analytic geometry
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use the determinant method i think.

given 3 points (x1,y1), (x2,y2), (x3,y3), then the area of the triangle is given by

0.5 |x1y2+x2y3+x3y1-x2y1-x3y2-x1y3|

sorry no latex knowledge.

[quote="236factorial"]The vertices of a triangle have cordinates (-3,1), (-2,-3), and (x,0). Find x given x is positive and the area of the triangle is 7.5 square units.

This is the last problem I don't get in the 1990-1992 sprints and targets!  :D

Does Heron's formula have any use?  :huh:[/quote]

Determinants are not required in MATHCOUNTS. It's messy anyways.

Was this a target though? Then it would be practical. :P Or you could use trig to find the included angle and use $A=\frac{ab \sin {C}}{2}$ :D

Ugh, this is going to be hard to explain without a diagram.  First, find the distance between the two known points.  I'll assume you know how to do that, and you get $\sqrt{17}$.  Use that as the base of the triangle.  So:

$\frac{\sqrt{17} \cdot h}{2} = \frac{15}{2} \rightarrow h = \frac{15}{\sqrt{17}}$

That means the point (x,0) is on a line parallel to $y = -4x - 11$, which is the base, and there is $\frac{15}{\sqrt{17}}$ units between those two parallel lines.  I hope you can visualize it.   ;)

So given that, you can draw a line perpendicular to both parallel lines, which is the height.  Since this line has a slope of $\frac{1}{4}$, the shift in x is 4 times the shift in y.  When I say shift, I mean the difference in coordinates of the base point and (x,0).  Draw a right triangle to help visualize it better.  Call the x shift $b$ and the y shift $a$.

$b = 4a$

$a^2 + b^2 = \frac{225}{17} \rightarrow a^2 + 16a^2 = 17a^2 = \frac{225}{17}$

$a = \frac{15}{17}$ and $b = \frac{60}{17}$

the base point is then $(x - \frac{60}{17}, -\frac{15}{17})$.  Plug that into $y = -4x - 11$ and solve for x.

$x = 1$

This is a sprint round question, so trig is invalid. Is there a shorter solution (dakyru's solution was nice)?

Here's an elementary solution.

Make a rectangle around the triangle, as shown in the figure below.
The area of the triangle is calculated by subtracting the areas of three right triangles from the area of the rectangle.

$4(x+3)-\frac{1}{2}(1\cdot 4+1(x+3)+3(x+2))=\frac{15}{2}$

$\Longleftrightarrow 4x+12-\frac{1}{2}(4x+13)=\frac{15}{2}$

$\Longleftrightarrow 2x+\frac{11}{2}=\frac{15}{2}$

$\Longleftrightarrow x=\boxed{1}$

That's the solution I would use (if I can think it up), dakyru's solution is the one that comes up first in my mind.


[img]http://bbs.pep.com.cn/attachments/a_n_mha2hZO3JAXv.jpg[/img]

Ask the formula of general term , but how to do it without using mathematical induction

[hide="Hint"]
$a_{n}(a_{n}-2)+1 =0$
$a_{n}=-\frac{1}{a_{n}-2}$
$a_{2}=-\frac{1}{-1.5}=-2/3$
a_3 = -3/8
a_4 = -8/19
Now look for patterns in the repeating fractions.
The limit of the sequence is zero, and it approaches from the left.
[/hide]

DFT, you're suffering from some typos in the subscripts and also from some sign errors, the result of which are that you've computed the sequence wrong.

Why would you want to prove a closed form for a recurrence relation [i]without[/i] using induction?  It's quick, natural and easy.

But how would you [b]derive[/b] the formula without knowing it beforehand? You couldn't use induction to do that...?

Something's screaming at me "factor it", but maybe that's just its superficial resemblence to $n^{2}-2n+1$. :(

There's a field of study in combinatorics (I'm pretty sure it's classified under combinatorics...) solely dedicated to solving recurrence relations (that is, exactly what we are trying to do here). This PDF shows a brief summary of some of the methods used: [url]http://www.cs.cmu.edu/~adamchik/21-127/lectures/recursion_4_print.pdf[/url].
As for this recurrence relation, I have no idea how you'd solve it without some clever guess work, it'd probably be ugly too.

[hide="this works out nicely..."]
We observe that the sequence $a_{n}$ might approach 1 (if it approaches anything, and if it does, it will probably do so from the left, as $a_{1}<1$), so it might help to consider $a_{n}=1-\frac{1}{b_{n}}$. After simplifying, the equation reduces to
\[b_{n+1}=b_{n}+1,\ b_{1}=2\]
Now it's simple; we get $b_{n}=n+1,\ a_{n}=\frac{n}{n+1}$.
[/hide]

If you calculate the recursion right you quickly discover that the series is this: [hide]$\frac{1}{2},\frac{2}{3},\frac{3}{4},\cdots$[/hide]

[hide="Sort of proof"]
\begin{eqnarray*}\frac{n}{n+1}(-\frac{n+1}{n})&=&-1\\ \frac{n}{n+1}(\frac{-n-1}{n})&=&-1\\ \frac{n}{n+1}(\frac{n-1-2n}{n})&=&-1\\ \frac{n}{n+1}(\frac{n-1}{n}-2)&=&-1\\ \frac{n}{n+1}(\frac{n-1}{n})-2\frac{n}{n+1}+1&=&0\\ a_{n}&=&\frac{n}{n+1}\end{eqnarray*}[/hide]


inequalities
summer program
Mathcamp
geometry
area of a triangle
Heron\u0027s formula

Hello, I was wondering if it were possible to use CS's extended version to prove Nesbitt's inequality?

CS Extended: $\sum^n_{i=1}\frac {{a_i}^2}{b_i} \geq \frac{({\sum^n_{i=1}a_i})^2}{\sum^n_{i=1}b_{i}}$

Nesbitt's: $\sum_{cyc} \frac a{b+c} \geq \frac 32$

Also, can somebody help explain homogenous, smoothing, and normalization as these terms seem to be employed a lot in inequalities? I am rather new in doing these proofs  :blush: . Thanks.

Yes, it is possible. Just apply it and a couple of AM-GM's should do it.

As for the other questions... someone else can handle those.

[quote="spang"]Also, can somebody help explain homogenous, smoothing, and normalization as these terms seem to be employed a lot in inequalities? I am rather new in doing these proofs  :blush: . Thanks.[/quote]

It would not help you by just telling you outright what those are. You need to see them in action, for example by reading the solutions of the inequalities posed on the forum.

But if you insist, I wrote a bit about it in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20553]this[/url] thread.

Lemma:
$\displaystyle\frac{a^2}{x}+\frac{b^2}{y} \geq \displaystyle\frac{(a+b)^2}{x+y}$
Start With Trivial ($x^2 \geq 0$):
$
\displaystyle (ay-bx)^2\geq 0
\\
\displaystyle (ay)^2+(bx)^2\geq 2(ay)(bx)
\\
\displaystyle (a^2)xy+(ay)^2+(bx)^2+(b^2)xy \geq (a^2)xy+2abxy + (b^2)xy
\\
\displaystyle (a^2)y(x+y)+(b^2)x(x+y) \geq xy(a+b)^2
\\
\displaystyle \frac{a^2}{x}+\frac{b^2}{y} \geq \frac{(a+b)^2}{x+y}$
So, for 2 variables it holds. Now for the inductive step. Assume:
$\displaystyle\sum_{i=1}^{n} \frac{t_{i}^{2}}{s_i} \geq \displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n} t_i\right)^2}{\displaystyle\sum_{i=1}^{n}{s_i}}
\\
\displaystyle\sum_{i=1}^{n} \frac{t_{i}^{2}}{s_i} + \displaystyle\frac{t_{n+1}^{2}}{s_{n+1}} \geq \displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n} t_i\right)^2}{\displaystyle\sum_{i=1}^{n}{s_i}} + \displaystyle\frac{t_{n+1}^{2}}{s_{n+1}}
$
Applying our Lemma:
$\displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n} t_i\right)^2}{\displaystyle\sum_{i=1}^{n}{s_i}} + \displaystyle\frac{t_{n+1}^{2}}{s_{n+1}} \geq \displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n} t_i + t_{n+1}\right)^2}{\displaystyle\sum_{i=1}^{n}{s_i} + s_{n+1}} = \displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n+1} t_i\right)^2}{\displaystyle\sum_{i=1}^{n+1}{s_i}}
\\
\displaystyle\sum_{i=1}^{n+1}\frac{t_{i}^{2}}{s_i} = \displaystyle\sum_{i=1}^{n} \frac{t_{i}^{2}}{s_i} + \displaystyle\frac{t_{n+1}^{2}}{s_{n+1}} \geq \displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n} t_i\right)^2}{\displaystyle\sum_{i=1}^{n}{s_i}} + \displaystyle\frac{t_{n+1}^{2}}{s_{n+1}} \geq \displaystyle\frac{\left(\displaystyle\sum_{i=1}^{n+1} t_i\right)^2}{\displaystyle\sum_{i=1}^{n+1}{s_i}}$
So, that version of Cauchy is proved.
Now, for Nesbitt:
$\displaystyle\frac{a}{b+c}+\displaystyle\frac{b}{a+c}+\displaystyle\frac{c}{a+b} = \displaystyle\frac{a^2}{ab+ac}+\displaystyle\frac{b^2}{ab+bc}+\displaystyle\frac{c^2}{ac+bc} \geq \displaystyle\frac{(a+b+c)^2}{2ab+2ac+2bc}$
Now, we use [hide="this"]By trivial,

$(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0
\\
2a^2 + 2b^2 + 2c^2 \geq 2ab + 2bc + 2ac
\\
a^2 + b^2 + c^2 \geq ab + bc + ac$[/hide] handy inequality:
$a^2+b^2+c^2 + 2ab+2ac+2bc \geq ab+ac+bc + 2ab+2ac+2bc
\\
(a+b+c)^2 \geq \frac{3}{2}(2ab+2ac+2bc)
\\
\displaystyle\frac{(a+b+c)^2}{2ab+2ac+2bc} \geq \frac{3}{2}$
And, connecting our chain:
$\displaystyle\frac{a}{b+c}+\displaystyle\frac{b}{a+c}+\displaystyle\frac{c}{a+b}  \geq \displaystyle\frac{(a+b+c)^2}{2ab+2ac+2bc}  \geq \frac{3}{2}$
And we are done.

[quote="spang"]Thanks!

Zach - This is John from Mathcamp; I'm trying to brush up on Dorin's Olympiad Problem Solving notes on inequalities  :lol:[/quote]

hi, maybe this request is a bit weird but do you mind telling me who dorin is and can you send me the notes on inequalities?

[quote="spang"]Hello, Dorin was the teacher of problem solving at Canada/USA mathcamp. The notes, unfortunately, are all handwritten - I don't have an electronic version. They are arranged similar to Engel's Problem-Solving Strategies with important theorems and then lots of problems.[/quote]

oh, never mind then..

[quote="zabelman"]Here's nesbitt by means of your version of Cauchy.

[hide]The key step is to make the numerator a square by doing the following:

$\sum_{cyc}\frac{a}{b+c} =\sum_{cyc}\frac{a^2}{ab+ac}\\
\ge \frac{(a+b+c)^2}{2(ab+bc+ca)}\\
=\frac{(a^2+b^2+c^2)+2(ab+bc+ca)}{2(ab+bc+ca)}\\
\ge\frac{(ab+bc+ca)+2(ab+bc+ca)}{2(ab+bc+ca)}\\
=\frac{3}{2}$[/hide][/quote]

there is an easier way to prove nesbitt, although it has nothing to do with Cauchy.

[hide="proof"]
$\sum_{cyc}\frac{a}{b+c} \ge \frac{3}{2}$
change this into this:
$\frac{a+b+c}{a+b} + \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} - 3 \ge \frac{3}{2}$
$(a+b+c)(\sum_{cyc}\frac{1}{a+b}) - 3 \ge \frac{3}{2}$
$\frac{1}{2} {(a+b)+(b+c)+(c+a)}(\sum_{cyc}\frac{1}{a+b}) - 3 \ge \frac{3}{2}$

now  we set a+b =x, b+c=y, a+c=z
$\frac{1}{2} (x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) - 3 \ge \frac{3}{2}$
$(x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) - 6 \ge 3$
$\frac{x}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{z}{y}+\frac{y}{z}+3-6 \ge 3$

Lemma: $\frac{a}{b} + \frac{b}{a} \ge 2$
$ \frac{a^2+b^2}{ab} \ge 2$
$ a^2+b^2 \ge 2ab$
$(a-b)^2 \ge 0$

So, back to the problem, we can group the six fractions into three groups where we can apply the Lemma. Thus the LHS becomes $\ge 3$, and we have finished the proof of Nesbitt. BTW, equality holds when $x=y=z$, or $a=b=c$
[/hide]

There are easily 10 million different ways to prove Nesbitt.  My personal choice is just an easy application of the rearrangement theorem.

What is the rearrangement theorem? I'm kinda new with inequalities. Thanks!

Also, to show that Nesbitt isn't very strong:

It's equivalent to $\sum a(a+b)(a+c)\geq \frac {3}{2}(a+b)(b+c)(c+a)$

which is just $\sum a^3+\sum a^2b + 3abc \geq \frac {3}{2}\left(\sum a^2b + 2abc\right)$

so all we need to show is $\sum a^3 \geq \sum a^2b$ which is obvious from $a^3+a^3+b^3\geq 3a^2b$ and so on.

[quote="Magnara"]$a_1b_1+a_2b_2+...+a_nb_n$ is maximized when a_i and b_i are similarly sorted, and minimized when they are oppositely sorted.

