algebra
polynomial
algorithm
number theory

So what are some common tips/techniques for proving primality? I am having a much easier time proving that stuff isn't prime that proving that stuff is prime. Thanks.

There isn't a really nice way of proving primality in general.  Recently, some mathematicians published a paper in which they gave a polynomial-time algorithm for doing it, but it was something like the order of x:^5: and in small cases that just isn't very good at all.  (That just means that the amount of time it takes to test a number goes up roughly equivalent to the size of the number to the 5th power, times some constant.  Earlier algorithms were exponential-time, meaning that doubling the size of your test number would square the time necessary, whereas this newer algorithm would just multiply the time by 32.)  But there still isn't a "nice" way of doing it.  Personally, I just stick with the standard check-every-prime-number-up-to-the-square-root.  What do you need to do this for?

It's much easier to show that a number isn't prime than to show that a number is prime. You can, for example, use Fermat's Little Theorem to disprove primality but not to prove primality. On the other hand, if your number is a Mersenne number (of the form 2^n-1, where n is prime), there's a simple algorithm for determining whether the number is prime that works in log(x) time. The number of iterations is n-2 for 2^n-1 I think.

Nono guys, you misunderstand me. I am not asking about algorithms used to prove that a number is prime, I am asking about general tactics for olympiad problems that ask you to prove that something (usually some expressions) is prime.

[quote]Olympiads don't ask you to prove that numbers or expressions are prime. They ask you to prove that numbers or expressions aren't prime.[/quote]

yes they do, they ask both. For example, a problem I am working on in Engel:

If p and p^2 + 2 are prime, then show that p^3 + 2 is also prime.

That's silly though. If p and p2+2 are both primes, then p=3. (Otherwise p2+2 is divisible by 3.)

[quote]If p and p^2+2 are both primes, then p=3. [/quote]

But how can you say that?

then p must be either 0 or 1. But how does that help?

clearly, for p>3, p cannot be 0 mod 3 (else it would be divisible by 3, and if it's greater than 3, then...well...it's not prime). however, p:^2::equiv:1 for p:equiv:1,2 mod 3. so p:^2:+2:equiv:0 mod 3, or it's a multiple of 3. this will not be composite unless it is equal to 3, which cannot happen. so therefore p=3.

By the way, what made you think of using mod 3?

I tried a bunch of other primes p, and p2+2 was always divisible by 3.

Yeah, I tried the same thing, except I was under the mistaken impression that 171 was a prime number . . . here's to math without thought!


function
integration
real analysis
real analysis unsolved

Give a function $ f : R \minus{} > R$ is bounded on $ ( \minus{} \infty,b]$ for all $ b \in R$
Prove that there a function $ g$ is convex and increasing, $ g \in C^{\infty}(R)$ satisfy : $ g(t) \geq f(t), g'(t) \geq f(t)$ for all $ t \in R$

Try this : $ h(t) \equal{} \sup_{x\leq t} |f(x)|$ - this function is non-decreasing, hence $ h'$ exists and is nonnegative a.e., next, define $ H(z) \equal{} h(0) \plus{} |\int_0^z h(t)dt|$ - it satisfies required conditions. You can make it $ C^{\infty}$ by standart procedure..


quadratics
geometry
modular arithmetic
number theory
relatively prime
prime factorization

Prove that for positive integers $ x,y,z$ the number $ x^2 \plus{} y^2 \plus{} z^2$  is not divisible by $ 3(xy \plus{} yz \plus{} zx)$.

Suppose that there exist numbers $ a,b,c,k$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3k(ab\plus{}bc\plus{}ca)$ .Then
$ (a\plus{}b\plus{}c)^2\equal{}(3k\plus{}2)(ab\plus{}bc\plus{}ca)$ . Now we choose a prime $ p\equiv 2(\mod 3)$ such that
$ p^{2m\minus{}1}\parallel{}3n\plus{}2$ . Such a prime exist because $ 3n\plus{}2$ cannot be a perfect square .
Now because the exponent of $ p$ at $ (a\plus{}b\plus{}c)^2$ will be an even number we have that
$ p|ab\plus{}bc\plus{}ca$ and subtituting the relation that $ c\equiv \minus{}a\minus{}b(\mod p)$ we have that
$ p|a^2\plus{}ab\plus{}b^2$ or $ p|(2a\plus{}b)^2\plus{}3b^2$ from which we conclude that $ \minus{}3$ is a quadratic residue mod p which is a contradiction since $ p\equiv 2(\mod 3)$

2.[u]Solution[/u]

Suppose that $ x,y,z$ and $ n$ are positive integers such that $ \frac {x^2 + y^2 + z^2}{3(xy + yz + zx)} = n$

$ \therefore x^2 + y^2 + z^2 = 3n(xy + yz + zx).$

Therefore  $ (x + y + z)^2 = (3n + 2)(xy + yz + zx)$

Choose prime number $ p \equiv 2 (mod 3)$ which divides $ 3n + 2$ with an odd exponent

such that $ p^{2i - 1} | 3n + 2$ and $ p^{2i} \nmid 3n + 2 \exists i \in \mathbb{N}$.Then $ p^i | x + y + z$

Therefore $ p | xy + yz + zx$. Substituting $ z \equiv - x - y (mod p)$

$ \therefore p | x^2 + xy + y^2 \therefore p | (2x + y)^2 + 3y^2$

$ \therefore (\frac { - 3}{p}) = 1$, Which is impossible because $ p \equiv 2 (mod 3)$.

Consider the following identity: \[ (x\plus{}y\plus{}z)^2 \equal{} x^2\plus{}y^2\plus{}z^2\plus{}2(xy\plus{}yz\plus{}xz)\] Now assume that $ 3(xy\plus{}yz\plus{}xz)|x^2\plus{}y^2\plus{}z^2$. This means $ x^2\plus{}y^2\plus{}z^2\equal{}3t(xy\plus{}yz\plus{}xz)$ for some positive integer $ t$. It follows that by combining the left side of the identity, \[ (x\plus{}y\plus{}z)^2\equal{}(3t\plus{}2)(xy\plus{}yz\plus{}xz)\] Now we will use congruences modulo $ 3(xy\plus{}yz\plus{}xz)$ (this will be assumed and not written every line).\\
$ (x\plus{}y\plus{}z)^2 \equiv (3t\plus{}2)(xy\plus{}yz\plus{}xz) \equiv 2(xy\plus{}yz\plus{}xz)$\\
$ \frac{(x\plus{}y\plus{}z)^2}{xy\plus{}yz\plus{}xz} \equiv 2$\\
Note $ \frac{(x\plus{}y\plus{}z)^2}{xy\plus{}yz\plus{}xz}$ must be an integer because if $ xy\plus{}yz\plus{}xz|x^2\plus{}y^2\plus{}z^2$, then it must also divide $ (x\plus{}y\plus{}z)^2$ because of the original identity. As a result of the modular algebra above, for some integer $ k$, \[ \frac{(x\plus{}y\plus{}z)^2}{xy\plus{}yz\plus{}xz} \equal{} 3k(xy\plus{}yz\plus{}xz) \plus{}2\]
Now we will divide both sides of the original identity by $ 3(xy\plus{}yz\plus{}xz)$. This gives \[ \frac{(x\plus{}y\plus{}z)^2}{3(xy\plus{}yz\plus{}xz)} \equal{} t \plus{}\frac23\]\\
\[ \frac{3k(xy\plus{}yz\plus{}xz) \plus{}2}{3} \equal{} t\plus{}\frac23\]\\
\[ k(xy\plus{}yz\plus{}xz) \plus{}\frac23 \equal{} t\plus{}\frac23\]\\
\[ k(xy\plus{}yz\plus{}xz) \equal{} t\]\\
Since $ x^2\plus{}y^2\plus{}z^2 \equal{} 3t(xy\plus{}yz\plus{}xz)$, our previous conclusion implies that $ x^2\plus{}y^2\plus{}z^2 \equal{} 3k(xy\plus{}yz\plus{}xz)^2$, which implies $ \frac{x^2\plus{}y^2\plus{}z^2}{(xy\plus{}yz\plus{}xz)^2} \equal{} 3k$, an integer. This is an absurdity because $ (xy\plus{}yz\plus{}xz)^2$ is strictly larger than $ x^2\plus{}y^2\plus{}z^2$, as shown below: \[ (xy\plus{}yz\plus{}xz)^2>x^2y^2\plus{}y^2z^2\plus{}z^2x^2\plus{}xy^2z>x^2\plus{}y^2\plus{}z^2\plus{}xy^2z>x^2\plus{}y^2\plus{}z^2\]

A contradiction has been reached, so $ 3(xy\plus{}yz\plus{}xz)$ never divides $ x^2\plus{}y^2\plus{}z^2$.

- assume the statement false. than there exists such a positive integer $ k$ that $ x^2 + y^2 + z^2 = 3k(xy + yz + xz)$. the expression is equivalent to $ (x + y + z)^2 = (3k + 2)(xy + yz + xz)$.
- since $ x^2 = 0,1 \mod 3$ for integer $ x$ we can conclude that $ 3k + 2$ is not a perfect square. which imples that there exists such a prime number $ p$ of a form $ p = 3l + 2$ (for a positive integer $ l$) that $ p$ devides $ xy + yz + zx$ (otherwise $ (3k + 2)(xy + yz + xz)$ can't be a perfect square). this also implies that $ p$ devides $ x + y + z$
- if $ p$ devides $ x$ than it has to devide $ yz$ and $ y + z$ implying $ p|y$ and $ p|z$. because $ x,y,z$ cannot be infinitely devisable by $ p$ we can assume without loss of generality that $ p$ doesn't devide any of $ x,y,z$ and is therefore relatively prime to each of them.
- obviously $ x = - (y + z) \mod p$ which by putting this into $ xy + yz + zx = 0 \mod p$ leads to $ x^2 = yz \mod p$. simmilary we conclude $ y^2 = xz \mod p$ and $ z^2 = xy \mod p$.
- now $ y^2 \cdot yz = x^2 \cdot xz \mod p$ and since $ p,z$ are relatively prime we conclude that $ x^3 = y^3 \mod p$
- if that is true than also $ x^{3l} = y^{3l} \mod p$. on the other hand since $ x,y$ are relatively prime to $ p$ ,${ x^{p - 1} = y^{p - 1} = 1 \mod p}$ or in other words $ x^{3l + 1} = y^{3l + 1} \mod p$ which combined with the former implies $ x = y \mod p$.
- using the same method as above but for a different pair of $ x,y,z$ we obtain $ x = y = z \mod p$. than $ x + y + z = 3x = 0 \mod p$ which implies $ x = 0 \mod p$ which is in contradiction with our assumption that $ x$ is relatively prime to $ p$
- our starting assumption must have been wrong, therefore such a $ k$ does not exist.

suppose that there are $ x,y,z$ that satisfy $ 3(xy \plus{} yz \plus{} zx)$ divides $ x^2 \plus{} y^2 \plus{} z^2$.
let $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3k(xy \plus{} yz \plus{} zx)$.
if $ x, y, z$ are all even numbers, divides them all with 2. then this case moves to other cases.
if two numbers of them are even, $ x^2 \plus{} y^2 \plus{} z^2$ is odd and $ xy \plus{} yz \plus{} zx$ is even so contradiction.
in the other hand, $ (x \plus{} y \plus{} z)^2 \equal{} (3k \plus{} 2)(xy \plus{} yz \plus{} zx)$, so there exists a prime number $ p$ so that is $ p \equiv 2(mod3)$, and the order of $ 3k \plus{} 2$ for $ p$ is an odd number..
if $ p$ only can be $ 2$, then $ 3k \plus{} 2 \equal{} 2^{(2l \minus{} 1)}$ for some $ l$. then $ 2$ divides $ xy \plus{} yz \plus{} zx$ because $ (x \plus{} y \plus{} z)^2$ is a square. but $ xy \plus{} yz \plus{} zx \equiv 1(mod2)$ so contradiction.
so there is $ p$ that is not $ 2$.
also, because $ (x \plus{} y \plus{} z)^2$ is a square number, $ p$ also divides $ xy \plus{} yz \plus{} zx$.
(the order of $ (x \plus{} y \plus{} z)^2$ for $ p$ is an even number)

meanwhile, because $ p$ divides $ x \plus{} y \plus{} z$, $ x \equiv \minus{} y \minus{} z(mod p)$, so $ x^2 \equiv y^2 \plus{} z^2 \plus{} 2yz(mod p)$.
therefore $ x^2 \plus{} y^2 \plus{} z^2 \equiv 2(y^2 \plus{} z^2 \plus{} yz) \equiv 0(mod p)$, so, $ 4y^2 \plus{} 4yz \plus{} z^2 \equiv \minus{} 3z^2(mod p)$.
this means $ \minus{} 3$ is quadratic residues of $ p$.(because $ 4y^2 \plus{} 4yz \plus{} z^2 \equal{} (2y \plus{} z)^2$).
but because $ p \equiv 2(mod 3)$, this is contradiction.

this means there are no $ x,y,z$ that satisfy $ 3(xy \plus{} yz \plus{} zx)$ divides $ x^2 \plus{} y^2 \plus{} z^2$.

