calculus
integration
trigonometry
calculus computations

[quote="Extremal"]Idea:
lets rewrite it as
$Rre(\int_{0}^{1}\frac{1-x^{n}}{e^{\frac{i\pi x}{2}}}dx)=Rre(\int_{0}^{1}e^{-\frac{i\pi x}{2}}dx+\int_{0}^{1}x^{n}e^{-\frac{i\pi x}{2}})$
and last integral you can calculate ( integration $n$ times by parts). I mean this :$\int_{0}^{1}x^{n}e^{-\frac{i\pi x}{2}}$[/quote]

Oh, you mean that

$\sec x = \mathfrak{Re}\left( \frac{1}{e^{ix}}\right) = \mathfrak{Re}(e^{-ix}) = \cos x$

holds?


search
inequalities
inequalities proposed

Prove for all real number $ a,b,c>0$:
\[ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \ge 9(ab\plus{}bc\plus{}ca)\]

That's APMO 2004 and has been posted many times before. Please use the search query before posting.  :)

[quote="hasan4444"]Prove for all real number $ a,b,c > 0$:
\[ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge 9(ab \plus{} bc \plus{} ca)\]
[/quote]
That's a well known inequality hassan ( From APMO 2004 ).Note that the stronger is also true : $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge 3(a\plus{}b\plus{}c)^2$ for $ a,b,c \geq 0$ and can be proved easily by Cauchy-Schwarz inequality ( See for example : http://www.mathlinks.ro/viewtopic.php?p=1613686#1613686 ).And the following stronger inequality also holds : $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 4(a^2\plus{}b^2\plus{}c^2)\plus{}5(ab\plus{}bc\plus{}ca)$ for $ a,b,c \geq 0$ ( See http://www.mathlinks.ro/viewtopic.php?p=1642907#1642907 )

Sorry guys but I wasn't know what competition was it and I don't know how can I search for such an inequality.

Thanks a lot guys


[quote="Klaas"]Does there exist such natural numbers a,b,c>1 for which all numbers

(a^2 -1)/b , (a^2 -1)/c , (b^2 -1)/a , (b^2 -1)/c, (c^2 -1)/a, (c^2 -1)/b
are natural?

 :football:[/quote]

Let $ a \leq b \leq c$ (*).

$ b$ and $ c$ are coprime because $ b$ divides $ c^2\minus{}1$. So $ bc$ divides $ a^2\minus{}1$.

It contradicts (*). There is no such $ (a,b,c)$.


integration
calculus
induction
inequalities
real analysis
real analysis unsolved

[quote="xirti"]Let $f: [0,\infty)\to[0,\infty)$ an increasing bijective function. Prove that for any$n\in \mathbb{N},a,b\in [0,\infty)$ we have

\[ a^{n-1}\int_0^a f^n(x)dx+b^{n-1}\int_a^b[f^{-1}(x)]^ndx\geq \frac{a^nb^n}{2^{n-1}}.  \]

this is exactly how is written. I think the the second integral must be from 0 to b.[/quote]
The correct statement is
\[ a^{n-1}\int_0^a f^n(x)dx+b^{n-1}\int_0^b[f^{-1}(x)]^ndx\geq \frac{a^nb^n}{2^{n-1}}.  \]
whenever $a > 0$ and $b=f(a)$. To prove this one first notes that
\[ ab = af(a) = \int_0^a f(x)dx + \int_0^{f(a)} f^{-1}(y)dy.  \]
Indeed, if $f$ is differentiable, this can be checked by differentiating with respect to $a$, and if $f$ is just continuous, it follows by approximation. [i]Drawing a picture shows why this is true.[/i]

Next, $\left(A+B \right)^n \le 2^{n-1}(A^n + B^n)$, as an induction argument shows. Therefore,
\[ a^nb^n \le 2^{n-1}\left(\left(\int_0^a f(x)dx\right)^n + \left(\int_0^b f^{-1}(y)dy\right)^n\right).  \]
Finally by Hoelder's inequality $\left(\int_0^a f(x) dx \right)^n \le a^{n-1}\int_0^af(x)^ndx$, and a similar estimate holds for the other integral. Putting it all together, the statement follows.

[quote="hpe"][quote="xirti"]Let $f: [0,\infty)\to[0,\infty)$ an increasing bijective function. Prove that for any$n\in \mathbb{N},a,b\in [0,\infty)$ we have

\[ a^{n-1}\int_0^a f^n(x)dx+b^{n-1}\int_a^b[f^{-1}(x)]^ndx\geq \frac{a^nb^n}{2^{n-1}}.  \]

this is exactly how is written. I think the the second integral must be from 0 to b.[/quote]
The correct statement is
\[ a^{n-1}\int_0^a f^n(x)dx+b^{n-1}\int_0^b[f^{-1}(x)]^ndx\geq \frac{a^nb^n}{2^{n-1}}.  \]
whenever $a > 0$ and $b=f(a)$. [/quote]
The last condition is't necesary for the rest of the proof.
[quote="hpe"]
To prove this one first notes that
\[ ab = af(a) = \int_0^a f(x)dx + \int_0^{f(a)} f^{-1}(y)dy.  \]
Indeed, if $f$ is differentiable, this can be checked by differentiating with respect to $a$, and if $f$ is just continuous, it follows by approximation. [i]Drawing a picture shows why this is true.[/i][/quote]
Actually a ineq holds:
\[ ab = af(a) \le \int_0^a f(x)dx + \int_0^{f(a)} f^{-1}(y)dy.  \]
[quote="hpe"]

Next, $\left(A+B \right)^n \le 2^{n-1}(A^n + B^n)$, as an induction argument shows. Therefore,
\[ a^nb^n = 2^{n-1}\left(\left(\int_0^a f(x)dx\right)^n + \left(\int_0^b f^{-1}(y)dy\right)^n\right).  \][/quote]

there must be
\[ a^nb^n \le 2^{n-1}\left(\left(\int_0^a f(x)dx\right)^n + \left(\int_0^b f^{-1}(y)dy\right)^n\right).  \]
[quote="hpe"]

Finally by Hoelder's inequality $\left(\int_0^a f(x) dx \right)^n \le a^{n-1}\int_0^af(x)^ndx$, and a similar estimate holds for the other integral. Putting it all together, the statement follows.[/quote]

Anyway, nice proof. Thank you.[/quote]


inequalities
algebra
system of equations

I am a bit confused about the definition of "cyclic" (when applied to simple algebra, not cyclic quadrilaterals or cyclic groups or anything). Is the following system of equations said to be cyclic?

x = Sqrt(y) - z
y = Sqrt(z) - x
z = Sqrt(x) - y

rrm.  Well.  Here is the context that I have seen cyclic used in most often.  Say you want the cyclic sum of five numbers, taken 2 at a time.  That would be (ab + bc + cd + de + ea).  If you wanted the symmetric sum of 5 numbers taken 2 at a time, it would be (ab + ac + ad + ae + ba + bc + bd + be + ca + cb + cd + ce + da + db + dc + de + ea + eb + ec + ed).  Generally speaking, something is symmetric if switching any two variables won't change it, and cyclic if rotating them cyclically (that is, each one goes only into the next one, so they make a full circuit) won't change it.  I'm not sure I've seen cyclic applied to systems of equations -- but if it is, then those definitely would be.

On further thought, those equations you gave are definitely symmetric.  I'm not sure I would call them cyclic.

On additional further thought, I have actually solved that system of equations:

[hide]So, to start off with, if x=y=z, we get solutions of (0, 0, 0) and (1/4, 1/4, 1/4).  If x, y, and z are not all equal, we can assume without loss of generality that x <= y <= z.  We can do this because the system is symmetric.  So we have

x + y =  :sqrt:z.

:sqrt:z < 2z

:sqrt:z > 1/2

z > 1/4

y + z = :sqrt:x

:sqrt:x > 2x

:sqrt:x < 1/2

x < 1/4

x + y = :sqrt:z

1/2 < :sqrt:z = x + y < 1/4 + y

y > 1/4

y + z = :sqrt:x

1/2 > :sqrt:x = y + z > y + 1/4

1/4 > y

Contradiction.

By the way, the reason all those >= became > is because we were assuming that not all three were equal, so one of the inequalities had to be strict, so when we added them, we had to come up with a strict inequality.[/hide]

Ok, I think I understand the definitions better now. Thanks.

By the way, that wasn't really a problem by design, I just listed it as an example of something symmetric.

Yeah, I figured, but it looked fun anyhow.   If the negative terms on the right had been, say, squared instead of first degree, then it wouldn't be symmetric anymore, just cyclic.

Wait, you mean like x = Sqrt(y) - z^2? Why would that not be cyclic anymore?

It wouldn't be symmetric.  The other one was symmetric -- it could be written as

x + z = :sqrt:y
z + y = :sqrt:x
y + x = :sqrt:z

Which is clearly symmetric -- if we switch any two of the variables, we still get a valid solution.  In fact, we get an identical system.  If we throw the squares in, though, we will not longer have a symmetric system -- it will only be cyclic, because we can cycle the variables to get other identical equations, but not just switch any pair.

[quote="Fierytycoon"]Is that because x^2 = n <=> +/- Sqrt(n)?[/quote]

Hmm.. I'm not sure exactly what you mean, but the reason it wouldn't be symmetric, for example, would be because there would be an equation x+z:^2: = :sqrt:y but no equation x:^2: + z = :sqrt:y. So it wouldn't be symmetric any more.


Hi,
Show that for any Positive Definte Matrix A nxn

 det(A^(1/n))<=1/n(tr(A))
Thanks

by the smaes arguments try to find an inequality between $\prod a_{i,i}$ and $det(A)$. Very classical.

[quote="Moubinool"]easy the matrix is diagonalisable its eigenvalues are positives, then AM-GM[/quote]
Hi,Thanks for your answer but it is very short and unclear to me.could you please give me more explanation
Cheers,

$A=P^tDP$ where $D$ is a diagonal matrix. But $tr(A)=tr(D)$, $det(A)=det(D)$. Then conclude with arithmetic-geometric mean inequality.


how to prove that:

1. A leg of a right triangle is the mean proportional between the hypotenuse and the projection of that leg on the hypotenuse.

2. The altitude drawn to the hypotenuse of a right traingle is the mean proportional between the segments of the hypotenuse.

3. In a right triangle, the product of the hypotenuse and the altitdue drawn to it is equal ot the product of the legs.

[hide="hint"]
construct the altitude. use similar triangles.[/hide]

all three could be solved using the same method.... draw an altitude to the hypotenuse....

EDIT: didnt see paladin8's post

Let the angle $\angle C = 90^o$ of a triangle $\triangle ABC$ be right and $H$ the foot of the altitude from the vertex $C$ to the side $AB = c$. Label the remaining triangle sides $BC = a, CA = b$, the altitude $CH = h$ and the segments $AH = y,, BH = x$.

The triangles $\triangle ACH \sim \triangle ABC$ are similar, because the angle $\angle A$ is common and the angles $\angle AHC = \angle C = 90^o$ are both right. Therefore

$(1a)\ \ \frac y b = \frac b c,\ \ b^2 = cy,\ \ b  = \sqrt{cy}$

$(2a)\ \ \frac h y = \frac a b$

$(3a)\ \ \frac h b = \frac a c,\ \ ch = ab$

The triangles $\triangle CBH \sim \triangle ABC$ are similar, because the angle $\angle B$ is common and the angles $\angle CHB = \angle C = 90^o$ are both right. Therefore

$(1b)\ \ \frac x a = \frac a c,\ \ a^2 = cx,\ \ a  = \sqrt{cx}$

$(2b)\ \ \frac x h= \frac a b$

$(3b)\ \ \frac h a = \frac b c,\ \ ch = ab$

Combining (2a) and (2b),

$(2c)\ \ \frac h y = \frac a b = \frac x h,\ \ h^2 = xy,\ \ h = \sqrt{xy}$

yes I do greatly appreciate your explanation, yetti.

If you guys think this is below intermediate, please move it to the Getting Started.

[quote="yetti"][quote="paladin8"]Come on, yetti. You have more advanced things to be working on than this!  ;)[/quote]

Done (the original poster) moves in the Getting Started section. This was his first post in the Intermediate Topics section. My guess is that he appreciated the detailed reply.

Yetti[/quote]

I was just joking around :D Good work, as usual.


induction
limit
algebra unsolved
algebra

[quote="Rust"]$ S_n \equal{} \sum_{k \equal{} 1}^n \frac {1}{4n^2 \minus{} 1} \equal{} \frac {1}{2}(1 \minus{} \frac {1}{2n \plus{} 1}) \equal{} \frac {n}{2n \plus{} 1}\to \frac 12$.[/quote]
My solution:
We have: $ \frac{1}{(2k\minus{}1)(2k\plus{}1)}\equal{}\frac{1}{2}(\frac{1}{2k\minus{}1}\minus{}\frac {1}{2k\plus{}1})$ $ \rightarrow$ \[ S_n\equal{}\frac{1}{2}(1\minus{}\frac{1}{3}\plus{}\frac{1}{3}\minus{}\frac{1}{5}\plus{}...\plus{}\frac{1}{2n\minus{}1}\minus{}\frac{1}{2n\minus{}1})\equal{}\frac{1}{2}(1\minus{}\frac{1}{2n\plus{}1})\]
$ \rightarrow limS_n\equal{}lim \frac{1}{2}(1\minus{} \frac{1}{2n\plus{}1})\equal{}\frac{1}{2}$.

[quote="Hesman"]The sequence $ (a_n), (n \equal{} 1,2,...)$ is defined by $ u_n \equal{} \frac {1}{4n^2 \minus{} 1}$. Institute $ (s_n)$ such that:
\[ s_1 \equal{} u_1 , s_2 \equal{} u_1 \plus{} u_2, s_3 \equal{} u_1 \plus{} u_2 \plus{} u_3,...,s_n \equal{} u_1 \plus{} u_2 \plus{} u_3 \plus{} ... \plus{} u_n.
\]
Find:
\[ limS_n
\]
[/quote]
We want to find a formula for $ s_n$. We know that $ s_{n\plus{}1} \minus{} s_n \equal{} \frac{1}{4(n\plus{}1)^2\minus{}1}$. Now instantly notice: $ 4(n\plus{}1)^2 \minus{} 1 \equal{} (2n\plus{}2)^2 \minus{} 1^2 \equal{} (2n\plus{}3)(2n\plus{}1)$. So $ s_{n\plus{}1} \minus{} s_n \equal{} \frac{1}{(2n\plus{}3)(2n\plus{}1)}$.
Notice the very useful "identity": $ \frac{1}{ab} \equal{} \frac{1}{b\minus{}a} ( \frac{1}{a} \minus{} \frac{1}{b})$. And it is easily proved by multiplying..
Plugging in gives us: $ \frac{1}{(2n\plus{}3)(2n\plus{}1)} \equal{} \frac{1}{(2n\plus{}1)\minus{}(2n\plus{}3)} ( \frac{1}{2n\plus{}3} \minus{} \frac{1}{2n\plus{}1} )$ $ \equal{} \frac{\minus{}1}{2}( \frac{1}{2n\plus{}3} \minus{} \frac{1}{2n\plus{}1} )$.
Thus if $ s_n \equal{} \frac{\minus{}1}{2(2n\plus{}1)} \plus{} k$. We get that $ s_{n\plus{}1} \minus{} s_n \equal{} \frac{1}{4(n\plus{}1)^2\minus{}1}$. Since $ s_1 \equal{} \frac{\minus{}1}{2(2 \cdot 1 \plus{} 1)} \plus{} k \equal{} \frac{1}{3}$ we get $ k \equal{} \frac{1}{2}$. So it should be easy to prove by induction that $ s_n \equal{} \frac{\minus{}1}{4n\plus{}2} \plus{} \frac{1}{2}$.

