Problem One: X + Y = 7; X + Z = 13; Y + Z = -34.

Find the value of X + Y + Z.

SOME GUIDELINES:

No spamming.  One problem at a time.  Not too hard.  Post a new problem after solving one.  Provide an explanation with your answer.

Suggestion: Read the sticky before posting. YOU CAN NOT POST HERE BEFORE READING THE STICKY.

Okay, I know no one's going to pay attention to what I just wrote, so I'll just solve ernie's problem.

Adding our equations yields

$ 2x \plus{} 2y \plus{} 2z \equal{} \minus{}14$

$ x \plus{} y \plus{} z \equal{} \boxed{\minus{}7}$

New Problem:

Cindy likes to sing. She practices singing for 30 minutes on even days and 1 hour on odd days. In March, in total, how many hours did she sing? If necessary, express your answer as a mixed number.

huh last one got locked...

there are 31 days in march.

therefore, 16*1+15*1/2=23 1/2

answer : 23 1/2 hours

<New Problem>

x+y=12, xy=13. find $ \frac {1}{x} \plus{} \frac {1}{y}$.

if we add unit fractions, the answer will always be $ \frac {\text{sum of numbers}} {\text{product of numbers}}$
so, we have our answer
$ \frac {12} {13}$
srry, no new problem

find smallest x that satisfies

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\equal{} x^2$ for some integers a,b,c,d, and x

pplz, plz try to keep these problems at a MOEMS level to prevent this one from being locked too

next problem
if Dan has 7 red marbles and 8 blue marbles, how many green marbles must he add in so that the probablity of selecting a green marble is 1/4?

r= # of red marbles
b= # of blue marbles
g= # of green marbles

$ r\plus{}b\equal{}15$

$ \dfrac{g}{r\plus{}b\plus{}g}\equal{}\dfrac{1}{4}$
$ \dfrac{g}{15\plus{}g}\equal{}\dfrac{1}{4}$
$ 4g\equal{}15\plus{}g$
$ 3g\equal{}15$
$ g\equal{}\boxed{5}$

<new problem>
Find the area of a triangle with sides 5,12,13.

[hide="trick problem"]
A boat in the middle of the ocean has a stair on the side with 8 steps. The water rises one step per hour. At 12 noon there are 3 steps underwater. How many steps are underwater at 3 pm?
[/hide]

5*12/2=30

answer : 30

<New Problem>

What is the hypotenuse of a triangle if other two sides of the triangle measure 7 and 24?


john nash i have seen it...

not hard for me... :P

here is the solution :

$X^3 + Y^3 + Z^3 = (X+Y+Z) (X^2+Y^2+Z^2 - XY-YZ-ZX)+3XYZ$

$=(X+Y+Z)^2 -3(XY+YZ+ZX)+3XYZ$

$=1-3(XY+YZ+ZX)+3XYZ$

then according to problem we should prove:

$(XY+YZ+ZX)-3XYZ\leq 1/4$

then:

$XY+YZ+ZX-3XYZ =XY(1-Z)+YZ(1-X)+ZX(1-Y)$

$=XY(X+Y)+YZ(Y+Z)+ZX(Z+X)$

$=X^2(Y+Z)+Y^2(Z+X)+Z^2(X+Y)$

$=X^2(1-X)+Y^2(1-Y)+Z^2(1-Z)\leq \frac{X+Y+Z}{4}\leq 1/4$

and in last inequality we used this fact:$X(1-X)\leq 1/4$

thanks ashegh ure very strong in algebra and geometry... :D

[quote="blahblahblah"]We have

$1=(x+y+z)^3=x^3+y^3+z^3+6xyz+3[xy(x+y)+yz(y+z)+zx(z+x)]$

By Schur's inequality, the term in square brackets is $\leq$

$\frac{2}{3}(x^3+y^3+z^3+6xyz)$

Hence it follows that

$1\leq 3(x^3+y^3+z^3+6xyz)$

which leads to

$\frac{1}{3}\leq x^3+y^3+z^3+6xyz$

which is a better result than ashegh's (since he uses the inequality $4x(1-x)\leq 1$, which cannot hold for all three variables simultaneously)[/quote]
By Schur's Inequality, $xy(x+y)+yz(y+z)+zx(z+x)\leq x^3+y^3+z^3+3xyz$. :(
To attain $\frac{1}{4}$, take $x=y\rightarrow\frac{1}{2}$ and $z\rightarrow 0$.

excuse me ...

did blahblahblah...  wrote some thing?

We have

$1=(x+y+z)^3=x^3+y^3+z^3+6xyz+3[xy(x+y)+yz(y+z)+zx(z+x)]$

By Schur's inequality, the term in square brackets is $\leq$

$x^3+y^3+z^3+3xyz$

Hence it follows that

$1\leq 4(x^3+y^3+z^3+6xyz) \ \ (*)$

which leads to

$\frac{1}{4}\leq x^3+y^3+z^3+6xyz$

Equality only holds in $(*)$ if one of the variables is zero and the other two are equal, so the equality case follows.

[quote="John Nash"]$X,Y,Z$ are real positive numbers.and we have :$X+Y+Z=1$

prove : $X^3 +Y^3 +Z^3 +6XYZ \geq 1/4$[/quote]

look at  Mehdi safa's book( page.306 ,No 7)

[quote="leepakhin"]Maybe he deleted it just before I posted my reply.[/quote]

Yes, there was a rather dumb mistake in the third line, or whatnot.  See my post before this one, which IS correct.


function
logarithms
geometric sequence
real analysis
real analysis unsolved

Find all continuous function $ f$ s.t for every geometric progression $ a_{1} , a_{2},....a_{n}$  have $ f(a_{1}) , f(a_{2}),...f(a_{n})$ are a geometric progression

We have: $ a_{n} \equal{} a_1 p^{n \minus{} 1}$ and $ f(a_n) \equal{} q^{n \minus{} 1}f(a_1)$ for every $ a_1$ and $ p$ and for some $ q$. Now we can write:
\[ f(a_1 p^{n \minus{} 1}) \equal{} q^{n \minus{} 1}f(a_1)
\]
Put here $ p \equal{} 1$ we get $ f(a_1) \equal{} q^{n \minus{} 1}f(a_1)$. Or $ f(a_1) \equal{} 0$ or $ q \equal{} 1$. If $ q \equal{} 1$ we get a constant function $ f(x) \equal{} c$. If $ f(a_1) \equal{} 0$ then all $ f(a_n) \equal{} 0$. So in this case $ f(x) \equal{} 0$. Now put $ a_1 \equal{} 1$: $ f(p^{n}) \equal{} q^n f(1)$. This gives $ f(x) \equal{} f(1)x^{\log_{p}{q}}$. $ \log_{p}{q}\in R$ so we can write $ f(x) \equal{} f(1)x^{\alpha}$. If $ \alpha$ is rational then solution exist for $ x\in R$ and if irrational then for $ x\geq0$.

Answer: $ f(x) \equal{} f(1)x^{\alpha}$ for $ x\in R$ if $ \alpha\in Q$ and $ f(x) \equal{} f(1)x^{\alpha}$ for $ x\geq 0$ if $ \alpha\in R/Q$


The lateral surface area of a cylindrical tube with a height of 6 cm is $ 48\pi$ square centimeters. In cubic centimeters, what is the tube's volume? Express your answer in terms of $ \pi$.

By logic and a little visual imagining, it is deducible (or more likely in cases, memorized) that the lateral surface area of a cylinder is the height times the circumference of a base circle. Thus, the circumference of the base circle in this case is $ \frac{48\pi}{6}\equal{}8\pi$, and the radius would be $ \frac{8\pi/2}{\pi}\equal{}4$. Since the volume of a cylinder is $ \pi r^2 h$, the volume of this cylinder is $ \pi(4^2)(6)\equal{}\boxed{96\pi}$.


logarithms
inequalities
inequalities proposed

[hide="a possible approach"]Try to prove that $ \frac{1}{x(1\plus{}x^3)}\plus{}\frac{5}{4}\ln x\minus{}\frac{1}{2} \geq 0$, $ \forall x>0$[/hide]

I try to find another way to approach and surprisingly, it came very smoothly. Now, getting a little change leads us to prove this inequality:
$ \frac{ab}{1\plus{}c^3}\plus{}\frac{bc}{1\plus{}a^3}\plus{}\frac{ca}{1\plus{}b^3}\ge \frac{3}{2}$
Rewrite it, we have to prove that:
$ \sum_{cyc} (ab\minus{} \frac{ab}{1\plus{}c^3}) \leq \sum_{cyc} (ab\minus{}\frac{3}{2})$
Now, using the well-know inequality called Cauchy- Schwartz, we have a range of following inequalities :
1)         $ \frac{c^2}{1\plus{}c^3} \leq \frac{\sqrt{c}}{2}$
2)         $ 2(ab\plus{}bc\plus{}ca) \geq \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\plus{}3$
Applying these inequalities, we can solve completely this problem!

Am-Gm kill it easily  :)
After using Am-Gm, it's becomes:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a} \geq\ \frac {9}{2}$
We can also use Am-Gm to complete the proof:
$ LHS \plus{} 3 \equal{} (\sum\ \frac {a \plus{} b}{4b} \plus{} \sum\ \frac {a}{a \plus{} b}) \plus{} \frac {3}{4} \sum\ \frac {a \plus{} b}{b} \geq\ 3 \plus{} \frac {9}{2} \equal{} RHS \Rightarrow q.e.d$
Besides, we can using Cauchy-Schwart to complete the proof.

[quote="Marius Mainea"]Let a,b,c positive numbers such that abc=1. Prove that
\[ \frac {ab}{1 \plus{} c^3} \plus{} \frac {bc}{1 \plus{} a^3} \plus{} \frac {ca}{1 \plus{} b^3}\ge \frac {3}{2}
\]
[/quote]

 Inequality is equivalent with $ \sum {\frac{1}{z(1\plus{}z^3)}}\ge \frac{3}{2}$

 Let $ f(x)\equal{}\frac{1}{x(1\plus{}x^3)}$ , then$ f_1(u)\equal{}f(e^u)$  is convex for $ u\ge \ln 1$ and $ f(x)\plus{}2f(y)\ge 3f(1)$  for all x,y such that $ \sqrt[3]{xy^2}\equal{}1$ and $ y\ge 1$

 Therefore follows the RCF-Corollary, $ \sum {f(x)}\ge \frac{3}{2}$ for all x,y,z such that $ \sqrt[3]{xyz}\ge 1$

[quote="Vasc"][quote="Marius Mainea"]Let a,b,c positive numbers such that abc=1. Prove that
\[ \frac {ab}{1 \plus{} c^3} \plus{} \frac {bc}{1 \plus{} a^3} \plus{} \frac {ca}{1 \plus{} b^3}\ge \frac {3}{2}
\]
[/quote]
I think the following nice inequality also holds for $ abc \equal{} 1$:
\[ \frac {ab}{2 \plus{} c^3} \plus{} \frac {bc}{2 \plus{} a^3} \plus{} \frac {ca}{2 \plus{} b^3}\ge 1
\]
[/quote]
Nice, VasC. :)
I have proved by Cauchy Schwarz the following inequalities
\[ \sum\frac{ab}{c^3\plus{}k} \ge \frac{3}{k\plus{}1}\]
\[ \sum \frac{ab}{c^4\plus{}k} \ge \frac{3}{k\plus{}1}\]
for all $ a,b,c>0,abc\equal{}1,k\ge 0$.
:)


How did all the Koreans in AoPS die??

by a brain burst?? XD

You could always become undead.. join me :D


logarithms
function
inequalities
limit
calculus
derivative
real analysis

[quote="Virgil Nicula"]
In a different way, if the initial function $F(x)=f(x)+f\left(\frac 1x\right),\ x>0$ has these properties, then $F(x)=F\left(\frac 1x\right)\Longrightarrow 2F(x)=F(x)+F\left(\frac 1x\right)\le 2F(1)\Longrightarrow F(x)\le F(1)$.
[/quote]
that is really a clever way! :roll:

[quote="Virgil Nicula"]$f(x),\ x>0$ is decreasing $(\ \searrow\ )$ and concave $(\ \frown\ )\Longrightarrow$ $f(x)+f\left(\frac 1x\right)\le 2f\left(\frac{x+\frac 1x}{2}\right)\le 2f(1)$. If the our function $f(x)=\frac{\ln (1+x)}{x},\ x>0$ has these properties, then the proposed inequality is proved. But I don't think, because  $\lim_{x\searrow 0} f(x)=1$ and $\lim_{x\to \infty } f(x)=0$.

In a different way, if the initial function $F(x)=f(x)+f\left(\frac 1x\right),\ x>0$ has these properties, then $F(x)=F\left(\frac 1x\right)\Longrightarrow 2F(x)=F(x)+F\left(\frac 1x\right)\le 2F(1)\Longrightarrow F(x)\le F(1)$.

And now you calculate the derivatives ![/quote]

Why $F(x),\ x>0$ is decreasing $(\ \searrow\ )$ and concave $(\ \frown\ )$

[quote="Virgil Nicula"]Please Jjchen611, read with greater attention what I wrote [u]clearly[/u] (two sentencies) :

$1).\ ....\ \boxed {\ \mathrm { if }\ }$ the initial function $F(x)=f(x)+f\left(\frac 1x\right),\ x>0$ has these properties, then .....

$2).\ \boxed {\ \mathrm {And\ now\ you\ calculate\ the\ derivatives\ !}\ }$[/quote]

1.Thank Virgil Nicula  very much

2.Please Virgil Nicula  , read with greater attention what I wrote :

  (a)$F(x),\ x>1$ is decreasing and concave $(\ \frown\ )$;
  (b)$F(x),\ 0<x<1$ is increasing and concave $(\ \frown\ )$;

[quote="Virgil Nicula"]O.K. What is the problem ? Then we together are think another solution ![/quote]

No problem,in fact you have solved it!

[quote="lccs"][quote="jjchen611"]
2.Please Virgil Nicula  , read with greater attention what I wrote :

  (a)$F(x),\ x>1$ is decreasing and concave $(\ \frown\ )$;
  (b)$F(x),\ 0<x<1$ is increasing and concave $(\ \frown\ )$;[/quote]
that is it,notice that$\frac{x+{1}/{x}}{2}\ge1$,so virgil's solution is correct.
but in fact if we have (a) and (b) ,it seems that Jensen Ineq. is not necessary. :?:
jjchen611, would you please post how you get (a) and (b)? ;) thanks[/quote]

ANOTHER PROOF!

