Let $ a,b,c>0$  Prove that:

\[ \frac{ab+c}{a+b+2c}+\frac{bc+a}{b+c+2a}+\frac{ca+b}{c+a+2b}\leq\frac{a+b+c+3}{4}\]

Since the inequality is inhomogeneous and the variables a, b, c are subject to no restrictions apart from the one that they are positive, it can only be true if the following two homogeneous inequalities are true:

$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\leq \frac{a+b+c}{4}$;

$\frac{c}{a+b+2c}+\frac{a}{b+c+2a}+\frac{b}{c+a+2b}\leq \frac{3}{4}$.

The second of these two inequalities is rather simple: Multiplying it with 4, we get

$\frac{4c}{a+b+2c}+\frac{4a}{b+c+2a}+\frac{4b}{c+a+2b}\leq 3$.

If we denote x = b + c + 2a, y = c + a + 2b, z = a + b + 2c, then we easily see that 3x - y - z = 4a, 3y - z - x = 4b, 3z - x - y = 4c, so that it remains to show

$\frac{3z-x-y}{z}+\frac{3x-y-z}{x}+\frac{3y-z-x}{y}\leq 3$,

or

$9-\left( \frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\right) \leq 3$.

This is equivalent to

$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\geq 6$,

what is clear since

$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\left( \frac{y}{z}+\frac{z}{y}\right) +\left( \frac{z}{x}+\frac{x}{z}\right) +\left( \frac{x}{y}+\frac{y}{x}\right) \geq 2+2+2=6$.

The first inequality seems to be very hard. The only proof I have is to write

$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}-\frac{a+b+c}{4}$
$=-\frac{2a^{4}+2b^{4}+2c^{4}+b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a-6b^{2}c^{2}-6c^{2}a^{2}-6a^{2}b^{2}+2a^{2}bc+2b^{2}ca+2c^{2}ab}{4\left( a+b+2c\right) \left( b+c+2a\right) \left( c+a+2b\right) }$.

So it remains to show

$2a^{4}+2b^{4}+2c^{4}+b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a-6b^{2}c^{2}-6c^{2}a^{2}-6a^{2}b^{2}+2a^{2}bc+2b^{2}ca+2c^{2}ab\geq 0$.

Now, from the Schur inequality,

$a^{2}\left( a-b\right) \left( a-c\right) +b^{2}\left( b-c\right) \left( b-a\right) +c^{2}\left( c-a\right) \left( c-b\right) \geq 0$,

so that

$a^{4}+b^{4}+c^{4}+a^{2}bc+b^{2}ca+c^{2}ab\geq b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a$,

and

$2a^{4}+2b^{4}+2c^{4}+b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a-6b^{2}c^{2}-6c^{2}a^{2}-6a^{2}b^{2}+2a^{2}bc+2b^{2}ca+2c^{2}ab$
$=2\left( a^{4}+b^{4}+c^{4}+a^{2}bc+b^{2}ca+c^{2}ab\right) +b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a-6b^{2}c^{2}-6c^{2}a^{2}-6a^{2}b^{2}$
$\geq 2\left( b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a\right) +b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a-6b^{2}c^{2}-6c^{2}a^{2}-6a^{2}b^{2}$
$=3\left( b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a\right) -6b^{2}c^{2}-6c^{2}a^{2}-6a^{2}b^{2}$
$=3\left( b^{3}c+c^{3}b+c^{3}a+a^{3}c+a^{3}b+b^{3}a-2b^{2}c^{2}-2c^{2}a^{2}-2a^{2}b^{2}\right) $
$=3\left( bc\left( b^{2}+c^{2}\right) +ca\left( c^{2}+a^{2}\right) +ab\left( a^{2}+b^{2}\right) -2b^{2}c^{2}-2c^{2}a^{2}-2a^{2}b^{2}\right) $
$\geq 3\left( bc\cdot 2bc+ca\cdot 2ca+ab\cdot 2ab-2b^{2}c^{2}-2c^{2}a^{2}-2a^{2}b^{2}\right) $
$=3\left( 2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-2b^{2}c^{2}-2c^{2}a^{2}-2a^{2}b^{2}\right) \geq 0$,

and the martyry is over.

  Darij

[quote="harazi"]Darij, why don't you just write $\frac{ab}{a+b+2c}\leq\frac{ab}{a+c}+\frac{ab}{b+c} $ and add up everything?[/quote]

I think you wanted to say

[tex]\frac{ab}{a+b+2c}\leq\frac14 \left( \frac{ab}{a+c}+\frac{ab}{b+c} \right) [/tex].

This is indeed simple, thanks a lot, I wonder how stupid I was :blush:  :blush:  :blush:

  dg


Show:
$ \sum_{cyc}\frac{z(x\plus{}y)}{(y\plus{}z)(x\plus{}z)} \leq \frac{3}{2}$ where x,y,z are positive reals.

[quote="kops723"]First of all, I don't see how those are equivalent at all, can you please explain?

Secondly, the bottom equality doesn't hold for x = y = z... unless you meant symmetric sum, but if you did I'm even more confused as to how you went from a cyclic to a symmetric, so I think your argument is flawed.[/quote]

This is not really a cyclic inequality, you can check that it is symmetric too.

I believe he just cleared the denominator and expanded everything.

Ah, I see what you mean, however I feel like there should be a nicer way to get from the start point to the end point of $ \sum x^2y \geq 6xyz$ than expanding.

And also, given that that is a symmetric sum, I think that inequality is true more directly by AM-GM than by Cauchy but to each his own I guess.

[quote="Hong Quy"]it's easy.
 it's equivalent  with  ineq :
 $ 2 \sum x^2y \plus{} 18 xyz \leq 3\sum x^2y \plus{} 12xyz$
  or $ \sum x^2y \geq 6xyz$  ( right  by cauchy's  ineq)[/quote]
$ 2 \sum (x\plus{}y)^2z \leq 3 (x\plus{}y)(y\plus{}z)(z\plus{}x)$
  or $ \sum x^2y \geq  6xyz$

 i have  a little  mistake.    but  it's terrible


inequalities
number theory unsolved
number theory

hello
i don't know if this problem was posted by someone else but i would like to know how to solve it:
find  all integers n  for which n+27 is a square
i tried n+27=(n+3)(n-3n+9)=x but..without results

It was posted before. Look up the topic "a perfect square" by Iris Aliaj in the Pre-Olympiad section.

Misha

Here I go:

(n+3)(n 2 -3n+9) is an square. suppose t divides n+3, then n=-3(modt) and n 2 -3n+9=27(t) so if d divides both factors, then d=1,3,9,27. Iff d=1 or 3 then both factors are perfect squares.
Case 1: n is positive. Iff n 2 -3n+9 is an square then m=4n 2 -12n+36 is an square but m=(2n-3) 2 +27. Now if (2n-2) 2 >(2n-3) 2 +27 then m cannot be a square. but the inequality is equivalent to:
4n-5>27 so n cannot be >5. this leaves n=0,1,2,3,4,5 and checking this values one finds out that n=3 is the only wich works but n+3=3+3 is not a square.
Case 2: n is negative. let k=|n| then: m=4k 2 +12k+36, but then m= (2k+3) 2 +27 and proceding as before we find that 4k+7<=27 so k=1,2,3,4,5. checking this values we find out also n+3>=0(it is an square) so we only try n=-1,-2,-3. n=-3 gives the solution n=-3, x=0, and the othr two values give no solutions.
Now if d=3, then n+3=3k and then n=3t but then n 2 -3n+9 is a multiple of 9, contradiction. So we just have to analize d=27. If d=27, n+3=27k 2  and also n 2 -3n+9=27t 2 . Now, n=3m, replacing this we find out the following: m+1=9k 2  and also
3m 2 -3m+3=9t 2  but then y=12m 2 -12m+12 is an square but y=(6m-1)2 +11. So we must have:
r 2 -(6m-1) 2 =11 so we get two options: r=6m and 12m-1=11 then m=1. Or r+6m-1=+-1 and r-6m=+-11 wich have no integer solutions. so if m=1, then n=3 and we find the solution we already had.

Sorry I know it was long and boring, but I think it is complete, if i forgot something just let me know.

thanx :)

you wrote that [b]
(n+3)(n 2 -3n+9) is an square. suppose t divides n+3, then n=-3(modt) and n 2 -3n+9=27(t) so if d divides both factors, then d=1,3,9,27. Iff d=1 or 3 then both factors are perfect squares[/b]
but is d equal to t ? or is it an other number?
i just wonder why n cant be greater than 5? the inequality gives n>8 so n can't be greater than 8?
sorry for my 'slow understanding' :blush:

First: d is the mcd of n+3 and n 2  -3n+9, sorry not for specifying it.
clearly d|27 so from it i got the expexted values.

Now i made a mistake here, I said that if 4n-5>27 then the number wasnt a square, so n cannot be >8, not 5! (sorry :blush: ) so we just had to analize n=1,2,3,4,5,6,7. because n+3 is a square also in this case, we just had to evaluate in 1 and 6. I dont know if n=6 drops a solution. Everithing clear now?

it's clear , thanks,
i have an other remark, for case2: n is negative
i think we can make it much easier because:

n+27 is a square so n>=-27 so n>=-3 for n=-3,-2,-1,0 there is no square.
am I right?

edit:  :blush:  why if  (2n-2) >(2n-3)  +27 then m cannot be a square?
is it because d<= 27?
if i get it right, only n=-3 is the solution?


search
symmetry
combinatorics unsolved
combinatorics

Within a group of $ 2009$ people, every two people has exactly one common friend. Find the least value of the difference between the person with maximum number of friends and the person with minimum number of friends.

Actually this one - as I also learned after the contest - is a special case of a known theorem which is called Friendship Theorem. Take a look at it from [url=http://www.mathlinks.ro/viewtopic.php?search_id=1643842659&t=48500]here[/url]

I got that the only possible graph is that one vertex has $ 2008$ edges,and other $ 2008$ vertices form $ 1004$ egdes.
So the answer is $ 2008\minus{}2\equal{}2006$
Is it right?

My solution:
Take the person with the maximal number of friends. Call him $ A$. Assume all of $ A$'s friends are $ A_{1},A_{2},\cdots,A_{m}$, respectively.
For $ 1\le i\le m$, since $ A$ and $ A_{i}$ have exactly one friend, which must be one of $ A_{1},A_{2},\cdots,A_{m}$. Therefore $ A_{i}$ has exactly one friend among $ A_{1},A_{2},\cdots,A_{m}$. So $ m$ is even. Assume $ m\equal{}2k$. W.l.o.g. we can assume that for $ 1\le i\le k$, $ A_{2i\minus{}1}$ and $ A_{2i}$ are friends.
Since $ A$ is chosen arbitrarily, we have that everyone has an even number of friends.

Now it suffices to prove that $ m\equal{}2008$, which is equivalent to that there are no more people other than $ A,A_{1},A_{2},\cdots,A_{m}$.
Assume the opposite,
for any person $ B$ other than $ A,A_{1},A_{2},\cdots,A_{m}$. Since $ A$ and $ B$ have exactly one friend, $ B$ has exactly one friend among $ A_{1},A_{2},\cdots,A_{m}$. So we can divide other persons into $ m\equal{}2k$ sets according to which of $ A_{1},A_{2},\cdots,A_{m}$ is his friend.
More specific, for $ 1\le i\le m$, denote set $ F_{i}\equal{}\{ x|x\notin \{A,A_{1},A_{2},\cdots,A_{m}\} \text{ and x is a friend of }A_{i} \}$ and $ S_{i}\equal{}|F_{i}|$.

It is trivial that $ k\ge 2$, so $ A_{3}$ exists. Due to the opposite assume $ F_{i}$ are not all empty, w.l.o.g. we assume $ F_{1}$ is not empty and take an element $ B_{1}$ from it. For $ 3\le i\le m$, $ B_{1}$ and $ A_{i}$ have exactly one friend, which belongs to $ F_{i}$. Therefore $ B_{1}$ has exactly one friend in $ F_{i}$ for $ 3\le i\le m$. So there exists a bijection between $ F_{1}$ and $ F_{i}$ for $ 3\le i\le m$, so $ S_{1}\equal{}S_{3}\equal{}\cdots \equal{}S_{m}$.
Due to the symmetry of all $ S_{i}$ we have all $ S_{i}$ are the same number. Assume it is $ x$. Because all of $ A_{1}$'s friends are $ A$, $ A_{2}$ and all $ x$ elements in $ F_{1}$, so  $ A_{1}$ has exactly $ x\plus{}2$ friends and $ x$ is an even number.
Therefore there are $ 1\plus{}m\plus{}mx\equal{}m(x\plus{}1)\plus{}1$ people in total. Also the condition told us that there are $ 2009$ people in total. So $ m(x\plus{}1)\plus{}1\equal{}2009$, $ m(x\plus{}1)\equal{}2008$, $ k(x\plus{}1)\equal{}1004$.

Now it is easy to get a contradiction. $ 1004\equal{}2^2*251$, and $ 251$ is a prime number. But $ x\plus{}1$ is an odd divisor of $ 1004$ that means it could only be either $ 1$ or $ 251$.
If $ x\plus{}1\equal{}251$, then $ x\equal{}250$ and $ k\equal{}4$, which makes $ A$ has less friends than $ A_{1}$, a contradiction.
If $ x\plus{}1\equal{}1$, then $ x\equal{}0$ also contradicts the hypothesis.
Therefore the hypothesis is wrong and the only possible graph is which I describe in the former post.
Q.E.D


group theory
superior algebra
superior algebra unsolved

Is the automorphism group of the permutation group S_n isomorphic to S_n if n>3??
In other words, is the permutation group S_n is a complete group if n>3?