Then $ \lim_{n \to \infty} s_n \equal{} \lim_{n \to \infty} \frac{\minus{}1}{4n\plus{}2} \plus{} \frac{1}{2} \equal{} \frac{1}{2}$.


algebra
polynomial
group theory
complex analysis
complex analysis unsolved

Suppose we know the location of zeros of $ f(z)\equal{}a_0\plus{}a_1z\plus{}\cdots\plus{}a_nz^n$.
Is there any result that tell us something about the zeros of
\[ a_{\sigma(0)}\plus{}a_{\sigma(1)}z\plus{}\cdots\plus{}a_{\sigma(n)}z^n ?\]

(here $ \sigma$ is an element of permutation group on $ n\plus{}1$ letters $ \{0,1,\cdots,n\}$)

Certainly if $ \sigma$ reverses the order of the indices, then roots $ r$ are taken to $ 1/r$. But if you're trying to find a general
formula, good luck...

It would make sense to consider the effect of a transposition $ \sigma$, since you could generate all permutations
through transpositions.


p is prime number and "m,n"are integer.find all(p,m,n)
p=m^2+n^2 and p|(m^3+n^3)-4

u shoudn't post this in this forum, it is obvious that it has a solution!


I'm having trouble making $ n$ the subject here:

$ x \equal{} \dfrac {m \minus{} n}{m \plus{} n}y$

I'd really appreciate it if someone could point out the steps please :)

I assume you want to solve for $ n$, so here it is.
\begin{align*}
x&=\frac{m-n}{m+n}y\\
\frac{x}{y}&=\frac{m-n}{m+n}\\
ym-yn&=xm+xn\\
-xn-yn&=xm-ym\\
n(-x-y)&=xm-ym\\
n&=-\frac{m(x-y)}{x+y}\\
\end{align*}

It's fairly straightforward: cross multiply, collect terms, factor, divide.

Thanks sunehra! It's the x-mult step I'd forgotten :(

It's just that process of "undoing the steps". The biggest problem people have when learning this is trying to get rid of the denominator. Also, when trying to solve for, um...let's say x, for example; when x is in the denom. people make the mistake of dividing by the numerator. This is fine if you take the reciprocal of both sides, but most people just panic and forget. Cross-multiplying is an important thing to learn when solving these equations.


function
limit
floor function
inequalities
topology
real analysis
real analysis unsolved

$ f: R - > R$
supose function $ f$ is monotonic increasing function and $ f - id$ is the periodical function  with period $ T$
then for all $ t\in R$and $ x\in R$ there exist limit  and ${{ \lim_{|n| - > + oo}\frac {f^{n}(t)}{n}} = \lim_{|n| - > + oo}\frac {f^{n}(x)}{n}}$
is  there some easy proof ? :maybe:

Did you mean that the limit exists and is independent of x?

I think you will have to add some add'l conditions, e.g., $ f(x)\equal{}[x]\equal{}\mbox{integer part of}\;x$ is not continuous, but monotonic increasing and $ f\minus{}id$ has period 1.

P.S. strictly monotonic increasing will not heal it, e.g., choose $ f(x)\equal{}0.5(x\plus{}[x])$

[quote="hjbrasch"]Did you mean that the limit exists and is independent of x?

I think you will have to add some add'l conditions, e.g., $ f(x) \equal{} [x] \equal{} \mbox{integer part of}\;x$ is not continuous, but monotonic increasing and $ f \minus{} id$ has period 1.

P.S. strictly monotonic increasing will not heal it, e.g., choose $ f(x) \equal{} 0.5(x \plus{} [x])$[/quote]
I mean  that the limit exists and is independent of x. It's didn't need add'l conditions. the statement is true .

This is an excellent result. I can restate it in a correct form as follows. Assume that $ f : \mathbb R \to \mathbb R$ is an increasing function and $ x \mapsto f(x) \minus{} x$ has the period 1. Denote by $ f^n$ the $ n$th iterate of $ f$, that is, $ f^1 \equal{} f$, $ f^n \equal{} f \circ f^{n \minus{} 1}$ for $ n \geq 2$. Then if
\[ \mathop {\lim }\limits_{n \to \infty } \frac {{f^n \left( 0 \right)}} {n}
\]
exists then for every $ x \in \mathbb R$
\[ \mathop {\lim }\limits_{n \to \infty } \frac {{f^n \left( x \right)}} {n} \equal{} \mathop {\lim }\limits_{n \to \infty } \frac {{f^n \left( 0 \right)}} {n}.
\]
Aha, we will show you that $ \mathop {\lim }\limits_{n \to \infty } \frac {{f^n \left( 0 \right)}} {n}$ exists apparently.

Now, to prove the statement, you must check the equality
\[ f\left( x \right) \equal{} f\left( r \right) \plus{} \left\lfloor x \right\rfloor
\]
where $ x \equal{} \left\lfloor x \right\rfloor \plus{} r$. Using the monotonicity of $ f$ we deduce that
\[ \frac {{f^n \left( 0 \right)}} {n} \leqslant \frac {{f^n \left( r \right)}} {n} \leqslant \frac {{f^n \left( 0 \right)}} {n} \plus{} \frac {1} {n}
\]
This implies
\[ \mathop {\lim }\limits_{n \to \infty } \frac {{f^n \left( x \right)}} {n}
\]
exists for every $ x \in [0,1]$. For an arbitrary numbre $ x \in \mathbb R$ we observe that
\[ f^n \left( x \right) \equal{} f^n \left( r \right) \plus{} \left\lfloor x \right\rfloor
\]
and then done.

The following inequalities play an important role in proving the existence of $ \mathop {\lim }\limits_{n \to \infty } \frac{{f^n \left( 0 \right)}}
{n}$
\[ \frac{p}
{{m_p }} \leqslant \mathop {\underline {\lim } }\limits_{n \to \infty } \frac{{f^n \left( 0 \right)}}
{n} \leqslant \mathop {\overline {\lim } }\limits_{n \to \infty } \frac{{f^n \left( 0 \right)}}
{n} \leqslant \frac{p}
{{m_p }} \plus{} \frac{{1 \plus{} f\left( 0 \right)}}
{{m_p }}
\]
if $ f(0)>0$ and $ m_p$ is the least positive integer such that $ {f^{m_p } \left( 0 \right) > p}$.

Now just consider the following 2 cases

[list]
$ \star$ If there exists $ p$ such that $ {f^m \left( 0 \right) < p}$ for every $ m$. In this situation, $ {\left\{ {f^m \left( 0 \right)} \right\}}$ is bounded.

$ \star$ If there does not exist such $ p$ then apply the above inequalities directly.
[/list]

thanks. this proof  is also in book about classical theory of dinamical system.
many aspect of this theory  find Danzua, also Van Kampenom, Koddington and Levinson. and other proof has Zigel.
this theorem is about homeomorphic of circle, which save orientation.
We consider two case:
if there exist periodical pouint, and if there not exist periodical point.

[quote="Extremal"]thanks. this proof  is also in book about classical theory of dinamical system.[/quote]

title of the book?

now I have only few papers from this book. I have no this book full.
and in this papers there is no title. but tommorow I will take this book


Show that $5^8 + 2^6$ is composite.

[hide="Hint"]Factor it.[/hide]


1.We say a number is ascending if its digits are strictly increasing. For example, 189 and 3468 are ascending while 142 and 466 are not. For which ascending 3-digit number n (between 100 and 999) is 6n also increasing
?

2. A regular octahedron has edges of length 6cm. If d cm is the shortest distance from the centre of one face to the centre of the opposite face measured around the surface of the octahedron, what is the value of d^2?

[hide="Problem 1"]Since $ n \ge 100$, $ n \equiv 3$ moulo $ 5$, and $ n < 117$. $ n \equal{} 113$ works using quick brute-force.[/hide]

unless I missed something 113 is not ascending.

This is not really a good way to do problems but it worked.
[hide="problem 1"]
n=def where d,e, and f are digits
I thought that f should be 8 since 6n will then give you abc8 where a,b, and c are digits
then since we carry a 4 over e should be 7 so we have  d78and ab68.
after this I just tried #'s until d=5 gave def=578=n and 3468=6n [/hide]


Two identical CDs regularly cost a total of $ \$28.$ What is the cost in dollars of five of these CDs?

2 CDs = 28 dollars
1 CD = 14 dollars
5 CDs = 14*5=70 dollars

answer : 70

Or you could use ratios.

$ \frac{28}{2} \equal{} \frac{x}{5}$

Cross multiply.

$ 140 \equal{} 2x$

$ \boxed{70} \equal{} x$

the problem says that 2 CD [color=#408000]COST[/color] $28  therefore you know that something plus something is 28 so lets call this [color=#404040]SOMETHING[/color] x so there are 2x so that equals to 28 so 2x=28 and each CD costs 14. and u want 5 CD so you multiply by 5 and get 70

so the answer is [size=200][color=#FF00FF][hide][b][i][u]70 [size=200][color=#BF8080][hide][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH !!!!!![/u][/i][/b][/hide][/color][/size][/u][/i][/b][/hide][/color][/size]

 :juggle:  :juggle:  :juggle:  :juggle:  I LIKE JUGGLING !!!!


MATHCOUNTS
quadratics
algebra
polynomial
geometry
area of a triangle
Heron\u0027s formula

Can any of you provide some programs for MATHCOUNTS target and team rounds?

Please write out the program.

Programs should be for either the TI-89 Titanium and TI-83 plus Silver Edition.

er
you can get many from http://www.ticalc.org
also, they are really not that useful.
The only calculator program/thingy i ever used in my whole life was a prime number list up to 1000

programmable calculators aren't allowed at mathcounts

not true.
The only calculators are those that have a QWERTY type keypad aren't or those that can communicate with each other.

Well in our chapter no one cared on which calculator.

Non QWERTY keypad calculators are allowed at MATHCOUNTS

[quote="Captain Sugar"]maybe not true in your state, but when they read out the rules, it says "no programmable calculators allowed."[/quote]

No.... TI calculators are programmable and at state and nationals they give out TI calculators to the competitors

Programmable calculators are allowed in every state, it could be that you are not allowed to have programs on your calculators.

not every TI calculator is programmable. my school has TI-29s, and they are not programmable.

I think you shouldn't use calculator programs beacuse its cheap, I think mathcounts should supply everyone with the same calculator so its fair

I was referring to the TI 82's and higher as being programmable, as those are the most common TI calculators. I agree with Naonoa though, it is unfair to have programs on your calculator. Also, almost every MC problem has a better and quicker solution than using a calculator for something such as primes under 1000. If you learn it instead on relying on your calculator, it will help your sprint score as well.

[quote="Captain Sugar"]ok, that makes sense.
have you noticed how expensive the TI graphing calculators are?
they're hideously expensive.[/quote]

Hideously expensive??? :P

Correct usage?

What do you mean by angle between two times? Are you talking about the angle between the hour hands?

Yeah. between the minute and hour havd is a bit easy

[quote="biffanddoc"]Heres what I think it would be.

Each hour adds 30 degrees.

Each minute adds 1/60 of the distance between the numbers.

Thus

30h+.5m
If h is the hour and m is the minute.
This works if it is the angle of the 12oclock and the time we gave.

Thus we can do
$|30(h_1-h_2)+.5(m_1-m_2)|$
to get the angle between the two times.

Here would be a calculator program (not sure if it works you should try)

[code]
ClrHome
Disp "Time 1"
Input "Hour",A
Input "Minute",B
Disp "Time 2"
Input "Hour",C
Input "Minute",D
abs(30!!!!!!!!!!!!!!(A-B)!!!!!!!!!!!!!!+.5!!!!!!!!!!!!(C-D)!!!!!!!!!)->E
If E>180
360-E->E
Disp E
[/code][/code][/quote]

Don't you mean $(A-C)$ and $(B-D)$?

Can you post a program that can calculate the number of degrees formed by the hour hand and minute hand of a clock?

Like: Find the number of degrees formed by an analog clock at 6:45.

the hour hand is at
$30h+.5m$
and the minute hand is at $6m$

so $|30h-5.5m|$

ok

[code]
ClrHome
Input A,"Hour "
Input B,"Minute "
abs(30A-5.5B)->C
If C>180
360-C->C
Disp C
[/code][/code]

[quote="ch1n353ch3s54a1l"]Can you use calcs for countdown? I would guess not, but...[/quote]
You cannot use calculators for the countdown or sprint round. They may be used only on the target and team rounds. That's the rules. :|

i wish you could for countdown. then it would be a race of who can enter into a calculator faster :rotfl:  :rotfl:

Or who can notice a non-calculator method faster.
Or who's memorized their factorials/pythag triplets/etc.

If you want an all-around good TI-83/84 program, try Allmath

http://62.65.69.4/pub/81/basic/math/allmath.zip

[quote="defiance"]My calculator has an abundance of programs, and here are the ones that I've found more useful.

BaseConverters
Clock Angles
Polynomial Manipulation
Prime Factorizer
Prime List
Triangle's Area SSS/SAS (Heron's Formula and something else)

Anything else can be done quickly without a specific program... During practice you should do the problems the hard way, and only use the programs during the real contest or to check.[/hide][/quote]

What do you mean by "Polynomial Manipulation"? and also, did you program these all yourself or did you download them?


Let $ A$ be an $ n\times n$ matrix of F such that for all nonzero $ n \times 1$ vectors $ B$ (where nonzero here means not all elements are zero), $ AB$ is nonzero. Is it true that $ A$ is invertible in F where F is $ \mathbb{Z}_p$? where F is finite field? Why?

