geometry
3D geometry
number theory
least common multiple
ratio
search

[quote=mathprincess09]
2) Find the next number in the sequence.
1,1,2,3,5,8,13,21,_____,_____,______
[/quote]

That seems to be a very famours sequence, what do you call it??

fibonacchi, even though I think I spelled it wrong...

[quote="mathprincess09"]1) Baskin-Robins advertises 31 flavors of ice cream. How many different double decker cones are possible if both the flavors have to be different? The order of the scoop doesnt matter,

2) Find the next number in the sequence.
1,1,2,3,5,8,13,21,_____,_____,______

3) The LCM of three consecutive one digit positive integers is 168. What are the integers?I

4) A window washer uses a vinegar and water solution in the ratio of one-half cup of vinegar to three cups of water. How much vinegar will be in two gallons of solution?
a)1 and 1/2     b) 5 and 1/3      c) 4 and 4/7      d) 12.0
 :D  :D  :lol:  :lol:  ;)[/quote]

1. [hide]it's gonna be the combination of choosing 2 out of 31.... so it's gonna be $(31*30)/(2*1) = 31 * 15 = 465... wow, lots of ice cream....[/hide]

2. [hide]Fibonacchi is the sequence name, and so it's gonna be 34... since it ask for the next number... but since you have 3 blanks, it's gonna be 34, 55,  and 89... unlesss I add it wrong... [/hide]

3. [hide]I'm not sure how you would do this... so I random thought of 3 numbers and got lucky... 6,7,and 8[/hide]

4.[hide] 0.5+3 = 3.5... and 2 gallons is equal to 8 quarts... which is 16 pints, which is 32 cups...     and the ratio of vinegar to water is equal to 0.5:3 or 1:6... so when you divide 32 by 7, you get 4 and 4/7 so it's gonna be c[/hide]

yes it is called Fibonnachi and it is very famous
and i dont no why but i just love that
sequence!
 :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

For the people who don't know the Fib. Sequence is a sequence where you can put a addition symbol inbetween any two numbers and the next number right after it, is the sum.

Fibonacci Sequence (for those who spelt it wrong, voila!) is what's called a recursive sequence. Recursive sequences are sequences where a term is formed in relation to the previous terms. Fibonacci Sequence is the most famous one.

for $n \ge 3$, it is stated that
$F_n = F_{n-1} + F_{n-2}$

so try this and verify yourself. good luck.

does it has any other bounds like $F_1=0$??

[quote="mathprincess09"]1) Baskin-Robins advertises 31 flavors of ice cream. How many different double decker cones are possible if both the flavors have to be different? The order of the scoop doesnt matter,

2) Find the next number in the sequence.
1,1,2,3,5,8,13,21,_____,_____,______

3) The LCM of three consecutive one digit positive integers is 168. What are the integers?

4) A window washer uses a vinegar and water solution in the ratio of one-half cup of vinegar to three cups of water. How much vinegar will be in two gallons of solution?
a)1 and 1/2     b) 5 and 1/3      c) 4 and 4/7      d) 12.0
 :D  :D  :lol:  :lol:  ;)[/quote]
[hide="Solutions"]
1) (31 x 30)/(2!) = [b]315[/b]

2) Fibbonaci sequence :D ! [b]34, 55, and 89.[/b]

3) 168 = 7 x 24. Let's try 5-6-7, 6-7-8, and 7-8-9.
5-6-7 - 210
[b]6-7-8[/b] - 168
7-8-9 - 504

4) I don't know these conversion thingies <_< . Most likely [b]c[/b], because the solution is 1/7 vinegar, and c involvs a sevenths term.
[/hide]

Alright here is number 4 with all conversions listed
[hide]
    16 tablespoons = 1 cup
    12 tablespoons = 3/4 cup
    10 tablespoons + 2 teaspoons = 2/3 cup
    8 tablespoons = 1/2 cup
    6 tablespoons = 3/8 cup
    5 tablespoons + 1 teaspoon = 1/3 cup
    4 tablespoons = 1/4 cup
    2 tablespoons = 1/8 cup
    2 tablespoons + 2 teaspoons = 1/6 cup
    1 tablespoon = 1/16 cup
    [b]2 cups = 1 pint[/b]
    [b]2 pints = 1 quart [/b]
    3 teaspoons = 1 tablespoon
    48 teaspoons = 1 cup
    [b]4 quarts = 1 gallon[/b]
(It is amazing how crappy the US system is...)

We only need to worry about the bold conversions.

So we have 1/2vin:3cups of water

If we reduce the rato we get 1cup:6cups

We are creating a 2 gallon solution which is converted to 32 cups.

1x+6x=32

x=32/7

Which turns to be 4 and 4/7 cups (c)

If you need a nice converter use google! You can say 2 gallons to cups or 3 quarts to gallons and anything else![/hide]

Google. The handy search engine usable whenever and whatever. Including math competitions :P .

[quote="mathprincess09"]3) The LCM of three consecutive one digit positive integers is 168. What are the integers?[/quote]

168=2*2*2*3*7

This means one of the numbers must have 23 as a factor, another has 3 as a factor, and tha third one has 7 as a factor. The only consecutive #'s that fit the requirements are 6,7,8.


analytic geometry
induction
limit
geometry proposed
geometry

Starting with the vertices $P_1(0,1)$, $P_2(1,1)$, $P_3(1,0)$, $P_4(0,0)$ of a square, we contstruct furthur points; $P_5$ is the midpoint of $P_1P_2$, $P_6$ is the midpoint of $P_2P_3$, $P_7$ is the midpoint of $P_3P_4$, $P_8$ is the midpoint of $P_4P_5$, and so on... Draw the figure for brevity's sake. Now, the polygonal spiral path $P_1P_2P_3P_4P_5P_6P_7...$ approaches a point $P$ inside the square.

(a) If the coordinates of $P_n$ are $(x_n, y_n)$, prove that $\frac{1}{2}x_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ and find a similar equation for the y-coordinates.

(b) Find the coordinates of $P$.

Masoud Zargar

Proposition:

$\frac{x_n}{2} + x_{n+1} + x_{n+2} + x_{n+3} = 2$

$\frac{y_n}{2} + y_{n+1} + y_{n+2} + y_{n+3} = \frac 3 2$

Proof is by induction with respect to n: Since, $x_1 = 0,\ x_2 = 1,\ x_3 = 1,\ x_4 = 0,$ the x-proposition holds for n = 1 and similarly, since $y_1 = 1,\ y_2 = 1,\ y_3 = 0,\ y_4 = 0,$ the y-proposition also holds for n = 1. Assume that the x- and y- propositions are true for some n. For n > 4, the point $P_n = (x_n, y_n)$ is obtained from the points $P_{n-4} = (x_{n-4},\ y_{n-4}),\ P_{n-3} = (x_{n-3},\ y_{n-3})$ as $P_n = \left(\frac{x_{n-4} + x_{n-3}}{2},\ \frac{y_{n-4} + y_{n-3}}{2}\right)$. Hence,

$\frac{x_{n+1}}{2} + x_{n+2} + x_{n+3} + x_{n+4} = \frac{x_{n+1}}{2} + x_{n+2} + x_{n+3} + \frac{x_n + x_{n+1}}{2} =$

$= \frac{x_n}{2} + x_{n+1} + x_{n+2} + x_{n+3} = 2$

The 2nd induction step for the y-proposition is exactly the same, except the rightmost side is $\frac 3 2$. In the limit transition $n \rightarrow \infty$, the x- and y- recurrence relations for the ${\{x_n\},\ \{y_n\}}$ sequences yield equations for their limits $L_x, L_y$, identical with the x-, y-coordinates of the point $P = (L_x, L_y)$:

$\lim_{n \rightarrow \infty} \left(\frac{x_n}{2} + x_{n+1} + x_{n+2} + x_{n+3}\right) = \frac{L_x}{2} + L_x + L_x + L_x = 2,\ \ L_x = \frac 4 7$

$\lim_{n \rightarrow \infty} \left(\frac{y_n}{2} + y_{n+1} + y_{n+2} + y_{n+3}\right) = \frac{L_y}{2} + L_y + L_y + L_y = \frac 3 2,\ \ L_y = \frac 3 7$


The circles $\mathcal C_1(O_1, r_1)$ and $\mathcal C_2(O_2, r_2)$, $r_2 > r_1$, intersect at $A$ and $B$ such that $\angle O_1AO_2 = 90^\circ$. The line $O_1O_2$ meets $\mathcal C_1$ at $C$ and $D$, and $\mathcal C_2$ at $E$ and $F$ (in the order $C$, $E$, $D$, $F$). The line $BE$ meets $\mathcal C_1$ at $K$ and $AC$ at $M$, and the line $BD$ meets $\mathcal C_2$ at $L$ and $AF$ at $N$. Prove that
\[ \frac{ r_2}{r_1} = \frac{KE}{KM} \cdot \frac{LN}{LD} . \]
[i]Greece[/i]

The claim is a consequence of the following lemma:

LemmaRL26042006.

Let  c1(O1,r1) and c2(O2,r2) be twoo orthogonal circles having twoo common points A and B. If D = c1/\O1O2  and F is the external (wrt c1) end of the diameter  of c2 through O1, and N and L are the intersection of BD with AF, then DN/LN = r1/r2.

Proof.

Later on ...

Manipulating angles, have to prove that K is the midpoint of the arc(CD) and AM is the angle bisector of angle(KAE).
So KM/ME=AK/AE (the interior angle bisector theorem)
But AK/AE=r_1/r_2 because if you aplicate a dilatative rotation with center = A, angle of rotation = O_1A O_2, quotient of homothecy = r2/r1, then C_1--> C_2 and K--> E.
Then, with a little of algebra KE/KM=(r_1+r_2)/r_1.
Similarly, LN/LD=r_2/(r_1+r_2), and the result follows.

[size=135][color=darkred][b]PP ([u]Balkan MO 1995[/u]).[/b] The circles $ \mathcal C_1(O_1, r_1)$ and $ \mathcal C_2(O_2, r_2)$ , $ r_2 > r_1$ , intersect at $ A$ and $ B$ such that $ AO_1\perp AO_2$ . The line $ O_1O_2$ meets $ \mathcal C_1$ at $ C$ and $ D$, and $ \mathcal C_2$ at $ E$ and $ F$ (in the order $ C$, $ E$, $ D$, $ F$).  The line $ BE$ meets $ \mathcal C_1$ at $ K$ and $ AC$ at $ M$ . The line $ BD$ meets $ \mathcal C_2$ at $ L$ and $ AF$ at $ N$ . Prove that $ \frac { r_2}{r_1} \equal{} \frac {KE}{KM} \cdot \frac {LN}{LD}$ .[/color] [/size]

[size=130][color=darkblue][b][u]Proof.[/u][/b] This problem is a nice application of the harmonical division. See the lemma from [/color] [url=http://www.artofproblemsolving.com/viewtopic.php?t=46146][color=darkred][b]here[/b][/color][/url].[/size]

[color=darkblue]$ 1\blacktriangleright$ The division $ (C,E,D,F)$ is harmonically\ ;

$2\blacktriangleright$ The angles $ \widehat {KAC}$ , $ \widehat {CAE}$ , $ \widehat{EAD}$ , $ \widehat {DAF}$ , $ \widehat {FAL}$ hava same measure, $ 45^{\circ}$ .

$ 3\blacktriangleright$ $ A\in KF\cap LC\ ;$

$4\blacktriangleright$ $ MN\parallel CF\ ;$

$5\blacktriangleright$ $ \frac {MK}{ME} \equal{} \frac {NL}{ND} \equal{} \frac {r_1}{r_2}$ (see [b]Camaronsou[/b]'s)  $\mathrm {\ \ OR\ \ }$ $MN\parallel CF$ $ \implies$ $ \left\{\begin{array}{cc} \triangle KEF\ : & \frac {KE}{KM} \equal{} \frac {EF}{MN} \\
 \\
\triangle LDC\ : & \frac {LN}{LD} \equal{} \frac {MN}{CD}\end{array}\right\|$ $ \implies$ $ \frac {KE}{KM}\cdot\frac {LN}{LD} \equal{} \frac {EF}{CD} \equal{} \frac {r_2}{r_1}$ .[/color][/size]

First it is easy to see that triangles $ ABC, AFB $ are isosceles, if $ X $ is the intersection of  $ AB $ and $ CF, AX=XB $ and $ \angle AXC=90 $, and also $ MN||CF $

Now, from the power of point theorem we have:
$ LN=\frac{AN \cdot NF}{BN}, LD=\frac{DE \cdot DF}{DB}, KM=\frac{MA \cdot MC}{MB}, KE=\frac{CE \cdot ED}{EB} $ so
$ \frac{LN}{LD} \frac{KE}{KM} =
\frac{CE}{EB} \frac{MB}{MC} \frac{BD}{DF} \frac{NF}{BN} \frac{AN}{AM} = \frac{sin CBE}{sin BCE} \frac{sin BCM}{sin CBM} \frac{sin BFD}{sin FBD} \frac{sin FBN}{sin BFN} \frac{AN}{AM} $
and now since
$ sin BCM=2\frac{AX}{AC}\frac{CX}{AC}, sin BCE =\frac{BX}{BC}, sin BFD=\frac{BX}{BF}, sin BFN=2\frac{AX}{AF}\frac{FX}{AF} $  and $ \frac{AN}{AM}=\frac{AF}{AC} $
 we have
$ \frac{LN}{LD}\frac{KE}{KM}=\frac{AF^2}{AC^2}\frac{CX}{FX} $ and we want to prove that
$ \frac{AF^2}{AC^2}\frac{CX}{FX}\frac{r_1}{r_2}=1 $ which is equivalent with
$ \frac{AF}{FX}\frac{AF}{O_2F}{\frac{CX}{AC}\frac{CO_1}{AC}=1 }$, note that
$ \frac{AF}{FX}\frac{AF}{O_2F}{\frac{CX}{AC}\frac{CO_1}{AC}=\frac{1}{cos DFA}\cdot 2cos DFA \cdot cos ECA \cdot \frac{1}{2cos ECA} =1 }$


I am already enrolled in the AMC-12 Problem Series, and I was considering taking the AIME Problem Series also.  How much more difficult is the AIME Problem Series?  Also, how much time will I need to get the most out of both classes?

The AIME Problem Series is definitely a step up from the AMC 12 Problem Series, though you have been in class for two of the harder days for the AMC 12 Problem Series.

