Start at M in the diagram and form a path by moving to an adjacent letter to the right, left, up or down. How many paths spell the word MATH?

[asy]for (int i=0; i<3; ++i) {
label("H",(-3+i, i),W);
label("H",(i,3-i),W);
label("H",(3-i,-i),W);
label("H",(-i,-3+i),W);
}
for (int i=0; i<2; ++i) {
label("T",(-2+i, i),W);
label("T",(i,2-i),W);
label("T",(2-i,-i),W);
label("T",(-i,-2+i),W);
}
label("A",(0,1),W);
label("A",(0,-1),W);
label("A",(1,0),W);
label("A",(-1,0),W);
label("M",(0,0),W);[/asy]

WLOG pick the top A (otherwise, rotate). Then there are two "side T"s that have 2 H's, and one "corner T" with 3 H's. 2*2+3=7, 7*4=28.

Note that there is only one path from the central M to the H's on the exact top, right, left, and bottom. Also, there are 3 paths to each of the remaining H's. Then the sum is 4*1 + 8*3 = 28.

hint: separate it into 4 quadrants and then count the number of ways in each quadrant

[quote=cowcow][quote=sap2018]... sorry dont understand. i in 5th grade, gifted[/quote]

what don't you understand? It seems simple enough to me.[/quote]

Do not discourage other learners.

[hide=Solution]The strategy we will use for this problem is multiplying individual cases of counting to find a group case. We start at $M$. Starting at $M$, there are $4$ possibilities of $A$'s we can move too. From each of those $A$'s, there are $3$ possibilities of $T$'s we can move to. From each of the $T$'s, there are $3$ possibilities of $T$ we can move to. Therefore, the solution is $4\times3\times3=48$ possible paths.

Try a harder problem [hide=here]look below[/hide][/hide]


The Sprint Round competition consists of 30 problems with a time limit of 40 minutes. You completed 20 Sprint Round problems in 25 minutes. On average, how many times longer will you be able to spend on each remaining problem than you did on each of the first 20? Express your answer as a common fraction.

[hide]My first 20 were done at 5/4 of a minute each.

Now I have 10 left, and 15 minutes.
So Now I can spend 3/2 of a minute each.
3/2 divided by 5/4 = 6/5[/hide]

[hide=" :nhl: "]
You spent 25 min on 20 problems

You have 15 min left for 10 problems.
That is 90 seconds per problem.

For the first 20, you spent 25 min=1500 seconds.
Per problem, that is 75 seconds.

$90/75=\frac 65$[/hide]

[hide]there's 15 minutes left
there are 10 problems left
15min/10problems= 3/2 minute per problem

originally you did 25min/20problems= 5/4 minutes per problem

3/2=5/4*x
3/2*4/5=x

x=6/5[/hide]


What/who do you guys think impacts math motivation the most? I mean like, do you think most people get motivated into math because of their parents, teachers, friends, or themselves?

Also, what kinds of factors might be the leading cause for losing that motivation? I'm wondering about this because I have a friend who used to always improve herself and incline herself to study for math contests when she was in her early middle school years, she was constantly doing very well. But now she's sorta having a burnout-losing her grip on practicing, and she doesn't really seem to be doing much math practice.

Any ideas of how to help someone like this?

Why do you think not being motivated to seriously invest oneself in math contests is something that needs fixing?

This is a very common and natural phenomenon, and I'll attribute it to two reasons:

[b](1) Increasing opportunity cost:[/b]
The direct cost of mathematics is paper and ink, but the opportunity cost is whatever you could be doing with the time you spend. In early middle school, this opportunity cost is very small -- there's not as much else you can be doing with your time. As you grow into high school, and then into college, the opportunity cost goes higher and higher. You could be working, interning, doing research/personal projects, exploring fields of potential interest, volunteering, building friendships, taking care of siblings, etc.

[b](2) Decreasing value:[/b]
In middle school, competition mathematics is basically right in line with any interests in math, science, and engineering. As you mature, your interests become more specific, and the fields themselves broaden at a rapid pace. By the time college comes around, a chemistry major won't find the same appeal in pure mathematics anymore. But they might've been in Kumon back in middle school.

I personally think that a really good/nice/awesome math teacher helps create a liking for math.  And yes, if you're reasonably good at math and all your friends are participating in competitive math, then that would also motivate you to do it as well.

Hmm.  There are (see previous post) quite a few reasons that some one would stop doing competitive math.  Another reason might be that the person lost interest in it.  I'm in middle school still, so I don't know that much about high school competitions.  But I'm getting the impression that they're EXTREMELY hard.  Failing stuff demotivates people a lot.

The thing that is motivating me to do math right now is the hunger to be able to participate in the International Mathematical Olympiad, and that is all I am looking forward to. (I have quite a way to go, I'm not even in the AIME level yet, but the feeling for the Olympiad has cornered me.) :)

I think one important motivation is the knowledge. As much as you know more, you like it more. You make associations between different topics, you understand what lies behind, you know where you stand and so on.
This is valid not only for mathematics, but for other disciplines.
 Another factor is the medium. In a medium that is predisposed to defeatism, or where knowledge/science/math is downplayed, the average person will be less likely to grow interested on these subjects.
These two factors are example of internal versus external motivations.
I find TZF's post quite interesting.

For me, I feel that having a peer is extremely important. When I run a race in track, if someone is in front of me, then I feel that I have to work harder to catch up, but when I can't see anyone else, I feel more lax. In a similar way, I believe having a person that you compete with will motivate you. This doesn't just apply to math, I think this applies to everything you do.


these are some exercises that my teacher has given me. however it's quite difficult and i want to see good solutions from you! ^^
1. Given $ f: R\rightarrow R$ that
$ f(u \plus{} v) \plus{} f(u \minus{} v) \equal{} 2f(u)f(v);$with any $ u,v\in R$
a) prove that f is
b) Quality $ f(0)$ to prove that $ f(x) \geq \minus{} 1;$ with any $ x\in R$
2. Given $ f: R\rightarrow R$ that
$ f(u \plus{} v) \plus{} f(u \minus{} v) \equal{} 2f(u)g(v);$with any $ u,v\in R$
$ f\not \equal{} 0; \left|f(x) \right|\leq 1$
prove that $ \left|g(x) \right|\leq 1$

This belongs in the Olympiad section. Maybe Pre-Olympiad, but certainly not High School Basics.

Hello!
1/a) there is no  a question  :?:

b/ $ (u,v) \equal{} (0,0) \Rightarrow f(0) \equal{} f(0)^2 \Rightarrow f(0) \equal{} 0$ or $ f(0) \equal{} 1$

case 1 : $ f(0) \equal{} 0$

$ (u,v) \equal{} (\frac {x}{2} ,\frac {x}{2}) \Rightarrow f(x) \equal{} 0 > \minus{} 1$

case 2 : f(0)=1

$ (u,v) \equal{} (\frac {x}{2} ,\frac {x}{2}) \Rightarrow f(x) \equal{} 2f^2(\frac {x}{2}) \minus{} 1 \geq    \minus{} 1$

Oh! I'm sorry. because there is a word in the first question that i can't translate into English.I am Vietnamese. I don't know exactly that word but it means bounded or round function. Do you know it?
 :?:  :!:


logarithms
function
calculus
calculus computations

$ e^{2\ln u} \equal{} u^{2}$, $ u > 0$
(Just to be picky)

$ 2\ln u \equal{} \ln u^2$ so it makes no difference if $ u > 0$ or $ u < 0$ so long as $ u\neq0$.

