calculus
geometry
3D geometry
symmetry
derivative
sphere
trigonometry

This was my classroom example today - I was trying to review the whole Calculus III course for the final which is next week.

Consider the hemisphere centered at the origin and of radius 1 with $z>0.$ Consider also the cone whose base is the circle of radius 1 in the $x$-$y$ plane centered at the origin and whose vertex is the point $(0,0,2).$

These two surfaces intersect along some curve with $z>0.$ (Note: I intend to ignore what's happening for $z=0.$) What angle do these surfaces make which each other where they intersect?

Since the problem have symmetry we can make simple things.

like. what's the angle that the cone makes with the plane $xy$?

suppose it is $\theta$.

we make a intersection of the 2 equations, of the cone and the sphere.
the intersection is a circle

for simplicity we can set $y=0$

so the problem is analagous to find a angle between a half circle and a triangle

we can just find one intersection, and calculate the derivative of the circle and after the angle $\phi$ that the tangent line makes with the $x$ axis.

i think this is not a hard problem, it's just a basica Calculus I problem, dont is?

I may have been using Calculus III vocabulary in my classroom discussion, but there are many ways to solve it, some involving no calculus at all. But I'd much rather see an actual solution rather than statements that you ought to be able to solve it some way or another.

${x^{2}+y^{2}+z^{2}=r^{2}}$  - A Sphere Equation
${\frac{x^{2}+y^{2}}{r^{2}}=\left(\frac{z-h}{h}\right)^{2}}$ - A Cone Equation

By symmetry we can deduce that the angle is equal, because the intersection is a circle, and the $3$ equations is symmetric to $z$ axis. so we can simplify
the problem setting $y=0$

${\left( \begin{array}{ll}z=0 & x=\pm r \end{array}\right)}$
${\left( \begin{array}{ll}z=\frac{2 h r^{2}}{h^{2}+r^{2}}& x=\pm \left(r-\frac{2 h^{2}r}{h^{2}+r^{2}}\right) \end{array}\right)}$

The first point we don't want, because $z>0$.
Let's derive the both equations. we now are just workin with the $zx$ plane

${\frac{d\left(x^{2}+y^{2}+z^{2}=r^{2}\right)}{dx}\Rightarrow 2 x+2 z \frac{dz}{dx}=0}$
${\frac{d\left(\frac{x^{2}+y^{2}}{r^{2}}=\left(\frac{z-h}{h}\right)^{2}\right)}{dx}\Rightarrow \frac{2 x}{r^{2}}=\frac{2 (z-h) \frac{dz}{dx}}{h^{2}}}$

We set

${\left( \begin{array}{ll}z=\frac{2 h r^{2}}{h^{2}+r^{2}}& x=+\left(r-\frac{2 h^{2}r}{h^{2}+r^{2}}\right) \end{array}\right)}$

the other value have the same effect, because the symmetry.
Now. the angle between the $2$ tangents is opposite the angle you want. if i understand your confuse statement

${\frac{dz}{dx}=-\frac{x}{z}=\left(\frac{h}{2 r}-\frac{r}{2 h}\right)}$
${\frac{dz}{dx}=-\frac{h^{2}x}{r^{2}(h-z)}=\left(\frac{h}{r}\right)}$

${\phi =\pi-\left|\tan^{-1}\left(\frac{h}{2 r}-\frac{r}{2 h}\right)-\tan^{-1}\left(\frac{h}{r}\right)\right|=\pi-\tan^{-1}\left(\frac{r}{h}\right)}$

${\{r=1,h=2\}}$
${\pi-\tan^{-1}\left(\frac{r}{h}\right)=153.43494882292202^{\circ}}$

since we have $2$ angles between the $2$ surfaces i choose the external angle for esthetical purpose.

Here's my answer, in the general case as yagaron was working it. We concentrate on finding normal vectors to the surfaces in question.

Let $\mathbf{a}$ be a normal to the sphere. The sphere is given by the equation $x^{2}+y^{2}+z^{2}=r^{2}.$ That's a level surface for the function $g(x,y,z)=x^{2}+y^{2}+z^{2}.$ To find the normal, we need any convenient scalar multiple of $\nabla g,$ so we take $\mathbf{a}=\langle x,y,z\rangle.$ Note that $\|\mathbf{a}\|=r.$

Let $\mathbf{b}$ be a normal to the cone. We can write the equation of the cone as $z=h-\frac{h}{r}\sqrt{x^{2}+y^{2}}.$ A general form for a normal to a surface given by $z$ as a function of $x$ and $y$ is $\left\langle-\frac{\partial z}{\partial x},-\frac{\partial z}{\partial y},1\right\rangle.$ Computing those derivatives and clearing the denominators, we get $\mathbb{b}=\left\langle hx,hy,r\sqrt{x^{2}+y^{2}}\right\rangle.$ In this case, $\|\mathbf{b}\|=\sqrt{h^{2}+r^{2}}\sqrt{x^{2}+y^{2}}.$

Our angle will be $\cos^{-1}\left(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}\right).$

Now we need to determine the intersection. From the sphere equation, we get that $\sqrt{x^{2}+y^{2}}=\sqrt{r^{2}-z^{2}}.$ Put that into the cone equation to get

$z=h-\frac{h}{r}\sqrt{r^{2}-z^{2}}$

$h-z=\frac{h}{r}\sqrt{r^{2}-z^{2}}$ (Note: I am assuming $h>r.$)

$h^{2}-2hz+z^{2}=h^{2}-\frac{h^{2}}{r^{2}}z^{2}$

$\left(1+\frac{h^{2}}{r^{2}}\right)z^{2}=2hz.$ Ignoring $z=0,$ we get

$z=\frac{2hr^{2}}{h^{2}+r^{2}}.$ From that, $\sqrt{x^{2}+y^{2}}=\frac{r(h^{2}-r^{2})}{h^{2}+r^{2}}.$

So now, $\mathbf{a}\cdot\mathbf{b}=h(x^{2}+y^{2})+rz\sqrt{x^{2}+y^{2}}.$

Inserting the above values for $z$ and $\sqrt{x^{2}+y^{2}},$ we get

$\mathbf{a}\cdot\mathbf{b}=\frac{r^{2}h(h^{2}-r^{2})}{h^{2}+r^{2}}.$

We also have that $\|\mathbf{b}\|=\frac{r(h^{2}-r^{2})}{\sqrt{h^{2}+r^{2}}}.$

Putting this all together, we have that $\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}$ simplifies down to $\frac{h}{\sqrt{h^{2}+r^{2}}}.$

Hence my answer is that the angle (the acute angle) between the surfaces is

$\cos^{-1}\left(\frac{h}{\sqrt{h^{2}+r^{2}}}\right).$

This is the same as yagaron's answer of $\tan^{-1}\left(\frac{r}{h}\right).$

i dont know why people rate you excelente and me below average since my proof is much more smaller simple and beautiful than yours. i used the symmetry of the system to simplificate it.
i made a lot of calculation to you understand what i was doing, but it can be did in only 3 or 4 lines.

dumpt. who understandl. tsk

Two points:
First, nobody has rated your solution yet- only the post you made indicating a method without following through.
Second, you haven't taught math yet. Simplification isn't always a good thing, especially when it comes without much explanation. How would you explain this to an ordinary student?

yeah. if a ordinary student dont understand my explanation they will never understand Kent solutions.

since it is much more harder than mine. uses gradient. etc

how should we call this? Kent paradox?

Easy things are not understood
Hard thigs is well understood

sorry jmerry. but you said some paradoxal things.

Kent talked just like "this problem is harder"

ai said "this problem is not harder"

blargh annoying

i'm just indignate with this. there is no answer for this. my solution was discarted to be too much easy?

im not offesing nobody.

I think we see one of the pitfalls of the post rating system, in that yagaron seems to be reacting as much to the ratings as to the content of the posts. In that, the ratings really added nothing to the discussion. yagaron's first post, the "it's an easy Calculus I problem, isn't it?" post didn't really need to be rated (and I didn't rate it), but it did need to be responded to, exactly as I did: "You say it's easy? Show me a solution." I don't dispute the contention that it could be solved using Calculus I methods. It could also be solved with geometry and trigonometry methods, with no mention of calculus at all.

yagaron then chose to make it a more difficult problem than the one I stated by going for the general case of radius $r$ and cone height $h.$ I had stated it in the specific: radius $1,$ cone height $2.$ The algebra is easier in the specific case, including that you get some 3-4-5 right triangles. This caused me to decide that I needed to solve the general case as well.

The answer yagaron posted is correct, and its methods mostly belong to a Calculus I course. That much is clear. I'll leave to others the question of how easy it was to follow. (I probably wouldn't have used my problem as a lecture example in a Calculus I course, as a matter of balancing the difficulty of the problem against the concepts it would illustrate.)

This caused me to post my solution. There is a matter of intended audience. I did say that I used this (well, the version with specific numbers) as a lecture example in an end-of-semester review for a Calculus III course. Hence the ingredients I chose to use belong to that specific course. In the context of that audience, these ingredients should be familiar, and, indeed, needed to be reviewed: normal vectors to surfaces, the relation between gradients and normal vectors to level surfaces, the cosine of the angle between vectors. If these are not things familiar to the intended audience, then you would need to find other methods and other language.

Kent. i am not fight with you or something else.

you made a really god problem. it's a pleasure for me to solve it.
wasnt easy to me to transforme a 3D problem in a 2D problem.
I'm a physics, so i ever try to use the symmetry of the problem to solve it more easy.

And for me seens that you was sarcastic on your second reply
[quote]
But I'd much rather see an actual solution rather than statements that you ought to be able to solve it some way or another.[/quote]

for me it was like "Oh, this boy is just saying random things"

on my post i didn't solve the problem, but i knew that i could solve in that manner, so i said it.

and jmerry said non-sense, since use Calculus I is more easy than Calculus III because I < III hhe

but let's stop this discution,
my solution is right, your too, and everybody is happy.

and we can generalize your problem :)

for you

Suppose we have a surface $S_{1}$ symmetrical to $z$ axis, we can write it in  simple cylindrical coordinates. and another curve $S_{2}$, in cylindrical coordinates too.

What's the anlge formed by the $2$ surfaces when they intersect?

I like general problems, and this is a good one. sorry for not making $r=1,h=2$ in the previous solution

c ya

[quote="yagaron"]on my post i didn't solve the problem, but i knew that i could solve in that manner, so i said it.[/quote]

And, obviously, whoever rated your post thought this made for a "below average" post.  Part of the problem is also that you were likely responding to a single rating: the problem with a lightly-used rating system is that it leaves the ratings posts in the hands of a very small number of raters.  I think it would be nice if the people who frequent each subforum most, and so really have the best idea of the expected quality of posts, took the time to rate much of what they read.  It would greatly increase the usefulness of the system.

Here's a bad solution with no reference to calculus whatsoever:

We might as well take $r = 1$, $h$ variable.  Take a cut through the plane $y = 0$ -- it's clear that the resulting 2-D problem gives the same result as the question actually asked.  So we have an intersection of the circle $x^{2}+z^{2}= 1$ with the line $hx+z = h$.  To find the point of intersection, we solve $x^{2}+(h-hx)^{2}= 1$, which gives $(h^{2}+1)x^{2}-2h^{2}x+(h^{2}-1) = 0$, which has roots $x = 1$ and $x = \frac{h^{2}-1}{h^{2}+1}$.  It's this second point we are interested in; the $z$-coordinate is $\frac{2h}{h^{2}+1}$.  The slope of the line at this point is the same as its slope at every point, $-h$.  The slope of the circle is the negative reciprocal of the slope of the radius, which is $\frac{-1}{\left(\frac{2h}{h^{2}+1}\right)/\left(\frac{h^{2}-1}{h^{2}+1}\right)}= \frac{1-h^{2}}{2h}$.  Thus we get the answer of $\arctan(-h)-\arctan \frac{1-h^{2}}{2h}$, which I believe is the same value as Kent's and yagaroon's (maybe up to sign?).

Here's a good solution with no calculus whatsoever:

Same set-up.  Draw the perpendicular $p$ from the origin to the line $\ell$ with equation $hx+z= h$.  These two lines together with the radius to the point of intersection define a right triangle, one of whose vertices is the origin.  The angle at this vertex is between the normals to the two curves we are interested in, i.e. it is exactly the angle we desire.  The hypotenuse of this triangle has length 1, and it is easy to see that the leg adjacent to this angle has length $\frac{h}{\sqrt{h^{2}+1}}$ (it's the altitude to the hypotenuse of the right triangle formed by $\ell$ and the two axes).  Then we have immediately that $\cos \theta = \frac{h}{\sqrt{h^{2}+1}}$.

Thank you, Joel. Your second solution is indeed the most elegant of the four that have appeared in this topic. That it wouldn't have served the lecture/review purpose for which I introduced the example in the first place is hardly relevant.


function
calculus
integration
modular arithmetic
quadratics
calculus computations

The Complex Variables response...

Write the sum as:

$ \sum_{n\equal{}\minus{}\infty}^{\infty}\frac{1}{(6n\plus{}1)^{3}}$.

and form the function:

$ f(z)\equal{}\frac{\pi\cot\pi z}{(6z\plus{}1)^{3}}$

and consider its contour integral over the square with vertices:

$ (N\plus{}\frac{1}{2})(\pm 1\pm i)$.

The only real calculation is determining that the residue of $ f(z)$ at $ z\equal{}\minus{}\frac{1}{6}$ is:

$ \minus{}\frac{\pi^{3}}{18\sqrt 3}$.

The rest is obvious.

Can someone come up with a solution using Fourier Analysis? I happen to know (seen many) similar looking sums are nicely solved this way.

All right, Fourier analysis. We'll work with functions periodic of period $ 1,$ and we'll use distributions and weak derivatives.

