geometry
3D geometry
algebra
difference of squares
special factorizations

Prove that the sum of $ n$ consecutive perfect cubes is never prime, for any $ n$.

[quote="#H34N1"][hide="Hint"]
$ n^3 \plus{} (n \plus{} 1)^3 \plus{} (n \plus{} 2)^3 \plus{} \dots \plus{} (n \plus{} k)^3 \equal{} (n \plus{} n \plus{} 1 \plus{} n \plus{} 2 \plus{} \dots \plus{} n \plus{} k)^2$[/hide][/quote]

What if $ n\equal{}2, k\equal{}1$?  :wink:

[quote="#H34N1"][hide="Hint"]
$ n^3 \plus{} (n \plus{} 1)^3 \plus{} (n \plus{} 2)^3 \plus{} \dots \plus{} (n \plus{} k)^3 \equal{} (n \plus{} n \plus{} 1 \plus{} n \plus{} 2 \plus{} \dots \plus{} n \plus{} k)^2$[/hide][/quote]

[quote="#H34N1"]oops $ n < k$[/quote]

How about $ n \equal{} 2$, $ k \equal{} 3$?

$ 2^3 \plus{} 3^3 \plus{} 4^3 \plus{} 5^3 \equal{} 224\neq n^2$ for any integer $ n$. And the result is true for any $ k$.

Okay forget about that I saw my flaw

[hide="Hint"]
$ 1^3\plus{}2^3\plus{}\dots\plus{}k^3\equal{}(1\plus{}2\plus{}\dots\plus{}k)^2$[/hide]

The sum of $ n$ cubes starting with $ k^3$ is the sum of $ n$ cubes starting with $ 1$ minus the sum of $ k\minus{}1$ cubes starting with $ 1$. Both of these are squares. So we have a difference of squares, which factors.

[quote="leoxnlin"]The sum of $ n$ cubes starting with $ k^3$ is the sum of $ n$ cubes starting with $ 1$ minus the sum of $ k \minus{} 1$ cubes starting with $ 1$. Both of these are squares. So we have a difference of squares, which factors.[/quote]

Correct!  :)

[hide="Alternately"] Gaussian pairing on $ n^3 \plus{} ... \plus{} m^3$ gives a factor of $ (n \plus{} m)$ in every pair.  If the number of terms is even, then $ n \plus{} m$ is a factor of the above sum.  If the number of terms is odd, then the middle term is $ \left( \frac{n \plus{} m}{2} \right)^3$ and every other term pairs, so $ \frac{n \plus{} m}{2}$ is a factor of the above sum. [/hide]


I am studying in the University of Macau and majoring in Education in Mathematics. I know that it is not famous (have you heard about it?), and I will only get a BEd in math, not BSc. Also, differential geometric, topology or other advanced things would not be taught. However, I want to go to a good graduate school after graduated in college. I am now a freshman in university, what should I do?

ps.: UM has a math dept but it only offers MSc and PhD in math. But the BEd course of math is also offered by math dept. We use English book, write in English but taught in Chinese. The thing taught to us is so simple, but so we have more spare time. (However, we need to spend time to take the education courses). Athough UM is not a famous in the world, it is the top college in Macao, and it is the only college in Macao offers the math program.

My undergrad university didn't offer the exact branch of linguistics I wanted to study in grad school, so I attended summer linguistics programs at other universities where I could take the classes I would need for graduate school.  It was useful not only for the subject material, but also for finding people in that area who could potentially write letters of recommendation for grad school for me.  Perhaps you could do the same for math?

You say that your university offers graduate degrees.  Could you also make up the deficiencies by taking graduate level courses?


You are presented with a ladder. At each stage, you may choose to advance either one rung or two rungs. How many different paths are there to climb to any particular rung; i.e. how many unique ways can you climb to rung "n"?

After you've solved that, generalize. At each stage, you can advance any number of rungs from 1 to K. How many ways are there to climb to rung "n"?

[hide="Part 1"] Let $ F_n$ denote the number of paths to rung $ n$.  Clearly $ F_1 \equal{} 1, F_2 \equal{} 2$.  Any particular path to rung $ n$ ends in either a one-rung advance (in which case it is a one-rung advance added onto a path to rung $ n \minus{} 1$) or a two-rung advance (in which case it is a two-rung advance added onto a path to rung $ n \minus{} 2$).  We therefore find that

$ F_n \equal{} F_{n \minus{} 1} \plus{} F_{n \minus{} 2}$

and so this sequence is the (index-shifted) Fibonacci sequence.  :) [/hide]

Here is a different approach...

[hide]The answer is Summation((n-m)!/(m!*(n-2m)!)) Where m = 0 to (n-2m) < or = 0.
And if we consider the steps could be 1 or 'k' steps it will be

Summation ( (n-(k-1)m)!/(m!*(n-km)!)) Where m = 0 to (n - km) < or = 0.[/hide]

The correct answer is
Summation((n-m)!/(m!*(n-2m)!))
Where m = 0, 1, 2,.. so on     until (n-2m) = 0 (for even 'n') and 1 (for odd 'n')


LaTeX
combinatorics unsolved
combinatorics

Suppose $X= \{1,...,n \}$
$(a_1,b_1),...,(a_k,b_k)$ are in $X^2$
$(c_1,d_1),...,(c_k,d_m)$ are in $X^2$
$(e_1,f_1),...,(e_k,f_n)$ are in $X^2$
Suppose $A,B,C$ are subsets of $X^3$ With this conditions:
$A=\{(a,b,c)\ |\ (a,b)=(a_i,b_i)\quad \forall\ 1\leq i \leq k\}$
$B=\{(a,b,c)\ |\ (a,c)=(c_i,d_i)\quad \forall\ 1\leq i \leq m\}$
$C=\{(a,b,c)\ |\ (b,c)=(e_i,f_i)\quad \forall\ 1\leq i \leq n\}$
Prove that if $D=A \cap B \cap C$ then $|D|^2 \leq mnk$.

Indeed, this seems very hard...to understand.

Pierre.

[quote="pbornsztein"]Indeed, this seems very hard...to understand.[/quote]
Your remark was too late ;)
Anyway, entire problem is not clear for me.

Well I give you some more geometric view.
A lattice line in $\mathbb{Z}^3$ is a set of points, for example $(x,y)=(a,b)$ that $a,b \in \mathbb{Z}$ (I mean all of points $\{ ...,(a,b,-1),(a,b,0),(a,b,1),... \}$)
We have three types of lattice lines.
Those that are parallel to line $x=y=0$
Those that are parallel to line $x=z=0$
Those that are parallel to line $y=z=0$
Now we have $m$ lines with the first type.$A$ is set of points of these lines.
Now we have $n$ lines with the first type.$B$ is set of points of these lines.
Now we have $k$ lines with the first type.$C$ is set of points of these lines.
Prove $|A \cap B \cap C|^2 \leq mnk$
I hope now the problem is clear.

It is IMO 1992-5, quite a good problem.
see http://www.kalva.demon.co.uk/imo/imo92.html


calculus
integration
geometry
rectangle
limit
complex analysis

using residue theorem :
$ \int_{\gamma}{\frac{z^2\plus{}5}{({z\minus{}1})^3{(z^3\plus{}8)}}dz}$
where $ \gamma$ is the rectangle $ 2\plus{}2i,\minus{}1\plus{}2i,\minus{}1\minus{}2i,2\minus{}2i$

That's neat!  This time the poles are on the contour but not the real axis . . . different from the text book but still ok.  See diagram below.  We still let the diameter, $ \rho$ of the indentations around the poles on the contour go to zero but this time I'm letting those poles remain INSIDE the contour so need to add them up when using the Residue Theorem.  Note also the orientation is positive so no negative sign.  Could have made the indentations the other way and excluded them though.  So using $ \text{Blue}$ below, I get:

\begin{align*} \mathop\lim\limits_{\rho\to 0}\mathop\oint\limits_{\text{Blue}}\frac {z^2 + 5}{(z - 1)^3(z^3 - 8)}dz & = \text{P.V.}\left\{\mathop\oint\limits_{\gamma} \frac {z^2 + 5}{(z - 1)^3(z^3 - 8)}dz\right\} \\
& + \pi i (r_1 + r_2 + r_3) \\
& = 2\pi i(r_0 + r_1 + r_2 + r_3) \end{align*}

With the Principal Value ending up being $ \approx - 2.48 i$

(Note:  I did actually calculate all the residues and calculate it algebraically and verify that result against the numerical result below)

Here's the Mathematica code to check it numerically:

[code]
f[z_] := (z^2 + 5)/((z - 1)^3*(z^3 - 8));
c1 = NIntegrate[f[z]*I /. z -> 2 + I*y,
       {y, -2, 0, 2}, Method ->
         "PrincipalValue"];
c2 = NIntegrate[f[z] /. z -> x + 2*I,
       {x, 2, -1}];
c3 = NIntegrate[f[z]*I /. z -> -1 + I*y,
       {y, 2, Sqrt[3], -Sqrt[3], -2},
       Method -> "PrincipalValue"];
c4 = NIntegrate[f[z] /. z -> x - 2*I,
       {x, -1, 2}];
c1 + c2 + c3 + c4

0. - 2.482132971203611*I
[/code]

I think you are solving something different

The integral over that contour does not converge absolutely. If you want it to be meaningful, you ask for the principal value, which looks like the integral over the path shown with the semicircles removed. Each of those semicircles contributes half the residue, so we can use the residue theorem on a path with them included, then subtract them off.


i don't live in NJ but i got 3rd in my chapter in IL. fail :(
But I got 2nd cd...happy though? :maybe:

I haven't had chapter yet :)

Although if you post the problems and answerse here, I'll probably win :)


[quote="pianoforte"]sorce: KCATM 2005 7-8

A math teacher at the local middle school likes to vary her homework assignments. She never assigns all of the problems in a set, and always picks between just 2 options. Some days she assigns just the odd problems while on other days she assigns every third problem. She always starts the assignment with problem 1.

One day there is a substitute and the teacher forgot to leave the problem assignment on the board, however the students know what page to work on. If there are 65 problems on the page, what is the total number of problems that the students must complete to be certain that they complete the assignment that the teacher forgot to write down?[/quote]
[hide]
The first option has problems 1, 3, 5, ..., 63, and 65, which has 33 numbers. The second option has problems 1, 4, 7, ..., 61, 64, which has 22 numbers, so there are 55 problems to work on, but you take away numbers $x\equiv 1 \bmod{6}$, which has 11 numbers, and $55-11=\boxed{44}$.[/hide]


geometry
3D geometry
quadratics
trigonometry
algebra
quadratic formula

I'm taking an analysis class and one of the harder problems on the homework is to find the cube roots of -1.  These include the complex roots.  I have used DeMoivre's theorem to find that one root is (obviously) -1 + 0i.  What method should I use to find the other roots of -1?  Please focus more on the concept rather than the actual answer.

Well you are looking for $ (\minus{}1\plus{}0i)^{1/3}$.

1. Convert this number to it's trigonometric form, which can also give it's polar form.

2. [i]Then[/i] apply DeMoivre's.

You dont need demoirvre's.

just x^3=-1.

so x^3+1=0.  then factor and solve

Well, I think DeMoivre's faster, but.

[hide="Quadratic"]
$ x^3 \plus{} 1 \equal{} 0$

$ (x \plus{} 1)(x^2 \minus{} x \plus{} 1) \equal{} 0$

Which yields,

$ \boxed{x \equal{} \minus{} 1}$ and

$ x^2 \minus{} x \plus{} 1 \equal{} 0$.

Using the Quadratic Formula, you get

$ x \equal{} \frac {1 \pm \sqrt {1 \minus{} 4(1)(1)}}{2}$

$ x \equal{} \frac {1 \pm \sqrt { \minus{} 3}}{2}$

$ \boxed{x \equal{} \frac {1 \pm i \sqrt {3}}{2}}$.
[/hide]

[hide="DeMoivres"]
Converting into polar form gives us $ (1 , 180^{\circ})$ and applying DeMoivre's to the trig form $ 1(\cos 180^{\circ} \plus{} \sin 180^{\circ})$, we get

$ (1^{1/3} , 180^{\circ} \plus{} (360^{\circ}k)/3)$

And using $ k \equal{} 0$, $ k \equal{} 1$, and $ k \equal{} 2$, we get

$ (1, 180^{\circ})$, $ (1, 300^{\circ})$, $ (1, 60^{\circ})$.
[/hide]


How many positive three-digit prime numbers less than 200 have two digits the same?

The number is either 11x, 1x1, or 1xx.

11x: 113
1x1: 101, 131, 151, 181, 191
1xx: 199

There are 7 primes total.


limit
function
Euler
calculus
calculus computations

Prove that $ f(x) \equal{} \sum_{1}^{\infty} \frac {x}{n(x \plus{} n)}$ is continuous on $ [0, \infty)$.

For $ x \ge 0$, $ |f(x)| \le \sum \frac{x}{n^2}\equal{}x\sum \frac{1}{n^2} < \infty$, so $ f$ is well-defined.
$ |f(x)\minus{}f(y)|\equal{}|\sum \frac{x\minus{}y}{(n\plus{}x)(n\plus{}y)}| \le |x\minus{}y| \sum \frac{1}{n^2}\equal{} |x\minus{}y|\frac{\pi^2}{6}$
Take $ x\equal{}y\plus{}h$ and let $ h \to 0$, and it follows that $ f$ is continuous in $ y$. $ y \ge 0$ is arbitrary, so $ f$ is continuous on $ [0, \infty)$.

bambaman has directly shown that $ f(x)$ is Lipschitz, and it won't take much more than that to show that $ f$ has a bounded derivative.

But what about asymptotic behavior? Is $ \lim_{x\to\infty}f(x)$ finite or infinite? And if infinite, how rapidly does $ f(x)$ grow?

We have $ f(x) \equal{} \sum_{n \equal{} 1}^\infty \frac {1}{n} \minus{} \frac {1}{x \plus{} n}$, so when $ x$ is an integer we have $ f(x) \equal{} H_x \equal{} \sum_{i \equal{} 1}^x \frac {1}{i}$.  So, this is some sort of continuous function interpolating the harmonic numbers and so growing about as fast as the logarithm.  (Mathematica tells me that this function is equal to $ \gamma \plus{} \psi(0, 1 \plus{} x)$, where $ \gamma$ is the Euler gamma and $ \psi$ is the digamma function (mathematica calls it "PolyGamma"), whatever that means :))


Solve the equation: \[ \frac{a^x\minus{}a\minus{}b}{b^{x\minus{}1}}\minus{}\frac{b^x\minus{}a\minus{}b}{a^{x\minus{}1}}\equal{}a\minus{}b,\] where $ a,b \in (0,1)$, and $ a \neq b$.