The proof:
a,b,c and 1/(b+c), 1/(a+c), and 1/(a+b) are similarly sorted (verify this!).  Therefore:
$a*\frac{1}{b+c}+b*\frac{1}{a+c}+c*\frac{1}{a+b}\geq{a*\frac{1}{a+c}+b*\frac{1}{a+b}+c*\frac{1}{b+c}}$

Similarly,
$a*\frac{1}{b+c}+b*\frac{1}{a+c}+c*\frac{1}{a+b}\geq{a*\frac{1}{a+b}+b*\frac{1}{b+c}+c*\frac{1}{a+c}}$

Summing these two, we get
$2*(a*\frac{1}{b+c}+b*\frac{1}{a+c}+c*\frac{1}{a+b})\geq{\frac{a+c}{a+c}+\frac{a+b}{a+b}+\frac{b+c}{b+c}}=3$.

Engel's book handles this inequality very well.  It's incredibly powerful, and not too well known.  I think it's one of those things where they don't tell you, because if you knew, you could trivialize a number of problems.  It's the same reason that "the man" doesn't teach you Heron's formula until the last week of your geometry class.[/quote]

In addition to this, Engel's book tries to, and very well does, explain the concept, by using what's called the Greedy Algorithm. Basically it explains how organized in same manner maximizes and organized in oppositely manner minimizes. The example that the book gives is as follows:

Four boxes contain 10, 20, 50, and 100 dollar bills. take as much as you want, but you can only take 3 of one, 4 of another, 5 of another, and 6 of another. Which way would give you the most amount? Which way the least?

to do this, we think of as if we were actually doing it. If you are greedy, you would want alot of money. Thus you would take 6 of 100 dollar bills, 5 of 50 dollar bills, 4 of 20 dollar bills, and 3 of 10 dollar bills. To minimize the amounrt, you do the opposite.

From the Rearrangement Inequality you can then derive Chebyshev's Inequality which is also quite powerful at times.


show that
(a+b)/(1+a^2  +b^2)lessthan or equal 1/sqrt(2)

[quote="bgbgbgbg"]show that
$ \frac {a \plus{} b}{1 \plus{} a^2 \plus{} b^2}\leq \frac 1{\sqrt 2}$[/quote]

$ \frac {a \plus{} b}{1 \plus{} a^2 \plus{} b^2}\leq \frac 1{\sqrt 2}$ $ \iff$ $ (a \plus{} b)\sqrt 2\leq 1 \plus{} a^2 \plus{} b^2$ $ \iff$ $ a^2 \minus{} a\sqrt 2 \plus{} \frac 12 \plus{} b^2 \minus{} b\sqrt 2 \plus{} \frac 12 \geq 0$ $ \iff$ $ (a \minus{} \frac 1{\sqrt 2})^2 \plus{} (b \minus{} \frac 1{\sqrt 2})^2 \geq 0$


solve the systeme in R:
[img]http://www.gnux.be/latex/data/995b313e1a27fbc206471b50ab7b32ec.png[/img]
[img]http://www.gnux.be/latex/data/2cb691924ebe8b99739ba8a5deea9715.png[/img][/img]

Are you sure this is right? Because it's kinda weird.

Solving for $ 2z$ and equating, we find that $ \frac{x\minus{}x^2}{y} \equal{} \frac{y\minus{}y^2}{x} \iff x^2 \minus{} x^3 \equal{} y^2\minus{}y^3$

Clearly, $ x\equal{}y$ solves this equation. Using this hint, write as $ x^3\minus{}x^2\minus{}y^3\plus{}y^2\equal{}0 \implies (x\minus{}y)(x^2\plus{}xy\plus{}y^2\minus{}x\minus{}y) \equal{} 0$

We can solve the latter factor using the quadratic formula: $ x \equal{} \frac{1\minus{}y \pm \sqrt{1\plus{}2y\minus{}3 y^2}}{2}$

And, of course, $ z \equal{} \frac{x\minus{}x^2}{2y}$

[quote]Are you sure this is right? Because it's kinda weird.[/quote]
of course this is right, that is my solution:
[code]x^2-y^2+2yz-2xz+x-y=0     <=>    (x-y)(x+y-2z+1)=0
<=>  x=y or x+y-2z+1=0 [/code]
[/code]

peine, the problem with your solution is that you haven't verified that the second constraint actually solves the [i]original[/i] system; in other words, $ x \plus{} y \minus{} 2z \plus{} 1 \equal{} 0$ is a necessary but not sufficient condition for $ (x, y, z)$ to be a solution to the system with $ x \neq y$, or in other other words, you've only shown that part of the solution lies on that plane but you haven't shown which part.

In other other other words, the original system is an intersection of quadric surfaces, so we generically expect it to be a curve.


limit
logarithms
calculus
calculus computations

$ \lim_{x \to \infty}\frac{x^{\ln x}}{(\ln x)^{x}}$

Let $ t = \ln x$. Then

$ \ln \left( \frac{x^{\ln x}}{(\ln x)^{x}}\right) \; = \; \ln^{2}x-x \ln \ln x \; = \; t^{2}-e^{t}\ln t.$

Now you'll see that it diverges to $ -\infty$ as $ t\to\infty$. So the answer is $ 0$.


induction
combinatorics unsolved
combinatorics

$ \binom{r}{0} \minus{} \binom{r}{1} \plus{} \binom{r}{2} \plus{} ... \plus{} (\minus{}1)^r\binom{r}{r} \equal{} 0$

I know this identity can be proved just by doing $ (1\minus{}1)^r \equal{} 0$

But I was wondering if anyone can show me how to prove it using induction and the use of the summation identity.

Thanks!

[quote="Kono"]$ \binom{r}{0} \minus{} \binom{r}{1} \plus{} \binom{r}{2} \plus{} ... \plus{} ( \minus{} 1)^r\binom{r}{r} \equal{} 0$

I know this identity can be proved just by doing $ (1 \minus{} 1)^r \equal{} 0$

But I was wondering if anyone can show me how to prove it using induction and the use of the summation identity.

Thanks![/quote]
$ \binom{r}{0} \minus{} \binom{r}{1} \plus{} \binom{r}{2} \plus{} ... \plus{} ( \minus{} 1)^r\binom{r}{r} \equal{} 0 \Rightarrow$

$ \binom{r}{0} \minus{} \left ( \binom{r}{1} \plus{} \binom{r}{0} \right ) \plus{} \left ( \binom{r}{2} \plus{} \binom{r}{1} \right ) \plus{} ... \plus{} ( \minus{} 1)^r \left ( \binom{r}{r} \plus{} \binom{r}{r\minus{}1} \right ) \plus{} (\minus{}1)^{r\plus{}1} \binom{r}{r} \equal{} 0 \iff$

$ \binom{r\plus{}1}{0} \minus{} \binom{r\plus{}1}{1} \plus{} \binom{r\plus{}1}{2} \plus{} .... \plus{} (\minus{}1)^{r\plus{}1} \binom{r\plus{}1}{r\plus{}1} \equal{} 0$

Here I'm using $ \binom{n}{r} \equal{} \binom{n\minus{}1}{r\minus{}1} \plus{} \binom{n\minus{}1}{r}$


Putnam
analytic geometry
calculus
integration
trigonometry
geometry
projective geometry

Find the volume of the region of points $(x,y,z)$ such that
\[\left(x^{2}+y^{2}+z^{2}+8\right)^{2}\le 36\left(x^{2}+y^{2}\right). \]

Sorry, kunny - I had a typo in the problem. Try again with the corrected problem.

That's pretty obvious, as it's a solid of revolution of some sort (being independent of $\theta$).  Of course on the Putnam you do whatever is most natural to you.

Certainly. I could use this in a second-quarter calculus class.

I did screw it up myself during the test, by making a mistake while solving the equation.

No calculus necessary.  :D
[hide="Solution"]Let $x=r\cos{\theta}$, $y=r\sin{\theta}$.

Our expression becomes:

$(r^{2}+z^{2}+8)^{2}\le 36r^{2}$

We can safely take the square root of both sides since we can let $6r$ be positive. So we have:

$r^{2}+z^{2}+8 \le 6r \Rightarrow (r-3)^{2}+z^{2}\le 1$.

So given an arbritary angle $\theta$, we want all points inside or on a vertical circle of radius $1$ centered at $(3\cos{\theta},3\sin{\theta},z)$. Since $\theta$ is arbritary, we have a torus centered at $(0,0,0)$ having a thickness of $2$ with the center of mass of an arbritary slice at $(3\cos{\theta},3\sin{\theta},z)$.

So the volume of a torus is the area of the cross section of a torus (${\pi}r^{2}$) times the circumfrence of the circle centered at $(0,0,0)$ and containing the center of the cross-section ($2{\pi}R$). In our case, $r$ is $1$ and $R$ is $3$. So:
$\text{Volume}=({\pi}r^{2})(2{\pi}R)=(\pi(1^{2}))(2(\pi)(3)))=\boxed{6{\pi}^{2}}$.[/hide]

[quote="jmerry"]I did screw it up myself during the test, by making a mistake while solving the equation.[/quote]

I forgot the $\theta$ integral :(

I really hope this knocks me down to 8 and not 2.

[quote="isaacchao"]just integrate disk of big circle-small circle over z: -1 -> 1
single integral =)[/quote]

Yes! So did I. :lol:

Cut this solid by the plane $z=k\ (-1\leq k\leq 1)$ or $z=\cos \theta.$

One of my students taking the Putnam is a first-semester freshman currently enrolled in Calculus II (my section, in fact.) He's pretty sure he got this one right - and he said that he felt comfortable with the problem, being that much closer to it.

I unfortunately didn't change to polar coordinates till the end, so I went through a fair amount of tedious algebra first. Even worse, when I wrote out the integral, I forgot that it's symmetric with respect to $z$, so I only found the volume of the top half of it. So I got $3\pi^{2}$....

As with blahblahblah I'm really hoping this is the kind of mistake that costs 1 or 2 points rather than 8 or 9...

[quote="blahblahblah"]In cylindrical coordinates, you want the points $(r,\theta,z)$ with $(r-3)^{2}+z^{2}\leq 1$.  This is straightforward to find (e.g. by a triple integral).[/quote]

what would the limits of your integrands be? :blush:

[quote="1234567890"]No calculus necessary.  :D
...
So the volume of a torus is the area of the cross section of a torus (${\pi}r^{2}$) times the circumfrence of the circle centered at $(0,0,0)$ and containing the center of the cross-section ($2{\pi}R$).
...[/quote]

This surely is a theorem of calculus.  Someone just did you the favor of already having computed the integral.

[quote="Kent Merryfield"]Well, the theorem does have a name, and the name belongs to someone who lived a millenium and a half before there was calculus:

[b]The Theorem of Pappus[/b]

The gist of the statement is this: the volume of the solid of revolution obtained by sweeping a two-dimensional region that lies on one side of a line through space around that line as an axis is equal to the area of the region times the circumference of the circle traveled by the center of mass of the two-dimensional region.

99% of modern mathematicians, if asked to prove this theorem, would use an integral. But it's an old theorem.[/quote]

i am familiar with the theorem but every proof i have seen involves calculus. do you know where i can find a non-calculus based proof?

[quote="kunny"]If my answer is right, Japanese high school students can solve this problem without using multi integral.  :lol:[/quote]

lol, when I saw the equation I recognized that it is a torus simply by inspection; I had memorized the equation of a torus before.  And of course it was darn easy to find the volume after that, no calculus required at all.  This was the only problem I got though.

Here is the solution for Japanese High School Students who don't know Pappus-Guildin's theorem.

Consider the cross section of the solid and the disk $x^2+y^2=r^2\ (r>0)$, we have
$(x^2+y^2+z^2+8)^2\leq 36(x^2+y^2)$
$\Longleftrightarrow (r^2+z^2+8)^2\leq 36r^2$
$\Longleftrightarrow -6r\leq r^2+z^2+8\leq 6r$
$\Longleftrightarrow (r+3)^2+z^2\geq 1\ and\ (r-3)^2+z^2\leq 1$,
thus desired volume can be obtained by finding that of a torus generated by rotating a circle with radius 1 centered on $(3,\ 0)$ in the $r$-$z$ plane about $z$ axis.

Thus desired volume $V=\int_{-1}^1 \pi\{(3+\sqrt{1-z^2})^2-(3-\sqrt{1-z^2})^2\}dz$

$=4\pi\int_{-1}^1 3\sqrt{1-z^2}dz=12\pi\ \text{times the area of semi circle with radius 1}$

$ =\boxed{6\pi ^ 2}.$

[hide="Solution With Calculus"]

First you find the limit of the figure in relation to the coordinate.

Fix $z=0$

$(x^{2}+y^{2}+0+8)^{2} \le 36(x^{2}+y^{2})$ Call it by $x^{2}+y^{2}=t$, we have

$(t+8)^{2} \le 36t$
$t^{2}+16t+64\le36t$
$t^{2}-20t+64\le0$
$(t-4)(t-16)\le0$
$t \le 4$

Knowing that the zone is being $4$ and $16$. Hence, on the $xy$ plane we have two circles.

$x^{2}+y^{2}=2^{2}$ and $x^{2}+y^{2}=4^{2}$

For $x=0$

$(0+y^{2}+z^{2}+8)^{2} \le 36(0+y^{2})$

Extracting the root of both members, we have

$y^{2}+z^{2}+8 \le 6y$
$y^{2}-6y+9+z^{2}+8-9 \le 0$
$(y-3)^{2}+(z-0)^{2}\le1$

So easy to see that we will have a circumference in $yz$, with center at the point $(3,0)$ with radius $1$.

For $y=0$

$(0+x^{2}+z^{2}+8)^{2} \le 36(0+x^{2})$

Extracting the root of both members, we have

$x^{2}+z^{2}+8 \le 6x$
$x^{2}-6x+9+z^{2}+8-9 \le 0$
$(x-3)^{2}+(z-0)^{2} \le 1$

Then the equation generates a torus.

