[quote="jimi"]Fraenkel

Are you expecting an exact answer to this?
Sorry, just saw the challenge symbol so obviously you are.[/quote]

yes.

i dont think you can do this w/o a calculator

The value of the sum is  irrational :huh:

just about all square roots are irrational

There are two possibilities;the answer is in the form $ a\sqrt{5}\plus{}b\sqrt{3}$ or$ a\plus{}b\sqrt{15}$.
Square those and set it equal to $ 53\minus{}8\sqrt{15}$.

one solution is:[hide]$ 4\sqrt3\minus{}\sqrt5$[/hide]

Actually you can.  This is an example of one.  Its $ \sqrt{49\minus{}20\sqrt{6}}$
Try solving it from there.  That should be a good exercise.  :yup:

are you sure that the square root of 49-20 times the square root of 6 is equal to square root 53-8 times the square root of 15? I think it is, put r u sure?javascript:emoticon(':?:')
Question

:maybe:
I'm pretty sure its correct.  Put it in the form of $ a\plus{}b\sqrt{15}$
 :)

Can you guys explain it from the beginning to the form of an answer?
(Sorry, I'm a visual learner...and also slow..)

I especially don't get what you guys mean by putting it as

$ \ a\plus{}b\sqrt{15}$ or $ \ a\sqrt{5}\plus{}b\sqrt{3}$....

[quote="xpmath"]There are two possibilities;the answer is in the form $ a\sqrt {5} \plus{} b\sqrt {3}$ or$ a \plus{} b\sqrt {15}$.
Square those and set it equal to $ 53 \minus{} 8\sqrt {15}$.[/quote]

do you mean the answer is $ 53\sqrt {5}\minus{}8\sqrt {3}$?

[quote="abacadaea"][quote="xpmath"]There are two possibilities;the answer is in the form $ a\sqrt {5} \plus{} b\sqrt {3}$ or$ a \plus{} b\sqrt {15}$.
Square those and set it equal to $ 53 \minus{} 8\sqrt {15}$.[/quote]

do you mean the answer is $ 53\sqrt {5} \minus{} 8\sqrt {3}$?[/quote]

I think so.. or it's $ \sqrt{49\minus{}20\sqrt{6}}$ according to minicon..

NO :!:
Thats not what I meant! I was putting an example for you guys.  That is [b][i][u]NOT[/u][/i][/b] the answer.
I said you could try to solve it step by step by looking at the attachment.
 :lol:

[quote="DonkeyKong"]So it's $ \ {53}\sqrt {5} \minus{} 8\sqrt {3}?$[/quote]

Obviously not (well, maybe not obviously).  Try squaring it to see if you get $ 53\minus{}8\sqrt{15}$.

well, if thats the case, then $ 8\sqrt {15}$ must equal 2ab. and $ 53\sqrt {5}*8\sqrt{3}*2$ clearly does not equal $ 8\sqrt {15}$

[quote="DonkeyKong"][quote="xpmath"][hide]
$ (a\sqrt {3} + b\sqt{5})^2 = 53 + 8\sqrt {15}$. Then,
$ 3a^2 + 5b^2 = 53$
and $ ab = 4$.
a=4,b=1 is a solution.
$ 4\sqrt {3} + \sqrt {5}$.[/hide][/quote]

Ah!!... I'm confused..
I don't get it..
Can you explain it with more details??
sorry...[/quote]

[hide]Basically what he did was let $ a\sqrt{3}+b\sqrt{5}=\sqrt{53+8\sqrt{15}}$.  Then, square both sides to get $ 3a^{2}+5b^{2}+2ab\sqrt{15}=53+8\sqrt{15}$.  Therefore, $ 3a^{2}+5b^{2}=53$ and $ 2ab=8$.  From there, you could solve as a quadratic or guess and check.  In this case, the guess and check is easier (in my opinion) since the second equation simplifies to $ ab=4$ and we try a=4, b=1 and it satisfies the first equation, so substituting $ a$ and $ b$ back into $ a\sqrt{3}+b\sqrt{5}=\sqrt{53+8\sqrt{15}}$, we get $ 4\sqrt{3}+\sqrt{5}$.[/hide]

You could also try to let the answer be in the form $ a+b\sqrt{15}$, but that gets very ugly results for $ a$ and $ b$.  Maybe not very, but it's not pretty.

[quote="alanchou"][quote="DonkeyKong"][quote="xpmath"][hide]
$ (a\sqrt {3} + b\sqt{5})^2 = 53 + 8\sqrt {15}$. Then,
$ 3a^2 + 5b^2 = 53$
and $ ab = 4$.
a=4,b=1 is a solution.
$ 4\sqrt {3} + \sqrt {5}$.[/hide][/quote]

Ah!!... I'm confused..
I don't get it..
Can you explain it with more details??
sorry...[/quote]

[hide]Basically what he did was let $ a\sqrt {3} + b\sqrt {5} = \sqrt {53 + 8\sqrt {15}}$.  Then, square both sides to get $ 3a^{2} + 5b^{2} + 2ab\sqrt {15} = 53 + 8\sqrt {15}$.  Therefore, $ 3a^{2} + 5b^{2} = 53$ and $ 2ab = 8$.  From there, you could solve as a quadratic or guess and check.  In this case, the guess and check is easier (in my opinion) since the second equation simplifies to $ ab = 4$ and we try a=4, b=1 and it satisfies the first equation, so substituting $ a$ and $ b$ back into $ a\sqrt {3} + b\sqrt {5} = \sqrt {53 + 8\sqrt {15}}$, we get $ 4\sqrt {3} + \sqrt {5}$.[/hide]

You could also try to let the answer be in the form $ a + b\sqrt {15}$, but that gets very ugly results for $ a$ and $ b$.  Maybe not very, but it's not pretty.[/quote]

Oh!! I see it now..

Thank you for your help!

Well, yeah, I did use trial and error first, but in some problems, it's hard to use trial and error, so I just wanted them to know you can also solve those.


conics
parabola
trigonometry
algebra proposed
algebra

Given a parabola passing through the two points $A\ \left(1,\frac{1-\cos \theta}{\sin \theta}\right),\ B\ \left(-1,\frac{1+\cos \theta}{\sin \theta}\right)$ and

with the directrix the $x-axis$.When $\theta$ varies, find the loci of the focus and vertex of this parabola.

i think this works.

the parabola is $y-k=1/(4a)(x-h)^2$.  The vertex will be $(h,k)$, directrix is $y=k-a)$ but the directrix is the x-axis so $k=a$.  The focus will be $(h,2k)$.

so we have

$\frac{1-\cos{\theta}}{\sin{\theta}}-k=\frac{1}{4a}(1-h)^2$ (i)
$\frac{1+\cos{\theta}}{\sin{\theta}}-k=\frac{1}{4a}(1+h)^2$ (ii)

(ii)-(i), and using the fact $h=k$ gives $2\cot{\theta}=h/a$ (iii)

(ii)+(i) gives $2\csc{\theta}-2a=\frac{1+h^2}{2a}$   (iv)

use (iii) to get $h=2a\cot{\theta}$, substitute into (iv) and use the fact 1+cot(x)^2=csc(x)^2,  it becomes $(2a\csc{\theta}-1)^2=0$ so $a=1/2 \cdot \sin{\theta}$.  so we can get $h$ from there and therefore the locii for vertex and focus.


Whats the evaluation of $I^{1066}$

a) 0
B)1
C) -3
D)-1

[hide]since $i^4=1$,

$i^{1066}=i^{4(264)}i^2=-1$.

Note that the 264 was not completely necessary. [/hide]


calculus inequalities vector quadratics algebra polynomial

[quote="hien"]Find all real numbers $a,b,c$ satify condition: $a-2b+3c-16=0$ such that expreesion $f=2a^{2}+2b^{2}+2c^{2}-4a-4b-4c+15$ has minimum[/quote]
This is a conditional extremum problem. With Lagrange's method, let $F(a,b,c,t)=2a^{2}+2b^{2}+2c^{2}-4a-4b-4c+15+t(a-2b+3c-16)$
$F'_{a}=0,F'_{b}=0,F'_{c}=0,F'_{t}=0 \implies~{\matrix{4a-4+t=0\cr, 4b-4-2t=0\cr, 4c-4+3t=0\cr, a-2b+3c-16=0\cr }$
Solving the system results:$a=2,b=-1,c=4 \implies f_{min}=37$

[i]Solution Using Calculus :([/i]:

We will use Lagrange multipliers.
Let
\[f(a,b,c)=2a^{2}+2b^{2}+2c^{2}-4a-4b-4c+15\wedge g(a,b,c)=a-2b+3c-16 \]
Hence,
\[\\ \frac{\partial f}{\partial a}=4a-4=\lambda\hspace{0.3 in}\text{(1)}\\ \frac{\partial f}{\partial b}=4b-4=-2\lambda\hspace{0.15 in}\text{(2)}\\ \frac{\partial f}{\partial c}=4c-4=3\lambda\hspace{0.25 in}\text{(3)}\]
Subtracting $(1)$ from $(2)$ and adding the result to $(3)$, we obtain $-a+b+c=1$.  Adding $-a+b+c=1$ to $g(a,b,c)=0$, we obtain $4c=17+b$.  Substituting this into $(3)$ and solving for $b$ using $(2)$, we obtain $b=-1$.  Using the obtained equations, we have $c=4$ and $a=2$.  $\min_{(a,b,c)\in\mathbb{R}^{3}}f(a,b,c)=f(2,-1,4)=37$, as required.

IF we don't use Lagrange multipliers then we can slove the problem?

Er, I'm not sure why you guys used Lagrange Multipliers, considering [hide="this method..."] $f = 2(a-1)^{2}+2(b-1)^{2}+2(c-1)^{2}+12$, applying Cauchy [/hide] works just as well, and is elementary.

Dear T0rajir0u: How did you applied Cauchy inequality?

The Cauchy-Schwarz Inequality, in vector form, states that

$| \vec{v}| | \vec{u}| \ge \vec{v}\cdot \vec{u}$

Let $\vec{v}= \left< a-1, b-1, c-1 \right>$ and let $\vec{u}= \left< 1,-2, 3 \right>$.  Then

$\sqrt{ (a-1)^{2}+(b-1)^{2}+(c-1)^{2}}\sqrt{ 1^{2}+(-2)^{2}+3^{2}}\ge \left( (a-1)-2(b-1)+3(c-1) \right)$

Which is the square root of the expression I gave.

Yeah, I think my solution can be applying for ganeral problem with condintion: $ma+nb+pc=Q$

[quote="cckek"]And 4 u this is elementaty?:)hmm, then also the Lagrange multiplicators are.[/quote]

The non-vector proof is quite elementary.

The quadratic polynomial

$P(x) = \sum_{k=1}^{n}(a_{k}x-b_{k})^{2}$

Has real roots if and only if there exists $x_{0}$ such that every term is zero; in other words, $\frac{b_{i}}{a_{i}}= x_{0}\forall i$.  This is the equality case of the Cauchy-Schwarz inequality.  In every other case, the polynomial has no real roots.

We can also determine whether it has real roots from its discriminant.  First, let's expand.

$P(x) = \left( \sum_{k=1}^{n}a_{k}^{2}\right) x^{2}-\left( \sum_{k=1}^{n}2a_{k}b_{k}\right) x+\left( \sum_{k=1}^{n}b_{k}^{2}\right)$

Hence the discriminant, which must be less than or equal to zero, is

$\left( \sum_{k=1}^{n}2a_{k}b_{k}\right)^{2}-4 \left( \sum_{k=1}^{n}a_{k}^{2}\right) \left( \sum_{k=1}^{n}b_{k}^{2}\right) \le 0$

Dividing by $4$ and moving to the other side,

$\left( \sum_{k=1}^{n}a_{k}b_{k}\right)^{2}\le \left( \sum_{k=1}^{n}a_{k}^{2}\right) \left( \sum_{k=1}^{n}b_{k}^{2}\right)$

Of course, letting

$\vec{v}= \left< a_{1}, a_{2}, ... a_{n}\right>$
$\vec{u}= \left< b_{1}, b_{2}, ... b_{n}\right>$

We have our previous vector result.  QED.


due to the fact that i've noticed math people are really good at identifying irrational numbers, i thought it would be cool to play "name that decimal".  so, you give the exact value for the decimal approximation that the previous person wrote, then post another decimal approximation.   example: 3.1415926535... (actually a calculator would round here, but you can't DO that!!! not to $\pi$, at least.)  so then the next person writes $\pi$, and then posts another decimal.

here we go, with an easyish one:

1.7320508...

That's $\sqrt{3}.$

This is actually easier for those of us who went to high school before (gasp!) there were calculators and who learned to rely on tables.

0.30103...

Wait Wait
Alan what problem are you solving...
Guys, we should start numbering the problem so we don't get confused between so many decimals lol

$\frac{1}{\sqrt[3]{2}}$  (i had to interrogate my calculator for like 4 minutes before i got that one  :lol: )

here's a good easyish one: 53.13

One angle (in degrees) of a 3-4-5 right triangle? (Your signature "we're number $sin^2\theta + cos^2 \theta$  " sort of gives it away.:)

Okay here are some which [b]everyone [/b] of my generation knew by heart - see how familiar are those: (Can you point why we knew them by heart)

(-1)  4.342944819  (with reciprocal 2.30258509) - Easy one..
(-1)  9.189385332

(BTW (-1) 4.342944 ... just means $4.34 \cdot 10^{-1}$ :)   IOW = .4342944... and
the other number is simply .9189385332 ....:) (Its just that , that's how we remembered them :)

(Hint:  You need not rememebrer these since the the age  of calculators!)

And the actual question
1.3541179

This one has been out there with no response for three days, so I'm putting a new one up:
1.2020569...

This is actually my favorite number. This one should be pretty tough.

