Find $ x,y,z,t \in R$ of this system: $ \begin {cases} (x\plus{}t\plus{}3y\plus{}4z)^2\equal{}27(x^2\plus{}y^2\plus{}z^2\plus{}t^2)\\ x^3\plus{}y^3\plus{}z^3\plus{}t^3 \equal{} 93\end {cases}$

[quote="queen"]hmm. Maybe cauchy can help :wink:[/quote]

Yes! I think so. But can you done this problem by it?

I dont know when does cauchy inequality holds.Do you?

The first equation is the equality case of Cauchy, which means that z/4=y/3=x=t, and that (64+27+1+1)t^3 = 93 and that the only solution is (z,y,x,t)=(4,3,1,1)

t=1, x=4, y=3, z=1 [url=http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality#Statement_of_the_inequality]see here[/url] for when inequality holds

At the Cauchy inequality , when does the equality hold?

[quote="thanhnam2902"]Find $ x,y,z,t \in R$ of this system: $ \begin {cases} (x \plus{} t \plus{} 3y \plus{} 4z)^2 \equal{} 27(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) \\
x^3 \plus{} y^3 \plus{} z^3 \plus{} t^3 \equal{} 93\end {cases}$[/quote]
That's seem verry difficult so...!

[quote="thanhnam2902"]Find $ x,y,z,t \in R$ of this system: $ \begin {cases} (x \plus{} t \plus{} 3y \plus{} 4z)^2 \equal{} 27(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) (1) \\
x^3 \plus{} y^3 \plus{} z^3 \plus{} t^3 \equal{} 93 (2)\end {cases}$[/quote]
we have
$ {\left( 1x \plus{} 1t \plus{} 3y \plus{} 4z\right)}^{2}\leq\left( {1}^{2} \plus{} {1}^{2} \plus{} {3}^{2} \plus{} {4}^{2}\right)\left( {x}^{2} \plus{} {y}^{2} \plus{} {z}^{2} \plus{} {t}^{2}\right) \equal{}$ $ 27\left( {x}^{2} \plus{} {y}^{2} \plus{} {z}^{2} \plus{} {t}^{2}\right)$
$ \left(1 \right)\Rightarrow \frac {x}{1} \equal{} \frac {t}{1} \equal{} \frac {y}{3} \equal{} \frac {z}{4} \equal{} k$
$ \left(2 \right)\Rightarrow 93{k}^{3} \equal{} 93\Rightarrow k \equal{} 1$
So $ x \equal{} 1,t \equal{} 1,y \equal{} 3,z \equal{} 4$


function
limit
real analysis
real analysis unsolved

Let $ f: (0,\infty) \rightarrow [0,\infty)$ be a bounded function. If $ \displaystyle\lim_{x\to 0}(f(x)\minus{}\frac{1}{2}\sqrt{f(\frac{1}{2}x)})\equal{}0$ and $ \displaystyle\lim_{x\to 0}(f(x)\minus{}2f^2(2x))\equal{}0$ then $ \displaystyle\lim_{x\to 0}f(x)\equal{}0$.

Let $ \Phi=\left\{\epsilon; \epsilon>0 \mathrm{\;such, that\; there\; does\; not\; exist\;} \delta>0,\; \mathrm{such \;that} \;f(x)<\epsilon, \;\mathrm{whenever}\; x<\delta \right\}$.

If $ \Phi$ is an empty set, then $ \lim_{x \to \infty}f(x)=0$, according to the definition of a limit.

Otherwise $ \Phi$ is not empty and is bounded with $ M$, where $ M=\sup{|f(x)|}$, which exists, since $ f(x)$ is bounded.

Therefore according to Dedekind axiom, there exists $ \sup{\Phi}$. Let us denote it with $ \epsilon_{0}.$

Now we construct a sequence $ x_{0}>x_{1}>x_{2}>\ldots$, which converges toward $ 0$, with the property, that $ f(x_i)\geq\epsilon_{0}.$

Now we plug $ 2x_i$ in the first condition to obtain $ \lim_{x_i \to 0}( f(2x_i)-\frac{1}{2}\sqrt{f(x_i)})=0.$
Similarly we plux $ x_i$ in the second condition to obtain $ \lim_{x_i \to 0}(f(x_i)-2f^{2}(2x_i))=0.$

$ \lim_{x_i \to 0}(f(x_i)-2f^{2}(2x_i))=0$ means, that for each $ \epsilon>0$, we have to find $ n_0$, such that for all $ i\geq n_0$ we have $ |(f(x_i)-2f^{2}(2x_i)|< \epsilon$. Therefore:
\[ -\epsilon<f(x_i)-2f^{2}(2x_i)<\epsilon,\]
which gives $ \frac{1}{2}(f(x_i)-\epsilon)<f^{2}(2x_i)<\frac{1}{2}(f(x_i)+\epsilon)$  (1)

Repeating the upper procedure on the first condition, we get:
${ \frac{1}{2}(\sqrt{f(x_i)}-\epsilon)<f(2x_i)<\frac{1}{2}( \sqrt{f(x_i)}}+\epsilon)$   (2)

Now we divide (1) by (2) (which we can do for $ \epsilon$ being small enough) and get:

$ \frac{f(x_i)-\epsilon}{\sqrt{f(x_i)-\epsilon}}\leq \frac{f^2(2x_i)}{f(2x_i)}=f(2x_i)<\frac{f(x_i)+\epsilon}{\sqrt{f(x_i)+\epsilon}}$ (3)

We get $ \frac{f(x_i)-\epsilon}{\sqrt{f(x_i)}-\epsilon}<\frac{f(x_i)+\epsilon}{\sqrt{f(x_i)}+\epsilon}$ and after multiplication it becomes: $ 0<2\epsilon(\sqrt{f(x_i)}-f(x_i))$, which means: $ \sqrt{f(x_i)}>f(x_i)$ for all $ i$ great enough. But that further implies: $ f(x_i)<1$ and hence $ \epsilon_0<1$.

Now, we again take a look at the LHS of (3) and send $ \epsilon \to 0$ and $ i \to \infty$ to obtain: $ \sqrt{\epsilon_0}\leq \lim_{i \to \infty}\sqrt{f(x_i)}\leq \lim_{i \to \infty}f(2x_i)$. But since $ \epsilon_0<1$, we get a sequence $ 2x_i>2x_{i+1}>\ldots$, which tends to $ 0$, for which $ f(2x_j)\geq \sqrt{\epsilon_0}$. But since $ \sqrt{\epsilon_0}>\epsilon_0$, that contradicts to the fact, that $ \epsilon_0$ is $ \sup{\Phi}$.

Therefore $ \Phi$ is an empty set and $ \lim_{x \to 0}f(x)=0$.


Let $U_n=$ the set of $n-$ieme roots of unity.Find the cardinal of the set $(z+z',$where $z,z'\in U_n)$.

Let $V_n$ denote $V_n = \{Z_a+Z_b\ |\ Z_a,Z_b\in U_n\}$
$Z_a = e^{i\frac{2a\pi}n},Z_b = e^{i\frac{2b\pi}n} , Z_a+Z_b=2cos(\frac{(b-a)\pi}n)e^{i\frac{(a+b)\pi}n}$
But $0\leq a \leq n-1,a\leq b\leq n-1$, so :

$n\ odd \Longrightarrow card(V_n)=\frac{n(n+1)}2$

$n\ even  \Longrightarrow card(V_n)=\frac{n^2}2+1$

 :cool:


In a trapezoid $ABCD$ , in which $AB$ i sparallel to $CD$ , $E$ is the mid point of $BC$ such that $AECD$ and $ABCD$ have inscribed circles. If $AB$ = $a$, $BC = b$ , $CD = c$ and $DA = d$ . prove that $a+c = \frac{b}{3} + d$ and $\frac{1}{a} + \frac{1}{c} = \frac{3}{d}$

After correcting the typo, this becomes http://www.mathlinks.ro/Forum/viewtopic.php?t=33041 .

  darij


geometry
complex numbers
geometry proposed

Assume that the circle with center $I$ touches the sides $BC,\ CA,\ AB$  in
 $D,\ E,\ F$ respectively .The lines AI and EF meet at K . Assume that the lines $ED$ and $KC$ meet at $L$ and the lines $DF$ and $KB$ at $M$ . Prove that $A$ belongs to $LM$ and $LM//EF$

I think there are some small typos in your problem.

The correct enuntiation should be: Let $ABC$ be a triangle and $w(I)$ its incircle for which $D$, $E$, $F$ are the points of tangenty with $BC$, $CA$, $AB$. Denote $K$ the midpoint of $(EF)$ and $X \in DF \cap KC$, $X \in DE \cap KB$. Prove that $A$, $L$, $M$ are collinear and $LM \| EF$.

(The thing I've slightly edited is that $X \in DF \cap KC$, i.e $Y \in DE \cap KB$, but maybe you sketched the points $B$ and $C$ in the opposite way)

[i]Proof:[/i]

Denote $T \in DE \cap AB$.

Because $AD$, $BE$, $CF$ are concurrent in the Gergonne point, the quadrupoint $(TAFB)$ is harmonic, i.e $(TAFB)=-1$.

Thus, $E(TAFB)=-1$, so by intersecting the pencil with $KB$, we obtain $(YGKB)=-1$, where $G \in AC \cap KB$.

Therefore $A(YGKB)=-1$ and now by intersecting it with $EF$, and denoting $R \in AY \cap EF$, we obtain that $(REKF)=-1$, and because $EK=EF$, $R =\infty$, i.e $AY \| EF$.

Similar $AX \| EF$, thus $A \in XY$ and $XY \| EF$.

Thank you for interesting nice proof .  :)
I have a proof with complex numbers and one synthetic one but a bit more complicated .

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/03.%20Parallele%20a%20un%20cote%20du%20triangle%20de%20contact%20passant%20par%20un%20sommet%20du%20triangle.pdf

p. 10-13.

Sincerely
Jean-Louis


geometry
perimeter
inequalities
geometric series

A lot of these APMO problems are surprisingly easy.  Here's a few more that you guys shouldn't have too much difficulty solving.

1.  Find all non-empty finite sets $S$ of positive integers such that if $m,n\in S$, then $\frac {m+n}{\gcd(m,n)}\in S$

2.  $\sum^8_{i=0} a_ix^i$ has $a_8=1, a_7=-4, a_6=7$ and all positive real roots.  Find $a_0$.

3.  A unit square lies across two parallel lines a unit distance apart, so that two triangular areas of the square lie outside the lines. Show that the sum of the perimeters of these two triangles is independent of how the square is placed.

[hide]Hint: Abuse the WLOG!.  Actually, it's easy enough that you don't need a hint.[/hide]

4.  If $n$ is a positive integer, let $d$ be the number of digits in $n$ (in base 10) and $s$ be the sum of the digits. Let $n(k)$ be the number formed by deleting the last $k$ digits of $n$. Prove that $n = s + 9 n(1) + 9 n(2) + \ldots + 9 n(d)$

5.  Positive reals $a,b,c$ satisfy $\frac {1}{a}+\frac {1}{b}+\frac {1}{c}=1$.  Prove that

$\sqrt {c+ab}+\sqrt {b+ac}+\sqrt {a+bc}\geq \sqrt {abc}+\sqrt{a}+\sqrt{b}+\sqrt{c}$

Of course, all of these problems and their solutions are on kalva.  But try and find your own solutions.

[hide="Hint for 5"]Square both sides. Then $ab + ac + bc = abc$ and $b^2 + c^2 \ge 2bc$.  Add a WLOG into the mix and you might just end up with an answer.[/hide]
[hide="Hint for 4"]It's a telescoping sum.  Sort of.[/hide]

BTW this is giving us a false sense of confidence about olympiads... at least they're fun problems ;)

#1

[hide]First, by taking $m=n$, we get $2\in S$.  Now, if $m$ is the largest odd element of $S$, and $n=2$, then we have $m+2\in S$, so contradiction.  Thus, all elements in $S$ are even.  Finally, if $m$ is the smallest even element of $S$ greater than $2$, then $\frac{m}{2}+1 \in S$ and $\frac{m}{2}+1 < m$, so we obtain another contradiction.  Thus $S = \{2\}$.[/hide]

A slightly different method for #5

[hide]Make the subsitutions $x=\frac{1}{a}$, etc, so $x+y+z=1$.  The inequality is equivalent to:

$\sum_{\text{cyc}} \sqrt{xy+z} \geq 1 + \sum_{\text{cyc}} \sqrt{xy}$, or

$\sum_{\text{cyc}} \sqrt{(x+y)(x+z)} \geq x+y+z + \sum_{\text{cyc}} \sqrt{xy}$, but this follows from Cauchy:

$(x+y)(x+z) \geq (x+\sqrt{yz})^2$, so square-rooting and adding this with two similar inequalities yields the desired result.[/hide]

And for 2:

[hide]Let the roots be $s_1,s_2,...,s_8$. Now we have that $s_1+s_2+...+s_8=4$ and $s_1s_2+...+s_7s_8=7$. So square the first equation and subtract twice the second. We get:

$s_1^2 + s_2^2 + ... + s_8^2=2$

Now multiply this by 7 and subtract twice the second equation above. We then factor and get

$(s_1-s_2)^2 + (s_1-s_3)^2 + ... + (s_7-s_8)^2 = 0$.

Thus, $s_1=s_2=...=s_8=\frac{1}{2}$. Then $a_8=s_1s_2...s_8=\frac{1}{256}$.[/hide]

Anyone want to come up with a solution to 3 that doesn't involve analytic geometry?

[hide="4."]

Let [tex]n=10^da_d+10^{d-1}a_{d-1}+...+10a_1+a_0[/tex] where [tex]a_i[/tex] is the digit in the [tex]10^i[/tex]th place.

Then [tex]s=a_d+a_{d-1}+...+a_1+a_0[/tex].

We can also write [tex]n(k) = 10^{d-k}a_{d}+10^{d-k-1}+...+10a_{k+1}+a_k[/tex].

Now, we wish to prove that if you sum up all the terms in [tex]\displaystyle n = s + 9 n(1) + 9 n(2) + ... + 9 n(d) [/tex] containing [tex]a_j[/tex] for any [tex]j[/tex] you will get [tex]10^ja_j[/tex]. We note that [tex]a_j[/tex] appears in terms where [tex]k \le j[/tex]. So if we take the term that has [tex]a_j[/tex] from the first [tex]j[/tex] terms of the sum and the [tex]s[/tex] term, we will get
[tex] 9(10^{j-1}a_j+10^{j-2}a_j+...+10a_j+a_j)+a_j[/tex],
which, by the sum of a finite geometric series, is
[tex] \displaystyle 9\left(\frac{10^j-1}{9}\right)a_j+a_j = 10^ja_j-a_j+a_j = 10^ja_j [/tex].

