K, well, Christmas is over, Hannukah and Kwanzaa are today, and here's the challenge:
Who got the geekiest gift?

Please provide pictures if necessary.
[i]Ex.

[Picture Necessary] I got a t-shirt that says "It's hip to b^2"
[Picture Unnecessary] I got a book of Mensa Sudoku puzzles.
[/i]

You can only list one item at a time- the goal is to top the "geekiness" of the person before you. If you got more than one geeky present, you can play again.

I'll start with something that's a little geeky.

[b][url=http://www.wbshop.com/catalog/detailed.xml?product_id=35811;option_id=;template=wbs_enlarge]I got a Harry Potter t-shirt[/url][/b]

I got a 4x4x4 AND 5x5x5 rubik's cubes :D

[quote="xxazurewrathxx"]I got an early New Year's gift, a coffee mug that had 2006 in binary.   :|

Yeah. It went around the cup.[/quote]

 :rotfl: that's AWESOME.

i got an albert einstein action figure.

I got a [url=http://images.amazon.com/images/P/B00006IFTK.01-A3KTM6ISXAEHAX._SCLZZZZZZZ_.jpg]Spanish-English Translator[/url]

I got a Santa candle, that's got to be geeky.

I got probability assignments to do now if that is not geeky then i dunno what is.

ipods and best buy giftcards are not geeky. anything that the average person would like as a present isnt geeky.

[url=http://www.mentalfloss.com/Merchant2/merchant.mvc?Screen=PROD&Product_Code=PYT&Category_Code=mer]However, this is. I mean, yes, a lot of people might wear it because it's "vintage" but how many people really "get it?"[/url]

[quote="math92"]my friend got something really weird and disturbing:
[hide]he got a Sports Illistrated Swimsuit Calendar[/hide][/quote]
That is strange... and rather disturbing, but not what I would call geeky...

I got nothing. :(

But our spring festival comes soon.

[quote="h_s_potter2002"][quote="math92"]my friend got something really weird and disturbing:
[hide]he got a Sports Illistrated Swimsuit Calendar[/hide][/quote]

Um, for most guys, thats definetley not weird.  And I wouldn't really call it disturbing.[/quote]
Some of us are less inured to others being so openly amoral than others. If one is not used to such disturbing and wierd gifts, the recipient is viewed as a rather warped person. And his parents don't do anything about it?!  :huh:

[quote="h_s_potter2002"][quote="math92"]my friend got something really weird and disturbing:
[hide]he got a Sports Illistrated Swimsuit Calendar[/hide][/quote]

Um, for most guys, thats definetley not weird.  And I wouldn't really call it disturbing.[/quote]
well since most of his posts are in the middle school forum, I'd assume that he's in middle school, and depending on what grade he's in, then it might be

If his parents or some other relative gave it to him, then I would find that a little weird.  Otherwise, I don't find it all that disturbing.


What is the sum of the next two terms in the geometric sequence 1, $ \frac{1}{2}$, $ \frac{1}{4}$, $ \frac{1}{8}$, $ \ldots$? Express your answer as a common fraction.

The next two terms in the sequence are 1/16 and 1/32, so the sum is 3/32.

The Reason Why:

The terms in geometric sequences are generated by multiplying what's known as the common ratio to each term. In this sequence, we see that the common ratio is $ \frac{1}{2}$. So we find the next two terms and add them, thus getting the answer dragon stated above.

The geometric sequence's common ratio is 1/2. 1/8*1/2=1/16; 1/16*1/2=1/32; their sum is 3/32


i was looking at this problem and i tried to do this problem using newtons sums but it didnt work out right for some reason

The roots of $x^4 - x^3 - x^2 - 1 = 0$ are $a, b, c, d$. Find $P(a) + P(b) + P(c) + P(d)$, where $P(x) = x^6 - x^5 - x^3 - x^2 - x$.

heres my work
[hide]

for roots $a,b,c,d$, your trying to find $a^6-a^5-a^3-a^2-a + b^6 - b^5 - ...$
which equals
$(a^6 + b^6 + c^6 + d^6) - (a^5 + b^5 + c^5 + d^5) - (a^3 + b^3 + ...  )$etc etc

so basically using newtons sums, you want:

$s_6 - s_5 - s_3 - s_2 - s_1$

knowing that $a+b+c+d = 1 = s_1$ you get

$(s_1) - 1 = 0 \Rightarrow s_1 = 1$
$s_2 - s_1 - 2 = 0 \Rightarrow s_2 = 3$
$s_3 - s_2 - s_1 - 3 = 0 \Rightarrow s_3 = 7$
$s_4 - s_3 - s_2 -s_1 = 0 \Rightarrow s_4 = 11$
$s_5 - s_4 - s_3 - s_2 = 0 \Rightarrow s_5 = 21$
$s_6 - s_5 - s_4 - s_3 = 0 \Rightarrow s_6 = 39$

so $39 - 21 - 7 - 3 -1 = 7$ which is one off from the answer... (answer is $6$)

can anyone tell me what i did wrong???

[/hide]

you don't have to do all of those newton sums - remember that from what they tell you, (let the first polynomial be Q(x)) Q(a)=Q(b)=Q(c)=Q(d)=0, so

P(a)=a^6-a^5-a^4-a^3-a^2-a+a^4
=a^4-a^3-a
=a^2-a+1, so really you just need s_2-s_1+4, which by your work turns out to be 6.

remember to use all of what you're given

hmm but how come my method didnt turn out right???
did i make an arithmatic error somewhere?

I think you forgot that the coefficient of x is 0.  If you take that into account:

$s_1 = 1$
$s_2 = 3$
$s_3 = 4$
$s_4 = 11$
$s_5 = 16$
$s_6 = 30$

So you have $30 - 16 - 4 - 3 -1 = 6$.


calculus
integration
probability
calculus computations

There are 10 cards, labeled from 1 to 10. Three cards denoted by $ a,\ b,\ c\ (a > b > c)$ are drawn from the cards at the same time.
Find the probability such that $ \int_0^a (x^2 \minus{} 2bx \plus{} 3c)\ dx \equal{} 0$.

Solving the integral we find that
$ \frac{a^3}{3}\minus{}(a^2b)\plus{}3ca\equal{}0$   Since $ a>0$ we can "cancel" it. Then

$ \frac{a^2}{3}\minus{}ab\plus{}3c\equal{}0$   Since $ \minus{}ab\plus{}3c$ is integer then we see that $ a\equal{}3k$ so that the expression is integer.
The posible values of $ a$ are $ 3,6,9$ Then doing some casework we find the options are
$ a\equal{} 9 , b\equal{}5 ,   c\equal{}4$
$ a\equal{}9 ,   b\equal{}4 ,   c\equal{}2$
$ a\equal{}6   , b\equal{} 5,   c\equal{}2$
$ a\equal{}3 ,   b\equal{}2 ,   c\equal{}1$
Hope I haven't missed anything :P
Then we have that the total cases are $ \binom{10}{3}\equal{}120$
And the cases we are looking for are $ 4$
So the probability is $ \frac{1}{30}$

Daniel

Oh :(
Could you tell me in which part did I made a mistake?
Is it when counting the total cases
Maybe I didn't take into account that choosing (a,b,c) is not the same as choosing (a,c,b)

Daniel

[quote="lambruscokid"]
$ a \equal{} 9 , b \equal{} 5 , c \equal{} 4$
$ a \equal{} 9 , b \equal{} 4 , c \equal{} 2$
$ a \equal{} 6 , b \equal{} 5, c \equal{} 2$
Daniel[/quote]

this part is incorrect.

Yes it sholud be
$ a\equal{}3, b\equal{}2, c\equal{}1$
$ a\equal{}6, b\equal{}3, c\equal{}2$
$ a\equal{}9, b\equal{}4, c\equal{}3$
I  don't know why I made those stupid calculation mistakes

So the probability should be $ \frac{1}{40}$

I expect it is correct now :)
Thank's Kunny

Daniel


calculus
integration
trigonometry
logarithms
calculus computations

here are some hints for the ones i can do in my head

#2) sub $u=ln(x)$ and you will have $du=\frac{dx}{x}$ so the new integral is $\int{\frac{du}{u^5}}$

#3)sub $u=e^{{-x}^{2}}$ and note that $ln(u)=-x^2$ and $du=-2xe^{{-x}^{2}}$ and u now have ${\frac{1}{2}}\int{ln(u)du}$ (check my work on this one i did it really quickly.)

#4) sub $x=sec(\theta)$

well for the first one just subsitute $u=\sin x - \cos x$ so $du =\cos x +\sin x$

[quote="maokid7"]

#3)sub $u=e^{{-x}^{2}}$ and note that $ln(u)=-x^2$ and $du=-2xe^{{-x}^{2}}$ and u now have ${\frac{1}{2}}\int{ln(u)du}$[/quote]
And how is this new integral ${\frac{1}{2}}\int{ln(u)du}$ solved?

Yes. :)

Hint : $ln(x) = ln(x) \times 1$.


geometry
3D geometry
pyramid
trigonometry
geometry proposed

[b]This was previously posted here and subsequently deleted. If that happens again, I would like to get a PM from a mod on the reasons. Thanks.[/b]

Prove that the volume of a triangular pyramid, in general case, is given by

$V={abc\over 3}\sqrt{\sin\phi\sin(\phi-\alpha)\sin(\phi-\beta)\sin(\phi-\gamma)}$

where $a,b,c$ are the side edges; $\alpha, \beta, \gamma$ are the angles between them; $\phi=\frac{\alpha+\beta+\gamma}{2}$

What are you talking about? This topic is already in Geometry Proposed & Own problems? I moved it there (Darij asked me).

It was cross-posted, now it's not and all is OK. Thanks.


ratio
geometry
circumcircle
geometry proposed

[color=darkred]Let $ ABC$ be a triangle with the incircle $ i \equal{} \mathcal C(I,r)$ and the circumcirle $ o \equal{} \mathcal C(O,R)$ .

Let $ l \equal{} \mathcal C(L,\rho)$ be the circle which is tangent to the sides $ (AB)$ , $ (AC)$ and (internal)

tangent to the circle $ o$ . Prove that $ OL\perp OI\Longleftrightarrow \frac {2}{\rho} \equal{} \frac {1}{R} \plus{} \frac {1}{R \minus{} r}$ .

[b]Remark.[/b] $ \left\{\begin{array}{c} T\in o\cap l \\
\ S\in AT\cap BC\end{array}\right\|\Longrightarrow\frac {SB}{SC}\cdot\frac {p \minus{} c}{p \minus{} b} \equal{} \frac {c^2}{b^2}$ . Interpret it syntheticallly ! [/color]

Denote by $ M$ and $ N$ two touch points of $ l$ to the sides $ (AB)$ and $ (AC)$, respectively.It is well known that $ I$ is the midpoint of $ MN$. $ o$ and $ l$ touch internally, so $ OL = |R - \rho|$ and from Euclid theorem $ IM^2 = AI \cdot IL = \frac {r}{sin \angle \frac {A}{2}} \cdot \rho sin \angle \frac {A}{2} = \rho\cdot r$
Hence

$ OL\perp OI\Longleftrightarrow OI^2 + OL^2 = IL^2 \Longleftrightarrow OI^2 + OL^2 = LM^2 - IM^2 \Longleftrightarrow$
 $ \Longleftrightarrow (R^2 - 2Rr) + (R - \rho)^2 = \rho^2 - %Error. "rhor" is a bad command.
\Longleftrightarrow 2R^2 - 2Rr = \rho(2R - r) \Longleftrightarrow$
$ \Longleftrightarrow\frac {2}{\rho} = \frac {2R - r}{R(R - r)}\Longleftrightarrow \frac {2}{\rho} = \frac {1}{R} + \frac {1}{R - r}$ .

Nice and [b]easy[/b], Cefer ! But the [b]remark[/b] ?!