As far as time goes, the AIME Problem Series probably takes a little more time than the AMC 12 as far as working message board problems and whatever time you want to spend reviewing the transcripts because of the step up in difficulty level.


Here is a very difficult problem.For given $n$ and $r$,what is the maximal dimension of a vectorial subspace consisting only of matrices $n*n$ with rank at most $r$?Great attention,the field is arbitrary.

You are right, Moubi. The intuition is important in this problem. Indeed, taking matrices with $n-r$ columns equal to 0 we find your answer. Proving this is the best dimension  is an absolutely non-trivial problem0, quite very difficult. Try to solve it first for the case of real numbers, it is so difficult even in this case. I will post some hints in the following days since it is clear that without hints this problem is discouraging.  :)


inequalities
logarithms
algebra proposed
algebra

Show that $ ln\frac{p}{q}\le\frac{p\minus{}q}{\sqrt{pq}},(0<q\leq{p})$

$ \ln x\leq \frac{1}{2}\left(x\minus{}\frac{1}{x}\right)\ (x\geq 1)$.


Let $ f(x)= a.cos(2x+\alpha) + b.cos(x+\beta) + 1 $ satisfying :
 $ f(x) \geq 0 $ with every $ x \in R $
 Prove that :
1) $ /a/ \leq 1 $ , $ /b/ \leq \sqrt2 $
2) $ f(x) \leq 3 $
3) If $ \alpha = \beta = 0 $ then $ /a/+/b/ \leq 2 $

This is my teacher 's problem !!!
 Here is a soluiton for 1) , 2)
 1) ugly computation
 2) It's easy to show $ f(x) + f(x+\frac{2\pi}{3}) + f(x+\frac{4\pi}{3}) =3 $
 ===> $ f(x) \leq 3 $

 3) It's still unsolved !

When is $ /a/ + /b/ =2 ??
 Let's solve it  :)


I found out the following question's answer is true. i need some explanation though. thanks.

If the sides of one triangle are congruent to the coreesponding sides of another triangle, then the corresponding angles must also be congruent.

I thought in order to use CPCTC, 2 triangles can only be congruent or similar by AA, SAS, SSS, and AAs

That is SSS... all the sides are congruent...


vector
function
linear algebra
linear algebra unsolved

I need on this question, I just can't see it...

Let V be the real vector space R^R and let V(even), V(odd) be the subsets of V consisting the even and the odd functions respectively. Show that V(even), V(odd) are subspaces of V and V=V(even) + V(odd).

From the previous question, give the decomposition of the function f(x) = e^x into its even and odd component functions.

Given any function $f$, the unique decomposition into even and odd parts is
$f(x)=\left(\frac12f(x)+\frac12f(-x)\right)+\left(\frac12f(x)-\frac12f(-x)\right)$

Thanks a lot, I still got another question...

If V is a vector space, prove or disprove the following statement:

If U[size=59]1[/size], U[size=59]2[/size], W are subspaces of V with U[size=59]1[/size] (O+) W = U[size=59]2[/size] (O+) W then U[size=59]1[/size] = U[size=59]2[/size]

Note: O+ means direct sum

That one's false. It should be pretty easy to construct a counterexample (try subspaces of dimension 1 in $\mathbb{R}^2$).


Which of the cones listed below can be formed from a $ 252^\circ$ sector of a circle of radius $ 10$ by aligning the two straight sides?
[asy]import graph;unitsize(1.5cm);defaultpen(fontsize(8pt));draw(Arc((0,0),1,-72,180),linewidth(.8pt));draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt));label("$10$",(-0.5,0),S);draw(Arc((0,0),0.1,-72,180));label("$252^{\circ}$",(0.05,0.05),NE);[/asy]
[asy]
import three;
picture mainframe;
defaultpen(fontsize(11pt));
picture conePic(picture pic, real r, real h, real sh)
{
	size(pic, 3cm);
    triple eye = (11, 0, 5);
	currentprojection = perspective(eye);
    real R = 1, y = 2;

	triple center = (0, 0, 0);
    triple radPt = (0, R, 0);
    triple negRadPt = (0, -R, 0);
    triple heightPt = (0, 0, y);

	draw(pic, arc(center, radPt, negRadPt, heightPt, CW));
	draw(pic, arc(center, radPt, negRadPt, heightPt, CCW), linetype("8 8"));
	draw(pic, center--radPt, linetype("8 8"));
	draw(pic, center--heightPt, linetype("8 8"));
	draw(pic, negRadPt--heightPt--radPt);

	label(pic, (string) r, center--radPt, dir(270));
    if (h != 0)
    {
		label(pic, (string) h, heightPt--center, dir(0));
    }
    if (sh != 0)
    {
		label(pic, (string) sh, heightPt--radPt, dir(0));
    }
    return pic;
}
picture pic1;
pic1 = conePic(pic1, 6, 0, 10);
picture pic2;
pic2 = conePic(pic2, 6, 10, 0);
picture pic3;
pic3 = conePic(pic3, 7, 0, 10);
picture pic4;
pic4 = conePic(pic4, 7, 10, 0);
picture pic5;
pic5 = conePic(pic5, 8, 0, 10);

picture aux1; picture aux2; picture aux3;
add(aux1, pic1.fit(), (0,0), W);
label(aux1, "$\textbf{(A)}$", (0,0), 22W, linewidth(4));
label(aux1, "$\textbf{(B)}$", (0,0), 3E);
add(aux1, pic2.fit(), (0,0), 35E);

add(aux2, aux1.fit(), (0,0), W);
label(aux2, "$\textbf{(C)}$", (0,0), 3E);
add(aux2, pic3.fit(), (0,0), 35E);

add(aux3, aux2.fit(), (0,0), W);
label(aux3, "$\textbf{(D)}$", (0,0), 3E);
add(aux3, pic4.fit(), (0,0), 35E);

add(mainframe, aux3.fit(), (0,0), W);
label(mainframe, "$\textbf{(E)}$", (0,0), 3E);
add(mainframe, pic5.fit(), (0,0), 35E);
add(mainframe.fit(), (0,0), N);
[/asy]

[hide="solution"]Note that the edge length of the sector is $(10)(7\pi/5) = 14 \pi$.  When the sector is folded into a cone, this length becomes the full circumference of the base circle, meaning $2 \pi r = 14 \pi \Rightarrow r = 7$.  It is obvious that the slant height is 10, since we have aligned the two edges of the sector to form part of the side, so the answer is $\textbf{(C)}$[/hide]

[hide="Answer"]$\frac{252}{360}=\frac{7}{10}$, and since the slant height must be $10$, then the radius is $7$, so the answer is $\boxed{C}$.[/hide]

The original contest pdf had the cones drawn as answer choices, so I'm adding these here:
[asy]
import three;
picture mainframe;
picture conePic(picture pic, real r, real h, real sh)
{
	size(pic, 3cm);
    triple eye = (11, 0, 5);
	currentprojection = perspective(eye);
    real R = 1, y = 2;

	triple center = (0, 0, 0);
    triple radPt = (0, R, 0);
    triple negRadPt = (0, -R, 0);
    triple heightPt = (0, 0, y);

	draw(pic, arc(center, radPt, negRadPt, heightPt, CW));
	draw(pic, arc(center, radPt, negRadPt, heightPt, CCW), linetype("8 8"));
	draw(pic, center--radPt, linetype("8 8"));
	draw(pic, center--heightPt, linetype("8 8"));
	draw(pic, negRadPt--heightPt--radPt);

	label(pic, (string) r, center--radPt, dir(270));
    if (h != 0)
    {
		label(pic, (string) h, heightPt--center, dir(0));
    }
    if (sh != 0)
    {
		label(pic, (string) sh, heightPt--radPt, dir(0));
    }
    return pic;
}
picture pic1;
pic1 = conePic(pic1, 6, 0, 10);
picture pic2;
pic2 = conePic(pic2, 6, 10, 0);
picture pic3;
pic3 = conePic(pic3, 7, 0, 10);
picture pic4;
pic4 = conePic(pic4, 7, 10, 0);
picture pic5;
pic5 = conePic(pic5, 8, 0, 10);

picture aux1; picture aux2; picture aux3;
add(aux1, pic1.fit(), (0,0), W);
label(aux1, "$\textbf{(A)}$", (0,0), 22W, linewidth(4));
label(aux1, "$\textbf{(B)}$", (0,0), 3E);
add(aux1, pic2.fit(), (0,0), 35E);

add(aux2, aux1.fit(), (0,0), W);
label(aux2, "$\textbf{(C)}$", (0,0), 3E);
add(aux2, pic3.fit(), (0,0), 35E);

add(aux3, aux2.fit(), (0,0), W);
label(aux3, "$\textbf{(D)}$", (0,0), 3E);
add(aux3, pic4.fit(), (0,0), 35E);

add(mainframe, aux3.fit(), (0,0), W);
label(mainframe, "$\textbf{(E)}$", (0,0), 3E);
add(mainframe, pic5.fit(), (0,0), 35E);
add(mainframe.fit(), (0,0), N);
[/asy]

What does it mean aligning two straight edges?

I think it means that you don't want any part of the sector to overlap. The edges should meet together, but not overlap. :)

The slant height is clearly $10$, and since the circumference is $10\pi \cdot 2 \cdot \frac{252}{360}=14\pi$, the radius is $7$. So, the answer is $\boxed{\text{C}}$.

[quote=RJ5303707]I don't really understand the question. Can someone please explain it to me?[/quote]

The sector is used to form a cone, it is done by aligning the straight sides.

[quote=RJ5303707]I don't really understand the question. Can someone please explain it to me?[/quote]

A cone is formed using the edges where it is cut, and you have to find how it looks

[quote=RJ5303707]I don't really understand the question. Can someone please explain it to me?[/quote]

[hide=Explanation][hide=1st PIECE OF WORK]Ok so the "perimeter" of the sector we have is $\frac{252}{360} \cdot 20\pi.$ Why? Well, a normal circle has a circumference of $\pi r^2$ and we want 252 degrees of this circle. Since a circle has $360$ degrees, we find that the circumference of the sector is $\frac{252}{360} \cdot 20\pi = 14\pi.$[/hide]

[hide=OBSERVATION]Now notice that when you curl the paper up you form a cone, with the base = to the circumference of the sector. Try playing around with this on paper. [/hide]

[hide=ANSWER]So after making that observation, you see that we just need to find the radius of a circle with a circumference of $14\pi.$ From here we can see that the radius is 7 which means the answer is C.
[/hide][/hide]


Ok I'm a complete noob and just started learning programming.

I'm using Dev-C++ compiler. Whenever I run a program a black box pops up and then immediately disappears. Is this supposed to happen? How can I make it stay so I can see my program?

Don't worry, it's normal. You can use [b]getch();[/b] or [b]system("pause");[/b] in the end of your program (before the [b]return(0);[/b]) to stop the closing of the window.

Ah yes, if you'll use [b]getch()[/b], include [b]<conio.h>[/b].

Many programmers simply hate [b]system("pause");[/b] and [b]getch();[/b] (and they're absolutely right-system commands such as pause are not designed to be used in such "trivial" matters, and they are not supported on all systems (same trouble with [b]getch() [/b]and its library [b]<conio.h>[/b]), so I don't recommend it... there's always [b]cin.get()[/b] with similar effect as [b]getch()[/b]. And no additional libraries needed  :wink:

Thanks, I'm actually learning C but I think [b]getch();[/b] is equivalent to [b]getchar();[/b] ?

As far as I know, [b]getchar();[/b] waits for the return key. [b]getch();[/b] doesn't. But [b]cin.get();[/b] or [b]getchar();[/b] will do the trick for your needs.

<conio.h> is not a part of the standard C or C++ and you don't have cin.get(); in plain C so I think that aspiringchemlete's best bet is getchar(); ;).

Dev-C++ is not a compiler, it's an IDE (Dev-C++ has gcc 3.4.2). I recommend you to use another editor, and compile your programs by command lines. If you do so, you will be able to use many compilers, know more about compiler directives and many things behind...
I have been using editor Notepad++ for a long time and it's my favourite editor. It has a plug-in call NppExec. With NppExec, you can call almost programs inside Notepad++. Moreover when you compile and run a program written in C++ (or any programming languages), you can directly see your results of your programs in a smalling docking window without doing something to pause your programs. Of course that's not all Notepad++ can do. It has many features that I'm sure you will like. Here is the link: [url]http://notepad-plus.sourceforge.net/[/url].
Try it out! I hope you will enjoy it.


[color=darkred]Let $ ABC$ be a triangle and $ H$ its orthocenter.
we draw a line $ l$ and denote its intersections with $ AB,AC,BC$ by $ C',B',A'$ respectively.
let $ d_a$ be a line which passes through $ A$ and is perpendicular to $ A'H$.
define similarly $ d_b,d_c$.
1)prove that $ d_a,d_b,d_c$ are concurrent.
2)call their point of intersection $ S$. prove that $ SH \perp l$.[/color]
(could u prove it with pole and polar?thx in advance :blush: )

http://www.mathlinks.ro/viewtopic.php?t=313257
This is only the reversed problem.

yes i know,could anyone give a proof using pole and polar(and without inversion plz,as i wanted )...
that would be kind...
thx in advance.... :)

[b]Result:[/b] Let $ A_1,A_2$ be the projection of $ H$ on $ AS, A'S$, similar define $ B_1,B_2,C_1,C_2$. Prove that $ A_1A_2,B_1B_2,C_1C_2$ concur at the polar of line $ A'B'C'$.


Also, much like the Fibonacci formula, this can be written in a somewhat nicer-looking form as $\left\lfloor \frac{(5+2\sqrt 6)^n}{4\sqrt6} \right\rfloor$.

Isn't that truncated form only for sufficiently large $n$ ?

Yes, but $5 - 2\sqrt6$ is very, very close to zero, and dividing it by $4\sqrt6$ doesn't hurt, either.  So it turns out in this case (as in the Fibonacci case, although there you round near rather than round down) that "sufficiently large" includes everything.


modular arithmetic
analytic geometry
combinatorics proposed
combinatorics

Suppose that we have put eight rooks on a chessboard so that none of them attack each other. Prove that the number of rooks that are on a black square is even.

Number the rows and columns of the board with numbers from $1$ to $8$ in such a way that the square $(i,j)$ is black iff $i\ne j\pmod 2$. The positions of the rooks are $(i,\sigma(i))$ for some permutation $\sigma$ of $\overline{1,n}$.