Therefore, [tex]\displaystyle n = s + 9 n(1) + 9 n(2) + ... + 9 n(d) [/tex].

QED.[/hide]


Um ... are you sure you want that x in there?  You definitely need either to replace the x with an n or to put in more information.

[hide]

n :ge: 9 from the problem.

25n + 8n + 3n can be rewritten as

2n +5 + 8 + 3 = x^2

2n + 16 = x^2

All perfect squares are 0 mod 4 or 1 mod 4, but 1 mod 4 is irrelevant. Since all even n will satisfy that statement, and there are an infinite number of perfect squares 0 mod 4, we conclude that there are infinite n. I'm not sure if you want a general pattern though...

[/hide]

Be careful there [hide]-- just because the result has to be congruent to 0 mod 4 doesn't mean that every number congruent to 0 mod 4 is a perfect square, or that because infinitely many n make it congruent to 0 mod 4 infinitely many n work.  (Although it does turn out that you are correct, there are infinitely many values.)[/hide]

We could say this n>8 and thus we know that x^2=3+8+5+20basen, this means 16+2*n=x^2, infinitly many solutions for n more than 9, for eample, 10 works, then next even square which will be at n=24, then at n=42, then n=64, then n=90, each time we get the last n+14 then 18, then 22 then 26 etc. so we have n is a set of (10+k*2) where the mth n will happen when k=(12+ 2(m-2)(m-1)).

I don't understand the flaw in my logic.

[hide]

If the general result is 16+2n, then for n=9,10,11,..., the result will be 34, 36, 38, 40, 42, 44, ...

This series includes all numbers 0 mod 4 that are greater than or equal to 36. Since there are an infinite number of perfect squares 0 mod 4 that are greater than or equal to 36, there are infinite solutions.

^Is there anything wrong with that? Or if there isn't, how is it different from what I posted earlier?

[/hide]

Thanks.

rlee wrote:[hide]This series includes all numbers 0 mod 4 that are greater than or equal to 36. Since there are an infinite number of perfect squares 0 mod 4 that are greater than or equal to 36, there are infinite solutions.[/hide]

This was the part that you didn't say.  I see now how you were trying to say it, but it wasn't clear to me.  I like your proof very much now, although the one downside to it is that it doesn't give the actual solutions.  But it's a very nice existence proof.


:mad:
actually, i participated in IOI 2004, but even now i don't know how to solve empodia problem.
you can see the problem at :: http://www.ioi2004.org/

the released solution explains about O(N^2) algorithm...
and it refers there is O(N) algorithm.

if you know O(N) algorithm or so, plz comment on!

huh, it's too difficult for me...

Your problem is interesting.
I have found an O(n) solution but I am not really sure it's correct.

Let's assume that the biosequence is a[n]
Firstly, i compute two additional array :

left[i] is the biggest index that is smaller than i and a[left[i]] > a[i]
It takes O(n) to compute left

right[i] is the smallest index that is bigger than i and a[right[i]] < a[i]
It also takes O(n) to compute right

Then i sort right using an O(n) sorting algo.

Finally, i iterate from 1 to m, mantaining linked lists of numbers having the same value - index difference(ie. a[i]-i). The number a[i] is inserted into the lists when i process to the index i and is removed when i process to the index right[i].
With these lists, it is very easy to obtain all the empodia.

Here is the most important part of my pascal code. I think explaining the code is more complicated than the code itself but i tried. ;)

[code]  {compute left}
  a[0]:=n;
  for i:=1 to n do
  begin
    j:=i-1;
    while a[i]>a[j] do j:=left[j];
    left[i]:=j;
  end;
  {compute right}
  a[n+1]:=-1;
  for i:=n downto 1 do
  begin
    j:=i+1;
    while a[i]<a[j] do j:=right[j];
    right[i]:=j;
  end;
  {sort right}
  fillchar(count,sizeof(count),0);
  for i:=1 to n do
    inc(count[right[i]]);
  for i:=2 to n+1 do
    inc(count[i],count[i-1]);
  for i:=1 to n do
  begin
    index[count[right[i]]]:=i;
    dec(count[right[i]]);
  end;
  {find the shortest framed interval for every right end point}
  max:=0;{max is the right end point of the latest empodio}
  j:=1;
  for i:=1 to n do
  begin
    if (last[a[i]-i] <> 0)and(last[a[i]-i]>left[i]) then {we have found a framed interval from last[a[i]-i] to i}
      if max <= last[a[i]-i] then {we have found a empodio}
      begin
        inc(h);
        empodio[i]:=last[a[i]-i];
        max:=i;
      end;
    {update last, prev, next}
    prev[i]:=last[a[i]-i];
    if prev[i] <> 0 then next[prev[i]]:=i;
    last[a[i]-i]:=i;
    while (j <= n)and(right[index[j]] = i) do
    begin
      j1:=index[j];
      last[a[j1]-j1]:=prev[j1];
      inc(j);
    end;
  end;[/code]
btw, do you know where i can get the problems in korean informatics olympiad. English version is best but chinese version is great, too.

actually, there is no web-sites or books that officially introduce Korean Informatics problems in English or any other languages....

there are only Korean Version problems..
:(  So if you want to find and solve Korean Inf. problems, you should know Korean..

i think it's too bad, but it is the case now.

and, i'm trying to understand your solution now..
i hope your solution is correct!
 :lol:

That looks similar to my solution (which I didn't find in the competition unfortunately but on the plane home).  I certainly used the same left and right arrays, and the value (sequence[i] - i) was very important.  I don't remember the details, but it was based around a stack.  I did once find one (and only one) website with google that contained this solution and explained it quite well.

--
Martin

I think the website you're referring to is the South African IOI forum. The page is located @ [url]http://olympiad.cs.uct.ac.za/pipermail/squad2004/2004-September.txt[/url].

The two last messages there are about the problem Empodia.


Let $a_n$ be  a sequence,  $a_1=1$, for  $n\geq 1$
$a_{n+1}=a_n+(1/a_n)$
Find the $[a_{1000}]$
[ ] denotes greatest integer function

Hint: Square it, you'll get $a_n^2=2.1000+a_1^2$...

Could you explain that a little more please, Lovasz?

Is there any closed form (i.e. non-recursive) definition for a_n?


superior algebra
superior algebra unsolved

here was a wrong problem. sorry!    :ninja:

I don't think it's true as stated here:
Take for example $G= \{ e \}$.

And setting $y=e$ gives $x^{p+r-q-1} = e$ and similar $y^{p+r-q+2}=e$ which gives $x^3=e$ for all $x$, so all nontrivial elements have order $3$ and $n$ is a power of $3$.

ok. i think u're right , my mistake in calculations. i'll delete it


geometry
incenter
circumcircle
perpendicular bisector

[quote="sam-n"]if the orthocenter H , the incenter I , and the circumcenter O of a traingle ABC  ,are the vertices of an equilateral traingle ,show that cosA+cosB+cosC=3/2[/quote]

 :)

The orthocenter H, circumcenter O and the incenter I of a triangle $\triangle ABC$ cannot form an equilateral triangle. The triangle $\triangle IOH$ is isosceles with IO = IH iff one of the the angles of the triangle $\triangle ABC$ is $60^\circ,$ say the angle $\angle A$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24538]www.mathlinks.ro/Forum/viewtopic.php?t=24538[/url]). Then $\angle BOC = \angle BHC = \angle BIC = 120^\circ$, so that the circumcenter of the triangle $\triangle IOH$ is the midpoint M of the minor arc BC of the circle (O) and the circumradii R = OM of the triangles $\triangle ABC, \triangle IOH$ are equal. Moreover, the orthocenter H and the incenter I always lie on the same side of the perpendicular bisector of BC passing through the circumcenter O, otherwise the triangle $\triangle IOH$ could not be isosceles with IO = IH. When $A \longrightarrow B$, OH is increasing and $OH \longrightarrow R \sqrt 3 = BC$, hence the angle $\angle OIH$ is decreasing and $\angle OIH > 120^\circ$.


conics
ellipse
geometry unsolved
geometry

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14698[/img]

[size=150][b]RULER  ONLY[/b][/size]

 :(

draw BT_2 and AT_1, call their intersection P, then draw $ T_3\in AB\cap CP$. (prove by carnot's theorem, its the same as that other problem basically)

[b]i need T4 and T5 to have 5 points of ellipse and could draw the ellipse[/b]

 :(


I didn't really get this question but I know it's not really hard question but I'm just posting the question as following what tokenadult asked.  I would not add any more comments but I'll keep adding problems that I sometimes understand or sometimes I don't understand. Also, people who want some challenges problems may visit topic Problem Solving Marathon which I posted to see more than easy questions.  This is question that's kinda easy but requires some understanding.

A plane flies 640 miles with the wind and 520 miles against the wind during the same time. Suppose that the speed of the wind was 15 miles per hour. Find the speed of the plane when the speed of the wind is zero.

I hope this is what you wanted tokenadult.

[hide](640 mi)(1 h / p+w mi) = (540 mi)( 1 h / p - w mi)

b/c time is equal

w=15

640 / p + 15 = 540 / p - 15

p=177

The v of plane = 177 mph.

[/hide]

If you need more explanation ask.  Apologies if this is wrong!

[hide]

(640 mi)(1 h / (p+w) mi) = (520 mi)( 1 h / (p - w) mi)

b/c time is equal

w=15

640 / (p + 15) = 520 / (p - 15)

p=145

The v of plane = 145 mph.

[/hide]

probably wrong.

Edited: Added parentheses.  Silver, was this a challenge or did you want an explanation.  I can give a better explanation if you need.

JS, parentheses are a wonderful thing and make math much more readable -- in standard notation, 640 / p - 15 = (640/p) - 15, not 640/(p - 15).

Sure thing.

A plane flies 640 miles with the wind and 520 miles against the wind during the same time. Suppose that the speed of the wind was 15 miles per hour. Find the speed of the plane when the speed of the wind is zero.

First thing we want to do is find the RATE of the plane when it is flying w/the wind and when it is flying against the wind.  Let p= the speed of the plane w/o anything going on (when speed of the wind=0).

Rate: p+15 mi/h  when plane and wind are working together
Rate: p - 15 mi/h when the plane and wind are working against each other

Now they both go a certain distance in the same time.  Since time for both is equal, you want to find time for both (in terms of p) and then solve for p.

(640 mi)(1 h / (p+15) mi) = 640/(p+15) hours

(520 mi)( 1 h / (p - 15) mi) = 520/(p-15) hours

b/c time is equal:

640 / (p + 15) = 520 / (p - 15)
Solve by cross multiplication.
p=145

So the speed of the plane when the wind=0 is 145 mi/h.

Ask if my explanation is confusing.


geometry
incenter
inradius
symmetry
geometry proposed

\[GI \perp BC \Leftrightarrow\]
\[GB^2 + IC^2 = GC^2 + IB^2 \Leftrightarrow\]
\[\frac{2a^2+2c^2-b^2}{9}+r^2+\left(\frac{a+b-c}{2}\right)^2 = \frac{2a^2+2b^2-c^2}{9}+r^2+\left(\frac{a+c-b}{2}\right)^2 \Leftrightarrow\]
\[12(b-c)(b+c-3a)=0.\]
Therefore $GI \perp BC$ if and only if $b=c$ or $b+c=3a$.

[quote="Valentin Vornicu"]Let $ABC$ be a triangle inscribed in a circle of center $O$ and radius $R$. Let $I$ be the incenter of $ABC$, and let $r$ be the inradius of the same triangle, $O\neq I$, and let $G$ be its centroid. Prove that $IG\perp BC$ if and only if $b=c$ or $b+c=3a$.[/quote]

Hmm, do you really need O and r here? (Actually, r is never used, and $O\neq I$ just means that the triangle ABC is not equilateral.) I would rewrite the problem as follows:

[color=blue]Let ABC be a non-equilateral triangle with centroid G and incenter I. Prove that $IG\perp BC$ if and only if b = c or b + c = 3a.[/color]

And, as usual, there is a synthetic solution:

At first, it is clear that if b = c, then $IG\perp BC$, since the line IG is then the symmetry axis of the isosceles triangle ABC and thus is perpendicular to its base BC. Hence, in the following, we will only consider the case $b\neq c$. Then, we have to show that $IG\perp BC$ if and only if b + c = 3a.

It is well-known that the incenter I, the centroid G and the Nagel point N of the triangle ABC are always collinear. Thus, the line IG coincides with the line IN. Hence, we have the following chain of equivalences:

{We have $IG\perp BC$.}
$\Longleftrightarrow$ {We have $IN\perp BC$.}
$\Longleftrightarrow$ {The point N lies on the perpendicular to the line BC through the point I.}

Now, the Nagel point N of triangle ABC is defined as the point of intersection of the lines AA', BB', CC', where A', B', C' are the points where the A-excircle, B-excircle, C-excircle of triangle ABC touch the sides BC, CA, AB, respectively. Let the incircle of triangle ABC touch the sides BC, CA, AB at the points D, E, F, respectively, and let D', E', F' be the points diametrically opposite to the points D, E, F on this incircle. Then, after Theorem 1 of post #6 in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=22461 , the points D', E', F' lie on the lines AA', BB', CC', respectively, and Theorem 4 of this post yields, with directed segments, that $\frac{NB^{\prime}}{E^{\prime}B^{\prime}}=\frac{s-b}{b}$, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Similarly, we can find $\frac{NA^{\prime}}{D^{\prime}A^{\prime}}=\frac{s-a}{a}$.

Since I is the center of the incircle of triangle ABC, and D is the point where this incircle touches the side BC, we have $ID\perp BC$. Thus, the line ID is the perpendicular to the line BC through the point I. Hence, we can continue our chain of equivalences:

{We have $IG\perp BC$.}
$\Longleftrightarrow$ {The point N lies on the perpendicular to the line BC through the point I.}
$\Longleftrightarrow$ {The point N lies on the line ID.}

Now, the point D' is the point diametrically opposite to the point D on the incircle of triangle ABC; hence, the segment DD' is a diameter of the incircle and thus passes through its center I. In other words, the line ID passes through the point D'. On the other hand, we know that the line AA' passes through the point D'. Hence, the point D' is a common point of the lines AA' and ID. Since we have $b\neq c$, the lines AA' and ID don't coincide (else, the points A, A', I and D would lie on one line, and since $ID\perp BC$, this line would be perpendicular to the line BC, so that we would have $AI\perp BC$, but this readily leads to b = c, contradicting with $b\neq c$). Hence, the point D' is the unique common point of the lines AA' and ID.

On the other hand, we know that the point N lies on the line AA'. Hence, the point N lies on the line ID if and only if the point N is the unique common point of the lines AA' and ID. Since we know that the point D' is the unique common point of the lines AA' and ID, we thus conclude: The point N lies on the line ID if and only if the point N coincides with the point D'.

