For $ a>0,\ b>0,\ c>0$ with $ a\plus{}b\plus{}c\equal{}1$, prove the following inequality.

\[ \frac{a^4}{\sqrt{b^4\plus{}c^4}}\plus{}\frac{b^4}{\sqrt{c^4\plus{}a^4}}\plus{}\frac{c^4}{\sqrt{a^4\plus{}b^4}}\geq \frac{1}{3\sqrt{2}}\]

From rearrangement,
\[ \sum\frac{a^4}{\sqrt{b^4\plus{}c^4}}\ge\sum\frac{b^4\plus{}c^4}{2\sqrt{b^4\plus{}c^4}}\equal{}\frac 12\sum\sqrt{b^4\plus{}c^4}\]
On the other hand,
\[ b^4\plus{}c^4\ge\frac{(b\plus{}c)^4}{8}\implies\sum\sqrt{b^4\plus{}c^4}\ge\frac{1}{2\sqrt 2}\sum (b\plus{}c)^2\]
Finally,
\[ \sum (b\plus{}c)^2\ge\frac 13\left(\sum (b\plus{}c)\right)^2\equal{}\frac 43\]
Hence we are done. :)

Maybe your solution is correct, although I can't understand rearrangement.

Could anyone make a confirmation of nayel solution?

OK I'll try to make it clearer.

Since the sequences $ (a^4,b^4,c^4)$ and $ \left(\frac{1}{\sqrt{b^4\plus{}c^4}},\frac{1}{\sqrt{c^4\plus{}a^4}},\frac{1}{\sqrt{a^4\plus{}b^4}}\right)$ are similarly sorted, rearrangement implies
\[ \sum\frac{a^4}{\sqrt{b^4\plus{}c^4}}\ge\sum\frac{b^4}{\sqrt{b^4\plus{}c^4}}\]
and
\[ \sum\frac{a^4}{\sqrt{b^4\plus{}c^4}}\ge\sum\frac{c^4}{\sqrt{b^4\plus{}c^4}}\]
Add the two inequalities and divide by $ 2$ to get what I wrote above. :)

kunny, nayel meant that if $ a\geq b\geq c$ then
$ a^4\geq b^4\geq c^4$ and $ \frac {1}{\sqrt {b^4 \plus{} c^4}}\geq\frac {1}{\sqrt {a^4 \plus{} c^4}}\geq\frac {1}{\sqrt {a^4 \plus{} b^4}}.$
Thus, $ \sum_{cyc}\frac {a^4}{\sqrt {b^4 \plus{} c^4}}\geq\sum_{cyc}\frac {b^4}{\sqrt {b^4 \plus{} c^4}}$ and $ \sum_{cyc}\frac {a^4}{\sqrt {b^4 \plus{} c^4}}\geq\sum_{cyc}\frac {c^4}{\sqrt {b^4 \plus{} c^4}}.$
After summing it gives $ \sum_{cyc}\frac {a^4}{\sqrt {b^4 \plus{} c^4}}\geq\sum_{cyc}\frac {b^4 \plus{} c^4}{2\sqrt {b^4 \plus{} c^4}}.$
My proof bases on Holder:
$ \left(\sum_{cyc}
\frac{a^4}{\sqrt{b^4\plus{}c^4}}\right)^2\sum_{cyc}a(b^4\plus{}c^4)\geq(a^3\plus{}b^3\plus{}c^3)^3.$
Hence, it remains to prove that
$ 18(a^3\plus{}b^3\plus{}c^3)^3\geq(a\plus{}b\plus{}c)^4\sum_{cyc}(a^4b\plus{}a^4c),$ which is true.

[quote="kunny"]Thank you! :)

This is how the problem is made.

$ \frac {1}{3}\left(\frac {a^4}{\sqrt {b^4 \plus{} c^4}} \plus{} \frac {b^4}{\sqrt {c^4 \plus{} a^4}} \plus{} \frac {c^4}{\sqrt {a^4 \plus{} b^4}}\right) \equal{} \frac {1}{3}\left(\sqrt {\frac {a^8}{b^4 \plus{} c^4}} \plus{} \sqrt {\frac {b^8}{c^4 \plus{} a^4}} \plus{} \sqrt {\frac {c^8}{a^4 \plus{} b^4}}\right)$

$ \geq \sqrt {\frac {1}{3}\left(\frac {a^8}{b^4 \plus{} c^4} \plus{} \frac {b^8}{c^4 \plus{} a^4} \plus{} \frac {c^8}{a^4 \plus{} b^4}\right)}$[/quote]

Are you sure? :maybe:

Ooops! $ y\equal{}\sqrt{x}$ was convex upward function. :blush:


How to find the general formula for
$ \sum_{i\equal{}1}^{n}i^{4}$

Assuming you know the formulas for
\[ \sum_{k\equal{}1}^{n}k^{r}\]
for $ r\equal{}0,1,2,3$, you can get the formula for $ r\equal{}4$ using
\[ n^{5}\equal{}\sum_{k\equal{}1}^{n}(k^{5}\minus{}(k\minus{}1)^{5})\equal{}\sum_{k\equal{}1}^{n}(5k^{4}\minus{}10k^{3}\plus{}10k^{2}\minus{}5k\plus{}1)\]
(Let $ S_{r}(n)\equal{}\sum_{k\equal{}1}^{n}k^{r}$, then we have $ n^{5}\equal{}5S_{4}(n)\minus{}10S_{3}(n)\plus{}10S_{2}(n)\minus{}5S_{1}(n)\plus{}S_{0}(n)$.)
More generally, if you have a formula for $ \sum_{k\equal{}1}^{n}k^{r}$ for all nonnegative integers $ r<R$, you can get a formula for $ r\equal{}R$.

[quote="Math^_^"]How to find the general formula for
$ \sum_{i \equal{} 1}^{n}i^{4}$[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167116

Please use the search key, or at least look around... this one was posted only 2 hours ago...  :ninja:

Consult [url=http://mathworld.wolfram.com/PowerSum.html]this link[/url].  An alternate technique is to write $ i^{k}$ as a sum of binomials $ {i\choose r}, r \equal{} 0, 1, 2, ... k$ and use Hockeystick.


a,b,c > 0
prove that : a*sqrt(a²+3bc)/(b+c) + b*sqrt(b²+3ac)/(a+c) + c*sqrt(c²+3ab)/(a+b) >= a + b + c

I think that you can use Murheid's and Schur's ineq. :)

I'm sorry. Certainly, this problem isn't solved by Murheid's and Schur's ineq. (I don't see "sqrt"). :blush:

[quote="vivemaths"]a,b,c > 0
prove that : $ a.\frac {\sqrt {a^2 \plus{} 3bc}}{b \plus{} c} \plus{} b.\frac {\sqrt {b^2 \plus{} 3ac}}{a \plus{} c} \plus{} c.\frac {\sqrt {c^2 \plus{} 3ab}}{a \plus{} b}\geq a \plus{} b \plus{} c$[/quote] or $ a.\sqrt{\frac{a^2 \plus{} 3bc}{b \plus{} c}} \plus{} b.\sqrt {\frac{b^2 \plus{} 3ac}{a \plus{} c}} \plus{} c.\sqrt {\frac{c^2 \plus{} 3ab}{a \plus{} b}}\geq a \plus{} b \plus{} c$
Dear vivemaths?

No reason, it's pretty obvious that if the inequaity is supposed to be true, it can't be the one you quoted, but it has to be the one you wrote yourself.


conics
ellipse
analytic geometry
function
inequalities
quadratics
algebra

In the Math Jam discussion of this problem (4/4/16), a minimum (38) and maximum (126) are found, but it isn't proven that all values from 38 to 126 can be achieved.  This seemed like the hardest part of the problem; why isn't it necessary to prove this?  (I did this part by casework.)

I reasoned it in the following way (roughly; take a look at my attached solutions for a more detailed account):

the intersection of the ellipsoid and the plane (assuming no change-of-variable substitutions) is a point or an ellipse when projected onto the xy-plane.  As long as part of the ellipse is in the first quadrant and "under" the line 23 = 3x + 4y, then there will be a solution in non-negative real numbers.  I showed that the minimum value of n lies in this region, so therefore all ellipses between the minimum and maximum value of n will lie at least partially in this region as well.  If you think of n roughly as the "size" of the ellipse, then as n gets bigger, the ellipse expands, but as long as it doesn't contain the triangular region, it will lie at least partially in it (if that makes any sense; playing around with a graphing program and a slider bar helped me to visualize the situation).

Once you prove there is a point with positive coordinates (x,y,z) that gives n less than 38 and a point that gives n greater than 126, you know n can assume all values in between because the function for n in terms of x, y, z is continuous.

I'm not sure that the fact that n = 2x^2 + 3y^2 + 6z^2 is continuous ensures that all values from 38 to 126 are possible.  But even if this is correct, my problem is that in the math jam, neither of your explanations are even given - it's simply assumed!

Good point, mathfanatic, we should have mentioned this in the Math Jam!  (Yes, you can use geometric argument to knock this part off.)

I just read over the Math Jam for this problem, and I think it's interesting that Cauchy-Schwartz was the only method used to prove the minimum.  I wasn't aware that this inequality could be applied to the problem, so I used a simple algebraic method (well, simple if you're willing to apply the quadratic formula to a very messy polynomial) to find the minimum - substitute the plane equation into the ellipsoid equation to get the "intersection" equation in two variables (plus the parameter $n$), solve for one variable, and see when that variable has real solutions; you'll find that if n is less than the minimum value, no real solutions exist (i.e. the part under the radical of the solution is always less than 0), so therefore we have our theoretical minimum.

[quote="mathfanatic"]I'm not sure that the fact that n = 2x^2 + 3y^2 + 6z^2 is continuous ensures that all values from 38 to 126 are possible.[/quote]

Construct a continuous curve between the minimal and maximal points in the domain (which you can do because the domain is connected), and then apply the intermediate value theorem to this curve.

I think there's an easier, elementary and rigorous way to do this:

We can turn the ellipsoid into a sphere (by multiplying each variable by an appropriate factor), and modifying the equation of the plane accordingly. Hence the original plane and the original ellipsoid intersect if and only if the new sphere and the new plane intersect. Now we have an octant of a sphere in the first octant (centered at the origin), and a plane in the first octant. The plane and the sphere intersect if and only if the radius of the sphere is between the smallest and largest distances from the origin to the plane.

To find the smallest distance we can use the standard "distance from a point to a plane" formula: http://mathworld.wolfram.com/Point-PlaneDistance.html (Or we can calculate the volume of the tetrahedron determined by the vertices of the plane and the origin in two ways). Now the largest distance is the largest distance from the origin to one of the vertices. Hence n = 38, 39, ... 126.

Here's my full solution if you want to read it...


i.e, prove that there is no way that [b]x^5 - y^2 = 4[/b] can be true with both x and y as integers?

-Isaac

It means there isn't a solution in which both are integers.

Guys,I know a solution to this problem but i want a nicer one from you

Problem: Show that for every positive integer n there exits a  pair of positive integers ([b]a[/b],[b]b[/b]) so that:

 a^2 + b
 ----------- = n
 a  + b^2

 For instance, for n = 4 we find a = 10,  b = 4

How often are you going to put up new problems? Once a month or something?

a^2+b=n(b^2+a)

a^2+b = nb^2 + na

a/n + b/na = nb^2 + 1

allright,,,I was trying to fake it and pretend I know how to do this but I dont...can I have u'r solutioon and then try to find a better one?

I'll provide a hint for one possible line of attack..

If the difference of two integers is even, then they must either both be even or both be odd. Assuming that they are even, you can set x = 2m and y = 2n. Our equation then reads:

32m^5 - 4n^2 = 4
8m^5 - n^2 = 1
m^5 = (n^2 + 1)/8

But can a square plus 1 ever be divisible by 8? Try a few numbers and see. What do you notice when you look at squares modulo 8. And for those who know about the theory of Quadratic Residues, how could you apply them to this problem?

Now what can you say about the case where both terms are odd?

TripleM - sorry I took so long to answer (I somehow missed your post earlier).

We expect to post new Challenge problems roughly once a week (though it may take longer if it takes the Community longer to solve a problem).

And if y'all don't knock off roman's problem soon, I may stick it up there...

Re Romans problem.

After working on this for aaaages I eventually came up with an almost answer. I can prove it has a solution for any [b]non-square[/b] n. The case for n square is most likely a similar idea but I just can't think any more.
Notice that the 'random' sort of as and bs you come up with. This gives a hint that it may be related in some way to Pells equation - this seems to produce 'random' sounding answers.

Anyway:
Let a = kc and b = kd. Cancelling ks we get (kc^2 + d)/(c+kd^2) = n. Multiply off the fraction and factorise out the k : k(c^2 - nd^2) = nc-d.
Now c^2 - nd^2 is a sort of Pells Equation, and we know for non square it has a (in fact infinitely many) solution for c,d where c^2 - nd^2 = 1. So set k as nc - d and we have a solution for any non square n.
For example,  n=7 gives us the equation c^2 - 7d^2 = 1. A solution to this is c=8, d=3, so k = 53 and the solution is a = 424, b = 159.

As I said, case for n square is likely to be similar and also involve Pells Equation but I just can't see it at the moment.

For those who haven't heard of Pells Equation, basically (and you've probably worked this out by now!), at simplest its the equation x^2 - ay^2 = 1 (it can also equal -1 and other things). If a is not a square then there are infinitely many integer solutions (x,y), and once we know the smallest solution there is a recursive formula which can be used to work out the others. Have a search for Pells Equation at a search engine.
If a is a square then we only get x=1, y=0, but this won't work since b has to be positive.
Can anyone finish this?

That's a very nice solution TripleM. One thing that you didn't discuss though was whether your solutions were always positive. Also a further question is does your method necessarily give all of the positive solutions when n is not a square?

