MATHCOUNTS
modular arithmetic
LaTeX
floor function
number theory
prime factorization

I have some questions I don't understand. Please help.

1.Find the value of x such that the four digit number x15x is divisible by 18.

2.How many 9 digit zip codes are possible if the first digit cannot be zero.

3.The walls of standard houses are constructed with 2 by 4 studs placed 16" apart, center to center. How many studs are needed for a wall 48 feet wide?

4.What is the largest perfect square that divides 64,800?

Please help me. Thanks, if you do.

[hide="1"]18=2*9
divisible by both

a15b
b=even
a+b=3 or 12
a+b=3

b=2, a=1 or
b=0, a=3

a+b=12
b=4, a=8
[b]b=6, a=6[/b]

that fills in for

x15x
if x=x

so its 6156[/hide]
[hide="2"]1st cant be 0. 9 other #s possible. all others: 10 possible #s
9x10x10x10x10x10x10x10x10=900000000 codes[/hide]

[hide="4"]prime factorization

64800=2^5*5^2*3^4

biggest square: (2^2*5*3^2)^2=2^4*5^2*3^4=64800/2=[b]32400[/b]
[/hide]
im thinking about #3

[quote="mathirons"]1.Find the value of x such that the four digit number x15x is divisible by 18.[/quote]
[hide]$ x$ must be even because the number is divisible by 2. Also, $ 2x\plus{}6\equal{}9k$ because the number is a multiple of 9. Quite clearly, $ \boxed{x\equal{}6}$ since no other even digits satisfy $ 2x\plus{}6\equal{}9k$.[/hide]

[quote="mathirons"]2.How many 9 digit zip codes are possible if the first digit cannot be zero.[/quote]
[hide]The first digit can have 9 possible values, and the other 8 can have 10. The answer is $ 9\cdot10^{8}$, or $ \boxed{900000000}$.[/hide]

[quote="mathirons"]4.What is the largest perfect square that divides 64,800?[/quote]
[hide]$ 64800\equal{}2^{5}\cdot3^{4}\cdot5^{2}$

The largest square that divides $ 64800$ is $ 2^{4}\cdot3^{4}\cdot5^{2}\equal{}\boxed{32400}$.[/hide]

[quote="mz94"][hide="1"]18=2*9
divisible by both

a15b
b=even
a+b=3 or 12
a+b=3

b=2, a=1 or
b=0, a=3

a+b=12
b=4, a=8
[b]b=6, a=6[/b]

that fills in for

x15x
if x=x

so its 6156[/hide]
[/quote]
WTH

hes saying the sum of all the digits is divisible by 9 and x is even...i think...atleast...thats what i think

so $ 2x\plus{}6$[u]=[/u]$ 0\pmod9$ and $ x$ [u]=[/u] $ 0\pmod2$

I'm not sure why he introduced two new variables, but what abacadaea said is correct.

@ abacadaea - The congruent symbol is written \equiv in $ \text{\LaTeX}$.

for 3, would it be rounded down(48*12/20). sorry i dodnt know how to do that cool boldy thingy.

"That cool boldy thingy" is called [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX[/url]. Simply put dollar signs around mathematical expressions, and it turns into a nice-looking image. Do not, however, type [i]everything[/i] in $ \text{\LaTeX}$ as this causes the site to be much slower for those without high speed internet connections.

As for #3, is the width of the studs 2" or 4"?

so, i only do it on the math part like 2x+3  making it look like $ 2x\plus{}3$ but not $ like this?$

Yes, unless you have a good reason (e.g. math symbols like $ \mathbb{N}$ and such, if they count as text at all).

what are those $ \mathbb{N}$'s used for?

N's i think are the set of all counting numbers?

i know Z is for integers
Q is for rationals
W i think is for whole numbers. ectr.

$ \mathbb{N}$ represents all natural numbers, in other words, all positive non-zero integers. Sometimes zero is included so that must be specified.

$ \mathbb{C}$ stands for complex. But we're getting of topic.

well arent all the problems solved?

i guess this topic should be locked

[/offtopic]

My question (Are the studs 2" or 4" wide?) hasn't been answered yet, so I can't answer #3.

usuallyits length by witdth so i think its 2 wide

Actually you don't need to know the width. [hide]*Center* to *center* means you would require $ \lfloor 48*12/16\rfloor\plus{}1$ studs, because it doesn't matter how wide the studs are you must have one at the beginning and end. Also, studs are always placed so the smaller dimension is facing the outside of the wall, since most stress on the studs will be up-down or into-out of the house, so you want to have as little flex in these dimensions as possible. You don't usually push parallel to the wall... :)[/hide] Having some construction experience helps with this question ;).

oh.....

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We got so wrapped up in this on the Intro to Geometry forum that I am moving it here.


integration
trigonometry
limit
function
abstract algebra
real analysis
real analysis unsolved

Set

f(x)= 0            if x<=0
        sin(pi/x)  if x>0

and

g(x)= 0           if x<=0
         1           if x>0

Prove that f has no antiderivative but g does not.

Thanks

$ f$ admits an antiderivative but $ g$ does not.

Letting $ F(x) \equal{} \int_0^x \sin {\pi\over t} dt,$ and $ F(0) \equal{} 0,$ $ F'(x) \equal{} f(x)$ by the
fundamental theorem at the points of continuity of $ f$ so $ x\ne0.$
If $ x \equal{} 0,$ $ F'(0) \equal{} 0$ holds. A proof runs as follows

$ F^\prime(0) \equal{} \lim_{h\to0} {1\over h}\int_0^h \sin{\pi\over z} dz.$
Changing variable $ \pi/z \equal{} y$ we have $ \lim_{h\to0} {\pi\over h} \int_{\pi\over h}^{ \plus{} \infty} dy {\sin y\over y^2} \equal{} \lim_{h\to0} {\pi\over h} \Bigl[ \minus{} {\cos z\over z^2}\Bigl\vert_{\pi\over h}^{ \plus{} \infty} \plus{} \int_{\pi\over h}^{ \plus{} \infty} dz { \minus{} 2\sin z\over z^3}\Bigr].$ The limit of the first part is clearly zero. As for the second contribution we can bound
$ \vert\int_{\pi\over h}^{ \plus{} \infty} dz { \minus{} 2\sin z\over z^3}\vert\le 2\int_{\pi\over h}^{ \plus{} \infty} dz {1\over z^3} \equal{} h^2/\pi^2$ and of course $ \lim_{h\to0} {\pi \over \vert h\vert} \vert\int_{\pi\over h}^{ \plus{} \infty} dz { \minus{} 2\sin z\over z^3}\vert\le \lim_{h\to0} {1\over \pi}{h^2\over \vert h\vert} \equal{} 0$ concluding the proof.

As for $ g(x),$ an easy and standard application of the Lagrange's theorem forbids
the functions having a jump or removable
discontinuity from being derivatives.

The application of Lagrange's theorem is the following

[i]
Let $ h\colon[a,b]\to {\bf R}$ be a differentiable function such that $ \lim_{x\to x_0^{\pm}} h'(x) \equal{} l^{\pm}\in {\bf R}\cup( \plus{} \infty)\cup( \minus{} \infty)$ and let $ x_0\in [a,b].$ Then $ \lim_{h\to0^{\pm}} {1\over h}(h(x_0 \plus{} h) \minus{} h(x_0)) \equal{} l^{\pm}$
(of course if $ x_0 \equal{} a$ we consider only right limits and left limits if $ x_0 \equal{} b$).
[/i]

Use Lagrange to prove the result above. The conclusions about jump or removable discontinuities are easy consequences.

your welcome

[quote="AlwaysMathLover"]Thank you Arrigo-Sacchi,

Very Thorough and helpful.

I appreciate the effort you put in helping me.[/quote]


I just discovered the MLG Gaming website today and became a user.

How do I join a video game competition there? (if anyone has experience)

What particular game are you trying to get into?

The most popular MLG game that exists is Halo 3, but there have been small factions of Call of Duty 4 or Gears of War, of even SSBB. To be identified as an MLG Professional player, you would have to dedicate your time to tournaments (paid entries) and winning them.


modular arithmetic
number theory unsolved
number theory

How many perfect squares are there  (mod $ 2^n$)  ?

For $ n$ even: The perfect squares are only $ 0,1,4,\ldots,(2^{n/2}\minus{}1)^2$, call this set $ S\equal{}\{0,1,4,\ldots,2^{n/2}\minus{}1\}$. Because if $ k\equiv r\pmod{2^n}$ ($ \minus{}2^{n/2}\le r\le2^{n/2}\minus{}1$), then $ k^2\equiv r^2\pmod{2^n}$. Since $ r^2\in S$, then $ k^2\in S$ too. We claim that these numbers are pairwise distinct. Assume $ a^2\equiv b^2\pmod{2^n}$ ($ 0\le a<b\le2^{n/2}\minus{}1$). Then $ (a\plus{}b)(a\minus{}b)\equiv0\pmod{2^n}$, which is clearly impossible (consider $ ord_2(a\plus{}b)\plus{}ord_2(a\minus{}b)$, it is clearly less than $ n$). So the answer is $ 2^{n/2}\minus{}1$.

For $ n$ odd, similar to the first case.

It is not true. Consider only odd integers $ x\le 2^{n\minus{}3}$, then $ x^2\not \equal{}y^2\mod 2^n$. Therefore we find $ 2^{n\minus{}3}$ solution for $ n\ge 3$ or 1 for  $ 1\le n\le 3$.
Then consider all numbers $ 2*x$, were $ x$ odd,... It give $ [2^{n\minus{}5}$ and so on.
Answer is $ 1\plus{}[\frac{2^{n\minus{}1}}{3}].$


Is the site down for some reason or can I just not access it?

I get the message:

Database currently unavailable. Please wait 5 minutes and try again. If problem persists, please contact webmaster@artofproblemsolving.com

So I don't believe it's just you.


Determine all finite nonempty sets S of positive integers satisfying

$ \frac{i\plus{}j}{gcd(i,j)}$

is an element of S for all i and j ( not necessarily distinct) in S

[hide="solution"]Since the definition of a set is that the elements ARE distinct, the (not necessarily distinct) means that we can choose the same element twice. Thus $ \frac {i \plus{} i}{gcd(i,i)} \equal{} \frac {2i}{i} \equal{} 2$ is always in $ S$.

If there is an odd number $ x$ in the set, then $ \frac {2 \plus{} x}{gcd(2,x)} \equal{} x \plus{} 2$ is in the set, and since that is odd, $ x \plus{} 2k$ is in the set for all naturals $ k$. But this means that $ S$ is infinite, so all elements must be even.

If there is an even number $ y\neq 2$ in the set, then $ \frac {2 \plus{} y}{gcd(2,y)} \equal{} 1 \plus{} \frac {y}{2}$ is in the set. Thus $ y$ cannot be a multiple of 4. Thus it is in the form $ 2y_1 \plus{} 1$. Also, $ y_1 \equal{} 2y_2 \plus{} 1$, and so on infinitely. But this is not possible, so there cannot be an even number $ y\neq 2$. Thus $ S \equal{} \{2\}$ is the only set that exists.[/hide]

EDIT: Sorry I didn't see your post.


I think the solution for this problem is incorrect. The problem is:

Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will 20! be the resulting product?

edit-nvm, I know where i went wrong.

Well here's the page in the resources section anyway: http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1991


A negatively charged particle is directed between two electrically charged plates. Why does the path of the charged particle bend? As the charge on the plates is increased, would you expect the bending to increase, decrease, or stay the same? As the mass of the particle is increased, would you expect the bending to increase, decrease, or stay the same?

The path of the particle bends because the particle is charged and there is an electric field present so the particle will be deflected to the positive source.  As the field gets stronger, the attraction is greater so the bending increases.