[quote="linker"]Solve the equation:
\[ \frac {a^x \minus{} a \minus{} b}{b^{x \minus{} 1}} \minus{} \frac {b^x \minus{} a \minus{} b}{a^{x \minus{} 1}} \equal{} a \minus{} b,
\]
where $ a,b \in (0,1)$, and $ a \neq b$.[/quote]

$ a^{x}\equal{}n;b^{x}\equal{}m$----> $ \frac{b(n\minus{}a\minus{}b)}{m}\minus{}\frac{a(m\minus{}a\minus{}b)}{n}\equal{}a\minus{}b$----->

$ bn^{2}\minus{}(a\plus{}b)bn\minus{}am^{2}\plus{}am(a\plus{}b)\equal{}(a\minus{}b)mn$------> $ bn^{2}\minus{}(ab\plus{}b^{2}\plus{}(a\minus{}b)m)n\minus{}am^{2}\plus{}a^{2}m\plus{}abm\equal{}0$----->

$ D\equal{}((a\plus{}b)b\plus{}(a\minus{}b)m)\minus{}4b(\minus{}am^{2}\plus{}a^{2}m\plus{}abm)\equal{}((a\plus{}b)b)^{2}\plus{}2bm(a^{2}\minus{}b^{2})\plus{}(a\minus{}b)^{2}m^{2}\plus{}4abm^{2}\minus{}4bm(a\plus{}b)\equal{}((a\plus{}b)b)^{2}\plus{}(a\plus{}b)^{2}m^{2}\minus{}2bm(a\plus{}b)^{2}\equal{}(a\plus{}b)^{2}(b\minus{}m)^{2}$----->

$ n_{1}\equal{}\frac{ab\plus{}b^{2}\plus{}am\minus{}bm\plus{}(a\plus{}b)(b\minus{}m)}{2b}\equal{}a\plus{}b\minus{}m$---->$ a^{x}\plus{}b^{x}\equal{}n\plus{}m\equal{}a\plus{}b$--->$ x\equal{}1$

$ n_{2}\equal{}\frac{ab\plus{}b^{2}\plus{}am\minus{}bm\minus{}(a\plus{}b)(b\minus{}m)}{2b}\equal{}\frac{am}{b}$------>$ \frac{a^{x}}{b^{x}}\equal{}\frac{n}{m}\equal{}\frac{a}{b}$----->$ x\equal{}1$


Triangle $ABC$ has a right angle at $A$. Among all points $P$ on the perimeter of the triangle, find the position of $P$ such that $AP+BP+CP$ is minimized.

[hide="Hint"] If $P$ is on a given side - say, $BC$ - then $AP+BP+CP = BC+AP$.  $BC$ is constant as $P$ slides around on a given side, so the only thing that changes is $AP$.  When is this a minimum?  On which side should $P$ be placed? [/hide]


Show that if $ p\equiv 1\mod 6$ then $ p \equal{} a^2 \minus{} ab \plus{} b^2$ with $ a,b\in \mathbb{Z}$.

[hide="if p is a rational prime..."]
Work in the ring $ \mathbb{Z}[\omega]: \equal{} \{ a\plus{}b\omega \mid \omega^3\equal{}1, a,b \in \mathbb{Z} \}$.
$ p\equiv 1\pmod{3}$ then $ \left(\frac{\minus{}3}{p}\right)\equal{}1$ so there is an integer $ n$ such that
$ p|n^2\plus{}3\equal{}(n\plus{}1\plus{}2\omega)(n\minus{}1\minus{}2\omega)$ hence $ p$ is not a prime in $ \mathbb{Z}[\,\omega\,]$
thus there exist a prime $ a\plus{}b\omega|p$ with $ a,b\in \mathbb{Z}$ and $ p\equal{}N(a\plus{}b\omega)\equal{}a^2\minus{}ab\plus{}b^2$.[/hide]


Let $l_a,l_b,l_c$ the three bisectors of a triangle $ABC$ and $O$ a point internal to $ABC$. Prove that

$(l_al_bl_c)^{\frac{1}{3}}<\frac{2}{3}(OA+OB+OC)$

I think this can be done orally.
Using well known facts:
1) La ≤ √p(p-a)
2) OA+OB+OC > p
3) (p-a)(p-b)(p-c)^1/3 ≤ p/3

(la*lb*lc)^[2/3] ≤ p*((p-a)(p-b)(p-c)^1/3) ≤ p2/3 < 4p2/9.
Check it.


trigonometry
ratio
function
geometric series
real analysis
real analysis solved

I need to find the sum of this series:

$\frac{1}{2}+\cos{x}+\cos{2x}+\cdots +\cos{nx}$

Using Euler's Identity, I rewrote it as:

$\frac{1}{2}\left(e^{-ixn}+e^{-ix\left(n-1\right)}+\cdots +e^{-ix}+1+e^{ix}+\cdots +e^{ix\left(n-1\right)}+e^{ixn}\right)$

This now looks like a geometric series, which I can write as:

$\frac{1}{2}\left(\sum_{k=0}^{n}e^{-ixk}+\sum_{k=0}^{n}e^{ix\left(k+1\right)}\right)$

My question is, before I do anything more, is this the right approach?  Also, did I make any mistakes above?  Thanks, I'd appreciate any help.

Yes, this is the right approach. I see one mistake: in your second sum, since you've got $(k+1)$ in the exponent, your upper limit should be $k=n-1,$ not $k=n.$

I'd be more inclined to write what you have as a single geometric series:

$\frac12\sum_{k=-n}^ne^{ikx}$

Making it one series like that will make your life simpler, and you can still use the formula for a finite geometric sum:

$\text{sum}=\frac{(\text{first term})-(\text{next term after the last})}{1-(\text{ratio})}$

A few things to note up front: you will get a real, continuous, even function of $x$, and the value at $x=0$ will be $n+\frac12.$

You're trying to compute a well-known object called the "Dirichlet kernel."


Hello,

I thought I understood the thing with URL and the brackets, but apparently I don't

It often happens than when the URL together with the brackets takes more than one line, something weird happens when I do preview.

I am trying to demonstrate but now it won't happen :mad: .

Can anyone enlighten me? :o
Thanks!

I don't know the problem, but here is a solution:
http://tinyurl.com/
;)
By the way, it often happens that the url won't work when there is some special character in it. I have had problems with wikipedia links occasionally because they have an apostrophe or something.

[url]http://verkiezingen2003.belgium.be/electionshome/nl/result/chamber/table_e3026.html[/url]
This was my problem. :(

EDIT : no fair, when i was writing my PM it didn't work :mad:

I usually use the alternate text code:
(url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633378)this thread(/url)
becomes
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633378]this thread[/url]
when I replace the parentheses with []. I've never had a problem with long urls using that.

As for the special characters- anything with dollar signs is really bad.


[quote="jdai_ming"]any free book for Mathematica book on the net ???
     Thank.[/quote]
The best Mathemetica book appears when you press F1. :D


You mean how do you enter the classroom? There's a button that says "classroom" on the left side of your screen. If you press it during class time, the classroom window should automatically appear.


[quote="Redriver"]Give $x,y,z >0$ such as : $xy+yz+zx+2xyz=1$
Prove that :
$\sqrt{x}+\sqrt{y}+\sqrt{z}\ge 2\sqrt{xyz+1}$[/quote]

From $xy+yz+zx+2xyz=1$, we can put $x=\frac{a}{b+c},y=\frac{b}{c+a},z=\frac{c}{a+b}$. Then the inequality is equivalent to
$\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq 2\sqrt{(a+b+c)(ab+bc+ca)}$,
and this is equivalent to
$\sum_{cyc}a(a+b)(a+c)+\sum_{cyc}2\sqrt{ab(a+c)(b+c)}\geq 4\sum_{sym}a^{2}b+12abc$, which can be given by squaring.

Now it is necessary to prove that
$a^{3}+b^{3}+c^{3}+2\sum_{cyc}\sqrt{ab(a+c)(b+c)}(a+b) \geq 3\sum_{sym}a^{2}b+9abc$. Here we use the following lemma:

$(a+c)(b+c)\geq(\sqrt{ab}+c)^{2}$

which is reached by a simple application of AM GM inequality. (note that the equality condition is $a=b$ or $c=0$.)

Then it is necessary to prove that
$a^{3}+b^{3}+c^{3}+2\sum_{cyc}2\sqrt{ab}c(a+b) \geq \sum_{sym}a^{2}b+9abc$.
Put $a=p^{2},b=q^{2},c=r^{2}$, then it is equivalent to
$\sum_{cyc}p^{6}+2\sum_{sym}p^{3}q^{2}r \geq \sum_{sym}p^{4}q^{2}+9p^{2}q^{2}r^{2}$,
and this is done by
$(p^{3}+q^{3}+r^{3}+3pqr-\sum_{sym}p^{2}q)(p^{3}+q^{3}+r^{3}-pqr+\sum_{sym}p^{2}q)\geq0$.
This completes the proof, leaving equality condition as $a=b=c$ or $a=b,c=0$, which means that $x=y=1,z=0$ or $x=y=z=\frac{1}{2}$.

by the way, is this your own inequality? i like this one as there are more than one equality conditions..

[quote="Redriver"]Give $x,y,z >0$ such as : $xy+yz+zx+2xyz=1$
Prove that :
$\sqrt{x}+\sqrt{y}+\sqrt{z}\ge 2\sqrt{xyz+1}$[/quote]
More way.
Since, $1=xy+xz+yz+2xyz\geq 3\sqrt[3]{x^{2}y^{2}z^{2}}+2xyz$ then $xyz\leq\frac{1}{8}.$
But $(\sqrt x+\sqrt y+\sqrt z)^{4}\geq16(xy+xz+yz)+11(\sqrt x+\sqrt y+\sqrt z)\sqrt{xyz}\geq$
$\geq16(xy+xz+yz)+33\sqrt[3]{x^{2}y^{2}z^{2}}=16+33\sqrt[3]{x^{2}y^{2}z^{2}}-32xyz.$
Id est, remain to prove that $16+33\sqrt[3]{x^{2}y^{2}z^{2}}-32xyz\geq16(xyz+1)^{2},$ which obviously. :)


geometry
geometric transformation
reflection
AMC

In triangle ABC, E is the midpoint of BC and D is a point on side AC. If AC = 1, angle BAC = 60 degrees, angle ACB = 20 degrees, and angle DEC = 80 degrees, then evaluate numerically the area of triangle ABC plus twice the area of triangle CDE. Do not include trigonometric expressions.

[hide](Sqrt(3))/8... draw 30-60-90 right triangle with hypotenuse AC and a right angle at vertex M. Reflect triangle BMC over MC and note that triangle BB'C is similar to triangle DEC, in fact since 2(BE)=BC and the triangles are isoceles, the area of BB'C is exactly four times that of DEC. Triangle BMC is half the area of triangle BB'C; therefore the area of right triangle AMC is equal to the area of triangle ABC+ 2(triangle DEC).

[/hide]

I've actually done this problem before, and when searching through old threads found this lol

Oldest thread in HSM, and it should have been in MSM  :oops_sign:

[quote=RedFireTruck][quote=Vince729]@SmartZX this thread is older than i am lol[/quote]

same lol[/quote]

same


Signups are open.

You cannot heal yourself.
You cannot hurt yourself.
You cannot double-post in this thread.
Once you hurt and heal, you must recharge (wait) for [b] 30 minutes [/b] (at least) before hurting & healing again.
You can hurt two people at once, two different, or two on the same person, and may heal one person at a time.

 :lol:

Each player will start with 25.

Check versions hurt and heal 1 and 2 for details about this game.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139053
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=136272


linear algebra
matrix
Ring Theory
superior algebra
superior algebra theorems

Let R be a ring, M - maximal ideal (all ideal, that contains M, is whether R or M itself). Why quotient ring R/M is a field?

Thank you very much.

If $ u$ is a noninvertible element of $ R/M$, then $ M \cup (u)$ is an ideal of $ R$ that properly contains $ M$.

Both of you seem to have tacitly assumed commutativity; in the absence of commutativity, you would have to specify whether you were talking about right ideals or left ideals, and t0rijir0u's argument would show $ R/M$ to be a division ring.

My apologies; as I understand it, "ideal" means "two-sided ideal."

Without commutativity, you need a two-sided ideal for the quotient operation to make sense: $ (a\plus{}M)\cdot(b\plus{}M)\equal{} ab\plus{}aM\plus{}Mb\plus{}M^2$, and we need both $ aM$ and $ Mb$ to be contained in $ M$ for quotient multiplication to work.

In that context, "maximal" also needs clarification. Even if we merely ask for the quotient to be a division ring, being maximal among two-sided ideals isn't enough. For example, take the matrix ring $ M_n(F)$ (with $ n>1$). The zero ideal is maximal among two-sided ideals, but there are nonzero elements that aren't invertible. In fact, even something like an element $ a\equal{}\begin{bmatrix}1&0&\cdots\\0&0&\cdots\\ \vdots&\vdots&\ddots\end{bmatrix}$ generates all of $ R\equal{}M_n(F)$: $ RaR$ spans $ R$.

The hypothesis we need here is that the two-sided ideal $ M$ is maximal either among left ideals or among right ideals (if one, then the other). The proof is almost the same. If $ u$ has no left inverse in $ R/M$, $ Ru\plus{}M$ is a left ideal properly between $ M$ and $ R$, and if $ u$ has no right inverse, $ uR\plus{}M$ is a right ideal properly between $ R$ and $ M$.
One detail: Suppose every nonzero element in a ring $ A$ has a left inverse. Choose a nonzero element $ a$. Let $ x$ be a left inverse of $ a$. Let $ b$ be a left inverse of $ x$. Then $ a\equal{}(bx)a\equal{}b(xa)\equal{}b$, and $ a$ and $ x$ are two-sided inverses of each other. The two-sided inverse must be unique (if $ x$ and $ y$ are inverses of $ a$, $ x\minus{}y\equal{}(x\minus{}y)ax\equal{}0\cdot x\equal{}0$), and $ A$ is a division ring.
Reversing the order of symbols, the same is true for right inverses.

This condition is very strong, and rarely occurs in the noncommutative case.


Does anyone have any good pdfs for complex numbers on the AIME level and beyond?

