Hi,

I am a sophomore and I want to take the Math IIC this summer to get it over with so I will have time to study for other SAT II tests next year. But will colleges accept scores of SAT II test taken in sophomore year?

Thanks.

Colleges generally accept SAT II scores from whenever you took them BUT because the SAT I math section has just changed its level of content, it is unclear whether or not there will be changes in the two SAT II math tests (including whether or not there will still two math tests in the SAT II series) by the time you apply for college. If you stay active in math problem-solving and studying math, your math level will stay up, and it might be prudent to hold off on taking an SAT II in math until you are sure what the latest requirements are at your favorite college.

For any other subject, take the SAT II test when you feel well prepared. Taking such a test "early" is fine--the score still counts.

i'm in 9th grade right now, and i will take level 1 this may and take level 2 next yr.
i don't think (hope not) the subj test will change too soon(only 2 more yrs before you go to college)
you should take it now because you will forget everything you learned next yr(i don't remember any thing about Geometry)
and if they test do change, well then, you are just unlucky(i'll be too)

Yes, take the test whenever you are able to.

I took it freshman year

I've since been to the College Board site and have seen an official statement that other than a name change to "Level 1" and "Level 2," there will still continue to be two SAT II math tests, with no particular change in content.

Yeah. Take SAT II's whenever you're ready.
Grouping tests together does save a few bucks, though...

[quote="juicybooty911"]i'm in 9th grade right now, and i will take level 1 this may and take level 2 next yr.[/quote]

NOOOOO
do NOT do this
the curve on the level 1 test is so harsh that if you make one stupid mistake and get the rest right you will probably NOT get an 800

however the curve on level 2 test is very generous... you might be able to leave 4-5 blank and still get an 800.

Also, colleges only care if you took one of these tests.
If you are advanced in math taking level 1 is a very bad idea.

[quote]however the curve on level 2 test is very generous... you might be able to leave 4-5 blank and still get an 800. [/quote]

It's more generous than that. If you get 4-5 [i]wrong[/i], you can still get a 800, but don't let that get into your head. I went in confident of an 800 and came out with 9 blank problems and 1 question wrong. I still got a 770 though  :)

I would definitely recommend taking the SAT II's during your sophomore year. Even if it's a little more expensive, it leaves you with extra time during your junior year so you won't have to stress out.


aca un problema, que aun no me sale.
--
sobre una cirunferencia hay puntos rojos y azules. Se puede adicionar un punto rojo el cual producira el cambio de los puntos vecinos(a otro color) o quitar un punto rojo el cual tambien producira el cambio de sus vecinos. Inicialmente hay dos puntos rojos. Probar que no existe secuencia de operaciones con la cual la configuracion final sea de dos puntos azules.
---
espero que lo resuelvan.

[quote="R.G.A.M."][color=blue]¿El cambio sólo se produce en los dos inmediatos vecinos del punto agregado o quitado?[/color][/quote]
SI, a sus vecinos inmediatos.

[color=blue]Encontré un problema que me hizo pensar en éste.

Sean $A$ y $B$ dos personas que, junto con otras $2001$ personas, se encuentran sobre una circunferencia. En el primer turno, $A$ toca a una persona que se encuentra a su lado, la cual sale del juego. Luego $B$ procede análogamente, y continúa el juego hasta que uno de los dos jugadores principales saca al otro del juego, y gana. Hay que encontrar la estrategia ganadora.

No creo que sea tan similar, a simple vista, pero la solución del mismo podría intentar aplicarse en el problema principal.[/color]

[quote="R.G.A.M."][color=blue]Encontré un problema que me hizo pensar en éste.

Sean $A$ y $B$ dos personas que, junto con otras $2001$ personas, se encuentran sobre una circunferencia. En el primer turno, $A$ toca a una persona que se encuentra a su lado, la cual sale del juego. Luego $B$ procede análogamente, y continúa el juego hasta que uno de los dos jugadores principales saca al otro del juego, y gana. Hay que encontrar la estrategia ganadora.

[/color][/quote]

SOLUCION.
Analisaramos la pariedad de la cantidad de personas que hay entre $A$ y $B$, notamos que cuando le toca el turno a el jugador $A$ hay una cantidad impar por un lado y par por el otro.entonces en algun momento llegara a que por un lado habra una persona y por el otro par, desps de una cantidad finita, llegaremos a que entre $A$ y $B$ hay 1 y 2 personas entonces $A$ toca a la persona que esta para el lado donde hay 2 personas y le quedara $B$ una persona por ambos costados, entonces toque al que toque, el jugador $A$ en su turno lo tocara.
entonce ganar el jugador $A$.

[color=blue]Correcta la solución. ¿Notas alguna similitud con relación al problema que enviaste? Trabajaré en ello, también.[/color]


AMC
AIME
percent
geometry
3D geometry
sphere
conics

Hello. The contest has ended. Results are posted. They seemed to be a bit lower than the results of the other AMCs.

Thank you for taking my contest. I hope you all enjoyed it!

Answers:

1.	C
2.	A
3.	E
4.	C
5.	C
6.	A
7.	E
8.	D
9.	C
10.	E
11.	E
12.	D
13.	D
14.	A
15.	B
16.	B
17.	A
18.	D
19.	D
20.	C
21.	A
22.	B
23.	D
24.	D
25.	B

Official Stats:

Number of participants: 17

Mean Score: 89.53

Median Score: 91

Mode Score: 73.5, 90, 93.5

Score Range: 65

Honors:

Gold Award: Teddy (126)

Silver Award: ginasboys (105.5)

Bronze Award: Rep123Max (100.5)

Other Mock AIME Qualifiers (rest of top half)
- interesting_move (95.5)
- pi^2 (94.5)
- beta (93.5)
- Nosepuppy (93.5)
- Optimosis (92)
- 0.5*joml88 (91) (just kidding, all of joml88 makes the AIME :))

Most Evil Problem(s) Award: 3, 8, 13 (hmm..., an arithmetic progression - too bad a lot of people got #18)

Well, if you are not listed here, do not feel bad. It seems this test was quite hard (compared to a real AMC). If you have any questions about your score or the answers, please private message me.

P.S. Do not forget to participate in the poll in the Mock AMC D thread!

What was the answer to 3?
The question was how many divisiors does 2^2004 have.  There's 2^1 all the way to 2^2004, plus 1 is a factor, so there are 2005?

negative numbers??? I got that one wrong also

Now that I think of it, it didn't specify positive divisors.  So the answer was 4010.  O well, tricky question.

How do you know that you guys got it wrong?

I got home from school at 6 today.  I knew I would be late so I took it Sunday.

how would you solve number 7?

how about 9, 10, 12, 16 ~ 25?
(I need those bad)
Should somebody open an official tread?

Just wondering...why was number 13 considered evil?  I got the other evil questions wrong!  EVIL INCHES TO FEET COVERSIONS!  HOW DARE YOU!   ;)  Can someone explain number 4 to me?  I know I must be really stupid or something...but it says PERCENT and how can 15r be a percent!

I wasn't able to take it, but I looked over the problems and answers.

Ultimatemathgeek, for number 7, you use Euler's formula. It says the for any polyhedron,

V + F - E = 2
V =vertices, F = faces, E = edges.

So if you set E = F+12, then plugging it into the equation, you get
V + F - F - 12 = 2
V - 12 = 2
[b]V = 14[/b]

For number 4, SingtheSorrow, a = .3r and b = .5r, so ab = .15r2
Then, since you want that as a percent of r, you make it .15r2 /r = .15r

But since that's the decimal value and you want percentage which is out of 100, you get .15r * 100 = [b]15r[/b]

I hope those explanations made sense  :)

How would you even approach number 9, 18, or 25?

[quote="Rep123max"]
#18. What is the coefficient of $x^3y^4z^6$ in $(x+y+z)^{13}$?

If we look at $(x+y+z)^{13}$ we have $(x+y+z)(x+y+z)\ldots(x+y+z)$ 13 times. The coefficient of $x^3y^4z^6$ will just be how many ways we can arrange xxxyyyyzzzzzz, which is 60060.

I probably didn't explain those good.[/quote]

I still don't get #18.  How did you get 60060?  When I was taking the test, I knew I could get it, but somehow I couldn't

So basically we want to know how many ways are there to rearrange the order of letters of the word xxxyyyyzzzzzz. There are [tex]\frac{13!}{3!4!6!} ways[/tex].

Yeah, I really like your solution for the (x+y+z)^13 one. I broke it up into ((x+y)+z)^13 and applied binomial theorem twice, getting (13 choose 7)*(7 choose 3) = 1716*35 = 60060. I think your way is more clever though.

16)Let the cord be AB, with B on the sphere, A on the ceiling. Let the eyes be point C Let O be the center of the sphere. Draw any tangent line CD, where D is the point of tangency. CD intersect the ceiling at E.  Let the radius of sphere be r. The area on the ceiling that is blocked is clearly a circle, so the radius of the circle is 10=AE. Since [tex]\Delta OCD \sim \Delta ECA[/tex]. [tex] AC=19+r, CD=\sqrt{13^2-r^2}. [/tex][tex]\frac{r}{10}=\frac{\sqrt{13^2-r^2}}{19+r}. [/tex]Solve for r to get r=5. The surface area of the sphere is  [tex]100 \pi[/tex]. So we need at least 315 gallons.

#17) Let the point be the origin and let equation of the line be y=-6. If (x, y) satisfies the condition then [tex]x^2+y^2=(-6-y)^2[/tex]. Simplify we get [tex]y=\frac{1}{12}x^2-3[/tex]. The graph of that is a parabola, which is the locus of points satisfy the condition. The lowest point C is (0, -3). It's clear that the slopes of CB and AC are 1 and -1. equations of AC and BC are y=x-3, y=-x-3. Sub in that for y and solves for x we get x=12, -12, y=9.
AB=24 and the altitude to AB=9+3=12. 24*12/2=144

the sum of the cubes one:
if you don't remember the formula, just look at smaller cases.
for 1^3+2^3+....n^3

n       sum        sqrt(sum)
1         1                 1
2         9                 3
3         36               6
4         100            10
5         225            15

the pattern seems to be quadratic.  so derive a formula for rad(sum) in terms of n, and voila :D

the edges of figures one:
look at smaller cases:
v=# of vertices
n=edges-faces

v        n
4        2
5        3
6        4

i didn't bother going any further because i felt this pattern was obvious enough

so if n=12, v=14

[quote="beta"]#25) A cheap trick(yes, I finally learned how to use them :) ). Let [tex]2a+3b+6c+24d=S[/tex]
[tex]a^2+b^2+c^2+d^2-\frac{S}{2}+(\frac{1}{4}+\frac{9}{16}+\frac{9}{4}+36)[/tex]
[tex]=(a-\frac{1}{2})^2+(b-\frac{3}{4})^2+(c-\frac{3}{2})^2+(d-6)^2=\frac{1201}{16}-\frac{S}{2} \geq 0[/tex]
[tex]S \leq 150.125[/tex]. That eliminates C, D, E. Now if [tex]d=6,a, b, c=0[/tex] then S=144>125, A is not the maximum. The only possible answer is B. :D[/quote]

I was wondering why you subtracted [tex]S/2[/tex] instead of [tex]S[/tex].
And also, I was wondering how you got [tex]\frac{1201}{16}-\frac{S}{2} \geq 0[/tex]

For number 10: A triangle has sides of length 5, 6, and 7. Find the greatest integer less than the area of its circumcircle.

The formula for the radius of a circumcircle is [tex]R = \frac{abc}{4A}[/tex]
a,b,c are the sides and A is the area of the triangle. Using Heron's formula, you get the area to be  6 :sqrt: 6 and abc = 210. Therefore the radius is 210/(4)(6 :sqrt:6) which equals about 3.57;  plug it into :pi: r 2 and you get 40.08, or about [b]40[/b].

FYI, for a proof of that circumcircle radius formula, go to  [url]http://www.bobbymcr.com/main/math/circumcircle.pdf[/url]

For the circumcircle one, I think that I used the law of cosines and the law of sines.

[quote="Rep123max"]The reason that when you have $(a+b)^n$ then the kth term will be $a^{n-k}b^k$ and we can arrange that in $\frac{n!}{k!(n-k)!}$.[/quote]

Wouldn't Kth term be [tex]a^{n-k+1}b^{k-1}[/tex] ?

It is like 1st term is [tex]a^nb^0[/tex], 2nd term [tex]a^{n-1}b^1[/tex] with their repective coefficients

[quote="interesting_move"]Well, if you had subtracted S, you would end up with S <= 112.5, which doesn't help at all.  The margin by which you subtracted is very significant.  So why did you subtract S/2, beta?

-interesting_move[/quote]
When I was doing the problem, I found that subtracting S wasn't too helpful. So I tried S/2.
That idea worked, here is my solution

If we divide S by k, then completing the square, we get
[tex]\frac{2^2+3^2+6^2+24^2}{4k^2}+36-\frac{S}{k} \geq 0[/tex]
simplify we get [tex]S \leq \frac{625+144k^2}{4k}=p[/tex], so we are trying to find the minimum value of p. simplify we get [tex]144k^2-4kp+625=0[/tex]

[tex]k=\frac{4p \pm \sqrt{16p^2-4(144)(625)}}{288}[/tex] So [tex]16p^2-4(144)(625) \geq 0[/tex]. [tex]p \leq 150[/tex], [tex]S \leq 150[/tex]  When S=150, a, b, c, d are[tex] \frac{12}{25}, \frac{18}{25}, \frac{36}{25}, \frac{144}{25}, a^2+b^2+c^2+d^2=36  [/tex] QED

[quote="JGeneson"]Did anyone use Cauchy for #25? The Cauchy solution is REALLY short.[/quote]

I can't seem to come up with this solution...would you mind posting it (or a hint)?

[quote="pkothari13"]WOW that was pretty cool and helpful Beta.

BTW, what's Cauchy? Is that some type of formula or method?[/quote]

Basically, Cauchy's Inequality (also the Cauchy-Schwarz Inequality) says that $(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$ for sequences $a_1, a_2, \ldots a_n$ and $b_1, b_2, \ldots b_n$.  More information can be found at [url]http://mathworld.wolfram.com/CauchysInequality.html[/url].


Let f(x)=x^2 when x is not equal to 0. How would you define the value of f at
x=0 so that the new function is continous at x=0. Justify your anwser.


