ratio
algorithm
number theory
Euclidean algorithm

Let $ P_1$ be a regular $ r$-gon and $ P_2$ be a regular $ s$-gon $ (r\geq s\geq 3)$ such that each interior angle of $ P_1$ is $ \frac {59}{58}$ as large as each interior angle of $ P_2$. What's the largest possible value of $ s$?

[hide]Let's set up the equations first. We're given that $\frac{(58)(180)(n-2)}{n}=\frac{(59)(180)(s-2)}{s}$.

Simplifying, $s(58n-116)=n(59s-118)$ and $n=\frac{116s}{118-s}$. We know that $n$ and $s$ must be positive, so the largest possible value of $s$ is $\boxed{117}$. [/hide]

The interior angle of $n$-gon is $\alpha_n=\frac{(n-2)\cdot 180^\circ}{n}$, and that of a $s$-gon is $\alpha_s=\frac{(s-2)\cdot 180^\circ}{s}$. Hence

\begin{eqnarray*}\frac{(n-2)s}{n(s-2)}={59\over 58} &\iff& 58ns-116s=59ns-118n\\ &\iff& -ns-116s=-118n\\ &\iff& s=\frac{118n}{n+116}\\ &\iff& s=\frac{118n+118\cdot 116-118\cdot 116}{n+116}\\ &\iff& s=118-\frac{118\cdot 116}{n+116}\end{eqnarray*}

For $s$ to be maximal, the expression $\frac{118\cdot 116}{n+116}\end{eqnarray*}$ must be minimal, which in turn means that $n+116$ must be maximal while in the same time it must divide $118\cdot 116.$ Obviously, $n+116$ must be equal to $118\cdot 116$, which gives $s_{max}=117.$

$ s\equal{}\frac{2*59r}{r\plus{}2*58}$

$ 2*59r\equal{}2*59(r\plus{}2*58)\minus{}4*59*58$

We must have $ (2*59r,r\plus{}2*58)\equal{}r\plus{}2*58$ so from the Euclidean Algorithm we have
$ r\plus{}2*58\equal{}4*59*58\implies 2*59r\equal{}2*59*2*58*117$
and $ r\plus{}2*58\equal{}2*58*118$

So the answer is the quotient of the two, which is $ 117$.

[hide=Solution]The interior angle of a regular $r$-gon is $\frac{180(r-2)}{r}$ and the interior angle of a regular $s$-gon is $\frac{180(s-2)}{s}$. So, $\frac{59}{58} \cdot \frac{180(s-2)}{s}=\frac{180(r-2)}{r}$, or $s(58n-116)=n(59s-118)$. Therefore we write $n=\frac{116s}{118-s}$, and given that $n,s$ are clearly positive the largest value is $s=\boxed{117}$.[/hide]

The interior angle of a regular $r$-gon is $\frac{180(r-2)}{r}$ and the interior angle of a regular $s$-gon is $\frac{180(s-2)}{s}.$ Hence, $s(58n-116)=n(59s-118)$, so $n=\frac{116s}{118-s}.$ Hence, the greatest value of $s$ that works is $\boxed{117}.$


conics
parabola
analytic geometry
graphing lines
slope
calculus
algebra

Find the eqn of the parabola, with vertex at the origin, the x-axis vertical and the line $x+y+2 = 0$ is a tangent to it

Show working please

[hide]A parabola has the form: $y=a(x-h)^{2}+k$.  Where $(h,k)$ is the vertex $\rightarrow (0,0)$ then plug the vertex into the equation $\implies y=ax^{2}$.  It is tangent to $y=-x-2$ which has a slope of $-1$.

$y'=2ax \implies-1=2ax \implies a=\frac{-1}{2x}$ and $ax^{2}=-x-2$ at this point.

$ax^{2}=-x-2$
$\frac{-1}{2x}\cdot x^{2}=-x-2$; This point is obviously not at $x=0$ because the slope at $x=0$ is $0$ on the parabola, not $-1$.  We may proceed.
$1x=2x+4$
$x=-4 \implies a=\frac{-1}{2(-4)}\implies a=\frac{1}{8}$

$\boxed{y=\frac{x^{2}}{8}}$[/hide]

Is there a way to do this without calculus?

We're talking lines with non-0 slope tangent to a parabola. I don't know how you'd deal with that without calculus.

Thought so, the tangent line part makes it almost impossible without calculus.

Here is the answer from the back of the book

[hide]$x^{2}= 8y$[/hide]

Yes how would I approach this question without calculus??

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

Here is an example without calculus approach.

Find the equation of parabola with vertex the origin, the x-axis is y=0 and the line x - y + 3 = 0 is a tangent to the curve

soln:

Let the curve be $y^{2}= 4ax$
Solving x - y + 3 = 0, $y^{2}= 4ax$ we have
$y^{2}= 4a(y-3)$ i.e. $y^{2}-4ay+12a$ = 0

This eqn gives the y-coordinates pr prdinates of the point of intersectio nof the line of the parabola.

[b]Since the line is a tangent to the curve, the roots of this eqn are equal to one another[/b]

For $ax^{2}+bx+c$ to have equal roots, we know that $b^{2}-4ac$ = 0. Applying this result to the eqn

$y^{2}-4ay+12a$ = 0 we have $(-4a)^{2}-4.1(12a)$ = 0

a=3

hence the eqn is
$y^{2}= 12x$

-----

I dont understand the step in bold

Also could someone use this example and try and solve the above

thanks

can this question be done without calculus?

[quote="bos1234"][b]Since the line is a tangent to the curve, the roots of this eqn are equal to one another[/b]

...

I dont understand the step in bold[/quote]

If a polynomial $P(x)$ is tangent to a line $y = mx+b$ at a point $x_{0}$, then

$P(x)-mx-b$

Is either only positive or only negative around the neighborhood of $x_{0}$, so $x_{0}$ is a double root of $P(x)$; in other words, $(x-x_{0})^{2}| P(x)$.

Example:

The parabola $y = x^{2}$ is tangent to the line $y = 2x-1$ at $x = 1$, so the polynomial

$x^{2}-(2x-1)$

Has a double root of $1$.  And in fact,

$x^{2}-(2x-1) = (x-1)^{2}$

Edit:  And of course this logic involves [b]no calculus.[/b]


Find the solutions, for $t \geq 0$, of the problem $x' = max(t,x); x(0)=0$.

Since the direction field is Lipschitz, the solution is unique: $x(t)=t^{2}/2$ for $0\le t\le 2$ and $x(t)=2e^{t-2}$ for $t\ge 2$. It may be helpful to draw a direction field on the $(t,x)$-plane to see how this works.


How many integers between 100 and 300 have both 11 and 8 as factors?

8 and 11 are relatively prime and 88 is the first number that has both as a factor.  Then all the numbers that contain both as factors must be multiples of 88.  That is, 88, 176, 264,... so there are only two in the range you specified.


Hi all,

From now on a dollar is going to be rendered as in normal LaTeX documents, namely with the code

$ \$ $ which gives a $ :)

PS Please don't quote this post, as it will ruin the thread :)

Test (you can quote this post :D) $ \$ $

[quote="Valentin Vornicu"]Test (you can quote this post :D) $\$$[/quote]

Awesome!!!! Good job, I've been waiting for that, particularly in my blog!

$[quote]$[/quote]

Can you fix this bug next?

It messes up the whole thread... :|

A picture of it messed up:

[hide="Click to Reveal Hidden Picture"]
[img]http://i96.photobucket.com/albums/l167/adasarathy/pictureofmesseduplayout.png[/img]
[/hide]

How'd you get the LaTeX not to compile in the first post?  :huh:

He checked the box next to "Disable BBCode in this post".  See, $ renders as normal with it checked.


Which natural numbers smaller than 100 000 are more in number-those,whose sum of digits is equal to 22 or those,whose sum of digits is equal to 23.

could you explain it more clearly because i can't seem to understand.

I think you mean: "How many more natural numbers less than 100,000 are there than numbers whose digit-sums are equal to 22 or 23?" :wink:

Which natural numbers smaller than 100 000 are more- with digit-sums equal to 22 or equal to 23.
Examples:
Which natural numbers smaller than 21 are more- with digit-sums equal to 2 or equal to 3.
natural numbers with digit-sums equal to 2: 2,11 and 20.
natural numbers with digit-sums equal to 3: 3 and 12.
So the answer is:natural numbers with digit-sums equal to 2.
I belive that now you can understand me.

[quote="AIME15"]It should be, "Are there more natural numbers less than 100,00 whose digit-sums are equal to 22, or are there more whose digit-sums are equal to 23?"[/quote]
Yes,that's my question,thank you AIME15.

[quote="barcelona"]Can anyone post a solution?[/quote]

Any natural number less than 100,00 whose digit-sum is equal to 22
corresponding to:
22=a+b+c
or
22=a+b+c+d
where a,b,c,d are single digit natural numbers with  non-zero a.

It is quite intuitive that
the number ways to write 22 = a+b+c
is less than the
the number ways to write 23 = a+b+c
and
the number ways to write 22 = a+b+c+d
is less than the
the number ways to write 23 = a+b+c+d

so there are more digit-sums are equal to 23 than digit-sums are equal to 22.

[quote="Math Champion"]Not necessarily the bigger the number, the more ways there are. For example, there are less ways to make a sum of 8 when rolling two dice than there are to make a sum of 7.[/quote]

I see.

The digit is limited to 9.
if you write 23 = a+b+c you should have more choices than 22=a+b+c.
However, if you change 23 to a bigger number say 35.
There is no way you can write
35=a+b+c
so you get less choices (actually none) when the number you pick is too big.

Interesting problem :-)

But the sum of the digits has a upper bound too...I think you mean that if you increase the number when it is below the average of the lower bound and the upper bound, you gain more combinations.

I think tiger21's solution is correct because as long as you increase the number, until it hits the maximum, there will be more ways to get it. Unless it gets to the point where the number has to have 2 nines or something like that. I don't know if this is clear.

What I'm trying to say is that it's not always the higher number. For example, with three digits, and you need the digits to add up to 27, there is only one way to do that. 999. But if u need the digits to add up to 26, there are more possibilities.

That's cause its 26 and 27 are larger than the average of the lower and upper bounds. See my previous post.

In fact, you are all wrong.As the nunmbers must be smaller than 100 000, therefore the largest possible must be 99 999.
We see that tha sum of the digits of 99 999 is equal to 45 (22 +23=45). Hence, if we take a number with digits' sum 22 from 99 999 , we will get a number with digits'  sum 23 and vice verse. It means that the numbers with digits' sum 22 and the numbers with the digits' sum 23 go in pairs.Then the amount of these numbers is equal.

I don't see why the numbers come in pairs just because the maximum sum possible is 45.

The numbers come in pairs not just because the maximum sum possible is 45 and because   if we take a number with digits' sum 22 from 99 999 , we will get a number with digits' sum 23 and vice verse.
Example:  let's  we take 679(6+7+9=22),then we have : 99 999-679=99320 (9+9+3+2+0=23)
Which it means  that the numbers with digits' sum 22 and the numbers with the digits' sum 23 go in pairs.Then the amount of these numbers is equal.

[quote="Math Champion"]Ah i see. But the minimum sum is not 0, its 1, so that screws up your count.[/quote]
I can't uderstand you,which minimum sum is not 0.

All of you are wrong. Some of you got the right answer, the wrong way. Others (barcelona) did it the right way but got the wrong answer.

Note there is a one-one correspondence between numbers with a given sum (i.e. there is 1 number with sum 0, 00000, and one with sum 45, 99999), so by all rights the number of each should be equal (as 22+23=45).

However, we need to include zeroes. Intuitively obvious, (and I will leave it for you to figure out why) the numbers with sum 22 will include more starting digits of 0 than the numbers with 23, so there are more numbers with sum 23.

...usually called "one-to-one correspondence," but the solution's still good.

^Mewto ...

I am pretty sure they are equal, I am not sure what you mean ... leading zeros are fine.   00499 is a natural number, it just isn't a 5-digit number.

Consider 5 urns, each containing 9 balls.  Remove balls and place them into a separate set of corresponding urns.  If you remove 22 balls there will be 23 left and each remaining set of urns corresponds to a number that is less than 100,000 (one with a digit sum of 22 and one with a digit sum of 23).

Can you come up with a number less than 100,000 whose digit sum is 23 and which will not yield a number whose digit sum is 22 when subtracted from 99,999 (or vice-versa?).

I am certain that there are an equal number of natural numbers less than 10 whose digit sum is 4 as there are numbers whose digit sum is 5.  I am confident that there are an equal number of natural numbers less than 1,000 whose digit sum is 13 as there are whose digit sum is 14 (there are 75 of each, figure that one out).  Why wouldn't there be an equal number of natural numbers less than 100,000 with digit sums 22 and 23?

Hm, he just misread the question and thought the numbers had to be five-digit...and I misinterpreted the question the same way only based on his solution...

And i think that the amount of these numbers is equal.


Let ABCD be a cyclic quadrilateral. Its diagonals AC and BD intersect at O. Let S1 and S2 be the circumcircles of ABO and CDO, respectively. The other intersection point of S1 and S2 is K. Through O we draw a parallel line to AB which intersects S1 again at L and a parallel line to CD which intersects S2 again at M. On segments [OL] and [OM] we choose points P and Q respectively s.t. OP/PL=QM/OQ. Prove that O, K, Q, P are on a circle.

[i]Russian olympiad 2003/11'th grade/problem 2[/i]

[quote="treegoner"]:D  I think this one is easy.
       First you should prove this lemma
    AOB is a triangle. (S) is a circle passing through A and B and intersects OA, OB respectively at C and D. Then SO is perpendicular to CD.
     Applying this lemma, we have OS1, OS2 is perpendicular to CD and AB. So OM and OL are respectively the tangent of (S1) and (S2) at O. This implies that  OLK are similar to MOK. We easily prove that OKP are similar to MQK we have done.  :D[/quote]

     I do not see SO  being perpendicular to CD.