Thanks,
bpms

DOes anyone have good pdfs on trig either? I'm also looking for Geo Problem sets....

there is two article about complexe number in mathematical excalibur :lol:

There is a sheet of inversion questions [url=http://mathcircle.berkeley.edu/BMC6/pdf0607/InversionI06.pdf]here[/url]; there is also a decent chapter on inversion in [url=http://www-math.mit.edu/~kedlaya/geometryunbound/gu-060118.pdf]Geometry Unbound[/url]


[quote="sneezywalrus"]
if a domain is given do you still find the absolute max and min within the domain or does the domain not apply when finding the absolute max and mins...

lost....[/quote]
[quote="paladin8"]To REAL moderators: why does it show that sneezywalrus is a moderator and has an absurd number of posts yet only registered yesterday?[/quote]
I deleted the entire original topic, to get rid of the ridiculously large picture file, and the entirely inappropriate poll.

I will let him ask the question, vague as it is - but it's so vague I won't bother to answer it myself.

To sneezywalrus: Don't post pictures that have nothing to do with your mathematical questions. And don't make polls. If you do either in any forum that I moderate, I'll delete your post.

To paladin8: a moderator wouldn't know the answer to that. Valentin Vornicu might.

[quote]has an absurd number of posts[/quote]
That absurd number is really -2 (right now). There's a bug right now in the post count calculation which decreases the post count twice for every deleted post. There's a thread on this in the support forum. False moderator status seems to be a side effect.

[quote="jmerry"]The picture is in your signature, and it shouldn't be. There are length limits on signatures for a reason.[/quote]

o.....sorry once again...

better?...better.


Hello,

This question may look silly, but I am reading a list of theorems that I did several years ago about the main theorems in Geometry needed in Olympiad of Mathematics. This [i]genius[/i] (yes, I am trying to be sarcastic) decided just to use the names of the theorems, and there is one that I don't remember and I can't find it on the web. Has anyone heard about the Pitot's Theorem?

Best,

[quote="darij grinberg"]I believe that the Pitot theorem is just the following fact:

A convex quadrilateral ABCD has an incircle if and only if AB + CD = BC + DA.

  darij[/quote]

Thank you Darij :first: , It may be, I don't have that theorem in my list, and I remember it.

Best regards,


find r in simplified terms.

$\sqrt{12(2+\sqrt{3}}=r$

i know my answer is wrong.  Can anyone tell me what i did wrong?  Thanks  :) [hide]
$r=a+b$ for some numbers a and b.
$a^{2}+2ab+b^{2}=24+12\sqrt{3}$

so $a^{2}+b^{2}=24$
$ab=12\sqrt{3}$ so  $a^{2}b^{2}=108$

so $a=\frac{12\sqrt{3}}{b}$

plugging this into $a^{2}+b^{2}=24$
i get  $\frac{432}{b^{2}}+b^{2}=24$

multiplying by $b^{2}$
$432+b^{4}=24b^{2}$

letting $p=b^{2}$
$p^{2}-24p+432=0$
$(p-36)(p-12)=0$

$b=\pm{6}$  or $b=\pm{2\sqrt{3}}$
$a=\frac{12\sqrt{3}}{6}=\pm{2\sqrt{3}}$ if $b=\pm{6}$

and  $a=\pm{6}$ if $b=\pm{2\sqrt{3}}$
therefore $r=2/sqrt{3}+6$

but the answer is $3\sqrt{2}+\sqrt{6}$
 :maybe:
[/hide]

[quote="myc"]$ab=12\sqrt{3}$ so  $a^{2}b^{2}=108$[/quote]
I'm assuming that it's supposed to be $2ab=12\sqrt{3}$....
anyways, I think you're assuming that $2ab$ is $12\sqrt{3}$.
You should start out with $r=a+b\sqrt{3}$ or $r=a+\sqrt{b}$ or something of that sort... :maybe:
I'm probably wrong so you can ignore me if you wish  :lol:

According to you:
$a^{2}+2ab+b^{2}=24-12\sqrt{3}$
so $a^{2}+b^{2}=24$
$ab=12\sqrt{3}so a^{2}b^{2}=108$

isn't $ab=-6\sqrt{3}$ then?

BTW, you said the answer is $3\sqrt2+\sqrt6$.
so I square it. And I get $24+12\sqrt3$. Why is $24-12\sqrt3$ being square-rooted? typo?

[quote="myc"]$r=a+b$ for some numbers a and b.
$a^{2}+2ab+b^{2}=24+12\sqrt{3}$

so $a^{2}+b^{2}=24$
$ab=12\sqrt{3}$ [/quote]

This step is incorrect.  You can't assume it splits like that.

Edit:  Actually,

[quote="myc"]$p^{2}-24p+432=0$
$(p-36)(p-12)=0$[/quote]

This factoring is incorrect.

You might want to try letting $r = \sqrt{a}+\sqrt{b}$ for $a, b$ integers.

[quote="t0rajir0u"][quote="myc"]$r=a+b$ for some numbers a and b.
$a^{2}+2ab+b^{2}=24+12\sqrt{3}$

so $a^{2}+b^{2}=24$
$ab=12\sqrt{3}$ [/quote]

This step is incorrect.  You can't assume it splits like that.
[/quote]

but can't you?  because $a^{2}+b^{2}$ must be an intergral value, so it has to be 24.

The problem is that you only specified $a, b$ were "numbers."  You didn't really specify what kind of numbers, and without doing so you certainly can't just assume that one part of the LHS is equal to another part of the RHS without specifying what you want $a, b$ to be.

That's why it would be best if you solved using $r = \sqrt{a}+\sqrt{b}$ where $a, b$ are [b]integers[/b], in which case your logic holds.


Find an example to prove that:
F, L are closed subspaces but F+L is not closed space in a normal space.

[quote="kingkong"]Find an example to prove that:
F, L are closed subspaces in a normed linear space but F+L is not closed subspace.[/quote]

I edited!


geometry
rotation
trigonometry
geometry unsolved

Draw square QRST and locate point U inside the square so that QU = 2, TU = 1, and UR = 3.  Then find the area of quadrilateral QTUR without finding the length of the side of the square.

Scale up the $ \triangle QTU$ by a factor $ \sqrt 2$ and rotate it around $ T$ by $ 45^\circ$ into a similar $ \triangle RTV.$
$ \angle UTV \equal{} 45^\circ,$ $ \frac {VT}{UT} \equal{} \sqrt {2}$ $ \Longrightarrow\ \triangle TUV$ is isosceles right with $ UT \equal{} UV$ and $ UT \perp UV.$
The $ \triangle UVR$ has sides $ UR \equal{} 3,\ UV \equal{} UT \equal{} 1,\ VR \equal{} UQ \sqrt {2} \equal{} 2 \sqrt 2,$ $ UR^2 \equal{} UV^2 \plus{} VR^2$ $ \Longrightarrow\ UV \perp VR$ and $ UT \parallel VR.$
The $ \triangle UTR$ has base $ UT \equal{} 1$ and altitude length $ UV \equal{} 1$ $ \Longrightarrow\ S(\triangle UTR) \equal{} \frac {UT \cdot UV}{2} \equal{} \frac {1}{2}.$
$ \angle QUT \equal{} \angle RVT \equal{} \angle RVU \plus{} \angle UVT \equal{} 90^\circ \plus{} 45^\circ \equal{} 135^\circ \Longrightarrow$ $ S(\triangle QRT) \equal{} \frac {QT^2}{2} \equal{} \frac {UQ^2 \plus{} UT^2 \minus{} 2\ UQ \cdot UT \cos \widehat{QUT}}{2} \equal{} \frac {5}{2} \plus{} \sqrt {2}.$
$ S(\square QRUT) \equal{} S(\triangle QRT) \minus{} S(\triangle UTR) \equal{} 2 \plus{} \sqrt 2.$


Hello, I am Damien Jiang.
I am a n00b. Please help me learn math, because I suck  :(
I also suck at StarCraft.
And life in general.
zoMG plz helps!  :|

THIS IS NOT HELPFUL, KEVIN HU

PLEASE DELETE THIS TOPIC PRETTY SOON OR I WILL BE FORCED TO EAT KEVIN HU

Huh?

The message is too small. Please make the message longer before submitting.

HAH I ACTUALLY FOUND THIS MILDLY AMUSING

I did not... you know the person posting as DJiang told me he was going to spam due to extreme boredom, and oh man he was pretty tricky.

dude you have a pretty good goal though

good job

Hmm, this is pretty amusing.

Still, impersonating others is bad, so I shall have to warn the offender.

Locked.


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

Prove that

$ a^4\plus{}b^4\plus{}c^4\ge 16A^2$

where $ a,b,c$ are three sides of a triangle and $ A$ is its area.

Applying Heron's formula gives:

$ a^4\plus{}b^4\plus{}c^4 \geq (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)$

or

$ a^4\plus{}b^4\plus{}c^4\geq a^2b^2\plus{}a^2 c^2\plus{}b^2 c^2$

which is true by AG or Muirhead.

[quote="hsiljak"]Applying Heron's formula gives:

$ a^4 \plus{} b^4 \plus{} c^4 \geq (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)$

or

$ a^4 \plus{} b^4 \plus{} c^4\geq a^2b^2 \plus{} a^2 c^2 \plus{} b^2 c^2$

which is true by AG or Muirhead.[/quote]

I know these two ineq., but how I get $ \ge 16$ in the above ineq.

[quote="hsiljak"]The $ 16$ disappears when you multiply it with $ A^2$:

$ RHS \equal{} 16\left(\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)} \right)^2$ where $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$.

Now, $ RHS \equal{} 16\left(\sqrt {\frac {a \plus{} b \plus{} c}{2}\frac {a \plus{} b \minus{} c}{2}\frac {a \minus{} b \plus{} c}{2}\frac { \minus{} a \plus{} b \plus{} c}{2}}\right)^2$

$ RHS \equal{} 16\frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)}{16} \\
 \\
\equal{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)$.[/quote]

thanks :D  :o

[quote="hsiljak"]Applying Heron's formula gives:

$ a^4 \plus{} b^4 \plus{} c^4 \geq (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)$

or

$ a^4 \plus{} b^4 \plus{} c^4\geq a^2b^2 \plus{} a^2 c^2 \plus{} b^2 c^2$

which is true by AG or Muirhead.[/quote]

I've found that
$ 16A^2 \equal{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c) \equal{} \cdots \equal{} 2\sum_{cyc}a^2b^2 \minus{} (a^4 \plus{} b^4 \plus{} c^4)$
 it implies that $ \sum_{cyc}a^2b^2 \ge 16A^2$
Conclude:
$ a^4 \plus{} b^4 \plus{} c^4\geq a^2b^2 \plus{} a^2 c^2 \plus{} b^2 c^2\ge 16A^2$

Is it right?

[quote="hsiljak"][quote="vinskman"]

I've found that
$ 16A^2 \equal{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c) \equal{} \cdots \equal{} 2\sum_{cyc}a^2b^2 \minus{} (a^4 \plus{} b^4 \plus{} c^4)$
 it implies that $ \sum_{cyc}a^2b^2 \ge 16A^2$
[/quote]
Yes, it is correct. I assume you've used

 $ \sum_{cyc}a^2b^2 \leq (a^4 \plus{} b^4 \plus{} c^4)$ so

$ 16A^2 \equal{} 2\sum_{cyc}a^2b^2 \minus{} (a^4 \plus{} b^4 \plus{} c^4)\leq \sum_{cyc}a^2b^2$.[/quote]

$ \mathbb {T}hanks!!$ :P


[quote="vishalarul"]Find the minimum value of $\sqrt{{a_{1}}^{2}+{a_{2}}^{2}+{a_{3}}^{2}+...+{a_{n}}^{2}}$ if $a_{1}+a_{2}+a_{3}+...+a_{n}= b$.

[hide="Great hint"]Use Cauchy or the fact that $RMS>AM$[/hide][/quote]
Your hint kills the problem immadietly: $\sqrt{\frac{\sum a_{i}^{2}}{n}}\geq\frac{\sum a_{i}}{n}=\frac{b}{n}$, so $\sqrt{{a_{1}}^{2}+{a_{2}}^{2}+{a_{3}}^{2}+...+{a_{n}}^{2}}\geq \frac{b\sqrt{n}}{n}$. The equality holds for $a_{1}=\ldots=a_{n}$.

[hide="And then the Cauchy way goes like this"]Let $a_{i}=a_{i}, b_{i}=1$: \[(\sum^{n}{a_{i}^{2}})(\sum^{n}{1^{2}})\geq (\sum^{n}a_{i})^{2}\implies \sqrt{(\sum^{n}{a_{i}^{2}})}{\sqrt{n}\geq b \implies \sqrt{\sum^{n}{a_{i}^{2}}}\geq \frac{b}{\sqrt{n}}=\frac{b\sqrt{n}}{n}}\] [/hide]
So Cauchy can let us prove RMS-AM, also AM-QM, can we prove generalized AM-GM with it or do we need a generalized Cauchy first?


I noticed that if you type the word t-r-i-o (without the hyphens), the system changes it to trio.  The same thing happens for the prefix t-r-i itself.

I noticed that the BBCode for our tri symbol is just "t-r-i" (without the hyphens), not ":t-r-i:".  Perhaps that explains the problem in the previous paragraph.

Yep - I forgot the colons.  Fixed now.  Thanks.