Let $ f(x)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}\frac{e^{2\pi i(6n\plus{}1)x}}{(2n\plus{}1)^{3}}$

Then $ \widehat{f}(k)\equal{}\begin{cases}\frac{1}{k^{3}},& k\equiv 1\pmod{6}\\0,&\text{otherwise}\end{cases}$

Differentiate three times to get

$ \widehat{f'''}(k)\equal{}\begin{cases}(2\pi i)^{3},& k\equiv 1\pmod{6}\\0,&\text{otherwise}\end{cases}$

Now let's work from the other side to identify $ f'''.$

Let $ g(x)\equal{}\frac{1}{6}\sum_{j\equal{}0}^{5}e^{2\pi ij/6}\delta\left(x\minus{}\frac{j}{6}\right).$

Then $ \widehat{g}(k)\equal{}\frac{1}{6}\sum_{j\equal{}0}^{5}e^{2\pi ij(1\minus{}k)/6}.$

This sum is zero unless $ \frac{1\minus{}k}{6}$ is an integer, in which case it is 1. So we now know that $ f'''(x)\equal{}\frac{1}{(2\pi i)^{3}}g(x).$

That makes it a matter of integrating back three times. $ f'''$ is a collection of point masses; $ f''$ will be a step function, $ f'$ will be a piecewise linear continuous function, and $ f$ will be a quadratic spline. At each stage there will be a constant of integration; at each stage we resolve that constant by the stipulation that the function has mean value zero over a period.


Dennis Phillips just got owned but hes still in 8th place! GO DENNIS!


[quote="maky"]nu inteleg cum vine la 3 acolo....[/quote]
Inegalitatea 3 este adevarata daca si numai daca este verificata pentru orice triplet (x,y,z) , cu x,y,z >= 1 . Observand ca membrul stang este simetric in a,b,c, iar cel drept nu, am scris inegalitatea pentru (c,b,a) si am adunat cele doua inegalitati pentru a obtine si membrul drept simetric in a,b,c .

[quote="maky"]pai nu prea merge asa :)
adica inegalitatea e falsa daca exista un triplet pentru care nu e adevarata. insa poate sa fie adevarata pentru restul tripletelor... daca nu`i simetrica, atunci ghinion :) am facut si eu "smecheria" asta intr`un concurs mai demuuult, si n`am fost prea fericit nici eu cand am aflat ca nu merge asa
ps, felicitari pentru solutiile de la 1 si 2. foarte elegante :)[/quote]
Multumesc  :) . Voi incerca sa gasesc alta solutie.

exact asa am facut si eu dupa ce am citit solutiile tale de la 1 si 2
si acum o mica generalizare
fie $\big(x_{i}\big)_{i=\overline{1,n}}$ numere reale cu proprietatea ca $x_{i}\geq 1$.
atunci are loc relatia :
$\sum\frac{1}{1+x_{i}}\geq\frac{n}{1+\sqrt[n]{x_{1}x_{2}\ldots x_{n}}}$.

[hide="demonstratie"]
pentru inceput voi demonstra pentru $n=2^{k}$, prin inductie.
pentru $n=2$ e mai sus.
apoi, de la $2^{k}$ la $2^{k+1}$ voi trece astfel :
notez $p=\prod_{i=1}^{2^{k}}x_{i}$ si $q=\prod_{i=2^{k}+1}^{2^{k+1}}x_{i}$.
$\sum_{i=1}^{2^{k}}\frac{1}{1+x_{i}}\geq \frac{2^{k}}{1+p}$.
$\sum_{i=2^{k}+1}^{2^{k+1}}\frac{1}{1+x_{i}}\geq \frac{2^{k}}{1+q}$.
$\frac{2^{k}}{1+p}+\frac{2^{k}}{1+q}\geq \frac{2^{k+1}}{1+\sqrt{pq}}$.
adunand primele 2 relatii, apoi analizand impreuna cu a treia, obtin exact ce vroiam.
acum fie $n$ natural, dar nu putere a lui $2$, si fie $k$ minim astfel incat $2^{k}>n$.
atunci aplic ipoteza de inductie pentru $2^{k}$ numere :
$y_{i}=x_{i}$ pt $1\leq i\leq n$.
$y_{i}=\sqrt[n]{x_{1}x_{2}\ldots x_{n}}$ pt $n+1\leq i \leq 2^{k}$.
se observa ca $\sqrt[n]{x_{1}x_{2}\ldots x_{n}}=\sqrt[2^{k}]{y_{1}y_{2}\ldots y_{2^{k}}}$. notez aceasta valoare cu $q$.
voi obtine
$\sum_{i=1}^{2^{k}}\frac{1}{1+y_{i}}\geq\frac{2^{k}}{1+q}$.
cum $y_{i}=q$ pt $n+1\leq i\leq 2^{k}$, trec acesti $y_{i}$ in membrul drept.
voi obtine exact
$\sum_{i=1}^{n}\frac{1}{1+x_{i}}\geq \frac{n}{1+q}$, adica ce trebuia de demonstrat[/hide]


There is a young boy who was collecting pebbles on a seashore.

Well, he collected a million pebbles and started playing this game.

He divided the pile into two (not necessarily of the same size) - let's say
with $ 7$ and $ 999993$ pebbles. He noted down the number $ 7 \times 999993$.

He then chose one of the piles, say the one with $ 7$ pebbles, and divided it into
two parts again, let's say he made piles of $ 2$ and $ 5$. He noted the product
$ 2 \times 5 \equal{} 10$.

He kept going like this, choosing one of the piles at random and dividing it into two
piles (could be of unequal sizes) and noting the product as described above.

He stopped when he had a million piles of one pebble each, and then he added up the
numbers that he noted at each stage.

If he repeats this whole process many many times over, what is the mean value of the
sum so derived, and what is the expected variance?

i am pretty sure the end result is the same no matter how the pebbles are split
lets prove using induction that if you start with 1 pile of n marbles, the end total will be $ \frac {n(n \minus{} 1)}{2}$
(that is the hypothesis)
base case: n=1
there is already 1 pile of 1 marble so the pile cannot be broken down anymore and the total is $ 0 \equal{} \frac {1(1 \minus{} 1)}{2}$

inductive step
assume that n=k and for all l such that l<k, the inductive hypothesis applies
then if the pile of k is broken into a pile of m and k-m, the amount $ m(k \minus{} m)$ is written and since both are positive, m<k and k-m<k, so for each of these piles, the total sum will be $ \frac {m(m \minus{} 1)}{2}$ and $ \frac {(k \minus{} m)(k \minus{} m \minus{} 1)}{2}$ for the piles of m and k-m respectively
we add all these and we get
$ m(k \minus{} m) \plus{} \frac {m^2 \minus{} m \plus{} k^2 \minus{} 2mk \minus{} k \plus{} m^2 \plus{} m}{2} \equal{} mk \minus{} m^2 \plus{} \frac {2m^2 \plus{} k^2 \minus{} 2mk \minus{} k}{2} \equal{} \frac {2mk \minus{} 2m^2 \plus{} 2m^2 \plus{} k^2 \minus{} 2mk \minus{} k}{2} \equal{} \frac {k^2 \minus{} k}{2} \equal{} \frac {k(k \minus{} 1)}{2}$

Q E Algebra bashed

so the value expected is 499999500000 and the variance is 0

aahhh....

non-bashing, more intuitive way:

Imagine there was an invisible string connecting every pair of two pebbles. Obviously there are $ n(n \minus{} 1)/2$ invisible strings. Each time we split a group of pebbles into two piles, any string that connects a pebble from one pile to the other pile breaks. And clearly, if the piles are of size $ a$ and $ b$, $ ab$ strings are thus broken. When all the pebbles are there own size-1 piles, every string will have been broken.

and blablablah you get your answer.

How happyful is that!  :D


I'm not sure what you mean as you have different numbering systems for the first and second sections. So assuming that you want the numbering to start again in each section then this will do it:
[code]\newcounter{problem}[section] %resets problem counter in each new section
\newcommand{\pno}{\stepcounter{problem}\thesection.\arabic{problem}}

\section{blabla}
problem \pno \\
problem \pno \\
....
\subsection{blablabla}
problem \pno \\
problem \pno \\
...
\section{kkakka}
problem \pno \\
problem \pno[/code]
You may have to adapt this if you want something different.

[quote="h.lam"][quote="nickolas"]Hello, probably this is easy for you, but for me its a problem:

I would like to enumerate my problems this way

\section{blabla}
problem 1
problem 2
....
\subsection{blablabla}
problem 1.1
problem 1.2
...
\section{kkakka}
problem 2.1
problem 2.2
....

if there is any solution, please tell me. Thank you[/quote]

Use the enumitem package:

[code]
\usepackage{enumitem}

\setenumerate[1]{label={\textbf{Problem \thesection.\arabic*}},labelsep=2em,align=left,ref={\thesection.\arabic*}}

\begin{document}
\begin{enumerate}
\item Write down Pythagoras' Theorem.
\item Write down Green's Theorem.
\end{enumerate}
\end{document}
[/code][/quote]

It works fine, but i need one more detail please. I would like enumerate to start from the beggining to each section:

\section{blabla}
1.1
1.2
....
....
\section{bloblo}
2.1
2.2
....
....

Thanks :)

You just need to use that same code and put in the relevant sections, otherwise the section counter will be "0", i.e.

0.1
0.2
0.3

will show up. That is,
[code]
\section{Hi}
\begin{enumerate}
\item Item One
\item Item Two
\end{enumerate}

\section{Hi again}
\begin{enumerate}
\item Item One
\item Item Two
\end{enumerate}[/code]


A square and a regular hexagon are from the same circle. If the side of the square measures $ 4\sqrt{2}$ cm, what's the measure of apothem of the regular hexagon?

I don't understand the answer. Is it $ 2\sqrt{2}$ ?

Hi Luis.

Since the side of the square is $ 4\sqrt{2}$, half its diagonal = the radius = $ 4$.

Then looking at one of the six equilateral triangles which make up the hexagon, if the length of the apothem is $ a$ , then $ \sin 60 \equal{} \frac{\sqrt{3}}{2} \equal{} \frac{a}{r}$, so $ a \equal{} \frac{r\sqrt{3}}{2} \equal{} \frac{4\sqrt{3}}{2}$ so $ a \equal{} 2\sqrt{3}$.


For $x,y,z$ positive reals such that $xyz=1$ prove that

$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geq 1$

something's not correct there, manlio...

The thing has got no maximum since when a,b -> 0 and c -> +oo (with abc constant) the LHS goes to infinity.

The thing's minimum is clealy 0 since when a,b -> +oo and c -> 0 (with abc constant) the LHS goes to 0.

I think this inequality misses something :)


When someone writes n!, what does he mean and what does ! stand for?

! stands for factorial. A factorial is an expression where you multiply all the numbers from 1 to that number.

(adding on to what monkeygirl13 said)
For example, $ 4!=4\cdot3\cdot2\cdot1$.
Likewise, $ 10!=10\cdot9...\2\cdot1$.

Also, this topic does not belong in this forum.

That depends. This n! you speak of can be anywhere. For example, it can be used in this sentence: 'I can use the variable n!' This means you would use [n] for something and be exclaiming it. The most common usage is factorial. If I have n! it means that the numbers from n to 1 are multiplied together. n can be any number EXCEPT A NEGATIVE NUMBER OR 0. -1! makes no sense. 0! makes no sense either. I could write n! like this: $ (n)*(n\minus{}1)*(n\minus{}2)*(n\minus{}3)*......*1$ Hope I could help!  :D

[quote="jwdkim"]0! makes no sense either. [/quote]  I have three complaints about your post:
1) It has irritating formatting.  In particular, it has SHOUTING and the use of * instead of \cdot for multiplication in LaTeX.
2) It added nothing new to the thread.
3) The quoted sentence is wrong: $ 0! \equal{} 1$.  This is not ambiguous or subject to debate, it's a definition that has an extremely good heuristic basis.  (In general, any empty product should be equal to 1: "not multiplying by anything" is the same as "multiplying by 1."  Also, there is exactly one way to rearrange zero objects: don't do anything at all.  Also, the factorial recursion $ n! \equal{} n \cdot (n \minus{} 1)!$ works perfectly well when $ n \equal{} 1$.  Also, this is the value you get from considering instead the gamma function, an important function in higher mathematics that can be thought of as the "right" way to extend the factorial function into a continuous function.)

Since the title of the thread has multiple exclamation points, I'll mention that $ n!!$ means not $ (n!)!$, but the product of all positive integers of the same parity as $ n$, up to and including $ n$.

Yes, you are right, abcak. For example, 2010!!!!!!! (that is 7) = 2010*2003*1996*1989...*1, while ((((((2010!)!)!)!)!)!)! is much larger than that.

BTW, why hasn't this thread been moved yet?


[quote="Joao Guerreiro"]For any positive reals $ a$, $ b$ and $ c$, show that:
$ \frac{a^{3}}{2a^{2}-ab+2b^{2}}+\frac{b^{3}}{2b^{2}-bc+2c^{2}}+\frac{c^{3}}{2c^{2}-ca+2a^{2}}\geq\frac{a+b+c}{3}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66271


The sequence $ a_n$ is defined by $ a_1 \equal{} 0, a_2 \equal{} 1, a_n \equal{}\frac {na_{n\minus{}1} \plus{} n(n\minus{}1) a_{n\minus{}2} \plus{} (\minus{}1)^{n\minus{}1}n}{2} \plus{}(\minus{}1)^n$.
Find $ a_n \plus{} 2C^1_n a_{n\minus{}1} \plus{} 3 C^2_n a_{n\minus{}2} \plus{} ... \plus{} nC^{n\minus{}1}_na_1$, where $ C^m_n$ is the binomial coefficient $ \frac{n!}{m! (n\minus{}m)!}$.

[quote="tdl"]The sequence $ a_n$ is defined by $ a_1 \equal{} 0, a_2 \equal{} 1, a_n \equal{} \frac {na_{n \minus{} 1} \plus{} n(n \minus{} 1) a_{n \minus{} 2} \plus{} ( \minus{} 1)^{n \minus{} 1}n}{2} \plus{} ( \minus{} 1)^n$.
Find $ a_n \plus{} 2C^1_n a_{n \minus{} 1} \plus{} 3 C^2_n a_{n \minus{} 2} \plus{} ... \plus{} nC^{n \minus{} 1}_na_1$, where $ C^m_n$ is the binomial coefficient $ \frac {n!}{m! (n \minus{} m)!}$.[/quote]
$ a_n\equal{}n!\minus{}\frac{n!}{1!}\plus{}\frac{n!}{2!}\minus{}\frac{n!}{3!}\plus{}...\plus{}(\minus{}1)^n \frac{n!}{n!}$


If p is prime and x,y,z are positive integers, then solve the following equation for (p,x,y,z):
p^x=y^2z+4.