    Lt. Pestich

[quote="pestich"]I do not see SO  being perpendicular to CD.

    Lt. Pestich[/quote]

$(S)$ is the circumcircle of $\triangle ABO$. Some other circle passes through A and B and intersects OA, OB, respectively, at C and D. Then S, O are collinear with the orthocenter of $\triangle CDO$.


The following one is a problem related to one of the problems given at Mathlinks contest (that one was much more difficult than this one). Prove that the limit of the sum of digits of n! goes to infinity when n goes to infinity. It is simple, but I don't find it trivial.

More generally :

Let S(x) be the sum of the (decimal) digits of the positive integer x.
Let v_2(x) be the 2-valuation of the positive integer x, and v_5(x) be its 5-valuation.

Then, you may prove (it is not so difficult, just a little boring to write here) that :
If (x_n) is an (infinite) increasing sequence of positive integers such that v_2(x_n) - v_5(x_n) tends to +oo or to -oo as n goes to +oo, then S(x_n) tends to +oo.

It may clearly be used to prove each of the two problems (and many more)

Pierre.

Would you be so kind to post a solution to this fact? I don't have any idea of proving it. Please!!

Ok... The following is due to Bruno Langlois :

[x] denotes the integer part of x.
d(n) denotes the number of non-zero digits of the integer n.

Lemma 1 :
Let (M_n) be an increasing sequence of positive integers such that the sequence (d(M_n)) is bounded from above by the integer N. Then there exists a subsequence (M_(n_k)) and a nonnegative integer m such that, for each k  \geq 0, we have M_(n_k) - m = 0 mod[10 k ].

Proof : Use induction on N. It is clear for N = 1.
Now suppose it is true for N, and let's give us a sequence (M_n) satisfying the statements for N+1.
Let (M'_n) be the sequence obtained from (M_n) by deleting the leftmost digit of each M_n.
- If (M'_n) is bounded, thene we may extract a constant subsequence, and choose m the constant value.
- Otherwise, we may extract an increasing subsequence (M'_(n_k)). The induction hypothesis applies to this subsequence, and we are done.

Remark :
Note that we may replaced the condition (M_n) is increasing by (M_n) tends to +oo.

Lemma 2 :
Let (U_n) be a sequence of positive integers.
If :
a) v_2(U_n) tends to +oo and none of the U_n is divisible by 5
or
b) v_5(U_n) tends to +oo and none of the U_n is even
then :
S(U_n) tends to +oo.

Proof :
Suppose a) (the proof is the same in the other case).
Suppose, for a contradiction, that S(U_n) does not tend to +oo.
Then, we may extract a subsequence (U'_n) from (U_n) such that (d(U'_n)) is bounded from above. Frome the lemma 1, we may find a nonnegative integer m and extract a subsequence (U_(n_k)) from (U_n) such that for each k, there exists a positive integer a_k satisfying :
U_(n_k) = (a_k)*10 k + m.

Note that 2 inf(k,v_2(U_(n_k)) divides U_(n_k) - (a_k)*10 k, which is not zero since 5 does not divides U_(n_k).
Then m  \geq 2 inf(k,v_2(U_(n_k)) . A contradiction.

Theorem :
If (x_n) is an (infinite) increasing sequence of positive integers such that v_2(x_n) - v_5(x_n) tends to +oo or to -oo as n goes to +oo, then S(x_n) tends to +oo.

Proof.
Suppose that v_2(x_n) - v_5(x_n) tends to +oo (the other case is proved in the same way).
If n  \geq n_0 is sufficiently large, then v_2(x_n)  \geq v_5(x_n). Then define U_n = (x_n)/10 v_2(x_n) . And applying lemma 2, we deduce that S(x_n) = S(U_n) tends to +oo.

Pierre.

And you call that "not so difficult to prove"? In fact, my solution was totally different. It uses a very nice result (prove it) that if a number is divisible by 10^n-1, then its sum of digits is at least n. So, s(n!) is at least int(logn).

>And you call that "not so difficult to prove"?

Hmmmm.....yes! :D

Pierre.

@hazari
could you please show your proof. im too dumb :( . also tell me guys , how do i make my brain cells more active ? :oops:


[quote="cezar lupu"]Let x(n) be a sequence such that x(1)>0 and x(n+1)=x(1)/n+1+x(2)/n+2+...+x(n)/n+n, for every n>=1.
  prove that x(n) is convergent to zero.[/quote]
You mean $ x_{n+1}=\frac{x_1}{n+1}+\frac{x_2}{n+2}+\cdots +\frac{x_n}{n+n}$?

I hope you understand what I wrote.I don't know how to write in Latex?

I can prove that $x_n$ is smaller than $x_{n-1}$, but I cannot prove that it converges to 0.

to prove that x(n-1)<x(n) is quite easy.

[quote="cezar lupu"]Let $x_n$ be a sequence such that $x_1>0$ and $x_n+1=x_1/n+1+x_2/n+2+...+x_n/n+n,$ for every $n>=1.$
  prove that &x_n$ is convergent to zero.[/quote]

it's easy to prove that $x_1 > x_i > 0$, then consider
$x_n=\frac{x_1}{n}+...+\frac{x_{n-1}}{2n-2}$
$\frac{x_1+...+x_{n-1}}{n} > x_n > \frac{x_1+...+x_{n-1}}{2n-2}$
Then take limit n-> inf, then since $x_i < x_1$ and x_1 is a constant, we have both sides tend to 0. Then by sandwich theorem we have the result.


Let $\{x_{n}\}_{n\ge0}$ and $\{y_{n}\}_{n\ge0}$ be two sequences defined recursively as follows \[x_{0}=1, \; x_{1}=4, \; x_{n+2}=3 x_{n+1}-x_{n},\] \[y_{0}=1, \; y_{1}=2, \; y_{n+2}=3 y_{n+1}-y_{n}.\] [list=a][*] Prove that ${x_{n}}^{2}-5{y_{n}}^{2}+4=0$ for all non-negative integers. [*] Suppose that $a$, $b$ are two positive integers such that $a^{2}-5b^{2}+4=0$. Prove that there exists a non-negative integer $k$ such that $a=x_{k}$ and $b=y_{k}$.[/list]


$a,b,c,d$ are positive prove that
$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\ge 4$

Well Mr payman_pm, you always post easy problems  :lol:
Our inequality is equivalent to : $(a+c)(\frac{1}{a+b}+\frac{1}{c+d})+(b+d)(\frac{1}{b+c}+\frac{1}{d+a}) \geq 4$
By $AM \geq HM$ we have $\frac{1}{a+b}+\frac{1}{c+d} \geq \frac{4}{a+b+c+d}$
and $\frac{1}{b+c}+\frac{1}{d+a} \geq \frac{4}{a+b+c+d}$
So : $LHS \geq (a+c)(\frac{4}{a+b+c+d})+(b+d)\frac{4}{a+b+c+d}=4$
And we are done   ;)


calculus
integration
logarithms
trigonometry
geometry
geometric transformation
reflection

It will take geniuses to figure this problem out (and im not kidding)

integral of (dx/(x^3+x))

I tried substitution, partial fractions but it just isn't working.

Partial fractions:

1/(x(x^2+1)) = A/x + (Bx+C)/(x^2+1)

1 = A(x^2+1) + (Bx+C)(x)

now you're supposed to find A, B, and C. you can find A by letting x = 0.

1 = A(1) + 0
A = 1.

1 = (x^2+1) + (Bx+C)(x)

I can't find C because the only value that would make the term Bx = 0 is letting x = 0. But if i do that, the whole thing (Bx+C) becomes 0.

Can anyone show me how to do this problem please? Show all the steps please.

I don't think anyone will be able to get it.  I'm pretty good in calculus but I can't figure it out.

I don't understand how u got from 1/(x(x^2+1)) to 1/x - x/(x^2+1)
can you please explain that

[quote="zubairkhan14"]
1 = (x^2+1) + (Bx+C)(x)
[/quote]

From there, we rearrange as $(1+B)x^2 + Cx = 0$ (for all $x$). So $B = -1$ and $C = 0$. Thus $\frac{1}{x^3+x} = \frac{1}{x} - \frac{x}{x^2+1}$

Btw, such posts should go under the forum [url=http://www.mathlinks.ro/Forum/index.php?f=296]Tutorials[/url].

[quote="billzhao"]It's straightforward partial fractions.

\[
\int \frac{dx}{x^3+x} = \int \left( \frac{1}{x} - \frac{x}{x^2+1} \right) \ dx = \int \frac{dx}{x} - \int \frac{x \ dx}{x^2+1} = \ln x - \frac{1}{2} \ln (x^2 + 1) + C
\][/quote]

$\ln |x|$

another question

integral of (dx/(4-x^2)^(3/2))

zubairkhan14- kunny is using a trigonomentric substitution, in this case $x = 2\sin{\theta}$

i don't get it...

can anyone solve this problem showing step by step what you did?

integral of (dx/(4-x^2)^(3/2))

Go read through the examples in the "trigonometric substitutions" section of your calculus textbook and give it another try yourself. You do have a calculus textbook?

You forgot to square $x$.

The whole point of a trigonometric substitution is to use a Pythagorean identity to turn a square root of a sum or difference of squares into a single term.

Suggestion: as soon as you decide that $x=2\sin\theta$ is the right substitution, draw a picture of a right triangle. Label one of the angles $\theta$, then label two of the sides in such a way as to reflect the substitution you have made. In this case, label the "side opposite" as $x$ and the hypotenuse as $2$. Use the Pythagorean theorem to fill in the remaining side. In this case, that's the "side opposite" and it is $\sqrt{4-x^2}.$

(We should worry about $\pm$ signs on the square root. In practice, I draw a first quadrant picture and assume everything is +; we can check quadrants at the end, if necessary.)

We see $\sqrt{4-x^2}$ as one of the sides of the triangle. We chose this substitution precisely for its power to simplyfy that. We can read off our our triangle that $\sqrt{4-x^2}=2\cos\theta.$

If the square root doesn't simplify like that, then you've used the wrong substitution.


Let two circles S and T touch (once) at X. Let AB be a common tangent with A on S and B on T. Let P be the point on S diametrically opposite A. Draw a tangent to T through P, and call the point where it touches T, Q.

Show that $AP=PQ$


inequalities
function
rearrangement inequality
inequalities proposed

If a,b,c positive integers proove that
   \[ \frac{a^a}{b}+\frac{b^b}{c}+\frac{c^c}{a}>a^{a-1}+b^{b-1}+c^{c-1} \]

$\frac{a^a}{b}+\frac{b^b}{c}+\frac{c^c}{a}\geq a{}^a{}^-{}^1+b{}^b{}^-{}^1+c{}^c{}^-{}^1$ Now to prove this inequality we use a rearrangement inequality.

WLOG : $a\geq b\geq c$ then, $\frac{1}{c}\geq\frac{1}{b}\geq\frac{1}{a}$ and also it is obvious that ...
$a^a\geq b^b\geq c^c$. Now we apply the a well know inequality to the last to inequalities. In this form
$Random Sum\geq Reverse Sum$ so we are done.

[quote="perfect_radio"]Take $a=0.6, \, b=0.1, \, c=0.3$ and see what will get.

$x \mapsto x^x$ is decreasing on one interval (the sign of the ineq is $\leq$), and increasing on the other (... is $\geq$). These results use the rearrangement inequality and some calculus.[/quote]

Perfect radio your counterexample is true but read the problem accurately then you will see that a,b,c are positive integers so you are wrong.

;)  [i]I don't understand what is going on there. Maruf had proven that inequality holds for every real numbers (since Rearrangemnet Inequality is true for every positive reals). But perfect radio is giving a counterexample for this case. That means the inequality is WRONG. [/i]

Davron

well firstly they are integers...not real
Secondly Maruf did proove it
SO IT'S OK :?  :mad:

Yes i am agree with Maruf's proof. But from his proof we can conclude that it is also true for +reals
but there is an counterexample, so i don't know what to say . Did you got me ?

[color=indigo][ If we show the minimum value of any function with the using the rearrangement's inequality that mean's the inequality holds for every positive reals] [/color]

Davron

Come on, Davron, didn't you note that the step $a\geq b\geq c\Longrightarrow a^a\geq b^b\geq c^c$ does not hold for every triple of positive reals a, b, c ? In fact, it holds if all three numbers a, b, c are $\geq\frac{1}{e}$, what is, of course, satisfied for positive integers.

Note that the inequality should be $\frac{a^a}{b}+\frac{b^b}{c}+\frac{c^c}{a}\geq a^{a-1}+b^{b-1}+c^{c-1}$ (mind the $\geq$ sign).

  darij

Ohhhhh...Yeah :blush: ..It's true if 0<a<1 and a>b>c then a^a>b^b>c^c isn't true...well...at least the problem is ok

[quote="luke999"]If a,b,c positive integers proove that
   \[ \frac{a^a}{b}+\frac{b^b}{c}+\frac{c^c}{a} \ge a^{a-1}+b^{b-1}+c^{c-1}\ \ (*)  \][/quote]
Assume $a\ge b \ge c$.
\[ (*) \iff (a-b)\frac{a^a-bc^{c-1}}{ab}+(b-c)\frac{ab^{b-1}-c^c}{bc} \ge 0  \]

You should see my generalization to this problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=86692


Hi Every Body !
I need an ebook about the apply of complex number in combinatorial and mumber theory !
Can you help me ! Thank Very Much  :)

Titu Andreescu, Dorin Andrica: "Complex numbers from A to Z", published by Birkhauser.


Question 1:
http://mathcounts.saab.org/mc.cgi/mc.cgi

This is one of the problems:

Positive integers  x, y  and  n  are each less than  8 . If  x^2 - y^3 = n  , find the median of all possible values for the sum  x + y + n .