Yes, if $ \text{nul}(A)\equal{}\{0\},$ then $ A$ is invertible. This result is true over any field. This is the general theory of dimension and rank, and all of the theorems can be proved by row elimination.


by schur, $a^{4}+b^{4}+c^{4}\geq \sum{a^{3}b}-3abc(a+b+c)$.

so, we only have to prove that $\sum{a^{3}b}\geq \sum{a^{2}b^{2}}$, true by Muirhead!


Let $ a,b,c$ be non-negative reals , prove that the following inequality holds good -
$ \sum_{cyclic}\frac{a^2}{a^2\plus{}2(a\plus{}b)^2} \ge \frac{1}{3}$

no onw interested in this nice inequality?
i am sorry if it is famous or has been posted before. If so please provide the link!

[quote="geniusbliss"]Let $ a,b,c$ be non-negative reals , prove that the following inequality holds good -
$ \sum_{cyclic}\frac {a^2}{a^2 \plus{} 2(a \plus{} b)^2} \ge \frac {1}{3}$[/quote]
Let $ \frac{b}{a}\equal{}x,$ $ \frac{c}{b}\equal{}y$ and $ \frac{a}{c}\equal{}z.$ Hence, $ xyz\equal{}1$ and your inequlity is equivalent to
$ \sum_{cyc}(9x^2\plus{}2x\minus{}8xy\minus{}3)\geq0,$ which is obviously true.

@arqady
i accept your solution every solution is equal but seriously whenever u conclude that sum big cyclic or symmetric is obviously non-negative and u indicate this with an expression inside the bracket which is repeated with cyclic terms, Now my question is the one in the bracket non-negative and the others similarly or is the whole expression getting factored into sum of squares or something?

anyway friends wat abt easier solutions(for me solution that doesnt involve big expressions is nice as i amnt gud enuf to compute them :rotfl: )

[quote="arqady"][quote="geniusbliss"]Let $ a,b,c$ be non-negative reals , prove that the following inequality holds good -
$ \sum_{cyclic}\frac {a^2}{a^2 \plus{} 2(a \plus{} b)^2} \ge \frac {1}{3}$[/quote]
Let $ \frac {b}{a} \equal{} x,$ $ \frac {c}{b} \equal{} y$ and $ \frac {a}{c} \equal{} z.$ Hence, $ xyz \equal{} 1$ and your inequlity is equivalent to
$ \sum_{cyc}(9x^2 \plus{} 2x \minus{} 8xy \minus{} 3)\geq0,$ which is obviously true.[/quote]
[quote="geniusbliss"]... Now my question is the one in the bracket non-negative and the others similarly or is the whole expression getting factored into sum of squares or something?
[/quote]
Because $ x^2\plus{}y^2\plus{}z^2\geq3,$ $ x\plus{}y\plus{}z\geq3$ and $ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}xz\plus{}yz.$

substitute $ a\plus{}b\equal{}p$ andsimilarly define q,r and the inequality becomes a direct,weak by cauchy-scwartz in the engel-form


[quote="ckck"][hide="Not sure"]Since the sign is to the right of 12 plug it in to the first one and get $-5$. Now the sign is left of it. So use the other equation to get $-12$.[/hide][/quote]
That's right, and this question is really easy because you don't need to do any difficult procedures...

[quote="tridecalogism_lover"]For any # x, we are  told that x& = 7-x and &x= x-7. What is the value of &(12&) ? I'm usually really good at logic problems, and problem solving stuff, but this one stumped me~[/quote]
[hide]$(7-12)-7=\boxed{-12}$[/hide]

This isn't really a logic problem...
more like those funny symbol problems.

[hide] 12&=7-12=-5. Now, &-5=-5-7=-12[/hide]

[quote="bpms"][quote="tridecalogism_lover"]For any # x, we are  told that x& = 7-x and &x= x-7. What is the value of &(12&) ? I'm usually really good at logic problems, and problem solving stuff, but this one stumped me~[/quote]
[hide=" a more general solution"]&(x&)=(7-x)-7=-x[/hide][/quote]
I never would have noticed that! That would have helped a lot...


logarithms
function
real analysis
real analysis unsolved

Evaluate

$ \sqrt {\frac {1}{2}\cdot\frac {3}{4}\cdot\frac {7}{8}\cdot\frac {15}{16}\cdot\frac {31}{32}\cdot\frac {63}{64}\cdot\cdots}$

as far as i know, infinite products like $ \prod_{k\equal{}1}^\infty\;\left(1\minus{}\frac{1}{n^k}\right)$
where $ n\equal{}2,3,4,5...$ are not known to have closed forms.

[quote="misan"]as far as i know, infinite products like $ \prod_{k \equal{} 1}^\infty\;\left(1 \minus{} \frac {1}{n^k}\right)$
where $ n \equal{} 2,3,4,5...$ are not known to have closed forms.[/quote]

http://mathworld.wolfram.com/InfiniteProduct.html
#49

Note the difference: the product in the original post is taken over $ k$ with $ n$ fixed while the product in bookworm_vn post is taken over $ n$ with $ k$ fixed  :!:

[quote="hadjimurad"][quote="bookworm_vn"]So that the result is
[img]http://mathworld.wolfram.com/images/equations/InfiniteProduct/equation9.gif[/img][/quote]

Could you show us a proof? I didn't see one on mathworld.[/quote]
for $ k = 2$
we have
$ \prod _{k = 2}^{N} = (\frac {2^{2} - 1}{2^{2}})...(\frac {N^{2} - 1}{N^{2}}) = \frac {1}{2}\frac {N + 1}{N}$
and when $ n - > + oo$ we have
$ \frac {1}{2}\frac {N + 1}{N} - > \frac {1}{2}$
and in this case :
[quote="misan"]as far as i know, infinite products like $ \prod_{k = 1}^\infty\;\left(1 - \frac {1}{n^k}\right)$
where $ n = 2,3,4,5...$ are not known to have closed forms.[/quote]
let $ n = 2$then
let's consider
$ \prod_{k = 1}^{N}(1 - \frac {1}{2^{k}}) = \frac {\prod_{k = 1}^{N}(2^{k} - 1)}{2^{\frac {N(N + 1)}{2}}}$
now let's consider
${ \prod_{k = 1}^{N}(2^{k} - 1)}$
 we have
${ \prod_{k = 1}^{N}(2^{k} - 1)} = \exp({\sum_{k = 1}^{N}\ln (2^{k} - 1)}) = \exp(\frac {N(N + 1)}{2}\ln 2 + \sum_{k = 1}^{N}\sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p2^{pk}})$ now let's consider sum
$ \sum_{k = 1}^{N}\sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p2^{pk}}$ when $ N - > + oo$
we have
$ \sum_{k = 1}^{ + oo}\sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p2^{pk}} = \sum_{p = 1}^{ + oo}\sum_{k = 1}^{ + oo}\frac {( - 1)^{p}}{p2^{pk}} = \sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p(2^{p} - 1)}$
(I have no idea how to calculate last sum )
in finally we have
$ \prod_{k = 1}^{N}(1 - \frac {1}{2^{k}}) = \exp(\sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p(2^{p} - 1)} + o(1))$ and when $ N - > + oo$ we have

$ \prod_{k = 1}^{ + oo}(1 - \frac {1}{2^{k}}) = \exp(\sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p(2^{p} - 1)})$
I try to consider function $ f(x)=\sum_{p = 1}^{ + oo}\frac {x^{p}}{p(2^{p} - 1)}$ when $ -1\leq x\leq 1$ but...
I think  is enough  to find this sum$ \sum_{k = 1}^{ + oo}\frac {1}{2^{k} - 1}$ and after we can find this sum $ \sum_{p = 1}^{ + oo}\frac {( - 1)^{p}}{p(2^{p} - 1)}$


geometry
perimeter
circumcircle
trigonometry
function
limit
congruent triangles

A rancher is adding a corral to his barn so that the barn opens directlyinto his corral. He has 195 feet of fencing left over from another project.  Find the greatest possible area for his corral using the length of the fence.

Circle has the largest area among all geometrical figures so he should have a circular fence.

I know that circle is the correct answer, but how can you prove it? Does anyone know?

[quote="Altheman"]consider an inscribed polygon

$A = 1/2 * a  * p$ where $a$ is the apothem
obviously the apothem maximizes when it becomes the length of the radius of the circle
the rest follows[/quote]

Be careful; the perimeter changes when you say "the rest follows". We're looking at the case where the perimeter is constant. Unless you meant something else. Please be more exact.

This is commonly known as the [url=http://en.wikipedia.org/wiki/Isoperimetry]isoperimetric problem[/url]

sorry i am a little lazy with the problems, but i suppose a more rigirous way to show it would be
let the perimeter be $P$, i will just do regular polygons... so a polygon has $n$ sides
dividing the polygon into congruent triangles by drawing the radii of the circumcircle
let the center be $O$ and 2 vertices of the polygon be $A$ and $B$
draw a perpenicular down from $O$ to $AB$ let that intersection be $D$
since there are $n$ sides, and $AD = BD$, $AD = \frac{P}{2n}$
Then $\angle AOD = \frac{360}{2n} = \frac{180}{n}$ because there are $n$ triangles that divide the 360 degrees of the circle, and $\angle AOD = \angle BOD$ (i did not show this but it is since $\triangle AOD = \triangle BOD$
then $\tan{\frac{180}{n}} = \frac{\frac{P}{2n}}{OD}$
$OD = \frac{P}{2n*\tan{\frac{180}{n}}}$
The area of the whole polygon then would be $n* \frac{1}{2}*OD*AB$ or $n*OD*AD$ since there are n congruent triangles
$A_{poly}=n*\frac{P}{2n*\tan{\frac{180}{n}}}*\frac{P}{2n}$
$A_{poly}=\frac{P^{2}}{4n*\tan{\frac{180}{n}}}$
and this function does infact maximize when $\lim_{n \rightarrow \infty}$, i have a really long way to show this, but i dont want to, it involves the fact that $x*\tan{\frac{180}{n}}$ approaces $\pi$ from above, and then switching it to the bottom, it would approach from below


How many different three-person committees can be formed from the 10 members of a club?

[quote="davidkim2106"][quote="firestar58"]Just do 10 C 3 which is 120.[/quote]

Um... Please be more specific for those people who are still learning basics... Please be a little more detailed.

Just saying "Just do 10 C 3 which is 120." does not, in any way, help a new student in math to learn.[/quote]

[hide="Solution"]
10C3 Simply means 10 Choose 3. This process is called combination.

$\frac{10\times 9\times 8}{3!}=\frac{10\times 72}{3\cdot 2\cdot 1}=10\times 12=\boxed{120}$

Formula:${{n}\choose{r}}=\frac{n!}{r!(n-r)!} $

An easy way to remember this is: There are $r$ number of terms in the numerator, going down by one each time, starting from $n$ because in the formula, it cancels out the factorial.

[/hide]

Just for clarification, we use combinations for problems that do not require a specific order, i.e. where order doesn't matter; in other words, a combination treats a order like ABC the same thing as CBA or BCA. This is why we have $(n-r)!$ in the denominator: to remove all redundant and equal orders/sequences.

If you come across a problem that treats a specific order of things differently from a different specific order of things, use a [url=http://mathworld.wolfram.com/Permutation.html]permutation[/url] instead.


trigonometry
pigeonhole principle
function

huh so use $ \tan(a\plus{}b) \equal{} \frac{1\minus{}\tan a \tan b}{\tan a\plus{}\tan b}$

dang I hope I remembered that right

anyway go do some real problems instead

Silly boy

$ \tan{\left(a\plus{}b\right)}\equal{}\frac{\tan a\plus{}\tan b}{1\minus{}\tan a\tan b}$

that sucks

that is why I'm "not_trig" :wink:

ok maybe I should remember that

yeah I would usually check it with easy values or just divide sin/cos ... w/e

yeah it takes like 15 seconds to derive it so remembering it is silly, and as long as you know the gist of the formula to avoid getting showed by problems that use it trickily...like

Prove that, among any seven real numbers , there exist two such that
\[ 0\le\frac {y_i \minus{} y_j}{1 \plus{} y_iy_j}\le\frac {1}{\sqrt {3}}.
\]

oh man the uber-famous pigeonhole+tangent formula one :wink:  :wink:

Canada MO like 1984 or something like that
i saw it in AoPS V2
basically what not_trig said: pigeonhole+tangent function

Find a closed-form formula for this recursion:
$ x_0 \equal{}2$
$ x_1 \equal{}x$
$ x_n \equal{}x_{n\minus{}1}(x_{n\minus{}2}^2\minus{}2)\minus{}x_1$
In terms of x and n.


[quote="dgreenb801"]Show that for x,z $ \ge 0$ (but it also holds for all reals)
$ 3 \sqrt {3} (x^2 \plus{} 1)(z^2 \plus{} 1) \ge 8(x \plus{} z)$
Is there an easier way without using the discriminant?[/quote]
I don't sure that the following way is easier.
$ (x^2 \plus{} 1)(z^2 \plus{} 1)\equal{}\left(x^2\plus{}\frac{1}{3}\plus{}\frac{2}{3}\right)\left(\frac{1}{3}\plus{}y^2\plus{}\frac{2}{3}\right)\geq$
$ \geq\left(\frac{x}{\sqrt3}\plus{}\frac{y}{\sqrt3}\plus{}\frac{2}{3}\right)^2\geq\frac{8(x\plus{}y)}{3\sqrt3},$ where the first inequality is Cauchy-Schwartz
and the last is equivalent to $ \left(\frac{x}{\sqrt3}\plus{}\frac{y}{\sqrt3}\minus{}\frac{2}{3}\right)^2\geq0.$

Yes, it is much easier! Of course, $ (a\minus{}b)^2 \ge 0$ is equivalent to $ (a\plus{}b)^2 \ge 4ab$, from which it is easier to see the last step. I got this problem from the more general problem (which was part of an even more general problem!)
Show that if $ xy\plus{}yz\plus{}xz\equal{}1$ and $ x,y,z \ge 0$, then
$ (x^2\plus{}1)(y^2\plus{}1)(z^2\plus{}1) \ge \frac{64}{27}$

Well, in the more general problem, we can just let $ x\equal{}cotA$ $ y\equal{}cotB$ $ z\equal{}cotC$ for some acute triangle ABC, with interior angles $ A,B,C$ then have to show, using the identity $ cot^2X\plus{}1\equal{}csc^2X$,

$ 27/64\ge sin^2(A)sin^2(B)sin^2(C)$

$ sin(A)sin(B)sin(C)\le \frac{3\sqrt{3}}{8}$

$ ln(sin(A))\plus{}ln(sin(B))\plus{}ln(sin(C))\le 3ln(\sqrt{3}{2})$

Which follows from the concavity $ f(x)\equal{}ln(sin(x))$

$ f'(x)\equal{}cot(x)$

$ f''(x)\equal{}\minus{}csc^2(x)\le 0$.