This remark !
[quote="Virgil Nicula"][color=darkred][b]Remark.[/b] $ \left\{\begin{array}{c} T\in o\cap l \\
\ S\in AT\cap BC\end{array}\right\|\Longrightarrow\frac {SB}{SC}\cdot\frac {p \minus{} c}{p \minus{} b} \equal{} \frac {c^2}{b^2}$ . Interpret it syntheticallly ! [/color][/quote]

[quote="Virgil Nicula"]This remark !
[quote="Virgil Nicula"][color=darkred][b]Remark.[/b] $ \left\{\begin{array}{c} T\in o\cap l \\
\ S\in AT\cap BC\end{array}\right\|\Longrightarrow\frac {SB}{SC}\cdot\frac {p - c}{p - b} = \frac {c^2}{b^2}$ . Interpret it syntheticallly ! [/color][/quote][/quote]

With some calculations we can find out that $ AM = \frac {cb}{p}$.
Let A ex-circle touch $ BC$ at $ V$. Let's prove that $ \bigtriangleup AVC \sim ABT$
Obviously $ \angle ACV = \angle ATB$ so it is enough to prove that $ \frac {AT}{BT} = \frac {AC}{CV}$
$ \Rightarrow \frac {AT}{BT} = \frac {AM}{BM} = \frac {\frac {bc}{p}}{c - \frac {bc}{p}} = \frac {b}{p - b} = \frac {AC}{CV}$
So $ \angle CAV = \angle SAB \Rightarrow \frac {CV}{VB}\frac {c}{b} = \frac {SB}{SC} \frac {b}{c}$
$ \Longrightarrow {\frac {SB}{SC}\cdot\frac {p - c}{p - b} = \frac {c^2}{b^2}}$

[hide]About calculation of $ AM$
$ AM=\frac{AI}{sin \angle\frac{A}{2}}=\frac{\frac{r}{cos\angle\frac{A}{2}}}{sin \angle\frac{A}{2}}=\frac{2r}{sin \angle A}=\frac{\frac{2S}{p}}{\frac{2S}{bc}}=\frac{bc}{p}$. That's all :wink:[/hide]

[quote="Virgil Nicula"][color=darkblue]That is no all, C[b]efer[/b] ! And the [b]synthetical interpretation[/b] ?! Thank you ! Happy new year ![/color][/quote]

OK.
Let $ U$ be the base of the perpendicular from $ I$ to $ AB$. Then $ IU^2=AU\cdot UM \Longrightarrow r^2=(p-a)(AM-(p-a))$
$ \Longrightarrow \frac{S^2}{p^2}=(p-a)(AM-(p-a)) \Longrightarrow \frac{(p-b)(p-c)}{p}=AM-(p-a)$
$ \Longrightarrow p-(a+b)+\frac{bc}{p}=AM-(p-a)\Longrightarrow AM=\frac{bc}{p}$

$ \frac{AT}{BT}=\frac{AM}{BM}=\frac{\frac{bc}{p}}{c-\frac{bc}{p}}=\frac{b}{p-b}=\frac{AC}{CV}$

$ \left \{\begin{array}{c}\angle ATB=\angle ACV \\frac{AT}{BT}=\frac{AC}{CV}\end{array} \right \| \Longrightarrow \bigtriangleup ACV \sim \bigtriangleup ATB \Longrightarrow \angle ACV=\angle TAB$
Let $ AV$ meet $ o$ again at $ K$. Then $ TK \parallel{} CB$ and $ CT=BK$

$ \bigtriangleup ATB \sim \bigtriangleup ACV  \sim \bigtriangleup BKV$

$ \Longrightarrow \frac{AB}{BV}=\frac{AT}{BK}$

$ \Longrightarrow\frac{c}{p-c}=\frac{AT}{CT}=\frac{AS}{CT}+\frac{ST}{CT}$

$ \Longrightarrow\frac{c}{p-c}=\frac{ST}{CT}(\frac{AS}{ST}+1)$

$ \Longrightarrow\frac{c}{p-c}=\frac{SB}{AB}(\frac{AV}{KV}+1)$

$ \Longrightarrow\boxed {\frac{c^2}{p-c}=SB(\frac{AV}{KV}+1)}$

From Stewart teorem $ AB^2 \cdot CV+AC^2 \cdot BV=CB \cdot (AV^2+CV \cdot BV)$

$ \Longrightarrow \frac{AB^2}{BV}+\frac{AC^2}{CV}=CB(\frac{AV^2}{CVBV}+1)$

$ \Longrightarrow \boxed{\frac{c^2}{p-c}+\frac{b^2}{p-b}=a(\frac{AV}{KV}+1)}$

$ \Longrightarrow \boxed{\frac{\frac{c^2}{p-c}}{\frac{c^2}{p-c}+\frac{b^2}{p-b}}=\frac{SB}{a}}$

$ SB+SC=a$
$ \Longrightarrow \boxed{ \frac{\frac{b^2}{p-b}}{\frac{c^2}{p-c}+\frac{b^2}{p-b}}=\frac{SC}{a}}$

$ \Longrightarrow\frac {SB}{SC}\cdot\frac {p - c}{p - b} = \frac {c^2}{b^2}$

I hope this is correct :)


I know that an admin had reccomended chapters from the second volume of AOPS that were most helpful for the AIME but I can't find it. I spent an hour searching for it but with no avail. I know I will get scolded for this post but I am very desperate to find that thread.

I apologize before hand.

To be honest, the whole book is going to be helpful.


Show us the progress you've made so far, [b]boohsunil[/b].

Just to clear up, should it be
$ ((((x^2\minus{}\frac{1}{2})^2\minus{}\frac{1}{3})^2\minus{}1)^2\minus{}\frac{1}{4}) ... \equal{}3$

I'm not sure where the 1 came from...

i'm sorry

i'll post the real question

evaluate

$ \sqrt{1\plus{}2\sqrt{1\plus{}3\sqrt{1\plus{}4\sqrt{1\plus{}5\sqrt{\cdots}}}}}$

till infinity


geometry
trigonometry
quadratics
analytic geometry
geometric transformation
rotation
algebra

Find the area of a square ABCD containing a point P such that PA=3, PB=7, PD=5.

[hide]we draw the lines through P parallel to the sides of the square. We let the distance from P to AB be w, from P to BC be x, to CD be y, and to AD be z. Now, we have w+y=x+z. We also have w:^2:+z:^2:=9, w:^2:+x:^2:=49, and y:^2:+z:^2:=25. It is easy to find that x:^2:+y:^2:=65.  We wish to find (w+y):^2:=(x+z):^2:=wz+wx+xy+yz. Hmmm...lemme think a bit longer.[/hide]

Ooh, this one is nice.  The key step is [hide]using the Law of Cosines.[/hide]

is the answer [hide]26 +2*sqrt(105)?

Edit: hmm that cant be right, i dont think, cause its less than 49.  But i just set up a system of equations like mystic + used law of cos.  i'll check my work to see where i messed up.

Edit again: yeah i misread a Z for a 2.[hide/][/hide]

actually Joel
u might not be accurate on this one

the key step is not Law of Cosines
makes the question messier than it needs to be, tho u do need it later on.

ANyway, THE key step is ROTATION.

And btw, cats...., ur answer is incorrect.

[color=cyan]Really?  Why don't you just apply LOC to the two triangles which have sides 3, 7, s  and  3, 5, s,  since they have to have their angles adding up to 90?  And then just solve a quadratic for s^2?  Does that not work, or can it be done more nicely?[/color]

[color=cyan]Because they together must be one of the angles of the square.[/color]

Okay, I'll give my solution, but Long, you have to post the nice one.

We've got PA = 3, PB = 7, and PD = 5.  Let AD = AB = s.  Then consider triangle APD.  Let us call angle PAD t.  We have by Law of Cosines
5:^2: = 3:^2: + s:^2: - 2*3*s*cos(t), rearanged to give us
cos(t) = (s:^2: - 16)/(6s)
In triangle APD, we have angle PAB = 90 - t.  So, lets do LOC again:
7:^2: = 3:^2: + s:^2: - 2*3*s*cos(90 - t).  Rearranging and replacing cos(90 - t) with sin(t), we get
sin(t) = (s:^2: - 40)/(6s).

Now, we have sine and cosine of an angle.  We can apply the identity sin:^2: + cos:^2: = 1 to get
[(s:^2: - 16)/(6s)]:^2: + [(s:^2: - 40)/(6s)]:^2: = 1.  Multiplying through and squaring out,
(s:^4: - 32s:^2: + 256) + (s:^4: - 80s:^2: + 1600) = 36s:^2:
2s:^4: - 148s:^2: + 1856 = 0
s:^4: - 74s:^2: + 928 = 0
s:^2: = [74 :pm: :sqrt:(74:^2: - 4*928)]/2
s:^2: = [74 :pm: 2:sqrt:(37:^2: - 928)]/2
s:^2: = 37 :pm: :sqrt:(1369 - 928)
s:^2: = 37 :pm: :sqrt:(441) = 37 :pm: 21

The negative case corresponds to the situation in which P is outside the square, while the plus case (which we want) is the case in which P is inside.  Thus, [area of square] = s:^2: = 58.

Thanks for the catch on that error, Ragingg.

One of my friends swears by the bible that the area is 58... LoL.  He said he put the square on the coordinate plane and then set the edges (0,x) and (x,0).  Then he used the distance formula for point P as (a,b).  3 equations, 3 variables. Ta da!

Yes, he was right.  You made an arithmetic mistake.  1856/2...  :lol:

thanks.  im also curious to see the other solution that darkquantum recommended.

well, my solution is slightly nicer
but Joel, i told u my solution
y don't u post it, seeing as how u have SOO much more time than i do.

My solution involves rotations, a very handy technique

I didn't post because I had forgotten it, but I believe I remember it now.

For any point X, let X' denote its image under rotation.

Rotate the entire diagram 90:^o: around point A so that B' = D.  Draw segment PP'.  Now, look at triangle APP'.  We see that the angle at A must be a right angle, since the two pieces that now make it up added up to 90:^o: in the original square.  Thus PP' = 3:rt2:.  Now, look at triangle PDP'.  Lets try Law of Cosines on angle P', because I said so.  Now
25 = 18 + 49 - 2*(3:rt2:)*7*cos:theta:
42:rt2:cos:theta: = 42
cos:theta: = 1/:rt2:
:theta: = 45:^o:

But wait!  Angle AP'P = 45:^o:  and  angle PP'D = 45:^o:, so that means angle AP'D = 90:^o:, so we have a right triangle, and s:^2: = 49 + 9 = 58.

yup
slightly easier than LOC, wouldn't u agree?


I've done a search, but I didn't see any posts that really explained how to spin a pencil and do beast tricks. Can anyone explain, please?  :(

[quote="Mewto55555"]Practice. One day at school I had like 3 tests in a row, so I finished early and had an hour of free time, so I just sat there until I figured it out.[/quote]

yes. exactly. do it every day at school. dont worry about your teachers and classmates in the background yelling at you telling you to stop dropping your pencil. it happened to me :P

also, use a pen. better grip

yeah... practice.
i mean. the videos online and stuff only help you so much.
i suggest starting out with a half spin. then moving onto a full one.
it takes a while. and patience. and a lot of picking up fallen pens/pencils ;]
but its worth it xD

Do you mean spining back and forth in between consecutive fingers?

or

Do you mean twirling hte pencil in between two fingers in a circular motion?

Thanks for the replies, everyone.  :)

Yeah, I do practice a lot, but I keep skewering people to the walls. I can make a circular motion between two fingers for all of ten seconds before killing someone (less, when my fingers are frozen, like now), but I can't do anything else.

^Yeah, that's usually the spin called the Charge. (Or Reversed Charge, depending on the direction you spin it in).

Generally, I develop a new spin every year (or at least, I have 4 spins right now and I hope to add more), and there all pretty much independent, meaning that I randomly screw around and find whatever works for me.

The videos on Pentrix don't really help you spin, they're only good for giving you ideas.

It's important to find a good pen/pencil. I use the PhD Ultra pencil, which is fatter than normal regular #2 pencils. It's also weighted, meaning when I spin it, the momentum of the spin is different because of the extra weight on one end of the pencil. This momentum is what helps keep the pencil in your hand and not on the floor. I first learned spinning with this kind of pencil, and whenever I try spinning #2 ones now, they always fall, because I'm not used to the balance.

We probably have better things to do now, but not after a test.

Warning: Spinning pencils can be awfully addicting.

[quote="nick42"]also, don't you guys have ANYTHING better to do?[/quote]

You can't practice pencil spinning; it's just something that you get used to over a long period of time. Sitting there for 5 hours spinning and dropping doesn't automatically make you a pro spinner. Eventually, you're going to get used to spinning so that you're not even consciously thinking about it when you are spinning. I've subconsciously spun pencils while walking, in the middle of tests, and even with my other hand.

So you hold the pencil between whatever which 2 fingers you use it with. So you have it know right? Put the end of the pencil facing towards the thumb and hook it in the web between your thumb and index. Now push your finger further away from the thumb that is holding the pencil towards the pencil and you should feel a slight force towards your thumb. When you want to release the pencil, pull your thumb back.

NOTE: Before spinning and you're getting ready to spin, your fingers should be bending just a bit. Now when you release the pencil, your fingers should straighten up. And when it's coming back to the thumb, bend a bit back into the position that you are getting ready to release the pencil.

It's hard to word it out. But I think that videos should help.

if you don't want to make noise than do what alex zhai does

practice in between your fingers at home for about 5 hours a day for 3 days and you're perfect.


The contest took place on 27 Jan. 2008, 10-13 (3 hours), the same problem set given in the 6 districts of Bucharest to determine the participants at the Regional Round (county level).
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185411]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185412]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185413]Problem 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185415]Problem 4[/url]

Well, PhilAndrew, could you post the remaining 2008 problems, as well as all the 2007 problems (since it appears they haven't been posted...)

Can you post 9th-grade      :blush:      please


For some reason, I cannot access junggi's shoutbox's bottom text input field. It used to be there on junggi's blog but just disappeared. Its there on everyone else's blog though. Can someone please look into this? Thanks.

edit: now 1337's also disappeared!

I see no problem with junggi's shoutbox  :maybe:

I think its a per user thing.

edit: Added screenshot(s):


function
limit
floor function
algebra
complex analysis

I have a series of questions that I suppose to be true, and if they are, I have a final question due to the answers I tend to get from you! If they turn out to be false, then my question doesn't hold.. So here it goes..

(1)

For function $ f$ defined on $ R$, it's the same thing to say it's defined on $ R$ or on the interval $ (\minus{}\infty,\infty)$?

(2)

If $ f: A \rightarrow R$ and $ \forall x \in A, f(x)\equal{}0$ then $ f(x)\equiv0$ on $ A$?

Now, final question:

Consider a function $ f: R \rightarrow [0,1)$ and $ lim_{x \rightarrow \infty}f(x)\equal{}1$

can we then say that $ f(x) \equiv 0$ ? I mean, it's true that $ \forall x \in (\minus{}\infty,\infty), f(x)\equal{}0$, but $ lim_{x \rightarrow \infty}0\equal{}1$ makes no sense!

The definitions (1) and (2) are correct. The final question is silly- of course $ f$ isn't identically zero. For that limit to be 1, it has to be nonzero somewhere- actually, everywhere once you're far enough.

This is not complex analysis.

Hm.. Yes, you're right, it's not complex analysis.. :( sorry.. Still, I need to ask.. What about $ f(x)\equal{}\lfloor1\minus{}2^{\minus{}x}\rfloor$, $ x>0$

$ \lim_{x \to \infty} f(x) \equal{} 0$, not $ 1$.  This is because $ f(x) \equal{} 0$ for all $ x > 0$.

Hm... I find this strange since $ \lim_{x \rightarrow \infty}(1\minus{}2^{\minus{}x})\equal{}1$ and $ \lfloor1\rfloor\equal{}1$...

