inequalities
function
inequalities unsolved

$ \textbf{(i)}$ Set $ a,b>0$ , prove : $ a^b\plus{}b^a>1$

$ \textbf{(ii)}$ Set $ a,b,c>0$ , and $ a\plus{}b\plus{}c\equal{}1$ , prove : $ \frac{a}{1\plus{}a^2}\plus{}\frac{b}{1\plus{}b^2}\plus{}\frac{c}{1\plus{}c^2}<\equal{}\frac{9}{10}$

false solution

Look in post#12 for the correct one  :wink:

[quote="FOURRIER"]for $ ii$:
Jensen give:
$ LHS\leq \frac {1}{1 \plus{} (a^2 \plus{} b^2 \plus{} c^2)^2}$
Thus It sufies to prove:
$ a^2 \plus{} b^2 \plus{} c^2\geq \frac {1}{3}$
which follows from CS  :wink:

@kg2100: It will be better if you change  the title :)
see here http://www.mathlinks.ro/Forum/viewtopic.php?t=135914[/quote]
But how did you prove that $ f(x)\equal{}\frac{x}{1\plus{}x^2}$ is concave?

[quote="Erken"]But how did you prove that $ f(x) \equal{} \frac {x}{1 \plus{} x^2}$ is concave?[/quote]

No, $ a\plus{}b\plus{}c\equal{}1$
and $ f(x)\equal{}\frac{1}{1\plus{}x^2}$ is concave  :wink:

SO,you need to prove that if $ 0\leq x\leq 1$ then $ f(x)\equal{}\frac{x}{x^2\plus{}1}$ is concave,show me your proof,cause i'm too lazy and it is too long to find $ f''$ and prove that $ f''>0$ for all $ x\in [0,1]$. :P

[quote="Erken"]SO,you need to prove that if $ 0\leq x\leq 1$ then $ f(x) \equal{} \frac {x}{x^2 \plus{} 1}$ is concave,show me your proof,cause i'm too lazy and it is too long to find $ f''$ and prove that $ f'' > 0$ for all $ x\in [0,1]$. :P[/quote]

I didnt consider $ f(x) \equal{} \frac {x}{x^2 \plus{} 1}$ at all !!
I considered $ f(x) \equal{} \frac {1}{x^2 \plus{} 1}$  :wink:

with $ a \plus{} b \plus{} c \equal{} 1$   :wink:  :lol:

Look!:

$ LHS \equal{} a\frac {1}{1 \plus{} a^2} \plus{} b\frac {1}{1 \plus{} b^2} \plus{} c\frac {1}{1 \plus{} c^2}$

with $ a \plus{} b \plus{} c \equal{} 1$  then apply Jensen to the concave function $ f(x) \equal{} \frac {1}{x^2 \plus{} 1}$

($ f' \equal{} \frac { \minus{} 2x}{(1 \plus{} x^2)^2}$) thats all

[quote="kg2100"]
$ \textbf{(ii)}$ Set $ a,b,c > 0$ , and $ a \plus{} b \plus{} c \equal{} 1$ , prove : $ \frac {a}{1 \plus{} a^2} \plus{} \frac {b}{1 \plus{} b^2} \plus{} \frac {c}{1 \plus{} c^2} < \equal{} \frac {9}{10}$[/quote]
We will prove
$ \frac{a}{a^2\plus{}1}\le \frac{18}{25}a\plus{}\frac{3}{50}$
Ineq <=> $ (4a\plus{}3)(3a\minus{}1)^2\ge 0$
So $ LHS\le \frac{18}{25}(a\plus{}b\plus{}c)\plus{}\frac{9}{50}\equal{}\frac{9}{10}$
I think only $ a,b,c\ge \frac{\minus{}3}{4}$
 :D

[quote="arqady"]
I think, you are wrong.
$ f''(x) \equal{} \frac {2(3x^2 \minus{} 1)}{(1 \plus{} x^2)^3}.$ :wink:
[/quote]

What a big mistake, I studied the sign of $ f'$ instead of $ f''$  :lol:

Sorry  :blush:

It is poland 1996 II round. But is extended. Compare all numbers with 0. There are different cases. You may use the result from poland or AM GM inequality after make all activities with fractions but it is too long.

I have found a solution
@arqady: dont tell me again its false  :wink:  :lol:

We have: $ \frac {a}{1 \plus{} a^2}\leq \frac {3}{10} \plus{} (a \minus{} \frac {1}{3})\frac {18}{25}$

How it cames :
[hide]I have founded it by lookin for k such that
$ \frac {a}{1 \plus{} a^2}\leq \frac {3}{10} \plus{} (a \minus{} \frac {1}{3})k$[/hide]

Now just sum them cycly to get the desired inequality  :wink:

[quote="arqady"][quote="FOURRIER"]

with $ a \plus{} b \plus{} c \equal{} 1$  then apply Jensen to the concave function $ f(x) \equal{} \frac {1}{x^2 \plus{} 1}$
[/quote]
I think, you are wrong.
$ f''(x) \equal{} \frac {2(3x^2 \minus{} 1)}{(1 \plus{} x^2)^3}.$ :wink:
By the way, for all non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 1$
 and for all $ k\geq\frac {1}{9}$ the following inequality holds:
$ \frac {a}{k \plus{} a^2} \plus{} \frac {b}{k \plus{} b^2} \plus{} \frac {c}{k \plus{} c^2}\leq\frac {9}{1 \plus{} 9k}.$ :wink:[/quote]
We will prove
$ \frac{a}{k\plus{}a^2}\le \frac{3}{9k\plus{}1}\plus{}\frac{27k\minus{}3}{(9k\plus{}1)^2}(3a\minus{}1)$
ineq $ \Leftrightarrow (3a\minus{}1)^2((9k\minus{}1)a\plus{}6k)\ge 0$
So $ LHS\le \frac{9}{9k\plus{}1}$
And I thing only $ a,b,c\ge \frac{\minus{}6k}{9k\minus{}1},k\ge \frac{1}{9}$
 :D

[quote="FOURRIER"]I have found a solution
@arqady: dont tell me again its false  :wink:  :lol:

We have: $ \frac {a}{1 \plus{} a^2}\leq \frac {3}{10} \plus{} (a \minus{} \frac {1}{3})\frac {18}{25}$
[/quote]
Yes, you are right! :lol:
[quote="FOURRIER"][quote="arqady"]
By the way, for all non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 1$
 and for all $ k\geq\frac {1}{9}$ the following inequality holds:
$ \frac {a}{k \plus{} a^2} \plus{} \frac {b}{k \plus{} b^2} \plus{} \frac {c}{k \plus{} c^2}\leq\frac {9}{1 \plus{} 9k}.$ :wink:[/quote]

arqady, how do find the generelisations, do you just try?[/quote]
See the last Mather's post. :wink:
By the way, it's true for $ k \equal{} \frac {1}{18}.$
Id est, for non-negative $ a,$ $ b$ and $ c$ such that $ a \plus{} b \plus{} c \equal{} 1$ the following inequality holds:
$ \frac {a}{1 \plus{} 18a^2} \plus{} \frac {b}{1 \plus{} 18b^2} \plus{} \frac {c}{1 \plus{} 18c^2}\leq\frac {1}{3}.$ :wink:
I have posted it here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170038

[quote="arqady"]
See the last Mather's post. :wink:
[/quote]

No, I asked about how do you find generlisations, not how do you prove them  :wink:
Because I saw you many times finding generelisations (with k) :thumbup:
Like this one http://www.mathlinks.ro/viewtopic.php?t=165948

Thanks  :laugh:

[quote="FOURRIER"]

No, I asked about how do you find generlisations, not how do you prove them  :wink:
[/quote]
OK. I just try to prove for all $ k$ and it gives the  condition  finding for $ k$.


The United States Postal Service charges an extra $ \$0.11$ in
postage if the length of an envelope, in inches, divided by its
height, in inches, is less than 1.3 or greater than 2.5. For how
many of these four envelopes must the extra $ \$0.11$ in postage
be paid?
\begin{tabular}[t]{ccc}
Envelope & Length in inches & Height in inches\\\hline
A &6 &4\\
B &9 &3\\
C &6 &6\\
D &11 &4
\end{tabular}

[quote="GameBot"]The United States Postal Service charges an extra $ \$0.11$ in
postage if the length of an envelope, in inches, divided by its
height, in inches, is less than 1.3 or greater than 2.5. For how
many of these four envelopes must the extra $ \$0.11$ in postage
be paid?
$ \begin{tabular}[t]{ccc}
Envelope & Length in inches & Height in inches\\\hline
A &6 &4\\
B &9 &3\\
C &6 &6\\
D &11 &4
\end{tabular}$[/quote]

6/4=1.5 (no extra)
9/3=3 (extra)
6/6=1 (extra)
11/4=2.75 (extra)

We must pay an extra 0.11 for 3 of them.


Show that an arbitary triangle can be dissected by straight line segments into three parts in three different ways so that each part has a line of symmetry.

Hint

[hide]For example, consider the segments connecting the vertices to the circumcentre...[/hide]


$x_1,..x_n$ are positive real numbers such that $x_ix_j \leq {\epsilon}^{|i-j|}$ for all $1 \leq i,j \leq n$, where $1> \epsilon > 0$ is a real.
Prove that : $x_1+x_2+..+x_n < \frac{1}{1-\sqrt{\epsilon}}$

Nice problem.
but sorry,I think it may be false  :P
for example:
let $x_k=1,x_{k-i}= \epsilon ^i$,and $x_{k+i}= \epsilon ^i$
let $n \to +\infty$ $k \to +\infty$
what will happen??
and I think we can change the problem to:
$x_1,x_2,\cdots,x_n$ are positive real numbers such that $x_i x_j \le \epsilon ^{|i-j|}$ for all $1 \le i,j \le n$, where  $0 < \epsilon < \frac 1 2$is a real.
Prove that : $x_1+x_2+..+x_n < \frac{1}{1-2\epsilon}$
and sorry $if$ I am wrong :P
what do you think?alekk.


Of the following numbers, what is the sum of the two smallest,
to the nearest thousandth: \\ $ 0.15 \hspace{.5cm} 0.42 \hspace{.5cm} 0.063 \hspace{.5cm} 0.1657$ ?

$ 0.063 \plus{} 0.15 \equal{} \boxed{0.213}$

[color=blue]What's easy for some people is hard for others, and vice versa. -Izzy[/color]


Can someone answer to my question please ? Google doesn't want to help me  :(

Only if it's cyclic, and that happens if and only if opposite angles are supplementary.  But opposite angles are also congruent.  Therefore, a rhombus is inscribable in a circle iff it's a square.


calculus
integration
trigonometry
logarithms
calculus computations

$ \int_{\minus{}\frac{\pi}{3}}^{\frac{\pi}{6}} \left|\frac{4\sin x}{\sqrt{3}\cos x\minus{}\sin x}\right|\ dx$.

$ \int_{\minus{}\frac{\pi }{3}}^{\frac{\pi }{6}}\left| \frac{4\sin x}{\sqrt{3}\cos x\minus{}\sin x}\right| dx$

$ \equal{}\int_{\minus{}\frac{\pi }{3}}^{0}\frac{4\sin x}{\sin x\minus{}\sqrt{3}\cos x}dx\plus{}\int_{0}^{\frac{\pi }{6}}\frac{4\sin x}{\sqrt{3}\cos x\minus{}\sin x}dx$

$ \equal{}\int_{\minus{}\frac{\pi }{3}}^{0}\frac{2(\sin x\minus{}\sqrt{3}\cos x)\plus{}2(\sin x\plus{}\sqrt{3}\cos x)}{\sin x\minus{}\sqrt{3}\cos x}dx\plus{}\int_{0}^{\frac{\pi }{6}}\frac{\minus{}(\sqrt{3}\cos x\minus{}\sin x)\minus{}\sqrt{3}(\minus{}\sqrt{3}\sin x\minus{}\cos x)}{\sqrt{3}\cos x\minus{}\sin x}dx$

$ \equal{}\allowbreak \frac{\pi }{6}\plus{}\sqrt{3}\ln \frac{3\left( 2\minus{}\sqrt{3}\right) }{2\sqrt{3}\minus{}3}$

Factoring out $ \sqrt{3}$ from $ 2\sqrt{3}\minus{}3$ gives simplest form.

hello, we have $ \frac{\pi}{6}\plus{}\sqrt{3}\ln\left(\frac{3(2\minus{}\sqrt{3})}{2\sqrt{3}\minus{}3}\right)\equal{}\frac{\pi}{6}\plus{}\frac{\sqrt{3}}{2}\ln(3)$.
Sonnhard.


geometry
circumcircle
symmetry
rectangle
analytic geometry
rotation
Pythagorean Theorem

Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$.
[asy]
size(200);
defaultpen(fontsize(10));
real x=22.61986495;
pair A=(0,26), B=(26,26), C=(26,0), D=origin, E=A+24*dir(x), F=C+24*dir(180+x);
draw(B--C--F--D--C^^D--A--E--B--A, linewidth(0.7));
dot(A^^B^^C^^D^^E^^F);
pair point=(13,13);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));[/asy]

I thought this problem was the easiest on the test once you saw the solution.

[hide="Solution"]Extend $FC$, $EB$, $EA$, and $FD$ through $C$, $B$, $A$, and $D$, respectively. The two newly formed triangles, using a tiny bit of angle chasing and ASA, are congruent to the original two (which are both right as they are 5-12-13). The extended sides form a $17\times 17$ square. $EF$ is simply a diagonal, and thus the length is $17\sqrt{2}$, so the answer is $\boxed{578}$.[/hide]

That's much better than the way I saw first looking at it... mine isn't too bad though.

[hide]
Draw the circumcircle around the triangles. If we let the point of tangency of the two circles be O, the distance we need is $FO+EO$. We find $AO=BO=CO=DO=\frac{13\sqrt{2}}{2}$. Then by Ptolemy:

$(5+12)\frac{13\sqrt{2}}{2}=13(FO)$
$\frac{17\sqrt{2}}{2}=FO$
So then double it and square it to get $\boxed{578}$
[/hide]

This one may not have been the easiest to see the solution, but once you include computation and all I thought it was the easiest on the test by far.

It took me under 30 seconds  :ninja:

[quote="dbell0971"]I drew an absolute blank on this problem.  I knew there was some simple solution, but I couldn't find it.  So, I drew it out, using a ruler :D .  It came out to be exactly 24 cm, so I put down 576  :( .  Dang it!!![/quote]

 :wink: Nice try

O is the center of the square..by symmetry, EOF is a line...AEB=AOB=90, so AEBO is cyclic...ptolemy: 12*13/root 2+5*13/root 2=EO*13...EF^2=(2EO)^2=578

I turned the thingy into a rectangle and then drew the heights of the two triangles to get a big right triangle that had nasty numbers but I didn't make a mistake in the algebra so I got this one pretty quickly.

I used analytic bashing, and then was checking and was like ...

oh

...

after I saw that you could extend the sides to form a square. I was sooooooooo mad... lol

Or you could draw it on a coordinate plane and realize that point E was at (0,0) and point F at (17,17)

[quote="SamE"]Ha ha...

I used Pythagorean Theorem on a triangle with legs parallel to the sides of the square. The sides of the right triangle where 119/13 and 289/13. Then I squared those 3-digit numbers, added them, and divided out 13 twice and still got the right answer.

Then I realized that both 119 and 289 were multiples of 17, so I checked my answer that way. But I never actually saw the obvious (kind of reminds me of the BEHOLD proof of Pythagorean Theorem) geometric solution.