Well c and d can be made positive as we're only worrying about their squares. That just leaves showing nc-d is positive. But c^2 - nd^2 = 1 >0 so c^2 > nd^2, c>d root n, and so nc-d > d(n root n - 1) is positive. This makes k, a, b, everything positive.
I doubt it gives all solutions - in fact of course it won't since we could set k to anything else. But luckily we didn't need to find all solutions, just prove one existed.
Still can't prove it for x a square though.

Very good proof that the solutions are positive. It's always important to check the small details like that though.

You're right on point that this has something to do with Pell's equation, but maybe there is a more general method that will give all solutions, including those for which n is a square. If you multiply all that stuff out in the original problem, what can you say about the discriminant?

I worked some on TripleM's equation, k(c^2 - nd^2) = nc-d.  I think that you can not find a solution for a perfect square, but am not sure.

k(c^2 - nd^2) = nc-d
factor in the k, and
k*c^2-k*n*d^2=n*c-d
multiplying by k yields:
k^2*c^2-k^2*d^2*n=k*c*n-k*d
you can factor that, getting:
n*(k*c-k*d)(k*c+k*d)=n(k*c-k*d)
dividing by n*(k*c-k*d) yields
k*c+k*d=1
Substituting a in for k*c, and b in for k*d, gives
a+b=1
a and b are both positive integers, and the sum of two positive integers is at least 2.

Does that work???

[quote="bookworm271828"]I worked some on TripleM's equation, k(c^2 - nd^2) = nc-d.  I think that you can not find a solution for a perfect square, but am not sure.

k(c^2 - nd^2) = nc-d
factor in the k, and
k*c^2-k*n*d^2=n*c-d
multiplying by k yields:
k^2*c^2-k^2*d^2*n=k*c*n-k*d
you can factor that, getting:
n*(k*c-k*d)(k*c+k*d)=n(k*c-k*d)
[/quote]

I don't think the last line is right, where does the k^2*c^2 and the -k*d terms go to?

I think you assumed the k^2*c^2 and -k*d terms had n's multiplied in. They don't. :twisted: MUAHAHAHA!

Its also not the best idea to try proving something false that appeared in a maths competition! Although in some cases, trying to prove it false ends up in you finding the solution, so maybe it does help. Probably the best way to check your proof is to check with small cases, as given n=4, a=10, b=4. This usually helps you to find your mistake.

[quote=integrated_JRC][quote=kevinmathz][quote=rrusczyk]Periodically we'll post difficult challenge problems, which will appear both in this forum and in the forum top bar above.

Here's the first one:

[img]http://www.artofproblemsolving.com/Community/Images/Problems/P1.gif[/img][/quote]

I somehow can't see the image.[/quote]

Me too ! Please make a PDF.[/quote]

Yes, I can't either.

Simply use a bit of modular arithmetic to obtain $y^2\equiv x\equiv 0,1 \mod 4$ (we toss out the $y\equiv 3\mod 4$ case because that wouldn’t work out in the initial equation. Plugging in $y=4a+1$ and $x=4b+1,3$ gives $v_2(LHS)$ is strictly less than or greater than $2$.

For all interested this is @rrusczyk's first post, and the question is asking to prove there are no integer solutions to $x^5-y^2=4$. How do you do these types of questions, by the way? Something with Pell's equation? What's that?


A fenced, rectangular field measures 24 meters by 52 meters.  An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots.  The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field.  What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence?

Let [i]s[/i] be the width of a single square in the described plot.  Then there must be whole numbers [i]m,n[/i] such that \[ s=24/m, s=52/n  \] By combining these, we get \[ 24/m=52/n \Leftrightarrow 24m=52n  \]
The interior fencing is \[ 52(n-1)+24(m-1) \le 1994  \] \[ 52n+24m \le 2070  \]
By substituting, \[ (24m)+24m \le 2070  \] \[ m \le \frac{345}{8} \approx 43.1  \]
By using our earlier equation \[ 24m=52n \Leftrightarrow 6m=13n  \]
Since [i]m,n[/i] are integers, [i]6m[/i] must be a multiple of 13.  Since 6 and 13 share no prime factors, [i]m[/i] must be a mulitple of 13.  Since it is less than or equal to 43, [i]m[/i] could be 13, 26, or 39.

I haven't tried any of these out yet, but I've left you with only 3 possiblities.

(Maybe I'll try them if I have time.)

[hide]
Call the side length of the squares $a$, we then have the inequality:

$\displaystyle \left(\frac{52}{a} - 1 \right) \times 24 + \displaystyle \left(\frac{24}{a} - 1 \right) \times 52 \leq 1994$

Simplifying the inequality, we have:

$a \geq \frac{416}{345}$

Because the resulting plots must all be identical, we have the restriction that $\frac{52}{a} = n$ and $\frac{24}{a} = m$ for some integers $n$ and $m$.

The ratio of $\frac{n}{m}$ must then equal $\frac{52}{24} = \frac{13}{6}$, so $n = 13k$ and $m=6k$ for some integer $k$.

Putting $13k$ into the expression $\frac{52}{n}$ (which equals $a$), we have $\frac{52}{13k} \geq \frac{416}{345}$. To minimize $a$, we want to maximize $k$, so $k=3$. The value of $a$ is then equal to $\displaystyle \frac{4}{3}$.

The total number of square plots is equal to $\displaystyle \frac{24 \times 52}{a^2} = \frac{24 \times 52 \times 9}{16} = \boxed{702}$
[/hide]


I guess you could say that this was a little off topic, but I was wondering what people use and whether there is any rhyme or reason to it.

I use "i". I've always used it, in my head I'm thinking "int" and it comes out rather naturally, I then follow with j,k etc,

Anyone else got any preferences, with reasons.

Thanks

[quote="paladin8"]I think most people use "i" as it has become rather standard. "i" is a standard indexing variable in mathematics too, so that might be why it has become so in programming. I also feel like "i" looks better as an array index, like array[i] as opposed to array[a] or array[x] or whatever.[/quote]

I use array[x][y] quite alot. Makes my think of lattice points in the plane. Things become clearer that way.

Yeah, it really depends on the context. If the loop variable has no real meaning other than the fact that its just looping a certain number of times, I'd use i,j,k. Anything else where it has some meaning, you could often use a much more descriptive (but still shortish) name, as it makes things far easier to keep track of.

I almost always use i and j. I use n and m iff i and j are taken for some other purpose (a special index, for example). If my loops are going more than 2 deep, I use k and l next. If it's going more than 4 deep it's probably time to swith to arrays :)

I always thought that i stood for "index", but apparently not. Supposedly the real reason everyone uses i and j is because in FORTRAN, variables starting with I and J were integers only. Of course, this derives from the fact that integer starts with i.

I use r and c where I'm looping over an array that means something like a map, for example. r stands for row and c for column. I almost never use x and y. This is because it's natural for you to write arr[x][y] rather than arr[y][x], but you almost always want the first index to be the row and the second to be the column (so you would write arr[y][x] instead of arr[x][y]). This makes it easier when it comes time to print the array :) The only time I'd use it is if I were literally working with lattice points in the plane. Wherever I'm imposing my own coordinate system, it's (r,c), not (x,y).

I use i,j,k too but when I get to l, things start to get really confusing :D. At one point in my life I were using i, ii, iii, iv, v, vi etc. but that didn't help too much. Sometimes I also use r, c or x, y but when it's x, y I always index arrays arr[x][y]. The code is much easier to read that way and seriously, who cares how's the data stored as long as you know which index to the array gets written first?


inequalities
number theory
prime numbers

Is there a general techniquse to solve the following problem?

[i]Find the smallest number with $ n$ factors, where $ n$ is any positive integer.[/i]

For prime values of n, the problem is easy. However, for composite values, distributing the exponents among the primes properly sounds like a hassle. How would you go about doing this?

EDIT: Fixed an egregious typo.

Are there any additional conditions?

If the prime factors have to be distinct, the answer is just the product of the first $ n$ primes. If not, then it's just $ 2^{n}$, because $ 2$ is the smallest prime.

You probably mean $ n$ factors, not $ n$ prime factors.

There isn't much of a technique beyond guess-and-check and keeping certain inequalities in mind.  For values of $ n$ without many divisors there aren't many possibilities, and in any case it's rare that you'll need to use more than the first few primes.

The factors have to be distinct, then, otherwise the answer would still be $ 2^{n}$.

Yeah, you're right, [b]t0rajir0u[/b]. :P

[quote="t0rajir0u"]You probably mean $ n$ factors, not $ n$ prime factors.[/quote]
Yes. One of my worse typos. <_< I'll edit.

Thanks for the responses.

[quote="metafor"]The factors have to be distinct, then, otherwise the answer would still be $ 2^{n}$. [/quote]
What does it even mean to count the non-distinct members of a set?  By all accounts, $ 2^n$ has $ n\plus{}1$ distinct factors and it's not very clear what "indistinct" factors is supposed to refer to.

The OP said nothing about primes, so I believe that the direction he was taking this was:

number:unique factors....
1:1
2:1,2
3:1,3
4:1,2,4
5:1,5
6:1,2,3,6
7:1,7
8:1,2,4,8
9:1,3,9
10:1,2,5,10

Note that 2^n is a much larger number, in general, than the smallest number with n unique factors.

To find the number of unique factors that a number has, reduce it to its prime factors:

n = 2^a * 3^b * 5^c * 7^d * 11^e * 13^f ...

...and the number of unique factors is (a+1)*(b+1)*(c+1)...

For 16 (2^4) this gives (4+1) unique factors, which is correct; 1,2,4,8,16.  For 48 (2^4 * 3^1) there are 10 unique factors; 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

Think of it this way: for any set of prime numbers, any unique combination of products of those primes is a unique factor of the product of all of them.  So to enumerate all the unique factors, all you have to do is figure out how many ways you can pull out the prime factors.  For each prime, you can pull out [0..n] of them, giving you n+1 choices.  The total number of combinations is therefore the product of (n+1)'s for each prime factor.

Clear as mud, right?

[quote="t0rajir0u"]What does it even mean to count the non-distinct members of a set?[/quote]  Well, in this case we have a quite natural multi-set (the unique unordered list of prime numbers whose product is equal to $ n$), and in such a case it is perfectly meaningful to talk about counting "non-distinct" members, though this is of course non-standard terminology.  (The usual thing to say is that we are counting prime factors [i]with multiplicity[/i], in the same way that we can say that $ (x \minus{} 1)^2$ has two roots counting multiplicities (or "one root with multiplicity two").)

Yes, but metafor's comment suggests that he's thinking of counting a multiset of [b]factors[/b], not of prime factors, and I can't imagine a sensible way of defining this.  Possibly the largest value of $ k$ such that $ d^k | n$, but I find this a much more artificial definition of the multiplicity of a divisor $ d$ than the corresponding definition for prime divisors, and if we're taking this definition then metafor's claim is probably still false.

Nevermind me, t0rajir0u, I was thinking nonsense. :P


Find $ \lim_{x\rightarrow \ 1} \frac {x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^n \minus{} n}{x \minus{} 1}$

Are you sure your question is well written?

Is this $ \lim_{x\rightarrow \ 1} \frac {x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^n \minus{} 5}{x \minus{} 1}$ your problem?

[quote="Meatman"]Sorry :blush:  Edited[/quote]

Still makes no sence! :)

Or:

$ n<5 \Rightarrow \lim \equal{} \infty$

$ n\equal{}5 \Rightarrow \lim \equal{} 15$

$ n>5 \Rightarrow \lim \equal{}\minus{}\infty$

hello, i think, it must be
$ \lim_{x\rightarrow \ 1} \frac {x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ...\plus{}x^n \minus{} n}{x \minus{} 1}$
Sonnhard.

Or written differently:

$ \lim_{x \rightarrow 1}\frac{\sum^{n}_{i\equal{}1}x^i\minus{}n}{x\minus{}1}$

Using L'Hopital's rule:

$ \lim_{x \rightarrow 1}\sum^{n}_{i\equal{}1}i x^{i\minus{}1}\equal{}\sum^{n}_{i\equal{}1}i\equal{}\frac{n(n\plus{}1)}{2}$

[quote="Dr Sonnhard Graubner"]hello, i think, it must be
$ \lim_{x\rightarrow \ 1} \frac {x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ... \plus{} x^n \minus{} n}{x \minus{} 1}$
Sonnhard.[/quote]

Maybe our problem is that.

[quote="milin"]Or written differently:

$ \lim_{x \rightarrow 1}\frac {\sum^{n}_{i \equal{} 1}x^i \minus{} n}{x \minus{} 1}$

Using L'Hopital's rule:

$ \lim_{x \rightarrow 1}\sum^{n}_{i \equal{} 1}i x^{i \minus{} 1} \equal{} \sum^{n}_{i \equal{} 1}i \equal{} \frac {n(n \plus{} 1)}{2}$[/quote]

$ \lim_{x\rightarrow 1} \sum_{i \equal{} 1}^n \frac {x^i \minus{} 1}{x \minus{} 1} \equal{} \lim_{x\rightarrow 1} \frac {(x \minus{} 1)(x^{i \minus{} 1} \plus{} x^{i \minus{} 2} \plus{} \cdots x \plus{} 1)}{x \minus{} 1}\equal{} \sum_{i \equal{} 1}^n i\equal{} \frac {n(n \plus{} 1)}{2}.$


modular arithmetic
calculus
integration
combinatorics proposed
combinatorics

I created the problem on my way home by train when we are playing cards.the problem is as following.
four students play cards,and in every time, one of them will be Landlord ,and if he win he will be +3points,and the other three will all -1.assume in some times over the four students' point will be (a,b,c,d).
question:what kind of (a,b,c,d) can be possibly appeared?

oh,grobber I don't want the form you write.
but I must say sorry because I didn't say an important to the problem:
if the Landlord doesn't win.than the Landlord will -3points and  the other three will +1point

[quote="Valiowk"]Side comment: Is this related to the game 斗地主 by any chance? :?[/quote]

what's that game?