1.2020569.= [hide]$\zeta(3)$ [/hide]
and the other looks lile[hide] $e \cdot \pi$ [/hide]

[quote="mathnerd314"]Try 8.539734221

[hide="Big hint"]
Favorite food...
[/hide][/quote]

pi and e HAHAHAHA  :rotfl:  :rotfl:

$\pi e$

BTW - Big Hint for 2.303 and  0.4343 (my previous post) ===>  are failry easy if you ask  [b] some  one who did Physics   or Math in old days[/b] ,  (Before the age of calculators) ...   I dare say these figure may be    just [i]slightly[/i] less familier than 3.14159 or 2.718   to them :)

[quote="frt"]$\ln 10\approx 2.302585093$ and $\frac{1}{\ln 10}=\log_{10} e\approx .4342944819$[/quote]
Of course!  Before  the age of calculators, for almost  a  few   hundred years,[i] everyone [/i] used  log tables (Like standar calculators we were allowed to use " standard log tables" in exam)   to do the calculations, and thus you start remembering log (based 10) of constants like 2, $\pi$ ( lucky that that was about (1/2) or 0.4971 to be more exact)  and e (.4343) and of course, the most important function was exponational - thus everyone  knew $log_{10} e$...

Similarly, if you wanted to know  natural log of a number, you look up the log (based 10) and multiply it by 2.3 (which may be little easier than dividing by 0..4343..)

We need another one! try this (its complex):

.31813+1.337i

this one is really tough. i dont even know if it has a closed form (sorry). but i will let you know that it is the smallest (in magnitude) solution to a quite simple equation.

Now, what I want to know is who (besides me) is cheating using the [url=http://www.research.att.com/~njas/sequences/]Online Encyclopaedia of Integer Sequences[/url]?

[quote="kokosai"][quote="Gyan"](-1)  9.189385332[/quote]$-\zeta'(0)=\frac{\ln2\pi}{2}$
[quote="Gyan"]and the actual question
1.3541179[/quote]$\Gamma(\frac{2}{3})$[/quote]
Correct! okay  $\Gamma(\frac{2}{3})$ was sort of sketchy but  $-\zeta'(0)=\frac{\ln2\pi}{2}$ which BTW is more familiarly written as
$\ln (\sqrt{(2 \pi)}$  was something nice to remember as this constant is used quite often in statistical  physics ,  many integrals. (eg stirling's formula)

Guess no one here is impressed  with my story when in our days we have to do all that calculation by log tables / bamboo slide rules,  and on top of that have to walk many kilometers uphill both ways in 2 meter of snow and  $50^o \ C$ temperature on our way to school :)

JBL - never!  :)


salam.biaid as hamin alan faalit dar inja ro bishtar karde va har kas masale baraye hal bezare :)
1.(easy) dar $ ABCD$ $ AD\equal{}AB\equal{}BC$ va $ DP\equal{}PC$ ke $ P$ taghatoe ghotrhaye $ AC,BD$ ast prvoe that $ ADC\plus{}DCB\equal{}120$

Soale khoobi bood :lol:  :lol:  .tanha chizi ke bayad estefade bokonid ghazie ye 2 zelo zavie nabein hast.
Problem.2
Let $ a,b,c$ azlaye yek mosallas be tori ke $ 2(a^2 b \plus{} b^2 c \plus{} c^2 a ) \equal{}a b^2 \plus{} b c^2 \plus{} ca^2  \plus{}3abc$
Prove  that $ a\equal{}b\equal{}c$

pas movafegh hastid inja ro marathon bokonim?
man ke kheyli dust daram intor bashe...
manzuram az marathon ine ke har kasi ham zamane ba inke soale ghabli ro javab mide ye soal ham bede va be hamin tartib ela akhar...(hamun chizi ke shoma farmudid)
agar movafegh hastid lotfan esme topic ro az 'hi' be 'marathon' taghyir bedid.
baraye soale mahanmath :
taghyire motaghayer bedid be dast miayad :
$ a \equal{} x \plus{} y \ , \ b \equal{} y \plus{} z \ , \ c \equal{} z \plus{} x$
$ \implies LHS \equal{} RHS$

$ \Longleftrightarrow 2 \left(\sum_{cyc} (x \plus{} y)^2(y \plus{} z) \right) \equal{} \left(\sum_{cyc} (x \plus{} y)(y \plus{} z)^2 \right) \plus{} 3\prod (x \plus{} y)$

$ \Longleftrightarrow 2 \left( \sum x^3 \plus{} 6 \prod x \plus{} 2\sum xy(x \plus{} y) \plus{} \sum_{cyc} xy^2 \right) \equal{} \left( \sum x^3 \plus{} 6 \prod x \plus{} 2\sum xy(x \plus{} y) \plus{} \sum_{cyc} x^2y \right) \plus{} 3 \left(\sum xy(x \plus{} y) \plus{} 2\prod x \right)$

$ \Longleftrightarrow \sum x^3 \plus{} \sum_{cyc} xy^2 \equal{} 2 \sum_{cyc} x^2y$

dar inja az AM-GM estefade mikonim darim $ x^3 \plus{} xy^2 \ge 2x^2y$
hal ba jame in ravanbet be dast miayad $ LHS \ge RHS$
va tasavi etefagh mioftad $ \Longleftrightarrow$ hameye AM-GM ha be tasavi tabdil shavad yani $ x \equal{} y \equal{} z$.
hal Masaleye 3
Problem 3-sabet konid hamneheshtiye $ x^{n \minus{} 1} \equiv 1 \pmod{n}$ tanha yek javab darad( yani haman adade $ 1$) agar va faghat agar $ \gcd(\phi(n),n \minus{} 1) \equal{} 1$.(marhale 3-vom)

Kheily mamnoon az halle zibatoon ,[b]shoki[/b].In soale TST Brazil bood vali too resources nabood.
Man shakhsan  :P  ba nazaretoon darbare marathon movafegham vali fek konam inja kheily khalvate !!!
(hala edame midim bebinim chi mishe  !! :wink: )
Baabat soalee ham ke gozashtin mamnoon

salam. manam be marathon peyvastam. :D
bara shoroo ye soale nazarie asoon : sabet konid in moadele binahaiat javabe tabiE dare : a^2+b^2+c^2+d^2=abcd
rasti soale akhare jabre marhale 3 sale 88 ro ke fek konam shoki zahmatesho keshide neveshte hame too dore 0 gereftan. age rahi peida shod benevisid

baraye soali ke goftid bayad moadele ro be surate chand-jomleiye daraje $ 2$ dar nazar begirim va baghish halee ...
vali khaheshan ru soali ke ferestadam fekr konid ...
[hide="rahnamayi"]
[hide="albate shayad rahnamayi nabashe :D"]
Fasle $ 9$ ketabe nazariye adade khanome maryam mirzakhani ro bekhunid...
[/hide]
[/hide]

tamame tavabe $ f: \mathbb{N} \to \mathbb{N}$ ra biabid betorike $ (m^2 \plus{} n)^2$ bar $ f(m)^2\plus{}f(n)$ bakhshpazir bashad..
I LIKE THIS PROB...

[b][i]Problem 3[/i][/b]:
[hide]Ba tavajjoh be rahnamaee haye Shoki   :lol:  :lol:
Ghazie $ 8.9$ va tamrin $ 11$ safhe ye badesh (dar ketab Nazarie adad) ro bekhoonid ,sepas  $ m$ ro adad dar tajziye $ n$ dar nazbegirid.
$ n$ ro ham $ n \minus{} 1$ dar nazar begirid.[/hide]

[b][i]Problem 4[/i][/b]:
[hide]Man too Mathlinks hamin moadelaro baraye halat 5 moteghaiere dide boodam , yani
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \equal{} abcde$ ke ba hamoon raveshi ke Shoki goft hal karde boodand , vali baraye 4 taee ravesh sade tar ham mishod.
$ a^2 \plus{} b^2 \plus{} 2^2 \plus{} 2^2 \equal{} 4ab$ $ \Leftrightarrow$ $ (2a \minus{} b)^2 \minus{} 3a^2 \equal{} \minus{} 8$
In moadeleye akhar ham ke [b]Pell[/b] hastesh ke toosh $ (10,6)$ yani $ a \equal{} 6 , b \equal{} 2$ kar mikone pas binahayat javab dare.[/hide]

Baraye soale jadid ham soale aghaye [b]sepehr ahmadzadeh[/b] hastesh.Vali lotfan aval in soal ro hal konid bad soale jadid bezarid.

In soal vaghan zibast hatman roosh fekr konid bad rahnamaee ro bekhoonid  :wink:
[hide="Hint"] Aval sabet konid $ f(1)\equal{}1$ . Bad az adad aval estefade konid.
[/hide]

Marhale avalam ke tamoom shod , ishalla hamatoon movaffagh boode bashin. :lol:
amma ....
In topic vaghan dare KAPAK mizane , haddeaghal ye nazari raje be soala bedin  :roll: ...

Pishnahad mikonam bishtar soal tarkibiat va hendese bezarin . age deghat karde bashin bishtar soal ha ee ke inja gozashte mishe jabr va nazariye hast . vali to marhale 2 , 1 jabr va 1 nazariye miad vali  2 tarkibiat va 2 hendese ! (Ke parsal 3 tarkibiat oomad  :mad: )
Pas Soal haye hendese va tarkibiat  bezarid ye kam .
Thanks  :) .

[quote="amin-nejatbakhsh"]
rasti soale akhare jabre marhale 3 sale 88 ro ke fek konam shoki zahmatesho keshide neveshte hame too dore 0 gereftan. age rahi peida shod benevisid[/quote]

motmaeni ? Akhe fekr konam ye tarafesh asoon bashe :
mikhahim sabet konim agar $ (\phi(n), n\minus{}1) \equal{} 1$ angah $ x \equiv 1 (mod n)$ .
------------------------
Lemma : agar $ f \equal{} ord(a,b)$ va dashte bashim $ a^g \equiv 1 (mod b)$ angah $ f | g$
ke esbatesh sade ast (farz konid $ g \equal{} fq \plus{} r$ angah ... ) .
--------------------------------
hal $ f \equal{} ord(x,n)$ . tebghe lem darim $ f | n \minus{} 1$ va hamchenin $ f | \phi(n)$ pas
$ f | (\phi(n), n\minus{}1) \equal{} 1$ pas $ f \equal{} 1$ va in yani $ x \equiv 1 (mod n)$ ke khaste mast .

Age ja e eshtebah kardam lotfan begid  :?:

[quote="navidj"][quote="amin-nejatbakhsh"]
rasti soale akhare jabre marhale 3 sale 88 ro ke fek konam shoki zahmatesho keshide neveshte hame too dore 0 gereftan. age rahi peida shod benevisid[/quote]

motmaeni ? Akhe fekr konam ye tarafesh asoon bashe :
mikhahim sabet konim agar $ (\phi(n), n \minus{} 1) \equal{} 1$ angah $ x \equiv 1 (mod n)$ .
------------------------
Lemma : agar $ f \equal{} ord(a,b)$ va dashte bashim $ a^g \equiv 1 (mod b)$ angah $ f | g$
ke esbatesh sade ast (farz konid $ g \equal{} fq \plus{} r$ angah ... ) .
--------------------------------
hal $ f \equal{} ord(x,n)$ . tebghe lem darim $ f | n \minus{} 1$ va hamchenin $ f | \phi(n)$ pas
$ f | (\phi(n), n \minus{} 1) \equal{} 1$ pas $ f \equal{} 1$ va in yani $ x \equiv 1 (mod n)$ ke khaste mast .

Age ja e eshtebah kardam lotfan begid  :?:[/quote]

1.Aghaye [b]amin-nejatbakhsh[/b] manzooreshoon ehtemalan in soal naboode chon in soal jabr nist.

2.Age mishe rahe hallaye manam too postaye ghabl check konid .(mamnoon)

3.Ye soale tarkibiat ziba ham bezoodi mifrestam , vali bayad soal haye ghabli halide shode bashe (marathon dige  :!: )...  :)

4.Bye.

2rost migi . manzooresh soal jabr boode .   :)   :)

Khob inam az ye soale zibaye tarkibiat (albate ye khorde sakht  :!: )

[b]Problem.6[/b] : Farz konid graph $ n \plus{} 2$ rasi kameli darim ke dar an yal ha ba tedadi rang tori rang shode and ke az har rang haddeaxar $ n$ bar estefade shode . Prove that mosallasi rangarang dar in graph mojood ast.

 :arrow:  :arrow:  Mosallas ke hamoon $ K_3$ hast.
Mosallas rangarang ham yani 3 yalesh ba rangaye mokhtalef rang shode bashand.
(Az khodam dar nayovordam !! estelahe too graph.)

DAR YEK GRAPH MIDANIM KE RAS AZ DARAJEYE 0 VA N-1 VOJOOD NADARAD AGAR ADAD-E-RANGI-E-IN GRAPH BARABAR-E-X BASHAD SABET KONID MITAVAN N-1 RAS AZ G ENTEKHAB KARD BE TORI KE HAR RAS-E- DIGAR AZ G HADEAGHAL BE YEKI AZ ANHA VASL BASHAD.

:blush: DAR SOAL-E-BALA BE ESHTEBAH BE JAYE N-X N-1 NEVESHTAM(GHLATE TYPI BOOD.)HOKM IN AST:...N-X RAS AZ G ENTEKHAB...

