[quote="236factorial"]Solve this equation in base 8.

2x + 16 = 4(x+2)[/quote]

[hide]2x + 16 <base 8> = 4 (x +2) <base 8>
2x+16 <base 8>= 4x + 10<base 8>
6 <base 8>= 2x <base 8>
x = 3 [/hide]

[quote="math92"]that looks exactly like B4k4ka's solution :agent:[/quote]

It is B4kaka's solution  ;)

let's stop copy and pasting  :(

[quote="risinglegend"]2x + 16 <base 8> = 4 (x +2) <base 8>
2x+16 <base 8>= 4x + 10<base 8>
6 <base 8>= 2x <base 8>
x = 3                      :winner_first:[/quote]

First of all, please hide your next solution.

Next, don't copy and paste. That's worse than spamming.

okay, why don't you solve the problem  :roll:

[hide][quote="236factorial"]Solve this equation in base 8.

2x + 16 = 4(x+2)

[/quote]

$ 2x + 16 = 4x + 10 \text{ since 8 in base eight is ten.}$

$ -2x = -6$

$x=3$[/hide]


calculus
number theory unsolved
number theory

Can any one help me  with this:

 x,y  \in  Z.Prove that the only solutions to

 x ^3 +y ^3 +xy 2  +x 2 y=8(x 2 +y 2 +xy+1)

are (2,8) and (8,2).

Here is a bad way,

Put u=x+y and v = x^2 +y^2 , then we have

     uv=8(v+1) + 4(u^2- v)
so
    uv=4(v+2+u^2)

    v(u-4)=4(2+u^2)

This implies u>4 , now note that 2v >= u^2 , then we get
    (u-4)u^2 <= 16+8u^2
so
    (u-12)u^2 <= 16
Then u<=12
Note that u and v are even , then we have u=6,8,10,12
Now calculus each case of u we hope to get your sol (8,2) ; (2,8) :D
[note 2v-u^2 is a perfect square]


Let $ABC$ be an equilateral triangle with side length $1$ and let $P$ be a point inside this triangle. What is the minimum value of $PA+PB+PC$?

[size=75][color=darkred]Added LaTeX coding
-nebula42[/color][/size]

[quote="nebula46"]Let ABC be an equaliteral triangle with side length 1 and P be a point inside this triangle. What is the minimum of PA+PB+PC?[/quote]
Delong, please stop using my name  :wink:  :mad: , and please stop posting problems that are too difficult for MathCounts in this forum. I'm moving it to Intermediate.  :wink:

this is the first fermat point, which happens to be the center of an equilateral triangle...it follows that the sum is $\sqrt{3}$

Oh, I was thinking of the sum of the perpendicular distances to the sides of the triangle which is constant. That's $\frac{\sqrt{3}}{2}$, right?


calculus
MATHCOUNTS
geometry
3D geometry
function
conics
hyperbola

Using calculus :

[hide]ln(x)/x has maximum for x=e and e^(1/e) < 1.45[/hide]

Challenge or request for help?

If you know calculus, belenos provides a very simple solution.  But there are non-calculus solutions to this problem.

Meanwhile, there's a closely related MATHCOUNTS problem, which was from the Target Round of the 2002 National Competition.  I'm paraphrasing the question:

Given a set of positive integers, whose sum is 20, what is the maximum possible product of these integers.

When x is negative (less than -1), imaginary numbers will result most of the time. You weren't specific.

When x is between -1 and 0, the number will be raised to a power; like 1/x = 1/-.25 = -4, so imaginaries are no problem. But the number will only get closer to zero because it is greater than -1. And THEN the number will be reciprocal'd to equal a largely negative number, like -.25^-4 = 1/-.25^4 =  -(4^4) = -256. So these will be less than 1.45.

x=0 is a double whammy paradox: 0^(1/0).

When x is between 0 and 1, the number will be raised to a power like before. Take x=1/3:
1/3^3 < 1.45?  1/729  :approx: 0.00137 <1.45.

Finally, when x is greater than 1, you basically say that the xth root of x is always less than 1.45. right?

2^(1/2) = square root of 2  :approx: 1.4142135
3^(1/3) = cube root of 3  :approx: 1.4422495
4^(1/4) = 4th root of 4  :approx: 1.4142135
5^(1/5) = 5th root of 5  :approx: 1.3797296
6^(1/6) = 6th root of 6   :approx: 1.3480061

And it goes down from there. The closest it gets to 1.45 is the cube root of 3. The graph of this function confirmed that.

Cool problem.

rcv wrote:Given a set of positive integers, whose sum is 20, what is the maximum possible product of these integers.

[hide]By trying different things I get 3^6 * 2, or 1458, but I don't know how to prove this other than listing out all the possible combinations.[/hide]

[quote="Hexaditidom"]And THEN the number will be reciprocal'd to equal a largely negative number, like -.25^-4 = 1/-.25^4 =  -(4^4) = -256. So these will be less than 1.45.[/quote]

Actually, I think we would have (-4)^4=+256.

So we have either trivially disproved the proposition, or we should assume the writer means "for positive x, prove that x^(1/x)<1.45".

Hexaditidom wrote:And it goes down from there. The closest it gets to 1.45 is the cube root of 3. The graph of this function confirmed that.

I suppose you are new here, based on your number of posts, so you really aren't expected to know this... I guess... anyway.

Unless otherwise stated, the use of a calculator or any computing device is not allowed... thus, a solution by decimals is not acceptable or useless, anyone can plug stuff into a calculator.

Also, the cube root of 3 is not the maximum value, it has already been posted that [hide]e^(1/e)[/hide] is the maximum.

I am sure the problem calls for x > 0... as this function becomes very messy when x < 0, jumping continuities all over the place...

[quote="rcv"]Challenge or request for help?.[/quote]
request.  Thought of it at school and tested for a lot of numbers.  Seemed to work.  Definetely worked for all integers (i think i have a proof for that).

[quote="Robert0201"]By trying different things I get 3^6 * 2, or 1458, but I don't know how to prove this other than listing out all the possible combinations.[/quote]

We are still lacking a (non-calculus) proof for the original problem.

Robert, your answer is correct.  Here is an outline of the proof applicable to the MATHCOUNTS problem:

By contradiction.  Suppose the optimal solution set, S, contains a "1".  The sum of the elements is known to be 20 and the product of the elements is given by P.  Create a set S', like the original.  But we remove the 1 and any other element, x, from the set and replace them with a single element (x+1).  The sum of the elements in S' is still 20, but the product of the elements in S' is [tex]P'=\frac{x+1}{x\times 1}P[/tex].  Since P'>P, we have a contradiction.  Therefore, the optimal solution set cannot contain a "1".

Suppose the optimal solution set, S, contains a "5".  We create set S' like S, but replace the "5" with two elements, "2" and "3".  The sum of the elements in S' is still 20, but the product of the elements in S' is [tex]P'=\frac{2\times 3}{5}P[/tex].  Since P'>P, we have a contradiction.  Therefore, the optimal solution set cannot contain a "5".

We can generalize the above by showing that [tex]2\times (x-2)>x, 2x-4>x, x>4[/tex].  Therefore, the optimal solution set cannot contain any element >= 5.

Suppose the optimal solution set contains a "2" and a "4".  We create set S' like S, but replace the "2" and "4" with two elements, "3" and "3".   [tex]P'=\frac{3\times 3}{2\times 4}P[/tex].  Since P'>P,, the optimal solution set cannot contain both a "2" and a "4".

Suppose the optimal solution set contains two "4"s.  We create set S' like S, but replace the {4,4} with {3,3,2}.  [tex]P'=\frac{3\times 3\times 2}{4\times 4}P[/tex].  Since P'>P, the optimal solution set cannot contain two (or more) "4"s.

Suppose the optimal solution set contains three "2"s.  We create set S' like S, but replace the {2,2,2} with {3,3}[tex]P'=\frac{3\times 3}{2\times 2\times 2}P[/tex].  Since P'>P, the optimal solution set cannot contain three (or more) "2"s.

By contradiction, we have eliminated all solutions except the following:  A solution set may contain up to two 2's and any number of 3's.  Or a solution set may contain up to one 4 and any number of 3's.  Since 20 modulo 3 = 2, the unique uncontradicted solution is the set containing exactly one 2 and enough 3's to bring the sum to 20.

[quote="Scrambled"][quote="rcv"]Challenge or request for help?.[/quote]
request.[/quote]

OK.  Here is my attempt at a non-calculus solution.

Lemma 1:  [tex]\ln(p) \le p-1$, where $p>0[/tex]

By definition, [tex]\ln(p)[/tex] is the area bounded by the hyperbola xy=1, the x-axis, and the vertical lines x=1 and x=p.  See illustration below and see [url=http://mathworld.wolfram.com/NaturalLogarithm.html]mathworld[/url] for more information on this definition.

When p=1, the value of the ln function is zero.  When p>1, the function is taken to be positive.  And when p<1, the function is taken to be negative.

Case 1:  Observe that when [tex]p>1[/tex], this area is contained within a rectangle bounded by the horizontal line y=1, the x-axis, and the vertical lines x=1 and x=p.  By the very definition of [tex]\ln(p)[/tex] (and the formula for the area of a rectangle) we see that:

[tex]\displaystyle\ln(p) < p-1$, where $p > 1[/tex]

Case 2:  Conversely, when [tex]p<1[/tex], the rectangle is contained within the area bounded by the hyperbola:

[tex]\displaystyle 1-p < -\ln(p)[/tex]
[tex]\displaystyle\ln(p) < p-1$, where $p < 1[/tex].

Case 3:  When p=1, the area is 0.

[tex]\displaystyle\ln(p) = 0 = p-1$, where $p = 1[/tex].

We have proven for p>1, p<1, and p=1 that [tex]\ln(p) \le p-1[/tex].
[End of Lemma 1]

Main theorem.  Now we intend to prove that [tex]e^{1/e} \ge x^{1/x}$, where $x>0[/tex]

[tex]\displaystyle e^{1/e} : x^{1/x}[/tex]

[tex]\displaystyle \frac{1}{e}\times\ln(e) : \frac{1}{x}\times\ln(x)[/tex]

[tex]\displaystyle \frac{1}{e}\times 1 : \frac{1}{x}\times\ln(x)[/tex]

[tex]\displaystyle \frac{x}{e} : \ln(x)[/tex]

[tex]\displaystyle \frac{x}{e}-1 : \ln(x)-1[/tex]

[tex]\displaystyle \frac{x}{e}-1 : \ln(x)-\ln(e)[/tex]

[tex]\displaystyle \frac{x}{e}-1 : \ln(\frac{x}{e})[/tex]

By Lemma 1:
[tex]\displaystyle \frac{x}{e}-1 \ge \ln(\frac{x}{e})[/tex]

Therefore,
[tex]\displaystyle e^{1/e} \ge x^{1/x}[/tex]

Finally, we evaluate [tex]e^{1/e} \approx 1.444668[/tex]

So:
[tex]\displaystyle x^{1/x} \le e^{1/e} \le 1.444668 < 1.45[/tex]

[tex]\displaystyle x^{1/x} < 1.45[/tex]

QED.

[quote="rcv"]there's a closely related MATHCOUNTS problem...

Given a set of positive integers, whose sum is 20, what is the maximum possible product of these integers.[/quote]

To tie these two together.  Suppose the MATHCOUNTS problem required all of the integers to be equal.  The possible products would have been 1^20, 2^10, 4^5, 5^4, 10^2, and 20^1.  [1, 1024, 1024, 625, 100, and 20]

The above are all of the form [tex]n^{20/n}[/tex].  We seek to maximize this value.

[tex]\displaystyle n^{20/n} = (n^{1/n})^{20} \approx 1.444668^{20} \approx 1568.05[/tex]

This provides an upper bound to the MATHCOUNTS problem, and suggests the solution set will have many integers near e :approx: 2.718.

>>>>>> a non-calculus solution.

I understand it may be interesting and challenging to find a non-calculus solution, but here I'm a bit confused !

First how do you define log and e^ without calculus (x^1/x = e^(log(x)/x))  by definition) ?? All the ones I know use some calculus (series, continuous morphisms, int, ...) at some point.
Or does the problem assumes that x is an integer (> 0) ??

How do you prove rigourously that log(x) <= x-1 without using some (perhaps hidden but still present) calculus ??

Don't take this post too seriously (there is no aggressivity and it is not a need to prove that calculus is the only way), it's just curiosity  :)

[quote="belenos"]>>>>>> a non-calculus solution.

I understand it may be interesting and challenging to find a non-calculus solution, but here I'm a bit confused !
...
Don't take this post too seriously (there is no aggressivity and it is not a need to prove that calculus is the only way), it's just curiosity  :)[/quote]

First of all, calculus is *clearly* the best way to solve this problem.  Your 1-line solution was great!  :)

But this sort of problem crops up from time-to-time on competitions that allegedly don't require any knowledge of calculus.  -- Perhaps the most famous such question is "which is larger, e^:pi: or :pi:^e".

If my solution used calculus, I apologize.

[quote="belenos"]How do you prove rigourously that log(x) <= x-1 without using some (perhaps hidden but still present) calculus ??[/quote]

According to Mathworld, ln(x) is "defined" as the area under the hyperbola.  To evaluate ln(x) requires calculus.  To see that ln(x) < x-1 is merely an observation based on simple properties of a hyperbola and a rectangle.  Is this calculus?

[If an ellipse fits entirely within a 1x2 rectangle, I think I can safely claim the perimeter of the ellipse is less than 6 and the area is less than 2.  Have I used calculus to make this claim?]