Sol:
[b]Lemma 1: If $ x,y,z$ are sides of one triangle.Then: $ x^2 \plus{} y^2 \plus{} z^2 < 2(xy \plus{} yz \plus{} zx)$.[/b]
dem: If $ 2P$: perimeter of the triangle $ ABC$ whit sides $ x,y,z$.
   Then: $ x \equal{} (P \minus{} y) \plus{} (P \minus{} z)$, $ y \equal{} (P \minus{} z) \plus{} (p \minus{} x)$, $ z \equal{} (P \minus{} x) \plus{} (P \minus{} y)$,
If. $ a \equal{} P \minus{} x$,$ b \equal{} P \minus{} y$,$ c \equal{} P \minus{} z$.
then : $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)$ , $ 2(xy \plus{} yz \plus{} zx) \equal{} 2(a^2 \plus{} ab \plus{} bc \plus{} ca \plus{} b^2 \plus{} bc \plus{} ba \plus{} ca \plus{} c^2 \plus{} ac \plus{} ab \plus{} cb) \equal{} 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca) \plus{} 4(ab \plus{} bc \plus{} ca)$[b]$ Q.E.D$[/b]
[b]Lemma 2: If $ x,y,z \in R^{ \plus{} }$ Then $ x^2 \plus{} y^2 \plus{} z^2 > \equal{} xy \plus{} yz \plus{} zx$.[/b]dem: $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2>\equal{}0$ where equality is given if only if $ x \equal{} y \equal{} z$ . [b]$ Q.E.D$[/b]
Now:
If $ K \equal{} (x^2 \plus{} y^2 \plus{} z^2)/(xy \plus{} yz \plus{} zx)$.Where $ x,y,z$ /$ K \in Z^{ \plus{} }$,
WLG $ x > \equal{} y > \equal{} z$.
$ I$ $ case$: $ x < y \plus{} z$ Then for lemma 1, $ K < 2$, But, for lemma 2. $ K > \equal{} 1$. how $ K$ is integer , $ K \equal{} 1$, then $ x \equal{} y \equal{} z$,
$ II$$ case$: $ x > \equal{} y \plus{} z$,
give the ecuation in $ x$.
$ x^2 \minus{} k(y \plus{} z)x \plus{} y^2 \plus{} z^2 \minus{} kyz \equal{} 0$ if $ x_0$ the other solution.For vietta.
$ x \plus{} x_0 \equal{} k(y \plus{} z)$,      $ x.x_0 \equal{} y^2 \plus{} z^2 \minus{} kyz$ ,Suppose that $ x \plus{} y \plus{} z$ is minimum.
 $ a)$ $ x_0 > \equal{} 0$.
Then: $ k(y \plus{} z) \equal{} x \plus{} x_0 > \equal{} 2x > \equal{} 2(y \plus{} z)$,-> $ k > \equal{} 2$..(1)
$ y^2 \plus{} z^2 \minus{} kyz \equal{} x.x_0 > \equal{} x^2 > \equal{} (y \plus{} z)^2 \equal{} y^2 \plus{} z^2 \plus{} 2yz$ -> $ k < \equal{} 2$..(2).
of (1) and (2). K=2.
 $ b)$ $ x_0 < 0$.

Suppose that a integer triple $ (a,b,c)$ satisfy the condition $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} k(ab \plus{} bc \plus{} ca)$ when $ k$ is multiple of 3.
Then $ (a \plus{} b \plus{} c)^{2} \equal{} (k \plus{} 2)(ab \plus{} bc \plus{} ca)$. (We can assume $ a,b,c$ are pairwise coprime. If $ p|a,b$ for prime $ p$, $ p|ab \plus{} bc \plus{} ca \Rightarrow p|a \plus{} b \plus{} c \Rightarrow p|a,b,c$. Now we can replace $ (a,b,c)$ into $ (\frac {a}{p},\frac {b}{p},\frac {c}{p})$.)
$ k \plus{} 2\equiv 2\pmod 3$. So it must has a prime factor in the form of $ 3t \plus{} 2$ which appears odd times in the prime factorization of $ k \plus{} 2$.
If there is a odd prime factor of $ k \plus{} 2$, $ p$ such that $ p\equiv 2\pmod 3$, $ p|(a \plus{} b \plus{} c)^{2} \Rightarrow p|a \plus{} b \plus{} c$.
The exponent of $ p$ in $ (a \plus{} b \plus{} c)^{2} \equal{} (k \plus{} 2)(ab \plus{} bc \plus{} ca)$ is even. Since that of $ k \plus{} 2$ is odd, $ p|ab \plus{} bc \plus{} ca$.
Applying $ p|a \plus{} b \plus{} c$, we get $ 0\equiv ab \plus{} bc \plus{} ca \equiv \minus{} (b \plus{} c)(b \plus{} c) \plus{} bc \equal{} \minus{} b^{2} \minus{} bc \minus{} c^{2}\pmod p$.
Hence, $ (2b \plus{} c)^{2}\equiv \minus{} 3c^{2}\pmod p$. $ 2b \plus{} c, c$ can't be a $ 0\pmod p$. ($ (b,c) \equal{} 1$)
Therefore $ \left( \frac { \minus{} 3}{p}\right) \equal{} 1$.
But $ \left(\frac { \minus{} 3}{p}\right) \equal{} \left(\frac { \minus{} 1}{p}\right)\left(\frac {3}{p}\right) \equal{} ( \minus{} 1)^{\frac {p \minus{} 1}{2}}( \minus{} 1)^{\frac {p \minus{} 1}{2}\frac {3 \minus{} 1}{2}}\left(\frac {p}{3}\right) \equal{} \minus{} 1$. Contradiction!

If exponent of $ 2$ in $ k \plus{} 2$ is odd, by the same way, $ a \plus{} b \plus{} c$ and $ ab \plus{} bc \plus{} ca$ are even. Both to be a even, $ a,b,c$ all should be a even. Contradiction!

Therefore $ 3(xy \plus{} yz \plus{} zx) \not| x^{2} \plus{} y^{2} \plus{} z^{2}$.


[quote="Franz Joseph"]How can you tell if a system of equations have one, none or infinite solutions?[/quote]
I think it usually depends on the number of equations and variables. But, lets say that even if you do have 3 variables and 3 equations, that doesn't mean you will have 3 solutions.

[quote="Franz Joseph"]How can you tell if a system of equations have one, none or infinite solutions?[/quote]
I hope you mean [u]linear equations[/u].  :maybe:
If you have a system \[{a_{11}x_{1}+a_{12}x_{2}+\ldots+a_{1n}x_{n}=b_{1}}\]  \[{a_{21}x_{1}+a_{22}x_{2}+\ldots+a_{2n}x_{n}=b_{2}}\]  \[{\vdots }\]  \[{a_{m1}x_{1}+a_{m2}x_{2}+\ldots+a_{mn}x_{n}=b_{m}}\] of $m$ equations with $n$ variables, put $\mathbf{A}= \left[ \begin{array}{*{20}c}{{a_{11}}}&{a_{12}}&{\cdots}&{{a_{1n}}}\\{{a_{21}}}&{a_{22}}&{\cdots}&{{a_{2n}}}\\{{\vdots}}&{\vdots}&{\ddots}&{{\vdots}}\\{{a_{m1}}}&{a_{m2}}&{\cdots}&{{a_{mn}}}\\ \end{array}\right]$ and $\mathbf{B}= \left[ \begin{array}{*{20}c}{a_{11}}&{a_{12}}&{\cdots}&{a_{1n}}&{b_{1}}\\{a_{21}}&{a_{22}}&{\cdots}&{a_{2n}}&{b_{2}}\\{\vdots}&{\vdots}&{\ddots}&{\vdots}&{\vdots}\\{a_{m1}}&{a_{m2}}&{\cdots}&{a_{mn}}&{b_{m}}\\ \end{array}\right]$. Then:
[b]1.[/b] If $\mathrm{rank}(\mathbf{A})\ne\mathrm{rank}(\mathbf{B})$, then the system has $0$ solutions.
[b]2.[/b] If $\mathrm{rank}(\mathbf{A})=\mathrm{rank}(\mathbf{B})=n$, then the system has $1$ solution.
[b]3.[/b] If $\mathrm{rank}(\mathbf{A})=\mathrm{rank}(\mathbf{B})<n$, then the system has $\infty$ solutions.

[quote="Franz Joseph"]What is $rank(A)$ :blush: ?[/quote]

you have to go into a little linear algebra. i'll start from the bottom and then tell you the definition.

A linear combination of the vectors $v_{1},v_{2},\ldots,v_{n}$ is a vector of the form $c_{1}v_{1}+c_{2}v_{2}+\ldots+c_{n}v_{n}$, where $c_{i}$ is a real number for all i.
The entries in a given column of a matrix can be represented by a column vector. The set Col(A) is the set of all linear combinations of the columns of of A.
We say that the columns of A [i]span[/i] Col(A), essentially meaning that you can produce any vector in Col(A) by choosing the appropriate coefficients $c_{1},\ldots,c_{n}$.
Associated with Col(A) is the [i]dimension [/i]of Col(A), dim Col(A). The dimension of Col(A) is the smallest number of vectors that can span Col(A). This may or may not equal to the number of column vectors that we originally had, which also span Col(A) (we can take some column vectors away and the remaining column vectors still span the original Col(A)).
So finally, we define rank(A)=dim Col(A).

rank(A) is not equal to rank(B) implies that adding the last column to A (the one with all the b's) changed the Col(A). This is the case iff the last column is not a linear combination of the preceding column vectors, which means that you have no solution if you consider the corresponding matrix equation.

rank(A) = rank(B)=n means that the last column is in fact a linear combination of the columns of A. Also, since rank(A)=n, the solution must be unique.

rank(A)=rank(B)<n implies that a solution exists (since rank(A)=rank(B)), but one column of A is a linear combination of the others (since rank(A)<n) which gives infinitely many solutions.

Hope that helps.  :lol:


Problem:Given triangle $ABC$ inscribed $(O)$ and $(C)$ is an arbitrary circle, $(C)$ doesn't cut $(O)$ let $AA',BB',CC'$ is three tangent from $A,B,C$ to $C$ prove that $aAA',bBB',cCC'$ are three side of a triangle and this triangle degrade to segment iff $C$ tangent to $O$.
Note:
$1/$ Ptoleme's inequality is a special case when $(C)$ is an point ($R_{c}=0$),
$2/$ When $ABC$ is equilateral triangle then $AA',BB',CC'$ are three side of a triangle we can calculate area of this triangle through side of $ABC$, $R_{c}$ and $OO'$ with $O'$ is center of $(C)$
$3/$ If $(C)$ cuts small arc $BC$ we will have the inverse inequality $aAA'\ge bBB'+cCC'$ similarly for $(C)$ cuts small arc $CA,AB$
$4/$ The case $(C)$ tangent $O$ we have the equality ex $(C)$ tangents small arc $BC$ we have $aAA'=bBB'+cCC'$, it is an extension of Ptoleme's theorem, too and it has nice solution base on Ptoleme's theorem. :)

In fact, this is a special case of Casey's Theorem.

Is this [url=http://mathworld.wolfram.com/CaseysTheorem.html]Casey's theorem[/url] ?, why don't I see there isn't any relation with my problem, note that my problem is inequality, or can you show clearly ?

I think it is really a nice extension of Ptoleme's inequality, actually I don't have any solution but I see it is true in computer, can anybody help me to show a solution for this nice problem ? :?:


trigonometry
function
real analysis
real analysis unsolved

Hello, I have a problem to compute the following serie :

\[ S(t,\delta)\equal{}\sum_{\substack {k, l\geq 0 \\ k\minus{}l > 0 }} {(1\minus{}t)^{k\plus{}l}\cdot\frac{\sin\big((k\minus{}l)\delta\big)}{k\minus{}l}\cdot\frac{kl}{k\plus{}l}}\equal{}\ ?\]

More precisly, my aim is to know if $ t^2 S(t,\delta)\stackrel{t\rightarrow0}{\rightarrow}0$ .

Thanks for this lecture.

No, it doesn't tend to $ 0$. It tends (pointwise, or in $ L^2$) to $ \frac 14 \sum_{m\equal{}1}^\infty \frac{\sin m\delta}m$ that is a piecewise linear function with a discontinuity at $ 0$ instead. The trick is to rearrange the sum as
\[ \sum_{m\ge 1}\left(\sum_{\ell>0}t^2(1\minus{}t)^{2\ell\plus{}m}\frac{(\ell\plus{}m)\ell}{2\ell\plus{}m}\right)\frac{\sin m\delta}{m}\]
and to note that the sums in the parentheses are uniformly bounded and are essentially equivalent to $ \sum_{\ell\ge 0}t^2(1\minus{}t)^{2\ell}\frac{\ell}2\to \frac 14$ as $ t\to 0\plus{}$, which is enough to draw the conclusion about the $ L^2$ convergence. The pointwise convergence is slightly more delicate but also not too hard.


Let 0<a,b,c$ \leq1$ Prove that   (2a+1)(2b+1)(2c+1) $ \geq 9(a \plus{} b \plus{} c)$

Hello, first man in space!

Well, this can't be right, because when a,b,c all go to 0, then (2a + 1)(2b + 1)(2c + 1) goes to 1,
while 9(a + b + c) goes to 0.

Now it's still not true.  When $ a\equal{}b\equal{}1$ and $ c\equal{}0$, $ (2a\plus{}1)(2b\plus{}1)(2c\plus{}1) \equal{} 9 < 18 \equal{} 9(a\plus{}b\plus{}c)$.

If $ a,b,c\ge1$, then the statement is true.


How many pairs of positive integers x, y are there, with x =< y, and such that gcd(x,y) = 5! and lcd(x,y) = 50!

True, but we must then divide by $2$ because we have the condition $x \le y$.

Because $x\not =y$, for all pairs (solutions) $(x_{2},y_{2})$ work $x=x_{2},y=y_{2}$ or $x=y_{2},y=x_{2}$, number of solutions $x\le y$ is $2^{15}=32768.$


trigonometry
geometry
geometric transformation
homothety
ratio
rectangle
analytic geometry

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$.  Suppose that the radius of the circle is 5, that $BC = 6$, and that $AD$ is bisected by $BC$.  Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$.  It follows that the sine of the minor arc $AB$ is a rational number.  If this fraction is expressed as a fraction $m/n$ in lowest terms, what is the product $mn$?
[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=dir(200), D=dir(95), M=midpoint(A--D), C=dir(30), BB=C+2*dir(C--M), B=intersectionpoint(M--BB, Circle(origin, 1));
draw(Circle(origin, 1)^^A--D^^B--C);
real r=0.05;
pair M1=midpoint(M--D), M2=midpoint(M--A);
draw((M1+0.1*dir(90)*dir(A--D))--(M1+0.1*dir(-90)*dir(A--D)));
draw((M2+0.1*dir(90)*dir(A--D))--(M2+0.1*dir(-90)*dir(A--D)));
pair point=origin;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));[/asy]

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=796[/img]

Let $A$ be any fixed point on circle $O$ and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AD$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to BC at point N.

Let M be the midpoint of the chord $BC$ such that $BC=3$. From right angle triangle $OMB$, $OM = \sqrt{OB^2 - BM^2} =4$. Thus, $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$.

Notice that the distance $OM$ equals $PN + PO \cos AOM = r(1 + \cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$. From $\cos \angle AOM$, we see that $\tan \angle AOM = \displaystyle \frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$

Next, notice that $\angle AOB = \angle AOM - \angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\tan AOM = \displaystyle \frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} = \displaystyle \frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$. It follows that $\sin AOM =\displaystyle \frac{7^2}{\sqrt{7^2+24^2}} = \frac{7}{25}$, resulting in an answer of $7 \cdot 25=\boxed{175}$.[/hide]

You have some of your labeling mixed up at the end...