Answer according to the site:
[hide][i]
 3.  (Sprint) Positive integers  x, y  and  n  are each less than  8 . If  x^2 - y^3 = n  , find the median of all possible values for the sum  x + y + n  .
    The right answer is 9
    Your answer was 6

Here is one way of doing this problem:
Here are all the triplets {x,y,n} such that
x2 - y3 = n:
{{3,2,1}, {2,1,3}, {3,1,8}, {4,2,8}}.
Here are their sums:
{6, 6, 12, 14}.
Here are their sorted sums:
{6, 6, 12, 14}.
The median of the last set is: 9[/i]

But the problem mentions that x,y, and n are each less than 8. So shouldn't the pairs just be 3,2,1 and 2,1,3?[/hide]

Question 2:
Here's a [math counts] question that I'm not sure about. I don't really know how to solve probability questions.

[i]In a school, student IDs are 2 letters followed by 2 digits. For the letters, O cannot be used and for the digits, 8 cannot be used. The letters must be in ascending order and the digits must increase in value. For example, AB15 is an acceptable ID while AA32 is not. How many possible IDs are there?[/i]

So can anybody help me with that problem?

for the first one, you're right. Nice catch.  :)

for the second one:

In a school, student IDs are 2 letters followed by 2 digits. For the letters, O cannot be used and for the digits, 8 cannot be used. The letters must be in ascending order and the digits must increase in value. For example, AB15 is an acceptable ID while AA32 is not. How many possible IDs are there?

so for the first letter choice, there are 25 letter choices. Say he starts by choosing an A. He has 24 choices after that. Say he starts by choosing a B, he has 23 choices after that. As you can see, it is 24+23+22....+1 so (24*25)/2
or 300. Do the same for the digits. If you choose 1 first, you have 8 choices, if you choose 2 first, you have 7 choices. As you can see, it is 8+7+6...+1 or (8*9)/2=36 so you do 300*36 since they are independent.

so 10800 ways.

[quote="d. philly"]Question 2:
Here's a [math counts] question that I'm not sure about. I don't really know how to solve probability questions.

[i]In a school, student IDs are 2 letters followed by 2 digits. For the letters, O cannot be used and for the digits, 8 cannot be used. The letters must be in ascending order and the digits must increase in value. For example, AB15 is an acceptable ID while AA32 is not. How many possible IDs are there?[/i]

So can anybody help me with that problem?[/quote]

[hide="Answer"]10800[/hide]

Here, it is clear that we should solve for the total combinations of letters and digits seperately, then multiply together to get answer.

Since O cannot be used for letters, then we have 25 possible letters for the first letter.
Let's do some casework:
If it is A, we have 24 possibilities for second letter, therefore 24 possible letter combinations.
If it is B, we have 23 possibilities for second letter, therefore 23 possible letter combinations.
We can easily see that the total number of combinations is $ 24 \plus{} 23 \plus{} 22 \plus{} ... \plus{} 1 \equal{} 12(25) \equal{} 300$

Similarly, we we see that
if the first number is 0, there are 8 other possibilities for the second number, etc.
$ 8 \plus{} 7 \plus{} 6 \plus{} ... 1 \equal{} 4(9) \equal{} 36$

So the answer is $ {10800}$
[/hide]

EDIT: Beaten to it


[b](1) How can I calculate the value of tan(22.5degree) and tan(15degree) with the help of Geometry.[/b]
plz give me full solution

[quote="Dr Sonnhard Graubner"]hello, see here
http://mathworld.wolfram.com/TrigonometryAnglesPi8.html
Sonnhard.[/quote]

Your comment isn't helping.
Because it does not do geometric solution.

[geogebra]b9cc693d94988614aa5c601d6bfd8fc334d304c9[/geogebra]

(let AB=1)

From the pic, AD is a bisector. Therefore, BD= 1/(1+√2)= √2 -1, and then tan(22.5)=BD/AB=√2 -1.

You can get tan(15) in similar way.

 [geogebra]a77ee8c7e73a457b9bdae1fbbaae6b37865141a6[/geogebra]

I'll just post a pic  :D

Use the fact that $ \tan 2x \equal{} \frac {2\tan x}{1 \plus{} \tan^2x}$, which can be proven geometrically as $ 0^\circ < x < 90^\circ$. For a picture, see http://www.jstor.org/pss/3620473 .


When 3-oxopentanoic acid is heated in the presence of bromine a compound A is obtained, which on treatment with sodium hydroxide, followed by acidification, yields 2-methylpropanoic acid.

$ CH_3CH_2COCH_2CO_2H \plus{} Br_2 \plus{} heat \longrightarrow A$

$ A \plus{} NaHO/then\, H^\plus{} \longrightarrow (CH_3)_2CHCO_2H$.

Find the structure of A and provide an explanation for these transformations.

A- Ch3ch2COCH2Br
then favorskii rearrangement.

A could also be Ch3Ch2COChBrCOOH
and probably CO2 is eliminated in the second step.

No CO2 is eliminated in the first step itself to get the carbanion(intermediate species) and then that attacks the Br2 to get A.
And then ofc Favorskii and then the ring opening of the cyclopropanone towards primary -CH2
side.

but then you wud be getting butanoic acid as another product too :ewpu:

[quote="VARUNARASU"]A- Ch3ch2COCH2Br
then favorskii rearrangement.[/quote]

Yes, that's right.

Yeah but major product will be carcul's product because it is formed via primary carbanion so I think that should be the major :maybe:  :maybe:


algebra
polynomial
advanced fields
advanced fields theorems

I saw a theorem in [url=http://math.arizona.edu/~ura/013/miller.justin/]this[/url] article and I can't seem to realize why it is true. The theorem states:

Let $ \mathbb{F}_q$ be a finite field with $ q\equal{}p^n$ elements, where $ p$ is prime, and let $ E$ be an elliptic curve over $ \mathbb{F}_q$. If $ E(\mathbb{F}_q)$ has order $ m$ then the group $ E(F_{q^k})$ has order $ 1\plus{}q^k\minus{}a^k\minus{}b^k$ where $ a$ and $ b$ are the reciprocals of the roots of the polynomial $ 1\minus{}(q\plus{}1\minus{}m)x\plus{}qx^2$.

Thanks.

This follows from the rationality part of the Weil conjectures. (You can generalise this to arbitrary curves and higher dimensions.)


I don't think a satisfactory solution has been provided for the following problem.

Compute the exact value of $ \sum_{k=1}^{1001}\,\log_{8}\left(\cos\frac{k\pi}{2003}\right)$.

[hide="Hint"]This (or very similar to this) has been posted a few times ..
see:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1857705171&t=147086[/hide]

I'll take a crack at this one.

[hide]$ \sum_{k=1}^{1001}{\log_{8}(\cos(\frac{k\pi}{2003}))}=\log_{8}(\prod_{k=1}^{1001}(\cos(\frac{k\pi}{2003})))$

So what we need to find is that product.  And from the link Gyan posted, we can use the formula:

$ \sin (mx) = 2^{m-1}\sin (x) \cdot sin (x+\pi/m) \cdot \sin (x+2\pi/m) \cdots \sin (x+(m-1)\pi/m)$

We can see that $ \prod_{k=1}^{1001}(\cos(\frac{k\pi}{2003}))=\sqrt{-\prod_{k=1}^{2002}(\cos(\frac{k\pi}{2003}))}$ as $ \cos(x)=-\cos(\pi-x)$.  And as we can see that $ \cos(x)=\sin(x+\frac{\pi}{2})$:

$ \sqrt{-\prod_{k=1}^{2002}(\cos(\frac{k\pi}{2003}))}=\sqrt{-\prod_{k=1}^{2002}(\sin(\frac{\pi}{2}+\frac{k\pi}{2003}))}$

Then from our formula above, we can plug in $ x=\frac{\pi}{2}$ and $ m=2003$ and find:

$ \sin{\frac{\pi}{2}*2003}=2^{2003-1}\prod_{k=0}^{2002}(\sin(\frac{\pi}{2}+\frac{k\pi}{2003}))$
$ -1=2^{2002}\prod_{k=0}^{2002}(\sin(\frac{\pi}{2}+\frac{k\pi}{2003}))$
$ -\frac{1}{2^{2002}}=\prod_{k=0}^{2002}(\sin(\frac{\pi}{2}+\frac{k\pi}{2003}))$

So that means:
$ \sqrt{-\prod_{k=1}^{2002}(\cos(\frac{k\pi}{2003}))}=$
$ \sqrt{\frac{1}{2^{2002}}=}$
$ \frac{1}{2^{1001}}$

So our final answer is $ \log_{8}(2^{-1001})=-\frac{1001}{3}$  And it seems that there is a solution with this formula, even if I might of messed up in the middle[/hide]

@Gyan or anybody else that might know: Do you know where the formula $ \sin (mx) = 2^{m-1}\sin (x) \cdot sin (x+\pi/m) \cdot \sin (x+2\pi/m) \cdots \sin (x+(m-1)\pi/m)$ comes from and how to prove it?  It is hard searching for it on the internet when you don't even know its name.


Let $ a,b,c \in R$. Prove that:
$ 3(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)\geq (ab\plus{}bc)^3\plus{}(bc\plus{}ca)^3\plus{}(ca\plus{}ab)^3$

[quote="No Reason"]Let $ a,b,c \in R$. Prove that:
$ 3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)\geq (ab \plus{} bc)^3 \plus{} (bc \plus{} ca)^3 \plus{} (ca \plus{} ab)^3$[/quote]
It is enough to prove  for $ a\geq0,$ $ b\geq0$ and $ c\geq0$ because $ |a|^3\left(|b| \plus{} |c|\right)^3\geq a^3(b \plus{} c)^3.$
We obtain:
$ 3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)\geq (ab \plus{} bc)^3 \plus{} (bc \plus{} ca)^3 \plus{} (ca \plus{} ab)^3\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(3a^4b^2 \minus{} 3a^3b^2c \minus{} a^3b^3 \plus{} a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow3\sum_{cyc}(a^3b^3 \minus{} a^3b^2c \minus{} a^3c^2b \plus{} a^2b^2c^2) \plus{} 3\sum_{cyc}(a \minus{} b)^2a^2b^2 \plus{} \sum_{cyc}(a^3b^3 \minus{} a^2b^2c^2)\geq0.$


In a party of $1000$ people, the number of people who have shaken hands with at most $962$ people is less than or equal to $37$. Show that we can find $27$ people in the party who have all shaken hands with each other.

This question was in the Singapore MO 2000, but I'm not sure if it has more ancient roots, which is perfectly possible. :)

Let $Y$ be the set of people at the party who have shaken hands with at least $963$ people.  Then $|Y| \geq 963$.  For any person $y$, let $S(y)$ be the set of all people who he has shaken hands with.  If $y \in Y$, then $|S(y)| \geq 963$.  Choose $y_1 \in Y$.  Then
\[|Y \cap S(y_1)| \geq |Y| + |S(y_1)| - 1000 > 0.\]

Choose $y_2 \in Y \cap S(y_1)$.  Then
\[|Y \cap S(y_1) \cap S(y_2)| \geq |Y \cap S(y_1)| + |S(y_2)| - 1000 > 0.\]

Proceeding similarly, we find that $27 \times 963 - 26 \times 1000 > 0$, so we can choose $y_1, y_2, \ldots, y_{26}$ so that $y_k \in Y \cap S(y_1) \cap \cdots \cap S(Y_{k-1})$ for $2 \leq k \leq 26$, and $Y \cap S(y_1) \cap \cdots \cap S(Y_{26}) \neq \emptyset$.  Choose $y_{27}$ in $Y \cap S(y_1) \cap \cdots \cap S(Y_{26})$, then $y_1, y_2, \ldots, y_{27}$ have all shaken hands with each other.

Don't know why but I have remembered the following problem http://www.mathlinks.ro/Forum/viewtopic.php?t=29343


1. Using the ten letters in the word MATHCOUNTS, what is the probability that any random arrangement of the letters will have the word MATCH in it?

2. Maria has four cookies and four chocolate chips. She randomly places the chocolate chips in any cookie. What is the probability that there are no more than two chips in any cookie?

For (1) I thought it should be: [5!*6/2]/[10!/2!], but that's wrong. Can someone explain what I'm doing wrong?

For (2) I tried the reverse by doing 3 and 4 chips, but I'm off by 3...

Thanks :)

[hide]There are 1*1*2*1*1 ways of making MATCH, and 5 possible positions of M, so there are 10 ways there. Now the different ways of arranging MATHCOUNTS is 10!/2! I think, since the 2 T's are indistinguishable. So we have 1/9*8*7*6*5*4*3 probability.

We have [code]
 _  _  _  _
(_)(_)(_)(_)
[/code]
(a graphical cookie diagram)
The cases that meet the condition are 1 in each and 0 in one and 0 in two.
1 in each occurrs 1 time
0inone occurrs 4 times
0 in two occurs 12 times
for a total of 17 that fit. Now can someone find all possible ones? I think it is 4^4 but am not sure.[/hide]

Billy

1
[hide]Ok, do it like so: MATCH: There are 1*1*2*1*1=2 ways of making it (since there are 2 Ts) and 5 differents places for the word to start, and there are 10!/2! ways to arrange them, so the equation is:
$\frac {10}{\frac {10!}{2!}}$
$\frac {10}{10*9*8*7*6*5*4*3*2*1}$
${\frac (1)(9*8*7*6*5*4*3*2}$
and I'm not bothering to multiply them out. (I also have a sneaking suspition I'm worng)[/hide]
2
[hide]I'm not as good at code as solafide, but anyways.
CASEWORK!
1 chip in each: 1 way
0 chips in one: 4*3-12 ways
0 chips in two: 6 ways
That's a total of 17 that work.