I hope the problem gurus on this site join the conversation.  What *is* the best pre-calc method for solving this class of problem?  Or should this class of problem never have appeared on high-school level competitions that claim calculus is not required.  Cheers!  :-)

[quote="rcv"]But this sort of problem crops up from time-to-time on competitions that allegedly don't require any knowledge of calculus.  -- Perhaps the most famous such question is "which is larger, e^:pi: or :pi:^e".[/quote]

How do you do that one without calculus? You can still solve it by finding the maximum of x^(1/x), and then you don't have to do any computations. (A friend gave me this problem once and told me to solve it with only basic calculus and no computation.)

belenos makes a good point. There is no purely "non-calculus" solution to this problem, because in essence it is a "calculus" result. But students start learning many subtle "calculus" concepts before they actually learn about derivatives and integrals. For example, the area of a circle is technically a result of calculus, but that doesn't mean students aren't expect to know it, or that questions involving the area of a circle can't be used on contests that claim to require no calculus. The natural logarithm is also a thing that technically belongs to calculus, but students learn about it long before then, and certain things about it are fair game to use on contests.

But the definition of the natural logarithm as the area under 1/x is a calculus topic for sure (notice that you will only see it defined that way in calculus books). Some teachers are starting to teach it in Pre-Calculus, but most still teach it as the inverse of the exponential function (without getting into the details of how the exponential is defined for irrational exponents). So this is a definition I would not expect students to know before taking Calculus.

So this question probably shouldn't appear on a contest that doesn't require calculus. rcv's solution is still quite excellent though! I like it very much.


[hide="solution"]
We see that there are four of $ 2^{50}.$ That means $ 4\left(2^{50}\right) \equal{} 2^{n}.$ We know $ 4$ can be expressed as $ 2^2,$ and by an exponent law, $ 2^2\cdot2^{50} \equal{} 2^{52}\Rightarrow n \equal{} \boxed{52}.$[/hide]

[quote="RoFlLoLcOpT"][hide="solution"]We know $ 4$ can be expressed as $ 2^2,$ and by an exponent law, $ 2^2\cdot2^{52} \equal{} 2^{52}\Rightarrow n \equal{} \boxed{52}.$[/hide][/quote]
A minor typo here, it should say $ 2^2\cdot2^{50} \equal{} 2^{52}\Rightarrow n \equal{} \boxed{52}.$ (typo was in the second exponent, should be 50 instead of 52)


Dear Forum  :)

I hope you could help me with these problems.

1.- The following equations have a straight line in commun. So the system is
a) Determined (consistent?)
b) indetermined
c) incopatible

2.- How can I solve this system, please?
 -x + 1/5y = 1
       x+2y = 4
     3x + y = 2

 3.- How can I solve this system with the Gauss Method, please?
x - y + 2z =1
  2x     - z = 2
 x + y     = -3

Many thanks
R097

hello, your first system has no real solutions. Considering the first and the second equations of your system
and adding both of them we have $ y\equal{}\frac{25}{11}$, substituting this in the first equaton we get $ x\equal{}\minus{}\frac{6}{11}$, substitung $ x$ and $ y$ in your third equation we get a contradiction.
Sonnhard.

3.

$ \left [ \begin{array}{ccccc} 1 & \minus{}1 & 2 & | & 1 \\ 2 & 0 & \minus{}1 & | & 2 \\ 1 & 1 & 0 & | & \minus{}3 \end{array} \right ] \longrightarrow \left [ \begin{array}{ccccc} 1 & \minus{}1 & 2 & | & 1 \\ 0 & 2 & \minus{}5 & | & 0 \\ 0 & 2 & \minus{}2 & | & \minus{}4 \end{array} \right ] \longrightarrow$

$ \left [ \begin{array}{ccccc} 1 & \minus{}1 & 2 & | & 1 \\ 0 & 2 & \minus{}5 & | & 0 \\ 0 & 0 & 3 & | & \minus{}4 \end{array} \right ] \longrightarrow \left [ \begin{array}{ccccc} 1 & \minus{}1 & 2 & | & 1 \\ 0 & 2 & \minus{}5 & | & 0 \\ 0 & 0 & 1 & | & \minus{}\frac43 \end{array} \right ] \longrightarrow$

$ \left [ \begin{array}{ccccc} 1 & \minus{}1 & 0 & | & \frac{11}{3} \\ 0 & 2 & 0 & | & \minus{}\frac{20}{3} \\ 0 & 0 & 1 & | & \minus{}\frac43 \end{array} \right ] \longrightarrow \left [ \begin{array}{ccccc} 1 & \minus{}1 & 0 & | & \frac{11}{3} \\ 0 & 1 & 0 & | & \minus{}\frac{10}{3} \\ 0 & 0 & 1 & | & \minus{}\frac43 \end{array} \right ] \longrightarrow$

$ \left [ \begin{array}{ccccc} 1 & 0 & 0 & | & \frac{1}{3} \\ 0 & 1 & 0 & | & \minus{}\frac{10}{3} \\ 0 & 0 & 1 & | & \minus{}\frac43 \end{array} \right ]$

Therefore, the solution is $ (x,y,z)^T \equal{} \left(\frac13, \minus{}\frac{10}{3}, \minus{}\frac{4}{3}\right)^T$.

2.

$ \left [ \begin{array}{cccc} \minus{}1 & \frac15 & | & 1 \\ 1 & 2 & | & 4 \\ 3 & 1 & | & 2 \end{array} \right] \longrightarrow \left [ \begin{array}{cccc} \minus{}1 & \frac15 & | & 1 \\ 0 & \frac{11}{5} & | & 5 \\ 0 & \frac85 & | & 5 \end{array} \right]$

and from here we can see that the system is inconsistent.


conics
ellipse
hyperbola
parabola
analytic geometry
rotation

1.With a given point and line as focus and directrix, a series of ellipses are described.The locus of the extremities of their minor axes is

A.an ellipse

B.a hyperbola

C.a pair of straight lines

D.a parabola

2.An ellipse of major axis $ 20\sqrt{3}$ and minor axis 20 slides along the coordinate axes and always remains confined in 1st quadrant.The locus of the ellipse therefore describes athe arc of a circle.Find the length of this arc and also the equation of the locus of the centre of the ellipse.

Please solve with appropriate explanation....

[hide="1"]
Consider some ellipse that has the given focus and directrix. Let the distance from the center be $ c$, the length of the major radius be $ a$ and the length of the minor be $ b$, so that the distance from the center to the directrix is $ \frac{a^2}c$, by definition, and the distance from the focus to the directrix is $ \frac{a^2}c\minus{}c\equal{}\frac{a^2\minus{}c^2}c\equal{}\frac{b^2}c\equal{}k$ for some constant $ k$. Now, rotate everything so that the directrix is parallel to the $ x$-axis and shift so that the origin is the focus. All points on the locus of endpoints of the minor axes can then be expressed as $ (\pm{ck},c)$; this corresponds to the parabola $ f(x)\equal{}\frac{x^2}k$.
[/hide]


function
inequalities
triangle inequality
real analysis
real analysis unsolved

let $ f(z)\equal{}\sum_{n\equal{}0}^{\infty}a_n.z^n$ . Assume that $ \sum|a_n|<\infty$ . then show that f is of  bounded variation on every radius of the unit circle.

It is just too easy. :) If you have any sequence of functions $ f_j$ on $ [0,1]$ such that $ \operatorname{Var}f_j\le 1$ for each $ j$, then for any sequence of numbers $ a_j$ such that $ \sum_j |a_j|<\plus{}\infty$, the function $ f\equal{}\sum_j a_jf_j$ also has bounded variation and, moreover, $ \operatorname{Var}f\le\sum_j|a_j|$. The proof is entirely straightforward: just write the definition of the variation, use the triangle inequality, and swap the sums.

i dont know how i miss that. i feel like a total idiot  :oops:


Suppose A is a subset of R^2. Show that A can contain at most one point p such that A is isometric to A\{p} with the usual metric.

Consider $ A\equal{}\{e^{i\sqrt{2}\pi k},\ k\in\mathbb{N}\}$. Then $ e^{i2\sqrt{2}\pi }\left(A\setminus\{e^{i\sqrt{2}\pi k},\ k \equal{}1,2\}\right) \equal{} A.$
so we have 2 such points

You're only allowed to remove one point at a time.  In other words, if removing a single point gets you an isometric space then removing any finite number of points does as well.  But the question is asking for points such that if you remove [i]only[/i] that point (and no others) the result is isometric.  $ k \equal{} 2$ in your example does not have this property.

It seems like the example $ A \equal{} \mathbb{N} \subset \mathbb{R}^2$ would have been more straightforward.


Each of four test scores in Connie's class is to be weighted equally. On the first three tests Connie scored $ 79\%$, $ 87\%$, and $ 98\%$. What percent must she score on her fourth tests to have an overall average of exactly $ 90\%$?

to get an average of $ 90\%$, she must get a sum of $ 360$

so, we subtract the scores that we do know
$ 360 \minus{} 79 \minus{} 87 \minus{} 98 \equal{} 96$

our answer is $ \boxed{96}$

Or do it my way:

$ 98$ is $ 8$ above $ 90$
$ 87$ is $ 3$ below $ 90$
$ 79$ is $ 11$ below $ 90$

So to compensate, the fourth number must be $ 11\plus{}3\minus{}8\equal{}6$ above $ 90$.

$ 90\plus{}6\equal{}96$

[hide="Or my way"]
If $90\%$ is her overall average, then her overall error must by $10\%$.  Let's assume that each test is $100$ points for a total of $400$ points.  $10\%$ of $400$ points is 40 points.  She has already missed $2+13+21=36$ points; therefore, she must lose $4$ more points or get a $\boxed{96\%}$ in order to get an overall average of $90\%$.
[/hide]


[quote="modx07"]i am currently in 9th grade (going into 10 for 09-10 school year)

i went to Mathcounts State competetion when i was a 7th and 8th grader, but havnt taken amc yet (hopefully good!)...

if i dont pass amc, will i still automatically qualify for MathZoom?[/quote]

Yes you qualify  :)


Let g be an even natural number and f(n) = g^n + 1 (n \in N).
Prove that the following properties are satisfied for all n \in N:

a.) f(n) is a divisor of each of the numbers f(3n), f(5n), f(7n),...
b.) f(n) is a divisor of none of the numbers f(2n), f(4n), f(6n),...

[quote="orl"]hey the problem is not difficult but you must be careful ... and you did not provide a solution ... some people might only prove part b.) and then they try deduce part a.) but that is not sufficient as part a.) is not the opposite of part b.)  :D[/quote]

Why do you say I didn't provide solution? Is there mistake?

a) g(2k+1)n+1=(gn+1)(g2kn-g(2k-1)n+...+1)

b) g2kn+1=(g2n)k-1+2=(g2n-1)(g2n(k-1)+...+g2n+1)+2=
(gn+1)(gn-1)(...)+2


I've been trying this but I'm having trouble:

Prove:

Matrix $ A$ is symmetric $ \Rightarrow$ $ A$ has real eigenvalues.


geometry
parallelogram
circumcircle
geometry unsolved

1.) In the plane let us consider a set $S$ consisting of $n \geq 3$ distinct points satisfying the following three conditions:

[b]I.[/b] The distance between any two points $\in S$ is not greater than 1.
[b]II.[/b] For every point $A \in S$, there are exactly two “neighbor” points, i.e. two points $X, Y \in S$ for which $AX = AY = 1$.
[b]III. [/b] For arbitrary two points $A, B \in S$, let $A', A''$ be the two neighbors of $A, B', B''$ the two neighbors of $B$, then $A'AA'' = B'BB''$.

Is there such a set $S$ if $n = 1991$? If $n = 2000$ ? Explain your answer.

The answer is yes $n=1991$ and no for $n=2000$.
For $n=1991$ we can take $S$ to be the set of vertices of a regular $1991$-gon $A_{1}A_{2}\dots A_{1991}$, such that $A_{i}A_{i+995}=A_{i}A_{i+996}=1$. Because $A_{i+995}A_{i+996}$ is opposite side of $A_{i}$, it is clear that $A_{i}A_{i+995}$ is the maximal diagonal, so $XY\leq 1$, $\forall X,Y\in S$. The conditions [b]II[/b] and [b]III[/b] are obviously satisfied as well.

Now we treat the case $n=2000$, or just $n$ even. Take $A\in S$. Let $A'$ and $A''$ be its [i]neighbors[/i]. We know that $A'$ has one more [i]neighbor[/i] Denote it $A'''$. If the points $A''$ and $A'''$ lie on different sides of $AA'$, then $AA''A'A'''$ is a parallelogram with greater diagonal $A''A'''>AA'=1$, which contradicts [b]I.[/b]
Thus $A'''$ lies on the same side of $AA'$ as $A''$. It's not difficult to prove that $A, A', A''$ and $A'''$ lie on a circle. Proceeding further for $A'A'''$ [as we did with $AA'$] and on, we obtain points all situated on the same circle, and we would either reach $A$ or obtain a point that lies on the chord $AA'''$ of the circle, that would be at a distance greater than $1$ from $A'$. Thus we must reach $A$. Then we obtain a regular polygon, with two maximal diagonals $AA'=AA''=1$, so it must have an odd number of vertices.
Because $n$ is even, $S$ can be decomposed in smaller subsets, each of which represents the set of the vertices of a regular $2k+1$-gon. Note that all these regular polygons must be congruent to each other. Consider one of them $P$, and its circumscribed circle $\Gamma$ with diameter $d$. Since $1<d$, we get that all vertices of other polygons must be inside each of the circles with radius $1$ centered at each vertex of $P$, or, as a weaker [but more needed] consequence, strictly inside $\Gamma$. But this is impossible, since all other regular polygons are congruent to $P$.


[b][size=150]Suppose f(x)=x^3 + ax^2 +bx + c satisfies f(-2)=-10, and takes the extreme value 50/27 when x=2/3. Find the values of a, b and c. [/b][/size]

Stop using this big fat text please!

Also don't post your school problems here, Getting started or so would be better (and when you want help on them show us what you already tried on it etc., e.g. show us your partial results).