How many ways are there to draw n-3  diagonals of regular n-gon such that none of them intersect with each other, and each of triangles have a common segment with n-gon.

No -- the Catalan numbers count triangulations without the restriction
[quote="Amir.S"]each of triangles have a common segment with n-gon.[/quote]

So, this sequence begins 1, 2, 5, 12, ... instead of 1, 2, 5, 14.  (We lose the two triangulations of a hexgon that together make the Star of David.)  I think it has appeared before on the forum, but I can't find it.  It's [url=http://www.research.att.com/~njas/sequences/A045623]A045623[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Edit:  Ah, here it is: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143776


logarithms
factorial
LaTeX
modular arithmetic
calculus
number theory
prime factorization

Here are problems from previous AOPS posts, I think they are from AOPS competitions where answers are sent privately so no solutions are available.  I have solved many problems but here are some which I had difficulty. I hope you can help me.

1. $\frac{4}{98} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ with $x$, $y$ and $z$ being positive integers.  Find $\sqrt\frac{yz}{z^2}$.

2. $(\log_2 128 + \log_2 64 + ... + \log_2 2)!$ has how many prime factors.

3.  The number $2769x4x$ is divisible by $7$. How many numbers in this format exist?

4. If $(-2,7)$ is the maximum point of the graph $y = -2x^2 + 4ax + k$, what is $k$?

5. Calculate for b if $a^6 - b^6 = 12746944$.

Thanks.

For the first question,since I feel lazy,i'll just give the idea of what I would have done if the L.H.S was convenient.
Bring the equation in the form:-

$\frac{1}{x}+\frac{1}{y}=\frac{a}{b}$

And then do casework. :roll:

[quote="hydro"][hide="4"]
$7=-8-8a+k \rightarrow k=15+8a$
$f'(x)=-4x+4a$ $f'(-2)=0 \rightarrow 8+4a=0 \rightarrow a=-2$
$k=-1$[/hide][/quote]

Since some of us don't know calculus... (that IS calculus, right?)
[hide="4"] The vertex is at $-\frac{4a}{-4}=-2$, so $a=-2$. Plugging $x=-2$, $y=7$, and $a=-2$ into the original equation, $7=-2(-2)^2+4(-2)(-2)+k$. Solving this, we get $k=\boxed{-1}$ :D [/hide]

[quote="guile"][quote="t0rajir0u"]Some of these questions are annoyingly tedious.

[hide="2"]$\log_2 128 + ... + \log_2 2 = 7 + ... + 1 = 28$
$\pi(28) = \boxed{9}$[/hide][/quote]

You're right, it will take a lot of time before you can solve it.  Problem no 2 has a factorial outside the parenthesis.[/quote]

I'm aware of that.

[hide="Explanation"]
$\pi(n)$ is the Prime Counting Function.  It counts the number of primes less than or equal to $n$ - precisely the number of distinct prime factors of $n!$.[/hide]

[quote="10000th User"]guile, isn't time for you to learn $\LaTeX$?

[hide="3"]$2769x4x\equiv 2(1)+7(-2)+6(-3)+9(-1)+2x+4(3)+x\equiv -27+3x\equiv 0\pmod{7}$

$x=9,2$[/hide]

For #1, there is/are missing condition(s), like $x<y<z$, otherwise you can get multiple values for it. Even with the condition I wrote, you still get 2 different values.[/quote]

I hope I can learn it during vacation.  I am not a student, so I don't have a lot of spare time like you guys.  If I only have time...

[b]HOW ABOUT NO. 5[/b]

[quote="surge"]but that's working with the answer....[/quote]

It's a terrible problem, but I'll try working from the factorization :P

[hide="Possible approach"]
Assume $769 = x^2 + xy + y^2$.  Note that $7 \cdot 37 = 259$.  Conveniently, if we set $259 = x^2 - xy + y^2$, their difference is $510 = 2xy = 2 \times 17 \times 15$, strongly suggesting $x = 17, y = 15$ (which happens to make $x^2 - y^2 = 64$).  :D
[/hide]

Anyone figured a sweet solution to number 1?


[hide]The radius is 21 inches so the area of the base is $441\pi$.  Multiply that be the height (40) and you get the volume.

15 tons is $2000\times 15=30000$.

Take the volume and divide that by 30000 and you will get the answer (hopefully).

[hide="Two reasons why I didnt calculate the answer"]1. I don't have my calculator.
2. I'm too lazy to go find one.  :D [/hide]
 [/hide]

Shouldn't there be a calculator on your computer? :D

[hide]
I have no idea what a representative column is but its volume would be $21^2\pi\cdot40=1764\pi$.  I think a ton is 2000 pounds, so the weight is $15\cdot2000=30000$ pounds.  Divide that by $1764\pi$ and you should get the answer.

So it's $5$, I think.
[/hide]

[hide]
It said 42 [i]inches[/i] so...

The volume would be $\left(\frac{21}{12}\right)^2\pi\cdot40=122.5\pi\approx384.845$.

Dividing 30000 by that, you get $77.953461783315360729644402291832\approx78$

Hope that's right.
[/hide]

[hide="just stuff all the numbers into your calculator"]Divide the number of pounds by the volume of the column and you'll get the answer.$30000/(21/12)^2*pi*40=about 78$.
[/hide]

Congrats! :thumbup: You guys got it right. I made the same mistake when I did that problem.

[quote="goldendomer"]A representative column has a diameter of 42 inches, height of 40 feet, and weighs 15 tons. What is the mean number of pounds per cubic foot? Round your answer to the nearest whole number.[/quote]

[hide]21^2 pi *40/15*2000 I think Im too lazy to figure it out[/hide]


[hide]
ok so l:w is 3:2
since l=12 then 2/3 of 12 is $\boxed{9}$
EDIT:What was i thinking??????????
[/hide]

[quote="pascal_1623"]The ratio of the length of a rectangular floor to its width is 3:2. If the length of the floor is 12 meters, what is the perimeter of the floor in meters?[/quote]

[hide]l to w is 3 to 2 the l is 12 so the w has to be 12/3=4*2=8
The perimeter is (2*12)+(2*8)=24+16=40.[/hide]


I found this problem on a website I cannot quite recal, anyways here it is.

At a party someone is curious to know what the total salary of eveyone in the room is. The problem is, no one wants to disclose their own salary to anyone else in the room. How do you solve this?

Does it have something to do with using modular arithmetic? I remember a similar problem that did.

I believe that this was a Mathpath problem of the day before.

I think the answer is:

[hide]Make everyone in the room write their salaries on a piece of paper and put it all in a hat. Then mix it all up and since you can't tell who is who's just add them all up!  :) [/hide]

Right?

[quote="jli"]drunken_math!!! how could you forget

solution:

[hide]ask everyone to write down 2 numbers on seperate papers.

the numbers will add up to he salary

put them in a hat and add them[/hide][/hide][/quote]

Ooopth!  :blush:

Well, mine's based on that answer....

Really? The one I heard was:

[hide]
Write down a big number on a piece of paper. Pass it to someone, tell them to add their salary to the number and write the new number on a new paper and pass it around telling the next person the same instructions. Once everyone has added, just subtract your big number and then you have the total salary
[/hide]

I guess there are lots of solutions  :D
I can think of a lot more...

[quote="jli"]ask everyone to write down 2 numbers on seperate papers.

the numbers will add up to he salary
[/quote]

How's that any different than Drunken_Math's solution?  ;)

based on this one:

a pilot needs to know the weight of 3 obese people.

they are extremely embarrased about their waits.

if they put only one number in, the captain can geuss their waits

however with 2 numbes each, the pilot will have no idea

Amputate various body parts so they become lighter.


I almost got the answer and ierno how to finish it..

Question is

For some reak k, the roots of 2x³ - 7x² + kx - 2 form a geometric progression. Find K

I'd be very thank you if u tell me how to solve it  :D

We can set the roots of the given cubic equation as $ a,\ ar,\ ar^{2}\ (a: first\ term\ , r: common\ ratio)$. Use the formula of Vieta, we have ........


[quote="xxxyyyy"]A $100 x 100$ square grid (with columns and rows labelled with letters) is filled with the numbers $1$ to $10000$ in the following way:
$1.$ The number $1$ is placed in any arbitary cell
$2.$ The number $2$ is placed in the row whose letter is the same as the column containing $1$
$3.$ The number $3$ is placed in the row whose letter is the same as the column containing $2$
$\cdot$
$\cdot$
In general, if $N$ is in the $Kth$ column then $N+1$ is in the $Kth$ row taking care that one does not go into an endless loop, it is always possible to fill in the grid

What is the difference between the sum of the numbers in all the cells of the row containing $1$ and the sum of the numbers in all the cells of the column containing $10000$?[/quote]

No attempts for this?


factorial
inequalities
inequalities proposed

[quote="kunny"]Prove the following inequality.

\[\left(\frac{n+1}{2}\right)^n>n!\ \ (n=2,3,\cdots)\][/quote]

AM-GM :D

$n!=1\times 2\times 3...\times n\leq \left(\frac{1+2+3+...+n}{n}\right)^n=\left(\frac{n+1}{2}\right)^n$

But since $n=2,3..$ we have the desired inequality .

 :)

Using $(x+k)(x-k)\leq x^2$,
$1\cdot n$, $2\cdot (n-1)$, ..., $n \cdot 1$ are all less than $\left( \frac{n+1}{2} \right )^2$

Then the result follows.


If A is an N x N matix and the determinant of A does not equal 0, assuming the existence of a left inverse, show that one exists.

Thanks for your help.

Should it be the case that $ det A \not\equal{} 0$ then $ A$ is invertible and hence, left inverse is also right inverse ?

[quote="mathangel"]assuming the existence of a left inverse, show that one exists.[/quote]That sounds like a tautology. "If P then P" is trivial.

Try fixing the statement.


modular arithmetic
inequalities unsolved
inequalities

prove that no square exists in the form of  11,111,1111,11111,.........

hey wrong place to post the question
just go thru the forum once since u r a newbie take care of that next tym.
[hide="Solution"]
all the numbers in the given series are of the form $ 4k \plus{} 3$ .Consider any number $ l$,$ l \equiv 0,1,2,3\pmod 4 \implies l^{2} \equiv 0,1 \pmod 4$  hence no number in teh series is a square since each of them is -1 mod 4[/hide]


I was wondering, what are the best prep books out there for the Math SAT IIC? I will be taking it "officialy" this May (as of yet I have only done several practice tests, getting around a 720-780 range). I really would like to get a perfect score on this test, and I was looking at different books, but I am sill not sure. One book that I really liked was the latest (Revised & Updated) [url=http://search.barnesandnoble.com/booksearch/isbninquiry.asp?userid=v12cnlgARz&cds2Pid=6323&isbn=1411402448]Sparknotes IIC Book[/url]. What other reccommendations are there?

i've always been a fan of princeton review but im gonna tell you, you probaly wont need a review. getting an 800 (perfect score) puts you in the 90th percentile range. which basically means, ten percent of the people that take it score a perfect. not a problem.

I assume you have the book "Real SAT II", which has one official test for Math 2C.  Two previous editions of that book each contain a different official test, but those editions are out-of-print and so are harder to find.  (You can find folks who sell them on Amazon.)

Yes, the SparkNotes book for Math 2C is quite good; the practice tests there are about the same level of difficulty as the real thing.  (The previous edition had several typos, but they seem to have fixed most in the revised edition.)  SparkNotes used to publish a book called "5 More Practice Tests for the SAT II Math IIC", but I think they may have stopped selling it.  If you sign up for the SparkNotes online course, I think you can get access to those extra tests.  Again, they are good tests at the right level.

Yep, they still sell the "5 More Practice Tests for the SAT II Math IIC" or whatever, because I ordered that book and the other Sparknotes book from Barnes and Noble. I am a sophmore right now, so I probably won't need the older version of the 10 Real SAT's (because of the new SAT), so I think I will need the new guide published from the College Board as well as the Real SAT II's, because I don't have that either...

Any other reccomendations that people have?

I am also looking to take the Math IIC. Here is a good/great book:

10 Realistic SAT II Math IIC Test Booklets

Its 29.99 on Amazon.


Hello all! I tried to solve the following problem:

dy/dt = -2y + 10

where y(0) = y_0

The book says that the correct answer is y = (y_0 - 5)e^(-2t)

But I got a different solution :maybe:

Please: See the steps I took:

http://www.mathbin.net/8266

What did I do wrong?

Thank you very much! Bye!

I usually don't check work posted off this site, so I don't know what you did.

This first-order differential equation is autonomous, which implies that it is separable, and linear. Since it is both separable and linear, it can be solved by two different methods: separation of variables, or by using the multiplying factor method approriate for  linear equations. When I teach this, I recommend that, given that choice, you use the multiplying factor method and solve it as a linear differential equation. It just seems, as a human matter, that fewer mistakes happen that way.

Since the equation is autonomous, we can examine it for stationary points and stability. We have $y'=-2y+10.$ If $-2y+10=0,$ then we have a constant solution. That happens only for $y=5.$ By the uniqueness portion of the existence-uniqueness theorem, other solutions cannot cross this one. Hence, if $y_{0}>5,$ then $y>5$ for all $t$ and if $y_{0}<5,$ then $y<5$ for all $t.$ If $y>5,$ then $y'=-2y+10<0,$ so $y$ is decreasing - and that has to be decreasing towards $5.$ If $y<5,$ then $y'=-2y+10>0,$ so $y$ is increasing - increasing towards $5.$ Put that together, and we get that for any initial condition, $\lim_{t\to\infty}y(t)=5.$

The "book" solution satisfies all of these conditions. Does yours?

[quote="nohope"]The book says that the correct answer is y = (y_0 - 5)e^(-2t).[/quote]

This is wrong. The correct answew is $y=(y_{0}-5)e^{-2t}+5.$


geometry
geometric transformation
reflection
circumcircle
Euler
trigonometry
projective geometry

[quote="jhaussmann5"]Let $A'$, $B'$, $C'$ be the feet of the altitudes of triangle $ABC$, let $D = A'B' \cap CC'$ and $E = A'C' \cap BB'$.  Let $O_a$ be the reflection of circumcentre $O$ in side $BC$.  Prove that $AO_a \perp DE$.[/quote]

This one is funny...