Let $F(x)=x\ln(1+\frac{1}{x})+\frac{ln(1+x)}{x}$ , it is easy to
prove that $F''(x)\le 0.$ For $x\in (0,+\infty),x\neq 1,$ using
Taylor's formula we have ${\xi} \in(1,x)$ or ${\xi} \in(x,1)$ such
that
$F(x)=F(1)+F'(1){(x-1)}+\frac{F''(\xi)}{2}\cdot{(x-1)^2} .$

Noting that
$F(1)=2\ln2,\qquad F'(1)=0.$
Therefore, $F(x)\le 2\ln2,$ .This completes the proof.


limit function integration algebra polynomial calculus derivative

[b]The 13th International Scientific Olympiad in Mathematics
for University Students[/b]
[i]15-18 July 2008
Shahid Beheshti University, Tehran, I.R. Iran[/i]

[b]First Day[/b]

[b]I. Mathematical Analysis (2 hours)[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216436]1.[/url] (25 points) Suppose $ \mathfrak{X}$ is a compact metric space and $ T: \mathfrak{X}\to\mathfrak{X}$ be a continuous map. For each $ x\in\mathfrak{X}$ let $ \omega(x)$ be the set of all limit points of $ \{T^nx: n \equal{} 0,1,2,\ldots\}$. Show that:
a. $ \omega(x)$ is a nonempty compact subset of $ \mathfrak{X}$.
b. $ T(\omega(x)) \equal{} \omega(x)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216437]2.[/url] (25 points) Let $ (c_n)$ be a sequence of strictly positive numbers such that $ \sum\limits_{n \equal{} 1}^{\infty}c_n \equal{} \infty$.
a. Show that, there exists a sequence $ \{x_n: n\in\mathbb{N}\}$ which is dense in $ \mathbb{R}$ and every point of $ \mathbb{R}$ belongs to an infinite number of $ I_n \equal{} (x_n \minus{} c_n,x_n \plus{} c_n)$, for $ n\in\mathbb{N}$.
b. Let $ f(x_n) \equal{} c_n$ and $ f(x) \equal{} 0$, if $ x$ is distinct from all the $ x_n$'s, and $ \lim\limits_{n\to\infty}c_n \equal{} 0$. Then $ f$ is differential at no point of $ \mathbb{R}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216438]3.[/url] (25 points) a. Let $ f$ be a piecewise continuously differentiable function on the interval $ [0,a]$ and $ f(0) \equal{} 0$, then
\[ \int_0^a\lvert f'(t)f(t)\rvert\,\mathrm{d}t\le\frac {a}{2}\int_0^a\lvert f'(t)\rvert^2\,\mathrm{d}t.
\]
b. When does equality hold?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216439]4.[/url] (25 points) Suppose $ f_n$ is a sequence of increasing functions on $ [0,1]$ (i.e., for all $ n$, if $ x_1 < x_2$ then $ f_n(x_1) < f_n(x_2)$) which converges pointwise to a continuous function $ g(x)$. Prove that the convergence ia actually uniform on $ [0,1]$.

[b]II. Numerical Analysis (2 hours)[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216442]1.[/url] (25 points) Let function $ f(x)$ be four times continuously differentiable and have a simple zero $ \xi$. Successive approximations $ x_n$, $ n \equal{} 1,2,\ldots$ to $ \xi$ are computed from
\[ x_{n \plus{} 1} \equal{} \frac {1}{2}(x_{n \plus{} 1}' \plus{} x_{n \plus{} 1}''),
\]
where $ x_{n \plus{} 1}' \equal{} x_n \minus{} \dfrac{f(x_n)}{f'(x_n)}$, $ x_{n \plus{} 1}'' \equal{} x_n \minus{} \dfrac{g(x_n)}{g'(x_n)}$, $ g(x) \equal{} \dfrac{f(x)}{f'(x)}$.
Prove that if the sequence $ \{x_n\}$ converges to $ \xi$, then the rate of convergence is cubic.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216443]2.[/url] (25 points) Suppose that $ p(x)$ is the interpolating polynomial of degree less than or equal to $ n$ to ($ n \plus{} 1$)-continuously differentiable function $ f(x)$ at distinct points $ x_i$, $ 0\le i\le n$. Prove that for any $ k$, $ 0\le k\le n$, there are points $ y_0,\ldots,y_{n \minus{} k}$ so that for any $ x$ there exists a corresponding $ \xi_x$ satisfying,
\[ f^{(k)}(x) \minus{} p^{(k)}(x) \equal{} \frac {f^{(n \plus{} 1)}(\xi_x)}{(n \minus{} k \plus{} 1)!}\prod_{i \equal{} 0}^{n \minus{} k}(x \minus{} y_i),
\]
where $ f^{(k)}$ and $ p^{(k)}$ denote the $ k$th derivative of $ f$ and $ p$ respectively.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216444]3.[/url] (25 points) Consider the following iteration formula for estimating $ A^{ \minus{} 1}$ of an $ n\times n$ symmetric positive definite matrix $ A$,
\[ X_{n \plus{} 1} \equal{} X_n \plus{} q(AX_n \minus{} I),\qquad(*)
\]
where $ q$ is a constant and $ I$ is the $ n\times n$ identity matrix.
i) Under what conditions on $ q$ (in terms of the eigenvalues of $ A$), the iteration formula $ (*)$ converges to $ A^{ \minus{} 1}$?
ii) What should the value of $ q$ be so that the convergence is as fast as possible?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216445]4.[/url] (25 points) Consider the following quadrature rule
\[ \int_0^{3h}f(x)\,\mathrm{d}x\approx\alpha_1f(0) \plus{} \alpha_2f(2h) \plus{} \alpha_3f(3h).\qquad(**)
\]
i) Find $ \alpha_1$, $ \alpha_2$ and $ \alpha_3$ such that the rule has highest order (or degree) of precision.
ii) Write the composite rule of $ (**)$ for $ \displaystyle\int_a^bf(x)\,\mathrm{d}x$ with $ h \equal{} \dfrac{b \minus{} a}{3n}$, $ x_i \equal{} a \plus{} ih$, $ i \equal{} 0,1,\ldots,3n$.
iii) Which one of the following integrals can be approximated by the composite rule of $ (**)$?
a) $ \displaystyle\int_0^1\frac {\tan x}{\sqrt {1 \minus{} x}}\,\mathrm{d}x$,
b) $ \displaystyle\int_{ \minus{} 1}^1\frac {\cos x}{\sqrt {1 \minus{} x^2}}\,\mathrm{d}x$,
c) $ \displaystyle\int_0^1\frac {\cos x}{\sqrt {x}}\,\mathrm{d}x$.

[b]Second Day[/b]

[b]III'. Algebra (2 hours)[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216446]1.[/url] (25 points) Prove that there is no ring epimorphism $ f: \mathcal{M}_2(\mathbb{R})\to\mathcal{M}_3(\mathbb{R})$, where $ \mathbb{R}$ is the field of real numbers.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216447]2.[/url] (25 points) Let $ R$ be an arbitrary ring and let $ Z(R)$ be the center of $ R$. Prove that: If $ x^2\in Z(R)$ for all $ x\in R$, then $ (xy)^2 \equal{} (yx)^2$ for all $ x,y\in R$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216448]3.[/url] (25 points) Let $ G$ be a finite group and for every subgroup $ H$ of $ G$ there is a homomorphism $ f: G\to H$ such that $ f(h) \equal{} h$ for all $ h\in H$. Prove that $ G$ is isomorphic to a direct product of cyclic groups of prime order.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216449]4.[/url] (25 points) Let $ G$ be a non-trivial group such that every normal subgroup of $ G$ is finitely generated. Prove that there is no non-trivial normal subgroup $ N$ of $ G$ such that $ G\cong\dfrac{G}{N}$.

[b]III''. Operations Research (2 hours)[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216450]1.[/url] (25 points) Consider the linear programming problem (P),
$ \min z \equal{} c^Tx$
$ \text{s.\,t. } Ax \equal{} b\qquad\text{(P)}$
$ \quad x\ge0.$
a) Prove that a feasible point for (P) is optimal if and only if there exist $ v$ and $ w$ so that:
$ c \minus{} A^Tw \minus{} v \equal{} 0$
$ v\ge0$
$ v^Tx \equal{} 0.$
b) Prove that if $ c\ge0$, then the problem (P) can not be infinite.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216451]2.[/url] (25 points) Prove that a network $ G \equal{} (V,A)$, with a finite number of vertices $ V$ and set of arcs $ A$ has infinite flow if and only if there exists a forward path from the source $ s$ to sink $ t$ with all its arcs having infinite capacity.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216452]3.[/url] (25 points) Consider the following primal and dual problems
$ \min c^Tx,\quad\max b^Ty$
$ x\in P\qquad(y,s)\in D$
where $ P \equal{} \{x\mid Ax \equal{} b,\ x\ge0\}$, $ D \equal{} \{y\mid A^Ty \plus{} s \equal{} c,\ s\ge0\}$, $ A$ is a full rank $ m\times n$ ($ m < n$) matrix, $ c,x,s\in\mathbb{R}^n$ and $ y,b\in\mathbb{R}^m$. Prove:
i) If $ \exists\overline{x}\in P$, $ \overline{x} > 0$ and $ \exists(\overline{y},\overline{s})\in D$, $ \overline{s} > 0$ then both primal and dual problems have optimal solutions.
ii) Let $ P^{*} \equal{} \{x\in P,$ $ x$ is optimal$ \}$ and $ D^{*} \equal{} \{(y,s)\in D,$ $ (y,s)$ is optimal$ \}$. Then $ P^{*}$ and $ D^{*}$ are bounded sets.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216453]4.[/url] (25 points) Assume $ S$ is a compact convex subset of $ \mathbb{R}^n$, and $ f: \mathbb{R}^n\to\mathbb{R}$ is a continuously differentiable convex function on $ S$. Show that for any two optimal points $ y$ and $ z$ for the problem,
$ \min f(x)$
$ \text{s.\,t. }x\in S,$
we have,
$ \nabla f(y) \equal{} \nabla f(z).$

[b]IV. Linear Algebra (2 hours)[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216454]1.[/url] (33.3 points) Let $ T_1,T_2,\ldots,T_m$ be linear transformations of an $ n$-dimensional vector space $ V$. Suppose that:
(i) $ \dim\mathrm{Im}(T_i) \equal{} 1$, for all $ i \equal{} 1,2,\ldots,m$.
(ii) $ T_i^2\ne0$ and $ T_iT_j \equal{} 0$ for each $ i\ne j$.
Prove that $ m\le n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216457]2.[/url] (33.3 points) Let $ A$, $ B$ and $ C$ be $ n\times n$ matrices over a field such that $ AB \plus{} C \equal{} I_n$. If there exists an $ n\times n$ matrix $ X$ such that $ A \plus{} CX$ is invertible, show that there exists an $ n\times n$ matrix $ Y$ such that $ B \plus{} YC$ is invertible.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=216458]3.[/url] (33.3 points) Let $ F$ be a field with at least three elements. If $ A\in\mathcal{M}_n(F)$, $ \mathrm{rank}(A) \equal{} n \minus{} 1$, and moreover $ A$ has no zero row, then prove that there exists $ X\in F^n$ such that no entry of the vector $ AX$ is zero.

[b]IV. 1[/b]

Let $ R(T) \equal{} Im(T)$.

$ R(T_i^2) \subseteq R(T_i)$

$ 0<rank(T_i^2) \leq rank(T_i) \equal{} 1$

Hence $ rank(T_i^2)\equal{}1$, and $ R(T_i^2)\equal{}R(T_i)$. Hence $ T_i^2 : R(T_i) \rightarrow R(T_i)$ is bijective.

Moreover $ T_iT_j(V) \equal{} T_i(R(T_j)) \equal{} \{0\}$ for $ j \neq i$. Hence $ R(T_j) \subseteq Kernel(T_i)$ for $ i \neq j$.

Now suppose we take a nonzero vector in each of $ R(T_1), ..., R(T_m)$; say we get $ x_1, ..., x_m$. Then if these vectors are linearly dependent, we can write

$ 0 \equal{} \sum_{j\equal{}1}^ma_jx_j$

where at least one of the $ a$'s is nonzero, say $ a_i$. Then

$ x_i \equal{} \frac{\minus{}1}{a_i}\sum_{j\equal{}1, \: j \neq i}^ma_jx_j$

$ T_i^2(x_i) \equal{} \frac{\minus{}1}{a_i}\sum_{j\equal{}1, \: j \neq i}^ma_jT_i^2(x_j)$

But $ x_j \in R(T_j) \subseteq Kernel(T_i)$  for $ i \neq j$, hence the right hand side is zero. But the left hand side is nonzero since $ x_i \neq 0$ and $ T_i^2 : R(T_i) \rightarrow R(T_i)$ is bijective. Contradiction; hence $ \{x_1, ..., x_m\}$ are linearly independent, and $ m \leq n$.

II.3)We have $ \forall m \in \mathbb{N^{*}}$
    $ X_{m\plus{}1}\equal{}X_n\plus{}q(AX_n\minus{}I)$ then $ X_{m\plus{}1}\minus{}A^{\minus{}1}\equal{}(I\plus{}qA)(X_m\minus{}A^{\minus{}1})$
By induction we have : $ X_m\minus{}A^{\minus{}1}\equal{}(I\plus{}qA)^{m\minus{}1}(X_1\minus{}A^{\minus{}1}),\forall k \geq 1$
1)Put $ \lambda_1,\lambda_2,..,\lambda_n$ are  positive eigenvalues of $ A$ and $ \lambda_s\equal{}Max_{1 \leq k \leq n}{\lambda_k}$ then conditions on $ q$ (in terms of the eigenvalues of $ A$ ), the iteration formula  converges to $ A^{\minus{}1}$
are $ |1\plus{}q\lambda_k|<1,\forall k\in \mathbb{N^{*}}$ or $ \frac{\minus{}2}{\lambda_k} <q<0,\forall k\in \mathbb{N^{*}}$ than if  $ \frac{\minus{}2}{\lambda_s} <q<0$ need find.
2)$ q\equal{}0$ so that the convergence is as fast as possible.$ X_m\equal{}X_1$


Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?

$\text{(A)}\ 72 \qquad \text{(B)}\ 73 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 76$

[hide]Completing the square we obtain $(x-7)^{2}+(y-3)^{2}=64$.  Let $4n$ be the maximum value we are looking for, so $3x+4y=4n$ and $y=-\frac{3}{4}+n$.  Plugging in $y$ we obtain $(x-7)^{2}+(-\frac{3x}{4}+n-3)^{2}=64$.  We want where these two graphs only have one intersection.  Since this equation is a quadratic, this will happen when the discriminant is zero.