Now, there are:
1 chip:1
1,0,2,1=12
1,0,3,0=12
0,0,4,0=4
2,2,0,0=6
Thats 35 ways total.

>>17/35<<
(Why do I think I'm doing this wrong?)[/hide]

[hide=" HINT for number 1"]The word MATCH once chosen must be arranged among the remaining letters OUNTS.  The letters OUNTS can be arranged 5! ways, and the word MATCH can be placed *O*U*N*T*S* in any of the starred spots, so there are 6 places MATCH can be put.  And as you all said, there are 2 ways to spell MATCH. [/hide]

There are no correct answers to #2 either.  Since I think this is a request for help, rather than a challenge, a methodical solution is contained below.  However, I would like to challenge members to devise a more efficient way of answering this problem.  [size=75][Note:  This would be a very challenging MATHCOUNTS question.  I cannot imagine that this would appear on anything below the level of a National Competition question.  However, the general techniques to solve this are within the skills expected of high-level MATHCOUNTS competitors.][/size]

[hide]We have a total of 4 chocolate chips.  There are 5 possible partitions of 4 chips:
$4+0+0+0 \\
3+1+0+0 \\
2+2+0+0 \\
2+1+1+0,\ {\rm and}\ \\
1+1+1+1.$

Let's examine these, case-by-case.

Case:  $4+0+0+0$
We have to choose a cookie to receive all 4 chips.  There are 4 ways to make this choice.  So there are a total of 4 combinations in which this partition will occur.

Case:  $3+1+0+0$
We have to choose a cookie to receive three chips.  (There are 4 ways to make this choice.  Then we must choose a cookie to receive one chip.  (There are 3 ways to make this choice.)  Then we must decide in what order the chips were given to these two cookies.  The choices are:  AAAB, AABA, ABAA, and BAAA.  There are 4 ways to make this choice.  So, there are a total of $4\times 3\times 4=48$ combinations in which this partition will occur.

[We could stop here, and subtract the choices with too many cookies from the total choices.  Or we could continue in order to check our work.]

Case:  $2+2+0+0$
We have to choose two cookies to receive two chips, each.  (There are ${4 \choose 2}=6$ ways to make this choice.  Then we must decide in what order the chips were given to these two cookies.  The choices are:  AABB, ABAB, ABBA, BAAB, BABA, BBAA.  There are ${4 \choose 2}=6$ ways to make this choice.  So, there are a total of $6\times 6=36$ combinations in which this partition will occur.

Case:  $2+1+1+0$
We have to choose one cookie to receive two chips.  (There are 4 ways to make this choice.)  Then we must choose two of the remaining cookies to receive one chip, each.  There are ${3\choose 2}=3$ ways to make this choice.  Then we must decide in what order the chips were given to these three cookies.  We call the cookie receiving two chips "A".  We call one of the cookies receiving one chip "B".  And we call the other cookie receiving one chip "C".  Now, cookie "B", might have received its chip first, second, third, or fourth.  (There are 4 ways to make this choice.)  Then, cookie "C", might have received its chip in any of the remaining positions.  (There are 3 ways to make this choice.)   So, there are a total of $4\times 3\times 4\times 3=144$ ways in which this partition will occur.

Case:  $1+1+1+1$
Every cookie receives one chip.  We just need to decide in what order the cookies received their chips.  There are $4!=24$ ways to make this choice.

Let's summarize this data in a table:

$\begin{array}{lccc}
{\rm Partition}&{\rm Choice\ of\ cookies}&{\rm Order\ of\ chip\ delivery}&{\rm Product} \\
\\
4+0+0+0&\phantom{0}4&\phantom{0}1&\phantom{00}4 \\
3+1+0+0&\phantom{}12&\phantom{0}4&\phantom{0}48 \\
2+2+0+0&\phantom{0}6&\phantom{0}6&\phantom{0}36 \\
2+1+1+0&\phantom{}12&\phantom{}12&\phantom{}144 \\
1+1+1+1&\phantom{0}1&\phantom{}24&\phantom{0}24 \\
\\
\quad\quad {\rm Total}&&&256
\end{array}$

Since each of 4 chips is randomly placed on one of four cookies, we should expect a total of $4^4=256$ choices from the above cases.  Since our work seems to check, all that's left is to tally the choices we want and divide by the total choices available.  $\frac{36+144+24}{256}=\frac{51}{64}$[/hide]

[quote="JBL"]No one has gotten number 1 right yet, although Josh's hint will help put you on the right track.  There's an additional mistake you've made, however:
[hide]You say that there are 2 ways to spell MATCH.  That means that the 2 T's are different from each other.  But then later, you say that there are 10!/2! ways to arrange MATHCOUNTS -- that's only true if the 2 T's are the same.  You have to pick one method over the other.[/hide][/quote]

yeah, that's what I thought, which is why I did [5!*6/2]/[10!/2!], but that becomes 1/5040 and the answer is 1/2520. :o is there a possibility that the answer is wrong...? It's a Mathcounts warmup question.

[quote="rcv"]There are no correct answers to #2 either.  Since I think this is a request for help, rather than a challenge, a methodical solution is contained below.  However, I would like to challenge members to devise a more efficient way of answering this problem.  [size=75][Note:  This would be a very challenging MATHCOUNTS question.  I cannot imagine that this would appear on anything below the level of a National Competition question.  However, the general techniques to solve this are within the skills expected of high-level MATHCOUNTS competitors.][/size]

[size=75][color=darkred][Edit:  Hidden text deleted by rcv][/color][/size][/quote]

thanks a lot! and btw these were from MATHCOUNTS warmups 13 and 18. :/

[quote="mathclass"]yeah, that's what I thought, which is why I did [5!*6/2]/[10!/2!], but that becomes 1/5040 and the answer is 1/2520. :o is there a possibility that the answer is wrong...? It's a Mathcounts warmup question.[/quote]
[hide="a solution"]I would solve this by considering the two T's as distinct.  The question then becomes:

[quote]Using the ten letters in the word MATHCOUNtS, what is the probability that any random arrangement of the letters will have either of the words MATCH or MAtCH in it?[/quote]

Now, I treat MATCH as one unit, and call it "X".  I will treat MAtCH as one unit, and call it "x".

How many ways can I arrange the letters in XOUNtS?  $6!$

How many ways can I arrange the letters in xTOUNS?  $6!$

How many ways can I arrange the letters in MATHCOUNtS?  $10!$

The final answer should be $\frac{6!+6!}{10!}=\frac{2}{10\times 9\times 8\times 7}=\frac{2}{5040}=\frac{1}{2520}$[/hide]

Mathclass, your intial division by 2 is wrong -- you haven't double-counted anything.  (That is, the 2! in the denominator of the denominator is correct if you wish to count the T's as interchangable, but the 2 in the denominator of the numerator doesn't account for anything at all.)

[quote="mathclass"]thanks a lot! and btw these were from MATHCOUNTS warmups 13 and 18. :/[/quote]

Thank you for identifying the source of the cookie problem.  I found the write-up in the School Handbook rather interesting. ...

The write-up in the Handbook begins "This is a difficult but rich problem that can lead to some great discussions!"  One of the solutions is by the MATHCOUNTS Staff.  Their solutions is very similar to the one I wrote above.  There are two additional solutions from National Competition students from Massachusetts.  These would meet the challenge I posed for more efficient solutions.  Any middle school students on AoPS care to give it a whirl?


vector
abstract algebra
linear algebra
matrix
linear algebra unsolved

Suppose $A\in M_n(F)$ and $A^t=-A$ and all elements of main diagnal of $A$ are zero. Prove that $Rank(A)$ is even

We look at $A$ as being an endomorphism of the vector space $V=F^n$ (over $F$). On $V$, consider the obvious symmetric bilinear form $(\cdot,\cdot)$, defined by $(x,y)=\sum_{i=1}^nx_iy^i$, where $x=(x_i)_{i=1}^n,y=(y_i)_{i=1}^n$. As usual, the orthogonal complement $W^\perp$ of a subspace $W\le V$ is defined as the set of those vectors $y$ which satisfy $(x,y)=0,\ \forall x\in W$.

The kernel $\ker A$ of $A$ is, of course, $A$-invariant. This means that its orthogonal complement is $A^t$ invariant. Indeed, let $y\in\ker A^\perp$. We then have $(x,A^ty)=(Ax,y)=0,\ \forall x\in \ker A$, so $A^ty\in\ker A^\perp,\ \forall y\in\ker A^\perp$. Since $A^t=-A,\ \ker A^\perp$ is $A$-invariant. Now, the restriction $\widetilde A=A|_{\ker A^\perp}: \ker A^\perp\to\ker A^\perp$ is a skew-symmetric [i]isomorphism[/i] of $\ker A^\perp$ onto itself, and now the problem reduces to this: given an invertible antisymmetric matrix $\widetilde A\in\mathcal M_k(F)$ (where $k$ is the dimension of $\ker A^\perp$), prove that $k$ is even. This follows at once from $0\ne\det \widetilde A=\det \widetilde A^t=\det(-\widetilde A)=(-1)^k\det \widetilde A$.


inequalities
function
inequalities proposed

Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. Prove that
$ \left( \frac{2a\plus{}b}{a\plus{}b\plus{}c}\right)^2 \plus{}\left( \frac{2b\plus{}c}{b\plus{}c\plus{}d}\right)^2\plus{}\left( \frac{2c\plus{}d}{c\plus{}d\plus{}a}\right)^2 \plus{}\left( \frac{2d\plus{}a}{d\plus{}a\plus{}b}\right)^2 \ge 4.$

I have proved now that: All values of $ k$ such that the above inequality holds is $ k \in (\minus{} \infty, \;0] \cup \left[\frac{1}{2}, \; 2\right].$

I wonder if anyone is interested in this problem?

ok here goes but i am not sure -
Assume $ d \ge c\ge b \ge a$ ,
Write the LHS as -
$ (1 \minus{} \frac {c \minus{} a}{a \plus{} b \plus{} c})^2 \plus{} (1 \minus{} \frac {a \minus{} c}{a \plus{} c \plus{} d})^2 \plus{} (1 \minus{} \frac {b \minus{} d}{b \plus{} c \plus{} d})^2 \plus{} (1 \minus{} \frac {d \minus{} b}{a \plus{} d \plus{} b})^2 \ge 4$

$ \Longrightarrow 4 \plus{} (\frac {c \minus{} a}{a \plus{} b \plus{} c})^2 \plus{} (\frac {c \minus{} a}{a \plus{} c \plus{} d})^2 \plus{} (\frac {d \minus{} b}{b \plus{} c \plus{} d})^2 \plus{}
(\frac {d \minus{} b}{a \plus{} d \plus{} b})^2 \ge \frac {2(c \minus{} a)(d \minus{} b)}{(a \plus{} c \plus{} d)(a \plus{} b \plus{} c)} \plus{} \frac {2(c \minus{} a)(d \minus{} b)}{(a \plus{} d \plus{} b)(b \plus{} c \plus{} d)} \plus{} 4$
applying AM-GM-
$ \frac {2(c \minus{} a)(d \minus{} b)}{(a \plus{} b \plus{} c)(a \plus{} c \plus{} d)(a \plus{} b \plus{} d)(b \plus{} c \plus{} d)} \cdot [ (a \plus{} c \plus{} d)(b \plus{} c \plus{} d) \plus{} (a \plus{} b \plus{} c)(a \plus{} b \plus{} d)]$ $ \ge \frac {2(c \minus{} a)(d \minus{} b)}{(a \plus{} b \plus{} c)(a \plus{} c \plus{} d)(a \plus{} b \plus{} d)(b \plus{} c \plus{} d)}\cdot [(a \plus{} b \plus{} c)(a \plus{} c \plus{} d) \plus{} (b \plus{} c \plus{} d)(a \plus{} d \plus{} b)]$
which re-arranges to -
$ (a \plus{} c \plus{} d)(d \minus{} a) \ge (a \plus{} b \plus{} d)(d \minus{} a)$
by assumption $ d \ge b$ and $ c \ge b$
so done..
is it correct?

[quote="Tourish"][quote="geniusbliss"][quote="can_hang2007"]Before checking your proof, I would like to ask you: Why can we assume that $ d \ge c \ge b \ge a$?[/quote]
 i hope u clarify this doubt - http://www.mathlinks.ro/viewtopic.php?p=1737570#1737570
.. i guess it was fundamental mistake in my notions.. so sry abt tht post just ignore it.. sry again cuz i cant delete my post.[/quote]
For $ n$ variables,we can assume $ a_1\geq a_2\geq ...\geq a_n$ only in the case $ F(a_1,a_2,...,a_n) \equal{} F(a_{i_1},a_{i_2},...,a_{i_n})$,where $ (i_1,i_2,...,i_n)$ is the permutation of $ (1,2,...,n)$.But if the function $ f$ only satisfy $ F(a_1,a_2,...,a_n) \equal{} F(a_2,a_3,...,a_n,a_1) \equal{} F(a_3,a_4,...,a_1,a_2) \equal{} ...$,we can only assume that $ a_k \equal{} max{\{a_1,a_2,...,a_n\}},k\in{1,2,...,n}$.I hope my comprehension aren't wrong.[/quote]
hey i am very sorry if i am acting like stupid... but you have only written my doubt mathematically ... my question was why???
why can we assume whatever we have assumed in both the cases .. someone please help me...

I am still waiting for a nice solution, especially a Cauchy-Schwarz solution.