But it's not true that $ \lfloor \lim f \rfloor \equal{} \lim \lfloor f \rfloor$.  The floor function is itself discontinuous, so there's no reason that it would treat limits so nicely.

Ah, yes.. That's the catch! Thanks for your replies..


function
search
quadratics
number theory proposed
number theory

Let $p$ is odd prime.Assume that the set ${1,2,...,p-1}$ can be expressed as the union of two nonempty subsets $S$ and $T$, $S\#T$,such that the product $(\mod p)$ of any two elements in the same subset lies in $S$,whereas the product $(\mod p)$ of any element in $S$ with any element in $T$ lies in $T$.Find $S$ and $T$

You give me reason to live!

This problem posted on forum, but i can not find it. You can use function search :)
[hide="Answer"] $S$ is set all quadratic residues of $p$ and $T$ is set all non-quadratic residues of $p$.[/hide]

:lol: I know it is not hard but i want to post good solution about it.


[hide="1"][hide="Method 1"]$ \frac{z_{2}\minus{}z_{1}}{z_{3}\minus{}z_{1}}\equal{}{1\over 2}\minus{}i{\sqrt{3}\over 2}\equal{} e^{\minus{}i\pi/3}$, hence $ z_{2}\minus{}z_{1}$ is a rotated image of $ z_{3}\minus{}z_{1}$ for $ 60^{\circ}$. Therefore, triangle $ (z_{1}, z_{2}, z_{3})$ is equilateral and the result follows.[/hide]

[hide="Method 2"]$ \left|\frac{z_{1}\minus{}z_{2}}{z_{3}\minus{}z_{1}}\right| \equal{}\left|\minus{}{1\over 2}\plus{}i{\sqrt{3}\over 2}\right| \equal{} 1\implies |z_{1}\minus{}z_{2}| \equal{} |z_{3}\minus{}z_{1}|$
$ \left|\frac{z_{3}\minus{}z_{2}}{z_{3}\minus{}z_{1}}\right| \equal{}\left|\frac{z_{1}\minus{}z_{2}}{z_{3}\minus{}z_{1}}\plus{}1\right| \equal{}\left|{1\over 2}\plus{}i{\sqrt{3}\over 2}\right| \equal{} 1\implies |z_{2}\minus{}z_{3}| \equal{} |z_{3}\minus{}z_{1}|$[/hide][/hide]

[hide="2"]By the result of 1, $ (z_{1}, z_{2}, z_{3})$ is an equilateral triangle, and it's inscribed in the unit circle centered at the origin. Hence the origin is its centroid and the result follows.[/hide]


vector
algebra
polynomial
linear algebra
linear algebra unsolved

if $T: P_3 \to P_3$  (polynomials in $x$ of degree less than or equal 3)

and $T(a+bx+cx^2+dx^3)=a(1+x^3)+b(x+x^3)+c(x^2+x^3)+d(1-x+x^2+x^3)$
What is the basis of quotient space $P_3/Ker(T)$ and $Im(T)$...
How do I treat a quotient space in general? What is it?

thanx

here material to understand quotient vector space

http://mathworld.wolfram.com/QuotientVectorSpace.html


[quote="nutz_for2.718281828"]Ky's awry is awry! The probability that he will catch him is .09
The probability that he will be awry after he catches his awry is .39
What is the probability that he will be awry, but he won't catch him?[/quote]

I don't think the question gives enough information, but I'd guess...
[hide]0?[/hide]

[quote="nutz_for2.718281828"]Ky's awry is awry! The probability that he will catch him is .09
The probability that he will be awry after he catches his awry is .39
What is the probability that he will be awry, but he won't catch him?[/quote]
Umm... there isn't enough information to solve this problem. However, you could solve the question "What is the probability that he will not be awry and catches his awry?" Is that what you meant?

[quote="nutz_for2.718281828"]Ky's awry is awry! The probability that he will catch him is .09
The probability that he will be awry after he catches his awry is .39
What is the probability that he will be awry, but he won't catch him?[/quote]
Probably if the second probability was given as "The probability that he will catch his awry when he is awry is 0.39", then we can say that the probability that he will be awry, but he won't catch him is $0.39 \times 0.91$. But it is not clear from the wording in the question.

u don t no what the probability is because the probability could change

Assuming that he is not already awry, then the probability is 0, because he cannot become awry without catching the awry.

kheg_k, what do you mean?  In a problem, it is given that the probabilities stay the same unless otherwise stated.

Oh, sorry about that. I meant "The probability that he will be awry after he chases his awry is .39"
mthiyaga is right.


A short, cute one.

Given triangle ABC, let P be an arbitrary pt on side BC. M is on AB and N on AC s.t PM and PN are parallel to AC and AB respectively. What's the locus of the orthocenter of AMN?

Answer. Let H be an ortocenter of ABC. K --- medium of AH. AF -- mediana
of triangle ABC. Then the locus is the line passing through the K and
perpendicular to AM.


From previous posts, I can see that there are quite a few pianists who frequent AOPS.  I was wondering if anybody (not necessarily just pianists either) was interested in Jazz, particularly improvisation.  Being a long time pure classical pianist, I was also wondering if any of the other pianists on the forum knew of any books that provide a good overview of basic Jazz piano.  Our high school recently started a Jazz band and making the transition from classical to Jazz (which is a ton of fun, yet very difficult), is proving to be more difficult than I thought.

One recommendation I can make right away is to watch the entire ten-video series Jazz by Ken Burns, originally broadcast on PBS. You'll learn a lot about jazz (and a lot about American history) by watching that, and that will give you more of a feel for the music and which performers you like. I'll ask my wife for other suggestions in the next few days. She is making that transition from classical pianist to jazz pianist these days.

I made that exact same transition myself, starting about two years ago. I had studied only classical piano, but I started getting interested in jazz and improvisization after joining my high school jazz band.

I couldn't find a good book that taught jazz improvisation. Mainly, I listend to A LOT of jazz, and continually attempted to improvise in the jazz style, though it sounded like crap for a long time. I bought [u]The Real Book[/u], a collection of several hundred jazz tunes, and tried improvising with those. Some of the fake books are probably just as good.

After a year of playing jazz on my own, I convinced my parents to let me take jazz lessons. Here are two of the things my teacher had me do that helped a lot:
1. Use a play-along CD (base and drum tracks). This helped me learn how to stay with the form and play the chord changes. I used the book [u]Body and Soul[/u], by Jamey Aebersold Jazz, Inc. Here is a link I just found http://www.jazzbooks.com/playalongs/Complete_Play-A-Long_List.htm
some of the other books might be useful to do.
2. Learn and memorize the piano solos of some of the masters, by listening to the recordings. This is very difficult; I had to slow the music down to figure out all the notes (using some freeware programs) but it's an extremely good exercise.

Good luck, and I'm interested in what music you listen to, and what you like to play, if you want to talk about it.

Try The Jazz Piano Book by Mark Levine. It's a really comprehensive and fun-to-read book.

Thanks for the input.

I like Chick Corea, Gary Burton, Lionel Hampton, Keith Jarrett, Pat Metheny, Milt Jackson (Modern Jazz Quartet), Dave Samuels (I play vibraphone too).

I played classical piano for 8 years and then switched to jazz.  I'm only in ninth grade and have only been playing jazz for a year so my perception may be naive.  Yet, I think that the real book idea was a great suggestion.  Also to improvise well you must listen a decent amount., learn from the best that ever lived.  Also, just fool around a lot with improvisation the more you practice the better you will get like all things, I'm sure that I didn't give you any new information there.

Haha, one more thing.  Once you do become proficient in jazz piano your classical background will help you greatly.  Your technique or as jazz muscians like to call it "chops" will allow you to pull of lightning fast licks and runs to the likes of Keith Jarrett :) .   Knowledge of the chords and various scale patterns is immensly helpful as well as knowledge of rhythym and time.

Hey have any of you'll learned "Take Five" by Dave Brubeck?  It's a really good song.  Also, I was wondering just how "hard" it is.  Could your average beginner play it?  Or would it take a couple of years of playing jazz before being able to learn it?

NO JAZZ!!! Classical is the way to success!!! Oh and if you're still wondering about Take Five, it's a pretty hard song... from what I've heard. Probably a intermediate student could handle it though if he/she works hard.

I'm making this transition after this summer.  I'm getting a jazz piano teacher at the community college after 7 or 8 years of classical playing.

My wife the classically trained pianist reports that learning jazz has added a new dimension to her classical playing and listening.

I played piano for about 7 years before switching to vibraphone.  I now play vibraphone in my school jazz band.  Improv isn't realy something that you can get from a book.  It would be an oxymoron if you could.  The point is to learn to be creative and expressive without having to memorize something.  I would recomend learning blues scales in the common keys.  This works very well for note selectoin because it's hard to play something that sounds wrong and you have more to choose from than if you went from the bass line or chord progression.  It also frees you from the chord pregression so you can try something different.  Hope this helps.


Prove that a prime of the form $2^{2^n}+1$ cannot be the difference of two fifth powers of two positive integers.

Pierre.

This seems a bit too straightforward but here goes:

2^(2^n) + 1 = a^5 - b^5 where a > b > 0.
a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).

So either a - b = prime and a^4 + a^3b + a^2b^2 + ab^3 + b^4 = 1
or a-b = 1 and a^4 + a^3b + a^2b^2 + ab^3 + b^4 = prime.

In the first case we cannot have because a^4 + a^3b + a^2b^2 + ab^3 + b^4 clearly > 1.  In the second case you have:
a-b=1 implies b = a-1.  So the factorization is:

2^(2^n) + 1 = 5a^4 -10a^3 + 10a^2 - 5a + 1 or

2^(2^n) = 5a^4 - 10a^3 + 10a^2 - 5a.

But 5|RHS but not the LHS, contradiction.  In fact it shows 2^k+1 is never a^5 - b^5.  This stronger result makes me think i made an error somewhere, any mistakes?

I agree this was a really straightforward problem...

Pierre.

Actually, if $p=a^5-b^5$, then $a,b$ are not multiles of $p$, so we can work in $\mathbb Z_p$ as follows: $(ab^{-1})^5-1=0\Rightarrow 5|p-1$. In our case this is $5|2^{2^n}$, clearly absurd.

[I needed the fact that $b$ is not a multiple of $p$ so that $b^{-1}$ exists in $\mathbb Z_p$]

Hmmm....strange...none of the contestants did that  :D

Pierre.


Ok. Fixed it so you can look at the pdf. Also can you post the code?

Also please say which part of the pdf you think has gone wrong.


What percent of the integers from 1 to 100 inclusive has at least one digit that is a 7?

There are $ 9 \times 9\equal{}81$ that don't include a 7, so answer is $ \frac{100\minus{}81}{100}\equal{}\boxed{19 \%}$.


Determine the area of the largest rectangle that can be inscribed in a circle of radius $ 4$.

You have to find the biggest xy such that x^2+y^2=64. This is done when x and y are the same. So x is 4sqrt2 so xy is 32.

[hide="proof"]
By AM-GM,
\[ \frac {x^2 \plus{} y^2}{2}\ge\sqrt {x^2y^2}\Rightarrow\frac {64}{2}\ge xy\Rightarrow32\ge xy.
\]
Therefore, the greatest area is is $ 32,$ and can be achieved when $ x$ and $ y$ are equal, or when they both equal $ 4\sqrt {2}.$[/hide]

@Ihatepie: try to delete your second post because double posting is not good :)

[quote="RoFlLoLcOpT"]

Therefore, the greatest area is is $ 32,$ and can be achieved when $ x$ and $ y$ are equal, or when they both equal $ 4\sqrt {3}.$

[/quote]

shouldn't that be $ 4\sqrt2$?


In a class of 62 students, 22 students do not like penguins, llamas, or giraffes.  The number of students who like penguins, llamas, and giraffes are 24, 25, and 20, respectively.  The number of students who like penguins and llamas is 15.  The number of students who like llamas and giraffes is 12.  The number of students who like giraffes and penguins is 6.  How many students like all 3, penguins, llamas, and giraffes?

i so wrote that. go penguins. and ninjas.  :ninja:  :ninja:

use pie.

[hide]so 24+25+20-15-12-6+x=40

so 36+x=40

x=[b]4[/b]

PIE, for those like some people in my math class, is (for a triple venn, although it can be infinite), is:

(the singles)-(the doubles)+(the triples)=all[/hide]

[hide] you see that there are 62 students, with 22 not liking anything. so...40 being the number to use. adding all the doubles together...we get 69. 69-40 is 29. thus we count 40 singles at least. we know that there are 29 doubles/triples. then, we subtract 15+12+6(the triples). this gives us 33-29, which is 4. so there are 4 students who like all three. Is this using PIE? [/hide]

yup. PIE is

principle of inclusion-exclusion
 :ninja:  :ninja:  :ninja:

you could also use a venn diagram  :D
though that is basically the same thing and it doesn't work for more than three animals  :(

Solve this: 22 of the last class has failed. Out of the remaining people in the class, 20 like llamas, and 35 like giraffes. At least 5 people like only llamas. What is the maximum possible number of people that do not like neither llamas nor giraffes, but penguins? What is my favorite color? What is the meaning of life? How big is the universe? WHERE IS CANADA!!!!????? Find these answers, and you are my new role model. :D Not very expectations, eh?