So expanding $(x-7)^{2}+(-\frac{3x}{4}+n-3)^{2}=64$, and (sparing a bunch of algebra) setting the discrimant (which will be in terms of $n$) equal to zero yields another quadratic, which has a solution $n=\frac{73}{4}$.  Therefore $4n=73$, and the answer is $\boxed{\text{B}}$.[/hide]

[quote="maokid7"]y did you choose to use $4n$. would it not have been almost as eays to use $n$?[/quote]Well, using $n$ would have given me $\frac{n}{4}$, which might have been harder to use because of the fraction.

But initially, using $4n$ was an accident.  When I divided the equation $4y=-3x+n$ by $4$, I accidentally forgot to divide the $n$ by $4$ also, which I only realized at the end of the problem.  :P

Nice solution djshowdown.  I totally did it eryaman's way... not good on AMC since you need speed.

nice solution.
but u have a small typo.
the last step should say $3x+4y$ and not $3x-4y$ :P

this type of problem has been posted many times, i am not sure if the numbers are the same for this one though, also there is also a good cauchy solution out there


Prove that there are infinitely many pairs $ (x,y)$ of different positive rational numbers such that both
$ \sqrt{x^2\plus{}y^3}$ and $ \sqrt{x^3\plus{}y^2}$ are rational


geometry trapezoid probability MATHCOUNTS

Here is Round 4...The people that may participate are:
JesusFreak197
nat mc
danne89
G-UNIT
Xantos C. Guin
colts18
mysmartmouth
Mo2(i did that with the help of nat mc.)
solafidefarms

Here are the problems...PM me the answers by two weeks from now. :)
In your PM you have to show your work!!!

1) Find the area of a trapezoid with sides of length 2, 3, 3, and 6, where the sides of length 3 and 6 are parallel.

--------------------------------------------------------------------
2)  I have 5 red pairs of gloves, 2 green pairs of gloves, and 4 yellow pairs of gloves.  What is the least mumber of gloves that i need to take out to ensure that i have a pair of gloves?

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

3)  What is the units digit of $1997^{1997}$?

---------------------------------------------------------------------
4)  A dog and cat together cost 25 dollars . If the dog had cost 5 dollars more and the cat’s price did not change, then the cat’s price would have been a third of the total. How much did the dog cost?

---------------------------------------------------------------------
5)  How many positive integers less than 600 are not a multiple of 2, 3, or 5?
-----------------------------------------------------------------------
6)  The cost of a widgit plus its box is 1 dollar and 10 cents, and the widgit  costs 1 dollar  more than its box. How much does the box cost?
------------------------------------------------------------------------
7)  Today is Saturday, April 26, 1997. The year 2000 is a leap  year. What day of the week will April 26, 2001 fall on?
----------------------------------------------------------------------
8) What is the probability of tossing 3 standard dice and getting a sum of 15?
------------------------------------------------------------------------
9) A and B each throw a die. What is the probability that A’s die shows a greater value than B’s die?
----------------------------------------------------------------
10)  Six people— Bob, Bobbie, Rob, Robbie, Robert and 31 Roberta— are to be divided into two study groups. The groups cannot have any person in common, and each group must contain at least one person. In how many ways can this be done?
--------------------------------------------------------------------------

GOOD LUCK!!!

heyyyy, and me?
 :blush: i had 5/5 on round 3  :roll:

oops, sorry i'll add your name.  I forget things frequently so don't think i hate you... :)

[quote="math92"]Mo$^2$(couldn't do the squared well)
[/quote]
try:
[code]2[/code]
for Mo2

thanks nat mc, i edited my post and i turned out well :lol:

How do you get an exponent on your name?

Do we have to show our work?

[quote="Xantos C. Guin"]
Do we have to show our work?[/quote]
i guess this time you should show your work. :D

[quote="Xantos C. Guin"]How do you get an exponent on your name?

[/quote]
I was wondering the same thing, thats really weird.  I wish mine would have an exponent

[quote="chess64"]Hey nat mc, do you know HTML?[/quote]
I know some, but I'm not like a master

[quote="G-UNIT"]math92, these problems are the same if not easier than the last round.  The point is to make problems more challenging as the rounds progress so we can eventually come to a winner....................... :)[/quote]
i don't know where to find challenging problems :?

[quote="G-UNIT"]math92, these problems are the same if not easier than the last round.  The point is to make problems more challenging as the rounds progress so we can eventually come to a winner....................... :)[/quote]

That's why he added the show your work part.

[quote="solafidefarms"]I do! And Javascript too.[/quote]

Me too!

#10 actually has some challenge to it, but it isn't too hard.

[b]EDIT:[/b] Nevermind, I was reading it as how many groups can they be divided into, not how many ways can they be divided into two groups.

The standings so far are:

1st place:  JesusFreak197 (18 points)
note: you get 1 point for showing work and 1 point for answer.

The new standings are:
JesusFreak197: 18 points
chess64 and solafidefarms: 17 points :)

Do you need to get a perfect score to advance on to the next round?

So what your saying is that chess64 and solafidearm cannot advance because jesusfreak has the highest score?

well, i don't know if there is going to be another round...how bout this:
[size=150]you have to be in the top 3 to make Round 5[/size]
there is 8 people and i am sure about 2 of them won't get their answers in so you have to be in the top 3. :)

The new standings are:
JesusFreak197, G-UNIT: 18 points
chess64, solafidefarms, Xantos C. Guin,: 17 points
okay, so i wanted three people...now what should i do.

If no one else sends in solutions, either try to decide who gave the best solutions out of those four, or put them in a separate round (maybe smaller, you decide) and take the top two from there.

Nope, I had 18. I missed #'s 1 and 2 - not sure how I missed #1, and I made a dumb mistake on #2. >_<

[quote="chess64"]Hey, didn't JF197 have like 17 points before? How come he has 18 now? :?[/quote]
no he had 18...(i know JF197 said this but i am just saying this because i run the contest and you may not beleive him)

The new standings are:
JesusFreak197, G-UNIT, solafidefarms: 18 points
chess64, Xantos C. Guin,: 17 points
solafidefarms had the correct answer but not simplified but i never said to simplify. So the top three are JesusFreak197, G-UNIT, solafidefarms :)

so when's round 5???  and what happened to round 1, and 2?????

Round 5 will be this saturday...
and Round 2 is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40656]here[/url]
and Round 1 is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Relay+Race&t=39498]here[/url]
 :)

Will it be PMed to us? Will we have but the one day to do it?

you will have one week to complete it
there will be one point awarded for answer and two for a full solution.
it will be posted today.

[quote="anirudh"]WHERE'S ROUND 3[/quote]

stop spamming :roll:

It is already unsticked.

[quote="anirudh"]WHERE'S ROUND 3[/quote]
its on page two of the mathcounts forum
round 2 is probably on page 3
 and round 1 is called "Relay Race" and its on page 4
 :)


find all natural $n$s such that:
$n!=\prod_{i=1}^{n-3}(x+i)$
$x \in \mathbb{N}$

We are asked to find all $n$ such that $n!$ can be expressed as the product of $(n-3)$ consecutive integers.

For small $n$, we get $\boxed{4!}= 24, \boxed{6!} = 8 \times 9 \times 10$.  Consider larger values of $n$, which have more consecutive integers in their expansions of $n!$ and are not so easily dealt with.

Clearly, $n! = 1 \times 2 \times 3 \times ... \times n = 2 \times 3 \times ... \times n$, which uses $(n-1)$ integers.  For large $n$, "most" of the terms in this product will remain, so we wish only to remove a few, putting them elsewhere in the product.  Removing $2 \times 3 \times 4$ and placing it at the head of the product (as $n+1$) gives us $n = \boxed{23}$.

An effort at showing there are no solutions $n > 23$:

We can express $n!$ as a product of $(n-1)$ consecutive integers, as shown.  In order to transform this into a product of $(n-3)$ consecutive integers, it is necessary to either remove three integers from the tail and add their product to the head, or remove four integers from the tail and add two products to the head (larger removals of $k$ integers require us to add $k-2$ integers to the head, so these are the only distinct scenarios for large $n$).  However, the four tailmost integers are $2, 3, 4, 5$, which cannot be multiplied to form two consecutive integers (a necessity for appending two products to the head).

if $n!=(k+1)...(k+n-3)$ then we call this $n$ a [b]good[/b] number and we show a [b]good[/b] number $n$ by $T(n,k)$
but $n!=(k+1)...(k+n-3)$=>$k!=(n+1)...(n+k-3)$
there for if $n$ can be shown by $T(n,k)$ then $k$ can be shown as $T(k,n)$
there for we have a one-by-one corespondence
know we show that if $n,n'$ are good numbers such that $n=T(n,k),n'=T(n',k'),n'>n$ then $k'<k$
assume that this is wrong
now assume that $k'\geq k$ and we want to claim a contradiction:
$n!=(k+1)...(k+n-3)$  (*)
,
$n'!=(k'+1)...(k'+n'-3)$  (**)
now by the assumption we have $n'>n$ there for $n'!>n!$
(**)=>$n!(n+1)...n'=(k'+1)...(k'+n'-3)$
(*)=>$(k+1)...(k+n-3)(n+1)...n'=(k'+1)...(k'+n'-3)$  (***)
but we have $n-3+n'-n=n'-3$ numbers at the LHS and also $n'-3$ numbers at the RHS
now for $k'>3$ we get that $k'+n'-3>n'$ and also by the assumption we have tha $k\leq k'$=>$k+1\leq k'+1$
but these conditions cause contradiction for (***)
there for $k>k'$ this means that if $\forall_{x,y \in good}x>y: f(x)<f(y)$
there for the maximum value $n$ such that $n$ is good happons when $k$ is minimum
but we have that $T(n,k)\to T(k,n)$ and there for $k$ must be good itself
but the least good number is $k=4\to k!=(n+1)...(n+k-3)=4!=24\to n=23$
there for we get that $4\leq n\leq 23,\forall_{n \in good}$
this is easy to check for $4\leq n\leq 23$ to find good numbers
it is trivial that $n \not= 5$ but if $n=6\to 6!=8*9*10\to k=7$
$n=7\to 7!=7*8*9*10\to k=6$ the next $k$ is $k=4\to n=23$
now our proof is completed
 ;)


analytic geometry
symmetry
advanced fields
advanced fields unsolved

Suppose $(U,\phi)$ is a normal chart at $p$. Show that the components of Riemannian metric with respect to the normal chart at $p$ are $g_{ij}(p)=\delta_{ij}$,$\partial_{k}g_{ij}(p)=0$, and $\Gamma_{ij}^{k}(p)=0$.

Notation: Denote by $(exp_{p}^{-1}, U)$ the normal chart with center $p$. Denote by $\Phi: T_{p}M\to\mathbb{R}^{d}$ the isomorphism $v=v^{i}e_{i}\mapsto (v^{1},...,v^{d})$ ($d=dimM$)where the $e_{i}$ form an orthonormal basis of $T_{p}M$ w.r.t. the Riemannian metric. In particular, $p$ is mapped to $0$. Denote $\partial_{k}g_{ij}$ by $g_{ij,k}$.

PROOF
($g_{ij}(0)=\delta_{ij}$) This follows from the definition of $\Phi$ (it maps an orthonormal basis of $T_{p}M$ w.r.t. the Riemannian metric onto an Euclidean orthonormal basis of $\mathbb{R}^{d}$.)

($\Gamma_{jk}^{i}(0)=0$) Since in normal coordinates the straight lines throught the origin of $\mathbb{R}^{d}$ (i.e. the image of $exp_{p}$) are geodesic you can insert $x(t)=tv$, $v\in\mathbb{R}^{d}$ into the geodesic equation, to obtain $\Gamma_{jk}^{i}(tv)v^{j}v^{k}=0$ (since $x''(t)=0$), and for $t=0$ $\Gamma_{jk}^{i}(0)v^{j}v^{k}=0$. Choosing $v=\frac{1}{2}(e_{l}+e_{m})$ and using the symmetry $\Gamma_{jk}^{i}=\Gamma_{kj}^{i}$ obtain $\Gamma_{lm}^{i}(0)=0$

($g_{ij,k}(0)=0$) By the above discussion, obtain (in 0) $g_{jl,k}+g_{kl,j}-g_{jk,l}=0$. After some manipulation (summing cyclic permutations of the indices, using $g_{kj}=g_{jk}$) get $g_{ij,k}(0)=0$.


Hello everyone,

I am a senior in high school planning to major in math and/or physics in college. My GPA is good but not godly like above 4.0 or anything like that. I am not planning on going to extremely competitive schools like caltech or MIT (i would be rejected in a heartbeat anyways). I plan on going to one of the cal poly schools. They are relatively close to home, and they are cal state schools (meaning they are easy to get into :D), AND they are known for their fine aerospace engineering programs, which is a field i am strongly considering. Now to get to the point.

Does anyone know the general prerequisites for a first year engineering physics course  (calculus based) and would i be able to take it as a freshman? As a senior i am taking AP Physics B , Trigonometry/Math analysis (fancy pants name for Precalculus), and AP statistics (if that is of any significance). Long story short, i didn't do so hot in honors pre-algebra in 7th grade otherwise i'd be in AP Calculus right now.

ANYWAYS, assuming i get a 4 on the Physics B exam (which gives me credit for general ed physics at cal poly), AND assuming i get a high enough MDPT score to place me into Calculus as a freshman, would i be able to go into engineering physics? and would it be a good idea to take it with Calculus 1, or should i wait until i have completed intro calculus?

One last note: All students in Trig/math analysis who keep up a B average (i have over 100% right now so i'm not worried about being able to earn a B) take a "Calculus readiness test". I'm not sure if these results are sent to colleges, but if they are, could i be placed into freshman calculus without the MDPT (mathematics diagnostic placement test)?

Sorry i know this is alot to ask but any answers or input would be greatly appreciated. Thanks for taking the time to read this.

There are two possibilities:
Calculus is a prerequisite for the physics course you're talking about, and you'll have to wait a semester.
or
Calculus is a corequisite for the physics course you're talking about, and you can take it right away if you place into calculus immediately.

For further details, you should ask someone at the school who might know these things.

AP statistics is not significant, and is nearly useless for you. AP Physics B doesn't do much for you either- you'll be taking more physics anyway. Precalculus is the most important piece of what you're doing now- math is the bottleneck which only moves one course at a time, and is required for everything else.

Introductory physics is always easier and clearer with more math, but you have to balance this against time. Waiting one semester even if you're allowed to take it right away might be a good idea, but waitng longer delays all later courses in physics.

If the physics class is taught with calculus, make sure you take calculus *before* the physics class.  In my first year physics class, the students who didn't have calculus before taking the physics class were in a world of trouble (even those taking physics concurrently with calculus).

So, in your case, I'd say take the calculus first, then later do the physics.  If they're making you take physics right off the bat, take calculus over the summer or get a book and teach it to yourself.  Don't go into calculus-based physics without already having the math unless the class is well-designed to teach the math as it's needed (this is almost never the case).