If it works, I guess...[/quote]

thats the way i did it (after i drew a diagram that was totally off). initially, my plan was to draw the diagram and do it mathematically to see if the answers i got somewhat clicked, but then i got 640 from my diagram and 578 from my calculations. i decided to go with the mathematical answer and got it right...trust ur math ppl! :lol:

I find it funny that if they had rotated just a couple degrees thae answer would be really obvious.

I really like this problem. From the position of the problem creator, this is very nice.

Except for two erased triangles, this is just the diagram for the pythagorean theorem proof given by archimedes.

[quote="SamE"]Ha ha...

I used Pythagorean Theorem on a triangle with legs parallel to the sides of the square. The sides of the right triangle where 119/13 and 289/13. Then I squared those 3-digit numbers, added them, and divided out 13 twice and still got the right answer.

Then I realized that both 119 and 289 were multiples of 17, so I checked my answer that way. But I never actually saw the obvious (kind of reminds me of the BEHOLD proof of Pythagorean Theorem) geometric solution.

If it works, I guess...[/quote]
That's exactly what I did... it doesn't take that long either...

[quote=peace09]Hello @Bove,
No need for all that. Simply extend $\overline{AE},$ $\overline{EB},$ $\overline{CF},$ and $\overline{FD}$ to form a square with sidelength 5 + 12 = 17, from which $EF^2,$ the square of the length of the diagonal, is $2 \cdot 17^2 = \boxed{578}.$[/quote]

omg that is so big brain :o

you really thought outside the box :rotfl:

(because the bigger square is outside the box)

[quote=sugar_rush][quote=brainfertilzer]Let $\angle CDF = x$. Then, let $O$ be the center of the square, so $EF = 2OF$. Note that $OD = 13/\sqrt{2}$.By the law of cosines on $\triangle ODF$, we have
\[ OF^2 = \frac{169}{2} + 25 - 65\sqrt{2}\cos(\angle ODF).\]
Note that $\cos\angle ODF = \cos(45^\circ + x) = \frac{\sqrt{2}}{2}(\cos x - \sin x) = \frac{\sqrt{2}}{2}(\frac{5}{13} - \frac{12}{13}) = -\frac{7\sqrt{2}}{26}$. Upon plugging this into our equation for $OF^2$, we get $OF^2 = 289/2$, so $EF^2 = 4OF^2 = \boxed{578}$[/quote]

That's kind of overkill actually, you can just realize that $EF$ is the diagonal of a square with side $12+5=17$ so $EF^{2}=2\cdot 17^{2}=\boxed{578}$[/quote]

nice you reached 1k posts congrats

[asy]
size(200);
defaultpen(fontsize(10));
real x=22.61986495;
pair A=(0,26), B=(26,26), C=(26,0), D=origin, E=A+24*dir(x), F=C+24*dir(180+x), G=(-24,0)*dir(-69),H=(35,0)*dir(6);
draw(B--C--F--D--C--H--B^^D--A--E--B--A--G--D, linewidth(0.7));
dot(A^^B^^C^^D^^E^^F);
pair point=(13,13);
label("$A$", A, dir(point--A));
label("$G$", G, dir(point--G));
label("$H$", H, dir(point--H));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));[/asy]
Extend $\overline{DF}$ and $\overline{AE}$ to intersection point $G$ and $\overline{EB}$ and $\overline{FC}$ to intersection point $H$. We now have square $GFHE$ with side lengths of $17$, so $\overline{EF}=17\sqrt{2}\implies \overline{EF}^2=\boxed{578}$.

[quote=sugar_rush][quote=brainfertilzer]Let $\angle CDF = x$. Then, let $O$ be the center of the square, so $EF = 2OF$. Note that $OD = 13/\sqrt{2}$.By the law of cosines on $\triangle ODF$, we have
\[ OF^2 = \frac{169}{2} + 25 - 65\sqrt{2}\cos(\angle ODF).\]
Note that $\cos\angle ODF = \cos(45^\circ + x) = \frac{\sqrt{2}}{2}(\cos x - \sin x) = \frac{\sqrt{2}}{2}(\frac{5}{13} - \frac{12}{13}) = -\frac{7\sqrt{2}}{26}$. Upon plugging this into our equation for $OF^2$, we get $OF^2 = 289/2$, so $EF^2 = 4OF^2 = \boxed{578}$[/quote]

That's kind of overkill actually, you can just realize that $EF$ is the diagonal of a square with side $12+5=17$ so $EF^{2}=2\cdot 17^{2}=\boxed{578}$[/quote]
Right, I read about that in the solutions, but I was just presenting the -1 iq solution that I used lol.


Determine the greatest $ k$ such that the inequality below
\[ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}abcd\plus{}1\geq k(abc\plus{}bcd\plus{}cda\plus{}dab)\]
i) is true for all non-negative real numbers $ a,b,c,d$.
ii) is true for all real numbers $ a,b,c,d$.

[quote="hitek"]Determine the greatest $ k$ such that the inequality below
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} abcd \plus{} 1\geq k(abc \plus{} bcd \plus{} cda \plus{} dab)
\]
i) is true for all non-negative real numbers $ a,b,c,d$.
[/quote]
If $ d\equal{}0$ and $ a\equal{}b\equal{}c$ then $ 3a^2\plus{}1\geq ka^3.$ Thus, the maximal $ k$ for which it's true for all $ a\geq0$ is $ k\equal{}0.$


$x,y,z>1$. Prove that $(log_{yz}x^4yz)(log_{zx}xy^4z)(log_{xy}xyz^4)>27$

I think it's false. If we let $a = log x$ and so forth, we have it equivalent to $(a+b-2c)(b+c-2a)(c+a-2b) \ge 0$, which may not be true..

if this is true then it is >=27 not >27(just look when x=y=z)

indeed the inequality is wrong :P .it took me some time to realize that :oops:


[quote="pii"]How do you do this problem?

When a and b are positive integers, a<b and ab[u]<[/u]a+3b, how many possible values are there for a?[/quote]
[hide]
$b(a-3) \leq a$
$b(a-3) < b$
$ab<4b$
$a<4$
3[/hide]

one question: why did you change the $a$ to a $b$ when you did the equation $b(a-3)\leq a$ to $b(a-3)\< b$?

[quote="tridecalogism_lover"]one question: why did you change the $a$ to a $b$ when you did the equation $b(a-3)\leq a$ to $b(a-3)\leq b$?[/quote]

so then both sides have a b, and it's easier to work with?

yeah but [b]how[/b] can you change from a to b? :| [/b]

Since $b(a-3)\le{a}$ and $a<b$, we have that $b(a-3)\le{a}<b\Rightarrow b(a-3)<b$.  This is how I would do it:
[hide] $ab\le{a+3b}\Rightarrow ab-a-3b\le{0}\Rightarrow ab-a-3b+3\le{3}\Rightarrow (a-3)(b-1)\le{3}$.  Since b must be greater than a and a is an integer greater than 0 (so, a is greater than or equal to one), b is greater than 1.  Thus, b is greater than or equal to 2.  So, b-1 is always positive.  Thus, a-3 must be negative or 0 unless b-1=3.  If b-1=3, then b=4.  So, a=(1,2,3).  Now, if $a-3\le{0}\Rightarrow a\le{3}\Rightarrow a=(1,2,3)$.  So there are $\boxed{3}$ values for a.  I would prefer bpms's way, though.  [/hide]


-------------------------------

(A union B union C)^c = (A^c intersection B^c intersection C^c)

Where ^c denotes the compliment, and A, B, and C denote subsets of a universe set U.

--------------------------------

Could I solve it this way?

(A union B union C)^c = (A^c intersection B^c intersection C^c)

U - (a union b union c) = U - a intersection U - b intersection U - c

U - (a + b + c) = U - a - b - c

U - a - b - c = U - a - b - c

I'm still a little confused, but I've been spending a long time on this problem and this was the best I could come up with. Could someone help me clarify the solution?

[b]Important:[/b] What is the notion of set subtraction that you're using? I won't try to guess, but all I can say is that if it is at all meaningful, then you'd anyway be better off by using conventional, algebraic means. There's a simple way to do this that will give you the generalization for your desired result in the same time it takes to get the proof for this problem.

[b]Hint:[/b] This generalizes to any integer number of sets. Prove it for 2 sets, then strong-induct over N, the number of sets.

I'm not sure I follow your hint... could you show me the algebraic way that you suggested?

I'm going to use ' to denote complement, + to denote union, and "multiplication" to denote intersection. The proof of the base case is a bit informal -- take a look at [url=http://en.wikipedia.org/wiki/Demorgan%27s_law]the DeMorgan's Law[/url] Wikipedia page to see a formal proof. The inductive step is the interesting part, and also the easier part.

[hide="Inductive Proof"]
[b]Base case (n=2):[/b]
Verify that AB, AB', A'B, A'B' are pairwise disjoint and cover U.
[color=darkblue](A+B)'
= (AU+BU)'
= (A(B+B')+B(A+A'))'
= (AB+AB'+AB+A'B)'
= (AB+AB'+A'B)'
= A'B'[/color]

[b]Inductive step:[/b]
[color=darkblue](A+B+....+N)'
= ((A+B+...+M)+N)'
= (A+B+...+M)'N'[/color] [color=blue][i][using the case of 2 sets, (A+B+...+M) and N][/i][/color]
[color=darkblue]= (A'B'...M')N'[/color] [color=blue][i][using the case of M sets, A, B, ..., M][/i][/color]
[color=darkblue]= A'B'...M'N'[/color][/hide]

absvalue,

the mathematical term you typed should be spelled "complements."

A "compliment" is something nice you say and is different from
a "complement."


Putnam
calculus
function
abstract algebra
algebra
polynomial
Stanford

This topic is for general discussion of the 2008 Putnam - how easy, how hard, general impressions, how you did, why they brought "determinant tic-tac-toe" from 2002 back for an encore, and so on.

Specific comments on problems should go on the 12 problem threads that I'm about to post.

Anyone else think it was a lot harder than last year? Especially the second set of 6.

While proctoring, I got A1, A3, A4, A5, B1, B2, B3, B5. Which isn't to say that I'd get 80 points if I turned in that work - I wouldn't. But it is a little better than I usually do while proctoring. I figure I'm a marker for the mid-range: how hard it is to get top 500 or top 200, not above or below that. On the other hand, I'm a calculus/analysis guy and maybe I appreciated that there were several calculus problems in there: A4 (which I enjoyed), B2, B3.

This was my first, but I'll agree that the second half was crazy hard; I didn't turn in anything on part B except a (probably wrong, and definitely incomplete) attempt at B3.

A2 stood out being the easiest problem among our group. We were talking about this question after session since it seemed so easy such that, it being a Putnam problem, there might be something sinister behind it or a catch. That, or our university shoved down a lot of linear algebra down our throats...

Can't believe I missed A2, but I think I got A1, A3, and some partial credit for B2 and B3.

I was just reading the solution to A2. Cant believe I missed it. Anyways, I think I got A1 and B2. I tried A3 but I am sure I will not get anything for that.

[quote="BirdKiller"]A2 stood out being the easiest problem among our group. We were talking about this question after session since it seemed so easy such that, it being a Putnam problem, there might be something sinister behind it or a catch.[/quote]

Considering that A2 was the only one I actually finished, I have to agree it was the easiest.
I got the answer really quickly, too.  Well, most of it.  I thought I had finished it about 40 minutes in, then looked at it again with about 45 left and realized I hadn't.  (Fortunately, it was pretty easy to finish.)  Our proctor agreed that it seemed very easy.  I think I've reread it at least five times since the exam to see if I missed anything.

*Looks again*

Still don't see anything I missed; maybe it really is that easy?

(On a related note, I [i]really[/i] hope Barbara knows this strategy.  Or, if she doesn't, that at least she and Alan have enough sense to put some zeroes in.  I don't even want to think about how long calculating that determinant would take . . .)

I think I might get some credit for A3; I'm pretty sure I was able to prove that the process has to stops.

I'm thinking maybe 11-12 points, at most.

I thought way too hard about A2 and missed it.  After our discussion about it at lunch, I was so annoyed that I spent all of my time working on B2 in the afternoon and kept messing up because I tried to be too tricky and introduced a generating function that had bad convergence properties.

In my opinion, A1 and A2 were awful problems. No one feels good about getting stupid tricks like that, and everyone who misses it feels like crap.

(for the record, I'm not the bitter one; I got A1 quickly, though I wasted a bit of time on A2 trying tricky stuff about the span of the rows before Barbara's last move)

I think this year's A exam was easier than last year's but part B was harder.
I think I got A1, A2, A3, A4, A5 and B1, B2, B4 and an incomplete B5 (time ran out, and I solved it only in the last 20 minutes..)

I think I got most guys except B3. As someone who doesn't know group theory, A6 was pretty easy after thinking about binary representations for abelian groups; the general case just requires you to double the bound by tacking on inverses in reverse order. B6 was really quite fun and it was very nice to see how the linear algebra naturally gave you a mod 2k+1 condition which you most likely would not have expected if it was not given in the problem statement! I think A5 was a fairly standard polynomial calculation with Lagrange interpolation that reduced to binomial formula and I don't remember what B5 was now. I found the problems very nice though.

This Putnam seemed about as hard or slightly easier than last year to me. The B half was difficult, but so was last year's A half.
Does anyone have guesses on what the cutoffs of the groups (fellow, group I, etc.) will be?

I found the A half much easier than last year, while the B half was probably about as hard as last year. I solved A1, A2 (took a long time to convince myself that such a simple solution was correct), A3, A4, B1, B2, B4. If I had time, I might have done one of A5/A6, but I took a bad approach to A4 that took too long to write. On the B half, I solved B1, B2, B4 in the first hour and then spent the last two hours banging my head on B3/B5 to no avail.

In the solution to A1, it's not just $ g(x)\equal{}f(x,0)\plus{}c$ that works but $ g(x)\equal{}f(x,k)\plus{}c\,\forall k,c\in\mathbb{R}$

Well, I think I might be the only person who actually did just as well on the B part than the A part. I ended up getting A1, A2, A3, A5, B1, B2, B3, B4. As happened in practice, it was the analysis that got me. I spent nearly an hour on A4 and even thought of the integral test... d'oh.

I would agree that A1 and A2 were bad problems, given the number of people I know who didn't get them and how stupid they felt afterwards. A3 was also quite tedious to write up, but most problems like that are so I was okay with that.

Also, was it just me or did we see a whole ton of natural logs on this Putnam?

Okay, on second thought, I'm thinking they might take off points on my A3.  I essentially stated the following without proving it:

Suppose we have a set of $ k$ sequences, each of length $ n$.  We perform an operation as follows.  Pick two different indices $ i$ and $ j$ between $ 1$ and $ n$, inclusive.  Take the $ i$th and $ j$th terms in each of the $ k$ sequences, and rearrange them in nondecreasing order.  This operation is only allowed if there is some sequence which is affected by this operation.  Then after some finite number of iterations of this operation, all of the sequences will be in nondecreasing order.

I basically figured proving this would disrupt the flow of the proof, and was completely unnecessary (you may do it by setting up a monovariant), but it seems that it was necessary to write up to get full credit.  If so, weak.  So totally weak.

this was my first putnam - i finished the first problem in about half an hour and was excited that i might finish all the problems.  then i read the rest of the problems. >_<

This was also my first Putnam.  I found A2 to be rather easy.  After that, A3 seemed rather easy to because it was so much like bubble sort and the fact that it resembled a problem from I think Ch. 3 or so from The Art and Craft of Problem Solving.