First, let me quote something from my website (knowing that it's not accessible from all over the world...):

[quote="Lemma 1 from [url=http://de.geocities.com/darij_grinberg/Schroeder/Schroeder.html#6]my Schröder points database, [Schroeder6][/url]"][b]Lemma 1.[/b] Let H be the orthocenter of a triangle ABC, and X, Y, Z be the feet of the altitudes from A, B, C. The sidelines YZ, ZX, XY of triangle XYZ intersect the sidelines BC, CA, AB of triangle ABC at the points O, P, Q, respectively. Then the points O, P, Q lie on one line - the so-called orthic axis of triangle ABC -, and this line is perpendicular to the Euler line of triangle ABC.

[i]Proof.[/i] Let o be the circumcircle and n the nine-point circle of triangle ABC. The circumcircle o passes through A, B, C, and the nine-point circle n passes through X, Y, Z.

Since the points Z and X lie on the circle with diameter CA, we have $PZ \cdot PX = PC \cdot PA$, and hence the point P has equal powers with respect to the circles n and o (in fact, $PZ \cdot PX$ is the power of P with respect to n, and $PC \cdot PA$ is the power of P with respect to o). Hence, P lies on the radical axis of n and o. Similarly, O and Q also lie on this radical axis; altogether, the points O, P, Q lie on one line, namely the radical axis of n and o. Moreover, this line is perpendicular to the Euler line of triangle ABC, since the radical axis of two circles is perpendicular to the central line, while the centers of n and o lie on the Euler line of triangle ABC. Lemma 1 is proven.[/quote]

Now, let H be the orthocenter of triangle ABC. Consider the triangle BHC.

- The altitudes of the triangle BHC are BC', HA' and CB' (since $BC^{\prime}\perp HC$, $HA^{\prime}\perp CB$ and $CB^{\prime}\perp BH$). Their feet are C', A' and B', respectively. The sidelines A'B' and C'A' of triangle C'A'B' intersect the sidelines HC and BH of triangle BHC at the points D and E, respectively. Thus, the points D and E lie on the orthic axis of triangle BHC. In other words, the line DE is the orthic axis of triangle BHC.

- Since the altitudes of the triangle BHC are BC', HA' and CB', the orthocenter of triangle BHC is their point of concurrence, i. e. the point A.

- The circumcenter of triangle BHC is $O_a$. In fact, it is well-known that the reflection of the orthocenter H of the triangle ABC in its side BC lies on the circumcircle of triangle ABC. Upon reflection with respect to the line BC, this becomes: The orthocenter H lies on the reflection of the circumcircle of triangle ABC in its side BC. Of course, the points B and C also lie on the reflection of the circumcircle of triangle ABC in its side BC. Thus, the circumcircle of triangle BHC is the reflection of the circumcircle of triangle ABC in its side BC. Hence, the circumcenter of triangle BHC is the reflection of the circumcenter of triangle ABC in its side BC, i. e. the reflection of the point O in the side BC, i. e. the point $O_a$.

- The Euler line of triangle BHC is the line joining its orthocenter A with its circumcenter $O_a$; in other words, it is the line $AO_a$.

So the orthic axis of triangle BHC is the line DE, and its Euler line is the line $AO_a$. Since the Lemma 1 quoted above states that the orthic axis of a triangle is perpendicular to its Euler line, we conclude that $DE\perp AO_a$. That's all.

  darij

I posted this one some time ago http://www.mathlinks.ro/Forum/viewtopic.php?t=50503 . (on the pre-olympiad board)

It was taken from my 9th grade textbook and to me the point (c) seemed almost impossible. Anyway, many of the  notions Darij used in his (nice) solution are not even taught in school. So it makes me wonder: Is there a more "elementary" solution?

Maybe someone will find it... To me it seems very weird that they would put a problem which requires serious geometry knowledge in a textbook.

Here is my solution (?) :

  O and H are isogonal conjugates.  Thus reflections in the sides of ABC of one

  lay on a circle centered in the other.

  Lines AOa, BOb, COc  intersect at 9 PC  center.  The DE-type lines are radical axis

  of 9 PC and circles which are reflections of circumcircle in the sides of ABC.

  Thank you.

   M.T.

[hide="Here is some details of the Darij-s proof and the his addressses."] [color=red]Lemma 1.[/color] Let $H$ be the orthocentre of the triangle $ABC$. For a point $L$ denote the points: $M\in BC$, $HM\perp AL$; $N\in CA$, $HN\perp BL$; $P\in AB$, $HN\perp CL$. Then the points $M$, $N$, $P$ are collinearly and $HL\perp \overline {MNP}$.
[color=green]Remark.[/color] The line $\overline {MNP}$ is the [u]trilinear axis[/u] of the point $L$ w.r.t. the triangle $ABC$.

[b]Proof.[/b] Denote: the circumcirle $w=C(O,R)$ of the triangle $ABC$; the power $p(X)$ of the point $X$ w.r.t. the circle $w$; the projections $D$, $E$, $F$ of the orthocentre $H$ on the sidelines $BC$, $CA$, $AB$ respectively; the reflections $D'$, $E'$, $F'$ of the orthocentre $H$ w.r.t. the lines $BC$, $CA$, $AB$ respectively. It is well-known that $\{D',E',F'\}\subset w$.
$MH\perp AL\Longleftrightarrow MA^2-ML^2=HA^2-HL^2\ \ (1).$
$MD\perp HA\Longleftrightarrow MH^2-MA^2=DH^2-DA^2\ \ (2).$
$(1)\ \wedge \ (2)\Longrightarrow MH^2-ML^2=(MH^2-MA^2)+(MA^2-ML^2)=$
$(DH^2-DA^2)+(HA^2-HL^2)=HA^2+HD^2-(HA+HD)^2-HL^2=$
$-2HA\cdot HD-HL^2=-HA\cdot HD'-HL^2=-p(H)-HL^2\Longrightarrow$
$MH^2-ML^2=R^2-OH^2-LH^2$ (a symmetrical expression w.r.t. the triangle $ABC$).
Therefore, $ML^2-MH^2=NL^2-NH^2=PL^2-PH^2=LH^2+OH^2-R^2$,
i.e. $P\in MN$ and $HL\perp \overline {MNP}.$

[color=red]Lemma 2.[/color] Denote: The centroid $G$ of the triangle $ABC$; the intersections $X\in BC\cap EF$, $Y\in CA\cap FD$, $Z\in AB\cap DE$. Then $XH\perp AG$, $YH\perp BG$, $ZH\perp CG$  and $HG\perp \overline {XYZ}.$

[b]Proof.[/b] Denote: the middlepoint $S$ of the side $[BC]$; $x=m(\widehat {HXD})$ and $y=m(\widehat {DAS})$.
Thus, $\tan x=\frac{HD}{XD}$, $\tan y=\frac{DS}{AD}$ and  $XH\perp AG\Longleftrightarrow \tan x=\tan y\Longleftrightarrow$
$DA\cdot DH=DX\cdot DS\Longleftrightarrow DA\cdot DD'=DX\cdot DS\Longleftrightarrow DB\cdot DC=DX\cdot DS$,
what is true, because $\frac{XB}{XC}=\frac{DB}{DC}$, i.e. the division $(B,C,X,D)$ is harmonically.
For $L: =G$ from the [color=red]Lemma 1[/color] results the [u]orthic axis[/u] $\overline {XYZ}$ is perpendicularly on the [u]Euler's line[/u] $HG$ of the triangle $ABC$.

[b]Remark.[/b] See the property $A4$ from my topic and reply
http://www.mathlinks.ro/Forum/viewtopic.php?t=46146[/hide]

armpist , your solution is correct, but explain more detailed why the $DE$ type lines are radical axis, because it doesn`t trivially result from what you said.

Let us to present another solution.

We denote as $P,$ the intersection point of $BC,$ $B'C'$ and it easy to show that the segment line $DE,$ passes through this point.

Because of $OO_{a}\parallel AA'$ and $ON = NO_{a},$ where $N,$ is the midpoint of the segment $BC,$  we conclude that the pencil $A.OO_{a}NA',$ is in harmonic conjugation.

Also we can easy to prove that the pencil $P.B'DHA',$ is in harmonic conjugation too.

We see now that $AO\perp PB',$ $($ because the segment line $B'C',$ is antiparallel to $BC,$ with respect to the angle $\angle A$ of $\bigtriangleup ABC$ $).$

Also is true that $AN\perp PH$ where $H,$ is the orthocenter of $\bigtriangleup ABC.$  $($ See at http://www.mathlinks.ro/Forum/viewtopic.php?t=110887 $).$

And of course we have that $AA'\perp BC.$

So, the pencils $A.OO_{a}NA',$ $P.B'DHA',$ both in harmonic conjugation, have three their homologous rays formed equal angles $($ easy to show from the above perpendicurarities $)$ and hence their homologous rays $AO_{a},$ $PD,$ forme also equal angles, with the other ones.

We conclude now, that $AOa\perp PD$ and the proof is completed.

Kostas Vittas.

Your proof, vittasko, is simply beautiful. Cheers!

[quote="Virgil Nicula"] [color=olive]Let $H$ be the orthocentre of the triangle $ABC$. For a point $L$ denote the points: $M\in BC$, $HM\perp AL$; $N\in CA$, $HN\perp BL$; $P\in AB$, $HN\perp CL$. Then the points $M$, $N$, $P$ are collinearly and $HL\perp \overline{MNP}$.[/color]
[/quote]
[color=indigo][b]Proof.[/b] Denote: $A_{1}=AH\cap BC,\,B_{1}=BH\cap CA,\,C_{1}=CH\cap AB.$
We have:
$\overrightarrow{MP}\cdot\overrightarrow{LH}=\left(\overrightarrow{HP}-\overrightarrow{HM}\right)\cdot\overrightarrow{LH}=\overrightarrow{HP}\cdot\overrightarrow{LH}-\overrightarrow{HM}\cdot\overrightarrow{LH}\\ =\overrightarrow{HP}\cdot\left(\overrightarrow{CH}-\overrightarrow{CL}\right)-\overrightarrow{HM}\left(\overrightarrow{AH}-\overrightarrow{AL}\right)=\overrightarrow{HP}\cdot\overrightarrow{CH}-\overrightarrow{HM}\cdot\overrightarrow{AH}\\ =\overline{HC_{1}}\cdot\overline{CH}-\overline{HA_{1}}\cdot\overline{AH}=0\Longrightarrow MP\perp LH$

Similar: $MN\perp LH\Longrightarrow N\in MP$ and $HL\perp\overline{MNP}$[/color]

[quote="SpongeBob"][quote="vittasko"]Let us to present another solution.

We denote as $P,$ the intersection point of $BC,$ $B'C'$ and it easy to show that the segment line $DE,$ passes through this point.
[/quote]

How exactly do you prove this, in a way that isn't hard? I proved this (and the problem from the beginning) using inversion, but it isn't so easy.[/quote]

It is easy to show that the segment line $DE,$ passes through the point $P,$ because of the triangles $\bigtriangleup AB'C'$ and $\bigtriangleup HDE$ are

perspective $($ $A'\equiv AH\cap B'D\cap C'E$ $)$ and hence, the conclusion it follows.

Kostas Vittas.

PS. Also, it is easy to show that the triangles $\bigtriangleup B'DC$ and $\bigtriangleup C'EB$ are persepective, because of the collinearity of $A,$ $H,$ $A'.$

$\bullet$ Another way is to show that the points $P,$ $P',$ are harmonic conjugates with respect to $B',$ $C',$ where $P'\equiv AH\cap B'C'$
$($ from the complete quadrilateral $AB'HC'BC$ $).$

So, the segment line $DE,$ as the diagonal of the complete quadrilateral $A'DHEB'C',$ passes through the point $P.$

Oh, yes, of course...
Anyway, Desargues theorem is not so "elementary" so I think it should not be skiped :)
Thanks!
Bye

[quote][color=darkred][b][u]Enunciation.[/u][/b] Let $A'$, $B'$, $C'$ be the feet of the altitudes of the triangle $ABC$.
Denote  the intersections $D\in A'B' \cap CC'$ and $E\in A'C' \cap BB'$.
Let $O_{a}$ be the reflection of the circumcenter $O$ in the line $BC$.  Prove that $AO_{a}\perp DE$.[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Prove easily that the point $O_{a}$ is the circumcenter of the triangle $BHC$ and the point $A$ is the orthocenter of the same triangle.

Denote the circumcircle $w=C(S)$ of the triangle $A'B'C'$ (the common Euler's circle of the triangles $ABC$, $BHC$, $CHA$, $AHB$)

and the intersection $\{U,V\}=DE\cap w$. Thus, $S\in AO_{a}$, i.e. the line $\overline{ASO_{a}}$ is the Euler's line of the triangle $BHC$ ($SA=SO_{a}$)

and $\left\{\begin{array}{ccc}EH\cdot EB=EC'\cdot EA'=EU\cdot EV & \Longrightarrow & EH\cdot EB=EU\cdot EV\ ,\ i.e.\ B,H,U,V\ are\ cyclically\\\ DH\cdot DC=DB'\cdot DA'=DU\cdot DV & \Longrightarrow & DH\cdot DC=DU\cdot DV\ ,\ i.e.\ C,H,U,V\ are\ cyclically\end{array}\right\|$ $\Longrightarrow$

the points $B,C,H,U,V$ are cyclically $\Longrightarrow$ the points $U,V$ belong to the circumcircles of the triangles $BHC$ and $A'B'C'$ $\Longrightarrow$

$UV\perp O_{a}S$ $\Longleftrightarrow$ $\overline{UEDV}\perp \overline{ASO_{a}}$ $\Longleftrightarrow$ $DE\perp AO_{a}$.[/color]

This is in fact armpist's solution, although he didn't prove that DE is radical axis for those circles :)


modular arithmetic
function
number theory proposed
number theory

Find all primes $ p,q $ such that $ \alpha^{3pq} -\alpha \equiv 0 \pmod {3pq} $ for all integers $ \alpha $.

Obviously $3$, $p$ and $q$ have to be pairwise distinct. In fact, assuming the contrary and letting $p = q$ (for example), it would be $\alpha^{3p^2} \equiv \alpha \bmod p^2$, for any $\alpha \in \mathbb{Z}$. Anyway this is trivially checked to be false for $\alpha = p$. So suppose $p < q$ by simmetry. Then $p \neq 2$ and $3 < p < q$. In fact, letting $p = 2$ and $\gcd(\alpha, 3) = 1$, you'd find out $\alpha \equiv \alpha^{3pq} \equiv 1 \bmod 3$, which fails to be true for $\alpha \equiv 2 \bmod 3$.