(Just to be more picky) :wink:

[quote="Rajiv"]$ 2\ln u \equal{} \ln u^2$[/quote]
Yes, but the LHS is only well-defined as a real function (and therefore this property is only true) when $ u > 0$.  ;) ;)

Okay!

[hide="And..."]$ 2\ln x$ exists only in first and fourth quadrant while $ \ln u^2$ exists in all four quadrants and is the mirror image of $ 2\ln x$ about the Y-Axis.[/hide]


This is the first year that my school is doing Mandelbrot (I got them into it  :D ), so we ordered a contest book. It says in the beginning that w/o expressed permission from the writers, we arent allowed to make photocopies, etc. Well, considering that you were one of hte writers, can we (pretty please  8-) ) make photocopies so that we dont have to use the book 1 at a time? Also, am I allowed to post select problems that I enjoy from the book onto this site?

You can post selected problems on this site (I doubt even copyright law could stop that even if problem authors wanted to).  As for photocopying, I'm not comfortable answering that question as that book is nearly entirely the work of Sam Vandervelde (Sandor and I only contributed to the first year or two of that spiral book).  You can email him at info@mandelbrot.org.  I suspect his policy will be much like the one we explicitly state in AoPS - photocopying of bits and pieces for educational purposes (as I'm guessing you're planning) is ok, but reproducing the entire book is not (and probably not economic, at any rate).


function
complex numbers
complex analysis

having trouble with this question:

given sinhz, show that u & v are harmonic conjugates.

u = real part of sinhz          = sinhxcosy
v = imaginary part of sinhz = icoshxsiny

how do i show that these are harmonic conjugates?

Harmonic conjugates get me going every time.

I can feel your pain!

 :o


Euler
algorithm
modular arithmetic
number theory
Euclidean algorithm
number theory proposed

What's a smart way to calculate $ 3^{15^{62944331} \plus{} 2} \mod 37$

[hide]
I set $ e \equal{} 15^{62944331} \plus{} 2$, then wanted to do $ e \mod \phi(37) \equal{} e \mod 36$, but then $ 15^{62944331} \plus{} 2 \mod 36$ is nasty again, so I looked for $ 15^k$ where $ k \equal{} 62944331 \mod \phi(36) \equal{} 62944331 \mod 12$, which is nasty as well (I can only use euclidean algorithm here)

[/hide]

[color=darkblue]I think the most important idea is $ 27 \cdot 37 \equal{} 999$
$ 3^{15^{62944331}\plus{}2} \equal{} 3^{9k \plus{} 2}$. Notice that $ k$ is odd.
But $ 3^{9k} \equal{} 27^{3k}\equiv ( \minus{} 10)^{3k} \equal{} ( \minus{} 1000)^k \equal{} \minus{} 1000^k\equiv \minus{} 1^k \pmod{37}$. So $ \boxed{3^{9k \plus{} 2}\equiv \minus{} 9 \equiv 28 \pmod{37}}$.[/color]


probability
linear algebra
matrix
symmetry
function
induction

The elements of a determinant are arbitrary integers. Determine the probability that the value of the determinant is odd.

can't you start with a 2 by 2 and prove it by induction?

[quote="me@home"]You mean the elements of a matrix (not a determinant) are integers...

anyway, the probability must be one half by symmetry; assume it isn't. The function of calculating a determinant is as symmetric as possible, except for one thing: the choice between positive and negative (i.e. letting the top element on the left be first, so positive when calculating the determinant of its mini matrix... as opposed to negative)
but negative/positive is symmetric with respect towards even/odd (except zero maybe-but that particular case has zero probability...)
so it must be symmetric

that was a really bad explanation, sorry, tell me if I'm wrong[/quote]

Er, you may be right, but I don't think that "the formula for a determinant is rather symmetric" is a sufficient solution for the problem. Anyhow, I just worded it as it is worded in the book.

EDIT: it's actually not true. The probability is $1/3$ for a 2 by 2 matrix.

[quote="Phelpedo"]EDIT: it's actually not true. The probability is $1/3$ for a 2 by 2 matrix.[/quote]

I got $\frac{3}{8}$ for a 2 by 2 :maybe:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=determinant+odd&t=43828]look here[/url]

e: Yes, the answer is $3/8$ for a matrix in $\mathcal{M}_{2}(\mathbb{Z}/2\mathbb{Z})$

Darn, I must have miscounted one case  :blush:


I've just begun studying a little bit of group theory and abstract algebra.  I was wondering if there are any math contest questions (at the high school level) where group theory would provide a simple, or at very least elegant, solution to the problem.

Not really. Good contest-writers want to avoid giving an advantage to people who have studied more advanced math, so if a problem can be solved very easily with group theory but not easily with other methods, then it isn't a very good problem for a high school contest.

Having said that, there's a problem from USAMTS 2001-2002 (the one about BAABAABAA) that I solved using, basically, group theory. I went overboard on that problem, using ninth roots of unity I think. I knew it could be done modulo 3, but I liked my solution more. That didn't stop the grader from saying something along the lines of ``You know, you could have done that by looking at it modulo 3.''


Many Math Counts and/or AMC problems I get wrong are problems of this type...

1. The median and the mean of a set of 5 integers is 8. What is the largest/smallest number that can be in this set?

2. The median of a set of 7 integers is 10. The mean is 13. What is the largest/smallest number that can be in this set?

Yup....

OK, don't worry, these problems can be really easily done if you know the technique. The thing u must understand is average multiplied by the number of integers in the set equals the sum of all the numbers in the set. I'll give you an example from your own problem. So for #1 you said the average is $ 8$ and the number of integers in the set is $ 5$. This must mean that the sum of all the integers in the set is $ 8*5$ or $ 40$. AFTER we know that, we can see what your problem is asking. First of all your problem looks like this {a,b,c,d,e} where $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}40$ and so we can assume a and b being $ 1$ and $ c\equal{}8$ because it said the median is 8 as well. This must mean $ d\equal{}9$ because you are trying to get "e" to be highest as possible, so u must make all the other variables to be small as possible. Sooo we have the equation $ 1\plus{}1\plus{}8\plus{}9\plus{}e\equal{}40$ and you solve for "e" to get $ e\equal{}21$. Thus, the greatest possible number that can be in such a set is 21!! Ask me if your unclear with anything. Using this technique, solve the next problem.

On the contrary, as the numbers need not be distinct, the maximum is 22 (1,1,8,8,22 achieves this).

If however they all had to be different, it would be (1,2,8,9,20)

Of course, this assumes the problem meant positive integers only.