Salam taha hala ke soalo hal kardi bayad ye soal bedi (maratone dige). montazerim,mamnoon :)

[quote="taha.hojjati"]SOAL SHOMA VAGEAN ZIBA BOOD HALA BE HAL E AN TAVAJOH KONID.
AVALAN MALOOM AST KE DARAJEYE HAR RAS BARABAR E n+1 AST.PAS AGAR YALHAYE KHAREJ SHODE AZ YEK RAS RA DAR NAZAR BEGIRIM HADEAGHAL DO RANG DAR ANHA MOSHAHEDE MIKONIM HALA HOKMI KOLITAR RA BA ESTEGHRA SABET MIKONIM:
AGAR DAR YEK GRAPH KAMEL RANG SHODE DAR BEIN YAL HAYE KHAREJ SHODE AZ HAR RAS HADE AGHAL DO RANG MOSHAHEDE SHAVAD ANGAH DAR GRAPH MOSALAS E RANGARANG VOJOOD DARAD.
PAYE ESTEGHRA BADIHI AST .PAS BE ESBAT-E-HOKM MIPARDAZIM BARAYE TAJASOM-E-RAHAT TAR FARZ KONID N=4(BEDOON E KAM SHODAN AZ KOLIAT E MASALE!)VA GRAPH KAMELI BA RASHAYE A,B,C,D,E,H RA DAR NAZAR BEGIRID.
HALA RAS-E-A RA HAZF MIKONIM AGAR  FARZ-E- ESTEGHRA DAR MORED GRAPH E BE DAST AMADE DOROST BASHAD KE MASALE HAL AST DAR GHEIR E IN SOORAT RASI KHAHIM DASHT KE YAL HAYASH FAGHAT AZ YEK RANG AND DAR GRAPH E ASLI IN RAS RA DAR NAZAR BEGIRID (RAS E B)TAMAME  YAL HAYE MOTASEL BE IN RAS BE JOZ YAL E AB GHEMEZ AND.HALA FARZ KONID MOSALAS E RANGARANGI VOJOOD NADARAD DAR IN SOORAT AD AC AH VA AE NIZ BAYAD ABI BASHAND KE IN BA FARZ E MASALE TANAGHOZ DARAD.[/quote]

Mamnoon az rahe halle zibatoon.Mikhastam begam ke in masale halate khasi az ye masale ye zibatar (vali asoon tar  :!:  :!: ) hast ke age har kasi khast mitoonam soalo post konam ...

Movaffagh bashin .

salam be rofagha...
mikhastam ye khaheshi bokonam...
age soali ro mikhahid ruye in topic bezarid soalate motavaset ru be bala(behtare begam marhale dovomi,ya nazdik be marhale 3 ) bezarid ke yehoyi in tori topic nakhabe(man ham ye jurayi shayad eshtebah kardam un soal ro andakhtam) ... :(
age har kasi soale sakht bezare,tori ke har chand ruz ye bar javab dade beshe,un vaght esme inja marathon nemishe(albate har chand hanuz aghaye Maijd (MJ GEO) esme topic ro marathon nakardand :D )...
haale marathon be ine ke kheyli zud javab dade beshe va kheyli ham soal ziyad(golchin shode va idedar) tush matrah beshe...
age ba hamin ravand pish berim khoda midune cheghadr soalate kami matrah mishe...
mamnunam az tavajohetun...
yademun ham nare ke marathon (hamunjuri ke az esmesh ham malume) yani ye jayi ke tond soal dade mishe va tond ham javab dade mishe,hamuntori ke goftam,va reghabat sare tond javab dadane...
albate soale aghaye hojjatti sare jaye khodesh baghiye va montazere rahe halim...
midunam kolan kheyliiiiiiii fak mizanam ... :oops:
sharmande,midunam saretuno dard avordam... :blush:
ba tashakore faravan :)

[quote="mohammad20"]salam.man ye soal  dashtam.mamnon misham javab bedid.
emsal hatman az tarkibiat va hendese ro ham rafte 4 ta miad???
be nazareton kodom mabhas vaghte kamtary bara motalee mibare(az mabahese marhale2)?
va mamolan sakht tarin soal male kodom mabhasas to marhale 2?
soale akhar:hendese khoshkhan ta kodom fasl bara marhale 2 kafie???
kheili mamoo :lol:  :)  :D n[/quote]
salam . hendese va tarkibiat az har kodoom 2 ta miad (har rooz ye hendese va ye tarkibiat) . albate goftam mamoolan . shayad didi emsal 5 ta tarkibiat dadan !!
raje be vaght gir boodan mabahes bastegi be khod shakhs dare va kolan bayad to har mabhas masale hal koni (tarkibiat bishtar ide yad gereftane vali hendese va jabr va nazariye bayad ghaziye ham bekhooni )
raje be sakht tarin soal ham nemidoonam chi begam ! toye har mabhas mishe ye soal ha ee dad ke KHEILI sakht bashan .
baraye hendese ham bayad hadeaghal ta fasle 4 khonde bashi va chizi ke bishtar moheme masale hal kardane .

bah bah che ajab tunestam ye soale tarkibiyat hal konam ...
tarif mikonim:
[b]Majmueye Rasiye Hambande Takrang (MRHT)[/b] :
be yek majmue rous az ras-haye $ G$ yek MRHT miguyim har gah beyne yalhaye vasl konandeye an majmue rous, rangi vojud dashte bashad ke agar tamami yalhaye beyne an majmue rous ,ke rangeshan be gheyr az an rang ast, ra hazf konim angah graphi ke padid miayad va tamami yalhaye an yekrangand, graphi hamband bashad.
pas in shod tarif MRHT.
hal bozorgtarin MRHT ra beyne tamami MRHT -ha dar nazar begirid (bozorgiye MRHT= manzur tedade rous MRHT hast) va an ra max-MRHT benamid.
badihist ke be dalile estefade shodane hadeaksar $ n$ bar az har rang , max-MRHT nemitavand hameye $ n\plus{}2$ ras ra dar bar dashte bashad chera ke sharte zir ra darad : $ e(max\minus{}MRHT) \le n$ va banabarin $ |V(max\minus{}MRHT)| \le n\plus{}1$.
pas yeki az an ras-ha ke dar max-MRHT nist ra dar nazar begirid va an ra $ v$ benamid.
farz konid $ V(max\minus{}MRHT)\equal{}\{t_1,...,t_k\}$.
tamami yalhaye $ vt_1,vt_2...,vt_k$ ra dar nazar begirid.badihist ke hich kodam az in yalha rangeshan ba range yalha dar max-MRHT yeksan nist.masalan farz konid $ t_1 \leftrightarrow t_2$.
chon $ vt_1t_2$ nabayad rangarang bashe pas range yalhaye $ vt_1$ va $ vt_2$ yeksan hast.
hala fekr konam fahmide bashid ke che etefaghi miofte...
are doroste tamami yalhaye $ vt_1,vt_2...,vt_k$ yek rang mishand va banabarin yek MRHT be vojud miarand .ama baraye in MRHT tedade rous $ k\plus{}1$ hast dar hali ke baraye max-MRHT chand bud ? $ k$ bud...
pas tanaghoz va HALE.
kolan rahe halam hamchin tulani nabud  bar khalafe chizi ke shayad be nazar biyad :huh: ...
soale aghaye taha ro aslan 0% ham nafahmidam ...
taze, CHAND ruze ke in 'marathon' khabide...
mamnun misham age khodeshun ya ye ensane digar be man befahmunan surate soal chiye ? :huh:
sorry for my stupidity... :(

[b]Shoki[/b] jan ye soal bezar felan , ta [b]taha[/b] biad javab bede bikar nabashim.  :lol:
rasti ino hal kardi ? $ 2x^2 \minus{} 73 y^2 \equal{}1$ (dar $ N$) ...

sharmande nemitunestam be internet val sham be khatere hamin ham hast ke in hame takhir dashtam ....
in soale shoma ro hanuz rush fekr nakardam...
khob soale jadid ...
hendese bashe chetore?
dar mosalase ABC sabet konid ke FF' movaziye khate eulere ABC hast .F va F' be tartib noghteye fermat va isogonal conjugate noghteye fermat mibashand.

[quote="shoki"]
dar mosalase ABC sabet konid ke FF' movaziye khate eulere ABC hast .F va F' be tartib noghteye fermat va isogonal conjugate noghteye fermat mibashand.[/quote]

اوههههههههه . shoki این سوال رو خودت حل کردی؟؟؟

ye rahnamayi  ...
inja ro bebinid ... dar akhare rahe halam eshare be ye morede kuchulu kardam ke kelide hale masale hast ... http://www.irysc.com/ftopict-1817.html


If $ A,B,C,D$ are four points in space, such that
\[ \angle ABC\equal{}\angle BCD\equal{}\angle CDA\equal{}\angle DAB\equal{}\pi/2,
\]
prove that $ A,B,C,D$ lie in a plane.

So, we have 4 right-angled triangles here, and applying Pythagoras we find

$ AB^2\plus{}BC^2 \equal{} AC^2 \equal{} CD^2\plus{}AD^2$

$ BC^2\plus{}CD^2 \equal{} BD^2 \equal{} AB^2\plus{}AD^2$

Subtracting, we get $ AB^2\minus{}CD^2 \equal{} CD^2\minus{}AB^2$ which implies $ AB^2\equal{}CD^2$ so $ AB\equal{}CD$ and it follows that $ AD\equal{}BC$ as well.

Consider now the planes in which $ BC$ and $ AD$ lie, calling them $ c$ and $ d$ respectively. They are both perpendicular to $ AB$, and the distance between them is thus $ AB$ as well. Drop a perpendicular from $ D$ to the plane $ c$. The length of this perpendicular is $ AB$, so if it doesn't hit $ C$ we would have $ CD > AB$ which isn't the case. Thus $ C$ lies on the plane perpendicular to $ d$ through $ B$ and $ D$, as does $ A$, and the points are coplanar.

Another good question : why is this problem in the number theory section?  :wink:  :wink:

Pierre.


Given triangle $ XYZ$ with vetices $ X(\minus{}2,0), Y(6,0),$ and $ Z(4,8)$, find the number of square units in its area.

The base is clearly $ 8$ (distance between $ (\minus{}2,0)$ and $ (6,0)$).  The height is $ 8\minus{}0$, or $ 8$.

Thus, the answer is $ \frac{8 \times 8}{2}$ or $ \boxed{32}$.


quadratics
parameterization
algebra
quadratic formula

For how many real values of $ \alpha$ , the quadratic equation

$ x^{2} \plus{} 2\alpha x \plus{} 2008 \equal{} 0$ has integer roots.

The discriminant is $ \Delta\equal{}4(\alpha^2\minus{}2008)$ and it must be a perfect square.

$ \alpha^2\minus{}2008\equal{}k^2\Rightarrow(\alpha\minus{}k)(\alpha\plus{}k)\equal{}2008$

$ \left\{\begin{array}{ll}\alpha\minus{}k\equal{}2\\\alpha\plus{}k\equal{}1004\end{array}\right.\Rightarrow \alpha\equal{}503$

$ \left\{\begin{array}{ll}\alpha\minus{}k\equal{}4\\\alpha\plus{}k\equal{}502\end{array}\right.\Rightarrow \alpha\equal{}253$

[quote="red_dog"]The discriminant is $ \Delta \equal{} 4(\alpha^2 \minus{} 2008)$ and it must be a perfect square.

$ \alpha^2 \minus{} 2008 \equal{} k^2\Rightarrow(\alpha \minus{} k)(\alpha \plus{} k) \equal{} 2008$

$ \left\{\begin{array}{ll}\alpha \minus{} k \equal{} 2 \\
\alpha \plus{} k \equal{} 1004\end{array}\right.\Rightarrow \alpha \equal{} 503$

$ \left\{\begin{array}{ll}\alpha \minus{} k \equal{} 4 \\
\alpha \plus{} k \equal{} 502\end{array}\right.\Rightarrow \alpha \equal{} 253$[/quote]

I am sorry if I am wrong.
I thought $ \alpha$ was a real number. So why should $ \Delta$ be a perfect square.

[hide]We can easily show that $ \alpha$ is rational: $ \alpha\equal{}\frac{2008\minus{}x^2}{2x}$, assuming that the roots are integers.

Using the quadratic formula and simplify, $ x\equal{}\minus{}\alpha\pm\sqrt{\alpha^2\minus{}2008}$.

Let $ r^2\equal{}\alpha^2\minus{}2008$. Then, $ (\alpha\minus{}r)(\alpha\plus{}r)\equal{}2008$. Since the roots are integers, $ r$ must also be rational.

Since $ \alpha$ is rational, we can let the rational parameters $ a\equal{}\alpha\minus{}r$ and $ b\equal{}\alpha\plus{}r$ so that $ ab\equal{}2008$. Solving for $ \alpha$ and $ r$ yields $ \alpha\equal{}\frac{b\plus{}a}2$ and $ r\equal{}\frac{b\minus{}a}2$.

$ x\equal{}\minus{}\alpha\pm r\equal{}\minus{}a\text{ or }\minus{}b$. The roots are integers, so $ a,b\in\mathbb{Z}$. The rest is straightforward.[/hide]


Well, its getting cold again. and that means that the football season is here.

get BOOM! Tough-actin' tinactin!- John Madden
 :huh:
I love that commercial :D [/hide]

Umm, you're about 3 months late.

I'm rooting for the Steelers even though they're 2-5  :roll:

Bears all the way! :football:  :winner_first:

Go Bears!!!

Chicago is my second favorite team, behind the Cleveland Browns.

You guys root for anything around the Great Lakes region, OSU, Bears, Browns, pfft.

my English teacher roots for the Bears. Whenever they win, he wears Bears stuff all week. Needless to say that he has worn nothing but Bears shirts all year! :D


function
geometry
inequalities
complex analysis
complex analysis unsolved

Can anyone help me out with any part of the following problem? I need to use Liouville's theorem to do them (can't use any knowledge about harmonic functions).

Thanks so much!

Let f:C-->C be an entire function. Let a be an arbitrary real-valued constant.

1)

Show that if [Re f(z)]^(2)[u]<[/u][Im f(z)]^(2)

for all z in the complex plane, then f is constant.

2)

Suppose h is another entire function and suppose there exists a real-valued constant a (a>0), such that

Re f(z)[u]<[/u]Re h(z) for all z in the complex plane.

Show that there exist complex constants c, d such that

f(z)=a*h(z)+d

for all z in the complex plane.

where $ a$ plays a rolle in your problems  :maybe:

1) means that all values of $ f$ lie in a union of two sectors. The shape does not really matter: the important thing is that $ f$ does not assume any values in some open disk $ \{w\colon |w\minus{}w_0|<r\}$. The function $ g(z)\equal{}\frac{1}{f(z)\minus{}w_0}$ is entire and is bounded by $ 1/r$. Hence  $ g$ is constant, and so is $ f$.