As a Cal State professor and academic advisor who posts regularly here, it looks like this one is for me.

I looked up the course descriptions at both Cal Poly Pomona and Cal Poly SLO. Both Cal Poly schools are on the quarter system. In both cases the first quarter of the physics sequence (the one that physics and engineering majors would take) has the first quarter of calculus as a prerequisite and the second quarter of calculus as a corequisite.

My own campus (CSULB) is on the semester system. Our first physics course (PHYS 151) has Calculus I as a prerequisite, and the second semester (PHYS 152) has Calculus II as a prerequisite.

All of these campuses have a different physics sequence, intended primarily for biological science majors. In our case, we accept AP Physics B as counting for this alternative sequence - which only does you any good if you are a biological science major. It doesn't count for engineering, physics, chemistry, or geology. (It is my impression, however, that students with this AP credit do generally do fairly well in the required calculus-based physics course.)

Most Cal State campuses, including both of the Cal Poly campuses and our campus (as of next year) administer a placement test called the MDPT and only allow students with certain scores on that test to take the first calculus course; a significant number of students are redirected to either precalculus or trigonometry courses. It is unlikely that any of our campuses would accept your high school's calculus readiness test in place of the MDPT. Assume that you will take that.

I concur wholeheartedly with jmerry when he said, "Precalculus is the most important piece of what you're doing now." (Actually, he knew that's what I'd say.) In particular, you want to be well prepared for that MDPT placement exam.

If you have any further questions about the Cal State system in general, feel free to contact me by private message.

Thanks for your input. I appreciate it.

And yeah about what professor maryfield said. I'm aware that physics B is only a survey course intended for the life sciences but i took it based on that same idea that it would provide a good physics foundation for engineering courses


inequalities
logarithms
inequalities proposed

Suppose that $0<a-b<1\leq a+b$. Prove that
\[0<a\ln\frac{2a}{a+b}+b\ln\frac{2b}{a+b}<\ln 2. \]

p.s.: I posted a similar problem on other forum, but still no one answered.

i think i'm wrong.[hide="solution"]we need to show that \[1<\left(\frac{2a}{a+b}\right)^{a}\left(\frac{2b}{a+b}\right)^{b}<2\]the left part is easy since $2a>a+b$.

for the right part, weighted AM-GM gives
$\left(\frac{2a}{a+b}\right)^{a}\left(\frac{2b}{a+b}\right)^{b}\le \frac{a\frac{2a}{a+b}+b\frac{2b}{a+b}}{a+b}$
$=\frac{2(a^{2}+b^{2})}{(a+b)^{2}}<2\iff 2ab>0$.[/hide]


The hypotenuse and a leg of a particular right triangle are
$ \sqrt{97}$ inches and 4 inches, respectively. The area of this triangle
is what common fraction of a square foot?

leg A^2 = hypotenuse^2 - leg B^2
leg A^2 = sqrt(97)^2 - 4 ^2
leg A^2 = 97 - 16
leg A = sqrt(81) or 9

A = 1/2bh
A = 1/2(9)(4)
A = 18 square inches

sq. in. = 1/144 sq. ft.
18 sq. in. = 1/8 sq. ft

answer = [b]1/8[/b]


Help me, please!
D=(x^4+2)/(x^4+4x^2+4)
find min of D :?:

Doesn't belong in Classroom Math...

Try posting this in High School Basics, or get a mod to move it.

hello, prove that
$\frac{x^{4}+2}{x^{4}+4x^{2}+4}\ge \frac{1}{3}$ holds for all real $x$. Sonnhard.


Let $ f: \mathbb{N}\mapsto \mathbb{R}$ be the function defined by $ f(1)\equal{}1, f(n)\equal{}n/10$ if $ n$ is a multiple of $ 10$ and $ f(n)\equal{}n\plus{}1$ otherwise. For each positive integer $ m$ define the sequence $ x_1, x_2, x_3, ...$ by $ x_1\equal{}m, f(x_{n\minus{}1})\equal{}x_n$. Let $ g(m)$ be the smallest $ n$ such that $ x_n\equal{}1$. (Examples: $ g(100)\equal{}3, g(87)\equal{}7$.) Denote by $ N$ be the number of positive integers $ m$ such that $ g(m)\equal{}20$. The number of distinct prime factors of $ N$ is equal to $ 2^u\cdot v$ for two non-negative integers $ u$ and $ v$ such that $ v$ is odd. Determine $ u\plus{}v$.


function
number theory
prime numbers
number theory proposed

Find all natural numbers $n$ for which $\phi (n) \cdot \tau (n)=6n$. Where
$\phi (n)$ is the Euler's function and $\tau (n)$ is the number of the of $n$'s divisors.

This is really a beautiful equation! :)

Let $n=\prod_{i=1}^{r} {p^{k_i}_i},$ where $p_i$ are the prime divisors of $n$ and $r$ is the number of the different prime divisors.

We will check the different cases for $r$

[b][hide="Intro"][/b]
It is known that $\boxed{\phi (n) = n \prod{(1-\frac{1}{p_i})}}$ and $\boxed{\tau (n) = \prod{(k_i+1)}}$

$($it is better to think that the prime divisors are ordered , $i<j\Rightarrow p_i<p_j)$

$\phi (n) \cdot \tau (n) = 6n \Rightarrow$

$\tau (n) = \frac{6n}{\phi (n)}$

$\blacksquare\ {}$ A first necessary condition is that the fraction $t=\frac{6n}{\phi (n)}$ is a natural number

$t=\frac{6n}{\phi (n)} = \frac{6n}{n \prod{(1-\frac{1}{p_i})}} = \frac{6\prod{p_i}}{\prod{p_i} \cdot \prod{(1-\frac{1}{p_i})}} = \frac{6 \prod{p_i}}{\prod{(p_i-1)}} = \frac{2 \cdot 3 \prod{p_i}}{\prod{(p_i-1)}}$

Once we have found a set of primes $\{p_i\}$ such that the above $t$ is a natural number,
the next step is to find the exponents $k_i>1$ such that $\prod{(k_i+1)=t}$

$\blacksquare\ {}$ So, another necessary condition (which will arise later) is that the natural number $t$ can be written as a product of $r$ [u]numbers[/u] (not necessary primes) greater than $1$.

For the number $n=\prod {p^{k_i}_i}$ we define $n_0 =\prod {p_i}$

(notice that $\frac{\phi(n)}{n}=\frac{\phi(n_0)}{n_0}$ and it depends only on the set $\{p_i\}$)

The solution depends only on the number $r$ of the prime divisors[/hide]
[list][b][hide="case-1, r=1"][/b]
if $r=1$, (that is, there is only one prime divisor of $n$) then $n=p^k$ and $n_0=p$

$t = \frac{6 \prod{p_i}}{\prod{(p_i-1)}}\Rightarrow t = \frac{6 p}{p-1}$

$t \in \mathbb{N} \Rightarrow p-1 \mid 6p$

if $p=2$ it is true
if $p>2$ we have $p-1\not{\mid} p \Rightarrow p-1 \mid 6$

The divisors of $6$ are $1,2,3,6$ (the case $p-1=1$ checked before) and we get $p=3,4$ or $7$.
From these values we don't accept $p=4$ because it's not a prime.

$\blacktriangle\ {}$ So, if $r=1$ we have $n_0=2,3,$ or $7$[/hide]
[/list]
[list][b][hide="case-2, r>1"][/b]
	[list][b][hide="Here we can see that the only possible values are r=2 and r=3"][/b]$r\geq 2$

$t = \frac{6 \prod{p_i}}{\prod{(p_i-1)}} = \frac{2 \cdot 3 \prod{p_i}}{\prod{(p_i-1)}}$

Suppose that $p_1 > 2 \ ($ then of course $p_i > 2, \forall i)$

Since every $p_i-1$ is an even number, we have (at least) $r$ 2s at the denominator, but only one at the numerator. So $t \not \in \mathbb{N}$
Hence, it is necessary that $p_1=2$.

$\Rightarrow p_1-1=1$, so we have $r-1$ 2s at the denominator and two 2s at the numerator. So, another necessary thing is $r-1\leq 2 \Rightarrow r\leq 3$.

So if $r\geq 2$ we have two sub-cases:[/hide][/list]
	[list][b][hide="case-2.1, r=2"][/b]
$r=2$

$n_0=p_1p_2,$ but $p_1=2$, so we can write $n_0=2p$, where $p>2$

$t = \frac{2 \cdot 3 \prod{p_i}}{\prod{(p_i-1)}} = \frac{2 \cdot 3 \cdot (2 \cdot p)}{1 \cdot (p-1)} = \frac{12 p}{p-1}$

Since $p>2$ we have $p-1 \not {\mid} p$, so $t \in \mathbb{N}$ gives $p-1 \mid 12.$
We get $p=3,5,7,$ or $13$

$\blacktriangle\ {}$ So, if $r=2$ we have $n_0 \in \{2 \cdot 3 ,\ \ 2 \cdot 5,\ \ 2 \cdot 7,\ \ 2 \cdot 13\}$[/hide]
[/list]
	[list][b][hide="case-2.2,  r=3"][/b]
$r=3$

$n_0 = 2pq,$ where $2<p<q$

We will show that $p=3$

$p,q>2 \Rightarrow$
$p-1=2a$
$q-1=2b.$
Notice that $b>a\geq 1$

$t = \frac{2 \cdot 3 \prod{p_i}}{\prod{(p_i-1)}} = \frac{2 \cdot 3 \cdot 2pq}{1 \cdot (p-1)(q-1)} = \frac{2 \cdot 3 \cdot 2 pq}{2a2b} = \frac{3 pq}{ab}$

Since $t \in \mathbb{N}$ we have $a \mid 3pq$

$a \not {\mid} p \Rightarrow a \mid 3q$

and $3,q$ are prime numbers, so we have that either $a \mid 3$ or $a \mid q$ (or both anyway)

But $a \mid q \Rightarrow a=q \Rightarrow p=2a+1=2q+1>q,$ contradiction.

So $a \mid 3 \Rightarrow a=1$ or $a=3$

If $a=3$ then $p=7$ and $t=\frac{3 pq}{ab}=\frac{pq}{b} \in \mathbb{N} \Rightarrow b \mid p \Rightarrow b=p=7\Rightarrow q=2b+1 = 15$, not a prime

So we have $a=1$ and consequently $p=3$

We got the conclusion that, if we have three prime divisors of $n \ ($so $r=3)$, then one of them is $2$ and another is $3$.

So $n_0=2 \cdot 3 \cdot q$

$t = \frac{3 pq}{ab} \Rightarrow t = \frac{9q}{b} \in \mathbb{N} \Rightarrow b \mid 9 \Rightarrow b=1,3,9$
$b=1$ is not accepted, because $a=1<b$

$b=3 \Rightarrow q=7$
$b=9 \Rightarrow q=19$

$\blacktriangle\ {}$ Finally, in this case $r=3$ we have two triples for the prime divisors: $(2,3,7)$ and $(2,3,19)$,
so $n_0 = 2 \cdot 3 \cdot 7$ or $n_0 = 2 \cdot 3 \cdot 19$[/hide][/list]
[/hide][/list]

[list][b][hide="So we have the solutions:"][/b]
	[list][b][hide="1"][/b]
$1.\ \ \boxed{r=1} \Rightarrow \boxed{n_0=p^k, t = \frac{6 \cdot p}{p-1}}$

$\tau(n) = k+1 \Rightarrow \boxed{k=\tau(n)-1}$

$n_0 = 2 \Rightarrow t = \frac{6 \cdot 2}{1} = 12 \Rightarrow k=11 \Rightarrow \boxed{n=2^{11}}$

$n_0 = 3 \Rightarrow t = \frac{6 \cdot 3}{2} = 9 \Rightarrow k=8 \Rightarrow \boxed{n=3^8}$

$n_0 = 7 \Rightarrow t = \frac{6 \cdot 7}{6} = 7 \Rightarrow k=6 \Rightarrow \boxed{n=7^6}$[/hide][/list]
	[list][b][hide="2.1"][/b]
$2.1 \ \ \boxed{r=2} \Rightarrow \boxed{n_0 = 2p, t = \frac{6 \cdot 2 \cdot p}{p-1} = \frac{12 \cdot p}{p-1}}$

$\tau(n) = (k+1)(m+1) \Rightarrow \boxed{(k,m)=\{(k,m) \mid k,m>1,(k+1)(m+1)=\tau(n) \}}$

$n_0 = 2 \cdot 3 \Rightarrow t = \frac{12 \cdot 3}{2} = 18 = 2 \cdot 3 \cdot 3\Rightarrow \boxed{n \in \{2^13^8,\ \ 2^23^5,\ \ 2^53^2,\ \ 2^83^1 \}}$

$n_0 = 2 \cdot 5 \Rightarrow t = \frac{12 \cdot 5}{4} = 15 = 3\cdot 5 \Rightarrow \boxed{n \in \{2^25^4,\ \ 2^45^2\} }$

$n_0 = 2 \cdot 7 \Rightarrow t = \frac{12 \cdot 7}{6} = 14 = 2 \cdot 7\Rightarrow \boxed{n \in \{2^17^6,\ \ 2^67^1 \} }$

$n_0 = 2 \cdot 13 \Rightarrow t = \frac{12 \cdot 13}{12} = 13 \Rightarrow \boxed{\text{no solution}}$

We have no solution in the last case because $13$ is prime, hence it can't be written as a product of two numbers greater than $1$.[/hide][/list]
	[list][b][hide="2.2"][/b]
$2.2\ \ \boxed{r=2} \Rightarrow \boxed{n_0 = 2 \cdot 3 \cdot q, t = \frac{6 \cdot 2 \cdot 3 \cdot q}{1 \cdot 2 \cdot (q-1)} = \frac{18 \cdot q}{q-1}}$

$\tau(n) = (k+1)(m+1)(s+1) \Rightarrow \boxed{(k,m,s)=\{(k,m,s) \mid k,m,s>1,(k+1)(m+1)(s+1)=\tau(n) \}}$

$n_0 = 2 \cdot 3 \cdot 7 \Rightarrow t = \frac{18 \cdot 7}{6} = 21 = 3 \cdot 7$

$n_0 = 2 \cdot 3 \cdot 19 \Rightarrow t = \frac{18 \cdot 19}{18} = 38 = 2 \cdot 19$

At either case, $n_0$ is a product of two primes, so we can't find a triple $(x,y,z)$ such that $x,y,z>1$ and $xyz=n_0.$
So we have not a solution in this case (the second necessary condition does not hold!)[/hide][/list]
[/hide]
[/list]

I read the intro (and plan to read the rest later).  This is such a cool problem.  Good work.