So we can continue our chain of equivalences again:

{We have $IG\perp BC$.}
$\Longleftrightarrow$ {The point N lies on the line ID.}
$\Longleftrightarrow$ {The point N coincides with the point D'.}

But clearly, the point N coincides with the point D' if and only if NA' = D'A'; since $\frac{NA^{\prime}}{D^{\prime}A^{\prime}}=\frac{s-a}{a}$, we have NA' = D'A' if and only if s - a = a. Thus we have:

{We have $IG\perp BC$.}
$\Longleftrightarrow$ {The point N coincides with the point D'.}
$\Longleftrightarrow$ {We have NA' = D'A'.}
$\Longleftrightarrow$ {We have s - a = a.}
$\Longleftrightarrow$ {We have s = 2a.}
$\Longleftrightarrow$ {We have 2s = 4a.}
$\Longleftrightarrow$ {We have a + b + c = 4a.}    (since 2s = a + b + c)
$\Longleftrightarrow$ {We have b + c = 3a.}

This completes the solution.

Sorry for this clumsy writeup of my solution, I am quite under time pressure and just wanted to complete it as quickly as possible.

  Darij

We can observe that this other proposition is worth:

[color=blue]Let $\triangle ABC$ be a non-equilateral triangle with centroid $G$ and incenter $I$. Prove that $GI\parallel BC$ if and only if $b + c = 2a$. [/color]

In fact, denoting with $M, r, p, \Delta$ the midpoint of $BC$, the inradius, the semiperimeter, the area of $\triangle ABC$, since  $GM = \frac{1}{3}AM$, we have:

$ GI\parallel BC \qquad \Leftrightarrow$

$ r = \frac{1}{3}h_a  \qquad \Leftrightarrow$

$ 3 \cdot \frac{\Delta }{p} = \frac{{2\Delta }}{a} \qquad \Leftrightarrow $

$ 3a = 2p \qquad \Leftrightarrow $

$ 2a = b + c \qquad \blacksquare$

I invite the other mathlinkers to find one synthetic solution !

[quote=Valentin Vornicu]Let $ABC$ be a triangle inscribed in a circle of center $O$ and radius $R$. Let $I$ be the incenter of $ABC$, and let $r$ be the inradius of the same triangle, $O\neq I$, and let $G$ be its centroid. Prove that $IG\perp BC$ if and only if $b=c$ or $b+c=3a$.[/quote]
$b=c$ case is obvious so assume $b$ and $c$ are not equal.

Consider the circumcircle of $ABC$ and let D be the midpoint of arc $BC$ not containing $A$.We have $A,I,D$ collinear.Let $M$ be the midpoint of side $BC$.So $DM\perp BC$.Note that $A,G,M$ collinear.
Now $IG\perp BC\qquad \Leftrightarrow$
$ GI\parallel DM \qquad \Leftrightarrow$
$\frac{DI}{DA}=\frac{MG}{MA}=\frac{1}{3}$

Now as $DI=DB=DC$ so by Ptolemy's theorem on $ABDC$
 $bDB+cDC=aAD$
$DI(b+c)=aDA$
$\frac{1}{3}=\frac{DI}{DA}=\frac{a}{b+c}$.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/AB%20+%20AC%20=%203.BC.pdf   p. 13-14.

Sincerely
Jean-Louis


1. Let $ f(x)\equal{}a_0\plus{}a_1x\plus{}....\plus{}a_nx^{n}$ where $ a_i \in Z$. If $ f(200)f(7)\equal{}2007$
then show that $ f(k)\neq200$ for any $ k \in Z$

2.Let $ f(x)$ be a polynomial with inetger coefficients and suppose that for 5 distinct integers $ a_1,a_2,....a_5$, we have $ f(a_i)\equal{}2000$. Show that there does not exist an integer $ b$ such that $ f(b)\equal{}2007$

[hide="Problem 1"]From $ f(200)f(7) \equal{} 2007$ and that $ f(200)$ and $ f(7)$ are integers, $ f(200)$ and $ f(7)$ are odd. Then $ f(200) \equal{} a_0 \plus{} 200a_1 \plus{} \ldots \plus{} 200^na_n$, so $ a_0$ is odd. Then $ 7a_1 \plus{} \ldots \plus{} 7^na_n$ is even. Then in $ \{a_1,\ldots,a_n\}$ there are an even number of odd $ a_i$.

If $ k$ is even, then $ a_1k \plus{} \ldots \plus{} a_nk^n$ is even and $ f(k)$ is odd.

If $ k$ is odd, then $ a_1k \plus{} \ldots \plus{} a_nk^n$ is even because in $ \{a_1,\ldots,a_n\}$ there are an even number of odd $ a_i$, and $ f(k)$ is odd again (because $ a_0$ is odd). [/hide]

[hide="Problem 2"]$ f(x)$ is at least of degree 5. Let $ f(x) \minus{} 2000 \equal{} g(x)$. Then $ g(x)$ has at least 5 roots, 5 of which are $ a_1, a_2,\ldots, a_5$. Suppose there is a $ b$ such that $ f(b) \equal{} 2007$. Then $ g(b) \equal{} 7$, and writing $ g(x) \equal{} a(x \minus{} a_1)(x \minus{} a_2)(x \minus{} a_3)(x \minus{} a_4)(x \minus{} a_5)\cdots$ where $ a$ is constant, then $ g(b)$ is a product of at least 5 integers, but $ g(b) \equal{} 7$ only has 4 integer factors, and $ (x \minus{} a_i)$ for $ i \equal{} 1,2,3,4,5$ are distinct. [/hide]


[url]http://ipho2008.hnue.edu.vn/Competition/Results/tabid/82/Default.aspx[/url]

Cestitke svima, a posebno Admiru na najvecem uspjehu BiH na IPhO svih vremena (srebro) !!!

 :winner_first:

I ja čestitam Admiru

Harune imam jednu veliku dilemu: Na koliko će dnevnika sutra Rajfa biti udarna ličnost??? 7 ili 8??  :D  :D  :D

Zanimljivo je to što je naša Nina Kamčev uzela medalju na fizici, a nije na matematici, iako je u matematici puuuno bolja i više radi. Govori li to što o slaboj konkurenciji na fizici? Pitam, jer nemam iskustva. Nikad se nisam natjecao iz fizike.

Prof. Dr. Musemic ce na ovom uspjehu zivjeti 10 godina. A bice gost i u Teleringu, i na 60 minuta, i u Mimohodu, mozda cak i u nekom kuharskom showu.

A sto se djurinog pitanja tice, mogu reci nesto iz vlastitog iskustva. 4 godine sam se takmicio iz matematike i fizike. Matematika je bila primarna, a fizika usput (sto bi rezultiralo trecim, cetvrtim mjestima na federalnim takimcenjima), a u cetvrtom razredu sam izabrao da se takmicim u D kategoriji (cjelokupna fizika), sto mi je omogucilo da se plasiram na drzavnu BHPhO-gdje sam sa minimalnim pripremama osvojio peto mjesto i upao u IPhO ekipu (jest da nisam otisao, ali to je vec druga prica). Mozda se ovo da dalje generalizirati na IPhO nivo?

My two cents.

Ma kod nas postoji u dijelu javnosti nekakvo "usađeno" mišljenje da je mnogo teže osvojiti medalju na fizici nego na matematici ali mislim da je to rezultat samo propagande društva fizičara tipa "mi nemamo sredstava za pripreme ne možemo do medalja" nego stanja u realnosti...

Mozda bi ispravnije bilo reci: mi ne znamo spremiti ekipu (bez uvrede profesoricama i profesorima, ali... )

BTW, Sa-x je vec objavio vijest. Mora se priznati da su brzi.

Zapanjen sam.... Rajfa nije ličnost dana u današnjem Avazu  :rotfl:  :rotfl:  morat ćemo se strpiti još koji dan  :D

Steta, Srbi su uzeli samo jednu bronzu... :(
Cestitam svim ekipama! :coolspeak:

Cestitke svima, posebno Nini, Petru i Veroniki.

Svaka čast Admiru Grelji
sa zakašnjenjem :D


I have two questions :

1.  is it true that in CW complex decomposition of X has no 1-cell then $ X$ is simply connected?
2. Let $ X$ be a manifold is it  $ H_n(X \# X) \equal{} H(X)\plus{}H(X)$ ?
thanks

1. yes, just apply van kampen (and induction).

2. yes, for every suitable dimension... it works in the general case of two manifolds $ M,N$ of the same dimension $ n$: just apply mayer-vietoris long exact sequence to obtain
$ H_k(S^{n\minus{}1})\equal{}0\to H_k(M)\oplus H_k(N) \to H_k(M\# N) \to 0\equal{}H_{k\minus{}1}(S^{n\minus{}1})$.
of course the trick doesn't work anymore (as expected) for $ k\equal{}0,n$, and, slightly more unexpectedly, for $ k\equal{}1,n\minus{}1$: i'll let you figure out how to proceed in these four cases...


algebra
polynomial
number theory theorems
number theory

Suppose we have numbers $ (a_{i}, b_{i})$ for $ i\equal{}1,2 ... n$. What is a necessary condition for the existence of a polynomial $ P$ to exist, with integer coefficients, such that $ P(a_{i}) \equal{} b_{i}$ for $ i\equal{}1,2 ... n$?
Is it this: $ a_{i} \minus{} a_{j} | b_{i} \minus{} b_{j}$, for all pairs $ (i,j)$. Clearly it is necessary, but is sufficient?.
I thought at first looking at this, Lagrange Interpolation Formula may be useful, but I couldn't see how it may kill this problem.

Try writing $ P(x) \equal{} c_0 \plus{} c_1 (x \minus{} a_1) \plus{} c_2 (x \minus{} a_1)(x \minus{} a_2) \plus{} c_3 (x \minus{} a_1)(x \minus{} a_2)(x \minus{} a_3) \plus{} ...$.  :)


inequalities
analytic geometry
linear algebra
matrix
geometric inequality
linear algebra unsolved

$A=(a(i,j)) \in M(n,R)$ such that

for any $1\leq i,j \leq n$ , $0\leq a(i,j)<1$

for any $1\leq i \leq n$ , $\sum_{j=1}^{n}a(i,j)\leq 1$

Prove that $|det(A)|\leq 1$

$\forall 1 \leq i \leq n, n (\prod_{j=1}^{n}{a_{i,j}})^{\frac{1}{n}} \leq \sum_{j=1}^{n}a_{i,j} \leq 1$ by the arithmetic-geometric inequality (which applies since the $a_{i,j} \geq 0$). Thus $\prod_{j=1}^{n}{a_{i,j}} \leq \frac{1}{n^{n}}$, and $|det(A)| \leq \prod_{i=1}^{n} (\sum_{j=1}^{n} a_{i,j}) \leq 1$.

That seems a bit obscure.

How about this: Let $v$ be a (complex) eigenvector $(x_1,x_2,\dots,x_n)^T$ with eigenvalue $\lambda$ and suppose $|x_m|\ge |x_j|$ for all $j$. Looking at the $m$th coordinate, $|\lambda x_m|=|\sum_j a_{mj}x_j|\le \sum_j |a_{mj}||x_j|\le\sum_j a_{mj}|x_m|=|x_m|\sum_j a_{mj}\le |x_m|$, and $|\lambda|\le 1$.
All eigenvalues are less than or equal to 1 in magnitude, and the matrix has determinant at most 1 in magnitude.


Let $a_{k},k=1,...,n$ be distinct natural numbers. Prove the inequality $\sum_{k=1}^{n}\frac{a_{k}}{k^{2}}\geq \sum_{k=1}^{n}\frac{1}{k}$


Some other nice pigeonholes. The first one :

Prove that for each positive integer n there exists a number of the form

111...111000...000

which is divisible by n.

[hide]Two of the (n+1) numbers

1, 11, ..., 1111...111 (with (n+1) '1').

must be equal modulo n. It means that their difference is divisible by n.

1111..111 (p '1') - 1111...111 (q '1') =  11.. 11000..000 (p - q '1', q '0')).  :mrgreen:

This one is not too difficult  [/hide]

I know, bit I like to start with some trivial problems ...

The fourth one is much harder.

The third one can be offputting if you don't look at it the right way, but if you realize its simplicity then it's quite trivial. Probably you already know a solution to this one ... I posted something similar before I think.

The second one is quite "intermediate".

hmm...they do get harder and harder...lol


inequalities
algebra
polynomial
inequalities proposed

Let $ a_1,a_{2},\ldots ,a_{100}$ be nonnegative integers satisfying the inequality
\[a_1\cdot (a_1-1)\cdot\ldots\cdot (a_1-20)+a_2\cdot (a_2-1)\cdot\ldots\cdot (a_2-20)+\\ \ldots+a_{100}\cdot (a_{100}-1)\cdot\ldots\cdot (a_{100}-20)\le 100\cdot 99\cdot 98\cdot\ldots\cdot 79.\]
Prove that $a_1+a_2+\ldots+a_{100}\le 9900$.

Let $ p(x)=x(x-1)\dots(x-20)$. Then we have to prove that if $ p(a_1)+p(a_2)+\dots+p(a_20) \leq 100.99.\dots.79$ then $ a_1+a_2+\dots a_20 \leq 9900$.

Now $ p(x)=0 \forall x=0,1,\dots,20$, and if $ a \geq b \geq 20$ then since $ a-i \geq b-i \geq 0 \forall 0 \leq i \leq 20$, $ p(a) \geq p(b) \geq 0 = p(20)=p(19)=\dots=p(0)$, and $ a(a-1)\dots(a-19) \geq b(b-1)\dots(b-19)$. Therefore $ p(x)$ is increasing on the nonnegative integers.

Also, if $ a>b$ are nonnegative integers, then $ p(a)+p(b)\geq p(a-1)+p(b+1)$
$ \iff (a-1)(a-2)\dots(a-20)[a-(a-21)] \geq b(b-1)\dots[b+1-(b-20)] %Error. "doubleleftarrow" is a bad command.
a-1 \geq b$ is true since $ a>b$. Hence we can always take the most and least of the $ a_i$ and move them closer together while keeping the sum of the $ a_i$ constant.
We continue doing this until the $ a_i$ consist of exactly two consecutive nonnegative integers: $ a$ with multiplicity $ x$ and $ a+1$ with multiplicity $ 20-x$. Then we have $ xp(a) +(20-x)p(a+1) \leq 100.99.\dots.79$.

[quote="enndb0x"]Suppose that there exist $ a_i \geq 100$ , then $ a_i(a_i \minus{} 1)(a_i \minus{} 2) ...(a_i \minus{} 20) > 100 \cdot 99 ...\cdot 79$ ,contradiction.
Thus all $ a_i$ are less or equal than $ 99$ .