[quote="grobber"]Um... Simply quadruples of the form $(3x-y-z-t,-x+3y-z-t,-x-y+3z-t,-x-y-z+3t)$, where $x,y,z,t$ are non-negative integers? :?[/quote]
Does it lead to an arbitrary quadruple with total sum is 0?

I don't know if any four integers which sum up to $0$ can be expressed like that. I have to add, however, that $x,y,z,t$ can be arbitrary integers, not just non-negative, because zhaobin added:

"if the Landlord doesn't win.than the Landlord will -3points and the other three will +1point".

[quote="grobber"]Actually, I think we can show that any quadruple $(a,b,c,d)$ where $a+b+c+d=0$ and $a\equiv b\equiv c\equiv d\pmod 4$ can be achieved, and these are all the quadruples we can achieve.[/quote]
yeh,it is the form I want. haha

I'm happy to hear that :).

To justify that (so that it's complete :)):

We attempt to solve th system $3x-y-z-t=a,-x+3y-z-t=b,-x-y+3z-t=c,-x-y-z+3t=d$. By subtracting the first equation from the second, the second from the third, and so on, we get $x-y=\frac{a-b}4,y-z=\frac{b-c}4,\ldots\ (*)$. It's clear now that $a,b,c,d$ are the same $\pmod 4$, and we knew from the beginning that their sum was $0$. On the other hand, if we are given $a,b,c,d$ satisfying the conditions, we can easily construct integral solutions $x,y,z,t$ for the system $(*)$.


I saw Dracula   :(

I am not the only one who saw Dracula, PiDeltaPhi also

well Moubinool, you should see Jesus, not Dracula... please try again and look more carefully! :)

I don't see anything when my eyes are closed  :D

Please follow the instructions carefully and TILTING your had back is the most important!

I saw Jesus! I saw Jesus! But... why does he have his moustache growing up his nose?  :D  :?

i saw jesus and dracula at the same time  :D

that must be some really awesome hybrid,PiDeltaPhi (jesus and dracula he!he! :D  :D )

take a photo editor (photoshop , gimp whatever)
invert the image
is that what you see when you close your eyes? :D

take your picture edit it a bit . vectorize it . and voila!
a stay-in-my-head read-to-go picture :D
shoud try it with your girlfreind's picture :D

this is incredible!!

AMAZING!! Best optical illusion I've seen so far :D

I need to use glass, my eyes are very ill :D

I saw Jesus, it's cool, i think i am gonna keep doing it all night long!

You know what freaked me out? I'm still seeing an outline of the picture even with my eyes open, while looking at the monitor! I hope it's not permanent.. :D:D

Oh, it went away :)

:? ...you had the chance to see Jesus every second  :D

Err... What exactly did the teacher hope to achieve by showing the illusion to the students? This really puzzles me :)

[quote="darij grinberg"]I just wanted to note that a [i]catholic religion teacher from our school[/i] really showed this picture to his class!! But even besides from such jokes, it is trivially clear that religious education at school lacks serious contents (it's partially for this reason I attend ethics instead)...

  Darij[/quote] just out of curiosity, did he get it from ML ? :)

[quote="grobber"]Err... What exactly did the teacher hope to achieve by showing the illusion to the students?[/quote]

I guess it was a desperate attempt to wake up a sleeping class... but I don't think it really helped.

[quote="Valentin Vornicu"]just out of curiosity, did he get it from ML ? :)[/quote]

Hmm, rather unlikely (this graphics really circulated through the whole web).

  Darij

First I see a a black rectangle. Then a circle appears in the middle of the rectangle, and a ghost appears inside the circle. Is that where Jesus is supposed to appear??!!  :maybe:  Or maybe I'm confusing the ghost with Jesus!

Well perhaps the mathematics have seriously damaged your brains :D but normally the ghost you see should look like Jezus :cool:

hmm i see exactly the image inverted  :?  :?
it looks like some scary black thing  :D

well if you've seen jezus on drawings, that's exactly what he looks like, the invert of the picture :D

just save picture, and invert colors in ms paint to see what you should have seen :)


This must have been asked before. When you want to view comments in a blog, the comments appear at the bottom of a very long page. Why is it like that? Is it being fixed?

I use IE7 on a Windows XP, and my board style is MathLinks. This is what I see on my blog:

EDIT: Apparently, this only happens to me.

It seems fine to me on my computer under all different browsers ... must be something wrong with your setup ... try using Firefox or Chrome first see what you get.

I tried it with Firefox. When I wasn't signed in, it worked fine. But when I signed in, it didn't work for any of the blogs. There was a long page again. Also, I tried it with both board styles and the same thing happened.

The board style doesn't affect the way your blog looks. That is weird that stuff happens when you log in ...

I just understood why this happens (thought it's from a long time ago). I had the same problem until today, when I changed my preferences. Instead of displaying posts in descending order, I set it to display them in ascending order (by post time). It is now fixed and works fine.

I am pretty sure the problem is in the order you set the comments to be displayed. Could/Should this be fixed?


geometry
circumcircle
parallelogram
geometry solved

In a convex quadrilateral $ABCD$ with $\angle ABC = \angle ADC = 135^\circ$, points $M$ and $N$ are taken on the rays $AB$ and $AD$ respectively such that $\angle MCD = \angle NCB = 90^\circ$. The circumcircles of triangles $AMN$ and $ABD$ intersect at $A$ and $K$. Prove that $AK \perp KC.$

Let $P, Q$ be the circumcenters of the triangles $\triangle ABD, \triangle AMN$ and $(O)$ a circle with diameter $AC$ centered at the midpoint $O$ of this diagonal of the quadrilateral $ABCD$. Let the lines $CB , CD, CM, CN$ intersect the circle $(O)$ at points $B', D', M', N'$ different from the point $C$. The angles $\angle AB'B \equiv \angle AB'C$, $\angle AD'D \equiv \angle AD'C$, $\angle AM'M \equiv \angle AM'C$, $\angle AN'N \equiv \angle AN'C$ are all right, because they span the diameter $AC$ of the circle $(O)$. The angles $\angle ABB', \angle ADD', \angle AMM', \angle ANN'$ are all equal to $45^o$:

$\angle ABB' = 180^o - \angle ABD = 180^o - 135^o = 45^o$

$\angle ADD' = 180^o - \angle ADC = 180^o - 135^o = 45^o$

$\angle AMM' = \angle ADC - \angle MCD = 135^o - 90^o = 45^o$

$\angle ANN' = \angle ABC - \angle NCB = 135^o - 90^o = 45^o$

Hence, the right angle triangles $\triangle ABB', \triangle ADD', \triangle AMM', \triangle ANN'$ are all isosceles and consequently, the perpendicular bisectors of the segments $AB, AD, AM, AN$ pass through the points $B', D', M', N'$. Let the perpendicular bisectors of the segments $AB, AD, AM, AN$ meet the circle $(O)$ at points $E, F, E', F'$. The angles $\angle AB'E = \angle AM'E' = 45^o$ are equal, hence, they span the same arc $AE \equiv AE'$ of the circle $(O)$ and the points $E \equiv E'$ are identical. Similarly, the angles $\angle AD'F = \angle AN'F' = 45^o$ are equal, hence, they span the same arc $AF \equiv AF'$ of the circle $(O)$ and the points $F \equiv F'$ are identical. Since these angles are all equal to $45^o$, the points $E, F$ are diametrally opposite points of the circle $(O)$, the points $E, O, F$ are collinear and $OE = OF$. The perpendicular bisectors $B'E \parallel N'F$ of the segments $AB, AN$ are parallel, being both perpendicular to the same line $AB \equiv AN$ and they pass through the circumcenters $P, Q$, respectively. Similarly, the perpendicular bisectors $D'F \parallel M'F$ of the segments $AD, AM$ are parallel, being both perpendicular to the same line $AD \equiv AM$ and they also pass through the circumcenters $P, Q$, respectively. The quadrilateral $PEQF$ is a parallelogram and the center $O$ of the circle $(O)$ is the intersection of its diagonals, being the midpoint of the diagonal $EF$. Consequently, the centers of the circles $(P), (O), (Q)$ are collinear. Since these 3 circles intersect at the point $A$, they belong to the same elliptic pencil and they all intersect at one other point, the intersection $K$ of the circles $(P), (Q)$. Since the angles $\angle AKC$ spans the diameter of the circle $(O)$, it is right and $AK \perp KC$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=16899

[color=red][Moderator edit: No, this is [i]not[/i] the same problem as the one discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 . In fact, in this problem, the points M and N lie on the extensions of the lines AD and AB and satisfy < MCD = < NCB = 90°, while in the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=16899 , the points M and N lie on the extensions of the lines AB and AD and satisfy < MCD = < NCB = 90°. Funnily enough, the two problems are different but both correct.][/color]

the quadraterial ABCD is a parallelogram. So, <BAC = <BCD = 45,
<DCN = <ADC - <DCN = 135 - 90 = 45
similarly, <MCB  = 45
<MCN = <MCB + <BCN = 45+90 = 135
<MCN + <MAN = 135 + 45 = 180,
so, A,K,N,M,C are concyclic.
<AMC = 180 - <MCD = 180 - 90 = 90,
<AKC = 180 - <AMC = 180 - 90 = 90

I added another solution  to [url=https://artofproblemsolving.com/community/c6h210284p1160288]here[/url] of this beautiful problem.

Let F and E be midpoints of BN and DM.
∠KBN = ∠KDM and ∠KMD = ∠KNB ---> KNB and KMD are similar ---> ∠KFA = ∠KEA ---> AKFE is cyclic.
∠CBN = 45 and ∠BCN = 90 ---> ∠CFA = 90
∠CDM = 45 and ∠DCM = 90 ---> ∠CEA = 90
so AFCE is cyclic as well and we have ∠AKC = ∠AFC = 90
we're Done.

[quote=liekkas]Let $P,Q$ be the midpoints of $BM,DN$, from $KBN \sim KDM$ we get $KBQ \sim KDP$, then $A,K,P,Q(,C)$ concyclic and we are done.[/quote]

Hello, shouldn't it be /{BN} and DQ?


Prove that \[ a^3\plus{}b^3\plus{}c^3\plus{}d^3\equal{}3\] has infinite solutions in intergers.

can someone give me a hint or something please   :D  :D  :D
thanks

[hide="hint"]
Try playing around with expressions like $ (m\plus{}1)^3$ and $ (m\minus{}1)^3$, see if you can find an explicit family of solutions.
[/hide]

[hide="solution"]
$ (6t^3\plus{}1)^3\minus{}(6t^3\minus{}1)^3\minus{}(6t^2)^3\plus{}1^3 \equal{} 3$.

Instead of pulling this out of thin air, a reasonable approach could go like this: working modulo $ 9$ you can see that $ a,b,c$ must be congruent to 1 modulo 3 and $ d$ must be divisible by 3 (or some other permutation, doesn't matter of course). You can then change variables to $ a\equal{}3k\plus{}1$, $ b\equal{}3l\plus{}1$, $ c\equal{}3m\plus{}1$ and $ d\equal{}3n$. Plugging this into the equation you get rid of the 3 and you have that a linear combination of $ k^3\plus{}l^3\plus{}m^3$, $ k^2\plus{}l^2\plus{}m^2$, $ k\plus{}l\plus{}m$ and $ n^3$ is 0. An easy way to make it 0 is to take $ m\equal{}0$ and $ k\equal{}\minus{}l$, since this immediately gets rid of the odd powers. You then fix the even powers with $ n$ and you're done.
[/hide]


function
real analysis
real analysis unsolved

$n$ an integer and $p(n)$ the number of partitions of $n$ as sum of integers.
e.g. $p(4)=5$ because $4=4=3+1=2+2=2+1+1=1+1+1+1$.
Find the radius of convergence of the power series $\sum_1^{\infty} p(n)z^n$.

Using Hardy-Ramanujan formul formula $p_n \~\ 1/(4n \sqrt{3})e^{\pi*\sqrt{2n/3}}$ ;)

$R = 1$


The expression $ \sqrt{25\minus{}t^2}\plus{}5$ equals zero for:

$ \textbf{(A)}\ \text{no real or imaginary values of }t \qquad
\textbf{(B)}\ \text{no real values of }t\text{ only} \\
\textbf{(C)}\ \text{no imaginary values of }t\text{ only} \qquad
\textbf{(D)}\ t\equal{}0 \qquad
\textbf{(E)}\ t\equal{}\pm 5$

[hide]

Since $ \sqrt{x}\ge 0$, $ \sqrt{25\minus{}t^2}\plus{}5\ge 5$, so $ \textbf{(A)}$

[/hide]


How many positive three-digit integers are multiples of 11?

The smallest one is $ 11\cdot 10$ and the largest is $ 11\cdot 90$ (use the fact that $ 7\cdot 11\cdot 13\equal{}1001$).  We need to count the number of numbers in the list \[ 10,11,12,...,89,90\]Subtracting 9 from each number gives us \[ 1,2,3,...,80,81\]This list obviously has $ \boxed{81}$ numbers.


What does math study?

Does it study the real world? or simply functions, sequences of symbols, and abstract things which only exist inside the human head?

Do you know any theories that evolved mearly from a study of the real world?

I had a friend who said exactly that: math does not study the real world, does not study the physic aspect of things.

What would you say to him? I presented many objects which math study like meteorology, creating models, design of buildings, etc.. but I would like to know the opinion of more people the deals with math everyday and likes the science.

Is he right or not?

Thanks to all

One must be somewhat careful to distinguish math as it [i]is[/i] and math as it is [i]applied.[/i]  Mathematics as it is applied has as its basis the notion of a model - a mathematical object that approximates the behavior of a real object to some desired degree of precision.  This is a question of science, not of mathematics.  Science applies mathematics as a tool.  Whether [url=http://www.dartmouth.edu/~matc/MathDrama/reading/Wigner.html]it is a good tool[/url] is, in my opinion, a question of philosophy.