[hide="ajai part 1"] We know that 35 people like giraffes (or giraffes and llamas), and at least an additional 5 people like only like llamas. So 40 people must like either giraffes, llamas, or both. 60-40 = 20. [/hide]
[hide="ajai part 2"] FIREY HOT PINK! [/hide]
[hide="ajai part 3"] Orange sodas. [/hide]
[hide="ajai part 4"] Who cares?? [/hide]
[hide="ajai part 5"] Canada?? Who's that? [/hide]


Find the smallest postive integer $a\geq 2$, so that there exists a prime $p$ and an integer $b\geq 2$ so that
\[ \frac{a^p-a}{p} = b^2  \]

Is it correct? a=p=2,b=1,so the minimum of a is 2

maybe a=50,p=2,b=35.
But i'm not sure if it's the smallest one

It's $a=9, p=2, b=6$

My solution:

First assume that $p\nmid a$.
We have $a(a^{p-1}-1)=pb^2$. Obviously $(a, a^{p-1}-1)=1$, so $a$ must be a perfect square. Let $a=x^2$.
Then we have $(a^{p-1}-1)=(x^{p-1}-1)(x^{p-1}+1)$.
If $a=4$ then we have $a^{p-1}-1 = (2^{p-1}-1)(2^{p-1}+1)$ Since they are coprime, one of them must be a perfect square. But that's impossible for $p\geq 3$ since then $p-1$ would be even. Hence $p=2$.
For $a=4, p=2$, we have $\frac{16-4}{2}=6$ which is not a square.
For $a=9, p=2$, we have $\frac{81-9}{2}=36$ which is a perfect square.
Since we must only find the smallest $a$, the case $p\nmid a$ is done.

Now let us assume that $p\mid a$. Let $a=pk$.
Then we have $k( (pk)^{p-1}-1)=b^2$. Hence $k$ and $( a^{p-1}-1)$ must be perfect squares since they are coprime. But if $p\geq 3$ then $a^{p-1}-1$ cannot be a square since $p-1$ is even. Hence $p=2$.
Since $k$ must be a square, we have $a\geq 8$.
For $a=8, p=2$ we have $\frac{64-8}{2}=28$ which is not a square.
Since we must only find the smallest $a$ and we already have found a solution for $a=9$, we are done.


How may different combinations of pennies, nickels , dimes, and quarters use 48 coins to total $1.00?

(A) 3      (B) 4   (C) 5   (D) 6     (E) 8

[quote="G-UNIT"]How may different combinations of pennies, nickels , dimes, and quarters use 48 coins to total 1.00?

(A) 3      (B) 4   (C) 5   (D) 6     (E) 8[/quote]

[hide]You need a number of pennies that is a multiple of 5 (for obvious reasons)...
This needs to be at least 35 because if you had 30, the minimum value you could make is 1.20 (30 pennies, 18 nickels)
If you have 35, you can make a dollar (35 pennies, 13 nickels)
If you have 40, you can also make a dollar (40 pennies, 4 nickels, 4 dimes) (40 pennies, 1 quarter, 7 nickels)
If you have 45, you can also make a dollar (45 pennies, 2 quarters, 1 nickel)

That makes a grand total of 4 ways, which is [b](B)[/b].[/hide]

stop doing that!  Why don't you actually work it out instead of just agreeing :o

[quote="chess64"]Wow, he has 12.15 posts per day! Why would someone want to get lots of posts? The AoPS forum is for learning math and practicing problems, not just getting lots of posts by saying random things.[/quote]
Good point, now lets get back to the problem and stop :spam: ming

nat mc I think you spam a lot also, because of all the times when you said : twin, and lol, not to mention the millions of polls you've done....and you have like 10 posts per day too....

[quote="GoBraves"]nat mc I think you spam a lot also, because of all the times when you said : twin, and lol....and you have like 10 posts per day too....[/quote]

You're  :spam: ming write now, :spam: er  :lol: . BTW, nat mc, we're done with this problem so let's go to some of the problems we haven't solved yet, like the [b]Complicated Lockers Problem[/b]...

I wasnt spamming...i was just making a point that some people shouldnt talk.

STOP WITH THE SPAMMING! WE'RE DONE IN THIS TOPIC! NOBODY POST HERE ANYMORE!  Sorry :blush: I had to say that.  :lol: [hide]Yes, now I got another post! Oh yeah! jk, jk, lol  :P [/hide]


For what value of $ x$ does $ 128^{x\plus{}2} \equal{} \left(\frac{1}{64}\right)^{x}$? Express your answer as a common fraction.

Convert to powers of 2
\begin{align*}
128^{x+2}=\left(\frac{1}{64}\right)^x &\Rightarrow \left(2^7\right)^{x+2}=\left(2^{-6}\right)^x \\
&\Rightarrow 2^{7x+14} = 2^{-6x} \\
&\Rightarrow 7x+14=-6x \\
&\Rightarrow x=\boxed{\frac{-14}{13}} \\
\end{align*}


An object is moving with velocity 16 ft/sec when it is suddenly subjected to a deceleration. The deceleration is proportional to the square root of the velocity, and the object comes to rest in 4 seconds.

a) How fast is the object moving 2 seconds after it begins decelerating?
b) How far does the object travel before coming to rest?


inequalities function calculus integration

Prove that

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{n^2}}\geq 2n+\frac{1}{2n}-\frac{3}{2}$ , $\forall n\in\mathbb{N}$

just integrate the left hand side and then you just have to prove 1/n<=1!

[quote="t0rajir0u"]Integrate a function defined on the naturals...

  I think you mean to bound it below with an integral, and I suspect there is an elementary solution anyway.[/quote]

there is one , using only AM-HM . :) Good Luck .

[quote="Shinta"]This is my solution. Are there any mistakes in it, mf? Thanks a lot. :)
[hide][b]First we have this inequality, using Cauchy theorem:[/b]
$\frac{1}{\sqrt{n^2+a}}+\frac{1}{\sqrt{n^2+2n-a}}\geq \frac{2}{\sqrt[4]{(n^2+a)(n^2+2n-a)}}\geq\frac{2}{\sqrt{n^2+n}}$

[b]Using this inequality, we have:[/b]
$\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+ \cdots +\frac{1}{\sqrt{n^2+2n}}$
$=(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+2n}})+(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2n-1}})+ \cdots + \frac{1}{\sqrt{n^2+n}}$
$\geq\frac{2}{\sqrt{n^2+n}}+\frac{2}{\sqrt{n^2+n}}+ \cdots +\frac{1}{\sqrt{n^2+n}}=\frac{2n+1}{\sqrt{n^2+n}}\geq 2$

[b]Now we have:[/b]
     $1\geq 1$
$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\geq 2$
$\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}+\cdots +\frac{1}{\sqrt{8}}\geq 2$

$\cdots$
$\cdots$

$\frac{1}{\sqrt{(n-1)^2}}+\frac{1}{\sqrt{(n-1)^2+1}}+\cdots +\frac{1}{\sqrt{(n-1)^2+2(n-1)}}\geq 2$
$\frac{1}{n}+\frac{1}{2}\geq\frac{1}{2n}$

[b]We now just add them together to have [color=blue]"Square root inequality"[/color][/b] :)[/hide][/quote]

thanks for replying Shinta . You solution is great !

Mine solution is using the idea of AM-HM and telescoping

$\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\geq \frac{4}{\sqrt{k+1}+\sqrt{k}}=4\left(\sqrt{k+1}-\sqrt{k}\right)$  :)

shinta shyong...
your methods are correct and intelligent
in fact, i also prefer elementry algebra to calculus!
never stop thinking


geometry
number theory
least common multiple

1. Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised. The remaining apples were perfect. How many perfect apples were harvested?

2000 small apples
+ 1500 green apples
+ 600 bruised apples
= 4100 bad

- 500 small, green apples
- 150 green, bruised apples
- 200 small, bruised apples
= 4100 - 850 = 3250

+ 50 small, green, bruised apples = 3300

...which leaves 2700 perfect apples. The author above me is right, just forgot to subtract. I still am wrong. What did I do wrong?

---

2. A circular garden is surrounded by a sidewalk with a uniform width of 25 ft. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.

Let R = the radius of the inner garden. The area of the sidewalk is therefore (pi)[50R + 625], and the area of the garden is equal. Therefore, R^2 - 50R - 625 = 0, (R - 25)^2 = 1250, R = 25 + 25 * root(2). D = 50 (1 + root(2)).

That is approximately equal to 120.710678, which rounds to 121 feet. Yayz! I R right this time!

[quote="jackdillon"]1.
2000 multiples of 3
1500 multiples of 4
600 multiples of 10

500 multiples of 3 and 4
150 multiples of 4 and 10
200 multiples of 3 and 10

50 multiples of 3,4 an 10

for some reason I keep geting 3200 as an answer....[/quote]

This is actually wrong since there are actually 100 multiples of 3,4,10 and 300 multiples of 4 and 10 because the
LCM of 4 and 10 is 20, not 40.

If you want to work with smaller numbers then:
[hide]Just look at the first 60 apples and multiply it by 100.
20 are multiples of 3.
15 are multiples of 4
6 are multiples of 10.

5 are multiples of 3,4
3 are multiples of 4,10
2 are multiples of 3,10

1 is a multiple of 3,4,10.
20+15+6-5-3-2+1= 20.
32*100=3200.

6000-3200=2800[/hide]

yes, i remember doing this problem yesterday, and i got 2800 also....

i think the main mistake is that the lcm of 10 and 4 is NOT 40

[quote="Bomboyage"]1. Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised. The remaining apples were perfect. How many perfect apples were harvested?
[hide="Answer"]2800 apples[/hide]
2. A circular garden is surrounded by a sidewalk with a uniform width of 25 ft. The total area of the sidewalk equals the total area of the garden. How many feet are in the diameter of the garden? Round your answer to the nearest whole number.
[hide="Answer"]121 ft[/hide][/quote]

[hide="1"]

2000 were too small
1500 were too green
600 were bruised.

500 were too small and green
300 were too green and bruised
200 were too small and bruised

100 were too small, green, and bruised.

so 2000+1500+600-500-300-200+100=3200

so there are 2800 perfect apples.[/hide]

[hide="2"]

pi*r^2=pi*(r+25)^2-pi*r^2
r^2=(r+25)^2-*r^2
r^2=r^2+50r+625-r^2
r^2=50r+625
r^2-50r=625
r^2-50r+625=1250
(r-25)^2=1250
r=sqrt1250+25
r=25sqrt2+25
D=50sqrt2+50
D is approximately 121.[/hide]


I have posted it in http://www.batdangthuc.net/showthread.php?p=2636#post2636
you can do that???

with x,y,z ≥ 0; x²+y²+z²=2 prove that:
1+ xyz(1+x+z)≥xyz([mat]\frac{1}{z}[/mat]+[mat]\frac{1}{x}[/mat] +1) +    [mat]\(xyz){3}[/mat]y;


A convex n-gon (n > 3) is divided into triangles by non-intersecting diagonals. Prove that one can mark n-1 segments among these diagonals and sides of the polygon so that no set of marked segments forms a closed polygon and no vertex belongs to exactly two segments.

Have fun...

What stops me from choosing all the segments connecting one vertex? i.e., if I have a polygon a_1,a_2,a_3,a_4,a_5,...a_n, couldn't I just choose a_1 to a_2, a_1 to a_3, a_1 to a_4, and so on to a_1 to a_n?  That would n-1 segments, and none of them form a closed polygon... Am I missing something?

the non-intersecting diagonals might not all come from a_1

Sure, they don't have to come from a_1, but why wouldn't we want them to, unless that counts as intersecting?

If they did, it would be nice, i agree. [i][b]However[/b][/i], we dont get to choose how they are!

[quote="probability1.01"]What stops me from choosing all the segments connecting one vertex?[/quote]

The fact that you don't get to choose how the segments are arranged.  You have to prove that it's true of any collection of diagonals that meets the other given requirements.


calculus
derivative
limit
inequalities unsolved
inequalities

Prove or disprove the following:

"For every $0<x\leq \frac{1}{8}$, $\frac{arcsin(x)}{arcsin(8x)}\leq \frac{1}{8}$."

So we want to show

$8 \text{ arcsin }x \le \text{arcsin }8x$

Obviously true for $x = \frac{1}{8}$, and the derivative will validate its truth for $x < \frac{1}{8}$.  There's also probably a nice geometric solution.

What I know, is that $\lim_{x \to 0}\frac{\arcsin x}{\arcsin 8x}=\frac{1}{8}$.
But I don't know how to prove that $f(x)=\frac{\arcsin x}{\arcsin 8x}$ is decreasing for $0<x \leq \frac{1}{8}$ [It is true at least when I try to graph it with a software]

I have tried to use derivative. But I can't see anything next  :oops:


Make an example.
Original: Height 6
Base 3
Area:1/2*6*3=9.

Enlarged triangle:
Height:18
Base 6
Area 1/2*6*18=54.

You need to multiply 6 by[b]9[/b]to get 54.

original triangle: bh/2
New triangle: (2b)(3h)/2=[b]6[/b]*(bh/2)

[quote="GameBot"]By what factor is the area of a triangle multiplied if the length of the base is doubled and the height is tripled?[/quote]

Just multiply 2 by 3...you don't need all those formulas.  Although james4l has a nice solution.  bh/2 becomes 6bh/2, shows that you just multiply 2*3.


Aaron has two bags of marbles, each with 2 blue marbles, 2 red marbles, and 3 green marbles.  He opens one bag, and puts all the marbles into a single pile.  He then removes one marble without replacement until there are only two left.  Then he opens the second bag, and adds these to the pile.  He repeats the process of removing marbles until there is now only one marble left.  What is the probability this marble is a red one from the first bag?


[quote="zhaobin"]Is $a_i=x_i$?
And the approach is the same as [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=422568#p422568]http://www.mathlinks.ro/Forum/viewtopic.php?p=422568#p422568[/url][/quote]
yes :blush:

i'm sorry to revive an old thread, but i couldn't figure out the above post.

how did he go from second to third step (bringing out the n in the front)?

can someone please post a more detailed solution to this problem? thanks!


So, like isabella said. You can do this because 100-1 equals 99, right? So (100-1)*(100-1) equals 10000-100-100+1. This equals [u][b]9801[/b][/u]


modular arithmetic
number theory proposed
number theory

$ a,b\in \mathbb{N}$  and $ a,b,n \geq 3$

If $ a^n \equiv 1 (mod 3)$ and $ b^n \equiv 1 (mod 3)$ ,is it posible that $ a^n \minus{} b^n \equiv 0 (mod {3^n})$ ?