This imply $ a_1 \plus{} a_2 \plus{} ... \plus{} a_{100} \leq 99 \cdot 100 \equal{} 9900$ ,as desired[/quote]
Definitely not.  :wink:

[quote=littletush]not so hard and dantern.
it can be second-killed by noting that polynomials $C_x^n$ are convex functions.[/quote]

How is taht solution with $C_x^n$ convex functions

If we know that Cnx is convex then f(x) = x!/(x-21)! Is convex meaning we know f(a1)+f(a2) + ... + f(a100) <= 100*f(99) and then it is easy to finish with Jensen's inequality.


The title question. I think I know who it is, but I don't know his AoPS name...

smartest? definetely not me

coolest? easy, hjoon

this is not true. i am definitely the coolest, because i am that cool.

[quote="hjoon0125"][quote="-$45!"][quote="hjoon0125"]no, im the coolest. and i actually live in va.[/quote]

Agreed. (Assuming you are who I think you are...)[/quote]

Is this Adam?[/quote]

Yup.

[quote="-$45!"][quote="hjoon0125"]Holy Smokes. I'm...lost for words. Adam... wow. I don't believe this (in a good way). You probably know me by my user name and signature. I'm having trouble writing sentences. Here goes:

How's summer?[/quote]

Why don't you believe this?

And my summer's been pretty good. I've been on vacation for a while of it. I am now, but have internet access.

I also know you from another thread in the VA forum...I think you know what I mean... :P

Don't really want to go quote it...

.......................................................................................................................................[/quote]

haha sorry.

Hmmm....
Does anybody go to math reasoning classes? (taught by Vern Williams)
There are some pretty smart kids in there.

[quote="PI-Dimension"][quote="Mewto55555"]No, its Max Schindler. Definitely the smartest in VA.[/quote]

QFT[/quote]

Quoting for Falseness.

Robin Park and Scott Becker are also pretty smart

This to me is not a valid question because you cant measure smartness.

[quote="r31415"]...Mitchell Lee actually made IMO; robinpark only made Nats CD...[/quote]
I know :(

I actually thought I would get a pretty decent JMO score this year. But as Abraham Lincoln once said, "The USAJMO is one dang hard test," a few months after his famous quote on Internet trustworthiness.

[quote="robinpark"]But as Abraham Lincoln once said, "The USAJMO is one dang hard test," a few months after his famous quote on Internet trustworthiness.[/quote]

whos Abraham Lincoln..

are you talking about the famous militia caption during the American revolution, a Kentucky pioneer?

lived from 1744 - 1786?

Mitchell Lee got accepted to MIT without completing 12th grade.

lol^ is the smartness we are talking about in math? If so, I think Robin Park is probably the smartest at math, but there are definitely many smart people in VA..some even posting on this forum! :)

aaditya is definitely the smartest. (middle-schooler)
he's not only amazing at math, but he goes to Langely for science, too

[quote="droid347"][quote="mathonator"]^wait I heard about that. how does he do that? i don't know anyone else who does that, and I didn't even know one [i]could[/i] do that...

[hide="..."]I met him once, and I was awestruck by his amazingness...[/hide][/quote]
uh wow fangirling over a math nerd lol jk[/quote]

Who doesn't love Aaditya?

plz charles wang is more op
I will provide proof if you desire.

[quote="mathonator"]^wait I heard about that. how does he do that? i don't know anyone else who does that, and I didn't even know one [i]could[/i] do that...

[hide="..."]I met him once, and I was awestruck by his amazingness...[/hide][/quote]

langley has a and b days, so 3 days a week he goes to langley for like 3 hours, and 2 days a week he's in the library for 3 hours.  idk if he has electives..

[quote="AkshajK"]

Mitchell Lee got accepted to MIT without completing 12th grade.[/quote]

I got accepted to MIT and I have not completed the 12th grade yet!


Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $ 1$. The polygons meet at a point $ A$ in such a way that the sum of the three interior angles at $ A$ is $ 360^\circ$. Thus the three polygons form a new polygon with $ A$ as an interior point. What is the largest possible perimeter that this polygon can have?

$ \textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 21\qquad \textbf{(E)}\ 24$

[hide="I did this at math camp"]
The interior angle of a regular $ n$-gon is $ 180 \left (1\minus{}\frac{2}{n} \right )$.  Let $ a$ be the number of sides of the congruent polygons and let $ b$ be the number of sides of the third polygon (which might be congruent to the first two polygons).  Then:

$ 2 \cdot 180 \left (1\minus{}\frac{2}{a} \right )\plus{}180 \left (1\minus{}\frac{2}{b} \right )\equal{}360$.

If we clear the denominators and factor, we get the equation

$ (a\minus{}4)(b\minus{}2)\equal{}8$

whose four positive integral solutions are $ (a,b)\equal{}(5,10),(6,6),(8,4),(12,3)$.  These four solutions give polygons with perimeters of $ 14$, $ 12$, $ 14$, and $ 21$, respectively, so the largest possible perimeter is $ 21$, or $ \boxed{\textbf{(D)}}$.
[/hide]


Find all $ f: \mathbb{R}^{ \plus{} } \rightarrow \mathbb{R}^{ \plus{} }$ such that
\[ f(f(x) \plus{} y) \equal{} xf(1 \plus{} xy)
\]

[i]Proposed by Dr. BJ Venkatachala, India[/i]

Could you please post the solution?  I tried it for quite long time but didn't get anything.  (Is $ f(x) \equal{} \frac{1}{x}$ the only answer?)

Yes it is the only solution.  I'll post the solution after a couple more days of letting people try it. BTW, the solution I know is so unusual that I see absolutely no way of motivating it. So it isn't all that bad if you can't solve it - nobody at the Indian TST's did!

If you want a hint, the first step is to prove that $ f(1) \equal{} 1$. Then you prove that $ f$ is always decreasing  and then finally you prove that $ f(x) \equal{} \frac{1}{x}$. This hint didn't help me one bit but maybe it'll help you!

Ashwath

Incomplete solution (and long)...

Let $ a: \equal{}f(1)$.  Then, for all $ y>0$,
\[ f(a\plus{}y) \equal{} f(1\plus{}y)\,.\]
Thus, if $ a < 1$, then $ f$ is periodic on $ (a,\infty)$ with period $ 1\minus{}a$.  If $ a>1$, then $ f$ is periodic on $ (1,\infty)$ with period $ a\minus{}1$.  Therefore, $ a \neq 1$ implies that $ f$ is periodic on $ (1,\infty)$ with period $ P: \equal{}|a\minus{}1|$.

Now, choose $ k \in \mathbb{N}$ so that $ kP>1$.  Then,
\[ f\left(f(kP)\plus{}1\right) \equal{} kPf(1\plus{}kP) \equal{} kPf(1\plus{}P)\,.\]
Also,
\[ f\left(f(kP\plus{}P)\plus{}1\right) \equal{}(kP\plus{}P)f\left(1\plus{}(kP\plus{}P)\right) \equal{} (k\plus{}1)Pf(1\plus{}P)\,.\]
However, $ f\left(f(kP)\plus{}1\right) \equal{} f\left(f(kP\plus{}P)\plus{}1\right)$.  Thus, $ kPf(1\plus{}P) \equal{} (k\plus{}1)Pf(1\plus{}P)$, which is a contradiction.  Therefore, $ f(1) \equal{} 1$.

Now, if we could prove that $ f$ is strictly decreasing, then $ f$ is continuous.  This is because
\[ f\left(f(x)\plus{}\frac{1}{x}\right) \equal{} xf(2)\]
implies that $ f$ is surjective.  Thus, $ f^{\minus{}1}\left((a,b)\right) \equal{} \left(f^{\minus{}1}(b),f^{\minus{}1}(a)\right)$ for all $ 0<a<b$.  This shows that $ f$ is continuous.  Now, by letting $ y \rightarrow 0$, we get
\[ f\left(f(x)\right) \equal{} xf(1) \equal{} x\,.\]
Now, suppose that $ x>1$.  Then, if $ \frac{f\left(\frac{1}{x}\right)\minus{}1}{x} > 0$.  Thus,
\[ f\left(f(x)\plus{}\frac{f\left(\frac{1}{x}\right)\minus{}1}{x}\right) \equal{} xf\left(f\left(\frac{1}{x}\right)\right) \equal{} 1 \equal{} f(1)\,.\]
Consequently, since $ f$ is strictly decreasing,
\[ f(x)\plus{}\frac{f\left(\frac{1}{x}\right)\minus{}1}{x} \equal{} 1\,.\]
The above equation also holds for $ x\equal{}1$.  Now, if $ x<1$, then we use the above equation with $ x$ being replaced by $ \frac{1}{x}$ and we still get the same relation.  Therefore, for all $ x>0$,
\[ f(x)\plus{}\frac{1}{x}f\left(\frac{1}{x}\right) \equal{} 1\plus{}\frac{1}{x}\,.\]
Now, with $ y: \equal{}\frac{1}{x}f\left(\frac{1}{x}\right)$, we get
\[ f\left(1\plus{}\frac{1}{x}\right) \equal{} f\left(f(x)\plus{}\frac{1}{x}f\left(\frac{1}{x}\right)\right) \equal{} xf\left(1\plus{}f\left(\frac{1}{x}\right)\right)\,.\]
Replace $ x$ by $ \frac{1}{f(x)}$ in the above equation, we get
\[ f\left(1\plus{}f(x)\right) \equal{} \frac{1}{f(x)}f(1\plus{}x) \,. \,\,...(*)\]
Replace $ x$ in (*) by $ f(x)$, we get
\[ f\left(1\plus{}x\right) \equal{} \frac{1}{x}f\left(1\plus{}f(x)\right) \,. \,\,...(**)\]
From (*) and (**), $ f(x) \equal{} \frac{1}{x}$ as desired.

NOW, I need to show that $ f$ is strictly decreasing.  But, how to do that???

here is a beauty (to some extent usual)
suppose both $ m, f(m) > 1$ or both $ m, f(m) < 1$
then we substitute $ x \equal{} m$ and and $ y \equal{} \frac {f(m) \minus{} 1}{m \minus{} 1}$
we get $ m \equal{} 1$
contadiction
therefore whenever $ f(x) > 1 , x \le 1$
and  whenever $ f(x) < 1 , x \ge 1$

for $ x > 1$
we put $ y \equal{} \frac {x \minus{} 1}{x}$
so $ xf(x) \equal{} f(f(x) \plus{} \frac {x \minus{} 1}{x})$
case $ 1$ $ xf(x) > 1$
  implies $ f(f(x) \plus{} \frac {x \minus{} 1}{x}) > 1$
  implies $ f(x) \plus{} \frac {x \minus{} 1}{x} \le 1$
  i.e. $ xf(x) \le 1$
  contradiction
case $ 2$ $ xf(x) < 1$
  implies  $ f(f(x) \plus{} \frac {x \minus{} 1}{x}) < 1$
  implies $ f(x) \plus{} \frac {x \minus{} 1}{x} \ge 1$
  i.e. $ xf(x) \ge 1$
  contradiction
therefore $ xf(x) \equal{} 1$ i.e. $ f(x) \equal{} \frac {1}{x}$ for $ x > 1$

for $ x\le 1$
we put $ y \equal{} 1$
so $ f(x) \plus{} 1$ , $ 1 \plus{} x > 1$ i.e. $ f(f(x) \plus{} 1) \equal{} \frac {1}{f(x) \plus{} 1}$
and $ f(1 \plus{} x) \equal{} \frac {1}{x \plus{} 1}$
substituting these we get $ f(x) \equal{} \frac {1}{x}$ for $ x \le 1$

QED
also now the function is decreasing as $ f(x) \equal{} \frac {1}{x}$ and hence Batominovski 's proof is completed
p.s.this question is neither own nor is it proposed it is used in the indian tst 2008 (creator of the problem is as aswath mentioned prof.BJV)

aswath could you post the solution given at the test
it wud be nice to have it


calculus
integration
trigonometry
email
function
limit
real analysis

Let s be real and > 1. Show that

[tex]\int_0^{\infty} \frac {dx}{x^s + 1} = \frac {\pi}{s\sin \frac {\pi}s}. [/tex]

Hey, Mindspa, do you have a solution for this one? I'd really love to see the solution. I like the problem but i can't solve it... I've tried writing 1/(1+x^s) as 1/(1-(-x^s))=1-x^s+x^2s-x^3s+... and I integrated this. I got something like this:

lim {x->00} (x-x^(s+1)/(s+1)+x^(2s+1)/(2s+1)-...) = Pi/(s*sinPi/s). Unfortunately, I have no idea how to prove this. I 've got the feeling that I'm not on the right track. Maybe it has something to do with cisPi/s = e^(i*Pi/s) being a root of x^s+1=0 and stuff like that, but I have no idea if this leads to the solution. I totally lost, so I would appreciate it if you sent me a solution by email or PM or something (if you have a solution, that is   :D ). Bye!

I haven't solved it either, but since for irrationall s it has infinitely many poles, I'll try today to show it for rational s with residues and do an continuation argument. I'll tell you if it worked or not later.

Hi grobber,

i use your idea if s=n a positive integer>1, the n-th root of unity
z_k =exp(i \pi /n+2k \pi /n) with 0 \leq k \leq n-1
x^n+1=(x-z_0)...(x-z_(n-1)) use the partial fraction technics
[tex] \frac 1{1+x^n}= \frac {A_0}{x-z_0}+\cdots +\frac {A_{n-1}}{x-z_{n-1}}[/tex] where  [tex] \ \ A_k=\frac  1{n(z_k^{n-1})}=-\frac {z_k}n [/tex] because [tex] z_k^n=-1 [/tex].

[tex] \int _0^y \frac {dx}{1+x^n} = -\frac1n \sum _{p=0}^{n-1}A_p \int _0^y \frac {dx} {x-z_p} \quad (1) [/tex]

use  primitive of [tex] \frac 1{x-z_k} ={ \left(x-\cos \left( \frac {\pi}n+\frac {2k \pi}n\right) \right) \over \left(x^2-2\cos \left( \frac {\pi}n+\frac {2k \pi}n\right)x+1\right)} + [/tex][tex]i {\sin\left( \frac {\pi}n+\frac {2k \pi}n\right)\over \left(x^2-2\cos\left( \frac {\pi }n + \frac {2k \pi }n\right )x+1\right)} [/tex] and plug in (1) and take the limit when [tex] y\to +\infty [/tex]
to get the  result when s=n integer >1
 \cap  \cap  \emptyset  \cup \beta  \int  \cap  \emptyset  \emptyset  \lambda

[quote="Mindspa"]Let s be real and > 1. Show that

\int_0^{\infty} dx/(x^s + 1) = pi/(s*sin(pi/s)).[/quote]
 have a look formulae (32)
http://mathworld.wolfram.com/GammaFunction.html
Could Gamma function help ?
 \cap  \cap  \emptyset  \cup \beta  \int

This problem can be easely solved with contour integration.
We must take contour \Gamma which include
(1) segment (r,R)
(2) arc of |z|=R
(3) segment (R.exp(2i\pi /s),r.exp(2i\pi /s))
(4) arc of |z|=r
Now R--> \infty, r-->0.
It is clear \int (3) = -exp(2i\pi /s) \int (1) for some determined C; \int (2) --> 0, \int (4) --> 0.
So on the one hand \int\Gamma --> (1-exp(2i\pi /s)).I and on the other hand \int\Gamma = 2i\pi Resx=exp(i\pi /s)(1/(xs+1)) !!!