The calculation of the volume can be done by a triple integral constant function $1$, for solids of revolution by a double integral or integral to the surface using the divergence theorem (Gauss theorem) in what is down there:
Note that its parameterization is:

$x=(3+cos(V)) \cos(u)$
$y=(3+cos(v)) \sin(u)$
$z=\sin(v)$ with $v,u$ between $[0,2\pi]$

$V=\int \int \frac{1}{3}( 3+con(v))^{2}.(\cos(u))^{2}.\cos(v) + \frac{1}{2}( 3+con(v))^{2}.(\sin(u))^{2}.\cos(v) + \frac{1}{3} \sin(v)( 3+con(v)).(sin(u))^{2}.\sin(v) + ( 3+con(v))^{2}.(\cos(u))^{2}. \sin(v)))du.dv$

We can also interpret the volume as the area under the circumference times greater.

$V=2\pi 3(\pi)$

So, finally we have the solution, that is $\boxed{V=6\pi^{2}}$[/hide]

Notice that setting $r^2=x^2+y^2$ we get that $r$ lies between $3-\sqrt{1-z^2}$ and $3+\sqrt{1-z^2}$. Hence, the volume of the solid of revolution is given by the definite integral $$\int^{+1}_{-1} \pi \left((3+\sqrt{1-z^2})^2-(3-\sqrt{1-z^2})^2\right) \, dz$$ which is equal to $6\pi^2$.

Setting $r=\sqrt{x^2+y^2} \implies r^2+z^2+8 \leq 6r,$ in cylindrical coordinates.
But this is just a solid torus, so we can apply Pappus' Theorem to get $\boxed{6\pi^2}.$


algebra
polynomial
number theory solved
number theory

It is well known that if the polynomial P with integer coefficients satisfies the property: P(n) is the square of an integer for each natural number n,then P is the square of a polynomial with integer coefficients. But is it true that if P(2^n) is the square of an integer for each natural number n, then P is the square of a polynomial with integer coeficients? If degP=1, we obtain a beautiful problem from polish olympiads. I have also solved the case degP=2, but I don't know how to solve the general problem.

[quote]P(n) is the square of an integer for each natural number n,then P is the square of a polynomial with integer coefficient [/quote]

It is not true for P(x)=x2(x+1)2/4.

Maybe I'm extremely stupid, but isn't your example wrong? The polynomial you indicated doesn't have integer coefficients. Please read the problem again.

Ok, you are right.

What polish problem you mean?

I don't know the year (I think 2001 or 2002), but the problem sounds like this: Let a,b natural numbers with the property that for any natural number n, the number a*2^n+b is the square of an integer. Prove that a=0. I have found a trully wonderful solution. The problem may seem simple, but it is in fact verry difficult. I know lots of people who tried in vain to solve it.

I think you mean following argumentation.

Suppose a<>0.
We have a2n+b=x2, a2n+2+b=y2 ==> 4x2 > y2 > (2x-1)2 for sufficently large n.

In fact, my solution is the following: put xn^2=a*2^n+b. Then , if a is non-zero, x(n+2)-2xn tends to zero when n tends to infinity. So from a certain point we have x(n+2)=2x(n) and from here we have that a and 2a are perfect squares, contradiction.

which is in fact almost the same solution :)


Express $ \frac12$ of $ \frac13$ of $ \frac18$ divided by $ \frac16$ as a decimal.

so this is basically $ \frac12*\frac13*\frac18*6$  and the $ \frac12*\frac13$ cancels out with the $ 6$ and you get simply $ \frac18\equal{}.125$


trigonometry inequalities algebra function domain

Show that for any a, b in (0,:pi:/2) :

sec a + sec b :ge: 2:sqrt:(tan a + tan b)

and determine the equality cases.

[hide]

LHS=1/cosa+1/cosb>=2sqrt(1/(cosacosb))

>=2sqrt(sin(a+b)/(cosacosb))  (since sin(a+b)>=0 in the given domain)

=2sqrt(tana+tanb)

[/hide]

forgot the inequalities cases

[hide]

1/cosa=1/cosb and sin(a+b)=1

hence cosa=cosb and a+b=pi/2

a=b=pi/4

[/hide]

Well done bugzpodder, quick and elegant solution

[hide]When I see a spoiler I can't resist looking in it instead of searching. But I'm sure I'm alone in this case   [/hide]


Billy Bob loves action figures. He gets action figures every day using his action figure making machine. The first day, Billy Bob gets one action figure from the machine. The second day, he gets two. The third day, he gets three, and so on. How many action figures will Billy Bob have on the 100th day?

 :ninja:

this kind of sum is calculated like:
1+2+......+n=n times (n+1)/2


linear algebra
matrix
linear algebra unsolved

Prove that the reduced row echelon form obtained by row reduction of a matrix A is uniquely determined by A.

i only need to prove rightmost single entry of it unique,then it follows by double induction.


Hello everyone ,

I think I will come soon in Norway ,since I won a scholarship at Red Cross Nordic United World College -RCNUWC.

I was just wondering if they will allow me to participate on Norway Mathematical Olympiad , even thought  I'm not a native ??

they would let u participate as far as you are in some officially recognised college of that country

I realize that this is an old topic, but yes, you would be able to participate in the Norwegian mathematical olympiad (Abelkonkurransen). In fact, the person who won the olympiad in 2009 attended a RCNUWC. :) However, you would probably not be eligible to join the Norwegian IMO team even if you did qualify because of IMO regulations.

Thank you, KarlErik. I came late for the first round of Contest, so I couldn't join next rounds. I hope next year I'll come to Trondheim.

P.S: As for IMO, I'm not sure if IMO requires to be native ? We have examples of people who participated at IMO, one year as a member of one country, and other year as a member of other country ! But still, I think it's fair enough not to allow non-natives to join IMO team.


I'm preparing for science olympiad this summer and was wondering what books and websites you find helpful for preparing. Are there books like the Art of Problem Solving books that are for science olympiad? Are there websites like the Art of Problem Solving website that are for science olympiad?

Try looking around the Arts & Sciences section of this website, and try searching for the words "science olympiad" on the forum.

The official Science Olympiad website is soinc.org  Under the events section, it has resources for all of the events.  If you manage to master all of the material they give you for one or more of the events, you should medal ant regionals and most likely at states as well.

You know Science Olympiad has almost an exact website like this one and looks almost exactly like Mathlinks with forums to discuss events and stuff.

Really? All I could find on the site was forums about rule clarifications and suggested events for 2007.

Edit: Never mind, I lied.

When I was preparing for SO, I googled "science olympiad" and got plenty of sites...


Find all function $f: \mathbb Q \to \mathbb Q$ such that for all rationals $x$ and $y$,
\[ \left(f(x)\right)^{f(y)}=\left(f(y)\right)^{f(x)}.\]

[quote="mathmen"]find all function $ f;\mathbb{Q} \to \mathbb{Q}$
such that

$ (\forall x,y \in \mathbb{Q})$; $ f(x)^{f(y)} = f(y)^{f(x)}$[/quote]
The equation $ x^y = y^x$ with $ x,y\in\mathbb Q$ is well known and has the general solution :

either $ x = y$

either $ x = \left(\frac {n + 1}n\right)^n$ and $ y = \left(\frac {n + 1}n\right)^{n + 1}$ where $ n\in\mathbb Z^*$ and, since $ g(n) = \left(\frac {n + 1}n\right)^n$ is injective on $ \mathbb Z^*$, we have the solutions :

1) $ f(x) = c$ constant

2) Let $ A$ and $ B$ a split of $ \mathbb Q$ :
$ f(x) = \left(\frac {n + 1}n\right)^n$ $ \forall x\in A$ and $ f(x) = \left(\frac {n + 1}n\right)^{n + 1}$ $ \forall x\in B$, for any $ n\in\mathbb Z^*$


Determine all solutions of the equation:

 4^x + 4^y + 4^z = u^2 with integers x,y,z and u !

I think the only solutions are : (k, n + k, (2n-1) + k, (1+2*4^(n-1)) * 2^k))  + permutations

Wolog 0 <= x <= y <= z

. If x > 0, then u is even : u = 2*U, then 4^(x-1) + 4^(y-1) + 4^(z-1) = (u/2)^2

So (x,y,z,u) is a solution with x > 0 iff (x-1,y-1,z-1,u/2) is a solution so we
can reduce the problem to x = 0.

. y = z = 0 is clearly not a solution so let z >= 1.

. 4*K + 2 is never a square so let y >= 1.

. Let's prove : if 1 + 4^y + 4^z1 = u1^2 and 1 + 4^y + 4^z2 = u2^2 then z1 = z2

u1 and u2 are odd : u1 = 2*U1+1, u2 = 2*U2+1
if z1 < z2 then 4^z1 (4^(z2-z1) - 1) = 4^z1 (2^(z2-z1) - 1)(2^(z2-z1) + 1) = 4 (U1+U2+1)(U2-U1)
which is impossible because (U1+U2+1) and (U2-U1) can't have the same parity.

. Now 1 + 4^n + 4^(2n-1) = (1 + 2*4^(n-1))^2 so the only solutions are y = n and z = 2n-1

QED (I think)

the problem told you to solve in integers so do not forget the other solution for u:

u = 2^n*(2(2r-1)+1) / -2^n*(2(2r-1)+1)

But nice solution anyway !  :)


There are 100 prisoners. They are told that on each of their heads, there will be either a black hat or a white hat. Each prisoner must in turn guess the color of their hat; if he is right, he is released, and if he is wrong, he must remain in prison. Every prisoner can see the other hats except his own. Before the game starts, they are allowed to work out a strategy.

What is the best strategy for the prisoners? (Note: You cannot use 'cheap' strategies like I scratch my nose means you have a white hat)

[hide="hint"]50 is too easy. Try for 99.[/hide]

classic? yeah, very classic. and the original problem had the prisoners executed unless they guessed right. i see you softened it a little :wink:

they could stand in alternating colors in a line.... maybe

How about everyone is told what their hat is in the plan making stage?

Without requiring them to talk after the plan making stage:

If they see an even number of white hats, go to the left and if they see an odd number go to the right. All the whites will be on one side and all the blacks on the other.

Have the players line up and guess sequentially.  The person who sees all of the other hats guesses "black" if he sees an even number of black hats, "white" otherwise.  The players ahead of him sum the number of "black" guesses they hear and the number of black hats they see, and if the sum's even, they guess black, and if not, they guess white (and they'll be right unless someone screwed up)  Of course, the last player can never be saved in general, so 99 is the best number that can be saved.

the same argument applies for any number of colors

There was a talk at the Las Vegas ARML site focusing on this very problem (with cannibals holding the captives, so losers get eaten).


Έστω τρίγωνο ΑΒΓ και έστω (Ι) ο εγγεγραμμένος κύκλος. Α', Β', Γ', τα σημεία επαφής του με τις πλευρές ΒΓ, ΓΑ, ΑΒ αντίστοιχα. Απο τυχαίο σημείο Κ του κύκλου φέρνουμε την εφαπτομένη του (ε). Οι παράλληλες ευθείες απο το Ι προς τις Β'Γ', Γ'Α', Α'Β' τέμνουν την (ε) στα Α'', Β'', Γ'' αντίστοιχα. Να δειχθεί οτι οι ΑΑ'', ΒΒ' και ΓΓ'' συντρέχουν.

Ρε Βαγγέλη, βλέπω μπήκες γρήγορα στα βαθειά νερά!!! Ωραίος :wink:

Πάντως το πρόβλημα είναι αρκετά δύσκολο (ακόμα περισσότερο αν δεν έλεγες ότι αφορά πολικές :wink: ) και ωραίο!

Θα δώσω απλά 2 hint:
[hide]1)Για να είναι τρεις ευθείες συντρέχουσες πρέπει οι πόλοι τους να είναι συνευθειακοί
2) Ευθεία Simpson[/hide]

Πολύ συνοπτικά η λύση. Με τα hint του mim είναι πανεύκολη...

$ A'C'$ is obviously the polar of $ B$. Let $ B1$ be the foot of the perpendicular from $ K$ to $ A'C'$. So $ B1$ belongs to the polar of $ B$.

Hence, also $ B$ belongs to the polar of $ B1$.

What's more, $ IB''$ is obviously perpendicular to $ KB1$, that means $ KB1$ is the polar of $ B''$. So, since $ B1$ belongs to the polar of $ B''$, then $ B''$ belongs to the polar of $ B1$.

We conclude that $ BB''$ is the polar of $ B1$ and similarly for the rest.

But $ A1, B1, C1$ belong to the Simpson line, hence the polars must coincide.

Έχει ξανασυζητηθεί  :wink:

http://www.mathlinks.ro/viewtopic.php?search_id=85815092&t=218065


function
geometry
algebra unsolved
algebra

(2005. 5) Find all functions $f: \mathbb R \to \mathbb R$ such that \[f(x+y)=f(x)f(y)f(xy)\] for all reals $x$ and $y$.

(2004.4) Find all $f: \mathbb R \to \mathbb R$ such that  \[f(x^2 + y) = 2 f(x) + f(y^2)\] for all reals $x$ and $y$.

hey someone answer these questions please~I would like to learn how to solve them too

it's really interesting that you achieved to find the most unrelated place to write this question. congratulations!

those are questions from the malaysian national mathematics olympiad so i do think its in the correct place to post?ain's it so?anyway how do we solve such questions?HELP!!!

You put problems in respective sections (Algebra,Geometry,...) and whole olympiads with links to problems in this section.

Let $ P(x,y)$ be the assertion that $ f(x\plus{}y)\equal{}f(x)f(y)f(xy)$

$ P(0,0)\rightarrow f(0)\equal{}f(0)^3$
So $ f(0)\equal{}{\minus{}1,0,1}$
$ P(0,x)\rightarrow f(x)\equal{}f(0)f(x)^2$

CASE 1:
$ f(0)\equal{}\minus{}1$
$ f(x)\equal{}\minus{}f(x)^2$
$ f(x)[f(x)\plus{}1]\equal{}0$
This gives us $ f(x)\equal{}0$ and $ f(x)\equal{}\minus{}1$

CASE 2:
$ f(0)\equal{}1$
$ f(x)\equal{}f(x)^2$
$ f(x)[f(x)\minus{}1]\equal{}0$
This gives us $ f(x)\equal{}1$

CASE 3:
$ f(0)\equal{}0$
$ P(0,x)\rightarrow f(x)\equal{}0$

So the only solutions are $ f(x)\equal{}0$, $ f(x)\equal{}1$, and $ f(x)\equal{}\minus{}1$

how would we understand that those are some malaysian olympiad problems if you don't write this? moreover, still the wrong place. just post those problems to algebra section and see how you get helped.