I did horribly on B.  I only took a stab at B3, but I definitely think I will get a 0 on it.  But since it is my first time taking the test, anything will really do. Lol.

[quote="Sly Si"]Consensus at Stanford was that A-side was the easiest half of a Putnam we've ever seen and will likely inflate scores quite a bit. [/quote]

I feel like that was just people exaggerating a lot based on A1/A2 being one trick ponies. Even still, they [i]did[/i] require a trick (that a number of people didn't see) and A3-A6 were far from trivial.

In my opinion, the only problems this year that were more difficult than their corresponding problems from last year were B1, B3, and possibly A5. Since last year's top 500 cutoff was >20 (if I remember correctly), I could definitely see really high cutoffs on the lower end this year.

Overall I felt like this was easier than last year's, but not by much. I would expect all cutoffs to be slightly higher. As usual my score will depend massively on generosity with partial credit -- I "solved" 7 problems but made omissions of varying degrees of severity on at least 4 of those. My past three ranks have been around 600, around 300, and just over 100; I'm hoping that I did better than last year but it's hard to tell. I'll be happy with top 100 but I'm really hoping for HM. It seems a bit optimistic at this point though.

Several people on this thread - plugh most recently - have made speculative statements comparing this to previous years. Here's my difficulty table (I've posted this before) - where do you think it fits?

[b]No. taking:[/b] The number of students in U.S. and Canadian universities that participated in the Putnam. Note that this number is generally trending upwards, although it has never been particularly large.

[b]Mean:[/b] The mean score among all takers on a 0-120 scale.

[b]Median:[/b] The median score among all takers. Mythology says it's always zero; as you can see that's a slight exaggeration.

[b]% zero:[/b] The percentage of all takers who did score zero.

[b]Top 500:[/b] The minimum score required to be placed on the list of the top 500 individuals. This is never precisely 500 names, because of broad-based ties. In 2007, it was the top 455.

$ \begin{array}{cccccc}\text{Year} & \text{No. taking} & \text{Mean} & \text{Median} & \text{\% zero} & \text{Top 500} \\
1998 & 2573 & 14.8 & 10 & 30.3 & 28 \\
1999 & 2900 & 6.3 & 0 & 60.2 & 11 \\
2000 & 2818 & 5.3 & 0 & 57.7 & 11 \\
2001 & 2954 & 8.9 & 1 & 44.9 & 20 \\
2002 & 3349 & 11.0 & 3 & 34.7 & 24 \\
2003 & 3615 & 7.1 & 1 & 27.8 & 18 \\
2004 & 3733 & 8.4 & 0 & 53.6 & 22 \\
2005 & 3545 & 7.9 & 1 & 46.7 & 20 \\
2006 & 3640 & 6.2 & 0 & 62.6 & 14 \\
2007 & 3753 & 7.7 & 1 & 42.5 & 21\end{array}$

This was my first time ever taking the putnam so I went in with the expectation of getting a zero.  I thought I had A1, but I guess in the process of worrying about proving a special case I managed to prove..a special case, I had an idea I just couldn't properly express it I guess.  I'm really hoping to pick up a point on A4, I used the integral test and showed that it diverged...I dunno, we'll see I guess.  I also submitted B1 but completely blew it on that one.

Since i'm new to this, how do we receive scores?  And when do they usually get posted/sent?

[quote="WelcomeGhosts"]Since i'm new to this, how do we receive scores?  And when do they usually get posted/sent?[/quote]
A mailing goes out in March - the actual date has varied quite a bit over the last several years - to the Putnam supervisor at each university. This mailing contains the numerical scores of each student at that university (but not Putnam Fellows), but not any question-by-question breakdown. If you're told that your score is 8, you've probably got a pretty good guess where you got that credit - but it's still a guess.

The mailing also contains the complete list of all of those anywhere in the top 500, broken down into categories: Fellows (top ~5); Group N1 (next ~10); Group N2 (next ~10); Honorable Mention (next ~50); Group I (rest of the top 100); Group II (rest of the top 200); and the rest of the top 500. There's also some statistical information about scores, from which it is possible to convert any raw score into a rank.

Nothing gets mailed to the competitors directly. How much information you ever get and how it gets shared with you is entirely up to the Putnam supervisor at your university.

When I get my copy of that mailing, I'll start a new topic here with the information I have. Look in this forum for versions of that particular topic from previous years (and note the date the first post of each such topic went up.) Don't bother looking on the official Putnam website for anything timely but do look at the Putnam archive on the AMC site - Kiran Kedlaya will update that as soon as he can.

Well, I'll go on record saying that I found this year to be harder than many previous years.  Sounds like I am a bit of an outlier though.  Usually when I sit down with a Putnam I can get one or two problems in a few hours time.  This test I failed miserably at getting anything more than a point.  Good times none-the-less.

I am really tired of Putnam by now.  I don't think I will get anything above 10.  I am hoping for a 10, but it may end up being a 1.  I got 0 last time.  I did few problems from previous years and it seems to me that there are really nothing new on this exam.  Almost all the problems have very similar techniques.  I guess if anyone is boring enough to do these kinds of problems over and over again, they will be good at it eventually.  Take b2, it's so very typical, work out the discrete form of the integral, and plug it in.  There's nothing creative about that. And who after topology and group theory still remember integrals!

[quote="sexynsmartjenny"]And who after topology and group theory still remember integrals![/quote]
Well, there are a few people out there who do :P. But you are right, the students who like hardcore analysis and are capable of doing it are quite rare nowdays :(.


Suppose $a,b\in\mathbb{N}$ s.t.
$p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}\in\mathbb{P}$;

find the maximum value of $p$.

$p=5$
obviously $b$ must be even, so let
$b=2c\Rightarrow p=\frac{c}{2}\sqrt{\frac{a-c}{a+c}}\Rightarrow \frac{4p^{2}}{c^{2}}=\frac{a-c}{a+c}$
assume $\frac{2p}{c}=\frac{m}{n}$ where $(m,n)=1$.
assume $k=(a-c,a+c)\Rightarrow a-c=km^{2}, a+c=kn^{2}$
$\Rightarrow 2c=k(n^{2}-m^{2})$
$\Rightarrow 4pn=km(n^{2}-m^{2})$
if both m,n are odd then $8|m^{2}-n^{2}$ which implies p is even $\Rightarrow p=2$.
if both m,n  are not odd then then $n^{2}-m^{2}$ is odd, so k is even.
$\Rightarrow k=2r$
$\Rightarrow 2pn=rm(n^{2}-m^{2})$
since $(n,m)=(n,n^{2}-m^{2})=1$ we have $n|r$.
$\Rightarrow r=ns$
$\Rightarrow 2p=s(n-m)(n+m)m$.

suppose $p>2$.  Since both $m+n$ and $n-m$ are odd, $m+n=p$ and $n-m=1$.  which implies that s and m are each at most 2. so now we have the following two possibilities:
$(m,n)=(1,2)$ or $(2,3)$.  in both cases, $p\leq 5$.  (note that for p=5, all we have to  is let:  m=2, n=3, s=1, r=3,k=6, c=15, b=30 and a=39).
and we are done. :)


There is a set of five positive integers whose average (mean) is 5,
whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?

We can write out the numbers, 5, 8, 8.

The other two numbers are 1 and 3, 25-8-8-5=4 (which can be written as 1 and 3). Thus our desired answer is $ 8\minus{}1\equal{}\boxed{7}$.


[u][b]Theorem:[/b][/u]Let $ ABC$ be a triangle .$ M;N;P$ are points on $ BC;CA;AB$,respectively.
Suppose that $ \Delta_1;\Delta_2;\Delta_3$ are three lines through $ M;N;P$ and $ \perp BC;CA;AB$,respectively.
Prove that $ \Delta_1;\Delta_2;\Delta_3$ concurrent  when $ MC^2\minus{}MB^2\plus{}NB^2\minus{}NA^2\plus{}PA^2\minus{}PC^2\equal{}0$
I need some manuals about this theorem.
Can you help me?
Thanks a lots!

This is Carnot theorem! It can be proved by use Pitago theorem!
But you have a mistake! It must be $ MB^2 \minus{} MC^2 \plus{} NC^2 \minus{} NA^2 \plus{} PA^2 \minus{} PB^2 \equal{} 0$
If $ \Delta_1\cap \Delta_2 \equal{} O$ then
$ MB^2 \minus{} MC^2 \plus{} NC^2 \minus{} NA^2 \plus{} PA^2 \minus{} PB^2 \equal{} 0$
$ \Leftrightarrow OB^2 \minus{} OC^2 \plus{} OC^2 \minus{} OA^2 \plus{} PA^2 \minus{} PB^2 \equal{} 0$
$ \Leftrightarrow PA^2 \minus{} PB^2 \equal{} OA^2 \minus{} OB^2$
It mean $ OP,AB$ are perpendicular!
Then $ O\in\Delta_3$

[b]A remarcable locus:[/b] [color=brown]Let $ A_1,A_2$ two given points and $ k,k_1,k_2$ real numbers.The locus for the points $ M$ with the property that $ k_1MA_1^2 + k_2MA_2^2 = k$ is:
[b]a)[/b] a circle,a point or nothing if $ k_1 + k_2$ diffrent from $ 0$
[b]b)[/b] a line perpendicular on $ A_1A_2$ if $ k_1 + k_2 = 0$.[/color]

[b][u]Proof[/u][/b] [color=brown]a).[/color]Consider the case when $ k_1,k_2 > 0$.Let $ D\in [A_1A_2]$ such as $ \frac {A_1D}{DA_2} = \frac {k_2}{k_1}$
From the cosine law in $ \{\begin{array}{cccc} \Delta MA_1D \\
 \\
\Delta MA_2D\end{array}$ $ \implies$  $ \{\begin{array}{cccc} MA_1^2 = MD^2 + A_1D^2 + 2A_1D\cdot DMcos \alpha \\
 \\
MA_2^2 = MD^2 + A_2D^2 - 2A_2D \cdot DMcos \alpha \end{array}$ $ \implies$

   $ k_1MA_1^2 + k_2MA_2^2 = (k_1 + k_2)MD^2 + k_1A_1D^2 + k_2A_2D^2$

If $ k_1 + k_2$ is different from $ 0$ we obtain $ MD^2 = \frac {k - k_1A_1D^2 - k_2A_2D^2}{k_1 + k_2} = K$(constant)

Therefore the locus for $ M$ is a circle with the center in $ D$ , a point (M=D) or nothing .
Note that in the case $ k_1 \cdot k_2 < 0$ the point $ D$ will not be situated on the segment  $ A_1A_2$ (the case $ k_1 < 0,k_2 < 0$ reduces to the first one by changing the signes of $ k_1,k_2,k$)
The reciprocal affirmation verifies imediately: any point $ M$  verifies the relation: $ k_1A_1M^2 + k_2A_2M^2 = k$

[b][u]Remark[/u][/b] [color=brown]The locus for the points $ M$ for who $ AM/MB = k$ is a circle with the center on the line $ AB$[/color] [color=brown][b](Apollonius)[/b][/color]

[color=brown]b).[/color]If $ k_1 + k_2 = 0$ $ \equiv$ $ k_1 = - k_2$,the problem reduces to the well know problem of determination of the locus for the points with the diferences of the square distances at two points given as constant.

And now, the famous Carnot theorem:
[color=brown]Let the triangle $ ABC$ and the points $ A_1,B_1,C_1 \in P$.The projections $ a_1,b_1,c_1$ from $ A_1,B_1,C_1$ on the lines $ BC,CA,AB$ concur only if takes place
\[ \boxed{BA_1^2 - CA_1^2 + CB_1^2 - AB_1^2 + AC_1^2 - BC_1^2 = 0}
\]
[/color]

[u][b]Proof[/b][/u] We will use the locus presented.
$ CM^2 - CA_1^2 = BM^2 - BA_1^2$ (because $ BC \bot MA_1$) therefore $ BA_1^2 - CA_1^2 = BM^2 - CM^2$.Writing the other analogously the conclusion results imediatly.

Here there are 3 aplications to this beautifull problem:
[b]1.[/b] [color=brown]Consider the triangles $ ABC$ and $ A_1B_1C_1$.The projections from $ A_1,B_1,C_1$ on $ BC,CA,AB$ concur only if the projections from $ A,B,C$ on $ B_1C_1,C_1A_1,A_1B_1$ concur .[/color]

[b]2.[/b][color=brown] Let a triangle $ ABC4$ and a line $ d$ as well as the points $ A',B',C'$ the projections of $ A,B,C$ on $ d$.The projections from $ A'$ on $ BC$ and the analogously concur in one point.[/color]

[b]3.[/b] [color=brown]On the sides of the triangle $ ABC$ construct the isoscelles triangles $ AB_1C,BA_1C,AC_1B$.Prove that the projections from $ A,B,C$ on the lines $ B_1C_1,C_1A_1,A_1B_1$ concur.[/color]

Hope this will help you!

[b]Aplication for:[/b] $ AC\perp BD \Longleftrightarrow |AD|^2 \plus{} |BC|^2 \equal{} |AB|^2 \plus{} |CD|^2.$

[b]Proof:[/b]
$ AC\perp BD \Longleftrightarrow \overrightarrow{AC}.\overrightarrow{BD} \equal{} 0 \Longleftrightarrow \overrightarrow{AC}.(\overrightarrow{AD} \minus{} \overrightarrow{AB}) \equal{} 0 \Longleftrightarrow \\
\Longleftrightarrow \overrightarrow{AC}.\overrightarrow{AD} \minus{} \overrightarrow{AC}.\overrightarrow{AB} \equal{} 0 \Longleftrightarrow \\
\Longleftrightarrow (|AC|^2 \plus{} |AD|^2 \minus{} |CD|^2) \minus{} (|AC|^2 \plus{} |AB|^2 \minus{} |BC|^2) \Longleftrightarrow \\
\Longleftrightarrow |AD|^2 \plus{} |BC|^2 \equal{} |AB|^2 \plus{} |CD|^2.$

In case the triangle ABC was an equilateral one the formula becomes:

[b]BM - MC + CN - AN + AP - BP = 0[/b] (*)

The proof is very simple:
BM^2 - MC^2 + CN^2 - AN^2 + AP^2 - BP^2 = 0  <==>
(BM - MC)(BM + MC) + (CN - AN)(CN + AN) + (AP - BP)(AP +BP) = 0 ( 1 ) but BM + MC = CN + AN = AP + BP, hence (1) becomes (*), after properly simplifying.

Best regards,
sunken rock


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Εδώ να μαζέψουμε θέματα από τα [color=red]Πολυώνυμα , Ακολουθίες , Συστήματα [/color]κλπ. Καλό είναι να συγκεντρώσουμε και να επιλέξουμε [b]20 ασκήσεις.[/b]
 Να βάλουμε πρώτα αυτές που ήδη έχουμε στείλλει και συζητήσει. Φυσικά αυτές που νομίζετε ότι ξεχωρίζουν.