Now let $\alpha_1$ and $\alpha_2$ be primitive roots $\bmod\;\! p$ and $\bmod\;\! q$, respectively, and argue $(q-1) \mid (3p - 1)$ and $(p-1) \mid (3q - 1)$, by properties of generators and multiplicative orders. Then let i) $3p - 1 = (q-1)u$ and ii) $3q - 1 = (p-1)v$, for suitable $u, v \in \mathbb{Z}^+$. Because $q \geq 7$, you need $1 \leq u \leq 3$, since otherwise $(q-1)u \geq 3q - 1 > 3p - 1$. But $u = 1$ and $u = 3$ are actually impossible values (that's trivial!), so necessarily $u = 2$ and $q = \frac{3p+1}{2}$. Subsequently, it follows $p = \frac{2v+1}{2v-9}$ from ii). You need $p \geq 5$ as well as $v \geq 5$. Moreover you know the function $p(\cdot): [5, +\infty[ \; \mapsto \mathbb{R}: v \mapsto \frac{2v+1}{2v-9}$ is strictly decreasing for $v \geq 5$. You check $p(5) = 11$ and $p(4) < 5$ by hand, concluding that $v = 5$ only fits. Hence you get $p = 11$ and $q = 17$ (or viceversa, by symmetrization) and everything works.

[quote="hitlEULER"]Obviously $ 3$, $ p$ and $ q$ have to be pairwise distinct. In fact, assuming the contrary and letting $ p \equal{} q$ (for example), it would be $ \alpha^{3p^2} \equiv \alpha \bmod p^2$, for any $ \alpha \in \mathbb{Z}$. Anyway this is trivially checked to be false for $ \alpha \equal{} p$. So suppose $ p < q$ by simmetry. Then $ p \neq 2$ and $ 3 < p < q$. In fact, letting $ p \equal{} 2$ and $ \gcd(\alpha, 3) \equal{} 1$, you'd find out $ \alpha \equiv \alpha^{3pq} \equiv 1 \bmod 3$, which fails to be true for $ \alpha \equiv 2 \bmod 3$.

Now let $ \alpha_1$ and $ \alpha_2$ be primitive roots $ \bmod\;\! p$ and $ \bmod\;\! q$, respectively, and argue $ (q \minus{} 1) \mid (3p \minus{} 1)$ and $ (p \minus{} 1) \mid (3q \minus{} 1)$, by properties of generators and multiplicative orders. Then let i) $ 3p \minus{} 1 \equal{} (q \minus{} 1)u$ and ii) $ 3q \minus{} 1 \equal{} (p \minus{} 1)v$, for suitable $ u, v \in \mathbb{Z}^ \plus{}$. Because $ q \geq 7$, you need $ 1 \leq u \leq 3$, since otherwise $ (q \minus{} 1)u \geq 3q \minus{} 1 > 3p \minus{} 1$. But $ u \equal{} 1$ and $ u \equal{} 3$ are actually impossible values (that's trivial!), so necessarily $ u \equal{} 2$ and $ q \equal{} \frac {3p \plus{} 1}{2}$. Subsequently, it follows $ p \equal{} \frac {2v \plus{} 1}{2v \minus{} 9}$ from ii). You need $ p \geq 5$ as well as $ v \geq 5$. Moreover you know the function $ p(\cdot): [5, \plus{} \infty[ \; \mapsto \mathbb{R}: v \mapsto \frac {2v \plus{} 1}{2v \minus{} 9}$ is strictly decreasing for $ v \geq 5$. You check $ p(5) \equal{} 11$ and $ p(4) < 5$ by hand, concluding that $ v \equal{} 5$ only fits. Hence you get $ p \equal{} 11$ and $ q \equal{} 17$ (or viceversa, by symmetrization) and everything works.[/quote]

Could somebody help me with the prove of (p - 1) / (3q - 1)?

let alpha be a primitive root mod q. then you do see why $ q\minus{}1|3pq\minus{}1$ right? (and then $ 3pq\minus{}1\equiv 3p\minus{}1 (\mod q\minus{}1)$)

[hide = Solution] First notice how $3, p, q$ are distinct. Assume $p = q$. Then $n^{3p^2} \equiv n \pmod{p^2}, \forall n \in \mathbb{Z^+}$. However if $n = p$, then we have that $p^{3p^2} \equiv p \pmod{p^2} \implies p^{3p^2 - 1} \equiv 1 \pmod{\frac{p^2}{\operatorname{gcd}(p, p^2)}} \implies 0 \equiv 1 \pmod p$, which is nonsense.

Now let $g_1, g_2$ be primitive roots modulo $p, q$ respectively. Then $g_1^{3qp} \equiv g_1 \pmod p \implies p - 1 \mid 3pq - 1$. Similarly $q -1 \mid 3pq - 1$. Thus we have that $p - 1 \mid 3q - 1$ and $q - 1 \mid 3p - 1$.
Let $3q - 1 = x(p - 1), 3p - 1 = y(q - 1)$, for some $x, y \in \mathbb N$. Now notice that $1 \leq x \leq 3$. If $x = 1$, we have that $3q - 1 = p - 1 \implies 3q = p$, which is absurd. If also $x = 3$, we have $3q - 1 = 3p - 3$. Reducing modulo $3$ gives us $2 \equiv 0 \pmod 3$. Another fallacy. Hence $x = 2$. Now then $3q - 1 = 2p - 1$ and $3p - 1 = y(q - 1)$. Solving for $q$ in the system of equations gives us $q = \frac{2y + 1}{2y - 9}$. Now $\frac{2y + 1}{2y - 9} = 1 + \frac{10}{2y - 9}$. Thus $y = 5, 7$. If $y = 5$ then $q = 11$ and $p = 17$. If $y = 7$, then $q = 3$ and $p = 5$. However $3, p, q$ are distinct, and so we can only have that $\boxed{(p, q) = (11, 17)}$ as the only solution. $\blacksquare$ [/hide]

WLOG $p\ge q,$ and note that if $q=2,$ then $3\nmid 2^{3\cdot 2\cdot p}-2.$ If $q=3,$ then $n^{9p}\equiv n\pmod{9}$ and setting $n=2$ yields $(-1)^p\equiv 2\pmod{9},$ which is absurd. Similarly, $p>3.$

We know $n^{3q}\equiv n^{3pq}\equiv n\pmod{p}$ so $g^{3q-1}\equiv 1\pmod{3}$ where $g$ is a primitive root modulo $p.$ Hence, $p-1\mid 3q-1$ since $\operatorname{ord}_p(g)=p-1.$ Similarly, we see $q-1\mid 3p-1$ so let $(p-1)\ell=3q-1$ and $(q-1)m=3p-1.$ Then, we have \[p\ell -\ell=3q-1\le 3p-1\implies \ell\le 3+\frac{2}{p-1}.\] Therefore, $\ell=1,$ $2,$ $3.$ If $\ell=1,$ then $p=3q,$ which is absurd. If $\ell=3,$ then we have a contradiction modulo $3$ so we need $\ell=2.$ Solving for $p$ in and substituting into the other equation, we have \[(q-1)m=3\cdot\frac{3q+1}{2}-1\implies q=\frac{2m+1}{2m-9}.\] In order for $2m+1\mid 2m-9,$ we need $2m-9\mid 10$ so $m=5,$ $7.$ Therefore, $q=3,$ $11$ and $p=5,$ $17,$ respectively. Since $p,q>3,$ our only solutions are $\boxed{(p,q)=(11,17),(17,11)}.$

Put $n=2$ in $3\mid n^{3pq}-n$ to get $p$, $q$ are both odd. Now if anyone of them $=3$ , say $p$, then $9\mid n^{9q}-n$, and choose a primitive root $g$ to get $g^{9q-1}\equiv 1\pmod{9}\implies 6\mid 9q-1$, contradiction. So, $p,q\not=2,3$. Now for a primitive root $g\pmod p$, we have $g^{3pq}\equiv g\pmod p\implies g^{3q-1}\equiv 1\pmod p\implies p-1\mid 3q-1$. Similarly choose a primitive root $\pmod q$ to get $q-1\mid 3p-1$.

So now we have $p-1\mid 3q-1$ and $q-1\mid 3p-1$. WLOG $p\le q$.

Now $\dfrac{3p-1}{q-1}\ge 5$ fails due to size reasons.

Now we use bounding from which we get for $\dfrac{3p-1}{q-1}=4$ that $q=3$, contradiction.

$\dfrac{3p-1}{q-1}=3$ fails $\pmod 3$.

$\dfrac{3p-1}{q-1}=1$ fails due to $q=3p$.

So we must have $\dfrac{3p-1}{q-1}=2\implies q=\dfrac{3p+1}{2}$. Put this in $p-1\mid 3q-1$ to get $p=11$ from which $q=17$. So our solutions for $(p,q)$ are $(11,17)$ and $(17,11)$.

The answer is $(11, 17)$ and $(17, 11)$ only, which work.

Pick some $n$ that is a primitive root modulo $p$ and $q$ which exists by CRT. Let $p, q$ be odd primes for now. Then, $pq \mid n^{3pq - 1} - 1$, thus $$p-1 \mid 3pq - 1 \iff p - 1 \mid 3q-1$$ and similarly $q-1 \mid 3p-1$. WLOG $p \leq q$, so we have $3q - 1 \geq 3p-1$; thus, we must have $3p-1 = 2q - 2$ is the only valid case. Plugging this back in, we have $$p-1 \mid \frac 92 p + \frac 12 \implies \frac 92 p + \frac 12 = 5p - 5 \iff p = 11.$$

On the other hand, if $2 \in \{p, q\}$, we have $6q \mid n^{6q} - 1$, so $q-1 \mid 6q-1$, and $q=2$. But this case fails too.

Solved with [b]dolphinday[/b]. It was pretty nice for such an old problem.

The answer is $(11,17)$ and $(17,11)$ only. (The confirmation of these solutions is at the bottom of the solution).

We first show the following key claim.

[color=#f00][b]Claim : [/b] The divisibilities $p-1 \mid 3q-1$ and $q-1 \mid 3p-1$ must hold.[/color]

[i]Proof: [/i] SImply note that by considering a primitive root $\pmod{p}$ for $n$ (which is known to exist), we have
\[n^{3pq-1}\equiv 1 \pmod{p} \implies \text{ord}_p(n) \mid 3pq-1 \]
combining this with the fact that $\text{ord}_p(n)=p-1$ we have that,
\[p-1 \mid 3q-1\]
as desired. Similarly, we obtain that (by considering a primitive root $\pmod{q}$ to be $n$) $q-1 \mid 3p-1$.

Now, we resort to size. WLOG, assume $p\geq q$.If $p=q$ then we have that $q-1 \mid 3q-1$ which implies that $p=q=2$ or $p=q=3$. These can be clearly seen to be impossible by considering $n=3$ and $n=2$ respectively.

Note that from $p-1\mid 3q-1$ we have that $3q-1=k(p-1)$. For all $k\geq3$, this implies that
\[k(p-1)\geq 3(p-1)=3p-3 > 3q-1\]
for all primes $p,q$ (since $p,q$ are distinct $p-q\geq 2$.)

Thus, $k=1$ or $k=2$. One can check that $k=1$ is impossible as it implies $p=3q$ which is clearly non-prime. Thus, we have that
\[p=\frac{3q+1}{2}\]
Now, plugging this into the other divisibility we obtain that
\begin{align*}
q-1 & \mid 3\left(\frac{3q+1}{2}\right)-1\\
q-1 & \mid \frac{9q+1}{2}\\
2q-2 & \mid 9p+1\\
2q-2 &\mid q+9
\end{align*}
which can then be seen to imply that $q\leq 11$ with some size bounding. Then, checking the values of $p$ for $q=2,3,5,7,11$ we see that the only possibilities are $(3,5),(7,11)$ and $(11,17)$. Of these, $(3,5)$ clearly fails when $n=3$. For, $(7,11)$ we have that,
\[3\times 7 \times 11 \mid n(n^{230}-1)\]
which clearly fails for $n\equiv 2 \pmod{7}$.
For, $(11,17)$ note that the original equation turns into
\[3\times 11 \times 17 \mid n(n^{560}-1)\]
which clearly holds for all $n$. (if $3,7,11\mid n$ it divides the first factor, if not it divides the second factor since $2,10,16\mid 560$). Thus, the only answer is the claimed one which has been checked to work.

I claim that the only solutions are $\boxed{(p, q) \in \{(11, 17), (17, 11)\}}$.

First observe that all of $3, p, q$ must be distinct and all odd. If say, $p$ were even, then $n^{6q} \equiv n \pmod 6$, so if $n = 5$ we would have $1 \equiv 5 \pmod 6$ which is a contradiction. Also note that $3, p, q$ must be distinct as if $p = q$, then $n^{3p^2 - 1} \equiv 1 \pmod{p^2}$, and so if $n = p$, we would derive a contradiction as $p$ is odd, and similar reasoning follows if either $p$ or $q$ are $3$. Therefore assume without loss of generality that $p > q > 3$. We now then make the following claim:

[color=#f00][b]Claim:[/b][/color] We have that $p - 1 \mid 3q - 1$ and $q - 1 \mid 3p - 1$.
[i]Proof:[/i] Let $g$ be a primitive root modulo $p$. Then \[g^{3pq} \equiv g \pmod p \implies g^{3q - 1} \equiv 1 \pmod p \implies p - 1 \mid 3q - 1.\] Similarly it follows that $q - 1 \mid 3p - 1$. $\square$

Hence we may write $3q - 1 = a(p - 1)$ and $3p - 1 = b(q - 1)$, and also observe $1 \leq a \leq 3$, as if $a > 3$, we must have $3q - 1 = a(p - 1) > 3p - 3 \geq 3q - 1$, as $p - q \geq 2$, which is a contradiction. Now $a = 1, 3$ gives us that $p \equiv 0 \pmod 3$ and $3q - 1 \equiv 0 \pmod 3$ respectively, both of which are impossible. Hence $a = 2$, and this gives us \[p = \frac{3q + 1}{2} \implies b = \frac{9q + 1}{2(q - 1)} \implies q - 1 \mid 10 \implies q = 2, 3, 6, 11.\] Of these only $\boxed{q = 11}$ works, which in turn provides $\boxed{p = 17}$, and vice versa. Therefore all of the solutions are $\boxed{(11, 17), (17, 11),}$ as was claimed. $\blacksquare$


Let $ m,n$ two distinct positive integers such that $ 10(m\minus{}n)| m^3\minus{}n^3$.

Show that $ 100(m\minus{}n) | m^3\minus{}n^3$.