Mathematics as it [i]is[/i] is the study of patterns.  Patterns can come from a variety of sources - the interplay between mathematics and physics or (more recently) mathematics and the other sciences (such as biology) is rich.  One of the most fascinating aspects of this interplay is when mathematicians develop an abstract tool based on one set of considerations that find unexpected applications in a physical science whose investigations were based on a totally different set of considerations.  I'm not good at examples, but rest assured that this happens and it is related to the subject of the link in the first paragraph.


Let the tangent to the circumcircle $O$ of $\triangle ABC$ at $A$ intersect side $BC$ at $X$. Point $P$ is an arbitrary point on side $BC$. The circumcircle of triangle $AXP$ intersect $O$ at $Z$. $ZP$ intersects $O$ again at $R$. Show that $AR\parallel BC$.

Because of the tangency of $AX$, we can easiley see that
\[\angle AXP = \angle C-\angle B = \pi-\angle AZP\]
But we also know that $\angle AZP = \pi-\angle C+\angle BZP$

Hence it follows that $\angle BZP = \angle B$.
This implies that $\angle RCB = \angle B$, equivalent with $AR \parallel BC$

\[\mathbb{QED}\]

[img]http://img501.imageshack.us/img501/7614/cancerparallelyc5.jpg[/img]


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Evaluate:
\[ \int_{\mathbb{R}}\frac{\sin(at)\sin(bt)}{t^2}dt\]

($ a,b$ - given reals)

Put (Sin[a x] Sin[b x]/(x^2)) into http://integrals.wolfram.com/index.jsp for your answer.

I know integrator :)
But here it doesnt't work:
I need a definite integral - over $ \mathbb{R}$.

The problem is how to achieve the result, not what the result is :)

As far as I know, it should be:
\[ \frac{\pi}{2}(|a\plus{}b|\minus{}|a\minus{}b|)\]

Write the product of two sines as the difference of two cosines and consider the integral $ \int_{\minus{}\infty}^\infty \frac{1\minus{}\cos t}{t^2}$, which can be found in several ways (probably, the easiest one is to recognize the integrand as the Fourier transform of some well-known function) ;)

[quote="fedja"]Write the product of two sines as the difference of two cosines and consider the integral $ \int_{ \minus{} \infty}^\infty \frac {1 \minus{} \cos t}{t^2}$, which can be found in several ways (probably, the easiest one is to recognize the integrand as the Fourier transform of some well-known function)[/quote]
Since the function is even

$ \int_{ \minus{} \infty }^{ \plus{} \infty } {\frac{{1 \minus{} \cos t}}{{t^2 }}\,dt}  \equal{} 2\int_0^\infty  {\int_0^\infty  {ue^{ \minus{} ut} (1 \minus{} \cos t)\,du} \,dt} .$

Or $ \int_{ \minus{} \infty }^{ \plus{} \infty } {\frac{{1 \minus{} \cos t}}{{t^2 }}\,dt}  \equal{} 2\int_0^\infty  {\int_0^1 {\frac{{\sin ut}}{t}\,du} \,dt} .$

Let's try it with properties of the Fourier transform. First, we've got to decide where to put the $ 2\pi.$ The notation I'm most comfortable with puts the $ 2\pi$ in the exponent:

$ \widehat{f}(s)\equal{}\int_{\minus{}\infty}^{\infty}f(x)e^{\minus{}2\pi isx}\,dx,$ $ f(x)\equal{}\int_{\minus{}\infty}^{\infty}\widehat{f}(s)e^{2\pi isx}\,ds.$

Define "box" by $ B(x)\equal{}\begin{cases}1,&|x|<\frac12\\0,&|x|>\frac12\end{cases}.$

We know that the Fourier transform of "box" is "sinc". That is,

$ \widehat{B}(s)\equal{}\text{sinc}(s)\equal{}\frac{\sin \pi s}{\pi s}$ (Made continuous at zero, of course.)

By the way scaling properties work, $ \frac{\sin a\pi s}{\pi s}\equal{}a\,\text{sinc}(as)$ is the Fourier transform of $ B\left(\frac{x}{a}\right).$

When I said that, I was tacitly assuming that $ a$ was positive. I will continue to do so.

Also, since these are all real, even functions, we can freely reverse the roles of function and transform.

Now we come to the most important manipulative relationship I know for the Fourier transform: convolution on one side gives pointwise multiplication on the other side. That is $ \widehat{f*g}(s)\equal{}\widehat{f}(s)\widehat{g}(s).$

Let $ f(x)\equal{}B\left(\frac{x}{a}\right)$ and $ g(x)\equal{}B\left(\frac{x}{b}\right).$

Then $ \widehat{f*g}(s)\equal{}\frac{\sin a\pi s}{\pi s}\cdot\frac{\sin b\pi s}{\pi s}\equal{}\frac{\sin(a\pi s)\sin(b\pi s)}{\pi^2s^2}$

Then, $ \int_{\minus{}\infty}^{\infty}\widehat{f*g}(s)\,ds\equal{}f*g(0)$

$ \int_{\minus{}\infty}^{\infty}\frac{\sin(a\pi s)\sin(b\pi s)}{\pi^2s^2}\,ds\equal{}f*g(0).$

Now we need to compute $ f*g(0).$ Assume WLOG that $ a\ge b>0.$

$ f*g(0)\equal{}\int_{\minus{}\infty}^{\infty}f(x)g(\minus{}x)\,dx\equal{}
\int_{\minus{}\frac{b}2}^{\frac{b}2}1\,dx\equal{}b.$

So $ \int_{\minus{}\infty}^{\infty}\frac{\sin(a\pi s)\sin(b\pi s)}{\pi^2s^2}\,ds\equal{}b,$ provided $ a\ge b>0.$

Substitute $ t\equal{}\pi s:$

$ \int_{\minus{}\infty}^{\infty}\frac{\sin(at)\sin(bt)}{t^2}\,\frac{dt}{\pi}\equal{}b$

$ \int_{\minus{}\infty}^{\infty}\frac{\sin(at)\sin(bt)}{t^2}\,dt\equal{}\pi b.$

So for $ a,b>0,$ the integral is $ \pi \min(a,b).$ The rest can be obtained by observing that the result is odd in $ a$ and odd in $ b.$


function
induction
algebra proposed
algebra

Find all the postive integer roots of equation :
 x^y -y^x=x+y
I'm waiting nice sol!

Well, I don't have a nice solution but I have one....

1) There is no solution of the form (a,a).
2) The function f(x) = ln(x)/x is decreasing on [e,+oo[ from 1/e to 0, then if 3  \leq  x < y we have x y > y x . (1)

In the same way, if 3  \leq y < x, we have x y > y x , thus there is no solution of 3  \leq  y < x.

3) Now we prove that there is no solution in the case where 3  \leq  x < y.
Let's x be fixed, with x  \geq  3. The function g(y) = x y  - y x  - x - y.
Then g'(y) = ln(x)*x y  - x*y x-1  - 1 > ln(x)*y x - x*y x-1  - 1 (from (1))
That leads to g'(y) > y x-1 [ln(x)*y - x - 1/y x-1 ] > y x-1 [ln(x)*x - x - 1/x 2 ] .
But (ln(x) - 1)*x - 1/x 2  > x - 1/x 2  > 0.
Then g is an increasing function, and we only have to prove that there g(x+1) > 0.
Let h(x) = g(x+1) = x x [x - (1 + 1/x) x ] - 2x - 1.
Let t(x) = (1 + 1/x) x . The function t is positive on [3,+oo[ and t'(x)/t(x) = ln(1+1/x) - 1/(x+1) = u(x).
We have u'(x) = - 1/[x(x+1) 2 ] < 0 then u is decreasing on [3, +oo[ from ln(4/3) - 1/4 to 0. It follows that t is increasing from (4/3) 3  to e.
Then for x  \geq  3, we have h(x) > x x  [x - e  - 2/9 - 1/27] > x x  (x - 2,978) > 0.

4) it only remains to look for the solutions with x or y less than 3.
It is easy to see that there is no solution for x = 0.
For y = 0, we find x = 1, and for x = 1 we find y = 0.
For y = 1, there is no solution.

For x = 2 the equation is now 2 y  = y 2  + y + 2. An easy induction leads to 2 y  > y 2  + y + 2 for y > 5.
And for y < 6, we find the solution x = 2 and y = 5.

For y = 2, the equation is 2 x  = x 2  - x - 2. As above, an easy induction leads to the fact that the LHS > RHS.

Then, the solutions are (x,y) = (1,0) and (2,5).

Pierre.

oh,I have same way! :)
It's difficult to find better sol! :(


calculus
derivative
algebra
function
domain
factorial
calculus computations

For a derivative to make sense, you need to have a domain like the real line- not just a bunch of integers.

Fortunately, there is an extension of the factorial to the real numbers (and more): the $ \Gamma$ function. Its derivative is certainly not zero.

Wait...
Isn't n! a constant, thus making the derivative 0?

I think I'm wrong just because it sounds so easy...

$ n!$ is not a constant. It is a sequence, meaning that it is a function whose domain is the integers. (The nonnegative integers, in this case.) You wouldn't graph it with a horizontal line, or even a succession of horizontal line segments; you'd graph it with dots. And for such a graph, the notion of "tangent line" is meaningless. This is what jmerry was saying above. For us to even attempt to compute the derivative of a function at a point, that function should be defined on an open neighborhood of that point.

Could you connect those dots with a reasonable curve? Absolutely yes, and that's jmerry's point about the Gamma function. $ \Gamma'(x)$ equals zero only at one particular point (the minimum value, somewhere in the interval $ (0,1)$ ) and not anywhere else for $ x > 0.$

-----

Reconsidered: the original question was ambiguous, because we didn't specify what variable we were trying to take the derivative with respect to. If $ n$ is considered the variable, then $ \frac{d}{dn}\,n!$ is meaningless, as I've discussed. If something else, say $ x$, is the variable, then it would be correct to call $ n!$ a constant, and $ \frac{d}{dx}\,n!\equal{}0.$

That's just a reminder of the need for precise language.


you know how to play the rule game, right?
scoring is
1.score=rule # times current score.
2.failure to follow a rule results in score reset and new ruleset,and everyones score doubles!
3.a rule must follow itself
4.only post your new rules.

lets start.

1 :D . use a number plus the correct smiley from the more emoticons box.(going across)
2  :lol: .  [b]F[/b]irst letter of any rule must be bold.

points:
asterday1:1*2=2

OK. This should either be locked or deleted, there is already a topic on this.


topology
function
real analysis
advanced fields
advanced fields unsolved

Let $X$ be a compact smooth manifold such that each smooth function $f: X\longrightarrow X$ has a fixed point. Prove that each continuous function $f: X\longrightarrow X$ has a fixed point.

Let F be a continuous function.
For any integer n>=1 there exists a smooth function Fn:X->X
s.t. norm(F-Fn)<1/n (density by local convolution!) where norm is the Sup-norm.
There exists Xn in X s.t. Fn(Xn)=Xn. We may extract from (Xn) a sequence which converges to U in X. Then F(U)=U.


LaTeX
number theory proposed
number theory

If $ F$ is Fibonacci sequnce, prove that

$ 2^{n\minus{}1}F_n\equal{}\Sigma_{k\equal{}0}^{[\frac{n\minus{}1}{2}]}C_{2k\plus{}1}^n5^k$

Sorry for LATEX  :huh:

$ C_0^n\plus{}C_{1}^{n\minus{}1}\plus{}C_{2}^{n\minus{}2}\plus{}...\equal{}F_{n\plus{}1}$

I posted this at pre-olympiad and a reply came from t0rajir0u

[quote="t0rajir0u"]Scroll down to "Excursion" [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].[/quote]

i found it nice and wanted to share with u  :)

So I will tell you right now that this is equivalent to Binet's formula, $ F_n \equal{} \frac{1}{ \sqrt{5} } \left( \left( \frac{1 \plus{} \sqrt{5}}{2} \right)^n \minus{} \left( \frac{1 \minus{} \sqrt{5}}{2} \right)^n \right)$, after an application of the binomial theorem.  What I'm wondering if there is an explicit combinatorial proof; I have not yet been able to find one.


The following inequality is stronger:
$\sqrt{\frac{ab+ac+bc+ad+bd+cd}{6}}\geq \sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$  :wink:


Determine all pairs of positive integers $(x\,,y)$ such that $x^2-y^2=2^{100}$ and $x$, $y$ are not divisible by any prime number greater than 5.


geometry
geometric transformation
reflection
3D geometry
tetrahedron
email

Just curious...what does everyone's username mean? Mine's pretty simple..."J" and "S" are my first two initials, "Rosen" is my last name, and "3" is the third-smallest positive integer.

Umm.. wasn't there another thread like this? I think it sort of got swept away because before there was only one page for fun and games and round table.. or maybe I'm hallucinating..

Anyway.. I like platypi.. so I think that takes care of the rest.

My username is a kind of circular parry in fencing.  Executing a countersixte parry involves making a clockwise circular motion with the blade.  If you're left-handed, though, it becomes a counterclockwise motion.  A parry, by the way, is an attempt made to block or deflect an opponent's blade so that it doesn't hit you.

Canadian pun (thats how canadians pronounce) of the character's name in the book "Who is Bugs Potter?" by Canadian author Gordon Korman

My username came about because I felt I was one of the few adults here. Actually, there are more adults (and very helpful adults, I might add) than I supposed at first.

I now use the same username on College Confidential, which gives me a much stronger feeling of there being too few grown-ups around.

I admire the young people here--there are people here one-third my age who could crush me in any math competition.

In elementary school, I thought Robin Hood was cool.  Still do.  Tried to sign up for something online, but found RobinHood was taken.  Added a 3000 to the end, and it's stuck to this day.

I like tetrahedrons. AIM sn tetrahedron was taken. Thus tetrahedr0n.

Someone decided that Frank the Tank would be a good nickname for me.  I have never seen [i]Old School[/i].  Maybe Stan the Van would be a better nick name, but my name isn't Stan.