 Μπάμπης

[size=150][color=red]ASKHSH 1[/color][/size]

Estw $S=1!2!3!\cdots 99!100!$. Apodeikste oti yparxei akeraios $k$, me $1\leq k \leq 100$ gia ton opoio to piliko $\frac{S}{k!}$ einai teleio tetragwno akeraioy. Einai o $k$ monadikos$;$

Alexandros

H Lisi episinaptetai sto arxeio .pdf

[quote="darksaga"]να θεσω κι εδω το αιτημα μου για το να συμπεριλαβουμε το 0,99..=1 και τουσ σχετικους τροπους...(νομιζω οτι βγαζει πολλους απο διαφορες στενομυαλες αποψεις...και πιστεψτε με ειναι παααααρα πολοι αυτοι)[/quote]

 Νομίζω ότι το ερώτημα αυτό έχει ενδιαφέρον , αλλά δεν άπτεται ολυμπιακού ενδιαφέροντος. Στείλτο βέβαια για να το δούμε, αλλά πιθανόν να μην κολάει πουθενά. Αν θέλεις , μπορώ να το αξιοποήσω διαφορετικά σε κάποιο άλλο χώρο. Όμως , έχεις τόσα άλλα ωραία θέματα να στείλλεις!  Στο 0 , 999... στάθηκες !:)
 Ξεκίνα λοιπόν με ολυμπιακά θέματα !

                   Μπάμπης

[b]ΑΣΚΗΣΗ 2[/b]

[b]ΕΚΦΩΝΗΣΗ[/b]

Η ακολουθία $a_{1}$ ,$a_{2}$,..... ορίζεται από τον τύπο :

$a_{1}=1$ , $a_{2}=a_{1}+\frac{1}{a_{1}}$ ,....., $a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$.

Να αποδείξετε ότι $a_{100}>14$

[b]ΛΥΣΗ
[/b]

Θα αποδείξουμε την ανισότητα $14<a_{100}<18$.Για $k>1$ έχουμε

$a^{2}_{k}=a^{2}_{k-1}+2+\frac{1}{a^{2}_{k-1}}$ με $a_{k}>1$

Για αυτό , $a^{2}_{k-1}+2 < a^{2}_{k}< a^{2}_{k-1}+3$

Προσθέντοντας αυτές τις ανισότητες για όλα τα $k$ με $1\leq  k\leq n$ ,θα έχουμε (λαμβάνοντας υπόψη ότι $a_{1}= 1$)

$2n-1<a^{2}_{n}<3n-2$ Αρα

$\sqrt{2n-1}< a_{n}<\sqrt{3n-2}$
Για  $n=100$ θα πάρουμε τις ζητούμενες ανισότητες $14<a_{100}<18$.

[color=red]ΔΙΟΡΘΩΣΕΙΣ:[/color]

(Για την 1 δεν έχω τις γνώσεις.)

[b]Άσκηση 2[/b]

[quote]Προσθέντοντας αυτές τις ανισότητες για όλα τα k με $1\leq k\leq n$ ,θα έχουμε (λαμβάνοντας υπόψη ότι$a_{1}= 1$)
[/quote]

Η σχέση θα πρέπει να είναι $1< k\leq n$ γιατί είπαμε αρχικά ότι $k>1$.

[b]Άσκηση 3[/b]

Σωστή  :D
Να μην αιτιολογήσουμε γιατί η αρχική σχέση δίνει άρρητη τιμή στο $P(1)$;  :roll:

ΑΣΚΗΣΗ 4-(Vietnam)
Να προσδιορίσετε το πολυώνυμο $P(x)$ ώστε $P(1960)=1992$ και $P(x)=\sqrt{P(x^{2}+1)-33}+32,$ για κάθε $x\geq0.$

ΛΥΣΗ(Γιαννης Παναγέας )
Παρατηρούμε ότι  $P(1960) = 1992 = 1960+32$
Θα δείξω ότι $P(x)= x+32$ για άπειρα $x$ και άρα θα είναι η μόνη λύση το $P(x)=x+32$.
Έχω ότι ισχύει για $x=1960$. Θέτοντας όπου x τo 1960 παρατηρώ ότι
$1960^{2}+33 = 1960^{2}+1+32 = P(1960^{2}+1)$ .
Άρα αν έχω μια λύση
$x_{i}$ και βρω μια $x_{i+1}>x_{i}$ τότε επαγωγικά θα υπάρχουν άπειρες.
Όμως αν $x_{i}$ μία τιμή για την οποία $P(x_{i}) = x_{i}+32$ τότε κάνοντας πράξεις προκύπτει ότι και για $x_{i+1}=x_{i}^{2}+1 > x_{i}$ ισχύει $P(x_{i+1}) = x_{i+1}+32$.

ΑΣΚΗΣΗ 5-(Προτάθηκε από τον Ράλλη Καραμιχάλη)
Για την ακολουθια $a_{n}$ ισχυει:

         $a_{1}=2$  και

         $a_{n+1}=\frac{2a_{n}}{1+\sqrt{{ \left( \frac{a_{n}}{2^{n}}\right)}^{2}+1}}$

        Νδο $a_{n}>\frac{\pi }{2}$   για καθε $n =1,2,...$.

ΛΥΣΗ -(Γιάννης Παναγέας)
Αρχική μου σκέψη , που παίζει το $a_{n}$ , παρατήρησα ότι κινείται μεταξύ 2(πάνω φράγμα) 0 κάτω χωρίς να είναι
καλό το κάτω φράγμα το αρχικό που βρίσκω. Μετά είδα κάτω από ρίζα το $x^{2}+1$ με $x= \frac{a_{n}}{2^{n}}$ και αμέσως
χωρίς σκέψη θέτω $\frac{a_{n}}{2^{n}}= tan(x_{n})$ ,$x_{n}$ μία ακολουθία με $x_{1}= \frac{\pi}{4}$. ($n$ εχω πάντα δείκτη, $\frac{\pi}{2}>x_{n}>0$ για κάθε $n$).

 μετά από πράξεις γίνεται.

  $\\ 2^{n+1}* tan(x_{n+1}) = 2^{n+1}\cdot \frac{ sin(x_{n})}{ 1+cos(x_{n}) }\Rightarrow \\ \\ sin(x_{n+1})+sin(x_{n+1})cos(x_{n})-cos(x_{n+1})sin(x_{n}) = 0 \Rightarrow \\ \\ sin(x_{n+1}) = sin(x_{n}-x_{n+1}) \Rightarrow x_{n}= 2 x_{n+1}\Rightarrow x_{n+1}= \frac{x_{1}}{2^{n}}= \frac{\pi}{ 2^{n+2}}$ .

 Όμως $tan(x)>x$ για κάθε $\frac{\pi}{2}>x >0$ και άρα $tan(x_{n+1})>x_{n+1}=\frac{\pi}{ 2^{n+2}}$ και άρα $a_{n+1}> \frac{\pi}{2}$

[quote="silouan"]Αυτή την άσκηση την είχε προτείνει ο Στέργιος παλιά .Πολύ ωραία άσκηση  .Αν σας φαίνεται συνθηματική η λύση προσθέστε σχόλια .

ΑΣΚΗΣΗ 3
Να βρεθούν όλα τα πολυόνυνα με πραγματικούς συντελεστές ώστε να ικανοποιούν την παρακάτω σχέση για όλα τα $x \in R$.
$P(2x^{2}-1)=\frac{(P(x))^{2}}{2}-1$

ΛΥΣΗ

Esto oti den einai stathero.  :o
Os polionimo to P einai paragogisimo. 'Esto oti $P'(x)=(x-1)^{r}\cdot q(x)$, opou r fisikos mporei kai miden,alla q(1) pote miden. Tote paragogizontas kai antikathistontas, exo oti : $4x(2x^{2}-2)^{r}q(2x^{2}-1)= P(x)(x-1)^{r}q(x)$. Alla apo tin ipothesi to q(x) den einai to mideniko kai epeidi milame gia polionima kai i pano sxesi isxuei gia apeira x. Pairno oti $2^{r+2}x(x+1)^{r}q(2x^{2}-1)=P(x)q(x)$ gia ola ta x. Tote omos to P(1) einai akeraios arithmos (apla balte stin teleutaia sxesi apou x to 1).
Antifasi dioti stin arxi sxesi an baloume opou x to 1, pairnoume oti to x einai arritos[/quote]

αλλη λυση: θεωρουμε την ακολουθια
$u_{n}=\sqrt(\frac{u_{n-1}+1}{2})$ for n>=3 and$u_{1}=1,u_{2}=2$
ισχύει  $u_{n}<u{n+1}<1$ εχουμε $P(u_{n})=(P(u_{n+1})^{2}/2-1$ επισης $P(u_{n})$ ειναι διάφορο του μηδεν , διαφορετικά τα $P(u_{n-1}),P(u_{n-2}),...$ θα ηταν ρητοι,ομως $P(1)=1+$ ή - $\sqrt3$
παραγωγιζωντας την αρχικη σχεση εχουμε $4P'(2x^{2}-1)=P(x)P'(x)$ για χ=1 παιρνουμε $P(1)=4$ ατοπο αρα $P'(1)=P'(u_{1})=0$
 αυτο συνεπαγεται $0=P'(u_{2})=P'(u_{3})=...$  αρα το $P'(x)$ ειναι το μηδενικο πολυονυμο , δηλ το   $P(x)$ ειναι σταθερο...

Στελιο το $u$ της προηγουμενης ασκησης ειναι το κρατουμενο της προσθεσης

Ωραία Κώστα . Την γράφω όπως την είχε γράψει ο galeos (Γαλάνης Ανδρέας ?) παλιά

ΑΣΚΗΣΗ 3-Λυση 2
Afu to $P(x)$ einai polionimo eukola pairnume apo ti dosmeni sxesi oti to $P$ eite einai artio eite einai peritto. An einai peritto, dld $P(x)=-P(-x)$ tote $P(0)=0$.
$x=0$ stin arxiki $\rightarrow P(-1)=-1$
$x=-1$ stin arxiki $\rightarrow P(1)=-\frac{1}{2}$
pu erxetai se antithesi me tin "peritotita" tu $P$.
Ara $P$ artio kai $P(1)=P(-1)$.
$x=-1$ stin arxiki $\rightarrow P(1)=1+\sqrt{3}=P(-1)$ h $P(1)=1-\sqrt{3}=P(-1)$.
Theoro tin proti periptosi pu einai akrivos idia me ti deuteri.
$x=0$ stin arxiki $\rightarrow (P(0))^{2}=(1+\sqrt{3})^{2}$
Paragogizo tin arxiki sxesi kai pairno $P'(2x^{2}-1)4x=P(x)P'(x)$(1), ap' opu $P'(0)=0$. Theoro akoluthia $a_{n}$
me $a_{1}=0$ kai $a_{n+1}=\sqrt{\frac{1+a_{n}}{2}}$. Einai apli epagogi na deiksei kaneis oti $a_{n}=\cos(\frac{\pi}{2^{n}})$ . Epomenos i $a_{n}$ einai gnisios monotoni kai diafori tu 0. Akoma apo tin (1) exume:
$P'(a_{n})4a_{n+1}=P(a_{n+1})P'(a_{n+1})$. An deiksume oti $P(a_{n+1})$ diaforo tu 0 exume teleiosei afu tote exume $P'(a_{n})=0$ gia ola ta $n$, pu simainei oti $P'(x)=0$ dld $P(x)$ stathero.
Esto oti $P(a_{n})=0$ gia kapoio $n$. Apo tin arxiki sxesi exume oti $P(a_{n})=\frac{(P(a_{n+1}))^{2}}{2}-1$. Tote $P(a_{n-1})=-1$, $P(a_{n-2})=-\frac{1}{2}$ kai profanos oso proxorame pros ta piso tha pairnume ritus, pragma pu simainei oti $P(a_{1})=P(0)$ ritos,atopo afu $(P(0))^{2}=(1+\sqrt{3})^{2}$

Telika $P$ stathero kai $P(x)=1+\sqrt{3}$ h $P(x)=1-\sqrt{3}$

[color=red]ΑΣΚΗΣΗ 7 - ΕΚΦΩΝΗΣΗ[/color] (cretanman)

Estw poliwnimo $P_{n}(x)=x^{n}-a_{n-1}x^{n-1}-\cdots-a_{1}x-a_{0}$ me $n\geq 2$, opou $a_{0},a_{1},\ldots,a_{n-1}>0$ kai $a_{0}+a_{1}+\cdots+a_{n-1}=1$. Na prosdiorisete oles tis rizes $\rho$ tou $P_{n}(x)$ me $|\rho|=1$, kathws kai ton vathmo pollaplotitas kathe tetoias rizas.

[color=red]ΑΣΚΗΣΗ 7 - ΛΥΣΗ[/color] (stedes)

Έστω $r$ μια ρίζα του $P_{n}(x)$ με $|r|=1$, τότε

$|P_{n}(r)| = |r^{n}-a_{n-1}r^{n-1}-a_{n-2}r^{n-2}+...-a_{0}| \geq$
$\geq |r^{n}|-|a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+...+a_{0}| \geq$
$\geq |r^{n}|-|a_{n-1}r^{n-1}|-|a_{n-2}r^{n-2}|-...-|a_{0}|=$
$=|r^{n}|-a_{n-1}|r^{n-1}|-a_{n-2}|r^{n-2}|-...-a_{0}=$
$= 1-a_{n-1}-a_{n-2}-...-a_{0}= 0$

Η ισότητα στην πρώτη ανισότητα ισχύει όταν $\overline{r}^{n}(a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+...+a_{0}) \in \mathbb R_{+}$ και στην δεύτερη όταν
$a_{n-1}r^{n-1},a_{n-2}r^{n-2},...,a_{1}r \in \mathbb R_{+}$ και επειδή το $r$ είναι ρίζα θα ισχύει η ισότητα και στις δυο περιπτώσεις.
Άρα, $r \in \mathbb R_{+}$ και επειδή $|r|=1$, θα ισχύει $r=1$.

Επίσης ισχύει ότι $P_{n}(x)=(x-1)[x^{n-1}+(1-a_{n-1})x^{n-2}+...+(1-a_{n-1}-a_{n-2}-...-a_{1})]$. Το πολυώνυμο όμως $t(x)=x^{n-1}+(1-a_{n-1})x^{n-2}+...+(1-a_{n-1}-a_{n-2}-...-a_{1})$ δεν έχει ρίζα το $1$, αφού $t(1)=1+(1-a_{n-1})+...+(1-a_{n-1}-a_{n-2}-...-a_{1}) > 0$.

Συνεπώς η $r=1$ είναι η μοναδική ρίζα του $P_{n}(x)$ για την οποία ισχύει $|r|=1$, και έχει ένα βαθμό πολλαπλότητας.

Στέλιος

[color=red]ASKHSH 8[/color] (cretanman)
Estw $p$ thetikos pragmatikos arithmos kai estw $|x_{0}|\leq 2p$. Gia $n\geq 1$ orizoume

$x_{n}=3x_{n-1}-\frac{1}{p^{2}}x_{n-1}^{3}$

Na ekfrasete tin $x_{n}$ ws sinartisi twn $n$ kai $x_{0}$.

[color=red]ΑΣΚΗΣΗ 8 - ΛΥΣΗ[/color] (pontios)
Έστω η συνάρτηση $f(x)=3x-\frac{x^{3}}{p^{2}}=x\left[3-\left(\frac{x}{p}\right)^{2}\right]$

Ο επόμενος όρος του $x$ στην ακολουθία θα είναι ο αριθμός $y=f(x)$.

Έστω $x$ κάποιος όρος της ακολουθίας τέτοιος ώστε $|x|\leq 2p$. Θα δούμε ότι και ο επόμενος όρος έχει την ίδια ιδιότητα.