EDITED NOW :

[quote="enndb0x"]$ a,b\in \mathbb{N}$  and $ a,b,n \geq 3$

If $ a^n \equiv 1 (mod 3)$ and $ b^n \equiv 1 (mod 3)$ ,is it posible that $ a^n \minus{} b^n \equiv 1 (mod {3^n})$ ?[/quote]
No because $ a^n\minus{}b^n \equiv 0 (\mod3)$

[quote="enndb0x"]Sorry ,it was a typo on the problem .

The problem is :

If $ a^n \equiv 1 (mod 3 )$ and $ b^n \equiv 1 (mod 3 )$,is it posible that $ a^n \minus{} b^n \equiv 0(mod 3^n)$[/quote]

It's always possible. Just choose $ a\equal{}b\equal{}1\pmod {3^n}$


algebra
polynomial
calculus
integration
abstract algebra
induction
linear algebra

Let $\alpha$ be a root of the polynomial $x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$, where each $a_i$ is an algebraic integer. Show that $\alpha$ is an algebraic integer.

Lemma: $b_1,b_2,...,b_n \in B$ are integral over $A$ iff $A[b_1,b_2,...,b_n]$ is a finitely generated $A$-module.

Then applying this to $b_i=a_i , b_n=\alpha$ gives the result by http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71108

Proof of the Lemma:

Let $b$ be integral over $A$ and $f(x)$ be a polynomial of degree $m$ such that $f(b)=0$. Since $b^m$ and all higher powers can, using $f(b)=0$, be expressed by the lower ones (with integral coefficients from $A$), $1,b,b^2,...,b^{m-1}$ generate the $A$-module $A[b]$.
Now when integral $b_1,b_2,...,b_n$ are given, $b_n$ is integral over the finitely generated (by induction hypothesis) ring/module $A[b_1,b_2,...,b_{n-1}]$, thus also $A[b_1,b_2,...,b_n]$ is finitely generated.

Converse:
Let $A[b_1,b_2,...,b_n]$ be finitely generated by $c_1,c_2,...,c_m$.
Take a $b \in A[b_1,b_2,...,b_n]$ and set $b c_i = \sum_j a_{i,j} c_j$ with $a_{i,j} \in A$.
Then we get by using $Mx = 0 \implies det(M) x =0$ ($M$ a matrix, $x$ a vector over $ A[b_1,b_2,...,b_n]$): $det(b E - (a_{i,j})) c_i = 0$
Since $1 \in A[b_1,b_2,...,b_n]$, by taking a representation of $1$, we get $det(b E - (a_{i,j}))=0$, thus a polynomial equation with leading coefficient $1$ for $b$ over $A$.


Using each of the digits 2 through 9, one per square, what is the maximum value of the following expression?

$\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}+(\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}\times\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}}\boxed{\text{\rule{0in}{0.1in}\rule{0.1in}{0in}}})$

Is there a systematic way to do this kind of problem?

[quote="xxm0rg07hxx"][hide]By guess and check I got [b]15932[/b].  Just multiply the highest numbers and add the next highest.[/hide][/quote]

The trick is: Can you prove that's the largest?

Well, to get 15932 you need 9632+(84*75). To get the maximum value, you need to put the largest digits into the highest place values, and from then on guess and check isn't too bad. It still doesn't prove anything, though...


The following equation holds for any real number $x$ and positive integer $n$:
\[ \lfloor nx \rfloor = \lfloor x \rfloor + \left \lfloor x + \frac{1}{n} \right \rfloor + \left \lfloor x + \frac{2}{n} \right \rfloor + \dots + \left \lfloor x + \frac{n - 1}{n} \right \rfloor. \]

I believe this is known as Hermite's Identity, but I can't find a reference for it.  Can anyone verify that this is indeed its name?

Yes, I could verify. Because I found this equality in my book titled MATHEMATICAL miniatures 43 MAA Page 41.

yeah,that's right ,this is hermite's identity ;)  (reference:"mathematical olympiad challenges")

true, that's Hermite's identity.
mathematical miniatures is an excellent book

Yes it is Hermite's. It's there in Mathematical Olympiad challenges I think.

the most beautiful part of it his proof using perodicity on e of those few really ingenious ye simple proof's i have ever seen.
 :omighty:

[quote="nsato"]Thanks for the reference, kunny.  I would feel more comfortable if there were some others.[/quote]

[url=http://en.wikipedia.org/wiki/Hermite%27s_identity]You can see here[/url]

another proof : Let $ f: R \minus{} > R$, $ f(x) \equal{} \sum_{k \equal{} 1}^{n \minus{} 1} \left[ x \plus{} \frac {k}{n}\right] \minus{} [nx]$. $ f(x \plus{} \frac {1}{n}) \equal{} f(x)$ we have to study $ f$ on $ [0,\frac {1}{n})$.For $ 0 < \equal{} x < 1/n , \left[x \plus{} \frac {k}{n}\right] \equal{} 0, [nx] \equal{} 0$ and therefore $ f(x) \equal{} 0$.


Demonstrati ca numerele OI,R-r si r pot fi laturile unui triunghi dreptunghic, unde notatiile sunt cele obisnuite intr-un triunghi oarecare.

Problema este usurica si imediata daca este cunoscuta relatia lui Euler: $OI^2=R^2-2Rr$,deoarece de aici putem scoate $(R-r)^2=OI^2+r^2$ si atunci laturile date in enunt sunt
$c_1=OI$;$c_2=r$;
ipotenuza$=\sqrt{OI^2+r^2}$ Si atunci triunghiul este dreptunghic de ipotenuza $R-r$.End of story

ce problema :) e un fel de ABC triunghiu , MN || BC...demonstratzi ca BM/AB = CN/AC...

dar e "own" nu e Thales :) ...in cazu tau euler :)

[quote="pureblack2003"]ia lasati spiritul creator al omului in pace . :) el propune d-astea pt brotaceii de a 5-a , ce stiti voi :P[/quote]
   Intr-un fel ai dreptate.Eu propun probleme grele rau de tot pentru astia micii.Pentru aia marii sa zicem ca nu am destula experienta si astfel, problemele mele sunt mai usoare.

Sorine, noi asteptam sa zici cum vine asta "own" la relatzia lui Euler :) (care e mort) :)) ( pt spix)

Kosmin, eu am crezut ca problema e cat decat originala.Imi cer scuze daca nu este(si cred ca nu e).

Nu e nici o problema mai :) am fost si eu rau un pik :)


geometry
complex numbers
algebra unsolved
algebra

Let $ A_1, A_2, ..., A_6$ the images of the complex numbers $ z_1, z_2 ,..., z_6$ respectively whith $ |z_1| \equal{} |z_2| \equal{} ... \equal{} |z_6| \equal{} \lambda > 0$ and $ w$ is a root of the equation $ z^2 \plus{} z \plus{} 1 \equal{} 0$. If

 $ z_1w^2 \plus{} z_3w \plus{} z_5 \equal{} 0$ and $ z_2w^2 \plus{} z_4w \plus{} z_6 \equal{} 0$

 prove that :

 a) triangle $ A_1A_3A_5$ is equilateral .

b)
$ |z_1 \minus{} z_2| \plus{} |z_2 \minus{} z_3| \plus{} |z_3 \minus{} z_4| \plus{} |z_4 \minus{} z_5| \plus{} |z_5 \minus{} z_6| \plus{} |z_6 \minus{} z_1| \equal{} 3 |z_1 \minus{} z_4| \equal{} 3|z_2 \minus{} z_5| \equal{} 3|z_3 \minus{} z_6|$

 Babis

a)
$ w$ is a root of the equation $ z^3 \minus{} 1 \equal{} (z \minus{} 1)(z^2 \plus{} z \plus{} 1) \equal{} 0$, so $ w^3 \equal{} 1$. $ w \neq 0$.
We start with the equation $ z_1w^2 \plus{} z_3w \plus{} z_5 \equal{} 0$.
We subtract $ z_1(w^2 \plus{} w \plus{} 1) \equal{} 0$ to get
$ (z_3 \minus{} z_1)w \plus{} (z_5 \minus{} z_1) \equal{} 0$
and solve for $ w$ to get
$ w \equal{} \dfrac{z_1 \minus{} z_5}{z_3 \minus{} z_1}$.
$ \dfrac{A_5A_1}{A_1A_3} \equal{} \dfrac{|z_1 \minus{} z_5|}{|z_3 \minus{} z_1|} \equal{} |w| \equal{} 1$, so
$ A_5A_1 \equal{} A_1A_3$.
Going back to the equation $ z_1w^2 \plus{} z_3w \plus{} z_5 \equal{} 0$,
we subtract $ z_3(w^2 \plus{} w \plus{} 1) \equal{} 0$ to get
$ (z_1 \minus{} z_3)w^2 \plus{} (z_5 \minus{} z_3) \equal{} 0$,
and since $ w \neq 0$, multiply by $ w$ to get
$ (z_5 \minus{} z_3)w \minus{} (z_1 \minus{} z_3) \equal{} 0$
and solve for $ w$ to get
$ w \equal{} \dfrac{z_3 \minus{} z_1}{z_5 \minus{} z_3}$.
$ \dfrac{A_1A_3}{A_3A_5} \equal{} \dfrac{|z_3 \minus{} z_1|}{|z_5 \minus{} z_3|} \equal{} |w| \equal{} 1$, so
$ A_1A_3 \equal{} A_3A_5$.
$ A_1A_3 \equal{} A_3A_5 \equal{} A_5A_1$, so triangle $ A_1A_3A_5$ is equilateral (similarly, so is triangle $ A_2A_4A_6$).

b)
$ A_1, A_2, A_3, \ldots, A_6$ lie on a circle of radius $ \lambda$, centered at $ (0,0)$ in the complex plane. Triangles $ A_1A_3A_5$ and $ A_2A_4A_6$ are both equilateral, and since they are inscribed in the same circle they are congruent. Their sides are chords of the circle that intercept arcs of $ 120^\circ$.
Angle chasing, we find that $ A_1A_2 \equal{} A_3A_4 \equal{} A_5A_6$, $ A_2A_3 \equal{} A_4A_5 \equal{} A_6A_1$, and $ A_1A_4 \equal{} A_2A_5 \equal{} A_3A_6$.
Extend each of $ A_2A_3$, $ A_4A_5$, and $ A_6A_1$ in both directions, so that $ A_2A_3$ and $ A_4A_5$ meet at $ P$, $ A_4A_5$ and $ A_6A_1$ meet at $ Q$, and $ A_6A_1$ and $ A_2A_3$ meet at $ R$.
$ \angle P \equal{} \dfrac{1}{2}(\widehat{A_3A_4} \plus{} \widehat{A_4A_5} \plus{} \widehat{A_5A_6} \minus{} \widehat{A_1A_2}) \equal{} \dfrac{1}{2}(\widehat{A_1A_2} \plus{} \widehat{A_2A_3} \plus{} \widehat{A_1A_2} \minus{} \widehat{A_1A_2}) \equal{} \dfrac{1}{2}(\widehat{A_1A_2} \plus{} \widehat{A_2A_3}) \equal{} \dfrac{1}{2}(120^\circ) \equal{} 60^\circ$.
Similarly, $ \angle Q \equal{} \angle R \equal{} 60^\circ$, so triangle $ PQR$ is equilateral.
$ \angle A_1PA_3 \plus{} \angle PA_3A_4 \equal{} 60^\circ \plus{} 120^\circ \equal{} 180^\circ$, so $ PA_1$ and $ A_3A_4$ are parallel. $ A_3A_4 \equal{} PA_1$, so $ A_1PA_3A_4$ is a parallelogram.
$ A_1A_4 \equal{} PA_2 \plus{} A_2A_3$
$ \angle A_1A_2A_3 \equal{} \angle A_6A_1A_2 \equal{} 120^\circ$ so
$ \angle PA_2A_1 \equal{} \angle PA_1A_2 \equal{} 60^\circ$ and triangle $ PA_2A_1$ is equilateral.
$ PA_2 \equal{} A_1A_2$, so
$ A_1A_4 \equal{} A_1A_2 \plus{} A_2A_3$.
Similarly,
$ A_2A_5 \equal{} A_2A_3 \plus{} A_3A_4$
and
$ A_3A_6 \equal{} A_3A_4 \plus{} A_4A_5$.
Adding the three equations we get
$ A_1A_4 \plus{} A_2A_5 \plus{} A_3A_6 \equal{} A_1A_2 \plus{} A_2A_3 \plus{} A_2A_3 \plus{} A_3A_4 \plus{} A_3A_4 \plus{} A_4A_5$.
$ 3A_1A_4 \equal{} 3A_2A_5 \equal{} 3A_3A_6 \equal{} A_1A_2 \plus{} A_2A_3 \plus{} A_3A_4 \plus{} A_4A_5 \plus{} A_5A_6 \plus{} A_6A_1$, so
$ |z_1 \minus{} z_2| \plus{} |z_2 \minus{} z_3| \plus{} |z_3 \minus{} z_4| \plus{} |z_4 \minus{} z_5| \plus{} |z_5 \minus{} z_6| \plus{} |z_6 \minus{} z_1| \equal{} 3 |z_1 \minus{} z_4| \equal{} 3|z_2 \minus{} z_5| \equal{} 3|z_3 \minus{} z_6|$, as desired.


trigonometry
real analysis
real analysis unsolved

Could we find exactly the maximum of $ |f^{(4)}(x)|$with $ f(x) = e^{5,25(sinx)^2}$ or the best upper bound
($ f^{k}(x)=f^{k-1}'(x)$ )

Is $ \sin x^2$ meant to be $ \sin(x^2)$ or $ (\sin x)^2$?