Let $ P(x,y)$ be the assertion that $ f(x^2 \plus{} y) \equal{} 2f(x) \plus{} f(y^2)$

$ P(0,0)\rightarrow f(0) \equal{} 3f(0)\rightarrow f(0) \equal{} 0$
$ P(x,0)\rightarrow f(x^2) \equal{} 2f(x)$
$ P(0,x)\rightarrow f(x) \equal{} f(x^2)$
$ 2f(x) \equal{} f(x)$
$ f(x) \equal{} 0$, $ \forall x\in \mathbb R$

Let $ Z(x,y)$ be the assertion that $ f(xy)\equal{}f(x)f\left (\frac{m}{y}\right )\plus{}f(y)f\left (\frac{m}{x}\right )$, $ \forall x,y\in\mathbb R^\plus{}$

$ Z(1,1)\rightarrow \frac{1}{2}\equal{}f(m)$

So that seems fairly simple, unless I'm missing something...


abstract algebra
group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

How many non-isomorphic finite abelian group are there of the following groups:

(a)5^6
(b)6^5

We don't need to find the groups, just find the number of partition

In part a, it is easy,  p(6) = 11 non-isomorphic abelian groups
part b, 6^5 = (2^5).(3^5), so the number of partition is: p(5)= 7, I am not sure about this, I think there are 7.7 = 49 non-isomorphic abelian groups (or p(5)+p(5)= 14 non-isomorphic groups). We add the number ofpartition together or multiply them

is that correct?

Many thanks

If $ m$ and $ n$ are relatively prime, any abelian group of order $ mn$ can be uniquely decomposed as the direct product of an abelian group of order $ m$ and an abelian group of order $ n$. You get to choose the group of order $ 2^5$, and the group of order $ 3^5$. Two independent choices leads to multiplication; sums in combinatorics come from choosing one from $ A$ [i]or[/i] one from $ B$.

thank you for your reply
so, in part b, the answer is 49 or 14?
when you say, "sums in combinatorics come from choosing one from A or one from B" .What do you mean? I dont understand it, please explain. Cheers


The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:
\[ \text{(A)}\ 1:2 \qquad \text{(B)}\ 2:3 \qquad \text{(C)}\ 2:5 \qquad \text{(D)}\ 3:4 \qquad \text{(E)}\ 3:5 \]

[hide="solution...that i got"]
i get $C$ let's say that we have a square of side $4x$ now, to get the radius of the semicircle, we find that we use half of a side and an entire side, using Pythagoran Theorem, we see that $\displaystyle\ \sqrt((2x)^2 + (4x)^2)\ $ this would give us the radius of $\displaystyle\ \sqrt (20x^2)\ $ then, we see that this would be half of the diagonal of the second square, so that gives us a $45-45-90 $ right triangle with side lengths of $\displaystyle\ \sqrt (20x^2)\ $ and a hypotenouse of $\displaystyle\ \sqrt (40x^2)\ $ so our original square has sides $4x $ and our new second square has sides $\displaystyle\ \sqrt (40x^2)\ $ so $\displaystyle\ \frac {16x^2} {40x^2} $ so that gives us $\displaystyle\ \frac 25\ $
[/hide]

[hide]If we let the the smaller square have side length 2, then its area is 4.  Drawing a line from the center of the circle, also on the midpoint of one of the edges of the square, to a vertex of the square which is on the circle, we find that this distance is $\sqrt{5}$, using pythagorean theorem. This is also equivalent to the radius of the circle.  Now we need the area of the larger square.  Since the diameter of the circle is equal to the diagonal of the large square, the area of the square is $\frac{(2\sqrt{5})^{2}}{2}=5$.  So the ratio is [b]C[/b], $2:5$.[/hide]


Let $ a,\ b,\ c$ be positive real numbers such that $ abc\equal{}1$. Find the possible greatest value of $ M$ satisfying the following inequality:

$ (a^2\plus{}b^2\plus{}c^2)(a^2\plus{}b^2\plus{}c^2\plus{}3)^2\left\{\frac{b}{(a^2\plus{}1)^2}\plus{}\frac{c}{(b^2\plus{}1)^2}\plus{}\frac{a}{(c^2\plus{}1)^2}\right\}\geq M.$

$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} 3)^2\left\{\frac {b}{(a^2 \plus{} 1)^2} \plus{} \frac {c}{(b^2 \plus{} 1)^2} \plus{} \frac {a}{(c^2 \plus{} 1)^2}\right\}$ $ \ge (a^2 \plus{} b^2 \plus{} c^2 \plus{} 3)^2\frac {\left(\frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1}\right)^2}{\frac {1}{a} \plus{} \frac {1}{c} \plus{} \frac {1}{b}}\ge \frac {81}{ab \plus{} bc \plus{} ca}$ (using two times Chaushy- Schwarz and $ \frac {1}{a} \plus{} \frac {1}{c} \plus{} \frac {1}{b} \equal{} ab \plus{} bc \plus{} ca$, when $ abc \equal{} 1$)

and since $ a^2 \plus{} b^2 \plus{} c^2\ge ab \plus{} bc \plus{} ca$, then $ M \equal{} 81$. Equality occurs when $ a \equal{} b \equal{} c \equal{} 1$. :read:

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers such that $ abc \equal{} 1$. Find the possible greatest value of $ M$ satisfying the following inequality:

$ (a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2 \plus{} 3)^2\left\{\frac {b}{(a^2 \plus{} 1)^2} \plus{} \frac {c}{(b^2 \plus{} 1)^2} \plus{} \frac {a}{(c^2 \plus{} 1)^2}\right\}\geq M.$[/quote]
BY AM-GM inequality,we have:
$ LHS \ge\ 81\sqrt[3]{((a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1))^2} \sqrt[3]{\frac{abc}{((a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1))^2}} \equal{}81$
then M=81  :)

Your answers are correct. :)

My solution is exactly same as BG Yoda's solution.

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers such that $ abc \equal{} 1$. Find the possible greatest value of $ M$ satisfying the following inequality:

$ (a^2 \plus{} b^2 \plus{} c^2)(a^2 \plus{} b^2 \plus{} c^2 \plus{} 3)^2\left\{\frac {b}{(a^2 \plus{} 1)^2} \plus{} \frac {c}{(b^2 \plus{} 1)^2} \plus{} \frac {a}{(c^2 \plus{} 1)^2}\right\}\geq M.$[/quote]
Hölder and AM-GM:
$ LHS \equal{}  \left ( \sum_{cyc} b^2 \right )  \left ( \sum_{cyc} a^2 \plus{} 1 \right )^2  \left ( \sum_{cyc} \frac{b}{(a^2\plus{}1)^2} \right ) \ge  \left ( b^{3/4} \right )^4 \ge 3^4 \equal{} 81$ :)

Hum, Hölder beats me! I can't use Hölder. :oops:

Respected kunny-san, Holder is a very useful and easy one.... :blush:
I use one form of this inequality quite a lot...
Let $ a_i, b_i$ be nonnegative reals($ i\equal{}1,2,\cdots n$) such that $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$, then we have:
$ \sum a_i b_i\le\left(\sum {a_i}^p\right)^{\frac{1}{p}}\left(\sum {b_i}^q\right)^{\frac{1}{q}}$
With equality iff there exists a constant $ k\in\mathbb{R}$ such that $ a_i\equal{}kb_i \forall i\equal{}1, 2, \cdots n$.
 :blush:
I think doing a lot of exercises will make one capable of applying this simple one......
@kunny-san[hide]私はあなたに失礼して申し訳ありません。[/hide] :(


geometry
search
geometric transformation
homothety

In the figure below point $D$ is a point on the first circle , $DC$ meets the other circle at $E$, and $AF \parallel BE$. Prove that $DF$ is a diametre of the left circle.

To clear the topic , just try to prove that $KD \parallel LE$ , where $L$ is the second center.Then try again to prove that $KF \parallel LE$ ect.

 babisl

stergiu: I like your geometry problems a lot! When I want a nice geometry problem, I search for your posts  :) .

I found this problem to be unsolved.

Solution:

Let $O_{1}$,$O_{2}$ be the center of the left ($C_{1}$), right($C_{2}$) circles.

Let $O_{1}O_{2}\cap AB=X$. Define a homothety center $X$, $\mathcal{H}_{1}$ that maps $B$ to $A$.  It is well known that $\mathcal{H}_{1}(C_{2})=C_{1}$.

Let $\mathcal{H}_{1}(E)=F'$. We know that $\mathcal{H}_{1}(B)=A$, so $\mathcal{H}_{1}(BE)=AF'$, so $BE//AF'$. But since $C_{2}$ maps to $C_{1}$, $F'$ lies on $C_{1}$, $F'\equiv F$

We also have that: $\mathcal{H}_{1}(O_{2})=O_{1}$, so $\mathcal{H}_{1}(O_{2}BE)=O_{1}AF$.

Hence, $O_{1}F//O_{2}E$

We also know that two tangent circles are homothetic wrt their point of tangency. i.e. Define $\mathcal{H}_{2}$ as a homothety center $C$ such that $\mathcal{H}_{2}(E)=D$. We also have that $\mathcal{H}_{2}(O_{2})=O_{1}$, so $\mathcal{H}_{2}(EO_{2})=DO_{1}$. Hence $EO_{2}//DO_{1}$.

But $O_{1}F//O_{2}E//DO_{1}$, so $D$, $F$, $O_{1}$ are collinear.

Comment: This is a "transformation" approach. I have not read Yaglom's book, but I presume that his solution is along these lines.

We have: $\triangle KAF\sim\triangle LBE\implies\begin{cases}\angle KFA=\angle LEB\\ AF\parallel BE\end{cases}\implies KF\parallel LE$
On the otherhand, $\triangle KCD\sim\triangle LCE\implies\angle KDC=\angle LEC\implies KD\parallel LE$
Hence: $D,\,K,\,F$ are collinear.


A square steel plate is 14.0 cm  on a side and 0.500 cm  thick. Assume that the plate is oriented with the square faces horizontal to the ground.

Find the shear strain that results if a force of magnitude 1.20×106  is applied to the top square face of the plate, parallel to the side, and a force of equal magnitude is applied in the opposite direction to the bottom face of the plate.Note that the shear modulus of steel in pascals is 7.50×10^10 Pa .

Find the displacement  delta x in centimeters

If I remember the definition correctly, the shear modulus is defined as $ \displaystyle S \equal{} \frac {F/A}{x/h}$, where $ x$ is the displacement (or alternatively, $ \tan\phi \equal{} x/h \approx \phi$ in the denominator).

Then you just plug in the given values for $ S$, $ F$, $ h$, and $ A \equal{} (14.0 \mbox{ cm})^2$ to solve for $ x$.  Really, the only hard part about this problem is remembering the definition.

EDIT: Oh wait, maybe that's not a question you need help on, but just something that you posted for other people to think about.


linear algebra
matrix
induction
complex numbers

Let A and B be M_n (C) so that AB=BA. Prove that A^k-B^k=(A-B)(A^(k-1)+A^(k-2)B+..+AB^(k-2)+b^(k-1)) for any k>0.

Obs:
A,B are matrices of order (n,n) of complex elements.

Please help me ! I would really love a detailed proof!!

Thank you very very much!!!!

I don't think it's that hard: the formula a^n+b^n=(a-b)(blablabla) is true for any 2 complex numbers, so if A and B commute any factorization which works for any 2 complex numbers a and b also works for A and B. For example, if A, B, C are in M_n(C) and each 2 commute, then A^#+B^3+C^#-3ABC=(A+B+C)(A^2+B^2+C^2-AB-BC-CA).


algebra
polynomial
trigonometry
algebra unsolved

Find (if there exists) the irreductable polynomial $f \in Q[X]$,  so that $f(cos 70)=0$.

From $\cos (9\cdot 70^\circ)=0$, by expanding $\cos 9x$: $\cos 9x=4\cos^33x-3\cos 3x=4(4\cos^3 x-3\cos x)^3-3(4\cos^3 x-3\cos x)$, we get $0=(4\cos^3 x-3\cos x)[4(4\cos^3 x-3\cos x)^2-3].$ Since the first factor is not zero for $x=70^\circ$ (it equals to $\cos (3\cdot 70^\circ)=-\frac{\sqrt{3}}{2}$), we can cancel it out, and what remains is $f(t)=4(4t^3-3t)^2-3=64t^6-96t^4+36t^2-3$

Note: Similar result can be obtained from $\cos(6\cdot 70^\circ)={1\over 2}$


calculus
integration
trigonometry
complex analysis
calculus computations

Evaluate
$ \int_{0}^{\infty}\cos x^{2}dx \ \ \ \int_{0}^{\infty}\sin x^{2}dx$

[hide]
Same answer to both of them: $ \sqrt{\frac{\pi}{8}}$
[/hide]

Note: I know how to do this with Complex Integration, so if a mod feels this doesnt belong here please move it to Complex Analysis sub-forum.

Your answers are incorrect.  The integrals are both equal to $ \frac{1}{2}\sqrt{\frac{\pi}{2}}$ and are known as Fresnel integrals.

The easy way to do these is with complex analysis, but since this is the advanced calculus forum, (and I know not many people are poking around the complex analysis forum), I will give a proof here for the cosine integral.

This may seem long and contrived, because it is.   :lol:

Let $ x^{2}=y$, then $ \int_{0}^{\infty}\cos x^{2}dx = \frac{1}{2}\int_{0}^{\infty}\frac{\cos y}{\sqrt y}dy$.