Ο x γράφεται $x=(2p)\sin\theta$ για κάποια γωνία $\theta$

$y = f(x) = x\left[3-\left(\frac{x}{p}\right)^{2}\right]\Rightarrow$

$y = p\cdot\left(\frac{x}{p}\right)\left[3-\left(\frac{x}{p}\right)^{2}\right]$

$= (2p)\sin \theta\left[3-(2\sin \theta)^{2}\right]$

$= (2p)\sin \theta (3-4\sin^{2}\theta)$

$= (2p)\cdot (3\sin \theta-4\sin^{3}\theta) \Rightarrow \boxed{y = (2p)\cdot \sin 3 \theta}$, που είναι $\leq 2p$

Έτσι, το $x=(2p)\sin\theta$ αντιστοιχίζεται στο $y=(2p)\sin 3\theta$

Έστω $\theta_{o}$ η αρχική γωνία. Τότε θα είναι $x_{n}=(2p)\sin(3^{n}\theta_{o})$.

ΑΣΚΗΣΗ 9-(Proposed by silouan)

Έστω $P$  ένα πολυώνυμο με πραγματικούς συντελεστές .Βρείτε όλες τις συναρτήσεις από το R->R έτσι ώστε να υπάρχει πραγματικός $t$  έτσι ώστε

$f(x+t)-f(x)=P(x)$ για κάθε πραγματικό αριθμό $x$

ΑΣΚΗΣΗ 9-ΛΥΣΗ (by silouan)
Προφανώς αν το $P=0$  τότε  για $t=0$  και για κάθε συνάρτηση από το R->R  ισχύει το ζητούμενο .έστω  $P(x)=\sum_{k=0}^{n}a_{k}x^{k}$  με $a_{n}\neq 0$  και  $t\in R^{*}$.Και ψάχνουμε για ένα πολυώνυμο $Q_{t}(x)=\sum_{k=1}^{n+1}b_{k}x^{k}$
τέτοιο ώστε $Q_{t}(x+t)-Q_{t}(x)=P(x)$  με x πραγματικό.
Κάνοντας εξίσωση συντελεστών έχουμε :
$\binom{n+1}{1}tb_{n+1}=a_{n}$
$\binom{n+1}{2}t^{2}b_{n+1}+\binom{n}{1}tb_{n}=a_{n-1}$
......

$\binom{n+1}{n+1}t^{n+1}b_{n+1}+......+\binom{1}{1}tb_{1}=a_{0}$
Παρατηρούμε ότι το σύστημα έχει μοναδική λύση , άρα υπάρχει μοναδικό πολυώνυμο
$Q_{t}$ ώστε $Q_{t}(x+t)-Q_{t}(x)=P(x)$  με  $Q_{t}(0)=0$  και
$deg\ Q_{t}=1+deq\ P$
Tέλος θέτουμε $g(x)=f(x)-Q_{t}(x)$ .
Τότε η $f$  ικανοποιεί το ζητούμενο  αν και μόνον αν $g(x+t)-g(x)=0$  για κάθε  πραγματικό x δηλαδή η $g$  να είναι περιοδική με περίοδο  $t$.
Άρα  $f(x)=Q_{t}(x)+g(x)$  όπου $g$  μια οποιαδήποτε περιοδική συνάρτηση με περίοδο $t$

Να και μια περίεργη
ΑΣΚΗΣΗ 10-(Θανάσης Κοντογεωργης)
Αν $a,b,c,d \in \mathbb{R}$ με

$a+b+c+d<-1$
$a-b+c-d<-1$
$\mid 4a+2c \mid=\mid 3b+d \mid$
$\mid b \mid \leq 1$

νδο $ac\leq \frac{3}{8}$

ΑΣΚΗΣΗ 10-ΛΥΣΗ -(Ramanujan)
Θεωρούμε το πολυώνυμο $p(x) = a x^{4}+b x^{3}+c x^{2}+d x+1$.

Έχουμε $p(1)=a+b+c+d+1<0$ και $p(-1)=a-b+c-d+1<0$ .

Αφού $p(0)=1>0$ το πολυώνυμο έχει δύο ρίζες $\rho_{1}\in (-1,0)$ και $\rho_{2}\in (0,1)$.

[u][b]Περίπτωση 1[/b][/u]

Άν $\boxed{a>0}$ αφού  $\lim_{x \to+\infty}p(x)=\lim_{x \to-\infty}p(x)=+\infty$ και $p(1), \, p(-1) \, <0$

το πολυώνυμο έχει άλλες δύο ρίζες $\rho_{3}\in (-\infty,-1)$ και $\rho_{4}\in (1, \infty)$ .

[u][b]Περίπτωση 2[/b][/u]

$\boxed{a<0}$

Έπίσης $p'(x)=4ax^{3}+3bx^{2}+2cx+d$ άρα από την συνθήκη

$\\ \mid 4a+2c \mid=\mid 3b+d \mid$ παίρνουμε $p'(1)=0$ ή $p'(-1)=0$ .

Υποθέτουμε ότι $p'(1)=4a+3b+2c+d=0$.

Αυτό μπορεί να συμβεί όταν $(4a+2c)(3b+d) \leq 0$ .

$\bullet$ Άν  $(4a+2c)(3b+d) =0$ τότε $p'(-1)$ συνεπώς το $p'(x)$ έχει τρεις ρίζες.

$\bullet$ Αν $4a+2c <0$ και $3b+d>0$ τότε $p'(-1)=(3b+d)-(4a+2c) >0$.
Σε αυτή την περίπτωση αν υποθέσουμε ότι $p'(x)>0$ για $x \in (-\infty ,1)$ καταλήγουμε σε
άτοπο γιατί το πολυώνυμο έχει ρίζα στο $(0,1)$ .
Άρα κάπου αλλάζει πρόσημο και έτσι παίρνουμε πάλι 3 ρίζες .

$\bullet$ Αν $4a+2c >0$ και $3b+d<0$ τότε $p'(-1)<0$ .
Τώρα αφού $\lim_{x \to-\infty}p'(x)=+\infty$ το $p'(x)$ έχει ρίζα $\xi \in (-\infty ,-1)$ άρα έχει 3 ρίζες .
[u][b]
Περίπτωση 3[/b][/u]

$\\ a=0$  :D

Από όλα τα παραπάνω το $p''(x)=12ax^{2}+6bx+2c$ έχει 2 ρίζες
άρα $D = 36b^{2}-96ac \geq 0 \Rightarrow \frac{3}{8}\cdot b^{2}\geq ac$

και αφού $|b| \leq 1$ παίρνουμε $\boxed{ac \leq \frac{3}{8}}$ .

ΑΣΚΗΣΗ 11- (ΙΜΟ 2006)
Έστω πολυώνυμο Ρ(x) βαθμού n>1 με ακεραίους συντελεστές και έστω $k$ θετικός ακέραιος.Θεωρούμε το πολυώνυμο $Q(x)=P(P(...P(x))...))$  όπου το Ρ  εμφανίζεται $k$ φορές .Αποδείξτε οτι υπάρχουν το πολύ $n$ ακέραιοι $t$ ώστε
$Q(t)=t$.

ΑΣΚΗΣΗ 11-ΛΥΣΗ (by silouan)
Αν $Q(t)=t$  όταν $P(t)=t$  το πρόβλημα γίνετα προφανές .
Θα εξετάσουμε την περίπτωση κατά την οποία
$P(x_{1})=x_{2}$ ,  $P(x_{2})=x_{3}$,...., $P(x_{n})=x_{1}$ (αυτές οι ισότητες προκύπτουν ώστε να έχουμε $Q(x_{1})=x_{1}$  χωρίς        $P(x_{1})=x_{1}$ ) kai $n\geq 3$.Υποθέτουμε ότι  $x_{1},x_{2},...,x_{n}$  διαφορετικοί μεταξύ τους .Πράγματι αν $x_{k}=x_{n}$  η περιοδικότητα απλα μειώνεται αλλά στην ουσία είναι σα να διαιρούμε το φαινόμενο σε δύο φάσεις άρα είναι το ίδιο πράγμα .Τότε θα εχουμε
$P(x)-x_{2}=(x-x_{1})P_{1}(x)$
$P(x)-x_{3}=(x-x_{2})P_{2}(x)$
....
....
$P(x)-x_{1}=(x-x_{n})P_{n}(x)$

Υποθέτουμε ότι η απόλυτη τιμή της διαφοράς $x_{1}-x_{n}$  είναι η μεγαλύτερη μεταξύ όλων των διαφορών που παρουσιάζονται στα γινόμενα του δευτέρου μέλους .Θέτουμε στην πρώτη σχέση όπου $x$  το $x_{n}$ .Τότε αυτή γίνεται $x_{1}-x_{2}=(x_{n}-x_{1})P_{1}(x)$  και αφού όλοι αυτοί είναι ακέραιοι η απόλυτη τιμή της διαφοράς $x_{1}-x_{2}$  είναι μεγαλύτεροι από την απόλυτη τιμή της διαφοράς  $x_{1}-x_{n}$ , άτοπο  άρα $n=k\leq 2$

Άρα αρκεί να λύσουμε το πρόβλημα για n=2 .Δηλαδή αν
$P(P(x))=Q(x)$ .Πάλι δεν εξετάζουμε την προφανή $P(t)=t$  και υποθέτουμε ότι $P(a)=b$  kai  $P(b)=a$
Πάλι  $P(x)-b=(x-a)P_{1}(x)$  και  $P(x)-a=(x-b)P_{2}(x)$
Βάζοντας όπου x  το b στην πρώτη παίρνουμε ότι $P_{1}(x)=-1$ kai όμοια $P_{2}(x)=-1$ .Kai exoume
$P(x)-b=a-x$  (1)
$P(x)-a=b-x$  (2)
(1)+(2) => $P(x)+x=a+b$  (3)
Έστω μια άλλη δυάδα τώρα ώστε $P(m)=n$  και $P(n)=m$
Πάλι τότε $P(x)+x=m+n$  . Άρα για κάθε τέτοιο ζεύγος θα συμβαίνει αυτό .Ή αφού $a+b=m+n$  μπορούμε την   (3) να την γράψουμε ως $a+b-m-P(m)=0$ (4) .Θέτουμε $a+b=C$  και θεωρούμε το πολυώνυμο
$F(x)=C-P(x)-x$ .Δείξαμε με την (4) ότι κάθε ζεύγος λύσης της
$Q(t)=t$ είναι ρίζα του πολυωνύμου F(x)
Αλλά το F(x) είναι νιοστού βαθμού (είναι ομοιόβαθμο με το P(x))
άρα έχει το πολύ $n$  ρίζες .Όμως αυτές είναι τα ζεύγη λύσεων της $Q(t)=t$  (5) (κάθε ζεύγος αντιστοιχεί σε μία ρίζα της (5)) άρα η (5)  έχει το πολύ   $n$  ρίζες , το ζητούμενο......

ΑΣΚΗΣΗ 12 -(proposed by Ράλλης Καραμιχάλης)
Ν.δ.ό. $\sqrt{1+\sqrt{2+\sqrt{3+.....+\sqrt{n}}}}<2$

για κάθε $n \in \mathbb N^{*}$

ΑΣΚΗΣΗ 12-ΛΥΣΗ  (Andrew Frimu )
θα αποδείξω και μια γενικότερη της (2) άρα και τη (2)
Θα δείξω [u]με αντίστροφη επαγωγή [/u]ότι
$\sqrt{m+\sqrt{(m+1)+\sqrt{(m+2)+.....+\sqrt{N}}}}<m+1$
όπου $m\geq 1$ ένας θετικός ακέραιος και $N$ ένας <<μεγάλος φυσικός αριθμός......
Προφανώς για $m=N$  θα έχουμε ότι $\sqrt{N}<N+1$  που προφανώς ισχύει .
Υποθέτουμε ότι
$\sqrt{m+1)\sqrt{(m+2)+\sqrt{(m+3)+.....+\sqrt{N}}}}<m+2$
και θα δείξουμε ότι
$\sqrt{m+\sqrt{(m+1)+\sqrt{(m+2)+.....+\sqrt{N}}}}<m+1$

Πράγματι αφού
$\sqrt{m+1)\sqrt{(m+2)+\sqrt{(m+3)+.....+\sqrt{N}}}}<m+2$
προσθέτωντας το $m$ και τετραγωνίζοντας έχουμε ότι
$\sqrt{m+\sqrt{(m+1)+\sqrt{(m+2)+.....+\sqrt{N}}}}<\sqrt{2m+2}<m+1$

άρα ολοκληρώνεται η αντίστροφη επαγωγή

ASKISI 15 (M.Klamkin)
Αν :$x+y+z=a+b+c$
      $x^{2}+y^{2}+z^{2}=a^{2}+b^{2}+c^{2}$
      $x^{3}+y^{3}+z^{3}=a^{3}+b^{3}+c^{3}$

$x,y,z,a,b,c$ πραγματικοι και $a,b,c$ δομενες σταθερες νδο :$(x,y,z)=(a,b,c)$

ΛΥΣΗ : εκ πρωτης οψεως η ασκηση ειναι δυσκολη,εχουμε αθροιματα  τετραγωνων ,κυβων που δεν οδηγουν πουθενα ,.οι παραστασεις που εχουμε στην διαθεση μας μας οδηγουν σε μια ποιο αλγεβρικη ιδεα :Εστω οτι τα $x,y,z$ ειναι ριζες της κυβικης εξισωσης $s^{3}-ks^{2}+ls-m=0$ $k=x+y+z,l=xy+yz+zx,m=xyz$.Απο εκει και περα τα πραγματα ειναι απλα :με απλες ταυτοτητες δειχνουμε οτι $l=ab+bc+ca,m=abc$ δηλαδη τα $a,b,c$ ειναι ριζες της ιδιας εξισωσης .Ομως εξισωση τριτου βαθμου $\Rightarrow$ το πολυ τρεις λυσεις.Εδω εχουμε 3+3=6 αρα σιμπειπτουν   αρα  αποδειχτηκε το ζητουμενο.

σχολια :το  αποτελεσμα γενικευεται για περισσοτερες απο τρεις μεταβλητες.ουσιαστικα το κλειδι ειναι οι συμετρικες συναρτησεις .στο βιβλιο των USAMO  υπαρχει μια ενδιαφερουσα αποδειξη για την γενικευση,οπου αν ζητηθει θα την βαλλω.
γενικα με αυτο τον τροπο μπορουμε να βρουμε παρατασεις που ειναι συμμετρικες συναρτησεις οταν μας δινονται αλλες στοιχειδης συναρτησεις.μια καταπληκτικη ειναι αυτη:
να βρεθει το $abc$ αν :
$a^{3}=3(b^{2}+c^{2})-25$
$b^{3}=3(c^{2}+a^{2})-25$
$c^{3}=3(a^{2}+b^{2})-25$
και αυτη λυνεται με ομοιο τροπο.

κανεις ? ειναι δυκολα?

apla tileskopika a8rismatakia... :P
LISIS 16

Isxii
${n}\cdot{n}!=(n+1-1)n!=(n+1)!-n!$...
diladi telika exume$(n+1)!-1$

Isxii
$\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$
opote tlk pernume $1-\frac{1}{(n+1)!}$

 :D

Btw gia to parapano evgala oti $abc=125$..ala yiloklevontas   :P :P
Exis kamia elegant solution?..