Although not exactly the same as your question, take a look at the Hall effect.  It is similar in nature.

1. I strongly recommend you post this topic in the physics forum. It would probably be more appropriate of a question there.
2. The answer to this (and a lot of life) can be abstracted from Coulomb's Law.
F = kqq/r^2

k =  electrostatic constant
q = charge
q = another charge
r = distance between charges

(assume each plate is a "charge")

Treating the plates as charges will lead to the correct qualitative answer to the questions posed, but it does not fit very well with applying Coulomb's Law.  Assuming (as one generally does) that the plates are large compared to the distance between them, the electron particle experiences a constant force, not one that varies as $ \frac{1}{r^2}$.

hey give the full version of the coulumbs law insluding the permittivity of constant!
i mean
 vaccum,!
if not air!


geometry
parallelogram
rectangle
Pythagorean Theorem

1.

[hide]For all paralellagrams,  One angle is the same measure of the opposite angle.  So since A = 90, C = 90.

B = D

The measure of the angles in the parallelagram add up to 360

B + D = 360-90-90 = 180

Since B=D, we can substitue B for D.

2B = 180

B = 90

So since B = D and B = 90, D = 90.[/hide]

2.[hide]

This parallelagram is a square, since all of the angles equal 90.

Let's make the value of one of the sides x.

Using what we know about 45:45:90 triangles, we can say that AC =  x :sqrt: 2

Since x = AB = BC = CD = AD, we can do the same to find out how long BD is.

So, AC = BD. [/hide]

Your #2 is flawed; having all 4 interior angles equal to 90o does not necessarily mean that the shape is a square.

When he said AB=CD, it does sound more if though the parallelogram is a rectangle instead.  Otherwise, he would probably ask to prove that all the sides are equal.

[quote="monotonic"]Let ABCD be a parallelogram and $\measuredangle A=90^0$.
    1) Prove that  $\measuredangle B=C=D=90^0$
    2) Prove that AC=BD[/quote]

If DAB = 90, DA and AB are perpendicular. As the shape, by defenition, has 4 sides with opposing sides parallel, the only possible positions of BC and CD are parallel to DA and AB, and perpendicular to AB and DC respectively...

this is so easy it's hard. I'm no good at geometry :(

[hide]1.  ABCD is a parallelogram with ang A = 90.

    That means ang C = 90.

    As interior angle measure = 360, (360 - 180)/2 = 90

    So, ang B = ang C = ang D

2.  1 proves that each ang = 90

    As ABCD is a parallelogram AB = CD and BC = AD

    Again, as each angle is 90, AB and CD are perpendicular to AD and BC.

    Pythagorean theorem (a^2 + b^2 = c^2) applies (Right triangles ABC and ACD)

    AB = CD = a; BC = AD = b; AC = BD = c

    Therefore, AC and BD are equal.

Not very direct, but everything is stated/implied.[/hide]

I'm going to promote this to Intermediate, because as usual problems involving geometry content and requiring the problem-solver to "prove .  .  . " are not at the getting-started level.

Thanks for posting the problem.

"Prove ... " is just a synonym of "Show that ... ", tokenadult.

As for intermediate, there's really no use for moving this to intermediate. Then it's just a waste of forum space to post it. This is really the basics of geometry, which you learn in the very first geometry lessons. These things we learnt in primary school (4th-5th grade) so if things for primary school are too hard for beginner's section, then what the heck do we have that section? Be honest, how many members from primary school do we have here?? :?


[quote="Flakky"]Does $ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2$ have solution???If yes please explain it?[/quote]

If you are talking about positive integer, I am sure that it has a lot...

for example $ 11^2\plus{}10^2\equal{}14^2\plus{}5^2\equal{}221$

You can consider $ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2\equal{}(ad\plus{}bc)^2\plus{}(ac\minus{}bd)^2$  :roll:


$\text{Find the largest area of the triangle whose three sides } a,b,c \text{ satisfy the conditions: }$

$0 < a \leq 2 \leq b \leq 4\leq c \leq 6$

Note that there are no specifications on the type of numbers a,b, and c must be, except that they must satisfy the above inequality.

i think the largest area will be a right triangle of 2,4, 2sqrt5 with an area of 4

how did you get that answer though? brute force or what?

One way the area of the triangle may be expressed is $\frac{1}{2}ab\sin{C}$.
Maximizing $a,b,\sin{C}$, the area is $\frac{1}{2}(2)(4)(1)=4$.

Since $\sin{C}=1$ implies $\angle ACB=90^\circ$, then $c=\sqrt{a^2+b^2}=2\sqrt{5}<6$, and this triangle does satisfy all requirements.

The smallest side must be maximum, $a = BC = 2$. The remaining vertex $A$ of the triangle $\triangle ABC$ must be within light yellow annulus $((B))$ centered at the vertex $B$ with radii $c_1 = 4, c_2 = 6$ and within the light green annulus $((C))$ centered at the vertex $C$ with radii $b_1 = 2, b_2 = 4$. For the maximum triangle area, the altitude from the vertex $A$ must be maximum, hence, $b = 4$ and the angle $\angle C = 90^o$ is right. Then $c^2 = a^2 + b^2 = 20,\ 16 < c^2 < 36,\ c = 2\sqrt 5$ and the triangle area is $|\triangle ABC| = 4$.

no...brutal forcing this problem would be....impossible....

i did it like Samson but as u should know i am Latex handicapped and lazy so i didn't give any work....

yes, you want it to be a right triangle, obviously, because this makes a maximum height of the other side, since $\sin{90}=1$, and if the angle is less or greater than 90, then $\sin{\theta}<1$ making the area and the height decrease


$ 2x\plus{}3\equal{}9\Rightarrow x\equal{}3$
$ 8y\minus{}1\equal{}7\Rightarrow y\equal{}1$

Then $ x\plus{}y\equal{}4$.


I wear glasses since the finish of 7th grade(now I'm 11th) :agent: . I could get rid of them next year if I make an surgical operation but I have to be 18 years old. How about you :P?

I've worn glasses since 3rd grade (I'm currently in 8th)

I heard that laser eye surgery wears off after a year or so...  :?

Wel, I had glasses(since around 3 rd grade), but  i got contacts a month ago. I like contacts much better. So should I vote yes or no?

I have glasses, but I dont wear them much, in fact, I havn't wore them ever since the start of summer break.

Since 5th grade. And I don't plan to make any surgical operations.

Glasses since 1st grade.  Contacts since 8th.

(Oh, and I'm in 12th now)

Since end of 7th, and I'm going into 8th. But, I barely wear them (only during a period in school where i need to see the board for a while, and thats it), and I'm gettign contacts soon.

technically, ive had glasses since 10th grade (now going into 12th), but i still dont really wear them that often.

So many have glasses.. O.o

I have 20/25 vision or something like that. My left eye is just a tiny bit below 20/20 alone, but my right eye makes up for the difference when I have both eyes open. Glasses look so inconvenient, I would go for the surgery or at least contacts.

I have glasses. I got them just a bit ago but I don't wear tham much...

glasses since 6th! w00t!
yay for 600/650 vision!

I don't know about measure system form US. or other countries. So I can't tell about my vision size.  :huh:
Left eye: -4.00
Right eye: -6.00

Both eyes are blue  :rotfl: !!

In american measure system might be -600 and -400... but I'm not sure of that.

[quote="236factorial"]I wear contact lenses, about -375 something like that.[/quote]
Don't you mean -3.75?
Both of my eyes are -5.25

I've worn glasses since 4th grade (I'm in 9th now)
I want to get contacts and when I turn 21, I want to get the eye surgery

Glasses for the last, oh, 10 years  (since 1st grade)
Got contacts two years ago for sports, but they are uncomfortable and don't properly correct for my astigmatism.  Oh well. Even the toric lenses they told me to use don't help much more (but it is a bit better)

I wear black framed rectangular glasses! They are so cool though, i'm proud of them! lol (I haven't been on in a while, school) Yeah, when I went to the place the lady at the counter looked at my prescription and said, "Do you wear contacts?"

Me: "No"
lady: "Do  you have another pair of glasses?"
Me: "yes, but they give me a headache because their too small, and the words are still a little blurry that I can't read them."
lady: "You'll probably want contacts, their better to drive with. What do you wear when driving."
Me: "I'm 11, I don't drive!"
lady: "YOU'RE 11?! Oh, well still how do you get around with this prescription?!"
Me: "i bump into every wall, it's not that bad."
1 week later
lady: "Here's your new glasses!"
*I put them on*
Me: "OH MY GOD!!!!!! I was blind! Those are bannas? I thought they were boxes of cereal!!!"
I don't know my prescription, but it was that bad... I heard a 7.75 or something with a 7, I have no idea... I don't even know where it is. For the longest time people's faces were little pale (tan, dark, whatever the case may be) ovals that had two black dots and a pink/red line. Yep, my sight was bad... Still is! But my glasses give me personality, and my eyes look so small without them, even though their not. Their not dork/geek glasses, they have an edge to them. Or at least I make them that way...  :lol:

wow glasses since 2nd grade, contacts since 3rd because of soccer... and im worse than -7.00 in both eyes.  :D

i'm 25-20 and the doctor says if i get to 30-20 i have to wear glasses :(

[quote="captcha000"]im a sophomore, worn glasses since about 1st grade. i wore contacts for two or three years but my parents have heard that they make your eyesight worse (?) so im back to glasses[/quote]

that's SO not true. i actually had to wear contacts for three years as part of the study to keep my eyesight from GETTING worse. and for the person with astigmitism: there are contacts for that now. i have it too, i used to have to wear hard contacts but now they make soft ones. they are the kind you keep for 3 months and then throw away. (much better than hard, BELIEVE ME).

[quote="sponge008"]I wear glasses, I've worn them since 6th grade (9th now). My right and left eyes are both -3, my middle eye is -3.5.[/quote]

middle-eye?????  :?
I have soft contacts too. I switch mine every 2 weeks.

[quote="Treething"]Glasses from 1st to 6th.

No glasses for me anymore.[/quote]

errrr contacts then? or do you just walk around blindly lol?
and why does everybody use the -5.00 or w/e system...what happend to like 600/20 and 550/20???

I just got contacts today.  But I voted for glasses, since  voted a week ago I think.

[quote="noneoftheabove"]how many people have lasik[/quote]

what's that?


Se da $f:[0,\infty ) \to R$,continua cu $\mathop {\lim }\limits_{x \to \infty } f(x) = 0$.
Sa se arate ca f este marginita.

Totusi, asta nu este o solutie. Ce se intelege dintr-un simplu "Weierstrass"? :?

Ca se aplica teorema lui Weierstrass, dupa ce ai "taiat jos" un interval catre $\infty$.

Si pun pariu ca aproape nimeni n-a inteles mai nimic cu "taiatul asta jos"? :P

Ideea este sa nu uitam totusi nivelul acestui sub-forum (bacalaureat) ;)

Fie. In primul rand nu este necesar ca limita sa fie $0$, ci doar sa fie finita.

Fie $\lim_{x\rightarrow\infty}f(x)=l\in{\bf R}$. Exista atunci $a>0$ astfel incat(apropo, cui ii place prescurtarea a.i.?) $f(x)\in[l-1,l+1],\,(\forall)x\geq a$. Deci $f$ este marginita pe $[a,\infty)$. Cum $f$ este continua, ea este marginita pe intervalul compact $[0,a]$. Prin urmare $f$ este marginita pe $[0,a]\cup[a,\infty)=[0,\infty)$.

Sper ca toata lumea este acum multumita.