We now add up the coordinates of the positions of the rooks, and look at the sum modulo $2$. On the one hand we get $2\sum_{i=1}^ni\pmod 2=0\pmod 2$, and on the other hand we get $N\pmod 2$, where $N$ is the number of rooks on positions $(i,\sigma(i))$ with $i+\sigma(i)=1\pmod 2$, i.e. the number of rooks on black squares. This shows that $N=0\pmod 2$, as desired.

Prove the stronger result:
Half of the rooks which stand on black squares stand on files a,c,e,g.


Find all positive integers $ a,b,c,d$, with $ a\ge b\ge c\ge0$ such that $ 2^a\plus{}2^b\plus{}2^c\equal{}d^2$.

For $ a\equal{}2n\plus{}2,b\equal{}m\plus{}2,c\equal{}0$, it becomes Zhautykov 2005: http://www.mathlinks.ro/viewtopic.php?t=245391
For $ a\equal{}2x\plus{}1,b\equal{}x,c\equal{}0$, it becomes IMO 2006: http://www.mathlinks.ro/viewtopic.php?t=101485
For $ a\equal{}2x,b\equal{}2y,c\equal{}0$, it becomes Korea 2007: http://www.mathlinks.ro/viewtopic.php?t=316742

[hide="What I have done"]If $ c$ is odd, $ 2^a\plus{}2^b\plus{}2^c\equal{}2^{c\minus{}1}(2^{a\minus{}c\plus{}1}\plus{}2^{b\minus{}c\plus{}1}\plus{}2)$. So $ 2^{a\minus{}c\plus{}1}\plus{}2^{b\minus{}c\plus{}1}\plus{}2$ is a perfect square which is impossible since it is 2 mod 4 (except for some trivial cases). So $ c$ is even. Let $ a\minus{}c\equal{}x,b\minus{}c\equal{}y$. It is enough to find $ x,y$ such that $ 1\plus{}2^y\plus{}2^x$ is a perfect square, let $ 1\plus{}2^y\plus{}2^x\equal{}(2k\plus{}1)^2$, we get $ 2^{y\minus{}2}(2^{x\minus{}y}\plus{}1)\equal{}k(k\plus{}1)$. Exactly one of $ k$ or $ k\plus{}1$ is divisible by $ 2^{y\minus{}2}$. So $ k\plus{}1\ge2^{y\minus{}2}$. But $ k\plus{}1\equal{}2^{y\minus{}2}$ is clearly impossible (except for some trivial cases maybe). Thus $ k\ge2^{y\minus{}2}$. This gives $ 2^{y\minus{}2}(2^{x\minus{}y}\plus{}1)\ge2^{y\minus{}2}(2^{y\minus{}2}\plus{}1)$, therefore $ x\minus{}y\ge y\minus{}2$. Let $ y\minus{}2\equal{}m,x\minus{}y\equal{}m\plus{}n$. Now it is enough to find $ m,n$ such that $ 2^m(2^{m\plus{}n}\plus{}1)\equal{}k(k\plus{}1)$... etc etc etc

I've made a lot of analysis, but it leads to nowhere.. For small $ n$, such as 1, 2, 3, we can simply do some caseworks. But I don't know how to solve the general case. Anyone can continue?[/hide]

I dint go through everything and you might have already got this, but anyway:

$ 2^{2a} \plus{} 2^{2b} \plus{} 2^{a \plus{} b \plus{} 1} \equal{} d^{2}$

$ 1 \plus{} 2^{a \plus{} 2} \plus{} 2^{2a \plus{} 2} \equal{} d^{2}$

and

$ 1 \plus{} 2^{4} \plus{} 2^{5} \equal{} 7^{2}$

I got something - but not sure whether corrct or wrong....unable to proceed further.

Assuming $ a\le b\le c$, we get

$ 1 \plus{} 2^{b \minus{} a} \plus{} 2^{c \minus{} a} \equal{} 2^a.k^2$

So that gives either $ (b \minus{} a) \equal{} 1$ or $ (c \minus{} a) \equal{} 1$ or $ a \equal{} 1$.

If $ a \equal{} 1$, we have

$ 1 \plus{} 2^{b \minus{} 1} \plus{} 2^{c \minus{} 1} \equal{} 2k^2 \\
\implies b$or $ c$ =1......But this is not giving all solutions!


rotation
counting
distinguishability
pigeonhole principle
geometry
geometric transformation
search

How many different ways are there to arrange six identical black chairs and six identical silver chairs around a large circular table in an evenly spaced manner?

[hide="attempt #1"]Set the black chairs in a line, and see how many ways you can arrange the silver chairs among the black chairs - 7 places for the first, 8 for the 2nd, and so on until 12 places for the last one, which can be described as $ \frac{12!}{6!}$ and they're identical so divide by 6!. Then divide by 12, to count for rotating this line. So $ \frac {12!}{12\cdot6!\cdot6!} \equal{} \boxed{77}$. [/hide]

[hide="attempt #2"]There are 12! ways to arrange the chairs in a line, then divide by 6! for the black chairs and another 6! for the silver chairs, and then by 12 to count for rotation. This'll give $ \frac {12!}{12\cdot6!\cdot6!} \equal{} \boxed{77}$[/hide]

I am not that good at counting (i.e. really bad) so this could be completely wrong. Anyways, here is a solution (I hope).

The number of permutations of the chairs in a straight line is equivalent to $ \frac{12!}{6!6!}\equal{}\binom{12}{6}$ (divided by number of ways to arrange identical objects). But when they are arranged in a circle, each permutation will occur $ 12$ times because of rotations. Therefore the answer is $ \frac{1}{12} \cdot \binom{12}{6}\equal{}77$.

[quote="SnowEverywhere"]each permutation will occur $ 12$ times because of rotations[/quote]  How many times does BSBSBSBSBSBS occur?

Hm so like

OK YAY WE FIX A BLACK CHAIR AT THE TOP.

Now we have 11 remaining slots, 6 of which must be silver. Pick these 6.

11c6 and we win. I think?

Btw, 11c6=11*10*9*8*7/120=11*6*7=432

However I am pretty sure this is wrong.

$ \binom{11}{6} \equal{} \frac {6}{12} \binom{12}{6}$, i.e., now you're arguing that every linear arrangement is counted twice.  But how many different ways are there to get something equivalent to BBBBBBSSSSSS from your procedure?

The key words here are "Polya theory."  However, one can solve this without any technology -- one just has to give up the idea that a simple formula is going to apply and instead do some necessary casework.

The problem has to do with the arranging on a circle
Suppose there are 4 chairs, two of each color
aabb=baab=bbaa=abba
abab=baba
We can see that some linear arrangement are repeated more than others.

I guess one can try all the combinations and find a pattern... I havent found one yet

[hide="Solution"]Let each setup have rotational period $ n$ so that if you rotate it $ n$ times it goes to an indistinguishable state from the first one. Then clearly $ n|12$ so $ n \equal{} 1,2,3,4,6,12$. In general if it has period $ n$ then we can divide the $ 12$ chairs into $ 12/n$ indistinguishable groups of $ n$ because it's periodic $ \bmod{n}$.

[i]Case: $ n \equal{} 1$.[/i] This means that there must be $ 12$ chairs of the same color, which is not possible since there must be $ 6$ of each, yielding $ \boxed{0}$ setups.
[i]Case: $ n \equal{} 2$.[/i] This means that the chairs must be alternating, yielding $ \boxed{1}$ setup.
[i]Case: $ n \equal{} 3$.[/i] This means that there are $ 4$ indistinguishable, equal groups of three chairs _ _ _, which means by Pigeonhole that there must be at least $ 2$ chairs of some color and hence at least $ 2\cdot4 \equal{} 8$ chairs of this color in all, which is not possible since there must be $ 6$ of each; $ \boxed{0}$ setups are allowed.
[i]Case: $ n \equal{} 4$.[/i] This means that there are $ 3$ groups of four chairs _ _ _ _. Note that we can make the groups such that they all start with B, so we have B _ _ _ with one remaining B and two remaining W's, so there are $ \binom31 \equal{} 3$ setups at most. BBWW and BWWB are the same by rotation, and have smallest period $ 4$, and BWBW has period $ 2 < 4$, so there's only $ \boxed{1}$ setup in this case.
[i]Case: $ n \equal{} 6$.[/i] This means that there are $ 2$ groups of six chairs _ _ _ _ _ _, which we can again let be B _ _ _ _ _ except with two remaining B's and three remaining W's for at most $ \binom52 \equal{} 10$ setups. One way is BWBWBW which has period $ 2 < 6$, so it doesn't work. And there are three distinct ways remaining. First, BBBWWW=BBWWWB=BWWWBB. Second, BWBBWW=BBWWBW=BWWBWB. Third, BWWBBW=BBWBWW=BWBWWB. Because $ 1 \plus{} 3 \plus{} 3 \plus{} 3 \equal{} 10$, we have found all possibilities, so there are $ \boxed{3}$ setups for this case.
[i]Case: $ n \equal{} 12$.[/i] Note that the count of $ \binom{12}6$ for the spots of the B chairs counts setups with period $ n$ precisely $ n$ times. Because we already know the number of setups with periods $ 1,2,3,4,6$, and the only other possibility is $ 12$, we can see that there are
\[ \binom{12}6 \minus{} (1)(0) \minus{} (2)(1) \minus{} (3)(0) \minus{} (4)(1) \minus{} (6)(3) \equal{} 900\]
ways with $ n \equal{} 12$ where we have counted each setup $ 12$ times. Hence $ 900/12 \equal{} 75$ setups are allowed for this case.

Finally, we have
\[ 0 \plus{} 1 \plus{} 0 \plus{} 1 \plus{} 3 \plus{} 75 \equal{} \boxed{80}.\]
[/hide]
For the same technique see this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2110930706&t=272283]MATHCOUNTS PROBLEM HA[/url].

(Whatever this is just equivalent to [url=http://en.wikipedia.org/wiki/Burnside%27s_lemma]Burnside's Lemma[/url] and stuff probably.)

Anyway, this is arguably harder for a 40 minute competition than any of the national problems, especially if you haven't seen something similar before.

Wow. I did not see that idea. Time to be more careful. :wallbash: (I always do this with counting problems).

I know this seems a little redundant, but what is the general strategy for counting rotations when we have something repeated like this?

I know that the (n-1)! formula depends on the items to be unique because there are varying rotational periods when we start to repeat items. I suppose my question is how can you determine which cases satisfy the given rotational period?

Consider another problem analogous to the posted one where we instead have four items AABB.

I drew it out and basically I found two cases:

*There is one case where the rotational period is 2. This occurs when we have ABAB (The alternating pattern makes it easy for us to see how this can quickly return to its original pattern.

*Now we need to find when the rotational period is 4. I drew it out and found the AABB pattern to give us the desired rotational period, but I can't intuitively explain why this is true.

let us consider chairs as people

no. of ways of arranging n persons in circular table is $ (n \minus{} 1)!$ ways ie $ 11!$ .$ since$ there are 6,6 different identical chairs.

so,no. of ways are

$ \frac {11!} {6!6!} \equal{} \boxed{77}$ ways

If you'll trouble yourself to read the thread, you'll notice that that incorrect answer has already been given, as well as an explanation of what's wrong with it.   :roll:

I found the phrase "evenly spaced" given in the problem to be a distraction
as it does not influence the number of orderings.  That is, it is immaterial
to solving the problem.

How close neighboring chairs are to each other, relative to other chairs
does not matter.


induction
number theory unsolved
number theory

Let $p(k)$ be the smallest prime not dividing $k$. Put $q(k) = 1$ if $p(k) = 2$, or the product of all primes $< p(k)$ if $p(k) > 2$. Define the sequence $x_0, x_1, x_2, ...$ by $x_0 = 1$, $x_{n+1} = \frac{x_np(x_n)}{q(x_n)}$. Find all $n$ such that $x_n = 111111$

here's a sol which i know: take $p_1=2<p_2=3<5=p_3<....$ are all primes. Then by induction show that if $n=1c_0+2c_1+4c_2+...$ in base $2$ then $x_n=\prod_{i}p_i^{c_i}$ for $n\geq1$.

To end [b]pluricomplex[/b]'s solution :
We have $x_0 = 1$, so, $n = 0$ does not work.
Also, we have : $111111 = 3 \cdot 7 \cdot 11 \cdot 13 \cdot 37 = p_2 \cdot p_4 \cdot p_5 \cdot p_6 \cdot p_{12}$.
Hence : $x_n = 111111 \iff n = 2^{11} + 2^5 + 2^4 + 2^3 + 2$, id est : $n = 2106$.

I am sorry .I know this problem is solved.but I have a question:
     I think if $x_{n}=p_{1}^{a_1}.p_{2}^{a_2}...p_{k}^{a_{k}}$ then $a_i=1$
  then $x_n=111111=3.13.7.37$ :arrow: there isn't $x_{n-1}$


write down as many! theorems or facts about sequences or series as possible.plz i need to find.

Maybe you could start here:

http://mathforum.org/library/drmath/sets/high_seq_series.html

[i]Before [/i]you go about memorizing formulas... you should try to understand their proof as well as [u]why [/u] the formula works.  If you do that, you can expand the formulas to improve your computing time, accuracy, and confidence.

Are you asking about finite geometric series, infinite geometric series, arithmetic progressions, or what? There are too many types of series to list; you should be more specific.


blogs
superior algebra
superior algebra unsolved

Show that every (finite?) group occurs as the Galois group of an extension of $ \mathbb{C}(X)$.

You say "difficulty unknown," but do you know if this is actually true or not?

I remember reading it (or a claim very like it) in a paper, but can't find it.

Here is a recent blog I did on a similar result (but I think you may already know this one)  :wink: [url=http://hilbertthm90.wordpress.com/2008/12/17/finite-groups-as-galois-groups/]link here[/url]

Aah, I didn't realize that about the finite case until now, thanks.  I think the problem for finite groups is open if the base field is fixed to be $ \mathbb{Q}$.