Is it something known about the number of nonquadratic residues smaller then $ \displaystyle \frac {p \minus{} 1}{2}$ for $ p$ prime with $ p \equiv 3(mod \, 4)$?? :?:

We have $ \sum_{a\equal{}1}^\frac{p\minus{}1}{2} \left( \frac ap \right) \equal{} \left(2\minus{}\left( \frac 2p \right) \right) \cdot h_{\minus{}p}$, where $ h_{\minus{}p}$ is the class number of $ \mathbb Q[\sqrt p]$.
From this you immediatly get the amount of nonquadratic residues to be $ \frac{p\minus{}1}4 \minus{} \left(2\minus{}\left( \frac 2p \right) \right) \cdot \frac{h_{\minus{}p}}{2}$.


So, what do we do now?  Should we introduce ourselves, or what?

Great idea. Shoulda thought of it sooner myself.


Dear friends,
i want 2 know how to solve 2nd degree or more diophantine equations.
I am sending an easy prob
1/x+1/y=1/z
plz help me to solute this nd plz send more probs nd solution
>moon

This is not an open question.  Could a moderator please move this?

[b]Solution.[/b] Let $\gcd (x,y)=d$.  Then $x=da,\ y=db$.
\[\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\implies z(x+y)=xy.\]
Substituting $x=da,\ y=db$, we have
\[z=\frac{dab}{a+b}\implies d=k(a+b).\]
As a result, $(x,y,z)=(ak(a+b),bk(a+b),kab)\in\mathbb{Z}^{3}$.$\blacksquare$

To Valentin: RSS makes this forum much better and less time consuming. :)


function
limit
LaTeX
real analysis
calculus
calculus computations

1)Let $f(x) = x$if x is rational and ${-}{x}$if x is irrational. Show that f is continuous at $x_{0}= 0$ but not continuous at $x \neq 0$.

2) Using $\epsilon-\delta$Show that $f(x) = \frac{1}{x}$ is continuous at $(0,1)$ but not uniformly continuous. Can you explain what is the difference between continuous and uniformly continuous?

1) For all $\epsilon > 0$
$|x-0| < \frac{ \epsilon }{ \max\{ 1, | a |\}}\Rightarrow | f(x)-f(0)| = |f(x)| \leq \max\{ 1, | a | \}\cdot |x| = \epsilon$.
Hence $f$ is continuous at $x_{0}= 0$.

If $x \not= 0, a \not= 1$, let $(a_{n})_{n \in \mathbb{N}}$ be a sequence of rational numbers with $\lim_{n \to \infty}a_{n}=x$ and $(b_{n})_{n \in \mathbb{N}}$ be a sequence of irrational numbers with $\lim_{n \to \infty}b_{n}=x$.
If $f$ was continuous at $x$, then
$x = \lim_{n \to \infty}f(a_{n}) = \lim_{n \to \infty}f(b_{n}) = ax$
Contradiction!

[quote="Ramanujan"]I think he want to write $-x$ for $x$ irrational .[/quote]

Although that is just a special case of what Pi-J has proved.

Guile, you use "long dashes" when you type, which Latex interprets incorrectly (that is, not as minus signs).  If you use "short dashes," it will work better.

[color=blue][Guile's post has now been edited by moderator][/color]


symmetry integration geometry calculus derivative

it's easy, for more info read "Simplifying the Laplace Transform and Fourier Transform and Winning 5nnnnnnnnn...n Dollars"

chapter 2, paragraph #3, the answer is there

I am not amused by your (singular) shenanigans.

Shenanigans, The Irish Pub, Restaurant & Bar?

google it

 :lol:

[quote="Thales418"]${f(x)=\sum_{1}^{\infty }x^{n}a_{n}}$
${g(x)=\int_{1}^{\infty }a_{n}(dx)^{n}}$

If $f = g$

${\sum_{1}^{\infty }x^{n}a_{n}==\int_{1}^{\infty }a_{n}(dx)^{n}}$

By symmetry

${dx =1}$
${\sum =\int{}}$

i will give 500000.....00 dollars for the first who simplify the laplace transforme and fourier transform[/quote]

it will be
$d = 1$

And what is there to prove in it? It is a direct consequence of the Poincare conjecture Grisha proved, with some differetnial geometry thrown in. On second thoughts, you need a spectral decomposition, but only once.

I know this buda

my conjecture prooves a lot of others conjectures and disproove a lot of anothers

but the question is

if a violin is made with Sigma Holes, the sound will be better?

since the f  is igual to sigma

you can't say that without checking the moore penrose inverse of the orthogonal linear transformation that takes the violin to a viola.

no no saxophones won't do - they are not topologically compact.

as for cellos, they are theoretically fine, but practically too huge to be of any use.

Hey, I play the cello! They are superior because they don't make a squeaky noise like violins, and their endpins can be taken out to whack [i]or[/i] stab people!

flute is the best!!!

i will prove this

suppose flute is not the best

so flute need to be the worst, but, everybody nows that the worst instrument is drums

so, by contradiction, flute is the best

why u make random posts?

are you bullying me?

[quote="randomdragoon"][quote="bubka"]ah but the endpins have a serious disadvantage, they have denumerably many points where the derivative is not continuous.[/quote]
That's why they hurt. Perfect for getting through mobs and playing triangle and stuff.[/quote]

That maybe true ... I've never used them.
And anyway why can't militia hoplites put up a fight against hastati?

this is my topic

someone can close this!?

the question have already been solved

dont make sense this idiot discussion


calculus
integration
logarithms
algebra open
algebra

I have a prolem so that.

if I=$\int_0^1 \frac{2+\sqrt{x+1}}{1+\sqrt{x+2}}\,dx$dx
I=?

Can you help me and reply soon?
can you do it by many way?
Thank you very much!

Please post in the correct (sub)forums!
This is neither Algebra nor an Open Problem!

[quote="Rust"]Let $1+\sqrt{x+2}=y,x=(y-1)^2-2,x+1=y^2-2y$. Then \[ I=2\int_{1+\sqrt 2}^{1+\sqrt 3} \frac{2+\sqrt{y^2-2y}}{y}dy=4\ln{\frac{1+\sqrt 3}{1+\sqrt 2}}+2\int_{1+\sqrt 2}^{1+\sqrt 3} \frac{\sqrt{y^2-2y}}{y}dy.  \][/quote]

Hey, Rust - and what about expressing $dx$ in terms of $dy$??? There should be $dx=2(y-1)dy$, but I don't see it anywhere.


What is the largest possible number of acute angles (angles less than 90 degrees) a 2001-gon (polygon with 2001 sides) can have if no 2 sides cross each other?

:roll: [hide]I think that is 3.Because we have the sum of the angles in 2001-gons is (2001-2).180=359820
Call the least possible number of the acute angles is x.
If $\ x\geq\ 4$ use Dirichle we have the wrong.
Try with 3 we have the right.[/hide]

Math is my life :!:

[hide="Guess"] $\boxed{1000}$? I don't know how to explain the shape. It's kind of like a zig-zag that eventually closes. Like how little kids draw suns: a circle, then zig zags around it. Of course, you could make this more "efficient" by having more acute angles for every reflex angle. The perfect efficiency would be at a ratio of $360: 90=4: 1$. So my next guess is $\boxed{1600}$??? :D [/hide]

[quote="aidan"] can have if no 2 sides cross each other?[/quote]
what do you mean by that :huh: if this is a convex polygon, then, the largest possible # of acute angles is 3. with 4, you would have a straight angle.

I think you guys misinterpreted the question. What it meant is the largest number of acute angles which can be [i]interior[/i] angles. Each vertice of the 2001-gon subtends an interior angle.  It definitely isn't a convex polygon if you want to maximize the number of acute angles.


geometry
3D geometry
tetrahedron
calculus
integration
analytic geometry
geometric transformation

A while ago I read in Hartshorne's "Beyond Euclid" and came to wonder, how you can prove rigorously, that (the inside of) a plane polygon has an area. One way to do this seems to be by proving, that you can triangulate the polygon and then add the areas together.

But how about polyhedrons in three-dimensional space. How does one prove, that a polyhedron has a volume? (I know that polyhedrons can be defined somewhat differently, but you are free to choose any convenient definition that includes the "classical" cases). Perhaps you can split the polyhedron into tetrahedrons (or something else) and add them up? Or is there an analytical way of  doing this?

One can define volume analytically (the value of some integral or other), and that works fine.  For rational polytopes (say, those whose vertices have integer coordinates) you can define volume combinatorially (in terms of the number of lattice points contained in dilations of the polytope -- the key words here are "Ehrhart polynomial"), but there is no combinatorial definition of volume for arbitrary polytopes because given any pair of tetrahedra, it is not necessarily possible to take a finite dissection of one and reassemble it into the other.  (One of Hilbert's problems?)

But how can you prove analytically that every polyhedron has a volume? In any given case, you can probably split the polyhedron into more manageable bits, whose volume you can compute (by an analytical method), but how can you prove, that this is always possible?

Instead perhaps the following idea would work:

Pick a point $ P$ outside the polyhedron and consider all pyramids with $ P$ as the apex and a face in the polyhedron as base. Each of these pyramids have a volume. Some of these pyramids have the property, that the points inside the pyramid "closest" to the base are inside the polyhedron, we add the volumes of these pyramids together. The rest of the pyramids have the property, that the points inside the pyramid "closest" to the base are outside the polyhedron, we subtract the volume of each of these pyramids from our sum.

This would yield the volume of the polyhedron, I think. Or is there a catch somewhere?

A polyhedron is by definition a closed (or if you want: an open) set. Thus it's lebesgue-measurable, thus it has a volume.
Alternatively, you should ask yourself what you want to express by saying "something has a volume".

Great, it was something like that, I was looking for.

What I was trying to do, was to find (or define) the volume by considering it is as a sum of simpler, known (or easily definable) volumes.

But perhaps the point of Dehn's result about not being able in general to dissect a polyhedron and reassemble it into another one is, that the analytical way to define the volume is actually much sounder than the other way, since if you define the volume as a sum of some simpler volumes, you could get into trouble proving, that the volume is well-defined, when you can't do the whole disassembly-reassembly thing.

Anyway, thanks for the input. I have one more question about this, but I think, it warrants a new thread.


Tomorrow, June 29th at 4:30PM, Apple stores will close early and will reopen at 6:00 to start selling their Iphones. Crowd control and security in malls and police in generall have been notified too. Sounds big.
Will you be getting one anytime soon, and do you think that the Iphone will live up to expectations?

Now there is the iPhone XS MAX...

back in the old days.....

revive...

lmao
lol

Right now there is a Iphone XS Max

Back in the old days everyone relys on nokia and old Iphones

but now we have Samsung s10, fold, and note 9, for iphone we have s10, s10 max, 10r

ok so im older than u auntievera cuz it was right when i was born

[quote=golden57y0]HOW DID YOU FIND THIS?!?!?![/quote]

Its because i'm genius and epic

[quote=douglasubella]Now there is a iPhone 11 Pro Max!![/quote]

A new iPhone may be coming soon!!!

[quote=babyzombievillager][quote=Valentin Vornicu]Iphone is [b]the most[/b] overrated phone product in the world. It was marketed as being "5 years ahead of the competition", when actually, if one is (even remotely) acquainted with mobile phones, it is quite the opposite.

Already, all windows mobile phones with touchscreen (and there are [b]HUNDREDS[/b] of them) can do all the "cool" finger stuff that the Iphone does (and it took 2 days of programming by an amateur to do that! :) ).

Furthermore, Nokia has already released on the market for a few months already the N95. Here's an image explaning the differences between the two.

As you can see the [b]only[/b] real advantage of the Iphone over the N95 is the larger screen. Period. Oh, did I mention that the N95 is cheaper than the Iphone, and works on all networks around the world (whereas Iphone will only work in Cingular)? Need I say more? :)[/quote]

I almost thought he/she was talking about the newer iphones. Since she was talking about the orginial one, this post did not age very well, did it? The iphone totally destroyed Nokia lol...[/quote]

The Nokia still destroyed the iPhone.
Just test it, drop both down a building :)

Ah yes, the revive of the "Indestructible Nokia" memes.

Thing is, I knew a friend who had his username as "NokiaBreaker".

His power was above godly.

[quote=Valentin Vornicu]As you can see the [b]only[/b] real advantage of the Iphone over the N95 is the larger screen.[/quote]

Oh boy did this age well

[quote=AuntieVera]OMG I JUST LOOKED AT THE TIME THIS WAS ORIGANALLY POSTED BRO THIS FROM B4 I WAS BORN  :omighty:[/quote]

same

[quote=Valentin Vornicu]Iphone is [b]the most[/b] overrated phone product in the world. It was marketed as being "5 years ahead of the competition", when actually, if one is (even remotely) acquainted with mobile phones, it is quite the opposite.

Already, all windows mobile phones with touchscreen (and there are [b]HUNDREDS[/b] of them) can do all the "cool" finger stuff that the Iphone does (and it took 2 days of programming by an amateur to do that! :) ).

Furthermore, Nokia has already released on the market for a few months already the N95. Here's an image explaning the differences between the two.

As you can see the [b]only[/b] real advantage of the Iphone over the N95 is the larger screen. Period. Oh, did I mention that the N95 is cheaper than the Iphone, and works on all networks around the world (whereas Iphone will only work in Cingular)? Need I say more? :)[/quote]

I just found these posts so many years later.

Fr (THREE DAY BUMP AIN'T NO WAY I'M A CERTIFIED NECROPOSTER NOW  :rotfl: )


This was in a Palo Alto Math Circle thing and the answer seemed trivial but I wasn't sure...

Find the sum of the digits of $ \Big(10^{4n^{2}+8}+1\Big)^{2}$.

[hide]
Isn't the answer $ 4$ always?
[/hide]

It looks easy $ 10^{m}$ is just $ 1$ with many zeros. Add 1 and you have $ 1$ a string of 0's and $ 1$. Now when you square that you get $ 1$ a string of zeros $ 2$ another string of zeros and $ 1$. So that $ 4$.