Hint: Take cases of x is odd and even.
$ p^x \equal{} y^{2z} \plus{} 4$. Solve for prime p and positive integers x,y and z.


I'm doing this NHD project on Chinese propaganda, and I would really appreciate it if you guys will give responses.

Note: Please, Chinese people, refrain from voting unless you can get rid of your bias (either way...).

Sorry, for convenience I couldn't add ambiguous choices.

Thanks, everyone!

well, non-chinese would also have biases


prove that $ \sqrt2+\sqrt3+....\sqrt{n}$is irrational for all $ n\in N$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1709488319&t=4496

Pierre.


[i]The real sequence $ \{ a_n \}_{n \equal{} 1 }^{ \plus{} \infty}$ satisfies the condition :

   The sequence : $ a_1  \plus{} 2a_2 \ ; \ a_2 \plus{} 2a_3 ; ..... ; a_n \plus{} 2a_{n\plus{}1} ;......$ is convergent

    Prove that the sequence  $ \{ a_n \}$ is also convergent  :lol: [/i]

$ a_{n\plus{}1}\plus{}a_n/2$ tends to $ l$. If we change $ a_n$ into $ a_n\minus{}2l/3$ then we may assume $ l\equal{}0$. If $ n>N$ then $ \minus{}\epsilon\minus{}a_n/2<a_{n\plus{}1}<\epsilon\minus{}a_n/2,\minus{}\epsilon(1\plus{}1/2)\plus{}a_{n\minus{}1}/2^2<a_{n\plus{}1}<\epsilon(1\plus{}1/2)\plus{}a_{n\minus{}1}/2^2$
and more generally $ \minus{}\epsilon(1\plus{}1/2\plus{}\cdots\plus{}1/2^k)\pm{a}_{n\minus{}k}/2^{k\plus{}1}<a_{n\plus{}1}<\epsilon(1\plus{}1/2\plus{}\cdots\plus{}1/2^k)\pm{a}_{n\minus{}k}/2^{k\plus{}1}$.
Finally $ \minus{}2\epsilon\pm{a_0}/2^{n\plus{}1}<a_{n\plus{}1}<2\epsilon\pm{a_0}/2^{n\plus{}1}$.
Conclusion: if $ n$ is sufficiently great then $ \minus{}3\epsilon<a_{n\plus{}1}<3\epsilon$ and $ a_n$ tends to $ 0$..

The last $ 2$ lines are false. I rewrite these lines.
$ \minus{}\epsilon(1\plus{}1/2\plus{}\cdots\plus{}1/2^{k})\pm{a}_{n\minus{}k}/2^{k\plus{}1}<a_{n\plus{}1}<\epsilon(1\plus{}1/2\plus{}\cdots\plus{}1/2^{k})\pm{a}_{n\minus{}k}/2^{k\plus{}1}$ is valid if $ n\minus{}k\geq{N}$.
Thus $ \minus{}2\epsilon\pm{a_{N}}/2^{k\plus{}1}<a_{N\plus{}1\plus{}k}< 2\epsilon\pm{a_{N}}/2^{k\plus{}1}$.
If $ k$ is sufficiently great then $ \minus{}3\epsilon<a_{N\plus{}k\plus{}1}<3\epsilon$ and $ a_n$ tends to $ 0$.


Let
\[ \prod_{n=1}^{1996}\left(1+nx^{3^{n}}\right)=1+a_1x^{k_1}+a_2x^{k_2}+...+a_mx^{k_m}  \]
where $a_1,a_2,...,a_m$ are nonzero and $k_1<k_2<...<k_m$. Find $a_{1996}$

I found this problem in ACoPS, and then checked the "Mathematical Contests from Around the World" (page 81) http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf . I don't think their answer is correct.

[hide]
(begin official solution)
Solution: Note that $k_i$ is the number obtained by writing $i$ in base 2 and reading the result as a number in base 3, and $a_i$ is the sum of the exponents of the powers of 3 used. In particular, $1996 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^3 + 2^2$, so $a_{1996} = 10 + 9 + 8 + 7 + 6 + 3 + 2 = 45$
--------------------
(end official solution)

The answer I think is 11*10*9*8*7*4*3.
Can someone explain why they add rather than multiply?
[/hide]

I think more than just that is wrong with the solution... because you don't have the $3^0$ power, it's more like reading the number in base 2, multiplying it by 2, and then reading it in base 3. In which case you get what you have. And it should be the product, because the coefficients multiply when the exponents add.

could someone exlain about the base 3/base 2 part? I didn't get that...

[hide]
If you multiply out the product, you will only get terms with powers that can be written only with 1's and 0's in base 3. This is because when you are multiplying to get an exponent, you are adding together single disctinct powers of 3. We want to find the 1996th smallest term. In comparing relative size, we can consider the string formed in base 3 to be in base 2, sinces there are no '2's in the string.
[/hide]

oh, I saw a problem that used this concept before. isn't it called bijection or something?
the question was something like:
all numbers that can be written as the sum of distinct powers of three are written in a list in increasing order. What is the 500th number written?
I can't remember trhe number, I just made it up...
anyway you solve it the same way, write 500 in base 2, then translate the answer from base 3 to base 10 (so $500_{10} = 11111010_2$ , and $11111010_3$ is $(\text{some big number})_{10}$ which is the answer.

do you remember what the original question was? it was an AIME, and I have seen you on those mesage boards..

No, unfortunately I do not remember seeing that problem, but it is a nice problem. :)

yes it is. how did you score on this years AIME by the way?

The problem was as follows:  A sequence of integers $1, 3, 4, ...$ consists of all the numbers expressible as the sum of distinct powers of $3$ (using no power more than once).  What is the $100^{th}$ term in the sequence?

It is a nice problem, I have solved it twice on the forums already, both times using a bijection in base $3$ to base-$2$ numerals.  Unfortunately, I do not remember what year or number it is.

I recall a similar problem from an AoPS book, does anyone remember than one in which you have 40 coins in 4 bags, all the coins in a bag wiegh the same. The weights possible are 1,2,3. If the total weight is 95 and there are 1,3,9,27 weights in each bag which bags has which weights?


This might be in the forum, don't know. Anyway it is a good excercise.

Let $o(G)$ be the number of components with an odd number of vertices of a graph $G$. Prove that $G$ has a perfect matching if and only if $o(G-S)\leq |S|$ for all $S$ subset of the set of vertices of G.

I am lazy to type a solution (I think it may be found if not in this site, then somewhere in web), but just want to mention a stronger result:

in any graph, $\max_S o(G-S)-|S|$ equals to the minimal number of vertices, which may not be simulateosly covered by disjoint edges. In other words, if for any$S$ we have $o(G-S)\le |S|+d$, then all but $d$ vertices may be covered by the set of disjoint edges. It is called Berges formula, I think.


calculus
integration
algebra
function
domain
abstract algebra
induction

Let $ R$ and $ S$ be integral domains and $ M$ be the $ S$-module of all maps from $ R$ to $ S$.
Show that every family $ \phi_1,\dots,\phi_n$ of distinct ring homomorphisms from $ R$ to $ S$ is linear independent in $ M$

note: in this problem all rings have unity.

How do you give the set of maps a module structure? If you mean ring homomorphisms, it doesn't have a module structure (that I know of). If you mean any map at all, then we can simply think of $ R$ as a set. But if $ 1 \neq s \in S$, $ \phi_1 \in M$, and we define $ \phi_2 \equal{} s \phi_1$, then $ \phi_1$ and $ \phi_2$ are not linearly dependent.

calc rulz: No. the ring homomorphisms are a subset of the maps from $ R$ to $ S$ (being a $ S$ module), and the claim is that as a subset of a module, they are linear independent.

Read the proof of Theorem 2.1 (Dedekind's Theorem on the linear independence of characters) in

[url=http://www.math.uconn.edu/~kconrad/math316/linearchar.pdf]Keith Conrad, [i]Linear independence of characters (Math 316, Spring 2006)[/i][/url].

Note that $ G$ needs not be a group, but can be any semigroup, and $ F$ needs not be a field, but can be any integral domain. Apply.

[b]EDIT:[/b] Alternatively, apply http://www.mathlinks.ro/viewtopic.php?t=89951 after replacing $ S$ by the quotient field of $ S$ (this is why $ S$ was required to be an integral - note that requiring $ R$ to be an integral domain is useless).

  darij

[quote="darij grinberg"]Read the proof of Theorem 2.1 (Dedekind's Theorem on the linear independence of characters) in

[url=http://www.math.uconn.edu/~kconrad/math316/linearchar.pdf]Keith Conrad, [i]Linear independence of characters (Math 316, Spring 2006)[/i][/url].

Note that $ G$ needs not be a group, but can be any semigroup, and $ F$ needs not be a field, but can be any integral domain. Apply.

  darij[/quote]

I came up with a similar solution, but I fail to see why $ (\phi_i\minus{}\phi_n)\phi_i$ needs to be a ring homomorphism if $ \phi_i$ and $ \phi_n$ are and frankly, I doubt that it is true. If it fails to be true, then I can't apply the induction hypothesis and the proof doesn't work. Am I missing something?
Thanks for the response.

Where do you need to consider $ \left(\phi_i\minus{}\phi_n\right)\phi_i$ ?
Re-read the proof of Theorem 2.1 in that note I linked to; read $ \phi_i$ for $ \chi_i$, read $ R$ for $ G$, and read $ S$ for $ F$.

  darij

[quote="darij grinberg"]Where do you need to consider $ \left(\phi_i \minus{} \phi_n\right)\phi_i$ ?
Re-read the proof of Theorem 2.1 in that note I linked to; read $ \phi_i$ for $ \chi_i$, read $ R$ for $ G$, and read $ S$ for $ F$.

  darij[/quote]

My apologies, I read  $ \left(\phi_i(r) \minus{} \phi_n(r)\right)\phi_i(r)$ instead of $ \left(\phi_i(r_0) \minus{} \phi_n(r_0)\right)\phi_i(r)$. I can see that it works now. Thanks.


[quote="Conjurer"]See attached file. Download and zoom for clarity. And of course do not mind my checks.[/quote]

Didnt you attend the session? You should have attended it, the answers were discussed! I'll post the solutions asap. Right now, its dinner time!!! :D

As I told you in the PM, I am not enrolled in it myself. :)

Plz post the solution with the minutest detail present :) since I think these two questions are really conceptual and even the mathematical aspects of it are very imp since many questions are framed on these concepts.

i also wrote this test,can you tell me how option <B> is correct

I don't know if you are taking about question 7 or 8 but <B> isn't correct in either...


Let $ (a_n)_{n\in\mathbb{N}}\subseteq  \Phi^{\mathbb{N}}$ such that $ \sum_{n\equal{}1}^{\plus{}\infty} a_n b_n$ converges  whenever $ \sum_{n\equal{}1}^{\plus{}\infty}|b_n|^p<\plus{}\infty,~1<p<\plus{}\infty$. Then $ \sum_{n\equal{}1}^{\plus{}\infty}|a_n|^{q}<\plus{}\infty,~\frac{1}{p}\plus{}\frac{1}{q}\equal{}1$.

thank you, I did not know that it was this Landau who proved that. Anyway, this result is a quite straightforward if one knows that the dual of $ l^p$ is $ l^q$. (see http://en.wikipedia.org/wiki/Lp_space)


設 N =13×17×41×829×56659712633。
已知N 是一個18 位數，而0 至9 十個數字,當中其中九個在N 的18
個數字裏各出現了兩次。求N 的數字之和

Translate:

Suppose $ N \equal{}13\times17\times41\times829\times56659712633$ is $ 18$ digits' number.
Now , if $ N$ is involved in 9 different digits only from {$ 0,1,2,3,4,5,6,7,8,9$} and each occurs twice. Find the sum of all digits of $ N$. :P

[hide="Hint"]Consider $ N\bmod 9$ to determine the number that is missing.[/hide]

EDIT: This is useless if you have read the above hidden solution (even though the above solution can be made slightly shorter).

[quote="ffao"][hide]
Let $ S$ denote the sum of digits of $ N$.

We know that $ S \equiv N \pmod{9}$
Furthermore, $ N \equal{} 13\times17\times41\times829\times56659712633 \equiv 4\times8\times5\times1\times8 \equiv 2 \pmod9$.

As we were told that N contains 9 digits from $ \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, we have that $ 72 \le S \le 90$. Also, as each digit appears exactly twice in the decimal representation of $ N$, then $ S$ is necessarily even.

Now, the only integers congruent to 2 (mod 9) between 72 and 90 are 74 and 83. But as S is even, then $ \boxed{S \equal{} 74}$.

[/hide][/quote]

$ \mathbb B$ingo!!nice work. :trampoline:


search
function
number theory proposed
number theory

Find all +ve integers a,b,c s.t.

bc+2ac+3ab=abc.

Bomb

it is easy to prove that c<37, a<37 and b<41. I have a nice proof of this, but i don't know how to write it down properly. Then you could perform a computer search on these numbers and it will give

{ [15,6,5],[36,9,4],[2,7,14],[2,5,30],[5,40,4],[2,6,18],[2,8,12
],[2,12,9],[8,16,4],[10,5,6],[3,4,18],[2,16,8],[6,6,6],[3,12,6],[6,4,9
],[5,10,5],[5,4,10],[3,30,5],[4,8,6],[3,6,9],[6,3,18],[2,10,10],[8,4,8
],[14,4,7],[2,28,7],[6,24,4],[30,3,10],[4,4,12],[4,3,36],[12,3,12],[20,10,4],[12,12,4] }

I tried to write down the proof that c<37 but it is very confusing, I hope someone will understand the idea and be able to write it down in a nice way.