This can be written as a double integral (ugh):

$ \frac{1}{2 \Gamma(\frac{1}{2})}\int_{0}^{\infty}\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-y u}\cos y \hspace{1 mm}du dy = \frac{1}{2 \sqrt \pi}\int_{0}^{\infty}\frac{u^{\frac{1}{2}}}{1+u^{2}}du$

Now let $ u^{2}=v$ to get $ \frac{1}{2 \sqrt \pi}\int_{0}^{\infty}\frac{u^{\frac{1}{2}}}{1+u^{2}}du = \frac{1}{4 \sqrt \pi}\int_{0}^{\infty}\frac{v^{-\frac{1}{4}}}{1+v}dv = \frac{1}{4 \sqrt \pi}\frac{\pi}{\sin(\frac{3 \pi}{4})}= \frac{1}{2}\sqrt{\frac{\pi}{2}}$

[quote="edoo"]Your answers are incorrect.  The integrals are both equal to $ \frac{1}{2}\sqrt{\frac{\pi}{2}}$ and are known as Fresnel integrals.[/quote]
Thanks I fixed it.

[quote]
The easy way to do these is with complex analysis, but since this is the advanced calculus forum, (and I know not many people are poking around the complex analysis forum), I will give a proof here for the cosine integral.
[/quote]
An off topic question, why do you think that the Complex Analysis forum is so small :(, since I started reading about it like a month ago it is awesomely fun.

[b]Sylow_Theory wrote:[/b]
[quote]

An off topic question, why do you think that the Complex Analysis forum is so small , since I started reading about it like a month ago it is awesomely fun.

[/quote]

I am not sure why it is so small either.  I thought maybe it had to do with summer vacation, but that effect is not true across the board.  I know I would love to get more practice with complex analysis problems, and I have some good problems to pose too -- unfortunately, I have failed to generate much response with my old problems  :(

That's really a shame too since complex analysis is so useful from an applied standpoint.


Suppose p is a prime.A set M has 2p-1 elements.
Prove that we can find a subset T of M.T has p elements ,the sum of which is a mutiple of p.

Here it is: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28929[/url].


Need some help with this one. Please be detailed and methodic in your understandable explanations because I find this very difficult.

http://a.imagehost.org/0436/Kop.jpg

[hide="Solution"]Ok, first let's try to find some angles, knowing that they add up to $ 180^{\circ}$ in each triangle.

$ \angle CAD \equal{} 180 \minus{} 100 \minus{} 60 \equal{} 20^{\circ}$

Since angles $ ADC$ and $ ABC$ share the same arc, they are congruent, so $ \angle ABC \equal{} 60^{\circ}$.

Since $ AB \equal{} AC$, we know $ \angle ACB \equal{} \angle ABC \equal{} 60^{\circ}$.

Now, we know that $ \angle CAB \equal{} 60$, so $ \angle BAD \equal{} 40$.

Now since $ v$ is the sum of the to angles $ ABC$ and $ BAD$, $ v \equal{} 60 \plus{} 40 \equal{} \boxed{100^{\circ}}$.[/hide]

Note: There are many, many ways to doing this problem, but the key step that was needed was using inscribed angles to find that $ \angle ADC \equal{} \angle ABC$.  After that step, there are lots of ways of finishing the problemIf you want, you can look that up, but it basically means that if two angles with their vertices on a circle share the same arc, they are congruent.

[hide]First of all, since the sum of the angles of a triangle is 180, $ \angle{CAD}\equal{}180\minus{}100\minus{}60\equal{}20^{\circ}$.
Also, ADC and ABC share (smaller) arc AC, so ADC=ABC=60.  However, we know that AB=BC.  Thus, ABC is an equilateral triangle, the only triangle that is isosceles with a 60 degree angle between them.

We have CAD=20 and ACB=60, so the angle made by A,the intersection of AD and BC, and C has degree 180-20-60=100.  Since v is this angle's vertical angle, they are equal, and $ v\equal{}\boxed{100^{\circ}}$[/hide]


Let $A,B\in\mathcal M_{2}(R)$ and $(A-B)^{2}=(A+B)^{2}=I_{2}$. Check if $A,B$ comute.

If they commute then it should turn up that $AB=O_n$..... :?


inequalities unsolved
inequalities
China

for $a,b,c>0$
prove
$1 \leq \sqrt{\frac{a^3}{a^3+(b+c)^3}}+ \sqrt{\frac{b^3}{b^3+(c+a)^3}}+\sqrt{\frac{c^3}{c^3+(a+b)^3}} < \sqrt 2$
maybe it is hard.because noneone have work it out in some month in chinese forum for the right side.

thanks.are you Mr jiang?
but you mistake your country...

[quote="leepakhin"]Why are you thanking youself?  :huh:[/quote]
make a mistake when thanks :P


For a concert, 72 fans attempted to get tickets. That was $ 60\%$
more than the number of tickets available. How many tickets
were available?

Let x be the number of tickets available.

1.6x=72
x=72/1.6=45 tickets available.


Correct me if I'm wrong, but isn't the number of partitions of a number just $\sum_{i=0}^{n-1}\binom{n-1}{i}$?

You mean $2^{n-1}$. No, there is no simple formula for partition of a number.

[quote="#H34N1"]Wait, so is it just $2^{n-1}$? :maybe:[/quote]

This is the answer if you count for example $1+3=4$ and $3+1=4$ as two distinct partitions.

This can be proven easily using strong induction.

[quote="#H34N1"]How do I use generating functions to find the partitions of a number?

1. Find the number of partitions in 7
2. Find the number of partitions in 8
3. Find the number of partitions in 9
4. Find the number of partitions in 10
5. Find the number of partitions in 100
6. Find the number of partitions in 200[/quote]
You can also use recursion by knowing that if $a_{n}$ represents the number of partitions of $n$, then:

$a_{n}=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+......+a_{2}+a_{1}+1$

This can also be proven easily using a constructive method.


Euler
number theory proposed
number theory

Find all integer solutions of $ p^3\equal{}p^2\plus{}q^2\plus{}r^2$ where $ p,q,r$ are primes.

[quote="feliz"] Answer: $ p \equal{} q \equal{} r \equal{} 2$.[/quote] Are you sure?  :lol:

The first if $ p=3$ we have $ (3,3,3)$ is a solution
Now if $ p>3$
If $ q,r\#3$ so $ p^2+q^2+r^2$ divides to $ 3$ thus $ p$ divides to $ 3$ it is contraction
Hence W.L.G we can suppose that $ q=3$
If $ r=3$ so $ p=3$
If $ r>3$
Our equation becomes $ p^2(p-1)=r^2+9$
Because $ p$ is a prime so we must have $ p-1$ divides to $ r^2+9$ or $ p=r^2+9$
And they follow us that $ p^2(p-1)>r^2+9$
We can finish our solution and the answer is $ (3,3,3)$
Am I wrong ?????

[quote="feliz"]Case $ p, q$ and $ r$ are odd, we have $ p^2p \equiv p^2 \plus{} q^2 \plus{} r^2 mod 4 \Rightarrow 1p \equiv 1 \plus{} 1 \plus{} 1 \equiv 3 mod 4$. But $ p|q^2 \plus{} r^2$, which implies $ p|q, r$ if $ p \equal{} 4k \plus{} 3$. Then $ p \equal{} q \equal{} r$. Now we just need parity arguments. Answer: $ p \equal{} q \equal{} r \equal{} 2$.[/quote]
a mistake.if p=q=r we must i.e p^3=3p^2
so p=q=r=3. a application of Fermat-euler theorem

$\pmod 3$ implies that one of $p,q,r=3$
Suppose that $p=3$ then $18=q^2+r^2$ with $(q,r)=(3,3)$
Now if either $q$ or $r=3$ then $p^3-p^2=q^2+9$
By $\pmod 4$ ,$p\equiv 3 \pmod 4$
Choose a prime divisor $p_i|p^3-p^2$ which is $\equiv 3 \pmod 4$
Since if $p|a^2+b^2$ with $p \equiv 3 \pmod 4$ it implies $p|a$ and $p|b$,and we are done since this implies $p^3-p^2=18$ or $p=3$ which is same as before so the only solution is $(p,q,r)=(3,3,3)$

The answer is $(3, 3, 3)$ only. Notice that if $3 \not \in \{p, q, r\}$, then we have mod $3$ issues.

Now if $p=3$, then $(q, r) = (3, 3)$ obviously. Thus let $ p > 3$ and $r = 3$, or $$p^2(p-1) = q^2 + 9.$$ For mod $4$ issues $p \equiv 3 \pmod 4$, then $p \mid q^2+9$ and by Fermat Christmas $p \mid q$, so we get $(3, 3, 3)$ again.


function
inequalities solved
inequalities

Show that $\forall x \in R$ and any $n \in N$  we have:
$ x^{2n}-2nx+2n-1 \geq 0 $.

cheers! :D :D :cool:

The easiest way:
  Take the function f(x)=x^2n-2nx+2n-1. Then f'(x)=2n(x^2n-1-1). So f is increasing for x>1 and decreasing for x<=1. The conclusion follows.


logarithms
inequalities
integration
inequalities proposed

Prove that for every natural number bigger than one like $n$, we have $\log n >\frac{n}{\sqrt[n]{n!}}-1$.

Oh sorry, don't check if the basis is $e$

I get a strange thought that is negative  :D


If $ n$ is an integer let $ f(n)$ be the sum of the square of the decimal digits of $ n$. For example, if  $ n \equal{} 37$ we have $ f(n) \equal{} 32 \plus{} 72 \equal{} 58$.
A integer a being choosen, we study the sequence $ (u(n)) n\geq0$ defined by
$ u_{0} \equal{} a$ and the recurrence relation $ u_{n \plus{} 1} \equal{} f(u_{n})$. Prove that this sequence is eventually periodic (this means that we can find $ T \geq 1$ and $ N \in \textbf{N}$ such
that $ u_{n \plus{} T} \equal{} u_{n}$ for $ n \geq N$).

we just need to find the permutations of the digits of a number so that f(n) remains the same

Consider a positive integer $ n$ having $ k$ decimal digits: $ n \equal{} \overline{d_1d_2\ldots d_k}$. Then $ f(n) \leq 81k < 10^{k\minus{}1} \leq n$ for $ k \geq 4$.

This means the (positive integer) terms of the sequence $ (u_n)_{n\geq 0}$ obey $ u_n \leq \max(a, 10^4)$, therefore a value will repeat, and the sequence turns periodic.


[color=darkblue][b]1.[/b] Consider the triangle $ ABC$ nonisoscelles.The exterior isector for the vertice $ A$ meets $ BC$ in point $ A'$.In the same way obtain the points $ B',C'$.Prove that the points $ A',B',C'$ are collinear.

[b]2.[/b] Let $ ABC$ be an nonisoscelles triangle and let $ A_1=pr_{BC}A$,$ C_1=pr_{AB}C$.$ B_1=pr_{AC}B$.
$ \{\begin{array}{cccc} A'\in BC\cap B_1C_1 \\
\\
B'\in AC\cap A_1C_1\\
\\
C' \in AB \cap A_1B_1 \end{array}$
Prove that the points $ A',B',C'$ are collinear.[/color]

[b]Remark:[/b] [color=brown]Note that $ A_1=pr_{BC}A$ means that $ A_1$ is the projection of the point $ A$ on $ BC$ ![/color]

1. It's know that $ \frac{\overline{BA_1}}{\overline{CA^1}} \equal{} \frac{c}{b}$ and cyclic, so appling menelaus to $ \triangle ABC$ and the trasversal $ A_1B_1C_1$ we are done.
2. True for Desargues on $ \triangle ABC$ and $ \triangle A_1B_1C_1$

very nice proof for the second one :wink:

Remark:
One can solve 1st problem by [b]Desargues[/b] as well: consider the triangle $\triangle I_aI_bI_c$ made by the excenters of $\triangle ABC$; because $AI_a, BI_b, CI_c$ concur at $I$, we may apply [b]Desargues[/b] to the two triangles and are done. ($I_b, I_c$  lay on the external bisector of $\angle BAC$, a.s.o.).

Best regards,
sunken rock


Yes Chatblazer is what the Virtual Classrooms use.

You do NOT need to download any software (provided that you have basic Flash).

Moved to Classes Information.

[quote="arbala"]Hi,

Do we need any software like ChatBlazer for attending a Math Jam?

Thanks in advance.[/quote]

No downloads necessary.  You click Classroom at the time of class, and the classroom loads.

I thought the latest version of Flash was required.

The vast majority of computers have that installed already, and those few that don't are prompted to install it the first time they visit our site.  (I don't think it requires the most recent version of Flash, either.)


[quote="Anood"]Hi,

Can you please help with the following problem:

In an arithmetic sequence the sum of the second term , the forth and the fifth term is 18.and the sum of the first therteen terms equals 13.What is the first term[/quote]

Hint:[hide]
Express the sum in terms of the first term and the common difference, set up the system of equation, and solve.
[/hide]

[hide="if you still dont get the hint"]
We are given:
$t_{2}+t_{4}+t_{5}= 18$
and
$t_{1}+t_{2}+\ldots+t_{1}3 = 13$

We can express each term in terms of the first since tehre is a common difference.
$t_{2}+t_{4}+t_{5}= (t_{1}+d)+(t_{1}+3d)+(t_{1}+4d) = 3t_{1}+8d = 18$

we can use the sum formula for the second since it includes all terms up to 13
$t_{1}+t_{2}+\ldots+t_{1}3 = \frac{13}{2}(2t_{1}+12d) = 13$
which can be simplified to $t_{1}+6d = 1$

now solve the system for both equations and you should find that

$\boxed{t_{1}= 10}$
[/hide]


Let $x,y\in\mathbb{N}$ such that $6 \cdot 3^x - 1 = 5^y$. Whether $x = 0$, then necessarily $y = 1$, making $(x,y) = (0,1)$ a solution. We want to prove it is the only one. So suppose $x > 0$ in the following. You need $5^y \equiv -1 \bmod 3$, hence $y = 2z +1$, for some $z\in\mathbb{N}$. Then $3^{4x} = \sum_{k=0}^{2z} (-1)^k \cdot 5^k$ and $0 \equiv 2z+1 \bmod 3$, i.e. $3 \mid y$. Subsequently $5^y \equiv \pm 1 \bmod 7$, and it's apparent a choice for the negative sign doesn't work. Therefore $3^{4x+1} \equiv 1 \bmod 7$, which is absurd, as $3$ is a primitive root $\bmod\;\! 7$. Now conclude...


linear algebra
matrix
complex numbers
absolute value

Show that a unitary matrix has det U = 1, but possibly det U is different from det U^H, where H is hermitian matrix

That wasn't stated very clearly. Here's what's really true:

Let $U^{*}$ be the complex conjugate of the transpose of $U.$ We call a square matrix $U$ unitary if and only if $U^{*}U=I.$

The problem appears to be to show that $|\det U|=1$ but that we need not have $\det U=\det U^{*}.$

What is it U^H? is UH?