... $\tan AOB =  \frac{\tan AOM-\tan BOM}{1+\tan AOM \cdot \tan AOM}=  \frac{\frac43-\frac34}{1+\frac43 \cdot \frac34}= \frac{7}{24}$. It follows that $\sin AOB = \frac{7^{2}}{\sqrt{7^{2}+24^{2}}}= \frac{7}{25}$, resulting in an answer of $7 \cdot 25=\boxed{175}$.

I also made a super accurate drawing :oops:
[img]7841[/img]
[img]7842[/img]

[quote="4everwise"][hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=796[/img]

Let $ A$ be any fixed point on circle $ O$ and let $ AD$ be a chord of circle $ O$. The locus of midpoints $ N$ of the chord $ AD$ is a circle $ P$, with diameter $ AD$. Generally, the circle $ P$ can intersect the chord $ BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $ P$ is tangent to BC at point N.

Let M be the midpoint of the chord $ BC$ such that $ BC \equal{} 3$. From right angle triangle $ OMB$, $ OM \equal{} \sqrt {OB^2 \minus{} BM^2} \equal{} 4$. Thus, $ \tan \angle BOM \equal{} \frac {BM}{OM} \equal{} \frac 3 4$.

Notice that the distance $ OM$ equals $ PN \plus{} PO \cos AOM \equal{} r(1 \plus{} \cos AOM)$ (Where $ r$ is the radius of circle P). Evaluating this, $ \cos \angle AOM \equal{} \frac {OM}{r} \minus{} 1 \equal{} \frac {2OM}{R} \minus{} 1 \equal{} \frac 8 5 \minus{} 1 \equal{} \frac 3 5$. From $ \cos \angle AOM$, we see that $ \tan \angle AOM \equal{} \displaystyle \frac {\sqrt {1 \minus{} \cos^2 \angle AOM}}{\cos \angle AOM} \equal{} \frac {\sqrt {5^2 \minus{} 3^2}}{3} \equal{} \frac 4 3$

Next, notice that $ \angle AOB \equal{} \angle AOM \minus{} \angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $ \tan AOM \equal{} \displaystyle \frac {\tan AOM \minus{} \tan BOM}{1 \plus{} \tan AOM \cdot \tan AOM} \equal{} \displaystyle \frac {\frac 4 3 \minus{} \frac 3 4}{1 \plus{} \frac 4 3 \cdot \frac 3 4} \equal{} \frac {7}{24}$. It follows that $ \sin AOM \equal{} \displaystyle \frac {7^2}{\sqrt {7^2 \plus{} 24^2}} \equal{} \frac {7}{25}$, resulting in an answer of $ 7 \cdot 25 \equal{} \boxed{175}$.[/hide][/quote]

The locus of midpoints  of the chord  is a circle , with diameter . Generally, the circle  can intersect the chord  at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle  is tangent to BC at point N. ???????????

What are you talking about?  I'm completely lost here.

He is using the condition that "$ AD$ is the only chord that is bisected by $ BC$".

[asy]pathpen = linewidth(0.5); pen d = linetype("2 2");
pair A = MP("A",expi(5*pi/4),SW); int n = 20;
D(MP("O",(0,0),SW)); D(unitcircle); D(expi(156*pi/180)--expi(340*pi/180)); D(A);
for(int i = 0; i < n; ++i){
 D(A--expi(2*i*pi/n),d);D((A+expi(2*i*pi/n))/2,black);
}
D(CP(A/2,A));[/asy]

Pick a point $ A$ on the circle and fix it. Let $ D$ travel across the circle, and consider the set of points (the locus) that the midpoint $ N$ takes. It forms a circle, the image of a homothety centered at $ A$ with ratio $ 1/2$, as shown in the diagram, though it has diameter $ AO$ [typo]. Since we are told that $ AD$ is unique, it follows that the line $ BC$ only intersects this circle at one point, eg $ BC$ is tangent.

[quote="4everwise"][hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=796[/img]

Let $ A$ be any fixed point on circle $ O$ and let $ AD$ be a chord of circle $ O$. The locus of midpoints $ N$ of the chord $ AD$ is a circle $ P$, with diameter $ AD$. Generally, the circle $ P$ can intersect the chord $ BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $ P$ is tangent to BC at point N.

Let M be the midpoint of the chord $ BC$ such that $ BC \equal{} 3$. From right angle triangle $ OMB$, $ OM \equal{} \sqrt {OB^2 \minus{} BM^2} \equal{} 4$. Thus, $ \tan \angle BOM \equal{} \frac {BM}{OM} \equal{} \frac 3 4$.

Notice that the distance $ OM$ equals $ PN \plus{} PO \cos AOM \equal{} r(1 \plus{} \cos AOM)$ (Where $ r$ is the radius of circle P). Evaluating this, $ \cos \angle AOM \equal{} \frac {OM}{r} \minus{} 1 \equal{} \frac {2OM}{R} \minus{} 1 \equal{} \frac 8 5 \minus{} 1 \equal{} \frac 3 5$. From $ \cos \angle AOM$, we see that $ \tan \angle AOM \equal{} \displaystyle \frac {\sqrt {1 \minus{} \cos^2 \angle AOM}}{\cos \angle AOM} \equal{} \frac {\sqrt {5^2 \minus{} 3^2}}{3} \equal{} \frac 4 3$

Next, notice that $ \angle AOB \equal{} \angle AOM \minus{} \angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $ \tan AOM \equal{} \displaystyle \frac {\tan AOM \minus{} \tan BOM}{1 \plus{} \tan AOM \cdot \tan AOM} \equal{} \displaystyle \frac {\frac 4 3 \minus{} \frac 3 4}{1 \plus{} \frac 4 3 \cdot \frac 3 4} \equal{} \frac {7}{24}$. It follows that $ \sin AOM \equal{} \displaystyle \frac {7^2}{\sqrt {7^2 \plus{} 24^2}} \equal{} \frac {7}{25}$, resulting in an answer of $ 7 \cdot 25 \equal{} \boxed{175}$.[/hide][/quote]

How did you get this $ OM$ equals $ PN \plus{} PO \cos AOM \equal{} r(1 \plus{} \cos AOM)$ (?
Since the problem is asking sine of arc, why are we computing sine of the angle?

He dropped an altitude from P to OM and noticed that it formed a rectangle with P, N, and M.  So the side of the rectangle that lies on OM is congruent to the radius of circle P, so its length is r.  The part of segment OM that's left is the leg of a right triangle with a hypoteneuse congruent to the radius of cirlce P, so it's length is rcos(AOM).

Hmm I definitely did something completely wrong on this problem can someone tell me what it was?

[hide="pseduo-solution"]
Call the center of the circle O.  $ \triangle AMO \cong \triangle DMO$ so $ \angle AMO \equal{} \angle DMO \equal{} 90$.  This means OM is the height from O to base BC of $ \triangle OBC$.  Then the area of $ \triangle OBC$ is $ 3OM$.  We can also find the area of triangle $ OBC$ using Heron's formula.  $ [OBC] \equal{} \sqrt{8(2)(3)(3)} \equal{} 12$.  Thus $ 3OM \equal{} 12 \Rightarrow OM \equal{} 4$.  $ AMO$ and $ DMO$ are 3-4-5 triangles.  So $ AM \equal{} DM \equal{} 3$  Call the intersection of $ OB$ and $ AD$ $ J$.  From right triangle $ MJO$ we find that $ JO \equal{} \sqrt{x^2 \plus{} 16}$.
By the law of sines $ \frac{3\minus{}x}{\sin\angle BOA} \equal{} \frac{\sqrt{x^2 \plus{} 16}}{\sin\angle MAO}$.  Rearranging gives us $ \sin \angle BOA \equal{} \frac{4(3\minus{}x)}{5\sqrt{x^2 \plus{} 16}}$

Using the law of cosines we get

$ x^2 \plus{} 16 \equal{} (3\minus{}x)^2 \plus{} (5)^2 \minus{} 2(3\minus{}x)(5)(\cos\angle JAO)$
$ x^2 \plus{} 16 \equal{} 9 \plus{} x^2 \minus{} 6x \plus{} 25 \minus{} 2(3\minus{}x)(5)(\frac{3}{5})$
$ \minus{}18 \equal{} \minus{}6x \minus{}2(3\minus{}x)$
$ \minus{}18 \equal{} \minus{}6x \minus{}6 \plus{} 2x$
$ \minus{}12 \equal{} \minus{}4x$
$ x \equal{} 3$

But this doesn't make any sense...!
[/hide]

[quote="Truffles"]
$ x^2 \plus{} 16 \equal{} 9 \plus{} x^2 \minus{} 6x \plus{} 25 \minus{} 2(3 \minus{} x)(5)(\frac {3}{5})$
$ \minus{} 18 \equal{} \minus{} 6x \minus{} 2(3 \minus{} x)$
[/quote]
If I'm not mistaken, your mistake occurred here.  From the top equation, you should simplify to get
\[ \minus{}18\equal{}\minus{}6x\minus{}2(3\minus{}x)(3)\]
(you forgot the $ (5)\left(\frac{3}{5}\right)$ part)
From this we get
\[ \minus{}18\equal{}\minus{}6x\minus{}6(3\minus{}x)\\
\minus{}18\equal{}\minus{}6x\minus{}18\plus{}6x\\
\minus{}18\equal{}\minus{}18\]
So that just leads you around in a circle...

Oh thanks, that was stupid I thought I was going crazy for a while.

[quote="Truffles"]Also is there any other way to do this problem than the way 4everwise posted?[/quote]

[hide]
First, we will show that radius OD is perpendicular to chord BC.

Draw a line through D, parallel to BC. Call this line $l$. If $l$ intersected the circle not only at point D but also at another point - call it E - then chord AE would also have to be bisected by BC (since a line parallel to one side of a triange and bisecting another side must also bisect the third side). That would contradict the statement "AD is the only chord starting at A which is bisected by BC". Therefore, D is the only point of intersection between line $l$ and the circle; therefore $l$ is tangent to the circle at point D, and therefore radius OD is perpendicular to both $l$ and BC.

Let M be the point of intersection of radius OD and chord BC. OD is perpendicular to BC, therefore in isosceles triangle OBC, OM is both altitude and median and BM=MC=3. From Pythegorean, we find OM=4 and DM=5-4=1.

Let N be the point of intersection of chords AD and BC. In isosceles triangle OAD, ON is median, altitude and angle bisector. ON is perpendicular to ND; in right triangle OND, OD is the hypotenuse and NM is an altitude; altitude NM divides triangle OND into similar triangles DNM and NOM; from similarity, $\frac{NM}{DM}=\frac{OM}{NM}$, so  $NM^2=4\cdot1$ and $NM=2$. Applying Pythagorean to triangle MND, we get $ND=\sqrt{5}$.

Now we can find that $sin\angle NOD = \frac{\sqrt{5}}{5}$ and $cos\angle NOD = \frac{2\sqrt{5}}{5}$ and, using double angle formulas, $sin\angle AOD = \frac{4}{5}$ and $cos\angle AOD = \frac{3}{5}$

Finally, $\angle AOB = \angle AOD - \angle BOD$ and using the the formula sin(x-y)=sin(x)cos(y)-cos(x)sin(y) together with sine and cosine of $\angle BOD=\angle BOM$ from the 3-4-5 triangle BOM, we find:
$sin\angle AOB = \frac{4}{5}\cdot\frac{4}{5} - \frac{3}{5}\cdot\frac{3}{5} = \frac{7}{25}$

[/hide]

[hide="Hybrid solution"] Note that there exists another chord through $A$ that is bisected by $BC$ iff there exists a chord through $D$ that is parallel to $BC$. This is easily proven as follows: suppose there exists another chord through $A$, say $AE$ that is bisected by $BC$. Then if we denote $AD \cap BC=F$ and $AE \cap BC=G$, it is easy to see by SAS that $\triangle AFG \sim \triangle ADE$. Thus $BC \parallel DE$. Similarly, if we take some point $E$ such that $DE \parallel BC$, we can see that $AE$ and $AD$ are both bisected by $BC$ by the same pair of similar triangles. Thus our claim is proven, and from it, we can deduce that $D$ must be the midpoint of arc $BC$.

Now if we let $O$ be the center of the circle, when we draw in $OD$, and let $OD \cap BC=H$, we see that since $HC=3$, $OH=4$. Then if we let $I$ be the projection of $A$ onto $OD$, we see that $DI$ must be equal to $2$, so $OI=3$.

Now we can use coordinates. We set $O$ to be the origin, so the equation of the circle is $x^2+y^2=25$. Then WLOG, we let $D=(0,5)$, so then $B=(-3,4)$ and $C=(3,4)$. Then from our second paragraph, we have proven that the y-coordinate of $A$ is $3$, in our configuration, the x-coordinate must be $-4$, so $A=(-4,3)$. At this point we can apply the shoelace theorem to $\triangle OAB$ to find its area to be $7/2$. However, this area is also equal to $1/2 \cdot (OA)(OB) \sin \theta$ where $\sin \theta$ is the answer we are looking for. Now obviously $OA=OB=5$, so solving, we get $\sin \theta=7/25$ so our final answer is $\boxed{175}$. [/hide]

[hide=New Solution]
As others noted, we find that $BC$ must be tangent to the circle with diameter $OA$. Let $M$ be the midpoint of $OA$, $OF\perp BC$ with $F$ on $BC$ and $OF$ defined similarly wrt the circle centered at $M$. We have that $MEFO$ is a trapezoid, and it can easily be found that $EF=2\implies EB=1$. Note that we also have that $AE=\sqrt5, OE=2\sqrt5$. From tangent-chord, $\angle BEA=\angle EOA$. We know that $\cos \angle EOA=\frac{2\sqrt5}{5}=\cos{\angle BEA}$. Using Law of Cosines on $\triangle ABE$, we have that $AB=\sqrt2$. It can easily be computed that $sin\angle AOB=\frac{7}{25}$, so our answer is $\boxed{175}$.
[/hide]

Let the center of the circle be $O$. Fix $B,C,$ and $A$. Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$, it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$.

Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$. Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$. Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$. Clearly, $KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\frac{5}{2}$. The same applies for $FO$, which also equals $\frac{5}{2}$. By the Pythagorean theorem, we deduce that $FL = 2$, so $EK = 2$. This is very important information! Now we know that $BE = 1$, so by Power of a Point, $AE = ED = \sqrt{5}$.