[hide="here it is"]
I will give I another try and prove that $c<37$. Consider the function $k_c$ defined as $k_c(a,b)=bc+2ac+3ab-abc$. (In fact $k$ kan be seen as a row of rows or an infinite matrix). We will prove that if $c>36$ then there are no $(m,n)$ so that $k(m,n)=0$ First of all, notice that the sequence $k_c(1,n)$ is always increasing: $k_c(1,n)-k_c(1,n+1)=3$. $k_{37}(1,1)=77$ and if $c$ increases then so will $k_c(1,1)$ . So for $m=1$ we will find no zero's for $k_c(m,n)$ if $c>36$. In the same way it can be proven that $n=1$ and $n=2$ will never provide solutions if $c>36$.  It may also be noticed that the element $k_c(2,3)$ will never provide a solution since $k_1(2,3)=19$ and $k_c(2,3)$ increases as $c$ increases. Furthermore $k_c(2,4)=24$ and $k_c(3,3)=27$ for all $c$. If $c>36$ then $k_c(2,n)$ is a decreasing sequence. It's first negative element is $k_c(2,5)$ because $k_c(2,4)$ was still positive, and all what's next will be negative too. As $c$ becomes larger, $k_c(2,5)$ becomes smaller, which proves that if $c>36$ there is no $n$ for wich $k_c(2,n)=0$. The same thing can be proven quite the same way with $k_c(3,n)$ and $k_c(m,3)$. Now suppose $m>3$. We saw that $k_c(m,3)$ will be negative then because it is decreasing as $m$ increases and $k_c(4,3)$ is alway negative when $c>36$. But if $m>3$ then will $k_c(m,n)>k_c(m,n+1)$ and this completes the proof.
[/hide]

I tried it again in an other way but  i failed:

$bc+2ac+3ab=abc$
then should $a|bc$

To make my life a little easier, let's say both $b$ and $c$ are primes.
This would mean that either $a=1$ (case I), $a=b$ (case II),$a=c$ (case III) or $a=bc$.

[b]case I[/b] if $a=1$ then we see that $bc+2c+3b=bc$ has no Solutions unless $b=c=0$. So $(1,0,0)$ is a first solution.

[b]case II[/b] if $a=b$ then $ac+2ac+3a^2=a^2c$ should be solved. We can assume $a\neq0$, having checked that $(0,b,c)$ provides a solution for every value of b and c.  Dividing by a, I get $3c+3a=ac$. This requires some time but it can be shown that its only solutions  are $a=c=0$ , $a=c=6$ , $a=12$ and $c=4$, $a=4$ and $c=12$.

[b]case III[/b] if $a=c$ then $ba+2aa+3ab=aba$ should be solved.  We already did the case where $a=0$ so we can divide by $a$ again  and solve $b+2a+3b=ba$. It can be shown again that its solutions are $(a,b) \in {(0,0),(6,6),(8,4),(5,10),(12,3)}$.

[b] case IV[/b]  $a=bc$ will give $bc+2bc^2+3cb^2=(bc)^2$.
Dealing with the cases $b=0$ and $c=0$ first, I see that $(a,0,b)$ and $(a,b,0)$ are solutions.
Now assuming that $b\neq0%Error. "neqc" is a bad command.
$, I can divide by $bc$ and get $1+2c+3b=bc$.
I found (b,c)=(9,4) and (b,c)=(3,10) as only solutions.

So all together the solutions for $(a,b,c)$ (exept from te solutions containing $0$): $(4,4,12)$, $(5,10,5)$, $(6,6,6)$, $(8,4,8)$, $(12,3,12)$, $(12,12,4)$, $(30,3,10)$ and
 $(36,9,4)$.

This is something weird: I started by saying that I would consider the cases where both
 $b$ and $c$ are primes, but I did [i]not[/i] find such solutions.

So actually i only proved that there are no solutions with $b$ and $c$ primes. :/


A computer is programmed to randomly generate a string of six symbols
using only the letters A,B,C. What is the probability that the string will not
contain three consecutive A’s?

The answer started: Let s_n denote the number of strings of length n with no three consecutive
A's. Thus $ s_1 = 3$, $ s_2 = 9$ and $ s_3 = 26$. We claim that we have the following
recurrence relation:
$ s_n = 2s_{n-1}+2s_{n-2}+2s_{n-3} (n ≥ 3)$.

How do they get that? can someone explain it

Consider how such a string can end.  If it ends in $ B$ or $ C$, deleting that letter gets you another string of length $ n\minus{}1$; this is the first term.  If it ends in $ BA$ or $ CA$, we can delete two letters; this is the second term.  Finally, if it ends in $ BAA$ or $ CAA$, we can delete three letters; this is the third term.

This is an instance of an extremely general technique for solving such problems.

dang, still dont get it. So what is that extremely general technique?

It might be easier if I name a few extra sequences.  Let $ a_n$ be the number of strings of length $ n$ satisfying the criterion that don't end in $ A$.  Let $ b_n$ be the number of such strings ending in exactly one $ A$.  Let $ c_n$ be the number of such strings ending in exactly two $ A$s.

By assumption, this partitions $ s_n$, hence $ s_n \equal{} a_n \plus{} b_n \plus{} c_n$.  The three explanations I gave are equivalent to the three recurrences

$ a_n \equal{} 2(a_n \plus{} b_n \plus{} c_n)$
$ b_n \equal{} a_{n\minus{}1}$
$ c_n \equal{} a_{n\minus{}2}$.

If you're still confused, work out $ s_4$ using this idea.

There are two general techniques for problems like this, both of which are beyond the scope of this forum.  The keywords are "regular languages" and "finite automata."

I understand up to s_n= a_n+b_n+c_n. why multiply the RHS of the first equation by 2. Then I get screwed by the n-1 and n-2 subscripts, just can't put it together. Also when you wrote a_n=2(a_n+b_n+c_n), are the a_n's the same, or did you mean to use another letter?

I worked s_4 but still don't get it

Sorry, I messed up the subscripts; I wanted $ a_n \equal{} 2(a_{n\minus{}1} \plus{} b_{n\minus{}1} \plus{} c_{n\minus{}1})$.

Maybe you should try understanding a simpler example.  Let $ F_n$ denote the number of strings of length $ n$ using the letters $ A, B$ such that there are no two consecutive $ A$s.  Do you see why

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

using a similar argument as the above?

ok. yes t0, I see it from the simpler example

When you say we can delete terms--- What if we delete terms and there's still an AAA in the ones we haven't deleted. Then what?


geometry
3D geometry
sphere
parameterization
calculus
integration
real analysis

Only by using the knowledge of general math II prove this theorem in differential geometry that ,by drawing big circles on a unit sphere, we make a triangle $ ABC$ with the angles $ \alpha , \beta$ and $ \gamma$, prove that $ \alpha\plus{}\beta\plus{}\gamma\minus{}\pi\equal{}S_{ABC}$.

PS. I know the proof that extend each arc of triangle and the area in each surface that has the angle $ \alpha$ is $ 2 \alpha$.

The area of two spherical sectors between lines AB and AC is $ 4\pi\frac{\alpha}{\pi}$; the area of two sectors between $ BA$ and $ BC$ is $ 4\pi\frac{\beta}{\pi}$; the area of two sectors between $ CA$ and $ CB$ is $ 4\pi\frac{\gamma}{\pi}$. These three sets (each consists of two sectors) cover every point outside of triangle $ ABC$ and its antipodal triangle $ A'B'C'$ once; they cover $ ABC$ and $ A'B'C'$ three times (draw a picture!). Therefore,
\[ 4(\alpha\plus{}\beta\plus{}\gamma) \equal{} 3 S(ABC) \plus{} 3 S(A'B'C') \plus{} S(\text{complement to }ABC\text{ and }A'B'C') \equal{} 2 S(ABC) \plus{} 2 S(A'B'C') \plus{} 4\pi.\]
Since $ S(ABC) \equal{} S(A'B'C')$,
\[ S(ABC) \equal{} \alpha \plus{} \beta \plus{} \gamma \minus{} \pi.\]

P.S. I don't know what General Math II is.

This is somewhat like I said, but can you give a good parameterization that by computing the surface integral we can get the result ?

PS. I think general math II in Iran is the second calculus course in USA that is taught in university.


limit
real analysis
real analysis solved

The limit is 00. I think you want lim[u(n)]2/n, which is 2 (just look at the expression of [u(n)]2 and use Cesaro-Stolz).

I think lephuocdaint1987's problem is :
u_(n+1)=u_n+1/(u_n) 2 .

Oh, in that case, u(n+1)3=u(n)3+3+3u(n)-3+u(n)-6, so u(n+1)sup>3-u(n)3->3 as n->00, so u(n)3/n->3 as n->00.

I used the fact that u(n) obviously tends to 00.


number theory
Diophantine equation
SFFT
special factorizations

8x+5y=600
is there a way to solve all solutions of this diophantine equation?
I tried using simon's favorite factoring trick, but I got into a muddle of mathematical mess....
 :?:  :)

there are two ways to do it.

one is to graph it and find all solutions.

the other is to find the first pair (x,y).

then each time you increase the value of of x by five you decrease y by 8 for example the first solution for
8x+5y=600. is (0,120) the next one is (5,112), etc

In the case you get a more subtle equation, which you can't see one solution immediately, you can use mods. For example, taking it mod 5, or mod 8.

uhh... i'm assuming you want non-negative integer values.

To solve Diophantine equations, you always solve for one variable in terms of the other, so let's take $ y$.

$ 5y\equal{}600\minus{}8x$
$ y\equal{}\frac {600\minus{}8x} {5}$
Notice that the smallest such pair that works is $ (0,120)$.
The next one will be when $ x$ is $ 5$.
It is then easy to see that for all $ x\equal{}5k$, where $ k\ge0$, both will be integer values.
The largest $ x$ that allows $ y\ge0$ is $ 75$, which makes $ y$ $ 0$.
If $ 600$ was not a multiple of $ 5$, it would be slightly different.

Well, you could take mod 5 of both sides to get $ 3x\equiv 0 (\mbox{mod }5)$, and multiplying both sides of the congruence tells you that there are integer solutions as long as $ x$ is a multiple of 5


equation
number theory
Divisibility
Cyclotomic Polynomials
modular arithmetic
IMO Shortlist

Find all integer solutions of the equation \[\frac {x^{7} \minus{} 1}{x \minus{} 1} \equal{} y^{5} \minus{} 1.\]

Lemma. If  $p, q$ are primes and $q|\frac{x^{p}-1}{x-1}$ then $q \equiv 1 \mod{p}$ or $q \equiv 0 \mod{p}$.
Proof: well-known, it was posted many times also.

Suppose that $x \not \equiv 1 \mod{7}$. Then of course $\frac{x^{7}-1}{x-1}\equiv \frac{x-1}{x-1}= 1 \mod{7}$. So $y^{5}-1 \equiv 1 \mod{7}$, it implies that $y \equiv 4 \mod{7}$, as one can easily check. But then: $y^{5}-1 = (y-1)\frac{y^{5}-1}{y-1}$ and $y-1 \equiv 3 \mod{7}$. It is contradiction with the lemma, since $y-1$ must have a prime divisor $q$ which is not congruent to $1$ or $0 \mod{7}$.

Suppose that $x \equiv 1 \mod{7}$. Then $\frac{x^{7}-1}{x-1}= x^{6}+...+1 \equiv 0 \mod{7}$. Then, we must have $y \equiv 1 \mod{7}$. But: $y^{5}-1=(y-1)(y^{4}+...+1)$ and $y^{4}+...+1 \equiv 5 \mod{7}$. One more time, a contradiction with the lemma.

[quote="TomciO"]It is contradiction with the lemma, since $y-1$ must have a prime divisor $q$ which is not congruent to $1$ or $0 \mod{7}$.
.[/quote]

I do not see the reason.  :(

It's clear that $y-1$ is not divisible by $7$. So, suppose that all prime divisors of $y-1$ are $1 \mod{7}$. Then of course $y-1 \equiv 1 \mod{7}$ as well, but we know that $y-1 \equiv 3 \mod{7}$.

[quote="TomciO"] So, suppose that all prime divisors of $y-1$ are $1 \mod{7}$.[/quote]

But how does this follow from the lemma?

This follows from the lemma for $p=7$ and the fact that $y-1|\frac{x^{7}-1}{x-1}$ (and of course $7 \not | y-1)$.

Fact (posted often enough before): $\frac{x^{7}-1}{x-1}$ has only prime divisors $\equiv 0$ or$1\mod 7$.

Then you want $\frac{x^{7}-1}{x-1}= (y-1)(y^{4}+y^{3}+y^{2}+y+1)$, thus $y \equiv 1,2 \mod 7$ by the fact. Then $y^{4}+y^{3}+y^{2}+y+1 \equiv 5,3 \mod 7$, contradicting the fact.

[Note: this post was merged from another topic]

[quote="e.lopes"](new topic, because our dear moderator ZetaX have ZERO TOLERANCE! :D )

Prove that the equation $\frac{x^{7}-1}{x-1}= y^{5}-1$ don't have integer solutions![/quote]
This problem also from Russian TST 2007(author - A. Efimov).
And for every prime $p=3k+2$ the equation $\frac{x^{7}-1}{x-1}=y^{p}-1$ hasn't solutions in $\mathbb{N}$. Funny, first we wanted to give this problem for $p=11$ but I find that this equation hasn't solutions modulo 23!

[quote="TomciO"]Lemma. If  $p, q$ are primes and $q|\frac{x^{p}-1}{x-1}$ then $q \equiv 1 \mod{p}$ or $q \equiv 0 \mod{p}$.
Proof: well-known, it was posted many times also.
[/quote]

When you say "posted many times", can you give a link then please? because new people like me do not know so many posts and i can not find this when i use search function  :blush:
this is not only here, it happend quite often to me when i looked at this forum in past some days since i found this forum, its a bit frustrating.

i couldn't see why the lemma is true ???

i didn't know the lemma and i tried modulo 11... the problem would hold if $y \neq 0$ (mod 11) or $x\neq 7,9$ (mod 11)... i would appretiate if somebody could be so gently to try to finish this up...