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. Prove that
$ \left( \frac {2a \plus{} b}{a \plus{} b \plus{} c}\right)^2 \plus{} \left( \frac {2b \plus{} c}{b \plus{} c \plus{} d}\right)^2 \plus{} \left( \frac {2c \plus{} d}{c \plus{} d \plus{} a}\right)^2 \plus{} \left( \frac {2d \plus{} a}{d \plus{} a \plus{} b}\right)^2 \ge 4.$[/quote]

I wonder $ \frac {2a \plus{} b}{a \plus{} b \plus{} c}\plus{}  \frac {2b \plus{} c}{b \plus{} c \plus{} d}\plus{}  \frac {2c \plus{} d}{c \plus{} d \plus{} a} \plus{}  \frac {2d \plus{} a}{d \plus{} a \plus{} b} \ge 4$    :?:

[quote="xyy"]I wonder $ \frac {2a \plus{} b}{a \plus{} b \plus{} c} \plus{} \frac {2b \plus{} c}{b \plus{} c \plus{} d} \plus{} \frac {2c \plus{} d}{c \plus{} d \plus{} a} \plus{} \frac {2d \plus{} a}{d \plus{} a \plus{} b} \ge 4$    :?:[/quote]
Try $ a\equal{}1,$ $ b\equal{}2$ and $ c\equal{}d\equal{}0.$

[quote="can_hang2007"][quote="xyy"]I wonder $ \frac {2a \plus{} b}{a \plus{} b \plus{} c} \plus{} \frac {2b \plus{} c}{b \plus{} c \plus{} d} \plus{} \frac {2c \plus{} d}{c \plus{} d \plus{} a} \plus{} \frac {2d \plus{} a}{d \plus{} a \plus{} b} \ge 4$    :?:[/quote]
Try $ a \equal{} 1,$ $ b \equal{} 2$ and $ c \equal{} d \equal{} 0.$[/quote]
But $ a,b,c,d>0$ because you wrote:
[quote="can_hang2007"]
Let $ a, b, c, d$ be [b]positive[/b] real numbers
[/quote]

[quote="xyy"][quote="can_hang2007"][quote="xyy"]I wonder $ \frac {2a \plus{} b}{a \plus{} b \plus{} c} \plus{} \frac {2b \plus{} c}{b \plus{} c \plus{} d} \plus{} \frac {2c \plus{} d}{c \plus{} d \plus{} a} \plus{} \frac {2d \plus{} a}{d \plus{} a \plus{} b} \ge 4$    :?:[/quote]
Try $ a \equal{} 1,$ $ b \equal{} 2$ and $ c \equal{} d \equal{} 0.$[/quote]
But $ a,b,c,d > 0$ because you wrote:
[quote="can_hang2007"]
Let $ a, b, c, d$ be [b]positive[/b] real numbers
[/quote][/quote]
$ a\equal{}1, b\equal{}2, c \rightarrow 0, d\rightarrow 0.$

[quote="xyy"][quote]
$ a \equal{} 1, b \equal{} 2, c \rightarrow 0, d\rightarrow 0.$[/quote]

it means? i don't really understand this.[/quote]

$ c\rightarrow 0$ means that $ c$ gets infinitely close to $ 0$, for example you can try $ c\equal{}d\equal{}10^{\minus{}2010}$

[quote="Pain rinnegan"][quote="xyy"][quote]
$ a \equal{} 1, b \equal{} 2, c \rightarrow 0, d\rightarrow 0.$[/quote]

it means? i don't really understand this.[/quote]

$ c\rightarrow 0$ means that $ c$ gets infinitely close to $ 0$, for example you can try $ c \equal{} d \equal{} 10^{ \minus{} 2010}$[/quote]
 thank a lot

If i am able to prove this then it will solve the sum i think -
$ \sum_{cyc} \frac{3a^2\plus{}2ab\plus{}c}{(a\plus{}b\plus{}c)^2} \ge 2\sum_{cyc} \frac{c}{a\plus{}b\plus{}c}$
But i am not able to prove that.

[quote="can_hang2007"]Actually, we can solve the more general problem:

Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. Determine all values of $ k\ge 0$ such that the following inequality holds
$ \left(\frac {a \plus{} kb}{a \plus{} b \plus{} c}\right)^2 \plus{} \left(\frac {b \plus{} kc}{b \plus{} c \plus{} d}\right)^2 \plus{} \left(\frac {c \plus{} kd}{c \plus{} d \plus{} a}\right)^2 \plus{} \left(\frac {d \plus{} ka}{d \plus{} a \plus{} b}\right)^2\ge \frac {4}{9}(k \plus{} 1)^2.$

The answer is $ k \in \{ 0\} \cup \left[ \frac {1}{2},\; 2\right].$[/quote]
See here for a super proof: http://can-hang2007.blogspot.com/2010/02/inequality-117-v-q-b-can.html


Hi,

i recently came across this problem. Its a Nim game. But this is misere play approach.(player who takes the lest stone(s) loses)

Here is the game description!

There are four stone piles with n1, n2, n3, n4 stones. One is permitted to take any number of stones from one pile for a move. The one who takes the last stone(s) LOOSES.

Ex:
 1 0 0 0 ---------- First Wins.
 3 2 1 0 ---------- Second wins.

Question is how can we determine who will win given these four number?
If it is normal play(the player who takes the last stone(s) WINS), the method is very simple!! jus take the XOR of the all the numbers and see if it is equal to zero. if it is then second wins, else first wins.

But in misere play i am stuck with this approach :( . but the method for misere play is also similar.
Please if anyone knows, help me.

Please, post your future questions just once. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=859291#859291[/url]


Given $ A(\minus{}6,\minus{}1)$ and $ B(9,4)$, find the coordinates of point $ C$ so that $ C$ is between $ A$ and $ B$ and $ \frac{AC}{BC}\equal{}\frac{3}{2}$.

I'm not sure what "between" means, so I'll find every point C without the "between" restriction.

[hide="solution"]
Let $ (x,y)$ be the coordinates of point $ C$

$ |AC| \equal{} \sqrt {(x \plus{} 6)^2 \plus{} (y \plus{} 1)^2}$

$ |BC| \equal{} \sqrt {(x \minus{} 9)^2 \plus{} (y \minus{} 4)^2}$

$ \frac {|AC|}{|BC|} \equal{} \frac {\sqrt {(x \plus{} 6)^2 \plus{} (y \plus{} 1)^2}}{\sqrt {(x \minus{} 9)^2 \plus{} (y \minus{} 4)^2}} \equal{} \frac {3}{2}$

$ 4(x \plus{} 6)^2 \plus{} 4(y \plus{} 1)^2 \equal{} 9(x \minus{} 9)^2 \plus{} 9(y \minus{} 4)^2$

after some "heavy work" we get:

$ \minus{} (x \minus{} 3)(x \minus{} 39) \equal{} (y \minus{} 2)(y \minus{} 14)$

$ \minus{} x^2 \plus{} 42x \minus{} 117 \equal{} y^2 \minus{} 16y \plus{} 28$

$ \minus{} (x^2 \minus{} 42x \plus{} 441) \plus{} 324 \equal{} y^2 \minus{} 16y \plus{} 64 \minus{} 36$

$ \minus{} (x \minus{} 21)^2 \plus{} 360 \equal{} (y \minus{} 8)^2$

$ (x \minus{} 21)^2 \plus{} (y \minus{} 8)^2 \equal{} 360$

so point $ C$ lies on a circle with coordinates of the centre $ (21,8)$ and $ r \equal{} 6\sqrt {10}$[/hide]

I assumed that by between it meant that the $ x$- and $ y$-coordinate of $ C$ lie between the $ x$- and $ y$-coordinates of $ A$ and $ B$.

Also, isn't the equation: $ 4(x\plus{}6)^2\plus{}4(y\plus{}1)^2\equal{}9(x\minus{}9)^2\plus{}9(y\minus{}4)^2$?

Yes, you are right, I missed it. I'll correct it.

[quote="sunehra"]I assumed that by between it meant that the $ x$- and $ y$-coordinate of $ C$ lie between the $ x$- and $ y$-coordinates of $ A$ and $ B$.

[/quote]
This is still ambiguous.  Suppose the coordinates of point A
are (0, 12), and the coordinates of point B are (6, -12).

Look at an x-coordinate that lies between the x-coordinates of
points A and B.  For instance, pick x = 5.

Look at a y-coordinate that lies between the y-coordinates of
points A and B.  For instance, pick y = -11.

Then this point is (5, -11), but it is not on the line segment AB.

[quote="sunehra"]Given $ A( \minus{} 6, \minus{} 1)$ and $ B(9,4)$, with point $ C$ on line $ AB$, find the coordinates of point $ C$ so that $ C$ is between $ A$ and $ B$ and $ \frac {AC}{BC} \equal{} \frac {3}{2}$.[/quote]
Try the above for the original statement.


Let $n$ be a five digit number (whose first digit is non-zero) and let $m$ be the four digit number formed from $n$ by removing its middle digit. Determine all $n$ such that $n/m$ is an integer.

[hide]

We know that $m$ has the same first two digits as $n$.  This means that if their ratio is an integer, it must be $\geq 10$.  This is true because $10m$ obviously has the same first two digits as $n$, and the subtraction of $m$ from $10m$ will change the first two digits since $m\geq 1000$.  Exactly the same argument shows that the ratio must be $\leq 10$.  So the ratio is equal to 10.

Say that $n=abcde$, with each letter representing a digit.  Them $m=abde$, and $10m=abde0$.  From this we get $e=0$, $d=e=0$, and $c=d=e=0$.  So $n$ must be a multiple of 1000. To show that all multiples of 1000 work, we simply note that removing the middle digit from $ab000$ gives $ab00$, and $10(ab00)=ab000$.

  [/hide]

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1971Problem7

Vo Duc Dien

Let n=10000a+1000b+100c+10d+e and let m=1000a+100b+10d+e.
Then if m|n, obviously $m|(n-10m)\Leftrightarrow{m|(100c-90d-9e)}$.

[b]Case 1:[/b]  The quotient k of n/m is 0.
In this case, $100c-90d-9e=0\Leftrightarrow{100c=9(10d+e)}$, which implies that $9|c$ (since $9\nmid{100}$), or c=0,9.
[i]Case 1.1:[/i] c=0.
Then $90d+9e=0\Leftrightarrow{10d+e=0}\Leftrightarrow{d=e=0}$ since $0\le{d,}e\le{9}$.
Thus, c=d=e=0 is a solution, i.e. all n of the form $n=\overline{ab000}$ work.
[i]Case 1.2:[/i] c=9.
Then $90d+9e=900\Leftrightarrow{10d+e=100}$, which is a contradiction since $d,e\le{9}\implies{10d+e\le{99}}<100$.

[b]Case 2:[/b] $k\ge{1}$.
Multiplying by the denominator, moving things to one side and factoring we get
100c=(1000k)a+(100k)b+(10k+90)d+(k+9)e.
But $c\le{9}\implies{LHS\le{900}}$ whereas $RHS\ge{1000}$ (this occurs when a,k are both minimal at 1), whence there are no solutions for this case.

To sum up, the only solution is of the form $\boxed{n=\overline{ab000}}$, i.e. m|n if n is a multiple of 1000.


linear algebra
matrix
trigonometry
geometry
geometric transformation
rotation
rotation matrix

yeah you are right, i did a stupid thing. thanks for pointing out the mistake

[quote="kouboy"]Its well known that $A$ and $A^{T}$ have the same eigenvalues. let $k$be an eigenvalue of $A$.now $A^{T}=A^{2}$ implies $k^{2}=k$. .[/quote]
We just know $A$ and $A^{T}$ have the same eigenvalues. But not have the same eigenvectors.So I can't know how do you get $k^{2}=k$?

[quote="who"]let A  be n*n matrix s.t.
 A traspose=A*A
  find all eigenvalues of  A.[/quote]$A^{T}= A^{2}= A \cdot A$ implies $A^{T}= (A^{T}\cdot A^{T})\cdot A = \left(A^{T}\right)^{4}$ and hence $A = A^{4}$. Thus all eigenvalues of $A$ satisfy $\lambda = \lambda^{4}$. Solutions: $\lambda = 0, \, \lambda = 1, \, \lambda = \cos 2 \pi / 3 \pm i \sin 2 \pi /3$. The example of a 2-d rotation matrix by an angle of $2 \pi / 3$ shows that the two latter cases can indeed occur.


modular arithmetic
number theory
least common multiple

[code]What are the last two digits of 3^1234[/code]

You should probably clarify where your premise comes from. $ 3^{20} \pmod{100}$ is 1 by Carmichael's theorem, as shown:

$ \lambda (100) \equal{} \lambda(2^2 5^2) \equal{} \text{LCM} [\lambda (2^2), \; \lambda (5^2)] \equal{}$ $ \text{LCM} [2^{2\minus{}2},(5\minus{}1)5^{2\minus{}1}]\equal{} \text{LCM}[1,20]\equal{}20$
Hence for all $ n$ coprime to $ 100$, $ n^{\lambda (100)} \equal{} n^{20} \equiv 1 \pmod{100}$

He then uses brute force to find the last part.

Just out of curiosity:

Is there an extension if $ n$ and $ m$ are not coprime?

[quote="The Zuton Force"]You should probably clarify where your premise comes from. $ 3^{20} \pmod{100}$ is 1 by Carmichael's theorem, as shown:

$ \lambda (100) \equal{} \lambda(2^2 5^2) \equal{} \text{LCM} [\lambda (2^2), \; \lambda (5^2)] \equal{}$ $ \text{LCM} [2^{2 \minus{} 2},(5 \minus{} 1)5^{2 \minus{} 1}] \equal{} \text{LCM}[1,20] \equal{} 20$
Hence for all $ n$ coprime to $ 100$, $ n^{\lambda (100)} \equal{} n^{20} \equiv 1 \pmod{100}$

He then uses brute force to find the last part.[/quote]

Could you give a link to the definition of Carmichael's theorem?