That's all.

It's all if s are integers, or I'm missing something. But s is real. I solved it btw using residue theory too, using a constructiobn very must similar to yours, but I don't have any "lower" arc, and I do a continuation argument at the end.

I take zs=|z|s(cos(s.arg(z))+i.sin(s.arg(z))) inside \Gamma, it will be analytic function.

[quote="Mindspa"]Let s be real and > 1. Show that
\int_0^{\infty} dx/(x^s + 1) = pi/(s*sin(pi/s)).[/quote]

Hi grobber,

Here a proof with high school mathematics, for grade  \leq 12th

\int _0^{\infty} dx/(x^s + 1)=\int_{0 to 1}dx(x^s+1) +
\int _{1 to +oo}dx/(x^s+1) in the second integral  the change variable
u=1/x  gives
\int _0^{\infty} dx/(x^s + 1)=\int_{0 to 1}dx(x^s+1) +
\int _{0 to 1}u^(s-2)/(u^s+1) du = \int_{0 to 1}(1+x^(s-2))/(1+x^s)dx
another change variable w=x^s

\int_ 0^{\infty} dx/(x^s + 1)=1/s\int_{0 to 1}(w^(1/s -1)+w^(-1/s))/(1+w)dw=q(A(q)+B(q))   (1)
with
A(q) =\int_{0 to 1}w^(q-1)/(1+w)dw, B(q)=\int_{0 to 1}w^(-q)/(1+w)dw , with    q= 1/s \in ]0;1[

Let I_n(q) and J_n(q) define in the following
|I_n(q)-A(q)| = |\sum_{k=0 to n}(-1)^k/(q+k) - A(q)|=
|\int _{0 to 1}t^(q-1)/(1+t).(1-(-t)^(n+1))/(1+t)dt -A(q)| \leq  \int _{0 to  1} t^(n+q)dt  \leq 1/n --->0 when n -->+oo

|Jn(q)-B(q)|=|-1/q + \sum_{k= 0 to n}(-1)^k/(q-k) -B(q)|=
|\int _{0 to 1} (1+(-1)^(n+1)t^n)/(t^q(1+t)) dt -B(q)| \leq
\int _{0 to 1}t^(n-q)dt  \leq 1/n -->0 when n-->+oo

 \lim I_n(q)=A(q) and  \lim J_n(q)=B(q)

Let T_n(q)=  \int _{0 to pi/2}((2sin((2n+1)x)/sin(x)).(-1+cos(2qx))dx
by Riemann-Lebesgue theorem  \lim T_n(q)=0

With some trigonometric formula
T_n(q)-T_(n-1)(q) = \int _{0 to pi/2}(-4cos(2nx)+2cos(2(n+q)x)+2cos(2(n-q)x))dx=
(-1)^n.sin(q.pi)(1/(n+q)-1/(n-q))   (2)

Let
C_n(q)= I_n(q) + J_n(q) - T_n(q)/sin(pi.q)
C_n(q) - C_(n-1)(q) =0 => C_n(q)=C_0(q)=pi/sin(pi.q)

I_n(q)+J_n(q) = pi/sin(pi.q) + T_n(q)/sin(pi.q) taking the limit when
n-->+oo

A(q)+B(q) = pi/sin(pi.q)   now with (1) we have
\int _ {0 to +oo} dx/(x^s+1) = q.pi/sin(pi.q)=pi/s.sin(pi/s)
 \cap  \cap  \emptyset  \cup \beta  \int  \cap  \emptyset  \emptyset  \lambda

use Beta function of euler:
Beta(p,q)= \int_0^1x^{p-1}*(1-x)^{q-1}dx.
with change of variable x=t/(1+t) we get
Beta (p,q)=\int_0^{oo}t^{p-1}/(1+t)^{p+q}dt
if, in the problem we note x^s=t,we obtain that
\int_0^1 1/(x^s+1)ds=1/s*Beta(1/s,2-1/s)=
1/s*Gama(1/s)*Gama(1-1/s)/Gama(1)=1/s*pi/sin(pi/s)

Here is an "easy" proof (without complex analysis)
Let f(s)=int(1/(1+x^s),x=0..+oo) and g(s)=(\pi/s)/sin(\pi/s)
Both functions are well defined and continuous on ]1,+oo[ (easy to prove)
Let p be even and q odd with p>q>0.
then f(p/q)=int(1/(1+x^(p/q),x=0..oo)=int(q*u^(q-1)/(1+u^p),u=0..oo)
=(q/2)*int(u^(q-1)/(1+u^p),u=-oo..oo)=(q/2)*lim(int(u^(q-1)/(1+u^p),u=-t..t),t->oo)
u^(q-1)/(1+u^p)=sum(a(k)/(u-b(k)),k=0..p-1) with b(k)=e^i(\pi(2k+1)/p)
and a(k)=-(b(k)^q)/p=-e^i((2k+1)\pi*(q/p))/p
Let x be a real number s.t. sin(x)<>0
Then 1/(u-e^ix)=(u-cos(x)+isin(x))/(u-2ucos(x)+1)
int((u-cos(x))/(u-2ucos(x)+1),u=-t..t)=ln((t-2tcos(x)+1)/(t+2tcos(x)+1))/2
(->0 when t->+oo)
int(sin(x)/(u+2ucos(x)+1))=arctan((t+cos(x))/sin(x))-arctan((-t+cos(x))/sin(x))
(->\pi if sin(x)>0 and -\pi if sin(x)<0)
and finally we get : lim(int(1/(u-e^ix),u=-t..t),t->+oo)=i*\pi (if sin(x)>0) or -i*\pi (if sin(x)<0). Finally we have :
(q/2)*lim(int(u^(q-1)/(1+u^p),u=-t..t),t->oo)=
(q*i*\pi)/2*(sum(a(k),k=0..p/2-1)-sum(a(k),k=p/2..p-1)=
(-q*\pi)*Im(sum(a(k),k=0..p/2-1))  (a(k)=conj(a(p-1-k)))
sum(a(k),k=0..p/2-1)=-(1/p)e^i(\pi*q/p)*(1-e^i(pi*q))(1-e^i(2*pi*q/p)=
-i/(psin(\pi*(q/p)))
So we finally get f(p/q)=-q*\pi*(-1/(psin(\pi*(q/p)))=(\pi/(p/q))/sin(\pi/(p/q))=g(p/q).
The subgroup of rationals p/q with p even and q odd is dense (since it's not of the aZ form). As f and g are continous, f=g.


This is from a regional exam monitored in early January. Only 1% out of 1200 solved the problem correctly.

Two points are randomly selected from inside the unit circle. Let $ X$ be a random variable representing the distance between them. $ (a)$ What is the expected value of $ X$? $ (b)$ What is the information entropy of $ X$ in nats?

Looks more like something that should go in the probability and statistics forum. I can't think of noncalculus methods of solving part a) and information entropy is definitely not a high school concept.


algebra
polynomial
modular arithmetic
Rational Root Theorem
number theory

Let f(x)=$px^5$+$qx^4$+$rx^3$+$sx^2$+$tx$+1 be a polynomial s.t. f(1)=4, f(2)=11 and all the numbers p,q,r,s,t are integers.

Prove that f(x) have no integer roots.

Hmmm....I tried to used "rational root theorem", meaning that the only possible integer solutions are 1 and -1, but i can't seem to be able to prove that f(-1) does not equal 0. If this way works, can someone please alert me.

However, i think this problem can be solved using contradiction, by letting $t$ be a root, then taking odd and even cases of $t$ to show both cases are not possible.

anyway, happy solving!!

[this problem comes from an australian source, an enrichment programs for year 10, or the equivalent of 15-16 year olds]

Your first approach is heading in the right direction.  There are a couple of ways to prove that -1 is not a root.

[hide="Hint"]One was is to assume that -1 is a root.  Then f(x) = (x + 1) g(x) for some polynomial g with integer coefficients.  Now see if you can reach a contradiction.[/hide]

There's an even easier way :D

[hide]
f(0) = 1 mod 3
f(1) = 1 mod 3
f(2) = 2 mod 3
Well-known and easily provable theorem: f(ab + c) = f(c) mod a
f(3a) = 1 mod 3
f(3a+1) = 1 mod 3
f(3a+2) = 2 mod 3
So f(any integer) = 1 or 2 mod 3.  However, 0 = 0 mod 3, so f(some integer) can't be zero.[/hide]

[quote="random_mathematician"]f(ab + c) = f(c) mod a

wats this theorem by its name??

(e.g. Pythag, or fermat's little theorem or sumthing)

i need to name it

thx[/quote]

Dunno if it really has a name.
On an olympiad, you would probably want to prove it.
It's really easy and straightfoward:
\begin{eqnarray*}
f(c) &\equiv& a_n c^n + a_{n-1} c^{n-1} + \cdots + a_0 \pmod a\\
f(ab+c) &\equiv& a_n (ab+c)^n + a_{n-1} (ab+c)^{n-1} + \cdots + a_0\\
&\equiv& a_n (0 + c)^n + a_{n-1} (0+c)^{n-1} +\cdots + a_0 \\
&\equiv& f(c) \pmod a
\end{eqnarray*}


[quote="4everwise"]A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to $(1,0)$. Then it makes a $90^\circ$ counterclockwise and travels $\frac 12$ a unit to $\left(1, \frac 12 \right)$. If it continues in this fashion, each time making a $90^\circ$ degree turn counterclockwise and traveling half as far as the previous move, to which of the following points will it come closest?

$\text{(A)} \ \left(\frac 23, \frac 23 \right) \qquad \text{(B)} \ \left( \frac 45, \frac 25 \right) \qquad \text{(C)} \ \left( \frac 23, \frac 45 \right) \qquad \text{(D)} \ \left(\frac 23, \frac 13 \right) \qquad \text{(E)} \ \left(\frac 25, \frac 45 \right)$[/quote]

[hide](B)
if you solve for the x coordinate, it's $1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}...$ so if you set that as X you get
$4x=4-1+\frac{1}{4}-\frac{1}{16}$
$-4x=-4+x$
$-5x=-4$
$x=\frac{4}{5}$
and you don't even have to solve for y because the only one with x-coord $\frac{4}{5}$ is (B)[/hide]

Yeah, you get the x-coord that way by doing infinite geometric series, and you do get $\frac25$ for the y-coord too, making B the answer, although you can't assume the answer just from the x-coordinate - what if the y-coordinate of B had been an outlier like $\frac{9}{10}$?  On AMC they probably wouldn't do this, but other tests might want you to calculate the distance from the point the bug approaches and find the smallest distance (I know my local tests would do that)

The problem states "To which of the following points will it come the closest", but how does it imply to find the point of convergence? I thought that we are supposed to find which of the points that is the closest to the path...

Well, if you draw all the points and look at them compared to the path, you can see that the intended answer can be arbitrarily close to the path, where as the other points are relatively much farther away from the path.


induction
logarithms
advanced fields
advanced fields solved

hi!
I would apprisiet some guidence for solving recursion
T(n) - T(n-1) = log (n), T(1) = 0

[quote="Babak"]T(n) = log(n!)[/quote]
s
so far I get it, but what kind of argument/proof is for this substitution.

There is a recursion. Induction is the [b]only[/b] way.

Sum $T(k)-T(k-1)=\log k$ for $k=2$ to $k=n$.  The LHS telescopes and you get
\begin{eqnarray*}T(n)-T(1) &=& \sum^{n}_{k=2}\log k \\ &=&\log\left(\prod^{n}_{k=2}k\right) \\ &=&\log k! \end{eqnarray*}


The decimal part of a real number $a$ is denoted by $[a]$. Find the smallest real number $x$ such that $x>5$ and $[\log(x+2)]+[\log x]=1$

Your answer is not correct, $[a]$ is [b]the decimal part[/b] of $a$, and so $[\log 8]$ is not 0, but $.903...$

$\left[ \log(x+2) \right] = 1 - \left[\log x \right]$

$m + \log(x+2) = 1 - (n + \log x)$, where $m, n$ are integers

$\log{((x + 2) \cdot 10^m)} = \log\frac{10}{x \cdot 10^n}$

$(x + 2) \cdot 10^m = \frac{10}{x \cdot 10^n}$

$x^2 + 2x - 10^{m + n - 1} = 0$

$x = \frac{-2 \pm \sqrt{4 + 4 \cdot 10^{m + n - 1}}}{2} = -1 \pm \sqrt{1 + 10^{m + n - 1}}$

The negative root is not acceptable, because $\log x$ would be undefined. In addition, $x > 5$, hence $m + n - 1 \ge 2$. Minimum $x$ occurs for $m + n - 1 = 2$, $x = \sqrt{101} - 1$.

Check:

$\sqrt{11} - 1 \doteq 2.32 < 5$

$\sqrt{101} - 1 \doteq 9.05 > 5$

$\log{(\sqrt{101} - 1)} \doteq 0.95664261$

$\log{(\sqrt{101} + 1)} \doteq 1.04335739$

The sum of fractional parts is clearly equal to 1.

x(x+2)=some multiple of 10  To minimize x, this multiple of 10 must be 100.  So, x(x+2)=100.  x^2+2x-100=0.  x= (-2[+/-]sqrt[404])/2, which is -1 +/- sqrt(101), as yetti said.


[quote="PenguinIntegral"]Let $ S$ be the set of reals of the form $ sin\left(\frac {2007\pi}{n}\right)$ where $ n$ is any positive integer greater than 2007. How many ordered pairs (a, b) exist such that a and b are both elements of $ S$, not necessarily distinct, and $ a^2 \plus{} b^2 \equal{} 1$?

Have fun next Tuesday.[/quote]
[hide]
We need \[ \sin^2\left(\frac{2007\pi}{n_1}\right) \plus{} \sin^2\left(\frac{2007\pi}{n_2}\right)\equal{}1\]
or \[ \cos^2\left(\frac{\pi}{2}\minus{}\frac{2007\pi}{n_1}\right)\plus{}\sin^2\left(\frac{2007\pi}{n_2}\right)\equal{}1\]

Setting the insides equal ... and getting rid of denominators and factoring, $ (n_1\minus{}4014)(n_2\minus{}4014)\equal{}4014^2\equal{}2^2\cdot 3^4\cdot 223^2$ which has $ 3\cdot 5\cdot 3 \equal{} 45$ positive factors and we can't have negative factors or else one of the $ n$ will fall below 2007...so $ n_1,n_2>4014$...