Here is a stronger one: http://www.mathlinks.ro/viewtopic.php?t=128529
I think the one above could be solved by the same technique

[quote="chien than"]Let $ a;b;c$ be positive real numbers satisfying $ abc \equal{} 1$
Prove that:
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \geq 4(a \plus{} b \plus{} c \minus{} 1)$[/quote]
$ f(a,b,c) \equal{} LHS \minus{} RHS;t \equal{} \sqrt {ab}$
$ \Rightarrow t^2c \equal{} 1$
$ f(a,b,c) \minus{} f(t,t,c) \equal{} (\sqrt {a} \minus{} \sqrt {b})^2(ab \plus{} c^2 \plus{} c(\sqrt {a} \plus{} \sqrt {b}) \minus{} 4)$
$ ab \plus{} c^2 \plus{} c\sqrt {a} \plus{} c\sqrt {b}\ge 4\sqrt [4]{a^3b^3c^3} \equal{} 4$
$ \Rightarrow f(a,b,c)\ge f(t,t,c)$
$ f(t,t,c) \equal{} 2t(t \plus{} \frac {1}{t^2})^2 \minus{} 4(2t \plus{} \frac {1}{t^2} \minus{} 1) \equal{} \frac {2(t \minus{} 1)^2(t^4 \plus{} 2t^3 \minus{} t^2 \plus{} 1)}{t^3}\ge 0$
So $ f(a,b,c)\ge 0$
 :blush:

[quote="chien than"]Let $ a;b;c$ be positive real numbers satisfying $ abc \equal{} 1$
Prove that:
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \geq 4(a \plus{} b \plus{} c \minus{} 1)$[/quote]
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 3k \minus{} 8\geq k(a \plus{} b \plus{} c)$ is true too
for all non-negative $ a,$ $ b$ and $ c$ such that $ abc \equal{} 1,$ where
 $ k_{best} \equal{} \min_{t\in(0,1)}\frac {2(t^2 \plus{} t \plus{} 1)^2}{t(2t \plus{} 1)} \equal{} 5.7028809...$ :wink:

[quote="FOURRIER"][quote="arqady"]
 $ k_{best} \equal{} \min_{t\in(0,1)}\frac {2(t^2 \plus{} t \plus{} 1)^2}{t(2t \plus{} 1)} \equal{} 5.7028809...$[/quote]

Could you tell me what is $ k$best and why its equal to this?[/quote]
Let $ f(a,b,c)\equal{}(a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 3k \minus{} 8\minus{}k(a \plus{} b \plus{} c)$ and $ a\equal{}\max\{a,b,c\}.$
Then $ a\geq1$ and $ f(a,b,c)\minus{}f\left(a,\sqrt{bc},\sqrt{bc}\right)\equal{}$
$ \equal{}\left(\sqrt b\minus{}\sqrt c\right)^2\left(a^2\plus{}bc\plus{}a\left(\sqrt b\plus{}\sqrt c\right)^2\minus{}k\right)\geq\left(\sqrt b\minus{}\sqrt c\right)^2\left(6a\sqrt{bc}\minus{}k\right)\equal{}$
$ \equal{}\left(\sqrt b\minus{}\sqrt c\right)^2\left(6\sqrt{a}\minus{}k\right)\geq0$ for $ k\leq6.$
Thus, it remains to check out when $ f(a,b,b)\geq0,$ where $ a\equal{}\frac{1}{b^2}$ and $ b\in(0,1].$
We obtain $ f(a,b,b)\geq0\Leftrightarrow2b^6\minus{}2kb^4\plus{}(3k\minus{}4)b^3\minus{}kb\plus{}2\geq0\Leftrightarrow$
$ \Leftrightarrow2(b^3\minus{}1)^2\minus{}kb(2b^3\minus{}3b^2\plus{}1)\geq0\Leftrightarrow$
$ \Leftrightarrow(b\minus{}1)^2\left(2(b^2\plus{}b\plus{}1)^2\minus{}(2b^2\plus{}b)k\right)\geq0.$
Id est, $ k\leq\frac{2(b^2\plus{}b\plus{}1)^2}{2b^2\plus{}b}.$ But $ \min\frac{2(b^2\plus{}b\plus{}1)^2}{2b^2\plus{}b}<6.$
Thus, $ k_{best}\equal{}\min\frac{2(b^2\plus{}b\plus{}1)^2}{2b^2\plus{}b}.$
If $ k>k_{best}$ then $ f(a,b,b)$ can be negative.
If $ k\leq k_{best}$ then $ f(a,b,c)\geq f\left(a,\sqrt{bc},\sqrt{bc}\right)$ and $ f(a,b,b)\geq0.$
Hence, $ f(a,b,c)\geq0$ is true.

[quote="arqady"]
Thus, it remains to check out when $ f(a,b,b)\geq0,$ where $ a \equal{} \frac {1}{b^2}$ and $ b\in(0,1].$[/quote]

First, thank you soooo much for your explanation!!

I think it remains to check $ f(a,\sqrt{ab},\sqrt{ab})\geq0,$
Why $ a \equal{} \frac {1}{b^2}$?

[quote="FOURRIER"][quote="arqady"]
Thus, it remains to check out when $ f(a,b,b)\geq0,$ where $ a \equal{} \frac {1}{b^2}$ and $ b\in(0,1].$[/quote]
Why $ a \equal{} \frac {1}{b^2}$?[/quote]
Because $ b \equal{} c$ and $ abc \equal{} 1.$ :wink:
[quote="FOURRIER"]
I think it remains to check $ f(a,\sqrt {ab},\sqrt {ab})\geq0,$[/quote]
We have proved that $ f(a,b,c)\geq f(a,\sqrt {bc},\sqrt {bc}).$
Hence, it remains to ptove that $ f(a,\sqrt {bc},\sqrt {bc})\geq0,$ where $ abc\equal{}1.$
Thus, it remains to ptove that $ f(a,b,b)\geq0,$ where $ abb\equal{}1$ since, $ a\cdot\sqrt{bc}\cdot\sqrt{bc}\equal{}abc\equal{}1.$ :wink:

[quote="arqady"]
Hence, it remains to ptove that $ f(a,\sqrt {bc},\sqrt {bc})\geq0,$ where $ abc \equal{} 1.$
Thus, it remains to ptove that $ f(a,b,b)\geq0,$ where $ abb \equal{} 1$ since, $ a\cdot\sqrt {bc}\cdot\sqrt {bc} \equal{} abc \equal{} 1.$ :wink:[/quote]

I think we put $ t\equal{}\sqrt{bc}$
It remains to prove $ f(a,t,t)\geq0$ where $ at^2\equal{}1$

Right?

[quote="FOURRIER"][quote="Mather"]
$ ab \plus{} c^2 \plus{} c\sqrt {a} \plus{} c\sqrt {b}\ge 4\sqrt [4]{a^3b^3c^3} \equal{} 4$
[/quote]

If you used $ AM$ $ GM$ here then Its false  :wink:
If you used something else please tell me about it[/quote]
I'm sorry  :blush:
Suppose $ c \equal{} Max(a,b,c)$
so $ 1 \equal{} abc\le c^3 \Rightarrow c\ge 1$
$ ab \plus{} c^2 \plus{} c\sqrt {a} \plus{} c\sqrt {b}\ge 4\sqrt [4]{\sqrt [2]{ab}abc^4}\ge 4\sqrt [12]{abc} \equal{} 4$
 :blush:

[quote="FOURRIER"][quote="arqady"]
Hence, it remains to ptove that $ f(a,\sqrt {bc},\sqrt {bc})\geq0,$ where $ abc \equal{} 1.$
Thus, it remains to ptove that $ f(a,b,b)\geq0,$ where $ abb \equal{} 1$ since, $ a\cdot\sqrt {bc}\cdot\sqrt {bc} \equal{} abc \equal{} 1.$ :wink:[/quote]

I think we put $ t \equal{} \sqrt {bc}$
It remains to prove $ f(a,t,t)\geq0$ where $ at^2 \equal{} 1$

Right?[/quote]
Yes!

[quote="Mather"]$ f(a,b,c) \minus{} f(t,t,c) \equal{} (\sqrt {a} \minus{} \sqrt {b})^2(ab \plus{} c^2 \plus{} c(\sqrt {a} \plus{} \sqrt {b}) \minus{} 4)$[/quote]

[color=red][b][size=150]This is FALSE[/size][/b][/color]

I Expanded it [b]many many times[/b]

Just give: $ a\equal{}4$ $ b\equal{}9$ $ c\equal{}\frac{1}{36}$ It will give you $ \frac{41653}{36^2}\equal{}\frac{42373}{36^2}$

[quote="FOURRIER"][quote="Mather"]$ f(a,b,c) \minus{} f(t,t,c) \equal{} (\sqrt {a} \minus{} \sqrt {b})^2(ab \plus{} c^2 \plus{} c(\sqrt {a} \plus{} \sqrt {b}) \minus{} 4)$[/quote]

[color=red][b][size=150]This is FALSE[/size][/b][/color]

I Expanded it [b]many many times[/b]

Just give: $ a \equal{} 4$ $ b \equal{} 9$ $ c \equal{} \frac {1}{36}$ It will give you $ \frac {41653}{36^2} \equal{} \frac {42373}{36^2}$[/quote]
Sorry
$ f(a,b,c) \minus{} f(t,t,c) \equal{} (\sqrt {a} \minus{} \sqrt {b})^2(ab \plus{} c^2 \plus{} c(\sqrt {a} \plus{} \sqrt {b})^2 \minus{} 4)$
 :blush:
Done !

[i]This problem is very easy :

  Note that : $ (ab \plus{} bc \plus{} ca )(a\plus{}b\plus{}c) \equal{} (a\plus{}b)(b\plus{}c)(c\plus{}a) \plus{} abc \equal{} (a\plus{}b)(b\plus{}c)(c\plus{}a) \plus{}1$

  so the initial ineq is equivalent to :

   $ (ab \plus{} bc \plus{} ca )(a\plus{}b\plus{}c) \plus{} 3 \ge 4 (a\plus{}b\plus{}c)$

   so we need to prove : $ \frac{3}{a\plus{}b\plus{}c} \plus{} (ab\plus{}bc\plus{}ca) \ge 4$

     from the well-known lemma : $ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)$ and the hypothesis $ abc\equal{}1$

  we deduce that : $ (ab\plus{}bc\plus{}ca)^2 \ge 3(a\plus{}b\plus{}c)$ and it's really trivial that : $ ab \plus{}bc \plus{} ca \ge 3$

  Now , applying $ AM\minus{}GM$ for $ 4$ positive real numbers

  $ \frac{3}{a\plus{}b\plus{}c} \plus{} (ab\plus{}bc\plus{}ca) \equal{}  \frac{3}{a\plus{}b\plus{}c} \plus{} \frac{ab\plus{}bc\plus{}ca}{3}  \plus{} \frac{ab\plus{}bc\plus{}ca}{3}  \plus{} \frac{ab\plus{}bc\plus{}ca}{3} \ge 4 \sqrt[4]{ \frac{ (ab\plus{}bc\plus{}ca)^3}{9(a\plus{}b\plus{}c)}} \equal{} 4 \sqrt[4]{ \frac{ (ab\plus{}bc\plus{}ca)^2}{3(a\plus{}b\plus{}c)}  \cdot \frac{(ab\plus{}bc\plus{}ca)}{3}} \ge 4$ , Done :D [/i]

Refer to Example 5 of [url]http://www.math.ust.hk/excalibur/v14_n1.pdf[/url] for another solution.  :lol:

[quote="sqing"][b]Generalization of  Ji Chen[/b]
Let $ a;b;c$ be positive real numbers .Prove that\[ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}\frac{1}{abc}).\][/quote]
Nice! :lol: There is a nice proof by AM-GM.

[quote="sqing"][b]Generalization of  Ji Chen[/b]
Let $ a;b;c$ be positive real numbers .Prove that\[ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}\frac{1}{abc}).\][/quote]
 There is a nice proof by AM-GM:
$\frac{ (a\plus{}b)(b\plus{}c)(c\plus{}a) }{4}+\frac{1}{abc}\ge \frac{2\sum a \sum bc}{9}+\frac{3\sum a}{( \sum bc)^2}$
$\ge 3\sqrt[3]{\left(\frac{\sum a \sum bc}{9}\right)^2\cdot \frac{3\sum a}{( \sum bc)^2}}=a+b+c.$

[quote=chien than]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}1)$[/quote]
The inequality is equivalent to
Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$. Prove that$$\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {3}{a+b+c}\geq 4.$$[url=http://www.artofproblemsolving.com/community/c6h623571p4627645]USA 2001[/url]
http://www.artofproblemsolving.com/community/c6h330978p1771117

[quote=chien than]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}1)$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1310567p7019979]We write inequality become[/url]
\[\frac{(a+b)(b+c)(c+a)}{4} + \frac{1}{abc} \geqslant a+b+c.\]Arcording to AM-GM Inequality, we have
\[\begin{aligned}\frac{(a+b)(b+c)(c+a)}{4}+\frac{1}{abc} & \ge \frac{2(a+b+c)(ab+bc+ca)}{9}+\frac{3(a+b+c)}{(ab+bc+ca)^2} \\ & \ge 3 \sqrt[3]{\left[\frac{(a+b+c)(ab+bc+ca)}{9}\right]^2 \cdot \frac{3(a+b+c)}{(ab+bc+ca)^2}} \\& =a+b+c. \end{aligned}\]

[quote=chien than]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}1)$[/quote]
https://artofproblemsolving.com/community/c4h1483883p8676302：
$(a+b)(b+c)(c+a)\geq4(a+b+c-1)$
$(a+b+c)(ab+bc+ca)-abc\geq 4(a+b+c)-4$
$(a+b+c)(ab+bc+ca-4)\geq-3$
By $AM-GM$
$a+b+c\geq3$
$ab+bc+ca\geq3$
$(a+b+c)(ab+bc+ca-4)\geq3(3-4)=-3$
[url=https://artofproblemsolving.com/community/c6h1310567p7020012]q.e.d[/url]

Let $a, b, c$ be positive real numbers satisfying $abc=1$. [url=https://artofproblemsolving.com/community/c6h1493502p8778039]Prove that [/url]$$(a+b+c)(a+b+c-3)
 \ge\frac{3+\sqrt{5}}{2}(ab+bc+ca-3)$$
$$\sum \frac{b+c}{a} \ge   2 (\sum a + \sum \frac{1}{a} - 3 )$$[url=https://artofproblemsolving.com/community/c6h575621p3392213]here[/url]

[quote=szl6208]The following inequality is also true!
Let $ a;b;c;d$ be positive real numbers satisfying $ abcd=1$
 Prove that:
$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-1)$[/quote]
Let $a+b+c+d=4u$, $abc+abd+acd+bcd=3w^3$ and $abcd=t^4$, where $t>0$.
Since $ (a+b)(b+c)(c+d)(d+a)\geq(a+b+c+d)(abc+abd+acd+bcd)$, it's enough to prove that
$$3uw^3+t^4\geq4ut^3.$$
Now, by AM-GM
$$3uw^3+t^4\geq4\sqrt[4]{u^3w^9t^4}.$$
Thus, it remains to prove that
$$w^9\geq ut^8.$$
Now, we'll replace $a\rightarrow\frac{1}{a}$... and similar.
Thus, we need to prove that
$$u^3\geq w^3,$$ which is obvious.
Done!