Thank you to you Tomic0.
campos, it will not work with modulo or they would have taken a different prime... read this:

[quote="pavel kozlov"] Funny, first we wanted to give this problem for $p=11$ but I find that this equation hasn't solutions modulo 23![/quote]

[quote="TomciO"]Lemma. If  $p, q$ are primes and $q|\frac{x^{p}-1}{x-1}$ then $q \equiv 1 \mod{p}$ or $q \equiv 0 \mod{p}$.
Proof: well-known, it was posted many times also. [/quote]
hi,
if p,q are primes and $q \equiv 0 \bmod{p}$ then $p=q$. :|

No solutions.
Obviously we are solving $\Phi_7(x)=y^5-1$. It is known that in general if some prime $p \mid \Phi_n(x)$ for some arbitrary $x$, then either $n \mid p-1$ or $p \mid n$. Since $7$ is prime, it follows that the divisors of the LHS are either $1 \pmod{7}$ or $7$ itself. It follows that the LHS is either congruent to $0$ or $1$ modulo $7$.
If $\Phi_7(x) \equiv 1 \pmod{7}$, then $y \equiv 4 \pmod{7}$. But then $y-1 \mid y^5-1$ and $y-1 \equiv 3 \pmod{7}$, so it must be divisible by some prime that's not $7$ nor $1 \pmod{7}$. But this prime can never divide $\Phi_7(x)$, contradiction.
If $\Phi_7(x) \equiv 0 \pmod{7}$, then clearly we require $y \equiv 1 \pmod{7}$ as well. But then $y^4+y^3+y^2+y+1 \mid y^5-1$ and is congruent to $5 \pmod{7}$, so it must be divisible by some prime that's not $7$ nor $1 \pmod{7}$ as well. Such prime can neger divide $\Phi_7(x)$, so we also get a contradiction.
In conclusion, we can always derive a contradiction, hence there is no solution as desired. $\blacksquare$

I claim that there are no solutions.

Consider a prime $p$ dividing both sides of the given equation. Then it must divide $x^7-1$. We can then casework on whether or not $p\mid x-1$.

Case 1, $p\mid x-1$: Then we need \[\nu_p(x^7-1)>\nu_p(x-1).\] By LTE, this means that $p=7$.
Case 2, $p\nmid x-1$: Then the order of $x$ mod $p$ is $7$, so $p\equiv 1\pmod 7$.

Now assume FTSOC that $7$ divides both sides. But then, $7\mid y^5-1$, so $7\mid y-1$. Then $7\nmid y^4+y^3+y^2+y+1$, so that must be $1$ mod $7$(since anything else that divides both sides must be $1$ mod $7$), absurd as it is clearly $5$ mod $7$ since $y\equiv 1\pmod 7$.

Now we can see that every divisor of both sides is $1$ mod $7$, so $y-1\equiv 1\pmod 7$, so $y\equiv 2\pmod 7$. But \[y^5-1\equiv 32-1\equiv 3\pmod 7,\] contradiction.

Note that if $p\mid y^5-1$ then $p\mid x^7-1$. If $p\mid x-1$ then $x\equiv 1\pmod p$ and so $x^6+\dots + x+1\equiv 7\pmod p\implies p=7$. Therefore, any factor of $y^5-1$ is either $0$ or $1$ $\pmod 7$. Thus, $y\equiv 1$ or $2\pmod p$, so $y^4+y^3+y^2+y+1\equiv 5$ or $3$ $\pmod p$, impossible!

By the condition, the order of $x$ mod $y^5-1$ equals 7. On the other hand, the order of $y$ mod $y^5 - 1$ equals $5$, so all primes $p$ that divide $y^5-1$ are 0 or 1 mod $35$. In particular, $5$ cannot divide $\frac{x^7-1}{x-1}$, thus $p = 7$ or $p \equiv 1 \pmod {35}$.

If $y^5 -1 \equiv 0 \pmod {35}$, then $y \equiv 1 \pmod {35}$. But then $$5 \equiv y^4+y^3+y^2+y+1 \mid y^5-1$$ under mod 7, contradiction.

If $y^5 -1 \equiv 1 \pmod {35}$, then $y \equiv 32 \pmod {35}$, so $$26 \equiv y^4+y^3+y^2+y+1 \mid y^5-1,$$ contradiction again.

There are no integer solutions.

First note that the LHS is always positive regardless of whether $x$ is negative, so $y$ is always positive. Also, plugging in $x=0$ and $y=0$ yield no integer solutions. Now we just focus on positive integer pairs $(x,y)$ since if $(x,y)$ works then $(-x,y)$ also works.

Let $p$ be a prime factor of $y^5-1$. It follows that $p \mid \frac{x^7-1}{x-1}$.

 If $x \equiv 1 \pmod p$, then $$\frac{x^7-1}{x-1} = x^6+x^5+x^4+x^3+x^2+x+1 \equiv 7 \pmod {p}$$ which implies $p = 7$. This means that $7 \mid y^5-1$ which implies $y \equiv 1 \pmod 7$. However, if we consider $y^5-1 = (y-1)(y^4+y^3+y^2+y+1)$, we see that $y^4+y^3+y^2+y+1 \equiv 5 \pmod 7$. This is bad because $y^5-1$ can only have $7$ as a prime factor.

If $x \not \equiv 1 \pmod p$, then we must have $x^7 \equiv 1 \pmod p$. The order of $x$ modulo $p$ is then $\gcd(7,p-1)$ which is either $1$ or $7$. If the order is $1$, then $x \equiv 1 \pmod p$, contradiction. So the order must be $7$, which implies that $p \equiv 1 \pmod 7$. However, if this is the case then $y^5 - 1 \equiv 1 \pmod 7$ which means $y \equiv 4 \pmod 7$. This is bad because again if we consider $y^5-1 = (y-1)(y^4+y^3+y^2+y+1)$ we see that $y-1 \equiv 3 \pmod 7$, implying that there exists a prime that is not $1 \pmod 7$ that divides $y^5-1$.

We've exhausted both cases and have proven that there are no solutions.

storage from a while ago

Take the equation mod $y^5-1$, which implies $k=\ord_{y^5-1}x|7$. If k=1, then $$x-1\equiv 0\pmod {y^5-1}\implies 1+x+\dots+x^6\equiv 7\pmod p\implies p=7\text{ }\forall \text{primes p}\implies y^5-1=7,$$ contradiction. So the order of x mod $y^5-1$ is 7. Then consider a prime p dividing both sides (a common strategy). Then $\ord_px=7|p-1\iff p\equiv 1\pmod 7$. This means all factors (because primes mod 7 make up all factors mod 7) of both sides are only 1 mod 7 or 0 mod 7. But then y is either 1 or 2 mod 7 (since y-1 is 0 or 1), hence $y^4+y^3+y^2+y+1$ is either 5 or 3 mod 7. But this is impossible, since it is a factor as well! $\blacksquare$

By cyclotmic polynomials we have that all prime factors are either $p = 7$ or $p \equiv 1 \pmod 7$. Note that this forces $y^5-1 \equiv 1 \pmod 7$ (If it were 0 mod 7, it would have to be a power of $7$ which is simple to show impossible). Then $y \equiv 4 \pmod 7$, so since $y-1 \equiv 3 \pmod 7$ is a divisor, it must contain a prime which isnt $1\pmod 7$ or $7$, a contradiction.

It is well known that prime factors of $\Phi_n$ are either $1\pmod{n}$ or $n$ is divisible by the prime factor $p$.

So, $p = 7$ or is $1\pmod{7}$. This forces all factors of $\Phi_7$ to be $0, 1\pmod{7}$.

$y^5 - 1 = (y - 1)(y^4 + y^3 + y^2 + y + 1)$, so $y - 1$ must be $0, 1\pmod{7}$. This also should apply to $y^4 + y^3 + y^2 + y + 1$.

This then makes $y \equiv 1, 2\pmod{7}$.Plugging in $y \equiv 1, 2\pmod{7}$ gives us $y^4 + y^3 + y^2 + y + 1 \equiv 3, 5\pmod{7}$. This is a contradiction, so we are done.

XOINK. Suppose $p$ is a prime dividing the left side. Then $x^7 \equiv 1 \pmod p.$ Then $x$ has order either $1$ or $7 \pmod p.$ If it has order $1$ then by LTE we get $\nu_p(x^7-1)=\nu_p(x-1)+\nu_p(7)$ so $\nu_p(7)\ge 1$ so $p=7.$ Otherwise $x$ has order $7$ implying that $7 \mid \varphi(p)=p-1,$ so $p \equiv 1 \pmod 7.$ Now factor the right side as $(y-1)(y^4+y^3+y^2+y+1).$ Now $y-1 \equiv 0,1 \pmod 7$ so $y \equiv 1,2 \pmod 7$ and we get $y^4+y^3+y^2+y+1 \equiv 5,3 \pmod 7,$ but it must also be $0,1 \pmod 7,$ contradiction.

We first present a brief lemma on cyclotomic polynomials, following from [b]v_enhance's[/b] "Orders Modulo A Prime".

[b][color=#f00]Lemma:[/color][/b] For any prime dividing $\Phi_7(a)$ we either have $p \equiv 1 \pmod{7}$ or $p \mid 7$.

[i]Proof.[/i] Take a prime $p$ dividing $\Phi_7(a)$. Note that clearly we must have $$a^7 \equiv 1 \pmod{p}$$ as $\Phi_7(X) \mid X^7 - 1$. However then $\text{ord}_p(a) \mid 7$. If $\text{ord}_p(a) = 1$ then we have $a \equiv 1\pmod{7}$, and hence $$\Phi_7(a) \equiv  7 \equiv 0 \pmod{p}$$ hence $p = 7$. Otherwise we have $\text{ord}_p(a) = 7$ and so $7 \mid p - 1$, which rearranges to $p \equiv 1 \pmod{7}$ as desired. $\blacksquare$

Now armed with the lemma we will show that this equation has no integer solutions. Note that $\Phi_7(x) = \frac{x^7-1}{x-1}$. Then from the lemma for any prime $p$ dividing $\Phi_7(x)$ for some $x$, we have one of the following:
[list]
[*] Either $x$ has order $7$ modulo $p$ and hence $p \equiv 1 \pmod{7}$
[*] Or we must have $p = 7$
[/list]
Now assume $7 \mid \Phi_7(x)$. Then we require $y \equiv 1 \pmod{7}$. Then clearly $\nu_7(y^5 - 1) = 1$ from LTE, and thus all other primes dividing $\Phi_7(x)$ satisfy the first condition. Then must have, $\frac{y^5-1}{y-1}$ congruent to $1$ modulo $7$, as it is only divisible by primes congruent to $1$ modulo $7$. However for $y \equiv 1 \pmod{7}$ this fails.

Then assume that $7 \nmid \Phi_7(x)$. From similar reasoning we also must have $y^5 - 1$ be congruent to $1$ modulo $7$ as it is only divisible by primes that are congruent to $1$ modulo $7$. Then we require $y \equiv 4 \pmod{7}$. However factoring the right hand side we can see the factor $(y-1)$ is then not congruent to $1$ modulo $7$ and hence there exists some prime that is not congruent to $1$ modulo $7$ dividing the left hand side, contradiction.

Thus there are: $\boxed{\text{No solutions}}$.

[b]Remark:[/b] I need to go over the general derivation of the lemma for $\Phi_n(X)$, rather than the specific case of $n$ prime, but it seems like it follows a similar idea.

It is well-known that if $\Phi_n(a) \equiv 0 \pmod p$, then either $n \equiv 0 \pmod p$ or $p \equiv 1 \pmod n$.

Let $p$ be a prime dividing $\frac{x^7-1}{x-1} = \Phi_7(x)$. Then either $p = 7$ or $p \equiv 1 \pmod 7$, so all factors of the LHS are either $0$ or $1$ $\pmod 7$. $y^5-1$ can be factored as $(y-1)(y^4+y^3+y^2+y+1)$, so we must have $y \equiv 1$ or $2 \pmod 7$.

If $y \equiv 1 \pmod 7$, then $y^4 + y^3 + y^2 + y + 1 \equiv 5 \pmod 7$, contradiction.

If $y \equiv 2 \pmod 7$, then $y^4 + y^3 + y^2 + y + 1 \equiv 3 \pmod 7$, also contradiction.

Therefore, there are no solutions.

We first gonna prove the following "well-known" lemma:
$$p\mid \Phi_n(a)\Rightarrow\begin{cases}
\operatorname{ord}_p(a)=n \rightarrow n\mid p-1 \hspace{0.3cm}\text{or} \\
p\mid n &
\end{cases}$$
as $a^n-1=\prod_{d\mid n} \Phi_d(a)$ then $r=\operatorname{ord}_p(a)\mid n$, we got two cases
[list]
[*] $r=n\Rightarrow n\mid \varphi(p)=p-1$

[*] $r<n \Rightarrow \exists \hspace{0.1cm} d\mid r \hspace{0.2cm} \text{such that}\hspace{0.2cm}  p\mid \Phi_d(a)$ because $$0\equiv a^r-1\equiv \prod_{d\mid r} \Phi_d(a)$$
Thus $a$ is double root of $x^n-1 =\prod_{d\mid n} \Phi_d(x) \pmod p$ so $a$ is a root of it´s derivative, this gives that $n\cdot a^{n-1}\equiv 0\pmod p$ so $p\mid n$ or $p\mid a$, the latter is impossible.
[/list]

$\textbf{Remark.}$ As a corollary of this we get that if $q,p$ are primes with $p\mid \Phi_q(a)$ then $p\equiv 1 \pmod q$ or $p=q$. (this is easier to prove) and also gives us immediately that in this case all divisor of $\Phi_q(a)$ is $1\hspace{0.1cm}\text{or} \hspace{0.1cm}0\pmod q$, because all divisor is a product of primes which are $1 \hspace{0.1cm}\text{or} \hspace{0.1cm}0\pmod q$.