We have $ \minus{}\frac{\pi}{2} < \frac{\pi}{2}\minus{}\frac{2007\pi}{n_1} < \frac{\pi}{2}$ and $ 0<\frac{2007\pi}{n_2}<\pi$ and only half the range in each interval is covered by the restriction $ n_1,n_2>4014$, so we multiply by 4 (taking into account when we made the assumption the insides of the sine and cosine were equal)...

But we also have the cases where $ \frac{\pi}{2}\minus{}\frac{2007\pi}{n_1}\equal{}0$ and $ \frac{2007\pi}{n_2}\equal{}0$ ... $ n_1\equal{}4014$ which gives $ n_2\equal{}2007$, which is illegal...

The second equation above gives $ n_2\Rightarrow \infty$...which is not possible, I guess.

So, uh, $ 45\cdot 4 \equal{} \boxed{180}$??? T_T[/hide]

[quote="ch1n353ch3s54a1l"]
\[ \cos^2\left(\frac {\pi}{2} \minus{} \frac {2007\pi}{n_1}\right) \plus{} \sin^2\left(\frac {2007\pi}{n_2}\right) \equal{} 1
\]
Setting the insides equal ... [/quote]

Unfortunately, all we know is that the difference of the two angles is a multiple of $ 2 \pi$, and that's without taking sign into consideration.

Edit:  I seem to have misread the question.  Which is strange; the last time I saw this question (not this thread), I could've sworn that condition wasn't there.

[hide]
The given condition, $ a^2 \plus{} b^2 \equal{} 1$, only occurs when (WLOG) $ a \equal{} \sin (x)$ and $ b \equal{} \cos (\frac {\pi}{2} \minus{} x)$. Which implies $ \$frac{\pi}{2} \minus{} \frac {2007\pi}{n} \equal{} \frac {2007\pi}{m}$ where $ m$ is a real, greater than 2007. Simplifying, we get $ \frac {mn}{m \plus{} n} \equal{} 4014$. Further simplifying, $ mn \minus{} 4014m \minus{} 4014n \equal{} 0$. Using SFFT, we get $ (m \minus{} 4014)(n \minus{} 4014) \minus{} 4014^2 \equal{} 0$. $ (m \minus{} 4014), (n \minus{} 4014)$ must both be factors of $ 4014^2$. $ 4014^2 \equal{} (2\cdot3^2\cdot223)^2$. $ \tau(4014^2) \equal{} (2 \plus{} 1)(4 \plus{} 1)(2 \plus{} 1) \equal{} 45$. Since it is impossible to get positive divisors of $ 4014^2$ of the form $ (x \minus{} 4014)$ where $ x < 2008$ (indeed, x must be at least 4015), the value from $ \tau(4014^2)$ stands. There are $ 45$ such pairs.
[/hide]

EDIT: [hide="Response to t0rajir0u"]
Doesn't matter. The angles we find (see solution above) are all between $ \frac {\pi}{2}$ and $ 0$.
[/hide]

I hate it when I'm not the first to post my solution because I went to do homework for a little bit instead of clicking post.


Let $m$ be a positive integer. Define the sequence $\{a_{n}\}_{n \ge 0}$ by \[a_{0}=0, \; a_{1}=m, \; a_{n+1}=m^{2}a_{n}-a_{n-1}.\] Prove that an ordered pair $(a, b)$ of non-negative integers, with $a \le b$, gives a solution to the equation \[\frac{a^{2}+b^{2}}{ab+1}= m^{2}\] if and only if $(a, b)$ is of the form $(a_{n}, a_{n+1})$ for some $n \ge 0$.


For $a,b,c,d$ positive reals such that $a+b+c+d =4$ prove

$(abc)^2+(bcd)^2+(cda)^2+(dab)^2 +abc+bcd+cda+dab \leq 8$

In diendantoanhoc forum , [b]Hatucdao[/b] said that it's stronger than this ineq :
$ \sum\frac{1}{5-abc} \leq 1 $.

OKie.It's not problem of Mine.
I had proved that if $\sum_{i=1,4}\frac{1}{5-b_i} \le 1$ then
$\sum_{i=1,n}\frac{1}{k-b_i} \le \frac{n}{k-1}$
For all $n\ge 4$ and $k \ge n+1$

Sorry, who is Hatucdao ?

Is he a member of this forum?

Very nice inequality.  :)
To prove it, it suffices to consider $b=c=d=1$. ;)

No $b=c=d=1$, only $b=c=d$.

Can you please explain better your idea, Vasc ?

$f(a)=(abc)^2+(bcd)^2+(cda)^2+(dab)^2 +abc+bcd+cda+dab$. I think Vasc meant that $f$ is convex, so it attain maximum for $a=1$ or $a=0$, with $a\le b \le c \le d$.

Thank you very much, Filip.

Can anybody to write a complete proof, please?

I used my $n-1$ Equal Variable Method.
Using this way, we can solve the following inequalities posted in this forum:

1. If x,y,z>0 and $\sum \frac 1{x_i} = n$ , then
$x_1\cdots x_n-1\geq \left(\frac{n-1}n \right)^{n-1} (x_1+...+x_n-n)$;

2. If x,y,z>0, then
 $(x_1+...+x_n-n)(\frac{1}{x_1}+...+\frac{1}{x_n}-n)+x_1...x_n+\frac{1}{x_1...x_n} -2 \ge 0$;

3. What are the maximal and minimal values of the expression $x_1+...+x_n-ax_1x_2...x_n$, where $x_1,...,x_n,a$ are real numbers and $ x_1^2+...+x_n^2=1$?

4. If x,y,z>0 and $x+y+z=3$, then
$\frac{12}{x^2+y^2+z^2}+\frac{1}{yz}+\frac{1}{xy}+\frac{1}{xz}\geq 7$;

5. If $a,b,c,d$ are positive reals and $a+b+c+d =4$, then
$(abc)^2+(bcd)^2+(cda)^2+(dab)^2 +abc+bcd+cda+dab \leq 8$.

6. If a,b,c are reals such that $a^2+b^2+c^2=1$, then
$a+b+c-2abc \le \sqrt{2}$.

I will present n-1EV Metod in my future book. You asked for it! ;)

When will you finish to write your book, Vasc?

Not so difficult,Manlio.
But I afraid my solution is same way as Vasc.

Assume that $a \ge b \ge c \ge d$ then

$f(a,b,c,d) \ge f(\frac{a+c}{2},\frac{a+c}{2},b,d)$

Sorry, but I cannot complete the solution. Can anybody try to complete mixing variables method, please?

To manlio:
I posted a fairly simple solution with Lagrange multipliers on [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=203524#p203524[/url]

[quote="hungkhtn"]
But I afraid my solution is same way as Vasc.[/quote]
To hungkhtn:
That is an excellent reason to post it!  :lol:


AMC
AIME
geometry
geometric transformation
reflection
floor function
3D geometry

I'm opening this thread to invite dicussion of the coolest AIME problems or the nicest solutions to them. To start, my favorite AIME problem of all time is from 1988 problem 13 (taken from kalva): $x^2-x-1$ is a factor of $ax^{17}+bx^{16}+1$ for some integers $a$, $b$. Find $a$.

The reason I like it is because it has so many solutions, all of them cool and none of them obvious. Here's mine, but feel free to add others.

[hide]If $z$ is one of the roots of $z^2=z+1$, then we must have $az^{17}+bz^{16}+1=0$. So, we continuously replace $z^2$ by $z+1$ in a careful manner:

$0=az^{17}+bz^{16}+1$
${}=az(z^2)^8+b(z^2)^8+1$
${}=(az+b)(z+1)^8+1$
${}=(az+b)(z^2+2z+1)^4+1$
${}=(az+b)(3z+2)^4+1$
${}=(az+b)(9z^2+12z+4)^2+1$
${}=(az+b)(21z+3)^2+1$
${}=(az+b)(441z^2+546z+169)+1$
${}=(az+b)(987z+610)+1$
${}=987az^2+(610a+987b)z+(610b+1)$
${}=(1597a+987b)z+(987a+610b+1)$.

Since $z$ is not rational, both of these numbers in parentheses must be zero. From $1597a+987b=0$ and the fact that $0<a<1000$ (you can check that $a\neq0$), we can already determine that $a=987$. Or, to avoid using assumptions, we can solve the system and verify our answer.[/hide]

My favorite one (out the few I've done) is #14 from 1994.  I didn't come up with the solution entirely on my own, however.  It was a board problem in the AIME PS and Mr. Crawford posted a picture after I tried, and failed, and the problem had been up for a while.  It took me a little bit to figure out why the picture helped but once I did I was amazed at the solution.
(I hope the instructors don't mind me borrowing their image for the problem statement)
[img]http://www.artofproblemsolving.com/Classes/AIME/Images/1994aime14ruti.gif[/img]

[hide="Solution"]
It's not hard to find how many times it reflects going toward the vertex because it turns around after the angle becomes greater than $90^\circ.$  You can use the equal angles and everything to figure that out.  Now, difficulty arises when coming out.  How do you know when it is 'out'?  At this point with this method you can only make an educated guess.  So we should try to find a better method.

We have all these equal angles and reflections etc.  Is there another way we can view this and still have equal angles and the relflections?  Aha! We could reflect $AB$ across $BC$.  The equal angles are preserved because they are now vertical angles.  This already looks better than the other method for we are know dealing with a straight line.  Imagine we did this every time there was a reflection.  Now we can think to draw a circle through the endpoints of these segments with center at $B$ (remember they are equal segments so this is perfect!).  Now how do we find out how many reflections it makes?  This is where the picture comes in.  The beam enters in at $C$ and leaves at $Z$ (I use $Z$ just to denote the point where it leaves, the fact that $Z$ is the 26th letter of the alphabet means nothing).  Connect the points to make $\triangle ABZ$.  This is an isosceles triangle (two radii) so thus $\angle BAZ=\angle BZA=19.94^\circ$ and $\angle ABZ =140.12^\circ.$  To find the number of reflections we just find $\left\lfloor\frac{140.12}{1.994}\right\rfloor+1=\fbox{71}.$  (the +1 is because we have to count the point where it enters the circle).[/hide]

[quote="zabelman"]I'm opening this thread to invite dicussion of the coolest AIME problems or the nicest solutions to them. To start, my favorite AIME problem of all time is from 1988 problem 13 (taken from kalva): $x^2-x-1$ is a factor of $ax^{17}+bx^{16}+1$ for some integers $a$, $b$. Find $a$.
[/quote]

I remember that problem, I just simply brute-force it.

What I did was:

[hide]$x=\frac{1+\sqrt{5}}{2}$

$x^2=x+1 $

$x^4= (x+1)^2 = x^2+2x+1=3x+2$

$x^8=(3x+2)^2 = 9x^2+12x+4 = 9(x+1)+12x+4 = 21x+13$

$x^{16}=(21x+13)^2= 441x^2 + 546x+169 = 441x+441+546x+169= 987x+610$

so

$a(987x+610)x + b(987x+610)+1=0$

$987ax^2 + (610a + 987b)x + (610b+1)=0$, which clearly must be a multiple of $x^2-x-1=0.$

so we have $987a=-(610a + 987b) = -(610b+1)$, or $1597a=-987b$, since $a$ has to be positive integer, clearly a is a multiple of 987, and by 3-digit AIME answer limit, we see that a=987[/hide]

Here's a cool (but easy) problem from the 1984 AIME

2.  Find the smallest positive integer n such that every digit of 15n is 0 or 8.

Not sure how many ways there are to solve this, but here's mine:

[hide]
Since 15 = 5 * 3, 15n must be divisible by 5 and 3.  Since 15n consists of only 0's or 8's, the units digit must be 0.  Then it's a matter of finding the smallest number of 8's such that 8x is divisible by 3.  x = 3 since that gives 24, so our number is 8880.  Then, we divide this by 15 to get n, and we get 592.
[/hide]

[quote="zabelman"]I'm opening this thread to invite dicussion of the coolest AIME problems or the nicest solutions to them. To start, my favorite AIME problem of all time is from 1988 problem 13 (taken from kalva): $x^2-x-1$ is a factor of $ax^{17}+bx^{16}+1$ for some integers $a$, $b$. Find $a$.[/quote]
What about a solution based on Fibonacci:
[hide]
Let $r_1,r_2$ be the two roots of $x^2-x-1=0$. Define a sequence $f_n=p r_1^n+q r_2^n$, then $f_{n+2}=f_{n+1}+f_n$ for $n\geq0$. So
$af_{17} + b f_{16} + (p+q) = 0$

If $f_0=0, f_1=1$, then it's Fibonacci, and $p+q = 0$, this gives $1597a+987b=0$.

By the way,  we could get $987a+610b+1=0$ if we let $f_0=1, f_1=0$, (Fibonacci shifted).  [/hide]

There are too many nice problems on AIMEs to pick one to be my favorite, but I like geometry and this one (AIME 2003 I, #15, as worded by Kalva) is really slick:

AB = 360, BC = 507, CA = 780. M is the midpoint of AC, D is the point on AC such that BD bisects ∠ABC. F is the point on BC such that BD and DF are perpendicular. The lines FD and BM meet at E. Find DE/EF.

A hint:
[hide]Work a generic triangle with $b > a > c$ and apply Menelaus to triangle $DFC$; now, the only tough part seems to be getting $FB$.[/hide]

A second hint:
[hide]Reflect $F$ over $\overline{BD}$ to $F'$ and apply Menelaus to $FBF'$.[/hide]

Here's a problem that I liked, it's from the 2002 AIME.

A cube with two faces ABCD, BCEF, has side 12. The point P is on the face BCEF a perpendicular distance 5 from the edge BC and from the edge CE. A beam of light leaves A and travels along AP, at P it is reflected inside the cube. Each time it strikes a face it is reflected. How far does it travel before it hits a vertex?

[quote=BooBooTM]2005 AIME II, Problem 15 was fun[/quote]

Hello,

Why did you bump this thread 12 years?

[quote=gogiants]
Hello,

Why did you bump this thread 12 years?[/quote]

What's wrong with him doing that?

tbh nobody on the history of the internet will ever top this display of forum etiquette I'm dying over here  :rotfl:

Personally I do not find any AIME problems particularly good, but if I were to choose, I'd probably say 2015 AIMEI #15. 3D geometry on the AIME is usually quite enjoyable.

My favorite AIME problems are the easiest because then I don't silly them and fail to make JMO.  :(

2014 AIME I #15 was somewhat easy but I kind of liked it.

[quote=BooBooTM]2005 AIME II, Problem 15 was fun[/quote]

Of course, that AIME hadn't even existed when this thread was started.

oops this thread is older than the average fifth grader


inequalities
function
logarithms
inequalities theorems

Let $ f: \left[\alpha,\beta\right]\to\mathbb{R}$ be a function such that for any two reals $ a$ and $ b$ from the interval $ \left[\alpha ,\beta\right]$, we have

$ f\left(a\right)f\left(b\right) \ge f\left(\frac{a+b}{2}\right)^{2}$.