Here is [b][u]Potentially Another Beautiful Equation[/u][/b]
Find all positive integer solutions in [i]n[/i] to
\[ \sigma(n)\tau(n)=6n \]

[quote="baz"]Here is [b][u]Potentially Another Beautiful Equation[/u][/b]
Find all positive integer solutions in [i]n[/i] to
\[ \sigma(n)\tau(n)=6n  \][/quote]

$\sigma(n)$ is the sum of the divisors of $n$ (including 1 and n)

$\tau (n) = \prod{(k_i+1)}$

$\sigma(n) = \prod {(1+p_i+ \cdots +p^{k_i}_i)}$

$\tau(n) \sigma(n) = \tau(n) \prod {(1+p_i+ \cdots +p^{k_i}_i)}$

$\prod {(1+p_i+ \cdots +p^{k_i}_i)}>\prod {p^{k_i}_i}=n$

So, if $\tau(n)\geq 6$, then $\tau(n) \sigma(n)>6n$

[u]1st Necessary Condition[/u] $\tau(n)<6 \Leftrightarrow \prod{(k_i+1)}<6$

Since $k_i+1\geq 2$, if we have $3$ factors then the product is $\geq 8$.

[u]2nd Necessary Condition[/u] $r\leq 2$.

If $r=2$ (two prime divisors of $n$) then $(k_1+1)(k_2+1)<6 \Rightarrow  k_1=1$ and $k_2=1$.

If $r=1$(so we have $n=p^k$) then the possible values are  $k+1=2,3,4,5 \Rightarrow  k=1,2,3,4$

For the first case it's easy too show that there are not solutions

$n=pq,$ with $p<q$
$\tau(n) = 4$
$\sigma(n) = (p+1)(q+1)$

$\tau(n) \sigma(n) = 6n \Leftrightarrow$
$4(p+1)(q+1) = 6n \Leftrightarrow$
$2(p+1)(q+1) = 3n \Leftrightarrow$
$2(p+q+pq+1) = 3pq \Leftrightarrow$
$2(p+q+1) = pq$

So $2 \mid pq \Rightarrow  p=2 \Rightarrow$

$2(2+q+1) = 2q \Rightarrow$
$3+q=q$
$3=0,$ contradiction

For the second case we have $n=p^k \Rightarrow$
$\tau(n)=k+1$
$\sigma(n) = 1+p+ \cdots +p^k = \frac{p^k-1}{p-1}$

$\tau(n) \sigma(n) = 6n \Leftrightarrow$
$(k+1)(1+p+ \cdots +p^k) = 6 p^k \ \ (1)$,where $1\leq k\leq 4$

I didn't prove it, but I noticed that there is not solution to $(1)$.

Because for $p\geq 7$ the function $f_p(k) = 6 p^k - (k+1)(1+p+ \cdots +p^k)$ seems to be an increasing function of  k
And it's not difficult to check all the cases for $p=2,3,5$ and $k=1,2,3,4$

If someone prove that the equation $(1)$ does not have solutions, it would be ok  :)


number theory
least common multiple
number theory proposed

Prove that for each $ n \geq  2$ then exist infinite n term $ (x_1,x_2,..,x_n)\in N$ such that
$ \sum_{i \equal{} 1}^n \frac {x_i \plus{} 1}{x_{i \plus{} 1}}$ is an integer .
Note $ x_{n \plus{} 1} \equal{} x_1$
a) (An easy question )
$ d \equal{} lcm (x_1,..,x_n)$
Prove that :
$ d^{n}\leq (\prod_{i \equal{} 1}^n x_i) (\sum _{i \equal{} 1}^n \frac {1}{x_i})$
This is an problem from a competition.

i don't understand. do you mean that there exists an infinite sequence of naturals $ (x_n)$ such that for every $ n$, $ \sum_{i\equal{}1}^n \frac{x_i\plus{}1}{x_{i\plus{}1}}$ is an integer? or just that for any $ n$ we can find some $ n$ integers that satisfy this property? please state your problems clearly.

Yes for each $ n\geq 2$ we can find infinite n tube $ (x_1,...,x_n)$

that doesn't make any sense. it's the same thing you said in your original post. and besides, if you mean what i think you mean, we can simply take the infinite sequence $ 1,2,3,4,...$. then $ \sum_{i\equal{}1}^n \frac{x_i\plus{}1}{x_{i\plus{}1}}\equal{}1\plus{}1\plus{}1\plus{}\cdots\plus{}1\plus{}n$, where $ 1$ appears $ n\minus{}1$ times. so the problem is trivial.

[quote="pleurestique"]that doesn't make any sense. it's the same thing you said in your original post. and besides, if you mean what i think you mean, we can simply take the infinite sequence $ 1,2,3,4,...$. then $ \sum_{i \equal{} 1}^n \frac {x_i \plus{} 1}{x_{i \plus{} 1}} \equal{} 1 \plus{} 1 \plus{} 1 \plus{} \cdots \plus{} 1 \plus{} n$, where $ 1$ appears $ n \minus{} 1$ times. so the problem is trivial.[/quote]
For each n there exist infinite tube $ (x_1,..,x_n)$ .
Your tube $ (1,..,n\minus{}1)$ is only one solution.
For example $ n\equal{}2$ we can find infinite pair of $ (x,y)$ such that
$ \frac{x\plus{}1}{y}\plus{}\frac{y\plus{}1}{x}$ is an integer.

okay. well then what you should have said was "infinitely many tuples". don't you know how to pluralize words?


calculus
integration
calculus computations

integral sign

(5r)(9^r) dr

how would i do this? :maybe:  :(

hello, do you mean $ \int5r9^r\,dr$?
Sonnhard.

hello, using integration by parts we get
$ \int5r9^r\,dr \equal{} \frac {5r}{\ln(9)}9^r \minus{} 5\int\frac {9^r}{\ln(9)}\,dr$
Sonnhard.

i tried

(5/ln9)*9^r  -  5(  (1/ln9)*(  (9^r)/ln9)  ) + C  a the final answer

but this is still wrong

hello, i have $ \frac{5r}{\ln(9)}9^r\minus{}5\frac{9^r}{(\ln(9))^2}\plus{}C$ as the answer.
Sonnhard.

i now tried that, but this wrong
this is what i got
http://i.imgur.com/zepO2.png


Points A and B line on a plane at distance 2 from one another. S is the set of points P such that (PA)^2 + (PB)^2 is less than or equal to 10.

Find area of S.

[hide="Solution"]In the coordinate system we can take $A(-1,0), B(1,0), P(x,y).$ Then

$(x+1)^2+y^2+(x-1)^2+y^2\leqslant 10$

After expanding and simplifying

$x^2+y^2\leqslant 4$

This is the circle with radius $2$, hence the desired area is $4\pi.$[/hide]

Define the midpoint $O$ of the fixed segment $AB$. From the median theorem $4OP^2+AB^2=2(PA^2+PB^2)\le 20$ results $4OP^2+4\le 20$, i.e. $0\le OP\le 2$. Therefore, the set of the points $P$ for which $PA^2+PB^2\le 10$ is the circle $C(O,2)$ with the area $4\pi$.

[color=blue]Try solve similarly (synteticaly) the following [b]generalization[/b] :
Let $a,b,c,d$ be four real numbers so that $a+b\ne 0$, $c>0$, $d>0$ and let $[AB]$ be a segment for which $AB=d$. Ascertain the area of the set $\Omega=\{ P| a\cdot PA^2+b\cdot PB^2\le c\}$. [u]Discussion.[/u][/color]

General question: Why analytical solutions are usually looked down upon? I don't understand that. Analytical geometry is a very powerful tool that can liquidate a number of problems in a few lines. Then why it is often insisted on "synthetical" solutions? Are they "prettier"? Do they do justice to the solvers, showing the elegance of their thinking? That's all very nice and necessary, but at the end of the day, a valid solution is a valid solution.

no they are fine, but on more difficult problems they don't just solve the problem instantly, and you loose that essential experience you could have gotten by finding clever solutions, and perhaps you could have solved that harder problem using ideas from a clever solution to another problem, not saying that you directly have these troubles, but it could increase your problem solving ability


inequalities
geometry
area of a triangle
Heron\u0027s formula
inequalities proposed

Prove that : In the triangle ABC :
 $ p^2+5r^2 \geq 16Rr $

Here  is  a  solution : Put  [tex] a=x+y , b=y+z , c=z+x  ,  x,y,z>0 . [/tex].
The  inital  inequality  becomes  [tex] (x+y+z)^3\geq 4(x+y)(y+z)(z+x)-5xyz [/tex].
This one  is  homogenous  so  consider  [tex] x+y+z=1 [/tex] So we  only must  prove  that  [tex] 1\geq 4(1-x)(1-y)(1-z)-5xyz \Leftrightarrow 1+9xyz\geq4(xy+yz+zx) [/tex]
which is  true  by Shur . Anyway  this  one is week , it also  follows  from  [tex] s^2\geq2R^2+8Rr+3r^2 [/tex]  which  is littlebit stronger.

[quote="romano"]$ 1+9xyz \geq 4(xy+yz+xz) $
 Why is it true by Schur ?[/quote]

Since we have normalised with $x + y + z = 1$, replace $1$ with $(x+y+z)^3$ and multiply RHS by $(x+y+z)$, expand and simplify and you will get Schur's: $ \sum_{cyc} (x^3 + xyz) \geq \sum_{sym} x^2y$

PS: I know this a bit dumb of me not to know, but could you explicitly state the formulae for r and R, and show a bit of how it reduces to the said inequality?

Use  Heron's  formula  for the  area  of  the triangle  and  find  [tex] [ABC]=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{xyz(x+y+z)} [/tex] .Also  we  have  [tex] [ABC]=sr=\frac{abc}{4R} [/tex]  so  we have  [tex]R=\frac{(x+y)(y+z)(z+x)}{4\sqrtxyz(x+y+z)} [/tex] and  [tex] r=\sqrt{\frac{xyz}{x+y+z}} [/tex] . Put  them  in the  initial  inequality  and  you will see  :)


linear algebra
matrix
limit
induction
linear algebra unsolved

The sequence of $m\times m$ real matrices $\{A_{n},\, n\ge0\}$ is defined as follows: $A_{0}=A,$ $A_{n+1}=A_{n}^{2}-A_{n}+\frac{3}{4}I,\, n\ge0,$ where $A$ is positively defined matrix such that $\mathrm{tr}(A)<1.$ Find $\lim_{n\to\infty}A_{n}.$

should be helpful to diagonalize it and study the sequence $u_{n+1}=u_{n}^{2}-u_{n}+\frac{3}{4}$ for $0 \leq u_{0}< 1$  :maybe:

In some books, 1) the condition that matrix be symmetric is included in the  definition of positive (semi)definite matrix; in others 2) not. If Moubinool meant 1), then here's my proof.

Set $f(u) = u^{2}-u+3/4$, then $A_{n+1}= f(A_{n})$. Define matrix norm as follows (spectral norm)
\[\|A\| = \sqrt{\max \sigma(A^{*}A)}. \]
Consider a closed subspace
\[X = \{ A \in M_{m}(\mathbf R) \mid A \text{ positive-semidefinite, }1/2 \le \|A\| \le 3/4 \}\]
of $M_{m}(\mathbf R)$, then $f \colon X \to X$ is a contraction. Indeed, the inclusion $f(X) \subset X$ follows from the fact that $\min f(\mathbf R)=1/2$ and $\max f([0,1])=3/4$. $f$ is a contraction, because $\|A+B-I\| \le 1/2$ so
\[\|f(A)-f(B)\| = \|(A+B-I)(A-B)\| \le \frac{1}{2}\|A-B\|. \]
Since $\mathrm{Tr}\, A_{0}< 1$ and $A_{0}$ is positive-definite, one deduces that $\|A_{0}\|<1$ and $A_{1}= f(A_{0}) \in X$. It now follows from Banach fixed point theorem that $B=\lim A_{n}$ exists and equals to the unique fixed point of $f$ in $X$:
\[B = B^{2}-B+\frac{3}{4}I, \]
or equivalently,
\[\left( B-\frac{1}{2}I \right) \left( B-\frac{3}{2}I \right) = 0. \]
Since $\|B\| \le 3/4$, the solution is $B = I/2$.

by induction $A_{p}$ are symetric real

for i=/=j  the matrices $A_{i}$ and $A_{j}$ commute

so they exist an orthogonal matrix $P$ such all $P^{-1}A_{j}P$ are diaogonal

if $a(p,k)$ are the eigenvaules of $A_{p}$    (k are integer from 1 to n)

we have

$a(p+1,k)=a(p,k)^{2}-a(p,k)+3/4$ with $0\leq a(p,k) <1$

the sequence p---> $a(p,k)$ is convergent

$A_{p}$ is convergent to $B$

$B^{2}-2B+3/4=0$

$(I_{n}-B)^{2}=I_{n}/4$ with $I_{n}-B$ symetric positive matrix

then I_n-B=I_n/2

B=I_n/2

a quick way

$C_{p}=A_{p}-\frac{I_{n}}{2}$

get a very simple recurrence relation with $C_{p+1}$ and $C_{p}$

and express $C_{p}$ with $C_{0}$ ...


algebra
polynomial
advanced fields
advanced fields unsolved

Let $ K$ be a number field. Let $ P \lhd \mathcal{O}_K$ be prime. Let $ a \in \mathcal{O}_K$.

Suppose that $ a^p \minus{} a \in P$. Prove that $ a \plus{} P \equal{} m \plus{} P$ for some $ m \in \mathbb{Z}$.

I suppose that $ p$ means the prime in Z with $ pZ\equal{}P\cap Z$ here. Then the field $ K(P): \equal{}\mathcal O_K/P$ has characteristic $ p$.

The case $ a\in P$ is trivial, so assume $ a\notin P$. Then $ \overline{a}\equal{}a\plus{}P\in K(P)$ is a zero of the polynomial $ f(X)\equal{}X^{p\minus{}1}\minus{}1$. But the zeros of $ f(X)$ in the field $ K(P)$ are precisely the non-zero elements of the field $ Z/pZ$ sitting in $ K(P)$, i.e. $ a\in Z/pZ$.


modular arithmetic
geometry
geometric transformation
reflection
number theory
relatively prime
Diophantine equation

A bad number is said to be "bad" if it has a multiple whose last three digits are "002". For example, 13 is "bad" since 2002 is a multiple of 13.

Find all positive integers which are "bad".