Like this:

[img]http://img229.imageshack.us/img229/1792/plotblj3.jpg[/img]

The shaded area is the image space of $ f(z)$  described above with $ |u|\leq |v|$ and $ |f(z)\minus{}w_0|$ shown by the line segment between the two points.  Since $ f(z)$ is constrained to this shaded area by design, we can see graphically that for any $ f(z)$, $ |f(z)\minus{}w_0|>r$ and thus:

$ \left|\frac{1}{f(z)\minus{}w_0}\right|\leq \frac{1}{r}$

for all z since by hypothesis, f(z) is entire.

Seems to me then if the image space of an entire function leaves ANY region in the u-v plane uncovered, then the function must necessarily be a constant function by the argument above.

That's true. See also: [url=http://en.wikipedia.org/wiki/Weierstrass%E2%80%93Casorati_theorem]Weierstrass–Casorati theorem[/url].
Part (2) can be solved by a similar argument applied to $ f\minus{}h$ (or $ f\minus{}ah$, if the inequality in (2) is supposed to contain $ a$.)

thanks a bunch! it helps a lot to think of it as a union of two sectors.


[quote="Valentin Vornicu"][quote="Peter VDD"]Well while we're talking about tags: it would be nice if [code]This is a $\LaTeX$ test.[/code] really ended up in code, so we could give latex codes in that way :)[/quote]This is a different topic, let's not mix them :D[/quote]

Ok, in a new topic then. :)

I really don't see how $\LaTeX$ in code can be useful.

That's exactly what I'm saying. Sometimes you want to put something in code, but you can't because it contains 2 dollar signs thus gets converted to latex. The code tag is invented to ignore any and all other tags, but now it loses that property as the tag can no longer deal with dollar signs.

For example, if I want to give some lines of php code, a lot of $-signs are needed. So there is currently no way I can write them.

Well I do not want php code on the board anyway, so that rules out a problem :)
It's not doable yet (without messing the other stuff).

Oh but php code is not dangerous at all :) phpBB is well secured for that matters. [and I can imagine it is often needed in the programming section]

Anyway I was just naming one of the dozens of applications... but it's not an urgent matter of course :)


[quote="biped51"]Given a chord with a length A and height B, find the radius.[/quote]
U should send this type of problems to High school basics.
[hide="solution"]consider a isosceles triangle the base of which is the chord.the sides  are the radious.now concider the height. and use pythagoras. u will get $r^{2}=B^{2}+(A/2)^{2}$ [/hide]


Rule 1: any two rows of the matrix can be interchanged.

Can anyone explain this rule to me specifically. Other than moving an equation with coefficient 1 in the first term to the top row; how do I decide when to switch the rows around? Is there something specific or a relation in the equations that I should be looking for to decide this?

Thanks for the help

You use this operation if you think it helps.  That simple.  For example, if you need to get a matrix into reduced row echelon form, you might want to do a row switch.  For example,
$ \begin{pmatrix}
1&0&0&0\\
0&0&1&0\\
0&1&0&0\\
0&0&0&1
\end{pmatrix}$
Now this is really simple but if you wanted it into row echelon form, you would switch the second and third rows.  Honestly, I use it only if I think it will help me but otherwise, I don't use it too much.

It's necessary if you're about to divide by zero if you don't do it. It's occasionally useful at other times.

[quote="jmerry"]It's occasionally useful at other times.[/quote]
Two categories:

1. Paper and pencil arithmetic, human motivations: Often it's convenient to move a $ 1$ into the pivot position, just to make the next round of arithmetic easier. Or if not a $ 1,$ then a $ 2$ or a $ 3.$

2. Floating point arithmetic, algorithmic decision making: The issue here is that computations are affected by roundoff error, and we would like to prevent roundoff error from causing more trouble than it has to. Generally speaking, it's bad to divide by small numbers (especially when such small numbers might themselves have been the result of the subtraction of larger numbers), so we try to make our pivot elements as large as we can. This falls under the general heading of "pivoting strategies." For more information, consult any reasonable numerical analysis textbook.

----

Note that if we can perform Gaussian elimination on $ A$ without using this kind of row operation, then we can factor $ A\equal{}LU$ where $ L$ is lower triangular and $ U$ is upper triangular. If we're forced to swap rows, then we'll have to write the factorization as $ A\equal{}PLU,$ where $ P$ is a permutation.


How many questions on the AMC 10 of the year 2008 did you get correct?

I did it unofficially, got a 25, anyways this isn't on mathcounts. Mods, plz move this topic.


trigonometry
real analysis
real analysis unsolved

Can anyone show me how to get the general solution for  this differential equation ?

$\dfrac{d^2y}{dx^2}+9y=\cos^2\dfrac{3x}{2}$

Start with some algebra; $\cos^2\dfrac{3x}2=\dfrac{1+\cos 3x}2$.

Now, we have a system driven at resonance; try a solution of the form $a+bx\cos 3x+cx\sin 3x+d\cos 3x+e\sin x$.

The last two coefficients will vary freely, as they give the general solution to the homogeneous equation.

For a more general version of this method, look up "undetermined coefficients"


trigonometry
geometry
power of a point
geometry unsolved

[geogebra]a5add7f8c737b034add0c7b29fcdea6418081f7f[/geogebra]
Let the center of the circle be O. Then AC = BC, and <ACO is 90 degrees.
Two arbitrary lines are constructed through A and B, and two other lines are constructed through the intersections of these lines with the circle as shown. Prove that LC = MC.

[geogebra]66702d0f9da7b89c78bebb8a74a34e5c04b65eb3[/geogebra]

(X=AQ∩BS)

From Menelaus,

MB/AM * RX/BR * QA/XQ =1
LA/BL * PX/AP * SB/XS =1

∴ MB/AM = LA/BL <=> RX/BR * QA/XQ * AP/PX * XS/SB =1

Then, because RX*XS = PX*XQ and BR*SB = BO^2 -R^2 = AO^2 -R^2 = AQ*AP, RX/BR * QA/XQ * AP/PX * XS/SB =1 is right, therefore MB/AM = LA/BL.

∴ because AM+MB = LA+BL = AB, AM=BL and then MC=LC.

I'll use the same names for the points as cadiTM.

As $ AO \equal{} BO$, from power of a point we get $ AP \cdot AQ \equal{} BR \cdot BS$. Then $ \frac {AP}{BR} \equal{} \frac {BS}{AQ}$

Now, using $ \angle APL \equal{} \angle MRB$ and $ \angle PSR \equal{} \angle PQR$ we get $ \frac {AL}{MB}\equal{} \frac {AP}{BR} \cdot \frac {\sin \angle BMR}{\sin \angle ALP} \equal{} \frac {BS}{AQ} \cdot \frac {\sin \angle BMR}{\sin \angle ALP} \equal{} \frac {LB}{AM}$

Then $ \frac {AL}{LB} \equal{} \frac {BM}{AM}$ and the result follows. (Notice $ M$ and $ L$ are both inside segment $ AB$ if $ RS$ and $ PQ$ intersect inside the circle, and they are both outside if $ RS$ and $ PQ$ intersect outside the circle).


The second ineguality is quite easy.We have \[\frac{x^{3}+y^{3}+z^{3}}{xyz}\leq\sum_{cyc}\frac{x^{3}}{y^{3}}\Leftrightarrow\sum_{cyc}\frac{x^{2}}{yz}\leq\sum_{cyc}\frac{x^{3}}{y^{3}}\]
where if we denote $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$(where we have that $abc=1$ )the ineguality is equivalent to \[\sum_{cyc}\frac{a}{c}\leq\sum_{cyc}a^{3}\Leftrightarrow a^{2}b+b^{2}c+c^{2}a\leq a^{3}+b^{3}+c^{3}\] wich is evidently true :) (i'm still meditating on the first one )

I was just going to write that same thing  :D


Let $ m$ and $ n$ be positive integers. Prove the identity

$ \sum_{1\leq i_1\leq i_2\leq ...\leq i_m\leq n}\frac{\left(\minus{}1\right)^{i_m\minus{}1}}{i_1i_2...i_m}\binom{n}{i_m}\equal{}\sum_{k\equal{}1}^n\frac{1}{k^m}$.

[i]Solution.[/i] [i]Preparations.[/i] As in http://www.mathlinks.ro/Forum/viewtopic.php?t=160313 , we notice that

[color=green][b](1)[/b][/color] $ \binom{a}{b} \equal{} \binom{a \minus{} 1}{b \minus{} 1} \plus{} \binom{a \minus{} 1}{b}$ for all integers $ a$ and $ b$;

[color=green][b](2)[/b][/color] $ \frac {1}{k \plus{} 1}\binom{i \minus{} 1}{k} \equal{} \frac {1}{i}\binom{i}{k \plus{} 1}$ for all integers $ i\neq0$ and $ k\neq \minus{} 1$;

[color=green][b](3)[/b][/color] $ \binom{n}{k} \equal{} \sum_{i \equal{} 1}^{n}\binom{i \minus{} 1}{k \minus{} 1}$ for all $ n\in\mathbb{N}$ and $ k\in\left\{ 1,2,...,n\right\}$.

We conclude that

[color=green][b](4)[/b][/color] $ \frac {1}{k}\binom{n}{k} \equal{} \sum_{i \equal{} 1}^{n}\frac {1}{i}\binom{i}{k}$ for all $ n\in\mathbb{N}$ and $ k\in\left\{ 1,2,...,n\right\}$,

since [color=green][b](3)[/b][/color] yields

$ \frac {1}{k}\binom{n}{k} \equal{} \frac {1}{k}\sum_{i \equal{} 1}^{n}\binom{i \minus{} 1}{k \minus{} 1} \equal{} \sum_{i \equal{} 1}^{n}\frac {1}{k}\binom{i \minus{} 1}{k \minus{} 1} \equal{} \sum_{i \equal{} 1}^{n}\frac {1}{i}\binom{i}{k}$

(since $ \frac {1}{k}\binom{i \minus{} 1}{k \minus{} 1} \equal{} \frac {1}{i}\binom{i}{k}$ follows from [color=green][b](2)[/b][/color] if we substitute $ k \minus{} 1$ for $ k$).

We will now show the identity

[color=green][b](5)[/b][/color] $ \sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_1\leq i_2\leq ...\leq i_m\leq k}\frac {1}{i_1i_2...i_m} \equal{} \frac {1}{i^m}$

for all integers $ i\geq 1$ and $ m\geq 0$. Here, for $ m \equal{} 0$, the sum $ \sum_{1\leq i_1\leq i_2\leq ...\leq i_m\leq k}\frac {1}{i_1i_2...i_m}$ has to be understood as $ 1$ (since this sum goes over $ m$-tuples, but there is only one $ 0$-tuple, namely the empty $ 0$-tuple, and the product of its elements is $ 1$).

We will prove [color=green][b](5)[/b][/color] by induction over $ m$:

[i]Induction base:[/i] For $ m \equal{} 0$, the identity [color=green][b](5)[/b][/color] states that

$ \sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\cdot 1 \equal{} \frac {1}{i^0}$.

But this follows from

$ \sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\cdot 1 \equal{} \sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}$
$ \equal{} \sum_{k \equal{} 0}^i\underbrace{\left( \minus{} 1\right)^{k \minus{} 1}}_{ \equal{} \minus{} \left( \minus{} 1\right)^k}\binom{i}{k} \minus{} \underbrace{\left( \minus{} 1\right)^{0 \minus{} 1}}_{ \equal{} \minus{} 1}\underbrace{\binom{i}{0}}_{ \equal{} 1}$
$ \equal{} \sum_{k \equal{} 0}^i\left( \minus{} \left( \minus{} 1\right)^k\right)\binom{i}{k} \minus{} \left( \minus{} 1\right) \equal{} \minus{} \underbrace{\sum_{k \equal{} 0}^{i}\left( \minus{} 1\right)^k\binom{i}{k}}_{\substack{ \equal{} \left(1 \plus{} \left( \minus{} 1\right)\right)^i\text{ by the} \\
\text{binomial formula}}} \plus{} 1$
$ \equal{} \minus{} \underbrace{\left(1 \plus{} \left( \minus{} 1\right)\right)^i}_{ \equal{} 0^i \equal{} 0} \plus{} 1$      (where $ 0^i \equal{} 0$ since $ i > 0$)
$ \equal{} 1 \equal{} \frac {1}{1} \equal{} \frac {1}{i^0}$.

Thus, [color=green][b](5)[/b][/color] is proven for $ m \equal{} 0$.

[i]Induction step:[/i] Let $ M\geq 1$ be an integer. Assume that [color=green][b](5)[/b][/color] holds for all integers $ i\geq 0$ if $ m \equal{} M \minus{} 1$. We want to show that [color=green][b](5)[/b][/color] holds for all integers $ i\geq 0$ if $ m \equal{} M$. In other words, we want to show that

$ \sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_1\leq i_2\leq ...\leq i_M\leq k}\frac {1}{i_1i_2...i_M} \equal{} \frac {1}{i^M}$

for all integers $ i\geq 0$.