Determine all function $ f: R \to R$ such that :
$ f(x\plus{}y)\plus{}f(y\plus{}z)\plus{}f(z\plus{}x) \ge 3f(x\plus{}2y\plus{}3z)$  ,$ \forall x,y,z \in R$

if $ y \equal{} z \equal{} 0$  $ \implies$  $ 2f(x) \plus{} f(0)\geq 3f(x)$  $ \implies$  $ f(0)\geq f(x)$

$ x \equal{} y \equal{} a,z \equal{} \minus{} a$ $ \implies$  $ f(2a) \plus{} 2f(0)\geq 3f(0)$  $ \implies$  $ f(0)\leq f(x)$

$ \implies$  $ f(x) \equal{} f(0) \equal{} c$  $ \blacksquare$


1. Find the best constant $ k$ such that
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}k\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge k\plus{}3
\]
for all $ a,b,c \ge 0.$
2. If $ a,b,c \ge 0,a\plus{}b\plus{}c\equal{}1$, prove that
\[ \minus{}\sqrt{\frac{71\minus{}17\sqrt{17}}{32}} \le \sum \frac{a\minus{}b}{\sqrt{a\plus{}b}} \le \sqrt{\frac{71\minus{}17\sqrt{17}}{32}}
\]
3. (IMO 2006) If $ a,b,c \in \mathbb R$, find the best constant $ k$ such that
\[ \vert ab(a^2\minus{}b^2)\plus{}bc(b^2\minus{}c^2)\plus{}ca(c^2\minus{}a^2) \vert \le k (a^2\plus{}b^2\plus{}c^2)^2
\]
4. If $ a,b,c >0$ then
\[ \sum \frac{a^2}{b} \ge 3\sqrt[6]{\frac{a^6\plus{}b^6\plus{}c^6}{3}}
\]
to be continued.....

I want to know what's your new technique?

Can you post the solution for the second?


geometry
parallelogram
circumcircle
geometric transformation
vector
geometry solved

the point O is situated in the parallelogram ABCD so that <AOB + <COD = 180
         prove R(OBC)=R(ODC)=R(AOB)=R(AOD)
 :)  :D  :?

[color=red][Moderator edit: For any triangle XYZ, the notation R(XYZ) denotes the circumradius of the triangle XYZ.][/color]

___________________________________________________________________

[color=red][b]Another posting by Thanliem, containing the same problem:[/b][/color]

the poin O is situated inside the parallelogram ABCD so that
<AOB + <COD = \pi
prove that R(OBC)=R(OAB)=R(OCD)=R(OAD)
(R is a radius of a circumcircle)

We obtain another point, O', from O by a translation of vector AD. It's easy to see that DOCO' is cyclic, so the circmradii of COD and CO'D are equal, but CO'D is congruent to BOA, so R(AOB)=R(COD). We obtain another point, O", from O by a translation of vector AB. By using some triangle congruences we can quickly show that quadrilaterals COBO" and CODO' are congruent, so the conclusion follows because R(AOB)=R(COD)=R(CODO')=R(COBO")=R(BOC)=R(AOD).


Given |a|+|b|+|c|>17

Prove that  |ax^2+bx+c|>1 has solution(s) belonging to  [0,1]

Have A Great Day!

[hide="fixed but still ugly"]Assume to the contrary that $ |ax^2 \plus{} bx \plus{} c|\le1\forall x\in[0,1]$. Therefore, $ |c|\le 1$ or $ \minus{} 1\le c\le1$ and $ \minus{} 1\le a \plus{} b \plus{} c\le1$. With $ x \equal{} \frac12$ we have $ \minus{} 4\le a \plus{} 2b \plus{} 4c\le4$. Now we can combine inequalities to receive $ \minus{} 5\le b \plus{} 3c\le5$ or $ |b|\le8$ and $ \minus{} 6\le \minus{} a \plus{} 2c\le6$ and $ |a|\le8$. Adding, we have $ |a| \plus{} |b| \plus{} |c|\le17$, a contradiction, so our assumption is false. $ \boxed{\text{QED}}$.[/hide]

[quote="worthawholebean"][hide="Slightly ugly solution"]Assume to the contrary that $ |ax^2 \plus{} bx \plus{} c|\le1\forall x\in[0,1]$. Therefore, $ |c|\le 1$ or $ \minus{} 1\le c\le1$ and $ \minus{} 1\le a \plus{} b \plus{} c\le1$, or $ \minus{} 2\le a \plus{} b\le2$. With $ x \equal{} \frac12$ we have $ \minus{} 4\le4a \plus{} 2b \plus{} c\le4$ or $ \minus{} 5\le4a \plus{} 2b\le5$. Now we can combine inequalities to receive $ \minus{} 9\le2a\le9$ or $ |a|\le\frac92$ and $ \minus{} 13\le \minus{} 2b\le13$ $ |b|\le\frac {13}{2}$. Adding, we have $ |a| \plus{} |b| \plus{} |c|\le12$, a contradiction, so our assumption is false. $ \boxed{\text{QED}}$.[/hide][/quote]

Nice! A small mistake in calculation as x= 1/2. The proof holds true though.

Have A Great Day!

I was wondering...
f(x)= ax^2 + bx+ c
f(1)= a + b + c
but a + b + c > 17 right??
 therefore , for f(x) to be greater than 1 its upper limit can be defined as 1 and it would be included in the solution set.
 Now , i dont know how to prove the lower limit as 0 because we cant c > 1 , we would only be able to conclude that c > 0 for this expression to have a postive value.

Any help?

You need to be very careful with the absolute values.

[quote="worthawholebean"][quote="Zed_darkx"]ohh darn..i took all to be postive.....
dat means that my trivial method would only work if (a,b,c)>0
which instantly satisfies c>0.
Well that was stoopid of me.[/quote]If you want you can take one of the three coefficients and make it positive as otherwise you can flip all the signs.[/quote]

Yea , well i got it now.

I just started with Inequalities a week ago.
:)


What's  the  differences  between   \pagenumbering{}  and  \setcounter{page}{} ?
Many thanks in advance!

\pagenumbering sets the style eg \pagenumbering{alph} gives a,b,c,... whereas \setcounter{page}{12} sets the page counter to whatever the current style is - in this case to m (12th letter of the alphabet).


a space $ X$ is  called limit point compact if every infinite subset of $ X$ has a limit point

My question is :  why $ [0,1]$ is not limit point compact as a subset of $ \mathbb{R}_l$ : with lower limit topology?

thanks

Try taking the infinite subset $ A \equal{} \{ 1 \minus{}  1/n : n\in \mathbb{N} \}$.

In the lower limit topology intervals of the form $ [a,b)$ are open; thus $ \{ 1 \}$ happens to be an open set in the subspace $ [0,1]$.

Now, since the lower limit topology is finer than the standard topology on $ \mathbb{R}$, you should have no trouble convincing yourself that the only possible limit point $ A$ is $ 1$.

However, as I've already mentioned $ \{1 \}$ is open and doesn't contain any point of $ A$.


search
combinatorics proposed
combinatorics

2009 boys and 2009 girls attend several groups. It is known that any of them is attending no more than groups and that any boy in a couple with any girl attend at least one common group. Prove that there is a group containing at least 11 boys and 11 girls.

This problem was posted in yesterday in this forum, with an easily identifiable subject name (and without missing a number in the text that renders the problem incomprehensible).  You wouldn't have even had to do a search to check whether you were making a redundant post!  I would think that "looking at the most recent five threads in a given forum to make sure my post is not this ridiculously redundant" should constitute a basic step when posting any new problem.  (Doing a basic search would have been nice, too.)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265865


inequalities
trigonometry
geometry proposed
geometry

[quote][color=darkred][b]Let $ABC$ be an $A$- isosceles triangle ($b=c$) with the circumcircle $w=C(O,R)$.
Denote the reflection $B'$ of the vertex $B$ w.r.t. the circumcenter $O$. Prove that $B'A+B'C\ge R\sqrt 2$.[/b][/color][/quote]

[quote]...... $B'C=BB'\sin(OBC)=2R\sin(\frac{\pi-A}{2})=2R\cos\frac{A}{2}$ ....[/quote] It is wrong. Correctly is $B'C=2R\sin\left(90^{\circ}-A\right)$ !

ok that's right thus
$B'A+B'C=2R(\cos A+\sin\frac{A}{2})=2R(1-2t^{2}+t)$ let $t=\sin\frac{A}{2}$ note that $A=\frac{180^{0}-B}{2}\le 90\Rightarrow \frac{A}{2}\le 45^{0}$ thus $t\in [0,\frac{1}{\sqrt{2}}]$ thus consider $f(t)=-2t^{2}+t+1\Rightarrow \min_{t\in[0,\frac{1}{\sqrt{2}}]}f(t)=f(\frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}$
Thus $B'A+B'C>\sqrt{2}R$ maybe there isn't equal.

[color=darkred][b]Sorry, again is wrong. I"ll use the your idea for finish correctly the your proof.[/b][/color]

[color=darkblue][b]Proof.[/b] Denote $A=2x$. Observe that $m(\widehat{ABB'})=x$ and $m(\widehat{CBB'})=\{\begin{array}{ccc}90-2x & , & 0<x\le 45\\\ 2x-90 & , & 45<x<90\end{array}$.
Thus, $\{\begin{array}{c}B'A=2R\sin x\\\ B'C=2R|\cos 2x|\end{array}$ and $B'A+B'C\ge R\sqrt 2$ $\Longleftrightarrow$ $2(\sin x+|\cos 2x|)\ge \sqrt 2\ \ (*)$.
$1\blacktriangleright$ If $45\le x<90$ then $\{\begin{array}{c}2|\cos 2x|\ge 0\\\ 2\sin x\ge \sqrt 2\end{array}$ and the relation $(*)$ is truly with the equality iff $x=45$, i.e. $A=90$.
$2\blacktriangleright$ If $0<x<45$ then denote $\sin x=t\in (0,\frac{\sqrt 2}{2})$, $\cos 2x=1-2\sin^{2}x\in (0,1)$ and
$2\sin x+2(1-2\sin^{2}x)\ge \sqrt 2$ $\Longleftrightarrow$ $4\sin^{2}x-2\sin x+\sqrt 2-2\le 0$ $\Longleftrightarrow$ $4t^{2}-2t+\sqrt 2-2\le 0$ $\Longleftrightarrow$
$(2t-\sqrt 2)(2t+\sqrt 2-1)\le 0$. But $2t+\sqrt 2-1>0$ and $2t-\sqrt 2<0$. Therefore, and in this case the relation $(*)$ is truly, even strictly.[/color]

ok thank you for your correct solution, I was so careless :)


ARML geometry rhombus rectangle trigonometry

Each side of a rhombus has length $12$ and one of its angles is $150^{\circ}$. External squares are drawn on each of the four sides of the rhombus. A point is marked at the center of each square and they are connected to form a quadrilateral. Find the area of this quadrilateral.

[quote="José"][quote="gnosis"]U can prove that the quadrilater is actually a rectangle.
After knowing this your goal is to find out the dimensions of the rectangle.
You can do that by first finding out the distance between the center of the rhombus and the center of one square (you will have that all 4 such distances are the same) $6\sqrt{3}$
Afterwards u get the dimensions of the rectangle to be $12\sqrt{2+\sqrt{3}}-8$ and $6\sqrt{6}$ which leads to the final result of $6(36+12\sqrt{3}-8\sqrt{6})$.
If I didn't do any silly mistake that should be it.[/quote]

Actually, i got that the cuadrilater is a square, with side equal to $6√2√3$. Then, the area is $216$. I hope it`s OK.[/quote]

Typing mistake. The square side is 6√2√3.

hmm im having a hard time finding the distance from the center of the rhombus to the center of the square...

By using the cosine theorem in $\Delta ABC$ we get that $AC^2=2\cdot12^2-2\cdot12^2\cos{150} \Rightarrow  OA=\frac{AC}{2}=6\sqrt{2+\sqrt{3}}$.
In the square with center $O_1$ we have that $O_1A=\frac{AB\sqrt{2}}{2}=6\sqrt{2}$. Since $\angle O_1AO=60 \Rightarrow  O_1A\cdot \sin60=h \Rightarrow  h=\frac{\sqrt{3}}{2}\cdot 6\sqrt{2} \Rightarrow  O_1O_4=2\cdot h=6\sqrt{6}$
For finding $\alpha=O_10_2$ we have that in the trapeze $AO_1O_2C$ by Phytagora $(\frac{AC-\alpha}{2})^2+h^2=O_1A^2$, hence $\frac{AC-\alpha}{2}=\sqrt{2\cdot 6^2-6\cdot 3^2}=3\sqrt{2} \Rightarrow  \alpha = 12\sqrt{2+\sqrt{3}}-6\sqrt{2}$.
In conclusion $\sigma(O_1O_2O_3O_4)=6\sqrt{6}\cdot(12\sqrt{2+\sqrt{3}}-6\sqrt{2})$.
I hope I didn't do any mistakes this time. :roll:

[quote="gotztahbeazn"]hmm... actually u couldve just stopped at $O_1O_4 = 6\sqrt{6}$ and square that to get $216$ which is the right answer

but can u prove that $O_1O_4$ is perpendicular to $AC$? maybe its trivial but ehh i ono[/quote]

I really don't think that $216$ is the right answer.Just check my calculations.Regarding the fact that $O_1O_4=2\cdot h$ that's due to the fact that $O_1$ and $O_4$ are symmetric by $AC$.

well thats the answer ARML gives on their answer sheet

[quote="gnosis"]$6\sqrt{6}\cdot(12\sqrt{2+\sqrt{3}}-6\sqrt{2})$.[/quote]
this can be simplified...
\begin{eqnarray*} 6\sqrt{6}\cdot(12\sqrt{2+\sqrt{3}}-6\sqrt{2}) &=& 6\sqrt{6}\cdot(12(\frac{\sqrt{6}+\sqrt{2}}{2})-6\sqrt{2})\\ &=& 6\sqrt{6}\cdot(6\sqrt{6})\\ &=& \boxed{216} \end{eqnarray*}

:D Sorry. They are right and I was wrong. I am not patient enough for so much calculus. That's why I found a shorter solution.
$O_1A=\frac{12\cdot\sqrt{2}}{2}=6\sqrt{2}$
$OA=\frac{\sqrt{2\cdot 12^2-2\cdot 12^2\cdot \cos{\frac{\pi}{3}}}}{2}=6\sqrt{2+\sqrt{3}}$
which in $\Delta O_1OA$ gives us $O_1O^2=2\cdot6^2+(2+\sqrt{3})6^2-2\cdot6^2\sqrt{4+2\sqrt{3}} \Rightarrow  O_1O=6\sqrt{3}$.
Next by using the sin theorem in $\Delta O_1OA$ we have $\frac{O_1A}{\sin(O_1OA)}=\frac{O_1O}{\sin\frac{\pi}{3}} \Rightarrow  \sin(O_1OA)=\frac{\frac{\sqrt{3}}{2}\cdot6\sqrt{2}}{6\sqrt{3}}=\frac{\sqrt{2}}{2}$, hence $\angle(AOO_1)=\frac{\pi}{4}$ and $\angle(O_1OO_4)=2\cdot\frac{\pi}{4}=\frac{\pi}{2}$, respectively $\angle(O_1OO_2)=2\pi-2\cdot\frac{\pi}{4}=\frac{\pi}{2}$.
This shows that $O_1O_2O_3O_4$ is a square and its diagonal is $2\cdot OO_1=12\sqrt{3}$, hence its side is $6\sqrt{6}$ and the area requested as they and you stated is $216$.