[quote=szl6208]The following inequality is also true!
Let $ a;b;c;d$ be positive real numbers satisfying $ abcd=1$
 Prove that:
$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-1)$[/quote]
Let $ a;b;c;d$ be positive real numbers . Prove that:
$$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-\frac{1}{abcd})$$

Let $ a;b;c;d$ be positive real numbers . Prove that:
$$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-\frac{1}{(abcd)^2}).$$
Let $ a;b;c;d$ be positive real numbers satisfying $ abcd=1.$ Prove that:
$$ (a+b)(b+c)(c+d)(d+a)+16\sqrt{2}-12 \geq (1+4\sqrt{2})(a+b+c+d).$$
[quote=szl6208]The following inequality is also true!
Let $ a;b;c;d$ be positive real numbers satisfying $ abcd=1$
 Prove that:
$$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-1)$$[/quote]
Proof of Zhangyunhua:

[quote=sqing]Let $ a;b;c;d$ be positive real numbers . Prove that:
$$ (a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-\frac{1}{(abcd)^2}).$$[/quote]

[quote=sqing]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$. Prove that$$\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {3}{a+b+c}\geq 4.$$[/quote]
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1a_2\cdots  a_n=1.$ Prove that$$\frac {1}{a_1} +\frac {1}{a_2} +\cdots+\frac {1}{a_n} +\frac {n}{a_1+a_2+\cdots+a_n}  \geq n+1 $$
Let $a,b>0$ and $ab=1.$  Prove that$$\frac{1}{a}+\frac{1}{b}+\frac{4}{a+b}\ge 4$$
Let $a,b,c>0$ and $abc=1.$  Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{7}{a+b+c}\ge\frac{16}{3}$$
Let $a,b,c,d>0$ and $abcd=1.$  Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7$$
Let $a,b,c,d,e>0$ and $abcde=1.$  Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{21}{a+b+c+d}\ge \frac{46}{5}$$

[quote=sqing][b]Generalization of  Ji Chen[/b]
Let $ a;b;c$ be positive real numbers .Prove that\[ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}\frac{1}{abc}).\][/quote]
Let $ a,b,c$ be positive real numbers .Prove that$$c(a^3+b^3+c^3-3-4c)> 7(abc-ab-c).$$
[size=50](p/6262708495)[/size]

[hide=Thanks.][quote=SBM][quote=sqing][b]Ji Chen[/b] (2019-09-21):
Let $ a,b,c$ be positive real numbers .Prove that$$c(a^3+b^3+c^3-3-4c)> 7(abc-ab-c).$$
[size=50](p/6262708495)[/size][/quote]

We have:
$$c(a^3+b^3+c^3-3-4c)-7(abc-ab-c)$$
$$= -\left[ 3ac+3bc+7c-7 \right] ab+c \left[  \left( a
+b \right) ^{3}+\left({c}^{3}-4c+4\right) \right]$$
\[ = \left[ {\frac{1}{4}{\mkern 1mu} {{\left( {a - b} \right)}^2} + \frac{{{{\left( {3{\mkern 1mu} c\left( {a + b} \right) - 14{\mkern 1mu} c + 14} \right)}^2}}}{{108{\mkern 1mu} {c^2}}}} \right]\left( {3{\mkern 1mu} ac + 3{\mkern 1mu} bc + 7{\mkern 1mu} c - 7} \right) + \frac{{27{\mkern 1mu} {c^6} - 108{\mkern 1mu} {c^4} - 235{\mkern 1mu} {c^3} + 1029{\mkern 1mu} {c^2} - 1029{\mkern 1mu} c + 343}}{{27{c^2}}}\]
Now$,$ easy prove.[/quote]

[/hide]
Let $ a,b,c$ be positive real numbers satisfying $a^2+b^2+c^2=3.$  Prove that$$4(a+b+c-2)< (a+b)(b+c)(c+a) \leq 4(a+b+c-1)$$

Let $ a,b,c$ be positive real numbers satisfying $ abc=1.$ Prove that
$$ (a+b)(b+c)(c+a) \geq 8(a+b+c-2)$$
Let $ a,b,c$ be non-negative numbers satisfying $ a(b+1)c=1.$ Prove that
$$ (a+b)(b+c)(c+a) \geq 2(a+b+c-1)$$
[hide=h]If $ a , b, c \ge 0  $ such that $  \boxed{ a^2+b^2+c^2+abc = 4}  $ prove that : $\boxed{8 (a+b+c-2)  \geq (a+b)(b+c)(c+a)}$ .
[url]https://artofproblemsolving.com/community/c6h506795p2846821[/url][/hide]

why is this thread still active from 2007? and more importantly why does sqing bump it every few years?

Why does sqing post a bunch of problems though on other peoples threads?

[quote=franzliszt]I believe it is because sqing posts problems related to the original, so that people can try to solve them.[/quote]

But why doesn't he make a thread of his own problems?

Because then there would be too many threads :P

You can get postbanned for creating too many threads.

[quote=chien than]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}1)$[/quote]

$\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )\geq 4\left ( a+b+c-1 \right ) $

$\Longleftrightarrow (a+b+c)(ab+bc+ca)-abc\geq 4\left ( a+b+c-1 \right )$

$\Longleftrightarrow (a+b+c)(ab+bc+ca)+3\geq 4(a+b+c).$

Proof :

$ab+bc+ca+\frac{3}{a+b+c}\geq \sqrt{3abc(a+b+c)}+\frac{3}{a+b+c}$

$=\sqrt{3}x+\frac{3}{x^2}$ where $x=\sqrt{a+b+c}\geq \sqrt{3}$ by AM-GM

$=\frac{(x-\sqrt{3})\{(x-\sqrt{3})(\sqrt{3}x+2)+\sqrt{3}\}}{x^2}+4\geq 4.$

Equality : $x = \sqrt{3}\Longleftrightarrow a = b = c = 1.$

$$ab+bc+ca +\frac{3}{a+b+c}\geq 4\sqrt[4]{\left(\frac{ab+bc+ca}{3}\right)^3\cdot \frac{3}{a+b+c} }\geq  4\sqrt[4]{\frac{ ab+bc+ca }{3 }}\geq 4$$

Let $ a,b,c$ be positive real numbers satisfying $ a^2+b^2+c^2+abc=4 $ .Show that $$\frac{a}{1+2b}+\frac{b}{1+2c}+\frac{c}{1+2a}\geq1$$
 $$\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\geq1$$
[quote=sqing]Let $ a,b,c$ be positive real numbers satisfying $ a^2+b^2+c^2+abc=4 $ .Show that $$ (b+c)(c+a)(a+b)\geq 4(a+b+c+abc-2) .$$[/quote]
[url=https://artofproblemsolving.com/community/c6h2903367p25891902]Solution of mudok:[/url]
Let $a+b+c=x, abc=z$. Then $ab+bc+ca=\frac{1}{2}(x^2+z-4)$

We need to prove: $x^3-12x+16\ge z(10-x)$

After using [url=https://artofproblemsolving.com/community/c6h1226093p6164570]this:[/url] $ \frac{(x-2)^2}{4-x}\ge z$

it is sufficient to prove:

$(x^3-12x+16)(4-x)\ge (x-2)^2(10-x)$ $\iff (3-x)(x-2)^2(x+2)\ge 0$

[quote=chien than]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \geq 4(a\plus{}b\plus{}c\minus{}1)$[/quote]

Very beautiful


Let a, b, c be distinct real numbers.
Find the minimum value of $ \frac {a^2}{(a \minus{} b)^2} \plus{} \frac {b^2}{(b \minus{} c)^2} \plus{} \frac {c^2}{(c \minus{} a)^2}$.

$ f(a,b,c)  \geq f(a\minus{}c,b\minus{}c,0)$
 hence  $ Min A \equal{}1$
  equal  hold when and only when $ (a,b,c)\equal{}(1,0,0)$

[quote="Hong Quy"]equal  hold when and only when $ (a,b,c) \equal{} (1,0,0)$[/quote]
a,b,c are distinct , Hong Quy , therefore , your equality doesn't occur  :)

[quote="Hong Quy"]$ f(a,b,c) \geq f(a \minus{} c,b \minus{} c,0)$ [/quote]
Why? a, b, c are not necessary positive.

[quote="apollo"]relate problem from Vasc's book
 If $ a, b, c$ are distinct real numbers, then
\[ \sum_{cyc}\frac {a^2}{(b \minus{} c)^2} \geq 2
\]
[/quote]
  it's   easy
 $ f(a,b,c) \geq f(a \minus{} c,b \minus{} c,0)$
 also seemmeriast's problem then  $ a \to 0$  then $ Min A \equal{}1$
 equal occur  then : $ a\to 0 , b \in R$ and $ c\equal{}0$

[b]My solution:[/b]

Let $ A \equal{} \frac {a}{a \minus{} b}, B \equal{} \frac {b}{b \minus{} c}, C \equal{} \frac {c}{c \minus{} a}$.

We have $ ABC \equal{} (A \minus{} 1)(B \minus{} 1)(C \minus{} 1)$,
that is $ AB \plus{} BC \plus{} CA \plus{} 1 \equal{} A \plus{} B \plus{} C$.

Then, $ (A \plus{} B \plus{} C \minus{} 1)^2 \equal{} A^2 \plus{} B^2 \plus{} C^2 \plus{} 2(AB \plus{} BC \plus{} CA \minus{} A \minus{} B \minus{} C) \plus{} 1$,
so $ (A \plus{} B \plus{} C \minus{} 1)^2 \equal{} A^2 \plus{} B^2 \plus{} C^2 \minus{} 1$.

Which means $ A^2 \plus{} B^2 \plus{} C^2 \equal{} (A \plus{} B \plus{} C \minus{} 1)^2 \plus{} 1 \geq 1$.

Equality can hold in infinitely many cases, e.g. $ (1, 0, \frac {1 \plus{} \sqrt5}{2})$.

[quote="Hong Quy"]equal occur  then : $ a\to 0 , b \in R$ and $ c \equal{} 0$[/quote]
The equality cannot hold, the value 1 can only be approched.


calculus
integration
calculus computations

I have just started differential equation and I want to know where do we put the constant while finding the solution in variable seperable form , somewhere i hav seen constant under log , pls explain how to put the constant after integration

Unless I've misunderestood the question, after you've separated the variables, you integrate both sides and so only need a constant on one side becuase if you wrote two, you could subtract one from the other. Also, it can be convenient to write the arbitrary constant as the log of a constant when, for example, the other terms are logs so you can use the rules of logs to simplify your answer. Similarly for exponentials.


Let $ ABC$ be acute angled triangle and $ AA_1,BB_1,CC_1$ are altitudes.Point $ K$ lies on $ BC$ also $ \angle BB_1K\equal{}\angle A$.$ A_1C_1$meets $ AC$ at $ L$.Prove that $ KL$ intersects $ CC_1$ at the middle.

Hmm...I don't really like trigonometry,but the proof of the main idea consists mostly from trigonometry,that's why I'll write down not the complete proof,but only a sketch of the solution:
[b]Sketch:[/b]
Suppose that perpendicular from $ L$ to $ AB$ intersects $ BC$ at $ P$.It is not hard to establish the length of $ A_1P$,and of $ A_1K$ as well.Indeed...(Maybe someone will be so kind to add this missing part of my solution and after that we would be able to name it  "complete solution")I do really believe that the things I will write down further are true,otherwise the problem would not be true.
After all calculations one will establish that division $ (A_1C,KP)$ is harmonic.Now it is enough to consider the pencil $ L(A_1C,KP)\cap CC_1$.It follows that $ CM \equal{} MC_1$,thus we are done.
$ \blacksquare$
P.S:
I've wasted five seconds of my life on reading your avatar but I believe that you will waste much more time on reading my post and on trying to finish my solution... :)

I've finally found out how this problem can be solved without using trigonometry(for "trigonometric" solution see my post above).
My solution based on harmonic fours and their properties.I would like to remind everyone one [b]well-known property[/b] of harmonic fours:
_________________________________________________________________________________________________________________
Suppose that $ X,Y,T,Z$ lies on the line in this order and division $ (XT,YZ)$ is harmonic.Also suppose that if $ E$ some point in the plane of
these points,such that $ \angle XET \equal{} 90^{\circ}$,then $ \angle YET \equal{} \angle TEZ$.
_________________________________________________________________________________________________________________

I guess it is not really hard to check out that this property is true indeed.
Also I will change notations as follows:
Instead of $ AA_1,BB_1,CC_1,K,L$ I will use $ AD,BE,CF,Z,X$.Let's return to the initial problem:

_________________________________________________________________________________________________________________

[b]Solution:[/b]
Suppose that $ XZ$ intersects $ AB$ at $ Y$ and $ BE$ intersects $ XZ$ at point $ T$.One can easily establish that $ (XT,YZ) \equal{} \minus{} 1$.Here we will use property of harmonic fours,which was mentioned above as well-known fact.(By the way notations are the same).
Therefore,we conclude that $ \boxed{\angle YEB \equal{} \angle BEZ \equal{} \angle BAC}$
It follows that $ EY\perp AB$,therefore:
\[ \frac {FY}{YA} \equal{} \frac {CE}{EA}
\]
On the other hand as we know division $ (XE,AC)$ is harmonic:
\[ \frac {CE}{EA} \equal{} \frac {CX}{XA}
\]
Combining previous two relations we finally conclude that:
\[ \frac {FY}{YA} \equal{} \frac {CX}{XA}
\]
Now it is enough to use Menelaus theorem for triangle $ \triangle AFC$ and transversal $ \overline{XYM}$,where $ M$ is the midpoint of $ CF$.
$ \blacksquare$.

Great solution!Thanks a lot. :lol:
I always enjoy your solutions :)


analytic geometry
geometry
Pythagorean Theorem
combinatorics proposed
combinatorics

Let $ n*n$ integer points with coordinate from $ (0;0)$ to $ (n \minus{} 1;n \minus{} 1)$. How many regular polygons with vertec are points from there?

what do you mean "with coordinates from $ (0,0)$ to $ (n\minus{}1,n\minus{}1)$"?

It mean $ (x;y)$ with $ 0\leq x,y \leq n\minus{}1$. :blush:

anyway, the answer is 0. if a regular n-gon has side length $ s$, then its area is $ \frac{ns^2}{4}\cot \frac{\pi}{n}$. so if there were a regular n-gon with vertices at lattice points and side length $ s$, $ s^2$ would be an integer (e.g., by the pythagorean theorem), so $ \frac{ns^2}{4}\cot \frac{\pi}{n}$ would be irrational (since $ \cot\frac{\pi}{n}$ is irrational). but at the same time, the polygon should have rational area, since its vertices are at lattice points, so we have a contradiction.

:wink:  But what about the squares?

Pierre.

well, the squares aren't so important  :D ... they have measure 0 with respect to the problem statement ;).