[i]Valentin. V.[/i], aici nu sunt de acord cu tine. Aceasta problema era in manualul vechi de analiza (XI), daca nu ma insel. Solutia era "cu taiatul jos" cum foarte bine s-a exprimat Mchis68, dar care a avut rabdarea (mai bine spus "darul") de a detalia ulterior riguros solutia problemei. De aici se poate trage o concluzie: daca postezi un reply cu indicatii, atunci ai bunavointa sa oferi cat de curand (daca solicitantul nu a inteles "indicatiile") si o solutie detaliata.

Imi permit sa mai fac in comentariu legat de cele mentionate mai sus. Un bun manual (sau culegere de probleme) trebuie sa contina solutiile tuturor problemelor si aceste solutii sa fie prezentate pe trei nivele: [u]solutie completa[/u]; [u]indicatii[/u]; [u]raspuns final[/u] (care corespund celor trei nivele ale dialogului profesor-elev:[u]meditatie[/u]; [u]consultatie[/u]; [u]cerc mate[/u]).

Iata un alt exemplu (problema de manual) la nivel [u]bac[/u] sau [u]admitere[/u]:
"o functie $f: R\rightarrow R$ continua si marginita are cel putin un punct fix."
[u]Indicatie:[/u] se considera functia $g: R\rightarrow R,\ g(x)=f(x)-x$ si se aplica o consecinta a proprietatii Darboux (extinsa la capete deschise) a unei functii continue: $\lim_{x\to -\infty} g(x)=\infty$ si $\lim_{x\to\infty} g(x)=-\infty$ etc.


Let M, N, P, Q be the midpoints of the sides AB, BC, CD and DA of the quadrilateral
    ABCD respectively. Assume that  X=(AX)∩(BQ), Y=(BQ)∩(CM), Z=(CM)∩(DN), T=(DN)∩(AP).
    Prove that S_{XYZT}=S_{AQX}+S_{BMY}+S_{CNZ}+S_{DPT}.

[quote="Sergey Ginsburg"]X=(AX)∩(BQ)[/quote]

do you mean X=(AP)∩(BQ)? if it's this the solution is simple:

$ [XYZT] \equal{} [ABCD] \minus{} [MBC]\minus{}[NCD]\minus{}[PDA]\minus{}[QAB]\plus{}[AQX]\plus{}[BMY]\plus{}[CNZ]\plus{}[DPT] \equal{} [ABCD] \minus{} \frac{[ACB]}{2} \minus{} \frac{[BCD]}{2} \minus{} \frac{[CDA]}{2} \minus{} \frac{[DAB]}{2} \plus{} [AQX]\plus{}[BMY]\plus{}[CNZ]\plus{}[DPT] \equal{} [ABCD] \minus{} \frac{[ABCD]}{2} \minus{} \frac{[ABCD]}{2} \plus{} [AQX]\plus{}[BMY]\plus{}[CNZ]\plus{}[DPT] \equal{} [AQX]\plus{}[BMY]\plus{}[CNZ]\plus{}[DPT]$


There are 3 groups.The 1st group contains 3 girls & and 1 boy.The 2nd group contains 2girls and 2 boys whereas the 3rd contain 1girl and 3 boys.Another group is formed by picking one student from the existing groups.What is the probability that the group formed will contain 1 boy & 2 girls.(in lowest form)

Group #1 - 3G, B
Group #2 - 2G, 2B
Group #3 - G, 3B

1) Boy comes from #1, and the girls come from #2 and #3.

$ \frac{1}{4}\times \frac{1}{2} \times \frac{1}{4}\equal{}\frac{1}{32}$

2) Boy comes from #2, and the girls come from #1 and #3.

$ \frac{3}{4}\times\frac{1}{2}\times\frac{1}{4}\equal{}\frac{3}{32}$

3) Boy comes from #3, and the girls come from #1 and #2.

$ \frac{3}{4}\times\frac{1}{2}\times\frac{3}{4}\equal{}\frac{9}{32}$

$ \frac{1}{32}\plus{}\frac{3}{32}\plus{}\frac{9}{32}\equal{}\frac{13}{32}$

answer : $ \frac{13}{32}$


Let $AB$ be a diameter of a circle; let $t_1$ and $t_2$ be the tangents at $A$ and $B$, respectively; let $C$ be any point other than $A$ on $t_1$; and let $D_1D_2. E_1E_2$ be arcs on the circle determined by two lines through $C$. Prove that the lines $AD_1$ and $AD_2$ determine a segment on $t_2$ equal in length to that of the segment on $t_2$ determined by $AE_1$ and $AE_2.$

The lines $AD_1$ and $AE_1$ determine an involution on the tangent from $C$ to the circle (the tangent different from $CA$). This means that they also determina an ivnolution on $t_2$, and since it's obvious that the infinity point of $t_2$ corresponds to itself in this involution, it means that the involution is a symmetry wrt a point (a well-known rersult), and from here the conclusion follows (we find the midpoint of $D_1'E_1'$ to be the same as that of $D_2'E_2'$, where, in general, $X'=AX\cap t_2$).

[hide="Solution"]Let $X_1$, $X_2$, $Y_1$, $Y_2$ be the intersections of lines $AD_1$, $AD_2$, $AE_1$, $AE_2$ with $t_2$. Furthermore, let $CT$ be the other tangent from $C$ to the circle. Finally, let $M=AT\cap t_2$. Since quadrilateral $AE_1TE_1$ is harmonic, the perspectivity at $A$ onto $t_2$ yields $-1=(A, D_1;T, E_1)=(P_{\infty}, X_1;M, Y_1)$, where $P_{\infty}$ is the point at infinity on $t_2$. It follows that $M$ is the midpoint of segment $X_1Y_1$. Similarly, $M$ is the midpoint of segment $X_2Y_2$, so $X_1X_2=Y_1Y_2$ and the result follows. $\blacksquare$[/hide]


function
quadratics
inequalities
limit
algebra
real analysis
real analysis unsolved

say that $ f(n) \equal{} 12n^2 \plus{} 6n$ and $ g(n) \equal{} n^3$

and $ f(n) < c * g(n)$, for all c > 0 and n > n0

there has to be exist an n0 that satisfies this statement

how do we find an no that is valid for this statement??

Clearly $ n_0$ must depend on $ c$. So all you have to do is put in those two functions and check.

We want $ 12n_0^2 \plus{} 6n_0 < cn_0^3$, or $ cn_0^2 \minus{} 12n_0 \minus{} 6 > 0$. The quadratic equation says that the larger root of this is
$ n_0 \equal{} \frac{6 \plus{} \sqrt{36 \plus{} 6c}}{c}$
and therefore if $ n$ is greater than that $ n_0$ then your inequality will hold.

If you don't want $ n_0$ to depend on $ c$, then you're going to have a hard time finding a single value that always works.

[quote="Xevarion"]Clearly $ n_0$ must depend on $ c$. So all you have to do is put in those two functions and check.

We want $ 12n_0^2 \plus{} 6n_0 < cn_0^3$, or $ cn_0^2 \minus{} 12n_0 \minus{} 6 > 0$. The quadratic equation says that the larger root of this is
$ n_0 \equal{} \frac {6 \plus{} \sqrt {36 \plus{} 6c}}{c}$
and therefore if $ n$ is greater than that $ n_0$ then your inequality will hold.

If you don't want $ n_0$ to depend on $ c$, then you're going to have a hard time finding a single value that always works.[/quote]

on the book it says that no = (12+6)/c

I am just confused where they find this equation?

[quote="Xevarion"]$ 12n^2 \plus{} 6n < 12n^2 \plus{} 6n^2 < cn^3$

if $ n > n_0$ where $ n_0 \equal{} 18/c$.

The idea is that in order to find a simpler expression for $ n_0$, you replace $ f$ by something that's pretty similar, but simpler and larger.

edit: obviously the $ n_0$ you get this way will be worse, but generally in this kind of situation you don't care too much.[/quote]

okay the book also quotes that too, but how do I get from

$ 12n^2 \plus{} 6n < 12n^2 \plus{} 6n^2$

sorry I am a little bit long at understanding something

Well, it should be obvious that $ 12n^2 \plus{} 6n < 12n^2 \plus{} 6n^2$ if $ n > 1$. So, if you have some inequality like
$ 12n^2 \plus{} 6n^2 < cn^3$ (1)
then you put that one and the previous one together and you get
$ 12n^2 \plus{} 6n < cn^3$ (2)
So, the objective was to find $ n_0$ such that (2) is true for all $ n > n_0$, right? The idea here is that all we need to do is find $ n_0$ such that (1) is true for all $ n > n_0$, and then that one will work for (2) as well!

But (1) is easy:
$ 18n^2 < cn^3$
$ 18 < cn$
$ 18/c < n$

So you just say $ n_0 \equal{} 18/c$, and the inequality holds for all $ n > n_0$. Since you've found $ n_0$ for (1), this also works for (2), and you are done!

In context, this is precisely about showing that $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{}0$, by the definition.

We have $ \frac{f(x)}{g(x)}\equal{}\frac{12}{x}\plus{}\frac{6}{x^2}$; each term clearly goes to zero, and they're easy to estimate separately. Now, how do we prove the addition part of the arithmetic theorem? The usual way is to estimate each piece to $ \frac{\epsilon}{2}$, and combine by the triangle inequality.

When is $ \frac{12}{n}<\frac c2$? We need $ n> \frac{24}{c}$.
When is $ \frac{6}{n^2}<\frac c2$? We need $ n> \sqrt{\frac{12}{c}}$

We can therefore choose $ n_0\equal{}\max\{\frac{24}{c},\sqrt{\frac{12}{c}}\}$. Since $ \frac{24}{c}\ge \sqrt{\frac{12}{c}}$ when $ c\ge \frac1{48}$, we could use the simpler $ n_0\equal{}\max\{\frac{24}{c},\frac12\}$.

There is no obligation to have the smallest $ n_0$ possible, or even close to it. Crude estimates of the error are perfectly fine.


trigonometry
geometry
geometric transformation
reflection
angle bisector
geometry proposed

In an isosceles triangle $ ABC , (AB \equal{}AC)$ we have $ \angle A \equal{} 100$. In the interior ot the triangle we take
 [b]a)[/b] a point $ P$ such that  $ \angle PAC\equal{} 20 , \angle PCA \equal{} 10$
 [b]b)[/b]a  point $ M$ such that $ \angle MAB \equal{} 20 , \angle MCB \equal{} 10$ .
 Prove that :
 [b]a)[/b] $ BA\equal{}BP$     [b]b)[/b] $ MA\equal{}MB$

[color=red][b]NOTE[/b][/color]

 The trigonometric form of Ceva's theorem gives simple solutions , to both questions, since it is sufficient to prove that $ P,M$  are on the bisector of angle $ B$.
[color=red][b] I'd like to see a synthetic solution![/b][/color]

but why serching for a syntetic solution when with ceva on 1 staps we conclude?
$ \frac {\sin \angle PBC}{\sin \angle PBA} \equal{} \frac {\sin20}{\sin80}\frac {\sin 30}{\sin{10}} \equal{} \frac {\sin 20}{2 \sin10 \cos10} \equal{} 1$ so P stay on the angle bisector of B and the same for M because P an M are isogonal cojugate.

$ AP$ meets the bisetor of $ \angle B$ at $ P'.$ $ \triangle ABP'$ is isosceles with $ BA \equal{} BP'$ base $ \angle P'AB \equal{} 80^\circ$ and vertex $ \angle ABP' \equal{} 40^\circ.$ Bisector of the $ \angle ABP'$ cuts bisector of the $ \angle CAB$ at $ Q.$ $ QA \equal{} QP',$ $ \triangle AQP'$ is isosceles with base $ \angle P'AQ \equal{} 30^\circ,$ vertex $ \angle AQP' \equal{} 120^\circ.$ $ QP'$ cuts $ AC$ at $ C'.$ $ \angle AC'Q \equal{} 180^\circ \minus{} (\angle CAQ \plus{} \angle AQP') \equal{} 10^\circ$ $ \Longrightarrow$ line $ QP'C'$ is reflection of line $ QB$ in $ AQ$ and $ C' \equiv C, P' \equiv P$ are identical.