But this follows from

$ \sum_{k \equal{} 1}^{i}\left( \minus{} 1\right) ^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \equal{} \sum_{k \equal{} 1}^{i}\left( \minus{} 1\right) ^{k \minus{} 1}\left( \binom{i \minus{} 1}{k \minus{} 1} \plus{} \binom{i \minus{} 1}{k}\right) \sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$ (since $ \binom{i}{k} \equal{} \binom{i \minus{} 1}{k \minus{} 1} \plus{} \binom{i \minus{} 1}{k}$ by [color=green][b](1)[/b][/color])
$ \equal{} \sum_{k \equal{} 1}^{i}\left( \minus{} 1\right) ^{k \minus{} 1}\binom{i \minus{} 1}{k \minus{} 1}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \plus{} \sum_{k \equal{} 1}^{i}\left( \minus{} 1\right) ^{k \minus{} 1}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \plus{} \sum_{k \equal{} 1}^{i}\underbrace{\left( \minus{} 1\right) ^{k \minus{} 1}}_{ \equal{} \minus{} \left( \minus{} 1\right) ^{k}}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
(here we substituted $ k$ for $ k \minus{} 1$ in the first sum)
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \minus{} \sum_{k \equal{} 1}^{i}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \minus{} \sum_{k \equal{} 0}^{i}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
(here we replaced the $ \sum_{k \equal{} 1}^{i}$ sign by a $ \sum_{k \equal{} 0}^{i}$, since the addend for $ k \equal{} 0$ is zero, since $ \sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}} \equal{} 0$ for $ k \equal{} 0$)
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \minus{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}$
(here we replaced the $ \sum_{k \equal{} 0}^{i}$ sign by a $ \sum_{k \equal{} 0}^{i \minus{} 1}$, since the addend for $ k \equal{} i$ is zero, since $ \binom{i \minus{} 1}{k} \equal{} 0$ for $ k \equal{} i$)
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\left( \sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M}} \minus{} \sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M}\leq k}\frac {1}{i_{1}i_{2}...i_{M}}\right)$
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M} \equal{} k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M}}$
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq i_{M} \equal{} k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}i_{M}}$
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}\left(k \plus{} 1\right) }$
(here, we used $ i_{M} \equal{} k \plus{} 1$ to replace $ i_{M}$ by $ k \plus{} 1$)
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\frac {1}{k \plus{} 1}\binom{i \minus{} 1}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}}$
$ \equal{} \sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\frac {1}{i}\binom{i}{k \plus{} 1}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}}$ (by [color=green][b](2)[/b][/color])
$ \equal{} \frac {1}{i}\cdot\sum_{k \equal{} 0}^{i \minus{} 1}\left( \minus{} 1\right) ^{k}\binom{i}{k \plus{} 1}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq k \plus{} 1}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}}$
$ \equal{} \frac {1}{i}\cdot\sum_{k \equal{} 1}^{i}\left( \minus{} 1\right) ^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq k}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}}$
(here we substituted $ k$ for $ k \minus{} 1$)
$ \equal{} \frac {1}{i}\cdot\frac {1}{i^{M \minus{} 1}}$ (since $ \sum_{k \equal{} 1}^{i}\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_{1}\leq i_{2}\leq...\leq i_{M \minus{} 1}\leq k}\frac {1}{i_{1}i_{2}...i_{M \minus{} 1}} \equal{} \frac {1}{i^{M \minus{} 1}}$ because of our assumption that [color=green][b](5)[/b][/color] holds for $ m \equal{} M \minus{} 1$)
$ \equal{} \frac {1}{i^{M}}$.

Hence, the induction step is complete. Thus, the proof of [color=green][b](5)[/b][/color] is complete.

If we substitute $ m \minus{} 1$ for $ m$ in [color=green][b](5)[/b][/color], we obtain

[color=green][b](6)[/b][/color] $ \sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq k}\frac {1}{i_1i_2...i_{m \minus{} 1}} \equal{} \frac {1}{i^{m \minus{} 1}}$

for all integers $ i\geq 1$ and $ m\geq 1$. Thus, for all integers $ m\geq 1$ and $ n\geq 1$, we have

$ \sum_{1\leq i_1\leq i_2\leq ...\leq i_m\leq n}\frac {\left( \minus{} 1\right)^{i_m \minus{} 1}}{i_1i_2...i_m}\binom{n}{i_m} \equal{} \sum_{i_{m} \equal{} 1}^{n}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq i_{m}}\frac {\left( \minus{} 1\right)^{i_{m} \minus{} 1}}{i_1i_2...i_{m}}\binom{n}{i_{m}}$
$ \equal{} \sum_{k \equal{} 1}^{n}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq k}\frac {\left( \minus{} 1\right)^{k \minus{} 1}}{i_1i_2...i_{m \minus{} 1}k}\binom{n}{k}$    (here we renamed $ i_{m}$ as $ k$)
$ \equal{} \sum_{k \equal{} 1}^{n}\frac {1}{k}\binom{n}{k}\left( \minus{} 1\right)^{k \minus{} 1}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq k}\frac {1}{i_1i_2...i_{m \minus{} 1}}$
$ \equal{} \sum_{k \equal{} 1}^{n}\sum_{i \equal{} 1}^{n}\frac {1}{i}\binom{i}{k}\left( \minus{} 1\right)^{k \minus{} 1}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq k}\frac {1}{i_1i_2...i_{m \minus{} 1}}$   (by [color=green][b](4)[/b][/color])
$ \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}\cdot\sum_{k \equal{} 1}^n\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq k}\frac {1}{i_1i_2...i_{m \minus{} 1}}$
$ \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}\cdot\sum_{k \equal{} 1}^i\left( \minus{} 1\right)^{k \minus{} 1}\binom{i}{k}\sum_{1\leq i_1\leq i_2\leq ...\leq i_{m \minus{} 1}\leq k}\frac {1}{i_1i_2...i_{m \minus{} 1}}$
(here, we replaced the $ \sum_{k \equal{} 1}^n$ sign by a $ \sum_{k \equal{} 1}^i$, since all terms for $ k > i$ are zero, since $ \binom{i}{k} \equal{} 0$ for $ k > i$)
$ \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}\cdot\frac {1}{i^{m \minus{} 1}}$    (by [color=green][b](6)[/b][/color])
$ \equal{} \sum_{i \equal{} 1}^n\frac {1}{i^m}$,

qed.

  darij


integration
calculus
trigonometry
calculus computations

This certainly cannot be done in closed form. Besides the usual means of numerical integration, power series are also quite effective in approximating it.

I happen to have the following result memorized, for reasons related to the Gibbs phenomenon in Fourier series:

$\frac2{\pi}\int_{0}^{\pi}\frac{\sin x}x\,dx\approx 1.18$


If Z = cos θ + i sin θ, (θ belongs to set of real numbers, i = √-1), then according to DeMoivre's theorem, Z^n = cos nθ + i sin nθ, (n belongs to set of natural numbers). Using this theorem alongwith binomial expansion, prove that:

(i) cos 2nθ =[size=150][b] ∑ [/b][/size][size=100]( r = 0 to n) (2n [i][b]C [/b][/i]2r) cos^(2(n - r)) θ sin^(2r) θ
(ii) cos 2nθ =[/size] [b][size=150]∑[/size][/b] [size=100](r = 0 to n) (2n [i][b]C[/b][/i] (2r + 1))cos^(2(n-r) - 1) θ sin^(2r + 1) θ [/size]

NOTE: 2n [i][b]C[/b][/i] 2r is 2n choose 2r.

I think you're wrong :D
$ (cos\theta \plus{} isin\theta)^{2n} \equal{} cos2n\theta \plus{} isin2n\theta$ by De Moivre's Theorem.

I will check if $ n$ is even.
Another case, you can get it same way.

by Binomial Expension,
$ (cos\theta \plus{} isin\theta)^{2n} \equal{} \sum_{r \equal{} 0}^{n}cos^{2r}\theta sin^{2n \minus{} 2r}\theta (\minus{}1)^{r} \plus{} i\sum_{r \equal{} 0}^{n \minus{} 1}cos^{2r \plus{} 1}\theta sin^{2n \minus{} 2r \minus{} 1}\theta (\minus{}1)^{r\plus{}1}$
And isology of real and ideal numbers,
$ cos2n\theta \equal{} \sum_{r \equal{} 0}^{n}cos^{2r}\theta sin^{2n \minus{} 2r}\theta(\minus{}1)^{r}$

Check out the general formulas:

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulas


function
AMC
AMC 10
AMC 10 A
algebra
polynomial
Rational Root Theorem

(Me) Function $ P(x)$ is defined as

$ P(x) \equal{} 2008 \plus{} \frac {2007}{x} \plus{} \frac {2006}{x^2} \plus{} \cdots \plus{} \frac {1}{x^{2007}}$

Find the sum of the reciprocal roots of $ P(x)$.

$ \text{(A)} \ \frac {2008!}{(2007!)^2} \ \qquad \text{(B)} \ \minus{} 2 \qquad \text{(C)} \ 1 \qquad$
$ \text{(D)} \frac {2006}{2005} \cdot P(2003) \ \qquad \text{(E)} \ \frac {1 \cdot 3 \cdot 5 \cdot \cdots 2007}{2 \cdot 4 \cdot 6 \cdot \cdots 2008}$

Inspiration: AMC10A 2003 #18
Level: First Round about #13-16

Am I missing something, or is P(x)... not a polynomial?

Hm...for a normal polynomial $ P(x) \equal{} a_0 \plus{} a_1x \plus{} a_2x^3 \plus{} ... \plus{} a_nx^n$, the sum of the reciprocals of the roots is equal to:$ \frac {a_1}{a_0}$,

Which can be derived pretty easily:

The sum of the reciprocals of the root, let's say $ r_1,r_2,...,r_n$, is:
$ \frac {1}{r_1} \plus{} \frac {1}{r_2} \plus{} ... \plus{} \frac {1}{r_n}$

$ \equal{} \frac {r_2r_3...r_n}{r_1r_2...r_n} \plus{} \frac {r_1r_3...r_n}{r_1r_2...r_n} \plus{}...\plus{} \frac {r_1r_2...r_{n \minus{} 1}}{r_1r_2...r_n}\equal{}\frac{a_1}{a_0}$

Well its not a polynomial but...
[hide]
Multiply by x^2007.
Flip the coefficients around for f(1/x) (easily proven), and then take the second coefficient, whatever that is.[/hide]

The function formed by multiplying $ P(x)$ by $ x^{2007}$ and reversing the coefficients has roots which are the reciprocal roots of $ P(x)$. Since the sum of the roots of a function $ f(x)$ is $ \frac{\minus{}a_1}{a_0}$, the answer is $ \frac{\minus{}2}{1}\equal{}\minus{}2 \Rightarrow \boxed{B}$.

For future reference, a Polynomial is an expression composed of one or more variables and constants using multiplication, addition, and subtraction. Division may also be within the expression as long as it does not involve division by variables. Exponents may also appear as long as the exponent is a positive integer. Also, the general form of a polynomial function is:

$ f(x)\equal{}a_nx^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_2x^{2}\plus{}a_1x\plus{}a_0$

See also: [url]http://en.wikipedia.org/wiki/Polynomial[/url]

Thanks. I need to prevent myself from making such mistakes ha.

Fixed.

[quote="t0rajir0u"]Now:  actually find the roots of $ P(x)$.  :)[/quote]

Hmm, how would one go about that? I know you can narrow some choices through RRT (Rational Root Theorem) but still, that's only [i]rational[/i] roots....

And yes, the answer is B.  :) Any input or comment about this problem?

[quote="t0rajir0u"]Do you remember how to find sums of the form $ \sum kx^k$?  ;)[/quote]

[hide="Attempt"]
$ \sum k x^k \equal{} x \plus{} 2x^2 \plus{} 3x^3 \plus{} \cdots \equal{} (x \plus{} x^2 \plus{} x^3 \plus{} \cdots) \plus{} (x^2 \plus{} x^3 \plus{} \cdots ) \plus{} (x^3 \plus{} \cdots ) \plus{} \cdots$

So, using this idea, it'll be possible to find their sum... But it looks bit annoyingly tedious though. Maybe I'm missing something?
[/hide]

The finite sum is more tedious, but yes.

[quote="t0rajir0u"]I was hoping it would simplify more than that.  Such is life  :|[/quote]

Oh, I thought you already solved it when you said, "Find the roots."  :D Nice job guys for finding the actual sum haha.  :P

[quote="iakovos1729"]i think the easiest approach to this one is to see that all roots of the function are negative. Now sum of negatives is negative, so the only possible option is B... right?[/quote]

No: many of the roots could be complex numbers.

[quote="mysmartmouth"][quote="iakovos1729"]i think the easiest approach to this one is to see that all roots of the function are negative. Now sum of negatives is negative, so the only possible option is B... right?[/quote]

Yep.  Something I learned from MIMC is that when you write a problem for a multiple choice test, part of the challenge is creating 5 plausible answer choices, because often there will be ways to eliminate certain answer choices that you won't see.[/quote]

Well in problems like this, I just make the ones that "look" similar or complicated. I don't really bother to check in depth.


calculus AMC USA(J)MO USAMO email trigonometry geometry

For clarification, see
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2816[/url]

[i]mod-edit: I'm adding "college" for the people who are no longer in highschool[/i]

[i][b]Writer Edit: Can a moderator put on an option that says 'beyond' for those freshmen taking BC or something like that?[/i][/b]

[i]mod-edit: done[/i]

I'm only in GT becuase of the school district, just like probably a whole lot of other people in here.  We were all probably ready for the math we're doing loooooooooong before right now.

I'm in high school, but taking classes at a local college.

8th grader taking Algebra 1 since thats as far as my school will let me go ahead.

I'll be taking multivariable calculus and possibly linear algebra as well when the semester starts.

linear algebra, 2nd semester differential equations

I took EQ and LA last year. And number Theory now.
Need advice: what would be a good course to take next? Why?
e.g.
Real analysis
modern algebra
or
....

(only one course)
Thanks!

[quote="peter"]freshman, trig/analysis, but you probably already knew that Dark_Shadow[/quote]

Actually, no. What is trig/analysis? Does it involve a lot of proofs?

Thanks!

I put "Beyond" because I actually am a freshman taking BC.

 :D

In some schools, if you're smart enough, you can take higher level courses.  I once heard of someone taking Calculus at grade 9, which may be the highest they will let him go at his school, while at other schools, the highest math you can take in grade 8 is Algebra 1.

But in high school, you get to pick what classes you take from a list they give you.  But to graduate, you have to take the rquired amount of classes in each subject.

[quote="Peter VDD"]I don't get this whole discussion? Can you guys pick what lessons you'll get at school??? Seems unbelievable to me...

Here you just go to school, and even on the biggest schools you take the only lessons pack they got :?

Can someone explain me what this is all about?[/quote]

In CA where we are, a junior high or high student may take college-level courses at community colleges or universities, IF he or she qualifies and gets the approval of the school principal.
To be qualified, one has to take assessment tests, SAT, or some other means depending on each university. In order to get your principal's signature, the course you will take has to be something that is not offered at your school. For home school students, this process is much easier and simpler for obvious reasons. Please check with your community college and your school.