I have a question, is the following problem true?
Problem: If the hexgon ABCDEF has AD,BE,CF are concurrent, then six segments of that hexagon all touch a conic.
Hope to receive your replies soon.


geometry
trigonometry
parallelogram
algebra
polynomial
perpendicular bisector
Pythagorean Theorem

Let $\triangle{ABC}$ be a right triangle with $\angle{C}=90^{\cir}$ and $D$ be the midpoint of $BC$. If $AC=BC=6$ and $E$ , $F$ are two points on sides $AC$ and $AB$,respectively, so that $\triangle{AEF}$ is congruent to $\triangle{DEF}$, find the area of quadrilateral $AEDF$. (without using trigonometrics )
[img]http://i5.photobucket.com/albums/y194/jelly119/001-1.jpg[/img]

See the attached diagram.

$AFDE$ can be either a parallelogram or a kite. If it's a parallelogram, then the problem is quite trivial and the answer is $[AFDE]={a^{2}\over 4}=9$

If it's a kite, then $EF$ is the perpendicular bisector of $AD$. Let $M$ be the midpoint of $AD$. Draw $DN\perp AB$. Then $EM: AM=CD: CA\implies EM={1\over 2}AM={1\over 4}AD$, and $FM: MA=DN: NA$, where $DN=\frac{DB}{\sqrt{2}}=\frac{a}{2\sqrt{2}}$ and $NA=AB-BN=a\sqrt{2}-\frac{a}{2\sqrt{2}}=\frac{3a}{2\sqrt{2}}$, hence $FM: MA=1: 3\implies FM={1\over 3}MA={1\over 6}AD$. Therefore $EF=EM+FM={5\over 12}AD$. On the other hand, Pythagorean theorem on $\triangle ACD$ gives $AD={a\sqrt{5}\over 2}\implies EF=\frac{5a\sqrt{5}}{24}$. Now $[AFDE]={1\over 2}AD\cdot EF={25a^{2}\over 96}={75\over 8}$

And can you guys believe that this is a problem in our midterm for 8 grades... :wacko:

a teenie tiny mistake: $[AEDF]=\frac{75}{4}$.

$\frac{25\cdot 36}{96}=\frac{25\cdot 6}{16}=\frac{25\cdot 3}{8}$

Besides that, ${75\over 4}$ exceeds the area of the triangle itself, which is $18$.

[quote="jelly119"][quote="Farenhajt"]Well, essentially it's not THAT complicated... Some basic similarity, Pythagorean Theorem... You just have to spot a thing or two - I would expect this to be solved by an 8-grader who knows his maths.[/quote]

But our 8-graders haven't learned the concepts and skills of similarity yet!  :(   Would it be possible that  there is any trick about this problem??[/quote]

all you do is use the pythagorean theorem a bunch of times...what is the big deal? FE is the perpendicular bisector of AD

[quote="Altheman"]all you do is use the pythagorean theorem a bunch of times...what is the big deal? FE is the perpendicular bisector of AD[/quote]

I would be grateful to know the details about your no-big-deal solutions.

[quote="Farenhajt"]There's another approach which circumvents similarity but relies on equations and basic polynomial skills.

Denote $x=AE$. Then $CE=a-x, ED=x$, hence by Pythagorean we have

\begin{eqnarray*}(a-x)^{2}+\left({a\over 2}\right)^{2}=x^{2}&\iff& a^{2}-2ax+x^{2}+{a^{2}\over 4}=x^{2}\\ &\iff&{5a^{2}\over 4}=2ax\\ &\iff& x={5a\over 8}\end{eqnarray*}

Then find $EM$ by applying Pythagorean on $\triangle AME$ - the answer is $\frac{a\sqrt{5}}{8}$ (keep in mind, of course, that $AM={AD\over 2}$ and $AD$ is calculated from $\triangle ACD$)

Similarly, denote $y=AF$. In the same manner as in my previous solution, find that $AN=\frac{3a}{2\sqrt{2}}, DN=\frac{a}{2\sqrt{2}}$, which gives

\begin{eqnarray*}\left(\frac{3a}{2\sqrt{2}}-y\right)^{2}+\left(\frac{a}{2\sqrt{2}}\right)^{2}=y^{2}&\iff&{9a^{2}\over 8}-{3a\over\sqrt{2}}y+y^{2}+{a^{2}\over 8}=y^{2}\\ &\iff&{5a^{2}\over 4}={3a\over\sqrt{2}}y\\ &\iff& y=\frac{5a\sqrt{2}}{12}\end{eqnarray*}

Then by Pythagorean on $\triangle AFM$ find $FM=\frac{a\sqrt{5}}{12}$.

From there proceed to finding $EF=EM+FM$, as above.[/quote]

Yep...that's the way I taught my 8-grades brother! But the equations really look terrible for an ordinary student...With the concepts of similariy in your previous solutions would be much better!!

[quote="Farenhajt"]Hey, are you sure that the creators of the problem didn't have a PARALLELOGRAM on their minds? :)[/quote]

They gave the correct answer $\frac{75}8$. So I don't think that's an "accident". :wink: As I know, no one got the right answer in that test!! After all, there are many other complicated problems -for an 8-grader- to be done in 45 minutes!!


Prove that: nonnegative real numbers $ x1 x2 ... xn$ satisfy the following inequality:

$ (x1x2...xn)(1/x1 \plus{} 1/x2 \plus{} ... \plus{} 1/xn)<\equal{}(x1)^(n\minus{}1)\plus{}(x2)^(n\minus{}1)\plus{}... \plus{}(xn)^(n\minus{}1)$

You mean:
$ (x_1\cdot x_2\cdot \ldots \cdot x_n)(\frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}\ldots \plus{}\frac{1}{x_n})\le (n\minus{}1)(x_1\plus{}x_2\plus{}\ldots \plus{}x_n)$?

[quote="Bugi"]You mean:
$ (x_1\cdot x_2\cdot \ldots \cdot x_n)(\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \ldots \plus{} \frac {1}{x_n})\le (n \minus{} 1)(x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n)$?[/quote]
That is wrong, so I think it is:

$ (x_1\cdot x_2\cdot \ldots \cdot x_n)(\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \ldots \plus{} \frac {1}{x_n})\le x_1^{n \minus{} 1} \plus{} x_2^{n \minus{} 1} \plus{} \ldots \plus{} x_n^{n \minus{} 1}$

Which is just rearrangement, muirhead or AM-GM :)

[quote="matex (L)(L)(L)"]Could anybody show how to prove this inequality directly from AM-GM inequality?[/quote]
$ \frac{x_1^{n\minus{}1} \plus{} .... \plus{} x_{n\minus{}1}^{n\minus{}1}}{n} \ge x_1x_2...x_{n\minus{}1}$. Just do this $ n$ times and add up!