[b]H. K. Iyer[/b], "On solving the equation Aut(X)=G", Rocky Mountain J. Math. 9 (1979), no. 4, 653--670.

This paper proves that for any finite group G, there are finitely many finite groups X with Aut(X)=G, and it explicitly solves the equation for some specific values of G. In particular, the complete solution when G is a symmetric group is worked in some detail.

Thanks!  That is a great paper that somehow I had never read and it pointed out a few other nice papers.

I have Miller's complete works in my office which I read through from time to time.  Each paper is short and covers something basically elementary, but usually something I had never seen in a textbook.  Unfortunately the language has become strange to modern eyes (and even stranger to non-native speakers; there are hardly any formulas, and the English is quaint). Miller has written so many papers, that it was not too surprising to see that he had looked at the smallest reasonable case in:

Miller, G. A. "On the groups which have the same group of isomorphisms."
Trans. Amer. Math. Soc. 1 (1900), no. 4, 395-401.
http://www.ams.org/mathscinet-getitem?mr=1500546
http://www.zentralblatt-math.org/zmath/en/search/?q=an:31.0135.01
http://dx.doi.org/10.2307/1986363

It is a nice paper, but again it has language like "It is well known that every simple isomorphism of a group G with itself may be obtained by transforming G by means of operators that transform it into itself." and this is followed by an insane description of an automorphism defined piecewise by multiplication.  If one just reproves the results one's self, then the paper is fairly easy to follow and has all sorts of interesting gems.  It also appears interested in solving Inn(X)=G, which has a more complicated answer but I think is implicitly treated in Iyer too.  Probably Miller's paper is much easier to read with Iyer alongside.

For easy reference, the great paper mentioned by coquito with links:
H. K. Iyer, "On solving the equation Aut(X)=G",
Rocky Mountain J. Math. 9 (1979), no. 4, 653-670.
http://www.ams.org/mathscinet-getitem?mr=560244
http://dx.doi.org/10.1216/RMJ-1979-9-4-653

For a counterexample at $ S_6$, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=118511#118511]this post[/url]. It's an explicit construction of an outer automorphism.

For the positive direction, here's an argument extending what grobber posted in the linked thread:
Any automorphism $ \phi$ of $ S_n$ is determined by the images of transpositions, and it's an inner automorphism (generated by conjugation with some element) if and only if every transposition is sent to a transposition.
Next, we note that the images of elements in the same conjugacy class must be in the same conjugacy class. Since this is also true for $ \phi^{\minus{}1}$, the transpositions are mapped bijectively to some conjugacy class. That class must generate all of $ S_n$, so it must consist of odd permutations. It must consist of involutions, so it's the products of some (odd) number $ k$ of disjoint transpositions.
How many elements are in that class? It's $ \binom{n}{2}$ for $ k\equal{}1$, $ \frac1{3!}\binom{n}{2}\cdot\binom{n\minus{}2}{2}\cdot\binom{n\minus{}4}{2}$ for $ k\equal{}3$, and so on. These are of course zero for $ n<2k$.
For $ n<6$, all of these classes except the transpositions $ k\equal{}1$ are empty, and all automorphisms are inner automorphisms. At $ n\equal{}6$, we get $ 20$ and $ \frac16(20\cdot 6\cdot 1)$, which are equal; this enables the construction above. For $ n\ge 2k\ge 6,$
$ \binom{n\minus{}2}{2}\cdot\binom{n\minus{}4}{2}\cdots \binom{n\minus{}2k}{2}\ge \binom{2k\minus{}2}{2}\cdot\binom{2k\minus{}4}{2}\cdots \binom{2}{2}$
$ \equal{}\frac{(2k\minus{}2)!}{2^{k\minus{}1}}\equal{}(k\minus{}1)!\cdot (2k\minus{}3)(2k\minus{}5)\cdots (1)\ge (k\minus{}1)!\cdot (2k\minus{}3)\ge k!$
with equality only if $ n\equal{}2k$ and $ k\equal{}3$. There are more products of $ k$ disjoint transpositions than there are transpositions, and we can't have the bijection for $ n>6$. Therefore, once again, all automorphisms are inner automorphisms in this case.

Finally, $ S_n$ has trivial center for $ n\ge 3$, making the group of inner automorphisms isomorphic to $ S_n$ itself.


Let$ a,b$ and $ c$ are positive real numbers. Prove that
$ (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2 \ge (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

[quote="Litlle 1000t"]Let$ a,b$ and $ c$ are positive real numbers. Prove that
$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})^2 \ge (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$[/quote]

\[ \left( {\sum {\frac {a}{b}} } \right)^2 \ge \left( {\sum a } \right)\left( {\sum {\frac {1}{a}} } \right)
\]

\[ \Leftrightarrow \left( {\sum {\frac {a}{b}} } \right)^2 \ge 3 \plus{} \sum {\frac {a}{b}} \plus{} \sum {\frac {b}{a}}
\]

\[ \Leftrightarrow \sum {\frac {{a^2 }}{{b^2 }}} \plus{} \sum {\frac {b}{a}} \ge 3 \plus{} \sum {\frac {a}{b}}
\]
which is true since
\[ \sum {\frac {b}{a}} \ge 3
\]

\[ \sum {\frac {{a^2 }}{{b^2 }}} \ge \frac {{\left( {\sum {\frac {a}{b}} } \right)^2 }}{3} \ge \sum {\frac {a}{b}}
\]

[quote="Litlle 1000t"]Let$ a,b$ and $ c$ are positive real numbers. Prove that
$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})^2 \ge (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49893

[quote="Litlle 1000t"]Let $ a,b$ and $ c$ be positive real numbers. Prove that
$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})^2 \ge (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$[/quote]

The following identity is true for all $ a,b,c$:

\[ {\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)^2 \minus{} (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \equal{}
\frac{a\plus{}b}{a\,b^{\,2}}\,(a\minus{}b)^2\plus{}\frac{b\plus{}c}{b\,c^{\,2}}\,(b\minus{}c)^2\plus{}\frac{c\plus{}a}{c\,a^{\,2}}\,(c\minus{}a)^2}\]

This clearly proves that the left part is non-negative.

Thank you for the nice inequality!

Sergey Markelov


Could somebody please show me a solution for question 4d) of the hypatia contest? The question was:
define $I(n) =\frac{\sigma (n)}{n}$, where $\sigma (n)$ (I think thats the symbol) denotes the sum of all positive divisors of $n$. Find the smallest odd positive integer $n$ such that $I(n) > 2$. I got, by pure trial and error, $945$, but I can't prove it.  :(

I think the answer is $n = 12$: $\sigma(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28 > 2 \cdot 12$.  Trial and error seems reasonable here.

If n must be odd, obviously that changes the answer.   :P

In that case, I agree the answer is 945, but I don't think there's anyway without some trial and error.

I agree. But it will take a short period of time to find the answer if you search in "a smart way"(not trying every odd integer from 1 to 945 ;) )

Sorry I forgot to specify that it's odd :blush: . I tried $3^2*5*7$ which didn't work, and I wanted more factors so I took another power of $3$. But my way was far from systematic, it was just trying random numbers.
Oh yes and rem, what is your smart way?  :)

I did it too and I got 945, basicly you just use the result from 9c to simplify a bit then you realizes you need small odd numbers. So by a little bit of trial and error you would get 945.

Crap. Due to some miscalculation I got 315.

I think you can use

$I(a_1a_2\cdots a_n) = I(a_1)I(a_2)\cdots I(a_n)$ if $gcd(a_i,a_j) = 1$ for $i \neq j$ and it's just a lot of cases. This actually helps you find the answer too.

Was there a way to prove that 945 was the answer?

[quote="rem"]Probably. But it's faster to write the traial and error way up than the proof I think.
btw do you guys find FGH kind of pointless? Because some problems are just stupid, you get a piece of paper if you win, and it has little weight on ur credentials.[/quote]

You say everything's stupid, but not everyone is good at Mathematics like you are Alex.  For AoPsers like us, yeah...I would say FGH is really easy but I wouldn't start bashing contests or going on an ego rant or whatever.

Besides, my question is, if it's so easy, why the **** did you even take the contest?

The point of FGH is to expose all students to Mathematics.  I know I got my start on the Fryer contest.  I know in Russia, coaches only recruit the brightest students, thereby effectively denying some kids that have a genuine passion for Math.  In North America, we do things differently:  We try to encourage everybody who's interested to do Mathematics.

What if some kid told you English was so easy, and that Ontario should get rid of English classes altogether?  You'd be screwed then wouldn't you?  Not so easy anymore is it?

Same thing with Math contests, be more considerate about others as well.

Credentials...jeez.  Is all you care about an IMO gold medal?  I should very much stress that that is secondary to what you're experiencing now...consider yourself fortunate.

Sorry. I seem to be making everyone mad on this forum and I am not meaning to do that. i took fgh because i automatically registeered for it when registering for fermat. sorry again :blush:
I should definitely not say my opinion about anything in here.

For what it's worth, alex isn't the first person to express an opinion like that.

In my view, it's a bit of an odd contest because it's supposedly for better students, but it's easier to get all the questions right on the FGH contests than the PCF contests.

(Here 'all questions right' =/= 'perfect score')

Hmm, its true, a lot of people felt that the Grade 10 and 11 Invitational exams (that FGH replaced) were more slightly more challenging in terms of problems.

i agree... i never did well in cayley and fermat contests, but hypatia 2003 i thought was not as challenging as i anticipated when i wrote it. i finished it with 15min left over. but i think hypatia is more accurate and better than pcf because it's full solutions. i think the point is to introduce stronger math students to write full solutions for prep for comc. plus they don't want to mark so many papers thats why they restrict it to stronger math students. north america is trying to educate the public rather than only elites get special previlages. of course, i've heard that some ppl complain that fgh contests r too hard. so seems like the difficulty is just right!  :D
changing the difficulty of contests because of AoPSers complain they r too easy is sampling biased!!

[quote="Sunny"][quote="J-I-M-B-O"]i agree... i never did well in cayley and fermat contests, but hypatia 2003 i thought was not as challenging as i anticipated when i wrote it. i finished it with 15min left over. but i think hypatia is more accurate and better than pcf because it's full solutions.[/quote]
Oh, the Grade 10 and 11 Invitational exams I was refering too is similar to Galois and Hypatia, full solutions.  You were invited to write based on your cayley or fermat score.  They were stopped and replaced with the FGH, which is open to everyone.  Those exams just seemed more challenging than hypatia or galois.[/quote]

yup! i remember those, those were challenging! comparison to fgh

I agree FGH contests are easier than PCF. But for some weird reason, some students in my school who did a good job on the PCF failed the FGH contest. Even I was expecting to get perfect on the Hypatia contest, but I skewed up on the first question and did not have enough time to finish 4d.

FGH is stupid.  Like Yufei Zhao said,

[quote="billzhao"]
Let's not kid ourselves here :D. FGH is a "non-competitve contest." And you can figure out what that means yourself.
[/quote]

I think that sums it up nicely.

[quote="d.lam.86"]Do schools that participated in the FGH contests receive the marked solutions that we wrote? I would like to find out how they marked my contest so I can learn from my mistakes. Hmm, I wonder if there is any bias in their marking?[/quote]

Occam's Razor; you probably lost marks because your solutions were wrong or you were unable to articulate yourself clearly.  Implying that there may be bias is just stupid.

[quote="d.lam.86"]Do schools that participated in the FGH contests receive the marked solutions that we wrote? I would like to find out how they marked my contest so I can learn from my mistakes. Hmm, I wonder if there is any bias in their marking?[/quote]

You didn't answer my original question though.

[quote="d.lam.86"]Do schools that participated in the FGH contests receive the marked solutions that we wrote? I would like to find out how they marked my contest so I can learn from my mistakes. Hmm, I wonder if there is any bias in their marking?[/quote]

Umm...i dont think so because when i went to see my results i just got a number out of 40 there was nothing else. However, you can check the solutions online (if u remember what u wrote) with ur own solutions. If you got the answer right and showed appropiate steps chances are u probably got 10/10


But that's different given situation...  :(

In fact $LHS=1$,

so we have to prove $RHS\leq 1$

$\leftrightarrow x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}\geq 4$

It is true, because

By Cauchy, we have $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2}=(4-xyz)^{2}$

And $(4-xyz)^{2}+3x^{2}y^{2}z^{2}\geq 12 \leftrightarrow (xyz-1)^{2}\geq 0$

Why $LHS = 1$ ??  :huh:

If we simplify the $LHS$, it will be like this..

$\frac{12+4(x+y+z)+(xy+xz+yz)}{8+4(x+y+z)+2(xy+xz+yz)+xyz}$

andi if $LHS = 1$, then $xyz+xy+xz+yz = 4$, which is not the given condition...

And I suggest for the rest (if that's right $1 \geq RHS$), use only $QM-AM$  :)


Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that $ f(1)\equal{}1$ and

$ f(f(x)y\plus{}\frac{x}{y})\equal{}xyf(x^2\plus{}y^2)$

for all real numbers $ x,y$ with $ y\neq 0$

Posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222523#1222523]before[/url] (but not solved, apparently). Also note the slight difference in requirements ($ f: \mathbb{R} \to \mathbb{R}$ vs $ f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\}$).

I'll give it a shot if I have some time....

I have very little experience with functional equations, so I may not have gotten the full answer here.
[hide]
If $ x \equal{} 0$, then $ f(f(0)y \plus{} \frac {0}{y}) \equal{} f(f(0)y) \equal{} 0yf(0^2 \plus{} y^2) \equal{} 0$ for all $ y$.
If $ f(0)\neq0$, then there exists $ y$ such that $ y \equal{} \frac {1}{f(0)}$, which would let $ f(f(0)y) \equal{} f(1) \equal{} 1 \equal{} 0$, a contradiction.  Therefore, $ f(0) \equal{} 0$.