[hide="Shortest solution"]Using the diagram from post #6, we have

$\angle O_1AO_4=30^\circ+2\cdot 45^\circ=120^\circ$ and $AO_1=AO_4$
$\angle O_1BO_2=360^\circ-2\cdot 45^\circ-150^\circ=120^\circ$ and $BO_1=BO_2$

Hence triangles $O_1AO_4$ and $O_1BO_2$ are isosceles and congruent because $AO_1=BO_1$. Hence

$1^\circ\quad O_1O_4=O_1O_2$
$2^\circ\quad\angle AO_1O_4=\angle BO_1O_2\implies$
$\implies\angle O_4O_1O_2=\angle AO_1B-\angle AO_1O_4+\angle BO_1O_2=\angle AO_1B=90^\circ$

Hence $O_1O_2O_3O_4$ is a square and $AO_1=6\sqrt{2}$

Applying Cosine Law to $\triangle AO_1O_4$ we get $O_1O_4^2=(6\sqrt{2})^2+(6\sqrt{2})^2-2\cdot 6\sqrt{2}\cdot 6\sqrt{2}\cdot\cos 120^\circ=216$, and in the same time that's the area of the square.[/hide]


given the array of integers  $ a_{n}$
$ a_{1}\equal{}a_{2}\equal{}1;a_{n\plus{}2}\equal{}7a_{n\plus{}1}\minus{}a_{n}\minus{}2$
prove that:
$ a_{n}$ is a square for every $ n\geq1$

Let $ b_1\equal{}1\equal{}b_2, b_{k\plus{}2}\equal{}3b_{k\plus{}1}\minus{}b_k$. By induction $ a_n\equal{}b_n^2$.


modular arithmetic
number theory
divisibility tests

1)Find a shortcut to test for divisibility by 5 with mods.

2)Find the shortcut to test for divisibility for 4, 8, and 20.

3) For the numbers 1717, 3451, and 173451, test for divisibility by 3 by a shortcut.

4) Which of the following are divisible by 11: 11, 111, 1111, 1716, 1761, 152637?

Not sure what you mean by "with mods" in #1.
[hide="#2"]If the last two digits of the number are divisible by 4, the number is divisible by 4, last three digits for 8, and for 20(?!), well, it has to be divisible by 5 and 4, doesn't it?[/hide]
[hide="#3"]$1717$: $1+7+1+7=16$, not divisible
$3451$: $3+4+5+1=13$, not divisible
$173451$: $1+7+3+4+5+1=21$, divsible.[/hide]
[hide="#4"]11: duh.
$111: 1+1\neq1$; nope.
$1111: 1+1=1+1$; yep.
$1716: 1+1\neq7+6$; nope.
$1761: 1+6\neq7+1$; nope.
$152637: 1+2+3\neq5+6+7$; nope.[/hide]

[hide="well for 20..."]Isn't it just if the last two digits are divisible by 20? Because then it is divisible by 4 of course, and also 5.[/hide]

Since $100\equiv 0\pmod{20}$, if the last two digits form a number divisible by 20, then the number is divisible by 20.

Edit: Darn.

OH

I thought that if it was in the mod chapter, it had something to do with mods

wow  :blush:

But how do you know that the sum of the digits can determine whether something is divisible by a number?  :maybe:

like 989 is divisible by 13 since its digit sum is 26...

and on number 4, how do you know the 1st half of the digits must equal to 2nd half of the digit sum to be divisible by 11?

[quote="davidyko"]For #4, it's actually just summing up every other digit and comparing them. I used to know why/how it worked, but I forgot, and I'm too lazy to find out. Maybe someone less lazy?  :P[/quote]
It's because $10\equiv-1\pmod{11}$.

[hide="1"]If a number $x$ is divisible by 5, we can write it as $x\equiv 0\pmod{5}\Longrightarrow x\equiv 0, 5\pmod{5}$. From this, we see that the only numbers divisible by 5 are ones that ends in 0 or 5. [/hide]

[quote="davidyko"][quote="cincodemayo5590"][quote="davidyko"]For #4, it's actually just summing up every other digit and comparing them. I used to know why/how it worked, but I forgot, and I'm too lazy to find out. Maybe someone less lazy?  :P[/quote]

Well the divisibility test for $11$ is adding every other digit, then subtracting the rest. If that sum is divisible by $11$, then the number is as well.[/quote]

Mm, then I guess my method isn't exactly accurate, then. I was assuming the difference had to be zero, but it could be any multiple of 11.[/quote]

So actually $1716$ is a multiple of $11$ since $13-2=11$ in #4, right?

Proving these useful divisibility properties are quite simple. Here are some examples and you might want to try the general case.

[b]Divisibility for 3 and 9[/b]

Let's say we want to test whether $126$ is divisible by $3$ or $9$

Since $126= 1 \times 100+2 \times 10+6= 1 \times (99+1)+2 \times (9+1)+6= 9(11+2)+(1+2+6)$

Now if $(1+2+6)$ is divisible by $3$ then $126$ is divisible by $3$ and if it is divisible by $9$ then $126$ is divisible by $9$.

Now this method works for proving divisibility for $11$, $4$, $8$, $7$, $13$, $37$. Try it.

[quote="kohjhsd"][/quote][quote="davidyko"][quote="cincodemayo5590"][quote="davidyko"]For #4, it's actually just summing up every other digit and comparing them. I used to know why/how it worked, but I forgot, and I'm too lazy to find out. Maybe someone less lazy?  :P[/quote]

Well the divisibility test for $11$ is adding every other digit, then subtracting the rest. If that sum is divisible by $11$, then the number is as well.[/quote]

Mm, then I guess my method isn't exactly accurate, then. I was assuming the difference had to be zero, but it could be any multiple of 11.[/quote][quote="kohjhsd"]

So actually $1716$ is a multiple of $11$ since $13-2=11$ in #4, right?[/quote]

Yep, that's right; I should definitely change that.


[hide]
By induction one can prove that $ \frac{1}{2}\cdot\frac{3}{4}\cdot\ldots\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}},\forall n\geq 1$
For $ n=50$ we have $ \frac{1}{2}\cdot\frac{3}{4}\cdot\ldots\cdot\frac{99}{100}<\frac{1}{\sqrt{101}}<\frac{1}{10}$
[/hide]


modular arithmetic
number theory
prime numbers

Let $n$ be a positive integer and $p_1, p_2, p_3, \ldots p_n$ be $n$ prime numbers all larger than $5$ such that $6$ divides $p_1 ^2 + p_2 ^2 + p_3 ^2 + \cdots p_n ^2$. prove that $6$ divides $n$.

[quote="paladin8"][hide]
If $p_i > 5$, then $p_i \equiv 1 \pmod{6}$ or $p_i \equiv 5 \pmod{6}$. Then $p_i^2 \equiv 1 \mod{6}$.

Hence $p_1^2+p_2^2+\cdots+p_n^2 \equiv 1+1+\cdots+1 \equiv n \equiv 0 \pmod{6}$.[/hide][/quote]
That is nice and simple solution, paladin8. Good job

[size=100]This also holds if the primes are larger than three [/size]

When RMO's were one liner...
Any $p$ is either $1,-1\pmod{6}$, so $p^2 \equiv 1 \pmod{6}$ Substituting immediately gives $n \equiv 0 \pmod{6}\,\,\square$


Solve the equation $ 2^x\plus{}x\equal{}\frac{x}{2^x}$

The equality occurs only if $ x(2^{\minus{}x} \minus{} 1) > 0$ or $ (x\minus{}0)(2^{\minus{}x} \minus{} 2^{\minus{}0})> 0$.

But the function $ f(x) \equal{} 2^{\minus{}x}$ is a decreasing function. Hence, $ x(2^{\minus{}x} \minus{} 1) \leq 0$.

So, the equation has no solution.


When you were born, your Grandmother gives you 100 dollars to put in the bank. Fortunately, at that time, Nick Bank is offering a 4% annual interest rate.
1. Identify the multiplier. _____________ Why?
2. Write an Exponential Model, V(t), for the given information.
Model: ________________________ Explain:
3. Identify the first 6 terms of the function to the nearest cent if t=0 in 1992.
t V(t)
t=0
t=1

4. Prove these values belong to the function V(t).

5. Algebraically determine the value of your [Unparseable or potentially dangerous latex formula.Error: 6: 1456x16]100 or less?

......................Does number seven make sense?

agreed. These problems seem suspiciously like homework questions.

I have no problem [i]explaining[/i] to you how to do the problem, but I refuse to just randomly give away answers.


geometry
perimeter
incenter
ratio
circumcircle
geometry solved

[u][b]The author of this posting is : Peter VDD[/b][/u]
____________________________________________________________________

most of us didn't really expect to get this, but here it goes (flanders mathematical olympiad 2004, today)

triangle with sides 501m, 668m, 835m

how many lines can be draws so that the line halves both area and circumference?

Consider a triangle with side lengths 501m, 668m, 835m. How many lines can be drawn with the property that such a line halves both area and perimeter?

This line must pass through the incenter of the triangle.

No, that's not correct!

I made the same mistake too on the contest, which costed me IMO.

[quote="Peter VDD"]No, that's not correct![/quote]

Why not? According to an old Russian book,

  D. O. Shklarsky, N. N. Chenzov, I. M. Yaglom, [i]Izbrannye zadachi i teoremy planimetrii[/i], 2nd edition, Moscow 1967,

problem 105, a line divides the area and the perimeter of a triangle in the same ratio if and only if this line passes through the incenter of the triangle. And hence your two lines pass through the incenter of the triangle, don't they?

Note that the same old book contains, under problem 79, a compass-and-ruler construction of all lines bisecting both the area and the perimeter of a triangle, and actually for an arbitrary triangle, not necessarily for a right-angled one with sidelengths 501m, 668m, 835m. However, I am not sure about whether the claim at the end of the solution is right that there are generally 3 such lines (the construction seems to yield 6 lines, 2 for every vertex, but some of them might work as solutions only if one allows areas and perimeters to be directed areas and algebraic sums of directed lengths, what, I think, is not the aim of the problem). So I am really unsure about the problem.

[b]PS.[/b] Or do you mean that the line should halve the [b]circumcircle[/b] of the triangle, not the [b]perimeter[/b]? I think the notion "circumference" is a bit misleading, but in case you mean the circumcircle, the problem would be a bit too trivial...

  Darij

Darij, in response to your PM:

For the case you hadn't noticed, 501m, 668m, 835m is just the same as 3,4,5. So I think it's easier for everyone if you'd work with those. And yes I mean perimeter with circumference.

Incenter can work, yes. I used the circumcenter (mass center) :blush:
But whatever you do, the solution must be 1. If you manage to find a proof saying it's 2 or more, please let me know :)

If you need the full solution, you can find it on kalva: it's from canada '85, question 1. (quite ironic, I tried all canadian olympiads back to 1986, my year of birth. Oh darn.)

[quote="Peter VDD"]For the case you hadn't noticed, 501m, 668m, 835m is just the same as 3,4,5. So I think it's easier for everyone if you'd work with those.[/quote]

Oh well, I hadn't noticed... since I was too lazy to calculate  :D

[quote="Peter VDD"]Incenter can work, yes. I used the circumcenter (mass center) :blush:[/quote]

Circumcenter or mass center? These are quite different things... (The circumcenter is the center of the circle through the vertices of the triangle. The mass center is the point where the medians meet. And the incenter is the center of the circle which touches the sides of the triangle. I think you just got these names confused.)

So the line actually passes through the incenter, doesn't it?

[quote="Peter VDD"]But whatever you do, the solution must be 1.[/quote]

I hadn't said that there are more than 1 such lines. Actually I've been speaking about a generic triangle when I quoted Shklarsky, Chenzov, Yaglom.

[quote="Peter VDD"]If you need the full solution, you can find it on kalva: it's from canada '85, question 1.[/quote]

Aha, thanks. http://www.kalva.demon.co.uk/canada/casoln/csol851.html . And actually the proposed solution shows that the unique line does pass through the incenter.

  Darij


If $x\in [0,\frac{\pi}{2}]$ prove that

$10x\cos x<65$

Well this is equivalent to proving:

$x cos(x) \le 6.5$ for x within those bounds.  This is pretty obvious.

We don't have $x \cos{x} \leq \frac{\pi}{2} $.

Must be something like this :

 10 x cos x < 65
<=>
 cos x < 6.5/x
<=>

 x>- 6.5/x + 2kpi V x<  6.5/x+2kpi,kEZ
<=>
x^2>-6.5+2kpi  V x^2 < 6.5 +2kpi,kEZ

 In .. [0,1.57]... => k=0

 we have :

 x <- + 2.54
 <=>
 x E ]-2.54,2.54[

nr 1337 may have your answer,i presume he did it with the Lagrange theorem.

You can also use your calculator, draw the graph,and then check if xcosx is minor than 6.5.
 In this terms:
  x=1.57,y=1.25 ;x=0,y=0.In the defined partition the function has no minimum,and it has a maximum equal to 0.56 for x equal to 0.86
And this can prove your expression.

"god must have emerged from the sea",<-homero :D  :D

[i]Dear Silouan[/i], with this problem you appear and at http://www.mathlinks.ro/Forum/viewtopic.php?t=52904 It isn't fair !

where I proved that for any $x\in \left( 0,\frac {\pi}{2}\right),\ x\cos x\le \frac {a^2}{\sqrt {1+a^2}}$, where $\tan a=\frac 1a,\ a\in \left(\frac {\pi}{4},\frac {\pi}{3}\right)$.

From this relations results easily the proposed problem.