We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 5$, we deduce that $EO = 2\sqrt{5}$. $EC=OC=5$, so $\sin (CEO) = \frac{2\sqrt{5}}{5}$. Furthermore, since $\sin (CEO) = \cos(DEC)$, we know that $\cos (DEC) = \frac{2\sqrt{5}}{5}$. By the law of cosines,
$$DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10$$Therefore, $DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}$. Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$. Then, by the Pythagorean theorem, $OZ = \frac{7\sqrt{2}}{2}$. Thus, $\sin (BOZ) = \frac{\sqrt{2}}{10}$ and $\cos (BOZ) = \frac{7\sqrt{2}}{10}$. As a result, $\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}$. $7 \cdot 25 = \boxed{125}$.

~AopsUser101

P.S. My solution is also in the wiki.

Let $O$ be the center of the circle. The condition is equivalent to $(AO)$ being tangent to $BC$. Since $O$ is $4$ away from $BC$ and the center of $(AO)$ is $\frac{5}{2}$ away from $BC$, $A$ must be $1$ away from BC, and hence $O$ is $3$ away from the line through $A$ parallel to $BC$. Then our answer is $\sin(\theta_1-\theta_2)$ where $\sin \theta_1 = \frac{4}{5}$ and $\sin \theta_2 = \frac{3}{5}$, giving $\frac{7}{25}\implies \boxed{175}$


linear algebra
matrix
linear algebra unsolved

Let $A,B \in {\mathbb R}^{n \times n}$ with $detB \neq 0$.
Show that $P=\left( \begin{array}{ccc} A&I_n\\I_n&B \end{array}\right)$ is positive semi-defined [b]if and only if[/b]$A-B^{-1}\geq0$ and $B>0$. (where $X\geq0$ means that the matrix $X$ is positive semi-definite and $X>0$ means $X$ is strictly positive defined)

Counterexample: $n=1$, $B=.1$, $A=.2$
The determinant of this matrix $P$ is negative.

In your example, isn't $A-B^{-1}$ negative definite?

[quote="jmerry"]No, $n=1$ works just fine; I misread that $A-B^{-1}$ condition as $A-B$.

Proving the $n=1$ case is an easy exercise.

Another clarification: Does "positive definite/semidefinite" include being symmetric?[/quote]

no, they are not symmetric....

ok......no one can prove this ?  :(

Give it some time. I had one direction in the symmetric case; that's why I asked for clarification.

[b]G E N E R A L I Z A T I O N[/b]:

Let $A,B,C \in {\mathbb R}^{n \times n}$ with $detC \neq 0$.
Show that symmetric matrix $P=\left( \begin{array}{ccc} A&B\\B^T&C \end{array}\right)$ is positive semi-defined [b]if and only if[/b]$A-BC^{-1}B^T\geq0$ and $C>0$. (where $X\geq0$ means that the matrix $X$ is positive semi-definite and $X>0$ means $X$ is strictly positive defined).


I have a slight problem with the proof provided for $ sin(a\plus{}b)\equal{}sin a cos b \plus{} sin b cos a$ in volume 2 on page 16.
I don't understand how they can let AH=1 without loss of generality. Can someone explain this to me?

It can be assigned an arbitrary value (here $ 1$ to make the calculations later on easier) because of similarity; as long as the angles remain the same (and here they remain $ 90\minus{}\beta$, $ \alpha\plus{}\beta$, and $ 90\minus{}\alpha$ whatever the altitude $ AH$ is), generality is not lost, as all triangles with those three angles are similar.


There are 26 students at Logic Camp; 18 of them always tell the truth and the other 8 always lie. Of course, everyone at the camp knows who is who. One day, n campers are sitting in a circle. Each of them says: "Exactly one of my two neighbors is a liar." For what values of n > 2 is this possible?

The first approach to this problem requires one to list our some small example and try to find a pattern. However we do know that any values n, for 9 > n > 2 satisfy our condition because if we consist the entire circle of liars, then each person’s statement’ “ Exactly one of my two neighbors is a liar”, would be false as expected due to the fact that everyone is a liar. We know this statement has to be false because each person has two neighbors and if exactly one is a liar, then that means that the other person must always tell the truth, which is false because no-one in the circle tells the truth. Therefore we know the values: 3, 4, 5, 6, 7, and 8 are all possibilities.

     From 9-26, just try to find possibilities using the fact that, if the person right next to a liar, is a truth-teller then the person to the other side must also be a truth-teller, and a truth-teller must be adjacent to exactly 1 liar and 1 truth-teller. The only way this can work is if the number of people is a multiple of 3. Therefore i get that all the possibilities are:

$ 3,4,5,6,7,8,9,12,15,18,21,24$


function
advanced fields
advanced fields unsolved

What is the laplace transform of exp(-t^2)? I couldn't find it in tables.

Thanks

[quote="Kent Merryfield"]Supported on what set?

$f(t)=e^{-t^{2}}$ on all of $\mathbb{R}$ has Laplace transform $\sqrt{\pi}e^{s^{2}/4}.$

However, it is not customary - at least not in a differential equations context - to take Laplace transforms of functions supported on $\mathbb{R}.$ More common is functions supported on $[0,\infty).$ The Laplace transform of the function that is $e^{-t^{2}}$ on $[0,\infty)$ and zero otherwise cannot be expressed in elementary terms.[/quote]

Thanks for the reply.
I need to solve this ODE using the laplace transform:
y" + y - exp (-t^2) = 0 , y(0) = y'(0) = 0

Can I use the transform that you suggested?

[quote="Kent Merryfield"]No, because the differential equation (with initial conditions) context implies that the function is supported on $[0,\infty),$ and then there is no elementary closed form.[/quote]

Do you have any suggestions on how to solve this problem using laplace transform?


Hi all,

After watching Criss Angel, I was at first totally amazed, but needed to dig deeper.  I taped the 2 shows back to back and watched over and over the levitations and the ring in the ice cube trick, and the queen of diamonds card trick and I see now how they're all done.

It appears that he's using a very strong magnetic device for starters.  If you tape the shows you can see this for yourself.

When he does the levitations he's wearing his heavey metal necklaces AND a necklace with a rectangular tag on it.  This tag must have some magnetic properties.  Of course, metal also conducts magnetic energy.  I also noted that as his subjects were arched over backwards (and as well both ladies were about the same height and weight: this probably only works with a certain weight for a person) their clothes in the center of their torso were lifting up instead of hanging off of them as they normally would need to do with gravity pulling down.  So clearly there was a magnetic field created between Criss and the levitation subject.  Just look very closely at the video and you'll see the fabric on each person rising where it should be just hanging down.

He also spends some time talking to the person prior to hypnotizing them, moving his hands around their energy field which is most likely creating the magnetic field at that very point.  He also wears in both instances long sleeved shirts with open cuffs, under which area (the cuff area) he must be wearing magnetic bracelets (the same ones perhaps that he had on when he did that Queen of Hearts card trick, where the "magnetic" deck of cards was controlled by his magnetic bracelet:  I only noticed the bracelets were there after I watched the video several times over.  You just don't pay attention to all those things right away, which is how he pulls this stuff off in the first place).

In both levitations of the women, he supports them from underneath their torsos and above the torso.  I believe at that time he's utilizing the magnetic charge that enables them each to stay "afloat" for that short time period.

After studying the ring in the ice trick I paid closer attention to those things that didn't quite look right, like the smaller slab of ice that seems to be attached to the larger cube where the ring is found.  Taking it back to the very beginning of the trick, he's already moved the ring from one hand to the other as SOON as he's got his hand over the ring.  Just tape it and watch carefully.  You'll see him moving his hand to where the ring can go up the sleeve, right after this he takes his other hand (left hand) reaches over and pulls up his right sleeve.  It's at this moment he's already transferred the ring over to the left hand.  He's also resting his left arm on some kind of a small notebook.  Whether or not the notebook plays a part in the trick, I'm not sure, accept that it probably serves to raise up the left elbow to a better angle where he can do the slight of hand easier.  When you study a video you have a chance to look at every single thing there in the picture.  At that point you start to question "Hey, what's that notebook doing there?  And why does that ice cube look so weird???".

When he goes over to the glass where the ice-cube with the ring supposedly is, he's already taken over this small ice-cube looking square that probably has some kind of a melting component in it. But in any case once he shows the ice with the ring in it, if you look down to the left you'll see a weird piece of "ice" that's black in appearance that must have been used for concealing the ring or burning a hole in a piece of ice or something.

Finally, whenever he levitates himself way up in the air, there's always something behind him, and when he does the super high levitations, there is never a crowd around him.  So he must be lifted up with strings/ invisible strings by high cranes that are out of the view of the camera at that point.

Lisa

with all seriousness now did u not see him lift himself up in the air at a street crossing.. there's nothing in front or behind him at the street crossing. and for the magnets i dont see how that could work for the women because u see the trick that he does where his knee is the only thing keeping him up how can magnets be doing that and then doing the same thing with the girl. it doesn't make that much sense.. and the cards with magnets have u ever seen a magnet bring a card closer to it honestly now i think you're trying a bit too hard here to find things that just arent possible

yeah....criss angel does not work with magnets at all...lets get that straight...no magnetic force is strong enough to hold a 150lb (well...im guessing) person...

i will give you this though..as a professional magician...

for the queen of diamonds card trick he is using influencing mentalist techniques...yes they do exist...and invisible elastic thread wrapped around his wrist. The thread is tied around the card as he slides them around on the floor, and as he gives the card to her...he simply rips it off

believe me..i own several spools of elastic IT...and with how much black he wears...you would never see it.


What is the smallest value of $x$ such that $|5x-1| = |3x+2|$?  Express your answer as a common fraction.

Nice work.

However, you only need 2 of those equations because the other 2 are just each of the first 2 times -1.

well, yeah//

but i'm becomin' a stickler for work

Gautam


faster : we have $a(x+2)(x-3)=ax^{2}+bx+1$ so immediately $a=-\frac{1}{6},\ b=\frac{1}{6}$


limit
logarithms
function
absolute value
real analysis
real analysis solved

This problem was given in the tests in 2003, which determined the team of Sofia University (Bulgaria) for the Bulgarian Maths Olympiad for Undergraduate Students. I found it interesting ... what's your opinion?

Given the real sequence $\{x_n\}_{n=0}^{\infty}$, such that $x_0 > 0$ and

$x_{n+1} = \frac{x_n^2}{e^{x_n} - 1}$  for  $n = 0,1,2,\ldots$

find

$\lim_{n\to\infty}\frac{n(2 - nx_n)}{\ln n}$.

The problem suggests that $x_n \sim 2/n + C{\ln n}/{n^2}$.   The first term is clear, the second would be interesting because the asymptotic expansion is not simply in powers of $x$.

$x_n$ positive decreasing so it's converge $L\geq 0$

$L=0$

$\frac{1}{x_{n+1}}-\frac{1}{x_n}$ tends to $1/2$  with Cesaro theorem

$x_n \sim \frac{2}{n}$

So, it gives first term of the asymptotic.

To: Myth and Moubinool: Obviously your observations are correct!

I don't know how [b]fleeting_guest [/b] got his/her expansion (he/she didn't post details), but it implies Moubinool's conclusion somehow.

Moubi's idea is a key.
We have $\frac{1}{x_{n+1}}-\frac{1}{x_n}=\frac{1}{2}+\frac{x_n}{3}+o(x_n)$.
We know that $x_n\sim \frac{2}{n}$. It gives $\frac{1}{x_n}\sim \frac{n}{2}+\frac{1}{3}(1/2+1/3+...+1/n)\sim \frac{n}{2}+\frac{\ln n}{3}$, therefore $x_n\sim \frac{2}{n}+\frac{2\ln n}{3n^2}$.

Not quite. Don't forget about $o(x_n)$ in Myth's formula, which is merely $o(1/n)$ and  sums just to $o(\ln n)$. Actually I suspect that the next term in the asymptotic decomposition depends on $x_0$ but I haven't checked it accurately ;)

I ran about 2500 terms in a spreadsheet. This suggests that $x_n<\frac2n$ always, and that whatever the next term is, convergence is incredibly slow. The asymptotic expansion isn't any good for actually computing the values without more terms.

[quote="fedja"] The next term is $\frac {C(x_0)}{n^2}$ where $C(x_0)$ is a constant depending on $x_0$, [/quote]

Can we get more information about the function $C(x_0)$ monotony ? continuity ? sign?

$C(x_0)$ behaves like $-\frac a{x_1}$ (I think) near both $x_0=0$ and $x_0=\infty$ and has a minimum (in absolute value) near $x_0=1.6$. The next term after $\frac{C(x_0)}{n^2}$ is positive.
All of this is based on simple numerical work. The calculation has roundoff problems for large or small values of $x_0$.

[quote="fedja"]Actually the last line in Myth's post is incorrect: it should be $\frac2n-\frac43\frac{\ln n}{n^2}$. The next term is $\frac {C(x_0)}{n^2}$ where $C(x_0)$ is a constant depending on $x_0$, so, indeed, you should go to really large values of $n$ to make the second term much larger than the third...[/quote] It seems you like to state that all my posts are incorrect... Though there are no mistakes, but small typos, and the right formulae is $\frac2n-\frac23\frac{\ln n}{n^2}$.

[quote="Myth"] the right formulae is $\frac2n-\frac23\frac{\ln n}{n^2}$.[/quote]
Look: $\frac1{\frac n2+\frac13\ln n}=\frac1{\frac n2\left(1+\frac23\frac{\ln n}n\right)}=\frac 2n \left(1-\frac 23\frac{\ln n}n+\dots\right)=\frac 2n-\frac43\frac{\ln n}{n^2}+\dots$.

As to the statement that "I like to state that all your posts are incorrect", I just like to read almost everything you post and thoroughly enjoy most of it. It is just customary to post a correction if you see a mistake and not to post a "WOW, how exciting!" comment if you enjoy the post (maybe we should propose to admin's to introduce something like "rate this post" button, then I'll gladly click on "I like it very much" option every time I see your good post, but until that is done I'll continue to admire them in silence :D). I have a few other favorite authors and my attitude towards them is the same.

Hi guys,

Thanks for your interest on that problem.
Yesterday I couldn't understand [b]Myth[/b]'s solution and prefered to leave more time to think about it or/and allow other people from the forum to comment it first before I post my solution.