[hide]we are trying to find all integers $ n$ for which there exists $ k$ such that $ nk\equiv 2\pmod{1000}$. clearly one of $ n,k$ must be even. if $ n$ is even, dividing by $ 2$ we get $ \left(\frac{n}{2}\right)\cdot k\equiv 1\pmod{500}$. This is possible only if $ \gcd\left(\frac{n}{2},500\right)=1$, that is, if $ \frac{n}{2}$ is not divisible by $ 2$ or $ 5$.

if $ n$ is odd, then $ k$ must be even. so $ n\cdot\left(\frac{k}{2}\right)\equiv 1\pmod{500}$. thus $ \gcd(n,500)=1$ i.e. $ n$ is relatively prime to $ 2$ and $ 5$.

the converse holds too: if $ n$ is relatively prime to $ 2$ and $ 5$, then there exists $ k$ such that $ nk\equiv 1\pmod{500}$, which is well known.

thus we conclude that $ n=m$ or $ 2m$, where $ m$ is an odd positive integer such that $ 5\not|\ m$.  [/hide]

[hide="One line"]$ n\text{ is bad}\Leftrightarrow \exists x : nx \equiv 2 \bmod 1000 \Leftrightarrow \exists x, y: nx+1000y = 2 \Leftrightarrow \gcd(n, 1000) | 2$[/hide]

$ P \Leftrightarrow Q$ means that $ P$ is true if and only if $ Q$ is true.  In other words, the two statements are equivalent.  (The last equivalence is an application of Bezout's Lemma.)

$ \exists$ means "there exists."

$ \gcd(n, 1000) | 2$ means $ \gcd(n, 1000) = 1$ or $ \gcd(n, 1000) = 2$ (since those are the only divisors of $ 2$).

[quote="0714446459923"]what does that mean exactly?[/quote]
Nice solution.That means that if there are solutions of that diophantine equation$ \in N_{+}$, then $ gcd(n,1000)$ divides $ 2$.

and I assume that the colons mean "such that"? It makes sense now

OMG! Thanks to all of you guys. What a solution! Thanks again  :D
_________________

Open your mind. Let the harmonious reflection of nature moves our hearts.

[url=http://www.printsplusposters.com]Art Prints & Art Posters[/url]


calculus
derivative
geometry
3D geometry
calculus computations

1. a snowball is melting such that the radius is shrinking at a rate of 5 in/hr. At what rate will the volume be decreasing at the moment when the radius of the snowball is 8 inches?

2. A cone shaped cup is 5 inches tall. The top of the cup has a radius of 3 inches. if the water is being poured into the cup at a rate of 2 cubic inches per sec, how fast will the water level be rising when the water is exactly 4 inches deep?

3. Suppose the base of a 10 foot ladder is being pushed toward the wall at a rate of 2 ft per sec. how fast will the top of the ladder be rising at the moment when the base of the ladder is 6 feet from the wall?

One guideline to related rates problems: carefully distinguish between equations or relationships that remain true even as things change, and equations which are "snapshots", true only for an instant. The former can be differentiated with respect to time, the latter cannot.

For example, in question 1, $V=\frac43\pi r^3$ remains true over time, so you may usefully take $\frac{d}{dt}$ of it. However, $r=8$ (inches) is a snapshot.


Prove that if $ gcd(a,p) \equal{} 1$, and the order of $ a (mod p)$ is $ 3$, then the order of $ (a \plus{} 1) mod p$ is $ 6$.

Note: Let gcd(a,m) = 1. Then the order of a (mod m) is k if and only if k is the smallest positive exponent such that $ a^k \equiv 1 (mod m).$

False when $ p$ is composite; consider $ a \equal{} 3, p \equal{} 26$.

[hide="Solution"] $ a^3 \equiv 1 \bmod p \implies p | (a \minus{} 1)(a^2 \plus{} a \plus{} 1)$.  Now, if $ p | a \minus{} 1$ then the order of $ a \bmod p$ is $ 1$, so $ p | a^2 \plus{} a \plus{} 1$.  Then $ (a \plus{} 1)^2 \equiv a \bmod p$, so $ (a \plus{} 1)^2$ has order $ 3$ and $ (a \plus{} 1)$ therefore has order $ 6$. [/hide]

I can understand up to the line that $ (a \plus{} 1)^2$ has order 3, but how did it follow that $ (a\plus{}1)$ has order $ 6$.

If $ (a \plus{} 1)^2$ has order $ 3$ then $ (a \plus{} 1)^6 \equiv 1 \bmod p$ so $ a \plus{} 1$ can only have orders $ 1, 2, 3, 6$.  And:

- Since $ a \not \equiv 0$ we know $ a \plus{} 1 \not \equiv 1$, so it does not have order $ 1$.
- Since $ a \not \equiv 1$ we know $ (a \plus{} 1)^2 \not \equiv 1$, so it does not have order $ 2$.
- $ (a \plus{} 1)^3 \equiv \minus{}1 \bmod p \not \equiv 1 \bmod p$, so it does not have order $ 3$.

Hence it must have order $ 6$.


calculus
derivative
function
real analysis
real analysis unsolved

Let a function $ f$  defined on $ [0,1]$, such that $ |f'(x)| \geq M >0$ for all $ x \in [0,1]$.Prove that there exists $ a \in [0 , \frac{3}{4}]$, such that $ |f(x)| \geq \frac{M}{4}$ for all $ x \in [a,a+\frac{1}{4}]$.

[b] In other words :[/b]

 There exists a subinterval $ I$ of $ [0,1]$ with length $ d = \frac{1}{4}$ , such that   $ |f(x)| \geq \frac{M}{4}$ for all $ x \in I$

[u] Babis[/u]

[color=red] EDITED[/color]

 I had edited one minute after the first message , but something happened and the change was canceled. Really sorry. :(

I think the goal is to prove that $ |f(x)|\ge \frac{M}{4}$ on a subinterval of length $ 1/4$.

[quote="jmerry"]This is utterly trivial as stated; $ |f'|>\frac14$ everywhere by hypothesis.

What question did you really mean to ask?[/quote]

 [color=red]Really sorry ! [/color]

 I have edited for second time. I hope now it is right ! ( $ |f(x)|\geq \frac{M}{4}$ )

 [u]Babis[/u]

As $ |f'| \geq M > 0$, either $ f' \geq M$ or $ f' \leq-M$ for all $ x$ since $ f'$ satisfies the intermediate value property. WLOG $ f' \geq M$, so $ f$ is increasing.

Assume the statement is false. By the Mean Value Theorem, $ f(\frac{3}{4})-f(0) = \frac{3}{4}f'(u) \geq \frac{3}{4}M$. Since $ f(\frac{3}{4}) < \frac{M}{4}$, we must have $ f(0) <-\frac{M}{2}$.

Let $ a = inf\{ \{1\}\cup \{x \in [0, 1] |f(x) =-\frac{M}{4}\}\}$. If $ a \geq \frac{1}{4}$, we're done since on $ [0, \frac{1}{4}]$ we have $ f \leq-\frac{M}{4}$. Thus $ a < \frac{1}{4}$. Applying MVT one more time gives $ f(\frac{3}{4})-f(a) = (\frac{3}{4}-a)f'(v) >\frac{1}{2}M$, a contradiction.

Edit note: I corrected the solution after I realized $ f$ itself doesn't need to be nonnegative.


superior algebra
superior algebra unsolved

Let $ K \subseteq E$ be an extension of odd degree. Suppose that $ E \equal{} K(\theta)$ for some $ \theta \in E$.

Is it true that $ E \equal{} K\left(\theta^2\right)$?

Yes, look at $ [K(\theta): K(\theta^2)]$.

:maybe: ... and then what? [sorry for being so slow today]

What can you say about it¿ And what does it imply for $ [K(\theta): K] \equal{} [K(\theta): K(\theta^2)] \cdot [K(\theta^2): K]$¿.

I want to show that it is equal to 1, of course. But I don't see how...

Oh, right. We have that $ \left[K(\theta) : K\left(\theta^2\right)\right] \leq 2$. However, it cannot be 2 since $ [K(\theta) : K]$ is odd. So we're done.


geometry
combinatorics solved
combinatorics

Determine all values of $n$ such that it is possible to divide a triangle in $n$ smaller triangles such that there are not three collinear vertices and such that each vertex belongs to the same number of segments.

The desired values are $n=1,3,7,19$.

Suppose $n$ is a good value, and consider the planar graph given by the configuration (that is whose vertices are the vertices of the triangles and the edges are the sides of the triangles). Since no three vertices in the configuration are collinear, we deduce that each of the faces of this graph are triangular (I omit the not bounded one) and even are the triangles themselves. Thus there are $f=n$ faces.

Now, each of these faces uses exactly $3$ edges, and each of these edges belong to exactly $2$ triangles except for the three of the initial triangles. It follows that there are exactly $e= \frac {3n-3} 2 + 3 =  \frac {3(n+3)} 2 $ edges.
Therefore $n$ has to be odd.

Let $d$ be the common degree of the vertices, and $v$ be the number of the vertices. Since the sum of the degree is twice the number of edges, we deduce that $ds = 3(n+1)$.

Now, using Euler's formula for planar graphs, that is $v+f-e=1$ (I still omit the not bounded face), we deduce that $d = \frac {6(n+1)} {n+5}$. (1)
Since $n$ is odd, let $n=2k+1$ where $k \geq 0$.
The relation (1) becomes $d= \frac {6(k+1)} {k+3}$ which has to be an integer.

- If $k$ is even, then $\gcd (k+1,k+3) = 1$ so that $k+3$ has to divide $6$, and $k+3$ is odd and greater than $1$. Thus, $k=0$ and $n=1$.
Conversely, $n=1$ is a solution. It suffices to consider the initial triangle itself.

- If $k=2q+1$ is odd. Then, the relation (1) becomes $d = \frac {6(q+1)} {q+2}$ where $\gcd (q+1,q+2)=1$. It follows that $q+2$ divides $6$ where $q+2 \geq 2$.
It leads to $q=0,1,4$ which in turn leads to $n=3,7,19$.

Conversely, $n=3$ is a solution. Just choose an interior point of the initial triangle and join it to the vertices.
$n=7$ is also a solution, as given by the well-known Morley's configuration (since here we do not care about the equilateral triangle, you just have to choose three interior point such that among the $6$ vertices there is no three collinear ones).

It remains to check $n=19$...
From above, if a configuration does exist it has $12$ vertices each with degree $5$ for a total of $30$ edges and $19$ triangles.
Start from a triangle $ABC$ and choose $6$ interior points, say $D,E,F,G,H,I$ in it which are the vertices of a convex hexagon, so that $A$ is joined with $D,E,F$, $B$ is joined with $F,G,H$ and $C$ is joined with $H,I,D$ (thus choose and label carefully these points...). Now, choose $3$ interior points of this hexagon, say $JKL$, and darw edges $JD,JE,JI$ and $KE,KF,KG$ and $LG,LH,LI$ and $JK,KL,LJ$. We are done...

Pierre.

We claim that $n \in \{1,3,7,19\}$.

Consider the obvious planar graph $G$ representation. Suppose that all the vertices have degree $k$. Hence, $$2A= \sum_{v \in G} deg v= kV=3F_3=3(n+1)$$
This implies that $n+1=\frac{2A}{3} \implies$ by Euler's relation, we know that $V+F=A+2 \implies V+ \frac{2A}{3}=A+2 \iff 3V=A+6 \implies 6A=k(A+6) \implies A+6|6A, 6A+36 \implies A+6|36$, so if we let $A=3a$, we have that $a+2|12$, so $a \in \{1,2,4,10\} \implies \frac{3(n+1)}{2}=A \in \{3,6,12,30\}$, which implies that $n \in \{1,3,7,19\}$, as claimed. For $n=1,n=3$ the example is trivial. For $n=7$, we can take the Morley configuration as said above. For $n=19$, we can consider the following example:


There are 25 students playing chess. Each of them plays one game with 3 other people (tries to). Does there have to be one person that plays with 4 people?

[quote="4everwise"]At least three other people? There is no forced play because everybody can just play each of the 24 others once.   ;)[/quote]
I think that the question is supposed to say "4 or more people" instead of "4 people".

Graph theory?

Viewing each player as a dot, there would be 25*3/2=75/2=37.5 lines. But because you cannot have fractional lines, one player must play at least 4 times.

PenguinIntegral has the right idea, although I've never heard of graph theory.

I've fixed it. does it become less ambigous (I'm trying to translate)?

Graph theory, we learned about that in MathPath. At least, I would've if I paid any attention at all during lectures...

I didn't go to mathpath. I learn't it from mathematical circles.

It's a great book :)

Ok, I get it.

My first post was inappropriate.  :blush:

[hide="My idea"]Group the people into 4 so that everyone can play against three, leaving one person left. He has to play with three people, and whoever plays against him plays four games (since they already played against three other people in their groups). [/hide]

The idea behind graph theory is to consider each player as a vertex of degree 3, but this gives you an odd number of odd vertices so it is not possible.

PenguinIntegral, did you just do the Independent Study?  (I also got this out of mathematical circles during the independent study, so just curious)

No. Sadly, my parents are internet-phobic. I have to teach myself.

I just hope im doing good for 12 1/2 years...


[quote="sweepsfairie"]I have some problems I am stuck on if anyone can help!!

A line perpendicular to 3x+4y=8 contains the point (6,-2) what is the equation?

I subtract 3x from both sides and that leaves me with
4y=-3x+8 and than I would divide by 4 and that gives me
y=-3/4x+2 and -3/4 is my slope so then I do y+2/x-6=-3/4 but how do I turn that into standard form??[/quote]
[hide]
The slope of a perpendicular line is the negative reciprocal of the line. Since the original equation's slope is -3/4, the slope of the perpendicular line is 4/3. Since y=mx+b and we have a point, we get $-2=8+b$ and b=-10, so the equation is $y=\frac{4}{3}x-10$. [/hide]


What is the least whole number value of x such that $ f(x)\equal{} x^2 \plus{} x \plus{} 11$ is not a prime?
(Without guess/check)

The astronomical unit (AU), the parsec, and the light-year are units used to express distance in astronomy. One light-year is equal to $ 6.33 \times 104$ AU's. And one AU is equal to $ 4.85 \times 10^{ \minus{} 6}$ parsecs. Use one significant digit approximations to estimate how many parsecs are in one light-year. Express your answer in scientific notation.

There are an equal number of pennies, nickels, dimes, and quarters in a bag. Four coins are pulled out, one at a time, and each coin is replaced before the next one is drawn. What is the probability that the total value of the four coins will be less than 20 cents? Express your answer as a common fraction.

For the first one, did you mean $ f(x) \equal{} x^2 \plus{}x \plus{}11$?