[i]Is this ever posted before. It must ! ? [/i]

If $a , b , c$  are positive numbers and $a+b+c+2= abc$ , prove that

 $ab+bc+ca \geq 2(a+b+c)$

 [u]Babis[/u]

   [b]Note[/b]
 [i]This problem is taken from a book(introductory Analysis) by  Prof Univ. Th. Rassias (in greek). His son is metallist in IMO (2003), when he was in class 10. Probably he has taken the problem from this forum, because I have seen that he has also problems by Vornicu, Dospinescu , Tetiva ect. [/i]

Well, this problem seems to be created for Lagrange multipliers   :P. Considering:

$f(a,b,c)=ab+bc+ca-2(a+b+c)-\lambda(abc-a-b-c-2)$

and proceeding as usual we get that $f' _a=f' _b=f' _c=0$ is equivalent to $\frac{b+c-2}{bc+1}=\frac{a+c-2}{ac+1}=\frac{b+a-2}{ba+1}$ which is equivalent to:

$(a-c)(b^2-2b-1)=0$
$(b-c)(a^2-2a-1)=0$
$(a-b)(c^2-2c-1)=0$

From which we just get some cases to check.

[quote="stergiu"][b]To post 2.[/b]

 Good , but is such a cyclic transformation allowed and why ?

 [u]babis[/u][/quote]
the initial condition is equivalent to $\sum \frac{1}{a+1} = 1$, i just let $\frac{1}{a+1} = \frac{x}{x+y+z}$, so I think there is no problem.

This problem is the key of the inequality:
For $a=x+\frac{1}{y}$, $b=y+\frac{1}{z}$, $c=z+\frac{1}{x}$, we have
$ab+bc+ca \ge 2(a+b+c)$.


Can anyone help me simplifying this?

Solve the following simultaneous equations:

$x^2+7y+2 = 2z + 4 \sqrt {7x-3}$
$y^2+7z+2 = 2x + 4 \sqrt {7y-3}$
$z^2+7x+2 = 2y + 4 \sqrt {7z-3}$

Maybe I should put this on Intermediate or something but this is more to hairy algebra.

I see that each coefficent is same and only change is x,y, and z but radical is the part I'm stuck at. I know if I square it, it would be a huge polynomial with degree four and then sum it?

Um.. This looks not nice. :(

just add it all up and note that it's equal to $\sum (x-1)^2 + (\sqrt{7x-3} - 2)^2 = 0$, which is trivial to solve.


Euler
geometry
circumcircle
conics
hyperbola
search
geometry proposed

Let  ABC a triangle  and let $O$ its circumcenter and $H$ its orthocenter.Prove that the circles of the nine points (Euler circles ) of the triangles  $OHA$  , $OHB$  and $OHC$  have two common points .

On the one hand, all nine-point circles pas through the midpoint of $OH$, which is also the nine-point center of $ABC$, and, on the second hand, since the nine-point circle is the locus of the centers of all rectangular hyperbolae passing through the vertices of a triangle, all three nine-point circles (and the nine-point circle of $ABC$) must pass through the center of the hyperbola passing through $A,B,C,O,H$.

Also my book has this very short and nice solution  :lol:
Great Grobber .But could anyone post a solution elementary ?.

Also discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=aoh&t=42838]here[/url] (just search "aoh").

The same solution was given. Also, there you can find a link to the synthetic proof of the fact that the nine point center lies on any rectangular hyperbola passing through $A,B,C$. Too bad it's in French.

[quote="perfect_radio"]Too bad it's in French.[/quote]

I know some french  :lol:  . Thank you perfect radio for the link .
As I saw at the other link the solution also used hyperbola so I think there is no other way to prove it.....
Unless someone will have another idea


Source: 2005 AMC-10B #9

One fair die has faces $1$, $1$, $2$, $2$, $3$, $3$ and another has faces $4$, $4$, $5$, $5$, $6$, $6$. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$\text{(A)}\ \frac13 \qquad\text{(B)}\ \frac49 \qquad\text{(C)}\ \frac12 \qquad\text{(D)}\ \frac59 \qquad\text{(E)}\ \frac23$

Yeah, I'm pretty sure it's 5/9 too by the same way (since 2/3 = odd (die 1) and 2/3 = even (die 2) put those together to get odd or 1/3 even (die 1) and 1/3 odd (die 2) put those together to get odd and then add the two probabilities together because it's OR).

[quote="breez"][hide="solution"]$2*\frac{1}{3}*\frac{2}{3} = \boxed{\frac{4}{9}}$[/hide][/quote]

I think you found the probability that the sum would be even.


calculus
integration
calculus computations

Evaluate $ \int_0^1 \sqrt{\frac{x\plus{}\sqrt{x^2\plus{}1}}{x^2\plus{}1}}\ dx$.

It's quite easy: put $ u\equal{}x\plus{}\sqrt{1\plus{}x^2}$ and the integral becomes $ \int_{0}^{1}{\sqrt{\frac{x\plus{}\sqrt{{{x}^{2}}\plus{}1}}{{{x}^{2}}\plus{}1}\,}\,dx}\equal{}\int_{1}^{1\plus{}\sqrt{2}}{\frac{du}{\sqrt{u}}}\equal{}2\left( \sqrt{1\plus{}\sqrt{2}}\minus{}1 \right).$


geometry calculus trigonometry probability percent ARML derivative

I took "Advanced Mathematics", Algebra 1, and Geometry in the 6th ,7th, and 8th grades respectively.

Grade 6 Math
Algebra 1
AP Calc AB(my school didn't allow students to take BC without finishing AB :( )

I took Pre-Algebra in 6th, Algebra in 7th, and Geometry in 8th. I could've been two grades ahead (I was up to geometry in 5th grade.)

Geometry in 6th, Algebra II in 7th, Pre-Calculus in 8th (now).

Algebra I in 6th, Geometry in 7th, Pre-Calculus in 8th (current).

Algebra 1 in 6th, Geometry in 7th, Advanced Algebra 2 now

I go to a PK3-12 school.

6th Grade- "Introduction to Math"
[skipped Pre-Algebra]
7th Grade- Algebra 1 Honors
[skipped Geometry]
8th Grade- Algebra 2 Honors

Right now, as a freshman, I'm taking Precalculus Honors.

6th and 7th grade was just the regular "advanced" math classes
and then 8th grade i took algebra 1 and geometry
now in algebra 2 (our school's math is a joke)

For me?

6th-Pre Algebra
7th-Algebra I(as in right now)
8th-Geometry(but I have to go to the high school for 1 period)

It's ok, compared to you guys, my school is almost laughable.

6th-(not really any name?)
7th-pre-algebra
8th-algebra I

For all you taking geometry during 8th grade, you're all a grade ahead of me.  :P

I had an interesting experience with my math classes:

[b]Mansfield Middle School, CT:[/b] [i]They had a cooperative program with JHU CTY's distance program (ie, they would pay for some people to substitute that for a school class).[/i]

[u]6th grade:[/u] H Algebra II and H Geometry

[b]Shrewsbury Middle School, MA:[/b] [i]Didn't want to accommodate my math requests. Looking back, this was a good thing, because I could do homework for other classes in math class, while focusing my math efforts on learning math on my own and getting into contest math.[/i]

[u]7th grade:[/u] Pre-Algebra
[u]8th grade:[/u] H Algebra I

[b]Shrewsbury High School, MA:[/b] [i]Automatically placed me in 9th grade. I personally raised hell until they let me place ahead for 10th grade. I didn't involve my parents, but I had to take 2 placement exams, make 2 petitions, and argue my case to the department head 3 times.[/i]

[u]9th grade:[/u] H Geometry
[u]10th grade:[/u] AP Calculus

Algebra I in sixth grade, Geometry and Algebra II in seventh, and Pre-Calculus in eighth (Currently).

pre algebra in 6th
algebra 1 in 7th
geometry 8th

Ooh!

umm....School phails....yay 4 aops/// U NEED to STUDY a LOT by URSELF..

1st- Like Counting (no, not combinatorics)
2nd- Operations
3rd- Computation
4th- Al 2/ Geo (hehe interesting story)
5th- Advanced Combinatorics and Stats I (hehe)
----then went to new school
6th- Pre Algebra
7th- Basic Algebra
8th- Al I/ Geo
9th- Trig/Pre-Cal (skipped Al II/Geo)

(will be doing l8r:
10th- Calc BC
11th- Diff Equations/ Calc III/ AP Stats mebe
12th- Advanced Class Study

By Myself- random proofs and books and reading and studying and competitions =D

6- algebra 2
7- analysis (pre-calc) and AP stat
8- calc bc

I know, unfortunately, my local community college offers none of those classes.  It's just Algebra 1/2, Geometry, Pre-Calculus, Calculus, Finite Math, Linear Algebra, and that weird teaching math

Math teaching is not a qualified mathematicians course.

That's semi-ironic: I took Calculus AB in 8th grade, yet most AoPS'ers who are in Algebra 2 for 8th grade are a heck of alot better than me. But the other half isn't so ironic: I skipped Pre-Calc and fell into Calc-Trap.

I am going into 6th grade. This is what I will take:

6: Geometry
7:Alg. 2
8: Pre Calc.
9: Calc BC
10,11,12: Whatever comes after calc. I guess stuff like D.E.

"D.E. Whazzat" is "Differential Equations" -- a first course in differential equations will often contain methods for solving some common forms of ordinary differential equations (ODEs) and partial differential equations (PDEs). Another important course often taken shortly after calculus is Linear Algebra (I could not possibly overestimate it's importance).

So maybe Linear Algebra. I just like DE because the name DE is mysterious.

Wow! It's interesting to hear other people share their experiences with taking math classes [i]just out[/i] of elementary. I'm wondering though, how in particular were you guys served when you did things such as take Precalculus in the eighth grade. Was there a class in the school, or were you accommodated alternatively, like being bussed to another school or maybe accessing an online class? I encourage other users to post their experiences as well.

My school in those days was very good, so it offered the class. However, I was taking it with much older kids. With some other classes, I was sent to the high school or university to take classes, or given a "free period" for that class and told to use it taking an online class.

In our school, the top class offered is Honors Algebra 1.  Fortunately our high school is right next door.  So for first period, my bus drops me off at the high school, I take Honors Geometry, and then walk back to the middle school and am 15 minutes late to my next class, We The People.  It's a decent system, except for the fact that sometimes adults think I'm a truant.

Edit to my last post: I am actually in BC Calc summer between 9th and 10th grade and in Multivarible and Stats in 10th grade.

I am in a bad school district even though we have a top 20 high school. Thus, math class is very boring.

6: Pre-Algebra
7: Algebra
8: Geometry

I thought my school had some advanced math...

6th Grade- Honors Algebra 1
7th Grade- Honors Geometry (Current)
8th Grade- Honors Algebra 2 (and kinda 1/2 of Precal)

6th: Algebra I
7th: Geometry [i]and[/i] Algebra II
8th: Pre-Calculus
9th: Calculus with Theory (Honors Calculus by most people) at University of Oregon

6th: preAlgebra

7th: Algebra 1

8th: Algebra 2 and Geometry (at the same time)

9th (right now): Precalc


1) Is my understanding correct?

Given the plane $ 2x \minus{} 3y \minus{} 5z \equal{} \minus{} 100$, give a point $ q$ not on the plane. Does this $ q$ and the origin lie on the same side of the plane?

[hide="My Answer"]The point $ (0,0,30)$ does not lie on the plane because it does not satisfy its equation. Since $ \minus{} 5(30) < \minus{} 100$, the point and the origin lie on the same side.[/hide]

2.) I already forgot this one: What do you know about the equation of the plane which is perpendicular to a given plane? Give an equation of a plane perpendicular to the given above/

[hide="My Answer"]
As far as i can remember, their dot product of their normals should be equal to zero. Am I correct? For example a plane with normal $ (a,b,c)$ is perpendicular to 2x +3y - 5z if $ 2a \plus{} 3b \minus{} 5c \equal{} 0$ [/hide]

The part about the origin in the first one isn't right. $ \minus{}150 < \minus{}100$ and $ 0 > \minus{}100$.

2nd one is right.

[quote="MellowMelon"]The part about the origin in the first one isn't right. $ \minus{} 150 < \minus{} 100$ and $ 0 > \minus{} 100$.

2nd one is right.[/quote]

is there anyway to generalize on how to determine which side of the plane is a point?

Sure. Call the positive z direction "up".

Say your point is $ (u, v, w)$. Now consider the line $ (x,y) \equal{} (u,v)$. This intersects the plane $ ax \plus{} by \plus{} cz \equal{} d$ when $ au \plus{} bv \plus{} cz \equal{} d \implies z \equal{} \frac {d \minus{} au \minus{} bv}{c} \; (*)$.

If $ w > \frac {d \minus{} au \minus{} bv}{c}$, then we are above the plane, and if $ w < \frac {d \minus{} au \minus{} bv}{c}$, we are below the plane. If they are equal then the point is on the plane.

$ (*)$ [i]The exception is when $ c \equal{} 0$, in which case we can just drop the $ z$ coordinate (ie, project onto the x-y plane), and do the same thing except with a point and a line.[/i]


Let $ABCD$ be convex quadrilatteral and let $AC$ and $BD$(diagonals) meet at $O$. If $P$ and $Q$ are circumcenters of $ABO$ and $CDO$ prove that \[ AB+CD \le 4PQ \]

[quote="X-man"]Let $ABCD$ be convex quadrilateral and let $AC$ and $BD$(diagonals) meet at $O$. If $P$ and $Q$ are circumcenters of $ABO$ and $CDO$ prove that \[ AB+CD \le 4PQ  \][/quote]

Consider $ABO$ and $CDO$ to be equilateral.
Contradiction.
It should be $AB+CD \le 2PQ$

[hide="Solution"]
In $ABP$
$AB < AP + BP$ i.e. $AB < 2OP$
Also In $CDQ$
$CD < CQ + DQ$ i.e. $CD < 2OQ$

But in $OPQ$
$OP + OQ \le PQ$                (Equality occurs if P,O,Q are collinear)

Thus $AB+CD<2OP+2OQ \le 2PQ$[/hide]


advanced fields
advanced fields unsolved

give an example of a regular language L1 and a context free language L2 such that L1  intersection L2 is not a regular language.