My opinion is that [b]fedja[/b]'s comment on [b]Myth[/b]'s solution is correct, but I feel a bit guilty as I gave a misleading "answer" to [b]Myth[/b] and I suspect that's the reason for his "error". Sorry about that, [b]Myth[/b]!

Anyways, today I decided to post my hidden solution, which to be honest is not entirely mine, and which I think differs from [b]Myth[/b]'s:

[hide][b]First step[/b]: is what Moubi already noticed, namely $\lim_{n\to\infty}nx_n = 2$;

[b]Second[/b]: Rearrange the terms in the limit in the following way:

$\frac{n(2 - nx_n)}{\ln n} = \frac{nx_n(\frac{2}{x_n} - n)}{\ln n}$ we can do that, since $x_0 > 0$ implies $x_n > 0\quad \forall n \in \mathbb{N}$;

[b]Third[/b] (already done in the postings above) Prove that $\lim_{n\to \infty}\frac{2}{x_n} - n = \infty$;

Edit2: We actually don't need this step!

[b]Fourth[/b] (the part, which I think is the most interesting) Using the discrete variant of L'hospital

$\lim_{n\to\infty}\frac{\frac{2}{x_n} - n}{\ln n} = \lim_{n\to\infty} \frac{\frac{x_n}{3} + o(x_n)}{\frac{1}{n}}= \frac{2}{3}$.

So, the finsl answer should be $\frac{4}{3}$.[/hide]

Edit1: OK, I must confess now that "my solution" only slightly differs from [b]Myth[/b]'s solution.

[quote="Moubinool"]
The problem is nice[/quote]

Thanks, Moubi!

[quote="Moubinool"]
Can we get more information about the function $C(x_0)$  monotony ? continuity ? sign?[/quote]

That's interesting ...

Regards,
jivko


[hide="hint"]
Let $S$ be the center of the smaller semicircle.
Then $\triangle QGS$ is... ;) [/hide]

[quote="pianoforte"]A circle named C1 has center Q. A semicircle is drawn inside C1 sharing it's diameter with the radius of C1. A second semicircle is drawn connected to the first semicircle shares it's diameter with radius of the first semicircle, on the side where it would be touching the edge of C1. A line segment extends from Q and is tangent to the second semicircle at point G. What is the ratio of the length of Line QG to the radius of C1?[/quote]

Thanks, frt for the hint. It really helped!  :)

[hide]Say that the radius of C1 is 1. then the radius of the first semicircle would be 0.5. The diameter of the second semicircle also equal to 0.5. The radius of the second semicircle would be 0.25. Using frt's hint, (having the center of the second semicircle S) $\triangle QGS$ would be a right triangle. Using the pythagorean theorum... we can find $QG$

$QG=\sqrt{\left(\frac{3}{4}\right)^2-\left(\frac{1}{4}\right)^2}$

$=\sqrt{\frac{9}{16}-\frac{1}{16}}$

$=\sqrt{\frac{1}{2}}$

$=\frac{\sqrt{2}}{2}$

So the ratio would be $1:\frac{\sqrt{2}}{2}$[/hide]


inequalities
blogs
LaTeX
inequalities proposed

$ 0\le a,b,c \le 1$ and $ a\plus{}b,b\plus{}c,c\plus{}a\ge 1$
Prove that $ \sum(1\minus{}a)^2 \plus{}\frac{2\sqrt{2}abc}{\sqrt{\sum a^2}}\ge 1$

[quote="hxy09"]$ 0\le a,b,c \le 1$ and $ a \plus{} b,b \plus{} c,c \plus{} a\ge 1$
Prove that $ \sum(1 \minus{} a)^2 \plus{} \frac {2\sqrt {2}abc}{\sqrt {\sum a^2}}\ge 1$[/quote]
I think this inequality holds for any nonnegative real numbers $ a,b,c.$ See here: http://canhang2007.wordpress.com/2009/11/23/inequality-76-unknown-author/

[quote="hxy09"]Hi,friend

Very sorry to disturb you again. :blush:
Would you please paste your solution here?
Though I can open your blog by zend2.com,Latex is replaced by blank :( Hardly can I read the formulas you wrote.

Thank you very much![/quote]
It's no problem, my friend. I must say sorry to you and you dont have to say sorry to me. I know this thing disturbs some Chinese members, and I dont know why Chinese members cannot open wordpress blog.

I did this thing because I want everything of me (files, problems, solutions) is easier for me to manage. And I can do everything I want (because this is my page). Very sorry.

And now, I paste my solution here:

[b]Problem. [/b]If $ \displaystyle a,b,c$ are nonnegative real numbers such that $ \displaystyle a^{2}\plus{}b^{2}\plus{}c^{2}>0,$ then

$ \displaystyle (1\minus{}a)^{2}\plus{}(1\minus{}b)^{2}\plus{}(1\minus{}c)^{2}\plus{}\frac{2\sqrt{2}abc}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}\geq 1.$

[b]Proof.[/b] By the AM-GM Inequality, we have

$ \begin{aligned} (1\minus{}a)^{2}\plus{}(1\minus{}b)^{2}\plus{}(1\minus{}c)^{2} &\equal{}3\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}2(a\plus{}b\plus{}c) \\ \displaystyle &\geq 3\plus{}a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}\frac{(a\plus{}b\plus{}c)^{2}\plus{}4}{2} \\ \displaystyle &\equal{}1\plus{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}2ab\minus{}2bc\minus{}2ca}{2}.\end{aligned}$

Therefore, it suffices to prove that

$ \displaystyle \frac{4\sqrt{2}abc}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}\geq 2(ab\plus{}bc\plus{}ca)\minus{}a^{2}\minus{}b^{2}\minus{}c^{2}.$

If $ \displaystyle a^{2}\plus{}b^{2}\plus{}c^{2}\geq 2(ab\plus{}bc\plus{}ca),$ then this is trivial. So let us assume that $ \displaystyle 2(ab\plus{}bc\plus{}ca)\geq a^{2}\plus{}b^{2}\plus{}c^{2}.$ From the fourth degree Schur's Inequality,

$ \displaystyle a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}abc(a\plus{}b\plus{}c)\geq ab(a^{2}\plus{}b^{2})\plus{}bc(b^{2}\plus{}c^{2})\plus{}ca(c^{2}\plus{}a^{2}),$

we get

$ \displaystyle 2(ab\plus{}bc\plus{}ca)\minus{}a^{2}\minus{}b^{2}\minus{}c^{2}\leq \frac{6abc(a\plus{}b\plus{}c)}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca}.$

So, it is enough to check that

$ \displaystyle \frac{4\sqrt{2}abc}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}\geq \frac{6abc(a\plus{}b\plus{}c)}{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca},$

or

$ \displaystyle 2(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)\geq 3\sqrt{\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(a\plus{}b\plus{}c)^{2} }{2}}.$

By the AM-GM Inequality, we have

$ \displaystyle \sqrt{\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(a\plus{}b\plus{}c)^{2}}{2}}\leq \frac{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\dfrac{(a\plus{}b\plus{}c)^{2}}{2}}{2}.$

Therefore,

$ \begin{aligned} &2(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)\minus{}3\sqrt{\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(a\plus{}b\plus{}c)^{2}}{2}} \ge \\ \displaystyle &\geq 2(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)\minus{}3\cdot \frac{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\dfrac{(a\plus{}b\plus{}c)^{2}}{2}}{2} \\ \displaystyle &\equal{}\frac{2(ab\plus{}bc\plus{}ca)\minus{}a^{2}\minus{}b^{2}\minus{}c^{2}}{4}\geq 0. \end{aligned}$

The proof is completed. On the assumption $ \displaystyle c\equal{}\min \{a,b,c\},$ we have equality for $ \displaystyle a\equal{}b\equal{}1$ and $ \displaystyle c\equal{}0.$

[quote="hxy09"]Nice solution,thank you very much :lol:
I knew two solutions for this nice problem,your solution is similar to the second's.
I will post them at this weekend :)[/quote]


This is similar to the last question, again, made by myself. I don't know how to solve it though.

Let $ a$ and $ b$ be positive integers greater than $ 1$ such that $ a^b<1000$, $ (a\plus{}1)^b>1000$, and $ a^{b\plus{}1}>1000$. Find the largest possible value of $ 1000\minus{}a^b$.

For example, a possible pair of $ a$ and $ b$ could be $ (5, 4)$, because $ 5^4<1000$, $ 6^4>1000$, and $ 5^5>1000$. The value $ 1000\minus{}a^b$ would then be $ 375$, but that is not the largest. What is the largest?

[quote="PowerOfPi"]This is similar to the last question, again, made by myself. I don't know how to solve it though.

Let $ a$ and $ b$ be positive integers greater than $ 1$ such that $ a^b < 1000$, $ (a \plus{} 1)^b > 1000$, and $ a^{b \plus{} 1} > 1000$. Find the largest possible value of $ 1000 \minus{} a^b$.

For example, a possible pair of $ a$ and $ b$ could be $ (5, 4)$, because $ 5^4 < 1000$, $ 6^4 > 1000$, and $ 5^5 > 1000$. The value $ 1000 \minus{} a^b$ would then be $ 375$, but that is not the largest. What is the largest?[/quote]

[hide="brutal force.........."]

Let condition $ (1)$ be $ \begin{cases}a^b<1000 \\ a^{b\plus{}1}>1000 \end{cases}$
codition $ (2)$ be $ (a\plus{}1)^b>1000$

Now by $ (1)$, $ a\equal{}2 \Rightarrow b\equal{}9 \Rightarrow a^b\equal{}512$, which satisfies $ (2)$
Similarly $ a\equal{}3 \Rightarrow b\equal{}6 \Rightarrow a^b\equal{}729$, which satisfies $ (2)$
$ a\equal{}4 \Rightarrow b\equal{}4$, which does not satisfies $ (2)$
$ a\equal{}5 \Rightarrow b\equal{}4 \Rightarrow a^b\equal{}625$, which satisfies $ (2)$
$ a \ge 6 \Rightarrow b \le 3$

[b]Case (a):[/b] $ b\equal{}3$, then $ \begin{cases}a^3<1000 \\ (a\plus{}1)^3 >1000 \end{cases} \Leftrightarrow \begin{cases} a<10 \\ a\plus{}1>10 \end{cases}$, contradiction.

[b]Case (b)[/b] $ b\equal{}2$, then $ \begin{cases}a^2<1000 \\ (a\plus{}1)^2 >1000 \end{cases} \Leftrightarrow \begin{cases} a \le 31 \\ a\plus{}1 \ge 32 \end{cases} \Leftrightarrow a\equal{}31$, which satisfies $ (2)$ also. And in this case $ a^b\equal{}961$

To maximize $ 1000\minus{}a^b$ is to minimize $ a^b$. And the minimum $ a^b$ is 512. So the largest possible value of $ 1000\minus{}a^b\equal{}488$
[/hide]


[quote="carlosbr"][color=blue]A rectangular grid is colored in checkerboard fashion, and each cell
contains an integer.

It is given that the sum of the numbers in each
row and the sum of the numbers in each column is even. Prove that
the sum of all numbers in black cells is even.  :o

Carlos Bravo
Lima -PERU[/color][/quote]

add up all the odd rows and columns, get the result by parity


What 5 digit number times 4 is its own digits backward?

[hide]

Let the number be $[abcde]$.

Clearly $4a < 10 \implies a = \{1, 2\}$. $a$ can't equal 1 since $4e = a \pmod{10}\therefore a = 2$.

Since $4e = a = 2 \pmod{10}, b = \{3, 8\}$. But $e$ cannot equal 3, since $3 < 4a \leq e \implies e = 8$. Continuing from here isn't hard...

$\boxed{21978}$

[/hide]

This is a very tricky question.

What do you mean by MOD 10?

$\text{4e = a (mod 10)}$ means that the remainder when $4e$ is divided by $10$ is $a$.


geometry
analytic geometry
trigonometry
function
LaTeX
symmetry
trig identities

...and in response to the request for solutions, here are some for week 7:

4. Drawing a diagram, we see that the resulting triangle is isosceles with a base of $4\sqrt{2}$ and a height of $(k-5)\sqrt{2}$, since the altitude from $(k,k)$ to the opposite side is the line from $(k,k)$ to (5,5).  So from the familiar formula area = 1/2*base*height, we have

\[\frac12 \left( (k-5)\sqrt{2} \right) \left( 4\sqrt{2} \right) = 600\]

Solving yields $k=155$.

5. From the given information, R = 1.6P and $\frac{P+R}{2} = 60$.  So $1.3P = 60$, or $P = \frac{600}{13}$.  Thus 613.

6. Since $100 \le n \le 909$, $1100 \le 11n \le 9999$.  Thus $11n$ is a four-digit number.  In order for $11n$ to contain two consecuitive zeroes, then, it must be of the form _00_ or _ _00.  For the first case, let such a number be a00b.  Recall that abcd is a multiple of 11 iff a - b + c - d is equivalent to 0 mod 11.  Thus a - b is a multiple of 11.  But this can only be the case if a = b, since a < b + 11 and a > b - 11.  So only 1001, 2002, ... 9009 could be values of 11n.  Of these, 2002, ... 9009 work for $11n$, but 1001 doesn't, because $\frac{1001}{11} = 91 < 100$.  Thus n = 182, ... 91*9 = 91(2, 3, ... 9).  Their sum is then 91*44 = 4004.  For the second case, let the number be ab00.  Again, a = b for the same reasons as before; this time, 1100, 2200, ... 9900 all work as values of 11n, i.e. 100, 200, ... 900 all work for n.  Their sum is 100(1+2+...9) = 100*45 = 4500.  Thus the total is 4004 + 4500, which is 504 mod 1000.

8. Coordinate geometry, ugh.  Anyone have a better way?

Ill just sketch the solution, and when I have time, I can fully write it out.

8: We can get $\sin MOA$ since we know $\sin MOB$. We know $PO$ and $BO$ so we apply the law of cosines to find $PM$. Now we use power of a point to find $P'P$. We see $P'M=P'P + PM$ and finally, by Pythagoras, $O'M^2-MP'^2=P'O'^2$.

Hmm.. for Individual #9, generating functions work, but i don't know any low-tech ways.

So basically you want the first 4 numbers not to have 0's and their sum needs to *not* be a power of three, since you can get 2^x (sum of 1st four numbers) for the rest of the numbers.