[quote="peter"]peter's my name[/quote]

so is mine :D

My mom signed up for me without consulting me for a creative username, so it's just my first and last name together. Had she asked me before signing up, I would probably have said 'bookaholic', my username on two other forums.

Lun Mu's my name... very creative eh? hehe

When I signed up for AoPS that was the first thing that popped into my head when I was deciding a username.

my scrambled brain, personality, and eggs

cold entre is a nickname i had at school for about a week.

I liked fairies and knitting a lot 2 years ago. "Knitted" was supposed to be "Knitting", that was a typo. Now, I don't like fairies, sorta like knitting, and wish I had a better username.

[quote=ld414think]My initials are L and D, 414 is my lucky number, my chinese name means think[/quote]
what's your chinese name?  :-D

Well,
My favourite animal is an owl

My house number [address] is 81

 (and it's also a square of 9, which was my sister's lucky number, my birthday month, and the digits of my birthday day added up (i.e. 1+8 [or 0+9 or 2+7]))

Anyway my parents made the username-

i like Australia, Albert Park is a track there, and then typo

[quote=mkhaim]My dad made up this name and it stuck. (mostly because I couldn't find out how to change it)[/quote]
lol

[quote=Idefix]Idefix is a dog in the Astérix comic book series. He is Obélix's dog.[/quote]
yes i read that in dutch i have a ton of them

Mine is antonyms (sorta) and my mom said i have 15 secs to make a username :rotfl:

I was born in the tiger year(2010) and my middle name is wonyong

I pulled mine from a random name generator, found that it's taken, and added a 1 on the end.

I like dungeons and dragons and dragonborn is a race.

I'm arrogant, and I want people to call me KingOfTheWorld in class XD.

When I made this username I wanted to be an inventor when I grow up and 6 is my fave number

@bove everyone does

mine is actually my moms cause she made htis account

[quote=HWenslawski]Mine is my first initial then my last name[/quote]

I guessed that lol, because there are 3 Wenlawki's starting with different letters

idk why i have a potato but 2017 is when i made my aops account... and now my name on literally everything is "Potato2017"


[quote="Alisher"][u]a,b,c,d-positive reals such that[/u]
\[ a^{2}+b^{2}+c^{2}+d^{2}= 1 \]
[b]prove [/b]
\[ a+b+c+d \ge a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+ac+bd \]
[/quote]

$ a+b+c+d \ge a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+ac+bd\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^{3}b^{3}-2a^{3}bcd+a^{2}b^{2}c^{2})\geq0\Leftrightarrow \sum_{sym}(a-b)^{2}S_{ab}\geq0,$
where $ S_{ab}=c^{3}a+c^{3}b+d^{3}a+d^{3}b+2c^{3}d+2d^{3}c-c^{2}ab-d^{2}ab.$
Let $ a\geq b\geq c\geq d.$
Hence, $ S_{ad},$ $ S_{bc},$ $ S_{bd}$ and $ S_{cd}$ are non-negative numbers and $ a-d\geq a-c\geq a-b.$
Thus, it remains to prove that $ S_{ad}+S_{ac}\geq0$ and $ S_{ab}+S_{ac}+S_{ad}\geq0,$ which easy.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135903


Prove that  tan70 = tan20 + 2tan40 + 4tan10, ( angles are in degrees ).

[hide]Using $ \tan\alpha+\tan\beta =\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}$ we get

$ \tan 20+\tan 10 ={1\over 2\cos 20\cos 10}$
$ 2(\tan 40+\tan 10) =\frac{2\sin 50}{\cos 50\cos 40}={2\over\cos 10}$
$ \tan 10 ={\sin 10\over\cos 10}$

Adding those up, using $ \tan 70 =\cot 20$ and multiplying by $ \cos 10$, the given identity becomes

$ \cot 20\cos 10 ={1\over 2\cos 20}+2+\sin 10$

$ {\cos 20\cos 10\over\sin 20}={1\over 2\cos 20}+2+\sin 10\qquad/\cdot\sin 20$

${ \cos 20\cos 10-\sin 20\sin 10 ={\sin 20\over 2\cos 20}+2\sin 20}$

${ {\sqrt{3}\over 2}={\sin 20\over 2\cos 20}+2\sin 20}\qquad/\cdot\cos 20$

$ {\sqrt{3}\over 2}\cos 20-{1\over 2}\sin 20 = 2\sin 20\cos 20$

$ \cos 30\cos 20-\sin 30\sin 20 =\sin 40$

$ \cos 50 =\sin 40$

Since all the steps are equivalences, the identity is proven.[/hide]


Let $a$ be a real number such that $\frac{1}{a}=a-\left[a\right]$. Prove that $a$ is irrational.

for sake of contradiction, let us suppose that $a=\frac{p}{q}$ with $\gcd(p,q)=1$, then

$\frac{q}{p}-\frac{p}{q}\in\mathbb{Z}$
$\frac{q^{2}-p^{2}}{pq}$
now we must have that $q|p^{2}$, contradiction, therefore if $a$ exists, then $a$ is irrational

If $[a]=n,\{a\}=\alpha$, then $(n+\alpha)\alpha=1\iff\alpha^{2}+n\alpha-1=0\iff \alpha=\frac{\sqrt{n^{2}+4}-n}{2}$ (the negative solution is discarded as $0\leqslant\alpha<1$)

[size=59]Note: It's also obvious that n can't be negative, since that would give the fractional part out of its legal range.[/size]

If $\alpha$ is rational, then $n^{2}+4=m^{2}\iff (m+n)(m-n)=4$. The only integer possibility for $n$ is $n=0$, but that gives $\alpha=1$ which is impossible. Hence $\alpha\notin\mathbb{Q}\implies a\notin\mathbb{Q}$.

NOTE: There exists infinitely many such $a=n+\alpha$, namely $a\in\{\frac{n+\sqrt{n^{2}+4}}{2}|n\in\mathbb{N}\}$ (it is understood here that $\mathbb{N}$ doesn't contain zero).


I've found one school in the vicinity that administers the AMC 12 B, but they're already full.  Is it possible that I can pay $ \$16$ to get another bundle of 10 tests, or will I have to pay another registration fee of $ \$60$?

This is probably not the place to ask this question.  The AMC Director occasionally checks this forum and answers questions (his username here is AMCDirector), but if you want your question answered you're best off contacting the AMC directly.  You should be able to find contact info at http://www.unl.edu/amc.

You can also send AMCDirector a PM, but as Matt said the best way is via the AMC website (and phone/email).


vector
linear algebra
linear algebra unsolved

If anybody could please help me out with this problem, I'd really appreciate it...

Show that the set of all (a+b+c, b-2c, 2a+2b-c, a+b) where a, b, and c are real numbers, is a three dimensional vector space.

That's $ a(1,0,1)\plus{}b(1,1,1)\plus{}c(1,\minus{}2,0)$- all linear combinations of three fixed vectors. How would you find the dimension of this space?


induction
inequalities
algebra
difference of squares
special factorizations
number theory proposed
number theory

Don't you just use difference of 2 squares.?

Bomb

Let n^2+3^n be = to a^2. Then, by difference of squares, we get:
2n=3^x-3^y where x+y=n. For n>3, we prove by induction that RHS>LHS, contradiction. So checkingg yields n=1 and 3.

Bomb

[quote="JL"]Sorry, this probably sounds really stupid :blush:  but how do you get from $n^{2}+3^{n}=a^{2}$ to $2n=3^{x}-3^{y}$, $x+y=n$? hope you could clarify. Thanks :)[/quote]

I guess he means $(a+n)(a-n)=3^n=3^x\cdot 3^y$ (letting $n=x+y$ of course). Then if say $x\geq y$ we must have $a+n=3^x$ , $a-n=3^y$ . So both of them gives us $2n=3^x-3^y$ .  :)

I'm sure shyong shall forgive me, whether it's me to reply your question, tiks! :oops: Well, it means the "right hand side" (RHS) is greater than the left hand one (LFS), sic et simpliciter. :D

[quote="Tiks"]Hi shyong,do you now what means RHS>LHS  :?:[/quote]

I think he means for $n>3$ , $2n<3^x-3^y$ is always true .

since we know that $min\{3^x-3^y\}=3^{p+1}-3^p=3^p\cdot 2$ where $2p+1=n$

But bernoulii inequality tell us that for $p\geq 1$

$3^p=(1+2)^p\geq 1+2p = n$ the equality holds when $p=1$ which is $n=3$ . So for any $n>3$ we always have $3^x-3^y>2n$ , and there is no solution for the equality to hold for $n>3$ . ;)

[quote=neverlose][quote=Valentin Vornicu]Find all positive integers $n\geq 1$ such that $n^2+3^n$ is the square of an integer.

[i]Bulgaria[/i][/quote]

Let $n^2 + 3^n = x^2$,we put $n^2$ at right and get $3^n = (x - n)(x+n)$.
Let $x - n = 3^y$ and $x + n = 3^{n-y}$,of course $n - y > y$,$2y< n$.$x = 3^{n-y} - n$ and we get that $n^2 + 3^n = (3^{n-y} - n)^2$ or $3^n + 2n3^{n-y} = 3^{2(n-y)}$ or $2n3^{n-y} = 3^n(3^{n-2y}-1)$,since $3^{n-2y} - 1 = 2(mod 3)$ we obtain $n = 3^y$ and $3^{n-2y} - 1 = 2$ we solve the system and obtain the equality $n = 3^y = 2y + 1$ $=>$ $y$ can be 0 or 1 and we finaly get     $n = 1$ or $n = 3$.[/quote]

Sorry for revival, but how do you go from $3^{n-2y}-1\equiv 2\mod3$ to $n=3^y?$

[b]Solution:[/b]

Let $3^n+n^2=k^2$, where $k$ is a positive integer. So, $3^n=k^2-n^2$. Let $d=\gcd(k,n). \Rightarrow d|3^n$. Let $d=3^e$. Let $n=3^e \cdot n', k=3^e \cdot k'$, where $\gcd(k',n')=1$. The equation becomes $3^{n-2e} = (k'-n')(k'+n')$.

If $k'-n'>1$, then clearly $k'+n'>1$, which will imply that both $k'$ and $n'$ are divisible by $3$, which contradicts the fact that $k'$ and $n'$ are co-prime. So, $k'-n'=1. \Rightarrow k'=n'+1$. So, $2n'+1=3^{3^e\cdot n' - 2e}$. We can notice that RHS grows very fast compared to LHS. Easy bounding gives us $n'=1$.

So, $3=3^{3^e-2e}$. So, $1=3^e-2e. \Rightarrow 2e+1=3^e. \Rightarrow e=0,1.$

So, $n=1\cdot 3^0, 1\cdot 3^1 = 1,3. \blacksquare$

After rearranging we get:
$(k-n)(k+n) = 3^n$

Let $k-n = 3^a, k+n = 3^{n-a}$

we get: $2n = 3^a(3^{n-2a} - 1)$
or, $(2n/(3^a)) + 1 = 3^{n-2a}$

Now, it is clear from above that $3^a$ divides $n$. so, $n \geq 3^a$

If $n = 3^a, n - 2a = 3^a - 2a \geq 1$
so $RHS \geq 3$
But $LHS = 3$

If $n > 3^a$ then $RHS$ increases exponentially compared to $LHS$ so $n$ cannot be $> 3^a$.

Thus $n = 3^a$.

Substituting value of $n$ above we get:

$3 = 3^{3^a - 2a}$

or $3^a - 2a = 1$ this results in only $a = 0$ or $a = 1$

Thus $n = 1$ or $3$.

Here's my work.

Find all positive integers $n \geq 1$ such that $n^2+3^n$ is the square of an integer.

[b]Solution -[/b] Note that $$n^2+3^n=k^2 \implies k^2-n^2=3^n \implies (k+n)(k-n)=3^n.$$ Let us experiment a bit with the problem. We can create a table showing values that work when $n=1,2,3,...,9.$ (We are only going to $9$ for a specific reason. Read until the end to find out.) We seek a pattern in the values that work. Here's the table: $$n=1; (k+1)(k-1)=3^1 \rightarrow k=2$$ $$n=2; (k+2)(k-2)=3^2 \rightarrow k \notin \mathbb{R}$$ $$n=3; (k+3)(k-3)=3^3 \rightarrow k=6$$ $$n=4; (k+1)(k-1)=3^4 \rightarrow k \notin \mathbb{R}$$ $$n=5; (k+1)(k-1)=3^5 \rightarrow k \notin \mathbb{R}$$ $$n=6; (k+1)(k-1)=3^6 \rightarrow k \notin \mathbb{R}$$ $$n=7; (k+1)(k-1)=3^7 \rightarrow k \notin \mathbb{R}$$ $$n=8; (k+1)(k-1)=3^8 \rightarrow k \notin \mathbb{R}$$ $$n=9; (k+1)(k-1)=3^9 \rightarrow k \notin \mathbb{R}.$$ Consider the following claim.

[b][color=#f00]Claim.[/color][/b] $n=1,3$ are the only solutions.

[i]Proof.[/i] Notice that these solutions claimed are in the form $3^k,$ where $k \in {\{0,1\}}.$ It suffices to show that $k>3$ is, indeed impossible. (Then, all we have to do is show that $n=2$ does not work, and we are done.)

Applying Bernoulli's Inequality, we obtain $$2n<3^x-3^y ~ \text{(As shown in Post 10)},$$ and the result follows. We now only use casework on $n=1,2,3$ to get that $n=1,3$ work, but $n=2$ does not. All cases have been exhausted, and thus the proof is complete. $~ \blacksquare$

[b]Remark.[/b] The table was nothing but motivation for the claim.


[quote="ComplexZeta"]When I work on hard geometry problems, I usually draw at least two diagrams. The first one will be nice (maybe an equilateral triangle or a right triangle or something), and the second will be a really weird one (definitely an obtuse triangle unless the problem forbids it). Anything that is the same in the two pictures will probably be the same in all triangles.[/quote]
the first picture is helpful when you want to solve some length or angles
the second picture is helpful when you want to draw a common conclusion, because sometimes it's correct in first pic but incorrect in second then it's incorrect.