EDIT: Never mind..read question wrong.....

mods..please delete this post

for the third one, there are 256 possible combinations (4*4*4*4)
the possible combination of coins less than 20 cents could be:
         different combinations
d n p p - 12
d p p p - 4
n n n p - 4
n n p p - 6
n p p p - 4
p p p p - 1
12+4+4+6+4+1=31

31/256

[quote="Luminescence"]For the first one, it's 11.
For it to not be prime, all the terms have to be divisible by the same number. Since the constant is 11, the smallest such number is 11.[/quote]

what about $ 10$? $ f(10)\equal{}10^2\plus{}10\plus{}11\equal{}100\plus{}21\equal{}121\equal{}11^2$. i think the answer is $ \boxed{10}$

f(n)=n(n+1)+11. n(n+1) has to be divisible by 11. so this is minimized when n+1=11 or n=10


Solve for $ x \in \mathbb{R}$,

 $ \sqrt{x^{2}\plus{}2x} \plus{} \sqrt{x\plus{}2} \plus{} \sqrt{x} \plus{} x \equal{} 3$.

$ (\sqrt x \plus{}1)(\sqrt{x\plus{}2}\plus{}\sqrt x)\equal{}3 $
obvious from here x=1/4

Thanks implex.

I can see that $ x\equal{}\frac{1}{4}$ is a good choice to try, but it wasn't immediately obvious to me. How do we know that this is the only solution? I think I can see it graphically.

hello, $ x\equal{}\frac{1}{4}$ is the only real solution.
Sonnhard.

Let $ z\equal{}\sqrt{x\plus{}2} \plus{} \sqrt{x}$, so

$ z^{2}\equal{} x\plus{}2 \plus{} 2\cdot\sqrt{x(x\plus{}2)} \plus{} x \equal{} 2(x\plus{}1\plus{}\sqrt{x(x\plus{}2)}$ and

$ \sqrt{x^{2}\plus{}2x} \plus{} \sqrt{x\plus{}2} \plus{} \sqrt{x} \plus{} x \equal{} 3$ becomes

$ \frac{1}{2}y^{2} \plus{}y\minus{}4\equal{}0$

$ y^{2} \plus{} 2y \minus{}8\equal{}0$

$ (y\minus{}2)(y\plus{}4) \equal{} 0$

But $ y >0$ so $ y\equal{}2$

Therefore $ \sqrt{x\plus{}2} \plus{} \sqrt{x} \equal{} 2$

$ x\plus{}2\equal{} 4\plus{}x\minus{}4\sqrt{x}$

$ 4\sqrt{x} \equal{} 2$

Therefore $ x \equal{} \frac{1}{4}$ is the unique solution.

[quote="implex"]$ (\sqrt x \plus{} 1)(\sqrt {x \plus{} 2} \plus{} \sqrt x) \equal{} 3$
obvious from here x=1/4[/quote]

There's nothing "obvious" about x = 1/4.  You happened to notice it.
Don't use a subjective word to describe something that is not
universal to all who look upon it by inspection.

Look at this difference in being reasonable:

"I noticed that x = 1/4 works."

However, getting a value by guessing is not a reliable method.


who ever u r  the answer i got is :-

However, I think the question is wrong.
a point needs to be added or I am wrong.

TRIED TO WRITE IT BIGGER .

ok. something wrong with the question? im not sure. i think thats wat the question said. im srry if its wrong. thnx tho :D

[quote="azn_invazn"]Four tennis players get together for a doubles match.  Two players are on the clubhouse side and two players on the side of the umpire's chair.

This is how it goes...

   1. Wang is to the left of the player with the big forehand.
   2. The player with a strong overhead has a Bonnie warm-up.
   3. The player wearing the Prince shirt is on Spaz's right.
   4. The player who has a good volley is the player with blisters.
   5. The player with a spin serve is across from Kye.
   6. Tex, in a Nike shirt, has Babolat tape on his racquet.
   7. Wang keeps looking at a girl in the clubhouse on his left.
   8. Who is wearing the Adidas shirt?[/quote]

Ok, so there are 4 people, Wang, Tex, Kye, Spaz.

_====club====_

_====umpi====_

These are the seats.

So Wang is in the Top Left. (#7)
The big forehand is the bottom right (#1)
Spaz doesn't have a prince shirt (#3)
Tex wears a Nike shirt, and has Babolet tape (#6)
Kye does not have a spin serve (#5)
Wang does not have a big forehand (#1)
I don't get #4. Is Blisters something to do with tennis?
Kye is on the Bottom left (#5), Spaz must be on the top right. Thus, Tex=Nike shirt, babolet tape, big forehand
What's a Bonnie warm-up? A shirt? If so, Spaz has Bonnie or Adidas.
If Spaz has Adidas, Wang has bonnie thingy, and thus a strong overhead.
But Wang has spin serve (#5), so Spaz has bonnie thingy, and [b]Wang has Adidas.[/b]

I  have cum up with another answer  :maybe: I thnik this is correct.  :maybe: .

THE DATA I GOT IS :-

WANG-  PRINCE SHIRT,VOLLEY.
   KYE-BONNIE SHIRT,OVERHEAD.
TEX-  NIKE SHIRT,BIG FORE HAND.
SPAZ-ADIDAS SHIRT,SPIN SERVE.

cant u reply ? r ur exams not yet over? please reply.


#7 on the State target was something like, point A makes triangles with line P. Points are line P are labeled P1, P2, etc. If a total of 120 triangles are made, what is the value of x in the last point Px.

Was the answer 16?

[quote="biffanddoc"]The key is that for $n$ base points, that makes $\binom n2$ triangles (overlapping ones are counted too!)
$\binom{16}{2}=120$, so $n=16$[/quote]
Remember that $\binom n2$ = n(n-1)/2

i found a pattern that goes like this:
2 points, 1 triangle
3 points, 3 triangles
4 points, 6 triangles
5 points, 10 triangles

it's faster than it looks.  :)
i just added them up on my calculator.
one of my teammates forgot to add the big triangle. he was mad!  :lol:

oh yeah, notice they're all triangle numbers. you can see lots of patterns and stuff just in the table i put up there.
if you put it like this:
2,1
3,3
4,6
5,10

i bet you could write a function for it if you wanted.
also, if you shifted the first column down one, you would get triangles and their bases.
2,3
3,6
4,10
5,15

you know, like 2 is the bottom number of a triangle and there are 3 points in all, 3 as the bottom number of points and 6 total points, so on.

[hide]You already know one of the three points of the triangle (point P).  You have to find the number n such that n C 2= 120 (you're choosing the other two vertices of the triangle).  So, you can figure out (I used the nCr function on my calculator) that n=16.[/hide]

You can also get an arithmetic sequence when you consider the number of additional trianges that are composed up to the point in question.
it goes P2->1, P3->2, P4->3.... Then all you need is to solve when you know the sum is 120.


With all of us math students, i was wondering what most of us got.

Im shooting for a 760 - 800.

[quote="trigonometry1.01"]NO YOU SHOULD SHOOT FOR 900.[/quote]

way to under-achieve...a 1337 is the ultimate goal

lolz you guys are funny

i just did a practice test(math 2) while listening to rhapsody(very good band btw) and got a 800 in 30 minutes
yea
i dont think aops is the right place to come to if you want to talk about math SATs since we're all overachievers

blahh if you don't get straight 800s on SAT IIs then you are like the majority of the people who take SAT IIs probably BUT HEY TAKING US IS PRETTY COOL

actually
if you don't get straight 800s you won't get accepted into the college that you want

and then you won't get a job

and you'll live on the streets

and die
doom

yeah, if you don't get 800 you'll get sucked in to a black hole and die.  :roll:
i got a 790 in 7th grade, btw, so technically i should be dead too.

Moved to college forum
Warning to those spamming in this thread.
I think the colleges only see the 800. (I'm not totally sure on what happens on tests with subscores, such as the language tests.

[quote="iamagenius"]I have a serious question.

do colleges only see the 800 score? because it is possible to get 800 even if you don't get every single question right. they don't see the score report/analysis do they?[/quote]

Thanks for asking a serious question. I'll rate your post accordingly.  :D  The answer is that on SAT II Subject Tests, the score report only shows the scaled score, not a count of items right and wrong. That's all the colleges want to know, so that is enough.

how sure are you that colleges only see the scaled score?

because if that is true, what a relief!!  :lol:

I have access to reported scores within my college's student information system. I can assure you that for SAT scores, the only thing there is the scaled score, with one exception: the essay score for the writing exam does get reported separately.

I dont think they have the time/care enough to see the scaled score because i think all they care about is you relative to others not your actual perfection.


If you missed the Math Jam, the transcript is now available.

[url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218[/url]

We discussed 21-25 on the 10A and 19-25 on the 12A.


Prove that for each $ a\in\mathbb{R}$ and each $ n\in\mathbb{N}$ there exists a rational $ r_n$ s.t $ |a\minus{}r_n|<1/n$

Thank You as I am slightly confused....

We can do much better:  there exists a rational $ r_n$ such that $ |a \minus{} r_n| \le \frac {1}{2^n}$.  Proof:  truncate the binary expansion of $ a$.


inequalities
function
LaTeX
inequalities proposed

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$\frac {b+c} {\sqrt a} + \frac {c+a} {\sqrt b } + \frac {a+b} {\sqrt c}  \geq \sqrt a + \sqrt b + \sqrt c +3$

I hope to see several solutions which are different from mine, but please comment my solution, as I'm not quite sure if I made a fool of myself or not.

And special thanks to Darij, who tought me this principle of solving inequalities yesterday.

[hide]
Let a function f(a,b,c) be defined as follows:
f(a,b,c) = (b+c)/sqrt a + (c+a)/sqrt b + (a+b)/sqrt c - sqrt a - sqrt b - sqrt c and I want to prove that its always >= 3
As f(ta,tb,tc) is sqrt t (f(a,b,c)) for all t in R and a,b,c are positive reals, f(a,b,c) is homogenous of degree 1/2.
So we may normalize the variables as follows:
a=1 b=1/c c=c
Then:
1/c +c + (c+1)/sqrt (1/c)+(1+1/c)/sqrt c - 1 - sqrt (1/c ) - sqrt c
=1/c+c+c(sqrt c)+1/c(sqrt 1/c) - 1

This is expression is  >= 0, as x+1/x >= 2 for all reals x, so the proof should be complete.[/hide]

[Edited by Pierre to be latex readable.]

Pierre, thank you for editing it to a latex version, could anyone compile my solution, too? I'm sorry, but I'm not able to handle Latex, I never tried and I have never installed it...

You can write the $3$ as $3(abc)^{1/6}$ to homogeneize your ineq, but the thingh I don't understand is that if you write

$f(a,b,c) = \frac{f(at,bt,ct)}{\sqrt{t}}$ and then you put $t=1/a$ you obtain

$\sqrt{a}f(1,b/a,c/a)$ so defining as new variables $B=b/a$ and $C=c/a$ you have not $BC=1$ as you said ???????? ;)

[quote="ikap"]Oki.So you should have observed that normalizing this can't be done because you omit the 3 on the other side of the inequality. When normalizing, you practically multiply all by $\sqrt{\frac 1a}$, right? So you also multiply 3 by $\sqrt{\frac 1a}$. BUT, allthough there is a way of normalizing this by choosing from $a,b,c$ one greter than 1, let's say $c$. Now choose $t$ to be $\sqrt {\frac 1c}$. You've got yourself $c=1,b=b,a=\frac 1b$ and the 3 replaced by something smaller than 3, and it continues as you wrote.[/quote]

I'm almost sure I don't understand you, but did you note that $abc=1$ ? This condition makes it impossible to multiply all variables with $\sqrt{\frac{1}{c}}$ without damaging the hypothesis, so I fear your proof is not really correct, is it?

[b]EDIT:[/b] The same objection was just posted by Manlio.

[quote="DanielKirchner"]Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$\frac {b+c} {\sqrt a} + \frac {c+a} {\sqrt b } + \frac {a+b} {\sqrt c}  \geq \sqrt a + \sqrt b + \sqrt c +3$

I hope to see several solutions which are different from mine, but please comment my solution, as I'm not quite sure if I made a fool of myself or not.

And special thanks to Darij, who tought me this principle of solving inequalities yesterday.

[hide]
Let a function f(a,b,c) be defined as follows:
f(a,b,c) = (b+c)/sqrt a + (c+a)/sqrt b + (a+b)/sqrt c - sqrt a - sqrt b - sqrt c and I want to prove that its always >= 3
As f(ta,tb,tc) is sqrt t (f(a,b,c)) for all t in R and a,b,c are positive reals, f(a,b,c) is homogenous of degree 1/2.
So we may normalize the variables as follows:
a=1 b=1/c c=c
Then:
1/c +c + (c+1)/sqrt (1/c)+(1+1/c)/sqrt c - 1 - sqrt (1/c ) - sqrt c
=1/c+c+c(sqrt c)+1/c(sqrt 1/c) - 1

This is expression is  >= 0, as x+1/x >= 2 for all reals x, so the proof should be complete.[/hide][/quote]

Well, to be honest, when explaining normalization, I simplified it too much... so there are indeed two errors in your above proof.  :(

[hide="First error"]
When talking about homogenization, I referred to inequalities of the form $f(a;\;b;\;c)\geq 0$ or $f(a;\;b;\;c)\leq 0$ for a certain function f. In order to bring your inequality into this form, we have to collect [b]all[/b] terms on the left hand side, i. e. we have to write

$f(a;\;b;\;c) = \frac {b+c} {\sqrt a} + \frac {c+a} {\sqrt b} + \frac {a+b} {\sqrt c} - \sqrt a - \sqrt b - \sqrt c - 3$,

and not

$f(a;\;b;\;c) = \frac {b+c} {\sqrt a} + \frac {c+a} {\sqrt b} + \frac {a+b} {\sqrt c} - \sqrt a - \sqrt b - \sqrt c$.

And the function

$f(a;\;b;\;c) = \frac {b+c} {\sqrt a} + \frac {c+a} {\sqrt b} + \frac {a+b} {\sqrt c} - \sqrt a - \sqrt b - \sqrt c - 3$

is, unfortunately, not homogeneous because of the 3.[/hide]

[hide="Second error"]
Even if the function f would be homogeneous, the condition abc = 1 wouldn't allow us to perform a normalization. In fact, speaking roughly, a normalization means to replace the numbers a, b, c by the numbers a' = ta, b' = tb, c' = tc with a certain t. Now, of course, except of the trivial case t = 1, the numbers a', b', c' don't obey the condition a'b'c' = 1 anymore, so you can't directly relate your inequality to a', b', c'.[/hide]

Anyway, your inequality is problem 3 in the famous book "Old and New Inequalities", one of whose authors is the abovementioned Mircea Lascu (= "Gil" on MathLinks), and the proposed solution goes as follows:

[hide="Proposed solution"]
From the AM-GM inequality, we have $b+c\geq 2\sqrt{bc}$, $c+a\geq 2\sqrt{ca}$ and $a+b\geq 2\sqrt{ab}$, so that

$\frac{b+c}{\sqrt a}+\frac{c+a}{\sqrt b}+\frac{a+b}{\sqrt c} \geq \frac{2\sqrt{bc}}{\sqrt a}+\frac{2\sqrt{ca}}{\sqrt b}+\frac{2\sqrt{ab}}{\sqrt c} = 2(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}})$
$= (\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}) + (\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}) + (\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}})$.