Δινονται οι μιγαδικοι $ x_1,x_2,...,x_n$
για τους οποιους για καθε φυσικο $ k$ ισχυει:

$ x_1^k\plus{}x_2^k\plus{}...\plus{}x_n^k\equal{}0$

Να αποδειξετε οτι

$ x_1\equal{}x_2\equal{}...\equal{}x_n\equal{}0$

[quote="AlexandrosG"]Δινονται οι μιγαδικοι $ x_1,x_2,...,x_n$
[b]για τους οποιους για καθε φυσικο $ k$ ισχυει[/b]:
[...][/quote]

Ίσως έχω κάνει πατάτα, αλλά μου φαίνεται ότι αρκεί να ισχύει για $ k\leq n$...

Cheerio,

Durandal 1707

PS: [hide]Βγαίνει ωραία και με επαγωγή ;)[/hide]

Εχω και εγω μια λυση με επαγωγη και με γραμμικη αλγεβρα

Βαζω τη βασικη ιδεα σε hide και αργοτερα ισως βαλω ολοκληρη τη λυση:

[hide]Η Οριζουσα του Vandermonde[/hide]

Νικο αν θες βαλε την λυση σου

Λοιπον θα δωσω λυση στο Γενικευμενο Προβλημα

Αν $ x_1, x_2, ... x_n$ μιγαδικοι και $ a_1, a_2, ... a_n \in R^{*}$
και για καθε $ k = 1,2,...n$
ισχυει $ {a_1}{x_1}^k + {a_2}{x_2}^k + ... + {a_n}{x_n}^k = 0$
Τοτε $ x_1 = x_2 = ... = x_n = 0$

Για $ n=1$ προφανως ισχυει
Εστω οτι ισχυει για $ n=m$
τοτε για $ n = m+1$ εχουμε
$ {a_{m+1}}{x_{m+1}}^k = -({a_1}{x_1}^k + {a_2}{x_2}^k + ... + {a_m}{x_m}^k)$ Για καθε $ k=1,2,...m+1$
Τοτε
${ }{det{\begin{pmatrix}{a_1}{x_1} & {a_1}{x_1}^2  & ... & {a_1}{x_1}^{m+1}\\ {a_2}{x_2} &{a_2}{x_2}^2& ... & {a_2}{x_2}^{m+1}\\ ... & ... & ... & ...\\ {a_{m+1}}{x_{m+1}} & {a_{m+1}}{x_{m+1}}^2 & ... &{a_{m+1}}{x_{m+1}}^{m+1}\end{pmatrix}}} = 0$
Αφου η τελευταια γραμμη ειναι γραμμικος συνδιασμος των προηγουμενων
Η τελευταια οριζουσα ομως ισουτε με $ a_1a_2...a_{m+1}x_1x_2...x_{m+1}V$ οπου $ V$ ειναι η οριζουσα Vandermonde των μιγαδικων $ x_i$, και για να γινεται 0 η προηγουμενη παρασταση πρεπει $ x_i = 0$ για καποιο $ i$ η $ V=0 \Rightarrow x_i = x_j$ για καποια $ i,j$.
Και στις 2 περιπτωσεις οδηγουμαστε στην σχεση που μας δινεται αρχικα για $ m$ μιγαδικους, απο την οποια χρησημοποιοντας την επαγωγικη υποθεση λαμβανουμε: $ x_1 = x_2 = ... = x_{m+1} = 0$. Q.E.D

Εσυ Αλεξανδρε θα κατεβεις την Παρασκευη στον Προκριματικο?

Ηθελα να δω την λυση σου γιατι ειχα βρει και εγω μια με γραμμικη αλγεβρα και
ηθελα να δω αν ειναι ιδιες. Οντως ειναι σχεδον ιδιες απλα  εγω πηρα τον πινακα $ A$
με γραμμες $ x_i,x_i^2,...,x_i^n$
για τον οποιο απο τη σχεση που δινεται προκυπτει οτι $ AA^t\equal{}0$
οποτε $ detA\equal{}0$ και μετα παρομοια(Vandermonde).

Ναι Νικο θα ερθω στον προκριματικο,θα τα πουμε εκει την Παρασκευη :)


If I get 32's on old states as is, how much would this help me?

-jorian

Just a question, although there is an obvious large difficulty gap between county and state level questions, I don't see too much of a difference between national and state level questions. To me, there is less of a difference between those two then the difference between the AMC 12 and 10. Why does everyone else seem to agree that nationals are MUCH harder than states?

Chapter, State, and national have pretty much the same kind of questions. National just requires a little more thinking with state, and the same with state and chapter.

Yeah...I'm saying, the TOPICS are the same. Like basic probability. This just takes very little thinking.

[quote="bpms"][quote="Phelpedo"]I beg your pardon? The following is a chapter question:

If the probability that it rains tomorrow is $\frac{3}{8}$, what is the probability that it does not rain tomorrow?[/quote]
Here is another-
What is six times one seventh?

But my point is that there is a HUGE difference between mastering chapter and state. But the difference between mastering state and nationals isn't that big.[/quote]

I think there is a HUGE difference between nationals and state.  I had trouble breaking 40 on nationals tests, while states and chapters both just tested me on dumb mistakes.  Reading AoPS book one was enough to get me past 40 on most states, while I got just a 29 or so on nationals...

thank you guys very much for this encouraging information!
-jorian

[quote="bpms"][quote="13375P34K43V312"][quote="kyyuanmathcount"]..., while I got just a 29 or so on nationals...[/quote]

Liar...you got a 27

Then again I averaged 36-37 on practice nationals and ended up with a 25, so yeah.[/quote]
I'm very strange this way but at states I lose 5+ points due to careless mistakes and not enough pressure. On nationals I lose like 2-5 points due to careless mistakes since I'm under a lot of pressure. That's th reason my nats score was only two points lower than states.[/quote]

My national score was only 16 points lower than my state score...

[quote="13375P34K43V312"][/quote][quote="bpms"][quote="13375P34K43V312"][quote="kyyuanmathcount"]..., while I got just a 29 or so on nationals...[/quote]

Liar...you got a 27

Then again I averaged 36-37 on practice nationals and ended up with a 25, so yeah.[/quote]
I'm very strange this way but at states I lose 5+ points due to careless mistakes and not enough pressure. On nationals I lose like 2-5 points due to careless mistakes since I'm under a lot of pressure. That's th reason my nats score was only two points lower than states.[/quote][quote="13375P34K43V312"]

My national score was only 16 points lower than my state score...[/quote]
Klebian's was as well...
You beat him though...
One of my teammates got a 23 on the sprint, but a 4 on the target...

SOMEBODY in this thread got 22 sprint, 3 target.

AMC 10 is basically just state-level, but AMC 12 is even above Nats-level.  I think Nats just demands a lot more speed and accuracy than states.

Yeah I just made up those phrases, most of them. But I managed to win both my state and chapter countdowns easily the Neal Wu way. That is, without cheating.

[quote="jhredsox"]it'd be fun to do the buzzer one in an unofficial state countdown!
-jorian[/quote]
It'd be fun to do the buzzer one at an official nationals one :P

if u luckily get into the top 12 :o

and ur against kevin chen, then u should use the buzzer one and say something random!

-jorian

i might use it at chapters...unless i want a plaque...i'm so happy my state countdown doesn't count, but who wants to take any chances

neal would pwn if he twitched :rotfl:

[quote="13375P34K43V312"]Kevin Chen isn't that good at countdown.[/quote]

You never know much he's practiced since then.

Also, I think either these posts should be either deleted or split into another topic for the MC forum.

yes, but dibbs not starting a topic called "twitching", although i would want to do it very bad

-jorian

Treething is the master of the twitch. He should be offended if he ever sees this topic.

[quote="jhredsox"]can u tell me exactly what treething did against nebula at nats?

he said something about tripping... :wink:

-jorian[/quote]
lmao
That was a joke.

Lol blockdude gts

WE ONLY PLAYED ON THE SUBWAY

[quote="funcia"]wow i want a free calc do you get them at nats?[/quote]
yes, a graphing

motivation is now luring funcia to her books

[quote="Treething"]ahhh twitching. mathematically inclined middle schoolers aren't always good at finding creative ways of bending the rules (on their own), so here i go

I've been doing it since I started MathCounts - I was always looking for ways to give myself an unfair advantage while still staying within the rules. I figure the 8,000-dollar-scholarship-and-a-laptop ends justified the means pretty good.

Basically what you do is buzz in before you have the question answered and finish up the remaining steps while the judge is calling your name. I believe you have three seconds (give or take a little bit, like you'd probably get a little bit more than three seconds if you're not instantly hateable, like Furby) after the judge calls your name to begin your answer, and three seconds from the start of your answer (again, give or take a little bit) to finish it.

What I started doing in 8th grade is an addition to twitching - glance at the problem and [i]instantly[/i] buzz in if the problem is one or two lines long, or if it looks familiar. Also, a lot of really long problems really only have a couple key sentences - learn to read quickly and stuff.

Twitching isn't for you if get nervous though. I guarantee you'll freeze up if you're one of those people who freak out. Think about it this way - even if you lose, what do you really lose? Nothing, and honestly, even if you lose, you still gain stuff, not necessarily monetary rewards, but experience, a free graphing calculator, and possibly lifelong enemies if you were a douche and sat in the corner playing Block Dude on the aforementioned graphing calculator.

Phoenix is okay though.[/quote]
Who's furby?
You mean the toy, or someone?


1.How do you make blog tags?

2.Can I organize all the entries I have so far into tags, or just new entries from the time I make tags?

3.I changed the font size of my blog to 12 pt font Times New Roman, but nothing happened. How do you change the font?

Thanks in advance.

When you post/edit an entry you have the ability to add tags (separated by a comma).

[quote="hurdler"]Ok, the font size has changed. But I still do not know how to make tags[/quote]

For me, there's a Tags link below the box where I type my entries.  Click that, and an input field will open into which you can type your tags.


Find all integer solutions of the equation $ (x^{7}-1)/(x-1)=y^{5}-1$.


calculus
function
limit
derivative
LaTeX
calculus computations

Hello. New to the boards. This is my first post. I'm enrolled in Calculus I and am having some difficulties seeing curve sketching when just the conditions are given. I could use some help and advice with these types of curve sketching:

Sketch the graph of a function satisfying the following conditions: f'(x)>0 on the interval (-2,2), f'(x)<0 on (-∞ ,-2)U(2,∞ ), f"(x)>0 on (3,+∞ ), f"(x)<0 on (-∞,  3), and lim f(x) = 1
                                                                           x > ∞
What would the graph look like?

Thanks in advance, --jeff[/code]

Are you sure you have transcribed this problem correctly?

The problem is that the function is concave down at x = -2 due to the conditions you have for f''(x) (if f''(x) < 0, the fcn is CD), however your conditions for f'(x) show that the function should go from decreasing to increasing at x = -2 (has a min there), implying that the function is concave up at that point.

Furthermore, for f(x) to approach +∞ at x = 1, it needs to be concave up.

You're reading it wrong- that last condition is that $\lim_{x\to\infty}f(x)=1$

On the other hand, the conditions on the first and second derivatives can't all be true.

Thanks for the replies. Yes, after double checking the problems reads just as I entered in the first post. My apologies about the latex, I didn't see this until after I posted. (my first post here.) Our assignment is to draw the curve according those given conditions.

Jeff

Well then I think you'll have to say that the curve can't exist. Like jmerry said you can't have a curve with the properties of the first and second derivatives you have listed. More explicitly: You can't have a min over a concave down region, which occurs at x=-2.


How does the size of a stringed instrument affect the characteristics of sound? Like, for violin, the size is small so the sound is higher, is it because the less surface area produces less resonance? But shouldn't that produce less sound. I don't understand how a bigger instrument produces a deeper, lower sound. Could somebody please explain? Thanks!

You know when you play on the instrument it oscillates and sound waves with the same frequencies spread which we are hearing. And these waves are quantised such that
$L=m\frac{\lambda}{2}$
$\nu=\frac{\upsilon}{\lambda}$
$J \sim \nu^2 \sim \lambda^{-2}$
So for the bigger instruments intensity is less.

There are two issues. The first is the pitch produced by the string itself, which depends on the length, mass per unit length, and tension of the string. I gave a talk on this a couple of years ago, and I'll attach the file for the overheads for that presentation. It gives a good overview of the relationship of pitch to those factors.

But there's a second issue: the string itself doesn't move very much air, and without the assistance of either a sounding box or an electric amplifier, would be almost inaudible. All non-electrified string instruments have a sounding box of some kind or another. The wavelength of a sound wave in air would be $\frac{s}{\nu}$ where $s$ is the speed of sound. An estimate for $s$ is about $330$ m/sec at sea level. The lowest pitch on a violin is about 196 Hz, which is a wavelength of 1.7 m, but the instrument spends much more time on notes with fundamental frequencies in the 400-1000 range (wavelengths of 30 to 80 cm) with overtones that are multiples of that (with even shorter wavelength). The sounding box - the body of the instrument - for a violin is perhaps 30 cm long. The bridge of the violin transmits the vibrations of the string to the wooden lid of the sounding box, and that movements of that lid in turn move the air inside the box, making a reasonably strong sound.

A lower instrument produces sound with lower frequencies, hence longer wavelengths. That's one reason it tends to have a larger sounding box. But that's only part of it: a lower instrument has bigger, longer strings, which take more force to put into motion in the first place. Those strings are then capable of transmitting more force to the sounding box and thus getting good motion out of a larger box. The extreme would be a grand piano: enormous strings under very high tension, moving a sounding box that may be as big as 1 meter by 3 meters - and that one piano can make itself heard above a 100 piece orchestra playing fortissimo.

There are various aspects to the sound: like thwe amplitude and quality of sound, mellowness, and given a violin the quality of sound varies with factors such as bowing pressure, bowing speed etc. I gave a presentation on this at an elementary level, I will attach a word document of that presentation, also web references are provided in the document.


ratio
geometric sequence
calculus
calculus computations

Hello.  I am stuck on a problem.

I need to rewrite the sums using notation.

The sums are: (1/2)-(1/4)+(1/8)-(1/16)+(1/32)-(1/64)+(1/128)

How would I go about writing that in the sigma notation? How would it look like?  Thank You

That is the sum of the geometric sequence with the first term $ \frac {1}{2},$ common ratio $ \minus{} \frac {1}{2}$ and the number of the term $ 7,$ so we can write the sum with sum notation, $ \sum_{k \equal{} 1}^7 \frac {1}{2}\left( \minus{} \frac {1}{2}\right)^{k \minus{} 1} \equal{} \minus{} \sum_{k \equal{} 1}^7 \left(\frac {1}{2}\right)^k$

ahh i see....I was doing it wrong.

Thank you very much.