[quote="maky"][hide="solution"]with am-gm, i get :
$a^{4}+a^{4}+b^{4}+c^{4}\geq 4a^{2}bc$. then i sum up cyclically this inequality, and the problem is solved.[/hide][/quote]thank you for your answer you are so smart :wink:  :P
please email me abook about calcules if you have.
if you have. :wink:


Recently, my post rating has decreased dramatically, but I don't know which of my posts are being rated as spam. I have never intentionally spammed and always keep my posts as high-quality as possible. Thus, if I really am spamming, may somebody please be kind enough to mention to me what is wrong with my posts? I have never intended to do anything harmful; I love this site.  :)  But, really, I wish to know if I'm doing something wrong. Sorry for any inconveniences I may have caused if I have really spammed. Thanks in advance.

[b] @r15s11z55y89w21 [/b] : Hello, What is post rating? How can it decrease/increase? What was your post rating before, and what is it now?

I'm new to this site, so I want to know about posts and post rating.

Thank you.

Post rating information can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659]here[/url]. Remember to read the stickies for the forum before posting on any forum.

OK, post ratings are somekind of important, but if you scroll down in this forum, you'll find several topics related to post ratings. Why are people so obsessed? Of course, r15... is right, when he's intentionally falling because of someone, that should be reported... But there are dozens of topics on this subject, and just PM an admin...

And it would be nice (I have no skills to tell the simplicity of that) that the poster can see who rated him and how...

We will enable the ability to see which of your posts were rated in the future. However it is very unlikely we will give you the ability to [b]see who[/b] rated your posts and how he rated them (it would be a flame war ignition sequence)

[b] @ Valentine Vornicu [/b] :  :D

Yes, I think that should be fine. I also have the same problem as the [b] OP [/b], many of my good (I mean-- posts pertaining to providing solutions to abstruse puzzles, technical posts etc) posts aren't rated at all or worse they are rated as being "spam". I think a high percentage of my posts ( I mean posts that were rated) were intentionally being rated as spam.

I had observed that most (almost 95%) of my posts weren't rated. (I do acknowledge some of my earliest/earlier posts were silly); I think life would be much better for me and the [b] OP [/b] (and people like us) if people found some time to dispassionately 'rate' posts otherwise we have to bear the burden of being labeled as "worthless/spam posters" ([b] which in fact, we're not [/b]) and receive brickbats from petty moderator(s) and other pesky posters.

[b] P.S.: [/b] By petty moderator(s) I meant, As far as I'm concerned there was only one petty modertor whom I know,who has  indulged in some gross over-simplification regarding my post rating.

Thanks!

@Gen8:

http://www.mathlinks.ro/viewtopic.php?p=1516485#1516485

http://www.mathlinks.ro/viewtopic.php?p=1515172#1515172

These aren't so early posts...

@Valentin:

Well, someone rated r15... a 1 again... And I don't see why his post is so bad.

My rating has also dropped from 3.8 to 3.1 in the past few days. (a lot considering i've been here for 3 years) And I don't think I've spammed.

The problem in general for low ratings is that most people rate spam, but not good posts. So someone may have 95 good posts and 5 spam, and have all the spam rated 1's and have 3 good post rated 5's. This brings the average to 2.5. BTW, I have a rating of 2.2, an improvement from 1.9. I have only been here a month and a half, though.

Another quesiton: Why can only the user see their own rating?

Mods can see all the ratings... And I think it's better to keep ratings to yourself (not allow others to see)


Let u''(w) < 0, show that E(u(X)) <= u(E(X)), by expressing u(w) as a series around the point w = E(X) and terminate the expansion with an error term involving the second derivative.

Let $ w_0\equal{}E[X]$. For any $ w\in \mathbb R$ we have $ u(w)\equal{}u(w_0)\plus{}u'(w_0)(w\minus{}w_0)\plus{}\frac{u''(s)}{2}(w\minus{}w_0)^2$ where $ s$ is between $ w$ and $ w_0$. The last term is nonpositive, therefore...

Since the last term is non-positive, therefore

u(w) <= u(w0) + u'(w0)(w - w0)
u(X) <= u(E(X)) + u'(E(X))(X- E(X))
E(u(X)) <= u(E(X))   Taking expectation on both sides

However, from the Taylor series, why the first equation is true?
u(w) = u(w0) + u'(w0)(w-w0) + (u''(s)/2)(w-w0)^2 ?

ok, it is the Taylor's series with Lagrange remainder


[quote="tcjy"]How many problems are supposed to be done before you master one of the tasks ex: counting independent events. I did 160 problems on alcumus and never mastered any  :([/quote]

You can master and unmaster it in one question, it's all about the percentile, I think it has to be at lest 87.5% on the subject, and have done at least 10 questions (maybe more, like 15) to master it.

i don't think it's based solely on the percentile
it's based on how others do at the topic as well

say you have 70% questions right
if the bulk of AoPS have on 60% right, alcumus will consider you to have "master" the topic

however, if you have 70% right while the bulk of AoPS have 80% right, you will not "master" the topic


This is a poll for the people that are competing in the Problem Solving Tournament. How far do you think you are going to make it?

I voted top 6.... but I am being optimistic today! :P

I wonder what BOGTRO will vote......

Is it too late to join the tournament????

Well, we are currently near the end of the 2nd round...

it is too late to join the tournament, but I have started another one called "Another Tournament"

When the one going on currently is over, I will be starting another one. There are currently fourteen people left, and on August 11th, I will be narrowing it down to the top 12.


rotation
geometry
3D geometry
prism
analytic geometry
ratio

Let A(0,0), B(18,0), and C(10,6) be the vertices of a triangular region that is rotated about the x-axis.  What is the volume, in cubic units, of the solid that is formed.  Express your answer in terms of pi.
I know how to it but I was wondering why find the are of the triangle and mutiplying it by the circumfirenz with radius 6.
Can someone please tell me why?

I don't quite understand what you're asking  ;)

Well I don't know why that would work anyway. The way I'd do it is take the volume of a frustrum with radii 18 and 6 and then subtract the upside-down cone.

$\frac 13 \cdot \pi \cdot 10 \cdot (324 + 36 + 288) - 120 \pi = 2040 \pi$

your answer is wrong . it is 216pi, which I know how to get. I just won't use my way from now on.

[quote="kchande"]Find the area of the triangle and take the furthest point, 10,6, and find the circumfrence with radius 6.  Multiply those.  Why does it not come out to the right answer.[/quote]

Visualize the solid that is created when you rotate the triangle about the x-axis.  You get two cones with a circular base of radius 6 joined at the bases.  What you are finding using the method above is the volume of a triangular prism with base having area of the triangle and height equal to the circumference of a circle with radius 6.

Naga

I got my x coordinates mixed up with my y-coordinates :(
I thought 18,0 was graphed like 0, 18

I no that way with the two cones, but thanks anyway

6 is not the height of the center of mass of the triangle.  What you're looking for is the height of the intersection of medians.  Since medians cut other medians into 2:1 ratios, you can easiliy see that the height of the center of mass is actually 2, since the total height is 6.  Then you get 2*2*pi*area = 4pi*54 = 216pi as required.