Note that $ f(f(x)y \plus{} \frac {x}{y}) \equal{} xyf(x^2 \plus{} y^2) \equal{} yxf(y^2 \plus{} x^2) \equal{} f(f(y)x \plus{} \frac {y}{x})$ for $ x\neq0$.
We examine the function when $ f(x)y \plus{} \frac {x}{y} \equal{} f(y)x \plus{} \frac {y}{x}$.  Letting $ y \equal{} 1$, we have

$ f(x)\cdot1 \plus{} \frac {x}{1} \equal{} f(1)x \plus{} \frac {1}{x}$
$ f(x) \plus{} x \equal{} x \plus{} \frac {1}{x}$
$ f(x) \equal{} \frac {1}{x}$ for $ x\neq0$, provided that $ f(x) \equal{} f(y) \Rightarrow x \equal{} y$ for $ x\neq0$.  Let $ g(x) \equal{} 1/x$ for $ x\neq0$, and consider $ x_1, x_2 \in \mathbb{R} \setminus \{0\}$ such that $ g(x_1) \equal{} g(x_2)$.  Then, since the multiplicative inverse of $ g(x_1) \equal{} g(x_2)$ is unique, $ g(x_1) \equal{} g(x_2)$ and $ x_1g(x_1) \equal{} 1 \equal{} x_2g(x_2) \equal{} 1 \Rightarrow$ $ x_1 \equal{} x_2$.

Combining our results, we have $ f(x) \equal{} \begin{cases}0 & x \equal{} 0 \\
\frac {1}{x} & x\neq0.\end{cases}$[/hide]

[quote="syckls"]
$ f(x) \equal{} \frac {1}{x}$ for $ x\neq0$, provided that $ f(x) \equal{} f(y) \Rightarrow x \equal{} y$.  This is indeed the case; since $ g(x) \equal{} \frac {1}{x}$ is an injection from $ \mathbb{R} \setminus \{0\}$ to $ \mathbb{R} \setminus \{0\}$, there exist no $ x_1, x_2 \in \mathbb{R} \setminus \{0\}$ such that $ x_1 \neq x_2$ and $ g(x_1) \equal{} g(x_2)$
[/quote]
Circular reasoning?

As you showed $ f$ injective ($ f(x) \equal{} f(y) \Rightarrow x \equal{} y$) would solve the problem, but I think it's harder to do that than to solve the problem.

[quote="MellowMelon"]
Circular reasoning?

As you showed $ f$ injective ($ f(x) \equal{} f(y) \Rightarrow x \equal{} y$) would solve the problem, but I think it's harder to do that than to solve the problem.[/quote]

Fixed.  I was just lazy and assumed that $ f$ injective was already known.

[quote="syckls"]
Fixed.  I was just lazy and assumed that $ f$ injective was already known.[/quote]

Sorry, but I don't see how you fixed it. Presumably $ g(x)\equal{}1/x$? (you don't define it in your fixed post). How does proving anything about $ g$ help in showing that $ f$ is injective?

[hide="Not sure if it works..."]
I'll solve the [url=http://www.mathlinks.ro/viewtopic.php?p=1222523#1222523]previous problem[/url] coz it looks more natural and doesn't involve cases.

[color=blue]For all $ x\in \mathbb R/\{0\}$ there exists some $ y\in \mathbb R/\{0\}$ satisfying
\[ f(x)y \plus{} \frac {x}{y} \equal{} x^2 \plus{} y^2
\]
because that is equivalent to this cubic equation of $ y$ with real coefficients:
\[ y^3 \minus{} f(x)y^2 \plus{} x^2y \minus{} x \equal{} 0
\]
Also that real root can't be $ 0$ because $ x\neq 0$

So plugging in that real root $ y_x$ for $ y$ in the main equation:
$ f\left(f(x)y_x \plus{} \frac {x}{y_x}\right)$ and $ f(x^2 \plus{} y_x^2)$ cancel out because
they are equal but aren't $ 0$ because $ f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\}$

So it remains: $ y_xx \equal{} 1$, or:
\[ y_x \equal{} \frac {1}{x}
\]
But that $ y_x$ also satisfies:
\[ y^3 \minus{} f(x)y^2 \plus{} x^2y \minus{} x \equal{} 0
\]
Putting $ \frac {1}{x}$ for $ y$ in that:
$ f(x) \equal{} \frac {1}{x}\ \forall x$[/color]

P.S. It has a good chance of being wrong coz It doesn't use $ f(1) \equal{} 1$[/hide]

[quote="randomgraph"][quote="syckls"]
Fixed.  I was just lazy and assumed that $ f$ injective was already known.[/quote]

Sorry, but I don't see how you fixed it. Presumably $ g(x) \equal{} 1/x$? (you don't define it in your fixed post). How does proving anything about $ g$ help in showing that $ f$ is injective?[/quote]

Sorry, it should work now.  I forgot to define g(x) after my first edit.

[hide="Current condition of the 2 times edited post"][quote="syckls"]
If $ x \equal{} 0$, then $ f(f(0)y \plus{} \frac {0}{y}) \equal{} f(f(0)y) \equal{} 0yf(0^2 \plus{} y^2) \equal{} 0$ for all $ y$.
If $ f(0)\neq0$, then there exists $ y$ such that $ y \equal{} \frac {1}{f(0)}$, which would let $ f(f(0)y) \equal{} f(1) \equal{} 1 \equal{} 0$, a contradiction.  Therefore, $ f(0) \equal{} 0$.

Note that $ f(f(x)y \plus{} \frac {x}{y}) \equal{} xyf(x^2 \plus{} y^2) \equal{} yxf(y^2 \plus{} x^2) \equal{} f(f(y)x \plus{} \frac {y}{x})$ for $ x\neq0$.
We examine the function when $ f(x)y \plus{} \frac {x}{y} \equal{} f(y)x \plus{} \frac {y}{x}$.  Letting $ y \equal{} 1$, we have

$ f(x)\cdot1 \plus{} \frac {x}{1} \equal{} f(1)x \plus{} \frac {1}{x}$
$ f(x) \plus{} x \equal{} x \plus{} \frac {1}{x}$
$ f(x) \equal{} \frac {1}{x}$ for $ x\neq0$, provided that $ f(x) \equal{} f(y) \Rightarrow x \equal{} y$ for $ x\neq0$.  Let $ g(x) \equal{} 1/x$ for $ x\neq0$, and consider $ x_1, x_2 \in \mathbb{R} \setminus \{0\}$ such that $ g(x_1) \equal{} g(x_2)$.  Then, since the multiplicative inverse of $ g(x_1) \equal{} g(x_2)$ is unique, $ g(x_1) \equal{} g(x_2)$ and $ x_1g(x_1) \equal{} 1 \equal{} x_2g(x_2) \equal{} 1 \Rightarrow$ $ x_1 \equal{} x_2$.

Combining our results, we have $ f(x) \equal{} \begin{cases}0 & x \equal{} 0 \\
\frac {1}{x} & x\neq0.\end{cases}$[/quote][/hide]
randomgraph already assumed what you edited. Look closely at the last sentence of his last post. Your solution is still not clear (at least to me)

Our issue is not with your proof that $ g(x) \equal{} 1/x$ is injective. We know that. But that doesn't do anything to help solve the problem. You proved that if $ f$ is injective on a set of points, then $ f(x) \equal{} 1/x$. There is absolutely no reason your proof discounts functions like $ f(x) \equal{} x^2$, where $ f$ is injective nowhere. To prove $ f$ is injective, you need to go back to the original equation, and proving injectivity from that is, as I said above, probably on the same level as solving the problem itself.

Akashnil, your proof looks fine to me. It works for $ x\equal{}1$ as well and shows that $ f(1)\equal{}1$ without needing to assume this at the outset. Here is how this works for this particular instance:

Like you, I'm assuming that $ f$ is defined on non-zero reals and takes non-zero values. Let $ a\equal{}f(1)$. The cubic equation

$ y^3\minus{}ay^2\plus{}y\minus{}1\equal{}0$

has at least one real root. It may have more but we pick one and call it $ y_1$, noting that $ y_1 \neq 0$. In the main equation we now set $ x\equal{}1$ and $ y\equal{}y_1$ to get

$ f(ay_1\plus{}\frac{1}{y_1}) \equal{} y_1f(1\plus{}y_1^2)$

but due to the choice of $ y_1$, $ ay_1\plus{}\frac{1}{y_1} \equal{} 1\plus{}y_1^2$ so this equation reduced to $ y_1\equal{}1$. It follows that the cubic above actually does have a unique real root at $ y\equal{}1$, and furthermore by substituting $ y\equal{}1$ into that cubic we get

$ 1\minus{}a\plus{}1\minus{}1\equal{}0$

so $ a\equal{}1$, that is, $ f(1)\equal{}1$. (Note that the cubic is now seen to be $ y^3\minus{}y^2\plus{}y\minus{}1 \equal{} (y\minus{}1)(y^2\plus{}1)$ which indeed has a unique real root at $ y\equal{}1$).


Find the product of the solutions of: $ |y|\equal{}2(|y|\minus{}1)$.

From $ |y|\equal{}2(|y|\minus{}1)$, we can divide both sides by $ 2$.  This will give us $ \frac{|y|}{2}\equal{}|y|\minus{}1$.  If we then add $ 1$ to both sides and subtract $ \frac{|y|}{2}$ from both sides, we get $ 1\equal{}\frac{|y|}{2}$  Then, we multiply both sides by $ 2$, and get $ |y|\equal{}2$.  Therefore, our answers for $ y$ are $ 2$ and $ \minus{}2$, so our answer is $ \boxed {\minus{}4}$


I post a question this morning asking "xxx xxx xxxxx xx xxxxx xxxxxxxxxxx"? but was deleted by somebody. May I ask why you did this?  [color=darkblue] [size=75]<Edited by moderator rcv>[/size][/color]

Mathcounts web site "message board" has a screening system, i.e. they read your message first then decide to post it or not. They accepted a lot of of similar questions post there like mine. What is wrong to post here?

Also a question post by Silverfalcon and answered by me  was deleted several days ago. The question was about the angle bisector formula. That question was very similar to a warm up question in the 2005 mathcounts handbook. May I know who deleted that one and why?

I believe your inquiry and remarks fall outside the guidelines of this forum.  PM me if you want further discussion.

Your response to an angle bisector problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=168787#p168787]here[/url].


A sculpture for a math classroom is a figure composed of a l'' cube stacked upon a 3'' cube which in turn is stacked upon a 5'' cube. The entire sculpture is to be painted. Find the number of square inches of the surface that will be painted.

First, you add the surface areas of the three cubes:
$ 1^2*6 \plus{} 3^2*6 \plus{} 5^2*6 \equal{} 210$

Then, you subtract the overlapping area * 2, because there are two sides not painted for each overlap.

Subtracting $ 1^2 * 2 \plus{} 3^2 * 2 \equal{} 20$, the answer becomes $ 210 \minus{} 20 \equal{} 190$ sq in.


Need to have an explanation of how to figure this out.
The question is : A children's store sells a toy for $94.25. The owner uses a 45% mark-up based on cost. What was the wholesale cost of the toy? :(  :blush:

guess: 94.25/1.45. do you have the answer?

i think you take the wholesale cost and apply the markup to get the 94.25
i'm too lazy...to work it out


Use limits to compute f' (x)

38. f(x)= -x+11

f-(x+h)+11- f(-x+h)/h

-x-h+11+x-h/h

-2h+11/h=9

and since if f(x)=mx+b and f'(x)=m I have the limit at -1.

Is that right or am I totally off track. Your response will be greatly appreciated because I definitely need all the help I can get in this subject.

I couldn't make sense of any single line in your solution :(, so let's start with this. Give the general definition of $ f'(x)$ as a limit.

:rotfl:  I wrote this problem down completely wrong. My bad. It's definitely time for me to put the books down and go watch tv.

f(x)= -x+11

f-(x+h)+11 - f(-x+11)/h

f(-x-h+11+x-11)/h

x cancels and the 11 cancels leaving -h/h which gives me

-1.

Is that right?

That was, if anything, more incomprehensible than the first effort.  Why not do what fedja suggested, first, before moving on to a second step?


Let the sequence $(x_{n})_{n\in\mathbb{N}}$ of real strictly pozitive numbers and $a,b>0$. If $\frac{1}{x_{n}}\leq a+bx_{n+1}\leq \frac{1}{x_{n+3}}$, for any natural $n$ then we have:

(a)  $x_{n}=x_{n+3}$ for any $n\in\mathbb{N}^{*}$

(b) The sequence $(x_{n})_{n\in\mathbb{N}^*}$ is a constant sequence.

clearly we have $x_{n+3}\leq x_n$  from $\frac{1}{x_{n+3}}\leq a+bx_{n+4}\leq \frac{1}{x_{n+6}}$ we get $a+bx_{n+1}\leq \frac{1}{x_{n+3}}\leq a+bx_{n+4}\leq a+bx_{n+1}$ so $x_n=x_{n+3}$ .
For point b after writting the 3 equalities obtained from the fact that$x_n=x_{n+3}$ we will have a system and by substituing we get a square equation with only one positive solution thus we must have $x_n$ constant.

Thanks! I know you'd solve it ...  :roll:

this one' s easy... the point a) is quite obvious whereas b) can be solved using a) and the given recurrence...you should obtain a second degree ecuation with a single positive solution


A sheet of 8-inch by 10-inch paper is placed on top of a sheet of $ 8 \frac{1}{2}$-inch by 11-inch paper, as shown. What is the area of the region of overlap?