It is very easy to check that: only square is satisfy! Now we have to count how many square in it. Now please do this, it is not so hard!

every square in question has a unique square circumscribed around it whose sides are parallel to the coordinate axes; moreover, this circumscribed square is contained within the large $ n\times n$ square. a square with its sides parallel to the axes and of side length $ s$ has $ s$ squares (including itself) inscribed in it. the number of squares of side length $ s$ with sides parallel to the axes is $ (n \minus{} s \plus{} 1)^2$, for $ 1\leq s\leq n$. so the total number of squares is

\[ \sum_{s \equal{} 1}^n s(n \minus{} s \plus{} 1)^2 \equal{} \sum_{s \equal{} 1}^n (sn^2 \plus{} s^3 \plus{} s \minus{} 2s^2n \plus{} 2sn \minus{} 2s^2) \equal{}
\]

\[ \equal{} \sum_{s \equal{} 1}^n s(n^2 \plus{} 2n \plus{} 1) \minus{} \sum_{s \equal{} 1}^n s^2(2n \minus{} 2) \plus{} \sum_{s \equal{} 1}^n s^3 \equal{}
\]

\[ \frac {n(n \plus{} 1)^3}{2} \minus{} \frac {n(n^2 \minus{} 1)(2n \plus{} 1)}{3} \plus{} \frac {n^2(n \plus{} 1)^2}{4}
\]


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $ a\in F_p$ with $ a\neq 0$, and let $ K$ be a splitting field of the polynomial $ x^p \minus{} x \minus{} a$ over $ F_p$.
Prove that $ [K: F_p] \equal{} p$ and that the Galois group $ G(K/F_p)$ is cyclic.

Presumably you know how to show that groups of prime order are cyclic. :)

The idea here is that once you have one of the roots, (degree p extension) you have all of them.  Looking at some examples will help you see the general pattern quite quickly, let us know when you have a guess...


trigonometry
logarithms
geometry
geometric transformation
rotation

This version of the four bugs problem, three snails stand 60 cm apart, and crawl twords the one clockwise to it, at a speed of 5cm/min. The questions are,

(1) How long do the snails take to reach each other, and
(Harder) How mant revolutions round the centre do they make before they meet?

So the snails are in a circle 60 cm apart? I do not understand the clockwise part.

Sorry, ill explain better....

The three snails are at the vertexs of an equilateral triangle. On a signal, they crawl twourds their clockwise neighbour, how long do the take to collide and what angle do the trace out at the centre?

This is the famous problem for any n-gon.  The curve is always a logarithmic  spiral, meeting at the center.

[hide]Easiest is to consider the   relative velocity (= $v+v\cos 60^o = 3v/2$  .. so answer would be 8 minutes.

[img]http://mathworld.wolfram.com/images/gifs/mice3.gif[/img][/hide]

Im assuming since you know everything that you have the rotation. But could you find the roatation on terms of time? :)

The total number of revolutions is , of course,  infinite.

You may like to look at [url=http://mathworld.wolfram.com/Whirl.html]Wirl[/url]

The curve, of course is given by $r = a(1-kt)$

  $a$  = distance from the center to the starting position,  $ak$ = (speed)(cos (angle)) (Or componet of velocity in the radial direction, in the case of a triangle it is $\cos 30^o$  (Let us call it $\alpha$ (For triangle $\alpha= 30^o$ for square it is $45^o$ etc   ($t$, of course is time)

And thus $\theta = \ln (\frac r a) \cdot \tan \alpha$

 ( log blows up when $r$ tends to zero - But you need not worry - even if the bugs are small, they would be squished  much before number of revolutions reach too high ...Even one rotation would be more than enough to squish them! - One revolution and you are about 10,000 of the distance you started from !  )

Hey, isn't this the first problem to "200 puzzling physics problems"?

I remember finding the whirl before, tis quite cool. Are you sure you answer (the 8 min one) is right? Not that im doubting you, but i got a different answer using Paul Zeitz method...

I might have made a mistake, but let me double check ... seems right.

Another way to think is $a= 20\sqrt 3$   (Each side is 60, so this is the distance from center to   corner)
and radial compnent =  $5 \cos 30^o = 5(\sqrt 3 / 2)$
dividing these one gets $8$ .. what answer do you get?

Odd, i must have divided those two wrong!


If magnitudes of vectors a and b are 5 and 12 respectively, then the magnitude of vector (b-a) could NOT be:

A. 5
B. 7
C. 10
D. 12
E. 17

[hide]THis is basically asking which number does not satisfy the triangle inequality when degenerate triangles (lines) are allowed.  THat would be A): 5[/hide]


superior algebra
superior algebra unsolved

Why is it true that every every epimorphism* of groups is surjective?

I read (in Freyd maybe?) that this proof requires something called "the amalgamation lemma," but I could not find a good reference.

* Definition for those curious: by epimorphism I mean [i]right-cancelable[/i] homomorphism, so a map $ f: G\to H$ such that whenever $ \alpha, \beta: H\to K$ with $ \alpha\circ f\equal{}\beta\circ f$, then $ \alpha\equal{}\beta$.

http://katmat.math.uni-bremen.de/acc/ , chapter II, ex. 7H

Thanks for the link.  It will take me a little while to get an intuition for this construction, but it's good to have a proof handy.


You have to unscramble the names of ten ports in which the sailor has been around the world.

1. GLSOLESANE
2. SODNLER
3. IEDERIJOONAR
4. SAILOB
5. LHANCTORSE
6. LMRSHAEA
7. BONRGTEAEDÍMNEAO
8. DVNUMIOETÉ
9. SRETB
10. NDNGELRIAO

1. Los Angeles
2. London
3. Rio de Janeiro
4. Lisbon
5. Charleston
6. Marseilles
7. Montego Bay
8. Montevideo
9. Brest
10. Leningrad


This one shows up in one form or another, but do you know how to solve it?

The Problem

This question is often posed as starting with 100 open lockers, numbered 1 through 100. A student passes by, closing only the even-numbered lockers. The next student "toggles" every 3rd locker. If it's open, she closes it, and if it's closed, she opens it. The next student toggles every 4th locker, and so on through the 100th student, who only toggles the 100th locker.

The question: after these 99 students are finished, how many lockers are open? What are their numbers? If there were 1000 lockers, and the same process was used, what is the number of the last open locker?

[hide="The Solution"]

The beauty of such problems is that, while they can be solved through a brute force approach, a key insight usually unravels the entire problem. For this problem, the insight involves "factors". Consider locker number 12. Student 2 will toggle it. So will students 3, 4, 6, and 12. The number 12 is a multiple of each of those numbers. In other words, with the addition of the number "1", those numbers are the factors of 12.

Someone must have opened all those lockers. We'll call him "student 1". For locker 12, the sequence is: 1 opens, 2 closes. 3 opens, 4 closes. 6 opens, 12 closes. In fact, for every locker number with an even number of factors, only the first and last students matter: the first one opens it and the last one closes it. Since factors always come in pairs the implication is that all the lockers will be closed. Not quite. Consider locker "4". 1 opens, 2 closes, 4 opens. 4 is different because it uses "2" twice: 2x2. It has an odd number of factors. However, the student doesn't make two passes through the lockers, so the door will be left open.

The only numbers with an odd number of factors are numbers that use a factor twice, and the only numbers that do that, are the squares. Thus, our answer to which lockers will be open is the series: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Isn't that just like kids? Always picking on the squares!

What if there are 1000 lockers? Another trick with squares makes determining the answer trivial. When figuring successive squares, the next answer is always the last answer, plus the next terms added together, minus 1. So if you know 12-squared is 144, 13-squared is 144+13+13-1: 169. With this insight in mind, we instantly see that 30x30 is 900, making 31x31= 961 (900+31+31-1). 32x32 would be 63 more - too much! So the last open locker would be 961.
http://www.daniweb.com/blogs/entry387.html[/hide]

Similar Problem: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166665]Click[/url]

@danb80(or Chinaboy): What is brute force?[/url]

It's where you actually do the work instead of doing cool math stuff to solve.
Like, if you were asked in how many ways you could arrange 1,2,3,4 in a line, you would actually list all the ways.


eqnarray is now obsolete (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url]) but both eqnarray and its replacement align use a counter called, yes, [i]equation[/i].

Thanks Stevem!
I read that "Avoid Eqnarray!" article, so suffice to say I'll be switching to the align environment in future.

Cheers.


Cone $ A$ has a volume of $ 144\pi$ and a radius of 3.  What is the height of the tallest cylinder with radius 2 that fits inside of cone $ A$?

Sorry about unpretty numbers, the author of this problem is kinda dumb.
I think i know what the answer is but have no idea how to solve. Thanks

[hide="hint"]
Remember that a cone's volume is $ \frac{1}{3}$ of the volume of a cylinder.
[/hide]

[hide]Well since the volume is $ 144\pi$, and we know the radius is $ 3$, we just solve a simple equation to get the height.

$ \frac{\pi r^2 h}{3}$ = $ 144\pi$.

$ 3h\pi \equal{} 144\pi$

$ h \equal{} 48$,

now we look at the similar right triangles with short legs $ 2$ and $ 3$ (drawing the picture always helps). Since they are similar ( Angle-Angle theorem/potualte idk), and the corresponding legs are in the ratio $ \frac{2}{3}$, we know the heights are in the ratio $ \frac{2}{3}$. We know that the height of the triangle with base 3 has height 48, so the height of smaller one must be 32. Now you just do 48-32 and you get $ \fbox{16}$.[/hide]

[quote="cognos599"][hide]Well since the volume is $ 144\pi$, and we know the radius is $ 3$, we just solve a simple equation to get the height.

$ \frac {\pi r^2 h}{3}$ = $ 144\pi$.

$ 3h\pi \equal{} 144\pi$

$ h \equal{} 48$,

now we look at the similar right triangles with short legs $ 2$ and $ 3$ (drawing the picture always helps). Since they are similar ( Angle-Angle theorem/potualte idk), and the corresponding legs are in the ratio $ \frac {2}{3}$, we know the heights are in the ratio $ \frac {2}{3}$. We know that the height of the triangle with base 3 has height 48, so the height of smaller one must be 32. Now you just do 48-32 and you get $ \fbox{16}$.[/hide][/quote]

I think you have the idea right but you forgot to square 3.

then the height is 16

then you can solve a proportion:

    16/3 = x/(3-2)

at least i think that will work

then the height of the new cylinder is 5 1/3

Guys, he did square three, but you divide by three too so it cancels out.
I agree, the height of the new cylinder would be 16. You basically just use similar triangles.

[quote="cognos599"][hide]Well since the volume is $ 144\pi$, and we know the radius is $ 3$, we just solve a simple equation to get the height.

$ \frac {\pi r^2 h}{3}$ = $ 144\pi$.

$ 3h\pi \equal{} 144\pi$

$ h \equal{} 48$,

now we look at the similar right triangles with short legs $ 2$ and $ 3$ (drawing the picture always helps). Since they are similar ( Angle-Angle theorem/potualte idk), and the corresponding legs are in the ratio $ \frac {2}{3}$, we know the heights are in the ratio $ \frac {2}{3}$. We know that the height of the triangle with base 3 has height 48, so the height of smaller one must be 32. Now you just do 48-32 and you get $ \fbox{16}$.[/hide][/quote]

what right triangle with legs 2 and 3?

I think he means two similar triangles that have corresponding legs of length 2 and 3.

Cone has a height of 48, you guys should be able to figure this out by yourselves.
Therefore, a right triangle is formed, 3 as the base, and 48 as the height.
That triangle is similar to another triangle with 3 as the base, so 32 would be the height.
The cylinder would be right under the triangle under the triangle with base 2, so the cylinder will have a radius of 2 and a height of 16.


http://www.dump.ro/download.php?id=YjBiYWVlOWQyNzlkMzRmYTFkZmQ3MWFhZGI5MDhjM2Y=&download=true

in that link you can find the 5 exercises? can someone help me with them? maybe 1 or 2 ?

Maybe I'm just spamming, but let me tell you this is a forum for Spanish-speaking people, while you want us to solve problems in Romanian... :huh: If you want your problems to be solved quickly, you should post them in the calculus forum, or in the Romanian forum, too:

[url]http://www.mathlinks.ro/Forum/index.php?f=250[/url]


On a recent trip, Mary Ann drove 189 miles in 4.5 hours. At this rate, how many hours will it take her to drive 294 miles?

We know that $ rt \equal{} d$, where $ r$ is the rate, $ t$ is the time, and $ d$ is the distance.  Plugging in the information given, we have $ r(4.5) \equal{} 189$, so $ r \equal{} \frac {189}{4.5} \equal{} \frac {189}{\frac {9}{2}} \equal{} \frac {189\cdot2}{9} \equal{} 21\cdot2 \equal{} 42$.  Knowing that, we can set up a new equation to solve for how long she takes to drive $ 294$ miles:
\[ 42(t) \equal{} 294\implies t \equal{} \frac {294}{42} \equal{} \boxed{7}\]


Let $0<x_n<\frac{1}{n}$ such that $\sum_{n\geq 1}{x_n}$ converges. Then $\sum_{n\geq 1}{\frac{x_n}{1-nx_n}}$ also converges.

This is false without some other condition on $x_n$, such as perhaps monotonicity.

Let $x_{2^k}=2^{-k}(1-2^{-k})$, with $x_n$ negligibly small for other values of $n$. (By that, I mean the terms of some convergent series that adds up to a small number.)

Then $\sum_{n=1}^{\infty}x_n$ converges but $\sum_{n=1}^{\infty}\frac{x_n}{1-nx_n} > \sum_{k=0}^{\infty}(1-2^{-k})$ which diverges.

? take $x_n=\frac{1}{n^2}$ if $n$ is not a square, $x_{n^2}=\frac{1}{n^2+1}$... it seems like this gives a counterexample.

Well, that's still not monotonic, and Kent had already given a counterexample.

Maybe the extra condition of monotonicity does it, but I still can't see an immediate solution. Maybe it's something similar to the condensation criterion, but that would only work out if we also knew that $(\frac{x_n}{1-nx_n})$ was decreasing, which we don't.

Sorry, guys for the useless post, it seems that the friend who gave me the problem forgot some little details.

sorry for my post, i didn't read kent's post before i wrote mine...you don't have to understand why ;)


[b]Fifteen[/b] tickets are numbered $ 1,2,3....,15$ respectively.[b]Seven[/b] tickets are selected at random one at a time with replacement.Find the probability that the largest number appearing on a selected ticket is $ 9$.

The number of ways to select seven tickets among which the largest is numbered 9 is ____.*  The number of ways to select seven tickets with no restrictions is ____.  Therefore, the probability that the largest ticket selected is numbered 9 is ____.

* You may wish to break this into two steps:  the number of ways of selecting seven tickets of numbered less than or equal to 9 is ____.  The number of ways of selecting seven tickets numbered less than 9 is ____.  Therefore, the number of ways of selecting seven tickets among which the largest is numbered exactly 9 is ____.


Solve

a # b= 1/a + 1/b
6 # k= 1/4

Find K.

What does the number sign means in this problem.