What you would do, kchande, is find the area of the triangle, multiply by the circumference of the circle and then divide by three to get the volume of the two cones which is ${216}\pi$.

By the way, this concept is called Pappus' Centroid Theorem or some variant of that name.

when I did it, I thought the problem was easy, but now it seems so hard.

Actually, you don't multiply the area of the tirangle by the circumference of a circle with radius six to get the answer. You actually need to find the centroid of the triangle, find its distance x from the x-axis, and multiply the area of the triangle by the circumference of a circle with radius x. This is actually a theorem proved by some ancient mathematician about 2000 years ago.

Well, whenever you draw the heighth of a triangle to the base and then the centroid to the base, the distance from the centroid to the base is always one third of the altitude to the base because the ratio of the distance of the midpoint to the centroid to the distance of the centroid to the vertex is always one to two. Also if you have a triangle with coordinates $(x_1,y_1), (x_2,y_2),$ and $(x_3,y_3)$, the center of mass is at point $(\displaystyle \frac{x_1+x_2+x_3}{3},\displaystyle \frac{y_1+y_2+y_3}{3})$

I found it very simple o_O ... add the volumes of the two cones together, and voila...

[quote="kchande"]well, finding the areas of the cones is a bit easier[/quote]

How do find you the area of a cone?


LaTeX
geometry
3D geometry
prism
Support
Asymptote
ratio

[size=200][color=green]Introduction[/color][/size]

AoPSia is certainly an interesting planet. It is known for being an incredible outlier on the positive end of the plot of average intelligence of populations of selected planets. Of course, AoPSians do not view their status as “outlier” at all, and rather as quite a normal state, and often exclaim a “phailure” when they make a stupid mistake that would definitely be noted a brilliant statement in another world. The planet and its surrounding moons are a very peaceful, gorgeous place, abundant in mathematical exhibits and monuments and thousands of mathematical tournaments taking place. Everything always goes well, as the president of AoPSia, rrusczyk, as well as all subordinates, and heck, just about all of the citizens, are always able to use the awesome art of problem solving to pretty much solve all problems that may arise. Quite a while ago, the population of AoPSia decided for a revolution for the good, remodeling all mathematical structures to the next degree of technology, practically proclaiming to all of the worlds of life that AoPSia is to be the new superpower, above all others.

Of course, the other words out there are often jealous of AoPSian intelligence, and thus, tease the inhabitants of AoPSia as “nerds.” The people of AoPSia are perfectly fine with that label; in fact, the majority is proud of it, and, to cap stuff off, have something that actually hurts the other parties to shoot back at them: calling them “÷0 brains.” Most other worlds’ inhabitants are devastated by such a comeback, and it is probably of the better of them to just give up right there. Of course, most people of AoPSia know of once when an Minicephleran challenged an AoPSian to prove the brainless status of those people. The AoPSian, being a good AoPSian, took out a piece of paper and neatly made a convincing two-column proof.

Given: Diagram of Minicephleran Brain as Labeled -->
Prove: Minicephleran$ \cong$brainless

Statements ------------------- Reasons ------------------
1. Minicehpleran$ \cong$brainless | 1. Reflexive Property

The poor guy couldn’t believe his eyes. The AoPSian has proved his intelligence level right in front of his eyes. He advised others of his type to try not to bother the “amazingly smart” population of AoPS.

AoPSia is certainly a wonderful prosperous world. Recently, however, much attention has shifted to the rapidly-advancing moon of Forthewinia, the site of most math tournaments. Colonization of that moon just started a while ago by levans, who became, as he should, the president of Forthewinia. Many loved the idea, and quickly followed, causing the population of Forthewinia to grow practically exponentially. Before long, Forthewinia was the brightest spot of AoPSia’s night sky. Lights are abundant, and there are often celebrations of great Forthewinian legends, like AIME15, Mewto55555, Jongao, stevenmeow, and BOGTRO, as well as newly-rising stars, like UberYoda and miller4math. It was pretty much, the moon that never sleeps, the moon of joyful competition and paradise. After all, what type of AoPSian won’t like to show off their elite math skills?

Forthewinia is certainly unique, and the members of Forthewinia highlight this uniqueness with a slightly unusual system. For example, they commonly write in LaTeX, and some solve equations in base 42. Forthewinia, as well as the rest of AoPSia, often taunts at foreign units of measurement and methods of communication because many don’t make logical sense.

One foreign unit most laughed at is the hit-point, or HP. It is often jokingly referred to as “headless pigs” and others of the sort that HP [i]might[/i] stand for. The Forthewinian equivalent of the HP, the MRU, stands for mathpower resistance unit, and has no funny alternative acronym comprehension. One benefit of using this unit is that Forthewinians, being AoPSians, can gauge the ability of resistance of an outsider in a flash, given their HP, but those outsiders fry their brains just trying to comprehend the unbelievably complicated formula for converting MRU to HP:

1 MRU = 1 HP.

Of course, this unit of MRU is hardly ever used, as Forthewinia is usually very peaceful; only at very special times, usually in contact with other worlds, is the unit noticed; usually, it’s not even worried about, as mathematical attacks would most likely overpower all other attacks.  But, yes, it is not a surprise that AoPSians often laugh uncontrollably when in discourse about other worlds.

Albeit the great ascending development of Forthewinia, the competitive nature of the planet has triggered a few heated arguments. One such was revolved around a decision levans made near the genesis of Forthewinia. levans selected advisors, popularly known as moderators or just “mods,” to aid him in the protection of the greatness and wealth of Forthewinia. So far, only four have been appointed to hold the special title and honor, and one retired. Many Forthewinians, being quite complacent in character from the experience of many tournaments, have questioned exactly what quality of them marks them as inferior to the ones that were selected as “mods,” and indeed, many have petitioned the king to gain moderator status and have been rigidly denied. Mods lived in the highest of possible luxury in Forthewinia, each with their own huge mansion, and a fourth building where they actually spend most of their time discussing organization in Forthewinia and loquaciously rambling about math. This sparked much jealousy, and in some instances, hate. In one worst case, an all-out war nearly commenced between one of the great Forthewinian legends and one of levans’s advisors. Usually, peace is retained, and most Forthewinians realize the importance of that. But just once in a while, some unrest arises.

As anyone can see, Forthewinia is certainly an interesting moon. It is now by far the most prosperous moon existent. However, the guys from the other moons of other planets...are still jealous, and one, that has actually prepared for an invasion for many years, now makes an unexpected strike...Forthewinia has been invaded. All heptagonal prism will break loose. Thus, we commence with Chapter 1, and on a dark note.

[size=200]Chapter 1[/size]

“Hmmmmm. It must be night in Forthewinia,” remarked james4l to himself. (A fair amount of light is radiated from AoPSia, but only for a certain sector of the surrounding space, giving moons day when it enters the radiation zone and night when it is away.) “This is the very moon that is becoming the apex of wealth of all of the worlds of space. Well, I’ll be there soon.”

