}{b^{3}} \equal{} \frac {c^{4}a \plus{} b^{5}}{c^{3}}$
can be satisfied only when $ a \equal{} b \equal{} c$ which unfortunately isn't as easy as the previous one, so I conclude that my proof will not be possible to use ;)

Great inequality arqady .Here is my solution (hope you will like it)
By AM-GM we have $ a^5b \plus{} b^5c \plus{} c^5a\leq \frac {a^6b^2 \plus{} a^4 \plus{} b^6c^2 \plus{} b^4 \plus{} c^6a^2\plus{}c^4}{2}$

So we have to prove that $ a^6b^2 \plus{} b^6c^2 \plus{} c^6a^2\leq 3$ or we have to prove that

$ (a^4 \plus{} b^4 \plus{} c^4)^2\geq 3(a^6b^2 \plus{} b^6c^2 \plus{} c^6a^2)$ which is the famous Vasc's inequality

$ (a^2 \plus{} b^2 \plus{} c^2)^2\geq 3(a^3b \plus{} b^3c \plus{} c^3a)$    :D