[hide="Solution"]
If $ 10(m\minus{}n)|m^3\minus{}n^3\equal{}(m\minus{}n)(m^2\plus{}mn\plus{}n^2)\implies 10|m^2\plus{}mn\plus{}n^2$.

The possible quadratic resdiues in $ \pmod {10}$ are: $ 0,1,4,5,6,9$

If $ m^2\equiv 0\pmod{10}$, then $ n^2$ has to be $ 0\pmod{10}$, and since the only time that $ x^2\equiv 0\pmod{10}$ is when $ x\equiv 0\pmod{10}$, so $ m^2\plus{}mn\plus{}n^2\equiv 0\pmod{100}$.

If $ m^2\equiv 1\pmod{10}$, then $ 1\plus{}n\plus{}n^2$ has to be $ 0\pmod{10}$, but this has no solutions for $ n$.

If $ m^2\equiv 4\pmod{10}$, then $ 6\plus{}4n\plus{}n^2$ has to be $ 0\pmod{10}$, but again this has no solutions for $ n$.

If $ m^2\equiv 5\pmod{10}$, then $ 5\plus{}5n\plus{}n^2$ has to be $ 0\pmod{10}$, and this has no solutions for $ n$.

If $ m^2\equiv 6\pmod{10}$, then $ 6\plus{}6n\plus{}n^2$ has to be $ 0\pmod{10}$, and this also has no solutions for $ n$.

If $ m^2\equiv 9\pmod{10}$, then $ 1\plus{}9n\plus{}n^2$ has to be $ 0\pmod{10}$, and this has no solutions for $ n$.

So the only solution is when $ m,n\equiv 0\pmod{10}\implies m^2\plus{}mn\plus{}n^2\equiv 0\pmod{100}$

... there's probably a faster way to do it.
[/hide]

[quote="tonypr"]
... there's probably a faster way to do it.
[/quote]

My solution:
m,n cannot be both odd, so at least one is even, but if wlog 2|m so 2|n, it would implies 4|m²+mn+n². Now 5|(2m+n)²+3n², but (-3/5)=-1, so 5|m and 5|n, that is our aim.


Parity
combinatorics solved
combinatorics
graph theory
vector

There are $2n$ people in a party, prove that there are 2 of them such that they have an even number of common friends.

hint:
[hide]use the following problem:
[b]there are $n$ people in a party s.t every two of them have odd number of common friends,prove that:
$i)$ every person has even friends in this party...
$ii)$ $n$ is odd...[/b][/hide]

BaBaK Ghalebi, can yoy finish the solution?

yeah sure,first we need to prove the problem that I gave:
[hide]for part $i)$,assume that person $v$ is friend with persons $v_{1},...,v_{k}$ so except $v$, every other person (including $v_{1},...,v_{k}$) must be friend with odd number of $v_{1},...,v_{k}$ because of the hypothesis...
now consider the induced subgraph with vertices $v_{1},...,v_{k}$ in this graph, as said above, the degree of every vertex is odd, so the number of vertices must be even (because the sum of degrees in a graph is even)...
so we get that $k$ is even, so the number of friends of every person in this party is even...
---------------------------------------------------------
for part $ii)$ name the set of friends of $v_{i}$ other than $v$ as $D_{i}$ ($i=1,...,k$) so according to $i)$ we get that $D_{i}$ has odd members...
now assume the other vertices of the graph (other than $v,v_{1},...,v_{k}$) and name them $v_{k+1},...,v_{n-1}$...
now for every $i=1,...,n-1$ denote the number of friends of $v_{i}$ between vertices $v_{1},...,v_{k}$ as $d_{i}$...
so acording to the hypothesis of the problem we get that $d_{i}$ is odd (because of vertex $v$ and common friends)...
but we know that:
$|D_{1}|+...+|D_{k}|=d_{1}+...+d_{n-1}$
(because each side of the equality counts the number of edges between vertices $v_{1},...,v_{k}$ and the rest of the graph except $v$)
now in $i)$ we proved that $k$ is even so the left hand side of the above equation is even...
and also we proved that every $d_{i}$ is odd, so the number of $d_{i}$s must be even thus we get that $n-1$ is even, so $n$ is odd...
--------------------------------------------------------
[/hide]
now according to this problem the rest is easy to prove:
for the original problem,we`ll prove it using contradiction,we assume that the number of common friends of every two person is odd,then according to the above problem we get that the number of persons in the party must be odd,which is a contradiction,this contradiction shows that there exists two persons s.t they have even number of common friends... :!:

http://www.mathlinks.ro/Forum/viewtopic.php?t=68109 as well.

  darij

One can also see http://www.mathlinks.ro/Forum/viewtopic.php?t=83868 for an algebraic approach.
(The two problems are easily seen to be equivalent.)

Something similar :)
In the case one person has an odd number of friends proceed as in Darij's solution. NOw assume that each person has even

number of friends. For each take a 2n-vector with 0 and 1 (1 on the jth position if the jth person is his friend or 0 otherwise;

 moreover, the ith person has 0 on the ith position). Let $x_{1},...x_{2n}$ the vectors. We know that

$x_{i}\cdot x_{j}\equiv 1-\delta_{i,j} (mod 2)$ for any $i$ and $j$, where $\delta_{i,j}$ means $1$ if $i=j$ and $0$ if $i\neq j$. Assume we can find $X$ a nonempty subset of $\left\{1,2,...,2n\right\}$ such that

$\sum_{i\in X}x_{i}\equiv (0,0,....0) (mod 2)$. For any $i\in X$ we have $x_{i}\cdot\sum_{j\in X}x_{j}\equiv 0(mod 2)$.

But the sum is also congruent to $|X|-1$. therefore $|X|$ is odd, so $X$ is a proper subset of $\left\{1,2,...,2n\right\}$. Now for $k\notin X$ we have $x_{k}\cdot\sum_{j\in X}x_{j}\equiv 0(mod 2)$ so $|X|$ is even, a contradiction. Therefore we can not find

such a  subset $X$. Thus, for each of the $2^{2n}$ subsets $Y$ of $\left\{1,2,...,2n\right\}$ the sum $\sum_{i\in Y}x_{i}$ is different $mod 2$.

On the other hand, for each sum, the sum of all the coordinates is even (each person has an even number of friends).

Therefore the first $2n-1$ coordinates uniquely determine the parity of the last coordinate. In conclusion we can have only

$2^{2n-1}$ possibilities $mod 2$, contradiction.

Let $G$ be the desired graph. Choose an arbitrary vertex $v$ from $G$. Remove $v$ from $G$ and all of its neighboring edges. The result is an induced subgraph $G'$ with an even number of odd vertices and an odd number of even vertices.

[b]Claim:[/b] An even vertex in $G'$ is adjacent to at least two odd vertices (in particular, some even vertex is adjacent to an even number of odd vertices $>0$).

[b]Proof:[/b] Assume no even vertices are connected to at least two odd vertices in $G'$. Then $G'$ is disconnected when we introduce vertex $v$, which is a clear contradiction. $\square$

Thus there must exist an even vertex in $G'$ (call it $u$) that is connected to at least two odd vertices $v_1, v_2$ in $G'$. When we add $v$ back into $G'$, we find that $v$ is connected to $v_1, v_2$, and so is $u$. Thus there are two people that are acquainted with the same even number of people. $\blacksquare$


algorithm
algebra
polynomial
Vieta
search
number theory
Euclidean algorithm

Suppose that $ a$, $ b$ are positive integers for which $ A \equal{}\frac{a\plus{}1}{b}\plus{}\frac{b}{a}$is an
integer.

Prove that $ A \equal{} 3$.

This problem looks kinda like number 2 on the BMO 2007 i think

[deleted] oops, read it wrong. :blush:  Yeah then I found the solution that scorpius119 has.

Here's how I did it:
[hide]From $ \frac{b}{a}\in\mathbb{Z}$ we know that $ b\equal{}ka$ for some number $ k$. Then $ \frac{a\plus{}1}{b}\equal{}\frac{a\plus{}1}{ka}$, so $ ka|a\plus{}1$ therefore $ a|a\plus{}1$.

By the Euclidean Algorithm, we know that $ (a, a\plus{}1)$ has to be $ (1, 2)$, leaving $ a\equal{}1$. Then $ A \equal{}\frac{2}{b}\plus{}b$ and since only $ (1, 2) | 2$, $ b\in (1, 2)$, in which either case leaves $ A\equal{}3$[/hide]

Are you both sure that $ a$ divides $ b$? I see $ (4,2)$ as a valid solution.  :wink:
[hide]
Suppose there exists a solution $ (a,b)$ with $ A\neq 3$. The consider one with least value of $ ab$.
From $ A \equal{}\frac{a\plus{}1}{b}\plus{}\frac{b}{a}\equal{}\frac{a^{2}\plus{}a\plus{}b^{2}}{ab}$, we get
\[ a^{2}\plus{}(1\minus{}Ab)a\plus{}b^{2}\equal{}0\]
The polynomial $ x^{2}\plus{}(1\minus{}Ab)x\plus{}b^{2}$ has as its roots $ a$ and $ c \equal{} Ab\minus{}1\minus{}a \equal{}\frac{b^{2}}{a}$. This implies $ c$ is a positive integer, and $ \frac{c\plus{}1}{b}\plus{}\frac{b}{c}\equal{} A\neq 3$ so that $ c\geq a$, which results in $ b\geq a$ (from $ c \equal{}\frac{b^{2}}{a}$).

Similarly, $ b^{2}\minus{}Aab\plus{}a^{2}\plus{}a\equal{}0$, so letting $ d\equal{}Aa\minus{}b\equal{}\frac{a^{2}\plus{}a}{b}$, $ d$ is a positive integer with $ \frac{a\plus{}1}{d}\plus{}\frac{d}{a}\equal{}A\neq 3$ so that $ d\geq b$, or
\[ b^{2}\leq a^{2}\plus{}a<(a\plus{}1)^{2}\Rightarrow b<a\plus{}1\]
It follows that $ a\equal{}b$, $ A\equal{}2\plus{}\frac{1}{a}$ so that $ 2<A\leq 3$. Therefore, $ A\equal{}3$, a contradiction.
[/hide]

[quote="gotztahbeazn"]first wlog $ b\ge a > 0$[/quote]

Unfortunately, you do lose generality here; the expression isn't symmetric.

[quote="gotztahbeazn"]that's an interesting solution scorpius... what was the inspiration behind that? and does my solution work?

[hide]

first wlog $ b\ge a > 0$

then we can quickly set a lower bound by using AM-GM:

$ \frac{a\plus{}1}{b}\plus{}\frac{b}{a}>\frac{a}{b}\plus{}\frac{b}{a}\ge 2$

then we set the given expression equal to $ 4$ (i was not able to find a way to obtain an upper bound) and so we have:

$ 4 \equal{}\frac{a^{2}\plus{}a\plus{}b^{2}}{ab}\implies a(2b\minus{}1) \equal{} (a\minus{}b)^{2}$

however, since $ a > a\minus{}b$ and $ 2b\minus{}1 > a\minus{}b$, we know $ a(2b\minus{}1) > (a\minus{}b)^{2}$ and that:

$ 4 >\frac{a\plus{}1}{b}\plus{}\frac{b}{a}$

hence the given expression must equal $ 3$

[/hide][/quote]

i think the solution of you is not general. because the equation is not symmetric and the choosing a and b as you choose will get lost the generality..

[quote="gotztahbeazn"]that's an interesting solution scorpius... what was the inspiration behind that? and does my solution work?

[hide]

first wlog $ b\ge a > 0$

then we can quickly set a lower bound by using AM-GM:

$ \frac{a\plus{}1}{b}\plus{}\frac{b}{a}>\frac{a}{b}\plus{}\frac{b}{a}\ge 2$

then we set the given expression equal to $ 4$ (i was not able to find a way to obtain an upper bound) and so we have:

$ 4 \equal{}\frac{a^{2}\plus{}a\plus{}b^{2}}{ab}\implies a(2b\minus{}1) \equal{} (a\minus{}b)^{2}$

however, since $ a > a\minus{}b$ and $ 2b\minus{}1 > a\minus{}b$, we know $ a(2b\minus{}1) > (a\minus{}b)^{2}$ and that:

$ 4 >\frac{a\plus{}1}{b}\plus{}\frac{b}{a}$

hence the given expression must equal $ 3$

[/hide][/quote]

You are wrong  :(

[quote="gotztahbeazn"]that's an interesting solution scorpius... what was the inspiration behind that?[/quote]
The method is a somewhat well-known number theory technique called root flipping or Vieta jumping. There are some recent topics asking about it, so you can do a search if you want to know more about it.


(a) If $ x \neq 0$, and $ xy \equal{} xz$, then $ y \equal{} z$.
(b) If $ x \neq 0$, and $ xy \equal{} x$, then $ y \equal{} 1$.
(c) If $ x \neq 0$, and $ xy \equal{} 1$, then $ y \equal{} 1/x$.
(d) If $ x \neq 0$, then $ 1/(1/x) \equal{} x$.

[b] Axioms [/b]
(1) If $ x \in F$ and $ y \in F$, then their product $ xy$ is in $ F$.
(2) $ xy \equal{} yx$, for all $ x,y \in F$.
(3) $ (xy)z \equal{} x(yz)$ for all $ x,y,z \in F$.
(4) $ F$ contains an element $ 1 \neq 0$, such that $ 1x \equal{} x$ for every $ x \in F$.
(5) If $ x \in F$ and $ x \neq 0$, then there exists an element $ 1/x \in F$ such that $ x \cdot (1/x) \equal{} 1$.

So (c) and (d) follows from Axiom (5), and (b) follows from Axiom (2). How about (1)?

(1) is just the "this operation exists" axiom. In fact, the only time it makes sense to state it in that form is when $ F$ is a subset of some larger set that already has an operation $ \cdot$.


Prove that if $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1,$ then $ \minus{} \frac {1}{2}\leq ab \plus{} ac \plus{} bc\leq 1.$

[color=darkblue][size=150]Oh! You must post again your problem. I don't understand your post:
"Prove that if $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, then $ \minus{} \frac {1}{2}\le a^2 \plus{} b^2 \plus{} c^2\le 1$" ?[/size][/color]

Well, umm yeah i think its a mistake. if it's not, then to prove it all you need is knowledge of what "less than or equal to" means...