Let $L1 = 0^{*}1^{*}$, a regular expression, hence regular.  Let $L2 = \{0^{n}1^{n}: n \geq 0\}$, this can be solved using a Push Down Automata, which shows that it is context free.  $L1 \cap L2 = L2$ and we just said that $L2$ is context free, not regular.


\[ \left( \begin{gathered}
  n \hfill \\
  0 \hfill \\
\end{gathered}  \right) \plus{} \left( \begin{gathered}
  n \hfill \\
  3 \hfill \\
\end{gathered}  \right) \plus{} \left( \begin{gathered}
  n \hfill \\
  6 \hfill \\
\end{gathered}  \right) \plus{} ... \plus{} \left( \begin{gathered}
  n \hfill \\
  3k \hfill \\
\end{gathered}  \right) \leqslant \frac{1}
{3}\left( {2^n  \plus{} 2} \right)
\]

Maybe these problems should go in the Pre-Olympiad section? They are not as difficult as, say, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=271350

In this case, we can use a classic technique called the "Roots of Unity Filter". Set our desired sum as $ S$.

$ (1 \plus{} 1)^n \equal{} {{n}\choose{0}} \plus{} {{n}\choose{1}} \plus{} {{n}\choose{2}} \plus{} {{n}\choose{3}} \plus{} \cdots$
$ (1 \plus{} \omega)^n \equal{} {{n}\choose{0}} \plus{} {{n}\choose{1}}\omega \plus{} {{n}\choose{2}}\omega^2 \plus{} {{n}\choose{3}} \plus{} \cdots$
$ (1 \plus{} \omega^2)^n \equal{} {{n}\choose{0}} \plus{} {{n}\choose{1}}\omega^2 \plus{} {{n}\choose{2}}\omega \plus{} {{n}\choose{3}} \plus{} \cdots$

Adding, we get $ 3S \equal{} 2^n \plus{} (\minus{}\omega)^n \plus{} (\minus{}\omega^2)^n$, which gives us our result.


geometry
3D geometry
vector
complex numbers

If $z_1, z_2$ and $z_3$ lie on the unit circle in the complex plane and $z_1 + z_2 + z_3 = 0$, prove that they are the vertices of a regular triangle.

since they all lie on the unit circle and have a sum of o they are the point of the cube roots of 1. $x^3=1$ and as we can see there is no $x^2$ term so the sum is truely 0, as i had previously stated. also it is know that when you connect the roots of unity you will get a regular polygon. since you only have 3 points (verticies) you have a regular triangle.

[hide]You have three vectors of magnitude one which sum to 0.  This is equivalent to making a triangle with three segments of length one.  There is only one triangle with all of its sides equal, and it is the regular/equilateral/equiangular triangle.[/hide]

maokid7: I don't think that proof works. Surely there are other possibilities than the three points being the solutions to $x^3 = 1$. You're showing that if they are the roots of $x^3 = 1$ then they fulfill the conditions given in the question, but that is not equivalent to that the conditions are only fulfilled when they are the roots of $x^3 = 1$. Note that $x^3 = \alpha$ where $|\alpha| = 1$ will result in an equilateral  triangle on the unit circle. However, this doesn't show that this is the only possibility for an equilateral triangle, not as far as I know anyway.

jomi88: I don't know how vectors work, so ok. How do you show the first equivalence though? The first equivalence is like the gist of this entire question and you just brush it off as trivial. is it?

[hide]let $z_n = a_n + b_ni$
by definition, $a_1+b_1i+a_2+b_2i+a_3+b_3i=0+0i$
therefore, $a_1+a_2+a_3=0, b_1+b_2+b_3=0$
square both equations:
$a_1^2+a_2^2+a_3^2+2(a_1a_2+a_1a_3+a_2a_3)=0$
$b_1^2+b_2^2+b_3^2+2(b_1b_2+b_1b_3+b_2b_3)=0$
add both sides together for one big fatty equation:
$a_1^2+a_2^2+a_3^2+2(a_1a_2+a_1a_3+a_2a_3)+b_1^2+b_2^2+b_3^2+2(b_1b_2+b_1b_3+b_2b_3)=0$
given $\sqrt{a_n^2+b_n^2}=1$
we can make things less tedious (kind of):
$3+2(a_1a_2+a_1a_3+a_2a_3)+2(b_1b_2+b_1b_3+b_2b_3)=0$
$a_1a_2+a_1a_3+a_2a_3+b_1b_2+b_1b_3+b_2b_3=-\frac{3}{2}$
since $a_1+a_2+a_3=0$
then $a_3=-a_1-a_2$
plugging back in to the big fatty equation again:
$-a_1^2-a_2^2-a_1a_2-b_1^2-b_2^2-b_1b_2=-\frac{3}{2}$
that turns out very sexy
$-a_1a_2-b_1b_2=\frac{1}{2}$
$a_1a_2+b_1b_2=-\frac{1}{2}$

now... finding distance between 2 points
$\sqrt{(a_1-a_2)^2+(b_1-b_2)^2}$
$=\sqrt{a_1^2-2a_1a_2+a_2^2+b_1^2-2b_1b_2+b_2^2}$
$=\sqrt{2-2a_1a_2-2b_1b_2}$
$=\sqrt{3}$
this is exactly what the side of an equilateral triangle should be if the segment from the middle to the vertice has length one[/hide]

tell me if i'm wrong

Oh, sorry...vectors and complex numbers are treated pretty much the same way.

The reason it is a triangle with three sides is: when adding complex numbers you can do this by tip to tail.  That is, you start with $z_1$, then you put the tail of $z_2$ at the tip of $z_1$, and finally the tail of $z_3$ goes at the tip of $z_2$.  $z_1+z_2+z_3$'s tip is where the tip of $z_3$ ends up.  But this is just at the origin.  The origin is also where $z_1$ started.  Thus, we have a triangle.  And $z_1=z_2=z_3=1$ and therefore it is equaliateral.


algorithm
induction
geometry proposed
geometry

Prove that in a plane, arbitrary $ n$ points can be overlapped by discs that the sum of all the diameters is less than $ n$, and the distances between arbitrary two are greater than $ 1$. (where the distances between two discs that have no common points are defined as that the distances between its centers subtract the sum of its radii; the distances between two discs that have  common points are zero)

[hide="Solution"]
Let any set of $ n$ points and $ k$ circles, with the property that the $ k$ circles cover all $ n$ points, and that the sum of the diameters of the $ k$ circles is less than $ n \minus{} k \plus{} 1$, be an "$ (n, k)$ covering."  Let a "complete $ (n,k)$ covering" be an $ (n, k)$ covering so that none of the $ k$ circles have a distance of $ x\le 1$.  We will prove that for any set of $ n$ points, there exists a $ k$ so that the set has an $ (n, k)$ complete covering.  Consider a set of $ n$ points.  Cover each of the $ n$ points with a circle of diameter $ \epsilon < \frac {1}{n}$.  These $ n$ circles form an $ (n,n)$ covering since the sum of the diameters is less than $ 1$.  We now present an algorithm that makes this covering complete:

[b]Step 1:[/b] Say that there are two circles that have a distance of $ x\le 1$.

Let these circles be $ (P)$ and $ (Q)$, having diameters $ D_P$ and $ D_Q$ respectively.  Let $ PQ$ hit $ (P)$ at $ P_1$ and $ P_2$ and $ (Q)$ at $ Q_1$ and $ Q_2$ so that $ PQ_2 > PQ$ and $ QP_2 > QP$.  Then, take the midpoint, $ R$ of $ P_2Q_2$ and draw the circle, $ (R)$, centered at $ R$ with radius $ P_2R \equal{} Q_2R$.  $ (R)$ has diameter $ D_R \equal{} D_P \plus{} D_Q \plus{} x \le D_P \plus{} D_Q \plus{} 1$.  Now, since $ (R)$ is internally tangent to $ (P)$ and $ (Q)$ (indeed, $ P$, $ Q$, and $ R$ are collinear and $ D_R > D_P, D_Q$),$ (R)$ contains any point that is either in $ (P)$ or $ (Q)$.  Furthermore, this step increases the sum of the diameters of the circles by at most one, but decreases the number of circles by precisely one.  Therefore, if the input into this step is a covering, the output of this step is also complete a covering.

[b]Step 2:[/b] Repeat Step 1 until it is impossble to do so anymore.

It is clear that we cannot do step one indefinitely since step one decreases the number of circles by precisely one each time.

By induction, repeating step 1 always yields coverings.  Yet, after doing step 2, there are no circles that have a distance of $ x\le 1$, implying that the covering produced after step 2 is complete, which means that our algorithm works.  Since there is at least one circle, the total sum of the diameters is less than $ n$. [/hide]

The process is simple. We can provide an algorithm to construct the circles.
The algorithm start by $n$ circles, whose centres are the original $n$ points and diameters are sufficently small.
If the distance of two circles is smaller than 1, then "combine" them together.
When the algorithm is done, we get a proper constructio !


Euler
search
combinatorics unsolved
combinatorics

was searching for some books on theory of graph in [color=red]PDF[/color]

can anyone send theme to me

[color=red]mathematicien@naseej.com[/color]

Well, I'll simply put the links here so that anyone can benefit of it :) :
http://www.sics.se/~jan/rapporter/diestel/diestel.pdf
http://www.southernct.edu/~fields/TeX-PDF/GraphTheory.pdf
http://www.mcb.mcgill.ca/~bryant/MS240/Notes/5_1_GraphsDigraphs.pdf
http://poppy.snu.ac.kr/~kchoi/class/cad_intro/graph.pdf

If you want more, just let me know  :P

thank you my friend

merci beaucoup [color=red]mathmanman[/color]

No problem Fermat - Euler :)
Hmm, in french, let me check lol

Yeah, I have some :) :
http://www.eduscol.education.fr/D0015/graphes.pdf
www-lmah.univ-lehavre.fr/gueye/enseignements_fichiers/graphes.pdf

Voilà!

(I'll have to search for more)


What is the largest value of $ x$, if $ \frac{x}{5} \plus{} \frac{1}{5x} \equal{} \frac{1}{2}$?

Since I know that x^2/5x + 1/5x = 1/2, i just used Guess and Check to find that 2 is the greatest possible value for x.

We can also multiply both sides by $ 10x$ to eliminate the denominator.

This yields:

$ 2x^2 \plus{} 2 \equal{} 5x$

Rearranging gives us the quadratic:

$ 2x^2 \minus{} 5x \plus{} 2 \equal{} 0$

We find the larger solution is $ \boxed{2}$.

Note: If you do not know how to solve quadratics, read Quadratics Part 1 ~ 5 in my blog. Link's in my signature.


The quadrilateral $ABCD$ is inscribed in a circle. The diagonals $AC,BD$ meet at $Q$. The sides $DA$, extended beyond $A$, and $CB$, extended beyond $B$, meet at $P$.
Given that $CD=CP=DQ$, prove that $\angle CAD=60$.

[hide="Proof"]Angle chasing. Denote $\phi=\angle QCD,\theta=\angle QCB$

\begin{eqnarray*}DQ=DC &\implies& \angle DQC=\phi\\ &\implies& \angle BQC=\pi-\phi\\ &\implies& \angle QBC=\phi-\theta\\ &\implies& \angle CAD=\phi-\theta\end{eqnarray*}

$CD=CP \&\implies\& \angle CPD=\frac{\pi-\theta-\phi}{2}$

$\begin{eqnarray*}\angle DAC=\angle APC+\angle ACP &\implies& \phi-\theta=\frac{\pi-\theta-\phi}{2}+\theta\\ &\implies& 2\phi-2\theta=\pi-\theta-\phi+2\theta\\ &\implies& 3\phi-3\theta=\pi\\ &\implies& \phi-\theta={\pi\over 3}$

$\angle CAD=\phi-\theta\implies\angle CAD={\pi\over 3}$[/hide]

[hide="Question about Proof"]How did you get $\angle QBC=\phi-\theta$?[/hide]

[hide="Answer"]By that point I found two angles in $\triangle QBC$, hence the third just complements those two to $\pi$. (Sketch a picture to see that clearly.)[/hide]

And I just have spotted a typo: In the beginning it should've been $\theta=\angle QCB$ (not $\angle QDB$, which is zero). Edited, with apologies.