The number of ways the sum can be a power of 3 is the sum of the coefficients of x^3k in f(x) = (1 + 2x + 2x^2)^4.  This is equal to (f(1) + f(omega) + f(omega^2))/3, where omega is a cubic root of unity, since the non-x^3k terms will all cancel out.  Evaluating, you get 209.  So the answer is 5^4 - 209 = 416.

and *not* 209... grr.

Sorry about the lack of LaTeX in this post -- I'm not very proficient with it

#8
[hide="Solution"]

Let the circle have radius $\frac{1}{2}$ (This is where i messed up). Let point $D$ be where $MD$ is perpendicular to $AB$. Thus, all you need to find is $\sin \angle P'MO'$ since that will give you $O'P'$. First we know that $OP=\frac{7}{10}-\frac{1}{2}=\frac{1}{5}$ and $OM=\frac{1}{2}$ since it is a radius. Then we find that $MD=\frac{2}{5}$ after using the sine formula to express the area of $\triangle MOB$ in two ways. Using Pythagorean Theorem we find $MP=\sqrt{0.41}$ and $\sin \angle MPD=\frac{0.4}{\sqrt{0.41}}$. Hence

\[\frac{\sin \angle P'MO'}{OP}=\frac{\sin \angle MPD}{OM}\]
\[\frac{\sin \angle P'MO'}{0.2}=\frac{\sin \angle MPD}{0.5}\]
\[\sin \angle P'MO'=\frac{2}{5} \cdot \sin \angle MPD\]
\[\sin \angle P'MO'=\frac{2}{5} \cdot \frac{0.4}{\sqrt{.41}}\]
\[\sin \angle P'MO'=\frac{8 \sqrt{41}}{205}\]

However, since I said that the circle had radius $\frac{1}{2}$ we multiply our answer by 2 since the circle actually has radius 1. Thus,$O'P'=\frac{16 \sqrt{41}}{205}$ so the answer is $\fbox{262}$

[/hide]

Alternate method for number 4:

If the vertices of a triangle are at $(x_1, y_1), (x_2, y_2), (x_3, y_3)$, then the area is the determinant of:
 \[
\left(
\begin{array}{ccc}
x_1 & y_1 & \frac{1}{2} \\
x_2 & y_2 & \frac{1}{2}  \\
x_3 & y_3 & \frac{1}{2}  \end{array}
\right)\]

In this case:
\[
\left(
\begin{array}{ccc}
3 & 7 & \frac{1}{2} \\
7 & 3 & \frac{1}{2}  \\
k & k & \frac{1}{2}  \end{array}
\right)\]

$1200=(4k)-(-4k)-40$
$k=155$

Probably slower, but I didn't see the symmetry along $y=x$ immediately.

Here's how I did #8

Drop a perpendicular from M to AB, say the foot is M'. MM'=4/5, M'O=3/5, AP=3/5, PO=2/5, PM'=1, MP=1^2+(4/5)^2=\sqrt{41}/5.

Power of a Point,

PP'(\sqrt{41}/5)=AP(PB)=3/5(7/5)=21/25

PP'=21rt(41)/205

P'M=62rt(41)/205, MO'P' is a right triangle
pythagoras,
O'P'= 16\sqrt{41}/205, add!

I used Law of Cosines many times on #8.

For #9, I wrote the terms as

$a, b, c, d, a+b+c+d, 2a+2b+2c+2d, 4a+4b+4c+4d, 2a+2b+2c+2d,\ldots$.

Therefore the sequence contains no zeros if the first four terms are nonzero and their sum is not divisible by 3. From there it's easy.

[quote="Simon"]Therefore the sequence contains no zeros if the first four terms are nonzero and their sum is not divisible by 3. From there it's easy.[/quote]

Thats how I did the problem. Unfortunately, when it came to counting the sequences of first four terms, I had to do it by brute force and I made a calculation error. Whats the easy way?

I got that the sum of the first 4 terms must be 1, 2(mod 3) but I don't know how to continue from there since it gets really messy.

I wrote a list of all the possibilities for the first four terms modulo 3 (without taking order into account) and checked which of those had sum divisible by 3. There weren't that many cases:

(1,1,1,1) no
(1,1,1,2) no
(1,1,1,3) yes
(1,1,2,2) yes
(1,1,2,3) no
(1,1,3,3) no
(1,2,2,2) no
(1,2,2,3) no
(1,2,3,3) yes
(1,3,3,3) no
(2,2,2,2) no
(2,2,2,3) yes
(2,2,3,3) no
(2,3,3,3) no
(3,3,3,3) yes

Now we count how many ways there are to do each of the ones with sum divisible by 3:

(1,1,1,3): 4*8=32
(1,1,2,2): 6*16=96
(1,2,3,3): 12*4=48
(3,3,3,3): 1*1=1

Sum is 209.

Now we subtract that from $5^4=625$ to get 416.


Let $X_1$, $Z_2$, $Y_1$, $X_2$, $Z_1$, $Y_2$ be six points lying on the periphery of a circle (in this order).

Let the chords $Y_1Y_2$ and $Z_1Z_2$ meet at a point $A$; let the chords $Z_1Z_2$ and $X_1X_2$ meet at a point $B$; let the chords $X_1X_2$ and $Y_1Y_2$ meet at a point $C$.

Prove that

$\left( BX_2-CX_1\right) \cdot BC+\left( CY_2-AY_1\right) \cdot CA+\left( AZ_2-BZ_1\right) \cdot AB=0$.

[i]Comment on the source.[/i] The problem is inspired by Stergiu's proof in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=326112#p326112]http://www.mathlinks.ro/Forum/viewtopic.php?t=50262 post #5[/url].

  Darij


AoPS Wiki seems unable to parse <math>\displaystyle C </math>, for some reason.  It's pretty easy to avert (just put curly braces around C), but I thought I'd make it known.

[quote="joml88"]It's already very well known that the wiki doesn't parse a lot of things.  Just find ways to work around it whenever it wont parse.[/quote]Okay; thanks.  I didn't know. :)


let a, b, c be sides of a triangle, prove
\[ \frac{3abc}{a\plus{}b\plus{}c} \leq \frac{((a\plus{}b)^{2}\plus{}(b\plus{}c)^{2}\plus{}(c\plus{}a)^{2} )^{2}}{9}\]

[quote="doicu"]let a, b, c be sides of a triangle, prove
\[ \frac {3abc}{a \plus{} b \plus{} c} \leq \frac {((a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \plus{} (c \plus{} a)^{2} )^{2}}{9}
\]
[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{5}.$ :wink:

sorry i forgot to mention $ a,b,c > 1$ ... i think this shuld work  :blush:

[quote="doicu"]let a, b, c be sides of a triangle, prove
\[ \frac {3abc}{a \plus{} b \plus{} c} \leq \frac {((a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \plus{} (c \plus{} a)^{2} )^{2}}{9}
\]
[/quote]please tell me i'm correct... :maybe:
$ \frac{((a\plus{}b)^2\plus{}(b\plus{}c)^2\plus{}(c\plus{}a)^2)^2}{9}\ge \frac{(4ab\plus{}4bc\plus{}4ca)^2}{9}$, and $ \frac{3abc}{a\plus{}b\plus{}c}\equal{}\frac{3}{\frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}}$
the inequality is equivalent to $ \left(\frac{ab\plus{}bc\plus{}ca}{3}\right)^2\left(\frac{\frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}}{3}\right)\ge \frac{1}{16}$
$ \left(\frac{ab\plus{}bc\plus{}ca}{3}\right)^2\left(\frac{\frac{1}{ab}\plus{}\frac{1}{bc}\plus{}\frac{1}{ca}}{3}\right)\ge \sqrt[3]{a^2b^2c^2}$ by AM-GM and $ abc\ge \frac{1}{64}$ clearly. i think the bound should be $ a,b,c\ge \frac{1}{4}$?

[quote="minsoens"][quote="doicu"]let a, b, c be sides of a triangle, prove
\[ \frac {3abc}{a \plus{} b \plus{} c} \leq \frac {((a \plus{} b)^{2} \plus{} (b \plus{} c)^{2} \plus{} (c \plus{} a)^{2} )^{2}}{9}
\]
[/quote]please tell me i'm correct... :maybe:
$ \frac {((a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2)^2}{9}\ge \frac {(4ab \plus{} 4bc \plus{} 4ca)^2}{9}$, and $ \frac {3abc}{a \plus{} b \plus{} c} \equal{} \frac {3}{\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}}$
the inequality is equivalent to $ \left(\frac {ab \plus{} bc \plus{} ca}{3}\right)^2\left(\frac {\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}}{3}\right)\ge \frac {1}{16}$
$ \left(\frac {ab \plus{} bc \plus{} ca}{3}\right)^2\left(\frac {\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}}{3}\right)\ge \sqrt [3]{a^2b^2c^2}$ by AM-GM and $ abc\ge \frac {1}{64}$ clearly. i think the bound should be $ a,b,c\ge \frac {1}{4}$?[/quote]

yes it seems correct. :wink:  (when i posted the inequality i have done some mistakes with the bounds)


1.(Russia 2001)
Find all $ x,y\in N$satisfying $ \frac{x^{2}\plus{}y}{xy\plus{}1}\in Z$
2. Consider $ x\in R$ and $ x^{3}\minus{}x,x^{4}\plus{}x\in Z$. Prove that: $ x\in Z$

for the first question, let $ k \equal{} \frac {x^{2} \plus{} 1}{xy \plus{} 1}$ with $ k\in\mathbb{Z}$.
this is equilalent to $ x^{2} \minus{} kxy \plus{} (y \minus{} k) \equal{} 0$
the determinent must be positive,so $ k^{2}y^{2} \plus{} 4k \minus{} 4y > 0.$
so $ 1 \minus{} k^{3}\geq\ 0$.
if $ k \equal{} 0$ so $ x \equal{} y \equal{} 0.$
if $ k \equal{} 1$ so $ x \equal{} y \equal{} 1$ or  $ x \equal{} 0$ and $ y \equal{} 1$.

[quote] so $ 1\minus{}k^3 \ge 0$ [/quote] Sorry but how does this follow?

[quote="duythuc_lqd"]1.(Russia 2001)
Find all $ x,y\in N$satisfying $ \frac {x^{2} \plus{} y}{xy \plus{} 1}\in Z$
2. Consider $ x\in R$ and $ x^{3} \minus{} x,x^{4} \plus{} x\in Z$. Prove that: $ x\in Z$[/quote]
Let $ x^4\plus{}x\equal{}a \in \mathbb{Z}, x^3\minus{}x \equal{} b \in \mathbb{Z}$, and denote:
$ f(t) \equal{} t^4\plus{}t\minus{}a, g(t) \equal{} t^3\minus{}t\minus{}b$, then both $ f(t)$ and $ g(t)$ has the root $ x$.
Then $ x$ is also a root of $ h(t) \equal{} f(t) \minus{} t \cdot g(t) \equal{} t^4\plus{}t\minus{}a\minus{}t^4\plus{}t^2\minus{}tb \equal{} t^2\plus{}t(1\minus{}b)\minus{}a$, and of:
$ j(t) \equal{} g(t) \minus{} t \cdot h(t) \equal{} t^3\minus{}t\minus{}b \minus{} t^3\minus{}t^2\plus{}bt^2\plus{}at \equal{} (b\minus{}1)t^2\plus{}(a\minus{}1)t\minus{}b \equal{} 0$, and of:
$ k(t) \equal{} j(t)\minus{}(b\minus{}1)h(t) \equal{} (b\minus{}1)t^2\plus{}(a\minus{}1)t\minus{}b\minus{}(b\minus{}1)t^2\plus{}t(1\minus{}b)^2\plus{}a(b\minus{}1) \equal{} (a\minus{}1\plus{}(b\minus{}1)^2)t\minus{}b\plus{}ab\minus{}a$
If $ a\minus{}1\plus{}(b\minus{}1)^2 \neq 0$ we see that $ x \equal{} \frac{\minus{}a\plus{}ab\minus{}b}{a\minus{}1\plus{}(b\minus{}1)^2}$, and hence $ x$ is rational. Let $ x \equal{} \frac{p}{q}$ then $ x^3\minus{}x \equal{} \frac{p^3\minus{}pq^2}{q^3} \in \mathbb{Z} \Rightarrow q^3 \mid p^3\minus{}pq^2 \Rightarrow q \mid p^3\minus{}pq^2 \iff q \mid p^3 \iff q \mid 1$, so $ q \equal{} \pm 1$, and hence $ x \in \mathbb{Z}$
Now assume $ a\minus{}1\plus{}(b\minus{}1)^2 \equal{} 0 \iff a \equal{} 1 \minus{} (b\minus{}1)^2$. Since $ k(x) \equal{} 0$ and $ k(t) \equal{} \minus{}a\plus{}ab\minus{}b$ we must also assume $ \minus{}a\plus{}ab\minus{}b \equal{} 0 \iff a\equal{}b\equal{}0 \text{  or  } a \equal{} \frac{b}{b\minus{}1} \equal{} \frac{1}{b\minus{}1}\minus{}1$.
First asssume $ (a,b) \neq 0$. Since $ a \in \mathbb{Z}$ we see that $ b\minus{}1 \equal{} \pm 1 \iff b \equal{} 0 \text{  or  } b \equal{} 2 \iff b \equal{} 2$ (Because $ (a,b) \neq (0,0)$. Then $ a \equal{} 2$. So we have $ x^3\minus{}x\equal{}2,x^4\plus{}x\equal{}2$.
Assume $ x^4\plus{}x\minus{}2\equal{}0,x^3\minus{}x\minus{}2\equal{}0$ and $ x \notin \mathbb{Z}$
$ x^4\plus{}x\minus{}2 \equal{} (x\minus{}1)(x^3\plus{}x^2\plus{}x\plus{}2) \equal{} 0 \iff x^3\plus{}x^2\plus{}x\plus{}2 \equal{} 0$ (Because $ x \notin \mathbb{Z}$). Using the same trick as i've used many times now: $ x^3\plus{}x^2\plus{}x\plus{}2\minus{}x(x^2\minus{}x\minus{}2) \equal{} 2x^2\plus{}3x\plus{}2 \equal{} 0$ and $ 2(x^3\minus{}x\minus{}2)\minus{}x(2x^2\plus{}3x\plus{}2) \equal{} \minus{}2(3x^2\plus{}2x\plus{}2) \equal{} 0$. And thus:
$ 3(2x^2\plus{}3x\plus{}2) \minus{} 2(3x^2\plus{}2x\plus{}2) \equal{} 5x\plus{}1$, and hence $ x \equal{} \frac{\minus{}1}{5}$, which clearly doesn't work. So we are left with the last case:
$ (a,b)\equal{}(0,0)$: $ x^3\minus{}x\equal{}x^4\plus{}x\equal{}0$. $ x^3\minus{}x \equal{} 0 \iff x(x\minus{}1)(x\plus{}1)\equal{}0$, and hence $ x \in \mathbb{Z}$.