For reading easiness, it is a good idea to combine the [b][size=100]yetti’s[/size][/b] solution, with the configuration of a regular [b][size=100]9-gon[/size][/b].

I think, after the result $ \angle AQP' \equal{} 120^{o},$ we can also say that the line segment $ QP',$ as the reflection of $ QB$ with respect to $ AQ,$ passes through the point $ C,$ as the reflection of $ B,$ wrt $ AQ.$

So, because of $ \angle ACQ \equal{} 10^{o},$ from the triangle $ \bigtriangleup AQC,$ where $ \angle AQC \equal{} 120^{o}$ and $ \angle CAQ \equal{} 50^{o},$ we conclude that $ CP'\equiv CP$ $ \Longrightarrow$ $ P'\equiv P$ and we are done.

Kostas vittas.

We can construct P' is the intersection of the bisector of $ \hat{B}$ with the ray CP and prove P=P' by proving $ \hat{P'AC}$ =20.
Construct MBC is the equilateral triangle, so $ \bigtriangleup$BMA= $ \bigtriangleup$BCP' then BA=BP'. So we have $ \hat{BAP'}$ =80 then $ \hat{P'AC}$=20, result is P'=P so BA=BP.

[quote="stergiu"][quote="plane geometry"]Construct equilateral triangle DPC, D and A are on different sides of BC
We have PD ⊥ BC , △PBC and △DBC are congruence  => PB=BD
∠CPQ=∠PAC + ∠PCA =20+10=30     =>  ∠DPQ=60-∠CPQ=30  △APD and △APC are congruence   =>  AD=AC  ∠BAD=100-2*20=60  => △ABD is equilateral triangle   => AB=BP[/quote]

[color=red][b] YES ! [/b][/color]

 I was seeking excactly such a solution ! Could someone try for a similar solution to question b) ?

[b] Note[/b]

 Thanks to all solvers for their nice solutions.

 Babis[/quote]
Exactly  the same way

For the result $ MA \equal{} MB,$ we can also say that we have $ M'A \equal{} M'B$ $ ($ $ \angle ABM' \equal{} \angle BAM' \equal{} 20^{o}$ $ ),$ where $ M'$ is the intersection point of the line segment $ AM,$ from the line segment $ BP,$ as the angle bisector of $ \angle B.$

Then, because of the point $ M'$ is the isogonal conjugate of $ P,$ with respect to the given triangle $ \bigtriangleup ABC$ ( easy to prove ), we conclude that $ \angle BCM' \equal{} \angle ACP \equal{} 10^{o}$ and so, we have $ CM'\equiv CM.$

Hence, we conclude that $ M'\equiv M$ and we are done.

Kostas Vittas.

There are many problem, which are similarity with this problem. The main idea often be used is construct a suitable equilateral triangle.


Try this problem on your own. (Note: This is not a trick question of any sort like the turtle or the circle problem I posted earlier)

Prove that there exist two positive prime integers with at least 10,000 composite integers in between them, and give two possible numbers.

I think this may be slightly misworded.. as I think 2 and any huge prime would satisfy this condition :).. perhaps you mean to say prove there exists 10000 consecutive composite numbers?

Arne posted this problem earlier. I think it was on the Advanced board although it probably should have been Intermediate.

o, ok, i didn't know somebody posted that already. i saw it at a talk at presentation high.

Oops. I was wrong. Arne's problem was different. Sorry. It's just that I used the same argument to solve that problem as I would to solve this one. That's why I was confused.

SPIKEzZ, try not to use too much SFBA terminology. I know what you mean when you say you saw it at a talk at Presentation High because I am from SFBA and went to those talks, but it's just gibberish to most others.

Is this the argument?

We take 2+10000!, 3+10000!, ..., 10000+10000!. These are all composite.

Pretty cute. :D

Well, that's only 9999, but it's the right idea.


Euler
number theory proposed
number theory

Find $n$ positive integer so that

        $n = 2 \phi(n)+150$

where $\phi$ is Euler number.


Support geometry calculus integration Yale college articles

[quote="hello"]I know this is getting way off topic and what I will say might seem really crazy but I think that the reason that some people get more then others and not necessarily that, like how can you tell how good is something? you probably cant exactly, so there is a problem in quantification. I think that the reason that most problems exist and are very difficult to solve is becuase it is impossible to determine their exact value, I call this my theory of quantification problems. LOL, I am just kidding[/quote]

You make a good point.  This is part of the reason we need more problem solvers to go around in fields outside of mathematics.  Usually when we quantify problems correctly, the solution either becomes apparent to the public or is accepted by the public over time due to confidence from the academic or expert communities.

There are many problems that have not been quantified or quantified well that society will battle with for years to come.  Teacher productivity is one of them.

[quote="tokenadult"]GOOD teachers might make a whole lot more money if parents, the clients of the school system, could shop readily for teachers. But the schoolteacher unions oppose voucher plans and most other policy proposals that help parents shop for teachers because many union members have a rational fear that their salaries would drop if clients could choose teachers on the basis of quality of teaching.[/quote]

Tokenadult, there are two assumptions embedded in your statement which I'd like to call into question.
(1) School systems have clients.
The use of the word "client" assumes a corporate model for school systems.  A corporate model is inappropriate here; school systems are a public service, not a corporation.  Similarly, public universities provide a public service, and both public and private universities provide educational opportunities (not guarantees of diplomas for class attendance).
(2) Parents are able to tell which teachers are "good."
I totally believe that you personally can assess the quality of a school teacher, but I don't think that a very high percentage of the general parent population has the experience or insight to judge this.  There are gross misperceptions as to what teachers' workloads and responsibilities and training and support actually are...

Also, I doubt that knowing which teachers are "good" would make a difference in salaries or in overall educational quality.  We don't have enough public funding to change the former or enough well-trained teachers to change the latter.

I don't see how the points you made affect tokenadult's argument.  It doesn't matter whether or not the corporate analogy is literally accurate or not, and as long as you can judge the teacher for your own needs, the whole workload/training and such thing would evolve to meet that.  It's selfish, but that's how America works in general.

[quote="smbelcas"]Tokenadult, there are two assumptions embedded in your statement which I'd like to call into question.
(1) School systems have clients.
The use of the word "client" assumes a corporate model for school systems.  A corporate model is inappropriate here; school systems are a public service, not a corporation.[/quote]

Could you lay out the additional major premises you need here to tighten up your argument? I have a feeling I disagree with them, but let's be sure before I further discuss the point.

Just as one example, no one disputes that lawyers have clients, and a few lawyers even think that the practice of law is a public service   ;)  but hardly any law firm is organized as a corporation.

And even if we agree about what is, couldn't we also agree that what should be is different from what now is?

[quote="smbelcas"]
Tokenadult, there are two assumptions embedded in your statement which I'd like to call into question.
(1) School systems have clients.
The use of the word "client" assumes a corporate model for school systems.  A corporate model is inappropriate here; school systems are a public service, not a corporation.  Similarly, public universities provide a public service, and both public and private universities provide educational opportunities (not guarantees of diplomas for class attendance).[/quote]

You assert by fiat that schools are a public service, but that's not only untrue in many observable cases, but it is false by definition.  You cannot have a public service of a scarce resource and if you've ever walked down the halls of a public school that isn't top-notch, you would know that education is still a scarce resource.  In fact, there is no such thing as a company that profits from any activity if there are no scarce resources.

Economics is the study of scarce resources.  To me this means that private education models are extremely appropriate -- and even necessary.  Besides, where does the "magic line" begin and end that allows for private university but not private elementary or secondary education?  Why the double-standard?

I know that many people have a knee-jerk reaction to corporate [insert anything here], but unless we're specifically critiquing poor management of individual leaders, all that getting corporations involved in schooling does is creates a fiduciary duty for educators to educate (hence the term "client").  As it is, teachers in most school systems have little incentive to do their job.  Two summers ago I played ultimate frisbee with a math teacher from Chula Vista who stated quite frankly that she hates math and works just to pick up the paycheck because she can't imagine she can work a better job.  Unions are strong and it's difficult to filter those kinds of teachers out of the system.

Additionally, every major study indicates that even as we strip out socioeconomic variables, [url=http://news.yahoo.com/news?tmpl=story&u=/ap/20050525/ap_on_re_us/charter_school_study_1]charter schools[/url] and [url=http://aspe.os.dhhs.gov/hsp/97trends/ea2-3.htm]private schools[/url] are doing much better jobs than public schools with secular private schools leading the way.  Public schools even outspend their private counterparts (something like 7k to 4.5k per student as of just a few years ago).

Having been through a public school system that is often ranked very high nationally among public schools in a variety of statistics, I didn't feel like a took a single course that challenged me or interested most of my peers.  Many of the top students whose 1500+ SAT scores and 8ish AP exams push up national rankings are taking summer college classes, attending corporate academic programs, or have private tutors.

Even at the top public schools in California, motivated students are bored and frustrated as evidenced by what we see on AoPS blogs.  It saddens me the number of times I read something along the lines of "I am so bored in school."

It is very very hard for me to imagine a pretty picture of public education.  Even our leaders of industry such as Bill Gates are telling us very frankly that our kids are not being well enough educated to hire.

Beyond that, I think that the assertion that corporations should stay out of education is offensive.  What is AoPS?  I would gladly contrast our model for private math education with that of most any public school.

One last question: If public schools are doing an adequate job, why do [url=http://www.faegre.com/articles/article_1601.aspx]H-1B visas for foreign workers without post-graduate degrees run out on the very first day of the fiscal year[/url]?

If you want a technology job in this country, you might still be better off with an education from a struggling, resource-poor, financially strapped former Soviet satellite than from an ordinary, well-funded American public school.

My answer is that the status quo system of education has dropped the ball in preparing students for the technology age.  Schools botched the job so badly, that I cannot even fathom how we could call it a service, much less a public service.

(note: that's not to say I don't consider the great teachers at all schools, public and private, among the heroes of our culture.  I hold them in high esteem as they struggle through a difficult system.)

[quote="smbelcas"]
(2) Parents are able to tell which teachers are "good."
I totally believe that you personally can assess the quality of a school teacher, but I don't think that a very high percentage of the general parent population has the experience or insight to judge this.  There are gross misperceptions as to what teachers' workloads and responsibilities and training and support actually are...

Also, I doubt that knowing which teachers are "good" would make a difference in salaries or in overall educational quality.  We don't have enough public funding to change the former or enough well-trained teachers to change the latter.[/quote]

So long as we stick to a public system, salaries will more more about politics than merit.  But I disagree that it doesn't affect salaries.  It will as education becomes more of an enterprise.  Besides that, giving up on the idea of a meritocracy in public education seems far more like an indictment of the system than anything else.

You're right that most parents are not involved and knowledgeable enough to make judgements about good teachers.  This seems like a good reason to judge based on what the students are capable of accomplishing.

Doesn't that at least beg us to experiment more to see if non-public forms of education get the job done better?

[quote="probability1.01"]It's selfish, but that's how America works in general.[/quote]

Fortunately, unless a situation is a prisoners dilemma, self interest results in meeting mutual needs.  This drives all cooperative economics.  Certainly education can and should be a win-win game.

I hope the two general economic "camps" (corporate and anti-corporate) can come to grips with the fact that their differences are mostly a matter of viewing only one side or the other of game theoretic situations.

Cooperation can consistently take place to the mutual benefit of large groups of people so long as people are only working with others that they trust.

On the other hand, the existence of scarce resources in a world in which some people do not act in trustworthy ways means that prisoners dilemmas necessarily exist.