[quote="1=2"]Could you give a link to the definition of Carmichael's theorem?[/quote]

Would using Google have been so hard?

http://mathworld.wolfram.com/CarmichaelsTheorem.html
http://mathworld.wolfram.com/CarmichaelFunction.html

[quote="Fraenkel"]Is there an extension if $ n$ and $ m$ are not coprime?[/quote]

Not one that is easy to state.  There is a [i]method:[/i] write the modulus $ m \equal{} ab$ where $ gcd(n, b) \equal{} 1$, calculate the remainder $ \bmod b$ by Carmichael and the remainder $ \bmod a$ by inspection, and use the Chinese Remainder Theorem.


Suppose $ a_{1}\equal{}1$. Assume that there isn't such $ k$. First we prove that all $ S_{i}\equal{}a_{1}\plus{}...\plus{}ia_{i}$ have the same sign. If not, we would have an $ i$ so that $ S_{i}\ge [\frac{i}{2}]\plus{}1$ and $ S_{i\plus{}1}\le\minus{}[\frac{i\plus{}1}{2}]\minus{}1$ but for this we must have $ [\frac{i}{2}]\plus{}1\minus{}(i\plus{}1)\le\minus{}[\frac{i\plus{}1}{2}]\minus{}1\rightarrow i\le i\minus{}1$ false.So all $ S_{i}>0$ ($ S_{1}>0$). Let $ t$ be minimal so that $ a_{1}\plus{}...\plus{}a_{t}\equal{}0$, so $ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{i}>0$ for all $ i<t$.
So we have:
$ a_{1}>0\rightarrow a_{2}\plus{}a_{3}\plus{}a_{4}\plus{}...\plus{}a_{t}<0$
$ a_{1}\plus{}a_{2}>0\rightarrow a_{3}\plus{}a_{4}\plus{}...\plus{}a_{t}<0$
$ a_{1}\plus{}a_{2}\plus{}a_{3}>0\rightarrow a_{4}\plus{}...\plus{}a_{t}<0$
$ \vdots$
$ a_{1}\plus{}...\plus{}a_{t}\equal{}0$
Summing these up we get $ S_{t}<0$ false.


analytic geometry
geometry
geometric transformation
number theory solved
number theory

Let $n$ points with integer coordinates be given in the $xy$-plane. What is the minimum value of $n$ which will ensure that three of the points are the vertices of a triangel with integer (possibly, 0) area?

Hmm...5?

Wlog, we may assume (use a translation) that one of the points is $O(0,0)$.
If $A(a,b)$ ad $B(x,y)$ are two of the points, the area of $OAB$ is $\dfrac{|ay-bx|} 2.$
Thus, if one of the points other than $O$, say $A$, has its two coordinates which are even numbers then with $O$ and any of the other points, they form a triangle with integer area.
If two of the points, say $A$ and $B$, have all their coordinates odd, the same is true for $OAB.$
The same is true for two points with coordinates of the form $(o,e)$ or two of the points with coordiates of the form $(e,o)$, where $e$ is for even and $o$ is for odd.

It follows that if a set of integer points does not contain three points which are the vertices of a triangle with integer area, then this set has at most $4$ elements.

Conversely, considering the four vertices of an unit square shows that such a set with $4$ elements does exist.

Pierre.


What is the probability that an integer in the set $ \{1,2,3,\ldots,100\}$
 is divisible by 2 and not divisible by 3? Express your answer as a common fraction.

Out of the 100 integers in the set, 50 are even, and 16 of those 50 are divisible by both 2 and 3.
So 34 integers are divisible by 2 but not 3.
34/100=17/50

Here's a solution that starts with the initial set of $\{1,2,3,\ldots,99,100\}$ and shows how you reduce it step by step by manipulating the members of the set as a sequence.

[hide="The 'weed-out' method"]We have 100 integers in the set $\{1,2,3,\ldots,99,100\}$. First, we only want the ones that are divisible by $2$, so we can kick every odd number out of the set.

Now we have the set $\{2,4,6,\ldots,98,100\}$. If we divide each member of this set by $2$, we get $\{1,2,3,\ldots,49,50\}$, and can easily see that this set has $50$ members.

Next we want to take our set of even numbers and throw out the members that are also divisible by $3$. Any even number that is divisible by $3$ is also a multiple of $6$. The multiples of $6$ in this set are $\{6,12,18,\ldots,90,96\}$. If we divide each member of this set by $6$, we get $\{1,2,3,\ldots,15,16\}$, and can easily see that this set has $16$ members.

We are now left with the set $\{2,4,8,10,14,16,\ldots,92,94,98,100\}$, which must have $50-16=34$ members. Therefore, the probability that an integer in the set  $\{1,2,3,\ldots,99,100\}$ is divisible by $2$ but not by $3$ is $\frac{34}{100}=\boxed{\frac{17}{50}=0.34}$.[/hide]

[hide="Why the answer is not 1/3, and under what circumstances 1/3 is the correct answer"]When I first set out to solve this problem, my answer was incorrect, because I didn't stop to consider that I was working with a finite set of numbers. It should be obvious that over the entire set of positive integers, the probability that any given integer is even is $\frac{1}{2}$. Of those even numbers, $\frac{1}{3}$ of them will be multiples of $6$ and therefore excluded from our answer. This means that over the entire set of positive integers, the probability that an integer is divisible by $2$ but not by $3$ is $\frac{1}{2} \times \frac{2}{3}= \frac{1}{3}$.

You see, for any sequence of $6$ consecutive integers, $3$ of them will be odd, $3$ will be even, and only $1$ will be a multiple of $6$ (divisible by both $2$ and $3$). For example, consider the sequence $\{9,10,11,12,13,14\}$. We can throw out the odd numbers $9,11,$ and $13$, leaving us with $\{10,12,14\}$. When we kick out the lone multiple of $6$, which in this case is $\{12\}$, we are left with $\{10,14\}$, which is only $\frac{1}{3}$ the size of our original set. Therefore, so long as you are dealing with an infinite set, or a set whose total population is a multiple of $6$, the correct probability is $\frac{1}{3}$. It is only when the number of elements in the set is not a multiple of $6$, for example, $100$, $500$, $1000$, etc. that the probability deviates from $\frac{1}{3}$.[/hide]

Tim

So I wish that it counted as right, because, I accidentally wrote 34 instead of 34/100, then to 17/50, which is the correct answer.

I think I know how to solve this, but instead, I made a program to do this. [code=javascript]for(var i = 1; i < 101; i++){
    if(i / 2 === round(i / 2) && i / 3 !== round(i / 3)){
        println(i);
    }
}[/code]

[hide=My solution]There are 50 integers divisible by two, and 16 even multiples of three. 50 - 16 = 34. So $\dfrac{34}{100}$ = $\boxed{\dfrac{17}{50}}$.[/hide]

haha waster my first attempt by writing 34 :cry:


Hello there,

I am having trouble understanding why this equation has two solutions. My work is shown below.

Thanks!

---

1. Solve the equation $ \sin 2x \plus{} 2 \sin x \equal{} 0$ and provide all solutions on the interval $ [0, 2\pi)$.

$ \sin 2x \plus{} 2 \sin x \equal{} 0$

$ 2\sin x \cos x \plus{} 2 \sin x \equal{} 0$

$ 2\sin x (\cos x \plus{} 1) \equal{} 0$

Therefore:

$ \sin x \equal{} 0$ and $ \cos x \equal{} \minus{} 1$

Since (-1, 0) occurs on 180° on the unit circle, $ x \equal{} \pi$.

However, the second answer given is $ x \equal{} 0$, could anyone please explain this?

is this homework?
Firstly what is the range of the sine fn?  :D

How did you take sinx common?  :huh:

Hello madness,

No it is not. I am reading ahead and I am attempting to learn this material on my own. I have also added a line in my original post to clarify how I factored the equation.

In any case, I figured it out. Since the graph of the sine function shows that $ 0 \equal{} \sin \pi \equal{} \sin 2\pi$, $ x \equal{} 0$ would be a solution as well.

But how did you take sinx out common? Or have you made a mistake in the question?  :maybe:

I apologise. Yes I did, I typed in \sinx so it didn't show up; it had to be \sin x.

Thanks for letting me know.

[quote="catalan"]$ \sin x \equal{} 0$ and $ \cos x \equal{} \minus{} 1$[/quote]
Isn't it $ \sin x \equal{} 0$ [b]or[/b] $ \cos x \equal{} \minus{} 1$?

Even though in this case a certain $ x$ makes the "and" statement true it should be "or" as i_like_pie said.
[hide]
$ x \equal{} \arcsin 0$ so, for this interval, $ x\equal{}0 , \pi$.
$ x \equal{} \arccos \minus{}1$ so $ x \equal{} \pi$.
[/hide]

Why should it be "or"? Since the question asks for all the values in the range $ [0, 2\pi)$, wouldn't both values be the answer?

a statement "p or q" means at least one of the statements has to be true, while "p and q" means that both statements must be true

You misunderstand our use of the word "or."  The complete statement is

"If $ x$ is a solution to this equation, then either $ \sin x \equal{} 0$ [b]or[/b] $ \cos x \equal{} \minus{}1$."

Thus,

"The set of all $ x$ in $ [0, 2\pi)$ satisfying this equation is $ x \equal{} 0$ [b]and[/b] $ x \equal{} \pi$."

i presume or means union here i.e. the mathematical or  :)


calculus
integration
limit
logarithms
real analysis
real analysis solved

Evaluate

\[\lim_{n\to\infty}n\int^e_1x^2(\ln x)^n dx (n=1,2,\cdot\cdot\cdot)\]

Let $J_n=n\int_1^ex^2(\ln x)^ndx$.
$J_n=\frac n3\int_1^e(x^3)'(\ln x)^ndx=\frac n3 e^3-\frac{n^2}3\int_1^ex^2(\ln x)^{n-1}=\frac n3 e^3-\frac{n^2}{3(n-1)}J_{n-1}$
And now if we denote $I_n=\frac {J_n}n$ we get that $I_n=\frac {e^3}3-\frac n3I_{n-1}$ (1) . Denote by $L=\lim_{n\to\infty}I_n\ge 0$.
Now (1)  becomes when $n\to\infty$, $L=\frac{e^3}3-\frac L3\lim_{n\to\infty}n$ (2).
If $L>0$ results from (2) that $L=-\infty$ which is absurd. So $L=0$ and now we have from (1) that $0=\frac {e^3}3-\frac 13\lim_{n\to\infty}nI_{n-1}\Leftrightarrow \lim_{n\to\infty}nI_{n-1}=\lim_{n\to\infty}J_{n-1}=e^3$.
Hope it's correct.


geometry
parallelogram
rectangle
geometry proposed

Let $ ABCD$ be a convex cuadrilateral inscribed in a circumference centered at $ O$ such that $ AC$ is a diameter. Pararellograms $ DAOE$ and $ BCOF$ are constructed. Show that if $ E$ and $ F$ lie on the circumference then $ ABCD$ is a rectangle.

If so, then $ BC\equal{}AD\equal{}R$ $ \Rightarrow$ $ \angle{BAC}\equal{}\angle{ACD}$ then $ ABCD$ is a parallelogram. $ \angle{B}\equal{}\angle{D}\equal{}\frac{\pi}{2}$$ \Rightarrow$ $ ABCD$ is a rectangle.

I forgot to say this problem is due to Aarón Ramírez (El Salvador)...  :D You can ask me if you are interested in problems related to the Central American competition (shortlists, proposed problems and the like).  :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

$ |H| \equal{} p$ because $ H\neq 0$ and $ p^i$ do not divide $ |S_p|$ if $ i > 1$.

[b]feniks[/b], who told you that $ \langle a, b \rangle$ is a $ p$-group?  :)
Believe me, this problem is much harder than you think.

[quote="Jan"]Let $ G \equal{} S_p$, where $ p$ is prime. Let $ a,b \in G$ be two cycles of length $ p$. Show that the group $ H \equal{} \langle a, b \rangle$ is simple.[/quote]

I have an hypothesis: if $ b \notin \langle a \rangle$ then $ H \equal{} \langle a, b \rangle$ is the alternating group of degree $ p$.

For example in $ S_5$, $ \langle (12345), (15423) \rangle \equal{} A_5$ because $ (12345)(15423) \equal{} (243)$ and $ A_5 \equal{} \langle (243), (12345) \rangle$.

Anyway I think it isn't easy to prove that...

To be honest, I'm having trouble even showing that [i]all[/i] the p-cycles generate $ A_p$.

Here's a proof for $ p\equal{}5$, at least: our group $ \langle a,b\rangle$ has a number of Sylow 5-subgroups dividing $ (5\minus{}2)!\equal{}6$ and congruent to $ 1\ (\rm{mod}\ 5)$, so it must contain all the 5-cycles, giving us 24 elements of order 5.  So our group has order 30 or 60, but in the former case, it would be a normal subgroup of $ A_5$.

Do we think p must be prime?

Yeah I got that proof for $ p\equal{}5$ as well. I'm wondering if we can prove the case $ p\equal{}7$ then we might be able to shed some more light on the general case. The sylow analysis is not as easy.

I admit this problem is really hard. It's from an UGent exam from $ 2007$.

Here's a hint: Assume that there exists a normal divisor $ N \trianglelefteq H$. Now investigate both cases where $ p$ divides $ \left|N\right|$ and where it doesn't.

First assume $ p\mid\mid{N}\mid.$ Then we must have $ a,b\in{N}$ since p-Sylows are conjugate and $ N$ is closed under conjugation.. But  $ H \equal{} < a,b >$ thus $ N \equal{} H$.

I'm still working on the other case but for now I have to go.

Edit: I changed the proof above because I found an error, but this proof looks good. However, the below implication still troubles me.

Ok now I'm a little skeptical of my above proof of the first case for the following reason.
Consider $ C_H(a)\equal{}\{{h\in{H}\mid{hah^{\minus{}1}\equal{}a}}\}$ Note that $ C_H(a)\lhd{H}$ and that $ p$ divides $ \mid{C_H(a)}\mid$
since $ a$ has order $ p$. Then using the same proof above we get that $ C_H(a) \equal{} H$.
But this is clearly ridiculous since this would imply all p-cycles commute.