[asy]draw((0,0)--(10,0)--(10,8)--(0,8)--(0,0)--cycle,linewidth(2));
draw((0,8)--(8.5,8)--(8.5,11.5)--(0,11.5)--(0,8)--cycle,linewidth(2));

draw((8.5,0)--(8.5,8),dashed);[/asy]

one rectangle has a side 8, and the other has a side 8.5
we can see that in the region of overlap, these are the dimensions

therefore, we multiply
$ 8\times8.5\equal{}68$

correct me if i made a blunder (we learned the difference between blunders and errors in science :D )


trigonometry
geometry
geometric transformation
reflection
geometry proposed

let $ ABC$ is a triangle such that $ \sqrt{(b\plus{}c)(b\minus{}c)} \equal{} a \sqrt {2}$
prove that :
$ \sin^2((A\minus{}B)/2)\plus{} \sin A \sin B\plus{}\sin^2(C/2)\equal{}1$

[quote="greatestmaths"][color=darkred]Let $ ABC$ be a triangle such that $ \sqrt {b^2 \minus{} c^2} \equal{} a \sqrt {2}$ .

Prove that $ \sin^2\frac {A \minus{} B}{2} \plus{} \sin A\cdot \sin B \plus{} \sin^2\frac C2 \equal{} 1$ .[/color] [/quote]

[color=darkblue][b]Remark I.[/b] The upper trigonometric relation is a identity, i.e. it is truly in any triangle $ ABC\ !$

Indeed, $ \left\|\ \sin^2\frac {A \minus{} B}{2} \plus{} \sin A\cdot \sin B \plus{} \sin^2\frac C2 \equal{} 1\ \right\|\ \bigodot\ 2$ $ \Longleftrightarrow$

$ 1 \minus{} \cos (A \minus{} B) \plus{} \cos (A \minus{} B) \minus{} \cos (A \plus{} B) \plus{} 1 \minus{} \cos C \equal{} 2$ $ \Longleftrightarrow$

$ \cos (A \plus{} B) \plus{} \cos C \equal{} 0$ , what is truly because $ A \plus{} B \plus{} C \equal{} \pi$ .

[b]Remark II.[/b] Denote the reflection $ D$ of the point $ A$ w.r.t. the point $ B$ .

Then $ \sqrt {b^2 \minus{} c^2} \equal{} a \sqrt {2}$ $ \Longleftrightarrow$ $ CD \equal{} c\ !$[/color]


calculus
integration
limit
real analysis
calculus computations

1) Find $\lim_{n\rightarrow\infty}$$\{\frac{1}{n+1}+\frac{1}{n+2}+.......+\frac{1}{pn}\}$ for $p\geq2$
2) Evaluate $\int(1-x^2)^ndx$ for $x=-a$ to $x=a$
3) Without L'Hospital technique, can we prove that $\lim_{n\rightarrow0}\frac{n}{1-e^{-n}}=1$

Hints and questions:

1. Factor $\frac1n$ out of each term, and see that it is a Riemann sum for a particular definite integral.

2. Are you sure that's not the integral from $-1$ to $1?$ As the integral from $-1$ to $1$ it has a nice general answer in $n.$ As the integral from $-a$ to $a$ for some $a\ne1$ it's just ugly.

3. Take the reciprocal, and use the definition of the derivative.

Or just use power series.


Given distinct straight lines $ OA$ and $ OB$.  From a point in $ OA$ a perpendicular is drawn to $ OB$; from the foot of this perpendicular a line is drawn perpendicular to $ OA$.  From the foot of this second perpendicular a line is drawn perpendicular to $ OB$; and so on indefinitely.  The lengths of the first and second perpendiculars are $ a$ and $ b$, respectively.  Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds.  This limit is:

$ \textbf{(A)}\ \frac {b}{a \minus{} b} \qquad \textbf{(B)}\ \frac {a}{a \minus{} b} \qquad \textbf{(C)}\ \frac {ab}{a \minus{} b} \qquad \textbf{(D)}\ \frac {b^2}{a \minus{} b} \qquad \textbf{(E)}\ \frac {a^2}{a \minus{} b}$

[hide]The length of a perpendicular is always the length of the previous perpendicular times $ \cos \theta$ where $ \theta$ is the angle $ O$. It's easy to see that $ \cos \theta \equal{} \frac{b}{a}$. So, we have a geometric sequence with first term $ a$ and ratio $ \frac{b}{a}$. Hence, the limit of the sum is $ \frac{a}{1 \minus{} \frac{b}{a}} \equal{} \frac{a^2}{a\minus{}b}$[/hide]


Let $a\geq0, b\geq0, c\geq0.$ Prove the follow inequality
$3(a+b+c)\geq 7\sqrt[3]{abc} +2 \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$

[quote="Lovasz"]It looks SOS is a good solution.[/quote]
Try. I am not sure, that SOS here it is good. ;)

I tried it. But it looks very messy.
\[ S_a=\frac32 (a+b+c)+\frac{4}{\sqrt[3]{3}}\frac{(a^2+b^2+c^2)^3...}{a^2b^2c^2+abc\sqrt{...}}.  \]
I'm in a net cafe, so, I can't remember it. But the denominator is very bad.
Can you find something between $S_a,S_b,S_c$, a.g.?


geometry
power of a point
radical axis
projective geometry
geometry unsolved

Let ${A,B,C,D,E,F}$ be six points of the circle $\omega$ with center ${O}$, such that ${AD,BE,CF}$ are concurrent at a point ${X}$ (with ${X\neq{O}}$). Let ${P= BC\cap{EF}}, {Q= AC\cap{DF}}$, and ${R= AB\cap{DE}}$. Show that ${P,Q}$ and ${R}$ lie on the radical axis of $\omega$ and the circle with diameter ${OX}$.

The radical axis of the two circles is the polar line of $X$ with respect to $w$

For the point $R=AB \cap DE$:

it is known that $R$ is on the polar line of $X=AD \cap BE$
(see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25036]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316994#p316994]here[/url])

Similarly for the points $P,Q$


trigonometry
real analysis
real analysis unsolved

Let $X: =\exp{\left(\pi\sqrt{163}\right)}$ and $Y: =1.000045\times (e^{\pi}-\pi) .$ If possible,  by means of  a computer find \[ A: =\sin{\left(\frac{\pi X}{8}\right)} \; \; \; \mbox{and}\; \; \; B: =\sin{(\pi Y)} \; .  \]
[b]It's true that $\; A=B=0 \;$ ? [/b]

$X$ is an old chestnut.  It is 'well-known' that $X$ is very close to an integer, and indeed the error is of order $1/X$.

[quote="rgep"]... $X$ is very close to an integer, and ...the error is of order $1/X$.[/quote]
Indeed , this is the "trick" (the key of solution):  $ \exp\left(\pi\sqrt{163}\right)\approx 262537412640768744- 0.75\times {10}^{-12}. $ Thank for interest.


calculus
derivative
logarithms
calculus computations

i need to take the derivative of y=2^x
now if it were the other way I could do it:
y=x^2
y'=2x

Do you do it the same way with the other one too?
y=2^x
y'=x2^X+1???

Thanks!

[quote="Nené"]$ y' = 2^{x} \cdot log_e 2. $[/quote]
If still not clear , for  above, write $2^x$ as  $e^{x (log_e 2)} $  and you get the result quoted above..

[hide=" details"] $y' =  (log_e 2) \cdot e^{x (log_e 2)} =  (log_e 2) (2^x) $[/hide]

For the derivative of $y=x2^{x+1}$, use the product rule and the chain rule with a u-substitution of $x+1$.
The answer is $\frac{dy}{dx} = 2^x(2x\ln{2}+2)$.

After you've done these, see if you can take the derivative of $y=x^x$.
[hide]The answer is $x^x(\ln{x}+1)$.[/hide]


The diagnols of a qaudrilateral separate the latter into four triangles having equal perimeters. Prove that the quadrilateral is a rhombus.

Hello.
Let ABCD be the quadrilateral and denote by K the intersection of its diagonals.

Assume that AK>KC and BK>KD. Take C' and D' such that OC=OC' and OD=OD' on AK and BK respectively. Denote by {XYZ} the perimeter of triangle XYZ. Then {AKB} > {C'KD'} = {DKC} since triangle(C'KP') = triangle(DKC) (by SAS :D), contradiction!

If AK=KC and BK=KD then AD=BC and afterwards AD=DC=AB.

Giwrgos

kalantzis It would  be nice if you add a figure or atleast describe where the points are. Simply you have just daid O, X,Y etc. So please do the needful to make your solution comprhendable. Please describe the points or add the diagram.

Ooops  :blush:, It's the third time i answer to a problem, i'm sory for the misunderstanding. Let's take it again...

Let ABCD be the quadrilateral and denote by K the intersection of its diagonals.

Assume that AK>KC and BK>KD. Take C' and D' such that KC=KC' and KD=KD' on AK and BK respectively. Denote by {XYZ} the perimeter of triangle XYZ. Then {AKB} > {C'KD'} = {DKC} since triangle(C'KP') = triangle(DKC) (by SAS :D), contradiction!

If AK=KC and BK=KD then AD=BC and afterwards AD=DC=AB.

Giwrgos


Κάθε χρόνο διοργανώνεται περί τα τέλη Ιουλίου/αρχές Αυγούστου ένας μαθηματικός διαγωνισμός για φοιτητές που δεν έχουν ξεπεράσει τα 4 χρόνια σπουδών.Η θεματολογία του είναι από τις περιοχές της Ανάλυσης (κυρίως),Μιγαδικής Ανάλυσης,Γραμμικής Άλγεβρας,Άλγεβρας και Συνδυαστικής.Ο διαγωνισμός διοργανώνεται σε χώρες της Ευρώπης (φέτος στη Βουλγαρία 3-9 Αυγούστου) από το University College London ,το Bremen University και το American University in Bulgaria. Φέτος,για πρώτη φορά, μια ομάδα φοιτητών ξεκινάμε κάποια βήματα ώστε και το ΕΜΠ να στείλει μια ομάδα στο διαγωνισμό. Γι αυτό επικοινωνήσαμε με κάποιους καθηγητές που γνωρίζουμε ότι έχουν ενδιαφέρον και εμπειρία σε τέτοιου είδους διοργανώσεις.Χωρίς ακόμα να έχουμε αποφασίσει κάτι,σκεφτόμαστε τη διοργάνωση κάποιου προκριματικού διαγωνισμού αλλά θα θέλαμε να γνωρίζουμε την απήχηση αυτού του διαβήματος σε ευρύτερο κοινό.Όποιος ενδιαφέρεται,ας ρίξει μια ματιά εδώ:[url]http://www.imc-math.org/[/url]  και ας ποστάρει ή ας μου στείλει pm για να δούμε περίπου τι κάνουμε (σαν αριθμός).Αυτό δεν συνεπάγεται δέσμευση για συμμετοχή ιδίως τώρα που τα πράγματα με τα εξάμηνα και τις εξεταστικές είναι ρευστά.Διαδώστε το και σε γνωστούς σας!Περιμένω απαντήσεις!

Υπάρχουν λυμένα θέματα από το 1994 ως το 2006. Αυτό για να ξέρουμε που βαδίζουμε και να τονωθεί και το ενδιαφέρον
R.Boris

kalo akougetai an k prepei na paiksei arketh prwtoboulia an einai na ginei kati tetoio. proswpika pantws endiaferomai..

Λοιπόν,μετά από πολλές συνεννοήσεις,αναιρέσεις και ματαιώσεις συναντήσεων κατάφερα σήμερα (αν και μόνος λόγω διαφόρων κωλυμάτων και "συμβάντων" στα άλλα μέλη) να μιλήσω με τον κο Φελλούρη σχετικά με το θέμα της αποστολής ομάδας του ΕΜΠ στον IMC.Βέβαια,τα δεδομένα είχαν αλλάξει μετά το Πάσχα και τις συνελεύσεις των σχολών καθώς πολλοί από εμάς δεν είμασταν πλέον διαθέσιμοι είτε λόγω υποχρεώσεων που μετατέθηκαν είτε λόγω του ό,τι δεν προλαβαίνουμε πλέον να ετοιμαστούμε για διαγωνισμό.Φυσικά,φαινόταν ότι -λόγω χρόνου- ούτε συζήτηση για διαγωνισμό επιλογής δεν μπορούσε να γίνει (και δεν έγινε).Ο κος Φελλούρης είναι πολύ συνεργάσιμος και έδειξε ιδιαίτερο ενδιαφέρον γαι την πρωτοβουλία μας την οποία είχε και εκείνος παλαιότερα (πέρυσι συγκεκριμένα).Καταλήξαμε ότι φέτος δεν υπάρχει δυνατότητα να σταλεί ΟΜΑΔΑ στο διαγωνισμό (δεν αποκλείονται,βέβαια, οι ατομικές συμμετοχές).Μιλήσαμε λίγο για του χρόνου και απ' ότι μου είπε είναι υπέρ ενός διαγωνισμού επιλογής ο οποίος θα γίνει στο εαρινό εξάμηνο (ώστε να μπορούν και οι πρωτοετείς να έχουν ελπίδες).Πιστεύει ότι κατά 99% θα μπορέσουμε να εξασφαλίσουμε χρηματοδότηση από τη Φοιτητική Μέριμνα για μια ομάδα 5-7 ατόμων.Ο ίδιος,λόγω των υποχρεώσεών του στις ΙΜΟ,θα είναι δύσκολο να συνοδεύσει και -αν και απ' ό,τι είπε δεν υπήρξε ιδιαίτερο ενδιαφέρον από άλλους καθηγητές- θα βρεθεί κάποιος συνοδός.Για προετοιμασία προτείνει,βέβαια,θέματα από IMC και Putnam.Α,μου ζήτησε να τονίσω ότι ο διαγωνισμός που έγινε φέτος στην Κύπρο (Νοτιοανατολικής Ευρώπης) θα γίνει του χρόνου το Μάρτιο στην Ελλάδα (είναι για 1οετείς και 2οετείς μόνο).Αν θυμηθώ κάτι αργότερα θα ξαναποστάρω.Αν έχετε ερωτήσεις που μπορώ να απαντήσω,ευχαρίστως.Προσωπική αίσθηση ότι η συμμετοχή του ΕΜΠ σαν ομάδα στο διαγωνισμό έχει μπει σε καλό δρόμο.Θα δείξει το μέλλον...


calculus
derivative
function
logarithms
calculus computations

For the function $ f(x)\equal{}log (x)$, estimate $ f'(1)$.

hello
$ f'(x)\equal{}\frac{1}{x}$
$ f'(1)\equal{}1$
thank u

I assume that by $ \log$ you mean the natural logarithm $ \ln$. If not, it just differs by a scale factor of $ \ln 10$.