A distinctively intelligent man, even in the AoPSian world, james4l is probably one of the most welcomed tourists in Forthewinia, highly famous for his good judgment in decisions, mathematical genius, and displays of nerdiness. Previously, james4l lived in Forthewinia himself, but a while ago moved back to the main world and planet of AoPSia. He occasionally comes to Forthewinia to visit, one such instance being now, as he travels in his private spaceship, the 4l-42 (pronounced “four-ell-forty-two,” not “negative one”). Hilariously, he came to visit just in time for a very big occurrence, which of course, he doesn’t know of yet.

--

Forthewinia boasts its strong 24/7 protection against outside invasion. From a very well-built guard tower, anyone can see anything coming. r15s11z55y89w21 runs the night shift every 10th day. However, lazy as he is, r15s11z55y89w21 usually dozes off halfway through his shift and pays his assistant, dragon96, to do the rest of the shift for him. Even though he was still skeptical about it, levans decided to approve what is going on, as the guarding work does get done correctly, and so, half of the time dragon96 is looking through the specially-manufactured Forthewinian telescopes that spot distant approaching spaceships even though r15s11z55y89w21 should be doing it.

“Hey r15s11z55y89w21,” says dragon96, “come here!”
Hesitant from laziness, r15s11z55y89w21 comes over.
“Look! Someone’s coming! And look closer! It says 4l-42!”
“It’s our old friend, james4l!”
“Well he’s coming for a visit. How nice. We’ll give him a warm welcome to the new Forthewinia that is so much different from what it was the last time he’s been here.”
“Allrighty! He’s almost landing.”

--

*DING*
The elevator door to the third floor opens, and james4l enters, but to r15s11z55y89w21’s and dragon96’s astonishment, bearing a melancholy expression on his face. r15s11z55y89w21 and dragon96 suddenly realize that james4l is holding in his hand a weird sheet of foreign-looking paper. Simple staring was all that happened for a short period of silence, and dragon96 finally enunciated, “Uh...what’s...wrong?”
“This,” said james4l, handing the obviously non-AoPSian-manufactured paper over.
“Oh gosh!” exclaimed r15s11z55y89w21, peering over dragon96 as he reads the paper. It was a poem. Short, but definitely not sweet.

[i]You evil nerds of Forthewinia,
So arrogantly bright for the universe to see,
To force you to realize your wrongness,
We will deprive a group of three. [/i]

“Gosh, so threatening. Geez, at this guard station, we could easily spot any intruders and tell everyone of the invasion to defend,” said dragon96. r15s11z55y89w21 reinforced dragon96’s statement, “Exactly, and who could possibly succeed against us?”
Deep in thought, james4l finally commented, “Exactly who is this group of three? Us?”
Using his elite logic skills, dragon96 explained, “Obviously not. They didn’t even know you were coming. What other groups of three are there?”
r15s11z55y89w21’s face turned slightly red. “...mods.”
dragon96, unaware of that possibility, exclaimed, “holy calculator.”
james4l advised, “Well, in that case  we better go and check the mods’ mansions.”

--

james4l came back from the mansions. “It happened. No one is answering the door; they always do.” r15s11z55y89w21 and dragon96 looked at each other. r15s11z55y89w21 questioned, “What puzzles me is exactly how they did it. Tonight we only found two spaceships coming, yours and levans’s, as he returns from his business trip to main-planet AoPSia.”

dragon96 reminded r15s11z55y89w21, “We don’t really have time to think. We got to plan something to do for this case we’ve got.”

--

Immediately the next morning all were informed of the mods’ disappearance. r15s11z55y89w21, dragon96, and james4l discovered that intruders hijacked levans’s plane and came sneakily via that. levans escaped, but the hijackers went on for more. They captured the three mods and then used an inflatable spaceship they had carried with them (by far the most advanced invention outside of AoPSia) to leave the planet.

To levans’s dismay, the lack of mods wreaked havoc in Forthewinia. Discourse immediately took a big turn for the bad, and fights and arguments arose, many concerning who the new mods to be replaced are; in such the heat, some citizens were so enveloped in jealousy as to think of this as a priority before saving the mods.

One day, r15s11z55y89w21 discussed with levans that he would go on a mission to try to find where the mods were, and levans approved. r15s11z55y89w21 deduced that the paper the horrid poem was written on was paper of Eipia (Ee-igh-PIE-ya), a planet that is becoming a rival to Forthewinia. After informing his trusted friend, james4l, of where the mods are being held and what he plans to do, r15s11z55y89w21 plans a trip to Eipia. He decided that dragon96 should go with him.

“Bring lots of any helpful stuff, we might have a lot to do, especially food,” he said to dragon96.
“Okay. I’ll help with the stuff.”

r15s11z55y89w21 brought 4200 LaTeX-dwollars (Forthewinian currency), and dragon96 brought some weapons, a lot a food, and a bunch of other miscellaneous items. And so they left for Eipia, and the rest of Forthewinia awaited their return with the mods, some mischievous citizens taking the last opportunity to commit nefarious actions before the policing system comes back. The only problem is, r15s11z55y89w21 and dragon96 did not come back.

james4l now became quite popular on Forthewinia, and people often asked him what was really going on. james4l just said that what is going on is really going on, r15s11z55y89w21 and dragon96 left and for some reason aren’t coming back.

Unrest has now became a huge crisis on Forthewinia, making the moon a terror of a place. Evenually, levans jumped on his spaceship and fled to nearby Alcumia, noting that even though he is king, he only has 196 MRU, and thus isn’t really in a very good position.

It took a long time for Forthewinia to realize what problem this fighting could become if it continues. At last, the current greatest citizen in accomplishment, AIME15, decided that he will support the cause to go get isabella2296, jjx1, vallon22 (, r15s11z55y89w21, and dragon96) back. james4l, now much more than a tourist, is also going. Sixteen other Forthewinians agreed to help the cause. AIME15 designated lotsofmath to take half of the crew, thus making a split in case an occasion calls for two tasks to be done. He still decided however, to have one group bigger than the other so that more can be done with this group. The whole crew of eighteen thus boarded the ASYMPTOTE XLII, and headed for Eipia.

"Surely with this many people we can conquer those Eipians," proclaimed AIME15. "However, note that they are getting smarter as well. We’re going to be going against people of our own kind. We should supposedly, be more powerful than them, but they’ve got their whole planet there and we have eighteen, possibly twenty if we find r15s11z55y89w21 and dragon96 somewhere and they weren't also captured. "

The crew:
AIME15, leader
BOGTRO, funtwo, Meta_Knight, pythag011, AIME15's top advisors
mz94, mathblitz, vahalla, eggylv999, rest of crew
lotsofmath, leader of secondary group
coolts, Joe10112, 1=2, Dojo, Garyzx, secondary group crew

Two days into their journey, AIME15 realized that he probably should have brought some money.

"Uh-oh, we might need to buy stuff out there, considering we don't have that many weapons, plus other stuff."
"Hmmmmmmm, I have 3500 LaTeX-dwollars with me," commented BOGTRO.