[quote="RAKAR"]
Btw gia to parapano evgala oti $abc=125$..ala yiloklevontas   :P :P
Exis kamia elegant solution?..[/quote]

Prosthetwntas diadoxika stis $3$ eksiswseis ta $3a^{2}, 3b^{2}, 3c^{2}$, pairnoume oti

$a^{3}+3a^{2}=3(a^{2}+b^{2}+c^{2})-25$
$b^{3}+3b^{2}=3(a^{2}+b^{2}+c^{2})-25$
$c^{3}+3c^{2}=3(a^{2}+b^{2}+c^{2})-25$

diladi oi arithmoi $a^{3}+3a^{2}, b^{3}+3b^{2}, c^{3}+3c^{2}$ einai metaksi tous isoi, as poume isoi me $t$.

ara oi $a,b,c$ einai rizes tis eksiswsis $x^{3}+3x^{2}-t=0$. Sinepws apo tous tipous Vieta exoume

$a+b+c=-3$
$ab+bc+ac=0$
$abc=t$

diladi $abc=3(a^{2}+b^{2}+c^{2})-25=3\left((a+b+c)^{2}-2(ab+bc+ac)\right)-25=3\cdot 9-25$

Ara $abc=2$

Exw xasei kamia praksi pouthena?

Alexandros

[quote="kostas"]πραγματι $abc=2$. η λυση του Αλεξανδρου μαλλον ειναι η μοναδικη για αυτη την ασκηση( να θεωρησεις οτι τα $a,b,c$ εινaι λυσεις μιας κυβικης εξισωσης)
αληθεια πως ψιλοεκλεψες Ραλλη? :P[/quote]
Xm..... :huh:  :huh: kati vromai edo....
ontos sosti fenet i lisi tu cretanman gt dini $(-1,-1-\sqrt{3},-1+\sqrt{3})$
+ontos telika $abc=2$
exo vrika tis $(5,5,5)$ pu dini $abc=125$
$(\frac{1-\sqrt{21}}{2},\frac{1-\sqrt{21}}{2},\frac{1-\sqrt{21}}{2})$
pu dini $abc=8-3\sqrt{21}$
$(\frac{1+\sqrt{21}}{2},\frac{1+\sqrt{21}}{2},\frac{1+\sqrt{21}}{2})$
pu dini $abc=8+3\sqrt{21}$

To periergo tis ipo8esis ine pos prokiptun toses lisis..(pos dikeologite to gegonos oti eprepe na exume teseris lisis?... :huh: )

Kosta,legontas ekleya..enousa oti eyaksa mono tis lisis me $a=b=c$...+gia ayto dn vrika tin lisi $abc=2$ :wink:

 :D

Δεν βρωμαει κατι.απλα το προβλημα λεει να βρεις το $abc$ with $a,b,c$ be distinct real numbers , sorry δεν το αναφερω στην εκφωνηση της ασκησης .

[quote="stedes"][color=red]ASKHSH 8[/color] (cretanman)
Estw $p$ thetikos pragmatikos arithmos kai estw $|x_{0}|\leq 2p$. Gia $n\geq 1$ orizoume

$x_{n}=3x_{n-1}-\frac{1}{p^{2}}x_{n-1}^{3}$

Na ekfrasete tin $x_{n}$ ws sinartisi twn $n$ kai $x_{0}$.

[color=red]ΑΣΚΗΣΗ 8 - ΛΥΣΗ[/color] (pontios) ...
[/quote]

Πιστεύω ότι είναι άδικο να αναφέρεται μόνο το δικό μου όνομα στη λύση.
Θα πρέπει να αναφερθεί και ο [color=blue][b]nikos[/b][/color] γιατί την είχαμε λύσει ταυτόχρονα
http://www.mathlinks.ro/Forum/viewtopic.php?p=698904#698904

Ναι Λουκά, έχεις δίκιο. Παράληψή μου. Απλά επειδή οι λύσεις είναι παρόμοιες ,έβαλα την πρώτη που είδα. Βάζω λοιπόν και τη λύση του nikou για να είναι εδώ όλες συγκεντρωμένες.

[color=red]ΑΣΚΗΣΗ 8 - Λύση 2η[/color] (nikos)

ΑΣΚΗΣΗ 17:(αρκετά απαιτητική θα έλεγα) :wink:

Να βρεθεί χωρίς τη χρήση ανάλυσης ένα πολυώνυμο 5ου βαθμού τέτοιο ώστε το P(x)+1 να διαιρείται από το (χ-1)^3 και το Ρ(χ)-1 να διαιρείται από το (χ+1)^3.

Λύση:

Επειδή (χ+1)^3|P(x)-1, αν θέσουμε όπου χ το -χ παίρνουμε τελικά ότι (χ-1)^3|Ρ(-χ)-1.

Ομοίως (χ+1)^3|Ρ(-χ)+1.

Έχουμε δηλαδή σύμφωνα με τις παραπάνω παρατηρήσεις και την υπόθεση ότι:

(χ+1)^3|Ρ(χ)-Ρ(-χ) και (χ-1)^3|Ρ(χ)-Ρ(-χ).

Άρα [(χ+1)^3][(χ-1)^3]|Ρ(χ)-Ρ(-χ).

Όμως επειδή το αριστερό μέλος είναι 6ου βαθμού και το δεξί το πολύ 5ου πρέπει:

Ρ(χ)-Ρ(-χ)=0 για κάθε χ.

Άρα οι συντελεστές των χ^4 και χ^2 πρέπει να είναι 0.

Παρατηρούμε ότι Ρ(χ)+1=[(χ-1)^3](Αχ^2+Βχ-1).

Αναπτύσσοντας και παρατηρώντας τους συντελεστές του χ^4 και χ^2 πρέπει Β-3Α=0 και 3+3Β-Α=0.

Λύνοντας τώρα το σύστημα παίρνουμε ότι Α=-3/8 και Β=-9/8.

Οπότε τελικά Ρ(χ)=-3χ^5/8 + 5χ^3/4 - 15χ/8

Στέλνω το ακόλουθο πρόβλημα το οποίο ανήκει στον τομέα Άλγεβρας και συγκεκριμένα στην Τριγωνομετρία. Επειδή δεν έχουμε αντίστοιχο topic θα πρότεινα είτε να μετονομαστει το υπάρχον topic σε  ΒΙΒΛΙΟ-ΑΛΓΕΒΡΑ , ΕΝΟΤΗΤΑ 2 : Πολυώνυμα - Ακολουθίες (αυτό ίσως να μπορεί να το κάνει ο κος Στεργίου) είτε να φτιάξουμε ένα άλλο topic με τίτλο "ΒΙΒΛΙΟ-ΑΛΓΕΒΡΑ , ΕΝΟΤΗΤΑ 3 : Τριγωνομετρία"

[color=Red]ΑΣΚΗΣΗ 19[/color]

Μετά το μάθημα στην τάξη έμεινε γραμμένο στον πίνακα η εξής άσκηση. (Αντί για τα νούμερα γράφω *)

"Να βρείτε το άθροισμα
$\begin{equation}t(0)-t\left(\frac{\pi}{5}\right)+t\left(\frac{2\pi}{5}\right)-t\left(\frac{3\pi}{5}\right)+\cdots+t\left(\frac{8\pi}{5}\right)-t\left(\frac{9\pi}{5}\right) \ \ (1) \nonumber \end{equation}$

όπου

$\begin{equation}t(x)=\cos{5x}+*\cos{4x}+*\cos{3x}+*\cos{2x}+*\cos{x}+*\cos{0}\ \ (2) \nonumber \end{equation}$ "

Ένας μαθητής είπε σε μία φίλη του ότι μπορούσε να βρεί το άθροισμα $(1)$ χωρίς να ξέρει τους συντελεστές που ήταν σβησμένοι στον πίνακα από την $(2)$. Έλεγε αλήθεια;

Φυσικά οι συντελεστές δεν είναι ανάγκη να είναι ίδιοι (Εξάλλου ήταν σβησμένοι στον πίνακα και δεν μπορούμε να το υποθέσουμε )

Αν δεν έχει δοθεί λύση σε $2-3$ ημέρες θα την βάλω.

Αλέξανδρος

[b][color=red]Άσκηση 19 - Λύση[/color][/b]

Μια όχι κομψή λύση :   :oops:

Το άθροισμα γράφεται σαν $S = A_{5}+* A_{4}+* A_{3}+* A_{2}+* A_{1}+* A_{0}$ όπου

$\\ A_{m}= \sum_{k=0}^{4}\left\{\cos \left( \frac{2k \pi m}{5}\right)-\cos \left[ \frac{(2k+1) \pi m}{5}\right] \right\}= 2 \sin\left( \frac{ \pi m}{10}\right) \sum_{k=0}^{4}\sin \left( \frac{4k \pi m }{10}+\frac{ \pi m }{10}\right)$

για $m \in \{0,1,2,3,4,5\}$

Αν $m \neq 5 ,0$ τότε από τον τύπο

$\\ \sin{\varphi}+\sin{(\varphi+\alpha)}+\cdots+\sin{(\varphi+n\alpha)}=\frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)}\cdot \sin{(\varphi+\frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}$

για $\varphi = \frac{ \pi m }{10}$ και $\alpha = \frac{ 4 \pi m }{10}$

παίρνουμε $A_{m}= 2 \sin\left( \frac{ \pi m}{10}\right) \cdot \frac{ \sin ( \pi m ) \cdot \sin \left( \frac{9 \pi m }{10}\right) }{ \sin \left( \frac{ \pi m }{5}\right) }= 0$

Αν $m= 0$ προφανώς $A_{0}= 0$ και αν $m = 5$ παίρνουμε $A_{5}= 2 \sum_{k=0}^{4}\sin \left( 2 k \pi+\frac{ \pi }{2}\right) = 10$

Άρα $S = 10$ .   :)

Sorry για τυχόν λάθη  και γιατί δεν είναι λεπτομερής η λύση . :D

Είναι απόλυτα σωστή η λύση σου Ramanujan! Δεν νομίζω να υπάρχει πολύ καλύτερη λύση εκτός από αυτή. Κι εγώ έχω την ίδια στο μυαλό μου. Μία άλλη η οποία αν δεν λυθεί σε 2-3 ημέρες θα βάλω την λύση.

[color=red]
[b]ΑΣΚΗΣΗ 20[/b][/color]

Να λυθεί η εξίσωση $6^{x}+1=8^{x}-27^{x-1}$ στο σύνολο των πραγματικών αριθμών.

Αλέξανδρος

[color=blue][b]'Ασκηση 20 - Λύση[/b] [/color]

Θέτουμε $2^{x}= \alpha , \,\, 3^{x}= \beta$ και η εξίσωση γίνεται :

$\\ \alpha \cdot \beta+1 = \alpha^{3}-\frac{ \beta^{3}}{27}\Leftrightarrow 27 \alpha \cdot \beta+27-27 \alpha^{3}+\beta^{3}= 0 \Leftrightarrow \\ \\ (\beta-3 \alpha+3) [ \beta^{2}+(3 \alpha-3) \beta+9 \alpha^{2}+9 \alpha+9 ] = 0 \Leftrightarrow \\ \\ \beta-3 \alpha+3 = 0 \Leftrightarrow 3^{x}+3-3 \cdot 2^{x}= 0$

η οποία έχει $2$ προφανείς λύσεις τις $\{1,2\}$

Θεωρούμε την $f(x) = 1+\frac{3}{3^{x}}-3 \left( \frac{2}{3}\right)^{x}$.

H $f'(x) = 3^{1-x}\left[2^{x}\cdot ( \log 3-\log 2)-\log 3 \right]$ μηδενίζεται σε

μοναδικό $x_{0}$ και είναι θετική για $x > x_{0}$ και αρνητική για $x < x_{0}$ .

Άρα η $f$ έχει λόγω μονοτονίας ακριβώς $2$ ρίζες , τις  $\{1,2\}$.

Sorry αν υπάρχει λάθος   :)

Eiναι απόλυτα σωστή η λύση σου Ramanujan. Να αναφέρω μόνο το εξής:

Η σχέση $27 \alpha \cdot \beta+27-27 \alpha^{3}+\beta^{3}= 0$ καταλήγει στην παραγοντοποίηση που γράφεις, λόγω της γνωστής ταυτότητας του Euler. Συνεπώς, η δεύτερη παρένθεση είναι μηδέν μόνο εαν ισχύει $\b=3a=3$ το οποίο προφανώς είναι αδύνατο. Αυτό μας εξασφαλίζει τον μηδενισμό της πρώτης παρένθεσης απ'όπου εξάγουμε και τις (σωστές) λύσεις σου.

Αλέξανδρος

[b][size=150][color=red]ΑΣΚΗΣΗ 21[/color][/size][/b]

Να υπολογίσετε την τιμή της παράστασης

$ \eta\mu5^\circ\cdot\eta\mu15^\circ\cdot\eta\mu25^\circ\cdot\eta\mu35^\circ\cdot\eta\mu45^\circ\cdot\eta\mu55^\circ\cdot\eta\mu65^\circ\cdot\eta\mu75^\circ\cdot\eta\mu85^\circ$

Αλέξανδρος

Σωστό το αποτέλεσμα! Εαν μπορέσεις γράψε την λύση σου αναλυτικά για να την έχουμε έτοιμη για το βιβλίο.

Αλέξανδρος

Προφανώς η παράσταση γράφεται:

$ \sin5^\circ\cos5^\circ\sin15^\circ\cos15^\circ\sin25^\circ\cos25^\circ\sin35^\circ\cos35^\circ\sin45^\circ$

Διαδοχικά έχουμε:

$ \frac{1}{2}2\sin5^\circ\cos5^\circ\frac{1}{2}2\sin15^\circ\cos15^\circ\frac{1}{2}2\sin25^\circ\cos25^\circ\frac{1}{2}2\sin35^\circ\cos35^\circ\frac{\sqrt{2}}{2}$.

Από τον τύπο $ \sin2a=2\sin a\cos a$, θα πάρουμε:

$ \frac{\sqrt{2}}{2^{5}}\sin10^\circ\sin30^\circ\sin50^\circ\sin70^\circ=\frac{\sqrt{2}}{2^{6}}\sin10^\circ\sin50^\circ\sin70^\circ$.

Δηλαδή μας μένει να υπολογίσουμε το $ \sin10^\circ\sin 50^\circ\sin 70^\circ$.

Όμως:

$ \sin10^\circ\sin 50^\circ\sin 70^\circ=\cos80^\circ\cos40^\circ\cos20^\circ\frac{\sin20^\circ}{\sin20^\circ}=\frac{1}{2\sin20^\circ}2\sin20^\circ\cos20\circ\cos80^\circ\cos40^\circ= \frac{1}{2\sin20^\circ}\sin40^\circ\cos80^\circ\cos40^\circ=\frac{1}{2}\frac{1}{2\sin20^\circ}2\sin40^\circ\cos40^\circ\cos80^\circ=\frac{1}{2}\frac{1}{2\sin20^\circ}\sin80^\circ\cos80^\circ=\frac{1}{8\sin20^\circ}2\cos80^\circ\sin80^\circ=\frac{1}{8\sin20^\circ}\sin160^\circ=\frac{1}{8}=\frac{1}{2^{3}}$

Έτσι:

$ \frac{\sqrt{2}}{2^{6}}\frac{1}{2^{3}}=\frac{\sqrt{2}}{2^{9}}$.

Πολύ ωραία Στέλιο!