[quote="mathkid"]When solving geometry problems, it often takes a fairly accurate drawing to finally 'see' it.

however, its so easy to draw diagrams that appear to have right angles where they don't belong, and parallel lines when they arent actually parallel. this can usually screw people up and distract them into thinking something about the problem that isn't necessarily true.

what do you guys do about this? i often waste too much time trying to draw a close to accurate picture when problems get more complex. and when drawing three dimensional solids like tetrahedrons, i always have a hard time drawing it.. and when i do, its so hard to work with a flat drawing of a 3-D object.

any tips?[/quote]
2-D pictures are not difficult to draw, you can find bunch of softwares to help this.
when it comes to 3-D pictures, hmm, that's why I come to this thread. I used forum search to find all solid geometry topics. I am also looking for a good software to draw solid figures.

I notice that sometimes simplifying the figure and just focusing on one part, then redrawing it for each other seemingly important part really helps me break the problem down and not get lost in a mess of lines.

[quote="paganinio"]
when it comes to 3-D pictures, hmm, that's why I come to this thread. I used forum search to find all solid geometry topics. I am also looking for a good software to draw solid figures.[/quote]
Maple 9.5 is amazing at plotting 3D objects, and if you're not experienced with it, its help file is really easy to read and follow.  I think it has some built in packages like plot3D to help with the programming too.

that's cool, a version 9.5 software must have been improved for numerous times. :D

[quote="mathkid"]When solving geometry problems, it often takes a fairly accurate drawing to finally 'see' it.

however, its so easy to draw diagrams that appear to have right angles where they don't belong, and parallel lines when they arent actually parallel. this can usually screw people up and distract them into thinking something about the problem that isn't necessarily true.[/quote]

Here are two suggestions:

1. Draw a big diagram.  Someone once told me if you can cover your diagram with a postage stamp, it's too small.
2. Use ruler and compass.  Use graph paper if you need to.

there're also many professional drawing tools, you can try them

I would like to add that you should bring a protractor to olympiads. Sometimes pure straightedge/compass constructions are impossible or difficult.

I've seen someone bring a 'mega-protractor' to exam...but we didn't have to use that during that exam at all!


function
number theory proposed
number theory

a)find all surjective functions $f:N^{*}->N^{*}$ such that:
                $f(m) \equiv 0(mod f(n)) <-> m \equiv 0(m0d n)$  for all $m,n \in N^{*}$
b)find all functions $f:N^{*}->N^{*}$ such that:
                 $m*f(n)+n*f(m)=(m+n)*f(m^2+n^2)$ for all $m,n \in N^{*}$

I think if $f(x)=x$ is one of the functions which satisfy the property.


I already posted a similar thread in the iTest forum...
I am continuing an iTest team that is part of the high school. Is there anyone who does not have a team and would like to join??? I expect that we should do fairly well this year.

Indianamath  (yuquing zhang) would like to join.

I hope this isn't an attempt to exacerbate tensions between you and indianamath lol.

[quote="indianamath"]what the heck vikas, what have i done to you.[/quote]
You called me names behind my back.

right...

are you sure this isn't udayan and shawn's idea of making you mad at me?

it could be...because shawn and udayan (and jimmy for that matter) do make fun of people alot

I've heard of that test, I'll join if its after college apps i.e. second semester. And I'll try to convince the Benjamins or Phil to join. Did you guys know that David is on the 2007 IOI team?

@ fibowade1123581321
What school are you going to?

I want to join if you could have more info (like how the problems are presented, dikfficulty, format) that would be nice.  btw I go to Carmel high school

well the rewards are nice for a teenager...

but i dont have five people that would do it..so probalby not.

[quote="hunter34"][url=http://en.wikipedia.org/wiki/Itest]To teh internet![/url][/quote]
doesn't even say what the problems are like...

fibo i already pmed you but yea what school do you go to...i might join.

[quote="hunter34"]TO the internet
[/quote]

[quote="hunter34"]TO GOOGLE!
[/quote]

lol


maybe im just stupid but how can 26 people use base 10 if what the second one says implies there are 25 residents?

They speak in their respective bases, so you have to figure it out using the fact that they will say stuff in their base.

There are at least 26 people because the first one said there are 26 people that speak my language, base 10. So 26 in base 10 is 26, and I think you can cross out answers and b.

Not quite true. In fact, no matter what base you speak in, you would always say "I speak in base 10". If you spoke in base 16, then 16 in base 16 is 10 :)

This problem isn't that hard once you put everything in one base.

  Let's name the two bases b and c.  If you translate all of the numbers into a base ten expression using the definition of bases, you get this.

Resident one:
2b + 6 speakers of base b
2b + 2 speakers of base (b + 4)=c

Resident two:
2c + 5 total residents
c + 3 bilingual speakers
1 non-speaking resident

Now convert all of resident two's statements into base b.  To do this, just substitute (b + 4) for c.  This gives you

Resident two:
2b + 13 total residents
b + 7 bilingual speakers
1 non-speaking resident

Now comes the part that's a little tricky:  To connect the two
speakers, we must find the total number of residents using Resident one's statements.  To do this, we add (2b + 6) + (2b + 2) and then subtract (b +  7) because we counted the bilingual speakers twice.  Now we 3b + 1 total  [i]speakers[/i].  There is one more resident!  He's mute, but he's still human, so we'll add him in, giving us 3b + 2 total residents.

To solve for b, we now solve (3b + 2)=(2b + 13) for b.  The answer is 11.  Then we plug 11 back in for b in the left side of the equation, giving us 3*11 + 2= [b]35 [/b] total residents!

mathclass wrote:In the small hamlet of Abzae, two base systems are in common us. Also, everyone speaks the truth. One resdient said, "26 people use my base, base 10, and only 22 people speak base 14." Another said, "Of the 25 residents, 13 are bilingual and 1 can't use either base." How many residents are there? (Use base 10, please).

A) 15 B) 25 C) 27 D) 35 E) 36

This problem is one of the dropped AHSME problems, and it's kinda confusing.

Can anyone help?

Spoiler (highlight to view):

[hide]Say that resident 1 speaks in base b.  Thus, 2b + 6 people use base b and 2b + 2 people speak base b+4.

If resident 2 speaks in base b also, then there are 2b+5 residents in all.  But 2b+6 people speak in base b, so this is impossible.  Thus, resident 2 must be speaking in base b+4, so there are 2(b+4) + 5 = 2b + 13 people in all.  (b+4) + 3 = b+7 people are bilingual, and 1 person cannot speak in either base.

Since 1 person cannot speak in either base, the remaining 2b + 12 people can speak one or both bases.  So (2b + 6) + (2b + 2) - (b+7) = 2b + 12.  (We subtracted the number of bilingual people because we counted then twice otherwise.)  Therefore 3b + 1 = 2b + 12, or b = 11.  Thus, there are 2*11 + 13 = 35 people in all.  (D)[/hide]

Fun problem!

thanks. haha, i figured out the base 10 thing after a bit more thinking


geometry
circumcircle
absolute value
national olympiad

[b]1)[/b] Find the number of the connected graphs with 6 vertices. (Vertices are considered to be different)

[b]2)[/b] Two different points $A$ and $B$ and a circle $w$ that passes through  $A$ and $B$ are given. $P$ is a variable point on $w$ (different from $A$ and $B$. $M$ is a point such that $MP$ is the bisector of the angle $<APB$ ($M$ lies outside of $w$) and $MP=AP+PB$. Find the geometrical locus of $M$.

[b]3)[/b] $a, b, c$ are positive reals such that their sum is $1$. Prove that
$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ac+2b^{2}+2b}$ is more then or equal to $\frac{1}{ab+bc+ac}$.

[b]4)[/b] $ABC$ is an acute angled triangle, $B_{1}, C_{1}, A_{1}$ are points respectively on $[AC, [BA, [CB$ such that $ABC$ is similar to $A_{1}B_{1}C_{1}$. Prove that orthocenter of $A_{1}B_{1}C_{1}$ coincides with circumcenter of $ABC$.

[b]5)[/b] $n$ is satisfying the above conditions
i) $n$ is a positive odd integer.
ii) There are some odd integers such that their squares' sum is equal to $n^{4}$.
Find all $n$ numbers.

[b]6)[/b] We write $1$ or $-1$ on each unit square of a $2007x2007$ board. Find the number of writings such that for every square on the board the absolute value of the sum of numbers on the square is less then or equal to $1$.

#3: Follows from recently proposed problem http://www.mathlinks.ro/Forum/viewtopic.php?t=141336

http://www.mathlinks.ro/Forum/viewtopic.php?t=143080
Some moderator: please close this one.

1  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143101[/url]
2  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143103[/url]
3  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143106[/url]
4  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143090[/url]
5  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143105[/url]
6  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=143102[/url]
Many thanks to Little1000t

Who are going to IMO this year?I heard that there are 3 student from Samanyolu liseym.

cafer tayyar yıldırım
melih üçer
süreyya emre kurt
burak sağlam
semih yavuz
tuba uzluer
(first three from samanyolu and the others from yamanlar)
are going to imo this year.

i don't know exactly, but i think 25 points is necessary.


function
invariant
real analysis
real analysis unsolved

$f:R\rightarrow R$ injective , suppose for all x in R

$f(x^3-9x^2-20x+1)=f^3(x)-9f^2(x)-20f(x)+1$

Prove that there exist $x_1,x_2,x_3$ "diferite (not equal) reals numbers such that

$f(x_1)+f(x_2)+f(x_3)=9$

Posted by Silviu on Romanian forum

Look at the fix points of $ x^3-9x^2-20x-1$ and observe they are let invariant by f.


[quote="fuzzylogic"]
For those of you who would like to practice the similar idea of problem solving, here is a straight extension of the problem posted in the thread help- ugly minimum.

Let $x_1, x_2,\ldots,x_n (n\ge 2)$ be positive real numbers. Show that

$\displaystyle \sum^n_{i=1} \left(\frac{x_i}{\sum^n_{k=1} x_k - x_i}\right)^{\frac{n-1}{n}} \geq n \left(\frac{1}{n-1}\right)^{\frac{n-1}{n}}$.[/quote]

[quote="fuzzylogic"]For those of you who would like to practice the similar idea of problem solving, here is a straight extension of the problem posted in the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=96527#96527]help- ugly minimum[/url].

Let [tex]x_1, x_2,\ldots,x_n (n\ge 2)[/tex] be positive real numbers. Show that

[tex]\displaystyle \sum^n_{i=1} \left(\frac{x_i}{\sum^n_{k=1} x_k - x_i}\right)^{\frac{n-1}{n}} \geq n \left(\frac{1}{n-1}\right)^{\frac{n-1}{n}}[/tex].[/quote]

Instead of [tex] / [/tex] (that is used on some other boards) , use the symbol dollar [b]\$[/b] for you latex expression to be compiled and displayed correctly.  :-)
[b]Example :[/b]
with tex command     : [tex] x^2 [/tex]
with dollar command : $ x^2$ (whose code is : \$ x^2 \$)

I like how my post gets deleted and then the only follow up post is regarding the use of [tex]/[/tex] and not the problem itself.

Hold on....please don't follow Sonnhards path. He revived a 3year old and 2year old thread once but this is like 8 years old. Pease don't revive things.


Let three congruent circles with one common point meet again in three points, $ A,B,C$. Then the common radius of the three given circles is equal to the circumradius of triangle $ ABC$, and their common point is its orthocenter.

The problem can be generalized:

[b]Three circles having a common pont and intersecting again at three points, $A, B, C$. Then $R^3=R_1\cdot R_2\cdot R_3$, where $R$ means the circumradius of $\triangle ABC$, while $R_i$ are the other three circles' radii.[/b]
I saw it in a Romanian Mathematics Magazine.

Best regards,
sunken rock

[quote="sunken rock"]Three circles having a common pont and intersecting again at three points, $A, B, C$. Then $R^3=R_1\cdot R_2\cdot R_3$, where $R$ means the circumradius of $\triangle ABC$, while $R_i$ are the other three circles' radii.[/quote]

I tried some examples, and this formula doesn't work.  You can also see this as follows: Let $A$, $B$, and $C$ be fixed points (which means $R$ is fixed), and let $O$, the common point, vary.  If $O$ approaches one of the side lengths of triangle $ABC$, then one of the $R_i$ goes to infinity.

[asy]
unitsize(1 cm);

pair A, B, C;
pair[] O;
real[] R;

A = (1,2);
B = (0,0);
C = (3,0);
O[0] = (1.5,1);
O[1] = circumcenter(O[0],B,C);
O[2] = circumcenter(O[0],C,A);
O[3] = circumcenter(O[0],A,B);
R[1] = circumradius(O[1],B,C);
R[2] = circumradius(O[0],C,A);
R[3] = circumradius(O[0],A,B);

draw(circumcircle(A,B,C),dashed);
draw(Circle(O[1],abs(O[1] - O[0])));
draw(Circle(O[2],abs(O[2] - O[0])));
draw(Circle(O[3],abs(O[3] - O[0])));

dot("$A$", A, NW);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$O$", O[0], NE);
dot("$O_1$", O[1], S);
dot("$O_2$", O[2], NE);
dot("$O_3$", O[3], NW);
[/asy]


calculus
integration
trigonometry
logarithms
function
calculus computations

Find $ \int_0^1 2 \cdot (tan^{\minus{}1}x)^2 dx$  :)

This sum came in the public exam!!!  :o

let me add some more  spice it was a 6 mark question in a 100 mark paper of three hours :rotfl:  :rotfl:  :huh:

Did any of you guys get the sum in the exam???  :huh:   :rotfl:  :rotfl:

yes i solved it before i had a look at it  :P  :rotfl:  :rotfl:

Can some one who's good at calculus try this one. Pls let us know even if u don't get the answer or if u think that it is not integrable.