One can use GAGA/SGA1 (Riemann's existence theorem suffices): $ \pi_1^{\mathrm{\acute{e}t}}(\mathbf{CP}^1 \minus{} \{x_0,\ldots,x_n\}) \equal{} \langle x_0,\ldots,x_n \mid x_0x_1\cdots x_n \equal{} 1\rangle\hat{}$ (by Seifert-van Kampen), the pro-free group on $ n$ generators. (According to Pop, an algebraic proof of this for $ n \equal{} 3$ is an open problem. One can even show that $ G_{\mathbf{C}(X)}$ is the free profinite group on $ |\mathbf{C}|$ generators.) A similar result holds for $ \mathbf{Q}_p(X)$.


algebra
pen
Vieta
induction
number theory
relatively prime
Divisibility Theory

Prove that there exist infinitely many pairs $(a, b)$ of relatively prime positive integers such that \[\frac{a^{2}-5}{b}\;\; \text{and}\;\; \frac{b^{2}-5}{a}\] are both positive integers.

Let $ b\equal{}2y,a\equal{}3y\plus{}x$, then $ a^2\plus{}b^2\minus{}5\minus{}3ab\equal{}x^2\minus{}5y^2\minus{}5\equal{}0$ give solution.
$ x\plus{}y\sqrt 5 \equal{}(5\plus{}2\sqrt 5)(9\plus{}4\sqrt 5 )^k$ is solution.

http://www.mathlinks.ro/Forum/viewtopic.php?t=170040
see more here.

Once you verified that for $a_0 = -1$, $a_1 = 1$, $a_{n+2} = 3a_{n+1} - a_n$ (careful, you have a typo there!) you have $a_n \mid a_{n+1}^2 - 5$ and $a_{n+1} \mid a_{n}^2 - 5$, it is clear we have $\gcd(a_n, a_{n+1}) \mid 5$. But modulo $5$ the sequence $(a_n)_{n\geq 0}$ is periodic, with period $-1,1$, and so no term is divisible by $5$, therefore $\gcd(a_n, a_{n+1}) = 1$.

Also, by the Vieta jumping method, it is proved [b]more[/b] than asked, namely that the solutions above are the [b]only[/b] ones. One correction though; it is asked that both $\dfrac {a^2-5} {b}$ and $\dfrac {b^2-5} {a}$ be positive integers, for $a,b$ positive integers, thus $a,b > 2$, so the solutions start with $(4,11)\mapsto (11,29)\mapsto (29, 76)\mapsto \cdots$. Otherwise we may take terms $\pm a_n, \pm a_{n+1}$, [b]and[/b] we also have the special solutions $(a,b) \in \{(1,2), (2,1)\}$, missing from your list, but verifying $ab \mid a^2 + b^2 - 5$.

If true, then $\tfrac{a^2-5}{b},\tfrac{b^2-5}{a}\in \bf Z$, so their product $\tfrac{(a^2-5)(b^2-5)}{ab}=\tfrac{a^2b^2-5a^2-5b^2+25}{ab}=ab-5\left(\tfrac{a^2+b^2-5}{ab}\right)$ is also integer. Hence, it suffices to show that there are infinitely many integers $a,b$ for which $ab\mid (a^2+b^2-5)$.

Consider the infinite sequence: -1,1,4,11,29,76,199,..., defined by $a_0=-1$, $a_1=1$, and $a_{n+2}=3a_{n+1}-a_n\ \forall{n\ge{0}}$.

We now quickly prove that $\gcd(a_n,a_{n+1})=1$ before moving onto the main proof. Note that we have $a_n \mid a_{n+1}^2 - 5$ and $a_{n+1} \mid a_{n}^2 - 5$, so it is clear that $\gcd(a_n, a_{n+1}) \mid 5$. But modulo $5$ the sequence $(a_n)_{n\geq 0}$ is periodic, with period $-1,1$, and so no term is divisible by $5$. Hence $\gcd(a_n, a_{n+1}) = 1$, as desired.

I claim that $a_n^2+a_{n+1}^2=3a_na_{n+1}+5$.

[b]Proof:[/b] It is clearly true when $n=0$, for we get $a_0^2+a_1^2=(-1)^2+1^2=2=3(-1)(1)+5=3a_0a_1+5$.
Suppose the proposition works for $n=k$, where $k\ge{0}$, i.e. $a_k^2+a_{k+1}^2=3a_ka_{k+1}+5$.

By definition, $a_{k+2}=3a_{k+1}-a_k$. Adding $2a_{k+2}$ to both sides yields \[3a_{k+1}2a_{k+2}-a_k.\] Multiplying both sides by $a_{k+1}$ yields \[3a_{k+1}a_{k+2}=3a_{k+1}^2+2a_{k+1}a_{k+2}-a_ka_{k+1}.\] Adding 5 to both sides yields
\begin{align*}
3a_{k+1}a_{k+2}+5 & =3a_{k+1}^2+2a_{k+1}a_{k+2}-a_ka_{k+1}+5\\
&=3a_{k+1}^2+2a_{k+1}a_{k+2}+3a_ka_{k+1}-4a_ka_{k+1}+5.
\end{align*}
By the induction hypothesis, $3a_ka_{k+1}+5=a_k^2+a_{k+1}^2$, thus
\begin{align*}
& 3a_{k+1}^2+2a_{k+1}a_{k+2}+3a_ka_{k+1}-4a_ka_{k+1}+5 \\ &=a_k^2+a_{k+1}^2+3a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}\\ &=a_k^2+a_{k+1}^2+3a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}\\ &=a_k^2+4a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}.\ (\clubsuit)
\end{align*}

But we can also get $(\clubsuit)$ as follows.

By definition, $a_{k+2}=3a_{k+1}-a_k$. Subtracting $a_{k+1}$ from both sides, squaring, then adding $2a_ka_{k+1}$ to both sides yields \[a_k^2+a_{k+1}^2=a_k^2+4a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}.\] But this is just $(\clubsuit)$, hence \[\boxed{a_k^2+a_{k+1}^2=3a_ka_{k+1}+5}\] $\bf Q.E.D.$

Well, then, let $(a,b)=(a_{n+1},a_{n+2})=(a_{n+1},3a_{n+1}-a_n)$. (Do you see the Lemma coming?) Then,
\begin{eqnarray*}\frac{a^2+b^2-5}{ab}&=&\frac{a_{n+1}^2+(3a_{n+1}-a_n)^2-5}{a_{n+1}(3a_{n+1}-a_n)}\\ &=&\frac{a_{n+1}^2+9a_{n+1}^2-6a_na_{n+1}+a_n^2-5}{a_{n+1}(3a_{n+1}-a_n)}\\ &=&\frac{3a_{n+1}(3a_{n+1}-a_n)+a_n^2+a_{n+1}^2-(3a_na_{n+1}+5)}{a_{n+1}(3a_{n+1}-a_n)}.
\end{eqnarray*}
But by the Lemma, \[a_n^2+a_{n+1}^2=3a_na_{n+1}+5\Leftrightarrow a_n^2+a_{n+1}^2-(3a_na_{n+1}+5)=0.\] Therefore,
\begin{align*}
& \frac{3a_{n+1}(3a_{n+1}-a_n)+a_n^2+a_{n+1}^2-(3a_na_{n+1}+5)}{a_{n+1}(3a_{n+1}-a_n)} \\ &=\frac{3a_{n+1}(3a_{n+1}-a_n)}{a_{n+1}(3a_{n+1}-a_n)}\\
&=\boxed{3}.
\end{align*}
Or, in other words, $\boxed{(a,b)=(a_{n+1},a_{n+2})}$ generates infinitely many pairs $(a,b)\in \bf Z$ for which $b\mid{a^2-5}$ and $a\mid{b^2-5}$, and we're done.

Multiply the two fractions:
$$\frac{(a^2 - 5)(b^2 - 5)}{ab} = ab + \frac{-5a^2 - 5b^2 + 25}{ab} = ab - 5\left(\frac{a^2 + b^2 + 5}{ab}\right)$$
Now I claim that
$\frac{a^2 + b^2 + 5}{ab} = 7$ has infinitely many positive integer solutions. Rearrange the equation and view it as a quadratic in terms of $a$:
$$a^2 - 7ba + b^2 + 5 = 0$$
The discriminant must be a perfect square so $49b^2 - 4(b^2 + 5) = k^2$ or $k^2 - 45b^2 = -20$ which has a solution $k = 5$, $b = 1$. And by well-known results, $k^2 - 45b^2 = 1$ has a solution, so there are infinitely many solutions to $k^2 - 45b^2 = -20$. If $b$ to be positive, $a$ must also be positive (otherwise, the left hand side of the quadratic would be positive). Thus, there are infinitely many $(a, b)$ such that both are positive integers that satisfy $\frac{a^2 + b^2 + 5}{ab} = 7$. Now restrict $b > 35$. Therefore, $ab > 35$ so
$$\frac{(a^2 - 5)(b^2 - 5)}{ab} = an - 35$$ is positive. Since $b > 35$, $b$ must $b^2 > 5$, so therefore $a^2 > 5$, otherwise, $ab - 35$ would not be positive.

My idea is sort of like what we do with vita jumping. let $(a,b)$ be a pair that have the property and $a>b$ now we shall assert

$(a,\frac{a^2-5}{b})$ also have the property and we can also show that under some circumstances $\frac{a^2-5}{b}>a$, if $a > 5$:

 $\frac{a^2-5}{b}>a \iff \frac{a^2-5}{a}=a -\frac{5}{a}>a-1\ge b$ so if $\max(a,b)\ge 5$ then $\frac{a^2-5}{b}>a$

now suppose that $(a,b)$ have the property and now we show $(a,\frac{a^2-5}{b})$ have the property as well

let $k=\frac{a^2-5}{b}$ then $\frac{k^2-5}{a} =\frac{(\frac{a^2-5}{b})^2-5}{a}=\frac{(a^2-5)^2-5b^2}{ab^2}$ and now it only take to prove:

$a \mid (a^2-5)^2-5b^2 , b^2 \mid (a^2-5)^2-5b^2$ that both are straightforward to prove.

we can also see $\frac{a^2-5}{k}=\frac{a^2-5}{\frac{a^2-5}{b}}=b \implies \frac{a^2-5}{k} \in N$

and the part is proving that $\gcd(k,a)=1$ we know that $\gcd(a,k)=\gcd(a,\frac{a^2-5}{b})$ now assume for some prime $p$:

$p \mid a,p \mid \frac{a^2-5}{b} \implies p \mid a^2-5 \implies p \mid 5$ which a contradiction because $a\mid b^2 -5 \implies 5 \mid b$

so $\gcd(a,b) >5$

so we have proven that $(a,k)$ have the properties we need and also $a+k>a+b$ if $\max(a,b)>5$

now we put two initial values of $b=4,a=11$ and we see they have all the properties that we want and we can

construct $k=29,a=11$ which has the

property as well. and we know the method never get recurring because the sum of the new pair is always

re-strictly greater than the sum of the previous pair.


If $ a,b,c\in\mathbb{R}^ \plus{}$, prove that:
$ \left(\frac {x^3 \plus{} y^3}{4} \plus{} 18\right)^3\geq(x^2 \plus{} 8y)(2y^2 \plus{} 2x)$
and determine when the equality happens.


Can you prove that If $ ax^{2}\plus{}bx\plus{}c >0$ for each x in real than  $ b^{2}\minus{}4ac<0$

[hide="hint"]
Graphically if we say a$ x^2$ + bx + c>0 it means the eq. will never touch x-axis or it would have only complex roots for which discriminant should be less than zero i.e. $ b^2$ - 4ac < 0.[/hide]
Tell me if u need some other proof.
Thank u.

So if a is negative than $ b^{2}\minus{}4ac>0$

if u r saying the eq is +ve in the whole period of real no. then it can't be -ve as -ve a signifies that parabola will face downward It will touch x-axis which is in the contrast of ur condition.

[quote="Elio (n)"]What does the discriminant represent grafically?[/quote]

How many roots the graph has

$ b^2 \minus{} 4ac \equal{} 0$ implies that for $ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$, $ ax^2 \plus{} bx \plus{} c \equal{} 0$. Therefore $ b^2 \minus{} 4ac < 0$.

The converse holds true if a is positive, that is, if $ a > 0$ and $ b^2 \minus{} 4ac < 0$, then $ ax^2 \plus{} bx \plus{} c > 0$ for all real $ x$.

Suppose $ a$ is positive.

$ ax^2\plus{}bx\plus{}c\equal{} a \left(x \plus{} \frac{b}{2a} \right)^2 \minus{} \frac{b^2\minus{}4ac}{4a}$

Thus, $ ax^2\plus{}bx\plus{}c > 0$ is equivalent to $ a \left(x \plus{} \frac{b}{2a} \right)^2 > \frac{b^2\minus{}4ac}{4a}$

Plug in $ x \equal{} \minus{} \frac{b}{2a}$ into the above to get $ 0 > \frac{b^2\minus{}4ac}{4a} \implies 0 > b^2\minus{}4ac$

So I got a conclusion : Since $ ax^{2} \plus{} bx \plus{} c > 0$  this means that the equation has no real roots . Since it has no real roots this means that its discriminant is negative . Since $ b^{2} \minus{} 4ac < 0$ this means that a>0 and c>0 and sure $ 4ac > b^{2}$

If $ ax^{2} \plus{} bx \plus{} c < 0$ this means that the equation has no real roots , so its discriminant is negative and  a,c<0 .

Supose a>0...