I had a hard time visualizing that, so here's a diagram for everyone:


A princess is in love with a dashing knight. Unfortunately, the King perfers another suitor for his daughter. The King has locked her in the castle tower. The castle is square with a moat that is 10 yards wide. The knight is attempting to cross the moat, but only has two 9.75 yard planks, and no way to fasten them together. How can the brave knight bridge the moat?

he could chuck the planks to the edge of the moat

like that:
_                                                 ___
 |eeeeeeeeeeeeeemmmmmm  |
 |eeeeeeemmmmmmm eeeeee|
 |mmmmmeeeeeeeeeeeeeeee|
 ~~~~~~~~~~~~~~~~~~~~

oooh... interesting problem, but i can't find the solution  :(

oh i got it now.  the knight puts one plank on the corner perp. to the diagonal of the square.  then he puts the other plank on the plank and the castle.

----------
|---------|
|----p-\--|/
|--------\/
---------/-

[hide]
hint: the first one should cross the cross the corner of the moat square... try bringing the start closer to the finish than the finish closer to the start  :) [/hide]

the knight swims across the moat. cool oh, it says *bridge* the moat. [hide]the knight covers the area of a corner with plank #1, walks onto it, and conects the castle to plank #1 with plank #2.
  :surf:[/hide]

This problem is very creative. NOw, try this:

There is a person with a dog, a sheep, and a basket of vegetables that wants to cross a river with a boat. however, due to the size of the boat, the person can only bring one thing on with him at a time. When left unattended, the dog will eat the sheep, and the sheep will eat the vegetables. How can the person carry all three things successfullly across the river without having anything eaten?

[hide]Let us designate 2 sides:

side1=original side with the sheep,dog & bag
side2=side on the other part of the river

Step1. sheep-->side2
Step2. dog-->side2
          sheep-->side1
Step3. bag-->side2
step4. sheep-->side2    :D  [/hide]

There are actually several answers to this one.
[hide="another way"]
Bring the sheep and the dog over, take the sheep back, leave the sheep there and take the vegetables over, come back empty, and finally take sheep back again. basically the same as jsc. [/hide]

[hide]first puzzle:the knight puts the plank on the corner of the moat, then walks across and connects the second to the castle.
second puzzle:
sheep :arrow: side 2
dog :arrow: side 2
sheep :arrow: side 1
veggies :arrow: side 2
sheep :arrow: side 2 :idea: [/hide]

ive seen this problem many times and figured it out alot but would it really work? I mean according to the pythagoream theorum the corner would be 14.1 ft away from the other side. And I've done some extra calculations and I might be wrong but I think it doesn't work.

edit: actually I think it would make it by about 7 in.

2nd problem:  Originally posed as "The Wolf, The Goat, and The Cabbage"

"There are actually several answers to this one."

No, there are just the two solutions I know of that have already been given in this thread.


Nice problem...

Let $a=a_1+a_2+...+a_n$, and $b=a_1^2+a_2^2+...+a_n^2$

By AM-GM,

$\sum \frac{a_k}{g+a-a_k} \geq \sum \frac{a_k}{\frac{a}{n}+a-a_k} = \sum \frac{na_k}{(n+1)a-na_k}$

But from Cauchy, we have

$\left(\sum \frac{na_k}{(n+1)a-na_k}\right) \left(\sum a_k[(n+1)a-na_k]\right) \geq na^2$

Applying Cauchy again,

$\sum a_k[(n+1)a-na_k] = (n+1)a^2-nb \leq (n+1)a^2-a^2$

So it follows that

$\sum \frac{a_k}{g+a-a_k} \geq \sum \frac{na_k}{(n+1)a-na_k}$, and

$\sum \frac{na_k}{(n+1)a-na_k} \geq \frac{na^2}{(n+1)a^2-nb} \geq \frac{na^2}{(n+1)a^2-a^2} = 1$,

as desired. $\blacksquare$


inequalities
probability
inequalities proposed

if you have a deep view in inequality solve this: :D
suppose the set A consists of 1,2,...,n  .we choose randomly an element of A ,so that the probability of i is p_i
now we choose subsets randomly if a_k be the probability of the event that the subset has k elements,anf if f(x)=-xlog2(x),prove that f(a_1)+f(a_2)+...+f(a_n) is maximum when p_i=0.5 for every i.

No solution?
i have a solution for that somehow incomplete but it is ugly,can anyone solve that in a nice way?

1. Learn LaTeX.
2. There are $ n$ elements and the probability $ p_i\equal{}0.5$ for every (!) $ i$. Thus, $ n\equal{}2$, as $ 1\equal{} \sum p_i$.

No,
every element in the set $ A$ is chosen independently of the others, for example $ 1$ is chosen by probability $ p_1$ and is not chosen by probability $ 1 \minus{} p_1$.
The problem is really hard.
I restate the problem.
Suppose the set $ A$ consists of $ 1,2,...,n$ .we denote disjoint probability to elements of $ A$ ,so that the probability of $ i$ is $ p_i$.
Now we choose subsets randomly if $ a_k$ be the probability of the event that the subset has $ k$ elements,and if $ f(x) \equal{} \minus{} xlog(x)$, prove that  $ f(a_1) \plus{} f(a_2) \plus{} ... \plus{} f(a_n)$  is maximum when $ p_i \equal{} 0.5$ for every $ i$.

I think Vasc or argady or someone specialist in something like majorization can help. :)


Could someone give me all the rules for congruent triangles? Thank you for helping me!

I think this link should help you : http://argyll.epsb.ca/jreed/math9/strand3/3203.htm  :)


Let $x,y,z$ be positive real numbers. Prove the following inequality: \[ \frac 1{x^2+yz} + \frac 1{y^2+zx } + \frac 1{z^2+xy} \leq \frac 12 \left( \frac 1{xy} + \frac 1{yz} + \frac 1{zx} \right). \]

I think this works

[hide]

By AM-GM, we have

$\frac{x^2 + yz}{2} \geq \sqrt{x^2yz}$

If we repeat this for the other two denominators and combine the results, we have
\[ \frac{1}{2}(\frac{1}{x\sqrt{yz}} + \frac{1}{y\sqrt{xz}} + \frac{1}{z\sqrt{xy}}) \geq \frac 1{x^2+yz} + \frac 1{y^2+zx } + \frac 1{z^2+xy} \]

So all that remains to be proven is that
\[ \frac1{2}(\frac{1}{xy} + \frac{1}{yz} + \frac1{xz}) \geq \frac1{2}(\frac{1}{x\sqrt{yz}} + \frac{1}{y\sqrt{xz}} + \frac{1}{z\sqrt{xy}}) \]

because the function is cyclic in nature, we can assume $x \geq y \geq z$. This implies that $1/x \leq 1/y \leq 1/z$ and $1/\sqrt{x} \leq 1/\sqrt{y} \leq 1/\sqrt{z}$. These two sequences are sorted the same way, so by rearrangement:

\[ \frac1{x} \frac1{\sqrt{y}} \frac1{\sqrt{y}} + \frac1{y} \frac1{\sqrt{z}} \frac1{\sqrt{z}} + \frac1{x} \frac1{\sqrt{z}} \frac1{\sqrt{z}} \geq \frac1{x} \frac1{\sqrt{y}} \frac1{\sqrt{z}} + \frac1{y} \frac1{\sqrt{z}} \frac1{\sqrt{x}} + \frac1{z} \frac1{\sqrt{x}} \frac1{\sqrt{y}} \]

and we are done[/hide]

Once again, I'm very surprised, but this problem had been proposed in GM before district olympiad. :?

Proposed by whom? And in what number? ...  :|

This is problem C:2971 from GM, no.1/2006 proposed by Traian Tamaian. :)

If is so, then this stage of the olympiad was prepared negligently or more correctly, it wasn't prepare ! By the way, see and another some problems from this competition.

Well, I belive it's mostly the author's fault, if it was proposed for the Olympiad by the same person. You cannot propose a problem that you have previously sent to the Mathematical Gazette ...

Anyway, but what do you think of problem 2, X grade. This was proposed in 2003, too. :)

Very nice ! This proof show many imagination. I like it very much ! Congratulations, Paladin8 !

[quote="blahblahblah"][quote="paladin8"]A little simpler...

[hide]
We have by AM-GM

$4xyz \le zx^2+yz^2+yx^2+zy^2 = (y+z)(x^2+yz)$, so

$\frac{1}{x^2+yz} \le \frac{1}{4}\left(\frac{1}{xy}+\frac{1}{zx}\right)$.

Summing up similar inequalities, we have the desired inequality.[/hide][/quote]

I claim priority! [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=a%5E2%2Bbc&t=69177]see here[/url] :)[/quote]

Fair enough. :P

Multiply $xyz$ to the inequality, we get:
\[ xyz \left( \sum\frac{1}{x^2+yz} \right) \leq \frac{1}{2}(x+y+z) \]
Use AM-GM then Cauchy:
\[ \sum\frac{xyz}{x^2+yz}=\sum\frac{1}{\frac{x}{yz}+\frac{1}{x}}\leq\sum\frac{\sqrt{yz}}{2}\leq\frac{1}{2}(x+y+z) \]
QED.

We can prove an strong inequality:$ \displaystyle\sum_{cyc}\frac{1}{x^2\plus{}yz}\le 2\displaystyle\sum_{cyc}\frac{1}{(x\plus{}y)^2}$

We'll prove the following inequality:$ \frac{1}{x^2\plus{}yz}\le \frac{1}{(x\plus{}y)^2}\plus{}\frac{1}{(x\plus{}z)^2}$

Using Cauchy-Schwartz inequality we obtain that: $ z(x\plus{}y)^2\equal{}(\sqrt{z}x\plus{}\sqrt{y}\sqrt{yz})^2\le (z\plus{}y)(x^2\plus{}yz)$

$ \Rightarrow$ $ \frac{1}{(x\plus{}y)^2}\ge \frac{z}{(y\plus{}z)(x^2\plus{}yz)}$.Similarly: $ \frac{1}{(x\plus{}z)^2}\ge \frac{y}{(y\plus{}z)(x^2\plus{}yz)}$

By adding last two relations,we get the result!

How old are the ninth graders in Romania? I am 14, which is the normal age for a ninth grader in Indonesia, but I think this is too difficult to be the first problem in a district olympiad.

[quote="micf"][quote="shobber"]
$ \sum\frac {\sqrt {yz}}{2}\leq\frac {1}{2}(x \plus{} y \plus{} z)$
[/quote]
How to use Cauchy Schwartz here?[/quote]I think it's just rearrangement: $ \sqrt{x^2}\plus{}\sqrt{y^2}\plus{}\sqrt{z^2}\ge\sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx}$.


hello, we get $ 42x^3\plus{}68x^2\plus{}16x\equal{}2x(4\plus{}3x)(2\plus{}7x)$.
Sonnhard.

uh..
btw u didn't show how
and he already knew  :)

[quote="eli140"]How would you factor a cubic polynomial, like $ 42x^3 \plus{} 68x^2 \plus{} 16x$?

EDIT: Never mind, I got it. Ooops
 :wallbash_red:

Can you delete new topics?[/quote]

You can delete new topics but you have to ask a moderator moderating the forum.


integration
geometry
rectangle
limit
calculus
function
real analysis

Show that $ \int_{C_1} e^{ \minus{} u^{2} \minus{} v^{2}} \ du \ dv \leq \int_{R} e^{ \minus{} u^{2} \minus{} v^{2}} \ du \ dv \leq \int_{C_{2}} e^{ \minus{} u^{2} \minus{} v^{2}} \ du \ dv$ where $ R$ is the rectangle $ \{ (u,v) | \minus{} x \leq u,v \leq x \}$, $ C_1$ is the largest circle that lies inside $ R$ and $ C_2$ is the smallest circle that contains $ R$.

This makes sense intuitively, but how would I show this rigorously? Eventually we show that $ \lim_{x \to \infty} \int_{R} e^{ \minus{} u^{2} \minus{} v^{2}} \ du \ dv \equal{} \pi$.

One way I tried: $ \int_{0}^{2 \pi} \int_{0}^{\infty} e^{\minus{}r^{2}} r \ dr \ d \theta \equal{} \pi$. But thats not really method of exhaustion.

The integral of non-negtive function is non-negtive.

What is your definition of the integral where this is not totally obvious?

Its a Riemann integral (the standard definition).

Any lower riemann sum (lower darboux sum?) for the integral over $ R$  is also a lower riemann sum for the integral of $ C_2$. It follows that the first integral is less than the second.


1-chloro 2,3-dimethyl but-2-ene on treatment with HCl gives 1,3-dichloro 2,3 dimethyl butane

[hide="Answer"]Chlorine is more electronegative than carbon, so it have an electrowithdrawing inductive effect. The carbocation with the positive charge beta to the chlorine would thus be more destabilized (because of the enhanced positive charged character of the carbon bearing the charge created by the attractive effect of the chlorine) than the carbocation with the charge more far away from the halogen: the proton thus attacks carbon 2 of the substract in order to form the less destabilized cation, and then chloride forms a bond with carbon 3.[/hide]

A very interesting question would be the following: what product are we expected to obtain by treatment of [b]2[/b]-chloro-2,3-dimethyl-but-2-ene with HCl?

can u explain again why the the carbocation beta to it is relatively unstable  :maybe:

The beta carbocation (with the positive charge in carbon 2) is more unstable than the carbocation with the charge in carbon 3 due to the electrowithdrawing effect of chlorine, since it is more electronegative than carbon. This is the same effect as the increased adicidity of chloroacetic acid as compared with acetic acid.

also what u would expect of treatment of 2-bromo but-1-ene with HBr.
also is the compound u give right?