Then prove that for any $ n$ reals $ a_{1}$, $ a_{2}$, ..., $ a_{n}$ from the interval $ \left[\alpha,\beta\right]$, we have

$ f\left(a_{1}\right)f\left(a_{2}\right) ... f\left(a_{n}\right) \ge f\left(\frac{a_{1}+a_{2}+...+a_{n}}{n}\right)^{n}$.

If $f\geq 0$, induct on $g=\log\circ f$, if  $f<0$, induct on $g=\log\circ(-f)$.


function trigonometry calculus derivative

Here's a hint: any bijection has an inverse.

There are, of course, infinitely many such bijections; this hint goes to the first one that popped into my mind, but I don't know if it will be the one that first pops into your mind.  The thing to think about is what makes it non-trivial to map one of these sets onto the other, and then think about what sorts of things you know how to do that get around that difficulty.

[hide]A bijection $f: (0,1)\rightarrow \mathbb{R}$ is given by $f(x)= \tan\left(\frac{x-\frac{1}{2}}{\pi}\right)$. [/hide]

[quote="Blackadder"]Could anyone help me to find a bijection that maps R onto (0,1) ?

My teacher gives me problems about finding functions satisfying some particular (sometimes weird) properties and i usually don't know where to begin. Is there any hint for this type of questions ?

Thank you in advance.[/quote]

 Do you like :

 $f(x) = \frac{e^{x}}{1+e^{x}}$ ? I think it is a good function.Use derivative to see that it is strict increasing and limits to $+\infty ,-\infty$ to find the range is the open interval  $(0,1)$.

[u] Babis[/u]

Thank you all very much.

 :) The function exp(x)/(exp(x)+1) is really nice and simple. Why can't i think of it ?  :(

To Rzeszut, i don't understand your example. Correct me if i am wrong but your function is bounded, so it can't map (0,1) onto R.

He meant to multiply by $\pi$, not divide by it.  (Incidentally, that was the function that I was thinking of.)

[hide]This is probably a restatement of someone's but:

Take the open interval (0,1) and "bend" it into a semicircle. Then project the points onto the real line (i.e. draw a line through the center of the semicircle and each point on it, and see where your line intersects the real line).[/hide]

That the first one that popped into my head aswell


[quote="ckck"][quote="slamchess"]wat are they asking for? :huh:[/quote]
They're basically asking for the perimeter of the entire object.[/quote]
well, try it and see wat you get, cuz when i did that, it didnt work... the answer is [hide]17[/hide]

Does the strippimg go between the semicircle and the rectangle also?  How do they expect you to know what weather stripping is...

you  have three sides of the rectangle and the periphery of the semicircle, so it is $2(4)+3.5+\pi*\frac{3.5}{2}$

i somehow stumbled upon this topic, and noticed that no one had explained how the answer could be

[hide]17[/hide]

hint:

[hide]read what it asks for[/hide]


Functional Analysis
real analysis
real analysis unsolved

Let $X$ be a Banach space and let $M$ be a closed subset of $X$. One can define a norm on $X/M$ by defining $\|x+M\|=\inf_{y\in M}\|x+y\|$.
Show that
(a) For every $\epsilon >0$ there is $x\in M$ such that $\|x\|=1$ and $\|x+M\|\geq 1-\epsilon$.
(b) Let $f$ be a linear functional on $X$. Use (a) to show that if $f^{-1}(0)$ is closed then $f$ is bounded.

(a)It is not need $ X$ be a banach space,in this part we concider $ X$ as a normed space
and $ M$ must be a subspace of $ X$(otherwise $ X/M$ is not well defined)

 suppose $ \epsilon > 0$ is given,
since $ M$ is proper subset of $ X$ so there exist $ x\in M^c$,
let $ d \equal{} \parallel{}x \plus{} M\parallel{}$ since $ M$ is closed so $ d\neq0$
with according to the properties of infimum there is a $ y\in M$ such that
 $ d\leq \parallel{}x \minus{} y\parallel{}\leq d(1 \plus{} \epsilon)$,
now let $ u \equal{} (x \minus{} y)/\parallel{}x \minus{} y\parallel{}$ so $ \parallel{}u\parallel{} \equal{} 1$ and for every $ m\in M$ we have

$ \parallel{}u \minus{} m\parallel{} \equal{} \parallel{}(x \minus{} y)/\parallel{}x \minus{} y\parallel{} \minus{} m\parallel{}$

$ \equal{} 1/\parallel{}x \minus{} y\parallel{}(\parallel{}x \minus{} (\parallel{}x \minus{} y\parallel{}m \plus{} y)\parallel{})\geq d/d(1 \plus{} \epsilon)$

$ \equal{} 1/(1 \plus{} \epsilon)\geq 1 \minus{} \epsilon$  $ \Rightarrow$   $ \parallel{}u \plus{} M\parallel{}\geq(1 \minus{} \epsilon)$

b)If ker$ f \equal{} X$ then obviously $ f$ is bounded
 Suppose ker$ f$ is a proper subset of $ X$ and $ f$ is not bounded so there is a sequnce
 $ (x_n)$ such that $ \parallel{}x_n\parallel{} \equal{} 1$ and $ |f(x_n)| > n$ for every $ n\in N$
since  ker$ f$ is a proper subset of $ X$  so there is a $ x\in$ (ker$ f$)^C
WLOG we can assume $ f(x) \equal{} 1$
$ x_n \minus{} f(x_n)x\in$ker$ f$ therefore there is a sequnce $ (z_n)$ of element ker$ f$
such that $ x_n \minus{} f(x_n)x \equal{} z_n$ ,let $ k_n \equal{} f(x_n)$ therefore we have

$ n\parallel{}x_n\parallel{} \equal{} n\leq|f(x_n)| \equal{} |k_n\parallel{}|f(x)\parallel{}$,let $ l_n \equal{} ( \minus{} 1/k_n). z_n\in$ $ \Rightarrow$

$ l_n\in$ker$ f$ and we have

$ x_n \equal{} z_n \plus{} k_nx \equal{} k_n(x \minus{} I_n )$ $ \Rightarrow$

 ||$ x_n\parallel{} \equal{} \parallel{}z_n \plus{} k_nx\parallel{} \equal{} |k_n\parallel{}|(x \minus{} l_n )\parallel{}\leq |k_n\parallel{}|f(x)\parallel{}/n$ $ \Rightarrow$

$ \parallel{}(x \minus{} l_n )\parallel{}\leq |f(x)|/n$  $ \Rightarrow$ $ l_n\to x$  $ \Rightarrow$

$ x\in$ ker$ f$ that is a contradiction.

[irrelevant non-mathematical things deleted by moderator]


algebra polynomial quadratics calculus integration function

If a and b are integers s.t. (a,b)=1 then there are infinitely many primes of the form ak+b, with k integer.

I don't know a proof (I don't think I'd understand it anyway :)), but some cases are easier:

a=2n, b=1;
a=p (prime), b=1;
the well known a=4, b=3; a=6, b=5 etc.

I think there's an elementary proof for the case b=1, much easier than the general case, but I don't know that one either.

That's just WAY over me, so I'll download it and read it in a few years' time :). Until then, here's the only case (or so I've heard) of the theorem which can be shown in an.. almost elementary way: try it for b=1.

Hint: it uses cyclotomic polynomials (it's not really elementary, it just doesn't use any analysis)..

Consider cyclotomic polynomial

 \emptyset _1(x) = x-1

 \emptyset _n(x) = prod_(1 \leq k \leq n, gcd(k,n) = 1)(x - exp(2ik \pi /n) )

1/Prove that the coefficients of  \emptyset _n are integers
2/ Let p a prime number wich divides  \emptyset _n(a) where a \in Z
but p does not divides  \emptyset _d(a) for any d strict divisor of n
Prove that p = kn+1
3/For n \geq 1 fixed deduced there exist an infinity prime number of the
form kn+1 with k integer

It is a subject of Caminha's article (in Portuguese), in the page

http://www.teorema.mat.br/ciclotomico.pdf

does anyone know of a generalization of Dirichlet's theorom?
maybe this is true? ax^2+b^x+c passes through infinitely many primes iff (a,b)=1, (a,c)=1,(b,c)=1?

There are generalizations by means of binary quadratic forms, a good reference is Serre's text "A Course in Arithmetic".

sorry i dont have access to that book  :D
could you please post the generalizations if they are not too complicated? thx

I do not quite understand which statements in Serre's book should be called generalizations of Dirichlet' theorem; there are indeed some conditions for a prime to be represented by a quadratic form, but these are really about forms representing all the primes in an arithmetical progression (so these theorems can be either deduced from Dirichlet's theorem in elementary way or used to prove it in elementary way :), as Selberg did it).
   I think "natural" generalization of Dirichlet's theorem is Buniakowski conjecture:
      If all the values of a polynomial with integral coefficients at integral points do not have a common divisor, then there are infinitely many primes among these values.
   The trouble is, the conjecture is proved to be wrong. Namely, for any $a$ there are only finitely many primes of the form $x^3+x+a$ (Sprindzhuk, 1986).
   The question about the second degree polynomials, to the best of my knowledge, is still open.

How it can be proved that x^3+x+a is prime only a finite number of times.......i mena form which point is this solved ? from an algebraic point or something else....
  I have no idea of this result an is really nice......... where can i found a proof? (since i have nearly no acces to books....do you know some internet link where i can found somethig?)

[quote="Charlydif"]anyway.....i heard several times that there is an elementary proof of the case b=-1 but i have never seen it.
 On tyhe other hand does anybody knows any elemnatry proof for b=1 which is not the cyclo.. polynomial solution. Or any quiet elementary approach not being that of he cyclotomic poly?[/quote]

This is a very beautiful solution (by Ivan Niven):

"For any positive integer $m$, prove that the arithmetic progression

\[ 1+m, 1+2m, 1+3m, \ldots\]

contains infinitely many primes."

That is a outline of the proof.

(i) Prove that it suffices to show that for any $m$ the arithmetic progression have at least one prime.
The cases $m=1$ and $m=2$ are trivial.

We show that for $m \geq 3$  the number $m^m - 1$ contains at least one prime divisor $q$ such that $q \equiv 1 (mod \ m)$.
Suppose that this $q$ does not exist.

(ii) Let $q$ be any prime divisor of $m^m - 1$, so that $q \not\equiv 1 (mod \ m)$. Let $h$ denote the order of $m$ $mod q$, so that $m^h \equiv 1 (mod \ q)$ and moreover $(m^d \equiv 1 (mod \ q) \Leftrightarrow h|d)$ (this is a famous theorem. If you have a "Mathlink" about this, post it here!).
Verify that $ h|(q-1)  \  , \ h|m $. Prove that $h<m$, so that $m=hc$ with $c>1$.

(iii)Let $q^r$ the highest power of $q$ dividing $m^m-1$; thus $q^r || (m^m-1)$. Prove that $q^r || (m^d-1)$ and $q^r || (m^h-1)$

These properties of $q$ hold for any prime divisor of $m^m-1$. Of course the values of $h,c,r$ are dependendent on $q$. To finish this post we need one additional concept. Consider the set of integers of the form $\frac{m}{s}$, where $s \not\eq 1$ is any square-free divisor of $m$. We partition this set in two disjoint subsets $E$ and $O$ accordind as the number of primes  dividing s is even or odd respectively.

Put \[\displaystyle{ Q=(\prod_{d \in O} (m^d-1) ) \cdot (\prod_{d \in E} (m^d-1) )^{-1}}\].

(iv)Let $q$ the prime factor of $m^m-1$ under consideration., and let $m^d-1$ be a factor that occurs in one of the products displayed. Use (ii) to show that $s|c$ if and only if $q|m^d-1$.

(v) Let $k$ denote the number of prime divisors of $c$. Show that the number of $d \in O$ for which $q|(m^d-1)$  is

\[
\left( \begin{array}{c}
k\\
1\\
\end{array} \right)
+
\left( \begin{array}{c}
k\\
3\\
\end{array} \right)
+
\left( \begin{array}{c}
k\\
5\\
\end{array} \right)
+
\left( \begin{array}{c}
k\\
7\\
\end{array} \right)
+ \ldots
=2^{k-1}
\]

Analogously show that the number of $ d \in E$ for which $q|(m^d-1)$  is

\[
\left( \begin{array}{c}
k\\
2\\
\end{array} \right)
+
\left( \begin{array}{c}
k\\
4\\
\end{array} \right)
+
\left( \begin{array}{c}
k\\
6\\
\end{array} \right)
+
\left( \begin{array}{c}
k\\
8\\
\end{array} \right)
+ \ldots
=2^{k-1}-1
\].
Use  (iii) to show that $q^r||Q$ for all prime divisor $q$ of $m^m-1$. Prove that $Q=m^m-1$.

(vi) Show that if $b$ is a positive integer and $m \geq 3$ then $m^b-1 \not\equiv \pm 1(mod \ m^{b+1})$

(vii)Prove that the equation $Q=m^m-1$ is impossible, by writing the equation as $$, and evaluating both sides $(mod \ m^{b+1})$ where $b$ is the least integer of type d that appears in the definition of $Q$.

Do you understand?

Can someone break down this theorem into a simple or a few simple statements. I lose the essence of the theorem when a problem is being solved. Thank you.

The one sensible generalization is the Chebotarev density theorem. One could restate that in terms of polynomials and get something like the Buniakowski theorem, but I doubt it would be anything of use.

The reason why I say that this is the only sensible generalization is that both the Dirichlet theorem and the Chebotarev theorem use $L$ and $\zeta$ functions and their analytic continuations (on the real line of Dirichlet, and Tate's theory for adeles for the Chebotarev) to get information about the density of primes.

Andrei

Can anyone provide a proof when a = 6 and b = 5 ?

Assume that there are only finitely many primes $p_1,p_2,...,p_n \equiv 5 \mod 6$. Then consider $n=6\cdot p_1p_2p_3\cdot...\cdot p_n -1 ( \equiv 5 \mod 6)$. This number has no prime divisors equal to $2$ or $3$, but can't have all it's prime divisors $\equiv 1 \mod 6$, so there is some prime $q , q|n , q \equiv 5 \mod 6$. But $q$ can't be equal to one of the $p_i$, so it's a new prime with the desired property, so contradiction.

A generalization of Dirichlet.
Let a,b,c be pairwise coprime positive integers. Then ax^2+bxy+cy^2 represents infinitely many primes as x,y vary in the positive integers.