Others also presents bits of money they found in their pocket or packages they brought. Collecting all of the money they had, AIME15 instantly summed in his head that in total they had 4725 LaTeX-dwollars to spend. "We're sort of short on supplies. So we'll take a stop at the AoPSian system-boundary store and try to see what we can buy there. Hopefully something useful." Thus, they will stop at the store on their trip.

Hmm... so what do we do?

[b]Buy Combinatorial Nullstellensatz from the store.[/b]

The first thing that you would do, before embarking on the journey to Eipia, is to decide what you will buy at the AoPSian system-boundary shop. That post will come soon.

This is too long to read. A summary please?

Blah blah blah you are on a planet with a store, you have 4725 dollars, what do you want to buy.

Realize how excellent that summary was.

[b]I buy the AoPS website[/b]

one question

how did you pick the groups and leaders and such?

by mru?

anyway, where is the store list?

did he pm it to you?

the store list will be up soon. be patient. He's recovering from Carpal Tunnel.

[quote="Joe10112"]sorry

whats the definition of "carpal tunnel?"

i know, im a n00b. o well

and i assume n00b is a very inexperianced player.[/quote]

Google is your friend.

but, i cannot go on google!

i have blocked internet  :(

r15 is an excellent writer.  How much must I pay you to write my essay for me?

Meh, so the store inventory list/prices is coming soon?

[quote="wiki"]Carpal tunnel syndrome (CTS), or median neuropathy at the wrist, is a medical condition in which the median nerve is compressed at the wrist, leading to paresthesias, numbness and muscle weakness in the hand. The diagnosis of CTS is often misapplied to patients who have activity-related arm pain.

Most cases of CTS are idiopathic (without known cause), genetic factors determine most of the risk, and the role of arm use and other environmental factors is disputed.

Night symptoms and waking at night--the hallmark of this illness--can be managed effectively with night-time wrist splinting in most patients. The role of medications, including corticosteroid injection into the carpal canal, is unclear. Surgery to cut the transverse carpal ligament is effective at relieving symptoms and preventing ongoing nerve damage, but established nerve dysfunction in the form of static (constant) numbness, atrophy, or weakness are usually permanent and do not respond predictably to surgery.

[/quote]

ohhhhhhh

thanks dojo

i hope this post isnt too short

I don't know. Announce if you're still here. If at least half of the people respond within 24 hours, the game shall continue.

Heh, I must be pro at indirectly reviving games.

I am here.

While I'm here, might as well advertise:

Go to http://cyneer.com, host of the first National Internet Math Olympiad!

bool(mehere)= 1

/me chuckles at C++ ref.

Yes, I'm here.

i am here but i dont come on very much so youll have to count me out

noooooooooo................... anyone want to send prods?

I'm still here. AIME15 decide what we do.

It appears that LTM has failed.

=(

Anyway, if you like LTM-ish stuff, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288353]here[/url].

[hide=" "]360-->3G0
363-->3G3
384-->3B4
NGG: 14:42-->coordinate strategy
E6--><6
Doctor, Silver, Oxygen, Nitrogen-->Dr., Ag, O, N
512 was 11th-->15n
r15's transmitter code-->785-IY-452
Eipian003: 110
Eipian004: 110
Eipian013: 125
Eipian014: 125
Eipian015: 145
Eipian016: 145
Eipian017: 90
Eipian018: 90
Eipian019: 100
Eipian020: 150
Eipian021: 150
Eipian022: 150
Eipian023: 175
Eipian024: 430
Eipian025: 300
Eipian026: 125
Eipian027: 125
Eipian028: 125
Eipian029: 200
Eipian030: 200
pispeakerfreaker-->202-->382-->411
(x,y)-->xth letter in yth line<--5
13-->lotsofmath
15-->BOGTRO
22-->AIME15[/hide]

You mean this topic is officially done with??? :?:

Do not post on this thread if you're not in the game. This rule still applies, even though at least this part of the game ended. Additionally, do not post unless you have something meaningful to say.

[quote="r15s11z55y89w21"]It appears that LTM has failed.

=(

Anyway, if you like LTM-ish stuff, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288353]here[/url].

360-->3G0
363-->3G3
384-->3B4
NGG: 14:42-->coordinate strategy
E6--><6
Doctor, Silver, Oxygen, Nitrogen-->Dr., Ag, O, N
512 was 11th-->15n
r15's transmitter code-->785-IY-452
Eipian003: 110
Eipian004: 110
Eipian013: 125
Eipian014: 125
Eipian015: 145
Eipian016: 145
Eipian017: 90
Eipian018: 90
Eipian019: 100
Eipian020: 150
Eipian021: 150
Eipian022: 150
Eipian023: 175
Eipian024: 430
Eipian025: 300
Eipian026: 125
Eipian027: 125
Eipian028: 125
Eipian029: 200
Eipian030: 200
pispeakerfreaker-->202-->382-->411
(x,y)-->xth letter in yth line<--5
13-->lotsofmath
15-->BOGTRO
22-->AIME15[/quote]

How were we supposed to figure this out?


AMC
AIME
geometry
inequalities
triangle inequality

A set of positive numbers has the triangle property if it has three distinct elements that are the lengths of the sides of a triangle whose area is positive. Consider sets $ \{4, 5, 6, \ldots, n\}$ of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of n?

Hints only, please. Thanks in advance!

I haven't tried this yet, but I would work backwards. Also, look for the smallest value of n that does not have the property rather than the largest value that does.

[hide="Hint"]Try to find the 10-element subset that meets this property, making the largest element in that subset as small as possible.
[hide="Little More"]Use 4 and 5, they are the smallest elements to start out with.
[hide="Complete Spoiler"] Triangle Inequality to find the next smallest-as-possible element that is viable.
[/hide][/hide][/hide]

[hide="mah hintz"]Consider the two smallest elements of one of the ten-element sets.
[hide="More"]What must all of the other elements be bigger than?[/hide][/hide]

Thanks for the hints! When I get a chance, I'll post the answer I get (because I don't have the key), and hopefully then someone who does have the answers can please let me know if it's right.

Nope, sorry.

Edit: You basically got it. Reread the question and see if you answered it correctly. You found the first set that does not have the property. That's not what the question is asking.

Oh, I think I see what I did! It's 253, right?


induction
combinatorics unsolved
combinatorics

How many ways can we select n-2 subset A_1...A_{n-2} of S={1;2...n} satisfying for i<j $|A_i|<|A_j|$ and $A_i\notin A_j$

There cannot be a set of size 1 and n-1 at the same time. Let $a_n$ be the number of set systems which contain sets of sizes 2 to n-1 and $b_n$  the number of set systems which contain sets of sizes 1 to n-2. Let $c_n$ be the total number of the systems.
If we wanna construct one of the sets in $b_n$, we can choose the set $A_1$ of size 1 in n ways. The other sets are of sizes 2 to n-2 and are subsets of $\{1,2,...,n\}-A_1$, so for each $A_1$ there are $a_{n-1}$ such subsets, so
(1) $b_n=n.a_{n-1}$.
If we wanna construct one of the sets in $a_n$, we can choose $A_1$ of size n-1 in n ways. Denote x the number not being in $A_1$. The other sets consist of the number x and a subset of size 1 to n-1. There are $b_{n-1}$ such subsets for each $A_1$, so
(2) $a_n=n.b_{n-1}$
From the equations (1) and (2) follows
(3)$c_n=n.c_{n-1}$.
As $c_3=6$, from (3) by induction  $c_n=n!$.