And we are done :) $ x^4\plus{}x, x^3\minus{}x \in \mathbb{Z} \iff x \in \mathbb{Z}$ :)

[quote="duythuc_lqd"]1.(Russia 2001)
Find all $ x,y\in N$satisfying $ \frac {x^{2} \plus{} y}{xy \plus{} 1}\in Z$
2. Consider $ x\in R$ and $ x^{3} \minus{} x,x^{4} \plus{} x\in Z$. Prove that: $ x\in Z$[/quote]

[b]1)[/b][b]Case 1:[/b]  $ y \ge x \plus{} 1$ , then
$ xy \plus{} 1 \equal{} (x \minus{} 1)y \plus{} y \plus{} 1 \ge (x \minus{} 1)(x \plus{} 1) \plus{} y \plus{} 1 \equal{} x^2 \plus{} y$
Equality holds $ \Leftrightarrow y \equal{} x \plus{} 1$ is a solution. If equality doesn't hold then $ denominator > nominator$, contradiction.

[b]Case 2:[/b] $ y \le x$, then
$ \frac {x^{2} \plus{} y}{xy \plus{} 1}\in Z \Rightarrow \frac {x^{2}y \plus{} y^2}{xy \plus{} 1}\in Z \Leftrightarrow \frac {y^{2} \minus{} x}{xy \plus{} 1}\in Z$

But $ xy \plus{} 1 \ge y^2 \plus{} 1 > y^2 \minus{} 1$, so $ \frac {y^{2} \minus{} x}{xy \plus{} 1} \equal{} 0$

Put $ x \equal{} y^2$ we get $ \frac {y^4 \plus{} y}{y^2 \plus{} 1}\in Z$
$ \Leftrightarrow \frac { \minus{} y^2 \plus{} y}{y^2 \plus{} 1}\in Z$

But $ y^2 \minus{} y < y^2 \plus{} 1$ so $ \frac { \minus{} y^2 \plus{} y}{y^2 \plus{} 1} \equal{} 0 \Leftrightarrow (x,y) \equal{} (1,1)$ as solution for this case.

So $ (x,y) \equal{} (1,1), (k,k \plus{} 1)$ , where $ k$ are any positive integers, are the solutions.
For either one of $ x,y$ be $ 0$, $ (x,y) \equal{} (0,k), (k,0)$ are solutions.

[b]2)[/b] Clearly for $ x^{3} \minus{} x$ or $ x^{4} \plus{} x$ being zero, the statement is true.

Now for $ x^{3} \minus{} x,x^{4} \plus{} x \neq 0$, we have

$ \frac {x^4 \plus{} x}{x^3 \minus{} x} \in Q$
$ \Leftrightarrow \frac {x^2 \minus{} x \plus{} 1}{x \minus{} 1} \in Q$
$ \Leftrightarrow x^2 \equal{} k(x \minus{} 1)$ for $ k \in Q$

Now $ (x^4 \plus{} x) \plus{} (x^3 \minus{} x) \in Q \Leftrightarrow x^4 \plus{} x^3 \in Q \Leftrightarrow [k(x \minus{} 1)]^2 \plus{} k(x \minus{} 1)x\in Q$
$ \Leftrightarrow k^2[k(x \minus{} 1) \minus{} 2x \plus{} 1] \plus{} k[k(x \minus{} 1) \minus{} x] \in Q$
$ \Leftrightarrow x \in Q$

Now let $ x \equal{} \frac {q}{p}$ in simplest form.
$ x^4 \plus{} x \in Z \Leftrightarrow \frac {q^4 \plus{} p^3q}{p^4} \Rightarrow p^4|(q^4 \plus{} p^3q) \Rightarrow p^3|q^4 \Rightarrow p \equal{} 1$,

$ \Leftrightarrow x \in Z$

[quote="pco"][quote="Mathias_DK"] ...$ f(t) \equal{} t^4 \plus{} t \minus{} a, g(t) \equal{} t^3 \minus{} t \minus{} b$, then both $ f(t)$ and $ g(t)$ has the root $ x$.
Then $ x$ is also a root of $ h(t) \equal{} f(t) \minus{} t \cdot g(t) \equal{} t^4 \plus{} t \minus{} a \minus{} t^4 \plus{} t^2 \minus{} tb \equal{} t^2 \plus{} t(1 \minus{} b) \minus{} a$, ...[/quote]

I think this is  $ t^2 \plus{} t(1 \plus{} b) \minus{} a$ here, instead of $ t^2 \plus{} t(1 \minus{} b) \minus{} a$[/quote]
Thanks. I'm not sure if i'm going to edit it, since the idea will be the same, but thanks :)

[quote="stephencheng"]
$ \frac {x^{2} \plus{} y}{xy \plus{} 1}\in Z \Rightarrow \frac {x^{2}y \plus{} y^2}{xy \plus{} 1}\in Z \Leftrightarrow \frac {y^{2} \minus{} x}{xy \plus{} 1}\in Z$

But $ xy \plus{} 1 \ge y^2 \plus{} 1 > y^2 \minus{} 1$, so $ \frac {y^{2} \minus{} x}{xy \plus{} 1} \equal{} 0$
[/quote]
I think there is a mistake.
$ \frac {y^{2} \minus{} x}{xy \plus{} 1}\in Z$
and $ \frac {y^{2} \minus{} x}{xy \plus{} 1}< 1$
$ \Rightarrow\frac {y^{2} \minus{} x}{xy \plus{} 1} \equal{} 0$ is incorrect

[quote="duythuc_lqd"][quote="stephencheng"]
$ \frac {x^{2} \plus{} y}{xy \plus{} 1}\in Z \Rightarrow \frac {x^{2}y \plus{} y^2}{xy \plus{} 1}\in Z \Leftrightarrow \frac {y^{2} \minus{} x}{xy \plus{} 1}\in Z$

But $ xy \plus{} 1 \ge y^2 \plus{} 1 > y^2 \minus{} 1$, so $ \frac {y^{2} \minus{} x}{xy \plus{} 1} \equal{} 0$
[/quote]
I think there is a mistake.
$ \frac {y^{2} \minus{} x}{xy \plus{} 1}\in Z$
and $ \frac {y^{2} \minus{} x}{xy \plus{} 1} < 1$
$ \Rightarrow\frac {y^{2} \minus{} x}{xy \plus{} 1} \equal{} 0$ is incorrect[/quote]

Sorry , I omitted a small case.

[b]Correction:[/b]
It's easy to see that $ \minus{} 1 < \frac {y^{2} \minus{} x}{xy \plus{} 1} \Leftrightarrow x < y^2 \plus{} xy \plus{} 1$ which is true for non-zero $ y$

Together with the above result ($ \frac {y^{2} \minus{} x}{xy \plus{} 1} < 1$) gives $ \frac {y^{2} \minus{} x}{xy \plus{} 1} \equal{} 0$


Of a group of boys and girls at Central Middle School's after-school party, 15 girls left early to play in a volleyball game. The ratio of boys to girls then remaining was 18 to 5. Later, 45 boys left for a football game. The ratio of girls to boys was then 5 to 9. How many students attended the party?

[hide]Call the number of girls at the party g and the number of boys at the party b. Then, after the fifteen girls left, it could be said that 18(g-15)=5(b). After the 45 boys left, then 9(g-15)=5(b-45). Distributing and solving these two equations yields g=40, b=90, or g+b=[b]130 students[/b].[/hide]


symmetry
geometry
trapezoid
geometry unsolved

[color=darkblue]Let $ ABC$ is a triangle, let $ E$ is a change point, know that $ E$ lie on the symmetry ray of $ CB$ ray. Let $ (C_1)$ and $ (C_2)$ are incribed circles of $ \triangle ABE$ and $ \triangle ACE$ respectively. Let $ (C_1)$ meet $ (C_2)$ at $ M$ and $ N$. Prove that the line $ MN$ always pass a fixed point when $ E$ change.[/color]

See the figure below.
$ BI_1\cap EF\equal{}\{M\}$ then $ \angle BMA\equal{}90^o$, $ \angle ABC\equal{}constant$ therefore $ M$ is fixed point.
Similarly, $ N\equal{}CI_2\cap HG$ then $ N$ is fixed point.
$ IJ$ passes through the midpoint $ P$ of $ EH$. because $ PE^2\equal{}PI.PJ\equal{}PH^2$
It's easy to see that $ HG//IJ//EM$ ($ \perp I_1D$) therefore $ IJ$ is the midline of trapezoid $ MNHE$. So $ IJ$ must passes through the midpoint $ Q$ of $ MN. Q$ is a fixed point.


combinatorial geometry
linear algebra
linear algebra unsolved

for $A\in \mathbb{R}^{m}$ define $A^{(n)}=\cup_{B\subset A,|B|=n}\overline{B}$ ,$\overline{B}$ is convex hull of $B$.prove that if $A\subset \mathbb{R}^{m}$ and $\overline{A}\neq A^{(m)}$ then $A$ has at least $m+1$ connected component.

This is a strenghening of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=convex&t=55852]this[/url], but that proof can easily be modified to yield the stronger result.

Basically, it works like this: take a point $P$ in the convex hull of $\mathcal A$. By Caratheodory's Theorem, it's a convex combination of $m+1$ points in $A$, say $A_{1},\ldots,A_{m+1}$, and we want to prove that if it's not a convex combination of $m$ points of $\mathcal A$, then $\mathcal A$ has at least $m+1$ connected components.

By examining the proof in the link, we see that if $P$ is not a convex combination of $m$ points in $\mathcal A$, then the connected component of $A_{m+1}$, for instance, cannot contain any of the $A_{i},\ i=\overline{1,m}$. Of course, this works for the other points as well, so each of the $A_{i},\ i=\overline{1,m+1}$ must lie in a different connected component, and we're done.


geometry
3D geometry
modular arithmetic
Support
algebra
polynomial
complex numbers

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is 7 and the sum of the cubes is 10.  What is the largest real value that $x + y$ can have?

[hide="Solution"]The best way to solve this problem seems to be by brute force.

$w^2+z^2=(w+z)^2-2wz=7$ and
$w^3+z^3=(w+z)(w^2+z^2)-wz(w+z)=(w+z)(7)-wz(w+z)=(7-wz)(w+z)=10$

Because we are only left with $w+z$ and $wz$, substitution won't be too bad. Let $x=w+z$ and $y=wz$.

We get $x^2-2y=7$ and
$x(7-y)=10$

Because we want the largest possible $x$, let's find an expression for $y$ in terms of $x$. $x^2-7=2y \implies y=\frac{x^2-7}{2}$.

Substituting, $x^3-21x+20=0$. Factored, $(x-1)(x+5)(x-4)=0$

The largest possible solution is therefore $4$.[/hide]

I got little different solutions.
Lets say $ x$ and $ y$ are complex numbers.
$ x \equal{} a \plus{} bi$ and $ y \equal{} c \plus{} di$
First we note that if we want $ x \plus{} y \in R$ we need to have $ \minus{} d \equal{} b$
So the complex numbers are really
$ x \equal{} a \plus{} bi$
$ y \equal{} c \minus{} bi$

$ x^2 \plus{} y^2 \equal{} a^2 \plus{} c^2 \minus{} 2b^2 \plus{} 2bi(a \minus{} c)$
Since $ x^2 \plus{} y^2 \in R$ and $ b\neq0$ then $ a \equal{} c$

$ x^3 \plus{} y^3 \equal{} 2a^3 \minus{} 6ab^2 \equal{} 10$
$ a^3 \minus{} 3ab^2 \equal{} 5$
$ a^3 \minus{} 5 \equal{} 3ab^2$

Note that $ a^3 \equiv 2$ $ mod 3$
By Fermat Little theorem $ a$ has to be $ 2$

Therefore $ x \plus{} y \equal{} (a \plus{} bi) \plus{} (a \minus{} bi) \equal{} 2a \equal{} 4$

If $ x$ and $ y$ are real numbers they would form circle with radius $ \sqrt{7}$ so I think the sum would be around $ \sqrt{7}$ too.

[quote="NapoleonXIV"]Note that $ a^3 \equiv 2$ $ mod 3$
By Fermat Little theorem $ a$ has to be $ 2$[/quote]

While Fermat's Little theorem is true, it's converse is not true. In fact, in this case, any number that is $ \equiv 2 \pmod{3}$ will satisfy that condition.

Oh yeah, I was thinking about that too.
So I came with something to support it.

$ x^2 \plus{} y^2$.... and $ a \equal{} c$ we get $ 2a^2 \minus{} 2b^2 \equal{} 7$ therefore $ 2b^2 \equal{} 2a^2 \minus{} 7$

Then we plug it in into $ a^3 \minus{} 5 \equal{} 3ab^2 /*2$
and we get $ 4a^3 \minus{} 21a \plus{} 10 \equal{} 0$
We can use synthetic division to show 2 is a root. And the remaining polynomial is $ 4x^2 \plus{} 8x \minus{} 5$
We can find other roots. I think there are $ 0.5$ and and $ \minus{}2.5$, clearly none of those give remainder 2 when divisible by 3.
Therefore $ a$ is $ 2$

Try this website: google.com. You should've done that before you revived an 8 year old thread.

Sorry I revived this thread - I already had tried googling it, and I still can't figure out how synthetic division had anything to do with the last steps in this solution (from above):

[quote=osdes][hide]

$w^2 + z^2 = (w + z)^2 - 2wz = 7$
$w^3 + z^3 = (w + z)(w^2 - wz + z^2) = 10$

$(w + z)(7 - wz) = 10$
$(w + z)^2 - 2wz = 7$

Let $a = w + z$ and $b = wz$, then
$a(7-b) = 7a - ab = 10$
$a^2 - 2b = 7$ ==> $b = \frac{a^2 - 7}{2}$

Substitute $b$ in:
$7a - ab = 10$
$7a - a(\frac{a^2 - 7}{2}) = 10$ and simplifies to
$a^3 - 21a + 20 = 0$ and guess that $a=1$ is a root, then by synthetic division, $a^3 - 21a + 20 = (a-1)(a^2 + a - 20) = (a-1)(a+5)(a-4)$.