Αλέξανδρος

[b][color=red][size=150]ΑΣΚΗΣΗ 23[/size][/color][/b]
Nα βρεθεί το υπόλοιπο της διαίρεσης του $ P(x) \equal{} (cos(\frac{\pi}{16})\plus{}xsin(\frac{\pi}{16}))^{4}$ με το πολυώνυμο $ x^{2}\plus{}1$

Το φέρνουμε στη μορφή ρ*ημ(χ+φ) και μετά γίνονται απλούστατα τα πράγματα.

Hint
Οι ρίζες του $ x^{2}\plus{}1\equal{}0$ είναι οι $ x_{1}\equal{}i$ και $ x_{2}\equal{}\minus{}i$
Είναι $ P(i)\equal{}(\cos\ \frac{\pi}{16}\plus{}i\sin\ \frac{\pi}{16})^{4}\equal{}\cos\ \frac{\pi}{4}\plus{}i\sin\ \frac{\pi}{4}\equal{}\frac{\sqrt 2}{2}\plus{}i\frac{\sqrt 2}{2}$

όμοια $ P(\minus{}i)\equal{}\frac{\sqrt 2}{2}\minus{}i\frac{\sqrt 2}{2}$


If hence ${a_{n}}$ have conditions:
$a_{n+2}a_{n}=a_{n+1}^{2}+8$ and $a_{1}=4$, $a_{2}=6$. Prove that $9{a_{n}}^{2}-128$ is perfect  of rationals numbers.

 :wink:

[quote="Fitim"]If hence ${a_{n}}$ have conditions:
$a_{n+2}a_{n}=a_{n+1}^{2}+8$ and $a_{1}=4$, $a_{2}=6$. Prove that $9{a_{n}}^{2}-128$ is perfect square.

 :wink:[/quote]
Interesting problem. What approach should we take here? The standard rearrange and solve?

There's some mistake there, since neither $a_{n}$ nor $9a_{n}^{2}$ are integer for $n\geqslant 4$

well the terms are not negative but its true that some are not integers
we can prove that :
$a_{n+1}+a_{n-1}=\frac{5}{2}a_{n}$ and $9a_{n}^{2}-128=(4a_{n-1}-5a_{n})^{2}$
so the asked quantity is indeed the square of a rational number...

[quote="Albanian Eagle"]well the terms are not negative[/quote]

Yup, that was a moment of stupidity. Deleted.

And generally - does "perfect square" imply "a square of an integer"? or "a square of a rational"?

Fitim you did the same thing as me above but notice that $a_{4}=21,5$ etc so not all terms are integers. I think you meant the square of a rational number...? :)

[quote="i_like_pie"][quote="Farenhajt"]And generally - does "perfect square" imply "a square of an integer"? or "a square of a rational"?[/quote]
"Perfect square" implies the square of an integer.[/quote]

I thought so too...


inequalities
function
inequalities proposed

Give three positive real number $a,b,c$ satisfying $a+b+c=3$.
Prove that:
$48abc + (ab+bc+ca)^3 +16 \geq 12(ab+bc+ca)+18abc(ab+bc+ca)$  :)

set $a=1$ and let $bc=k$ where $0<k\leq 1$ and you will see that the inequality become

$(k+3)^2-15\geq 0$ which is not generally correct ....

Uhm, i don't think so since the degree of $bc$ is $3$ . This one is quite interesting, the equality occurs only when $a=1,b=2,c=0$ and the permutation.

It's exciting. The equality occurs if $a=2,b=1,c=0$. It must be prove by changing the expression to a sum of squares, if we don't want to touch in ABC. But $3^6=729$. :(

Would you post a more concrete solution ? Indeed this was made less strict, the original form can't be solved by ABC theorem.

again we see the case  when two variables are equal to get an inequality in one variable which can be then solved by calculus.(I can't find other ways to this problem! :D )

Opp, but this one doesn't occur when two variables are equal, would you post a full solution ?

[quote="Anh Cuong"]Give three positive real number $a,b,c$ satisfying $a+b+c=3$.
Prove that:
$48abc + (ab+bc+ca)^3 +16 \geq 12(ab+bc+ca)+18abc(a+b+c)$  :)[/quote]If $a+b+c=3$, why did you write $abc(a+b+c$? It is strange!
Replacing $a+b+c$ by 3 the inequality becomes
$(ab+bc+ca)^3 +16 \geq 12(ab+bc+ca)+16abc$.
For $ab+bc+ca=2$  the inequality becomes
$0 \geq 16abc$,
which is not true. :oops:

hahaha that's the reason why it does'nt work

Yes, but in fact, the real inequality is :
$48abc + (ab+bc+ca)^3 +16 \geq 12(ab+bc+ca)+18abc(ab+bc+ca)$. ;)

Opp, i terrribly sorry, thanks mathmanman for correcting it.

Let $x = ab+bc+ca$. We need to show that $x^3 - 6(3abc+2)x + 16(3abc+1) \geq 0$.

Consider the function $f(x) = x^3 - 6(3abc+2)x + 16(3abc+1)$ on $R^{+}$. Then $f'(x) = 3x^2 -6(3abc+2) = 0$ if and only if $x = x_0 = \sqrt{2(3abc+2)}$. Thus, if $x$ lies on the $[0, x_0]$, $f(x) \geq f(x_0)$.

By plugging $x =x_0$ into the function $f(x)$, we obtain

$f(x_0) = -4\sqrt{2(3abc+2)^3} + 16(3abc+1)$.

Let $u=3abc$. Then $0 < u \leq 3$. It is not hard to prove that $g(u) = -4\sqrt{2(u+2)^3} + 16(u+1) \geq 0$.

Hence $f(x_0) \geq 0$.

It is now sufficient to show that $0 <ab+bc+ca < x_0$. The left inequality is obvious. To prove the right one, we have

$(ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2 + 2abc(a+b+c)$

$= a^2b^2+b^2c^2+c^2a^2+6abc \leq 6abc + \frac{(a+b+c)^4}{27} < 2(3abc+2) = x_0^2$.

Thus the problem is solved.

I 'm sorry the final step $a^2b^2+b^2c^2+c^2a^2 \leq \frac{(a+b+c)^4}{27}$ is not true. However, it isn't really a big trouble to the solution.

We know that $f(x_0)$ is the minimal value of $f(x)$ on $R^{+}$. Hence, since we have proved that $f(x_0) \geq 0$, we obtain $f(x) \geq 0$ for every positive $x$. Thus, since $ab+bc+ca > 0$, we obtain $f(ab+bc+ca) \geq 0$. Therefore the problem is solved.


Hi everyone!  There is this proof that I'm having trouble with, and I was wondering if anyone knew how to do this:

A perpendicular matrix is one such that A transpose = A^-1

Prove that if a matrix, namely A, is perpendicular, then any two distinct columsn of A have dot product zero.  Is the converse of this statement true?  Justify your answer.

[quote="jmerry"]That's an orthogonal matrix. I've never heard "perpendicular" in this context.

$A^{T}A=I$. What are the entries of $A^{T}A$ in terms of the columns of $A$?[/quote]

Sorry, I'm really confused as to how to find that.  Like, I know the identity matrix is I = A^(T)A =

(1 0 0)
(0 1 0)
(0 0 1)

I'm not sure how to find the what the columns of A are, though.

[quote="jmerry"]You're misunderstanding the question- I didn't ask you to find the columns. Given $A=\begin{bmatrix}A_{1}&A_{2}&\cdots&A_{n}\end{bmatrix}$, with $A_{1},\dots,A_{n}$ column vectors, what is the $ij$-entry of $A^{T}A$?[/quote]

A = [A1, A2,...An}

So, A^T is

[A11, A21..., Ai1]
[A12, A22,...,Ai2]
[A1j, A2j,....,Aij]

Therefore the ij of A^T * A

is

(A1 * A1j) + (A2 * A2j) + (An * Aij)

No, that's not it. Don't confuse vectors with scalars; my $A_{i}$ were meant to be vectors, which means your notation for $A^{T}$ and $A$ don't match.

ok, well let's say the matrix A is:

[a11, a12, a1j]
[a21, a22, a2j]
[ai1, ai2, aij]

and let's say it's transpose is

[a11, a21, ai1]
[a12, a22, ai2]
[a1j, a2j, aij]

therefore, to find the aij entry of A times A transpose, we would need the dot product of:

the ith row of A, which is [ai1, ai2, aij] with the ith column of A transpose, meaning:

(ai1 * ai1) + (ai2 * ai2) + (aij * aij)

Now what does that mean?
You should also look at $A^{T}A$- that's the one that's relevant to your original question.

But Perpendicular matrix, it sounds cool.

$A^{T}A$ has dot products of all columns A as entries (elements). For ortogonal (perpendicular in chosen termins) matrices transposed is the same, like inverted. $A^{-1}A=I$. You have really - distinct dot products of any two (different with eath other) columns outside of diagonal $I$.

Oh!  I get it.

So if we have two columns - named i and j such that i isn't equal to j (because they are distinct), the dot product of the 2 columns is the product of the ith row of A transpose with the jth column of A.

A^T (A) is the identity matrix.  If i and j aren't equal, then the (i,j) entry of A^T(A) is not 1, but 0.

I was wondering - is the converse true?  Meaning, if any two distinct columns of A have dot product zero, is the matrix A orthogonal?

The key additional point: the $i$th row of $A^{T}$ is the $i$th column of $A$.

As for the converse- what about the diagonal?

>Oh!  I get it.
Good :)

>if any two distinct columns of A have dot product zero, is the matrix A orthogonal?
Yes.. (but u've to norm columns of A to get 'pure' 1/0, orthogonal matrix has normalized columns)


number theory
prime factorization
relatively prime
superior algebra
superior algebra unsolved

Let m be a positive integer and let $ n \equal{} \phi(m) \equal{} \prod_{i\equal{}1}^{r}p_{i}^{e_{i}}$ be the the prime factorization of n. Let g be an integer relatively prime to m. Show that g is a primitive root modulo m iff $ g^{\frac{n}{p_{i}}}$ is not congruent to 1 modulo m for each $ i \equal{} \{1, ..., r\}$.

[quote="akech"]Let m be a positive integer and let $ n \equal{} \phi(m) \equal{} \prod_{i \equal{} 1}^{r}p_{i}^{e_{i}}$ be the the prime factorization of n. Let g be an integer relatively prime to m. Show that g is a primitive root modulo m iff $ g^{\frac {n}{p_{i}}}$ is not congruent to 1 modulo m for each $ i \equal{} \{1, ..., r\}$.[/quote]

Let d be the order of g.  g is a primitive root iff d = n.  It is clearly necessary that $ g^{n/p_i}$ not be congruent to 1 or else g would have order dividing n/p_i which is strictly less than n.  For sufficiency, the order d divides n (why?), so we can write $ d \equal{} n/(p_i q)$ for some q and some $ p_i.$  Now $ 1 \equiv g^{dq} \equiv g^{n/p_i}$ the latter not equivalent to 1 which gives a contradiction.

-evo


calculus
integration
limit
trigonometry
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calculus computations

Find $ \lim_{n\to\infty} \int_0^{\pi} e^x|\sin nx|dx.$

Let we devide interval $ (0,\pi)$ on $ n$ parts. Let $ \frac {n}{2}$ is an integer. So, the value of integral will be:
$ I_n \equal{} \int_{0}^{\pi}{e^x|sinnx|dx} \equal{} \sum_{k \equal{} 0}^{\frac {n}{2} \minus{} 1}\left(\int_{\frac {2\pi k}{n}}^{\frac {(2k \plus{} 1)\pi}{n}}{e^x\sin nx dx} \minus{} \int_{\frac {(2k \plus{} 1)\pi}{n}}^{\frac {(2k \plus{} 2)\pi}{n}}{e^x\sin nx dx}\right)$
We could write: $ \sin nx \equal{} Im\left(e^{nxi}\right)$
So: $ I_n \equal{} Im\left(\sum_{k \equal{} 0}^{\frac {n}{2} \minus{} 1}\left(\int_{\frac {2\pi k}{n}}^{\frac {(2k \plus{} 1)\pi}{n}}{e^{(ni \plus{} 1)x} dx} \minus{} \int_{\frac {(2k \plus{} 1)\pi}{n}}^{\frac {(2k \plus{} 2)\pi}{n}}{e^{(ni \plus{} 1)x} dx}\right)\right)$
We have:
$ I_n \equal{} Im\left(\frac {1}{ni \plus{} 1}\cdot\sum_{k \equal{} 0}^{\frac {n}{2} \minus{} 1}\left(2e^{\frac {(2k \plus{} 1)\pi}{n}(ni \plus{} 1)} \minus{} e^{\frac {2k\pi}{n}(ni \plus{} 1)} \minus{} e^{\frac {(2k \plus{} 2)\pi}{n}(ni \plus{} 1)}\right)\right)$
$ I_n \equal{} Im\left(\frac {1}{ni \plus{} 1}\cdot\sum_{k \equal{} 0}^{\frac {n}{2} \minus{} 1}e^{\frac {2k\pi}{n}(ni \plus{} 1)}\cdot\left(2e^{\frac {\pi}{n}(ni \plus{} 1)} \minus{} 1 \minus{} e^{\frac {2\pi}{n}(ni \plus{} 1)}\right)\right)$
Then:
$ I_n \equal{} Im\left(\frac {2e^{\frac {\pi}{n}(ni \plus{} 1)} \minus{} 1 \minus{} e^{\frac {2\pi}{n}(ni \plus{} 1)}}{ni \plus{} 1}\cdot\sum_{k \equal{} 0}^{\frac {n}{2} \minus{} 1}{e^{\frac {2k\pi}{n}(ni \plus{} 1)}}\right)$
$ I_n \equal{} Im\left(\frac {2e^{\frac {\pi}{n}(ni \plus{} 1)} \minus{} 1 \minus{} e^{\frac {2\pi}{n}(ni \plus{} 1)}}{ni \plus{} 1}\cdot\frac {e^{\pi (ni \plus{} 1)} \minus{} 1}{e^{\frac {2\pi}{n}(ni \plus{} 1)} \minus{} 1}\right)$
or
$ I_n \equal{} Im\left(\frac {2e^{\pi i \plus{} \frac {\pi}{n}} \minus{} 1 \minus{} e^{2\pi i \plus{} \frac {2\pi}{n}}}{ni \plus{} 1}\cdot\frac {e^{\pi (ni \plus{} 1)} \minus{} 1}{e^{\frac {2\pi}{n}(ni \plus{} 1)} \minus{} 1}\right)$
$ I_n \equal{} Im\left(\frac {2e^{\pi i}e^{\frac {\pi}{n}} \minus{} 1 \minus{} e^{2\pi i}e^{\frac {2\pi}{n}}}{ni \plus{} 1}\cdot\frac {e^{\pi}e^{\pi ni} \minus{} 1}{e^{\frac {2\pi}{n}}e^{2\pi i} \minus{} 1}\right)$
We have here: $ e^{\pi i} \equal{} \minus{} 1$, $ e^{2\pi i} \equal{} 1$
So:
$ I_n \equal{} Im\left(\frac { \minus{} 2e^{\frac {\pi}{n}} \minus{} 1 \minus{} e^{\frac {2\pi}{n}}}{ni \plus{} 1}\cdot\frac {e^{\pi}e^{\pi ni} \minus{} 1}{e^{\frac {2\pi}{n}} \minus{} 1}\right)$
$ I_n \equal{} \minus{} Im\left(\frac {e^{\frac {\pi}{n}} \plus{} 1}{e^{\frac {\pi}{n}} \minus{} 1}\cdot\frac {e^{\pi}e^{\pi ni} \minus{} 1}{ni \plus{} 1}\right)$
$ I_n \equal{} \minus{} Im\left(\frac {e^{\frac {\pi}{n}} \plus{} 1}{e^{\frac {\pi}{n}} \minus{} 1}\cdot\frac {\left(e^{\pi}e^{\pi ni} \minus{} 1\right)\cdot \left(1 \minus{} ni\right)}{n^2 \plus{} 1}\right)$
$ I_n \equal{} \minus{} Im\left(\frac {e^{\frac {\pi}{n}} \plus{} 1}{e^{\frac {\pi}{n}} \minus{} 1}\cdot\frac {\left(e^{\pi}(\cos\pi n \plus{} i\sin\pi n) \minus{} 1\right)\cdot \left(1 \minus{} ni\right)}{n^2 \plus{} 1}\right)$