There are many HS students that I know of who are taking college-level math and science courses (beyond AP) after 3pm. Some JH students are able to do this as well. There are also EPGY and other distant learning programs (expensive) that offers universities level courses.

If your own school does not offer courses that meet your needs, you may look into this concurrent enrollment at a community college or university.

Hope this helps.

Why don't you stop caring about the length of time in between posts, and start caring about more relevant things instead? Math is important, and just because nobody bothered to discuss on this topic for such a long time doesn't mean that we should proceed to ignore it as long as the universe exists. I see people bumping old threads all the time. Some old threads can contain really interesting pieces of information. We shouldn't ignore the things that interest us to satisfy these nonexistent "rules".

If bumping old threads weren't allowed, this thread would have been locked.

[quote=tigeryong]the 0.75% of AoPS that has a moderator position in an official forum?
EDIT: Sniped[/quote]

I have no idea where you pulled that statistic but if that's correct then yes. Typically if you report a thread that has been bumped after a long time moderators will delete the message that bumped the thread and lock the thread.

[quote name="AlienGirl05" url="/community/p29035038"]
Oh my, did I just summon my Xarcade! clan?
[/quote]

I think you did...

There is a difference between necro-posting, and general bumping. All I did was answer the EXACT question that I was asked to answer in the OP. The only difference between my response, and other responses is the fact that there is a $19$-year gap between my response, and the response that came before mine. Stop being annoyed by the disobedience of irrelevant etiquette that nobody really cares about, admit that you do not have an argument, and move on.

Anyway, guys, let's discuss our math levels. Who here is also doing calculus? :)

[quote=AlienGirl05][quote=tigeryong]Me obviously. Class #3611[/quote]

Nice! I'm in Class $\#3,556$. :)[/quote]

same here

Diff Eq // Lin Alg [first one] and Discrete Math [second one] [I'm a sophomore]

Next year I'm taking Analysis. Senior year abstract algebra.

9th grade, precalc
is that advanced? (or smth else)

8th grade unofficially taking calc, officially taking alg 2


Let $ x_1, x_2,...,x_n$ is a set of real numbers such that $ \sum^n x_i \equal{} n$, where $ n > 0$. Prove that for all $ i \ge 1$,

$ \left( \sum^n \frac {x_i}{x_i \plus{} 1} \right)^2 \plus{} 2 \left(\sum^n \frac {x_i^{3/2}}{x_i \plus{} 1} \right)^2 \plus{} \left(\sum^n \frac {x_i^2}{x_i \plus{} 1} \right)^2 \le n^2$

What do you mean "for all $ i\geq 1$"? Isn't $ i$ running from 1 to $ n$ in each summation?

And is this really so easy that it should be in the Pre-Olympiad section? hmm...

Well, the first term, when simplified, is less than 1.
The second team, when simplified, is less the $ 2\sqrt n$
The last term is less than n.
So, $ LHS<(\sqrt{n}+1)^2$
That's gotta be useful...argh.

[quote="pluristiq"]What do you mean "for all "? Isn't  running from 1 to  in each summation?

And is this really so easy that it should be in the Pre-Olympiad section? hmm...[/quote]

Of course, each summations run from 1 to n.

By the way, I just use [b]only[/b] CS inequality to prove this inequality............ :lol: (i think it is suitable for pre-olympiad.....)

[quote="abcak"]Well, the first term, when simplified, is less than 1.
The second team, when simplified, is less the $ 2\sqrt n$
The last term is less than n.
So, $ LHS < (\sqrt {n} + 1)^2$
That's gotta be useful...argh.[/quote]

try n=2,

$ (\sqrt {2} + 1)^2 \approx 5.828427...... > 2^2$ :wink:

[quote="Rofler"]$ \left(\sum\frac {x_i^2}{x_i \plus{} 1} \plus{} \sum\frac {x_i}{x_i \plus{} 1}\right)^2 \le n^2$

$ \sum\frac {x_i^2}{x_i \plus{} 1} \plus{} \sum\frac {x_i}{x_i \plus{} 1} \le n$

$ \sum \frac {x_i^2 \plus{} x_i}{x_i \plus{} 1}\le n$

$ \sum x_i \le n$

$ n \le n$

$ 0 \le 0$

Not that it matters, but this is actually an identity... CS might be overkill :P

Cheers,

Rofler[/quote]

that's right. thx. Actually by CS, easy to prove $ (\sum \frac{x^{3/2}}{x\plus{}1} )^2\le \sum \frac{x^2}{x\plus{}1} \sum \frac{x}{x\plus{}1}$


For a rising 8th grader who has pretty good experience with Algebra I and Algebra II, would you recommend your Introduction to Algebra book (I think I heard somewhere there would be a course eventually also...)? I have worked in the Introduction to Number Theory book and found that while most of the material would seem elementary, it and its great problems are very good. I was thinking that the Introduction to Algebra would be similar in this. Also, would a good indication of the level of the book be the Intermediate Algebra class "Are You Ready" PDF? If I happened to take that course, would I be missing out or behind if I never read "Introduction to Algebra"?

If you've already gone through Algebra 1 and Algebra 2, you're probably ready to dive into Intermediate Algebra.  The Intermediate Algebra text will likely be finished around year end.


rotation
blogs
LaTeX
number theory
prime factorization

1. In the sequence 13, 10, 1, 3, ... each term, after the first, is found by adding the
tens digit and three times the units digit of the previous term. How many distinct
perfect squares will appear in this sequence?

2. N is an integer greater than 5, and when 205 is divided by N, the remainder is 5.
For how many distinct positive integers N is this statement true?

And also, if they give you an equilateral triangle, divided into sections of equilateral triangles, how many triangles will there be?

[hide="1)"]We find the sequence $ 13$, $ 10$, $ 1,3,9,27,23,11,4,12,7,21,5,15,16,19,28,26,20,2,6,18,25,17,22,8,24,14,13,\dots$ repeats every $ 27$ terms. Hence, we only need to find the number of distinct perfect squares in the first $ 27$ terms. Counting, we find there to be $ \boxed{4}$ such terms namely $ 1,4,9,$ and $ 25$.[/hide]

[hide="2)"]If the remainder is $ 5$, when $ 205$ is divided by $ N$, then $ N$ must divide $ 200$ evenly. Furthermore, we have $ N > 5$, so we need to find all factors of $ 200$ greater than $ 5$.

We note that the prime factorization of $ 200$ is $ 2^3\cdot5^2$. Hence, there are $ (3 \plus{} 1)(2 \plus{} 1) \equal{} 12$ factors of $ 200$ because we have $ 4$ choices of what power of $ 2$ we want in the factor namely $ 0$, $ 1$, $ 2$, and $ 3$ choices as to what power of $ 5$ we want in the factor namely $ 0,1,$ and $ 2$. Now, we need to find the number of positive factors less than or equal to $ 5$.

We know that $ 1,2,4,$ and $ 5$ are factors, so there are $ 4$ factors of $ 200$ less than or equal to $ 5$. Hence, our answer is $ 12 \minus{} 4 \equal{} \boxed{8}$[/hide]

[hide="3)"]There will be $ \boxed{5}$ because there are $ 4$ little ones and $ 1$ big one.[/hide]

1. In the sequence 13, 10, 1, 3, ... each term, after the first, is found by adding the
tens digit and three times the units digit of the previous term. How many distinct
perfect squares will appear in this sequence?

The best way to solve this is to list everything out . I did this problem a while ago, I think the answer is 5 or something, but I'm not sure.......

2. N is an integer greater than 5, and when 205 is divided by N, the remainder is 5.
For how many distinct positive integers N is this statement true?

Just factor 200, and all of the divisors that are greater than 5, and don't divide 205 work.

If they give you an equilateral triangle, divided into sections of equilateral triangles, how many triangles will there be?  (Do you want to know the least number that you can divide it into?)

I'm pretty sure this is 5...

[quote="pytheagle"]We find the sequence $ 13$, $ 10$, $ 1,3,9,27,23,11,4,12,7,21,5,15,16,19,28,26,20,2,6,18,25,17,22,8,24,14,13,\dots$ repeats every $ 27$ terms. Hence, we only need to find the number of distinct perfect squares in the first $ 27$ terms. Counting, we find there to be $ \boxed{4}$ such terms namely $ 1,4,9,$ and $ 25$.
[/quote]

Isn't $ 16$ a perfect square?

[quote="randomguy64"]I'm pretty sure that the third part is N^2, where N can be any positive integer, and the side lengths of the triangles equals 1/N of the original side length.

If you divide the sides with N evenly spread dots per side, and connect the dots the correct way, you will get it. The correct way is to connect the dots horizontally (lines parallel to the base), and rotate the triangle so that another vertex is pointing up, repeat, rotate, repeat.

I concur with the other two answers.[/quote]

This only accounts for unit triangles - what about ones with, say, side lengths 2/N, 3/N, ...?

For the triangle problem (in general, with the sides of the smallest triangles being $ \frac{1}{n}$ times the sides of the largest triangle), we have to consider "upside-down" triangles too.

[hide="Solution/General Formula"]
We will begin by looking at the first few cases:

Let $ n\equal{}1$. It is obvious that we only have $ \boxed{1}$ triangle.

Let $ n\equal{}2$. As pytheagle said, we have $ 4$ small triangles and $ 1$ big one, for a total of $ \boxed{5}$.

Let $ n\equal{}3$. Let us look at the upright and upside-down triangles separately. There are $ 1\plus{}2\plus{}3$ upright small triangles because there is $ 1$ in the first "level", $ 2$ in the next "level", and $ 3$ in the last. Similarly, there are $ 1\plus{}2$ upright triangles of the next size, and just $ 1$ upright triangle which is the same size as the biggest triangle. For the upside-down triangles, we have $ 1\plus{}2$ small triangles (again, going by "level"), and only $ 1$ upside-down triangle of the next level. We add these to get a total of $ (1\plus{}2\plus{}3)\plus{}(1\plus{}2)\plus{}(1)\plus{}(1\plus{}2)\plus{}(1)\equal{}\boxed{14}$.

Let $ n\equal{}4$. Similar to the above case, we have $ 1\plus{}2\plus{}3\plus{}4$ small upright triangles, $ 1\plus{}2\plus{}3$ upright triangles of the next size, $ 1\plus{}2$ for the next, and $ 1$ for the biggest, all going by "levels". Next is the smallest upside-down triangles, and there are $ 1\plus{}2\plus{}3$ of those. There are $ 1\plus{}2$ of the next level and finally $ 1$ upside-down triangle of the biggest size that fits. We add these for a total of $ (1\plus{}2\plus{}3\plus{}4)\plus{}(1\plus{}2\plus{}3)\plus{}(1\plus{}2)\plus{}(1)\plus{}(1\plus{}2\plus{}3)\plus{}(1\plus{}2)\plus{}(1)\equal{}\boxed{30}$.

We can start to see a pattern. The number of triangles is the sum of the first $ n$ triangular numbers plus the sum of the first $ n\minus{}1$ triangular numbers. We can also derive a formula for the sum of the first $ n$ triangular numbers, see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=306743]my blog[/url]. Finally, we put it all together to get a formula:
\[ \; (1\plus{}2\plus{} \cdots\plus{}n)\plus{}\cdots\plus{}(1) \plus{} (1\plus{}2\plus{} \cdots \plus{}(n\minus{}1))\plus{}(1) \\ \\
\equal{} \frac{n(n\plus{}1)}{2} \plus{} \cdots \plus{} \frac{1(1\plus{}1)}{2} \plus{}\frac{(n\minus{}1)((n\minus{}1)\plus{}1)}{2} \plus{} \cdots \plus{} \frac{1(1\plus{}1)}{2} \\ \\
\equal{} \frac{n(n\plus{}1)(n\plus{}2)}{6} \plus{} \frac{(n\minus{}1)n(n\plus{}1)}{6} \\ \\
\equal{} \frac{ n(n\plus{}1)(n\plus{}2)\plus{}(n\minus{}1)n(n\plus{}1)}{6} \\ \\
\equal{} \frac{ n(n\plus{}1)(n\plus{}2\plus{}n\minus{}1)}{6} \\ \\
\equal{} \boxed{\frac{n(n\plus{}1)(2n\plus{}1)}{6}}\]
This is the same as the sum of the first $ n$ squares. Why? Because the sum of any two consecutive triangular numbers is a square...
[/hide]

So it is just the sum of the first $ n$ squares.

1. If you think about it you should find 28 is the largest number you can get, 25 is the largest perfect square therefore 1-5, so you should get an answer of "5"
2. find all the factors of 200 and is greater than 5, so you find the prime factorization which is 2^3x3^2, so you raise the amount of factors so its 4x3 which is 12, then you find the factors that are lessthan or equal to 5 which is 1,2,4,5. so it is 12-4,  so the answer is "8"
3.and last one i go with latex i think it was, i though it was "5", can you explain how its 3?

I don't believe you know exactly how many sections of equilateral triangles are, i.e. how small the unit triangles are in relation to the biggest one, so you can't just say 5.


USAMTS email LaTeX geometry AMC AIME USA(J)MO

We are now accepting online registrations for Year 17 of the USA Mathematical Search.  If you participated in Year 16, or have otherwise already registered online, all you have to do is log in to [url]http://www.usamts.org[/url], then click on My USAMTS, then Profile, then set your grade for the 2005-6 school year.  That's it!  If you register for Year 17, you'll receive an email in late August or early September when the first round of Year 17 is available.

[quote="rrusczyk"]We are now accepting online registrations for Year 17 of the USA Mathematical Search.  If you participated in Year 16, or have otherwise already registered online, all you have to do is log in to http://www.usamts.org, then click on My USAMTS, then Profile, then set your grade for the 2005-6 school year.  That's it!  If you register for Year 17, you'll receive an email in late August or early September when the first round of Year 17 is available.[/quote]

You have an extra comma there ;) [url=http://www.usamts.org]www.usamts.org[/url].