Then $a = 4$ is the largest solution, so the largest value for $w + z$ is $4$.[/hide][/quote]

PS:  Why is reviving discussions so bad?  I don't get why we can't ask questions about the solutions on old threads...

synthetic division is a way to divide 2 polynomials

@mathcrazymj Reviving discussions [i]isn't[/i] bad...
even @rrusczyk supported asking questions and learning more about old threads

[quote=osdes][hide]

$w^2 + z^2 = (w + z)^2 - 2wz = 7$
$w^3 + z^3 = (w + z)(w^2 - wz + z^2) = 10$

$(w + z)(7 - wz) = 10$
$(w + z)^2 - 2wz = 7$

Let $a = w + z$ and $b = wz$, then
$a(7-b) = 7a - ab = 10$
$a^2 - 2b = 7$ ==> $b = \frac{a^2 - 7}{2}$

Substitute $b$ in:
$7a - ab = 10$
$7a - a(\frac{a^2 - 7}{2}) = 10$ and simplifies to
$a^3 - 21a + 20 = 0$ and guess that $a=1$ is a root, then by synthetic division, $a^3 - 21a + 20 = (a-1)(a^2 + a - 20) = (a-1)(a+5)(a-4)$.

Then $a = 4$ is the largest solution, so the largest value for $w + z$ is $4$.[/hide][/quote]

Same solution

$a^3–21a+20=0$ can be found by vieta also :)

We have $x^2+y^2=7, x^3+y^3=10$. Denote $x+y=a, xy=b$, so $a^2-2b = 7, a(a^2-3b) = 10$. Multiplying the first equation by $a$ gives $a^3-2ab = 7a$ and we also have $a^3 - 3ab = 10$, so $ab = 7a-10$. Then $a^3-3(7a-10) = 10$, or $a^3 - 21a + 30 = 10 \implies a^3 - 21a + 20 = 0$. Obviously $1$ is a root, so factoring $a-1$ gives $(a-1)(a-4)(a+5) \implies \boxed{4}$.

Let $s=x+y$ and $p=xy$, so that $x^2+y^2=s^2-2p=7$ and $x^3+y^3=s(s^2-3p)=10$, or, equivalently, $s^2-3p=\tfrac{10}{s}$. Seeking to create elimination in $p$, we multiply the two equations by $3$ and $2$ in that order to obtain $3s^2-6p=21$ and $2s^2-6p=\tfrac{20}{s}$, which after subtracting yields $s^2=21-\tfrac{20}{s}$ or $s^3-21s+20=0$. Evidently the LHS is divisible by $s-1$, and dividing by $s-1$ gives $s^2+s-20$, which factors as $(s+5)(s-4)$. Therefore, the solutions to the equation are $s=-5$, $s=1$, and $s=4$, of which the largest real value is $\boxed{4}$, the requested answer.

Let $x+y=a$, and $xy=b$, hence, $a^2-2b=7$, and $a(a^2-3b)=10$, or $a^3-2ab=7a$, so we have $7a-10=ab$, so $a^3-21a+30=10$, by further computations, factoring this gives $(a-1)(a-4)(a+5)$, hence, the answer is $\boxed{4}.$

Let $s=x+y$ and $p=xy$. We have

\[s^2-2p=7,\]
\[s(s^2-3p) = 10.\]

Solving via substitution yields $s=\boxed{4}$ as our largest solution.

[hide]Given that we have two equations and two variables, we will probably be able to solve for the possibilities for $(x,y)$. Next, we note that we can simplify our systems of equations by substituting $x+y=a$ and $xy=b$; plus, we just have to solve for $a$ now to find the largest real value of $x+y$ instead of having to solve for all pairs $(x, y)$ (which are most likely complex) and adding the two up. $x^2+y^2=7$ becomes $a^2-2b=7$ and $x^3+y^3=10$ becomes $a(a^2-3b)=10$[. We can solve the first equation for $b$ and substitute that into the second, getting $b=\frac{a^2-7}{2}$ and then $a(\frac{21-a^2}{2})=10$. We can expand and factor the cubic to get $(a-4)(a+5)(a-1)=0$, so the largest real value of $x+y$ aka $a$ is $4$.[/hide]


Find the length of rectangle ABCD if the width is 3 inches and its diagonal is 20 inches.

[hide]Using Pythag. the length of the rectangle is $400-9=391$.
So its $\sqrt{391}$.[/hide]


Prove that every positive integer can be expressed as a sum of not more than five cubes.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=60224 (there is at least one more, answered by Myth I think, but that one suffices).

Here's a closely related problem.

Show that every integer $k$ divisible by $3$ can be written as the sum of four cubes.

Solution
[hide]It is well known that
$6n = (-n)^{3}+(-n)^{3}+(n+1)^{3}+(n-1)^{3}$
But we also have
$6n+3 = (n)^{3}+(-n+4)^{3}+(2n-5)^{3}+(-2n+4)^{3}$[/hide]

Isn't this one of the online crux problem and solutions?


calculus
topology
function
Functional Analysis

When learning math, do you use textbooks or online notes/textbooks? For every page that you read, do you take 3-5 pages of notes? Would it really matter if you used the internet or textbooks to learn advanced math (i.e. multivariable calculus, topology, functional analysis,etc..)?

I will use  textbooks if I can. I only use internet when learning problem.

I mostly use handbooks; maybe it's kind of "old-school" fbut for me, it's definitely the best way to practice something.

[quote="new riemann_albanian"]i use only textbooks, i don't like to use the internet[/quote]
At this time. If use internet, you can find all books rank Graduate :P

I used to use EPGY but it was pretty bad. I just read out of textbooks and if I need extra practice, I use the AoPS books

I use the internet, but I can't always find what I'm looking for.

More people should publish online!!! (goshdarn stable roommates problem grumble grumble elliptic functions grumble grumble grumble) :mad:

I agree with Zarathustra. I can't get access to JSTOR and I'm really pissed.
  AR


A set S of district  >=  2 points is given on a plane.Show that there are two district points P[i],P[j] in S such that the circle with diameter p[i]p[j] contains at lest n/3 of the poits from S

I'm not very sure, but I think this works. Take $\cal C$ to be the smallest circle such that $S \subset Int(C)$. It is should be clear this smallest circle exists. It is not hard to prove using the minimality that there are either 3 points $A,B,C$ on the circle which form an acute triangle or 2 points $D$ and $D$ diametrally opposite on ${\cal C}$. In this first case, the discs of diameters $AB$, $BC$ and $CA$ cover $Int(C)$, hence, one of them contains at least one third of the points. In the latter, the circle of diamater $DE$ contains all points.

I think a made a small mistake. I edited the proof.

I think you meant $S \supset \textrm{Int} \left( \mathcal C \right)$. But how do you prove that [quote]there are either 3 points A,B,C on the circle which form an acute triangle or 2 points D and D diametrally opposite on $\mathcal C$.[/quote]

No, I mean $S \subset Int({ \cal C})$ (or rather, $S \subset Int({\cal C}) \cup {\cal C}$). For the second, I'll try to explain though it's not very "scientifical". It is clear that there are at least 2 points on the circle. If the claim is false, then there is an arc $XY<\pi$ which contains all points from ${\cal C} \cap S$. Move the circle such that $XY$ remains a chord and in such way to decrease its radius. On a small distance, since $S$ is finite, the effect is the following: the points originally on the circle are now inside it and the others also remain inside. We have a smaller circle with that property, a contradiction.

Yes, you're right. I confused the notations, but I was thinking of the same thing.

And the non-scientifical proof makes sense.


What do we mean by :

Find the representation for bounded linear functionals on $ \ell^p, 1\leq p < \infty$ .

I know how to prove the dual space of $ \ell^p\equal{}\ell^q$ . During the proof we define for any $ a_n \in \ell^q$ the linear functional by :

$ F(x)\equal{}\sum\limits_{i\equal{}1}^{\infty} x_i a_i , \forall x\in \ell^p$.

$ F$ is Linear ,Bounded , and it's an isometry for sure .

Is that what the question need ????

$ F$ is most assuredly not an isometry. It's not even a mapping from a space into itself.

$ F$ is defined  : $ F: C_0 \rightarrow  \mathbb R$ .

$ F$ is an [b]isometry[/b] for sure.

But , what is the representation the question asked about it???

Please write out what you think the word "isometry" means.

Sorry , I meant $ F: \ell^p\rightarrow \mathbb R$

Isometry : if $ T : X \rightarrow Y$ , we have $ \parallel{}Tx\parallel{}\equal{}\parallel{}x\parallel{} , \forall x\in X$

To be more clear :

To prove that $ (\ell^p)*\equal{}\ell^q$ : (over real numbers field)

1st of all , we need to do :

1) for every sequence in $ a\in \ell^q$ , there is $ F: \ell^p\rightarrow \mathbb R$  such that $ \parallel{}F\parallel{}\equal{}\parallel{}a\parallel{}$.

the $ F$ I defined is just satisfies this condition .

That's to say , $ \ell^q$ is just a copy from the dual of $ \ell^p$

Correct me If I'm wrong , that's what I could figure out from books .

Thx


calculus
integration
trigonometry
Putnam
calculus computations

Evaluate:

$\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}$

I tried to make the substitution $x= \tan \theta$ but I got stuck half-way through.

How are you getting stuck? subbing for tan, dont you see that the denominator is secant^2?


a,b,c,d strictly positive real numbers
prove that   a/(b+c)+b/(c+d)+c/(d+a)+d/(a+b) superior to 2

i think it should be in inequality unsolved problem

the ineq you ask for is called Nesbitt's ineq for the case n = 4.if you go to ineq section i am sure some one can point you to the right post that you can get your answer.
m.c-


Hello, rrusczyk!

I'll try to explain this without a diagram . . .

[quote]A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle.
What is the ratio of the area of the triangle to the area of the square?[/quote]
Let the side of the equilateral triangle be $S$.
Its area is: $\frac{\sqrt{3}}{4}S^2$.

The altitude of the triangle is $\frac{\sqrt{3}}{2}S$.
It can be shown that the incenter is one-third of the way "up" the altitude.
Hence, the radius of the inscribed circle is: $r =\frac{\sqrt{3}}{6}S$.

Sketch a square inside this circle.
We can see that the side of the square is the hypotenuse of a 45-45-90 right triangle
where $r$ is the legs of the right triangle.
Hence, the side of the square is: $\sqrt{2}\times \frac{\sqrt{3}}{6}S \;= \;\frac{S}{\sqrt{6}} $
And the area of the square is: $\left(\frac{S}{\sqrt{6}}\right)^2 \;= \;\frac{S^2}{6} $

The ratio is: $\displaystyle{\frac{A_{tr}}{A_{squ}} \;= \;\frac{ \frac{\sqrt{3}}{4}S^2}{\frac{S^2}{\sqrt{6}}} \;= \;\frac{3\sqrt{2}}{4}  }$

[hide]1. Let us call the radius of the circle r.

2. 1/2 the length of the inscribed square's diagnol is also r, so the length of the square's diagnol is 2r.

3. Through the Pythagoreom Theorem, 1/2 of one of the square's side is ( :sqrt:  2)r.

4.  The area of the square is thus (( :sqrt:  2)r) 2 or 2r 2.

5. Drawing lines from the center of the circle to each of the triangle's vertices, we have divided the triangle into three 120 degree sections.

6. The larger triangle now consists of 6 triangles of are (r 2  :sqrt:3))/2. Combined, the total area is r 2 3 :sqrt: 3).

7. Thus, the ratio of the triangle's area to the square's area is (3 :sqrt: 3)/2.  [/hide]

A different answer?

You made a slight error at the end, Soroban. For the area of the square, you used $\frac{S^2}{\sqrt{6}}$, when you had stated previously that the area of the square is $\frac{S^2}{6}$ (the latter being correct).


function
ratio
limit
calculus
calculus computations

Find the interval of convergence of serie:
Sum(n=0->infinity)n!*x^n/n^n.
I'm stucking at checking whether it converges or not at x=e.Anyone can help me?

I'd try applying the ratio test, ie finding all x such that

$ \lim_{n \rightarrow \infty} a_{n} \equal{} \lim_{n \rightarrow \infty} \frac { (n \plus{} 1)!\left(\frac {x}{n \plus{} 1}\right)^{n \plus{} 1} }{n!\left(\frac {x}{n}\right)^{n}} < 1$

$ \lim_{n \to \infty} \frac{e^n n!}{n^n \sqrt n} \equal{} \sqrt {2\pi}$

this should tell you whether or not you have convergence, right?

this has been asked numerous times. look for the latest question on some gre problem (posted a few days ago). you'll find it there.

I know the interval of convergence is (-e,e).But i can not find the limit of n!*e^n/n^n   (n->infinity)
I also check some gre problems but its not like this question.
Anyway thanks for your help

I don't know what happens at that boundary point. Mathematica doesn't give an answer. Perhaps someone more skilled can help you.

Ya,actually Maple Mathematica and Matlab dont give the answer for this limit,but you can use for loop to calculate this limit and I aprroximate the limit is infinity

kenn4000 gave the answer already.  It's based on [url=http://en.wikipedia.org/wiki/Stirling%27s_approximation]Stirling's approximation[/url], which states that for large $ n$ we have asymptotically

$ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$.

Once you know this, you can verify that the series diverges at both $ e$ and $ \minus{}e$.

So   n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n => \left(\frac{e}{n} \right)^n*n! ~~ \sqrt{2pi*n} =>limit ->infinity   ,rite?But How to prove Stirling's approximation?^^,its my assignment and actually,in one book(Advanced Calculus of Richardson),the author suggest that we should use the expansion of e^n and use the n-th term test.

Sorry,I have problem in typing latex...I will edit my post later.

Oh,thank you very much.Yet I have had my own solution already by using n-th term test.If I use Stirling's approximation in my assignment my Prof will kill me.

[hide="Without Stirling Approximation"]Let $ a_{n} \equal{} \frac {n!e^{n}}{n^{n}}$. We have: $ \frac {a_{n \plus{} 1}}{a_{n}} \equal{} \frac {e}{(1 \plus{} \frac {1}{n})^{n}} > 1$, since denominator is increasing and converges to $ e$. Therefore $ a_{n}$ is increasing, and there can't be $ \lim_{n\to\infty}a_{n} \equal{} 0$. Similar argument shows that the series diverges in $ x \equal{} \minus{} e$.[/hide]