If we supose $ b^2 \minus{} 4ac \ge 0$, we get $ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$ are real roots of $ ax^2 \plus{} bx \plus{} c$, which contradicts the fact that $ ax^2 \plus{} bx \plus{} c > 0$. So, $ b^2 \minus{} 4ac < 0$.


calculus
integration
trigonometry
logarithms
calculus computations

I have to know how to resolve some integrals but im a bit stucked, please if someone can give me some tips :lol:

[img]http://img204.imageshack.us/img204/2355/dibujone0.jpg[/img]

hello, for $ \int\sin(\ln(x))\,dx$ we have with $ \ln(x)\equal{}t$ the integral $ \int e^t\sin(t)\,dt$. By partial integration we get
$ \int e^t\sin(t)\,dt\equal{}e^t\sin(t)\minus{}\int e^t\cos(t)\,dt$
and
$ \int e^t\cos(t)\,dt\equal{}e^t\cos(t)\plus{}\int e^t\sin(t)\,dt$
plugging the second integral into the first we have the result
$ \int e^t\sin(t)\,dt\equal{}\frac{1}{2}\left(\sin(t)\minus{}\cos(t)\right)\plus{}C$ with $ t\equal{}\ln(x)$.
Sonnhard.

hello
$ \int xtan^2xdx$
=$ \int x(sec^2\minus{}1)dx$
=$ xtan(x)\plus{}ln(cos(x)\minus{}\frac{x^2}{2}$
I m posting other in few mins
thank u

$ \int xsec^{ - 1}xdx$
=$ sec^{ - 1}x\frac {x^2}{2} - \int{\frac {1}{x\sqrt {x^2 - 1}}\frac {x^2dx}{2}}$
for second integral put $ x^2 -1= t$
=$ \frac {x^2}{2}sec^{ - 1}x - \frac {\sqrt {x^2 - 1}}{2x}$
thank u

hello, to $ \int e^{\sqrt {3x \plus{} 9}}\,dx$, setting $ t \equal{} \sqrt {3x \plus{} 9}$ we get
$ x \equal{} \frac {t^2 \minus{} 9}{3}$ and $ dx \equal{} \frac {2}{3}t\,dt$ and our integral is $ \frac {2}{3}\int te^t\,dt$, by partial integration we have $ \int te^t\,dt \equal{} e^t(t \minus{} 1) \plus{} C$ with $ t \equal{} \sqrt {3x \plus{} 9}$.
Sonnhard.

hello Dr.
$ \int \frac {dx}{e^{x} \plus{} e^{ \minus{} x}}$
=$ \int \frac {e^x}{e^{2x} \plus{} 1}$
put $ e^x \equal{} t$
Answer=$ tan^{ \minus{} 1}(e^x)$

hello, to $ \int \frac{1}{e^x\plus{}e^{\minus{}x}}\,dx\equal{}\int \frac{e^x}{e^{2x}\plus{}1}\,dx$, setting $ t\equal{}e^x$ and $ dt\equal{}e^x\,dx$ so we have $ \int\frac{1}{t^2\plus{}1}\,dt\equal{}\arctan(t)\plus{}C$ with $ t\equal{}e^x$.
Sonnhard.

hello, note that $ \tan(x)^7\equal{}\minus{}\tan(x)\plus{}3\sec(x)^2\tan(x)\minus{}3\sec(x)^4\tan(x)\plus{}\sec(x)^6\tan(x)$, so we have $ \int\tan(x)^7\,dx\equal{}\log(\cos(x))\plus{}\frac{3}{2}\sec(x)^2\minus{}\frac{3}{4}\sec(x)^6\plus{}\frac{\sec(x)^6}{6}\plus{}C$.
Sonnhard.

hello
for $ \int tan^{7}xdx$ u can use reduction formula of $ \int \tan^{n}xdx$

$ \int tan^{n}(x)dx \equal{} \frac {tan^{n \minus{} 1}x}{n \minus{} 1} \minus{} \int tan^{n \minus{} 2}xdx$
thank u

hello kabi, here we say $ \tan^{\minus{}1}(x)\equal{}\arctan(x)$.
wishing you a nice day,Sonnhard.

[quote="Dr Sonnhard Graubner"]hello kabi, here we say $ \tan^{ \minus{} 1}(x) \equal{} \arctan(x)$.
wishing you a nice day,Sonnhard.[/quote]
ok Dr ,I will remember that.
Have a nice day to u too.

hello, to $ \int \csc(x)\sin(3x)\,dx$, note that $ \csc(x)\sin(3x)\equal{}1\plus{}2\cos(2x)$ and so our integral is $ x\plus{}\sin(2x)\plus{}C$
Sonnhard.

Thanks for your help Kabi and thanks again Sonnhard, I have a final test the next week and this is very helpful

hello, to $ \int \frac{7}{(x\minus{}1)\sqrt{x^2\minus{}2x\minus{}48}}\,dx$ let $ t\equal{}\sqrt{x^2\minus{}2x\minus{}48}$,
then we have $ dt\equal{}\frac{\sqrt{t^2\plus{}49}}{t}\,dx$ and so our integral is
$ 7\int\frac{1}{t^2\plus{}49}\,dt\equal{}\minus{}\arctan\left(\frac{7}{\sqrt{(x\minus{}1)^2\minus{}49}}\right)\plus{}C$.
Sonnhard.

Hello
For $ \int \frac {7dx}{(x \minus{} 1)\sqrt {x^2 \minus{} 2x \minus{} 48}}$
u can do this by just putting  $ (x \minus{} 1) \equal{} \frac {1}{t}$

thank u


Έστω $a,b,c>0$ ώστε $ab+bc+ca+2abc=1$
Να δείξετε ότι $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq \frac{3}{2}$

καλή ασκση αγαπητέ σιλουανέ αλλά ευκολάκι τρελό :D
χαιρετω

Παιδιά θα ήθελα να με συγχωρήσετε για την λάθος λύση της άσκησης της ανισότητας κι θα ήθελα  να σας γράψω την σωστή (σημείωση όποιος θέλει ας την γράψει σε latex γιατί δεν γνωρίζω καθόλου λοιπόν χρησιμοποιώντας την Cauchy-Schwarz έχουμε
(αβ+βγ+γα+2αβγ)(3/8+3/8+3/8+3/8)>= (αβ+βγ+γα+αβγ)(3/8+3/8+3/8+3/8)>= ((αβ3/8)^1/2+( γβ3/8)^1/2+( αγ3/8)^1/2+( αβγ3/8)^1/2)^2
Άρα μετά από λίγες πράξεις
2-( αβγ)^1/2>= ((αβ3/8)^1/2+( γβ3/8)^1/2+( αγ3/8)^1/2) άρα
Για να ισχύει το ζητούμενο πρέπει κι αρκεί
2-( βαγ)^1/2>=3/2 ή
1/4>=αβγ
Όμως αβγ από ανισότητα Cauchy στην δοσμένη ισότητα
Ισχύει
1/4>/1/8>=αβγ  άρα η ανισότητα ισχύει
Ευχαριστώ χαιρετώ

Η ισότητα πότε ισχύει  Αμβρόσιε??

Συγνώμη αγαπητέ Σιλουανέ αλλά την είχα γράψει στο αρχικό μου πόστ κι κατα λάθος την ξεγραψα :blush: αφούμ συγχώρησες (πιστεύω :D )η ισότητα ισχύει αν κι μόνο αν α=β=γ=1/2 οπότε κι η  δοσμένη ισότητα ισχύει κι η ανισότητα ισχύει ως ισότητα
ελπίζω να σε ικανοποίησηα Σιλουανέ κι ελπίζω ακόμα να σου άρεσε η λύση μου κι να είναι κι σωστή γιατί ομολογώ ότι η άσκηση θέλει φαντασία
χαιρετώ μετά τιμής
Υ.Γ. αν θέλει κι ευκαιρεί κάποιος για την διευκόλυνση σας ας γράψει την λύση μου σε latex  ευχαριστώ

[quote="ambros"]
(αβ+βγ+γα+2αβγ)(3/8+3/8+3/8+3/8)>= (αβ+βγ+γα+αβγ[/quote]

Η ισότητα όμως στο πρώτο βήμα που χρησιμοποιείς ισχύει όταν
$2abc=abc$ δηλαδή όταν ένας τουλάχιστον είναι μηδέν .Άρα κάπου τα μπέρδεψες τα πράγματα αφού κατά τη διάρκεια όλης της λύσης η ισότητα πρέπει να ισχύει για τις ίδιες <<σταθερές>>

ΣΙΛΟΥΑΝΕ ΕΧΕΙΣ ΔΙΚΙΟ ΑΛΛΑ ΤΟ ΠΡΩΤΟ ΒΗΜΑ ΤΗΣ ΑΝΙΣΟΤΗΤΑΣ ΕΙΝΑΙ ΛΟΓΙΚΟ ΚΙ ΟΧΙ ΠΡΑΓΜΑ ΠΟΥ ΑΠΟΡΡΕΙ ΑΠΟ ΚΑΠΟΙΑ ΑΝΙΣΟΤΗΤΑ ΚΙ ΕΙΝΑΙ ΠΡΟΦΑΝΕΣ ΟΤΙ ΟΤΙ ΠΟΤΕ ΔΕΝ ΙΣΧΥΕΙ Η ΙΣΟΤΗΤΑ ΟΠΟΤΕ ΑΝ ΘΕΣ ΕΚΑΝΑ ΛΑΘΟΣ ΣΤΟ ΠΡΩΤΟ ΒΗΜΑ ΤΗς ΑΝΙΣΟΤΗΤΑΣ ΠΟΥ ΕΒΑΛΑ ΚΙ ΤΟ ΙΣΟΝ ΟΚ?

Ναι αυτό σου λέω .......Με τον τρόπο σου αποδεικνύεις ότι
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}<\frac{3}{2}$
που είναι λάθος αφού η ισότητα ισχύει ......
Επίσης αυτό το πρέπει και αρκεί δεν το έχω δει σε καμία λύση ανισότητας   :o
Προσπάθησε για μια άλλη λύση .

η ανισότητα είναι σωστά λυμένη καθότι ισχύει η ανισότητα σε όλα τα σημεία απλώς για την ισότητα ακολουθούμε αλλη διαδικασία συνηθως επίσης το πρεπει κι αρκει μου το εμαθε ο κ Τασσόπουλος ως μια εκφραση της αναγκαιας κι ικανης συνθηκης

Λοιπόν αυτό είναι το τελευταίο ποστ που γράφω για συζήτηση για τη λύση σου .
Εκεί που γράφεις πρέπει και αρκεί κάνεις λάθος ......
Αν θες να με ακούσεις άκουσέ με . Αν όχι... πίστευε ότι η λύση σου είναι σωστη . Δικαίωμά σου .....

Λοιπον, αμβροσιε νομιζω κανεις λαθος...

Δινω δυο λυσεις σε μορφη hint:

Α) θετουμε $\sqrt{ab}=cosA$ κτλ Α,Β,C γωνιες τριγωνου

Β) θετουμε $\frac{x}{y+z}=a$ και κυκλικως....

 :D

Σιλουάνε παιδί μου απλά δεν χρειάζεται να θυμώνεις προφανώς κι είναι δικαίωμα μου να πιστεύω οτι είναι σωστό οκ? κι αν εχω λάθος κι το δώ θα το παραδεχτώ δημόσια οκ? είμαστε φίλοι νομίζω κι αν θες διαψευσε μας κι δεν χρειαζεται να αναβουν οι αντιπαραθεσεις για τις απόψεις μας πέραν του θεμιτού δεν νομίζεις ?
χαιρετω

Σίγουρα είμαστε φίλοι οκ ?Απλά στο λέω αυτό ώστε όταν σου λέει κάποιος ότι κάνεις λάθος να το ψάχνεις λίγο το ζήτημα...

Επίσης η πρώτη λύση του Socrates είναι παραπλήσια με την δική μου απλά εγώ έθεσα όπου
$\sqrt{ab}=sin\frac{A}{2}$ .
Kai μετά είναι απλή εφαρμογή της ανισότητας Jensen


Euler
geometry
projective geometry
geometry proposed

Let $ ABC$ be a triangle and $ BE$, $ CF$ be its altitudes. Let $ M$, $ N$ be the mid-points of the sides $ CA$, $ AB$, respectively, $ K$ the intersection of $ MN$ and $ EF$. Show that $ AK$ is perpendicular to the Euler line of triangle $ ABC$.

For $ \gamma \equal{} \beta$ is obvious, then w.l.o.g $ \gamma > \beta$. Let D the Feet of the altitude from A in $ \triangle ABC$, $ J: DF \cap MN$ and $ P : CF \cap MN$. The perpendicolar line from J to MN meet AB on Q and PF on T. Then T is the orthocenter of $ \triangle NPQ$ and so $ QP \parallel AC$. Now call $ S: FE \cap QP$. Then cause the triangle $ \triangle NPQ$ and $ \triangle ABC$ are homothetic with center F, S is the feet of the altitude from F to PQ. Then JS is the symmetric of JD w.r.t. MN and so meets DH on A. Then The line AK is the perspectrix of the orthic triangle of $ \triangle NPQ$ and $ \triangle NPQ$ and so it's perpendicolar to the Euler line of $ \triangle NPQ$ that is parallel to the Euler line of $ \triangle ABC$.

Dear Mathlinkers,
for a nice synthetic proof, you can see
http://perso.orange.fr/jl.ayme  vol. 2  Les deux points de Schroeter p. 4-6.
Sincerely
Jean-louis

I think the following is a known proof.

Let $MF\cap NE=L$ , apply [i][b]Pappus[/b][/i] to $\square BMFCNE$ and then $L$,$G$,$H$ are collinear.
If $N_{9}$ is the center of the [i][b]Euler[/b][/i] circle of $\triangle ABC$, then $O$,$N_{9}$ are on $\overline{LGH}$
Since $\square MFNE$ is cyclic on the [i][b]Euler[/b][/i] circle of $\triangle ABC$, $N_{9}$ is the orthocenter of $\triangle AKL$ ([i][b]Brocard[/b][/i])


Αρκετά  καλή  άσκηση  θα την  προσπαθήσω .... :lol:

ενα hint......!

και την source....!

 :)

Η λύση μου. H άσκηση είναι του  Vasc γιατό είναι δύσκολη


The [i]Euler line[/i] of a triangle is the line which connexting its centroid and itsorthocenter.
Let $ P$ be the intersection of the diagolnals of an inscribed quadrilateral $ ABCD$. Prove that all Euler lines of the triangles $ PAB,PBC,PCA,PAD$ are concurrent.


geometry
analytic geometry
vector
calculus
inequalities
derivative
function

Does anyone have any recommendations for Maths( Algebra , geometry , Coordinate geo, vectors, calculus ...anything) Problems which are [b]interesting[/b] and [b]thought-provoking[/b] and which have [b]ingenious solutions [/b]that do not use [i]any[/i] advanced concepts. These solutions should be built up from the [b]absolute basics[/b]( sort of like Maths commonsense) without any shortcuts but must be absolutely ingenious and those which clearly indicate that the solver is a genious and has a deep understanding. These solutins shouldn't require too much concepts of e.g. number theory etc. For eg, no Graph theory problem in Combinatorics. Basically problems which can't be solved easily with advanced concpts but remarkably with simple,basic solutions. Any suggestions??