Sorry,I've edited my first post.Noone care about it ?


We have a war with images found on Google Images. A very standard duel could take place completely from images found on staples.com (paper, scissors, rock, paper, ...) but be more creative than that.

Basically, search for something that beats the last image.

http://images.google.com

Oh yeah, rules:
1) No Chuck Norris.
2) Black holes, supernovas, and nuclear weapons can only be posted once.
3) Try not to post high-resolution images. The image size is given at the bottom of each search result.

[img]http://www.cow-sensation.de/MuKuh.jpg[/img]
I start the war with my cow army.

I attack your cows with my war elephants!

[img]http://www.worth1000.com/entries/209500/209926lQsp_w.jpg[/img]

:rotfl:

lololololololololollololol  :rotfl:

that picture just makes me laugh dunno y

maybe cuz its cool! unless i get attacked  :blush:

[quote="randomdragoon"]
[img]http://www.cow-sensation.de/MuKuh.jpg[/img]
I start the war with my cow army.[/quote]

[img]http://www.tvdance.com/shop/-00-images/halloween-costumes/12917.jpg[/img]
that beats all

[img]http://www.allergysa.org/assets/images/cockroach.jpg[/img]
An army of Cockroaches survive the nuke and start invasion of the world. :D

[img]http://z.about.com/d/politicalhumor/1/0/2/-/bush_debate_poland.jpg[/img]
you forgot poland!

[img]http://www.explodingdog.com/dumbpict51/onedaygodlookeddownandsaid.gif[/img]

but god did not

[img]http://school.discovery.com/clipart/images/erase.gif[/img]

Erase the existance of the "god" :D

[img]http://www.asianfoodgrocer.com/stationery/eraser/burger.jpg[/img]

It's a hamburger! No! It's an eraser! It looks good enough to eat! I think I will. The eraser is eaten by.......

[img]http://www.hedonistica.com/archives/hugeburger-thumb.jpg[/img]

A COMPETITIVE EATER! Who will eat anything that is thrown at him :D

[quote="bubka"]This looks familiar. Where have I seen it? Oh yes, I know:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=84573[/url]

I have half a mind to bump that up. In fact I will.[/quote]
But this one has rules that fixes the old version! For example, no one can post nukes anymore.

Will this man really eat [i]anything[/i]?
[img]http://content.answers.com/main/content/wp/en/thumb/1/15/200px-Sulfuric_acid_chemical_structure.png[/img]

He ate too many Big Macs, and got diabetes. *insert diabetes germ here*

[img]http://flint.cs.yale.edu/spam/spam.gif[/img]
this topic just got spammed

[quote]*insert diabetes germ here*[/quote]
I hope you're kidding.

[img]http://www.hubris.net/images/illustrations/spam_filter.gif[/img]
Love the name they gave themselves...

[img]http://livefromcern.web.cern.ch/livefromcern/antimatter/imageHomeEn/EtoM.jpg[/img]

Antimatter. :harhar:

But the antimatter was killed by [img]http://upload.wikimedia.org/wikipedia/commons/thumb/b/be/Cyanide-montage.png/200px-Cyanide-montage.png[/img]

Won't help if you're a dummy!
[img]http://www.hometrainingtools.com/tbimages/1706.477.600.jpg[/img]

[quote="Klebian"][img]http://www.thinkgeek.com/images/products/front/largevalues.jpg[/img][/quote]

$\frac{d}{dx}$

(is that right?)

Nothing a derivative can't take care of...

[img]http://img187.imageshack.us/img187/7348/schrodingerscattv8.jpg[/img]

For physics lovers like myself  :lol:

[img]http://www.pbs.org/wgbh/nova/origins/images/fabe-sombrero-l.jpg[/img]

ooh blam! what now? not even physics can stop the mysteries of space  :lol:

These geeks can.

[img]http://flashforservergeeks.com/fsg/images/frontcoversmaller.jpg[/img]

[img]http://vandamere.com/grades.jpg[/img]

geeks get killed

[img]http://nerdparadise.com/images/20060307.png[/img]

 :rotfl:

[quote="nutz_for2.718281828"][img]http://adrian.gimp.org/lava.jpg[/img]

P.S. NO immortal sources allowed!

P.P.S. There can be a Chuck Norris, because he is MORTAL.[/quote]

it looks to me like that's fire, but idk...here is my solution:

[img]http://www.hln-store.com/catalog/water.gif[/img]

[quote="nutz_for2.718281828"]

P.P.S. There can be a Chuck Norris, because he is MORTAL.[/quote]
sorry, but chuck norris is GOD, aka immortal

[img]http://www.freakshop.com/band-pics/spongebob.jpg[/img]

Spongebob soaks up water!  :roll:

[quote="nutz_for2.718281828"][img]http://www.pl-259.com/Hole%20Plug.JPG[/img]

No explination needed.[/quote]
Is that supposed to be a black hole? If so, it's against the rules. :roll:


integration
trigonometry
calculus
calculus computations

Hey all

This is a suggested exercise in my textbook, and I know how to do it, I just get stuck on an integral. Here is what I have so far:

Question - A particle of mass m is released from rest a distance b away from an origin of attractive force that attracts accoring to

$ F(x) = - kx^{ - 2}$

Show that the time required to reach the origin is

$ \pi\left(\frac {mb^3}{8k}\right)^{1/2}$

Here is what I have done:

$ ma = m\frac {dv}{dt} = m\frac {dv}{dx}\frac {dx}{dt} = mv\frac {dv}{dx}$

So $ mv\frac {dv}{dx} = - kx^{ - 2}$
$ m\int_0^vvdv = - k\int_b^x\frac {1}{x^2}dx$
$ \frac {1}{2}mv^2 = - k\left[\frac {1}{x} - \frac {1}{b}\right]$

$ v = + - \sqrt {\left(\frac {2k}{m}\left[\frac {1}{b} - \frac {1}{x}\right]\right)} = \frac {dx}{dt}$ and since the motion toward the origin will be all in the same direction, I take the positive root. (Or would it be the negative root, since it is motion away from b, a positive constant?)

$ \int_0^tdt = \sqrt {\frac {m}{2k}}\int_b^0\frac {1}{\sqrt\left(\frac {1}{b} - \frac {1}{x}\right)}}dx$

Here I get stuck. I don't know how to do that last integral. Since there is a $ \pi$ in the answer it's a good bet that there is a trig substitution that will solve all my problems, but I don't know what. Could someone tell me what substitution to use and I'll go from there?

Also, did I set up the limits of integration right in the context of this problem?

I .You could have avoided the fist Integral by just using energy conservation,

II . Then, probably you wouldn't forget the sign of $ \int \frac{1}{x^{2}}dx \equal{} \minus{} {1\over x}$, getting a [b] positive kinetic energy [/b]...

III. In the last integral you can make , for instance, $ x \equal{} b. cos^2 \theta$.


answer is $ C$

this is because $ \sqrt {a^2 \plus{} b^2 \plus{} c^2} < a \plus{} b \plus{} c$
and this holds true only in $ C$

:)

your answer is perfectly right n that is the best method but the point why i gave the sum is for a beautiful observation  :D  Try to find it  :D

1056 = 32 x 33.

Basically we have

$ a^2 \plus{} b^2 \plus{} a^2b^2 \equal{} (ab \plus{} 1)^2.$

perfect da but just one mistake b necessarily = a+1 bcos its $ h^2 \plus{} (h\plus{}1)^2 \plus{} (h \cdot (h\plus{}1))^2 \equal{} (h \cdot (h\plus{}1) \plus{} 1)^2$  :)

I wanted to finish the observation but got a phone call.

Anyways, if you solve the above eqn of mine

$ (a\minus{}b)^2 \equal{} 1$

is what we get

so w/o LOG a = b+1


Suppose the numbers $a_{0},a_{1},a_{2},\dots,a_{n}$ satisfy the following conditions:
\[ a_{0}=\frac{1}{2}, \ \ a_{k+1}=a_{k}+\frac{1}{n}a_{k}^{2} \ \ \mbox{for}\ \ k=0,1,\dots,n-1. \]
Prove that ${\displaystyle 1-\frac{1}{n} < a_{n} < 1}$.

It is not correct. $a_k\ge a_1= \frac 12 \Longrightarrow a_n\ge \frac 12 +\frac 14 H_{n-1}>\frac 12 +\frac{\ln(n-1)}{4} \to \infty, n\to \infty$.

[quote="Rust"]I am sorry. I consider $a_{n+1}=a_n+\frac{a_n^2}{n}.$[/quote]

Ok ,it doesn't mind as you have shown that you are <<a metr>> on sequences  :) .

[quote="silouan"]The problem is true I think .
$\frac{1}{a_{k+1}}=\frac{1}{a_k}-\frac{1}{a_k+n}$

so $\sum_{k=0}^{n-1}(\frac{1}{a_k}-\frac{1}{a_{k+1}})=\sum_{k=0}^{n-1}\frac{1}{n+a_k}<1$  and from this
$a_n<1$. Now because it is increasing we have
$\sum_{k=0}^{n-1}\frac{1}{n+a_k}>\frac{n}{n+1}$
 and the result follows[/quote]
Nice solution.And the official solution is very long.about four pages.


LaTeX
algorithm
analytic geometry
graphing lines
slope
arithmetic sequence

[quote="nano"]Fill in the magic square:

                2005 **** 2004

                **** 2003 ****

                **** **** ****

Note: In a magic square the sum of the numbers in each column, row and diagonal is the same.[/quote]

[hide]2005     2000     2004

2002     2003     2004

2002     2006     2001[/hide]

hint:set up some variables

I don't think that's right. In a magic sqare all the numbers are supposed to be different.

[hide="Hint"]
Try taking the 2000 part of the number off.  It should make it easier to work with the variables then.[/hide]

ummm... do the numbers have to be different nano?

the numbers don't have to be different? then

1 1 1
1 1 1
1 1 1

is a magic square???

[quote="Xantos C. Guin"]816
357
492

is a magic square
now add 1999 to each square to get the answer.[/quote]

1999+5=2004, not 2003

Here is another way you can do it in a few seconds with one variable :
[hide]2005______2004
_____2003_____
__________[i]x[/i]

Then the sum is 4008 +[i]x[/i], which gives us

2005______2004
_____2003 2004
[i]x[/i]+1_______[i]x[/i]

and therefore

2005______2004
2002  2003  2004
[i]x[/i]+1_______[i]x[/i]

Thus, the magic sum is 6009, and therefore,

[i]x[/i]=2001

From this, we can swiftly produce the following magic square :

2005  2000 2004
2002  2003  2004
2002  2006  2001
                                                                                Q.E.D.[/hide]How do you do that fancy mathematical notation?  Those lines get annoying, and the spacing is difficult.

I don't think this has a solution in which all numbers are distinct

Boy Soprano, are you referring to $\LaTeX$?

Do you guys know the algorithm for generating a 5X5 magic square?

No  :? I think I read it in Mathematical Mosaics, but I didn't pay much attention. It was boring  :D

I think that we need some clarification here.  I always thought that a magic square had to be different numbers, but I can't do it without doing the same numbers.

[quote="mathgeek2006"]There would be tons of possibilities, but it would take too long to find out if there really was a correct answer.[/quote]

Give me one magic square where 3 times the middle number isn't the sum of any of the rows, columns, or diagonals.

Sorry about that.  It's finished now.
Thank you, NoSoupForYou.
A magic square [i]can[/i] have repeated numbers in it.  The only trouble is that if one tries to set up a 3 × 3 magic square with a number present three times, most of the time, one is forced to fill the entire square with that number, which makes it degenerate and uninteresting.  If you look at the solutions given by chess64 and myself, you will see that in this case the solution given is the [i]only[/i] solution.
bob123, there is no solution to the problem you posed :
[hide] Let us have a 3 × 3 magic square with the numbers of the top row being $x$, $y$, and $z$, respectively, making the magic sum $x+y+z$.  Let us call the number below $\\z \quad m\\$ and solve for $m$ :
$\\x \quad y\ \quad z\\$
$\\? \quad ?\ \quad m\\$
$\\? \quad ?\ \quad -m+x+y\\$

Thus :

$\\x \qquad y\ \qquad\ \quad z\\$
$\\? \quad m-x+z\ \quad m\\$
$\\? \quad -m+2x\ \quad -m+x+y\\$

Using the diagonal with the positive slope and the bottom row,
\begin{eqnarray*}
m-x+2z &=& -2m+3x+y\\
3m &=& 4x+y-2z\\
m &=& \displaystyle\frac{4x+y-2z}{3}\\
\end{eqnarray*}
Substituting for $m$ in the center term, we find that the center term is equal to
$\displaystyle\frac{x+y+z}{3}\\$,
or the one-third of the magic sum.
Q.E.D.[/hide]

[quote="Boy Soprano"]Sorry about that, people.  I accidentally hit the wrong button.  I am still working on this.  Please don't read further!
[hide]Thank you, NoSoupForYou.
A magic square [i]can[/i] have repeated numbers in it.  The only trouble is that if one tries to set up a 3 × 3 magic square with a number present three times, most of the time, one is forced to fill the entire square with that number, which makes it degenerate and uninteresting.  If you look at the solutions given by chess64 and myself, you will see that in this case the solution given is the [i]only[/i] solution.
bo123, there is no solution to the problem you posed :
[hide] Let us have a 3 × 3 magic square with the numbers of the top row being $x$, $y$, and $z$, respectively, making the magic sum $x+y+z$.  Let us call the number below $\\z \quad m\\$ and solve for $m$ :
$\\x \quad y\ \quad z\\$
$\\? \quad ?\ \quad m\\$
$\\? \quad ?\ \quad -m+x+y\\$

Thus :

$\\x \qquad y\ \qquad\ \quad z\\$
$\\? \quad m-x+z\ \quad m\\$
$\\? \quad -m+2x\ \quad -m+x+y\\$

Using the diagonal with the positive slope and the bottom row,
$-2m+3x+y=m-x-2z$
$3m=4x+y-2z$[/hide][/hide][/quote]

I know there's no solution to the question I posed, that's exactly why I asked it...