[quote="Pardesi"]also is the compound u give right?[/quote]

Obviously not. Just remove the 2-methyl:

"A very interesting question would be the following: what product are we expected to obtain by treatment of [b]2[/b]-chloro-3-methyl-but-2-ene with HCl?"

[quote="Pardesi"]also what u would expect of treatment of 2-bromo but-1-ene with HBr.[/quote]

Solve my previous problem and you have the answer. However, if you get stuck, [hide="See this hint"]What is the orientation observed in electrophilic aromatic substitutions of halobenzenes? How can we explain it?[/hide]

so is the answer 2,3 dichloro 2 methyl butane


calculus
derivative
function
real analysis
real analysis theorems

Hi.
can someone provide me with the example of function $ f: \mathbb{R}^2 \rightarrow \mathbb{R}$ whose directional derivative exists at $ p$ (for any direction) but it is not differentiable at $ p$.

I want a simpler than this one
\[ f(x,y) \equal{} \begin{cases}1  & 0<y<x^2 \\ 0 & \text{otherwise} \end{cases}\]

the above function has an "ugly" $ \epsilon\minus{}\delta$ proof  :( .

thanks

[quote="Beginner"]
the above function has an "ugly" $ \epsilon \minus{} \delta$ proof  :( .
[/quote]
What "ugly proof"? :? All you need is to notice that every line through the origin contains a small open interval containing $ 0$ on which $ f$ vanishes identically and that $ f$ is discontinuous at $ 0$.

Another example of a slightly different nature is
\[ f(x,y)\equal{}\begin{cases}x, & x\equal{}y\\0, &x\ne y\end{cases}\]

Here's a continuous example: $ f(x,y)\equal{}\frac{x^2y}{x^2\plus{}y^2}$ (and zero at zero). It's homogeneous of degree one, so the directional derivative in the $ u$ direction is $ f(u)$; these aren't identically zero. The partial derivatives are both zero, so it can't be differentiable.

Hi all Thanks for your replies...
Fedja, i mean the rigirous $ \epsilon\minus{}\delta$ proof produces some cases.
Actually, i already noticed something similar with your argument (line through origin and open set). I was trying to give a simple example for a second-semester students, who only have the definition in hands (they might wondering when they'll move to the next section and find some handy directional derivative theorem to use  :P ) . As for this matter, i must give an example which have a friendly definition proof, rather than an intuitive one.  hmm i think i can use your example Fedja, Thanks.

@Jmerry
I've seen that example as an exercise in some of calculus book. thank you very much to remind me again.
As I said above, i think the proof of this example by definition is not smooth-straightfoward, excuse if i'm wrong.

Thanks.


linear algebra
matrix
vector
Galois Theory
superior algebra
superior algebra unsolved

Prove that the Galois group of $ x^p\minus{}2$ for $ p$ a prime is isomorphic to the group of matrices $ \begin{pmatrix} a&b\\0&1 \end{pmatrix}$ where $ a,b \in \mathbb{F}_p, a \ne 0$.

Both are the semidirect product of $ \mathbb F_p$ with $ \mathbb F_p^\times$ (the latter acting by multiplication).

Why the group of matrices $ \begin{pmatrix} a&b\\0&1 \end{pmatrix}$ where  $ a,b \in \mathbb{F}_p, a\ne0$
is the semidirect product of $ \mathbb{F}_p$ with $ \mathbb{F}_p^\times$?
Please give me more...

[quote="Bradley"]Why the group of matrices $ \begin{pmatrix} a & b \\
0 & 1 \end{pmatrix}$ where  $ a,b \in \mathbb{F}_p, a\ne0$
is the semidirect product of $ \mathbb{F}_p$ with $ \mathbb{F}_p^\times$?
Please give me more...[/quote]

Another way to view the semi-direct product of $ \mathbb{F}_p$ with $ \mathbb{F}_p^\times$ is as the group of permutations on a $ p$-element set induced by addition by an element or multiplication by nonzero element of $ \mathbb{F}_p$.

To see that this semi-direct product is isomorphic to the matrix group given, we represent an element $ c \in \mathbb{F}_p$ by the column vector $ \begin{pmatrix} c \\
1 \end{pmatrix}$. Then let this matrix group act on these column vectors in the normal way, so that multiplication by $ \begin{pmatrix} a & b \\
0 & 1 \end{pmatrix}$ corresponds to multiplying by $ a$ and then adding $ b$. Check that this action is faithful, and that shows that the two are isomorphic  :)


if gcd(a,p) = 1 where p is odd prime and a has order 2k (mod p), then $ a^k \equiv \minus{}1 (mod p)$

[quote="guile"]if gcd(a,p) = 1 where p is odd prime and a has order 2k (mod p), then $ a^k \equiv \minus{} 1 (mod p)$[/quote]

[hide]We have $ a^{2k}\equiv 1\mod p$, so that $ a^{2k}\minus{}1\equiv 0 \mod p$, or $ (a^k)^2 \minus{} 1\equiv 0 \mod p$, which has a maximum of two roots for $ a^k\mod p$. Roots are $ 1,\minus{}1$ but if $ a^k\equiv 1\mod p$ then that contradicts the given that $ 2k$ is the order of $ a\mod p$ so $ a^k\equiv \minus{}1\mod p$[/hide]


[quote="57_factorial"]Prove that:

m=n if the following is true:

x^n + x^(n+a) = x^m + x^(m+a)[/quote]
[hide]$x^n+x^{n+a}=x^m+x^{m+a}$. Factor: $x^n(x^a+1)=x^m(x^a+1)$, since $m=n$,  $x^m(x^a+1)=x^m(x^a+1)$.[/hide]

Ah, so simple  :blush:

Thank you, I see it now.

I just wanted to make sure you guys know there are many assumptions in this problem:

1) $x^a\ne-1$

2) $x\ne0,1$

3) $a\le\min\{m,n\}$

With these conditions, we can safely get $m=n$ as shown by ch1n353ch3s54a1l (with one slight error :D )

[quote="ch1n353ch3s54a1l"]Wait...
with those assumptions :dry: you could just plug in m for n and see that they are equal...[/quote]
That's because he asked for a 'proof'. A 'proof' consists more than mere plug-ins. Plugging in values is easy, proofs are never easy. If you want to prove something like $x^2\ge0$, it is not an easy task. Of course, it is so easy to understand that square of any real number is always nonnegative, but to prove it rigorously, it is not as easy as you think it is (unless you have seen it proven of course). In the same manner, 'seeing that they are equal' is easy, you can call this truly obvious, yes obvious, but to show the steps with sound and persuasive reasoning, I see that your solution is flawed at various steps, but I am lenient enough to only point out 1 error in your solution.

[EDIT]I'm not trying to make this complicated but the nature of the question can be very easy if certain assumptions are made but it can be a very hard problem if some of these assumptions are not made.


inequalities
trigonometry
inequalities unsolved

let $ x_1$ , $ x_2$ ,...., $ x_n$  ,$ y_1$ , $ y_2$ ,...., $ y_n$ $ \in \mathbb R$ and $ n$ $ \in {\mathbb N }^{*}$ ,prove that :

$ n\sqrt {2} \geq \sum_{i \equal{} 1}^{n} \cos(x_i) \plus{} \sum_{i \equal{} 1}^{n} \sin(y_i)$

when iquality occure

first question: why do you define the $ y_i$ if you do not use them?
Second: did you notice that the "general case" of $ n\in \mathbb{N}$ follows trivially from $ n\equal{}1$

Cauchy gives $ \sqrt{2}\equal{}\sqrt{2(\cos^2 x\plus{}\sin^2 x)}\geq \cos x\plus{}\sin x$  :wink:


trigonometry
inequalities unsolved
inequalities

Prove that for all x,y,z-real
 (x+y+z)[ :sqrt: (1-x 2 ) +  :sqrt: (4-y 2 ) +  :sqrt: (9-z 2 )] <= 18

Am I right to say that $|x| \leq 1, |y| \leq 2, |z| \leq 3$?

If so, we can sub $x = \sin a, y = 2\sin b, z = 3\sin c$,
yielding $LHS = (\sin{a} + 2\sin{b} + 3\sin{c})(\cos{a} + 2\cos{b} + 3\cos{c}) $$= \frac{\sin{2a} + 4\sin{2b} + 9\sin{3c} + 4\sin(a+b) + 6\sin(a+c) + 12\sin(b+c)}{2} \leq 18$

Equality obviousholds when $a = b = c = \frac{\pi}{4}$, or $x = \frac{\sqrt{2}}{2}, y = \sqrt{2}, z = \frac{3\sqrt{2}}{2}$


inequalities
induction
inequalities open

Find all positive integers $n$ such that for all positive real numbers $a,b,c$, the following inequality holds:
$$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq 3 \frac{a^n+b^n+c^n}{a^{n-1}+b^{n-1}+c^{n-1}}.$$

The inequality is false for say $(a, b, c, n) = (2, 1, 1, 5)$.

oh, I'm sorry the question is find $n$
for example in the case of $n=1$,it's satisfied ,because when $n=1$
the question is $a ^2/b+b ^2/c+c ^2/a>=a+b+c$                           (1)
and we have $a ^2/b+b>=2a$,
                   $b ^2/c+c>=2b$,
                    $c ^2/a+a>=2c$
sum them we can got (1)                        so $n=1$is right
but what about n>=2?I don't know
and as far as I know it's right for $n=2$

[quote="hungkhtn"]I think when $n=3$,it's true and will false if $n=4$.[/quote]

I agree with you

[quote="Fiachra"]As far as I can tell, this is true for $n=2$ but no further.
For the $n=2$ case, multiplying out, we need $\sum_{cyc} a^2b^3+ab^4 \geq 2(a^3bc+ab^3c+abc^3)$. This follows from
(a) $a^2b^3+b^2c^3+ab^4 \geq ab^3c$ (AM-GM) and hence $2\sum_{cyc} a^2b^3+\sum_{cyc} ab^4 \geq 3\sum_{cyc} a^3bc$.
(b) $b^2c^3+3ab^4 \geq 4a^{\frac{3}{4}}b^{\frac{7}{2}}c^{\frac{3}{4}}$ (AM-GM) and hence $\sum_{cyc} a^2b^3 + 3\sum_{cyc} ab^4 \geq 4\sum_{cyc} a^3bc$ (using Muirhead).

For $n=3$, $\sqrt{2}, 1, 1$ is a counter-example. By Cauchy-Schwarz we have $(a^{n+1}+b^{n+1}+c^{n+1})(a^{n-1}+b^{n-1}+c^{n-1}) \geq (a^n+b^n+c^n)^2$, so by induction we have that this is a counter-example for all $n \geq 3$.[/quote]

there is my solution
add they:
$a^3/b+ab>=2a^2$
$b^3/c+cb>=2b^2$
$c^3/a+ac>=2c^2$
$ab^2/c+ac>=2ab$
$bc^2/a+ab>=2bc$
$ca^2/b+bc>=2ac$

With n=2,it's really easy because you can expand to:
$\sum\frac{(a-b)^2}{a} \ge \frac{\sum(a-b)^2}{\sum{a}}$

[quote="hungkhtn"][quote="zxingtan"]
there is my solution
add they:
$a^3/b+ab>=2a^2$
$b^3/c+cb>=2b^2$
$c^3/a+ac>=2c^2$
$ab^2/c+ac>=2ab$
$bc^2/a+ab>=2bc$
$ca^2/b+bc>=2ac$[/quote]

Sorry,but I can't understand it?
My solution is expand:
$\sum\frac{a^2}{b}-\sum{a}\ge 3\sum\frac{a^3+b^3+c^3}{a^2+b^2+c^2}-\sum{a}$
become $\sum{k_a(a-b)^2}$[/quote]

when $n=2$ we should prove $a^2/b$+…>=$3(a^2+b^2+c^2)/(a+b+c)$
iff (a+b+c)($a^2/b$+…)>=$3(a^2+b^2+c^2)
expand it and use $a^3/b+ab>=2a^2$
$b^3/c+cb>=2b^2$
$c^3/a+ac>=2c^2$
$ab^2/c+ac>=2ab$
$bc^2/a+ab>=2bc$
$ca^2/b+bc>=2ac$

I cannot prove the case $n=3$ using AM-GM. :(

[quote="hungkhtn"]Oh,I see!
 :)
Where is your solution when $n=3$?[/quote]

I can't prove when $n=3$
but I guess it still right when $n=3$
just my guesses :)

who are interested when $n=3$(prove it or disprove it)

[quote="hungkhtn"]Yes, for $n=3$ this ineq still true.
You can check for it in my solution:
$S_a(a-b)^2+S_b(b-c)^2+S_c(c-a)^2 \ge 0$
When $S_a=\frac{a^2+c^2}{b}-a$ ...[/quote]

excuse me, how to prove $S_a(a-b)^2+S_b(b-c)^2+S_c(c-a)^2 \ge 0$
Thank you

OK,in two case:
If $a\ge b \ge c$ you need to prove:
$S_a+2S_c \ge 0$
$S_b+2S_c \ge 0$
If $a\ge c\ge b$
It's easy because $S_a+S_c \ge 0,S_b \ge 0$

Dear Hungkhtn
Please, rewrite your solution in more details.

Thank you very much.

Ok,you can expand the inequality to:
$S_a(a-b)^2+S_b(b-c)^2+S_c(c-a)^2 \ge 0$ in above.

Two case:
$a \ge b \ge c$,$S_a \ge 0,S_b \ge 0$.If S_c \ge 0,we are done.Else:
$S_a+2S_c \ge 0$
$S_b+2S_c \ge 0$
Because $(a-c)^2 \le 2(b-c)^2+2(a-c)^2$
Then $S_a(a-b)^2+S_b(b-c)^2+S_c(c-a)^2 \ge )S_a+2S_c)(a-b)^2+(S_b+2S_c)(b-c)^2 \ge 0$.OK?

If $a\ge c\ge b$,its easy because:
$S_a+S_c \ge 0,S_b \ge 0$
Then
$S_a(a-b)^2+S_b(b-c)^2+S_c(c-a)^2 \ge (S_a+S_c)(a-c)^2+S_b(b-c)^2 \ge 0$
And we are done!

Okie. Now I want we consider a new problem :
 [i]Find all n is positive integer that for all (positive real number):
 $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \sqrt[n]{3^{n-1}(a^n+b^n+c^n)}$[/i]

When n=1, 2, 3 it's true. I thinks it's true when n=4.

But i have checked many times that are all true for n=4,can you give an example which is wrong for n=4? :oops:

:lol: This inequality holds for n=1,2,3,4,5,6, but does not for n=7.

Really?How to prove it without computers?

Can you show how to prove it for n=2,3,5..


[quote="learningfish"]A Product costing 60 cents per unit to produce had been selling at the rate of 1,200,000 units per month. After the product was improved, sales increased to 2,000,000 units per month.
However, the new product costs 5 percent more to produce. If the manufacturer's in each instance was 75 cents per unit, what was the manufacturer's added profit per month with the new product?[/quote]