Unitary is $U^{*}=U^{-1}$, det$U$=1, hermitian $H^{*}=H$.

Different matrices with different properties det($UH$)=det($H$), it is different with det($U$)=1 if $H^{*}<>H^{-1}$ :)

A unitary matrix with determinant 1 is a "special unitary matrix". One cannot and should not say that a unitary matrix has determinant 1; the correct statement is that the absolute value of its determinant is 1.

Let $\omega=e^{2\pi i/3}=-\frac12+\frac{i\sqrt{3}}2,$ and let

$U=\frac1{\sqrt{3}}\begin{pmatrix}1&1&1\\1&\omega&\omega^{2}\\1&\omega^{2}&\omega\end{pmatrix}$

Then $U$ is certainly unitary, but $\det U=-i.$

Thanks, by absolute meaning (well known eigen values are 1, by abs. m.,  follows, besides more simple presenter is U=-1) and could be written abs(det($A$)) or |det($A$)|, it's too long, seems observed "special case" was put in statement, shortly. Attention shifted to multiplication or?, less to common look of  factors.


geometry
rectangle
trigonometry
calculus
rotation
cyclic quadrilateral

[quote="Chalao"]Please solve this two problems step-by-step, and very,very,very clear using geometry. I solved them with algebra but it was a very large procedure.

1) Find the lenght of the sides of an equilateral triangle which has its vertex in a circle with radius 1.

2) Find the maximum area of a rectangle which has its vertex in a circle with radius 5.

It is very important to me that you explain it very clear, and step-by-step.

Thank you!! :lol:  :lol:[/quote]

Im am unsure i understand your terminology. Do you mean that the triangle and the rectangle are circumscribed in a circle with aforementioned radii? I will proceed as such.

q1) Construct a triangle with a circumscribed equalateral triangle. Connect each vertex with the centre of the circle (ie the circumcentre). Notice that the 3 newly formed triangles are congruent and hence we have each of the triangles original angles bisected, ie the 60 degree angles will be split up into two 30 degree angles. Taking any one of the three triangles note that due to two sides being radii, then their bases are equal ie 30 degrees. Hence the remaining angle is 120 degrees.

Using the cosine rule,

$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos C$

$ AB^2 \equal{} 1^2 \plus{} 1^2 \minus{} 2\cdot1\cdot1\cdot \cos 120$

$ \equal{} 2 \minus{} 2 \cdot \cos 120$

$ \equal{} 2 \plus{} 2 \cos(180 \minus{} 120)$

$ \equal{} 2 \plus{} 2 \times \frac{1}{2}$

$ \equal{} 3$

Therefore $ AB \equal{} \boxed{\sqrt{3}}$

2) One particular way to solve this problem is to acknowledge the theorem that states the maximum area of a circumscribed rectangle occurs when the sides are equal ie when its a square. This can be proved via calculus. So to solve this problem construct a circle with a circumscribed square and construct the diagonal ie the diameter. Noting that the triangle formed by the diagonal of the square is a 45-45-90 triangle and the hypotenuse is 10,

$ \sin(45) \equal{} \frac{AB}{10}$

$ \therefore AB \equal{} 10 \times \sin(45) \equal{} 5\sqrt{2}$

Hence the side of the square is $ 5\sqrt{2}$ and its area is $ (5\sqrt{2})^2 \equal{} \boxed{50}$

[quote="10000th User"][quote="kurt.math"]This can be proved via calculus.[/quote]I challenge you, or any AoPSer, to prove it without calculus (yes it is possible).[/quote]

We prove that the maximum area of a cyclic quadrilateral given a circle with fixed radius is when said quadrilateral is a square.

The area of a quadrilateral is given by $ \frac{1}{2} pq \sin A$ where $ p$ and $ q$ are the lengths of the diagonals and $ A$ is an angle between them. In a cyclic quadrialteral, each diagonal is a chord of the circle. Since we want the maximal area, from the formula evidently we want the diagonals to be as long as possible, and this is when they are both diameters.

Since $ 0 \le \sin A \le 1$ we have $ 0 \le \frac{1}{2} \sin A\le \frac{1}{2}$ and from the formula we want this part as big as possible, ie $ \frac{1}{2}$ which occurs when $ A\equal{}90^{\circ}$.

So the diagonals of the cyclic quadrilateral of maximal area are perpendicular diameters. Joining the vertices to form the quadrialteral, it is clear from the four congruent 90-45-45 triangles we have that the sides of the quadrilateral are equal and each angle is $ 90^{\circ}$. So it is a square.

[quote="10000th User"][quote="kurt.math"]This can be proved via calculus.[/quote]I challenge you, or any AoPSer, to prove it without calculus (yes it is possible).[/quote] and i will do it without trig too  ;)   :lol:

Let $ BC$ be the diameter of a semicircle with a point $ A$ anywhere on the circumference. To prove that the area of an inscribed circle is greatest when its a rectangle it is sufficient to prove that a triangle in a semicircle (hence right angled triangle) has a maximum area when its sides are equal (besides the hypotenuse), this is due to the fact that we can rotate the semicircle $ 180$ degrees to achieve a circle with an inscribed rectangle.

From the point $ A$ drop a perpendicular $ P$ to the diameter $ BC$. As the point $ A$ (and hence the point $ P$ such as to retain the perpendicularity) moves along the circumference it is obvious that the largest value of $ AP$ occurs when $ P$ coincides on $ O$ (the center of the circle). Hence as $ BC$ is constant the maximum area occurs when $ AP$ is greatest which is when it is the length of the radius (at the point $ O$). [at that point] due to congruency of the triangles $ BAP$ and $ CAP$, $ AB \equal{} AC$, as required.


calculus
integration
algebra theorems
algebra

I've heard that Abel Summation is the discrete analogue to integration by parts, but I've never seen it in a form that recalls the relation, and google gives absolutely nothing.

Try this: If $ A_n\equal{}\sum_{k\equal{}1}^na_k,$ then:

$ \sum_{k\equal{}1}^na_kb_k\equal{}
A_nb_n\plus{}\sum_{k\equal{}1}^{n\minus{}1}A_k(b_k\minus{}b_{k\plus{}1})$

There are several minor tweaks you can give to this formula, and the starting index could vary. The analogy to integration by parts is there: $ a_n,$ which we sum, is analogous to $ dv,$ and the part of $ u,$ which we would have differentiated, is taken by $ b_n,$ which we difference. We soak up the minus sign intrinsic to integration by parts by writing the difference backwards, as $ b_k\minus{}b_{k\plus{}1}.$

Check that this works for $ n\equal{}1$ or $ n\equal{}2.$ If you're trying to reconstruct this from having forgotten it, that's what you should focus on. The hard part is getting the right formula; once you have that, the inductive proof that it is correct writes itself.


Dear Mathlinkers,
remark that the third circle is a Morley's circle.
 http://perso.orange.fr/jl.ayme  vol. 2 Les cercles de Morley...
Sincerely
Jean-Louis

[quote=Potla][quote="shobber"]$ \odot O_1$ and $ \odot O_2$ meet at points $ P$ and $ Q$. The circle through $ P$, $ O_1$ and $ O_2$ meets $ \odot O_1$ and $ \odot O_2$ at points $ A$ and $ B$. Prove that the distance from $ Q$ to the lines $ PA$, $ PB$ and $ AB$ are equal.

(Prove the following three cases: $ O_1$ and $ O_2$ are in the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ and $ O_2$ are out of the common space of $ \odot O_1$ and $ \odot O_2$; $ O_1$ is in the common space of $ \odot O_1$ and $ \odot O_2$, $ O_2$ is out of the common space of $ \odot O_1$ and $ \odot O_2$.[/quote]
This one is quite easy compared to the other China TST problems.
Note that $\triangle O_1PO_2\cong\triangle O_1QO_2\implies \angle O_2O_1P=\angle O_2O_1B.$ Also, $\angle PO_1O_2=\angle PAO_2,$ and $\angle QO_1O_2=\angle BAO_2,$ leading to $\angle PAQ=\angle BAQ.$
So $AQ$ bisects $\angle PAB.$ Similarly $BQ$ bisects $\angle PBA,$ so $Q$ is the incentre of $\triangle PAB.$
So it is equidistant from $PA,PB$ and $AB.\Box$
:)[/quote]

I'm missing something in the solution, why is $ O_1QB $ a straight line?

[b]One line proof:[/b]
It is well known that $A,Q,C_2$ and $C_1,QB$ are collinear and thus, $C_1C_2BA$ is cyclic. Thus, $2\angle APQ=\angle AC_1Q=\angle QC_2B=2\angle QPB$ and $\angle QAB=\angle BC_1C_2=\angle PC_1C_2=\angle PAQ$ completing the proof.


Solve the following equation:
$f'(x)=2xf(2x),~f: R\to R$

[quote="cckek"]It's trivial, only f=0 is solution.[/quote]

$f=\sum_{n=0}^\infty \frac{(2n-1)!!}{2n!}2^{n^{2}}x^{2n}$

V.V., what is the radius of convergence of that power series? [This affects whether you can quote it as a solution to a differential equation.]

Yeah, quite this was my Taylor serie too, unfortunatelly has radius 0.


Can anyone do diabolo?  Refer to http://www.artofdiabolo.com and go see some of the movies.  Or you can go to http://www.diabolotricks.com to see different tricks.  I wanna see how many can do diabolo and give me some tips on like doicides or infinite suicides.  Thanks.

i'm a decent beginner at diaboloing (sp.?) i can do grinds, climb the rope, throws, catches, all that simple stuff. i'm never good at knots and cradles though. :( i always end up inflicting myself with lots of pain :rotfl:

ur happy that you're inflicting urself with pain?  :huh: I can do waterfall, inverse trapezee, right/left/infinite trapezees, chinese suicide, 3 point sun, spagetti, half spagetti into whip catch, trapezee cradlee (once.....) I'm working on my around the leg orbit and the suicide attempts (they HURT :( )  Anyone else do diabolo?

elvenchamp777

no, i find inflicting myself with pain to be amusing


actually, this is a problem from a book
i have slightly modified it
evaluate:
$ (2(2(2(2(...2(2(k \minus{} 1) \minus{} 2) \minus{} 4) \minus{} 8) \minus{} ...2^{n \minus{} 1})) \minus{} 2^n \equal{} 0$
find n

(n and k are positive integers)

[hide="Solution"]
Define a sequence for the inner term in the parenthesis. $ a_1 \equal{} 2(k\minus{}2)$. And then, $ a_d \equal{} 2(a_{d\minus{}1}) \minus{} 2^d$. So, I claim that $ a_d \equal{} 2^d(k\minus{}d\minus{}1)$. We'll prove that using induction. The base case (d=1) obviously works. Now let's look at the induction step. Assume it works for $ a_{d\minus{}1}$. Therefore,

\[ a_d \equal{} 2(2^{d\minus{}1}(k\minus{}d)) \minus{} 2^d\]
\[ a_d \equal{} 2^d(k\minus{}d) \minus{} 2^d\]
\[ a_d \equal{} 2^d(k\minus{}d\minus{}1)\]
And the induction is done. Using that fact, we can see that the stuff in the parenthesis on the LHS is $ a_{n\minus{}1}$. So we now know that:

\[ 2^{n\minus{}1}(k\minus{}n\minus{}1) \minus{} 2^n \equal{} 0\]

Let us divide both sides by $ 2^{n\minus{}1}$, which is valid because that can never be equal to 0. We see that:

\[ k\minus{}n\minus{}3 \equal{} 0\]

And we see that n = k-3.

[/hide]


function
analytic geometry
graphing lines
slope

The function $ f$ is linear and satisfies $ f(d\plus{}1)\minus{}f(d) \equal{} 3$ for all real numbers $ d$. What is $ f(3)\minus{}f(5)$?


abstract algebra
Euler
number theory unsolved
number theory

get the last three digits of $ {{2009}^{2008}}^{2007}$

plz keep it simple explain elaborately...not very well versed with congruences just have a basic idea

Note that $ 2009^{50} \equiv 1 (1000)$, then $ 2009^{2008^{2007}} \equiv 2009^{2008*2007} \equiv 2009^{56} \equiv 2009^{6} \equiv 9^{6} \equiv 441 (1000)$.
A question, how is related your post with the chinese remainder theorem?

have a look at this....

then tell me who's right...

u'll get the answer here...
http://forum.100percentile.com/iitjee_aieee_forums/jforum.htm;jsessionid=0401B17DBC622C360BFC7046D6C819EC.100p_jserv1?module=search&action=search&match_type=all&search_keywords=chinese+remainder+theorem

I have two things to say:
1) I thought the problem was with $ (2009^{2008})^{2007}$ not $ 2009^{(2008^{2007})}$ because of that the result is different, have you even thought about what I wrote?
2) This is not about who is right and who is not, I was just making a question, did you get angry? it was not my intention, I'm sorry

Well, I'm gonna explain it better know and I apologize.....
$ \phi(1000) \equal{} \phi (2^{3}5^{3}) \equal{} \phi(2^3)\phi(5^3) \equal{} 400$, Well, now what do we know? By Fermat-Euler we know that $ a^{\phi(n)} \equiv 1 (modn)$ and then $ 2009^{\phi(1000)} \equal{} 2009^{400} \equiv 1(mod 1000)$. By the other hand we have seen that $ 2008^{2007} \equal{} k.400 \plus{} 352$ for some $ k \in Z$.
Then we have $ 2009^{2008^{2007}} \equal{} 2009^{k.400 \plus{} 352} \equal{} 2009^{k400}2009^{352} \equiv 2009^{352} \equiv 9^{352} (mod 1000)$. But we know form past posts that $ 9^{50} \equiv 1(mod1000)$ and then $ 9^{352} \equal{} 9^{7*50}9^2 \equiv 9^{2} \equiv 81 (mod 1000)$.