$ f'(1) \equal{} \frac {1}{g'(f(1))} \equal{} \frac {1}{g'(0)}$, where $ g(x) \equal{} e^x$

How you approximate the derivative of that depends on how you choose to define $ e^x$. But it's probably easier than dealing with the logarithm directly (unless you define $ e^x$ as the inverse of the $ \ln$ function).

[quote="Kent Merryfield"]ben b's other posts strongly suggest that he is in the first week or two of a Calculus I course using some "mainstream" calculus textbook. My suspicion that this was an exercise for which the intended "method" was the computation of several difference quotients with a spread of possible values of $ \Delta x,$ using a calculator. (Followed by an "eyeball" estimate based on those results.)

ben b's other posts also suggest that he came here trying to get other people to do his homework for him.[/quote]

In no way shape or form do I want other people to do my homework for me. I have modified the numbers in nearly every post so that they do not match numbers in the homework. I am clear just looking for a little assistance. I thought that was your guys goal here but if that's not the case I will go elsewhere.

[quote="jmerry"]As a general principle, "estimate _" is an incomplete question. There are many different techniques, and many different levels of accuracy we could be looking for.[/quote]

We are looking for accuracy to 4 decimal places.

... which is the less important half of the two things jmerry mentioned.


probability
geometry
3D geometry
expected value

I was doing practice test for a competition that is over school stuff and I came across a few problems which I didn't think were right:
1. Three six sided dice are rolled.  What is the probability that they all show the same number?
[hide="Answer"]The answer key claims it's $ \frac{1}{36}$ but I think it's $ \frac{1}{216}$ because $ \frac16\bullet\frac16\bullet\frac16\equal{}\frac{1}{216}$[/hide]
2.  Three six sided dice are rolled.  What is the probability that at least one five is showing?
[hide="Answer"]  The answer key says $ \frac{91}{216}$.  I took this into cases and got $ \frac{43}{216}$.  So, what is right?[/hide]
3.  Three six sided dice are rolled.  What is the probability that the sim of the numbers is 5 or less?
[hide="Answer"]  The answer key says $ \frac{5}{108}$ but I think it's $ \frac{1}{12}$  I'm not sure...[/hide]
4.  In a lottery, 1000 people pay \$1 for a ticket.  One grand prize winner will recieve \$300 and nine runner ups will recieve \$50 each.  What is the expected value of playing in this lottery?
[hide="Answer"]  I don't get "expected value" part.  The answer is $ \minus{}\$0.25$.  I need help...[/hide]

[quote="ajain"]1. Three six sided dice are rolled.  What is the probability that they all show the same number?
[hide="Answer"]The answer key claims it's $ \frac {1}{36}$ but I think it's $ \frac {1}{216}$ because $ \frac16\bullet\frac16\bullet\frac16 \equal{} \frac {1}{216}$[/hide][/quote]
[hide]What the first cube is does not matter; what matters is that the second and third cubes are the same as the first.

$ P \equal{} \frac {1}{6}\cdot\frac {1}{6} \equal{} \boxed{\frac {1}{36}}$[/hide]

[quote="ajain"]2.  Three six sided dice are rolled.  What is the probability that at least one five is showing?
[hide="Answer"]  The answer key says $ \frac {91}{216}$.  I took this into cases and got $ \frac {43}{216}$.  So, what is right?[/hide][/quote]
[hide]Case 1: Exactly one 5 is showing. Probability: $ 3\cdot\frac {1}{6}\cdot\frac {5}{6}\cdot\frac {5}{6} \equal{} \frac {25}{72}$

Case 2: Exactly two 5's are showing. Probability: $ 3\cdot\frac {1}{6}\cdot\frac {1}{6}\cdot\frac {5}{6} \equal{} \frac {5}{72}$

Case 3: Exactly three 5's are showing. Probability: $ \frac {1}{6}\cdot\frac {1}{6}\cdot\frac {1}{6} \equal{} \frac {1}{216}$

$ P \equal{} \frac {25}{72} \plus{} \frac {5}{216} \plus{} \frac {1}{216} \equal{} \boxed{\frac {91}{216}}$[/hide]

[quote="ajain"]3.  Three six sided dice are rolled.  What is the probability that the sim of the numbers is 5 or less?
[hide="Answer"]  The answer key says $ \frac {5}{108}$ but I think it's $ \frac {1}{12}$  I'm not sure...[/hide][/quote]
[hide]There are only $ 10$ cases where the sum of the $ 3$ numbers is $ 5$ or less: (1,1,1); (1,1,2); (1,2,1); (2,1,1); (1,1,3); (1,3,1); (3,1,1); (1,2,2); (2,1,2); (2,2,1).

There are $ 216$ possible outcomes when $ 3$ dice are rolled, so $ P \equal{} \frac {10}{216} \equal{} \boxed{\frac {5}{108}}$.[/hide]

[quote="ajain"]4.  In a lottery, 1000 people pay \$1 for a ticket.  One grand prize winner will recieve \$300 and nine runner ups will recieve \$50 each.  What is the expected value of playing in this lottery?
[hide="Answer"]  I don't get "expected value" part.  The answer is $ \minus{} \$0.25$.  I need help...[/hide][/quote]
[hide]"Expected value" is the theoretical weighted average of all possible outcomes. Here, it means how much each person will get on average.

In this case, you first find the gross earnings, $ \$750$. Then, you subtract the expenses, $ \$1000$, from this to get the net earnings, $ \minus{} \$250$. Finally, divide by the number of people there are, $ 1000$, to find the expected value for each person.

$ E \equal{} \frac {\$750 \minus{} \$1000}{1000} \equal{} \minus{} \frac {\$250}{1000} \equal{} \boxed{ \minus{} \$0.25}$[/hide]

If you understand why my solutions give the correct answer but don't understand why yours don't (this has happened to me before), you can post your solutions. :)

[quote="7h3.D3m0n.117"][quote="ajain"]Why doesn't the first one matter?  I understand everything but don't get why the first roll does not matter, shouldn't it?[/quote]

The first one matters because it sets the precedent for the other dice. If you roll a 5 on the first di, you have to make sure the others are also 5. If you roll a 2 on the first di, the others have to be 2.[/quote]

That makes a lot more sense.  Thanks!

I still don't know why he multiplied by three though...

[quote="ajain"]I still don't know why he multiplied by three though...[/quote]

Well each die is a seperate and independent event, meaning that the outcome of one roll does not effect another. Therefore, the probability of rolling a 5 on three [i]seperate[/i] dice would be $ \left( \frac{1}{6} \right)^3$. He didn't really multiply by 3, he raised it to the 3rd power instead.

I think they're asking about why they multiplied by 3 in say, part 1 of problem 2.
Reason: There are 3 different ways to arrange the dice and still have one 5. With dice, order is (almost) always important.

Logical explanation:

For each combination of 3 rolls (example: 1-2-5), there are 3 permutations (only the position of the 5 is changed): 1-2-5, 1-5-2, 5-1-2. Notice that 2-1-5 does not belong in this set; it is accounted for by the $ \left(\frac {5}{6}\right)^2$ part.

Algebraic explanation:

The probability of there being exactly one 5 is actually $ \frac {1}{6}\cdot\frac {5}{6}\cdot\frac {5}{6} \plus{} \frac {5}{6}\cdot\frac {1}{6}\cdot\frac {5}{6} \plus{} \frac {5}{6}\cdot\frac {5}{6}\cdot\frac {1}{6}$, but I just wrote $ 3\cdot\frac {1}{6}\cdot\frac {5}{6}\cdot\frac {5}{6}$.

[quote="perfect628"]Well, for  that one, there is a much better way.

[hide]Try to find the probability that [b]no[/b] 5 is rolled.[/hide][/quote]
Thats exactly what I did. The probability that [b]no[/b] 5 is rolled is $ ({\frac{5}{6}})^3$, or $ \frac{125}{216}$. The probability that you get at least one five is 1-this, or $ \frac{91}{216}$

Just to make sure I understand this, if I rolled four dice, the probability that they all show the same number would be $ \frac{1}{216}$?

So the probability of at least one five showing with four dice would be $ \frac{23}{48}$

The probability that the sum is five or less with four dice would be $ \frac{5}{1296}$

And in problem 4, instead of 9 runner ups, say we only had 4, so the expected value would be $ \$$-.50.

Am I right, though I am not sure on a few...

All of those are correct except for the second one.

The probability of at least one five showing on four dice is $ 1\minus{}\left(\frac{5}{6}\right)^4\equal{}\frac{671}{1296}$.

[quote="dreambold22"]can 671/1296 be reduced at all?[/quote]

nada.

remember, 5/6 is simplified, so 5^4/6^4 is also in simplest form. they have no factors in common. since that is true, 6^4-5^4/6^4 must also be in simplest form

[quote="ajain"][quote]What the first cube is does not matter; what matters is that the second and third cubes are the same as the first.
[/quote]

Why doesn't the first one matter?  I understand everything but don't get why the first roll does not matter, shouldn't it?
[quote]
Case 1: Exactly one 5 is showing. Probability: 25/72

Case 2: Exactly two 5's are showing. Probability:  5/72

Case 3: Exactly three 5's are showing. Probability:  1/216
[/quote]

Okay...why did you multiply by 3?[/quote]
The first number could be any number, while the next two have to be the same as the first number, thus, the first number doesn't matter

I believe he multiplied by three because there are three dice, so the probability of any roll on one of the dice could happen on either of the other 2... :|

BTW: ilovepi, you're new here, so I would like to remind you to try not to revive old threads... :P


geometry
3D geometry
pyramid
combinatorics proposed
combinatorics

The base of a pyramid is a convex p[olygon with 9 sides

Each diagonal of the base and each edge on the lateral surface is coloured either black or white.Both colours are used. But the edges of the base are not coloured.

Prove that there are three segments of the same colour that form a triangle.

In the latteral edges, at least $5$ have the same colour, WLOG white ones.
Let their endpoints in the base be $A_1,A_2,A_3,A_4,A_5$ in that order. When any coloured diagonal between these points is white, we are done, so consider them all black.
Now there are two consecutive vertices $A_i, A_{i+1}$ (indices seen $\mod 5$) that don't share a common side of the $9$-gon, WLOG $A_1,A_5$.
Then also $A_3$ doesn't share a common side with one of them (only $A_2,A_4$ can be direct neighbours) and our black triangle is $A_1A_3A_5$.


In triangle $ ABC$, the measure of angle $ A$ is $ x$ degrees, the measure of angle $ B$ is $ 2x$ degrees and the measure of angle $ C$ is $ 5x$ degrees. What is the value of $ x$? Express your answer as a decimal to the nearest tenth.

hello, we have $ x\plus{}2x\plus{}5x\equal{}180^{\circ}$ and from here we get $ x\equal{}\frac{180^{\circ}}{8}\equal{}22.5^{\circ}$.
Sonnhard.


probability
algorithm
limit
real analysis
real analysis unsolved

I wanted to ask a question on Real analysis but I haven't found the right forum. So I'm sorry for posting it here, but I really want to ask.
Can you prove ( or give a counter example) to this proposition:
In a sigma Algebra M, let (E_i) , 1 < i < infinity, be a sequence such that measure(E_i) -> some number c. Then there exists a set in M (say E) such that measure(E) = c.

Let's try this. Let $ M$ be the power set of $ \mathbb{N},$ and assign a measure such that $ m(\{n\})\equal{}1\plus{}\frac{1}n.$ Then we have a sequence of sets whose measures tend to $ 1$ but no single set of measure $ 1,$ since any nonempty set has measure greater than $ 1.$

The essential problem I want to prove is in a measure space $ \{M,X,\mu\},$ suppose $ \mu(X) \equal{} 1$ and $ \forall E\in M$, there exists $ F\subset E$ such that $ 0 <\mu(F) <\mu (E)$. Show that $ \{\mu(E) | E\in M\}\equal{} [0,1]$.
My approach: Show that $ \forall\epsilon > 0$, there exists $ E\in M$ such that $ \mu(E) <\epsilon$. (Easy to do). Then given $ 0 < c < 1$ I look at $ d \equal{} sup\{\mu(E) | E\in M,\mu(E) < c\}$. I want to show 2 things:(i) There exists $ F\subset X$ such that $ \mu(F) \equal{} d$ and (ii) $ d \equal{} c$. I couldn't prove (i) and as Dr. Merryfield pointed out, it's not necessarily true. So is there any other approach ?

No pmtriet00, I think your conclusion (i) is also true because the space $ X$ which you consider has mesure equals $ 1$, ie.. finitely. I's not hard to prove this fact, but it's quite hard to find a complete proof to your problem.  :D

Such measures are called non-atomic measures.