[color=blue][b]Solution to the problem [/color]
The problems follows almost immediately with this, as we need $\Phi_7(x)=(y-1)(y^4+y^3+y^2+y+1)$, so $y-1$ is $1 \hspace{0.1cm}\text{or} \hspace{0.1cm}0\pmod 7$ thus $y^4+y^3+y^2+y+1$ is $5 \hspace{0.1cm}\text{or} \hspace{0.1cm}3\pmod 7$ which is impossible.

Notice the LHS is positive, so we only have to consider $y>0$.

It is well known that the prime factors of $\Phi_p$ are either $p$ or $1 \pmod{p}$. Applying this to $p=7$ as the LHS is $\Phi_7$, we see that any prime dividing the RHS follows this format as well.

If $y \not \equiv 1 \pmod{7}$, $y-1 \not \equiv 0 \pmod{7}$, so we must instead have $y \equiv 2 \pmod{7}$, which gives $y^4+y^3+y^2+y+1 \equiv 3 \pmod{7}$.

If $y \equiv 1 \pmod{7}$, we have $y^4+y^3+y^2+y+1 \equiv 5 \pmod{7}$.

Neither of those cases yields $y^4+y^3+y^2+y+1 \equiv 0,1 \pmod{7}$, so there are no solutions. $\square$

I claim that there are no solutions. We first use a small lemma:

[color=#ff9a00][b]Lemma:[/b][/color] If $p, q$ are primes so that $p \mid \frac{x^q - 1}{x - 1}$, then $q \mid p - 1$ or $q = p$.
[i]Proof:[/i] Let $o(x)_p$ denote the order of $x$ modulo $p$. Then if $p \mid (x^q - 1)/(x - 1)$ then $p \mid x^q - 1$ so that $o(x)_p \mid q$ and $o(x)_p \mid p - 1$, so that $o(x)_p \mid \gcd(q, p - 1) \in \{1, q\}$.

[b][color=#00f]Case 1:[/color][/b] $\left(o(x)_p = q \right).$ Then trivially $q \mid p - 1$.

[b][color=#00f]Case 2:[/color][/b] $\left(o(x)_p = 1 \right).$ Then $p \mid x - 1$, so that $x^{q - 1} + x^{q - 2} + \cdots + x + 1 \equiv q \equiv 0 \pmod p$. Hence $q = p$. $\square$

Now for $q = 7$, we find that $7 \mid p - 1$ or $p = 7$, in particular, every factor of $(x^7 - 1)/(x - 1)$ is either $0, 1$ modulo $7$. Therefore \[y^5 - 1 \equiv 0, 1 \pmod 7 \implies y^5 \equiv 1, 2 \pmod 7 \implies y \equiv 1, 4 \pmod 7. \] However, if $y \equiv 4 \pmod 7$, then $y - 1 \equiv 3 \pmod 7$, which is a contradiction. Similarly if $y \equiv 1 \pmod 7$, then $y^4 + y^3 + y^2 + y + 1 \equiv 5 \pmod 7$, which is another contradiction. Hence no solution exists, as initially claimed. $\blacksquare$

Assume FTSOC that there does exist an integer solution $(x,y)$ to the equation. Note that $(x,y)$ is a solution if and only if $(-x,y)$ is a solution, so we can assume WLOG that $x$ is a positive integer ($x=0$ clearly has no solutions).

[color=#9a00ff][b]1. If a prime $p\neq 7$ divides $\frac{x^7-1}{x-1}$, then $ord_p(x)=7$.[/b][/color]
Since $7$ is prime, this is basically equivalent to proving that $x$ cannot be $1$ mod $p$ if $p$ divides $\frac{x^7-1}{x-1}$. Assume FTSOC that $x$ is $1$ mod $p$, meaning that $p\mid x-1$. Then by LTE, we have that
\[\nu_p(x^7-1)=\nu_p(x-1)+\nu_p(7)=\nu_p(x-1),\]
since $p\neq 7$. Therefore, if $p\mid x-1$, that means that $p$ does not divide $\frac{x^7-1}{x-1}$, a contradiction. Therefore we must have that $ord_p(x)=7$. Additionally, extending this claim a bit further, we can also find that since $ord_p(x)=7$, we have that if $p\mid \frac{x^7-1}{x-1}$, then $p$ must be $1$ mod $7$, since $ord_p(x)=7$ must divide $\phi(p)=p-1$.

[color=#9a00ff][b]2. If $7\mid \frac{x^7-1}{x-1}$?[/b][/color]
Since $7\mid \frac{x^7-1}{x-1}$, we must have that $7\mid x^7-1$. By FLT we have that
\[x^7-1 \equiv x-1 \mod 7,\]
meaning that in order for $7$ to divide $x^7-1$, we must have that $x$ is $1$ mod $7$. This implies that $7\mid x-1$, and by LTE we have that
\[\nu_7(x^7-1)=\nu_7(x-1)+\nu_7(7)=\nu_7(x-1)+1,\]
which implies that
\[1=\nu_7\left(\frac{x^7-1}{x-1}\right)=\nu_7(y^5-1).\]
Using this, we get that we must have $7\mid y^5-1$. Running through all of the mods mod $7$, we get that this means that $y$ must be $1$ mod $7$. Therefore we have
\[y^5-1=(y-1)(1+y+y^2+y^3+y^4),\]
and since $7\mid y-1$ and $\nu_y(y^5-1)=1$, this means that $7$ cannot divide $1+y+y^2+y^3+y^4$. However, we have from [color=#9a00ff][b](1)[/b][/color] that all prime factors other than $7$ of $\frac{x^7-1}{x-1}=y^5-1$ are $1$ mod $7$, implying that $1+y+y^2+y^3+y^4$ must only have prime factors that are $1$ mod $7$. This means that
\[1+y+y^2+y^3+y^4\equiv 1\mod 7 \iff y(1+y+y^2+y^3)\equiv 0\mod 7.\]
However, running through all mods mod $7$, we get that $1+y+y^2+y^3$ is never $0$ mod $7$, meaning that we must have $y$ be $0$ mod $7$, a contradiction to our previous claim that $y$ must be $1$ mod $7$. Therefore we have that $7$ cannot divide $\frac{x^7-1}{x-1}$, meaning that this case has no solutions.

[color=#9a00ff][b]3. If $\frac{x^7-1}{x-1}$ is not a multiple of $7$?[/b][/color]
Since $7$ is not a prime factor of this number, we have that the prime factors of this number must all be $1$ mod $7$ by [color=#9a00ff][b](1)[/b][/color]. This means that the number $\frac{x^7-1}{x-1}=y^5-1$ and all its factors must be $1$ mod $7$. Factoring $y^5-1$, we get
\[y^5-1=(y-1)(1+y+y^2+y^3+y^4),\]
meaning that
\[1+y+y^2+y^3+y^4\equiv 1\mod 7 \iff y(1+y+y^2+y^3)\equiv 0\mod 7,\]
which we already established in [color=#9a00ff][b](2)[/b][/color] only works if $y$ is $0$ mod $7$. However, this implies that $(y-1)$, a factor of $y^5-1$, is $6$ mod $7$. This is a contradiction to our earlier claim that all factors of $\frac{x^7-1}{x-1}=y^5-1$ are $1$ mod $7$. Therefore there are no solutions in the case if $\frac{x^7-1}{x-1}$ is not a multiple of $7$. Since we have exhausted all possible cases, this implies that there are no integer solutions to this equation, finishing the problem.


analytic geometry
geometry
incenter
vector
inradius
angle bisector

[quote="iamagenius"]1. given triangle RST with R(-3,8), S(3,0), T(12,12). find the y-coordinate of the point at which the bisector of angle RST intersect RT.

2. triangle ABC has vertices (0,0), (10,0), and (6,6). find the ordered pair (x,y) of the incenter of the triangle. round to 4 decimal places.[/quote]
[hide="1"]
The triangle is a 10, 15, $\sqrt{241}$
Just use the angle bisector theorem to find the segment breakup in the triangle.
Then use stewart's theorem.
Then use some analytic.[/hide]

[quote="iamagenius"]wouldn't that be very messy. i'm trying to find a "elegant" solution.[/quote]

 Notice that the point you are looking for in #1 is just $\frac{2}{5}$ of the way between R and T so the answer is just $\frac{48}{5}$.

EDIT: I suck at calculating stuff.

[quote="iamagenius"]this is probably a dumb question but how did you get 48/5?[/quote]

x is the distance from R to the point where the angle bisector and RT meet

Angle Bisector theorem:

$\frac{10}{15}=\frac{x}{\sqrt{241}-x}$

so from that you get that $x=\frac{2\sqrt{241}}{5}$ and that is $\frac{2}{5}$ of the way up RT therefore the y coordiante is 2/5 the way up 8 and 12 which is 48/5

EDIT:Error in calculation
EDIT: First answer was right :D

You can use vectors to get 48/5 as the answer.

[quote="iamagenius"]2. triangle ABC has vertices (0,0), (10,0), and (6,6). find the ordered pair (x,y) of the incenter of the triangle. round to 4 decimal places.[/quote]

[hide]Let $A(0,0)$, $B(10,0)$, and $C(6,6)$.

Let the incenter be $I$. Let the altitude drawn from $I$ to $AB$, $BC$, and $CA$ be $D$, $E$, and $F$, respectively...

Then by using distance formula, we get $AB=10$, $BC=2\sqrt{13}$, and $CA=6\sqrt{2}$. The semiperimeter $s$ equals $5+\sqrt{13}+3\sqrt{2}$.

The length of the inradius is then calculated, which equals $r=\frac{[ABC]}{s}$ $=\frac{10 \cdot 6 \cdot \frac{1}{2}}{s}$ $=\frac{30}{s}$ $\approx 2.33496$.

Then $AD=s-2\sqrt{13}=5-\sqrt{13}+3\sqrt{2}$ $\approx 5.63709$.
$ID=r \approx 2.33496$, so we have the coordinate of the incenter.
$I$ has same $x$-coordinate as $D$ and $y$-coordinate $r$.

Then the coordinate is approximately $(5.6371,2.3350)$.[/hide]


linear algebra
matrix
algebra
polynomial
function
linear algebra unsolved

Let $A \in M_n(C)$.Show that:
1)$A$ is diagonalizable if and only if its class of similarity it's closed.
2)$A$ is nilpotent if and only if $0$ it's adherent to its class of similarity.Here the ring can be also $R$.

In (1), I believe you mean diagonalizable in $M_n(\mathbb C)$, not $M_n(\mathbb R)$, right? Otherwise, I think it's wrong.

Suppose first that the similarity class of $A$ is closed. All the matrices obtained from $A$'s rational canonical form by replacing the $1$'s underneath (or above; it depends on the way you prefer to visualize the rational canonical form) the block diagonal with some arbitrary $\varepsilon$ belong to $A$'s similarity class, and since $\varepsilon$ can be made arbitrarily close to $0$, it means that the block diagonal matrix belongs to the closure of the similarity class, and thus to the similarity class itself. This implies that the rational canonical form itself must be block diagonal with blocks of dimension $1$ (corresponding to real eigenvalues) or $2$ (corresponding to irreducible factors $x^2+bx+c$ of the characteristic polynomial), which is just another way of saying that $A$ is diagonalizable in $M_n(\mathbb C)$.

If, on the other hand, $A$ is diagonalizable in $M_n(\mathbb C)$, suppose we show that its similarity class in $M_n(\mathbb C)$ is closed (also in $M_n(\mathbb C)$). Then, since it's well-known that two real matrices are similar in $M_n(\mathbb C)$ iff they are similar in $M_n(\mathbb R)$, we find that the similarity class of $A$ in $M_n(\mathbb R)$ is closed in $M_n(\mathbb R)$, as desired. This means that what we want to prove now is:

If $A\in M_n(\mathbb C)$ is diagonalizable, then its similarity class is closed.

Take any matrix $M$, and assume $S^{-1}MS$ is $M$’s Jordan form. Then, $M\mapsto SNS^{-1}$ is a continuous function $\varphi: M_n(\mathbb C)\to M_n(\mathbb C)$, where $N$ is the nilpotent matrix having $1$ below the main diagonal wherever $S^{-1}MS$ does, and $0$ everywhere else. The set of diagonalizable matrices is $\varphi^{-1}(O)$, so it’s closed, and the conclusion follows.


3 points are chosen at random along the square's outline. They then combine to form a triangle.
What is the probability that the centre of the square will be 'engulfed' by this triangle?

this seems far too simple to be correct,  but i don't see any flaws in it.  then again, it is past midnight.

first, pick apoint.  any point.  as long as it's on the square

then take a line, pass it through that point and the center of the square.

this line divides the square into two regions of equal area/perimeter.  we'll name these two regions Left and Right.

so now picka  second point anywhere.

now the third point:
if the second point was in Left, the third point must be in Right.
if the second point was in Right, the third point must be in Left.

either, way there's a 1/2 probability of picking a third point that works.

the picking of the first two points are arbitrary, and the probabilty of getting a final point that works is 1/2, so the probability is 1/2.

I dont think that way works cuz u may take the first point, then have the second and third points on the sides adjacent to the first, but very close to it, and the center is not included...

More specifically you may draw a line from 2nd pnt thru center.  Let 2nd pnt be in left, this new line divides right, into 2 parts as well, right-top, right-bottom  :P .  And all points in right-top would counter the arguement...

would it be the same answer if square is changed to circle?

To answer my own question, I don't think so, and I think the answer is 13/48 but I could be totally wrong...for the circle it's 1/4 so pretty close

yea, that solution's so wrong it's not even funny.

thought i could get away without having to do case analysis >_<

ah, i gues that's what i get for staying up so late.


Prove for $ a,b,c\ge0$ that $ a^3/b\plus{}b^3/c\plus{}c^3/a\ge a^2\plus{}b^2\plus{}c^2\plus{}3/2(a\minus{}b)^2$

hello, do you mean
$ \frac{a^3}{b}\plus{}\frac{b^3}{c}\plus{}\frac{c^3}{a}\geq a^2\plus{}b^2\plus{}c^2\plus{}\frac{3}{2(a\minus{}b)^2}$ $ \forall a,b,c>0$?
Sonnhard.