Apart I explain the two conguences you asked,

$ 2008^{2007} \equiv 8^{2007} \equiv 2097142 \equiv 352 (mod 400)$.

By the other hand, $ 9^{400} \equiv 1 (mod 1000)$ (By Fermat-Euler again), then $ (9^{50})^{8} \equiv 1 (mod 1000)$, but the only solution of the equation $ x^{8} \equiv 1 (mod 1000)$ is $ x \equiv 1 (mod 1000)$ and the result follows.

well i'm not angry in any manner

u shud hav read this...

[quote]plz keep it simple explain elaborately...not very well versed with congruences just have a basic idea[/quote]

anyways...few things

first plz check that <totient> 1000=400 nd not 800
second u shudn't have expected ME to think so wisely to point out ur mistake in understanding the problem
third now since i know almost nothing...so my next question ought to be to explain the things which U can SEE "easily"

...how did u get ...(i mean claculate)

$ {2008}^{2007}\equiv352(mod\;1000)\\and\;similarly\\
9^{50}\equiv1(mod\;1000)$

get me the full steps plz

sorry for asking another silly question...but unfortunately u hav tried to explain me the things which are the only ones i hav understood in congruences...

its my mistake for asking the question so subjectively i shud hav been more to the point...

anyways thanks for spending so much time trying to explain me the whole thing..it has helped me to strengthen my concept...

what i wanted to ask is how did u calculate "352 " and how did u get that $ 9^{50}$ yes..i mean how "50" is 1 mod 1000
thanks in advance..hope this won't take much of ur time...

[b]sfsreporter[/b] no problem friend  :D but don't you have a msn mail? so we cant talk there, because it is getting to long, and I don't understand your last question, we can talk better in that way

only yahoo and gmail..  same id for both contact me on any i'm online on both

let's talk by gmail, alejnaser@gmail.com  :lol:


A dam is holding back a lake 175 m deep. What pressure must the dam be able to withstand? (Answers are given in atm)

Is the answer supposed to be the hydrostatic pressure at the bottom of the lake?


quadratics
algebra
function
domain
polynomial
Gauss
complex numbers

Help me with this:

$ \begin{array}{l}
 10x^2  \plus{} 7x \plus{} 4 \\
 \\
  \equal{} \frac{{ \minus{} b \pm \sqrt {b^2  \minus{} 4ac} }}{{2a}} \\
\\
  \equal{} \frac{{ \minus{} 7 \pm \sqrt {7^2  \minus{} 4(10)(4)} }}{{2(10)}} \\
\\
  \equal{} \frac{{ \minus{} 7 \pm \sqrt { \minus{} 111} }}{{20}} \\
 \end{array}$
In the past I was thinking this has no solution.
But now I see a strange solution that I never see in fact.
$ \frac{{ \minus{} 7 \pm i\sqrt {111} }}{{20}}$
How this come and what is $ i$?

If you were looking for real solutions, then you're right when you say there's no solutions. But if you extend the domain to so-called complex numbers, two solutions do exist.

So the solution you have is an example of a complex number

http://en.wikipedia.org/wiki/Complex_number

where i denotes the imaginary unit

http://en.wikipedia.org/wiki/Imaginary_unit

$ i$ is defined as the non-real number that is the solution to the equation $ x^2\equal{}\minus{}1$. (To put it simply, $ i\equal{}\sqrt{\minus{}1}$).
From this, we get that $ i^2\equal{}\minus{}1$, and therefore negative square roots become acceptable (they're obviously not defined in the real numbers).

Thanking you all guys.
Best wishes to you.  :)

[quote="KMST"][quote="hasan4444"]So we can simply say that there is no real solution but there is an imaginary one.  :?:   :D[/quote]There is no real solution but there [b][i]are two[/i][/b] imaginary one[i][b]s[/b][/i].[/quote]
Oh yes I forget to say it.

[quote="hasan4444"]Help me with this:

$ \begin{array}{l} 10x^2 + 7x + 4 \\
 \\
= \frac {{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
 \\
= \frac {{ - 7 \pm \sqrt {7^2 - 4(10)(4)} }}{{2(10)}} \\
 \\
= \frac {{ - 7 \pm \sqrt { - 111} }}{{20}} \\
\end{array}$
In the past I was thinking this has no solution.
But now I see a strange solution that I never see in fact.
$ \frac {{ - 7 \pm i\sqrt {111} }}{{20}}$
How this come and what is $ i$?[/quote]

Your polynomial, $ 10x^2 + 7x + 4$, is what is known as an irreducible quadratic polynomial. An irreducible quadratic is irreducible because it has no real roots, and therefore it cannot be factored. It cannot be factored because negative numbers do not have square roots, i.e. no (real) number when multiplied by itself results in a number less than zero. Thus an expression like $ \sqrt{-111}$ makes no sense.

That numbers such as $ \sqrt{-111}$ do not exist on the real number line notwithstanding, such objects do occur frequently in practice. Since they do indeed occur, Carl Friedrich Gauss decided to make a study of them, just as he made a study of almost every other thing in mathematics. In order to study these roots of negative numbers, Gauss invented a number $ i$, that he defined as ${ i\stackrel{\mathrm{def}}=\sqrt{-1}}$. He called it $ i$ because $ \sqrt{-1}$ is [i]imaginary[/i].

All irreducible quadratics do have roots, but as you discovered in your example where $ x =\frac{-7\pm i\sqrt{111}}{20}$, the roots are partly real and partly imaginary. It turns out that numbers that are partly real and partly imaginary are themselves a field of numbers. This field is called the [i]complex numbers[/i] because such numbers are neither entirely real nor entirely imaginary but an admixture of both.

You can't place an imaginary number on the real number line, but you can plot a complex number on a plane. A complex number can be represented by an ordered pair. In other words, if you have a complex number
\[ z = x + iy\]
you can represent it by the ordered pair $ (x, y)$. Then obviously you can plot the point $ (x, y)$ on a graph.

I see.
So the complex number is the number that half real and half imaginary.  :lol:
Thanks a lot "Dr. No".


Let $ p$ be a prime number greater than $ 5$ for any integer $ x$ defined :
          $ f_p(x) \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}{\frac {1}{(px \plus{} k)^2}}$
Prove that for all positive integers $ x$ and $ y$ the numerator of $ f_p(x) \minus{} f_q(y)$ , when written in lowest terms , is divisible by $ p^3$

Just a generaliation (don't know how to proove)

Let $ f_p^n(x)\equal{}\sum _{k\equal{}1}^{p\minus{}1} \frac{1}{(k\plus{}p x)^n}$

Then $ \left(Numerator\left( \sum _{k\equal{}1}^{p\minus{}1} \left(\frac{1}{(k\plus{}p x)^n}\minus{}\frac{1}{(k\plus{}p y)^n}\right)\right)\equiv 0\right) \bmod p^{\frac{1}{2} \left((\minus{}1)^n\plus{}5\right)}$

And the minimum $ p$ for a given $ n$ varies randomly

[i]You can find the solution of this problem and many other nice ones in MOSP in that file[/i] :lol:


[quote="amirhtlusa"]$a=\frac{13}{4}$
$b=\frac{10}{3}$[/quote]
Solution?
I am puzzled by this question, logs scare me :?

[quote="shyong"]What do you mean by log10(2) ? Is it $\log_{10}2$ or $\log 10\times 2$ ? :?[/quote]$\log_{10}2$

first i'm assuming by log10(2) you mean $\log_2{10}$.(if not, then the same solution applies).

notice that
$\log_2{10}=\frac{1}{\log{2}}$.
$1024>1000\Rightarrow 10\log{2}>3\Rightarrow \log{2}>\frac{3}{10}\Rightarrow \frac{1}{\log{2}}<\frac{10}{3}=b$

$2^{13}<10^4\Rightarrow  13\log{2}<4\Rightarrow \log{2}<\frac{4}{13}\Rightarrow  \frac{1}{\log{2}}>\frac{13}{4}=a$

well if you mean $\log{2}$
then its easier.

$a=\frac{3}{10}$
$b=\frac{4}{13}$

[quote="AntonioMainenti"][quote="PenguinIntegral"]If one uses only the information that 10^3=1000, 10^4=10000, 2^10=1024, 2^11=2048, 2^12=4096, and 2^13=8192, what is the largest a and smallest b such that a<log10(2)<b[/quote]There is technically no largest $a$ or smallest $b$, which is why the actual problem asks for the largest $a$ and smallest $b$ that you can prove with the given information. [hide]You have \[10^3<2^{10}<2^{11}<2^{12}<2^{13}<10^4,\] so, since $\text{log}$ is strictly increasing, \[\log10^3<\log2^{10}<\log2^{11}<\log2^{12}<\log2^{13}<\log10^4\] \[\Longrightarrow3<10\log2<11\log2<12\log2<13\log2<4.\] Then $\frac{3}{10}<\log2<\frac{4}{13}$.[/hide][/quote]
Thank you!


abstract algebra
category theory
advanced fields
advanced fields theorems

let $ X$ be a scheme; a $ \mathcal{O}_X$-module $ M$ is called quasi-coherent if it is a sheaf and for every open affine $ U \subseteq X$ there is a $ \mathcal{O}_X(U)$-module $ N(U)$, whose associated $ \mathcal{O}_U$-module is isomorphic to $ M|_U$, that is $ M|_U \cong \widetilde{N(U)}$. obviously, we may assume $ N(U) = M(U)$.

I want to show that if $ U$ is quasi-coherent $ \mathcal{O}_X$-submodule of $ M$, then $ M/U$ is also a quasi-coherent $ \mathcal{O}_X$-module (which is easy according to the author of my book). the problem is that the presheaf $ (M/U)_{pre}$, defined by $ V \mapsto M(V)/U(V)$, does not seem to be a sheaf in general, so that we have to consider $ M/U$ as the garbification of $ (M/U)_{pre}$. since garbification is a functor which preserves products, one easily extends the module structure $ \mathcal{O}_X \times (M/U)_{pre} \to (M/U)_{pre}$ to a module structure ${ \mathcal{O}_X \times M/U \to M/U}$. but why should it be quasi-coherent? this is only clear to me if $ (M/U)_{pre}$ happens to be a sheaf on affine parts of $ X$.

cleared (since $ (\widetilde{P} / \widetilde{Q})_{pre} \equal{} \widetilde{P} / \widetilde{Q} \cong \widetilde{P/Q}$ is known).


I read somewhere before that in an outline, each "sub-level" must have at least two points.  Like this:

[code]I. Text
     a. Text
          i. Text 1
          ii. Text 2
[/code]

And that something like this:

[code]I. Text
     a. Text
          i. Text 1
[/code]

Is wrong.  The difference is that the third level, [b]i[/b]/[b]ii[/b], has two points in the first example but only one in the example.  Can anyone verify that the first example is correct and that the second is wrong?  Thanks!

Nothing is ever "wrong."  The point of this convention is that if you only have one point in a sub-level, there's no reason not to make that point one level higher.  And outlines just look nicer if the sub-levels (or sub-sub-levels, etc.) are filled out.

Edit:  Unless you mean "wrong with respect to a specific convention," in which case, what convention?

[quote="t0rajir0u"]The point of this convention is that if you only have one point in a sub-level, there's no reason not to make that point one level higher.[/quote]

Would it be a good reason to have one point in a sub-level if I'm using it to give one example of a statement of a higher level?

[quote="t0rajir0u"]Edit:  Unless you mean "wrong with respect to a specific convention," in which case, what convention?[/quote]

No, I wasn't referring to any specific convention.

Thanks!

[quote="elcric"]Would it be a good reason to have one point in a sub-level if I'm using it to give one example of a statement of a higher level?[/quote]
Try to give two examples if you can.  It just looks better, at least in my opinion.


induction
combinatorics unsolved
combinatorics

A graph has $n$ vertix .Prove that:$\chi(G)+\chi(\overline{G}) \leq n+1$

induct on $n$. If $G=K_n$ then the result is best possible.if $G\neq K_n$ then in a minimal coloring of $G$,some monochrome class has atleast $2$ vertices.So there exists some vertex $v$ such that $\chi(G)=\chi(G\setminus\{v\})$.
By induction $\chi(G\setminus\{v\})+\chi(\overline{G\setminus\{v\}})\leq n$, but $\chi(\overline{G})\leq \chi(\overline{G\setminus\{v\}})+1$ and so we are through.


Let $m \ge 2$ be an integer. Find the smallest integer $n>m$ such that for any partition of the set $\{m,m+1,\cdots,n\}$ into two subsets, at least one subset contains three numbers $a, b, c$ such that $c=a^{b}$.

The answer is $ m^{m^{m\plus{}2}}$.

Consider the set $ S\equal{}\{m,m\plus{}1,\ldots,m^{m^{m\plus{}2}}\}$. Assume that there is a partition of $ S$ into two subsets $ A$ and $ B$ such that no subset contains three number $ a,b,c$ for which $ c\equal{}a^b$. Without loss of generality, assume $ m\in A$, thus $ m^m\in B$, $ (m^m)^{m^m}\equal{}m^{m^{m\plus{}1}}\in A$, $ \left(m^{m^{m\plus{}1}}\right)^m\equal{}m^{m^{m\plus{}2}}\in B$. Note that $ \left(m^m\right)^{m^{m\plus{}1}}\equal{}m^{m^{m\plus{}2}}$ so $ m^{m\plus{}1}\in A$. But then $ m,m^{m\plus{}1},m^{m^{m\plus{}1}}\in A$, contradiction.