I found a new solution (not answer) based on the following facts:

(1) The sum of the rows, columns, and diagonals are 3 times the middle number, because

(2) The rows, columns, and diagonals form an arithmetic sequence.

Now we see
\[5 \ \ a \ \ 4\]
\[b \ \ 3 \ \ c\]
\[d \ \ e \ \ f\]

Because of (1), $3 \cdot 3 = 5+3+f$ so that $f=1 \Rightarrow c=4 \Rightarrow b=2 \Rightarrow d=2 \Rightarrow a=0 \Rightarrow e=6$.

\[2005 \ \ 2000 \ \ 2004\]
\[2002 \ \ 2003 \ \ 2004\]
\[2002 \ \ 2006 \ \ 2001\]

An easy problem:

$\text {What is the minimum number of numbers necessary to complete a magic square?}$

[quote="math92"][hide]1?[/hide][/quote]

There can only be 1 possible magic square made from it.

chess64,
"The rows, columns, and diagonals form an arithmetic sequence"?  Only if they pass through the middle number.  2004, 2004, 2001 is hardly an arithmetic sequence.
[hide]Do you mean "magic square in which it is possible to have distinct terms"?  If so, the answer is 3 × 3, because a 1 × 1 square has only one term and a 2 × 2 square has no distinct terms, either.  If we call the top left term $x$ and the bottom right term $m$ :
\[\begin{tabular}[b]{ll}\hline
x&?\\
?&m\\\hline
\end{tabular}\]
\[\begin{tabular}[b]{ll}\hline
x&?\\
x&m\\\hline
\end{tabular}\]
The magic sum is thus $2x$, so every term must be $x$.
On a 3 × 3 square, as has been demonstrated, it is possible to have distinct terms.[/hide]


For each positive integer $n$, prove that there are two consecutive positive integers each of which is the product of $n$ positive integers greater than $1$.

We has a result:
If $ \tau(k)\geq n \plus{} 1$ then k is product of $ n$ positive integer great than 1.
Now we prove that for each n there exist an $ k$ such that $ \tau(k \plus{} 1),\tau(k)\geq {n \plus{} 1}$
Consider the reside
$ x\equiv 0(\mod p_i),\forall i \equal{} 1,..,n \plus{} 1$
$ x\equiv \minus{} 1(\mod p_i),\forall i \equal{} n \plus{} 2,..,2n \plus{} 2$
Where $ p_i$ is distint prime .
More we can prove for a sequence contain k term with the same method.

$ \tau(k)$ is the number of different prime divisors of k.


I need a good cryptogram deocder that will decipher the following:

abcdefeghijghkcmncopkqarsedieghkhtfjohfhrdircoathsfeghieifhhtkcmncopkqarshkankfnafhfqhicrkhciaffrfkfectphfuhhtfqho

If anyone can do this for me or knows of a good (but easy) decoder, please let me know. Thanks!

I worked on it for a long time, supposing that this used monoalphabetic substitution. But, I couldn't get it :( .

It's called a single displacment cryptogram. Maybe that will help.

[quote="fd9"]It's called a single displacment cryptogram.[/quote]
which is another word for 'monoalphabetic substitution'. :D

[quote="fd9"]Nevermind, I got the answer on another forum. But I still don't know how to go about solving something like this. Can you give me an explanation of how to decode this, and which decoder to use?[/quote]
I'm not sure about decorder, but you can always use statistics. Go to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36443&postorder=asc&start=4]here[/url], and look at post #5 and #7.

BTW, what was the answer?


topology
real analysis
real analysis unsolved

Let $S \in \mathbb{R}^n$ be a subset which is uncountable. Prove that there is a sequence of distinct points in $S$ converging to a point of $S$

We can prove a lot more:

Any uncountable set $\mathcal S$ in a separable metric space can be written as the union of a set which is at most countable and a set consisting only of condensation points of $\mathcal S$ (condensation point of $\mathcal S$ means point for which any neighbourhood contains uncountably many points of $\mathcal S$).

Every point which is not a condensation point of $\mathcal S$ is covered by a basis element which contains at most countably many points of $\mathcal S$. If we eliminate all the points of $\mathcal S$ contained in such basis elements (from a countable basis of the space) we are left with condensation points of $\mathcal S$ only, and we have eliminated a set which is at most countable.


Let $k, m, n$ be integers such that $1<n\le m-1 \le k$. Determine the maximum size of a subset $S$ of the set $\{ 1,2, \cdots, k \}$ such that no $n$ distinct elements of $S$ add up to $m$.

I have a conjecture:

Let x be the minimum natural number such that:

x + (x+1) + (x+2).... + (x+n-1)>m

Then let S={x,x+1...k}

Notice that it has the desirable (but inconclusive) property that if I add in another element to this set, it no longer satisfies the condition. It can be proven easily for n=2, and probably not too hard for n=3 either. (n=1 it is not true).

Any ideas? very curious about this.

Someone reminded me of this problem and so I have given it another shot  :)

Let $x(n,m,k)$ be the number $x$ as given in the above post. Let $f(n,m,k)$ be the size of the set $S$ in the above post.

Notice: $x(n,m,k)<x(n',m',k) <=> f(n,m,k)>f(n',m',k)$

Now suppose $N>2$ we make the inductive hypothesis that $f(n,m,k)$ is indeed the answer to the problem for $n=2,3...N-1$ (the base case $n=2$ is easy and I will omit it).

Suppose we have chosen $f(N,m,k)+1$ numbers. Then we have some number $y$ that is less than $x(N,m,k)$. Let $m'=m-y$. Now remove $y$ from our set. We have $f(N,m,k)$ numbers in our set. Now we want to show that $f(N,m,k)>f(N-1,m',k)$ and we will do this if we can show that $x(N,m,k)<x(N-1,m',k)$.

Let $x=x(N,m,k)$ then we have:

$(x-1)+x+(x+1)+...+(x+N-2)\leq m$

And so:

$x+(x+1)...+(x+N-2) \leq m-(x-1) \leq m-y=m'$ (since $y<x$ as mentioned above)

And so this shows that $x(N-1,m',k)>x(N,m,k)$ Hence $f(N-1,m',k)<f(N,m,k)$.

Now we use the induction hypothesis: We have $f(N,m,k)$ numbers in our set hence we have more than $f(N-1,m',k)$ numbers in our set. That means by the induction hypothesis that we can choose $N-1$ of them such that their sum is $m'$, but now we can add in the number $y$ to this set and the sum is exactly $m$.

So in summary $f(n,m,k)$ is the answer to the problem for $n>1$, while for $n=1$ it is $k-1$ (choose every number other than $m$)

This number is easy to compute hence I declare it a suitable solution to the problem :P


A perfect power is a number that can be written as a positive integer raised to an integer power greater than one. For example, 125 is a perfect power because it is equal to $ 5^3$. The list 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, $ \ldots$ contains every positive integer less than 1000 that is not a perfect power. How many integers are in the list?

[quote="leewongjao"]What is wrong with this?

9 powers of 2
6 powers of 3
4 powers of 5
3 powers of 7
2 powers of 11,13,17,19,23,29,31

3 powers of 6
2 powers of 10,14,22,26,15,21

999-9-6-4-3-2(7)-3-2(6)=948[/quote]

Because we overcount powers of 6,15,14,21, etc.

You're counting numbers such as $ 2^1\equal{}2$ and $ 6^1\equal{}6$.

Also, you forgot about $ 1^{\text{large number}}\equal{}1$ :wink:


What is the volume of a regular octahedron whose vertices are the centers of the faces of a cube whose edge has length 6?

The height ofhalf of the octahedron is 3. And the base edge is $3\sqrt2$. The base  is a square with area $18$. So $\frac12V=\frac{18\cdot3}3=18$. $\boxed{V=36}$


function calculus integration limit quadratics algebra polynomial

$f \in C^{2}[0,1]$, how to show that we can extend $f$ to a periodic $g$ on $R$, which is $C^{2}$, with period 3.
Also we need $max(\parallel g\parallel _{\infty},\parallel g'\parallel _{\infty},\parallel g''\parallel _{\infty})\leq \parallel f\parallel _{\infty}+\parallel f''\parallel _{\infty}$.

I think there should be some constant in the bound:

[quote="kalazealot"]$f \in C^{2}[0,1]$, how to show that we can extend $f$ to a periodic $g$ on $R$, which is $C^{2}$, with period 3.
Also we need $max(\parallel g\parallel_{\infty},\parallel g'\parallel_{\infty},\parallel g''\parallel_{\infty})\leq C(\parallel f\parallel_{\infty}+\parallel f'\parallel_{\infty}+\parallel f''\parallel_{\infty})$. C an absolute constant. [/quote]

What I got so far: to use indefinate integration to define extension, so that it is differentiable, and
hopefully by MVT, it also satisfy the boundness condition.
consider extention to $[1,2]$ (extension to $[-1,0]$ similiar), for $x\in [1,2]$, define $f(x)=\frac{1}{x-1}\int_{2-x}^{1}f(t)dt$, easy to see $|f|$ is boudned since
this intgration is averaging process,
then it is easy to see that, by L'Hospital's rule, $\lim_{x\to 1}f(x)=\lim_{x\to 1}\frac{f(2-x)}{1}=f(1)$, so at least $f$ is continuous at $1$ now.

Using chain rule, calculate
$f'(x)=\frac{1}{(x-1)^{2}}((x-1)f(2-x)-\int_{2-x}^{1}f(t)dt) =\frac{1}{(x-1)^{2}}(\int_{2-x}^{1}f(2-x)-f(t)dt) \leq \frac{1}{(x-1)^{2}}\parallel f'(x)\parallel_{\infty}\int_{2-x}^{1}(2-x-t)dt \leq \frac{1}{(x-1)^{2}}\parallel f'(x)\parallel_{\infty}(x-1)^{2}=\parallel f'(x)\parallel_{\infty}$
so $|f'|$ is bounded in the extension.

But problems appear in continuity of $f'$ at 1, by L'Hospital's rule:
$\lim_{x\to 1}f'(x)=\lim_{x\to 1}\frac{-f'(2-x)}{2}=-f'(1)/2$, there is a factor -1/2.

More problems is $f''(x)=\frac{1}{(x-1)^{3}}(2\int_{2-x}^{1}f(t)dt-2(x-1)f(2-x)-(x-1)^{2}f'(2-x))$,
this one is difficult to bound using MVT...

Also continuity of f'' at 1 is not quite right either: by L'Hospital's rule:
$\lim_{x\to 1}f''(x)=\lim_{x\to 1}f''(2-x)/3=f''(1)/3$, a factor of 1/3.

One might suggest to use $3f(1)-2f(x)$ so that continuity of $f$ at 1 is kept and continuity of $f'$ at 1
is rescued. But $f''$ still won't be continuous at 1, and how to bound $|f''|$ anyway?

Can someone gives hint?

I would do the following: (1) extend $f$ to $f_{1}\in C^{2}[-0.5,1.5]$ by quadratic polynomials defined on $[-0.5,0]$ and $[1,1.5]$. (You can easily find a quadratic polynomial $p$ with given $p(0)$, $p'(0)$, and $p''(0)$.)
(2) Construct a $C^{2}$-smooth bump function $\phi$ such that $\phi=1$ on $[0,1]$ and $\phi=0$ outside of $[-0.5,1.5]$. You can do it piecewise with polynomials.
(3) The product $f_{2}=f_{1}\phi$ is $C^{2}$, with estimates for derivatives, and is supported on $[-0.5,1.5]$. Extend it by periodicity.


inequalities
LaTeX
inequalities proposed

(a^2+b^2+c^2)/a+b+c>=(a+b+c)/3 and similar such inequalities .
adding them gives us the result.Although the idea is same as cauchy ,a shorter proof!i am too lazy to use latex! :oops:

By Chebyshev Inequality. ( Too easy problem... )

I never said it was a hard problem  :roll: But maybe for newbies that's  :)

What's "Engel form" ?

This problem is very famous! This has posed in Japanese university entrance exam.

[quote]What's "Engel form" ?[/quote]
$\sum_{i=1}^{n}\frac{x_{i}^{m}}{a_{i}}\geq \frac{(\sum_{i=1}^{n}x_{i})^{m}}{n^{m-2}\sum_{i=1}^{n}a_{i}}$
$n,m \in \mathbb N^{*}$; $m\geq 2$; $a_{1},a_{2},...,a_{n}>0$

For $m=2$ you get Cauchy in Engel form.

I propose you the following stronger generalization:

$\sum_{i=1}^{n}\frac{x_{i}^{p}}{a_{i}^{q}}\geq ?$
$n,p,q \in \mathbb N^{*}$; $a_{1},a_{2},...,a_{n}>0$, $p\geq q$

The general case is just a particular form of holder...


Can we see the source code for the Round 2 Problem sheet?  Or, at least, the source code for the Problem 1 diagram?

Thanks.

uh....I don't think they would give you the source code, but I use the tabular command to make the picture.  You can learn it from the latex tab on the left

Yeah, me too, except I tried to use double lines and it looked really bad...  how'd you do it?


Disclaimer:  Sitan is a totally awesome person.    :surf:

Sitan has decided that he likes basketball and wants to improve.  He does so by playing a game of around the world.  This is done by shooting from eight different points.  If Sitan makes a shot, he can move on to the next place.  If he misses, he has two options:  he can choose to take a "risk," where he shoots again, and if he makes it, he continues as usual, but if he misses it, he must start at the beginning again.  Or, he can choose to "not shoot" a turn and continue shooting from his spot (this would be like a golfer hitting a ball into water, where the ball is basically placed nearby and a stroke is added).