Prove that there are an infinite number of primes

[quote="h_s_potter2002"][hide]Let's say that there are $k$ primes, where ${p_n}$ is prime.  Then ${p_1}{p_2}{p_3}\ldots{p_k}+1$ wouldn't be divisible by any of the primes (since ${p_1}{p_2}{p_3}\ldots{p_k}+1 \equiv 1 \mod{p_n}$), creating a new prime.[/hide][/quote]
correct  :roll:

I would just like to remind everyone to watch the level of problems that are posted in the Getting Started forum.  Most proofs should be posted in the Intermediate forum or above.  This is a famous proof and is often times used as an introduction to the method of proof by contradiction.  This proof is actually an ingenious work of Euclid.

[quote="fourierseries"]There is an interesting proof that involves point set topology:

Given a,b integers

S(a,b) := {a+nb|n integer}

A topology on the integers can be defined by:

A subset N of the integers is a neighborhood of an integer a iff there exists a b>0 s.t. S(a,b) is a subset of N

S(a,b) is both an open and a closed set by definition.

Now we note that

Z \ {-1,1} = union    S(0,p)
                  primes p

If there are finitely many primes, then there is a contradiction.[/quote]
What are you talking about? This is Intermediate Topics. You can mention it can be proved with something advanced like point set topology? But don't post the whole proof whatever your intent is. Sorry, but I did not understand your proof.

[Edit]Please read my signature if you don't understand what I mean


Whats the...
      Sin
    /
60--Tan
    \
      Cos

and

      Sin
    /
90--Tan
    \
      Cos

$ \sin 60 \equal{} \frac{\sqrt{3}}{2}$
$ \cos 60 \equal{} \frac{1}{2}$
$ \tan 60 \equal{} \sqrt{3}$

$ \sin 90 \equal{} 1$
$ \cos 90 \equal{} 0$

Is this what you are looking for?

$ \sin 30 \equal{} \frac {1}{2}$
$ \cos 30 \equal{} \frac {\sqrt {3}}{2}$
$ \tan 30 \equal{} \frac {\sqrt {3}}{3}$

To help remember these think of a 30 - 60 - 90 triangle.

Another helpful hint may be $ \cos (90 \minus{} x) \equal{} \sin x$ and $ \sin (90 \minus{} x) \equal{} \cos x$

These should be in your textbook and you should memorize them.

An easy way to memorize $ \sin$.

[hide]
$ \sin 0^{\circ} \equal{} \frac {\sqrt {0}}{2}$.

$ \sin 30^{\circ} \equal{} \frac {\sqrt {1}}{2}$

$ \sin 45^{\circ} \equal{} \frac {\sqrt {2}}{2}$

$ \sin 60^{\circ} \equal{} \frac {\sqrt {3}}{2}$

$ \sin 90^{\circ} \equal{} \frac {\sqrt {4}}{2}$
[/hide]

Similarly we can have the same thing, except backwards, for $ \cos$.

[hide]
$ \cos 0^{\circ} \equal{} \frac {\sqrt {4}}{2}$.

$ \cos 30^{\circ} \equal{} \frac {\sqrt {3}}{2}$

$ \cos 45^{\circ} \equal{} \frac {\sqrt {2}}{2}$

$ \cos 60^{\circ} \equal{} \frac {\sqrt {1}}{2}$

$ \cos 90^{\circ} \equal{} \frac {\sqrt {0}}{2}$
[/hide]

Now once you have those memorized (which are really easy - just look at the numerators) you can perform $ \frac {\sin \theta}{\cos \theta} \equal{} \tan \theta$. Where the denominators should cancel out (seeing as they're all $ 2$), leaving the numerators.  :)

I would actually use the unit circle:

[img]http://dcr.csusb.edu/LearningCenter/LCimages/UnitCircle.gif[/img]

[quote="airyork6=)"]oh and also do 30 sin,cos,tan....[/quote]

That, I think is what you're supposed to see by doing the problem.

For any angle, $ x$ the following identities hold:

\begin{eqnarray} \cos\left(x\right) =\sin\left(x + 90^{\circ}\right) \\ \sin \left(x\right) = \cos\left(x - 90^{\circ}\right) \end{eqnarray}

So,

\[ \begin{array}{lc} \sin 60^{\circ} = \cos \left(-30^{\circ}\right) = \frac{\sqrt{3}}{2} & \cos 60^{\circ} =\sin\left(150^{\circ}\right) =\frac{1}{2} \\ \sin 90^{\circ} =\cos 0^{\circ} = 1 & \cos 90^{\circ} =\sin 180^{\circ} = 0\end{array}\]

The other thing to know is: $ \begin{array}{lc} \cos\left(-x\right) =\cos\left(x\right) & \sin\left(-x\right) = -\sin\left(x\right)\end{array}$

From this you can tell that    $ \cos\left(-30^{\circ}\right) =\cos\left(30^{\circ}\right) =\frac{\sqrt{3}}{2}$

Also by definition, $ \tan x =\frac{\sin x}{\cos x}$.

$ \tan 60^{\circ} =\frac{\sin 60^{\circ}}{\cos 60^{\circ}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$

$ \tan 90^{\circ} =\frac{\sin 90^{\circ}}{\cos 90^{\circ}} = \frac{1}{0}$ -- undefined.

$ \tan 30^{\circ} =\frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} =\frac{1}{\sqrt{3}}$


Bill walks $ \frac{1}{2}$ mile south, then $ \frac{3}{4}$ mile east, and finally $ \frac{1}{2}$ mile south. How many miles is he, in a direct line, from his starting point?  Express your answer as a decimal to the nearest hundredth.

You can draw a picture, to get a sort of zigzag line. Now extend the lines to get a right triangle with longer leg 1 and shorter leg $ \frac{3}{4}$. By the pythagorean theorem, our answer is $ 1.25$.

We construct two triangles by connecting the end point to the starting point. By $SAS$ Theorem, we know the triangles are congruent.

Dividing $\frac{3}{4}$ by $2$, we get a leg lengths of $\frac{3}{8}$ and the other leg lengths is given to us as $\frac{1}{2}$.

Using the Pythagorean Theorem, we get a hypotenuse lengths of $\frac{5}{8}$ but because we want the total lengths from the starting point to the ending point, we must multiply that number by $2$.

Hence $\frac{5}{4}=\boxed{1.25}$.


Δίνεται τρίγωνο $ ABC$, η διχοτόμος $ AD$ και η διάμεσος $ AM$. Ο κύκλος που γράφεται περί των $ A,D,M$ τέμνει την $ AB$ στο $ P$, την $ AC$ στο $ Q$ και τον περιγεγραμμένο κύκλο του $ ABC$ στο $ E$. Η $ BQ$ και η $ CP$ τέμνονται στο $ K$. Να αποδειχθεί ότι η κάθετη από το $ M$ στην $ AE$ διχοτομεί το τμήμα $ AK$.

Εστω $ L$ το μεσο της $ AK$ και $ G,O$ τα περικεντρα των $ ADM,ABC$ αντιστοιχως.
Αρκει να δειξουμε οτι $ ML$ ειναι παραλληλη στην $ GO$.

Το $ APKQ$ ειναι πληρες τετραπλευρο οποτε εφοσον τα $ M,L$ ειναι τα μεσα των δυο διαγωνιων του, η $ ML$ τεμνει την $ PQ$ στο $ H$ που ειναι το μεσο της.
Οι $ AD,OM$ συντρεχουν πανω στον περιγεγραμμενο κυκλο του $ ABC$ στο $ I$.
Εχουμε οτι $ \angle{HGQ} \equal{} \angle{MOC} \equal{} \frac {a}{2}$. Αρα το τριγωνα $ HGQ$ και $ MOC$ ειναι ομοια.
Αρα $ \frac {OM}{GH} \equal{} \frac {R}{r} \equal{} \frac {OI}{GD}$.
Απο θεωρημα θαλη εχουμε οτι $ OG//ML//DI$. Αρα $ ML$ καθετη στην $ AE$.  [geogebra]2d435e2e32f38ce69f998568864605dffd827046[/geogebra]

$ \bullet$ Από $ (BP)\cdot (BA) \equal{} (BD)\cdot (BM)$ $ ,(1)$ και $ (CQ)\cdot (CA) \equal{} (CM)\cdot (CD)$ $ ,(2)$ προκύπτει ότι $ \frac {(BP)\cdot (BA)}{(CQ)\cdot (CA)} \equal{} \frac {(BD)\cdot (BM)}{(CM)\cdot (CD)}$ $ \Longrightarrow$ $ BP \equal{} CQ$ $ ,(3)$ γιατί $ BM \equal{} CM$ και $ \frac {BA}{CA} \equal{} \frac {BD}{CD}$ λόγω της διχοτόμου $ AD$ του $ \bigtriangleup ABC.$

Από την $ (3)$, σύμφωνα με το παρακάτω [b]Λήμμα[/b], συμπεραίνουμε ότι $ A'K\parallel AD,$ όπου $ A'$ είναι το συμμετρικό του $ A$ ως προς το μέσον $ M$ της πλευράς $ BC$ του δοσμένου τριγώνου $ \bigtriangleup ABC.$

$ \bullet$ Έστω $ Z$  προβολή του $ M$ επί της ευθείας $ AE$ και εύκολα αποδεικνύεται ότι $ MZ\parallel AD,$ γιατί $ AD\perp AE,$ λόγω του $ E$ ως το αντιδιαμετρικό του $ F\equiv AD\cap OM$ ( αντί του $ I$ στη λύση του [b]Σάκη[/b] ), όπου $ O$ είναι το περίκεντρο του $ \bigtriangleup ABC.$

Άρα, στο τρίγωνο $ \bigtriangleup KAA',$ η δια του μέσου $ M$ της πλευράς του $ AA'$ παράλληλη ευθεία προς την $ KA'\parallel AD\perp AE,$ περνάει από το μέσον $ L$ της $ KA$ και η πρόταση έχει αποδειχθεί.

[b][size=100][color=DarkBlue]ΛΗΜΜΑ. - Επί των πλευρών $ AB,\ AC$ τριγώνου $ \bigtriangleup ABC,$ λαμβάνουμε τα σημεία $ P,\ Q$ αντιστοίχως, ώστε να είναι $ BP \equal{} CQ.$ Αποδείξτε ότι το σημείο $ K\equiv BQ\cap CP$ ανήκει στην παράλληλη ευθεία προς την διχοτόμο της γωνίας $ \angle A,$ από το σημείο $ A',$ συμμετρικό του $ A$ ως προς το μέσον $ M$ της πλευράς του $ BC.$[/color][/size][/b]

Εύχομαι σε όλους Καλή Χρονιά, με Υγεία, Ειρήνη, Δημιουργικότητα και Προκοπή.

Κώστας Βήττας.

ΥΓ(1). Θα βάλω αργότερα την απόδειξη που έχω υπόψη μου για το παραπάνω [b]Λήμμα.[/b]

ΥΓ(2). Μήπως μπορεί κάποιος να μου πει γιατί δεν μπορώ να ανοίξω την εικόνα στη λύση του Σάκη ; Ευχαριστώ.

[quote="Sakis"]
Αρα $ \frac {OM}{GH} \equal{} \frac {R}{r} \equal{} \frac {OI}{GD}$.
Απο θεωρημα θαλη εχουμε οτι $ OG//ML//DI$. Αρα $ ML$ καθετη στην $ AE$.  [/quote]
[quote="Protonios"]That is so true  :)[/quote]

Ρε παιδιά δεν ξέρω αν λόγω ξενυχτιού θα βλέπω κάτι προφανές αλλά δεν μου φαίνεται σωστό!
Από τους λόγους και το θεώρημα Θαλή ΔΕΝ συνεπάγεται ότι οι 3 ευθείες είναι παράλληλες! Θα' πρεπε να χεις ήδη 2 απ' τις 3 παράλληλες για να καταλήξεις ότι θα ναι και η τρίτη παράλληλη  :wink:
Για να βγάλεις $ ML//DI$ χρησιμοποιήστε ότι:
[hide]$ BP\equal{}CQ\iff MH//AD$[/hide]

Καλή και δημιουργική χρονιά σε όλους!

Νομίζω ότι σωστά από την ισότητα $ \frac {OM}{GH} \equal{} \frac {R}{r} \equal{} \frac {OI}{GD}$ συμπεραίνεται ότι $ OG\parallel ML\parallel DI,$ γιατί ισχύει $ OG\parallel DI$, από $ AD\perp AE$ και $ OG\perp AE$ ( $ OG$ διάκεντρος και $ AE$ κοινή χορδή των δύο κύκλων ).

[quote="bilot4"]Για να βγάλεις $ ML//DI$ χρησιμοποιήστε ότι: $ BP \equal{} CQ\iff MH//AD$[/quote]
Αυτό είναι ένα άλλο χρήσιμο [b]Λήμμα[/b], που έχει δημοσιευτεί [url=http://www.mathematica.gr/forum/viewtopic.php?f=50&t=1813][b][color=Blue]Εδώ[/color][/b][/url] .

$ \bullet$ Αν και προσεγγίζεται απλούστερα το πρόβλημα που έχει τεθεί, χρησιμοποιώντας την ευθεία [b]Gaus[/b] στο πλήρες τετράπλευρο, σκέφτηκα ότι ήταν μία καλή ευκαιρία να δούμε το γνωστό [b]Λήμμα[/b] στο οποίο αναφέρθηκα πιο πάνω, χρήσιμο πιστεύω για το μέλλον.

Κώστας Βήττας.

Ναι πράγματι με αυτήν την δικιολόγηση ισχύει! Αλλά επειδή κατ' εμένα χρειάζεται δικιολόγηση ότι OG//DI για αυτό έκανα την παραπάνω παρατήρηση (όταν είπα λάθος πριν εννοούσα ατελές )

Υ.Γ. Για το σχήμα του Σάκη που ρωτήσατε μήπως δεν έχετε κατεβάσει το Java?

[quote="bilot4"]Υ.Γ. Για το σχήμα του Σάκη που ρωτήσατε μήπως δεν έχετε κατεβάσει το Java?[/quote]
Δημήτρη σ΄ευχαριστώ. Πράγματι δεν είχα το Java, αλλά τώρα είναι εντάξει.

Κώστας Βήττας.


Coplanar squares $ ABGH$ and $ BCDF$ are adjacent, with $ CD \equal{} 10$ units and $ AH \equal{} 5$ units. Point $ E$ is on segments $ AD$ and $ GB$. What is the area of triangle $ ABE$, in square units? Express your answer as a common fraction.