[hide="answer"]
Original profit: $ (75 \minus{} 60)(1\;200\;000)\;\text{cents}$
New profit: $ (75 \minus{} 1.05\cdot 60)(2\;000\;000)\;\text{cents}$
added profit: $ (75 \minus{} 63)(20)(100\;000) \minus{} (15)(12)(100\;000)\;\text{cents}$
$ \equal{} (20)(12)(100\;000) \minus{} (15)(12)(100\;000)$
$ \equal{} (5)(12)(100\;000)$
$ \equal{} 6\;000\;000\;\text{cents}$
$ \equal{} \$60\;000$
[/hide]


The Pony Express carried mail 2000 miles between St. Joseph, MO, and Sacramento, CA, in 10 days. Given that it costs $ \$5.00$ for each $ \frac12$ ounce of mail carried by the Pony Express,  how many dollars were charged for a rider to carry a 15-pound sachel of mail?

$ 15 \text{ pounds}\equal{}480 \text{ half\minus{}ounces} \implies 480 \times \$5\equal{}\boxed{\$2400}$.


Euler
geometry
circumcircle
incenter
geometry proposed

Let $ABC$ be a triangle with all angles $\leq 120^{\circ}$. Let $F$ be the Fermat point of triangle $ABC$, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$. For each one of the three triangles $BFC$, $CFA$ and $AFB$, draw its Euler line - that is, the line connecting its circumcenter and its centroid.

Prove that these three Euler lines pass through one common point.

[i]Remark.[/i] The Fermat point $F$ is also known as the [b]first Fermat point[/b] or the [b]first Toricelli point[/b] of triangle $ABC$.

[i]Floor van Lamoen[/i]

Because $\angle BFC=120^{\circ}$, it's easy to show that the Euler line of $BFC$ is parallel to the bisector of $\angle BFC$, which is $FA'\perp B_1C_1$. We do the same, of course, for $CFA,\ AFB$. This means that the three Euler lines concur at the center of the equilateral triangle $A_1B_1C_1$ ($A_1$ is the center of $BCA'$ and $B_1,\ C_1$ are defined similarly).

It's also problem A.323 from Komal, having appeared in September 2003.

Yes, this is one possible way. Sorry for posting a Kömal problem, I didn't know it had appeared there.

My proof goes as follows: The circumcenter $A_1$ of triangle BFC is simultaneously the circumcenter of triangle BA'C; hence, $A_1$ lies on the line A'D, where D is the midpoint of BC, and we have $A'A_1 : A_1 D = 2$. On the other hand, the centroid X of triangle BFC lies on the line FD, and we have FX : XD = 2. Finally, the centroid G of triangle ABC lies on the line AD, and we have AG : GD = 2. Since the points A, F and A' are collinear, and AG : GD = FX : XD = $A'A_1 : A_1 D$ holds, the points G, X and $A_1$ lie on one line. In other words, the line $X A_1$ passes through G. But this line is the Euler line of triangle BFC. Similarly, the Euler lines of triangle CFA and AFB pass through G. And perhaps I need not explain why the Euler line of triangle ABC does as well  :)

  Darij

[hide="Some hints"]Prove that the Euler's line of the triangle $CFB$ is parallel to $AF$.

You also can prove(if you wish :) ) that if $O_{A},O_{B}$ and $O_{C}$ are the circumcenters of the triangles $BCF,ACF$ and $ABF$ respectively,then the point of intersection of the Euler's lines is the center of the equilateral triangle $O_{A}O_{B}O_{C}$ :wink: .
[/hide]

[quote="darij grinberg"]Let $ABC$ be a triangle with all angles $\leq 120^{\circ}$. Let $F$ be the Fermat point of triangle $ABC$, that is, the interior point of $ABC$ such that $\angle AFB = \angle BFC = \angle CFA = 120^\circ$.

Prove that the Euler lines of the triangles $BCF$, $CAF$ and $ABF$ pass through one common point.[/quote]

[hide="hint"]I know they are pass centroid of triangle $ABC$, but I can't remember prove now  :( [/hide]

Let $A_{1}, B_{1}, C_{1}$ are midpoints of $BC,CA,AB$ . Construct three equilateral triangles $BCA_{2}, CAB_{2}, ABC_{2}$ with center $O_{1}, O_{2},O_{3}$ out of the triangle $ABC$.
Because $\hat{BFC}+\hat{BA_{2}C}= 120^{o}+60^{o}= 180^{o}$ so $B,F,C,A_{2}$ are lie on a circle have $O_{1}$ for center.
Moreover, note that the centroid $G_{a}$ of $\Delta BFC$ are lie in $A_{1}F$ and $A_{1}G_{a}= \frac{1}{3}A_{1}F$, so $O_{1}G_{a}$ - the euler line of $\Delta BFC$ is parallel with $FA_{2}$. We have $\hat{BFA_{2}}= \hat{BCA_{2}}= 60^{o}= 180^{o}-\hat{BFA}$ so $A,F,A_{2}$ are colinear. So using $O_{1}G_{a}$ and $AA_{2}$ are parallel with the respect, easy to prove that $O_{1}G_{a}$ is joining the centroid $G$ of $\Delta ABC$. So the Euler lines of $\Delta BFC, \Delta CFA, \Delta AFB$ are concurrent at the centroid $G$ of $\Delta ABC$

P/s: The problem is still true and more interesting if we replace the Toricelli point $F$ with the incenter $I$, I remember that the Euler lines of $\Delta BIC , \Delta CIA, \Delta AIB, \Delta BAC$ are concurent at a point called "Schiffler point" :wink:


algebra
polynomial
linear algebra
matrix
vector
Galois Theory
superior algebra

Let $ E$ be a commutative $ F\minus{}$algebra of finite dimension over $ F$. The norm map $ N_{E/F}: E\to F$ maps $ a\in E$ to the determinant of the $ F\minus{}$linear endomorphism $ : x\mapsto xa$ of $ E$.

a. Show that the norm induces a homomorphism $ E^*\to F^*$.

b. Let $ f\in F[X]$ be of degree $ d\geq 1$ and put$ E\equal{}F[X]/(f)$. Suppose that $ f$ is seperable and factors over an algebraic closure $ \bar F$ of $ F$ as $ f\equal{}(X\minus{}a_1)\cdots (X\minus{}a_d)$ with $ a_1,\cdots ,a_d\in \bar F$. Prove that for $ f\in F[X]$ we have
$ N_{E/F}(g mod f)\equal{}g(a_1)\cdots g(a_d)$

Thks in advance.

I assume that $ F$ denotes a field.

a - Define the $ F$-endomorphism of $ E$ $ L_a(x) = ax$ for every $ a\in E$. Then $ N_{E/F}(a) = {\det}_E(L_a)$. Since $ L_{ab} = L_a\circ L_b$ for $ a,b\in E$, you immediately get $ N_{E/F}(ab) = N_{E/F}(a)\cdot N_{E/F}(b)$.

Note that the norm does not necessarily take values in $ F^*$ only. Here a counterexample: $ E = F[X]/(X^2)$ and take $ L_X$ which is not injective (because $ L_X(X) = 0$), hence its determinant is zero. But if we start with a unit $ a\in E$, i.e. $ a\in E^*$, then $ N_{E/F}(a)\in F^*$.

b - Since $ f$ is separable, all the $ a_i$ are pairwise different. Thus there are $ \overline{F}$-algebra isomorphisms
\[ E': = E\otimes_F\overline{F}\cong_{\overline{F}}\prod_{i = 1}^d\overline{F}[X]/(X - a_i)\cong_{\overline{F}}\prod_{i = 1}^d\overline{F}\]
where the first isomorphism is induced by the Chinese remainder theorem, the second by evaluating the polynomials at $ a_i$ in each component. Given an element $ \overline{g} = g(X)\mod f(X)\in E$ - here's a slight typo is the original statement - we get
\[ \begin{matrix} N_{E/F}(\overline{g}) & = & \det_E(L_{\overline{g}}^{(E)}) \\
& = & \det_{E'}(L_{\overline{g}\otimes 1}^{(E')}) \\
& = & \prod_{i_ = 1}^d \det_{\overline{F}}L_{g(a_i)}^{(\overline{F})) }\\
& = & \prod_{i_ = 1}^d g(a_i) \end{matrix}\]
Here it is used
- that the determinant does not change if the field is extended by tensoring and
- that, given endomorphisms $ \varphi_i\in End(V_i)$ on two vector spaces $ V_1,V_2$, then for the endomorphism $ (\varphi_1,\varphi_2)$ on $ V_1\oplus V_2$ defined by $ (\varphi_1,\varphi_2)(v_1,v_2) = (\varphi_1(v_1),\varphi_2(v_2))$ it holds
\[ {\det}_{V_1\oplus V_2}(\varphi_1,\varphi_2) = {\det}_{V_1}(\varphi_1)\cdot {\det}_{V_2}(\varphi_2)\,.\]

BTW: After reducing to the case that $ f$ is irreducible over $ F$ another approach could use Galois theory of course.


How many positive integers not exceeding 2001 are multiple of 3 or 4 but not 5?

$ \textbf{(A)} \ 768 \qquad \textbf{(B)} \ 801 \qquad \textbf{(C)} \ 934 \qquad \textbf{(D)} \ 1067 \qquad \textbf{(E)} \ 1167$

[hide] By dividing $2001$ by $3$ and $4$ and taking the integer part of the quotient we get $667$ and $500$ respectively. By the principle of inclusion and exclusion, we have to subtract the quotients when $2001$ is divided by $12$, $15$, and $20$. Then add the quotient when $2001$ is divided by $60$. Thus we get $667+500-166-133-100+33=801$ or $\boxed{B}$[/hide]

[hide]There are 667 multiples of 3 and 500 multiples of 4 from 0 - 2001, or 1167 total. The multiples of both three and four (or multiples of 12) are counted twice, so 1167 must be reduced by the number of multiples of 12 from 0 - 2001 or 166 numbers, leaving 1001 numbers. The problem states that multiples of 5 must be excluded, so we must eliminate the numbers that are multiples of both 3 and 5 (or multiples of 15), and numbers that are multiples of both 4 and 5 (or multiples of 20). After subtracting, there are 1001 - 133 - 100 = 768 numbers. We also eliminated the numbers that are multiples of 3, 4, and 5 (or multiples of 60) twice. We must add the number of multiples of 60 from 0 - 2001. There are 33, and adding this to 768 gives us 801 numbers. [/hide]

[hide="Answer"]We take multiples of $3$ and multiples of $4$, and subtract the overlap, so we have $667+500-166=1001$. However, we must remove multiples of $5$. Every $60$, we have $15$, $20$, $30$, $40$, $45$, $60$ that overlap, so we remove $6$ numbers per $60$. There are 33 $60$'s before $2001$, so we subtract $198$, and finally the $15$ and $20$ of the next $60$, so we have $1001-200=801\Rightarrow \boxed{B}$.[/hide]

[quote="Silverfalcon"]How many positive integers not exceeding 2001 are multiple of 3 or 4 but not 5?

$\text{(A)} \ 768 \qquad \text{(B)} \ 801 \qquad \text{(C)} \ 934 \qquad \text{(D)} \ 1067 \qquad \text{(E)} \ 1167$[/quote]
[hide]
Take the # of multiples of 4 and add the # of multiples of 3 subtract the # of multiples of 3 and 4 (12) and you get the 3 or 4 part to get the non 5's you subtract the # of multiples of 3 and 5 (15) and the multiples of 4 and 5 (20) and finally add the # of multiples of all 3 numbers (60) (because we got rid of them twice).

So then you get .....

500+667-166-133-100+33=801.
SO the answer is[size=150] B 801[/size]
Sometimes using a venn diagram helps me when I do these kind of problems.[/hide]


limit
real analysis
real analysis unsolved

Let $ x_n, y_n, z_n$ be three sequences of positive real numbers, so that the sequences defined as

$ a_n\equal{} x_n\plus{}y_n\plus{}z_n$

$ b_n\equal{}x_ny_n\plus{}y_nz_n\plus{}z_nx_n$, and

$ c_n\equal{}x_ny_nz_n$

converge to the same limit. Prove that the sequences $ x_n, y_n, z_n$ converge to the same limit.

First, consider cubics of the form $ G_r(u) \equal{} u^3 \plus{} r(u^2 \plus{} u \plus{} 1)$ with $ r\in\mathbb{R}$. The discrimant is given by $ \Delta_r \equal{} \minus{} \frac {r^2}{3}[(3r \plus{} 7)^2 \plus{} 32]\leq 0$ with equality, iff $ r \equal{} 0$, i.e. in all cases $ G_r$ has exactly one real root $ R(r)$.

Second, denote the common limit of the series $ a_n,b_n,c_n$ by $ L$. Squaring $ a_n$ gives $ 0\leq x_n^2 \plus{} y_n^2 \plus{} z_n^2 \equal{} a_n^2 \minus{} 2b_n\rightarrow L^2 \minus{} 2L$. In particular, $ x_n,y_n,z_n$ are bounded. Multiplication of $ b_n$ with $ x_n$ (resp. $ y_n,z_n$) yields: $ \minus{} w_n^3 \plus{} a_nw_n^2 \minus{} b_nw_n \plus{} c_n \equal{} 0$ with $ w_n$ being one of the sequences $ x_n,y_n,z_n$. Applying limits and the boundedness(!) of $ w_n$ we get

$ 0 \equal{} \lim_{n\rightarrow\infty} \left[G_L( \minus{} w_n) \plus{} (a_n \minus{} L)w_n^2 \plus{} (L \minus{} b_n)w_n \plus{} (c_n \minus{} L)\right] \equal{} \lim_{n\rightarrow\infty} G_L( \minus{} w_n)$.

Since $ G_L$ has a unique real root $ R(L)$, we get $ w_n\rightarrow \minus{} R(L)$. Substitution into $ a_n,b_n,c_n$ yields $ L \equal{} \minus{}3R(L) \equal{} 3R(L)^2 \equal{} \minus{}R(L)^3$ and the only possible solution is $ R(L) \equal{} 0 \equal{} L$.

Let $ s$ be the limit of $ a_n$,
$ \lim x_n^2y_n^2\plus{}y_n^2z_n^2\plus{}z_n^2x_n^2\equal{}\lim(x_ny_n\plus{}y_nz_n\plus{}z_nx_n)^2\minus{}2x_ny_nz_n(y_n\plus{}z_n\plus{}x_n)\equal{}s^2\minus{}2s*s\equal{}\minus{}s^2$
But $ LHS\ge0$, we have $ s\equal{}0$.
Then $ \lim x_n^2\plus{}y_n^2\plus{}z_n^2\equal{}\lim(x_n\plus{}y_n\plus{}z_n)^2\minus{}2(x_ny_n\plus{}y_nz_n\plus{}z_nx_n)\equal{}0$ implies $ \lim x_n\equal{}\lim y_n\equal{}\lim z_n\equal{}0$.

We can see that $ x_n$ being positive is redundant.