Suppose Sitan is a $ 50$% shooter from all places of the basketball court.  Is there a strategy for him to reduce the number of "shots taken" in his practice?

Oh, and I don't have an answer for this.

I believe such optimization problems are solved using [url=http://en.wikipedia.org/wiki/Stochastic_process]stochastic processes[/url], but I could be wrong.

Well, after a point, a risk of going to the beginning will have a much greater expected cost than skipping a turn and trying again.

[hide]To be precise, let $ S(n)$ be the expected number of shots that Sitan will make using his optimal strategy on $ n$ baskets.

We can construct Sitan's best strategy inductively.  For 1 basket, we can see that it is clearly optimal for him to go back to the beginning, since this way he gets to try again without penalty.  The expected number of shots it will take him to finish is
\[ \sum_{k=1}^\infty \frac{1}{2^k}\cdot k = \frac{1/2}{(1 - 1/2)^2} = 2 . \]

Now, suppose that Sitan has $ n$ baskets, for $ n>1$.  He should clearly use the best strategy for $ n-1$ baskets to get through the first $ n-1$ baskets whenever he has to do this.  Now, if he is at the $ n$th place, and he uses the conservative (penalty) strategy, then the expected number of turns it will take him to shoot again from spot $ n$ is clearly 1.  But if he chooses the risky strategy, then with probability 1/2, he will have to go back to the start, where it will take him 2 expected turns just to get past the first basket!  Therefore after the first basket, the conservative strategy is always at least as good as the risky one.  So for $ n>1$, we have
\begin{align*}
S(n) &= S(n-1) + \sum_{k=1}^\infty \frac{1}{2^k} (2k-1) \\
&= S(n-1) + 2 \cdot \sum_{k=1}^{\infty} \frac{k}{2^k} - \sum_{k=1}^\infty \frac{1}{2^k} \\
&= S(n-1) + 2 \cdot 2 - 1 = S(n-1) + 3 .
\end{align*}

So by induction, $ S(n) = 3n - 1$, for $ n>0$.[/hide]

Thank you for all the responses!

As an extension, how would Sitan's strategy change if his shooting probability was changed?  Say, if he was an $ 80$% shooter?

It's pretty much the same method regardless of the probability-just change the numbers around.


vector
geometry
geometric transformation
rotation
trigonometry
combinatorics proposed
combinatorics

[quote="gb2124"]Given 2n+1 vectors in the plane such that all the vectors are of unit length and the sum of these vectors are 0
Prove that one can always choose n vectors from these such that the sum of these vectors have the length less than 1.[/quote]

Consider roots of unity.

[quote="gb2124"]Dear cautiouscauchy
Sorry,I don't know what you mean.
It is given 2n+1 vectors,not what you can choose.(such as unity roots).
Anyone wants to try it?I hope there's someone who can show me the solution.[/quote]

WLOG we have all 2n+1 vectors at the origin. Either these represent the 2n+1 roots of unity [and we are don] or they do not. If the do not can you see how considering roots of unity will still help [if they are not then at least one pair of vectors exist that are opposite in direction. But then we are left with 2k+1 vectors that are not paired in such a way and these must be some kind of rotation of the 2k+1 roots of unity.

Dear CautiousCauchy
I still don't understand.
How can you get the conclusion?
At least one pair of vectors exist that are opposite in direction.
I don't think it is true. :P

[quote="gb2124"]At least one pair of vectors exist that are opposite in direction.
I don't think it is true. :P[/quote]

Rotation of all the vectors will not alter the corresponding angles between corresponding vectors. So if none of the 2n+1 pairs of vectors had a partner that was opposite in direction then the vectors must represent some rotation of the vectors we can usee to represent the 2n+1 roots of unity [as all the vectors have magnitude 1]. If not, then at least one pair of vectors exist with the above condition.

Your claim is plainly ridiculous.  For $ n \equal{} 2$ and any $ \alpha \in \left(\frac {\pi}{3}, \frac {\pi}{2}\right)$, let $ \beta \equal{} \arccos\left( \cos \alpha \plus{} \frac {1}{2}\right)$, choose the vectors $ 1, e^{\pm i \alpha}$ and $ e^{\pm i \beta}$.  Of these uncountably many arrangements, exactly one has the form you describe  :roll:

I vaguely remember that this problem may be solved by the probabilistic method, but I'll have to look through my notes next week to see if I'm thinking of the right problem.

Dear JBL and mathlinkers:
Can anyone help me to solve this?
I'm waiting for a so long time.There must be some method I don't know.
To dear CautiousCauchy
What you say is probably wrong.You can't rotate only one vector one time since it has the condition those vectors have the sum 0. :P


function
calculus
derivative
calculus computations

Supppose you have a function given by a fraction in which the whole denominator is raised to a fraction other than $ 1.$ For instance,

$ f(x)\equal{}\frac{2x}{(x^2\plus{}4)^{3/2}}$  or  $ g(x)\equal{}\frac{x^2\minus{}a^2}{(x^2\plus{}a^2)^2}$

and we want to find, and simplify, the derivative of this function.

(I should add that these often occur as 2nd or higher order derivatives; I am teaching a multivariable calculus course right now and just covered the notion of partial derivative, in which case many problems like this occurred as computational exercises.)

The question: would you explicitly use the quotient rule? Or would your write these functions as products, use the product rule, and then find a common denominator as the last step? That is, write

$ f(x)\equal{}2x(x^2\plus{}4)^{\minus{}3/2}$  or  $ g(x)\equal{}(x^2\minus{}a^2)(x^2\plus{}a^2)^{\minus{}2}.$

I should add that for my own sake, I nearly always use the second option - product rule followed by common denominator. To what extent would you steer a class into doing it that way?

Logarithmic derivatives.  :P  But that's just because I can never remember the quotient rule myself.

LOdHI-HIdLO over LO squared is how I remember it. :lol:

I usually use the quotient rule, but that's just because I'm too lazy to put it into the form Kent wrote.  If you're trying to get the class to understand the quotient rule, I would have them use the quotient rule-if you want them to understand the power rule, then re-write it for the power rule.  I think the product rule form is easier to follow since the expression is linear but I don't prefer one way over another.

I'm not talking about promoting either understanding of the quotient rule or understanding of the product rule: I'm talking about computational efficiency and getting the right answer. (Which includes being able to locate critical points and other algebraic questions about the derivative.) Just for an example, I'll attack the first one in two different ways:

$ f(x)\equal{}\frac{2x}{(x^2\plus{}4)^{3/2}}$

$ f'(x)\equal{}\frac{(x^2\plus{}4)^{3/2}\cdot 2\minus{}2x\cdot\frac32(x^2\plus{}4)^{1/2}(2x)}{(x^2\plus{}4)^3}$

$ \equal{}\frac{2(x^2\plus{}4)^{3/2}\minus{}6x^2(x^2\plus{}4)^{1/2}}{(x^2\plus{}4)^3}\equal{} \frac{2(x^2\plus{}4)^{1/2}(x^2\plus{}4\minus{}3x^2)}{(x^2\plus{}4)^3}$

$ \equal{}\frac{4(2\minus{}x^2)}{(x^2\plus{}4)^{5/2}}$

versus the other pathway:

$ f(x)\equal{}2x(x^2\plus{}4)^{\minus{}3/2}$

$ f'(x)\equal{}2(x^2\plus{}4)^{\minus{}3/2}\plus{}2x\left(\minus{}\frac32\right)(x^2\plus{}4)^{\minus{}5/2}(2x)$

$ \equal{}\frac{2}{(x^2\plus{}4)^{3/2}}\minus{}\frac{6x^2}{(x^2\plus{}4)^{5/2}}\equal{}\frac{2(x^2\plus{}4)\minus{}6x^2}{(x^2\plus{}4)^{5/2}}$

$ \equal{}\frac{4(2\minus{}x^2)}{(x^2\plus{}4)^{5/2}}$

The number of steps is roughly the same. The question is which one represents more economy of thought and hence less chance of computational error? The primary advantage of the quotient rule is that it proceed directly to a common denominator, but the particular issue in this class of problem is that that denominator is the wrong power and must be further adjusted.

I sometimes go on to say that the quotient rule is not a "primary" rule of differentiation. The product rule is a primary rule; the quotient rule is a secondary rule derived from the product rule. It's very useful - in the right place. I happen to think that this class of problems isn't that right place.

I taught my Calc 1 recitation the second method, and a nontrivial fraction of them seem to use it in practice (and it seems to reduce their sign errors).  At a minimum I think that it's worthwhile to point it out to them a couple of times.

One of my students told me that in the past, she'd learned an additional rule we hadn't covered: the differentiation rule for $ \frac {1}{f(x)}$.  I think reducing the number of "things people just memorize for particular tasks" in favor of "tools that might be used for more than just one task" is definitely a good thing.

I also think that calc 1 students tend to be weak at algebra, and so (for example) might not recognize the last simplification step when using the quotient rule.  (This was certainly true of those who've done similar problems in the class I TA.)  So, to the extent that "fully simplified" answers are good, I think they're more likely to end up in the right place using the product rule.

I find that the algebra of simplifying the quotient rule is much easier for students than the algebra of finding the common denominator in the product rule.

So in my calculus classes I encourage students to use the quotient rule if the numerator is not a constant and the denominator is not a monomial -- as is the case in your two examples.

I think the quotient rule takes some of the work out but in order to get there, you have to be more careful.  I think that if you feel you're good at the quotient rule, there's no reason to discourage its use.

I think the quotient rule is more efficient IF you can avoid the computational errors and forgetting the order of everything.  I think the product rule is easier to use because the steps are more straightforward and there are fewer moving parts.


induction
number theory proposed
number theory
strong induction

$k$ is a positive integer. The sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = k+1$, $a_{n+1} = a_n ^2 - ka_n + k$. Show that $a_m$ and $a_n$ are coprime (for $m \not = n$).

We have :
 $a_{n+1}-k=a_n(a_n-k)=a_na_{n-1}...a_1$ (1)
 Suppose $n >m$ . Put $d =gcd(a_m,a_n)$
  Since (1) , we have $d|k$
 Because $a_n-k=a_{n-1}(a_{n-1}-k)$ , we infer $d|a_{n-1}$
 By induction , we get $d|a_1=1+k$ . This infers $d|1$ . So $d=1$
 --> Q.E.D

[quote="nttu"]
 Because $ a_n \minus{} k \equal{} a_{n \minus{} 1}(a_{n \minus{} 1} \minus{} k)$ , we infer $ d|a_{n \minus{} 1}$
 By induction , we get $ d|a_1 \equal{} 1 \plus{} k$ . This infers $ d|1$ . So $ d \equal{} 1$
 --> Q.E.D[/quote]

I don´t see that, because we infer $ d|a_{n \minus{} 1}^2$ and not $ a_{n \minus{} 1}$
It would be: Let $ p$ a prime number suh that $ p|d$. Because $ a_n \minus{} k \equal{} a_{n \minus{} 1}(a_{n \minus{} 1} \minus{} k)$ , we infer $ p|a_{n \minus{} 1}$
and the result follows.
Thanks campos.

Polish Math Olympiad 2002 final round and TST at the same time.


The number $ 25^{64}\cdot64^{25}$ is the square of a positive integer $ N$. In decimal representation, the sum of the digits of $ N$ is

$ \textbf{(A)}\ 7 \qquad
\textbf{(B)}\ 14 \qquad
\textbf{(C)}\ 21 \qquad
\textbf{(D)}\ 28 \qquad
\textbf{(E)}\ 35$

[quote=SnipedYou][hide="solution"]
$ N \equal{} \sqrt{25^{64} \cdot 64^{25}}$
$ N \equal{} \sqrt{5^{128} \cdot 8^{50}}$
$ N \equal{} 5^{64} \cdot 8^{25}$
$ N \equal{} 5^{64} \cdot 2^{75}$
$ N \equal{} 10^{64} \cdot 2^{11}$
Now sum up the digits of $ 2^{11}$, and the answer is $ 14$, $ B$
[/hide][/quote]

Shouldn't the answer be $15$ because you forgot to add up the $1$s digit from the $10^{64}$

I used a slightly different solution.
[hide = Solution] All we need to do is find $N$ in mod $9$ and compare it to the answers. $$N^2=25^{64}\times64^{25} \implies N = 5^{64}\times2^{75} \equiv 2^{11} \equiv 2048 \equiv 5 (mod 9)$$ Since the only answer $5$ modulo $9$ is (B)$14$, $\boxed{(B)14}$ is the answer. [/hide]


At first, please write your problem in plain text and explain the definitions of all the points involved. Of course, you can attach a picture, but the points should be explicitely defined in text form. This makes everything clearer. It is particularily important to know what assumptions you do and what you want to be proven. For instance, the perpendicularity of AC and DB is required not only for the last part of your problem (to show that PMNRLSTK is a cyclic octagon), but also for the first part (the equations).

Then, please use some more standard notations. I can only hope that I got you right in translating "IDNI" as "|DN|" (length of the segment DN). And please don't tell me that you don't have those | signs on your keyboard.

Finally, please try avoiding stupid errors. Look at your picture: can you really mean |LC| = |CB| ? Maybe you wanted to say |LC| = |LB| instead? And are you sure that |MA| = |DN| shouldn't be |MA| = |DM| instead?

Ah yes, and I've almost forgotten, if the source is called "internet"... well, the internet is big, could you be more precise?

PS. The second problem, which doesn't require the cyclicity of the quadrilateral ABCD, by the way, is briefly solved on http://www.cut-the-knot.org/Curriculum/Geometry/EightPointCircle.shtml

  Darij


divide by $\sqrt[n]{1-x^2}$. Denote $\sqrt[n]{\frac{1+x}{1-x}}=y$ and    obtain $y=\frac{1+\sqrt5}{2} , y=\frac{1-\sqrt5}{2}$
take $y=\frac{1+\sqrt5}{2}$ if n=2k and both y if n=2k+1.