[hide="Answer"]Make $ \arctan x \equal{} u$:

$ I \equal{} 2 \int_0^1(\arctan x)^2\,dx \equal{} 2 \int_0^{\pi/4}u^2 \sec^2u\,du \equal{} 2[u^2 \tan u]_0^{\pi/4} \minus{}$

$ \minus{} 4 \int_0^{\pi/4}u \tan u\,du \equal{} \frac{\pi^2}{8} \plus{} 4 [\ln(\cos u) \cdot u]_0^{\pi/4} \minus{} 4 \int_0^{\pi/4}\ln(\cos u)\,du$

$ \equal{} \frac{\pi^2}{8} \minus{} \frac{\pi}{2} \ln 2 \minus{} 4 \int_0^{\pi/2}\ln(\sin t)\,dt \plus{} 4 \int_{0}^{\pi/4}\ln(\sin t)\,dt$

(where the substitution $ u \equal{} \frac{\pi}{2} \minus{} t$ was made)

$ \equal{}> I \equal{} \frac{\pi^2}{8} \plus{} \frac{\pi}{2} \ln 2 \minus{} 2G$

where G is Catalan's constant and where I made use of the facts that $ \int_0^{\pi/2}\ln(\sin t)\,dt \equal{} \minus{}\frac{\pi}{2} \ln 2$ and $ \int_{0}^{\pi/4}\ln(\sin t)\,dt \equal{} \minus{} \frac{\pi}{4} \ln 2 \minus{} \frac{G}{2}$. In these two last integrals, the first is easily calculated, and the second can be computed, for example, by using the Fourier series of the function $ \ln\left(2 \sin \frac{x}{2}\right)$.[/hide]

wow! hat's off carcul!  :)
but do you think that this is undergrad level? none of us have even heard of fourier series or catalan's constant before.  :oops:

Wow. Great, Carcul.

I think looking at the soln, it definitely means 6 marks for all of those who have attempted. This is even beyond JEE level. :D But how can the board compensate for the mental torture the students have undergone.

[quote="Kalle"]Of course it's undergrad level, although it is an incredibly dumb problem. Absolutely no reason to include something with catalan's constant when people might not know what that is, or that this constant might be considered in elementary form already. I think you're being a bit whiny though, this problem isn't that hard.

What kind of exam was this on? Public exam, what is that?[/quote]

Class XII Public Maths exam. :D:D

Oh, so you're not university students? Then yes, it's very difficult. Maybe they hired misan to construct the problems  :D

Maybe they made a mistake and this problem was supposed to be something else. Certainly seems like it.

No misan wouldnt have given such an easy sum n all :D :D that would have been solved by his 5th std students :D

[quote="Kalle"]

Maybe they made a mistake and this problem was supposed to be something else. Certainly seems like it.[/quote]

We are all in grade 12.  :rotfl:  :rotfl: The question was probably meant for gifted students. :D I guess this was a typo or they meant something much much easier.

GOOD ONE .... Everybody would be getting their 6 marks.....
Can u tell what the Catalan's Constant is. Is it a value

Obviously since its a constant. ;) http://en.wikipedia.org/wiki/Catalan%27s_constant

[quote="madness"]Find $ \int_0^1 2 \cdot (tan^{ \minus{} 1}x)^2 dx$  :)

This sum came in the public exam!!!  :o[/quote]
but i beleive that 2 was a typo...one who can integrate
$ \int_{0}^{1}(\tan^{\minus{}1} x)^{2} dx$ can integrate the above one so what's the use of having a $ 2$

[quote="pardesi"]but i beleive that 2 was a typo...one who can integrate
$ \int_{0}^{1}(\tan^{ \minus{} 1} x)^{2} dx$ can integrate the above one so what's the use of having a $ 2$[/quote]
yeah.
i think the correct sum is :
$ \int_{0}^{1}x(\tan^{ \minus{} 1} x)^{2} dx$ , which i presume is simpler


[quote="sos440"]Solving $ k^{4}= Ak(k-1)(k-2)(k-3)+Bk(k-1)(k-2)+Ck(k-1)+Dk+E$ gives $ A=1, B=6, C=7, D=1, E=0$. So

$ \sum_{k=0}^{\infty}\frac{k^{4}x^{k}}{k!}\; = \; (Ax^{4}e^{x}+Bx^{3}+Cx^{2}+Dx+E)e^{x}\; = \; (x^{4}+6x^{3}+7x^{2}+x)e^{x}$

Plugging $ x = 1$ to this equation gives the answer $ 15e$[/quote]

Maybe i`ve done some mistakes in my calculations :(

[quote="misan"]i have asked the question on this forum before, as a matter of fact,
it can be shown that the [i]generic[/i] series

$ \frac{1}{e}\cdot\sum_{k=1}^\infty\;\frac{k^{p}}{k!}\;\;\;\quad (\;\;where\;\;\;p\in\mathbb N\;\; )$

forms a sequence known as the [b]bell [/b]numbers, i.e. $ \{\;1,2,5,15,52,203,877\cdots\;\}$[/quote]
Can you give the link? Can`t find...  :maybe:


geometry
perimeter
area of a triangle
Heron\u0027s formula

[quote="Icosahedroid"]The three edge-lengths of a triangle are all the square root of a prime number. Prove that the area of that triangle cannot be an integer.[/quote]
[hide="Solution...maybe?"]Denote the sides of the triangle $ a,b,c$, knowing that since there are no perfect prime squares, $ a,b,c$ are all non-integer numbers. The semi-perimeter is therefore $ \frac{a\plus{}b\plus{}c}{2}$, and using Heron's Formula, we get that the area of the triangle is $ \sqrt{\frac{a\plus{}b\plus{}c}{2}\cdot\frac{b\plus{}c\minus{}a}{2}\cdot\frac{a\minus{}b\plus{}c}{2}\cdot\frac{a\plus{}b\minus{}c}{2}}$. Simplifying the inside results in $ \sqrt{\frac{a^2\plus{}b^2\minus{}c^2\plus{}2ab}{4}\cdot\frac{b^2\plus{}c^2\minus{}a^2\plus{}2bc}{4}}$. Now $ a^2, b^2, c^2$ are all integers, but on the other hand we still have the terms $ 2ab$ and $ 2bc$, and multiplying even further results in more irrational numbers. And even if the possibility of them canceling out, the term $ 4ab^{2}c$ is irrational because $ ac$ is not a perfect square, therefore the area is not an integer.[/hide]

[quote="Icosahedroid"]That proof is invalid. If you multiply out all of the terms under the square root you will get the square root of:
[2(a^2)(b^2) +2(b^2)(c^2)+2(a^2)(c^2)-(a^4)-(b^4)-(c^4)]/16

btw, how do you make a square root sign?[/quote]
[code]$\sqrt{blah}$[/code][/code]
Are you sure that's the result of multiplying everything out under the radical? Shouldn't the $ b^4$ be positive?

I'm pretty sure. I worked it out by hand and wikipedia agrees with me.

http://en.wikipedia.org/wiki/Heron's_formula


calculus
integration
modular arithmetic
number theory unsolved
number theory

am not sure whether this is the right forum(am a new user).hope youll can help me.

the problem is

solve for integral x,y,z,w

  x^4 + y^4 + z^4 - w^4=1995

[b][u]Hint:[/u][/b] Use the fact that $n^{4}\equiv 0,1 \pmod{16}$ for every integer $n$.


modular arithmetic
number theory
number theory proposed

Let $ p$ be an odd prime. Prove that $ \sum_{k \equal{} 1}^{p \minus{} 1}(2^k.k^{p \minus{} 2}) \minus{} \sum_{k \equal{} 1}^{\frac {p \minus{} 1}{2}}(k^{p \minus{} 2})$ is divisible by $ p$.

[color=blue][b]Problem.[/b] Let $ p$ be an odd prime. Prove that

$ \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} k^{p\minus{}2} \equiv \sum_{k\equal{}1}^{p\minus{}1} 2^k k^{p\minus{}2} \mod p$.[/color]

[i]Solution.[/i] For every $ k\in\left\{1,2,...,p\minus{}1\right\}$, we have $ \binom{p\minus{}1}{k\minus{}1}\equiv\left(\minus{}1\right)^{k\minus{}1}\mod p$ (by Lemma 1 in [url=http://www.mathlinks.ro/viewtopic.php?p=849241#849241]post #2 of http://www.mathlinks.ro/viewtopic.php?t=150392[/url]). But

$ \binom{p\minus{}1}{k\minus{}1}\equal{}\frac{\left(p\minus{}1\right)!}{\left(k\minus{}1\right)!\left(\left(p\minus{}1\right)\minus{}\left(k\minus{}1\right)\right)!}\equal{}\frac{\left(p\minus{}1\right)!}{\left(k\minus{}1\right)!\left(p\minus{}k\right)!}$
$ \equal{}\frac{k}{p}\cdot\frac{\left(p\minus{}1\right)!\cdot p}{\left(k\minus{}1\right)!\cdot k\cdot\left(p\minus{}k\right)!} \equal{} \frac{k}{p}\cdot\frac{p!}{k!\left(p\minus{}k\right)!}\equal{}\frac{k}{p}\cdot\binom{p}{k}$,

so that $ \binom{p\minus{}1}{k\minus{}1}\equiv\left(\minus{}1\right)^{k\minus{}1}\mod p$ becomes $ \frac{k}{p}\cdot\binom{p}{k}\equiv\left(\minus{}1\right)^{k\minus{}1}\mod p$. Multiplying this by $ \left(\minus{}1\right)^{k\minus{}1}$, we get $ \left(\minus{}1\right)^{k\minus{}1}\frac{k}{p}\cdot\binom{p}{k}\equiv\left(\minus{}1\right)^{2\left(k\minus{}1\right)}\mod p$. But $ \left(\minus{}1\right)^{2\left(k\minus{}1\right)}\equal{}1\equiv k^{p\minus{}1}\mod p$ by Fermat's Little Theorem, so $ \left(\minus{}1\right)^{k\minus{}1}\frac{k}{p}\cdot\binom{p}{k}\equiv k^{p\minus{}1}\mod p$. We can divide this congruence by $ k$ (since $ p$ is prime and $ k\in\left\{1,2,...,p\minus{}1\right\}$, and thus $ k$ is coprime to $ p$), and obtain

[color=green][b](1)[/b][/color] $ \left(\minus{}1\right)^{k\minus{}1}\frac{1}{p}\cdot\binom{p}{k}\equiv k^{p\minus{}2}\mod p$.

(Note that $ \frac{1}{p}\cdot\binom{p}{k}$ is an integer, since $ p$ is prime and $ k\in\left\{1,2,...,p\minus{}1\right\}$ and therefore $ p\mid\binom{p}{k}$.)

Now, [color=green][b](1)[/b][/color] yields

$ \sum_{k\equal{}1}^{p\minus{}1} 2^k k^{p\minus{}2} \equiv \sum_{k\equal{}1}^{p\minus{}1} 2^k \cdot \left(\minus{}1\right)^{k\minus{}1}\frac{1}{p}\cdot\binom{p}{k}$
$ \equal{}\sum_{k\equal{}1}^{p\minus{}1} 2^k\cdot\left(\minus{}\left(\minus{}1\right)^k\right)\frac{1}{p}\cdot\binom{p}{k}\equal{}\minus{}\frac{1}{p}\cdot\sum_{k\equal{}1}^{p\minus{}1} 2^k\left(\minus{}1\right)^k\binom{p}{k}$
$ \equal{}\minus{}\frac{1}{p}\cdot\left(\sum_{k\equal{}0}^p 2^k\left(\minus{}1\right)^k\binom{p}{k}\minus{}\underbrace{2^0}_{\equal{}1}\underbrace{\left(\minus{}1\right)^0}_{\equal{}1}\underbrace{\binom{p}{0}}_{\equal{}1}\minus{}2^p\underbrace{\left(\minus{}1\right)^p}_{\equal{}\minus{}1}\underbrace{\binom{p}{p}}_{\equal{}1}\right)$       (here, $ \left(\minus{}1\right)^p\equal{}\minus{}1$ since $ p$ is odd)
$ \equal{}\minus{}\frac{1}{p}\cdot\left(\sum_{k\equal{}0}^p\left(\minus{}2\right)^k\binom{p}{k}\minus{}1\minus{}\left(\minus{}2^p\right)\right)$
$ \equal{}\minus{}\frac{1}{p}\cdot\left(\minus{}1\minus{}1\minus{}\left(\minus{}2^p\right)\right)$           (since the binomial formula yields $ \sum_{k\equal{}0}^p\left(\minus{}2\right)^k\binom{p}{k}\equal{}\left(1\plus{}\left(\minus{}2\right)\right)^p\equal{}\left(\minus{}1\right)^p\equal{}\minus{}1$, where we again use that $ p$ is odd)
$ \equal{}\minus{}\frac{1}{p}\cdot\left(\minus{}2\plus{}2^p\right)\equal{}\minus{}\frac{2^p\minus{}2}{p}\mod p$.

On the other hand, by Fermat's Little Theorem, $ 1\equiv k^{p\minus{}1}\mod p$ for every $ k\in\left\{1,2,...,p\minus{}1\right\}$, while Wilson's Theorem says that $ \left(p\minus{}1\right)!\equiv \minus{}1\mod p$. Thus, $ \left(p\minus{}1\right)!\equiv \minus{}1\equiv \minus{}k^{p\minus{}1}\mod p$, so that $ k^{p\minus{}1}\equiv \minus{}\left(p\minus{}1\right)!\mod p$. We can divide this congruence by $ k$ (since $ k$ is coprime to $ p$, as before) and get $ k^{p\minus{}2}\equiv \minus{}\frac{\left(p\minus{}1\right)!}{k}\mod p$ (notice that $ \frac{\left(p\minus{}1\right)!}{k}$ is an integer, since $ 1\leq k\leq p\minus{}1$ and thus $ k\mid\left(p\minus{}1\right)!$). Hence,

$ \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} k^{p\minus{}2}\equiv \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} \left(\minus{}\frac{\left(p\minus{}1\right)!}{k}\right)\equal{}\minus{}\sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} \frac{\left(p\minus{}1\right)!}{k}\mod p$.