If $ m$ is a non-atomic probability measure on $ X$, for each $ 0 < c < 1$ there is $ E\subset X$ with $ m(E) \equal{} c$.
A short proof can be obtained using greedy algorithm as in [url=http://www.everything2.com/index.pl?node_id=1773590]the proof of Vitali's covering lemma[/url].

proof)

We want to build a sequence $ \{E_{n}\}_{1}^{\infty}\subset X$ of disjoint subsets such that the measure of its union is $ c$.

Suppose $ E_{1}, E_{2},\cdots, E_{n}$ are already constructed so that they are disjoint and that $ m(F_{n}) < c$ where $ F_{n}\equal{}\cup_{1}^{n}E_{k}$. Consider the family $ \mathcal{C}_{n}$ of all sets $ E$ such that $ E\subset F_{n}^{c}$ and $ 0 < m(E) < c\minus{}m(F_{n})$. this family is nonempty. Therefore there exists $ E_{n\plus{}1}\in\mathcal{C}_{n}$ such that $ m(E_{n\plus{}1}) >\frac{1}{2} m(E)$ for all $ E\in\mathcal{C}_{n}$. Continue this process to infinity then we have a disjoint sequence of sets $ \{E_{n}\}_{1}^{\infty}$.

The next step is to show that the measure of $ F \equal{}\cup_{1}^{\infty}E_{n}$ is $ c$. If it is not $ c$, there exists $ A\subset F^{c}$ with $ 0 < m(A) < c\minus{}m(F)$. Then there is $ n$ such that $ m(E_{n}) <\frac{1}{2} m(A)$. But this contradict the greedy choice of $ E_{n}$. Therefore we must have $ m(F) \equal{} c$.

Thank you redkimchi for giving us a lovely proof. It's nice surprise to know the same elegant arguments in proof of Vitali covering theorem could be use for this question.

By the way, i don't understand why (the last ) "Then there is $ n$ such that $ m(E_{n})<\frac{1}{2}m(A)$". May be i'm missing some trivial arguments?  :blush:

[quote="pluricomplex"]"Then there is $ n$ such that $ m(E_{n}) <\frac{1}{2}m(A)$". May be i'm missing some trivial arguments?  :blush:[/quote]

$ \sum_{1}^{\infty}m(E_{n})\le 1$ implies $ \lim_{n\to\infty}m(E_{n}) \equal{} 0$.

[quote="redkimchi"]
$ \sum_{1}^{\infty}m(E_{n})\le 1$ implies $ \lim_{n\to\infty}m(E_{n}) \equal{} 0$.[/quote]

Ah I see, thanks again.


hello, from the third equation we get
$ z\equal{}\minus{}\frac{4y(y^2\minus{}1)}{1\plus{}y^4\minus{}6y^2}$ setting this in the first equation and factoring we get
$ \minus{}28y^6\plus{}70y^4\minus{}28y^2\plus{}56y^4x\plus{}8y^8x\minus{}56y^6x\minus{}8y^2x\plus{}1\plus{}y^8\equal{}0$
solving this for $ x$ we get
$ x\equal{}\minus{}\frac{1}{8}\frac{\minus{}28y^6\plus{}70y^4\minus{}28y^2\plus{}1\plus{}y^8}{y^2(7y^2\plus{}y^6\minus{}7y^4\minus{}1)}$
setting this in the seconde equation and factoring we get
$ (1\plus{}y^2)(y^{22}\minus{}253y^{20}\plus{}8855y^{18}\minus{}100947y^{16}\plus{}490314y^{14}\minus{}1144066y^{12}\plus{}1352078y^{10}\minus{}817190y^8\plus{}245157y^6\minus{}33649y^4\plus{}1771y^2\minus{}23)\equal{}0$
Sonnhard.


geometry
analytic geometry
graphing lines
slope
algebra
system of equations
algebra unsolved

Find values for k and d so that there is no real solutions (x,y) for the eqautions.

x^2+y^2 = 2
y = k*x+d

[hide="Answer."]
Using geometric arguments about the relationship between a line and a circle, I found:

\[ \left\{(d,k): |d| > \sqrt {2},\ |k| < \frac {\sqrt {2d^2 \minus{} 4}}{2}\right\}\]
[/hide]

First of all, thx for your solution.
I would like to know how you found out that solution. Could you elaborate more on your solution?

Thank You!

Not sure if it's correct, but I'll explain what I did. Note that my solution makes it look more like a geometry problem than an algebra one:

[hide="My solution."]
The first equation represent a circle with center at the origin and radius $ \sqrt{2}$. The second equation is a line with slope $ k$ and y-intercept $ d$. Thus the system of equations will have no real solutions if and only if the circle and line do not intersect. Obviously, if the y-intercept is inside the circle, the line will always intersect the circle, so we must have:

 $ |d|>\sqrt{2}$

Now consider the line $ y\equal{}kx\plus{}d$ with $ |d|>\sqrt{2}$ such that the line is tangent to the circle at some point $ (a,b)$. Then the line is perpendicular to the radius of the circle drawn to $ (a,b)$, so:

$ \frac{b\minus{}d}{a}\equal{}\frac{\minus{}a}{b}$ (Draw a picture to see this.)

$ \implies a^2\plus{}b^2\equal{}bd$

But since $ (a,b)$ is on the circle, $ a^2\plus{}b^2\equal{}2$, and $ bd\equal{}2 \implies b\equal{}\frac{2}{d} \implies a\equal{}\pm \sqrt{2\minus{}\frac{4}{d^2}}$. Now we can find the slope of our line in terms of $ a$ and $ b$:

$ k\in\left\{\frac{d\minus{}b}{a}\right\}$
$ k\in\left\{\frac{d\minus{}\frac{2}{d}}{\pm \sqrt{2\minus{}\frac{4}{d^2}}}\right\}$ (Now some algebra.)
$ \implies k\in\left\{\pm \frac{\sqrt{2d^2\minus{}4}}{2}\right\}$ (Note that there are two values of $ k$ depending on what side of the circle the line is tangent to.)

$ \implies |k|\equal{}\frac{\sqrt{2d^2\minus{}4}}{2}$

Now we find that if $ |k|$ is greater than $ \frac{\sqrt{2d^2\minus{}4}}{2}$, then the line will intersect the circle twice, so $ |k|$ must be less than $ \frac{\sqrt{2d^2\minus{}4}}{2}$, and:

$ |k|<\frac{\sqrt{2d^2\minus{}4}}{2}$

Putting it all together, our solutions become:

$ \left\{(d,k): |d| > \sqrt {2},\ |k| < \frac {\sqrt {2d^2 \minus{} 4}}{2}\right\}$

[/hide]


Let $ n$ be a positive integer, and let $ a_1, a_2, a_3, \ldots, a_n$ be $ n$ positive real numbers such that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots \plus{} a_n \equal{} 1$. Is it true that
\[ \frac {a_1^4}{a_1^2\plus{}a_2^2} \plus{} \frac {a_2^4}{a_2^2\plus{}a_3^2} \plus{} \frac {a_3^4}{a_3^2\plus{}a_4^2} \plus{} \cdots \plus{} \frac {a_{n\minus{}1}^4}{a_{n\minus{}1}^2\plus{}a_n^2} \plus{} \frac {a_n^4}{a_n^2\plus{}a_1^2} \ge \frac {1}{2n}?
\] Justify your answer.


[quote="socrates"]If $a,b \in [1,4]$
show that $2\leq\frac{a}{b}+\frac{b}{a}\leq\frac{17}{4}$

 :)[/quote]

The left side is true for all $a,b \in R^+$
The right reduces to $(4a-b)(a-4b) \leq 0$  :)


number theory
greatest common divisor
inequalities
number theory proposed

Prove that for all integers $ m$ and $ n$, the inequality
\[ \dfrac{\phi(\gcd(2^m \plus{} 1,2^n \plus{} 1))}{\gcd(\phi(2^m \plus{} 1),\phi(2^n \plus{} 1))} \ge \dfrac{2\gcd(m,n)}{2^{\gcd(m,n)}}\]
holds.
[i]Nanang Susyanto, Jogjakarta [/i]

I spent a lot of hours on this problem, up to realize that the text was wrong  :ninja:
After some heavy trying, I am sure that the right statement is "$ \displaystyle \dfrac{\gcd(\varphi(2^m \plus{} 1),\varphi(2^n \plus{} 1))}{\varphi(\gcd(2^m \plus{} 1,2^n \plus{} 1))} \ge \gcd(m,n)2^{1\minus{}\gcd(m,n)}$ for all $ (m,n) \in \mathbb{N}_0^2$."
If someone is interested in my solution can look [url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]here[/url] on Problem5 :P
PaoloL.

oh my god.

bboypa, that's  totally my fault, really.
thanks for correcting.  :oops:

i'm very sorry for this.

Raja.

Edit: if the moderator does not mind, please edit the problem. i can not edit the problem. thank u very much.


I was using Latex to submit my answers to USAMTS, and knowing I had little time, I had to hurry. ok. I already had TeXnicCenter downloaded. I went to this site [url]http://www.usamts.org/MyUSAMTS/U_MyForms.php[/url] I downloaded the three things at the bottom and opened the solutions page in TeXnicCenter. When I press Ctrl+F7, it says there are two errors and it doesn't run smoothly, so I can't get my solution onto pdf. It says in Error #1,
[quote]Latex Error: File 'fancyhdr.sty' not found.[/quote]
and the second one says
[quote]Emergency Stop.[/quote] Any help is appreciated. If I don't get it to work, I guess I could just fax my solutions.

They are the same error, telling you that LaTeX couldn't find fancyhdr.sty so had to stop. Are you sure that the fancyhdr.sty that you downloaded is in the same folder as your tex file?

If you can't get it to work, go into Miktex Package Manager, look for fancyhdr. Rightclick on it and click on install. That will put a copy of fancyhdr in the LaTeX system so it can be used from anywhere.

Maybe it's the downloading. on the site i mentioned earlier, first I clicked on the link that says, "Click here to download fancyhdr.sty" Then it opens up something else with just text. So, I go to file, save as, and save it in the Latex folder where I have everything else. When I open the folder, it's saved there as fancyhdr_sty as a Text document. what's wrong with this?

where's the Miktex Package Manager? thanks.

I see why things have gone wrong now. The problem is the way that the browser does things. To correct it, go to the fancyhdr link but [b]Right[/b] click on it and choose [i]Save Target As...[/i] (in Firefox it is [i]Save Link As...[/i]) In the [i]Save as type[/i] dropdown box choose [i]All Files[/i] and use fancyhdr.sty as the file name.
(if you don't use [i]All Files[/i] then the file will be saved as fancyhdr.sty.txt and you would have to change its name.)
It should work now.

The MiKTeX Package Manager (which allows you to easily install all sorts of packages) is found in Start->All Programs->Miktex menu.

[color=darkred]To DPatrick: Perhaps the instructions for downloading fancyhdr.sty could be amplified as both IE & Firefox open the file by default.[/color]

Thanks, but now, when I press F5, it says "There was an error opening the document. The file cannot be found." when I'm trying to convert it to PDF.

Have you put any of your solutions into the file? If you don't then it won't create the pdf. To test this just put $\$1+1=2\$$ on the line after
%% Solution to Problem 1 goes here
and it should then produce a pdf.

thank you stevem for answering that, I had the same problem as well.

As far as I know, that doesn't do anything. When I try to compile, I get the same message, but I am still able to view the PDF.

Thanks!

I tried that already, though, and it didn't work for me, neither in PDF or DVI.

Thanks anyways!

[quote="astrangequark"]I have a different problem:

When I try to compile, it says (in the errors section):

I found no \citation commands---while reading file C:\Documents                 and Settings\Owner\Desktop\a\solutions.aux

I found no \bibdata command---while reading file C:\Documents and Settings\Owner\Desktop\a\solutions.aux

I found no \bibstyle command---while reading file C:\Documents and Settings\Owner\Desktop\a\solutions.aux

I downloaded all three files correctly into a folder 'a' on the Desktop.  I tried deleting and trying again, three times.  The first time I ever tried compiling was about two weeks ago, and it worked fine that time.  I tried compiling the file when it was empty of all but the form (ie, no name, etc.) and also with my name, username, etc.  It displayed the same error every time.

Can anyone help me?[/quote]It's not really an error as such - all it's telling you is that there's no bibliography which I guess is correct, so you can ignore it.

The line you need to keep an eye on for errors is the last one which says something like:
LaTeX-Result: 0 Error(s), 0 Warning(s), 0 Bad Box(es), 1 Page(s)

A while ago the link to the fancyhdr file would actually start downloading the file and not just open the file.  I'm guessing that this is a result of the server switch AoPS underwent that hasn't been corrected for yet.

[quote]
When I try to compile, it says (in the errors section):

I found no \citation commands---while reading file C:\Documents and Settings\Owner\Desktop\a\solutions.aux

I found no \bibdata command---while reading file C:\Documents and Settings\Owner\Desktop\a\solutions.aux

I found no \bibstyle command---while reading file C:\Documents and Settings\Owner\Desktop\a\solutions.aux [/quote]

I saw these errors, and your replies have been helpful, but I wasn't able to make it into a PDF.  I am still not able to.

[quote]Are there any other errors? What does the log file say? Are any files created or is it just the pdf that's missing? What happens if you create a DVI file instead, does that work?[/quote]

No, these are the only three errors.  I don't know how to access the log file; I think some files [i]are[/i] created, but I can't do a DVI, either.

Thanks!

The log file, solutions.log, can be found in the same directory as solutions.tex. It is a text file so can be opened in Notepad for example.

Are you sure you have added your own content to solutions.tex, as you'd get the same problems if you haven't?