When I stand back and evaluate what kind of system would allow for Nash equilibriums of cooperative scenerios and promote iterated cooperation in regards to scarce resources, I cannot imagine how this could be achieved without the option of families to choose those with whom they trust and cooperate.

Forced cooperation does not promote tit-for-tat, an evolutionarily robust strategy.  It's a quick fix that makes the voicing of self-interest look greedy when it's exactly what makes our species strong.

We just need to get down to the job of educating people so that they understand that building a [i]better world is a matter of self-interest[/i].

I recently read a few books by Ayn Rand such as Atlas Shrugged and The Fountainhead.  MCrawford, you seem to have very similar views.  Do you know anything about Objectivism?  I agree with you entirely, by the way.

We're pretty off topic now but I'd like to comment on the charter school issue.  Since 1993, over 100 charter schools have opened in Colorado.  The results have been mixed.  The worst schools suffer from poor management and lack of focus.  The best charter schools outperform the neighborhood schools by a wide margin.

My kids have benefited from attending one of these fine charter schools, which is not only one of the best performing middle schools in the state but has repeatedly won national recognition as one of the finest schools in the country.  Does our community applaud this school and use it as a shining example for other schools in the district?  On the contrary, our charter school is repeatedly attacked for elitism in spite of a lottery-based admissions policy.  The local school board (and teachers union) is very antagonistic toward our school and would love to see it disappear.  Fortunately the state of Colorado currently supports charter schools and encourages their continuation.  Unfortunately our local school board has almost complete control over charter school funding!  This means our school receives 70% of the per-pupil funding awarded to non-charter schools.  And it still outperforms year after year, as do other charter schools in our district.

Since the introduction of charter schools to our community, the overall performance of all the schools has improved because they're now competing for students.  Some of this competition has had negative consequences but in general I think our community and our children have benefited from the increased number of choices available.  We no longer accept the old "one size fits all" policy.  There are now Montessori schools, Core Knowledge schools, arts schools, etc., to choose from.  Parents now look at standardized test results and other performance data before selecting a school for their children.

One of the biggest problems with the introduction of charter schools has been an increase in ethnic and economic segregation.  The district could handle this problem by providing free busing for low-income students, making it easier for them to attend charter schools.  Instead they're proposing a takeover of the charter schools admissions process in order to discourage families from choosing non-neighborhood schools.  It's been very frustrating watching the local school board repeatedly place handcuffs on our charter schools, hoping they will fail.

Because of the lower per-pupil funding, charter schools in our district cannot match the teacher salaries offered by the regular public schools.  Unfortunately this has resulted in high teacher turnover.  Fortunately at our school we've had a loyal group of excellent teachers who have stayed for years because they believe in the mission of our school (to challenge students with high academic standards) and because they appreciate having more autonomy in their classrooms.  It's a struggle to manage costs on such small budgets but our local charter schools have succeeded, mostly by reducing administrative overhead.  If charter schools were able to match regular teacher salaries, I think they would be able to attract more of the best teachers.

Some charter schools in other communities have suffered from management problems but in our community I think charter schools have been very successful in providing alternatives for families dissatisfied with their neighborhood schools and raising academic standards across the entire school district.

Thank you for posting your experiences.

[quote="wild octagon"]
One of the biggest problems with the introduction of charter schools has been an increase in ethnic and economic segregation.[/quote]

It seems to me that this is a problem in schools in most areas of the country no matter what system is used (unless integration is forced).  I remember the bus students at my own high school being lost in the process of fitting in.  For the most part it was like there were two distinct cultures in the same school -- the larger, wealthier local culture, and the less affluent bus-student culture.  While I'm sure that helped some of the students, there was no cultural integration.

I think cultural integration occurs more naturally when better schools simply elevate more students of different race backgrounds.  Perhaps the first step is on the local level where standards would be raised for low-income schools leading to great economic development.  While the benefits of cultural integration would not be immediate, there may be no good immediate solution (even if many well-meaning people want there to be one).

I would love to see more charter schools established near low income neighborhoods, which would reduce white flight and segregation.  The problem is the parents who take the initiative to start new schools tend to live in wealthier, more educated, neighborhoods.   Maybe some of these successful charter school founders should share their experiences with parents in low income neighborhoods.

In Colorado, one simple way to improve the performance of low-performing schools would be to eliminate ESL and bilingual programs or offer them as afterschool options only.  As the child of immigrant parents who did not speak English, I sympathize with the desire to raise bilingual and bicultural children but I firmly believe that bilingual education should take place outside of school.  All young Americans deserve to learn English in school and to learn it well.  How else will they be able to compete for jobs when they're adults?  The last time I read a debate about bilingual education in the local paper, the Hispanic parents were against it while the English-speaking parents were in favor because they wanted their children to be exposed to multiculturalism!

[quote="wild octagon"]I would love to see more charter schools established near low income neighborhoods, which would reduce white flight and segregation.  The problem is the parents who take the initiative to start new schools tend to live in wealthier, more educated, neighborhoods.   Maybe some of these successful charter school founders should share their experiences with parents in low income neighborhoods.
[/quote]

Unfortunately, lower income parents are not nearly as likely to have the time and resources (including know-how and social connections) to initiate such action.

I often wonder if a company might develop that specializes in providing the know-how and a plan for such schools.  States could hire them to help clean up the problem.  This might solve the problem of states spending money that isn't used well in some schools due to the level of poverty and chaos.

[quote]I often wonder if a company might develop that specializes in providing the know-how and a plan for such schools. States could hire them to help clean up the problem. [/quote]

Such companies exist already---notably Edison Schools, which made the flamboyant move of hiring the president of Yale, Benno Schmidt, to head the company over a decade ago.   The company currently runs charter schools serving a quarter million kids in 20 states.   http://www.edisonschools.com/overview/ov0.html

There's a number of other companies doing the same sort of thing on various scales around the country.

However,  such businesses face lots of obstacles--to begin with, big poitical opposotion from teachers unions and government employee unions generally.

In NY, teacher unions and allies are the single biggest political donor block--in terms of their  contributions to state legislative campaigns.   Most of the very large troubled urban school systems are located in states that happen to have heavy union organization.   And privately operated charter schools rarely save the district money in the short run, because the district is typically still stuck with heavy fixed costs for bond issues, teacher pensions, etc     So when some kids go off to privately operated charters  (followed by their share of the tax dollars), the total school district budget often rises and the local taxpayers often end up paying more, at least in the short run.

In states where teacher & government employee unions are not so strongly organized, it's a bit easier, but there are stilll difficulties.  How to measure value-added is controversial, especially if there is a lot of student/family turnover (which often happens in low-income areas.)

There was a lot of excitement and sizzle back when the Edison Schools project was launched in the early 90s.   It was being sold as  "the next big hot thing" to investors.    I can imagine investors must be far more jaded now.

It's hard to do a great job of educating kids who come from very troubled and chaotic family backgrounds, who may have a lot of health problems, who move around a lot, who have parents with alcohol and drug problems, domestic violence, etc.  Just keeping some semblance of order in the classroom isn't easy.

 And if your company's schools  manage to educate some kids, it's often hard to prove or establish the value-added to a skeptical public constituency--you may have had a real impact on kids who have  moved out of your school by the time the test comes around.    And high-stakes standardized tests present enormous incentives and opportunities for cheating/integrity issues, especially if corporate profits are at stake.

Different scholars have looked at the same data from charter schools and tuition voucher plans and come up with diametrically opposite conclusions as to whether those schools do in fact do a better job of educating kids than the regular public schools.  After all, it's necessary to adjust for the fact that the sample of students whose parents choose to enroll them may not be representative.    Reasonable people can differ about how to do the statistical adjustments necessary--and the public winds up confused in the end.

(I'm struck by the parallels to the drug tests run by the big pharmaceutical companies to support the introduction of the new medicines they've invented.   There seems to be a lot of concern about the integrity of the research protocols, given that the drug companies fund and sponsor the testing and have an incentive to make sure that the studies come out in a way that will present their new drug in the most favorable possible light.   And yet, drug testing can be a lot more straightforward than educational testing.  You can do double-blind placebo-controlled studies for drug tests.  You can't really do random-assignment double-blind placebo-controlled studies for educational interventions.  Whatever the flaws in drug-testing methodology, I suspect the flaws in educational-program--testing methodology are far greater. )

In the end, I'm pessimistic about the idea that investor-backed charter schools are going to be wildly successful on a large scale in turning out incredibly well-educated kids.

The discipline problems in school are huge.   And school administrators have a limited amount of ability to control those discipline problems if they have to take "all comers,"  as charter schools are generally required to do.   The liability exposure if bad stuff happens to a kid in a school is big too.

I'm not against market solutions in general, but as a a potential investor looking for a place to put my money, I'd cast a jaundiced eye on private companies that think they can make a quick buck delivering high quality education in troubled neighborhoods.

The big success story, in my eyes, are the struggling Catholic schools that manage to operate on a shoestring, with dedicated and caring teachers who work for a pittance (lower salaries and MUCH lower benefits), but who manage to create an orderly environment in which kids who want to learn can do so--because they have the power to expel disruptive kids.  A big part of the reason for their success is the fact that the kids' parents are committed to the enterprise---they may not be able to pay a lot of tuition, but whatever they do pay is a real sacrifice and means that they have a commitment and stake in their kids' education.   They are going to make sure their kid makes the most of the opportunity.   If the kid gets in trouble at school, the kid knows he's going to be in trouble at home too.  Increasingly, a majority of the kids attending urban Catholic schools are not Catholic---but parents of the enrolled kids see the school as a haven of safety and order and caring  in which their kids can learn.   But the impetus for those schools doesn't come from a profit motive--it comes from a sense of a larger mission of service and dedication to the community.  And the success of those schools comes from the fact that the sense of mission is shared by the parents as well as the school staff.

I'm no longer Catholic, but I'm awed by the personal charisma and dedication of some of the teachers I know who work in inner city schools (both public schools and Catholic schools.)  It is possible for dedicated and caring public school teachers to make a huge difference in urban schools--but too many of the best burn out.   The daunting and silly educational bureaucracy frustrates a lot of the best public school teachers.    Susan Ohanian, a former Teacher of the Year from a troubled urban school district in NY, writes eloquently of the idiocy of the bureaucratic constraints she faced as a teacher.

sophia,

I realized right after I posted that I had heard of some companies like that, but had never been in an area where they operated (at least to the extent that it was noticeable).

I agree with what you say.  Dedicated staff are the key, not profit motive.  Unfortunately I think Catholic schools are not easily replicated in creating a missionary environment, so I keep looking for somebody who will devise a solution that scales.  It seems difficult to create a large scale solution without profit motive.  At the very least, an economically robust system is likely to stand the test of time to provide worthy experimentation in education, particularly when the obstacles in experimentation as so great as you mention.

In addition to the problem of selecting an appropriate vehicle for action, you hint at the problem of a nonexistent incentive structure.  Unions and therefore government have lower incentive for supporting the creation of alternative environments, particularly those that create competition.  That's a shame given that evolution in virtually any environment is dictated by competition for resources (and I think that students are often implicitly viewed as "resources" to teachers unions and therefore to politicians).

Additionally, the incentives for parents to enroll students in experimental or new schools may be skewed for many of the reasons you mention.

A problem solving approach might be a matter of providing a new, more sane incentive structure, but it's hard for me to imagine that happening within the status quo system.  And it's a mind-bender to formulate a solution from outside the system which often views itself as both a paternal entity and with great defensiveness.

[quote="JBL"]Wildoctagon, what schools in NYC did you attend?[/quote]Uh ... I'd rather not say, if you don't mind.  Few have ever heard of my elementary school and junior high.  My magnet HS ([i]not[/i] Stuyvesant, Bronx Science, Brooklyn Tech, or Hunter HS) was a very good school with an excellent math dept.  (btw - this was eons ago, probably before you were born.)