[asy]draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)--cycle,linewidth(2));
draw((5,0)--(15,0)--(15,10)--(5,10)--(5,0)--cycle,linewidth(2));

draw((0,0)--(15,10),linewidth(2));

filldraw((0,0)--(5,0)--(5,10/3)--(0,0)--cycle,gray,linewidth(2));

label("A",(0,0),SW);
label("B",(5,0),S);
label("C",(15,0),SE);
label("D",(15,10),NE);
label("E",(5,10/3),SE);
label("F",(5,10),NW);
label("G",(5,5),NW);
label("H",(0,5),NW);[/asy]

we can see that triangle ABE is similar to triangle ACD

now we fill in lengths

AB=5
AC=15
CD=10

$ \frac{15}{5}\equal{}\frac{10}{x}$ where x=BE

so $ x\equal{}\frac{10}{3}$

$ [ABE]\equal{}\frac{1}{2}\cdot\frac{10}{3}\cdot5\equal{}\frac{25}{3}$


induction
combinatorics proposed
combinatorics

Fifty teams participate in a round robin competition over 50 days. Moreover, all the teams (at least two) that show up in any day must play against each other. Prove that on every pair of consecutive days, there is a team that has to play on those two days.

Really wonderful problem!
I hope I have a solution, I will just state the main idea here.
We have to prove that if $ K_n$ is decomposed into $ n$ complete graphs then no 2 of them
are disjoint.
We can prove a stronger statement:
"If $ K_n$ is decomposed into less than $ n\plus{}k\minus{}1$ complete graphs, then no $ k$ of them
are pairwise disjoint (this holds for $ k\geq2$). Furthermore, there is no decomposition
into less than $ n$ graphs, besides the trivial one."
This statement can be proved by induction on $ n$, pretty straightforward.

Sorry for my poor combinatorics, shasha said using induction could prove the stronger statement , but I'm still puzzled.Could anyone say it more clearly? Many thanks.

This is nothing but a particular case of the de Bruijn-Erdös incidence theorem, stating that given a set of $n$ elements $A = \{a_1,a_2,\ldots,a_n\}$, and $m>1$ subsets of it, such that each pair of elements in $A$ is contained in one and only one subset, then $m\geq n$, with equality only in the cases when (up to a reindexing) the subsets are $\{a_1,a_n\}, \{a_2,a_n\}, \ldots, \{ a_{n-1}, a_n\}$ and $\{a_1,a_2,\ldots,a_{n-1}\}$, or when $n=k^2+k+1$ for some integer $k$, each subset contains $k+1$ elements, each element is contained in $k+1$ subsets, and any two subsets intersect (in precisely one element, of course). See [url]http://www.renyi.hu/~p_erdos/1948-01.pdf[/url].

The elements are the teams, and the subsets are the days, made of the players that showed up that day.

Clearly this is also saying things about finite projective planes. For the last case, which is of most interest, it is known a model exists when $k$ is a power of a prime, and no other possibility is yet known to occur.

Unfortunately, the problem setters either had no knowledge of this, or botched it! First, they only talk about consecutive days; in fact this is true for any two days. Second, $50$ is not of the form $k^2+k+1$ (so the issue of $k$ being a power of a prime is moot); therefore the only possibility is the first one mentioned - in one day all teams but one showed up, and in any other day, that last team showed up, with just, each day, another of the first teams. Therefore complete knowledge is acquired on the possible round-robin. It was not so difficult for $50$ to have been replaced by $57 = 7^2 + 7 + 1$, eh?


Last time I checked this was not a place for people to post a single post advertising something.

[quote="TheMage"]Mage's Maze is more than a riddle game. Besides solving the regular enigmas, you have to find your way through the maze, in order to face The Mage and become his apprentice.

Good Luck.

http://magesmaze.50webs.com/[/quote]
yet another riddle site...

anyways I'm already stuck on the first on what does it mean by fridays' friend?


[quote="lipako"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {a^2 \plus{} b}{b(a^2 \plus{} ab \plus{} b^2)} \plus{} \frac {b^2 \plus{} c}{c(b^2 \plus{} bc \plus{} c^2)} \plus{} \frac {c^2 \plus{} a}{a(c^2 \plus{} ca \plus{} a^2)}\geq2$[/quote]

My solution  :oops:
$ LHS \equal{} \sum {\frac{{a^2 }}{{b(a^2  \plus{} ab \plus{} b^2 )}}}  \plus{} \sum {\frac{1}{{a^2  \plus{} ab \plus{} b^2 }}}$
Setting $ ab \plus{} bc \plus{} ac \equal{} X$
By Cauchy-Schwarz
$ \sum {\frac{{a^2 }}{{b(a^2  \plus{} ab \plus{} b^2 )}}}  \ge \frac{{\left( {\sum a } \right)^2 }}{{\sum {ab(a \plus{} b)}  \plus{} \sum {a^3 } }} \equal{} \frac{{\left( {\sum a } \right)^2 }}{{\left( {\sum a } \right)\left( {\sum {ab} } \right) \plus{} \left( {\sum a } \right)\left( {\left( {\sum a } \right)^2  \minus{} 3\left( {\sum {ab} } \right)} \right)}} \equal{} \frac{9}{{3X \plus{} 3(9 \minus{} 3X)}} \equal{} \frac{3}{{9 \minus{} 2X}}$
$ \sum {\frac{1}{{a^2  \plus{} ab \plus{} b^2 }}}  \ge \frac{9}{{2\sum {a^2 }  \plus{} \sum {ab} }} \equal{} \frac{9}{{2\left( {\sum a } \right)^2  \minus{} 3\left( {\sum {ab} } \right)}} \equal{} \frac{9}{{18 \minus{} 3X}} \equal{} \frac{3}{{6 \minus{} X}}$
We must to prove this ineq
$ \frac{3}{{9 \minus{} 2X}} \plus{} \frac{3}{{6 \minus{} X}} \ge 2$
$ \Leftrightarrow (X \minus{} 3)(21X \minus{} 4) \ge 0$
which is true because by Cauchy-Schwarz
$ 3X \le \left( {\sum a } \right)^2  \equal{} 9$
$ \Rightarrow X \le 3$

[quote="lipako"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {a^2 \plus{} b}{b(a^2 \plus{} ab \plus{} b^2)} \plus{} \frac {b^2 \plus{} c}{c(b^2 \plus{} bc \plus{} c^2)} \plus{} \frac {c^2 \plus{} a}{a(c^2 \plus{} ca \plus{} a^2)}\geq2$[/quote]
This is my solution!  :D
By Cauchy-Schwarz we have:
$ 3LHS \equal{} \sum \frac {3a^2 \plus{} b(a \plus{} b \plus{} c)}{b(a^2 \plus{} b^2 \plus{} c^2)} \equal{} \sum \frac {1}{b} \plus{} \sum \frac {2a^2}{b(a^2 \plus{} ab \plus{} b^2)} \plus{} \sum \frac {c}{a^2 \plus{} ab \plus{} b^2} \ge \sum \frac{1}{b} \plus{} \frac {2(a \plus{} b \plus{} c)^2}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {(a \plus{} b \plus{} c)^2}{(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)} \equal{} \sum \frac {1}{b} \plus{} \frac {6}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {3}{ab \plus{} bc \plus{} ca}$
Beside we also have:
$ (ab \plus{} bc \plus{} ca)^2 \ge 3abc(a \plus{} b \plus{} c) \Rightarrow \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge \frac {9}{ab \plus{} bc \plus{} ca}$
So we only need to improve:
$ \frac {6}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {12}{ab \plus{} bc \plus{} ca} \ge 6$
It's true by Cauchy-Schwarz

[quote="mathstarofvn"][quote="Lanyes"]
We must to prove this ineq
$ \frac {3}{{9 \minus{} 2X}} \plus{} \frac {3}{{6 \minus{} X}} \ge 2$
$ \Leftrightarrow (X \minus{} 3)(21X \minus{} 4) \ge 0$
which is true because by Cauchy-Schwarz
$ 3X \le \left( {\sum a } \right)^2 \equal{} 9$
$ \Rightarrow X \le 3$[/quote]
I think you were wrong![/quote]

Sorry. I've had some mistakes :blush:  :blush:  :blush:

Hi,

Can someone please expand how this is true ?
By Cauchy-Schwarz we have:

$ 3LHS = \sum \frac {3a^2 + b(a + b + c)}{b(a^2 + b^2 + c^2)} = \sum \frac {1}{b} + \sum \frac {2a^2}{b(a^2 + ab + b^2)} + \sum \frac {c}{a^2 + ab + b^2} \ge \sum \frac{1}{b} + \frac {2(a + b + c)^2}{(a + b + c)(a^2 + b^2 + c^2)} + \frac {(a + b + c)^2}{(ab + bc + ca)(a + b + c)} = \sum \frac {1}{b} + \frac {6}{a^2 + b^2 + c^2} + \frac {3}{ab + bc + ca}$

[quote=lipako]If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$, then

$ \frac{a^2\plus{}b}{b(a^2\plus{}ab\plus{}b^2)}\plus{}\frac{b^2\plus{}c}{c(b^2\plus{}bc\plus{}c^2)}\plus{}\frac{c^2\plus{}a}{a(c^2\plus{}ca\plus{}a^2)}\geq2$[/quote]

After homogeneous,the inequality becomes
\[\sum{\frac{3a^2+b(a+b+c)}{b(a^2+ab+b^2)}}\cdot (a+b+c)\ge{18}\]
\[<=>LHS-RHS=\frac{1}{2}\sum{\frac{(a^3b+13a^2b^2+6a^2c^2+ab^3+24abc^2+6b^2c^2+12c^4)(a-b)^2}{ab(b^2+bc+c^2)(a^2+ac+c^2)}}\ge{0}\]

[quote=szl6208][quote=luofangxiang]The following inequality is also true
[quote=luofangxiang][img]http://s14.sinaimg.cn/middle/0018zOAxgy71QX7erCdcd&690[/img][/quote]
[/quote]
We have
\[\sum{\frac{3a^2+kb(a+b+c)}{3b(a^2+ab+b^2)}}-\frac{3(1+k)}{a+b+c}\]
\[=\frac{1}{6(a+b+c)}\sum{\frac{(a^3bk+7a^2b^2k+ab^3k+6a^2bc+6a^2c^2+6ab^2c+18abc^2+6b^2c^2+12c^4)(a-b)^2}{ab(b^2+bc+c^2)(a^2+ac+c^2)}}\ge{0}\][/quote]

展开关于k单调，可直接分解。


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

$ \int{\sqrt{\frac{x}{Ax\plus{}B}}}\,\mathrm{d}x\quad\mbox{where }A$ and $ B$ are constants and $ B>0$.

[hide="This might help"]Make $ \frac {x}{ax \plus{} b} \equal{} t^2 \equal{} > x \equal{} \frac {bt^2}{1 \minus{} at^2} \equal{} > dx \equal{} \left(\frac {2bt}{1 \minus{} at^2} \plus{} \frac {2abt^3}{(1 \minus{} at^2)^2}\right)dt$, and so

$ P \equal{} \int \sqrt {\frac {x}{ax \plus{} b}}\,dx \equal{} \underbrace{\int\frac {2bt^2}{1 \minus{} at^2}\,dt}_{ \equal{} \,P_1} \plus{} \underbrace{\int\frac {2abt^4}{(1 \minus{} at^2)^2}\,dt}_{ \equal{} \, P_2}$.

[hide="P1"]$ P_1 \equal{} \minus{} \frac {2b}{a} \int\frac {at^2}{at^2 \minus{} 1}\,dt \equal{} \minus{} \frac {2b}{a} \left(t \plus{} \int\frac {1}{at^2 \minus{} 1}\,dt\right)$

$ \equal{} \minus{} \frac {2b}{a} t \minus{} \minus{} \frac {2b}{a\sqrt {a}} \ln \left|\frac {\sqrt {a}t \plus{} 1}{\sqrt {at^2 \minus{} 1}}\right| \plus{} C$.[/hide]

[hide="P2"]$ P_2 \equal{} 2ab \int\frac {t^4}{(1 \minus{} at^2)^2}\,dt \underbrace{ \equal{} }_{t \equal{} \frac {1}{\sqrt {a}} \sin u} \frac {2b}{a\sqrt {a}} \int \frac {\sin^4u}{\cos^3 u}\,du$

$ \equal{} \frac {2b}{a\sqrt {a}} \int\frac{(1 \minus{} \cos^2 u)^2}{\cos^3u}\,du \equal{} \frac {2b}{a\sqrt {a}} \int\frac{1 \minus{} 2 \cos^2u \plus{} \cos^4u}{\cos^3u}\,du$

$ \equal{} \frac {2b}{a\sqrt {a}} \left(\int \sec^3u\,du \minus{} 2 \int \sec u\,du \plus{} \sin u\right)$

and you can finish from here.[/hide][/hide]

hello, by substituting $ t=\sqrt{\frac{x}{Ax+B}}$ you get $ x=\frac{Bt^2}{1-At^2}$ and it follows $ dx=\frac{2tB}{(t^2A-1)^2}dt$.
Your integral becomes
$ \int\frac{2Bt^2}{(t^2A-1)^2}dt$
Sonnhard.


[quote="GameBot"]A cube has a surface area of $ 900 \text{cm} ^{2}$. What is the number of cubic centimeters in the volume of the cube? Express your answer in simplest radical form.[/quote]

$ 6x^2\equal{}900 \implies x^2\equal{}150 \implies x^3\equal{}150\sqrt{150}\equal{}750\sqrt{6}$


What is the sum of the number of faces, vertices and edges in a triangular prism?

[hide="Solution"]There are 5 faces, two bases and three faces connecting those bases.
There are 6 vertices, 3 for each base.
There are 9 edges, 3 for each base plus three connecting the vertices of the bases.
5+6+9=20.[/hide]

[quote="236factorial"]this is boring, how about the sum of the number of faces, vertices and edges in a rectangular prism?[/quote]
[hide]
$6+8+12=26$
[/hide]


Solve for $x$ in the following equation:

$2^{x^{2^{x^{2^{x \cdots}}}}}= 3^{x}$

Notice that we are to solve for $x$ in $2^{x^{3^{x}}}=3^{x}$.  For $x<0$, $2^{x^{3^{x}}}>3^{x}$. Clearly, a solution is $x_{1}=0$.  For $0<x\leq 1$, $3^{x}>2^{x^{3^{x}}}$.  Using a calculator, I found that another zero is $x_{2}\approx 1.187145$.  Notice that $2^{x^{3^{x}}}>3^{x}$ for all real $x>x_{2}$, so there are only two solutions.

Yes, that is why your answer is not what was required; I never said you were not making an assumption.  By the problem's [i]obvious[/i] requirement, your solution is incorrect.  However, I completely understand that that was a misunderstanding on your side.