$ I_n \equal{} \frac {e^{\frac {\pi}{n}} \plus{} 1}{e^{\frac {\pi}{n}} \minus{} 1}\cdot\frac { \minus{} e^{\pi}\sin\pi n \plus{} n e^\pi\cos\pi n \minus{} n}{n^2 \plus{} 1}$

But we have: $ \sin\pi n \equal{} 0$ and $ \cos\pi n \equal{} 1$. So:

$ I_n \equal{} \frac {e^{\frac {\pi}{n}} \plus{} 1}{e^{\frac {\pi}{n}} \minus{} 1}\cdot\frac {n e^\pi \minus{} n}{n^2 \plus{} 1}$

If we take $ n\to \infty$ we will have:
$ e^{\frac {\pi}{n}} \equal{} 1 \plus{} \frac {\pi}{n} \plus{} O\left(\frac {1}{n^2}\right)$ and
$ \lim_{n\to \infty}{I_n} \equal{} \frac {2}{\pi}\left(e^\pi \minus{} 1\right)$
Maybe I`m wrong. I`m not shure. Interesting to see your solution.  :ninja:

Note that the average value of $ |\sin nx|$ is $ \frac2{\pi}$ and the result is this average value times $ \int_0^{\pi}e^{x}\,dx.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=648136194&t=21933]Here[/url] is one old topic about this; there are probably others to be found.


analytic geometry
number theory
prime numbers
greatest common divisor
geometry unsolved
geometry

The circle $x^2+ y^2 = r^2$ meets the coordinate axis at $A = (r,0), B = (-r,0), C = (0,r)$ and $D = (0,-r).$ Let $P = (u,v)$ and $Q = (-u,v)$ be two points on the circumference of the circle. Let $N$ be the point of intersection of $PQ$ and the $y$-axis, and $M$ be the foot of the perpendicular drawn from $P$ to the $x$-axis. If $r^2$ is odd, $u = p^m > q^n = v,$ where $p$ and $q$ are prime numbers and $m$ and $n$ are natural numbers, show that

\[ |AM| = 1, |BM| = 9, |DN| = 8, |PQ| = 8. \]

It suffices to prove that the only solution of the equation $x^2+y^2=r^2$ in integers such that $r$ is odd, and $x$, $y$ are powers of prime numbers is $4^2+3^2=5^2$.
First, as $r$ is odd, one of $x$, $y$ should be even, so we can suppose $x=2^n$ for some positive integer $n$. Now we have $y^2=(r-2^n)(r+2^n)$ so both $r-2^n$ and $r+2^n$ should be powers of the same prime, but being their greatest common divisor $(r-2^n,r+2^n)=(2r,r+2^n)=(r,r+2^n)=(r,2^n)=1$,  we conclude that $r-2^n=1$.
Now we have $y^2=(r-2^n)(r+2^n)=2^{n+1}+1$ and also that $2^{n+1}=y^2-1=(y-1)(y+1)$ and from this we conclude that $y-1$ and $y+1$ are two powers of two that are two units apart, so it should be $y-1=2$ and $y+1=4$, that is, $y=3$, $n=2$, $x=4$ and $r=5$.

QED


source:IRMO 1997, second round.
this a pretty good problem, (it's not hard!)
if $n, r$ are positive integers, and $2\le n$ and $r\not\equiv {0}(mod n)$ and $g=(n,r)$ prove:
$\sum_{i=1}^{n-1}{\{\frac{ir}{n}\}}=\frac{1}{2}(n-g)$
where $\{x\}$ is the fractional part of $x$

[hide]
Let $n=kg$.

Notice that the numerator of the LHS cycles through $g, 2g,..., kg \mod{n}$ (unless $k=n$, in which case it doesn't ever reach $kg$ but that doesn't matter because it will result in a $0$ fractional part anyway).

We note that each term appears $g$ times except $kg$ which appears $g-1$ times (but again it doesn't matter).

So the resulting sum is $\frac{g^2+2g^2+...+(k-1)g^2}{n} = \frac{g^2\frac{k(k-1)}{2}}{n}$. But $n=kg$ so we have $\frac{1}{2}g(k-1) = \frac{1}{2}(kg-g) = \frac{1}{2}(n-g)$.

Note that is does work when $r \equiv 0 \mod{n}$ as well because then $g=n$ and all the terms of the LHS are $0$.

Sorry I didn't do a great job of explaining that first part, but I'll leave it to be explored by yourselves.[/hide]

i proved it, another way,
i proved two cases:
 if $g=(n,r)=1$, i proved that the series is $\frac{n-1}{2}$, then i proved that if $(n,r)=1, r'=gr, n'=gn$. then the sum for r' and n' is true... i can post the proof if anyone is intrested... ;)

What exactly do you mean by g=(n,r)?  Doesn't that mean that g is not a number?

(n,r) is a mathematical notation for the greatest common divisor of n and r.


geometry
geometric transformation
reflection
trapezoid
symmetry
ratio
cyclic quadrilateral

Let $\tau$ be a circumference [i]inscribed[/i] in a circumscribed quadrilateral $ABCD$. Let $X$ and $Y$ be the points where $BD$ intersects $\tau$, and $M$ be the midpoint of $XY$. Let $P$ and $Q$ be the points of tangency of $DA$ and $DC$ with $\tau$. Prove $\angle AMP = \angle CMQ$.

Daniel

Let R, S be the tangency points of BC, BA with the circle $\tau$. The opposite sides PQ, RS of the contact cyclic quadrilateral PQRS meet at the pole K of the diagonal BD of the tangential quadrilateral ABCD with respect to its incircle $\tau$ and this pole lies on the other diagonal AC. Since the normal to $XY \equiv BD$ at the midpoint M of XY passes through the center of $\tau,$ it also passes through the pole K of BD. WLOG, assume that K lies on the same side of BD as C. Let P', Q' be the reflections of P, Q in MK. PQQ'P' is an isosceles trapezoid with PQ, P'Q' intersecting at K. Its diagonals PQ', QP' meet at the polar BD of K with respect to $\tau$ and by symmetry, also on the line MK, hence $PQ' \cap QP' \equiv BD \cap MK \equiv M.$ Since MK bisects $\angle PMP',\ \angle QMQ',$ its perpendicular $MD \equiv BD$ bisects the complementary angle $\angle PMQ$ and $\angle DMP = \angle DMQ$ are equal. In exactly the same way, if R', S' are reflections of R, S in MK, then $RS', SR'$ meet at M and MB bisects the angle $\angle RMS.$ Consider the cyclic quadrilateral QRP'S' with the diagonal intersection $M \equiv QP' \cap RS'$ and let the tangents to the circle $\tau$ at P', S' meet at A', let the tangent at P', R meet at B', and let the tangents at S', Q meet at D'. For notation purpose, let $N \in MK$ be the intersection of SP, S'P'. Since $\triangle S'P'M$ is a reflection of $\triangle SPM$ in MK and $\tau$ is reflected into itself, A' is a reflection of A in $MK \equiv MN$, hence $\angle A'MN = \angle AMN$ are equal. A'C is a diagonal of the tangential quadrilateral A'B'CD' with the contact quadrilateral QRP'S', hence they have the same diagonal intersection, $M \in A'C.$ Thus the angles $\angle CMK = \angle A'MN = \angle AMN$ are equal. As a result,

$\angle AMP = 90^\circ-(\angle AMN+\angle DMP) = 90^\circ-(\angle CMK+\angle DMQ) = \angle CMQ$

[quote="yetti"]Since MK bisects $\angle PMP',\ \angle QMQ',$ its perpendicular $MD \equiv BD$ bisects the complementary angle $\angle PMQ$ and $\angle DMP = \angle DMQ$ are equal.[/quote]

But this assumes that $P'$, $M$, and $Q$ are collinear, which is not necessarily the case.

[quote="nsato"]But this assumes that $P'$, $M$, and $Q$ are collinear, which is not necessarily the case.[/quote]

You are mistaken. Not only are P', M, Q necessarily collinear, it is easy to see why, and I gave that reason.

You're right, I was looking at my diagram incorrectly.  :(

[quote="dhernandez"]Let $ \tau$ be a circumference [i]inscribed[/i] in a circumscribed quadrilateral $ ABCD$. Let $ X$ and $ Y$ be the points where $ BD$ intersects $ \tau$, and $ M$ be the midpoint of $ XY$. Let $ P$ and $ Q$ be the points of tangency of $ DA$ and $ DC$ with $ \tau$. Prove $ \angle AMP \equal{} \angle CMQ$.

Daniel[/quote]

[b][size=100]PROOF[/size][/b]. – It is easy to show that the points $ P,$ $ D,$ $ Q,$ $ O,$ $ M,$ where $ O,$ is the center of the incircle $ (O)$ of $ ABCD$ $ ($ instead of $ \tau$ $ ),$ are concyclic $ ($ from $ \angle DMO \equal{} \angle DPO \equal{} \angle DQO \equal{} 90^{o}$ $ ).$

So, we have $ \angle DMP \equal{} \angle DOP \equal{} \angle DOQ \equal{} \angle DMQ$ $ \Longrightarrow$ $ \angle DMP \equal{} \angle DMQ$ $ ,(1)$

We denote as $ R,$ $ S,$ the tangency points of $ (O),$ to the sidelines $ BC,$ $ AD$ respectively and we will prove that the line segments $ PQ,$ $ AC,$ $ OM,$ $ RS,$ are concurrent at one point.

Let $ N,$ $ N'$ be, the intersection points of the line segment $ AC,$ from the line segments $ PQ,$ $ RS$ respectively and we will prove that $ N'\equiv N.$

Applying the [b][size=100]Menelaus’s theorem[/size][/b] in the triangle $ \bigtriangleup ADC,$ taken the line segment $ PQ$ as its transversal, we have $ \frac {NA}{NC}\cdot \frac {QC}{QD}\cdot \frac {PD}{PA} \equal{} 1$ $ \Longrightarrow$ $ \frac {NA}{NC} \equal{} \frac {QD}{QC}\cdot \frac {PA}{PD}$ $ ,(2)$

Similarly from the triangle $ \bigtriangleup ABC,$ with its transversal the line segment $ RS,$ we have $ \frac {N'A}{N'C}\cdot \frac {RC}{RB}\cdot \frac {SB}{SA} \equal{} 1$ $ \Longrightarrow$ $ \frac {N'A}{N'C} \equal{} \frac {RB}{RC}\cdot \frac {SA}{SB}$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {NA}{NC} \equal{} \frac {N'A}{N'C}$ $ \Longrightarrow$ $ N'\equiv N.$

It is easy now, to show that the line segment $ OM,$ passes through the point $ N,$ based on the polar theory. Because of the line segments $ PQ,$ $ RS,$ as the polars of the points $ D,$ $ B$ respectively, with respect to the circle $ (O),$ we conclude that the line segment $ BD,$ connecting these points, is the polar of $ N,$ with respect to $ (O)$ and so, we have that $ BD\perp ON.$

That is the line segment $ OM\perp BD,$ passes through the point $ N.$

We denote as $ E,$ $ F,$ the intersection points of the line segment $ OM,$ from the line sidelines $ AD,$ $ CD$ respectively.

Because of $ \angle DMP \equal{} \angle DMQ,$ we have also $ \angle EMP \equal{} \angle FMQ$ $ ,(4)$

Because of the collinearity of the points $ A,$ $ P,$ $ D,$ we conclude that the pencils $ N.EAPD$ and $ M.EAPD,$ have congruent double ratios $ ($ because they have also the line segment $ MN,$ as their common ray $ ).$

That is, we have  $ (N.EAPD) \equal{} (M.EAPD$ $ ,(5)$ and similarly $ (N.FCQD) \equal{} (M.FCQD)$ $ ,(6)$

$ ($ from the collinearity of the points $ C,$ $ Q,$ $ D$ and the line segment $ NF,$ as the common ray of the pencils $ N.FCQD,$ $ M.FCQD$ $ ).$

From $ (5),$ $ (6)$ $ \Longrightarrow$ $ (M.EAPD) \equal{} (M.FCQD)$ $ ,(7)$ because of $ (N.EAPD)\equiv (M.FCQD).$

From $ (7),$ because of $ \angle DMP \equal{} \angle DMQ$ and $ \angle EMP \equal{} \angle FMQ,$ we conclude that $ \angle AMP \equal{} \angle CMQ$ and the proof is completed.

$ ($ because of in the pencils $ M.EAPD$ and $ M.FCQD,$ with congruent double ratios, three of their homologous rays form equal angles with each other, it is easy to show that  the same result holds, as for the rest of their homologous rays $ ).$

Kostas Vittas.

hint.
Let $ R,S$ be tangency of $ BC,AB$ with $ \tau$. Denote $ E\equal{}AB\cap CD, F\equal{}AD\cap BC, G\equal{}CA\cap EF,H\equal{}CA\cap BD$. Then $ C,H,A,G$ form harmonic division. Show $ \angle GMH\equal{}90$(not hard). Then $ \angle AMD\equal{}\angle DMC$. So it is suffice to show $ \angle PMD\equal{}\angle DMQ$. This is true since $ D,P,M,Q$ are cyclic and $ DP\equal{}DQ$.


[b]Problem:[/b]
Let $ ABC$ be a triangle where sides $ AB$, $ BC$ and $ AC$ satisfy that,
$ \frac {AB^{3} \plus{} BC^{3} \plus{} AC^{3}}{AB \plus{} BC \plus{} AC} \equal{} BC^{2}$
If $ n \equal{} BC$, prove that the maximum possible value for $ AB$ is $ \frac{2}{3}n\sqrt{3}$.

I think it's not true.

Let $ AB\equal{}c, \ BC\equal{}a, \ AC\equal{}b$

Then $ \frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c}\equal{}a^2\Rightarrow a^3\plus{}b^3\plus{}c^3\equal{}a^3\plus{}a^2b\plus{}a^2c\Rightarrow$

$ \Rightarrow b^3\plus{}c^3\equal{}a^2(b\plus{}c)\Rightarrow(b\plus{}c)(b^2\minus{}bc\plus{}c^2)\equal{}a^2(b\plus{}c)\Rightarrow$

$ \Rightarrow b^2\minus{}bc\plus{}c^2\equal{}a^2\Rightarrow \frac{b^2\plus{}c^2\minus{}a^2}{bc}\equal{}1\Rightarrow \cos A\equal{}\frac{1}{2}\Rightarrow A\equal{}60^{\circ}$

Then $ a\equal{}2R\sin A\equal{}2n\cdot\frac{\sqrt{3}}{2}\equal{}n\sqrt{3}$

Sorry for the mistake. I forgot which angle I had given what cosine, and it was a mess. Anyways, I have edited in the actual problem.