[quote="JBL"]Fair enough.  Just as an aside, I don't think your suggestion for parental involvement in the poorest city areas is realistic unless we implement a far superior social safety net.  And until we do that, schools will continue getting stuck with issues that they really have no business or expertice dealing with and that undoubtedly take away from the academic programs.[/quote]

I think admitting that we don't have a system for dealing with those locales means admitting that we can't school them.

I'm not saying I disagree.  Unfortunately I agree with you.

My thought is that continued economic improvement will hopefully generate more jobs to get such locales more integrated into the social fabric.

[quote="JBL"]Fair enough.  Just as an aside, I don't think your suggestion for parental involvement in the poorest city areas is realistic unless we implement a far superior social safety net.  And until we do that, schools will continue getting stuck with issues that they really have no business or expertice dealing with and that undoubtedly take away from the academic programs.[/quote]Unfortunately inner city schools have to make it their business to deal with social issues as well as academic ones.  I realize there are no easy solutions to inner city education problems but there have been a few success stories like East Harlem's [url=http://www.thenotebook.org/editions/2003/fall/harlem.htm][u]District 4[/u][/url] that have demonstrated the progress that can be made when bureaucratic micromanagement is eliminated and school choice is offered to families, teachers and principals.

I think we should also keep in mind that inner city schools aren't the only ones with major problems. When I hear discussions about bad schools I always hear people bring up inner city schools as examples. But it may be that, as a whole, poor rural schools are even worse off academically.

South Carolina for instance has the lowest average Math SAT score in the country (I think around 475), and a high school graduation rate just barely over 50%.

School choice can only do so much for people in rural areas because there may only be one or two high schools within a gazzillion miles of them. Plus, its harder to get good teachers to come to rural areas than to come to schools in urban areas.

When you look nation wide I bet there are just as many kids who live in these type of areas than who in poor urban environments - but no one ever sees or talks about them. This is just an example of the scale and diversity of educational situations. So even in reform tactics, one size doesn't fit all.

The fact that school choice only works when there are choices seems like a great reason to continue to develop better internet resources.  At the very least the internet is likely to begin supplementing the education of more and more of those students.

For students with high speed connections, the Internet is a great resource.  I recently discovered MIT's OpenCourseWare [url=http://ocw.mit.edu/OcwWeb/Global/OCWHelp/avocw.htm][u]video lectures[/u][/url] in physics, linear algebra, and a few other subjects.  We can now audit, for free, courses taught by top MIT professors.  Imagine if all our top universities followed MIT's example and offered free video lectures on the Internet.

Yeah, I love MIT Open Sourceware.  I've probably spent 100 hours reading from their lecture notes over the past couple of years when I've had the itch to learn about something from their huge online course selection.

The Intro to Bio course includes lectures by Eric Lander, former US Math Olympiad team member.

[quote="gauss202"]I think we should also keep in mind that inner city schools aren't the only ones with major problems. When I hear discussions about bad schools I always hear people bring up inner city schools as examples. But it may be that, as a whole, poor rural schools are even worse off academically.[/quote]As a native New Yorker, I don't know much about rural schools but I'm interested in learning more.  I assume they don't have to deal with the crime problems prevalent in urban ghettoes.  Are there problems specific to rural schools that are not found in urban and suburban settings?

[quote="wild octagon"]Are there problems specific to rural schools that are not found in urban and suburban settings?[/quote]

It's damn hard to attract teachers.  Students must travel over great distances.  Funding can be very hard to come by.  Uneducated parents.  Students needed to work at home (on farm, for example).
Not an exhaustive list, of course :) -- urban and rural schools have a lot of overlapping problems and a lot of great differences (especially racial, at least in the northeast).

[quote="JBL"]It's - hard to attract teachers.  Students must travel over great distances.  Funding can be very hard to come by.  Uneducated parents.  Students needed to work at home (on farm, for example).
Not an exhaustive list, of course :) -- urban and rural schools have a lot of overlapping problems and a lot of great differences (especially racial, at least in the northeast).[/quote]The working on the farm issue is the only one that seems unique to rural schools.  If kids have to spend hours each day on farm chores then it would be difficult for them to get homework done.  On the other hand, they're learning about hard work and discipline instead of watching TV or playing video games so that should help them in the long run.

Urban schools also have difficulty hiring teachers, for different reasons obviously.  (I remember one NYC substitute teacher who clutched her purse tightly the whole time she was writing on the blackboard.  I guess she thought we would steal her money if she put her purse down.  :lol:  My school must have neglected to inform her that our class was one of the good ones.)  Commute time is also an issue in urban areas.  In NYC, for example, it's not uncommon for middle and high school students to commute an hour or more each way on buses and trains to reach magnet schools.  I did.  I guess the difference with rural districts is the little kids also have to commute long distances.  And of course the issue of uneducated parents is common in many areas of the country.

I'm not sure what funding problem you're referring to.  Are rural districts allocated less money than urban districts?  In New York State the opposite has been true.  For many years the state government siphoned off funds that should have gone to NYC and allocated them to wealthier districts instead.

Many of the problems rural districts face are similar but take on a different form.  For example, in New York City, those students (and I too was among them) who travel long distances to school have a good transit network to travel in and could have chosen a nearer school, had they so desired.  In rural districts, there is very often one single school, with only small numbers (30-100) of students per class, but covering very large geographic regions.  Also, rural districts suffer brain drain to a much, much worse degree than urban districts do.

[quote="wild octagon"]I'm not sure what funding problem you're referring to.  Are rural districts allocated less money than urban districts?  In New York State the opposite has been true.  For many years the state government siphoned off funds that should have gone to NYC and allocated them to wealthier districts instead.[/quote]

That's not the opposite effect, it's just a different effect.  Rural districts aren't wealthy, just as urban districts aren't.  Property values are low, as is the ability of residents to pay property taxes.  It's the suburban districts that have constantly been funded at higher levels than urban districts in New York, but rural districts get screwed by the system as well.  I don't know that they are allocated less money on a per-pupil basis than NYC and New York's other urban areas, but like the inner city, they require more state funding.  You can't generate such convenient numbers (like the 5 or 6 billion dollars the state should be paying the city but isn't) for rural districts because they are much more spread out and serve many fewer students and have different needs, but it's clear in New York that the suburbs have a plum situation while much of the rest of the state, city and rural, does not.

[quote="qvc122"]If you go back 100 years, most people were illiterate.[/quote]

That statement is not factually true about the United States. It surely is true about, for example, China. Horace Mann, when he promoted compulsory school attendance laws, published literacy figures for Massachusetts in his Common School Journal. Those literacy figures, from the pen of an advocate of compulsory school attendance, were above 90 percent. It's rather doubtful that literacy is much higher today than it was when Mann was writing. Mann's desire was not to bring an illiterate society into literacy, but rather to bring an increasingly diverse society into uniformity. He wanted to standardize religion and political attitudes at a time when immigration from various countries (particularly Ireland) was populating Massachusetts with people who didn't agree with Horace Mann on those issues.

Much more about the history of compulsory school attendance can be found in an [url=http://learninfreedom.org/school_state.html]online bibliography about school and state issues[/url] that I refer to from time to time.


[b]ROUND 1 HAS BEGUN[/b]

Players:
Klebian
pianoforte
AnonAmoz
azhang
Mustafa

Here are the current powers:
[list]
[b]Cloned Sidekick[/b]
	Cloned Sidekick has 50 Energy and a Power of your choice from those in the current game.

[b]Curse[/b]
 The lowest bid(s) is stuck with this curse. This curse cannot be discarded. The highest bidder pays for it. You can only use one activated power per turn. You can only attack one person per turn. All normal attacks you make deal 10 less damage than normal. (THIS IS BAD, IN CASE YOU DIDN'T REALIZE IT. IF YOU DON'T BID, YOU BID 0 ON THIS, MEANING YOU AND ANY OTHER PEOPLE WHO BID 0 WIN THIS CURSE)

[b]Sniper[/b]
 Global Effect (This Power may not be bid on. It affects all Heroes): Each round, Sniper makes a 10 Point Attack against the opponent with the lowest total Defense , after all other Powers are applied.

[b]Assassin[/b]
 Choose one player in the game. Choice is not revealed to others until end of round. That player's attack and defense is lowered to zero for one turn. Max damage to that player's energy that turn is 50. Can only be used once.

[b]Army of Noobs[/b]
Summons army of 10 noobs who can each do 2 damage. They are controlled by the owner of this power. Once they attack, they are removed from the game. Their damage is unblockable, but no more than 5 noobs can be used on the same person. If you are attacked by them, you cannot attack next turn.

[b]High-Powered Energy Shield[/b]
	Burn X: Add 10X to your Defense.

[b]Masochist[/b]
Each time you take Damage, gain a permanent +1 to your regular Attacks.

[b]I think I can...[/b]
You can use one other hero's power (of your choice) each turn. (You can pick each turn)

[b]Regeneration[/b]
Gain 3 Energy at the beginning of each round.

[b]Glass Swords[/b]
Add 40 to each of your Attacks. Lose this power the first time you sustain any damage.
[/list]

Please nobody take Pseudonite, that one is just cruel as a cucumber.

Wait what the heck, you chose the powers??????

Yes, I wanted to choose just to see how it would work out, since some people may have not played before.

Ok... so... should I get rid of the list and allow everyone to send in two?

EDIT: I will, since you two seem unhappy.

[quote="mustafa"]Just because someone hasn't played doesn't mean that they are helpless, plus Klebian already made a game of this a while ago.[/quote]
Oh, huh, I actually did run it. I forgot I had.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113443 is the link.

Some people have submitted one power, one has submitted none, please submit ASAP. There will be 10 powers at the end. If I have less than 7 at the deadline(6:35 PM PST), it will be extended another day. If I have 7 or more, the remaining powers will be selected by me.

So for Army of Noobs, at the beginning of the game, a player gets 10 noobs, and can have them attack when he wants to. Max damage allowed using noobs the entire game is 10. Can noobs be attacked?

I'd have to say no. I never really thought of that, because this power was made by a player, but since they don't really have energy, they just do damage, you cannot attack them.

One person has not submitted.

EDIT: The deadline has passed, I will make two powers.

Regeneration
	Gain 3 Energy at the beginning of each round.

Glass Swords
	Add 40 to each of your Attacks. Lose this power the first time you sustain any damage.

Klebian, could you please add these to the list and change the bold text to say "Send in your bids within 48 hours"?

EDIT: I don't think you'll see this, so I'll just PM you.

Oooohh... bold text. :lol:  It looks better like that, thanks!

How many people still haven't sent in bids?

Deadline is in like 10 minutes, but I'll extend it another day, if the people who have submitted agree. If not, all the people who didn't submit use half their energy on defense(the other half is not used, this is your "punishment" for not submitting).

It's ok with me if you extend the deadline, this game is pretty pointless if only half the people play. You might want to send out reminder pms, in case the missing people forgot about this

AnonAmoz can't play anymore, and I think it's kinda late for a replacement, so his energy will all be allocated to defense every turn, unless that makes it impossible for the game to end.

Man, I think this game is failing. :o

It's a fun game, we just need more committed people.

Wow, this sucks. azhang has forfeited.  :(

Do you guys still wanna continue, or should I try to start a new game with more people? :o

Let's continue, the way the game was going, it was basically going to turn into a showdown between me and Pulak anyways

Alright. Round 2 results coming soon in this post. azhang's hero will become a time bomb. Every turn he doesn't take damage(his defense will be set at the highest possible, and the curse will be removed.), the counter goes up by 1. Every turn he does take damage, the counter goes up by 2. Once the counter goes past 5, he will make an all out attack on one surviving person the next turn, determined by http://www.random.org, and will continue doing so until he dies.