It means to be
Prove for $ a,b,c\ge0$ that $ a^3/b \plus{} b^3/c \plus{} c^3/a\ge a^2 \plus{} b^2 \plus{} c^2 \plus{} (3(a \minus{} b)^2$)/2

[quote="Inequalities Master"]I think is highly likely:

$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {2(a \minus{} b)^2}{3}$ or

$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3(a \minus{} b)^2}{2}$

because the inequality:
$ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3}{2(a \minus{} b)^2}$

is clearly wrong if $ a \minus{} b \rightarrow 0$

[b]dima ukraine[/b], please say us what is the correct inequality.[/quote]
I think $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3(a \minus{} b)^2}{2}$
And I think it can be kill by Cauchy schawrt ineq
Because i have a  similar problem
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\geq a \plus{} b \plus{} c \plus{} \frac {4(a \minus{} b)^2}{a\plus{}b\plus{}c}$

Well, allnames, your second problem is very known.
It's from Balkan MO 2005.

The inequality is true if $ b \equal{} min  [a,b,c]$
We use the similar method like the inequality that appeared in Balkan Olimpiad 2005.  :wink:

This is $ \frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3(a \minus{} b)^2}{2}$ for all $ a,b,c\ge 0$ I have only one ugly solution and I am interesting in others


I just installed asymptote 1.55, gsview 4.8, and ghostscript 8.63.  I just bought a new laptop and it came with a 64-bit Vista OS.  I was careful to get the 64-bit versions of all the above programs.  Whenever I do anything in asymptote, it gives me "cannot write out.eps".  What should I do?

EDIT: I could not find a 64 bit version of asymtote.  Could this be the problem?

Modify BOTH shortcuts (start menu AND desktop) to start in "C:\Users\funmath\Documents\Asymptote" or any other directory which you have permission to write to. The current start directory is in "C:\Program Files\Asymptote" which is clearly not writable by users because of Vista's architecture.

(Obviously, replace funmath with your real Windows username.)


A circular table is tangent to two adjacent walls of a rectangular room. Point P, on the edge of the table, is 12 inches from one wall and 16 inches from the other wall as shown to the right. What is the number of inches in the diameter of the table? Express your answer to the nearest whole number.

[hide]Subtract 12 and 16 from side with the radius(r) perpendicular to the side.
Get r^2 = (r-12)^2 + (r-16)^2 which simplifies to r^2 - 56r +400 = 0
solve with the quadratic formula to get 47.5 and 8.4,rounded. 8.4 won't work so 47.5 is the radius.  So the diameter is [b]95[/b][/hide]

I would recomemd a picture.

I'll add one. Just pretend it's a good picture.


[quote="chien than"]Let $a;b;c>0$.Prove that:
$\sum \frac{a^{2}}{a^{2}+b^{2}+ab}+\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\le 2$[/quote]
That's problem of Mr. Thuan, not yours :P. The solution is very simple
\[2-LHS=\frac{(a+b+c)(ab+bc+ca)(ab^{5}+bc^{5}+ca^{5}-ab^{2}c^{3}-bc^{2}a^{3}-ca^{2}b^{3})}{(a^{2}+b^{2}+c^{2})(a^{2}+ab+b^{2})(b^{2}+bc+c^{2})(c^{2}+ca+a^{2})}\ge 0 \]
:)

[hide="To Chienthan"]@chienthan: why do you always post the old problems, most of them is not yours. [/hide]

[quote="Zamfirmihai"]$a,b,c$ are >0[/quote]
Zamfirmihai, I speak about identity. :wink:

maybe the identity works for a,b,c>0 :maybe:

Sorry, I have some mistakes in typing. :(


if f(x)= (4x+5)/(2x-3), find the inverse of f(x).

im having trouble figuring this one out


show that f:Z\to Z does not exist for:
f(x+f(y))=f(x)-y \formal x,y\in Z

Using the information:
f(f(-x)) = x      ,     f(-x)=-f(x)  ,   f(m*x) =m*f(x)   ,   f(n) = k*n   , x, m ,n  and k (constant ) are integers .
 f(f(-n)) = f(-k*n) = (-k)*f(n) = -k^2 * n  =  n   then  k^2  =  -1 ,a contradiction .

How did you get $ f( \minus{} x) \equal{} \minus{} f(x)$?  I got $ f(f( \minus{} x) \plus{} f(x)) \equal{} 0$ instead by replacing $ x$ by $ f(x)$ and $ y$ by $ \minus{} x$.
Or instead, replacing $ y$ by $ \minus{} f( \minus{} x)$ yields $ f(x \plus{} f( \minus{} f( \minus{} x))) \equal{} f(x) \plus{} f( \minus{} x)$, but we don't know if the LHS is 0 yet.

For $ f(f(\minus{}x)\plus{}f(x))\equal{}0$, I think we should have $ f(x)\equal{}0 \iff x\equal{}0$.

When you see that $ f(f(\minus{}x)\plus{}f(x))\equal{}0$ that means that there exists an integer $ k$ such that $ f(k)\equal{}0$ Pugging $ y\equal{}k$ you get that $ k\equal{}0$ so $ f(0)\equal{}0$
Then plugging $ x\equal{}0$ we obtain 1) $ f(f(y)\equal{}\minus{}y$ Then it is obvious that $ f(x)$ must be biyective.  Then we can plug in the original equation $ f(y)\equal{}\minus{}x$ and we conclude that f(-x)=y (using 1) and making the substitution f(-x)+f(x)=0 and then you can continue as they have done before.

Daniel

Yeah, I think the other way to get $ f(0)\equal{}0$ is to replace $ y$ by $ 0$ in $ f(x\plus{}f(y))\equal{}f(x)\minus{}y$ to get $ f(x\plus{}f(0))\equal{}f(x)$.  However, is it necessary to show that $ x\equal{}0$ is the only value for $ f(x)$ to be $ 0$ (or showing f is bijective)?
You said that from $ f(f(y)) \equal{} \minus{}y$, $ f$ must be bijective, but could you offer a proof please?

Yes of course... let $ f(y)\equal{}k$  so $ f(k)\equal{}\minus{}y$ so for every $ x$ there exists $ k$ such that $ f(k)\equal{}x$ then $ f(x)$ is suryective
If $ f(y)\equal{}f(z)\equal{}m$ then $ f(m)\equal{}\minus{}z\equal{}\minus{}y$ and $ z\equal{}y$ so $ f(x)$ is inyective. Since it is both suryective and inyective it follows that it is biyective
Daniel


geometry
MATHCOUNTS
ARML
3D geometry
sphere
trigonometry
AMC 8

Here's a question I put on our local high school contest. It was supposed to be easy, but you'd be surprised how many people didn't get it.

On a very clear night with a very smooth, calm ocean, if you stand on top of a coastal bluff, you can just barely see the blinking light at the top of an offshore oil rig. The light seems to be right on the horizon. If the bluff is 100 meters tall and the oil rig is 60 meters tall, how far away from the shore is the oil rig? Assume that the radius of the earth is 6360 kilometers. Round your answer to the nearest whole kilometer.

Nice question.  I'm not surprised people found it difficult, though -- how long did they have to solve it in?

Joel: Team round, 6 member teams, 8 questions total, 25 minutes for the 8 problems, calculators permitted.

[quote="Kent Merryfield"]Here's a question I put on our local high school contest. It was supposed to be easy, but you'd be surprised how many people didn't get it.[/quote]

You are talking about a senior high school contest, right? That would be a contest for young people too old to take the AMC 8 or to participate in Mathcounts. I take the AMC 8 and Mathcounts implicit syllabuses (STRICTLY prealgebra for AMC 8, mostly "algebra 1" and "middle school geometry" for Mathcounts, as detailed in the Mathcounts School Handbook) as the level of this Getting Started Forum. If you have a challenge question for senior high students, which most of them didn't tackle successfully, you have at least an Intermediate Forum challenge question, it seems to me. I'm quite empirical about gauging difficulty levels.

High school contest, yes, but the question itself fits comfortably within the scope of Mathcounts subjects - as you put it, "algebra 1" and "middle school geometry." Let's see what the readers of this forum can do.

we're assuming that the person is 0 ft tall, right?

well, if im interpreting the question correctly...[hide]

We can construct two right triangles, each of which has one leg as a radius of the earth (which is assumed to be a circle for simplicity):

hyp: 6360.1 km, side: 6360 km and x

hyp: 6360.06 km, side: 6360 km and y

Add x and y to get the total length between the two points, which, using the Pythagorean Theorem and a calculator, appears to be 63 km rounded.[/hide]

I can see some reasons why a lot of students might have missed this problem. First of all - the interpretation of the problem. Students notoriously have a hard time with "word problems." If the same problem had been presented purely geometrically, more students may have gotten it.

The second reason is with the conversion of units between meters and kilometers. Conversions of units always give students a hard time, and is a potential cause of a lot of careless errors.

I assume students were allowed to use a calculator on this question? If so, then I do like it though. It involves some skills that students aren't usually asked to exhibit on a contest problem.

I have to agree with gauss202.  This looks like it could be a hard problem.  You know this was intended to be an easy problem.  You told the readers of this forum this was intended to be an easy problem.  But did you tell your ARML high school students this was intended to be an easy problem?

I happen to have seen many problems of this general sort.  G. Bomford's "Geodesy" and John P. Snyder's various USGS works contain the derivation of solutions to problems like this.  Some are easy.  Some are extraordinarily difficult.

If I'm expecting a hard problem, I think I'd save this problem for last.

If you tell me it's intended to be an easy problem, I'm sure I'd look look at it.

- The km vs meters issue will trip up a lot of people.

- The word problem itself will trip up (or confuse) a lot of people.  "What does he mean by 'seems to be on the horizon'?  Is this some sort of a trick question?"

- A student might wonder, "I know they correct the time of sunrise and sunset for atmospheric refraction.  Do I have to correct for atmospheric refraction on this problem?  I don't know the correction, so why bother."

- A student might wonder, "do I have to correct for tides?  I don't know the correction, so why bother."

- A student may wonder, "do I have to correct for the earth being an ellipsoid, rather than a sphere?"

- As someone else asked, a student might wonder, "do I have to correct for the height of the observer?"

- When I solved this problem earlier today, I used arccos.  With the method I used, I don't think I could have solved this without a calculator.  I don't think I could have solved this with a slide rule.  If I had written intermediate answers to paper, I would have needed to record about 6 digits of precision, so I may have lost something in copying intermediate results.

- It's also quite unnatural to work with such disparate figures (60 meters vs ~6000000 meters).  If I'm a visual person, how do I make a meaningful sketch?

I know the underlying math behind this problem is really quite simple.  I also admit many of the above concerns are red herrings.  But all these unusual factors do conspire to make this problem intimidating.

It's a nice problem.  It should be solvable by a bright high school student.  But I'm not surprised that few students solved the problem.

is this right?:

[hide]

since the light appears to be on the horizon, it's actually on the "other" side of the curvature of the earth, so that everything on the rig below the light is hidden (blah I can't draw it here). So we have two right triangles, drawing a line from the center of the earth to the top of the bluff and to the top of the oil rig (the lines are 6360.1 km and 6360.06 km, respectively).

Then, find the two angles of the right triangles:

cos :theta: = (6360)/(6360.1) ...  :theta: = 0.005607 radians

cos :theta: = (6360)/(6360.06) ...  :theta: = 0.00434 radians

add them together to find the angle of the larger triangle, which isn't right anymore, and we find that the larger angle is 0.0099514 radians.

using s = r :theta: , we have s = 6360 * 0.0099514 radians  :approx: 63.2908 km  :approx: 63 km

this is the arclength distance between them, because if you were "measuring distance" by traversing the surface of the earth, this is what you would get.

alternatively, you could find the "straight distance" between the top of the bluff and the rig light bulb, not the distance of the round earth below them, simply by making another triangle with angle 0.0099514 radians and using the law of cosines to find:

x :^2: = a :^2: + b :^2: - 2ab*cos :theta:

but I'm guessing that the question wants the first answer, the curved distance of the earth below the bluff and the rig. either way however, you would get around 63 km if you rounded to the nearest integer b/c  :theta: is so small in this question.

[/hide]

paladin8 and VeCTer are both correct. My, rcv, that's an entire school of red herring. At that you missed the one question I was asked, which is the one VeCTer had: did I mean the straight line distance of the distance along the curve of the earth? I actually meant the straight line distance, but it makes about 2 cm worth of difference in a question that called for an answer rounded to the nearest km, so it doesn't matter.

As long as you take it to be the straight line distance, it shouldn't be necessary to use trigonometry.

I did see one particular wrong answer - it's > 2000 km - that indicates a failure to convert between meters and kilometers.

The fact that the heights of the bluff and the oil rig are much less than the radius of the earth (by a factor of  :sim:10^5) is a feature, not a bug. It can be exploited to give a simple approximating formula that might prove useful to you the next time you're in a tall building, or in a car looking a a distant mountain.

One lesson to the problem solver. Of course you're going to draw a picture. But do NOT attempt to draw it accurately or to scale, because you'd never be able to see anything. Draw it schematically, in such a way that you can see what is going on.


What integer is closest to $ \frac{100}7$?

The two nearest multiples of $ 7$ to $ 100$ are $ 14(7)\equal{}98$ and $ 15(7)\equal{}105$. $ 100$ is closer to $ 98$ than $ 105$, and so $ \frac{98}{7}\equal{}\fbox{14}$ is the nearest integer to $ \frac{100}{7}$.


inequalities
rearrangement inequality
inequalities proposed

[quote="Nebraska boy"]I've got several days trying to prove this inequality  :blush: , could someone help me?

$ x^2y \plus{} y^2z \plus{} z^2x \le x^3 \plus{} y^3 \plus{} z^3 \le 1 \plus{} \frac {1}{2}(x^4 \plus{} y^4 \plus{} z^4)$

where $ x, y, z$ are real non-negative numbers such that $ x \plus{} y \plus{} z \equal{} 2.$[/quote]
The left inequality follows directly from rearrangement inequality or AM-GM.