It was just defining an operation. So when it says a # b, it means $ \frac {1}{a} \plus{} \frac {1}{b}$, just like it defined in the first equation. For instance, 6 # 8 = $ \frac {1}{6} \plus{} \frac {1}{8}$.

So to find $ k$, solve:

$ \frac {1}{6} \plus{} \frac {1}{k} \equal{} \frac{1}{4}$

Hint: Multiply both sides by 24k.

EDIT - Oops, fixed.

[quote="helijet"]Then 1/6 + 1/k should equal 1/4, but how come you make it equal 4.[/quote]
So, 1/6 + 1/k = 1/4.
Then, 2 + 12/k = 3 (Multiply by 12)
Next, 12/k = 1
So we have k = 12


1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]


Was anyone at Seton Hall for the Andrushkiw competition today? I thought the test was a lot easier this year than it has been in previous years. They also took two identical problems from 2007, and one where the changed the numbers a little bit but forgot adjust the answer. They need to come up with new and more challenging problems for next year.

Anyways if you were there I was the one in fifth with a score of 29.

Yes, the problems were indeed similar.  The trig problem on #9 was challenging, and I felt that the 3 point problems were harder than the 4 point ones.  I was also there, and I got a 33 (3rd).

Yea, I ran out of time and couldn't simplify it. For those that weren't there:

9) Find the exact value of

$ \frac {\frac {2\sin 140\sec 280}{\sec 220} \plus{} \frac {\sec 340}{\csc 20}}{\frac {\cot 200 \minus{} \tan280}{\cot 200}}$

I got a score of 25.

I messed up on some really simple stuff, made various mistakes everywhere.

I completely screwed up the combinatorics problem on the first page somehow. The problem with the chairs I got 40 because I didn't realize that my method only applied to odd rows of chairs. The gym floor triangle question was just very stupid computational errors.

Assuming I got those 3 questions, I would've gotten 25+2+3+3 = 33. On the bright side though, I scored JUST enough to bring our school up to overall 5th place, so it's not that bad.


Problema 1

--------------------------------------------------------------------------------
in triunghiul MBE avem unghiurile MBE si EMB congruenter deci triunghiul este isoscel cu laturile congruente (MB) si (EB) deci avem in triunghiul MBC (ME) mediana si jumatate din BC deci triunghiul este dreptunghic in M. Cum unghiul AMB este drept, rezulta ca A, M, C coliniare si atunci unghiul MAB este de 45 de grade, deci x=45.

scuzati-ma ca am trimis gresit rezolvarea problemei 1 propusa de razvan balan la subiectul 3 pr de geometrie


Let $ A$ and $ B$ be positive integers satisfying $ B\equal{}3A$. Assume that the digits of $ B$ are a permutation of the digits of $ A$. Prove that $ A$ is divisible by $ 27$.

That's not true. A counterexample: $ A \equal{} 1035$, $ B \equal{} 3A \equal{} 3105$ where $ A$ is not divisible by $ 27$.

However, $ B$ is always divisible by 27 (or equivalently $ A$ is divisible by $ 9$).
Let $ a_0, \dots, a_n$ be digits of $ A$ and $ B$ so that $ A \equal{} \sum_{i \equal{} 0}^n a_i\cdot 10^i$ and $ B \equal{} \sum_{i \equal{} 0}^n a_i 10^{p_i}$ for some permutation $ p$.
Then
\[ 2A \equal{} B \minus{} A \equal{} \sum_{i \equal{} 0}^n a_i\cdot (10^{p_i} \minus{} 10^i)\]
is divisible by 9, since every difference $ 10^{p_i} \minus{} 10^i$ is. Therefore, $ A$ is divisible by 9 and $ B$ is divisible by 27.


abstract algebra
algebra
polynomial
vector
linear algebra

Find the kernel of the following linear transformation :

$T : P_n = \{ \text{all polynomials in} \ x \ \text{of degree} \ \leq n \} \rightarrow \Re \ \ \ \text{where} \ \ \ T(p(x))=p(2)$

Thanks a lot!
Mark.

The kernel is simply $\{p\in P_n: p(2)=0.\}$ There are several other ways one could express this. You could write out what $p(2)=0$ means in terms of the coefficients of $p,$ or you could also write the kernel as $\{(x-2)q(x): q\in P_{n-1}\}.$ This last formulation makes it clear that the kernel is an $n$-dimensional subspace of the $(n+1)$-dimensional vector space $P_n,$ and could be used to help us find a basis for the kernel.

Thanks a lot.

And could you tell me what a base for the kernel would be please?

You could try to start from a base for $P_{n - 1}$ and transform it into a base for $\ker T$. Do you see how to do that?

actually no, i don't see how to transform a base for $P_{n-1}$ into a base for $\ker \ T$. Could you help me in this please?

Thank you.

[quote="headhunt"]actually no, i don't see how to transform a base for $P_{n-1}$ into a base for $\ker \ T$. Could you help me in this please?[/quote]
Just multiply each element of that basis for $P_{n-1}$ by $x-2.$


inequalities
trigonometry
inequalities unsolved

Prouve that all reels $x_1,x_2,.......,x_n$ satisfy:

$\sum_{n\geq{i}\geq{j}\geq1}\cos^2(x_i-x_j)\geq\frac{n(n-2)}{4}$

Thre is a typo. You should write $i>j$ or $\frac{n(n+2)}{4}$.

We are to maximize $\sum_{i\leq j} \sin^2(x_i-x_j)$. Consider $n$ points on the unit circles, those angles are $2x_1,2x_2,...2x_n$, then $\sin^2(x_i-x_j)$ is a quarter of the square of the distance between points $i$ and $j$, and the whole sum is the $n/4$ times moment of inertia of this system of points (it is a Jacobi's theorem). Due to Steiner's theorem we know that this moment of inertia is not greater than the moment of inertia wrt to $O$, and the last one is $n$. Thus $\sum_{i\leq j} \sin^2(x_i-x_j)\leq n\cdot n/4=\frac{n^2}{4}$. Therefore $\sum_{i\leq j} \cos^2(x_i-x_j)\geq \frac{n(n+1)}{2}-\frac{n^2}{4}=\frac{n(n+2)}{4}$.

[quote="Myth"]Thre is a typo. You should write $i>j$ or $\frac{n(n+2)}{4}$.

No i'm sure ${n\geq{i}\geq{j}\geq1}$ and $\frac{n(n-2)}{4}$

Then if you read my solution you see that I proved it for $\frac{n(n+2)}{4}$.


How would you simplify:

$ \sqrt {\frac { \minus{} 1 \pm i\sqrt {7}}{2}}$

[hide="HINT"]
Convert it to phasor form and then use DeMoivre's.
[/hide]

just write it as x+iy
square  and solve.

It equals $ x \plus{} yi$ for some real $ x$ and $ y$. Hence,

$ x^2 \minus{} y^2 \plus{} 2xyi \equal{} \frac{\minus{}1}{2} \pm i\frac{\sqrt{7}}{2}$.

Since if two complex numbers are equal, their real and imaginary parts must be equal, we equate the real and imaginary parts to get:

$ x^2 \minus{} y^2 \equal{} \frac{\minus{}1}{2}$
and
$ 2xy \equal{} \pm \frac{\sqrt{7}}{2}$.

From here, you could just solve for $ x$ and $ y$. However, you can take an even easier route: first, note that $ \sqrt{|z|} \equal{} |\sqrt{z}|$ for all complex numbers $ z$.

$ x^2 \plus{} y^2 \equal{} |x \plus{} yi| \equal{} |\sqrt{\frac{\minus{}1 \pm \sqrt{7}}{2}}| \equal{} \sqrt{|\frac{\minus{}1 \pm \sqrt{7}}{2}|}$.

Adding this equation to the first one in the system, you can find $ 2x^2$ in terms of a real number. Sorry for being too lazy to simplify that expression above.


I have solved this problem but not sure about my answer. Please help me!
    On a 20 mile bike ride, you ride the first 10 miles at an average speed of 24 mi/h. What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be 16 mi/h?

  I obtain the result is 12 mi/h but my friend said that there is no solution. So who is correct?

i agree with you:
[hide="solution"]
total time: $ \frac{10}{24}+\frac{10}{x}$ hr
total distance: $20$ mi
total speed: $ \frac{20}{\frac{10}{24}+\frac{10}{x}}=16$ mi/hr
solve this and you get $x=12$.
[/hide]

[hide]
The sum of the times for each leg must equal the total time for the entire ride.

$\frac{10}{24}+\frac{10}{v}=\frac{20}{16}$

$\Rightarrow \frac{10}{v}= \frac{20}{16}-\frac{10}{24}$

$\Rightarrow \frac{10}{v}= \frac{\left(60-20\right)}{48}$

$\Rightarrow \frac{v}{10}= \frac{48}{40}$

$\therefore v = 12\text{ mph}$[/hide]


calculus
integration
geometry
conics
parabola
real analysis
real analysis unsolved

I have tried twice but I am unable to get the correct numerical answer:

The closed curve $ C$ in the $ z \equal{} 0$ plane consists of the arc of the parabola $ y^2 \equal{} 4ax \textup{ (}a > 0)$ between the points $ (a, \pm 2a)$ [b]and the straight line joining[/b] $ (a, \mp 2a)$. The area enclosed by $ C$ is $ A$. Show, by calculating the integrals explicitly, that
\[ \int_C (x^2y \textup{ d}x \plus{} xy^2 \textup{ d}y) \equal{} \int_A (y^2 \minus{} x^2) \textup{ d}A \equal{} \frac {104}{105}a^4\]
where $ C$ is traversed anticlockwise.

I parameterized the first integral as $ y \equal{} t$, $ x \equal{} \frac {t^2}{4a}$ and substituted everything in: $ dy \equal{} dt$, $ dx \equal{} \frac {t}{2a} dt$.

Then I add the integrals over the 2 paths:
\[ \int_{2a}^{ \minus{} 2a}\frac {t^6}{32a^3} \plus{} \frac {t^4}{4a} \textup{ d}t \plus{} \int_{ \minus{} 2a}^{2a} \frac {t^4}{4a} \textup{ d}t\]
However, the second definite integral somehow cancels out with the second part of first and I get a negative answer. Where's the mistake?

Perhaps the curve $ C$ is the arc of the parabola together with a line segment?  (That is, I think you left out some words in your second paragraph.)  You might want to double-check in which direction you've traversed the line segment.

Yes, sorry, indeed I missed out some words (corrected in bold in original post).

I just realised my careless mistake for the second integral of the line integral - along the straight line it should be
\[ \int_{\minus{}2a}^{2a} at^2 \textup{ d}t\]However my area integral is still wrong

Well, everything you've written is fine, so probably you should check your arithmetic and other similar things.  For the double integral, I have $ \int_{\minus{}2a}^{2a}\int_{\frac{y^2}{4a}}^a y^2 \minus{} x^2\, dx\,dy$, which gives the same value (as it must).

There are some arithmetic errors.. such a pain. Thanks :)


In a remote island, a language in which every word can be written using only the letters $a$, $b$, $c$, $d$, $e$, $f$, $g$ is spoken. Let's say two words are [i]synonymous[/i] if we can transform one into the other according to the following rules:

i) Change a letter by another two in the following way: \[a \rightarrow bc,\ b \rightarrow cd,\ c \rightarrow de,\ d \rightarrow ef,\ e \rightarrow fg,\ f\rightarrow ga,\ g\rightarrow ab\]
ii) If a letter is between other two equal letters, these can be removed. For example, $dfd \rightarrow f$.

Show that all words in this language are synonymous.

Edit:  Hidden.

[hide="Solution"] First, consider the sequence of transformations

$aa \to bcbc \to bb$

Applied cyclically, every double letter is equivalent to every other double letter.  Next, consider the sequence of transformations

$a \to bc \to cdde \to ccce \to eece \to ec \to ede \to d$

Applied cyclically, every letter is equivalent to every other letter.  Then the algorithm is trivial; lengthen the shorter word until it is the same length as the longer word, then change the letters accordingly. [/hide]

This would be cool to do with graph theory.

Just a note, t0rajir0u's 2nd transformation finishes the problem only because we have 7 letters, it moves 3 to the right, and (7,3)=1, so we can reach each other letter.

If we had 6 letters, that process would not necessarily work.


Has anyone participated or know of somebody that has participated at a Mathnasium? Are they effective at math acceleration? They seem expensive. Are they worth the price?

TIA

If you're looking for competition math preparation, a brief glance at their website tells me they aren't what you're looking for.

[quote="Christiano89"]What is it?? :maybe:[/quote]

It's a chain/franchise of after school math academies. There is a lot of money to be made in math tutoring, but the best part is that teaching/tutoring is fun and rewarding.


real analysis
calculus
derivative
integration
function
limit
real analysis unsolved

let E be a subset of the reals with positive measure and $ c \in (0,1)$ constant

then show there exists an interval I with

$ \mu_L (E \cap I ) > c  \mu_L (I)$

i have a proof but im not really happy with it, so i would love to see others ideas

$ 0 < \mu (E) = \sum_{n \in Z} \mu (E \cap \left[n, n + 1\right]) \to \exists N$ s.t.  $ E_N = E \cap [N, N + 1]$, WLOG assume $ c > \mu (E_N) = \eta > 0$

then consider $ a_0 = N, b_0 = N + 1, I_k = (a_k, b_k) \forall k = 1, 2, ..$

where $ I_k = (a_k, \frac {a_k + b_k}{2}) \cup \{\frac {a_k + b_k}{2}\} \cup ( \frac {a_k + b_k}{2}, b_k) = A_k \cup \{p_k\} \cup B_k,$

and we pick $ I_{k + 1} \in \{A_k, B_k\}$ s.t. $ \mu (I_{k + 1}) = max( \mu (A_k) , \mu (B_k))$

note that since $ \mu (E \cap I_{k + 1}) = d_k \mu (E \cap I_k)$ with $ 1 \le 2d_k \le 2$ then ${ \mu (E \cap I_{k + 1}) = \left[\prod_0^k d_k \right]\mu (E \cap I_0)} = \eta \prod_0^k d_k$
then since $ \mu (I_n) = 2^{ - n}$ for each k, we get $ \mu (E \cap I_n) = \eta \prod_{k = 1}^n d_k = \eta D_n$ where $ 2^n D_n$ is a nondecreasing sequence.

here i do a hand waving argument stating that we must have infinitely many $ d_{k_j} > \frac {1}{2}$
 therefore there is a monotone increasing subsequence $ c_j = 2^{n_j}D_{n_j} \eta$ bounded by 1

letting j be sufficiently large, $ c_j > c$ for prescribed $ c \in (0,1)$

i feel not so good about this

Apply the Lebesgue theorem on the differentiation of the integral to the characteristic function of $ E.$ The result of that theorem then shows that

$ \lim_{\stackrel{\mu(I)\to0}{x\in I}}\frac{\mu(E\cap I)}{\mu(I)}\equal{}1$ for almost every $ x\in E.$

Hence for every $ c<1$ and almost every $ x\in E,$ we can find $ I$ such that $ x\in I$ and $ \mu(E\cap I)>c\mu(I).$

ah.. it was the chapter after we stopped having lecture, had to give it a quick skim.

so...

since $ \mu (E) > 0$ then we know $ \Delta E \ne \emptyset$

so let $ x \in \Delta E$ and let $ I_n \equal{} B(x, n^{\minus{}1})$

then we know $ \forall \epsilon > 0, \exists N$ s.t. $ n\ge N \to  1\minus{} \frac{\mu ( I_n)}{\mu (I)} < \epsilon$

letting $ \epsilon < 1 \minus{} c$ we win. thank you much km.