Bomb


Define a sequence :
$a_{0}$ is any real number.
$a_{n+1}$ $=$ $[a_{n}](a_{n})$
where $(a_{n})$ is fractional part of $a_{n}$

Prove that after some '$m$'
$a_{n}$ $=$ $a_{n+2}$ for all $n > m$


MATHCOUNTS
number theory
least common multiple
geometry
perimeter
3D geometry
prism

[quote="sirorange"]Everyday I (your new moderator) will be posting a problem. After a week, if enough people are responding, I'll give out points.

DAY 1:

What is:

100+99-98+97-96+95-94. . . . . +3-2+1?[/quote]
[hide]100 + 49 pairs of 1's and one extra one so 150[/hide]

It is so hard to hide your answer, clicking the hide button burns all my lunch calories  :roll:

A MOEMS thing?  :D  i wonder why no one is answer ing my Mathcounts one  :(

[hide]100+1+1+1+1+....+1+1 (50 1's) = 100+50=150. [/hide]

I'm trying, through this, and possibly a mock-MOEMS that I'll host, to bring some more action to this forum. 150 is the correct answer, and nice solutions everyone!

[quote="sirorange"]Alright, todays problem of the day. (Please hide your answers)
And from now on, I'll post what level of MOEMS this question is.

DAY 2: (Level 7-8)

A machine takes two consecutive integers, each 20 or under, multiplies them and adds 17. It then determines whether the output is prime or composite. If it is prime the printout reads P, if composite the printout reads C. There are two different sets of consecutive integers such that the printout will read C, what is the greater number in each set?[/quote][hide]16*17+17 works and so does  17*18+17
so the larger numbers in each set are 17, and 18
I had to guess and check  :oops:  [/hide]

Can anyone figure out a better solution than guess and check?

[hide="no"]

$\frac{n(n+1) + 17}{17}$

  Thus, $n(n+1)$ must be a multiple of 17.  Meaning either 17 is $n$ or $n+1$.  The larger numbers of the two cases is 17 and 18. [/hide]

[quote="sirorange"]Day 3: What is the largest whole number less than 1500 which leaves a remainder of one when it is divided by each of 5, 6, and 7.[/quote]
[hide="my answer"]
the lcm of 5,6,7 is 210 1500/210= 7 R 30 210*7= 1470 so it is 1471
 strange..no one has posted in my problem of the day (basics forum)[/hide]

When the fraction 9/14 is expressed as a decimal, what digit will be in the 75th decimal place to the right of the decimal point?

[hide="Click for answer"]
When you start dividing manually, you see that there is a repeating decimal .6  [u]428571[/u] [u]428571[/u]... 75/6=12 r3. The 75th digit is the same as the 3rd digit so the answer is ...

 [b]2[/b]
[/hide]

:starwars:

Day 5:

A large square region is divided into four congruent non overlapping square regions. If the semi-perimeter of one of these small squares is 30, what is the area of the large square?

Mark, Mike, Lawrence and Joshua bought a motorbike for 60.00 dollars. Mark paid one half of the sum of the amounts paid by the other boys, Mike paid one third of the sum of the amounts paid by the other boys, and Lawrence paid one fourth of the sum of the amounts paid by the other boys. How much did Joshua have to pay?

[hide]call mark $a$, mike $i$, lawrence $l$ and joshua $j$.  first, we have the system $a+i+j+l=60\\ a=\frac{i+j+l}{2}$.  we can solve for a by doing $2a=i+j+l$
$i+j+l=60-a$
$2a=60-a$
$a=20$
we can repeat this process for mike and lawrence, finding $i=15,l=12$.  so, $j=60-20-15-12=\boxed{13}$[/hide]

anybody?  this isn't too hard of a question :maybe:

[hide="hint"]cross-multiply
then [hide="another hint"]take the square root[/hide][/hide]

[quote="daermon"][quote="mimar"]NEXT PROBLEM!!!!!

 Plz :blush:[/quote]

 :spam:  :spam:
if u don't want to post the next problem dont say anything

Next problem:

if $\frac{x}{y}=\frac{4y}{x}=z$, find all possible values of $z$

(AoPS vol. 1)[/quote]

x^2=4y^2
x=2y
z can be anything?i know thats wrong

[quote="daermon"]find all possible values of $z$
[/quote] :wink:

i know. i cahnged it. gimme another hint? :D

[hide]$z=\frac{x}{y}$ and $x=\pm2y$.  :wink: [/hide]

[hide]Since z=(x/y) and x=+/- 2y, z must equal 2 or -2.[/hide] :huh: [/quote]

finally somebody answered!  :D
correct.  you can post the next question

Someone cough* hadasah *cough should post the next problem.


Ok I REALLY have absolutely NO clue where to post it (or if it's even a volume vs. area, etc type of question)... but since I remember doing problems like these when I was in highschool I figured I'd place it here.  If a mod thinks it would be better somewhere else, please move it :)

I majored in Art and 3D Animation so I haven't done math in a very... VERY... VEEEERY long time!  :rotfl:  Just looking for help on a present I'm trying to make for someone for Valentine's Day.

I have about 100 1/8dram glass bottles.

[img]http://i20.ebayimg.com/02/i/00/a8/35/ed_1.JPG[/img]
http://cgi.ebay.com/PATENT-LIP-VIALS-12-1-8-DRAM_W0QQitemZ7584713684QQcategoryZ26409QQcmdZViewItem#ebayphotohosting

Each one is [color=red][b]30mm in length[/b][/color] and [color=red][b]7.25mm in diameter[/b][/color].  I want to fit them all into some sort of casing for holding - like a wooded box (But I'm not sure exactly what size of box to buy since I don't have the bottles yet, and won't have them for about a month.  The box I can buy tomorrow.).  I don't really want them stacked perfectly, but rather mixed about and just kinda thrown in there.

I want to know (or how to figure out) whats the smallest box (lxwxh and depth) I can fit about 100 of these bottles in without really having to stack them neatly.

Also... I had an idea of using one of these...

[img]http://www.worldofboxes.com/images/unfinished/TW-32.jpg[/img]
http://www.worldofboxes.com/unfinished/wood-boxes.htm?group=25

It says the deminsions are as follows: 2.1" x 2.1" x 1.5" deep (upper chamber); 4.2" x 4.2" x 1.5" deep (middle chamber); 6.2" x 6.2" x 1.5" deep (lower chamber).

If anyone can, how many bottles will fit in each one of those chambers.

[quote="1234567890"]And please don't post homework questions..... ;) [/quote]

What?  This isnt homework - I graduated from college and haven't had a math class in about 4 years  :lol:
I honestly need to know because I'm trying to figure out which box to buy from this site: http://www.worldofboxes.com/unfinished/unfinished.htm :)

My bad, sorry. It looked like one, though.


Pro.1: Find f(x) that satisfy the conditions:
$f(x+y) \leq f(x)+f(y)$ and $f(x) \leq x \forall x \in R$
Pro.2: Let $f(x): (0,+\infty)\rightarrow(0,+\infty).$ and:
A. $f(x.f(y))=y.f(x) \forall x,y>0$
B.  $\{\begin{array}{l}f(xy)=f(x).f(y) \forall x,y>0\\f(f(x))=x\end{array}$
Prove that $A \Leftrightarrow B$
Pro.3: Find two functions f and g that satisfy the relation:
$f(x)-f(y)=(x+y).g(x-y) \forall x,y \in R$

Enjoy!  :D

Oh, they are not very difficult.
Who can solve these problems?

I'll solve the 2nd problem first :lol:
 A.f(xf(y))=yf(x)    x,y>0
 B.f(xy)=f(x)f(y)
    f(f(x))=x       x,y>0

[u][b]A->B[/b][/u]
    +)If y and y'are different,we have f(y) and f(y') are diffirent,too
          x=y=1: f(f(1))=f(1) so f(1)=1
          x=1:f(f(y))=y.f(1)=y
    +)We have f(y).f(x)=f(x.f(f(y)))=f(x.y)
[u][b]B->A[/b][/u]
    We have f(x.f(y))=f(x).f(f(y))=y.f(x) ,x,y>0

Finish! :D

and this is my solution for pro.1
$x=y=0\to f(0) \leq 2f(0) \to  f(0) \geq 0$ but from condition we have $f(0) \leq 0$ so that $f(0)=0$
$y=-x \to f(0) \leq f(x)+f(-x) \leq x+(-x)=0 \to  0 \leq f(x)+f(-x) \leq 0$ so $f(x)+f(-x)=0$ so $f(x)=-f(-x)$
we have $f(-x) \leq-x \to -f(-x) \geq x \to  f(x) \geq x$ but $f(x) \leq x$ so $f(x)=x$
Done!

I have just solved Pro.3  :blush:
And this is my solution:
$y=-x \to f(x)-f(-x)=0 \to f(x)=f(-x)$
$f(x)-f(-y)=(x-y)g(x+y)$
$f(x)-f(y)=(x+y)g(x-y)$
So we have $(x-y)g(x+y)=(x+y)g(x-y)$
from this we have (easily) $ag(b)=bg(a)$
$b=1\to ag(1)=g(a)\to g(x)=kx(k\in R)$
From the relation, $y=0 \to f(x)=xg(x)+f(0)$
so $f(x)=kx^{2}+b$
 :blush:


inequalities
vector
LaTeX
geometry
parallelogram
triangle inequality
geometry solved

A parallelopiped has vertices $A_1, A_2, ... , A_8$ and center $O$. Show that:
\[ 4 \sum_{i=1}^8 OA_i ^2 \leq \left(\sum_{i=1}^8 OA_i \right) ^2  \]

Suppose $A$ is the center of the face $A_1A_2A_3A_4$ of the parallelepiped. Now we work with vectors (I don't put the arrows above the vectors because I don't know the LaTeX command, could anybody post it, please?) and suppose that $AA_1=u, AA_2=v, AA_3=-u, AA_4=-v$, then the inequality becomes
\[ 8\|OA\|^2+4\|u\|^2+4\|v\|^2\leq(\|OA+u\|^2+\|OA-u\|^2+\|OA+v\|^2+\|OA-v\|^2)^2  \]
The remaining ingredient to solve the problem is the triangular inequality $\|OA+u\|+\|OA-u\|\geq 2\|OA\|, 2\|u\|$ and $\|OA+v\|+\|OA-v\|\geq 2\|OA\|, 2\|v\|$. There are three cases.
CASE 1: $\|v\|\leq\|OA\|\leq\|u\|$
Using the triangular inequalities above
\[ (\|OA+u\|^2+\|OA-u\|^2+\|OA+v\|^2+\|OA-v\|^2)^2\geq(2\|u\|+2\|OA\|)^2 \geq 8\|OA\|^2+4\|u\|^2+4\|v\|^2  \]
CASE 2: $\|v\|\leq\|u\|\leq\|OA\|$
Just follow the same pattern of proof above, using the adequate triangular inequalities.
CASE...(all the cases the same).

QED

Here is my solution.

First we will prove that $\sum_{i=1}^8 OA_i\geq 4\cdot OA_1$.

In fact, let's WLOG assume that the vertices $A_1$, $A_2$, ..., $A_8$ of the parallelepiped are labeled in such a way that the quadrilateral $A_1A_2A_3A_4$ is a parallelogram, and the segments $A_1A_5$, $A_2A_6$, $A_3A_7$ and $A_4A_8$ are the great diagonals of the parallelepiped (i. e., the diagonals which connect opposite vertices of the parallelepiped). Then, the center O of the parallelepiped must be the midpoint of each of these segments $A_1A_5$, $A_2A_6$, $A_3A_7$ and $A_4A_8$; hence, $OA_5=OA_1$, $OA_6=OA_2$, $OA_7=OA_3$ and $OA_8=OA_4$.

Since the quadrilateral $A_1A_2A_3A_4$ is a parallelogram, its diagonals $A_1A_3$ and $A_2A_4$ bisect each other; in other words, the segments $A_1A_3$ and $A_2A_4$ have a common midpoint. Denote this common midpoint by M. Then, since M is the midpoint of the segment $A_2A_4$, we have, using vectors, $\overrightarrow{OM}=\frac{\overrightarrow{OA_2}+\overrightarrow{OA_4}}{2}$, so that $\overrightarrow{OA_2}+\overrightarrow{OA_4}=2\cdot\overrightarrow{OM}$; thus, by the triangle inequality for vectors,

$\left|\overrightarrow{OA_2}\right|+\left|\overrightarrow{OA_4}\right|\geq\left|\overrightarrow{OA_2}+\overrightarrow{OA_4}\right|=\left|2\cdot\overrightarrow{OM}\right|$.

On the other hand, since M is the midpoint of the segment $A_1A_3$, we have $\overrightarrow{OM}=\frac{\overrightarrow{OA_1}+\overrightarrow{OA_3}}{2}$, so that $2\cdot\overrightarrow{OM}=\overrightarrow{OA_1}+\overrightarrow{OA_3}$, and thus, by the triangle inequality for vectors,

$\left|2\cdot\overrightarrow{OM}\right|=\left|\overrightarrow{OA_1}+\overrightarrow{OA_3}\right|\geq\left|\overrightarrow{OA_1}\right|-\left|\overrightarrow{OA_3}\right|$.

Hence,

$\sum_{i=1}^8 OA_i=OA_1+OA_2+OA_3+OA_4+OA_5+OA_6+OA_7+OA_8$
$=OA_1+OA_2+OA_3+OA_4+OA_1+OA_2+OA_3+OA_4$
$=2\left(OA_1+OA_2+OA_3+OA_4\right)$
$=2\left(OA_1+OA_3+\left(OA_2+OA_4\right)\right)=2\left(\left|\overrightarrow{OA_1}\right|+\left|\overrightarrow{OA_3}\right|+\left(\left|\overrightarrow{OA_2}\right|+\left|\overrightarrow{OA_4}\right|\right)\right)$
$\geq 2\left(\left|\overrightarrow{OA_1}\right|+\left|\overrightarrow{OA_3}\right|+\left|2\cdot\overrightarrow{OM}\right|\right)\geq 2\left(\left|\overrightarrow{OA_1}\right|+\left|\overrightarrow{OA_3}\right|+\left(\left|\overrightarrow{OA_1}\right|-\left|\overrightarrow{OA_3}\right|\right)\right)$
$=4\cdot\left|\overrightarrow{OA_1}\right|=4\cdot OA_1$.

Similarly, of course, $\sum_{i=1}^8 OA_i\geq 4\cdot OA_i$ for every other index i. Thus,

$\left(\sum_{i=1}^8 OA_i\right)^2=\sum_{i=1}^8 OA_i\cdot\sum_{i=1}^8 OA_i=\sum_{i=1}^8 \left(OA_i\cdot\sum_{i=1}^8 OA_i\right)$
$\geq \sum_{i=1}^8 \left(OA_i\cdot 4\cdot OA_i\right)=4\cdot\sum_{i=1}^8 OA_i^2$,

and the problem is solved.

  Darij