Please post the problem (and your solution in hidden) if you have any. I can clarify my question if you want!!!

if you want to make a line tangent to $y = x^2$, then you would think of calculus, but there is another way to do it,
the solution is basically when you have the interesection of $y = x^2$ and $y = mx +b$, you have a doulble soultion for x,i guess it is not easier, but it is more clever than just using the derivitive,
also if you want to find the minimum value of a function $f(x) = ax^n + \frac{b}{x^n}$, then you can just use the am-gm
$\frac{ax^n + \frac{b}{x^n}}{2} \ge ab$ since you want the minimum value, you just make it the minimum (an equals sign)
$ax^n + \frac{b}{x^n} = 2ab$
(this essentially comes from an aime)
i also am interested in ways that instead of using say calculus, you just use elementary concepts

[quote="Altheman"]
$\frac{ax^n + \frac{b}{x^n}}{2} \ge ab$[/quote]

I am sure you meant \[\frac{ax^n + \frac{b}{x^n}}{2} \geq \sqrt{ab}.\]

yes, that is correct, my bad, i seem to make that mistake often though

[quote="blahblahblah"]Given $a,b,c,d\geq 0$ such that $abcd=1$, prove that

$\frac {1}{(1+a)^2}+\frac {1}{(1+b)^2}+\frac {1}{(1+c)^2}+\frac {1}{(1+d)^2}\geq 1$

This is a brutally hard inequality (it was used in a Chinese TST!) but the solution is unbelievably simple.

[hide]

Prove the inequality $\frac {1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac {1}{1+ab}$ and sum.

[/hide][/quote]
edit: deleted
understand what you said blahblahblah

No.  In fact, I see no connection between what you wrote and the hint I posted.  You should cross multiply and cancel out terms, then try to factor or use AM-GM.

[quote="Bictor717"]Inscribe two congruent circles in a right triangle. (One is tangent to the hypothenuse and one leg, the other is tangent to the hypothenuse and the other leg, and they are externally tangent to each other.) Find the radius of these circles, for a general right triangle.
You can try it for specific right triangles first. Perhaps you can generalize to more congruent circles, all lined up along the hypothenuse.
I need to go now, I'll post a solution later.[/quote]

So the thing most people do would be to construct a lot of perpendiculars and parallels, connecting the centers and use the ice cream cone theorem a few times.
I fiddled around and found that drawing the circles' common tangent was the best approach. The tangent through the circles' common point is also perpendicular to the hypothenuse. When the shorter side is extended to meet the perpendicular, a triangle similar to the original is formed, with one of the circles inscribed in it. The other circle is inscribed in another similar triangle, so these two triangles are congruent.
Say the ratio of similitude between the original triangle and the newly formed triangles is $x$. Then the hypothenuse, $c$, is $ax+bx$. So $x=\frac{c}{a+b}$. Since the inradius of the original triangle is $\frac{ab}{a+b+c}$, the radius of the circles is $\frac{abc}{(a+b)(a+b+c)}$.
For $n$ circles, you would have $ax+(n-2)\cdot x\cdot\frac{2ab}{a+b+c}+bx=c$ and get $\frac{abc}{(a+b)(a+b+c)+2ab(n-2)}$.

Do you have suggestions for problems which require cross-application and correlation of different field in Math??/

A parallelogram lies inside a triangle.

Prove that the area of the parallelogram is not greater than half the area of the triangle.

I hope  this meets your requirement for a problem whose solution tests ingenuity wihout the need to apply advanced knowledge.

In answer to your post (No 10).,I find that problems in geometric combinatorics and  especially geometric inequalities often depend on theory from several different areas of maths.Assembling a collection of problems that meets all your requirements will take some time I would imagine. The Pigeon hole principle is often used as in this question:

Prove that every convex polygon with an even number of sides has a diagonal which is not parallel to one of its sides

Rushil,

1. Concerning your enquiry about diificult problems with elementary solutions have you seen this book:"Challenging Mathematical Problems With Elementary Solutions" written by A.M. Yaglom & I.M Yaglom. It is now a classic,only available second hand.

2.Concerning your other enquiry about problems requiring cross application of theory from different branches of math, these lecture notes from Berkeley Math Circle 1999-2000 may be useful to you:

"Combinatorial Number Theory" by Gabriel Carroll, "Complex Numbers in Geometry" by Zvezdelina Stankova-Frankel. You can find them at http://www.madras.fife.sch.uk/maths/enrichment.

Also there is the book "Complex Numbers in Geometry" by I.M.Yaglom. Again.it is another old  text that is  only available second hand.

A book which is in print ( published by Canadian Mathematical Society) is "Counting and Configurations-Problems in Combinatorics,Arithmetic  and Geometry" Authors are J.Herman, R Kucera, J.Simsa .This book has 392 pages ,problems( with solutions) in all areas of combinatorics  and a comprehensive bibliography.It is an ideal text for students trying to understand combinatorial problems of IMO standard. I strongly recommend it .

 Also you might like to check out the solutions at Kalva and elsewhere to these IMO questions on Combinatorial Geometry
1986/Q2, 1989/Q3, 1995/Q3, 1997/Q1, 1999/Q1, 2002/Q6

Thanks for taking out the time to suggest these invaluabe resoucres. Don't know which I can get in India!!!

check this out:

find all $(x;y)\in\mathbb R$ s.t.
\[x^4+y^3=9\\
x^2+y=3\]

[hide]
the solution is almost easy if u see that it is equvilent to the folowing:
\[y^3=9-x^4=(3+x^2)(3-x^2)\\
y=3-x^2\]

therefore either $y=3-x^2=0$ or $y^2=9-x^4=3+x^2$

the first becomes $x^2=3$ and $y=0$

and the second becomes with $y=3-x^2$: $y^2+y-6=0$
so $y=-3$ and $x^2=6$ or $y=2$ and $x^2=1$
[/hide]

or another problem would be:

find all $x\in\mathbb R$ for every $a,b\in\mathbb R$ s.t.
\[(2x+a+b)^3=(x+a)^3+(x+b)^3\]

[hide]
the solution for this a trick as well. just say $(2x+a+b)^3=((x+a)+(x+b))^3$

then u get the equation
\[3(x+a)^2(x+b)+3(x+a)(x+b)^2=0\]

so $x$ is either $-a$, $-b$ or $-\frac{a+b}2$
[/hide]

there are alot of nice simple but ingenious equations/ equ sets out there :)

[quote="Michael Niland"] 1. Concerning your enquiry about diificult problems with elementary solutions have you seen this book:"Challenging Mathematical Problems With Elementary Solutions" written by A.M. Yaglom & I.M Yaglom. It is now a classic,only available second hand. [/quote]

There is a Dover Publications reprint of the English translation.   Probably still in print.   Volume 2 exists in Russian but not English.

Somebody mentioned the "ice cream cone theorem";  what is that?

[quote="blahblahblah"]Given $a,b,c,d\geq 0$ such that $abcd=1$, prove that

$\frac {1}{(1+a)^2}+\frac {1}{(1+b)^2}+\frac {1}{(1+c)^2}+\frac {1}{(1+d)^2}\geq 1$

This is a brutally hard inequality (it was used in a Chinese TST!) but the solution is unbelievably simple.

[hide]

Prove the inequality $\frac {1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac {1}{1+ab}$ and sum.

[/hide][/quote]

I proved it on a different way, but I think solution is equally ingenious.. Or at least it would be if it isn't wrong. :D :( [hide]Induction over number of variables.. We'll prove that if $n\in N$ and $a_i\geq 0$ for every i, and $a_1a_2\ldots a_n=1$ then \[\sum_{i=1}^{n} \frac {1}{(1+a_i)^2} \geq \frac {n}{4}\]. For $n=1$ we have trivial inequality $\frac {1}{4} \geq \frac {1}{4}$. If we suppose that for $n=k-1$ we know that $\sum_{i=1}^{k-1} \frac {1}{(1+a_i)^2} \geq \frac {k-1}{4}$, then we have to prove that $\sum_{i=1}^{k} \frac {1}{(1+a_i)^2} \geq \frac {k}{4}$. Since inequality is symmetric, we may assume that $a_k \leq a_i$ for every i, and from that we get $a_k \leq 1$ (by contradiction). Now it's really easy to prove the induction step, just use the hypothesis on first $k-1$ members of LHS, and you'll get that it suffices to show that $4 \geq 1+a_k^2+2a_k$ which is obvious from $a_k \leq 1$ For $n=4$ we get what we wanted.[/hide]

Peeta, I don't find any of your problems ingenious, not even interesting, but who knows, maybe you'll say same thing for my problems (I want to say that ingenious is pretty relative term).. For example, here's my solution of first one, it's almost identical to the one you gave.. [hide]Simply square the second equation and put it in the first one (sorry about bad English), you'll get that $y(y^2-y-2x^2)=0$. First case $y=0$, we get solutions $(\sqrt 3, 0)$ and $(-\sqrt3, 0)$. Second case $y^2-y-2x^2=0$. From second equation we get that $2x^2=6-2y$, so we have to solve $y^2+y-6=0$, and that's easy.. [/hide]

I think this one is appropriate, so.. It's combinatorical.. We have square 10x10 divided into 100 unit squares, and some number of "infected" unit squares. Unit square will become infected if at least 2 of his neighbours (2 unit squares are neighbors if they share a side) are infected. Initially, there are 9 infected unit squares. Is it possible that at some moment, every unit square will be infected?
This problem was on Yugoslavian national competition when Yugoslavia still existed..

Then there's one from IMO 1975, so most of you probably saw it.. Let A be the sum of the decimal digits of $4444^{4444}$, and B the sum of the decimal digits of A. Find the sum of the decimal digits of B.

Here's one more, I saw it in school on preparations for competitions, I was astonished by idea of solution, but some of you will probably see it as well-known problem, but I think that that doesn't make its solution any less ingenious (same thing for problem from IMo 1975). Find the minimum of $(u-v)^2+(\sqrt{2-u^2} -\frac{9}{v})^2$ for $0 \leq u \leq \sqrt2$, $v>0$.

[quote="blahblahblah"]Your solution is wrong.  If you proceed by induction on the first $n-1$ variables, then your only choice for the $n$th variable is $1$.[/quote]

You're right. I didn't think about that, thanks. I'll edit my post now.. Too bad I checked the solution of person who wrote this prolem here, I can't try to solve it anymore..

[quote="idioteque"]Here's one more, I saw it in school on preparations for competitions, I was astonished by idea of solution, but some of you will probably see it as well-known problem, but I think that that doesn't make its solution any less ingenious (same thing for problem from IMo 1975). Find the minimum of $(u-v)^2+(\sqrt{2-u^2} -\frac{9}{v})^2$ for $0 \leq u \leq \sqrt2$, $v>0$.[/quote]

Are you talking about a geometric interpretation of the problem? :)

Draw a circle of radius $\sqrt{2}$, draw the curve $y = \frac{9}{x}$.  It becomes "obvious" what the minimum distance between two points on these two curves is.  It's a slight bit trickier to make it rigorous.

Yeah, I had exactly the same idea.

I don't think it's that hard to give a rigorous proof...

[In this "proof", all points $(x,y)$ satisfy $x > 0$ and $y > 0$.)

For example, consider some circle centered at the origin and hyperbola $xy = k > 0$ (such that the circle and the hyperbola do not intersect), and let $P$ and $Q$ be points on the circle and the hyperbola respectively. Let $O$ be the origin, and let $P'$ and $Q'$ be the points where the line with equation $y = x$ intersect the circle and the hyperbola respectively. Then, we get \[PQ \geq OQ - OP = OQ - OP' \geq OQ' - OP' = P'Q'\] where it is very easy to prove that $OQ \geq OQ'$, and we are done.

[quote="peeta"]
or another problem would be:

find all $x\in\mathbb R$ for every $a,b\in\mathbb R$ s.t.
\[(2x+a+b)^3=(x+a)^3+(x+b)^3\]

the solution for this a trick as well. just say $(2x+a+b)^3=((x+a)+(x+b))^3$

then u get the equation
\[3(x+a)^2(x+b)+3(x+a)(x+b)^2=0\]

so $x$ is either $-a$, $-b$ or $-\frac{a+b}2$

there are alot of nice simple but ingenious equations/ equ sets out there :)[/quote]

you proposed a problem and solved another??

anyway, i think that the problem you first proposed can be solved by $u=x+a$, $v=x+b$, $s=u+v$, $p=uv$. then compute...

[quote="Arne"][quote="ThAzN1"]
Let $p$ and $q$ be two consecutive odd primes.  Prove that $p+q$ has at least $3$ (not necessarily distinct) prime factors.

Hint:

[hide]Can $\frac{p+q}{2}$ be prime?[/hide][/quote]

I don't see why you need the hint. We need $p = -1$ and $q = 1$ (mod 6) so $p + q$ is divisible by 2, 3 and some other prime (possibly 2 or 3) (since $p + q \geq 12$).[/quote]

What about 23 and 29, these are consecutive primes and both $\equiv -1$ (mod 6)? (Unless you read 'consecutive primes' as 'twin primes', but I doubt this is the problem.)

Oh, sorry, I did read the question as "twin primes" indeed :blush:

Well, it's even simpler actually, since $p$ and $q$ are consecutive primes, $p < q$, we have $p < \frac{p + q}{2} < q$ so $\frac{p + q}{2}$ cannot be prime (this is the point of $p$ and $q$ being consecutive primes :)) and hence $p + q$ is divisible by 2 and at least two other positive primes (divisors of $\frac{p + q}{2}$).

[Edit: never mind, I see that a link to a solution has been given already]

[quote="ThAzN1"]Try this one...

Let $p$ and $q$ be two consecutive odd primes.  Prove that $p+q$ has at least $3$ (not necessarily distinct) prime factors.

Hint:

[hide]Can $\frac{p+q}{2}$ be prime?[/hide][/quote]
I found solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43361]here[/url]


let $a,b,c$ be natural numbers such that ${1 \over b}+{1 \over c}={1 \over a}$, and let $d=gcd(a,b,c)$. Prove that $abcd$ and $d(b-a)$ are both perfect squares.

Check previous posts. This was a former BMO question. You just factorise the equation.