But $ \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} \frac{\left(p\minus{}1\right)!}{k}\equiv \frac{2^p\minus{}2}{p}\mod p$ as proven in http://www.mathlinks.ro/viewtopic.php?t=38958 . Thus,

$ \sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} k^{p\minus{}2}\equiv \minus{}\sum_{k\equal{}1}^{\frac{p\minus{}1}{2}} \frac{\left(p\minus{}1\right)!}{k}\equiv \minus{}\frac{2^p\minus{}2}{p} \equiv \sum_{k\equal{}1}^{p\minus{}1} 2^k k^{p\minus{}2}\mod p$,

qed.

  darij

Hey, I am TeXing the solutions of some problems from Polish olympiad math camp and I have written down the solution of this problem just yesterday, so I will simply copy it here, but I feel that it is very similar to the darij's approach.

We use the fact that
\[ \sum_{i \equal{} 1}^{n} \binom{i}{j} \equal{} \binom{n \plus{} 1}{j \plus{} 1}
\]
and also
\[ \sum_{i \equal{} 1}^{p \minus{} 1} \frac {1}{i} \equiv 0 \pmod{p}.
\]
And then:
\[ \sum_{i \equal{} 1}^{p \minus{} 1} 2^i i^{p \minus{} 2} \equiv \sum_{i \equal{} 1}^{p \minus{} 1} \frac {(1 \plus{} 1)^i}{i} \equal{} \sum_{i \equal{} 1}^{p \minus{} 1} \sum_{j \equal{} 0}^{i} \frac {\binom{i}{j}}{i} \equal{} \sum_{j \equal{} 0}^{p \minus{} 1} \sum_{i \equal{} j}^{p \minus{} 1} \frac {\binom{i}{j}}{i} \equal{} \sum_{j \equal{} 1}^{p \minus{} 1} \sum_{i \equal{} j}^{p \minus{} 1} \frac {\binom{i}{j}}{i} \plus{} \sum_{i \equal{} 1}^{p \minus{} 1} \frac {1}{i} \equiv
\]

\[ \equiv \sum_{j \equal{} 1}^{p \minus{} 1} \sum_{i \equal{} j}^{p \minus{} 1} \frac {\frac {i!}{j!(i \minus{} j)!}}{i} \equal{} \sum_{j \equal{} 1}^{p \minus{} 1} \frac {1}{j} \sum_{i \equal{} j}^{p \minus{} 1} \binom{i \minus{} 1}{j \minus{} 1} \equal{} \sum_{j \equal{} 1}^{p \minus{} 1} \frac {1}{j} \binom{p \minus{} 1}{j} \equal{}
\]

\[ \equal{} \sum_{j \equal{} 1}^{p \minus{} 1} \frac {1}{j} \frac {(p \minus{} 1)(p \minus{} 2) \ldots (p \minus{} j)}{j!} \equiv \sum_{j \equal{} 1}^{p \minus{} 1} \frac {1}{j} \frac {( \minus{} 1)^{j}j!}{j!} \equal{} \sum_{j \equal{} 1}^{p \minus{} 1} \frac {( \minus{} 1)^{j}}{j} \equal{}
\]

\[ \equal{} \minus{} \sum_{j \equal{} 1}^{\frac {p \minus{} 1}{2} \minus{} 1} \frac {1}{2j \plus{} 1} \plus{} \sum_{j \equal{} 1}^{\frac {p \minus{} 1}{2}} \frac {1}{2j} \equiv 2\sum_{j \equal{} 1}^{\frac {p \minus{} 1}{2}} \frac {1}{2j} \equiv \sum_{j \equal{} 1}^{\frac {p \minus{} 1}{2}} j^{p \minus{} 2} \pmod{p}.
\]


trigonometry
inequalities unsolved
inequalities

let $A+B+C=\pi$ prove that:
$\cos(A-B).\cos(B-C).\cos(C-A)\ge \cos A.\cos B.\cos C$

Maybe $cos(A-B)\cdot cos(B-C)\cdot cos(C-A)\ge 8\cos A\cdot cos B\cdot cos C?$
If so then:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60450


The product of a number $ M$ and six less than $ M$ is $ \minus{}5$. What is the sum of all possible values of $ M$?

We have M(M-6)=-5. We expand and move over the 5 then fator again, giving us (M-1)(M-5)=0. Thus, the roots are 1 and 5, so the sum is 6.

An easier way:

$ m(m \minus{} 6) \equal{} \minus{}5$

$ m^2 \minus{} 6m \plus{} 5 \equal{} 0$

Vieta's formula says that the sum of the roots of a quadratic is $ \frac{\minus{}b}{a}$. (To see derivations go to my blog and find the Table of Contents.) That means our answer is $ \frac{\minus{}(\minus{}6)}{1} \equal{} \boxed{6}$.


A Mathlon is a competition in which there are $ M$ athletic events. Only teams $ A,B,C$ participate. In each event $ p_1$ points were awarded for first place, $ p_2$ for second and $ p_3$ for third, where $ p_1>p_2>p_3>0$ are integers. The final score for $ A$ was $ 22$, and for $ B$ and $ C$ was $ 9$. It is known that $ B$ won the $ 100$ meters. What is the value of $ M$ and who was the second in the high jump?

[hide="Solution by casework"]
We know that $ 3\leqq_{1}<9, p_{2}\geq2, p_{1}\geq1$ and that A has not won atleast one game.

So we get that $ 3\leq M \leq7$.

We can easily dismiss cases $ M=3$ and $ M=6$ and $ M=7$ (There are only unique combinations for these cases).

Hence $ 3<p_{1}\leq6$

Again we can dismiss the case of $ M=4$ and thus, $ p_{1}\neq6$.

Hence we get $ M=5$ and $ p_{1}=5$, $ p_{2}=2$ and $ p_{3}=1$.

C was second in the high jump.
[/hide]


I just solve for $a,b,c$ since $e$ and $d$ are similar to $a$ and $b$
We have:
[tex] 1 \geq a^2+\frac{(b+c+d+e)^2}{4}=\frac{5a^2}{4} \Rightarrow a\geq-0.4\sqrt{5} [/tex]
If there exist $n=4$ negative number then
[tex] 1=a^2+b^2+c^2+d^2+e^2 \leq (a+b+c+d)^2+e^2=2e^2 \Rightarrow e\geq \frac{\sqrt2}{2} [/tex]
Similar for $n=3,2,1$ and when $n=1$ we have $1 \leq 20e^2$ or $e\geq 0.1\sqrt{5}$ and this is the least value of $e$ in all case.
Assume that $b \leq 0$ then
[tex] 1 \geq a^2+b^2+\frac{(c+d+e)^2}{3}=a^2+b^2+\frac{(a+b)^2}{3}\geq \frac{10b^2}{3} \Rightarrow b\geq -\sqrt{0.3} [/tex]
Assume that $b \geq 0$ then
[tex] 1=(b+c+d+e)^2+b^2+c^2+d^2+e^2 \geq 20b^2 \Rightarrow b \leq 0.1\sqrt{5} [/tex]
Assume that $c \leq 0$ then
[tex] 1\leq a^2+b^2+c^2+\frac{(d+e)^2}{2} \leq \frac{15}{2}c^2[/tex] then [tex] c\geq-0.2\sqrt{\frac{10}{3}} [/tex] and similar for $c \geq 0$.
We have:[tex] (a+e)^2 \geq 0 \Rightarrow 4ae \geq -(a-c)^2 [/tex]
 [tex] 1\geq a^2+e^2=(a-e)^2+2ae\geq (a-e)^2 -\frac12(a-e)^2 [/tex] or
[tex] e-a \leq \sqrt{2} [/tex]
It still remains the case [tex] e-a \geq \sqrt{\frac56} [/tex].I think someone will finish it. :)


algebra
polynomial
quadratics
algebra unsolved

I guess this could kind of be NT, but mostly algebra. I don't know the answer. I couldn't find an example.

Does there exist a polynomial $Q(x) = (x-a)(x-b)(x-c)(x-d)$ where $a, b, c, d$ are distinct integers and there is a real quadratic $P(x)$ so that $P(P(x)) = Q(x)$?

because it means that $\ P(x)$ will have 4 roots $\ P(a),P(b),P(c)$ and $\ P(d)$. That could not be because P(x) is quadratic (only 2 roots)

It is not prove. Can be $P(a)=P(b),P(c)=P(d)$,

Nice. I think, that it is not true, but I don't find simple example.

Kolumbus: very nice, thanks! May I ask how you found it? Did you have an efficient method or was it just brute force?


I hope this has not been asked yet... since this seems like a new game...  :D

its a an MMORPG... simply an online game...  :D
this is the Philippine Site --> http://www.guilds.ph
though i think there are other sites.

the international site is http://www.guildwars.com


The (restricted access) "Teachers" forum and the (restricted access) "Parents" forum are both stone cold dead. It may not make any difference to anything if they sit there doing nothing, but if you need the space for anything ...

Does the subforum "Teachers" still exist? It doesn't show on my screen at all though I'm a member of the group :? I thought Valentin just deleted it during the time I didn't visit the forum...

On the other hand, as far as I remember, they used to occupy less space than a typical daily addition to "College Playground" so I do not see how removing them can help anything :P. Seriously, what is there depends on us only; I thought of posting a couple of questions on the "Teachers" subforum recently but I just couldn't find it :(.

P.S. I finally found it in Online School forum (had to look at the subfora tree to figure it out :lol:). Actually it shows up OK if you expand the Online School subforum but if you keep it compressed (as I did all the time) and just click on the name, you see only "Classes information" in the subfora list. That is what confused me when I tried to look for it before :blush:).

I moved those into the Archive a while back - you shouldn't see them where they used to be anymore.


The sum of three numbers is greater than or equal to their product. Prove that the sum of their squares also is.

Let's put it into $Latex$.You say that if $a+b+c \geq abc$ then $a^2+b^2+c^2 \geq abc $

 It can be proven very easily by using Cauchy that $a^2+b^2+c^2\geq abc\sqrt3 $ ;)

i'm reffering to the cauchy-Schwarz inequality.Anyway,I realised i don't need that to solve this problem.Just use
 the well known inequalities $a^2+b^2+c^2 \geq ab+bc+ca $ and $ab+bc+ca \geq \sqrt{3abc(a+b+c)} $ and we are done. ;)

could it be product of "squares" of numbers. I don't know if it can be proved but could that be what the question is asking?


Each of the squares shown is inscribed in a larger square so that the vertices of the inscribed square bisect the sides of the larger square. What fraction of the area of the largest square is shaded? Express your answer as a common fraction.

[asy]fill((-0.5,0.5)--(0,1)--(0.5,0.5)--cycle,gray(0.7));
fill((0.5,0.5)--(1,0)--(0.5,-0.5)--cycle,gray(0.7));
draw((-0.5,-0.5)--(-0.5,0.5)--(0.5,0.5)--(0.5,-0.5)--cycle,linewidth(0.7));
draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,linewidth(0.7));
draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,linewidth(0.7));
draw((-2,0)--(0,2)--(2,0)--(0,-2)--cycle,linewidth(0.7));
draw((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,linewidth(0.7));[/asy]

The shaded region consists of two right triangles.  In the second-smallest square there are $ 8$ such triangles.  The shaded region is $ \frac{1}{4}$ of the second-smallest square.  The second-smallest square is half of the third-smallest square, making the shaded region equal to $ \frac{1}{8}$ of the third smallest square.  This in turn is $ \frac{1}{2}$ of the fourth smallest square, making the shaded region $ \frac{1}{16}$ of the second largest square, which is half of the largest square.  The shaded region is $ \boxed{\frac{1}{32}}$ of the largest square.


integration
calculus
derivative
logarithms
calculus computations

$ \int \minus{} \frac{2}{x}\,dx \equal{} \minus{}2 \ln|x| \plus{} C$

Recall that the derivative $ \frac{d}{dx} \ln x \equal{} \frac{1}{x}$ for positive x can be found by implicitly integrating $ x \equal{} e^ {\ln x}$ and using the product rule on the right side. This covers positive x.

For negative x, $ \minus{}x \equal{} e^{\ln (\minus{}x)} \equal{} e^{\ln |x|}$ can be implicitly differentiated to achieve $ \frac{d}{dx} \ln (\minus{}x) \equal{} \frac{d}{dx} \ln |x| \equal{} \frac{1}{x}$.

Hence, in general, $ \frac{d}{dx} \ln |x| \equal{} \frac{1}{x}$.

You can use this in your antidifferentiation.

[quote="Endlessx89"]Please show me how to Integrate f(x)=-2/x ?
And please explain which formula you base your decisions on.[/quote]

$ \int \frac { \minus{} 2}{x}\, dx \equal{} \minus{} 2\int \frac {1}{x}\, dx$

The result is as several posters have pointed out: $ \minus{} 2\left(\ln\left|x\right| \plus{} C\right) \equal{} \ln\left(\frac {1}{x^2}\right) \plus{} C_1$. However, this is not at all obvious.

You might try this in order to prove it.

Let $ x \equal{} e^{y}$ then $ 1 \equal{} e^{y}\frac {dy}{dx}$.

Therefore $ \frac {dy}{dx} \equal{} \frac {1}{e^y} \equal{} \frac {1}{x}$

$ f\left(x\right) \equal{} \minus{} \frac {2}{x} \equal{} \minus{} 2\frac {dy}{dx}$

$ \int f\left(x\right)\, dx \equal{} \int \minus{} 2\frac {dy}{dx}\, dx \equal{} \minus{} 2 y \plus{} C_2$

If $ x \equal{} e^{y}$ then $ y \equal{} \ln x \Rightarrow \int f\left(x\right)\, dx \equal{} \minus{} 2\ln x \plus{} C_2 \equal{} \ln\left(\frac {1}{x^2}\right) \plus{} C_2$