[quote="miyomiyo"][quote="boxedexe"]$x_{2}\approx 1.187145$.[/quote]I found this root using a calculator also, but what is it in terms of $ln$'s and stuff?[/quote]

Equations of this form typically do not have elementary solutions, i.e. the inverse function of $x^{3^{x}-1}$ cannot actually be expressed in terms of logarithms, exponentials and polynomials.  It would be rather remarkable if this number could be so expressed.


Hi!

We assume that for every real number $x$ we have
\[\vert ax^{2}+bx+c\vert\leq \vert dx^{2}+ex+f\vert\]
where $a,b,c,d,e,f$ are real numbers. Prove that :
\[\vert b^{2}-4ac\vert\leq \vert e^{2}-4df\vert.\]

[quote="Moubinool"]putnam 2003 problem A4

I see you read RMS[/quote]

what is RMS? :maybe:


limit
trigonometry
calculus
calculus computations

[quote="misan"]prove:
$\lim_{k\to\infty}\left|\left( \tan{x}\;\cdot\;\sqrt{\tan{(2x)}}\;\cdot\;\sqrt[4]{\tan{(4x)}}\;\cdot\;\sqrt[8]{\tan{(8x)}}\;\cdot\;\;\;...\;\;\;...\;\;\;\cdot\;\sqrt[2^{k}]{\tan{(2^{k}x)}}\;\right) \right| \;\\ \\ \\ \hspace* \;\;\;\;\;\quad\;\;\;\;\quad\;\;\;\;\;\quad\;\;\;\;\;\quad=\;\boxed{4\;\sin^{2}{x}}$
(PS: you will need to give a criterion under which [u]each[/u] of the [b]individual[/b] terms of the infinite sequence is properly defined...)[/quote]

I need some references regarding [b]"Dobinski  identity"[/b]. His/her/ first name is [b]G.[/b] ?
Thanking in advance,Alex


geometry
calculus
integration
calculus computations

Hi all,

I know that in order to find the area between two curves I can just integrate both and take the difference, but I am unclear on how to find the bounds for the integration on this particular problem.

Here are the curves:

$y=2+e^{-x}$
$y=1+x^{2}$

I think I am supposed to do something like

$2+e^{-x}= 1+x^{2}$ then find out what values of x make sense, but I get stuck here.

Any help is appreciated!

Thanks a lot

Are you sure that's the right question? There's only one intersection point, and it can't be found exactly- you need a numerical method. I have no idea where the other limit is supposed to come from.

we cant find a explicit formula for x. just a aproximation

after this you can integrate with this limit.

but is just a aproximation. dont have a explicit formula

and i dont know if are complex roots

That's exactly what the problem states.  It must be a mistake.  I thought it was not right, but I wasn't sure.

Thanks a lot


floor function
logarithms
function
number theory proposed
number theory

hi!
$\omega (n) $ denote the numbers of distinct prime factors of $n$

show that :

$\sum_{n \leq x} 2^{\omega (n) } = \frac{6}{\pi^2}xlogx + O(x)$

wl

and... what is $O(x)$ supposed to be? :?

It's a notation regarding sizes of functions. $f(x)=O(g(x))$ means that $\frac{f(x)}{g(x)}$ is bounded.

I gave it a try, but I might be wrong. I'm very inexperienced in analyitc number theory.

$2^{\omega(n)}$ is the number of sets of prime divisors of $n$, so it's the number of square-free divisors of $n$. Let $\{a_k\}$ be the sequence of square-free numbers.

Each square-free number $a_k$ appears in the set $\{1,2,\ldots,\lfloor x\rfloor\}$ exactly $\lfloor\frac x{a_k} \rfloor$, so what we have to show is that $\displaystyle \frac{\sum_{a_k\le x}\lfloor\frac x{a_k} \rfloor}x-\frac 6{\pi^2}\ln x=O(1)$. Since $\sum_{a_k\le x}\frac 1{a_k}-\displaystyle \frac{\sum_{a_k\le x}\lfloor\frac x{a_k} \rfloor}x=O(1)$ and $\sum_{n\le x}\frac 1n-\ln x=O(1)$, what we need to show is $\frac{\pi^2}6\sum_{a_k\le x}\frac 1{a_k}-\sum_{n\le x}\frac 1n=O(1)$.

If $Q(x)$ is the number of square-free numbers which are not $>x$, then it's a well-known result that $Q(x)=\frac 6{\pi^2}x+O(\sqrt x)$. From this we can derive $a_k=\frac{\pi^2}6k+O(\sqrt k)$. I think it should pretty clear now that $\sum_{a_k\le x}\frac 1{k+O(\sqrt k)}-\sum_{n\le x}\frac 1n=O(1)$.

Think it works?

hi! :D
it works ( and I like very much your solution) ...

this is my solution ... ( sorry for my poor english)
 $\ast $ is the Dirichlet moltiplication , is easy to prove that  $2^{\omega (n)} = N_0 \ast  \mu ^2 (n) $ , but I think is the key of the problem.

so we have  : $\sum_{n \leq x } 2^{\omega (n)}  = \sum_{n \leq x } N_0 \ast \mu^2 (n)$ now we use  a  particular case of Dirichlet's hyperbola metod :

if  $f(n) , g(n) $ are arithmetic functions  $\sum_{n \leq x } (f \ast g )(n) = \sum_{n \leq x} f(n) G(\frac{x}{n})$ where  $G(y) = \sum_{n \leq y} g(n)$ .

so $ \sum_{n \leq x } N_0 \ast \mu^2 (n) = \sum_{n \leq x } \sum_{m \leq x/n } \mu^2(m) = \sum_{n \leq x } ( \frac{6}{\pi^2} \frac{x} {n} + O((\frac{x}{n})^{1/2}) )$

$x\sum_{n \leq x } ( \frac{6}{\pi^2 n} )+ O(x)  =  \frac{6}{\pi^2}x logx + O(x)$

wl


[hide="solution"]we have $\cos\frac{4\pi}{7}=-\cos(\pi-\frac{4\pi}{7})=-\cos \frac{3\pi}{7}$. now use $\cos2x=\cos^{2}x-\sin^{2}x$.[/hide]


Okay, I don't know if it's just me, but any website googley related is down (Gmail, Google, Froogle, Google Images, etc.) Everytime I try to load it I'm getting a "Cannot Display blah blah" page.
It was working fine last night; has this happened to anyone else?

Also, I tried Restarting my computer and modem, each of which didn't work.

[i]Edit:[/i] Plus, my ISP is Qwest.

Yeah, me too. That's weird.

EDIT: It's working for me, now.

Still not working for me. What did you do to get it to work?

I'm using google chrome, so i typed something in the search bar. That automatically searched it on google. That worked, so I clicked the google icon at the top of the search results page. That took me there.

Wow, that's strange. I just got on Chrome and google worked. Normally, I just randomly click one of the 3 browsers I have. :P Guess I'm going to have to start using Chrome.

What's Google Chrome? Is that the toolbar?

Google Chrome is a new broswer made by Google, just like Firefox.

I used to use Google Chrome before I switched to Linux. I found Google Chrome much faster than Firefox of IE. It is much more secure than other browsers because of its sandboxing.


Find all real values of t such the following sistem has real solutions:

\[\begin{cases}\left |x+\frac{1}{x}  \right |=\left | y+\frac{2}{y} \right |=\left | z+\frac{3}{z} \right |\\
x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} =t\end{cases}\]

I obtained an order 4 equation in z, with t parameter.
Nasty system, isn't it?

Someone told me that the problem is very simple, and the answer is that for every real t that system has solutions, but i cant prove this, so i leave it for someone smarter!


geometry conics parabola calculus integration

[quote="warut_suk"]Find the area enclosed by the parabola $x = y^2-4$ between $x=0$ and $x=1$.[/quote]
[hide]
using calculus, you can just integrate $\sqrt{x + 4} = y$ and multiply by 2.
$\displaystyle \int\sqrt{x+4}\,dx = \frac{2(x+4)^{\frac{3}{2}}}{3}$
$\displaystyle \frac{2(5)^{\frac{3}{2}}}{3} - \frac{2(4)^\frac{3}{2}}{3}$

so the answer is:
\[ \frac{4}{3}(5\sqrt{5} - 4\sqrt{4}) \]
[/hide]


ABC triangle Prove that:
 \sqrt a+ \sqrt b+ \sqrt c<2 \sqrt 3p

Is it blablabla < 2*(3p)^1/2 or blablabla < 2*3^1/2*p ? Because if it's the second one I think there's a problem concerning the degrees of the 2 terms, which are different (there goes my obsession about degrees again  :D )

Is p = a + b + c or 2p = a + b + c or something else ?

For all positive a, b, c we clearly have

\sqrta +  \sqrtb +  \sqrtc  \leq  \sqrt(3(a + b + c))

since the Cauchy - Schwarz - inequality gives

(\sqrta +  \sqrtb +  \sqrtc)2 \leq (1 + 1 + 1)(a + b + c).

So probably the solution has something to do with this.
Now the only thing that's annoying me is the factor 2 in 2\sqrt(3p).

(I hope I'm right about this because I'm not sure about the meaning of that "p".)

No matter what p is (a+b+c or (a+b+c)/2) the inequality is pretty weak. And, besides, we don't use the fact that a, b, c are the sides of a triangle..

So, Maverick, what do you mean with "p" ?

I think p=(a+b+c)/2. I'm not sure I have to check my materials again to be sure. I think this problem was proposed by Gh. Molea. I'll come back  to tell you for sure and also tell you the author.

p = (a+b+c)/2 and it is  \sqrt a+ \sqrt b+ \sqrt c<2(3p)^1/2
This problem was proposed by Gh.Molea in the MG in 1993.

Here's the solution:
 \sqrt a< \sqrt (b+c)
 \sqrt b< \sqrt (c+a)
 \sqrt c< \sqrt (a+b)

 \sqrt a+ \sqrt b+ \sqrt c< \sqrt (b+c)+ \sqrt (c+a)+ \sqrt (a+b) \leq  \sqrt (12p)=2 \sqrt (3p)


1. Find all polynomials $ P(x)$ satisfying $ P(x^2 \plus{} 1) \equal{} (P(x))^2 \plus{} 1$ for all $ x$
and $ P(0) \equal{} 0$

2. Find a polynomial of degree at most 3 such that $ p(2) \equal{} 3$, $ p(3)\equal{}5$, $ p(5) \equal{} 8$ and $ p(7) \equal{} 13$

3. Let $ a\equal{} 2^{1/3} \plus{} 5^{1/2}$. Find a polynomial $ p(x)$ with integer coefficients
satisfying $ p(a) \equal{} 0$.

1. If $ P\left( {{x_0}} \right) \equal{} {x_0}$, so $ P\left( {{x_0}^2 \plus{} 1} \right) \equal{} {x_0}^2 \plus{} 1$. This means the polynomial Q(x)=P(x)-x has infinite roots, which implies P(x)=x for all x
2. You can use Lagrange interpolation formula
3. $ a \equal{} \sqrt[3]{2} \plus{} \sqrt 5  \Rightarrow {\left( {a \minus{} \sqrt 5 } \right)^3} \equal{} 2 \Rightarrow \left( {{a^3} \plus{} 15a \minus{} 2} \right) \equal{} \sqrt 5 \left( {3{a^2} \plus{} 5} \right)$
Squaring both sides, we have the polynomial satisfy f(a)=0

For numer 1, how do we know that P(x)=x is the only solution. What's a rigorous proof?

[hide="This is what dvtrung meant"]
Let a sequence $ a_0, a_1, a_2, a_3, a_4, \ldots$ be $ 0, 1, 2, 5, 26, \ldots$ where $ a_0\equal{}0$ and $ a_{n\plus{}1} \equal{} a_n^2\plus{}1$ for all $ n\geq0$.
Then, by induction, $ P(a_n) \equal{} a_n$ for all $ n\geq0$.
The polynomial $ Q(x) \equal{} P(x)\minus{}x$ then has an infinite number of roots (at $ a_0, a_1, \ldots$), so $ Q(x) \equal{} 0 \implies P(x) \equal{} x$.[/hide]


What is the value of $ \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \ldots \times \frac{8}{9}$? Express your answer as a common fraction.

Note that everything cancels out except the $ 2$ and the $ 9$. Therefore, $ \frac{2}{3}\times{\frac{3}{4}}\times{\frac{4}{5}}\times\ldots\times{\frac{8}{9}}\equal{}\boxed{\frac{2}{9}}$.


Which 2 courses are you guys going to pick? Where do you check about what courses you take because I looked at the AMSP website and I did not see the link?
 :?:

The Academic Curriculum has the information you're looking for:

http://www.awesomemath.org/AMSP_dailysch.pdf

Don't be dependent on that table of courses in the link because they recently change it for people going to Dallas.

Yes, my friend Hassan - the ghost assistant (lol inside joke) is right. They recently sent out an email about a week ago to all people attending, and the course schedule looks EXTREMELY different from the one provided for 2008 (the link that someone posted).

BTW, about the courses themselves now.

I am unsure about exactly which courses I'm going to take - and sometimes you can't take all you like because some classes are morning and others are afternoon.

Some course plans that I like for myself:

Any combination of 1 each from morning and afternoon :

MORNING: Algebra 2.5 OR Counting Strategies (intermediate combinatorics)

AFTERNOON: Number Sense (beginning number theory) OR Modular Arithmetic (intermediate number theory)

Not sure about number theory - not too good at it either, so.

I don't know how advanced you are in your mathematical abilities, so I'm just informing you on what I am taking.

Hope I was helpful!

Sorry, I did not link to the correct academic schedule. The courses offered are the same as in 2008, but they're offered at different times.

Does anyone know if the Texas and California schedules are going to be the same (or very similar)?  If so, could one of the students going to Texas please post the course schedule?

No one can know right now. And remember if it was very similar one change may ruin your plan. So waiting is the best thing.

Here is the schedule for the Dallas site.

[quote="sjyoo93"]Yes, my friend Hassan - the ghost assistant (lol inside joke) is right. They recently sent out an email about a week ago to all people attending, and the course schedule looks EXTREMELY different from the one provided for 2008 (the link that someone posted).[/quote]

thats strange, i dont recall getting an email from them.

@iYOA: Are you attending in Texas or California? If Texas e-mail them because maybe your e-mail was mistakenly not included.

Anyway I will send it to you now.

im going to texas.

i think i should ask my parents because it was their email account but i dont think they got it or else they would have told me.

im not sure if this was a reason for not getting the email, but i mailed the consent forms on the 10th but i think it would have reached them on the 13th which would be late.

I don't believe this is the problem. Maybe they just forgot.