Your proof of the first case looks fine.

On your second post, why is $ C_G(a) \trianglelefteq H$? Of course $ C_G(a) \trianglelefteq \langle a \rangle$, maybe that's what you meant?

Oooops you are right. It is not necessarily normal. I got a little too messy in my scratch work lol. Thhe other case is admitedly harder and I have no solution yet.


Hola, aquí están los problemas de la olimpíada nacional que estamos celebrando por estos días. En realidad hay algunas curiosidades en cuanto a nuestro sistema :D Primero, la participación es abierta, los problemas se publican en el periódico y quien quiera participar manda por correo las soluciones de los problemas correspondientes a su grado escolar... Además (esto es lo más triste) no hay medallas!! :mad: Así que, si lo quieren ver de esta forma, este es un "examen de admisión" para nuestra Academia de Matemática. En todo caso siempre se insiste en llamar al examen olimpíada :)

Estos son los problemas de décimo grado. Por una u otra razón no hay prueba de undécimo grado, por eso me quedé sin participar jejeje :) No he puesto los problemas de grados más bajos porque creo que el nivel de dificultad no es muy significativo que digamos.

En fin, aquí están los enunciados.

[b]Problema 1.[/b] Dado el cuadrado $ABCD$, encontrar todos los puntos $P$ interiores al cuadrado tales que el área del cuadrilátero $ABCP$ es igual al triple del área del cuadrilátero $APCD$.

[b]Problema 2.[/b] Encontrar todos los posibles pares de enteros positivos $(x,\, y)$ tales que ni $x$ ni $y$ tienen un factor primo mayor que 5, y que además satisfacen $x^2-y^2=2^{100}$.

[b]Problema 3.[/b] En el plato derecho de una balanza hay una bolsa que pesa 11111 gramos. Una persona coloca pesas en cualquier plato de la balanza; la primera pesa es de un gramo, y cada pesa colocada tiene el doble de peso que la anterior; es decir, las pesas son sucesivamente de 1, 2, 4, 8,$\ldots$ gramos. En algún momento los platos de la balanza se equilibran. Determinar, justificando la respuesta, en qué plato está la pesa de 16 gramos, el derecho o el izquierdo.

[b]Problema 4.[/b] ¿Cuántos números enteros entre 1 y 1000 inclusive pueden escribirse como la suma de un múltiplo positivo de 7 y un múltiplo positivo de 4?

[b]Problema 5.[/b] Digamos que "concatenamos" dos números cuando escribimos uno a continuación del otro; por ejemplo, al concatenar 200 y 6 obtenemos 2006 y al concatener 6 y 200 obtenemos 6200. Encontrar dos números de 6 dígitos tales que el número que resulta al concatenarlos sea divisible por su producto.

[hide="MI Solución al prob. 1"]
(Denotaremos $[A_1A_2...A_n]$ al área del polígono $A_1A_2...A_n$. )
Sea $[ABCD]=4S$ ; claramente $P$ debe estar en él interior del triángulo $ACD$ , pues $[ABCP]=3S > 2S=[ABC]$.
Notemos también que $[APC]=S=\frac{1}{2}[ACD]$, luego, si llamamos $h$ a  la distancia de $D$ a $AC$, la distancia de $P$ a $AC$ será $\frac{h}{2}$ (esto debido a que si dos triángulos tienen un lado en común la relación de sus áreas será igual a la relación entre sus alturas), con esto concluimos que todos los puntos $P$ que consideraremos distan $\frac{h}{2}$ de $AC$, luego $P$ está en la recta paralela a $AC$ que dista $\frac{h}{2}$, es claro notar, también ,que dicha recta pasa por los puntos medios $M$ y $N$ de $AD$ y $CD$, respectivamente, es decir $P$ está en $MN$.
Recíprocamente cualquier punto $P'$ e $MN$ cumple $[AP'C]=\frac{1}{2}[ACD]=S$ luego $[ABCP']=[ABC]+[AP'C]=3S=3[AP'CD]$.
Por lo tanto todos los punto $P$ que buscamos están en el segmento (abierto) de extremos $M$ y $N$, donde $M$ y $N$ son puntos medios de $AD$ y $CD$ respectivamente.
  [/hide]

[quote="José"][quote="Jutaro"][b]Problema 2.[/b] Encontrar todos los posibles pares de enteros positivos $(x,\, y)$ tales que ni $x$ ni $y$ tienen un factor primo mayor que 5, y que además satisfacen $x^2-y^2=2^{100}$.[/quote]

Alguien puede postear la solución??? Gracias.[/quote]

Puedes encontrar una solución ( en inglés ) en :
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=75767&highlight=]www.mathlinks.ro/Forum/viewtopic.php?t=75767&highlight=[/url]

$Tipe$


[quote="cezar lupu"]Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3abc.$ Prove the following inequality:

                                           \[ \frac{a}{b^{2}c^{2}}+\frac{b}{c^{2}a^{2}}+\frac{c}{a^{2}b^{2}}\geq\frac{9}{a+b+c}  \][/quote]
$3=\frac{a^2+b^2+c^2}{abc}$ then $9=(\frac{a^2+b^2+c^2}{abc})^2$
$\frac{a}{b^{2}c^{2}}+\frac{b}{c^{2}a^{2}}+\frac{c}{a^{2}b^{2}}\geq \frac{9}{a+b+c}$ is the same as proving
$(abc)^2(a^3+b^3+c^3)(a+b+c) \geq 9(abc)^4$ then plugging 9 in $(a^3+b^3+c^3)(a+b+c) \geq \frac{(a^2+b^2+c^2)^2}{(abc)^2}(abc)^2$ which becomes $(a^3+b^3+c^3)(a+b+c) \geq (a^2+b^2+c^2)^2$ and it's true by Cauchy-Schwarz inequality.

[quote="sandumarin"]We have:
$ a,b,c>0,a^2+b^2+c^2=3abc\Leftrightarrow a,b,c>0,\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=3 $
$ \sum\frac{a}{b^2c^2}=\sum\frac{(\frac{a}{bc})^2}{a}\ge\frac{(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab})^2}{a+b+c} $
$ =\frac{9}{a+b+c}\Leftrightarrow QED $
_________
Sandu Marin[/quote]
$\frac{a}{b^{2}c^{2}}+\frac{b}{c^{2}a^{2}}+\frac{c}{a^{2}b^{2}}=\frac{(\frac{a}{bc})^2}{a}+\frac{(\frac{b}{ca})^2}{b}+\frac{(\frac{c}{ab})^2}{c}$$\ge\frac{(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab})^2}{a+b+c} =\frac{9}{a+b+c} .$

Very nice.

[quote="sandumarin"]If $ a,b,c>0,\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=3,n\ge 1 $ n integer, prove that:
$ \sum\frac{a^n}{b^{n+1}c^{n+1}}\ge\frac{9}{a+b+c} $
_____
Sandu Marin[/quote]
If $a,b,c>0,\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=3,n>m\ge 1$ , prove that:
$\sum{\frac{{{a}^{m}}}{{{b}^{n}}{{c}^{n}}}}\ge \frac{{{3}^{1+n-m}}}{{{\left( a+b+c \right)}^{n-m}}}$


if a>1 natural ,m,n natural and m,n=2k+1
gcd(m,n)=1 then prove gcd((a^m)+1,(a^n)-1) is 1 or 2.[/code]

Numbers $a$ and $d=gcd(a^m+1,a^n-1)$ are obviously coprime.
Let x be order of a mod d.
(1) $a^m\equiv -1(mod d)$ => $a^{2m}\equiv 1(mod d)$
$a^n\equiv 1 (mod d)$
so $x|2m$ and $x|n$. Because $gcd(m,n)=1$ we hve $x|2$,
$a^{2k+1}\equiv a^{2l+1} (mod d)$
$1\equiv a^m\equiv a^n\equiv -1 (mod d)$
so d=1 or 2. Both values can be reached, 1 for a even, 2 for a odd.

It is easy $d=gcd(a^m+1,a^n-1)|gcd(a^{2m}-1,a^n-1)=a^{gcd(2m,n)}-1\Longrightarrow d=1 \ or \ d=2.$


[quote="N.T.TUAN"]Chào cả nhà,

Chúc năm mới mạnh khoẻ và hạnh phúc. :) Sao cái box Việt Nam mình vắng nhể? :D[/quote]
Có lẽ người VN biết Mathlink nhưng chưa biết đến box này đó bác.


The letters of the alphabet are given numeric values based on
the two conditions below.\\
$ \bullet$ Only the numeric values of $ \minus{}2, \minus{}1, 0, 1 \text{ and } 2$ are used.\\
$ \bullet$ Starting with A and going through Z, a numeric value is
assigned to each letter according to the following pattern:\\
\indent $ 1, 2, 1, 0, \minus{}1, \minus{}2, \minus{}1, 0, 1, 2, 1, 0, \minus{}1, \minus{}2, \minus{}1, 0,$ ....\\
Two complete cycles of the pattern are shown above. The letter A has a value of 1, B has a value of 2, F has a value of $ \minus{}2$ and Z has a value of 2. What is the sum of the numeric values of the letters in the word ``numeric?''

I see no better way than to just evaluate...

We find that this evaluates to $ \boxed{\minus{}1}$; just find each individual value of each letter and add up.


ABC triangle. Prove that:

 \sum (p-a)/(2p-a) 2  \leq 9/8p

I think, you mean \sum (p-a)/(2p-a)2 \leq 9/16p.

Let a=x+y, b=y+z, c=z+x, where x,y,z > 0.

We must prove \sum z/(x+y+2z)2 \leq 9/16(x+y+z).

We have x+y+2z = z+3*(x+y+z)/3 \geq 4(z(x+y+z)3/27)1/4.
Thus initial inequality becomes \sqrt x + \sqrt y + \sqrt z \leq \sqrt (3(x+y+z)) - it is true.


[quote="SnowStorm"]Hey could someone help explain this algorithm to me?
http://mathworld.wolfram.com/Rabin-MillerStrongPseudoprimeTest.html
Or suggest a better one?
What doe the = sign with three lines mean?
Thanks[/quote]
is congruent with (used in modular arithmatic)


Hmm, would it be something like $ (5!)\plus{}\binom{5}{2}*\binom{4}{3}*(3!)\plus{}\binom{5}{3}*\binom{4}{2}*{2!}\plus{}5\plus{}1$ ?

(I'm too lazy to expand that out right now  :P )

Becuase you're splitting this up into cases: one 2, two 2's, three 2's, four 2's, five 2's.

Like in the term $ \binom{5}{3}*\binom{4}{2}*{2!}$,  you first choose the 3 places for the 2's to go, then out of the 2 places left, you choose 2 out of the 4 non-2 digits and then the 2! for order...  sorry for the confusing words

Basically, do casework based on the number of 2's in the number.

[quote="lotsofmath"]I took the 2007 Chapter Team Round and got an 8. I still can't figure out this problem:

Using the digits 2,3,4,7 and 8, Carlos will form 5-digit positive integers. Only the digit 2 can be used more than once in any of Carlos' 5-digit integers. How many distinct 5-digit integers are possible?[/quote]

Well as he said, divide it into cases.

So first, with all 2s, their is obviously only one way.

With 4 2s, do 5!/4!1! so it is 5. Multiply it by 4 because there are 4 ways that the other numbers could fill in the blanks. 20.

With 3 2s, do 5!/3!2! so it is 10. Multiply it by 4*3 which is the amount of ways it
can fill in the blanks. 120

With 2 2s it is 5!/2!3! so it is 10. Multiply it by 4*3*2 which is the amount of ways to fill in the blanks. 240.

With 1 2 it is 5!/1!4! so it is 5. Multiply is by 4*3*2*1 which is the amount of ways to fill in the blanks. 120.

120+240+120+20+1 is 501.


Let a,b,c nenegative numbers. Prove that :

 \[ \sqrt{a^2\plus{}b^2}\plus{}\sqrt{b^2\plus{}c^2}\plus{}\sqrt{c^2\plus{}a^2}\le a\plus{}b\plus{}c\plus{}\sqrt{a^2\plus{}b^2\plus{}c^2}\]

[quote="Marius Mainea"]Let a,b,c nenegative numbers. Prove that :
\[ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\]
[/quote]

without loss of generality, we can setting $ a \equal{} \sqrt {x},b \equal{} \sqrt {y},c \equal{} \sqrt {z},x,y,z\ge0$ and assume that $ x\le y \le z,x \plus{} y \plus{} z \equal{} 1$.
so
\[ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\]
\[ \Longleftrightarrow \sum{\left(\sqrt {1 \minus{} x} \minus{} \sqrt {x}\right)} \le 1.\]
let $ f(k) \equal{} \sqrt {1 \minus{} k} \minus{} \sqrt {k}$, easy to prove that when $ k \in \left[0,\frac {1}{2}\right],f''(k)\ge0$ and $ 0\le x \le y \le x \plus{} y \le \frac {1}{2}$, so we have
\[ \sum{\left(\sqrt {1 \minus{} x} \minus{} \sqrt {x}\right)}\]
\[ \equal{} f(x) \plus{} f(y) \plus{} f(z)\]
\[ \le f(0) \plus{} f(x \plus{} y) \plus{} f(z)\]
\[ \equal{} 1 \plus{} f(1 \minus{} z) \plus{} f(z)\]
\[ \equal{} 1.\]
done!