If you haven't added anything then just put $\$1+1=2\$$ on the line [i]after[/i]
%% Solution to Problem 1 goes here
and try again.

If you have put in your own content, then post your solutions.log and solutions.tex files (as attachments as they can get rather long) so we can have a look to see what the problem is.


Monica reduced 154/253 to a common fraction by cancelling the two middle 5's to get 14/23. This answer is correct, but the procedure is incorrect. Kia's example worked with Monica's incorrect procedure too; the fraction 242/x also reduced to its equivalent common fraction when the identical middle digits of the three-digit integers were cancelled out. What is the greatest possible value of x?

[hide]What I did was, let the digits of x be a, 4, and b in that order. So,
$\frac{242}{100a+40+b}= \frac{22}{10a+b}$ then cross multiply and reduce then a+b=4. So let b=0, a=4 and x=440. And 242/440 = 22/40. So why is the answer 341?[/hide]

I liked this problem, but I got it wrong.
Later, I redid the test and got it right.

[hide]
So first you realize what happened - you divided by 11.
So you just try a few multiples of 11 that have 4 as a middle digit and you suddenly realize that it's 341.[/hide]

hmm i got this question right but now i'm wondering why 891 doesn't work.

242/891=22/81


Prove that if $ x,y,z>0$, then \[ \sum{\frac{x^2z^2\plus{}y^4\plus{}y^2z^2}{yz(xz\plus{}y^2\plus{}yz)}} \geq 3.\]

[quote="ozgurkircak"]Prove that if $ x,y,z > 0$, then
\[ \sum{\frac {x^2z^2 \plus{} y^4 \plus{} y^2z^2}{yz(xz \plus{} y^2 \plus{} yz)}} \geq 3.
\]
[/quote]
$ \frac{x^2z^2 \plus{} y^4 \plus{} y^2z^2}{xz\plus{}y^2\plus{}yz} \geqslant \frac{xz\plus{}y^2\plus{}yz}{3}$. (By AM-QM or common sense)

Then we just have to prove:
$ \sum{\frac {xz \plus{} y^2 \plus{} yz}{yz}} \geqslant 9 \iff$

$ \sum{\frac {x}{y} \plus{} \frac{y}{z}} \geqslant 6$

Which is obvious by AM-GM.


A rectangular prism has a volume of 240 cubic inches. Each dimension is increased by a factor of 1.5. What is the number of cubic inches in the volume of the enlarged prism?

The factor 1.5 is multiplied on all 3 sides, so the volume is 240*1.5^3=360*1.5*1.5=540*1.5=[b]810[/b]


rotation
geometry
geometric transformation
3D geometry

Imagine you built a roof over all of the earth. How fast would the earth have to rotate to make the force outwards be equal to the force we currently feel towards the earth?

What do you mean by a roof?
This should also probably go into the Introductory Physics forum.
[hide="Centripetal Acceleration is given by"]
$ a = \frac{v^{2}}{r}$
$ v$ is tangental velocity

In this case
$ a=9.80$
$ r= \ \text{The Radius of the Earth}$
[/hide]

Answer

The questioner has fallen into the trap that a force is exerted outwards on an object is spinning in a circle: this is quite a common misconception. This "centrifugal" (centre-fleeing) force is a "virtual" force and only seems to occur because our frame of reference is also moving in a circle. We have all probably felt that when we are in a car, and we turn a corner, we get thrown to the outside of the bend...surely this feels like a real force? To answer this we must consider an independant observer, perhaps someone having a birds-eye view of the car going around the bend. What this observer will see is a passenger continuing in a straight line while the car turns until  the passenger reaches the inside of the car door. The car door exerts a central, centripetal (centre-seeking) force on the passenger causing them to deviate from the straight line path, and move in the same circular motion as the car. Hence when we are being "flung" to the outside of the bend, we are in fact continuing in a straight line at constant velocity until a force acts inwards (from the car door, or friction from the seat).

I assume that the question can be rephrased as: How fast would the earth have to rotate to produce no normal reaction from the Earth's surface acting on a person "standing" on the Earth such that the person does not "fly" off the Earth. I assume that the person is standing at the equator such that the gravitational and centripetal force act in the same direction (remember the centripetal force acts towards the axis of rotation, while the gravitational force acts towards the centre - It is in fact impossible for someone standing on the North or South poles to experiance any change in normal reaction since they experiance no centripetel force).

The only force acting on the person is gravitational and is given by the following equation:-

$ F = mg$

-$ m$ is the mass of the person and $ g$ is the acceleration due to gravity.

This force, and this force alone, must supply the centripetal force which is given by:-

$ F = mr{\omega}^{2}$

-where $ \omega$ is the Earth's angular velocity.

Setting these forces equal:

$ mg = mr{\omega}^{2}$

$ \Rightarrow g = r{\omega}^{2}$

$ \Rightarrow{\omega}^{2}= \frac{g}{r}$

Using the equation: $ \omega = 2\pi f$ where f is the frequency of the Earth's revolutions.

$ (2\pi f)^{2}= \frac{g}{r}$

$ f^{2}= \frac{g}{4{\pi}^{2}r}$

At this stage I would like to note that this is an equivalent expression for Kepler's third law of planetary motion: that the square of the orbital period of a planet is directly proportional to the cube of the mean distance of the planet from the sun.

Subbing $ g = 9.81 ms^{-2}$ and $ r = 6.3781 \times 10^{6}m$ into the equation gives:

$ f^{2}= 3.90 \times 10^{-8}\Rightarrow f = 1.97 \times 10^{-4}Hz$

Hence $ 1.97 \times 10^{-4}$ revolutions per second

= $ 1.18 \times 10^{-2}$ revolutions per minute
= $ 0.711$ revolutions per hour
= $ 17.1$ revolutions per day

(note the time taken for one revolution is approximately $ 84$ mins)


number theory
Diophantine equation
number theory unsolved

Show \[3x^3 + 4y^3 + 5z^3 = 0\] has nonzero solutions modulo every integer but not in $\mathbb{Z}$.

[i]Category: Diophantine equation[/i]

this is a well-known equation. the interesting thing about it is that has no nontrivial solutions in $\mathbb{Z}$, but it has nontrivial solutions in any $\mathbb{Z}_p$ or equivalent in $\mathbb{Z}/m\mathbb{Z}$ for any integer $m$. thus the local-global-principle of hasse doesn't work for curves of degree 3.

[quote="Peter Scholze"]the interesting thing about it is that has no nontrivial solutions in $\mathbb{Z}$, but it has nontrivial solutions in any $\mathbb{Z}_p$ or equivalent in $\mathbb{Z}/m\mathbb{Z}$ for any integer $m$.[/quote]

 :D
I think that it is what we are asking to prove Peter.

Pierre.

ah sorry, i didn't read it properly  :blush:

[quote="koo"]Show \[ 3x^3 + 4y^3 + 5z^3 = 0 \] has nonzero solutions modulo every integer but not in $\mathbb{Z}$.

[i]Category: Diophantine equation[/i][/quote]

:)

As much as I know, the (shortest/easiest¿) solution requires some work in $\mathbb{Z}[\sqrt[3]{60}]$ or something like that... cruel... very cruel...

:o , I see...
Well, I'll try to look for an easier solution, maybe in vain, but I'll try..  :P

I think you might want to read http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf and work out details.


Consider an experiment throwing 2 fair dice.  Let X be the random variable indicating the sum of the numbers that appear.

P(X=3)  ?

P(X <= 4) ?

P(3 < x <= 7)  ?

I found that $ P(\text{me doing your homework for you})\equal{}0$.

Just list the necessary summing pairs and divide them by the total number of ways you can get results for the throw. It's not too much casework so don't complain.


inequalities
inequalities solved
BritishMathematicalOlympiad

(a) Given reals a, b, c such that a + b + c = 0 prove that \sum a3 > 0 iff \sum a5 > 0.
(b) Given reals a, b, c, d such that a + b + c + d = 0 prove that \sum a3 > 0 iff \sum a5 > 0.

I think the second one can be proved in the following manner:
  Put d=-a-b-c. Then  \sum a^3=-3(a+b)(b+c)(c+a) and  \sum a^5=cst*(a+b)(b+c)(c+a)(f(a^2+b^2+c^2)+g(ab+bc+ca)), where f and g can be determined. The first one can be proved similarly.

for both the inequalities note that
6 (\suma^5)=5( \sum a^3)( \sum a 2 )
obvious enough from now


I am trying to put a "prove" problem in Getting Started forum. Feel free to move it if it shouldn't belong here.

This is one of the most basic theorem in mathematics. Can you prove it?

Theorem Given any two integers a, b where b>0, there exists a unique pair of integers q, r such that  a=qb+r and 0 :le:  r < b.

What are  the names for q and r? Answer in spoiler:

[hide]

q is called the quotient of a and b, and r is the remainder.

You might need The Well Ordering Principle  which states: Every non-empty subset of non-negative integers has a least element.

[/hide]

The name for q is [hide]quotient[/hide].

The name for r is [hide]remainder[/hide].

Sad experience has taught me that the people who will be able to provide the proof are best considered to be at Intermediate level or above, so I will move the thread.


geometry
circumcircle
rhombus
geometry proposed

Let D be the midpoint of the base AB of an isosceles acute triangle ABC. A point E is chosen on AB and O is the circumcenter of triangle ACE. Prove that the perp line from D to OD, the perp line from E to BC and the parallel line from B to AC all pass through a common point.

Dear Mathlinkers,
a rapid proof (sorry):

let X the point of intersection of the perp line from E to BC and the parallel line from B to AC,
1 the circumcircle of ACE,
and 1', C', E' the symmetrics of 1, C, E wrt D.
ACBC' is a "losange" (rhombus ?).
The perp line from D to OD is the radical axe of 1 and 1'.
Sincerely
Jean-Louis


greatest common divisor
AMC
USA(J)MO
USAMO
number theory
relatively prime

yes
proofy way:
suppose they are not relatively prime
there then exists f, a, and b (all integers, f>1) such that n can be rewritten as af and n+1 can be rewritten as bf
we have
n=af
n+1=bf
subtracting the first equation from the second gives
1=(b-a)f
b-a=1/f
b-a must be an integer but 1/f is never an integer for all f, f>2
therefore we have a contradiction

non proofy way
there is a theorem for finding the GCF of 2 numbers that states
the GCF of a and b (given they are distinct) is at most their positive difference
(but this ofcourse needs a proof...)
anyway the GCF of n and n+1 is at most 1, so they are relatively prime

i think it was a question on USAMO some time ago to prove or disprove that 4n and 4n+1 are always relatively prime.
EDIT: i think that same exact proof stevenmeow did is in Intro to NT ...

[quote="stevenmeow"]
non proofy way
there is a theorem for finding the GCF of 2 numbers that states
the GCF of a and b (given they are distinct) is at most their positive difference
(but this ofcourse needs a proof...)
anyway the GCF of n and n+1 is at most 1, so they are relatively prime[/quote]

I can simplify that.
non proofy way
DUH!

[quote="stevenmeow"]
non proofy way
there is a theorem for finding the GCF of 2 numbers that states
the GCF of a and b (given they are distinct) is at most their positive difference
(but this ofcourse needs a proof...)
anyway the GCF of n and n+1 is at most 1, so they are relatively prime[/quote]

Another ''non-proofy'' way similiar to yours.
We know that if $ a\ge b$ $ gcd(a,b)\equal{}gcd(a\minus{}b,b)$ so $ gcd(n\plus{}1,n)\equal{}gcd(1,n)\equal{}1$

Yep; $ kn$ and $ kn\plus{}1$ are always relatively prime
same proof as stevenmeow:

suppose they are not relatively prime
there then exists $ f$, $ a$, and $ b$ (all integers, $ f>1$) such that
$ kn\equal{}af$ and $ kn\plus{}1\equal{}bf$
after subtraction
$ 1\equal{}f(b\minus{}a)
b\minus{}a\equal{}\frac{1}{f}$
but $ f>2$...
contradiction

Do not revive old threads if you are not going to contribute anything valuable to the discussion.


The Archimedes principle states that the upward or buoyant force on an objet within a fluid is equal to the weight of the fluid that the object displaces.

For a partially submerged object, you can obtain information about the relative densities of the floating object and the fluid by observing how much of the object is above and below the surface.  You can also determine the size of a floating object if you know the amount that is above the surface and the relative densities.

You can see the top of a floating iceberg.
The density of ocean water is $ 1.03 \times 10^3 \frac{kg}{m^3}$.
The density of ice is $ .92 \times 10^3 \frac{kg}{m^3}$.

What percent of the total iceberg is below the surface?

[quote]The percentage is 10.68%[/quote]
@castigioni
well the questions asks for the %of iceberg below water,so u have calculated the % above it :).The answer therefore is 100-ur answer.U can also calculate the value by:
(92/103)*100.


trigonometry
inequalities
inequalities unsolved

Let $ a,b$ be positive numbers such that $ a^2\plus{}b^2\equal{}1$ and $ a>b$. Let $ n>m$ be natural numbers. Prove that

$ 2(a^n\minus{}b^n)\leq 3(a^m\minus{}b^m)$


For next time, problems like this are better off in the high school forum.

I'll remember that for next time.

Can you help me find the answer here?

pm a moderator to move your post
people in the high school forum are more likely to be able to answer this.

[quote="sharkman"]Gerald Carter and Marysia Wokawski bought a property by making semi-annual
payments of 2,500 dollars for seven years. If the first payment is due on the date
of purchase and interest is 9 percent compounded quarterly, what is the purchase
price of the property?

Do I use Quarterly = P (1 + r/4)^4 = (quarterly compounding)?

P is the principal (the initial amount you borrow or deposit)

r is the annual rate of interest (percentage)

Am I close?