[quote="varunrocks"]Is there a way I can do Algebra 2.5 or 3.5 and counting strategies even though the time does not allow it. Because I really set my goals to.[/quote]

If they're in the same time slot, i dont think so. sorry.
unless you can be in two places at once....


1. The average (arithmetic mean) of the test scores of a class of p students is 70, and the average of the test scores of a class of n students is 92. When the scores of both classes are combined, the average score is 86. What is the value of p/n ??

2. In a group study on nutrition, the average daily intake of calories per person was 8 percent higher in April than it was in March. If this average was 3200 calories in March, what was the average daily intae of calories per person in April?

3. Smith ran 10 mile distance at 5 miles per hour and then returned to the starting place at 3 miles per hour along the same route. What is the average speed rate for the round trip? I got 6 miles per hour. Is that correct???

[hide="number 2"]
the average in April was 8% higher than march. So 8% of 3200 is 256.  3200+256=$\boxed{3456}$[/hide]
[hide="number 3"]i think you do
$\frac{5+3}{2}$
that gives me an answer of 4 mph.
can you explain to me how you got 6?[/hide]

No. 3 is wrong for here it's not a time-average but a distance-average

[hide]answer :$ \frac{10+10}{ \frac{10}{5} +\frac{10}{3} } $ = $ \frac{15}{4} $[/hide]

[hide="Problem 1"]
The sum of the scores for the first class is $70p$.
The sum of the scores for the second class is $92n$.
$\frac{70p+92n}{p+n}=86$
$70p+92n=86p+86n$
$6n=16p$
$\frac{p}{n}=\frac{6}{16}=\boxed {\frac{3}{8}}$
[/hide]

Thank you so much.

For the "Smith ran 10 miles distance" I got 3.75 too. I know what I did wrong. Thank you to everyone else too^^


[quote="mayhemkerim"]Let x,y,z be distinct positive integers satisfying x - y + z = 1. Show that there are infinitely many triples (x,y,z) such that  xy is divisible by z or xz divisible y  or  yz  divisible x.[/quote]

$ x \equal{} ab \plus{} 1 \minus{} b$
$ y \equal{} ab$
$ z \equal{} b$

We have $ x \minus{} y \plus{} z \equal{}1$ and $ xy$ divisible by $ z$

Also if you mean "and" would be trivial again, just choose (x,y,z)=(1,y,y)  :huh:

[quote="mayhemkerim"]yz divisible  x?[/quote]

As bboypa said, your requirement contains "or" so $ xy$ divisible by $ z$ is enough.

[quote="mayhemkerim"]I must restatement my problem.Sorry...
Let x,y,z be distinct positive integers satisfying x - y + z = 1. Show that there are infinitely many triples (x,y,z) such that xy can divisible by z and xz divisible y and yz divisible x.[/quote]

Ok, then take :

Let $ a>1$ positive integer  :

$ x\equal{}a(a\plus{}1)$
$ y\equal{}(a\plus{}1)(a^2\plus{}a\minus{}1)$
$ z\equal{}a(a^2\plus{}a\minus{}1)$


Assume that we know $x^{3}+y^{3}=z^{3}$ just has trivial answers in the natural numbers (We are actually sure about it  :) ). Prove that answers of equation \[y^{2}=x^{3}-432\] in rational numbers , are $(x,y)=(12,36),(12,-36)$

It would be nice to see a proof using FLT is the case $n=3$. Anyway, the elliptic curve $y^{2}=x^{3}-432$ has rank 0 (according to MWRank) and torsion group of order 3, so there are no more integer solutions.

A solution using Fermat's last theorem: http://www.qbyte.org/puzzles/p078s.html


In 8th grade, my teacher told me an interesting problem:

In a party, the mother of a family talks to a math professor and asks him a math problem. She says, "I have 3 children, all of whom were born on the same day of the year, but not necessarily the same year. The product of their ages is 72 and the sum of their ages is their house number (the doctor knows their house number, but you guys don't). What are the ages of my children?"
The doctor ponders a minute, and then says, "That is not enough information."
Then the mother replies, "By the way, my oldest son has a pet iguana." And now, the doctor is able to answer the question.

The question (for you guys), is what is their house number?

We must find a house number $ n$ that has two possibilities with one being ruled out by the fact that there is an oldest child.
$ a\plus{}b\plus{}c\equal{}n$
$ abc\equal{}72$
$ 72\equal{}2^3\cdot3^2$
Our possible triples are:
$ (1,1,72)\equal{}74$
$ (1,2,36)\equal{}39$
$ (1,3,24)\equal{}28$
$ (1,4,18)\equal{}23$
$ (1,6,12)\equal{}19$
$ (1,8,9)\equal{}18$
$ (2,2,18)\equal{}22$
$ (2,3,12)\equal{}17$
$ (2,4,9)\equal{}15$
$ (2,6,6)\equal{}14$
$ (3,3,8)\equal{}14$
$ (3,4,6)\equal{}13$

The only pair with two matching house numbers are the ones with house number $ \boxed{14}$.

[hide]
This is a very popular problem that once wormed its way into a mock test! You have to list out of combinations of 3 numbers multiplying to 72 and listing out their sums. Then the sum with exactly TWO sets adding to that will be the answer, provided that one sum has a unique greatest term and the other doesn't. By the way, the answer is $ 3 \plus{} 3 \plus{} 8 \equal{} 14$.[/hide]


Here is another physics problem, sth went wrong with my solution though.

I don't get how the momentum conservation worked in the second case... If there were no cord, then $ v_f = v'_f = \frac12 v_0$, but there is a cord, so I'm not pretty sure. Anyway:

[hide]
1st case: conservation of momentum

$ mv_0 = 4mv_1$
$ v_1 = \frac {v_0}{4}$

Conservation of energy

$ \frac {mv_0^2}{8} = E_p$

2nd case: conservation of energy

$ \frac12 mv_0^2 - \frac {mv_0^2}{8} = \frac12 3mv_f^2 + \frac12 mv'_f^2$
$ \frac {3v_0^2}{4} = 3v_f^2 + v'_f^2$

If we consider that the smaller ball wasn't affected (as it went backwards after the collision, in which the cord had no effect), then we can say it achieved the speed it would normally, i.e. $ \frac12 v_0$, and that the bigger object was the one slowed down (is this correct?)

Then:

$ \frac {3v_0^2}{4} = 3v_f^2 + \frac {v_0^2}{4}$
$ v_0^2 = 6v_f^2$
$ \boxed{\frac {v_f}{v_0} = \frac {1}{\sqrt {6}}}$
[/hide]


integration
algebra
partial fractions
calculus
calculus computations

Primitivate:

\[ f(x) \equal{} \frac{x}{{\left( {x \minus{} 1} \right){{\left( {x \plus{} 1} \right)}^2}}},{\rm{ x }} \in {\rm{ }}\left] {1, \plus{} \infty } \right[\]

Thanks

Antonio

Use Partial Fractions

$ \frac{x}{(x \minus{} 1)(x \plus{} 1)^2} \equal{} \frac{A}{x \minus{} 1} \plus{} \frac{B}{x \plus{} 1} \plus{} \frac{C}{(x \plus{} 1)^2}$

comparing coefficients gives you

$ A \equal{} \frac{1}{4} , \ B \equal{} \frac{\minus{}1}{4} , \ C \equal{} \frac{1}{2}$

$ \int \frac{x}{(x \minus{} 1)(x \plus{} 1)^2} \equal{} \frac{1}{4} \int \frac{1}{x \minus{} 1} \plus{}  \frac{\minus{}1}{4} \int \frac{1}{x \plus{} 1} \plus{}   \frac{1}{2} \int \frac{1}{(x \plus{} 1)^2}$

$ \equal{} \frac{1}{4}ln(x \minus{} 1) \minus{} \frac{1}{4}ln(x\plus{}1) \minus{} \frac{1}{2} \frac{1}{(x\plus{}1)} \plus{} C$

$ \equal{} \frac{1}{4}ln(\frac{x \minus{} 1}{x\plus{}1})  \minus{} \frac{1}{2} \frac{1}{(x\plus{}1)} \plus{} C$

Thanks,

I understand the problem.

 :blush:  :roll:

[quote="Kabi"][hide="@Mirka"]
congrats for yesterday win[/hide][/quote]

[b]Thank You !!![/b]  :winner_first:


number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ E$ and $ F$ be fields with $ F \subset E$ and suppose that $ d \equal{} [E : F]$ is finite. Let $ h(X) \in F[X]$ be irreducible and suppose that $ \deg f$ is relatively prime to $ d$.

Show that $ h(X)$ is also irreducible in $ E[X]$.

[hide]
Let $ \lambda$ be any root of $ h$. Then clearly $ F(\lambda) \subset E(\lambda)$.

Let $ d_F(\lambda)$ be the degree of $ \lambda$ over $ F$, and define $ E$ accordingly. Then by the tower law it follows that

$ [E(\lambda): F(\lambda)] \equal{} [E: F]\frac{d_E(\lambda)}{d_F(\lambda)}$. This must be an integer, and since $ d_F(\lambda)$ is by assumption relatively prime to $ [E: F]$, we have $ d_F(\lambda) | d_E(\lambda)$. Since $ d_E(\lambda) \leq d_F(\lambda)$, it follows that $ d_F(\lambda) \equal{} d_E(\lambda)$, which is as desired.
[/hide]

Nice problem!


Counting with Multiplication Part 2

We learn how to count the number of ways a series of independent events can occur by using multiplication.

:| [quote="Butler"]These videos are great! They showed me a lot of stuff.
Personally i think they should put a 100000 more on here! :10:[/quote]

These videos are [b][u]AWESOME!!![/u][/b]
Also, the answer to the number of ways is:

[b]208,971,104,256,000,000,000[/b].

That is (in word form):

[b]208 quantillion, 971 quadrillion, 104 trillion, 256 billion ways.[/b]

That's alot of ways!!!

quintillion, not quantillion
$\phantom{sorry for necropost}$


logarithms
induction
inequalities
function
inequalities proposed

Sure, bacause
\[ f(x)=\ln(\frac{1-x}{1+x})=\ln(1-x)-\ln(1+x) \]
is concave function.

No,I think we should prove $\frac{1-x}{1+x}\frac{1-y}{1+y}\geq\frac{1-x-y}{1+x+y}$ for all $x,y>0,x+y<1$
And that's easy to prove so by induction we can finish it.
For f(x) is concave we can only have $\prod\frac{1-x_i}{1+x_i}\geq(\frac{n^2-1}{n^2+1})^n$

Sorry, I was confused with "concave" and what?I forget it again.

[quote="arqady"][quote="jin"]
For f(x) is concave we can only have $\prod\frac{1-x_i}{1+x_i}\geq(\frac{n^2-1}{n^2+1})^n$[/quote]
hungkhtn means:
for $f$ is concave we have
 $f(x_1)+f(x_2)+...+f(x_n)\geq(n-1)f(0)+f\left(\frac{1}{n}\right)=f\left(\frac{1}{n}\right)=ln\frac{n-1}{n+1}.$ ;)[/quote]

  Arqady! Are you using Karamata's inequality?
  Thanks!

[quote="Gibbenergy"]

  Arqady! Are you using Karamata's inequality?
  Thanks![/quote]
1.If you will draw graph of concave function will see, that it is obvious. ;)
2. It is possible, since $(x_1,x_2,...,x_n)\prec(\frac{1}{n},0,...,0).$


See a similar exercise at http://www.mathlinks.ro/Forum/viewtopic.php?t=111272

Ascertain the pairs $(x,y)$ of integer numbers for which $(2x-3)\cdot y^{2}+(3x-1)\cdot y+(x-2)=0\ .$

[color=darkblue][b]Very nice the both solutions ([u]Sinajackson's[/u] and [u]Farenhajt's[/u]) ![/b
But can solve with these methods (yours) and the exercise $(2x-3)y^{2}+(x-1)y+(3x-2)=0\ ?$
In this case the denominator cann't be decomposed as product of factors. [/color]

[hide]Well this one is different. We have $y=\frac{1-x\pm \sqrt{-23x^{2}+50x-23}}{2(2x-3)}$
thus for having answers, we must have : $-23x^{2}+50x-23 \geq 0$ $\Longrightarrow$ $-23(x-1)^{2}+4 \geq 0$ $\Longrightarrow$ $\frac{4}{23}\geq (x-1)^{2}$ $\Longrightarrow$ $x-1=0$ and $x=1$. But if $x=1$ we have : $-y^{2}+1=0$ and $y=\pm1$ so the only answers are $(1,1),(1,-1)$. !!! :!:[/hide]

[hide="Second problem"]$x=\frac{3y^{2}+y+2}{2y^{2}+y+3}=1+\frac{y^{2}-1}{2y^{2}+y+3}\implies\frac{y^{2}-1}{2y^{2}+y+3}\in\mathbb{Z}$. Inequalities $\frac{y^{2}-1}{2y^{2}+y+3}\geqslant 1$ and $\frac{y^{2}-1}{2y^{2}+y+3}\leqslant-1$ don't have solution, hence $y=\pm 1$ giving $x=1$. The solutions are $(1,1),(1,-1)$[/hide]

Beaten again...  :roll:

Another exercise : $(2x-3)\cdot y^{2}+(5x-2)\cdot y+(x-1)=0$ and ... good nignt !

Then $x=1+\frac{y^{2}-3y}{2y^{2}+5y+1}$.
This will certainly be integer if y=0 or y=3. This gives already the solutions (1,0) and (1,3).
Now the fraction above has negative denominator only if y=-1 or y=-2. These values give indeed new solutions: (-1,-1) and (-9,-2).
For all other values of y we now have:
1) $\frac{y^{2}-3y}{2y^{2}+5y+1}\geq 1 \Leftrightarrow y^{2}+8y+1\leq0$
  This only occurs for y=-7,-6,-5,-4,-3; none of them gives an integer value for x.
2) $\frac{y^{2}-3y}{2y^{2}+5y+1}\leq-1\Leftrightarrow 3y^{2}+2y+1\leq0$, which even never occurs.
So only four solutions: (1,0),(1,3),(-1,-1),(-9,-2)

edit: corrected a typo

I confess I could't solve the last exercise :D  :oops:
Nice solution B23

I had to think a little bit why it works, but this method is very good, Virgil! I like it. It is more elegant than the inequality-approach (which in my opinion -but I didn't think that over very well- is also general). However both methods will run into problems when the degree of x or y is increased, but that shouldn't surprise us...
 :lol:

this may be dumb but where do you get $3y^{2}+2y+1 |-2$

Here the |-symbol doesn't mean 'is divisor of' but is just a seperator. The number on the right says how many times the polynomial on the left is to be multiplied with.