[quote="djshowdown2"]I've never participated before, but I filled out the online application earlier this day (before this post actually, does that make any difference?).  So do I just wait till Round 1 (which is approximately when?) and send in the entry form then?[/quote]

Yes - send in the Entry Form with your Round 1 solutions if you like.  You can send it in now if you want, but you don't have to.  The first round problems will be up in late August or early September.  The first round will be due in October.

[quote="nr1337"]I registered for the USAMTS last year (but unfortunately only had time for round 1 :( ), so do I have to submit another entry form this year with my first solutions?[/quote]

Yep - we collect an Entry Form each year.

[quote="Mr.Ocax"]2 questions: If you're sending your solutions in Latex (well, PDF) then how do you send your entry form?

And, do you get counted off if your proof doesn't have pictures?  Because I don't know how to do pictures on Latex.  Should I learn how to?

Mr.Ocax[/quote]

If using LaTeX, you can mail your Entry Form separately.

You will make graders of most geometry problems sad if you don't include your pictures.  I recommend learning how to include pictures - it really doesn't take that long.  Follow the pictures guide in our LaTeX tutorial.

[quote="nr1337"]I guess if all else fails you can senf in your geometry graphics files not embedded in the LaTeX file... :)[/quote]No, please do not do this.  Include your graphics into your main solutions file -- they are much, much less likely to get lost this way.

Please take the time to learn how to embed them in your document -- it's really not that hard.

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]How to include graphics in a LaTeX document[/url] -- note that the examples on this page are for including pdf files, but jpeg's, png's, and eps's can be included just as easily.

[quote="xxreddevilzxx"]If we were to qualify for AIME through USAMTS, would it still be that maximum 100 again??? (because that wouldnt be fair)[/quote]I don't know if the AMC has yet set a policy for 2006 for this.

Why would that not be fair?

Personally I feel that USAMTS is much harder than an AMC test although it is based on the honor system. Therefore it really is a tough decision on how to grade it and I think the AMC administration will use their best judgement.

This is just me speculating, but I don't think that the AMC is using relative difficulty when they set the USAMTS->AIME qualifying score to 100. Also, for those who are not yet Juniors or Seniors, that 100 is as good as a 150.

I faxed my forms on July 18, but on the USAMTS site it says I haven't turned in my forms yet. Did they not get the fax, or did they just not update their site yet?

We haven't yet parsed all the forms we have received.  We will probably start doing that in mid-late August after we release the first set of problems.

[quote="JesusFreak197"]So, I printed out the forms from the USAMTS website, and I filled them out, but I was planning on sending them in with my answers for the first round... is that okay? I don't know if there are any forms besides the two(?) that I have, and some need to be sent in earlier, or what.[/quote]There are just the two forms, and yes, they can be sent with your first set of solutions.

Okay, thank you. So I don't have to worry about submitting anything beforehand, and I just have to keep myself updated on when the first round goes up. :)

If you registered online, you should get an email when the first round is up.  Also, we'll make an announcement on this site.

[quote="yunorman"]How many people participate in USAMTS last year?[/quote]

Around 800 sent in at least one round.

Is it extremely worthy to learn how to type up a latex document as opposed to sending handwritten solutions?

I was uh... :blush: ... lazy last year so I wrote all mine.

I think they're only admissable for being quoted in the solutions if they're TeXed.  They may still be commended, just not quoted.  And if I remember correctly, LaTeX makes for a happy grader. :P

But learning TeX isn't hard.  It took me about an hour or so to get the feel for it and another hour to learn more advanced things like tables/arrays, etc.  I'd recommend learning it, personally, but if you have some sort of aversion to TeX, you won't be penalized for it.

Everything no_one_2000 said is correct.  There is absolutely no penalty for handwriting your solutions, and you can still receive a commendation.  The only difference is that when we select students' solutions to publish on the USAMTS website, we will only consider those solutions that were written in LaTeX.

Okay... the rules say that I should not fax solutions in pencil. Does this mean that my solutions should be written in pen, or is that just referring to faxes, and I can send in solutions written in pencil by mail along with my permission stuff?

[b]EDIT:[/b] Also, can I assume obvious things? I have a specific question on that, so if you need to know, I can PM the question... I don't want to post it here.

[quote="JesusFreak197"]

[b]EDIT:[/b] Also, can I assume obvious things? I have a specific question on that, so if you need to know, I can PM the question... I don't want to post it here.[/quote]

yea i have a similar problem. however i dont think you can simply pm and ask them whether it is needed. i think its part of the comptetition to know whats trivial and whats not.

Okay... I'll do my best to determine that.

Indeed - please do not PM us asking what you can and cannot assume.  We cannot answer those sorts of questions until after the round is over.  (However, don't PM us then, either - post those questions here on the message board, but only after we have given permission to discuss the problems publicly.)

Okay... one other question that is not included in the rules. Am I allowed to write on the back of the paper, or should I start a new sheet if I run out of room on the front, even though it's the same problem? I'm not putting multiple problems on one page, but I have a long explanation.

[quote="joml88"]I can't remember whether or not I submitted an entry form yet...Should I just send another one in case?[/quote]Sure, you can send one with your Round 1 solutions.

I have never heard of the USAMTS, but I will try to spread the word and sign up. For what ages do this competition allow...?

usually high school, but if middle schools people want to, they can

can I sign up even though I live in Canada? :?

Are you a US citizen?  If so, yes; it not, then no, since the program is funded by the NSA and thus only open to US citizens (and possibly permanent residents - I'm not clear on that; but you don't seem to be one).

The contest is open to high school and middle school students who are US residents and/or US citizens.  So, an American living in Canada may participate.

AAAAuh :( Oh well. At least the Americans let Canadians write USAMO :)


function
limit
real analysis
real analysis unsolved

Let $f: R\to R$ be a continous funtion such that $\lim_{x\to\infty}f(x)$ and $\lim_{x\to -\infty}f(x)$ exist and are equal. Prove that $\forall d>0$ exists $x_1,x_2$ such that $x_1-x_2=d$ and $f(x_1)=f(x_2)$.

Since the limits are equal f(x) reaches it's maximum or his minimum in a real point $a$. Take $g(x)=f(x+d)-f(x)$. We have that $g(a)\cdot g(a-d)\le 0$ and the conclusion follows.

Let $g(x)=f(x+d)-f(x)$-contiious function.If we suppose the contary there exists $d>0$ s.t. $g(x)>0$ for all $x$, thus $f(x+kd)>f(x)>f(x-kd)$ (*), $k \in \mathbb{N}$ for all $x$.In (*) taking $k \to \infty$ we get that $f(x) \equiv const$, which contradicts our supposion.
The same could be done in the case $g(x)<0$.

I haven't seen xirti answered.

[quote="spix"]So if $\lim_{x\to\infty}f(x)$=$\lim_{x\to -\infty}f(x)$ the function is bounded ? $\lim_{x\to\infty}f(x)$ is necessary finite right?  :blush:[/quote]

The function may not be bounded. Think at $f(x)=|x|$. But the function have at least one of two: an upper bound or a lower bound.


Three friends have a total of 6 identical pencils, and each one has at
least one pencil. In how many ways can this happen?

the solution they gave was based on casework, but couldn't they have just simplified it to a simple distribution case? After each friend gets a pencil, there are 3 pencils that are left over that can be given to any one of them.  Thus, there are 3 *s and 2 |s (ex: *|**| would mean friend 1 got 1 pencil, friend 2 got 2, and friend 3 got none). there are 5 objects of which you are trying to pick 2 to be bars, so 5C2 = 10

[font=TIMES NEW ROMAN][hide=Solution][c]4   1   1                1
1   4   1                2
1   1   4                3

3   2   1                4
3   1   2                5
2   1   3                6
2   3   1                7
1   2   3                8
1   3   2                9

2   2   2                10[/c] $$\definecolor{color}{rgb}{0.6, 0.5, 0.4} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

[quote=dragon78][quote=Harry10][quote=charzar]I believe the easiest way is to do what pifinity did, list the solutions.[/quote]

Agreed[/quote]

Haha yeah totally.[/quote]

lol agreed!

Stars and bars, 3+3-1 choose 3 5 choose 3 10!!!!!!!!!!!!

Extra Reasoning on the Stars and Bars Solution:

We are given that each friend must have at least one pencil. So there are $6-3 = 3$ pencils that can be distributed in any way. There are 3 friends, meaning there are 2 bars when using stars and bars. There are 3 pencils left, so there are 3 stars.
We have: *** | |.
The number of ways to arrange this is $\frac{5!}{3!2!} = 10$.

Hence, there are $\boxed{10}$ ways this can happen.

[hide=Solution in Words] The friends can have one pencil each - so there are three pencils excess to give out. there are $10$ ways to do this.


Hello.  I wasn't sure where to ask this, so I'm posting here.  Could someone reset the post

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102491

so I can continue to edit it?  I periodically update the leader board in the post.  I've a new entry for a high-scoring student I would like to put in.

Thanks!

Best,
Dr. A.

It should be fixed for you now Catherine.


Give x+y+z=xyz for all x,y,z posivetive numbers
Prove that
$\sum_{x,y,z}(\sqrt{(1+x^2)(1+y^2)}-\sqrt{1+x^2}-\sqrt{1+x^2})/(xy)$=0

I think the last $\sqrt{1+x^2}$ should be $\sqrt{1+y^2}$.
And posted before.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=79661[/url]

This problem is for secondary school , so that solution is too complex
I always want to find an sencondary-school solution


I always saw someone said rank top 2 or top 5 in the class of ...
What score decide the rank?

Grade point average (GPA). Usually, if your grade in a course is on a scale of 0-100, the GPA is (grade - 60) / 10, or 0 if you are doing worse than 60. (Alternatively, you might be given letter grades for your courses, and each letter grade is assigned a value, e.g. A = 4.0, A- = 3.7, B+ = 3.3, B = 3.0, ... This is what usually happens in college). Your overall GPA is the average of your GPA for all of the classes you've taken. Some schools give you a bonus if you take honor or AP classes.

Thank you! If someone takes more courses, does that means he will have higher overall GPA and have higher rank? May the rank by average GPA?

The rank is by average. However, colleges will see your transcript in detail, so if they see that you took many challenging courses, they might view that more highly than a very good GPA. Some people at my high school took the easiest classes possible in order to get the best GPA, but they weren't the smartest people in the school. College admissions officers know that happens.


Woohoo!  Apparently I've won 10 bucks.  Still not a big amount, but I won indeed. :lol:

http://www.hitzradio.com/

Love how for a Canadian station, most of the winners are American.

Now, anyone know any good Math books with AIME-related material?  I'm thinking about Challenging Problems in Geometry...but I'm also looking for books with lots of good AIME-related Counting problems.  Agh...so indecisive!

Please Canadians...help yo brotha!

[quote="lowest_reynolds"]

Now, anyone know any good Math books with AIME-related material?  I'm thinking about Challenging Problems in Geometry...but I'm also looking for books with lots of good AIME-related Counting problems.  Agh...so indecisive!

[/quote]
Challenging Problems in Geometry, good book.  Have you already got the AoPS books? Maybe those might help.

I have the AoPs books thank.

Suggestions?

Geometry Revisited is a good book too.  And any of the books you get when writing the CMO, so In Polya's Footsteps, the CMO book, er...can't remember the other ones right now.


modular arithmetic
arithmetic sequence
algebra unsolved
algebra

[quote="Moonmathpi496"]What is the largest possible length of an arithmetic progression of positive integers $ a_{1}, a_{2},\cdots , a_{n}$ with difference $ 2$, such that $ {a_{k}}^{2}\plus{}1$ is prime for $ k \equal{} 1, 2, . . . , n$?[/quote]

Take any three consecutive terms $ a,a\plus{}2,a\plus{}4$ and look modulo 5 :
If $ a\equal{}0\pmod{5}$, then $ (a\plus{}2)^{2}\plus{}1\equal{}0\pmod{5}$ and the second is divisible by $ 5$.
If $ a\equal{}1\pmod{5}$, then $ (a\plus{}2)^{2}\plus{}1\equal{}0\pmod{5}$ and the second is divisible by $ 5$.
If $ a\equal{}2\pmod{5}$, then $ a^{2}\plus{}1\equal{}0\pmod{5}$ and the first is divisible by $ 5$.
If $ a\equal{}3\pmod{5}$, then $ a^{2}\plus{}1\equal{}0\pmod{5}$ and the first is divisible by $ 5$.
If $ a\equal{}4\pmod{5}$, then $ (a\plus{}4)^{2}\plus{}1\equal{}0\pmod{5}$ and the third is divisible by $ 5$.

So, among any three consecutive terms, one of the $ {a_{k}}^{2}\plus{}1$ is divisible by $ 5$, so is not prime (except if it is equal to 5).
So the maximum is two, except if the number $ 2$ is in the sequence. And in this case we can have a three-elements sequence :
$ 2,4,6$ which gives the three primes $ 5,17,37$.

The answer is $ 3$.

Answer:$ n \equal{} 3$.
Solution:
Denote $ p_{i}\equal{} a_{i}^{2}\plus{}1$.If $ a_{i}\equiv 2,\minus{}2(mod 5)$ then $ 5|a_{i}^{2}\plus{}1$.If $ a_{i}> 2$ and $ a_{i}\equiv 2,\minus{}2(mod 5)$ then $ p_{i}$ isn't prime number.So it is obviously that among any three numbers $ a,a\plus{}2,a\plus{}4$ exist one number that $ \equiv 2,\minus{}2(mod 5)$.Now if in the progression isn't exist $ 2$,then $ n\leq 2$.If $ a_{1}\equal{} 2$ then $ a_{2}\equal{} 4$,$ a_{3}\equal{} 6$.And $ p_{1}\equal{} 5,p_{2}\equal{} 17,p_{3}\equal{} 37$.
Edit:I was too slow.I'm sorry pco.

Anyone have the whole problem set(with the distribution for each grades) and solution for russia 2002 & 2003?
Thanks


Let $\sum_{k=1}^n{a_k^2}=2003, a_k\in\mathbb{N}$. And $a_i\neq a_j$ if $i\neq j$, ($1\leq i, j\leq n$). Find the maximum of n.