Klebian
Powers: Assassin, I think I can...(used for army of noobs)
Starting Energy: 61
Attacks: AnonAmoz-20, pianoforte- 6
Defense: 25
Attacked by: pianoforte- 4
Ending Energy: 61

Klebian's Sidekick
Powers: I think I can...(army of noobs)
Starting Energy: 40
Attacks: AnonAmoz- 30 + 2 noobs(4), azhang- 2 noobs(4)
Defense: 10
Attacked by: nothing
Ending Energy: 40

pianoforte
Powers: Army of Noobs
Starting Energy: 4
Attacks: Klebian-4
Defense: 0
Attacked by: Klebian- 6
Ending Energy: 0 DEAD(also hit by sniper)

azhang- TIME BOMB
Powers: none
Starting Energy: 50
Attacks: nothing
Defense: 50
Attacked by: Klebian- 2 noobs(4)
Ending Energy: 46
Other: Counter - 2

AnonAmoz
Powers: Masochist, Regeneration
Starting Energy: 36(regeneration)
Attacks: nothing
Defense: 36
Attacked by: Klebian- 20, Klebian's sidekick- 30 + 2 noobs(4)
Ending Energy: 18
Other: Masochist bonus total- 2

Mustafa
Powers: Cloned Sidekick, High-Powered Energy Shield,
Starting Energy: 44
Attacks: nothing
Defense: 43 + BURN 1 to use Energy Shield(10)= 53
Attacked by: nobody
Ending Energy: 43

Mustafa's Sidekick
Powers: I think I can...(Energy Shield)
Starting Energy: 50
Attacks: azhang- 30
Defense: 18 + BURN 2(20)= 38
Attacked by: nobody
Ending Energy: 48

Check to make sure it's right. I was too lazy to bold it and stuff. Send in round 3 moves ASAP.

This thing sure seems dead, I've only got moves from one person.... But then again, there only are 2 active ones left in the game.

Should I end this and start possibly start a new game?

If so, I'd like to join.  (As long as everybody is active.)

[quote="CheeseIsGood"]MUSTAFA! Where are you?[/quote]

ME?!?!?! I sent in my stuff a long time ago...

Really?  :huh:

[url=http://img207.imageshack.us/my.php?image=inboxan7.jpg][img]http://img207.imageshack.us/img207/7163/inboxan7.th.jpg[/img][/url]
I don't see it.

:o
Results will be up tomorrow... I'm kinda busy, and the interest in this game seems to have dropped to 0.

Argh... I don't think I can do this anymore, it takes about 5 minutes, but the interest in the game has died. Closed.  :(


otaghi darim ba iek lamp,va kharej az an 3 kelid .vali nemidanim ke kodam kelid baraye lamp ast,chegoone mitavan faghat ba 2 bar estefade az kelid ha va iek bar varede otagh shodan kelide lamp ra yaft?

[quote="ehsan2004"]Aval iek kelid ra be tasadof roshan mikonim , [i]bad az gozashte chand daghighe[/i] yeki az kelid haye baghimande ra roshan mikonim[u] va haman lahze varede otagh mishavim [/u], agar lamp roshan bood anvaght be lamp dast mizanim ,agar lamp garm bood mifamim ke kelid lamp kelid aval boode vali agar sard bood mifahmim ke kelid lamp kelide dovom boode (chon ke taze roshanash kardim)  vali agar roshan nabood maloom ast ke kelid sevom kelid lamp boode.[/quote]

ghabule azadeh? :maybe:
mitunim be lamp dast bezanim? :huh:


there is a competition here in the philippines that asked this question:

"A = 16^(0.249999999....)

Write A as a fraction in simplest form."

Some of the contestants answered 2, while others answered 2/1.

initially, the correct answer that flashed on the screen was 2. but of course, those students with 2/1 as an answer protested. in the end, the judges considered both answers as correct.

any opinions on this?

Every integer is also a rational number, and as 2/1=2 and 2/1 is reduced, both answered should be considered correct.

If we were to really seek for an argument, 2 is 'wrong' (notice that I've put it in single quotes). Why? Literally, assuming that we take word for word and following all instructions, 2 is not 'written' as a fraction; on the other hand, 2/1 is written as a fraction in simplest form (as a ratio of two integers). Thus, for the sake of winning an argument and biting at the problem committee, they were 'wrong', literally  :wink:

I have no idea if there is a formal definition of fraction that takes the case of integers into account. Just compute A, then say "Saying 'Express 2 as a common fraction' makes no sense. Hence my answer will be A=2".

Actually in retrospect I think this competition is multiple choice.

[quote="leprosy"]"A = 16^(0.249999999....)

Write A as a fraction in simplest form."[/quote]

I would have to say that both 2 and 2/1 are correct as well. One of them is simplest and the other is a "fraction". Take your pick.

To avoid confusion the next time around, I'd say "write A as a rational number in the simplest form". That way you don't have people protesting that it was supposed to be a fraction.

If it says A in a fraction, 2/1 would be correct. If its simplest form, 2 would be correct.

Stop the discussion, everything has been said, and you are just repeating.


[i][color=brown]This comes out from another problem [/color][/i]

Find $ m$ such that $ 2223-m^{2}\mid 14553$

[hide="I dont know if this is a hint"] $ 14553 = 21^{2}\cdot 33$

$ 2223 = 33^{2}+21^{2}+21\cdot 33$[/hide]

Why not just use that $ 2223=3^{2}\cdot 13 \cdot 19$?  Then, you are solving
\[ 3^{2}\cdot 13 \cdot 19-m^{2}\mid 3^{3}\cdot 7^{2}\cdot 11\] i.e. solve \[ 3^{3}\cdot 7^{2}\cdot 11=y(3^{2}\cdot 13 \cdot 19-m^{2}).\]  Notice that if $ 3$ does not divide $ y$, then there is no solution for $ m$ because at most $ 3^{2}\mid (3^{2}\cdot 13 \cdot 19-m^{2})$.  So, one only needs to check the cases where $ 3\mid y$.  Altogether, you just need to check:
$ (3^{2}\cdot 13 \cdot 19-m^{2}) = 3^{a}7^{b}11^{c}$ where $ a=0,1,$ or $ 2$, $ b=0,1,$ or $ 2$, and $ c=0$ or $ 1$ (18 equations).  For example, the case where $ a=b=c=0$: $ 2223-m^{2}= 3^{2}\cdot 13 \cdot 19-m^{2}= 1$ implies $ 2224=m^{2}$ which is not true.

This may not be the most elegant, but it gets the job done. :(   (For the moment I don't see anything easier).

[quote="pco"]So the only solutions are $ m=12$ and $ m=30$[/quote]
What about $ m=54$?  :)  I think you might want to check those again.  But, the ideas above should work (its just the computation that is a little annoying).

[quote="baz"] What about $ m=54$?  :)  [/quote]

I considered everything was in $ \mathbb N$, including $ 2223-m^{2}$  :blush:


geometry
perimeter
trigonometry
LaTeX
area of a triangle
trig identities
Law of Cosines

[hide="icky way"]
Area of a triangle is $ \frac{ab \sin C}{2}$, in this case, $ C\equal{}30$, so $ \frac{ab}{4} \equal{} 2$.
Obviously, $ ab\equal{}8$. Express all the sides in terms of $ a$. The sides of the triangle are $ a$,  $ \frac{8}{a}$,   $ 8\minus{}(a\plus{}\frac{8}{a})$
Plug these values into herons formula and set it equal to 2 and solve the equation.
There has to be an easier way though.
[/hide]

[quote="Lulze"]This was a problem from a team round that I wasn't able to figure out (source in the subject description):

A triangle has an area of 2, a perimeter of 8, and an angle of 30. What is the length of the side opposite the 30 degree angle?[/quote]

[hide="probably a tad advanced and long"]
call the sides a, b, and c
c is the side opposite the 30 degree angle
we know area is $ \frac {1}{2}ab\sin{C}$
which in this case leads to $ 8 \equal{} ab$
by law of cosines $ a^2 \plus{} b^2 \minus{} ab\sqrt {3} \equal{} c^2$
now add $ ab(2 \plus{} \sqrt {3})$
$ a^2 \plus{} b^2 \plus{} 2ab \equal{} c^2 \plus{} 16 \plus{} 8\sqrt {3}$
$ (8 \minus{} c)^2 \equal{} c^2 \plus{} 16 \plus{} 8\sqrt {3}$
$ 16c \equal{} 48 \minus{} 8\sqrt {3}$
$ c \equal{} 3 \minus{} \frac{\sqrt {3}}{2}$

i think thats wrong
looks too ugly

[/hide]

and it looks like mathcountsfreak had the same starting idea as me, ehe.

shouldn't the final answer be $ 3\minus{}\frac{\sqrt{3}}{2}$?

Might've been a typo.

[quote="mathcountsfreak"]shouldn't the final answer be $ 3 \minus{} \frac {\sqrt {3}}{2}$?

Might've been a typo.[/quote]

oh thanks.  that was probably a latex error on my part

[quote="Lulze"]This was a problem from a team round that I wasn't able to figure out (source in the subject description):

A triangle has an area of 2, a perimeter of 8, and an angle of 30. What is the length of the side opposite the 30 degree angle?[/quote]
[hide]Let $ c$ be the side opposite to angle 30. So $ \frac{1}{2} ab \sin 30 \equal{} 2$ or $ ab \equal{} 8$.

Using cosine rule, we have $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos 30$ or $ a^2 \plus{} b^2 \minus{} c^2 \equal{} 8\sqrt{3}$.

Since $ a\plus{}b\plus{}c \equal{} 8$, we have $ (8\minus{}c)^2 \equal{} a^2 \plus{} b^2 \plus{} 2ab$ or $ (8\minus{}c)^2 \equal{} a^2 \plus{} b^2 \plus{} 16$. Subtracting this from the previous eqn, and solving for $ c$, we obtain $ c \equal{} 3 \minus{} \frac{\sqrt{3}}{2}$.[/hide]

[quote="CircleSquared"][quote="Lulze"]Thank you very much! But I'm confused as to how you went from

$ a^2 \plus{} b^2 \plus{} 2ab \equal{} c^2 \plus{} 16 \plus{} 8\sqrt {3}$
to
$ (8 \minus{} c)^2 \equal{} c^2 \plus{} 16 \plus{} 8\sqrt {3}$[/quote]

perimeter is 8
and perimeter is a+b+c
so a+b+c=8
a+b=8-c
on the LHS on the top equation
you see (a+b)^2
so its (8-c)^2

too lazy to latexify[/quote]

Ahh, thank you!


I'm interested in how many schools offer computer science, and how many students take it.

If it is offered at your school, is it some sort of requirement for the students there? What language is it taught in? Does your school offer AP computer science?

It's an elective at my school. There is one teacher. There are between 10 and 20 kids in all of his classes. He offers regular classes (in basic computer literacy and BASIC), an honors class in BASIC and C++, and an AP class (in Java). The honors class is the prerequisite for the AP class.

My school only has AP computer science (Java) which is an elective.

nope.
I don't have one, that's probably cuz I'm only in middle school..

My school has

Programming for the Internet (half yr)
which is a recommended prerequisite for
AP Comp Sci (full yr)

I think there are some other courses as well that at least dabble in programming

I'm planning on taking both, first next yr, then AP the yr after.  We don't always have enough people sign up for the AP though...

We've got Basic, C++, and AP Java. I'll probably take advantage of this, but not in my schedule for next year.

My school used to offer one, but they closed it about 3 years ago.

at my school there were courses in grades 10, 11 and 12.  VB6, C++ and Java respectively.  no AP stuff in my high school at all.  really simplistic material.  no algorithms.  cool final projects though.  i prepared for the Canadian Computing Contest a lot by myself.