[hide="Answer"]
For $ ab\plus{}bc\plus{}ca \ge \frac{\minus{}1}{2}$.
We have
$ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca) \ge 0 \\
\Leftrightarrow 2(ab\plus{}bc\plus{}ca) \ge \minus{}1 \Rightarrow ab\plus{}bc\plus{}ca \ge \frac{\minus{}1}{2}$
For $ ab\plus{}bc\plus{}ca \le 1$
By Cauchy we have
$ (a\plus{}b\plus{}c)^2 \le 3(a^2\plus{}b^2\plus{}c^2) \\
\Leftrightarrow 1\plus{}2(ab\plus{}bc\plus{}ca) \le 3 \\
\Leftrightarrow ab\plus{}bc\plus{}ca \le 1$
[/hide]

[hide="Another prove for the right hand."]
$ 1 \equal{} a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$. Easy to prove.
[/hide]

Edited

[quote="mathson"][hide="Another prove for the right hand."]
$ 1 = a^2 + b^2 + c^2 \ge ab + bc + cd$. Easy to prove.
[/hide][/quote]

I think there is a mistake in your post, it should be $ ca$, not $ cd$

 :)

Another problem

Prove that if :
$ a^2 + b^2 + c^2 = 1$, then $ a^2 + b^2 + c^2 + \frac1{a^2} + \frac1{b^2} + \frac1{c^2}\ge\frac {9a^2 + 9b^2 + 9c^2 + 21ab + 21ac + 21bc}{(a + b + c)^2}$
$ a,b,c\in \mathbb{R^+}$

[quote="IW@IT"][quote="mathson"][hide="Another prove for the right hand."]
$ 1 \equal{} a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} cd$. Easy to prove.
[/hide][/quote]
I think there is a mistake in your post, it should be $ ca$, not $ cd$
 :)
[/quote]

Thanks. I edited it.

[quote="IW@IT"]
Prove that if :
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, then $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac1{a^2} \plus{} \frac1{b^2} \plus{} \frac1{c^2}\ge\frac {9a^2 \plus{} 9b^2 \plus{} 9c^2 \plus{} 12ab \plus{} 12ac \plus{} 12bc}{(a \plus{} b \plus{} c)^2}$[/quote]

This isn't true for a, b, c in R because we can make the denominator of the RHS as close to 0 as we wish while the numerator of the RHS and the entire LHS remains bounded.

@JavaMan: $ a,b,c \in \mathbb{R^\plus{}}$
I'm sorry, I didn't mention it before :blush:
And the coefficient of $ ab,bc,ac$ should be $ 21$, not $ 12$ :P
I've edited my problem

[quote="IW@IT"]
Another problem

Prove that if :
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, then $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac1{a^2} \plus{} \frac1{b^2} \plus{} \frac1{c^2}\ge\frac {9a^2 \plus{} 9b^2 \plus{} 9c^2 \plus{} 12ab \plus{} 12ac \plus{} 12bc}{(a \plus{} b \plus{} c)^2}$[/quote]

$ \frac {9a^2 \plus{} 9b^2 \plus{} 9c^2 \plus{} 12ab \plus{} 12ac \plus{} 12bc}{(a \plus{} b \plus{} c)^2}\equal{} \frac{3}{(a\plus{}b\plus{}c)^2} \plus{} 6$

From Cauchy $ \frac1{a^2} \plus{} \frac1{b^2} \plus{} \frac1{c^2}\ge 9$

So it suffices to proove
$ \Rightarrow 10 \ge \frac{3}{(a\plus{}b\plus{}c)^2} \plus{} 6$

$ \rightarrow (a\plus{}b\plus{}c)^2 \ge \frac{3}{4}$
And this is obvious...

[quote="IW@IT"]@JavaMan: $ a,b,c \in \mathbb{R^ \plus{} }$
I'm sorry, I didn't mention it before :blush:
And the coefficient of $ ab,bc,ac$ should be $ 21$, not $ 12$ :P
I've edited my problem[/quote]

In that case use the same method and the inequality reduces to
$ \dfrac{3(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \le 1$

Which is the same as
$ ab \plus{} bc \plus{} ca \le 1$
Which is true by rearrangement


linear algebra
matrix
algebra
polynomial
linear algebra solved

[b]Show that A and C^(-1)AC (where A is an n*n matrix and C is n*n invertible matrix), have the same eigenvalues. MUST they have the same eigenvectors??[/b]

Thanks for your help in advance!

They have the same eigenvalues if their characteristic polynomials have the same roots.

$\\ h_{C^{-1}AC}(x)=det(C^{-1}AC -xI)=det(C^{-1}AC -xC^{-1}C)=\\ \\ det(C^{-1}) \cdot det(A-xI)\cdot det(C) = det(A-xI) =h_{A}(x)$.

For the second question the answer in  [b]no[/b] . Check this $A=\left(\begin{array}{cc} 3&6\\\ 0&2\end{array}\right)$ and $C=\left(\begin{array}{cc} 1&-1\\\ 0& 1\end{array}\right)$.

 :)


analytic geometry
graphing lines
slope
rotation
geometry
geometric transformation
reflection

Find the number of finite sequences $ \{a_1,a_2,\ldots,a_{2n\plus{}1}\}$, formed with nonnegative integers, for which $ a_1\equal{}a_{2n\plus{}1}\equal{}0$ and $ |a_k \minus{}a_{k\plus{}1}|\equal{}1$, for all $ k\in\{1,2,\ldots,2n\}$.

Note that $ a_1 \equal{} a_{2n \plus{} 1} \equal{} 0$ then $ a_2 \equal{} a_{2n} \equal{} 1$
we investigate value can  have of $ a_i$
With jugde that : $ a_3 \equal{} 0$ or$ 2$  and $ a_{2n \minus{} 1} \equal{} 0$  or$ 2$.
Siliar completly, $ a_4,a_{2n \minus{} 2} \equal{} 1$ or$ 3$
Term right in the middle of sequence can receive maximum $ a_{n \plus{} 1} \equal{} n$
 We bring up 1 example about sequence  following conditons of problem  : $ 0,1,2,3,....,n,n \minus{} 1,...,0$
Conclude $ (a_1,a_2,...,a_i,...a_{2n \plus{} 1})$ contain maximum $ (n \plus{} 1)$ elements
 and number of finite  sequences is $ n$

Consider travelling from $ (0,0)$ by means of the moves $ A: (a,b) \mapsto (a\plus{}1,b)$ and $ B: (a,b) \mapsto (a,b\plus{}1)$ according to the following instructions, for a given sequence $ (a_i)$ with the properties required:
If you are at a point $ (a,b)$, look at $ b_{a\plus{}b}\equal{}a_{a\plus{}b\plus{}2}\minus{}a_{a\plus{}b\plus{}1}$ in the sequence. If $ b_{a\plus{}b}\equal{}1$, apply move $ A$. If $ b_{a\plus{}b}\equal{}\minus{}1$ apply move $ B$.
It's fairly easy to see that the result of this is that after $ 2k$ moves, you are at the point $ (k\plus{}a_{2k\plus{}1},k\minus{}a_{2k\plus{}1})$ (since $ a_1\equal{}0$)
By the condition that $ a_{2n\plus{}1}\equal{}0$, we must therefore end up at $ (n,n)$.
Also, since the sequence is nonnegative, any lattice point $ (a,b)$ we cross must have $ a\geq b$.

However, these are all the conditions taken account of, and by reversing this procedure it is clear that we can construct a valid sequence $ (a_i)$ from any path from $ (0,0)$ to $ (n,n)$ using moves $ A$ and $ B$ satisfying the condition that at any point $ (a,b)$ of the path $ a\geq b$.

Thus there is a bijection between paths from $ (0,0)$ to $ (n,n)$ using only points $ (a,b): a\geq b$ and sequences $ (a_i)$ having the desired properties.

But the number of such paths is well known (for example, see chapter 5 of Engel's Problem Solving Strategies) to be equal to the $ n$th Catalan number $ \frac{1}{n\plus{}1}\binom{2n}{n}$. So this is also the number of sequences satisfying the conditions mentioned in the question.

(Counting by the method of oriented broken line on coordinate plane)

The points $ \{(i, a_{i} \plus{} 1)\}$ forms a broken line with slopes of either 1 or -1, then by a corollary of a theorem on the number of finite sequence, the number of broken line that starts with $ (1, 1)$ and ends with $ (2n \plus{} 1, 1)$ is
$ C(2n, \frac{1}{2}(2n \plus{} 1 \minus{} 1))\minus{}C(2n, \frac{1}{2}(2n \plus{} 1 \plus{} 1))\equal{}\frac{(2n)!}{n!n!}\minus{}\frac{(2n)!}{(n\plus{}1)!(n\minus{}1)!}\equal{}\frac{(2n)!}{n!(n\minus{}1)!}$

The number of such sequences is the number of paths from the leftmost $ 0$ to the rightmost $ 0$ in the following figure:

[code]
The numbers in parentheses denote the columns' indexes

(1) (3) (5) (7)   . . .      (2n+1)
 0   0   0   0 . . . . .    0   0
   1   1   1  . . . . .   1   1
     2   2   2 . . . .  2   2
       3   3  . . .   3   3
         4   4. . . 4   4
           5  . . .   5
               . . .
                 n
[/code]

where from each of the entries we may reach the next column either to the top-right or to the bottom-right entry. Here the $ i$th column contains all the possible values of $ a_i$. Let us call such a figure with $ 2n + 1$ columns and $ n$ rows a $ n$-[i]triangle[/i].

Consider the following configuration of the $ n$-triangle:

[code]
                           0
                        0 1
                         .
                         .
       0  1  2 . . .  n-2
    0  1  2  3 . . . n-1
0  1  2  3  4 . . . n

[/code]

where each of the numbers denote a lattice point in the region $ 0\le y\le x\le n$. Hence our problem is equivalent to finding the number of paths from $ (0,0)$ to $ (n,n)$ which do not cross the line $ y = x$ and from a point we may go either up or right. Let $ k(n,r)$ be the number of ways to reach the point $ (n,r)$ from the point $ (0,0)$. Note that we have the following recursion:
\[ k(n,r) = k(n - 1,r) + k(n,r - 1)
\]
with the additional values $ k(n,0) = 1, k(n,1) = n, k(n,n) = k(n,n - 1)$ and $ k(n,r) = 0$ for $ r > n$. Using these, we can determine $ k(n,n)$ for any $ n$ for example:
\begin{align*}k(3,3) & = k(3,2) \\
& = k(2,2) + k(3,1) \\
& = k(2,1) + 1 \\
& = 2\end{align*}
And likewise $ k(4,4) = 14,k(5,5) = 42$ etc.

I haven't been able to find a closed form of $ k(n,n)$ yet. But if we consider this another way, $ k(n,n)$ is the number of integer solutions to the following equation:
\begin{align}x_1 + x_2 + \dots + x_n & = n\end{align}
where $ 0\le x_i\le i$ for all $ 1\le i\le n$. We call such a solution a [i]good[/i] solution. Since (1) has a total of $ \binom{n + n - 1}{n - 1} = \binom{2n - 1}{n - 1}$ nonnegative integer solutions, we can count the number of good solutions using the inclusion-exclusion principle. When one of the $ x_i$'s exceed $ i$ we consider the equation:
\[ x_1 + x_2 + \dots + (x_i - i - 1) + \dots + x_n = n - i - 1
\]
and the number of solutions is $ \sum_{i = 1}^n\binom{n - i - 1 + n - 1}{n - 1}$. Likewise, when two of the $ x_i$'s, say $ i_1,i_2$, exceed $ i_1$ and $ i_2$ respectively, the number of solutions is $ \binom{n - i_1 - 1 - i_2 - 1 + n - 1}{n - 1}$. If we continue like this, we find that the number of good solutions is
\[ \binom{2n - 1}{n - 1} - \sum_{1\le i_1\le n}\binom{n - i_1 - 1 + n - 1}{n - 1} + \sum_{1\le i_1 < i_2\le n}\binom{n - i_1 - 1 - i_2 - 1 + n - 1}{n - 1} - \dots
\]
And thus
\[ k(n,n) = \sum_{k = 0}^n\sum_{i_0 = 0,1\le i_1 < i_2 < \dots < i_k\le n}( - 1)^k\binom{2n - 1 - (i_0 + i_1 + i_2 + \dots + i_k + k)}{n - 1}
\]
which is our desired number.

We may visualise such a sequence a route through some object which is symmetrical about the centre and fans out from $ 0$ to $ \pm n$ and back to $ 0$ with any path traveling from $ k$ to $ k \pm 1$

However, there is a constraint that all the integers are non negative, so by constructing a "line" between the two ends, we need to count the number of paths through the object which are always above the dotted line.

By considering increasing steps as up and decreasing as right, we can rotate and transform our object into an n by n grid with the dotted line as y=x.

[b]Lemma 1[/b]: The number of illegal routes is $ 2n \choose {n-1}$

Consider an illegal route. It must cross y=x at least once. Consider the first time it does this. Now "reflect" the remaining paths in "y=x" (essentially rename "u" by "r" for remaining moves). Since at the point of crossing y=x there was one more r than u. The new path will have n+1, r's and n-1 u's. This is a route through a similar grid which is $ n+1$ by $ n-1$, which crosses $ y=x$. However, all the paths in this new grid cross $ y=x$. Therefore the number of illegal routes is the number of routes through this tree, which is $ 2n \choose {n-1}$

Therefore the number of legal routes is the complement of the number of illegal routes.

\begin{align*}
{2n \choose n} - {2n \choose {n-1}} &= \frac{2n!}{n!n!} - \frac{2n!}{(n-1)!(n+1)!} \\
&= \frac{2n! (n+1)^2 - 2n! (n+1)(n)}{(n+1)!(n+1)!} \\
&= \frac{2n!(n+1)}{(n+1)!(n+1)!} \\
&= \frac{2n!}{n!(n+1)!} \\
&= \boxed{\frac{1}{n+1} {2n \choose n}}
\end{align*}

just another standard problem on catalan numbers
we just interpret a path as a sequence

There are endless number of solutions.a(1)=0 , we have it as condition.Due to condition |a(k)-a(k-1)| a(2)=1 , a(3)=0 or a(3)=2.And for every integer k from 2 to end we have a(k+1)=a(k)+1 or a(k+1)=a(k)-1.And that means that for every integer k from 1 to any integer number which is bigger than two, we have sequence
{0,1,2, ... ,k-1,k,k-1, ... ,2,1,0} and all the conditions are held.And that means that we have endless number of consequenses of this type.

P.S I'm bad in English

Draw points $ P_{i}=(i,a_{i})$. and join $ P_{i}, P_{i+1}$. To count the number of such path, We will count the number of paths which meet $ y=-1$. For these path, reflect from $ P_{1}$ to the first point which meet $ y=-1$.
Then the start point of the path become a $ (1,-2)$. all of this path have a bijection, and the number of such path is $ {2n}\choose{n-1}$. So the answer is $ {2n}\choose {n}-{2n}\choose {n-1}$.