Again from AM-GM, we have $\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}) \geq 2\sqrt{\sqrt{\frac{ca}{b}}\cdot\sqrt{\frac{ab}{c}}) = 2\sqrt a}$ and similarly $\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}} \geq 2\sqrt b$ and $\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}} \geq 2\sqrt c$, so that

$\frac{b+c}{\sqrt a}+\frac{c+a}{\sqrt b}+\frac{a+b}{\sqrt c}$
$= (\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}) + (\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}) + (\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}})$
$\geq 2\sqrt a+2\sqrt b+2\sqrt c = (\sqrt a + \sqrt b + \sqrt c) + (\sqrt a + \sqrt b + \sqrt c)$.

Now, after AM-GM again, $\sqrt a + \sqrt b + \sqrt c \geq 3 \sqrt[3]{\sqrt a\cdot \sqrt b\cdot \sqrt c} = 3 \sqrt[6]{abc} = 3 \sqrt[6] 1 = 3$, so that

$\frac{b+c}{\sqrt a}+\frac{c+a}{\sqrt b}+\frac{a+b}{\sqrt c}$
$\geq (\sqrt a + \sqrt b + \sqrt c) + (\sqrt a + \sqrt b + \sqrt c) \geq (\sqrt a + \sqrt b + \sqrt c) + 3$.

Proof complete.[/hide]

Anyway, don't feel bad - my first proof of this inequality was a horror.

  Darij

Thank you Darij:
Do you remeber what Kiran Kedlaya wrote:
"It seems easier to fool oneself by constructing a false proof of an inequality than of any other type of mathematical assertion."
I think that is right! And my proof surely is a wreck.

I suppose I got it now with the normalization and homogenous functions.

I won't correct my answer above, as you wrote a nice proof here.

Thank you very much Darij for your perfect clarification and your clear remarks.
I think that it is very important to use correctly these powerful techniques and it is very easy to make wrong considerations if anybody uses them in a great hurry.

;) In fact! I used it in a [b]great[/b] hurry

You're right. I got tricked by the "normalize method". In fact as I knew this method is used to make $abc=1$ when you don't have it.

However. Here is a solution:

We will break the $\sum a(\frac 1{\sqrt{b}}+\frac 1{\sqrt{c}})\geq 3+\sum \sqrt {a}$ in $\sum \frac a{\sqrt{b}}\geq \sum \sqrt{a}$ and $\sum \frac a{\sqrt{c}}\geq 3$, cyclic sums.

For the first one use rearrangements, and for the second one the AM-GM inequality.

My solution:
[hide]
Let $x = \sqrt{a}, y = \sqrt{b}, z = \sqrt{c}$. The desired, after homogenization, is equivalent to

$\sum_{Sym} x^3y \ge \sum_{Sym} \frac{1}{2}x^2yz + \frac{1}{2}x^{\frac{4}{3}}y^{\frac{4}{3}}z^{\frac{4}{3}}$

which is a direct consequence of Muirhead's inequality.[/hide]

my solution is nearly the same as mildorf's one, except i didn't change the 3, because (1/2) sym x^3y >= 3 by amgm
anyway, happy new year!

Here is my proof.
After Am-Qm we get
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq{\frac{(\sqrt{b}+\sqrt{c})^{2}}{2\sqrt{a}}+\frac{(\sqrt{c}+\sqrt{a})^{2}}{2\sqrt{b}}+\frac{(\sqrt{a}+\sqrt{b})^{2}}{2\sqrt{c}}}$ Now after Causchy we obtain
$\frac{(\sqrt{b}+\sqrt{c})^{2}}{2\sqrt{a}}+\frac{(\sqrt{c}+\sqrt{a})^{2}}{2\sqrt{b}}+\frac{(\sqrt{a}+\sqrt{b})^{2}}{2\sqrt{c}}\geq{\frac{4(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}} .$So we have to prove that $\frac{4(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}\geq{\sqrt{a}+\sqrt{b}+\sqrt{c}+3} $
Which is sufficient to prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq{3}$ which easily follows from Am-Gm.
[b]Happy New Year guys[/b]!!!!!!!!!! :)  :)  :)  :)  :)  :)


calculus
integration
real analysis
real analysis unsolved

Suppose $ f \geq 0$, show that $ f \in L^p$ if and only if
\[ \sum\limits_{k \equal{}  \minus{} \infty }^{ \plus{} \infty } {2^{pk} \omega \left( {2^k } \right)}  <  \plus{} \infty
\]
where
\[ \omega \left( \alpha  \right) \equal{} \left| {\left\{ {x \in E: f\left( x \right) > \alpha } \right\}} \right|.
\]
Any hints?  :(

I would compare it to the integral:

$ \int f^p \, dx \equal{} \int \int_0^f p \lambda^{p \minus{} 1} \, d\lambda \, dx \equal{} \int_0^\infty p \lambda^{p \minus{} 1} \omega(\lambda) \, d\lambda$ (fubini)

[hide="one direction"]
$ A_k \equal{} f^{\minus{}1} ( [2^k,2^{k\plus{}1}))$

$ g(x) \equal{} \sum_{k \in Z} 2^{k\plus{}1} 1_{A_k}(x) \ge f(x)$

$ \int_0^\infty f^p(x) \le \int_0^\infty g^p(x)$

$ \int_0^\infty f^p(x) \le \int_0^\infty g^p(x)\equal{}\sum_{k \in Z} 2^{pk}[w(2^k)\minus{}w(2^{k\plus{}1})]$
$ \le \sum_{k \in Z} 2^{pk}w(2^k) < \infty$

so $ f \in L^p$[/hide]

Let $ E_k\equal{}\{x\in E: f(x)>2^k\}$

Let $ g(x)\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}2^{pk}\mathbf{1}_{E_k}$

Suppose $ 2^n<x\le 2^{n\plus{}1}.$ Then $ g(x)\equal{}\sum_{k\equal{}\minus{}\infty}^n2^{pk}\equal{}\frac{2^{pn}}{1\minus{}2^{\minus{}p}}.$

Hence there exist constants $ c_p,C_p$ such that $ c_pg(x)\le f(x)^p\le C_pg(x).$

Then there's a valid two-sided comparison between $ \int f^p$ and

$ \int g\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}2^{pk}\int \mathbf{1}_{E_k}\equal{} \sum_{k\equal{}\minus{}\infty}^{\infty}2^{pk}\omega(2^k).$

This works for all $ 0<p<\infty.$


Solve the equation:

[tex]\sin (x-y)+\sin y-\sin x=4\sin\frac{y}{2}[/tex]

Any ideas :?:

$\sin \left( {x - y} \right) + \sin y - \sin x = 4\sin \frac{y}{2}\mathop  \Leftrightarrow \limits_{\mathop {\sin u - \sin v = 2\sin \frac{{u - v}}{2}\cos \frac{{u + v}}{2}}\limits^ \uparrow  } $

$ \Leftrightarrow 2\sin \frac{{ - y}}{2}\cos \frac{{2x - y}}{2} + 2\sin \frac{y}{2}\cos \frac{y}{2} = 4\sin \frac{y}{2} \Leftrightarrow $

$ \Leftrightarrow \left( {\sin \frac{y}{2} = 0} \right) \vee \left( {\cos \frac{y}{2} - \cos \frac{{2x - y}}{2} = 2} \right) \Leftrightarrow $

$ \Leftrightarrow \left( {\left\{ \begin{array}{l}
 \frac{y}{2} = k\pi ,k \in \mathbb{Z} \\
 x = \alpha ,\alpha  \in \mathbb{R} \\
 \end{array} \right.} \right) \vee \left( {\left\{ \begin{array}{l}
 \cos \frac{y}{2} = 1 \\
 \cos \frac{{2x - y}}{2} =  - 1 \\
 \end{array} \right.} \right) \Leftrightarrow $

$ \Leftrightarrow \left( {\left\{ \begin{array}{l}
 x = \alpha ,\alpha  \in \mathbb{R} \\
 y = 2k\pi ,k \in \mathbb{Z} \\
 \end{array} \right.} \right) \vee \left( {\left\{ \begin{array}{l}
 \frac{y}{2} = 2k\pi ,k \in \mathbb{Z} \\
 \cos \frac{{2x - y}}{2} =  - 1 \\
 \end{array} \right.} \right) \Leftrightarrow $

$ \Leftrightarrow \left( {\left\{ \begin{array}{l}
 x = \alpha ,\alpha  \in \mathbb{R} \\
 y = 2k\pi ,k \in \mathbb{Z} \\
 \end{array} \right.} \right) \vee \left( {\left\{ \begin{array}{l}
 y = 4k\pi ,k \in \mathbb{Z} \\
 \cos x =  - 1 \\
 \end{array} \right.} \right) \Leftrightarrow $

$ \Leftrightarrow \left( {\left\{ \begin{array}{l}
 x = \alpha ,\alpha  \in \mathbb{R} \\
 y = 2k\pi ,k \in \mathbb{Z} \\
 \end{array} \right.} \right) \vee \left( {\left\{ \begin{array}{l}
 x = \pi  + 2p\pi ,p \in \mathbb{Z} \\
 y = 4k\pi ,k \in \mathbb{Z} \\
 \end{array} \right.} \right) \Leftrightarrow $

$ \Leftrightarrow \left( {x,y} \right) \in \left\{ {\left. {\left( {\alpha ,2k\pi } \right)} \right|\alpha  \in \mathbb{R},k \in \mathbb{Z}} \right\}$

 :?:


[quote="amirhtlusa"]move the bishob behind the knight.[/quote]

Can't the king then eat the rook?

What bishop?  :huh: You mean the pawn? I don't see any mate in 1, but I several future mates...

[quote="Valentin Vornicu"][hide="Solution"]Move the horse 2 in the right and 1 cell down. It's mate then :P[/hide]But I don't have another problem to post  :blush:[/quote]

Yes, that works [edit] or maybe it doesn't [/edit].  I was stuck on why the pawn couldn't advance to the end row until I realized that there was nothing to attack there...  :blush:

Anyway, a new puzzle (a1 is the bottom right-hand corner)

sorry guys i posetd the wrong picture (i have so many) here is the correct one:

find mate in one

[quote="bubala"][quote="Valentin Vornicu"][hide="Solution"]Move the horse 2 in the right and 1 cell down. It's mate then :P[/hide]But I don't have another problem to post  :blush:[/quote]

Wait; I don't think that works.  Then the black king can capture the rook.[/quote]Yup! :( You're right ...

YES!!! A CHESS MARATHON!! Dream Come true!!! Ok guys you'll see me post here a lot!

[quote="G-UNIT"]Well marathons seem to be the flavour of the month, so why not create a chess marathon?

  Heres the Rules:

  [b]  A problem will be posted where white must checkmate black in 1 move.  If you would like to make some variation of this, i.e ( checkmate in 2 moves, checkmate in 3 moves, etc)  please note that in your problem, otherwise checkmate in 1 move will be assumed.  Give your answer in chess notation (preferabally)  and then post the next problem.

  Ill start with a simple one.

  [i]Problem 1[/i]

Find mate in one.

[/b][/quote]

Well..I think mate in one is possible if its blacks turn to move.

Solution to green and yellow problem
[hide]fg8N++; same idea for first problem[/hide]
Here is a very very easy one, I couldn't find a better one quickly. (attached)

Jack

[quote="jmadsen"]g-unit's second problem is easy.
[hide]Nxf7[/hide]

I'm pretty sure danbert's problem doesn't have a solution. White has only four pieces, and we obviously cannot get mate with the king. The pawns can only put the king in check by moving c7xd8, but this is not mate. Also, the only way for the rook to get check is Rxd6, but this is also not mate.[/quote]

Wrong! I'm quite sure that there's a solution!  (Of course, I also thought that Valentin was right for a while on the first puzzle...  ;) )

Is the white's turn to move at the first one?  :?

Yup, there's only one possibility for the last five.

Potential solution to above?

[hide]
...f7 - f5 forced (blocking check from white bishop)

e5xf6+ (en passant)

...Ke6 - d6+ forced

c2 - c4, blocking check

...b4xc3+ (en passant)
[/hide]
And what position are you referring to regarding legality?

[quote="bakwardz"]Potential solution to above?
...f7 - f5 forced (blocking check from white bishop)[/quote]
But how did the white bishop come to h3? Answer to this question and you've solved the problem.

I know the solution but I won't post it, let somebody else solve this.

3. B*f2+ ke2 Nd4 mate OMG i did this in a game on chess live!
4. Qf7+ Kd6 Ne4++
6. I did this exact mate on OTB! number 3 is a famous trap but where did they get this one! f*g7+ kh8 r*f8 mate
number 5 i figured out and wrote down but i lost it so i will have to go back and redo it after this post

for number 7 a8=Q+ Ke2 Qe7 mate i wont post answers to any more until a new one is made too confusing

[quote="jhollenbeck"]:blush: i forgot that you can edit your posts im sorry guys.[/quote]

You can edit or delete them now.

did someone say chess? I'm a nerd at that.  Anyway, the answer to the first problem is wrong because the king can take the rook.

a good place to play chess free is US chesslive :)


absolute value
inequalities unsolved
inequalities

For $ x,y,z \in R$, prove that: $ |x| \plus{} |y| \plus{} |z| \plus{} |x\plus{}y\plus{}z| >\equal{} |x\plus{}y| \plus{} |y\plus{}z| \plus{} |z\plus{}x|.$

[quote="iTAQ"]For $ x,y,z \in R$, prove that: $ |x| \plus{} |y| \plus{} |z| \plus{} |x \plus{} y \plus{} z| \geq |x \plus{} y| \plus{} |y \plus{} z| \plus{} |z \plus{} x|.$[/quote]
Squaring of both sides kills it. :wink:

suppose x has the biggest absolute value of the three

then $ |x| \plus{} |x \plus{} y \plus{} z| \geq |2x \plus{} y \plus{} z| \equal{} |x \plus{} y| \plus{} |z \plus{} x|.$

the left is really simple.
[quote="iTAQ"]For $ x,y,z \in R$, prove that: $ |x| \plus{} |y| \plus{} |z| \plus{} |x \plus{} y \plus{} z| > \equal{} |x \plus{} y| \plus{} |y \plus{} z| \plus{} |z \plus{} x|.$[/quote]