Now consider the set $ \{m,m\plus{}1,\ldots,m^{m^{m\plus{}2}}\}$ which is partitioned into $ A\equal{}\{m,m\plus{}1,\ldots,m^m\minus{}1\}\cup\{m^{m^{m\plus{}1}},\ldots,m^{m^{m\plus{}2}}\minus{}1\}$ and $ B\equal{}\{m^m,\ldots,m^{m^{m\plus{}1}}\minus{}1\}$. We have $ m^m>m^m\minus{}1$, $ \left(m^m\minus{}1\right)^{m^m\minus{}1}<\left(m^m\right)^{m^m}\equal{}m^{m^{m\plus{}1}}$, $ \left(m^{m^{m\plus{}1}}\right)^{m^{m^{m\plus{}1}}}>m^{m^{m\plus{}2}}\minus{}1$, so there cannot be $ a,b\in A$ for which $ a^b\in A$. Also for all $ c,d\in B$, $ c^d\ge\left(m^m\right)^{m^m}>m^{m^{m\plus{}1}}\minus{}1$, so $ c^d\not\in B$. Our proof is now complete.


Four of the ordered pairs listed are solutions of the same linear equation and one is not. Give the letter of the one that is \textit{not}. \[ A(0,\minus{}5),\;B(3,\minus{}4),\;C(\minus{}6,\minus{}1),\;D(\minus{}9,\minus{}8),\;E(6,\minus{}3)\]

$ \fbox{C}( \minus{} 6, \minus{} 1)$ is not since it doesn't lie on the line $ y \equal{} \frac {1}{3}x \minus{} 5$.


Why is cyclopentadiene unstable? What compound does it form on long standing?(answer with stereochemistry ofcourse)

Can the answer be bcos its very basic and wants to form its anion?  :maybe:

Or maybe i am blabbering and the answer is a diels alder reaction  :maybe:

something like a dianion? Like in Ferrocene?  :maybe:

Like shridhar style:
[hide="W"][hide="R"][hide="O"]wrong answer[/hide][/hide][/hide]

Sorry couldnt resist but answer is here  :P   :D
http://en.wikipedia.org/wiki/Cyclopentadiene

or more precisely here http://en.wikipedia.org/wiki/Dicyclopentadiene

Cyclopentadiene is unstable in the sense that it is a very reactive diene in cycloaddition reactions: that's why in a bottle of this compound you don't have really cyclopentadiene, but instead the dimer resulting from Diels-Alder reaction with itself. Therefore, in order to use cyclopentadiene in synthesis one have to distill it first i  order to revert the dimer to the monomer. And by the way, cyclopentadiene is slightly acidic, not basic.

[quote="madness"]is it exo as the rn proceeds for a long time n hence thermodynamic prdt is formed  :maybe:[/quote]
actually i think it forms the endo product

see the link in wiki abt diels alder, it says long time rns will favour exo atleast that's what i interpreted it to say :maybe:

it doesnt say so and im pretty sure its endo, coz of the endo addition rule.

sorry fer double posting. actually your second link says that it is endo

i dont think so! i think it says on further dimerisation only, its endo  :)


Jack spent one-half of his money for a movie ticket, two-thirds of the remainder for popcorn, and the remaining $2 for a drink.  How many dollars did he spend for the movie ticket?

the main part was to read the question carefully and take half of his money for the movie ticket

Well, you don't need to take half the money really, if you do it the easiest way by working backwards (0 -> 2 -> 6)

yes, thats another way, i always get confused with working backwards even though all you have to do is execute the opposite. but meh

Well it does make the problem a lot easier, especially if its a lot longer than this one, and it means you don't need any algebra..
For example, if it said someone spent half their money on this then 1 dollar on that then 2/3 the remainder then 2 dollars then ....

Then backwards is the easiest :)

[color=cyan]indeed, but on the some problems it wouldnt take [i]that[/i] much longer, but still, backwards is goodly[/color]


The triangular numbers are the numbers 1,3,6,10, ..., where the nth
triangular number is the sum of the first n positive integers. The square
numbers are the numbers 1,4,9,16,..., where the Kth square number is the kth
positive integer squared.  What is the sum of the first three positive
numbers that are both triangular numbers and square numbers?

[hide]
Trianglular numbers are of the form $\frac{n(n+1)}{2}$ the first three values that this is a perfect square is n=1, 8, 49 which yields the triangular numbers 1, 36, and 1225.  The sum of these is 1262. :D
[/hide]

Good problem.

[hide]For $n(n-1)=2m^2$, and if m is to be even, than one of n and n-1 must be either divisible by 4 and the other divisible by 2, or one must be divisible by 8, since we have to have n(n-1) divisible by 2 and then be divisible by 4 so as to allow m to be even. The first is not possible since n and n-1 are consecutive integers.

Also, for n(n-1) to be a perfect square and m to be even, the member that is NOT divisible by 8 must be a perfect square. This works where n=9 and m=6, and probably some other cases also.

If m is odd, then one of n and n-1 must be even. But since they are consecutive integers, this is obviously true. Also, for n(n-1) to be a perfect square, then the one that is not divisible by 2 must be a perfect square, and the other(the one divisible by 2) must be a perfect square multiplied by 2. This works where n=1 and n=49, among others.

So the desired sum is (1+36+1225), which is 1262.[/hide]

Billy


It is not hard to prove by S.O.S method  :wink:
\[ \frac {a^2 \minus{} bc}{3a \plus{} b \plus{} c} \plus{} \frac {b^2 \minus{} ca}{a \plus{} 3b \plus{} c} \plus{} \frac {c^2 \minus{} ab}{a \plus{} b \plus{} 3c}\ge 0
\]

\[ \iff a^4 \plus{} b^4 \plus{} c^4\sum ab(a^2 \plus{} b^2) \plus{} abc(a \plus{} b \plus{} c)\ge 4(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)
\]

\[ \iff S_a(a \minus{} b)^2 \plus{} S_b(b \minus{} c)^2 \plus{} S_c(c \minus{} a)^2\ge 0
\]
which is:
\[ S_a \equal{} a^2 \plus{} b^2 \minus{} c^2 \plus{} 4ab
\]

\[ S_b \equal{} b^2 \plus{} c^2 \minus{} a^2 \plus{} 4bc
\]

\[ S_c \equal{} c^2 \plus{} a^2 \minus{} b^2 \plus{} 4bc
\]
Assume that $ a\ge b\ge c$, easy to check out that: $ S_a,S_c\ge 0,S_a \plus{} S_b\ge 0$. We are done
Another proof for:
\[ \frac {a \plus{} b \plus{} c}{2}\ge \sum \frac {2bc}{2a \plus{} b \plus{} c}
\]

\[ 2(a^4 \plus{} b^4 \plus{} c^4) \plus{} \sum ab(a^2 \plus{} b^2) \plus{} 2abc(a \plus{} b \plus{} c)\ge 6(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)
\]
S.O.S killed it again  :roll:

If use the way to prove by standardize method  of Inequality,   you'll prove it, will not?
1a/
 try make a+b= x, b+c =y, c+a=z
standardize general x+y+z=6
...........

[hide="a"]
The right inequality reduces to
\[ 0\leq \sum_\text{cyc} \frac{2a^2\plus{}2bc\minus{}a(2a\plus{}b\plus{}c)}{2a\plus{}b\plus{}c}\equal{}\sum_\text{cyc} \frac{b(c\minus{}a)\minus{}c(a\minus{}b)}{2a\plus{}b\plus{}c}\]
\[ \equal{}\sum_\text{cyc}c(a\minus{}b)\left( \frac{1}{2b\plus{}a\plus{}c}\minus{}\frac{1}{2a\plus{}b\plus{}c}\right)\equal{}\sum_\text{cyc} \frac{c(a\minus{}b)^2}{(2a\plus{}b\plus{}c)(2b\plus{}a\plus{}c)}\]
which works out.

For the left inequality, we want to show the nonnegativity of
\[ \sum_\text{cyc} \frac{(2a\plus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\minus{}4bc}{2a\plus{}b\plus{}c}\equal{}\sum_\text{cyc} \frac{(b\minus{}c)^2\plus{}a(c\minus{}a)\minus{}a(a\minus{}b)}{2a\plus{}b\plus{}c}\]
\[ \equal{}\sum_\text{cyc} \frac{(b\minus{}c)^2}{2a\plus{}b\plus{}c}\plus{}(b\minus{}c)\left( \frac{c}{2c\plus{}a\plus{}b}\minus{}\frac{b}{2b\plus{}c\plus{}a}\right)\]
\[ \equal{}\sum_\text{cyc} (b\minus{}c)^2\left( \frac{1}{2a\plus{}b\plus{}c}\minus{}\frac{a\plus{}b\plus{}c}{(2b\plus{}c\plus{}a)(2c\plus{}a\plus{}b)}\right)\]
After clearing denominators, it's equivalent to dealing with $ \sum_\text{cyc} (b\minus{}c)^2(b^2\plus{}3bc\plus{}c^2\minus{}a^2)$, which is easy enough to deal with by SOS.
[/hide]

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

(a)$ \sum\frac {2bc}{2a \plus{} b \plus{} c}\le \frac {a \plus{} b \plus{} c}{2} \le \sum\frac {a^2 \plus{} bc}{2a \plus{} b \plus{} c}$;
[/quote]

We have
\[RHS-MID=\frac{1}{2}\sum{\frac{c(a-b)^2}{(2b+c+a)(2a+b+c)}}\ge{0}\]

\[MID-LHS=\frac{3\sum{ab(a-b)^2}+\sum{(a-b)^2(a+b-c)^2}}{2(2a+b+c)(2b+c+a)(2c+a+b)}\ge{0}\]

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then
(b)$ \sum\frac {a^2\minus{}bc}{3a\plus{}b\plus{}c} \ge 0$.[/quote]

We have
\[LHS=\frac{2\sum{ab(a-b)^2}}{(3a+b+c)(3b+c+a)(3c+a+b)}+\frac{\sum{(a-b)^2(a+b-c)^2}}{2(3a+b+c)(3b+c+a)(3c+a+b)}\ge{0}\]

[quote=Vasc]If $ a,b,c$ are nonnegative real numbers, then

(a)$ \sum\frac {2bc}{2a \plus{} b \plus{} c}\le \frac {a \plus{} b \plus{} c}{2} \le \sum\frac {a^2 \plus{} bc}{2a \plus{} b \plus{} c}$;

(b)$ \sum\frac {a^2\minus{}bc}{3a\plus{}b\plus{}c} \ge 0$.[/quote]
We have
\[\frac{a+b+c}{2} - \sum \frac{2bc}{2a+b+c} = \frac{\displaystyle 3\sum ab(a-b)^2 + 2 \sum a^2(a-b)(a-c)}{2\displaystyle \prod (2a+b+c)} \geqslant 0,\]
and
\[\sum \frac{a^2-bc}{3a+b+c} = \frac{\displaystyle 2\sum ab(a-b)^2 + \sum a^2(a-b)(a-c)}{\displaystyle \prod (3a+b+c)} \geqslant 0.\]


I have a quick question.

Exactly how quick do the questions get harder/easier?

This is why I ask: In my diagnostic phase, some of the questions were "whats 7c3" so I started doing them in my head and I got a lot of stuff wrong because I was just making stupid mistakes because I was trying to do everything without pencil or paper. I was wondering after the diagnostic phase are you basically stuck in a general range of problems, or can they potentially get up to the "easy-medium AIME level" once you start doing better? I don't want to make a whole new AoPS account just so I can restart Alcumus.

If you begin do to better, you will get harder problems. You will not be stuck in a general range. However, Alcumus is geared towards Intro to Counting & Probability level. There aren't any AIME level problems in Alcumus.

Really?  Why do people claim that alcumus helped them get to an AIME level of c & p?

Often, knowledge becomes stronger and reaches further simply by reinforcement of the foundation.

Here, going through ample practice of slightly-easier-than-AIME-level problems helps promote knowledge of concepts further and highly quickens the thought process when encountering a harder-level AIME Counting & Probability problem. It's just one of those interesting things about learning.

I think I'm on Slightly easier than AIME problems.

I'm sure I saw an Admin post that the hardest problems on alcumus were easy-medium AIME range.


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let $ \omega \equal{}exp(\frac{2i \pi}{n})$

$ p , n \in \mathbb N^*$

prove that $ \sum_{k\equal{}0 } ^ {n\minus{}1} (1\plus{}\omega ^k )^p \equal{}n \sum_ {0 \le k \le \frac{p}{n}} \binom {p} {kn}$

i really need help  :(

A search for "roots of unity filter" on the forum will find much information.  The key idea: if $ \omega$ is a primitive $ n$-th root of unity, what do you know about $ \sum_{i \equal{} 0}^n \omega ^k$?


Let $ f(x) \equal{} a_nx^{n \plus{} 1} \plus{} a_{n \minus{} 1}x^n \plus{} \cdots \plus{} a_1x^2 \plus{} a_0x$.

we're given that $ f(1) \minus{} 1 \equal{} f(2) \minus{} 1 \equal{} \cdots \equal{} f(n \plus{} 1) \minus{} 1 \equal{} 0$. since $ f(x) \minus{} 1$ has degree at most $ n \plus{} 1$, we must have $ f(x) \minus{} 1 \equal{} a_n(x \minus{} 1)(x \minus{} 2)\cdots (x \minus{} n \minus{} 1)$. we also have $ f(0) \equal{} 0$, so $ a_n \equal{} \frac{( \minus{} 1)^n}{(n \plus{} 1)!}$.

now, the $ x$ coefficient of $ f(x)$ (=$ a_0$) is the same as the $ x$ coefficient of $ f(x) \minus{} 1$, which is $ ( \minus{} 1)^n a_n(n \plus{} 1)! \sum_{j \equal{} 1}^{n \plus{} 1} \frac {1}{j} \equal{} \sum_{j \equal{} 1}^{n \plus{} 1}\frac {1}{j}$.