Suppose that the sequence  ${(a_{n})}_{n \ge 0}$ satisfies \[a_{0}= 0,\ a_{1}= a_{2}= 1,\ \frac{a_{n+1}-3a_{n}+a_{n-1}}{2}={(-1)}^{n}\] Prove that $a_{n}$ is a perfect square.

Solved by induction [url=http://wangsblog.com/jeffrey/?p=123]here[/url].

And here is the practice problem you gave:
Practice Problem: (360 Problems For Mathematical Contests - 2.1.57) Let $a_{0}= a_{1}= 3$  and $a_{n+1}= 7a_{n}-a_{n-1}$  for $n \ge 1$ . Prove that $a_{n}-2$  is a perfect square for all $n \ge 1$ .
[hide="My Solution"]i'll post it tomorrow, but i found the sequence of squares: it  goes $b_{n+1}=3b_{n}-b{n-1}$ now only easy induction left...[/hide]


trigonometry
inequalities proposed
inequalities

$ o<x<90^{o}$ prove:
$ (tan)^{sinx}\plus{}(cot)^{cosx}\geq2$

[quote="zaya_yc"]$ o < x < 90^{o}$ prove:
$ (\tan x)^{\sin x}\plus{}(\cot x)^{\cos x}\geq2$[/quote]$ \left(\frac{\sin{x}}{\cos{x}}\right)^{\sin{x}}\plus{}\left(\frac{\cos{x}}{\sin{x}}\right)^{\cos{x}}\geq2$ $ \implies$ $ \frac{\frac{\sin{x}^{\sin{x}}}{\cos{x}^{\sin{x}}}\plus{}\frac{\cos{x}^{\cos{x}}}{\sin{x}^{\cos{x}}}}{2}\geq\sqrt{\frac{\sin{x}^{\sin{x}\minus{}\cos{x}}}{\cos{x}^{\sin{x}\minus{}\cos{x}}}}\equal{}\left(\frac{\sin{x}}{\cos{x}}\right)^{\frac{\sin{x}\minus{}\cos{x}}{2}}$
Now if $ \sin{x}>\cos{x}$, the quantity $ \left(\frac{\sin{x}}{\cos{x}}\right)^{\frac{\sin{x}\minus{}\cos{x}}{2}}>1$ because $ \frac{\sin{x}}{\cos{x}}>1$ and $ \frac{\sin{x}\minus{}\cos{x}}{2}>0$.
If $ \sin{x}<\cos{x}$, $ \left(\frac{\sin{x}}{\cos{x}}\right)^{\frac{\sin{x}\minus{}\cos{x}}{2}}>1$ because $ \frac{\sin{x}}{\cos{x}}<1$ and $ \frac{\sin{x}\minus{}\cos{x}}{2}<0$.
Equality is at $ x\equal{}45$.

if $ o<x\leq45^{o}$ ----> $ cosx\geq sinx$ and $ tanx\leq1$ ---> $ (tanx)^{sinx}\geq(tanx)^{cosx}$;
if $ 45^{o}\leq x<90^{o}$-----> $ sinx\geq cosx$ and $ tanx\geq1$---->$ (tanx)^{sinx}\geq(tanx)^{cosx}$
$ (tanx)^{sinx}\plus{}(cotx)^{cosx}\geq(tanx)^{cosx}\plus{}(cotx)^{cosx}\geq2\sqrt{(tanx)^{cosx}(cotx)^{cosx}}\equal{}2\sqrt{(tanxcotx)^{cosx}}\equal{}2$
$ x\equal{}45^{o}$


Ben was twenty years old in 1980 but only fifteen years old in 1985. How come ?

[quote="gaussintraining"][hide="Solution?"] Because Ben lived in B.C. times, meaning he was $ 15$ in $ 1985$ B.C, and $ 20$ in $ 1980$ B.C?[/hide][/quote]

Yes, right

[quote="batteredbutnotdefeated"]The curious case of Benjamin button! He ages in reverse! :rotfl:[/quote]

lol Benjamain Button.........or he could be living in B.C. times........or both! lol

 :rotfl:

Wouldn't be because he is Benjamin Button or something?

This could be he lived in the BC times or there simply was not even a real Ben!!!

why hasn't anyone sugjested a time machine
he could have gone through a time machine
or he had superpowers
or he was rank 20 than rank 15(military, games, etc)


bye, i am italian but i would like to go to university in england or in the united states, i wuold like to ask to some of the best universities in these countries.
i've seen watching on internet what i should do to be accepted.i'm going to do the SAT exam and for SAT 2 i've choosen 4 subjects(math,physics,chemistry and latin) and i am going to do also the AP exam of math (AP calculus) and the physics one.
i am going to do also STEP because i would like to ask also for english universities, but reading on this forum i've understood that is very difficoult to be accepted in important universities in USA, do you think that if i'do these exams good i'll have some possibilites to be accepted? there something else i should do?
ado you nthink is is easier to be accepted by cambridge or by an american famous universities(MIT,princeton....)?
P.S i''ve never gone to IMO.

[quote="tokenadult"]You can have some good choices of colleges without ever going to the IMO. Good luck in your applications.[/quote]
do ypu think there's something else i could do to have more choise of being accepted?


καλημερα
κοιταξτε τι εφτιαξα με mathematica!!(καλα ειμαι νεος ακομα μην περιμενετε κανενα τρομερο θαυμα)
[code]Do[ParametricPlot3D[{(Abs[5 - i]/3 + 4)/
    4*Cos[t1]Sin[t2], (Abs[5 - i]/3 + 4)/4*Sin[t1]Sin[t2], Cos[t2] + 3 - Abs[
      5 - i]/3}, {t1, -Pi, Pi}, {t2, -Pi, Pi},
          PlotRange -> {{-5, 5}, {-5, 5}, {-3, 5}}], {i, 11}][/code]

μολις το γρψετε επιλαξτε ολα τα σχηματα που θα φτιαχτουν απο το loop και πατηστε ctrl+Y για animation(υποθετω πως οσοι εχετε mathematica ξερετε τι παιζει)...

περιμενω σχολια :)

ΥΓγια οσους δεν εχουν mathematica και δεν εχουν καταλαβει τι ειναι ειναι ενα μπαλακι που χοροπηδαει και καθως πεφτει κατω πεπλατιζεται(δεν ξερω αν υπαρχει τετοια λεξη παντω το νοημα ειναι οτι προσομοιωνει ως ενα βαθμο την ελαστικη παραμορφωση που του προκαλει το πατωμα)

edit:οριστε και το επιπεδο(μαζι με το μαπλακι)
[code]Do[a = ParametricPlot3D[{(Abs[5 - i]/3 + 4)/4*Cos[t1]Sin[t2], (Abs[5 - i]/3 +
                4)/4*Sin[t1]Sin[t2], Cos[t2] + 3 - Abs[5 - i]/
        3}, {t1, -Pi, Pi}, {t2, -Pi, Pi}, DisplayFunction -> Identity,
            PlotRange -> {{-5, 5}, {-5, 5}, {-3, 5}}];
  c = ParametricPlot3D[{
      t1, t2, -1}, {t1, -5, 5}, {t2, -5, 5}, DisplayFunction ->
      Identity, PlotRange -> {{-5, 5}, {-5, 5}, {-3, 5}}];

  Show[{a, c}, Axes -> False, DisplayFunction -> $DisplayFunction], {i, 10}]
[/code]

parte ki ena mpalaki poy gyrizei gyrv apo ena allo!(kanei kabosi wra na fanei alla einai grafikotato)
[code]Do[
  a = ParametricPlot3D[{Cos[t1]Sin[t2] + 4Cos[i/3],
       Sin[t1]Sin[t2] + 4Sin[i/3], Cos[t2]}, {t1, -Pi, Pi}, {
        t2, -Pi, Pi}, DisplayFunction ->
          Identity, PlotRange -> {{-5, 5}, {-5, 5}, {-3,
       5}}]; b = ParametricPlot3D[{Cos[t1]Sin[t2], Sin[t1]Sin[t2],
       Cos[t2]}, {t1, -Pi, Pi}, {t2, -Pi,
        Pi}, DisplayFunction -> Identity, PlotRange -> {{-5, 5}, {-5, 5}, \
{-3, 5}}]; Show[{a,
          b}, Axes -> False, DisplayFunction -> $DisplayFunction], {i, 6Pi}][/code]


True or not:
if$a = b(mod k)$ then $a^2 = b^2 (mod k^2)$

All we can say is $a^2 = b^2 (k)$

if $k$ is prime, $gcd(a,k)=gcd(b,k)=1$, $k \neq 2$ this is true iff $a^2 = b^2 (k^2)$ from the beginning I think.

I didn't totally understand what you wanted to say, but as I understood it, it mustn't be so from beginning, e.g. set $a=1 , b=4 , k=3$ again.

But the following should hold: $a \equiv b \mod k^n \implies a^k \equiv b^k \mod k^{n+1}$


If $ a$ and $ n$ are positive integers, then compute $ (a^{8n^2\plus{}6n\plus{}1}\minus{}1,a^{4n^2\plus{}5n\plus{}1}\plus{}1)$.

Let's denote $ a^{4n\plus{}1}\equal{}m$. We have to find $ d\equal{}(m^{2n\plus{}1}\minus{}1,m^{n\plus{}1}\plus{}1)$. We have $ d|m^{2n\plus{}2}\minus{}1$
and so $ d|(m^{2n\plus{}1}\minus{}1,m^{2n\plus{}2}\minus{}1)\equal{}m\minus{}1$, but we also have $ d|m^{n\plus{}1}\plus{}1$ so $ d|2$.
It's clear that $ d\equal{}1$ if $ a$ is even and $ d\equal{}2$ if $ a$ is odd.


calculus
integration
trigonometry
logarithms
geometry
perimeter
calculus computations

I was doing some practice using Calculus BC APs from previous years, and I came across an integration I just couldn't solve...

\[ \int\limits_{\frac{5\pi}{4}}^{\frac{7\pi}{4}}{\sqrt{9+16\cos^2{x}}dx} \] was the specific one.  I reduced it to something of the form \[ \int\limits_{\frac{5\pi}{2}}^{\frac{7\pi}{2}}{\sqrt{17+8\cos{x}}dx} \] using some substitution, but then I realized: I don't even know how to evaluate something like that.  The AP is coming closer...and I'm getting worried.  Any help here?

By the way, I also found that I couldn't find an explicit solution for $x$ for \[ \ln{(x^2+1)}=\cos{x} \] although the AP seemed to expect me to be able to (or just wanted me to use a calculator).  Any help there?  Thanks.

Neither of these can be done exactly. The first is an elliptic integral- they're well-studied, but the antiderivative you're looking for is not an elementary function.

So in a problem like that on the AP, is a calculator the way to go? (yeah, it was asking me to find the perimeter (circumference?) of a section of a parametrized ellipse...)


limit
real analysis
real analysis unsolved

Explore lower & upper bounds. Besides find infimum and supremum of a set:

$ \lbrace \frac {|n \minus{} k|}{n \plus{} k}\rbrace$

$ n, k$ - natural numbers

$ n\equal{}rk$ for $ r\in Q$

$ \to \frac {|rk \minus{} k|}{rk \plus{} k}\equal{} \frac{|r\minus{}1|}{r\plus{}1}\equal{}f(r)$

for $ r\ge 1 \to 1 \ge f(r)\equal{}\frac{r\minus{}1}{r\plus{}1}\equal{}1\minus{}\frac{2}{r\plus{}1}\ge 0$

for $ r\in (0,1)\to 1 \ge  f(r) \equal{} \frac{2}{r\plus{}1} \minus{}1 \ge 0$

$ \cup_{n,k \in N} \frac {|n \minus{} k|}{n \plus{} k} \equal{} \cup_{r \in Q^\plus{}} f(r)$

$ f(r) \equal{} 0 \to r\equal{}1 \to n\equal{}k$ must be the min (and therefore inf)

$ \lim_{r \to 0^\plus{}} f(r) \equal{} 1 \to$ our sup must be 1


Prove that if $ \frac{2^n\minus{}2}{n} \in \mathbb{Z}$ then $ \frac{2^{2^n\minus{}1}\minus{}2}{2^n\minus{}1} \in \mathbb{Z}$, where $ n \in \mathbb{N}^*$.

[hide="Solution"]$ \frac {2^{n} \minus{} 2}{n}\equal{} k \Rightarrow 2^n \minus{} 2 \equal{} nk$

$ \frac {2^{2^{n} \minus{} 1} \minus{} 2}{2^{n} \minus{} 1} \equal{} \frac {2^{nk \plus{} 1} \minus{} 2}{2^{n} \minus{} 1} \equal{} \frac {2\left(2^{nk} \minus{} 1\right)}{2^{n} \minus{} 1} \in \mathbb{Z}$[/hide]


Find all integers $x$, such that:

$cos(\frac{\pi}{8}(3x-\sqrt{9x^{2}+160x+800}))=1$

[quote="matt276eagles"][quote="ch1n353ch3s54a1l"][hide]$x=\frac{8k}{3}-\frac{40}{3}+\frac{125}{3k+5}$

Since $k$ does not satisfy this equation for integers $x$ mod 3 then there are no solutions[/hide]

[/quote]

Could you explain this last step?[/quote]

Uhh...the mod 3 part?
I just took 0,1,2 and substituted them in for $k$ and none worked...is that legal?

[quote="ch1n353ch3s54a1l"]I just took 0,1,2 and substituted them in for $k$ and none worked...is that legal?[/quote]

Not with that division sign.  You need to do casework on all the possible values of $3k+5$.