Bomb


For $a,b,c$ real numbers with sum null prove the ineq

$\sum a^2b^2 +3 \geq 6abc$

We may assume $a,b<0$ and $c>0$. Then inequality can be transformed to
\[f(a,b)=a^2b^2+(a^2+b^2)(a+b)^2+3-6ab(a+b)\geq 0\]
(I have revert signs $a$ and $b$)
It is clear if $a+b\geq 3$ then $(a^2+b^2)(a+b)^2\geq 2ab\cdot 3(a+b)$, so inequality holds. Suppose $(a,b)$ is the optimal point for difference (it exists!). If $a=0$ or $b=0$ then we are done. If $a,b> 0$ then
\[0=\frac{1}{2}f'_a(a,b)=ab^2+a(a+b)^2+(a^2+b^2)(a+b)-6ab-3b,\]
\[0=\frac{1}{2}f'_b(a,b)=a^2b+b(a+b)^2+(a^2+b^2)(a+b)-6ab-3a.\]
Suppose $a\neq b$, then substracting and dividing by $a-b$ we obtain
\[-ab+(a+b)^2+3=0\]
- that is impossible.
Thus $a=b$. Then we obtain inequality $9a^4+3\geq 12a^3$ -- true, since $9a^4+3- 12a^3=3(3a^4+1-4a^3)=3(a-1)^2(3a^2+2a+1)$.

Recently the posting velocity in the inequalities section has become so big that I don't even find the time to read through all the topics. Anyway, this one is particularly nice. Here is my solution:

For any three real numbers a, b, c, we have

$\left(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3\right)-6abc$
$=\left(b+1\right)^{2}\left(c+1\right)^{2}+\left(c+1\right)^{2}\left(a+1\right)^{2}+\left(a+1\right)^{2}\left(b+1\right)^{2}-2\left(a+b+c\right)\left(a+b+c+bc+ca+ab+2\right)$,

so that, in the particular case when a + b + c = 0, we have

$\left(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3\right)-6abc=\left(b+1\right)^{2}\left(c+1\right)^{2}+\left(c+1\right)^{2}\left(a+1\right)^{2}+\left(a+1\right)^{2}\left(b+1\right)^{2}$,

and thus $\left(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3\right)-6abc\geq 0$, so that $b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3\geq 6abc$.

Done.

Note that this problem was problem 7 of 239MO 1995 for grades 8-9.

  darij

The secret is to write short proofs, Darij! Simple and efficient.  ;)

:D  :D  :D  :D   i think this pro very trivial:
we only must prove that: for every possitive real number x,y
then: $x^2y^2+(x^2+y^2)(x+y)^2+3\ge{6xy(x+y)}$
indeed:
$LHS=(x^2+y^2)(x^2+xy+y^2)+xy(x^2+y^2)+x^2y^2+3$
$\ge{3xy(x^2+y^2)+3(x^2y^2+1)}$
$\ge{3xy(x^2+y^2+2)}\ge{RHS}$

okie?????????!!!!!!!!

Okie!!!!
General method + 3 particular methods...


linear algebra
matrix
linear algebra unsolved

Prove that
$ det(A_{1}^tA_{1} \plus{} A_{2}^tA_{2} \plus{} ... + A_{k}^tA_{k}) \geq 0$
where $ A_i \in M_{n}(R)$ for each $i$.

For any real matrix $ A,$ the matrix $ A^TA$ is positive semidefinite, because $ x^TA^TAx \equal{} (Ax)^T(Ax) \equal{} \|Ax\|^2\ge0.$

In fact, for a real symmetric matrix $ B$, the following are equivalent:

(1) $ x^TBx\ge0\,\forall\,x\in\mathbb{R}^n.$
(2) For any $ \lambda$ an eigenvalue of $ B,$ $ \lambda\ge 0.$

Now for the problem in question:

$ x^T\left(\sum_{j \equal{} 1}^kA_j^TA_j\right)x \equal{} \sum_{j \equal{} 1}^k(x^TA_j^TA_jx)\ge0.$

Hence all of the eigenvalues of that matrix are nonnegative, hence the determinant, which is the product of the eigenvalues, is nonnegative.

Thx Kent Merryfield...now i know that is a positive-definite matrix  :D


MATHCOUNTS
ARML
AMC
AIME
trigonometry
geometry
analytic geometry
\/closed

I dont' think most of us are to the skills of studying specific topics of mathematics. It will take too long to carefully a topic. But I hope we might do some of the math circle stuff that AOPS offers in San Diego sometimes online.

[quote="churchilljrhigh"] But I hope we might do some of the math circle stuff that AOPS offers in San Diego sometimes online.[/quote]

What do you mean by math cirlce in online? I thought it was meeting in actually by person, or is it?

[quote="Silverfalcon"]I understand that MathJams only cover the math competition but in my opinion, I think if we have MathJam in certain branch of math, it could be pretty awesome, too.. Are we going to have one of those in future?[/quote]

er...correct me if I'm wrong, but isn't that what online classes are for?

er...correct me if I'm wrong, but isn't that what online classes are for?[/quote]

It is kinda true but for my knowledge, I think it's more specified in certain competition such as Mathcounts, ARML, AIME, and I don't think any for AMC... And it is true that we do cover some parts in math, but I don't think MathJams is for specified branch or part.  Maybe Math class is, but I do not take them any so..

I agree that it would be pretty cool to have a mathjam on specific branches of math (i.e. trig, algebra, basic number theory etc.) but I also believe that Mr. Rusczyk and Mr. Crawford are doing a great job as it is.  I appreciate having any mathjam at all.  Don't get me wrong.  It s okay to give your opinion.  I'm just saying don't take them for granted (which you aren't doing).

[quote="Silverfalcon"]It is kinda true but for my knowledge, I think it's more specified in certain competition such as Mathcounts, ARML, AIME, and I don't think any for AMC... And it is true that we do cover some parts in math, but I don't think MathJams is for specified branch or part.  Maybe Math class is, but I do not take them any so..[/quote]

Rest assured that when AMC season rolls around next year, there will be AMC Math Jams. If you'd like to get extra practice, the classes are designed for that. The independent study courses are also excellent! (although I'm still trying to see if one would fit into my schedule) And there are AMC/AIME Problem Series Classes...I'm sure there'll be one run next year.

I think that the idea of a student hosted course or whatever is a good idea.  I would help out as much as possible but the only problem is that I am definitely not knowledgeable enough to really teach anyone.

[quote]I'm sure it just wouldn't be the same to a parent paying and registering for a class taught by professional teachers, test-writers, and very respectable mathletes (when they were still in jr/sr high school) than some high school (or even college) students.[/quote]

You're right that it's not the same but I don't think that Magnara was suggesting that it be a cost kind of thing.  I interpreted Magnara to mean that it would be sort of like a joint effort.

[quote="MithsApprentice"][quote="Magnara"]Maybe, if we want to do a specialized sort of thing, we could organize a student teaching sort of thing.  Rather than waste time repeatedly of Mr. C and Mr. R, we could do it ourselves.  I probably don't know enough to teach anything too significant, but I'd be willing to do as much as I could, which would be something like "Intro to..." whatever.
[/quote]

I don't think that would be appropriate. The reason that they're getting another teacher is just that...so that more classes can be taught and Mr. Rusczyk and Crawford aren't working overtime and sometimes weekends (e.g. AIME Seminar) to keep up with the AoPS workload. The way this site has expanded is pretty unbelievable and we'll be seeing more increase as time goes on and AoPS is more well known.

I'm sure it just wouldn't be the same to a parent paying and registering for a class taught by professional teachers, test-writers, and very respectable mathletes (when they were still in jr/sr high school) than some high school (or even college) students. Not only that, to us high schoolers, it would be a huge committment on our time. It's not like we can just call in one day b/c we have too much hw and can't run the class. Also, it would be terribly difficult to teach kids about the same age as you.

If someone needs help in a particular area, then they can go to the forums to post questions and other students/parents/coaches/instructors can answer them.[/quote]

It seems as though your interpretation of my idea was the exact opposite of what I thought.

I meant by having student teachers Mr. Rusczyk and Mr. Crawford wouldn't have to come to the classes.  All that my plan potentially requires from them would be access to the math classroom, but even this isn't necessary, there are many other ways on the internet to access chat rooms.

I also meant that it would be volunteer work on the part of the student teachers.  Therefore, no money would be necessary on the part of any parents.

Edit: Just a little addition.  I think that if this was ever done, it would best be done if its goal was non-competetive mathematics.  There is plenty competetive mathematics out there.  Heck, you could just take one of the classes they offer here.  Rather, it would be just to know more stuff, which is always a good thing.

[quote="Magnara"]Maybe, if we want to do a specialized sort of thing, we could organize a student teaching sort of thing.  Rather than waste time repeatedly of Mr. C and Mr. R, we could do it ourselves.  I probably don't know enough to teach anything too significant, but I'd be willing to do as much as I could, which would be something like "Intro to..." whatever.

We could either do it on AIM, or possibly the teachers would allow us to use the Math Jam program.[/quote]

I personally don't suggest AIM. There were some last year and they were chaotic. Some very smart people (you know who you are) posted nonsense and swore. If it was with Digichat, then ok. But its how life is everywhere. When I first joined FICS there was all this inappropiate shouts, and the first thing I did was turn off two tell whatever.

Fine then.  Teachers?  Would it be possible to use the Math Classroom here?

Probably not, for the same reason we're not allowed to PM through the classroom... "legal reasons." But thats just speculation.

Yeah, zabelman's pretty much right - we'd have a tough time letting students use the classroom with out an AoPSer present for legal reasons (rough litigious world we live in, unfortunately).  We're happy to see you use the board to coordinate something for AIM or Yahoo or whatever.


Let p be a prime number ;$p>2$.Prove that:
  $\sum_{i=1}^{p-1} 2^i i^{p-2} \equiv \frac {2-2^p}{p}(modp)$
More difficult(I think):
$\sum_{i=1}^{p-1}  i^{p-2} \equiv \frac {2-2^p}{p}(modp)$
Sorry if this problem was posted before!

The second one is wrong. For $p=3$ the sum is $1+2=3 = 0 \mod [3]$ whereas $\frac {2-2^3} 3 = -2 = 1 \neq 0 \mod [3]$.
In fact for odd prime $p$, the second sum, say $S_2$,  is always $0 \mod [p]$ since, from Fermat's little theorem, for $i=1,2, \cdots, p-1$, we have $i^{p-2} = i^{-1} \mod [p]$.
Moeover, the numbers $1^{-1}, \cdots, (p-1)^{-1}$ form a permutation of $1, \cdots, p-1$ (they are pairwise distincts), so that $\dsp S_2 = \sum_{i=1}^{p-1} i = \frac {p(p-1)} 2 = 0 \md [p] .$

Pierre.

:? Hmm,... :blush:  :blush:  :blush:  :blush: .Yes ,it is wrong!I will correct it

For the first one:
p is an odd prime.
Lemma 1: pCk=p(p-1)(p-2)...(p-k+1)/k(k-1)......2.1.
we always have: pCk=[p*(-1)^(k-1)]/k modp^2. for1<k<p
It is easy to see (pCk)/p is an integer, so look it mod p and it is easy to see it equals (1/k)*(-1)(k-1) and the lemma is proved.

Now:
1=(2-1)^p= :Sigma: (-1)^k*2^(p-k)*pC(p-k)= 2^p-1+:Sigma:2^i/i (mod p^2)

And result follows, taking into acount thAt i^p-2=i^-1.

A similar problem:
Let $p\in P,p>2: 1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{p-1}=\frac{a}{(p-1)!},a\in Z$.
Prove that: $a\equiv \frac{2-2^{p}}{p}(\mod p)$

What pascual did in the post above sems to work pretty well on this one too.


David is checking the water bill for his laundry business.  The rate sheet below was enclosed with his bill.  He is an REU2, he must pay the basic facility charge, and he used 42,000 gallons of water this month.  How much should his bill be?
 \begin{tabular}{c|c|c|c}
  & Basic & Tier 1 & Tier 2 \\\cline{3-4}
  & Facility & $ \$5.20$ per $ 1000$ gal & $ \$7.90$ per $ 1000$ gal \\
  & Charge & up to ... & over ...  \\\hline
  REU1 & $ \$3.50$ & $ 18,\!000$ gal & $ 18,\!000$ gal \\\hline
  REU2 & $ \$7.00$ & $ 36,\!000$ gal & $ 36,\!000$ gal
 \end{tabular}

Fixing the latex again:
\[ \begin{tabular}{c|c|c|c} & Basic & Tier 1 & Tier 2 \\
\cline{3 \minus{} 4} & Facility & per gal & per gal \\
& Charge & up to ... & over ... \\
\hline REU1 &\$3.50 &18,000 gal &18,000 gal \\
\hline REU2 & \$7.00&36,000 gal &36,000 gal \end{tabular}
\]


Suppose that $a_{1},a_{2},\ldots,a_{n},r,s,t,\ell$ are positive real numbers such that $a_{1}^{s}+a_{2}^{s}+\cdots+a_{n}^{s}=c$ and $2s\geq r\geq s\geq t\geq 0$. Prove the following inequality \[\sum_{j=1}^{n}a_{j}^{r}+\ell\prod_{j=1}^{n}a_{j}^{t}\geq\min\left\{\frac{c^{r/s}}{(n-1)^{r/s-1}},\frac{c^{r/s}n^{(nt+s)/s}+\ell c^{nt/s}n^{r/s}}{n^{(nt+r)/s}}\right\}.\]

The weighted version also holds, just substitute $\sum a_{j}^{k}$ with $\sum w_{j}a_{j}^{k},$ $\prod a_{j}^{k}$ with $\prod a_{j}^{w_{j}k}$ and $n$ with $w=w_{1}+\cdots+w_{n}.$


Given:
i) $ AB\perp OY$
ii) $ AQ // OY$
iii) $ \overline{PQ} \equal{} 2\cdot\overline{OA}$
iv) $ P\hat{O}B \equal{} 26^{\circ}$

Find $ X\hat{O}Y$.

[hide="Answer"]

$ X\hat{O}Y\equal{}78^{\circ}$[/hide]

Let $ R$ be the midpoint of $ PQ$. Then $ AR\equal{}PR\equal{}RQ\equal{}AO$ and $ \triangle AOR$ is isosceles.
The triangle $ APR$ is also isosceles.
$ \widehat{AOR}\equal{}\widehat{ARO}\equal{}180^{\circ}\minus{}2\cdot\widehat{APR}\equal{}180^{\circ}\minus{}2\cdot 64^{\circ}\equal{}52^{\circ}$.
Then $ \widehat{AOY}\equal{}26^{\circ}\plus{}52^{\circ}\equal{}78^{\circ}$.


Connie can wash nine windows in one hour. Rick can wash seven windows in one hour. Connie has been washing windows for one hour, when Rick joins her to finish the job. The entire job consists of washing 57 windows. How many hours does Rick work?