The school I used to go to offers AP Computer Science, but from what I hear, it bairly even succeeds in getting students to AP level.

My current school offers ~1 year introduction of C++ programming which is self-paced, so you can finish the course in 3 weeks if you have the time.  After that, it offers an introdcution to Java, which essentially means reading a big book on Java.  The third course in the sequence it offers is AP Computer Science, which is fairly trivial if you've taken the first two (Intro to C++ and intro to Java).

Really, AP Computer Science is a nice introduction to ideas in programming, but that course alone won't get you far in programming. (...although, the same could be said for all other AP courses...)

AP CS used to be not too bad.
Then they cut the AB, bc not many people took it.
ARGH.

I was gonna take the AB too, this year.  Too bad for me.
So I'm taking CS A, but as the actual "main course," I'm reading Intro to Algorithms and doing usaco :D

I know that my school has Computer Science, and AP Computer Science. I've only seen Java textbooks lying around. Also, I accidentally walked in on a class ( :blush: ), not sure if it was AP or not, but it only had 6 people or so...

high school = AP comp sci.  It is an elective, but I basically HAVE to take it because we are required to take 2 math courses in High school.  By the end of the year, I'll have completed half of calc BC, so I was hoping that during the summer, I would study the rest of Calc BC and take the AP test.  But now I have to actually take the CLASS calc BC at the high school :(

We have two Computer Programming classes. We have Computer Programming I (the class I am in now), which teaches QBasic, and Computer Programming II (the class I want to take next year) which teaches C++ or Java. The teacher for that class goes super slow, and I want to be a computer programmer when I grow up, so  :dry: .

Fortunately, my school has a club, but no course. The courses on programming are only offered at the High School

We only have Java courses + general software/tech courses. The only classes that are geared specifically toward programming are Computer sciences I, II, III, IV and Advanced Object-Oriented Design.

My school(middle school) mandates "Computer Keyboarding" for all 6th graders.  There are also electives:  Computer Graphics, Digital Tools 1, Digital Tools 2, and Microsoft Office.  But there is no computer programming for the Munster School District.

Our computer science program has slowly dwindled to nonexistence. We used to offer one section of AP Computer Science A and at least two sections of Computer Science Programming Java (an introduction to coding in Java). This year, we're only offering the one section of AP Computer Science A, and we're having two seniors and one sophomore take an online class in Data Structures (which covers the now defunct AP Computer Science AB).


function
LaTeX
calculus
integration
real analysis
calculus computations

Find the laplace of the following function:

[color=white].........[/color]$ \{t$ $ 0{\le}t<1$
$ f(t)$=$ \{$
[color=white].........[/color]$ \{0$ $ t{\ge}1$

f(t) = t-U(t-1)?
is the equation of the above function?

similar qn

[color=white].........[/color]$ \{1\minus{}e^t$ $ 0<t<1$
$ f(t)$=$ \{$
[color=white].........[/color]$ \{0$ $ t>1$

find $ L{f(t)}$ using the unit step function

1. Isn't it $ t(U(t) - U(t - 1))$ ? The term in brackets is 0 for $ t < 0$, one for $ 0 < t < 1$ and zero again for $ t > 1$.

Applying similar logic, you can obtain an equation for the second.

$ \text{\LaTeX}$ help:

$ f(t) = \begin{cases}t & t\in[0,1) \\
0 & t\in[1,\infty)\end{cases}$

$ f(t) = \begin{cases}1 - e^t & t\in(0,1) \\
0 & t\in(1,\infty)\end{cases}$

PS: what's the value of $ f(1)$ in 2nd problem?  :)

[quote="hsiljak"]PS: what's the value of $ f(1)$ in 2nd problem?  :)[/quote]
Irrelevant. Changing the function on a set of measure zero does not change its Laplace transform. In practice, I tend to care about the value at a point of continuity, but not at a point of discontinuity.

One approach is simply to brute-force the appropriate integrals:

For the first example

$ F(s)\equal{}\int_0^1te^{\minus{}st}\,dt\equal{} \left.\minus{}\frac{t}{s}e^{\minus{}st}\right|_0^1\plus{} \frac1s\int_0^1e^{\minus{}st}\,dt$

$ \equal{}\minus{}\frac{e^{\minus{}s}}{s}\minus{}\frac{e^{\minus{}s}}{s^2}\plus{}\frac1{s^2}.$

If we now try to reverse-engineer a formula for $ f$ by running that through our table of inverse Laplace transforms, we get

$ f(t)\equal{}U(t\minus{}1)(\minus{}1\minus{}(t\minus{}1))\plus{}t\equal{}t\minus{}U(t\minus{}1)t.$

(I'm treating $ U(t)$ as understood in context. This is consistent with what hsiljak wrote.)

We could similarly get the second example just by doing the integral.


geometry
vector
trigonometry
geometric transformation
reflection
geometry proposed

Two circles $S_{1}$ and $S_{2}$ with centers $O_{1}$ and $O_{2}$, respectively, intersect at two points $A$ and $B$. The tangents to the circles $S_{1}$ and $S_{2}$ at the point $A$ intersect the segments $BO_{2}$ and $BO_{1}$ at the points $K$ and $L$, respectively. Prove that $KL \parallel O_{1}O_{2}$.

[i]Russian olympiad/2003/proposed by S.Berlov[/i]

Not really that hard, but I felt like posting something :D

well after 10 mins of trying the problem with my lucky pen (a blue parker vector :D ) i think i've got the problem.
first,let P_1= BO_1 \cap AK   and  P_2=BO_2 \cap AL.
<AO_1O_2= x  and <AO_2O_1 = y.
note that BK/BL=P_1B cos(2x)/P_2Bcos(2y)
and again by sine rule ABsin(x)=P_1Bcos(2x) and similarly
ABsin(y)=P_2Bcos(2y).this implies
 BK/BL=sin(x)/sin(y)=BO_2/BO_1.
HENCE KL ll O_1O_2

Another solution, with no trigonometry at all:

Let $D_1$ and $D_2$ be the points diametrically opposite to the point B on the circles $S_1$ and $S_2$, i. e. the reflections of the point B in the points $O_1$ and $O_2$. Since the segments $BD_1$ and $BD_2$ are diameters of the circles $S_1$ and $S_2$, we have $\measuredangle BAD_1 = 90^{\circ}$ and $\measuredangle BAD_2 = 90^{\circ}$, so that $\measuredangle BAD_1 + \measuredangle BAD_2 = 90^{\circ} + 90^{\circ} = 180^{\circ}$, what shows that the point A lies on the line $D_1D_2$. Since the segments $BD_1$ and $BD_2$ are diameters of the circles $S_1$ and $S_2$, the centers $O_1$ and $O_2$ of these circles are the midpoints of these segments, and thus $O_1O_2 \parallel D_1D_2$. Therefore, instead of proving $KL \parallel O_1O_2$, it will be enough to show $KL \parallel D_1D_2$.

Well, since the angle between a chord of a circle and a tangent to the circle at an endpoint of the chord equals the chordal angle of the chord, we have $\measuredangle LD_1A = \measuredangle KAB$ and $\measuredangle D_1AL = \measuredangle ABK$. Hence, the triangles $D_1AL$ and ABK are similar, and we get $\frac{D_1L}{AL} = \frac{AK}{BK}$. Equivalently, $D_1L \cdot BK = AL \cdot AK$. Similarly, $D_2K \cdot BL = AK \cdot AL$. Hence, $D_1L \cdot BK = D_2K \cdot BL$, so that $\frac{D_1L}{D_2K} = \frac{BL}{BK}$. This immediately implies $KL \parallel D_1D_2$, and the problem is solved.

  Darij

Just another way to say the same thing.
LBKA is ciclic. Indeed <O1AL = <O2AK = <O1AO2 - 90 = x. Then <LAK = 90 - x and <LBK = 90 + x.
Therefore <BLK = <BAK = <BO1A/2 = <BO1O2. Then the thesis.

Wow. I didn't think it was so simple!

  Darij


What is the sum of the first $ 60$ odd positive integers?

Method I:
Since the sum of the first $ n$ odd positive integers is $ n^2$, the answer is $ 60^2\equal{}\fbox{3600}$.

Method II:
The 30th odd positive integer is $ 2(30)\minus{}1\equal{}59$. The 31st is the next odd integer, $ 61$. Since odd positive integers form an arithmetic sequence, the mean of these two values will be the mean of the entire set. Thus, we multiply the mean of $ \frac{59\plus{}61}{2}\equal{}60$ by $ 60$, the number of terms, to get $ \fbox{3600}$.


calculus
derivative
real analysis
real analysis unsolved

Can anyone show me how we can use the following two conditions to show that $ (x\minus{}\sqrt{x^{2}\plus{}y^{2}})dx\plus{}y dy \equal{} 0)$ can have a solution? Note that the equation is not an exact equation.
1. $ \frac{(x dx\plus{}y dy)}{\sqrt{x^{2}\plus{}y^{2}}}\equal{}dx$
2. $ \frac{1}{2}d(x^{2}\plus{}y^{2}) \equal{} x dx\plus{}y dy$

Another question:

Consider the integrating factor $ \mu$, often used to make a nonexact equation exact. Are the two equations $ Mdx\plus{}Ndy \equal{} 0$ and $ \mu M dx\plus{}\mu N dy \equal{} 0$ necessarily equivalent in the sense that a solution of one is also a solution of the other?

I said no b/c there could be some $ \mu$ such that $ \mu N$ or $ \mu M$ could be defined at say $ x\equal{}0$, while the original $ M$ or $ N$ was undefined at $ x \equal{} 0$.

Any advice?

Thanks.

forget the 2 conditions you have. start with the equation itself. clearly, it is [b]inexact [/b]since $ \frac{\partial}{\partial x}\,(y)\;\not\equiv\;\frac{\partial}{\partial y}\,\left(x\minus{}\sqrt{x^{2}\plus{}y^{2}}\right)$

this means you have to multiply the given ODE by an integrating factor $ \frac{1}{\textbf g(x,y)}$ which you will have to determine, i.e. try and turn the $ \text{LHS}$ into the derivative of a product (or maybe even a sum, but that is generally considered to be a little harder).

now, your equation is easy, and anyone can see the integrating factor has to be $ \frac{1}{\sqrt{x^{2}\plus{}y^{2}}}$, and so, multiplying the given ODE by that gives

$ \frac{x\,\textbf dx\plus{}y\,\textbf dy}{\sqrt{x^{2}\plus{}y^{2}}}\minus{}\textbf dx\;\equal{}\;0$
$ \Rightarrow\;\frac{\textbf d\,(x^{2}\plus{}y^{2})}{2\sqrt{x^{2}\plus{}y^{2}}}\minus{}\textbf dx\;\equal{}\;0$

integrating then yields
$ \sqrt{x^{2}\plus{}y^{2}}\minus{}x\;\equal{}\;C$
where $ C$ is an arbitrary constant

[b]TIP:[/b] always recall these 2 identities:
$ \textbf d(x\,y)\;\equal{}\;x\,\textbf dy\plus{}y\,\textbf dx$
and
$ \frac{1}{2}\textbf d(x^{2}\plus{}y^{2})\;\equal{}\;x\,\textbf dx\plus{}y\,\textbf dy$

and considering your other question, yes, both $ M\;dx\plus{}N\;dy\;\equal{}\;0$ and $ \mu\;M\;dx\plus{}\mu\;N\;dy\;\equal{}\;0$ are equivalent... remember, the integrating factor does nothing more than facilitate the solving of your equation. so long as the integrating factor is [u]not[/u] undefined within the region of validity of the differential equation, that will always be true. refer to a standard text on ODE (e.g. boyce & diprima) for a rigorous proof.