Here is my work:

P[(1+9%/4)^4]-1
35000[1.0225^4]-1
35000(1.0930833187890625)-1
38256.92

Is this correct?[/quote]

But the purchase price should be Less than the amount paid (because of interest), not more. (amount paid = 35000)

Actually, this type of problem appears in many different forms, so they're actually pretty good to learn to do.  But maybe this should go in the MATHCOUNTS forum or something around that.  It's not exactly all the way at high school level.


prove that in the set {11,111,1111,....,111...1,....} there isn't a number who is a perffect square.

u are right. if k=2n it is imposible that (10^2n)-1 to be perffect squares, but what happened when k=2m+1?

[quote="ckck"]Um.. I don't think proofs are allowed in classroom math.[/quote]

yea.  this is in no way classroom math.  i would say hsb.

if you can go find me a regular middle school textbook that has that in it, wow.

If n=111...1=$x^{2}$, then the last digit of x must be either 1 or 9, so x=10s+1 or x=10s-1. Again, n=111...1=$100s^{2}+20s+1$ or n=111...1=$100s^{2}-20s+1$, hence the sequence 111...1=(n-1)/10 is equal to $10s^{2}+2s$  $10s^{2}-2s$ so it is an even number. Contradiction.

Or just use that every # in that set is 3 (mod 4), and all squares are 0 or 1.


I orginally had a problem but figured it out.

Put it back and let us figure it out too :wink:

Given equations of two non-intersecting circles, show that the graph of the difference of the equations consists of those points $T(x,y)$ for which tangents $\overline{TP}$ and $\overline{TQ}$ are congruent.

$P$ is a point on one of the circles and $Q$ is a point on the other.


inequalities
Vieta
calculus
integration
algebra
polynomial
IMO Shortlist

Ok I think that a problem from the 1999 IMO Shortlist breaks down to proving this .. any hints ?

Let f(x) = x^3 + x + 1 , and u,v,w be its roots . Prove the following inequality :

(u^p - 1)(v^p - 1)(w^p - 1) >= 3 , where p is a prime number which is neither 2 nor 3 .

A nice idea would be to use Vieta's formula, that is uvw=1, bla bla bla
Open the paranthesis and have fun :) trying to prove it. I think it should be quite easy. :)

Maybe you should try it yourself and also keep in mind that two of the numbers u,v are COMPLEX :=) Oh and by the way isn't that uvw = -1 ?:P

The problem can be solve by two step:

1) Show that (u^p-1)(v^p-1)(w^p-1) is an integer divided by 3. It easy, b'se (u-1)(v-1)(p-1)=-3 and (u^p-1)(v^p-1)(w^p-1)/(u-1)(v-1)(p-1) is integral (symmetric polynomial of roots).
2) Seconly, we will show that if p odd, p>3 then (u^p-1)(v^p-1)(w^p-1) > 0. Indeed, polynomial x^3 + x + 1 has one real sln u,  - 1 < u < 0. So, we have to show that     (v^p-1)(w^p-1) < 0. But
(v^p-1)(w^p-1) = (vw)^p - (v^+w^p) + 1 <= (-1/u)^p + (|v|^p+|w|^p) + 1 = (-1/u)^p + 2(-1/u)^(p/2) + 1 < 0
(Here, we use -1/u ~ -1.4!)

Namdung


If $p$ is a prime number and $a_1,...,a_n \in Z$ then:
$(a_1 + ... + a_n )^p \equiv a_1 ^p + ... + a_n ^p (\bmod p)$

In the expansion of $(a_1+a_2+\cdots +a_n)^p$, every term is in the form $\frac{p!}{x_1!x_2!\cdots x_n!}a_1^{x_1}a_2^{x_2}\cdots a_n^{x_n}$ where $x_1,x_2,\cdots ,x_n$ are positive integers with sum $p$.
It is clear that when none of $x_i$ is equal to $p$, $\frac{p!}{x_1!x_2!\cdots x_n!}$ is divisible by $p$ since $p$ is prime. Therefore, $(a_1+a_2+\cdots +a_n)^p\equiv a_1^p+a_2^p+\cdots +a_n^p \pmod{p}$

[quote="pontios"]Another way is

$(a_1 + ... + a_n )^p \equiv a_1 + ... + a_n (\bmod p)$

and $a_i \equiv a_i^p (\bmod p), \ \ \forall i$

because for every integer $a$, we have $a^p \equiv a (\bmod p)$[/quote]

the only problem with that is you do not know that $\sum a_i$ is coprime to $p$, so FLT might not be useable

Are you sure about that, altheman?

Think of what residues $a^{p-1}\equiv 1\bmod p$ is valid for, and think what residues $a^p\equiv a\bmod p$ is valid for.

$a^{p-1}\equiv 1\bmod p$ true if $(a,p)=1$

$a^p\equiv a\bmod p$ true $\forall a \in \mathbb{Z}$


[quote=georg]Prove that :

[tex]\displaystyle \sum_{k=1}^{n} \frac{1}{n-1+y_{k}} \leq 1[/tex] if:

1)all y_{k} are positive (>0)  and 2) [tex] \displaystyle \sum_{k=1}^{n} \frac{1}{y_{k}}=\sum_{k=1}^{n}  y_{k}[/tex][/quote]
Let $ y_1, y_2,\cdots, y_n  >0$  and  $ \displaystyle \sum_{k=1}^{n} \frac{1}{y_{k}}=\sum_{k=1}^{n}  y_{k}.$  Prove that :$$\displaystyle \sum_{k=1}^{n} \frac{1}{n-1+y_{k}} \leq 1$$

@sqing writing that doesnt really help @georg because the last time he came on aops was in 2007.....


trigonometry
function
algebra
domain
linear equation

This is a conceptual question: Take for example the graph of sin x or any graph of a trigonometric function for that matter. In your x-axis, you have your degrees in radians, while in your y-axis, you have just regular real numbers.
I'm not sure how to word this question, but what is the name for this type of graph? It can't really be a cartesian graph. In a cartesian graph, $ \pi$ and other expressions involving it would represent the actual value 3.14159..., right? I mean, I guess you can think of my question as a question of units. In a cartesian graph, you have no units on your x and y axes. When you graph a trigonometric function, your x-axis is in terms of radians. Can it therefore still be considered a cartesian graph? For example, in one trig problem, it says solve for all x, such that $ \frac{x}{100}\equal{}\sin x$. The simplest way to solve this would be to graph the linear equation $ \frac{x}{100}$ alongside the graph of $ \sin x$ and see where the two intersect. But, can you do that. I mean, the x in $ \sin x$ is in terms of radians while the x in $ \frac{x}{100}$ is just the real number x?
I truly apologize for the vague and unusual wording of my question, but this is a concept that doesn't seem to make sense to me. I hope in the jumble of phrases above, my question was understood.  :)

Radians aren't really units (when we're talking about the trigonometric functions defined on the real numbers).  They're a consistent way to interpret the input to sine and cosine.  This is just a normal graph.

Edit:  Let me be even clearer.  The trigonometric functions have domain $ \mathbb{R}$.


inequalities
calculus
derivative
logarithms
function
inequalities proposed

[quote="arqady"]Let $x,$ $y$ and $z$ are non-negative numbers. Prove that
$\left(\frac{x+y+z}{3}\right)^{54}\geq\left(\frac{xy+xz+yz}{3}\right)^{13}\left(\frac{x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}}{3}\right)^{7}.$[/quote]

Thanks for awesome gift! :)

Let $a=\frac{x+y+z}{3}, b=\sqrt{\frac{xy+yz+zx}{3}}, c=\sqrt[3]{xyz}$. Then the inequality is equivalent to $a^{54}\geq b^{26}(3b^{4}-2ac^{3})^{7}$. Now we know that $2ac^{3}\geq-9a^{4}+15a^{2}b^{2}-4b^{4}$ which is equivalent to $\sum_{cyc}x^{4}+\sum_{cyc}x^{2}yz \geq \sum_{sym}x^{3}y$, Schur's. So it's sufficient to prove that $a^{54}\geq b^{26}(9a^{4}-15a^{2}b^{2}+7b^{4})^{7}$. Now divide both sides by $b^{54}$, and let $t=\frac{a^{2}}{b^{2}}$. Since $a \geq b \iff (x+y+z)^{2}\geq 3(xy+yz+zx)$, $t \geq 1$, and it is sufficient to prove that $t^{27}\geq (9t^{2}-15t+7)^{7}$. Now let $s=t-1$, then it is equivalent to prove that $(s+1)^{27}\geq (9s^{2}+3s+1)^{7}$.

I proved the last ineqaulity by differentiation of $27 \ln (s+1)-7 \ln (9s^{2}+3s+1)$, but I couldn'y prove it without using differentiation. In fact, the function has minimum at $\frac{11+\sqrt{43}}{39}$, and at this point $(s+1)^{27}$ is near about 22828 while $(9s^{2}+3s+1)^{7}$ is near about 22096. I tried to find another way but failed. I found that $(s+1)^{4}\geq 9s^{2}+3s+1$, but it seems not so useful.

[quote="Sung-yoon Kim"] Now we know that $2ac^{3}\geq-9a^{4}+15a^{2}b^{2}-4b^{4}$
[/quote]
Yes! I meant it!
[quote="Sung-yoon Kim"]
...and it is sufficient to prove that $t^{27}\geq (9t^{2}-15t+7)^{7}$.[/quote]
I continued so:
$t^{27}\geq (9t^{2}-15t+7)^{7}\Leftrightarrow t^{\frac{27}{7}}-9t^{2}+15t-7\geq0...$ and differentiation.
[quote="Sung-yoon Kim"]
 Now let $s=t-1$, then it is equivalent to prove that $(s+1)^{27}\geq (9s^{2}+3s+1)^{7}$.
I proved the last ineqaulity by differentiation of $27 \ln (s+1)-7 \ln (9s^{2}+3s+1)$, but I couldn'y prove it without using differentiation. In fact, the function has minimum at $\frac{11+\sqrt{43}}{39}$, and at this point $(s+1)^{27}$ is near about 22828 while $(9s^{2}+3s+1)^{7}$ is near about 22096.[/quote]
It is nice!
My congratulations, Sung-yoon Kim! :)

[quote="Vasc"]Actually, the inequality holds
$(xy+yz+zx)^{r}(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}) \leq 3^{r+1}$
for $x+y+z=3$ and $r \geq \frac{ln27-ln16}{ln4-ln3}= 1.8188...$.[/quote]
Certainly,Vasc. I don't know to prove it, but so I have made the mine problem.
And only after that I have solved it (mine problem).
By the way, the Sung-yoon Kim's  method  transforms the following problem:
"Let $x,$ $y$ and $z$ are non-negative numbers such that $x+y+z=3.$
Prove that $\sqrt[3]x+\sqrt[3]y+\sqrt[3]z\geq xy+xz+yz.$ "
in an easy walk. :wink:

[quote="Vasc"]I proved it by EV-Theorem.
On the other hand, the problem 28 from chapter 5 of Algebraic Inequality states the followind:
[i]Let $x,y,z$ be non-negative numbers such that $x+y+z=3$. Then,
$x^{p}+y^{p}+z^{p}\geq xy+yz+zx$
for $p \geq \frac{ln9-ln8}{ln3-ln2}=0.29...$.[/i][/quote]

Yes, it's aslo in my book, chapter II :)


linear algebra
matrix
algorithm
complex numbers

Let $ T$ be a self-adjoin linear operator on an $ n$-dimensional inner product space $ V$, and let $ A=[T]_\beta$, where $ \beta$ is an orthonormal basis for $ V$. Prove that $ T$ is positive semidefinite if and only if $ A=B^{*}B$ for some square matrix $ B$.

I have managed to show that if $ A=B^{*}B$ for some $ B$ then $ T$ is positive semidefinite but an not able to find a proof for the reverse direction.

Just in case you do not know:
positive semidefinite means that $ <T(x),x>$ is nonnegative for all nonzero $ x$, and self-adjoint means that $ T=T^{*}$ and $ A=A^{*}$

What does $ A=[T]_{\beta}$ mean?
By the way, positive semi definite means $ x^{*}Tx$ for at least one nonzero $ x$, and non-negative for all $ x$. What you defined was non negative definite, and it may include positive definite as well.

$ [T]_\beta$ is the matrix representation of $ T$ with respect to an ordered basis $ \beta$. Also I was just using the definitions provided to my by my book.

See the following link:

[url]http://en.wikipedia.org/wiki/Cholesky_decomposition[/url]

On this webpage, they give the proof showing Cholesky can be extended to semi-positive definite matrices.  For a proof of Cholesky itself and a suitable discussion, the usual sources should have it:

-  Matrix Computations  by Golub and Van Loan
-  Numerical Linear Algebra  by Trefethen and Bau
-  Accuracy and Stability of Numerical Algorithms  by Higham

etc.

Note that you do need some algebraic properties of the field here; every positive number must have a square root, at least.

What's being asked here is simpler than Cholesky, since there's no demand that $ B$ have any particular structure. As such, it has a very simple proof.

Let's say up front that we're working over the complex numbers, and that $ A=A^{*}$ is a Hermitian nonnegative matrix. (That is, either positive definite or positive semidefinite.) It is well known that Hermitian matrices can be unitarily diagonalized and that all of their eigenvalues are real. Hence we can write $ A=Q^{*}DQ$ where $ D$ is real, diagonal, and nonnegative. Then we have an unambiguously defined $ D^{\frac12}$ which is also real, diagonal, and nonnegative. Hence $ A=\left(Q^{*}D^{\frac12}\right)\left(D^{\frac12}Q\right)=B^{*}B$ where $ B=D^{\frac12}Q.$