[hide=" :nhl: "]
After an hour, Connie washed 9 of the windows.
That leaves 48 left.
combined they can do 16 per hour.
Thus Rick works 3 hours.[/hide]

[hide]Connie's washed nine windows, when Rick joins her for the remaining 48. They wash 16 windows an hour together, so Rick will work for 3 hours with Connie.[/hide]


As the title, what IMO that you like its question the most? There are some judgment of this IMo 2007 but not for previous IMOs. So we can discuss here.

Personally, I like questions of IMO 2004 the most. Not because I take part in this IMO, but I think all its questions are in Olympiad Style,  very suitable for IMO.

Give your opinion.

Not the same as you, I like the 47th IMO, though I did very bad at that IMO. In my opinion, except the problem 1, the others were very good for IMO. Besides, I like problem 2 and 6.

I also like 45th IMO. however, problem 5 was not suitable for that IMO, it was too easy.


Since you have said that the function is analytic it means that the function is differentiable in all neighborhood  of points in the left hand plane. Infact the function is infinitely differentiable in all the points in the left hand plane and the Taylor's series $ \sum_{n \equal{} 0}^{\infty}\frac {f^{n}(z_{0})}{n!}(z \minus{} z_{0})^{n}$ converges to $ f(z)$.

Poles are certain types of singular points. Analyticity fail at those points.

that means i'm correct when i say that all the poles are in right hand plane. Thus no pole exists in the left hand plan and because of this it is analytic in left hand plane as there are no poles on that side?

Thanks for ur reply

[quote="aitezaz.abd"]When working with signal estimation problem using wiener filtering, there was written something like this
"Suppose H(s) is analytic in left-hand plane"
what does that mean and tell me about H(s). What i figured out that it means that H(s) has all the poles on right hand plane? Is that correct?
Thanks[/quote]

No. H may only be defined in the left-hand plane, so what sense does its behavior in the right hand plane make? Look up the definition of "analytic in a region".


How big is the math department at the aforementioned universities? i.e. How close is the student-faculty relationship?

A good way to find out about the "feel" of their departments is to visit them and to try talking to a few professors.  :)

How about if you can't visit them (like if you are an international student)?

Digging up some info on the MIT website: http://web.mit.edu/annualreports/pres07/# .  In particular, here's the statistical breakdown of students by major: http://web.mit.edu/annualreports/pres07/26.01.pdf

So there's about 200 undergrad math majors (177+31 = 208 to be exact).  However, this is only considering those people that declared math as their primary major (I believe).  I found somewhere in one of those documents that about 100 people per year graduate with a math degree (many have it as a double degree) so that's about 400 math majors at MIT.  Anyway, the department is pretty sizable.

From the classes I've taken, the professors are very accessible.  They all have office hours and if that time doesn't work out for you, they'll schedule appointments (same with TA's).

Of course, I expect that Harvard and Princeton have just as good faculty so you should definitely consider other things.  I believe their departments are smaller.  One pro of this is smaller classes.  Though you should definitely consider other things.  For one, there is definitely a huge problem solving / math culture at MIT (e.g. look at the Putnam results from the recent past...).  But then again, maybe the Ivy League culture is better suited to you.  Or maybe you'd much rather be in a more rural setting.  Plenty of things to consider.

Princeton's math department is quite large. We have about 60 or so profs ([url]http://www.math.princeton.edu/directory/[/url]) and generally about 20 students majoring in math each year (class of 2009 seems to be an exception, which might continue - there are about 35 math majors).

Some professors make more effort to talk to students than others, but if a student goes to a professor, the professor will almost always be friendly and helpful (I have never heard of a complaint that a professor refused to talk to students). A thing to keep in mind is that they probably won't like it if you randomly visit them every day or something like that since they also have busy schedules (so either make an appointment or come to office hours).

Most profs probably don't teach. There are only so many courses offered, and usually at most a few profs per course (I hear you can also make up your own course, however, and get a professor to teach/supervise you - something like a reading course). Intro classes (applied calc, lin alg) have large numbers of students, but then meet in small groups called precepts with a grad student. The only complaints I have heard from students about math so far are from those intro classes when some grad student's proficiency in English is not to their liking. Some also complain that tests are hard, but that's irrelevant since all scores are always curved...

My classes so far had these numbers: MAT 215 (about 25 or so students in the class, a handful of questions from students during class), MAT 217 (20 or so students in the class, probably no more than 15 show up on any given day, very few questions since everyone is quite lost and the text is quite good), MAT 330 (8 people, one guy almost never showed up, prof asks questions and we ask questions, very good amount of interaction during class).

As far as p-sets go, there are always groups collaborating on p-sets, sometimes even with a grad student. I didn't work with others very much on MAT 215 or MAT 217, but I probably would not have survived MAT 330 all by myself. Unless you're a genius or something, you'll be forced eventually to work in groups since for most math major classes (300 and 400 level), hw goes around 70% of your grade, so it is quite substantial and will probably take you more than 10 hrs a week even when working in a group.

[quote="The Visionary"]I heard collaboration is unique in MIT compared to other colleges in the sense that one can just walk downstairs and if you meet several people that you want to discuss something and do hw/psets with, you can just ask them and they will say yes immediately. Is this true? Anyway, I want to hear from you about the collaboration thingi at MIT? Thanks![/quote]

In my experience, yes, I've definitely found that to be true.  Personally, I like to work on assignments independently if possible since that's the way I learn best.  However, I've been over my head in a couple of classes in which case I would find a classmate or two to work with and regularly go to office hours for help.  But even beyond classes, I've found the environment to be pretty amazing.

Sorry for the short answer...there's so much at MIT that it really takes more than a post to explain.

@ blah...blah
I actually think this is a very good question. It's proly one of the things I considered the most when making my choice. Dept X could have 200 profs, none of whom want to talk to you whereas dept Y could have 25 profs, many of whom could make a difference in your undergrad years. Also, it matters a lot what the profs are doing. It's good to have a balance in profs' research interests because it gives you more opportunities come jr. or sr. year when you have to do a research project in something to get to grad school. The size of the dept does not necessarily translate to the number of profs or number of students, but to the relative numbers and the way students and profs interact.

Continuing on math at Princeton, it does seem that the number of majors is increasing somewhat, but a lot of the upper classes stlll have solid student-teacher levels, with class going from single-digits to low twenties.  If the class size is a little larger, it's possible several (or a lot) of students got together and planned to take some tough course together to make it more bearable, or just because the teacher's supposed to be really good.

In terms of student-student interactions, there appears to be a sort of bond between the math majors even in separate classes; my first semester, there were several instances when I'd be discussing some problems, at a problem session or more casually, and an upperclassmen asked if we were doing MAT 215.  Then they'd say something like, Yeah, I did that my first semester.  Fun class.


I have a question regarding how you interpret a power tower.

Does $3^{3^3}$ mean $3^{27}$ or $27 ^3$ ? If both interpretations are used, which is more prevalent. Also, I have noticed that 2 has the special property that $2^{(2^2)}=(2^2)^{2}$. Why would this only be for 2 or are there other numbers that work as well?

Usually the problem statement should clarify which definition should be used for the power tower.

For the power tower of length three, the identity \[ x^{(x^x)}=(x^x)^x \] holds for $x\in\{-1,1,2\}$. This can be found by solving the general equation using logarithms and factoring/simplifying.


When I just started with TeXnicCenter, it asked for the executable. At that time I didn't have Asymptote. Is Asymptote an executable? Because if it is, what do I do? The wiki says nothing about importing Asymptote into TeXnicCenter, unless I'm not looking hard enough :maybe:

Read [url=http://www.mathlinks.ro/Wiki/index.php/Asymptote:_Getting_Started/Windows/Downloads_and_Installation]this[/url] for asymptote installation information.

Also, I just downloaded Asymptote v 1.69 to get geometry_dev working, but now I cannot output files in pdf. I believe the error lies in this command:

[code]
-batchView -tex "pdflatex" %tc
[/code]

Removing pdf outputs the pictures just fine in GSview.

Anyone? The problem still persists in 1.70.

You need to call asy.bat, not asy.exe now. Check that your Asymptote tool in TeXnic center is configured properly. Also asy.bat may not take the options into account, moreover, they may just confuse it...

In Texnic Centre, I go tools -> cutsomize -> tools -> command

and replaced "C:\Program Files\Asymptote\asy.exe" with "C:\Program Files\Asymptote\asy.bat". However, when I hit Alt+A (hotkey for asymptote), Asymptote just opens as though I directly clicked the Asymptote shortcut on my desktop.

Well, you still need to pass the file name (encoded by %tc) to asy.bat as the [b]only[/b] parameter. So, your command in the tool should be

C:\Asymptote\asy.bat %tc

(or something like that). Another thing is that cd %userprofile% line in asy.bat may result in placing the output files into the root directiory instead of the directory with your tex file, so I recommend commenting that line out.

Still nothing. Currently, I have

Command: C:\Asymptote\asy.bat %tc

Argument: -batchView -tex "pdflatex" %tc

Initial Directory: %dc

What is your current set-up?

[quote]Another thing is that cd %userprofile% line in asy.bat may result in placing the output files into the root directiory instead of the directory with your tex file, so I recommend commenting that line out.[/quote]

What?

Try to change command to C:\Asymptote\asy.bat and argument to %tc, leaving the rest as it was. I'm not 100% sure about my current setup because I'm now 500 miles away from my home computer. I'll return today in the late evening and look at it then.

I managed to fix the problem. It turns out it was a problem with Adobe Reader itself, and was fixed either by upgrading from version 8 to 9, or having restarted my computer.


Hope this is the reight forum  :maybe: Well anyways I'm just starting grade 11 and the kind of stuff we learn is nothing to me since all my tests and exam last year was perfect. Now around this forum I see people that are in grade 9 that solve very complex problems that I cant even understand. I was wondering if it is too late for me to start learning above what i learn in school and get to the level of understanding that some people here have?

If its not too late what things could I do to expand my knowledge and understanding?

It is most certainly not too late.  The main thing you have to do is to swallow your pride and start with material that those younger students are working on.  So what if they're 2-3 years younger than you - that may only mean they were exposed to challenging material 5 years before you were.  If you're willing to work and challenge yourself, you'll be fine.  Life's not a race - so even if you're not where some 9th graders are now, it's no big deal.

The first step is to determine at what level you should be working.  Poke around different forums, get practice contests from a variety of contests.  You want to find resources/a place in the forum where you can consistently get 50-75% of the problems.  Once you have found a few resources like that, report back here and the community can give you guidance about what to study next.

Thank you for the wonderful support. I will do as suggested and report back. Thank you very much!!!


Let $a,b,c$ be three real numbers such that $1 \geq a \geq b \geq c \geq 0$. prove that if $\lambda$ is a root of the cubic equation $x^3 + ax^2 + bx + c = 0$ (real or complex), then $| \lambda | \leq 1.$

I think that it will help
[hide]Use Bolzano theorem :) [/hide]

Can you please tell waht that is? Please post the whole solution for the Resoures Section!

can anyone post a solution without using complex numbers????????
It's getting complex using them.............

Here's a long and class-10-student-like solution :D
Firstly we prove this for a real root $\lambda$
Case 1.$|\lambda|=\lambda$
Assume that $|\lambda|>1$
This implies $|\lambda|^3+a|\lambda|^2+b|\lambda|+c>1+a+b+c\ge 1$ which is a contradiction
Case 2. $|\lambda|=-\lambda$
Assume that $|\lambda|>1$
This implies $|\lambda|^2+b\ge a|\lambda|^2+c=|\lambda|(|\lambda|^2+b)>|\lambda|^2+b$ .Contradiction !

Nextly
Let the roots of the equation be $\lambda,re^{i\theta}$ and $re^{-i\theta}$
Since $\lambda r^2=-c<0$ we have $|\lambda|=-\lambda$
We also have $\lambda+r(e^{i\theta}+e^{-i\theta})=-a$ and $\lambda r(e^{i\theta}+e^{-i\theta})+r^2=b$.
Combining these two we get
$r^2=|\lambda|^2-a|\lambda|+b$
Here we again consider 2 cases
case(I)
$|\lambda|\le a$.This implies $r^2=|\lambda|^2-a|\lambda|+b\le a|\lambda|-a|\lambda|+b=b\le 1$.
case(II)
$|\lambda|>a$.
$r^2=|\lambda|^2-a|\lambda|+b=|\lambda|(|\lambda|-a)+b\le|\lambda|-a+b=|\lambda|-(a-b)\le |\lambda|\le 1$
QED

The generalisation to higher degree polynomials is also true

I think here is a simple proof.
Since  is a root of the equation x3 + ax2 + bx + c = 0, we have
y^3 = −ay^2 − by − c.
This implies that
y^4 = −ay^3 − by^2 − cy
= (1 − a)y^3 + (a − b)y^2 + (b − c)y + c
where we have used again
−y^3 − ay^2 − by − c = 0.
Suppose |y| ≥ 1. Then we obtain
|y|^4 ≤ (1 − a)|y^|3 + (a − b)|y|^2 + (b − c)|y| + c
≤ (1 − a)|y|^3 + (a − b)|y|^3 + (b − c)|y|^3 + c|y|^3
≤ |y|^3.
This shows that |y| ≤ 1. Hence the only possibility in this case is |y| = 1. We conclude
that |y| ≤ 1 is always true.

[quote=DeepanshuPrasad]$0 \le c \le b \le a \le 1$
therefore $a \le 1 , b \le 1, c \le 1$ so,
$|\lambda^3+a\lambda^2+b\lambda+c|=0\le|\lambda|^3+|\lambda|^2+|\lambda|+1 \le 1+|\lambda|+|\lambda|^2+|\lambda|^3+|\lambda|^4+.....$ upto infinity $=\frac{1}{1-|\lambda|}$
i.e. $\frac{1}{1-|\lambda|}\ge 0$  i.e. $1-|\lambda| \ge 0$
hence, $|\lambda| \le 1$[/quote]

But in this solution ....to sum it to infinity $|\lambda| \le 1$ if its not then you cannot sum it to infinity....so you are assuming the same thing which you have to prove.

First prove that it has only one real root (since f(x) is <=0 for all x<=-1 and >=0 for all x>=1).And the only real root say "£" lies between -1 and 0.

Now let the roots be £,m+in,m-in
£*(m+in)*(m-in)= (-c)
m^2+n^2= (-c)/£ <=1      since £>=-1
So √(m^2+n^2) <=1, |£|<=1 and we are done.

Somebody please convert it into LaTeX