For the right one :

We have to prove that :

$ x^4 \plus{} y^4 \plus{} z^4 \geq 2(x^3 \plus{} y^3 \plus{} z^3) \minus{} 2$

$ \Leftrightarrow x^4 \plus{} y^4 \plus{} z^4 \geq (x \plus{} y \plus{} z)(x^3 \plus{} y^3 \plus{} z^3) \minus{} 2$

$ \Leftrightarrow (xy \plus{} yz \plus{} zx)(x^2 \plus{} y^2 \plus{} z^2) \minus{} 2xyz \leq 2$

But,this is clearly true,because from AM-GM's inequality we have:

$ (xy \plus{} yz \plus{} zx)(x^2 \plus{} y^2 \plus{} z^2) \minus{} 2xyz \leq \frac { (2xy \plus{} 2yz \plus{} 2zx)(x^2 \plus{} y^2 \plus{} z^2) }{2} \leq \frac { (x \plus{} y \plus{} z)^4 }{8} \equal{} 2$

Done !

 :)


Hi everyone!
Does someone know tutorials for solving linear programming problems? I am looking for tricks to help me.

is linear programming like.. graphing out the region in which something may occur. and finding the maximum and minimum?

umm.. just read the question and write the inequalities and graph them

[url=http://www.google.com/search?hl=en&hs=6Yf&client=firefox-a&rls=org.mozilla:en-US:official&q=linear+programming+exercises&spell=1]Here's a link[/url]
Try googling around. You'll find as much as you need to do exersises on your own.


AMC
AIME
algebra
polynomial
AMC 12
AMC 12 B
AMC 12 A

So far...

[url=http://mathideas.org/public/math/mock2007A_ans.pdf]Answer Key[/url]

[hide="Top three scorers"]

1. mathgeniuse^ln(x) (135)
2. matt276eagles (120)
3T. Xanthos C. Guin (118.5)
3T. deej21 (118.5)

[/hide]

[hide="Number of participants"]
24
[/hide]

[hide="All scores"]
76.5
81
87
88.5
90
93
97.5
99
100.5
102
103.5
103.5
105
108
108
108
111
112.5
117
117
118.5
118.5
120
135
[/hide]

[hide="Number of AIME qualifiers"]
16
[/hide]

[hide="Average score"]
104.1875
[/hide]

[hide="Minimum score"]
76.5
[/hide]

[hide="Median"]
103.5
[/hide]

[hide="Highest number of correct answers"]
22 (mathgeniuse^ln(x))
[/hide]

Nice job to the outright (and unexpected, IMHO) winner - mathgeniuse^ln(x). :first:

Note: nr1337 got a perfect score, but he wasn't able to submit his answers on time because he was at MOP :(

Yay!  Thanks for posting the results!

Man, I barely qualified, and I got below average, but I qualified  :lol:  :D

I got first with a 117, Wow that's surprising, (I should have left one more blank)

[quote="eryaman"]Yarr 105...that was pretty hard.[/quote]

I agree, harder than most AMC 12s I have taken.

[quote="eryaman"]Yarr 105...that was pretty hard.[/quote]

I agree.

The only thing I noticed was that there seemed to be a lot of problems about solutions to equations.

Chess64, can you tell me which questions I missed?

chess64, when are the solutions going to be posted or are we suppose to post the solutions?

Not yet, because answers aren't due until tomorrow at 7:00.

Did anyone else think that there were a lot of problems on solving equations?

[quote="b-flat"]Did anyone else think that there were a lot of problems on solving equations?[/quote]

Yeah, I did.

Does anyone who got 13 right want to pm me the solution, please?

[quote="solafidefarms"]Eh, I got that 102. Hurrah for being, at least at one time, the median score :D.[/quote]

Isn't the median technically 102.75?

[quote="eryaman"]Splash, I sent you a PM with a solution to #13.[/quote]

Really, I didn't get it yet?

By the way, is jclarke's name before mine because he sent in his answers earlier than me?

I got 102, 15 right, 2 wrong, 8 blank (I missed #3  :wallbash: ). Yay for median, and >100 score! That was a fun test (but hard).

Wow, 97.5, that's really bad.  It was much harder than a real AMC 12, but still I made a lot of silly mistakes.  Oh, well, at least it doesn't count . . .

[quote="mathgeniuse^ln(x)"]I just turned in my answers, the maximum score I could get is 141.0.[/quote]

That's what my max score could be but look what I got. If you got less than 4 questions wrong, then you beat me. Then again, you took the test after all the typos were fixed and you had a whole day.

111

uh...17 right...2 wrong...6 blank...I think...

Can someone PM me the solution to #3? It's kinda bugging me :/

Haha...for one of them...I got the right answer but looked at the wrong letter...>.>. Oh well, I would still like the solution to #3.

The [url=http://mathideas.org/public/math/mock2007A_ans.pdf]Answer key[/url] has been posted, but I still have a few more tests to grade...

(you can discuss solutions now)

AAAAAAARG!!!!

I had printed the test out and done the problems to submit when I got back from my MOP test, but then I got back and it was too late :(  Oh well, can I still take the Mock AIME?

Sorry :( Well, how'd you do? (BTW, how is MOP going? :))

145.5.  I could swear the answer to #23 was 1, which wasn't one of the choices, so I left it blank (it's the f(k) = 1 + 1/k one).  MOP's going fine, though it's the second to last day.... MOP tests are fun though :)

Sequence $\{ x_n \}^\infty_{n=1}$ is defined as $x_1 = 1$,

\[ x_{2n} = 1 + x_n \text{ and } x_{2n+1} = \frac{1}{x_{2n}} \]

for every positive integer $n$.  Prove that each positive rational number appears in the sequence exactly once.

^--- our very last MOP test problem.  It was quite fun; after I proved it I even wrote neat little algorithm for finding p/q in the sequence :)

Nice job! You should definitely have won, then, and for #23 I made an error when I solved it myself, so the answer actually was 1. :blush:

If I solve that problem, I'll PM you the solution. :)

[quote="SplashD"]nr1337, what grade are you in?[/quote]

9th (Red MOP)

I guess I don't know the difference between the sum of numbers in an interval and the number of numbers...  I swear I screw up that kind of problem on every contest that ever appears on.  Nice job mathgenius x (I simplified your name  :D ) and nr1337.

I really liked #25 because it took me like 5 seconds to solve.  #24 was funny because I kept debating with myself whether I should answer 48 or leave it blank, but in the end I answered 48.  Nice test, chess64 - It was close in difficulty to a real AMC, except that there weren't as many extremely easy problems (even #1 I misinterpreted at first because there was a word missing).  Thanks!

Thanks

I still don't know which question I got wrong.  But, for the last 2 I actually found the quadruple and the triple, respectively then I was able to solve the questions.

Wow I guess some people didn't realize:

[quote="Sample person's answers"]
1. A
2. B
3. C
4. D
5. E
6. F [b]-A[/b]
...
[/quote]

Here, the person put F for #6 but it was actually A. Notice the text which I have bolded here. I added that next to your answers, so you should be able to see which ones you got wrong. :roll:


geometry
circumcircle
parallelogram
ratio
quadratics
power of a point
radical axis

Given a non-isoceles triangle $ABC$ inscribes a circle $(O,R)$ (center $O$, radius $R$). Consider a varying line $l$ such that $l\perp OA$ and $l$ always intersects the rays $AB,AC$ and these intersectional points are called $M,N$. Suppose that the lines $BN$ and $CM$ intersect, and if the intersectional point is called $K$ then the lines $AK$ and $BC$ intersect.
$1$, Assume that  $P$ is the intersectional point of $AK$ and $BC$. Show that the circumcircle of the triangle $MNP$ is always through a fixed point.
$2$, Assume that $H$ is the orthocentre of the triangle $AMN$. Denote $BC=a$, and $d$ is the distance between $A$ and the line $HK$. Prove that $d\leq\sqrt{4R^2-a^2}$ and the equality occurs iff the line $l$ is through the intersectional point of two lines $AO$ and $BC$.

a) Let $MN\cap{BC}=\{T\}$ and $C(\triangle{MNP})\cap{BC}=\{P,Q\}$.Then \[ TB\cdot{TC}=TP\cdot{TQ}  \] and \[ \frac{CP}{BP}=\frac{CT}{BT}.  \]
Suppose that $TB=x$, $BP=y$,$PQ=z$, $QC=t$ ($T,B,P,Q,C$ are collinear in this order). Then
\[ (z+t)x=y(x+y+z+t)  \] and
\[ xt=y(x+y+z).  \] Thus $yt=zx$, or
\[ t=\frac{y(x+y+z)}{x}=\frac{zx}{y}.  \] It yields that $y^2(x+y+z)=x^2z$, consequently $y(y+z)=xz$. In conclusion $Q$ is the midpoint of $BC$, which is a fixed point.

b) If $H'$ is the ortocenter of $\triangle{ABC}$, then it is well-known that the points $H,H',K$ are collinear. Therefore:
\[ d\leq{AH'}=\sqrt{4R^2-a^2},  \] and the equality cases can be obtained quite easily.

hey could someone explain 'the easy' bit at the end of part ( b ) ? i cant quite get it so that would be appreciated.
btw, the 'well-known result' sailor quoted is from a previous ISL. i'll prove it now:
Question: B'C'BC is concyclic with B' on AC and C' on AB. Prove the orthocentre H' of AB'C', the orthocentre H of ABC and T = BB' intersect CC' are collinear.
Proof: define B'' on AC s.t. BB'' is perp to AC. Similarly define C'' on AB. Consider a circle C1 with diameter BB' and circle C2 with diameter CC'. It suffices to prove that the H', H and T all lie on the radical axis of C1 and C2. Define P(x, T) = power of X wrt circle T.
p ( H, C1) =  HB * HB '' = HC * HC'' ( BCB''C" is cyclic)
               = p (H, C2)
p ( H',C1 ) = p (H',C2) by exactly same argument.
p ( T, C1) = TB * T B' = T C * T C'  ( BCB'C' is cyclic)
                = p ( T, C2)
so we are done.

[quote="Sailor"]

$\frac{CP}{BP}=\frac{CT}{BT}$.

 consequently $y(y+z)=xz$. In conclusion $Q$ is the midpoint of $BC$, .
[/quote]

How to prove these two facts?

The lines AB, AC, BN, CM form a complete quadrilateral AMKN with diagonals AK, MN, BC. The 2 diagonals MN, AK cut the 3rd diagonal BC at T, P harmonically. You can project the line AT to infinity, the quadrilateral BCNM becomes a parallelogram B'C'N'M' and P goes into the midpoint P' of B'C'. Cross ratio is preserved in a central projection, $\frac{CP}{BP}\cdot \frac{BT}{CT}= \frac{C'P'}{B'P'}\cdot \frac{B'T'}{C'T'}=-1,$ because C'P' = -B'P' (P' is the midpoint of B'C') and B'T' = C'T' (T' is at infinity).

$MN \perp AO$ means MN is antiparallel to BC with respect to the $\angle CAB,$ quadrilateral BCNM is cyclic, MN is radical axis of the circumcircles of the $\triangle MNP$ and of the quadrilateral BCNM. $T \in MN \Longrightarrow TB \cdot TC = TP \cdot TQ.$

The quadratic equation $x^{2}z = y^{2}(x+y+z)$ for x has 2 roots $x = \frac{y(y+z)}{z}$ and $x =-y.$ Ignoring the negative root, $xz = y(y+z) = yt,$ $y+z = t,$ $BQ = BP+PQ = QC.$

(a) We claim that the fixed point is just the midpoint $X$ of $BC.$ Let $T = MN \cap BC.$ Observe that by Ceva/Menelaus Duality, we have that $(B, C, R;, T) = -1.$ Hence, it's well-known that $TR \cdot TX = TB \cdot TC.$ Since $MNCB$ is cyclic, we have that $TB \cdot TC = TM \cdot TN$ by Power of the Point, and so combining the two previous results gives $TM \cdot TN = TR \cdot TX$. By Power of the Point, this implies that $MNXR$ is cyclic, and so $X$ is the desired fixed point.

(b) We claim that $HK$ passes through the orthocenter $H'$ of $\triangle ABC.$ This would imply the first part, because $AH' = \sqrt{4R^2-a^2}$. In other words, we have to show that $HH', BN, CM$ are concurrent. By Desargues' Theorem, we just need that $T, BH' \cap NH, CH' \cap MH$ are collinear. This is saying that $T$ is collinear with the orthocenters of $\triangle ABN, \triangle ACM.$ However, this is obvious because they lie on the radical axis of the circles with diameters $MN$ and $BC.$

As for the second part, it's clear that equality holds iff $AH' \perp HH'.$ This means that equality holds iff $HH' || BC$, and this is easily seen to be equivalent to $\ell$ passing through $AO \cap BC.$

$\square$


limit
integration
calculus
real analysis
real analysis theorems

So say we're trying to evaluate

$ \lim_{k\to 0} \frac{1}{k} \int_0^k f(x) dx$.

Are we allowed to say that $\lim_{k\to 0} \int_0^k f(x) dx$ is always equal to zero if $f(x)$ does not tend to infinity as $x$ tends to zero (and thus we can allow L'Hopital's rule) and have our argument be mathematically rigorous? I know this argument makes sense, but I just wanted to be sure that it is rigorous enough without needing any complicated analytical proof.

Any confirmation would be nice, thanks.

Yes. If $\int_0^k f(x)dx$ is a proper integral, then $f$ is bounded on $[0,k]$, say $|f|\leq M (M>0)$. Thus given $\epsilon > 0$, let $\delta = \epsilon / M$, then for $0<h<\delta$, we have $|\int_0^\delta f(x)dx |\leq M\delta < \epsilon$, which means that $\lim_{k\to 0} \int_0^k f(x)dx \to 0$ as $k\to 0$.


floor function
number theory proposed
number theory

Prove that we can find infinitely many $n\in N$ s.t. $\lfloor \sqrt{2} n \rfloor$ is a square number.

I found a solution of this problem using unelementary method.

I hope I can see elementary solution here!  :D

I've seen previously a solution (which I found amazing) which uses the binary expansion of $\sqrt{2}$ Is'it yours also ?