BTW, this inequality's geometric meaning:

[quote="Marius Mainea"]Let a,b,c nenegative numbers. Prove that :
\[ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\]
[/quote]
$ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\Leftrightarrow$
$ \Leftrightarrow\left(\sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\right)^2\le \left(a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\right)^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a\sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} bc \minus{} \sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{2abc\sqrt{a^2\plus{}b^2\plus{}c^2}}{a\sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} bc \plus{} \sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)}}\geq0.$

[quote="arqady"][quote="Marius Mainea"]Let a,b,c nenegative numbers. Prove that :
\[ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\]
[/quote]
$ \sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\le a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\Leftrightarrow$
$ \Leftrightarrow\left(\sqrt {a^2 \plus{} b^2} \plus{} \sqrt {b^2 \plus{} c^2} \plus{} \sqrt {c^2 \plus{} a^2}\right)^2\le \left(a \plus{} b \plus{} c \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\right)^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a\sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} bc \minus{} \sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {2abc\sqrt {a^2 \plus{} b^2 \plus{} c^2}}{a\sqrt {a^2 \plus{} b^2 \plus{} c^2} \plus{} bc \plus{} \sqrt {(a^2 \plus{} b^2)(a^2 \plus{} c^2)}}\geq0.$[/quote]

very nice, learning... :)

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200987

Case 1. $ a^2 \plus{} b^2 \minus{} c^2\ge 0,b^2 \plus{} c^2 \minus{} a^2\ge 0,c^2 \plus{} a^2 \minus{} b^2\ge 0$

 Then inequality is echivalent with
\[ \sum\frac {a^2 \plus{} b^2 \minus{} c^2}{\sqrt {a^2 \plus{} b^2} \plus{} c^2}\le\sqrt {a^2 \plus{} b^2 \plus{} c^2}\]
wich is true because
\[ \sum\frac {a^2 \plus{} b^2 \minus{} c^2}{\sqrt {a^2 \plus{} b^2} \plus{} c^2}\le\frac {a^2 \plus{} b^2 \minus{} c^2 \plus{} b^2 \plus{} c^2 \minus{} a^2 \plus{} c^2 \plus{} a^2 \minus{} b^2}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \equal{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\]
Case 2. $ a^2 \plus{} b^2 \minus{} c^2 < 0,b^2 \plus{} c^2 \minus{} a^2\ge 0,c^2 \plus{} a^2 \minus{} b^2\ge 0$ then presume $ a\le b\le c$

$ \sum\frac {a^2 \plus{} b^2 \minus{} c^2}{\sqrt {a^2 \plus{} b^2} \plus{} c^2}\le\frac {a^2 \plus{} b^2 \minus{} c^2 \plus{} b^2 \plus{} c^2 \minus{} a^2 \plus{} c^2 \plus{} a^2 \minus{} b^2}{\sqrt {b^2 \plus{} c^2} \plus{} a}\le\sqrt {a^2 \plus{} b^2 \plus{} c^2}$

 because  $ \sqrt{b^2\plus{}c^2}\plus{}a\le \sqrt{c^2\plus{}a^2}\plus{}b\le\sqrt{a^2\plus{}b^2}\plus{}c$

[quote="Marius Mainea"]$ \sum\frac {a^2 \plus{} b^2 \minus{} c^2}{\sqrt {a^2 \plus{} b^2} \plus{} c^2}\le\frac {a^2 \plus{} b^2 \minus{} c^2 \plus{} b^2 \plus{} c^2 \minus{} a^2 \plus{} c^2 \plus{} a^2 \minus{} b^2}{\sqrt {b^2 \plus{} c^2} \plus{} a}\le\sqrt {a^2 \plus{} b^2 \plus{} c^2}$[/quote]why? $ \frac {a^2 \plus{} b^2 \minus{} c^2}{\sqrt {a^2 \plus{} b^2} \plus{} c^2}>\frac {a^2 \plus{} b^2 \minus{} c^2}{\sqrt {a^2 \plus{} b^2} \plus{} c^2}$

for the case abc>0 another solution uses Jensen for $ f(x)\equal{}\sqrt{1\plus{}x^2}$


vector
trigonometry
Pythagorean Theorem
geometry

[quote="pkothari13"]What angle does the vector $A=\hat{i}+\hat{j}+\hat{k}$ make with the $x,y$ and $z$ axes?[/quote]
$\vec{A}=<1,1,1>$, $|\vec{A}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$
$\vec{i}=<1,0,0>$, $|\vec{i}|=\sqrt{1^2+0^2+0^2}=1$.
$\vec{A}\cdot\vec{i}=1*1+1*0+1*0=1=|\vec{A}|*|\vec{i}|*\cos\theta$, where $\theta$ is the angle between $\vec{A}$ and the $x$-axis.
$\cos\theta=\frac{\sqrt{3}}{3}$

Ack, oops it was supposed to be What angle does the vector $A=\hat{i}+2\hat{j}+3\hat{k}$ make with the $x,y$ and $z$ axes?

Well, from your explanation, is this how it should work?

$\vec{A} \cdot \hat{i} = 1(1)+2(0)+3(0) = 1$

$1 = \sqrt{14}\cos(\theta)$, so $\theta = \cos^{-1}(\frac{\sqrt{14}}{14})$

Maybe Intermediate Maths would have been better for this!   :D

I don't know, I was thinking of maybe putting it there, but I had this problem on my Physics test the other day and I had no clue how to do it, so I figured the Physics forum would be my best bet.  :P

Does anyone have any comments on that problem?

Well, you can use Pythagorean theorem if you're not familiar with scalar multiplication of vectors or if you can't figure it out. The angle between $A$ and $x$ axis is the angle in right triangle between sides $1$ and $\sqrt{1+4+9}$, similarly for the other axes.


[quote="jinmadaxia"]0<x<1,0<y<1,proof
x^y+y^x>1[/quote]

we can prove that
$ x^y>\frac {x}{x \plus{} y}$
and
$ y^x>\frac {y}{x \plus{} y}$


By CSI, $ (\sqrt {1 - a^2} + \sqrt {1 - a} + \sqrt {1 + a})^2 \leq 3(3 - a^2) \leq 9$, implying $ \sqrt {1 - a^2} + \sqrt {1 - a} + \sqrt {1 + a} \leq 3$.

Then, again by CSI, ${ (\sqrt [4]{1 - a^{2}} + \sqrt [4]{1 - a} + \sqrt [4]{1 + a}})^2 \leq 3(\sqrt {1 - a^2} + \sqrt {1 - a} + \sqrt {1 + a}) \leq 9.$

Taking the square root gives the desired inequality. Equality holds iff $ a = 0$.


quadratics
number theory unsolved
number theory

Let n be a positive integer.Then there exist integers x and y such that $ n\equal{}x^2\plus{}y^2$ iff each prime factor of n of the form 4k+3 appears an even number of times.

[quote="lajanugen"]Let p be a prime divisor of n. If (p,x)=1, then -1 is a quadratic residue mod p and hence p=-1(mod4)

If (p,x)>1 , then it follows that $ p^{2}$ divides n.

If there existed a prime divisor =3(mod4) of n , then it would divide both x and y. Now, applying the same argument to $ n/p^{2}$, and so on, we will have to stop at a certain situation and we have the result.[/quote]
Thanks but I don't understand your solution very clearly.Can you or anyone else explain more to me please?
[quote]If (p,x)=1, then -1 is a quadratic residue mod p [/quote]


[b]Let F(x) = [x^2 + bx + c]/[x^2 + b1x + c1]
 if the eqns x^2 + bx + c = 0 and x^2 + b1x + c1 = 0 don't have    real roots, then
(a) F'(x) = 0 has real and distinct roots.
(b) F'(x) = 0 has real and equal roots
(c) F'(x) = 0 has imaginary roots.
(d) nothing can be said.[b][/b][/b]

$ F(x)$ is always positive since $ x^2 \plus{} bx \plus{} c$  and $ x^2 \plus{} b_1x \plus{} c_1$ are always positive for real $ x$.
$ F(x)$ has no vertical asymptotas.
$ F(x)$ takes $ 1$.
$ F(x) \equal{} 1 \plus{} \frac {x(b \minus{} b_1) \plus{} (c \minus{} c_1)}{x^2 \plus{} b_{1}x \plus{} c_1}$ has a horizontal asympotata $ y \equal{} 1$ and as $ \frac {x(b \minus{} b_1) \plus{} (c \minus{} c_1) }{x^2 \plus{} b_{1}x \plus{} c_1}$ can take positive and negative values $ F$ will reach the asymptota from below and above: this means $ F'(x)$ has at least two different real zeros.(I think)

 :)


[quote="Mateescu Constantin"]$ a,b,c > 0,\ ab \plus{} bc \plus{} ca \equal{} 3\ \implies\ \frac {a^3(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^3(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^3(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2}\ge 2$[/quote]

The above inequality is equivalent to $ \sum a^{2}\geq \sum \frac {a^2b^2}{a^2 \plus{} ab \plus{} b^2} \plus{} 2$. But from AM-GM we get that $ \sum \frac {a^2b^2}{a^2 \plus{} ab \plus{} b^2} \plus{} 2\leq 2 \plus{} \sum \frac {ab}{3} \equal{} 3$. So we only need to prove that $ 3\leq \sum a^2$ which is true from hypothesis

[quote="Mateescu Constantin"]$ a,b,c > 0,\ ab \plus{} bc \plus{} ca \equal{} 3\ \implies\ \frac {a^3(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^3(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^3(c \plus{} a)}{c^2 \plus{} ca \plus{} a^2}\ge 2$[/quote]

first, we prove that

$ \sum{\frac {a^4}{a^2 \plus{} ab \plus{} b^2}} \ge \sum{\frac {ab^3}{a^2 \plus{} ab \plus{} b^2}}.$

actually, $ \frac {a^4}{a^2 \plus{} ab \plus{} b^2} \minus{} \frac {ab^3}{a^2 \plus{} ab \plus{} b^2} \equal{} a(a \minus{} b)$, so

$ \sum{\frac {a^4}{a^2 \plus{} ab \plus{} b^2}} \ge \sum{\frac {ab^3}{a^2 \plus{} ab \plus{} b^2}} \Longleftrightarrow \sum{a^2} \minus{} \sum{ab}\ge0,$

obvious true.

thus, we have

$ \sum{\frac {a^3(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}} \ge \sum{\frac {ab(a^2 \plus{} b^2)}{a^2 \plus{} ab \plus{} b^2}},$

so, we just need to prove that

$ \sum{\frac {ab(a^2 \plus{} b^2)}{a^2 \plus{} ab \plus{} b^2}} \ge 2$

$ \Longleftrightarrow \sum{\frac {ab(a^2 \plus{} b^2)}{a^2 \plus{} ab \plus{} b^2}} \ge \frac {2}{3}(ab \plus{} bc \plus{} ca)$

$ \Longleftrightarrow \sum{\left(\frac {ab(a^2 \plus{} b^2)}{a^2 \plus{} ab \plus{} b^2} \minus{} \frac {2}{3}ab\right)} \ge 0$

$ \Longleftrightarrow \sum{\frac {ab(a \minus{} b)^2}{a^2 \plus{} ab \plus{} b^2}} \ge 0,$

obvious true, so ineq holds.

[quote="Pain rinnegan"]
\[ LHS \equal{} \frac {1}{2}\sum_{cyc}\frac {(a \plus{} b)^2(a^2 \minus{} ab \plus{} b^2)}{a^2 \plus{} ab \plus{} b^2}\ge \frac {2}{3}\sum_{cyc}(a^2 \minus{} ab \plus{} b^2)\ge RHS\]
[/quote]

I don't think it is true. :wink:

You are right manlio, but the proof can be repared, because we have $ 3(a\plus{}b)^2(a^2\minus{}ab\plus{}b^2)\ge 4(a^2\plus{}ab\plus{}b^2)ab$.


i was hoping for top 3 or so.....btw, does anyone know the individual placings by the 4 people??

first place written = david yang.
2nd written = max schindler
3rd =Bobby shen
4th = calvin deng

Or after cd

1st: bobby shen
2nd: David yang
3rd: Max schindler

And someone else....

Technically 3rd and 4th after countdown are considered "semifinalists" since they never go against each other.

Why don't the MATHCOUNTS people make the semifinalist go each other?


Suppose that $ x \equal{} 2\minus{}t$ and $ y \equal{} 4t \plus{} 7$.  If $ x$ = -3, what is $ y$?

If x=-3 that means that t is 5 so the answer is just 20+7=27.


This is a generalization of an AIME problem:

Let us define a cyclic k-set of teams to be a set S of k teams such that S(0) beats S(1), S(1) beats S(2), S(2) beats S(3), ..., S(k-1) beats S(k), S(k) beats S(1). If n teams play a round robin tournament, what is the maximum number of k-sets that can occur?

Fierytycoon

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I believe the answer is nCk*(k-1)! = nPk/k.

First out of those n teams you have to choose which k belong to your cyclic k-set. Then once they are chosen, you can imagine them being put as the vertices of a k-sided polygon to represent their cyclic order. There are nCk way to pick the k teams, and then (k-1)! ways to order them cyclically.

[/hide]

That doesn't look right to me. Let's get rid of the S(0) because otherwise it just doesn't make sense. So it's S(1) B S(2), S(2) B S(3), etc, S(k) B S(1), where x B y denotes x beats y. In Gauss's solution, let's let n=4 and k=2. There should be 0 because if team X beats team Y, then team Y cannot beat team X also since they only play once. But Gauss's solution gives 6.

I think it's quite a tough problem, and I'll probably have to work out a bunch of cases and get a conjecture before I can prove it. But I'll see what I can do. Thanks for the interesting problem.

Oh yeah. my solution only works when k /= 0 or 2. Otherwise it is zero because you can't have a cyclic polygon with only 0 or 2 vertices. It does happen to work for k = 1 though.

[quote]Let's get rid of the S(0) because otherwise it just doesn't make sense. [/quote]

Sorry, my bad, I sorta included that extra win by accident. Was typing up the problem sorta quickly and didn't catch the error.

Fierytycoon