Answer the poll above. I did not put ZzZz... on the list as he will clearly pwn all of us   :o

You clearly should put me on the poll.

I vote myself.  :P

Nice though... 3 from middle school, 1 from high school in missouri... nice

Aww! Everyone says math will fail. I voted tiny!

[quote="Mewto55555"][quote="ZzZzZzZzZzZz"][quote="Mewto55555"]I did not put ZzZz... on the list as he will clearly pwn all of us   :o[/quote]

You can count on it.  I've got my lucky streak going.  :D

Most recent demonstration of luckiness:  Qualifying for USAMO despite being put in a room with a teacher who, despite my polite request, wouldn't stop laughing on the phone [b]AND[/b] was ignorant enough to tell her next class, "I had this arrogant kid working on some math thing last period.  This is my room, and I'll talk whenever I want."

Most recent demonstration of unluckiness: Not going to Ladue.  Your schools have too much money.  I go to school in a cardboard box compared to you guys.

Anyways, back on topic, I voted for sjlite assuming his real name is S## Lite, and he's in high school.  Otherwise I don't know him.[/quote]

Yeah clearly who else would it be?

Btw, our school teaches us very little contest math. We had announcements going on for like 30 minutes durin gAIME.[/quote]

I wasn't sure about the "j" in the middle.  You never know how many people have similar names.  I was expecting "slite"

Yeah, no school teaches contest math, but it's the overall atmosphere of acknowledging achievement that is lacking in my school district, and I guess the lack of smart people who will actually qualify for AIME.  A boost in intelligent people will typically make a school care more about them.  And the lack of money (though our school would rather install huge TVs in the cafeteria instead of focusing on academics.  hmmm...)

30 minute announcements?   :huh:  I would love to have to sit through that everyday.   :lol:   And I mean that seriously; announcements are infinitely better than classes at my school.

5 minutes of distractions feels like 30 when you have to bash EVERYTHING because you suck that much.

And no, Jeff Boyd's schools have contest classes.

[quote="ZzZzZzZzZzZz"][quote="Mewto55555"]And no, Jeff Boyd's schools have contest classes.[/quote]

No, Jeff Boyd's schools have Jeff Boyd.[/quote]

And contest classes.

I'd vote for Ravi F.

//but it's the overall atmosphere of acknowledging achievement that is lacking in my school district//

That's quite true here as well, [color=red]everyone in my school district including myself is terrible at math, absolutely terrible in all respects in all things in all ways possible and impossible and somewhat non-possible, at absolutely everything from mathcounts to geographic bee to spelling bee and any other academic thing you can think of.[/color]

[quote="serialk11r"]Darn I take it only 5 people in Missouri have a chance at making USAMO since you made a poll with 5 voting options?[/quote]

Replace "have a chance at making" with "are taking."


Hi, I have a question is about $ \sigma$ finite meaure space, please help. Thank you very much.

If $ (X, M,\mu)$ is a $ \sigma$ finte measure space, then show that there are at most countably many $ a\in X$ for which $ \mu({a})>0.$

sigma finite means we really gotta just prove it for the finite measure space case

so if $ \mu (X)  <\infty$

then let $ A_k\equal{} \{a\in X: \frac{1}{k} > \mu (a) \ge \frac{1}{k\plus{}1} \}$ for k=1,2,3..

$ A_0 \equal{} \{ a\in X: \mu(a) \ge 1\}$

so the set we care about $ A\equal{} \{ a\in X: \mu(a) > 0 \} \equal{} \cup_{k\equal{}0}^\infty A_k$

and A uncountable iff $ \exists A_k$ uncountable

so it remains to show that each $ A_k$ is countable...

assume $ A_k$ uncountable.. then there exists an infinite countable subset $ A_k'$

but $ \mu(A) \ge \mu (A_k) \ge \mu (A_k') \ge \sum_{n\equal{}1}^\infty \frac{1}{k\plus{}1}\equal{} \infty$

obvious contradiction


Suppose that circles $ O_{1}$ and $ O_{2}$ intersect at points $ A$ and $ B$,and the tangent line of circle $ O_{2}$ at point $ C$, passing through point $ O_{1}$,intersects with circle $ O_{1}$ at point $ D$.
let $ AE$ be a chord of circle $ O_{1}$ which is perpendicular to line $ DC$.Suppose $ AF$ is perpendicular to $ DE$ at point $ F$.Denote by $ G$ the intersection point of lines $ BD$ and $ AF$.Prove that if line $ AC$ is tangent to a circle $ O_{1}$,then $ AG=FG$.

$ M \in AE \cap DC, N \in O_{1}B \cap (O_{2})$. We can prove $ NC \parallel O_{1}O_{2}$ easily.
$ \overline{O_{1}B}.\overline{O_{1}N}= O_{1}C^{2}= O_{1}A^{2}+AC^{2}\Rightarrow \overline{O_{1}B}.\overline{BN}=AC^{2}$
$ \Rightarrow \frac{O_{1}M}{MC}= \left( \frac{O_{1}A}{AC}\right)^{2}= \frac{O_{1}B}{BN}\Rightarrow MB \parallel CN \perp AB$
$ \angle GAM = 90^{o}-\angle AED = \angle ABM-\angle ABD = \angle GBM \Rightarrow M,G,A,B$ are concyclic. So $ MG \parallel EF (\perp GA)$ and $ G$ is the midpoint of $ AF$.


An open question: Take a cube and decompose it into 27 smaller cubes in the typical way. Put a base 3 digit where each little cube is (0, 1, or 2). Let us then read the sequences of the cube as follows: we pick any of the smaller cubes that show on the front, left, or top face. We then go straight through the cube from that cube. So for example, if it was the very top left front cube that we chose, relative to the top face, we would count first that cube, then the one directly below it, then the one below that. Thus we have 27 total possible strings. Now the question: can we put our base-3 digits in this such that every base-3 string from 000 to 222 occurs exactly once?

Can we generalize to an n dimensional cube with digits in base n with strings ranging from 000 to $n^{n}-1$?

*bump* Anyone? Is the problem statement unclear or something?

It is possible:

top layer:
012
002
011
middle layer:
221
202
211
bottom layer:
100
010
212

Of course, given this solution there are 5 more by permuting 0,1 and 2. But there may still be other solutions (but not too many).


γεια σε ολους. εστω p ενας πρωτος >2 και k ενας φυσικος αριθμος ωστε 1<=k<=2(p+1) και p δεν διαιρει  k. εστω N=2kp+1 τοτε τα παρακατω ειναι ισοδυναμα :1) Ν ειναι πρωτος ,2) υπαρχει α ακεραιος>0 ωστε ΜΚΔ(α^κ+1,Ν)=1

συγνωμη αλλα μαλλον ισχυει οτι 1) συνεπαγεται 2). το αντιστροφο δεν ισχυει. και επισης εκει που λεει a^κ+1 εννοω a^k+1  :)

Φίλε κώστα τα δύο αυτά που έγρταψες στο δεύτερο πόστ είναι ιδια εννοεις α^(κ+1) η (α^κ)+1?

Μια παρατηρηση.Το a>1.Φιλε Αμβροσιε μαλλον το δευτερο ειναι το σωστο ετσι Κωστα?

Αμβροσιε ρωτησα τελικα τον Κωστα και επειδη αυτος δεν μπορει να απαντησει μου ειπε να σας ενημερωσω οτι ηθελε να γραψει a^k+1 (με το αγγλικο k).Αν και νομιζω οτι κατι τρεχει με την ασκηση.Εχω βρει μια παρα πολυ απλη λυση και αναρωτιεμαι αν ειναι λαθος η λυση μου η τελικα η ασκηση του Κωστα. :huh:

παιδια δεν πρεπει να ειναι δυσκολο,απλα ειναι ενα επωνυμο κριτηριο για το αν ενας αριθμος ειναι πρωτος.για το αντιστροφο δεν ειμαι σιγουρος αφου η λεξη που ελεγε ηταν:τα 1),2) ειναι equivalent. παρακαλω αν καποιος ξερει τι σημαινει να το πει. :oops:  χρονια πολλα σε οσους γιορταζουν  :lol:

Κωστα equivalent σημαινει ισοδυναμος.Παντα υπαρχει a ωστε (a^k+1,N)=1. Αν a=0modN τοτε a^k+1=1modN αρα (a^k+1,N)=1.....

ε τοτε ισχυει και το αντιστροφο της ασκησης...α και κατι αλλο ποτε θα μαθεις μπιλιαρδο ρε μητσο? ολο διπλα παιρνω :rotfl: ελα πλακα κανω μπραβο ομως βρηκες τη φορμα σου ε?

Ητεχνη δεν ξεχνιεται! :rotfl:

οκ η ρεβανς σημερα αν προλαβω να βγαλω τα γενικευμενα ολοκληρωματα και της συναρτησεις β,γ.αληθεια δωσε την αποδειξη του αντιστροφου :D

Διαλεξε εναν πρωτο....ας πουμε ο 3.Εστω οτι ο κ=4.τοτε Ν=2κρ+1=25 και ομως (5^4+1,25)=1 Αρα 0 25 ειναι πρωτος!Προφανως το αντιστροφο δεν ισχυει και επειδη η λυση μου για το ευθη ηταν πολυ περιεργη(δεν χρησιμοπ. καν την βασικη συνθηκη οτι Ν-πρωτος) ισως το προβλημα σου να ειναι λαθος φιλε μου. :rotfl:

αγαπητοί φίλοι πολύ ομορφη  η "κόντρα" σας πάντως οσον αφορά την ασκηση δεν την κοίταξα δυστυχώς αλλά θα βάλω εγώ κάποια
καλη συνέχεια
Χρόνια Πολλα

Αμβροσιε δεν ξερω αλλα ειναι λιγο μπερδεμενο το κριτηριο ετσι οπως το διατυπωσα(κρ.του Pochlington,νομιζω). αυτο ειναι αποτελεσμα ενος  ποιο γενικου θεωρηματος το οποιο (ειναι σωστο) λεει:Εστω  α,ν ακεραιοι, ν>1,    αν α^(ν-1)=1modν και α^d οχι ισο με το 1 modν για καθε φυσικο διαιρετη d του ν-1 με d<n-1, τοτε ο ν ειναι πρωτος. ελπιζω να μην σε μπερδεψα.  :)

[b]Kritirio Pocklington[/b]

Estw $n \in \mathbb{N}, n>0$. Ipothetoume oti iparxei $q\in \mathbb{P}$ (sinolo Prwtwn arithmwn) me $q|n-1$ tetoio wste $q>\sqrt{n}-1$. Ean iparxei akeraios arithmos $a$ tetoios wste:
i)$a^{n-1}\equiv 1 \pmod{n}$ kai
ii)$\big(a^{\frac{n-1}{q}},n\big)=1$,
tote o $n$ einai prwtos.

To kritirio auto genikeuetai kai sti Thewria twn Elleiptikwn kampilwn, poli xrisimwn stin Kriptografia...

Grafw episis $3$ isodinames protaseis.

Stis parakatw periptwseis o arithmos $n>2$ einai prwtos:

i)Yparxei $a\in \mathbb{Z}$, tetoios wste $a^{n-1}\equiv 1 \pmod{n}$, kai isxiei $a^{k}\not\equiv 1 \pmod{n}$, gia kathe $k=1,\ldots,n-2$.

ii)(Veltiwsi tou (i)) Yparxei $a\in \mathbb{Z}$, tetoios wste $a^{n-1}\equiv 1 \pmod{n}$, kai gia kathe thetiko diaireti $k$ tou $n-1$, o opoios einai gnisiws mikroteros tou $n-1$, isxiei $a^{k}\not\equiv 1 \pmod{n}$.

iii) (Veltiwsi tou (ii)- Kritirio Procklington) Yparxei $a\in\mathbb{Z}$, tetoios wste $a^{n-1}\equiv 1 \pmod{n}$, kai gia kathe prwto diaireti $q$ tou $n-1$, isxiei $a^{\frac{n-1}{q}}\not\equiv 1 \pmod{n}$.

Alexandros

Y.G. Yparxoun arketes diaforetikes diatipwseis tou thewrimatos tou kritiriou tou Poclington (to opoio sixna anaferetai kai san thewrima twn Poclington-Lehmer).


analytic geometry
geometry
trapezoid
absolute value

Did you put them in the right order?

They need to be ordered, but it doesn't matter if it's a triangle.

i put them in order and NO it is not a triangle it is a trapazoid did i do something wrong

[quote="shybox"]When plotted in the standard rectangular coordinate system trapezoid ABCD has vertices A(1,-2), B(1,1), C(5,7), and D(5,1). What is the area of trapazoid ABCD.

I am confused cause i posted a question like this and people told me to use the shoelace theorm. I used it on this question but got the wrong answer. Can someone please explain. I think i did the formula wrong.

Here is what i did.

1,-2
1,1
5,7
5,1
cross multiply both ways add and i got the absolute value of 38-12 so my answer was 26

Answer
 [hide]18[/hide][/quote][hide]the mod is right. Also you need to put 1,-2 at the end again and divide the difference in half.
1,-2
1,1
5,7
5,1
1,-2
1+7+5-10=3
-2+5+35+1=39
39-3=36
Divide that by 2 to get 18[/hide]

Thank you soo sooo much you do not know how much that will help me cause i didn't know you had to put the starting number back on the bottom and then i forgot to divide by two :D  :)

Yeah, I was a bit reckless at my first few sad attempts at Shoestringing.

So:

1. In order.

2. Repeat first one.

3. Divide by 2.

does anyone know why multiplying random numbers and subtracting crazily can give you the area of a figure? I know it works and all it just seems really out of the blue.

[quote="kyyuanmathcount"]It WAS some really ugly formula that a smart guy derived, then figured how he could make it easier on people.[/quote] :rotfl:  :rotfl:

I googled and yahooed and searched a bunch of the other places for the guy, but he doesn't seem to exist. He doesn't get enough for his job. :(