A cone and a cylinder have the same height and a base with the
same diameter. What is the ratio of the volume of the cone to
the volume of the cylinder? Express your answer as a common
fraction.

Let r be the radius of the cone and cylinder, and h the height. Since the diameters are the same, the radii are the same.

Volume of cone= (1/3)$ \pi$*$ r^2$*$ h$.

Volume of cylinder= $ \pi$*$ r^2$*$ h$.

Ratio is $ \boxed{\frac {1}{3}}$


geometry
geometric transformation
rotation

there are 16 books on a shelf.in how many ways can these be arrangedif 12 of them are volumes on history and must be kept together in order.( i will like to see yours)my anwser was 1 less than the real anwser

I am definitely NOT even close to gifted, I estimate my iq somewhere around 60-70
    [hide="bogus solution"] since the 12 history books have to always be together and in the same order then just imagine that all 12 are one block so then there would be 5 blocks which can be rearranged 120 ways. [/hide]

Just imagine that the group of 12 is only 1 book.

Then there are 5 books, and 5 spaces, leaving 5! as the correct solution.

If some items must be kept together in the same order, you might as well treat 'em as one object.

The 12 books are one "book"
There are 5 spaces.
So:
$ 5\cdot4\cdot3\cdot2\cdot1\equal{}120$
or
$ 5!\equal{}120$

This line of thinking may be useful in contest problems when you have a situation similar to the following:

[b]Problem:[/b] [i]Seven cheap cars and seven expensive cars are parked in a parking lot with 32 spaces, arranged in one row. Noone wants to risk scratching an expensive car while opening his/her door, so there can't be any cars in the two spaces adjacent to an expensive car. In how many ways can this be done?
[/i]
[b]Solution:[/b] [i]There are four cases: no expensive cars in the leftmost or rightmost slot, expensive car in only the leftmost slot, expensive car in only the rightmost slot, and expensive car in the leftmost and rightmost slots.

In the first case, we group each expensive car with the 2 adjacent spaces to form "units", burning up 14 of the 32 spaces. So now, out of effectively 18 spaces, we pick 7 for cheap cars, and 7 for expensive car units. This gives 18!/(7! 7! 4!) combos.

In both the second and third cases, the expensive car on the edge is grouped with the single adjacent slot, reducing the number of slots to 31. Now we want to fit 7 regular cars, and 6 expensive car units (for which we burn 12 slots). Now, we have 18!/(7!6!5!) ways of doing this for each of these two cases (we multiply that number by 2).

In the fourth case, place an expensive car on each end, leaving 28 slots. Now, group each of the 5 remaining expensive cars with the two adjacent spaces, burning another 10 spaces. There are 18!/(7!5!6!) ways of arranging these.

The total number of arrangements is 42007680[/i]

Obviously you wouldn't see those exact numbers on a contest since the numbers are huge. But it's an example of a slightly less obvious situation where grouping might help.

I don't see where the IQ of 129 comes in...

[url]http://en.wikipedia.org/wiki/IQ_reference_chart[/url]
I'm sorry, but above 129 is only moderately gifted. :P

And by Zuton Force's poke-hint, you should deduce that the answer is [i]not[/i] 120.

well, above 129 is where they start calling people gifted, moderately or not, i guess

Oh, well I was assuming the books were marked and distinct from each other.

[quote="kysuke"]If it's on a shelf, why can't it be 2*5!
The volumes can be counting up or counting down right? They just have to be in numerical order?

edit: They're unmarked so I'm sure the answer is just 5![/quote]

Well then you have to think about flipping books upside down and stuff. Well I don't suppose flipping a book upside down is a different arrangement

ALL OF YOU AER WRONG!!!!!! THe history books are not distinct, so its $ 5!\cdot12!$ because yo can rearrange the history books!

[quote="Euler221"]I don't understand The Zuton Force's post.  Specifically, if each expensive car is grouped with a space on each of its sides, then how is it possible for there to be an expensive car followed by a space followed by another expensive car?
The arrangement "expensive car, space, expensive car" is not possible.  Instead, the arrangement would have to be "expensive car, space, space, expensive car."  So isn't this undercounting the true count by not including arrangements in which there is only one space between expensive cars?[/quote]

Yeah, I was smoking... and then they wouldn't let me delete it. But the point was, that grouping is a useful technique :P

Why do so many people not read the original post?

It clearly says that the 16 history books must be kept in order. (So they are distinct, mewto.)

I am almost certain that rotations do not count in these types of problems, but then again, the problem should have specified.

[quote="metafor"][quote="Euler221"]I don't understand The Zuton Force's post.  Specifically, if each expensive car is grouped with a space on each of its sides, then how is it possible for there to be an expensive car followed by a space followed by another expensive car?
The arrangement "expensive car, space, expensive car" is not possible.  Instead, the arrangement would have to be "expensive car, space, space, expensive car."[/quote]

Actually, no: the problem only requires that the spaces [i]adjacent[/i] to the expensive car (i.e., the space to the right of the car and the space to the left of the car) be left empty, so the arrangement "expensive car, space, expensive car" is indeed possible (as is the arrangement you suggested). Perhaps a clearer problem statement would be as follows:

[i]An employee of the Legendre Hotel has been ordered to park seven cheap cars and seven expensive cars in a parking lot. The parking lot consists of a single row of 32 parking spaces. The manager of the hotel doesn't want to risk damaging any of the expensive cars and, therefore, has told the employee that there must be at least one empty parking space between an expensive car, and any other car. If each car must be parked in a parking space, in how many ways can the cars be arranged? [/i][/quote]

I am aware that the arrangement "expensive car, space, expensive car" is possible, as you stated.  My complaint is that the Zuton Force's solution does not allow for this arrangement, and therefore undercounts the total number of possible arrangements.  In his solution, the Zuton Force groups each expensive car with a space on both of its sides, forming what he calls "units."  Let's say that two of these "units" are placed next to one another.  Then the arrangement would be "space, expensive car, space, space, expensive car, space."  As you can see, there are two spaces between the expensive cars.  Furthermore, there is no way to get just one space between two expensive cars if they are grouped into the "units."  Therefore, the Zuton Force's solution is incorrect.

Ah — apologize for the misunderstanding, then. -_-;

I think that this isn't an IQ problem. Its just counting here is the solution : consider the 12 history books as just one book
then at first you have 4 choices then 3 choices... so the are $ 4!$ solutions read Intro C and P you will never have a question again  :D

When he says "in order" maybe he means that you can also reverse the order of history books. The answer would then be $ 2\cdot 5!$ . You may say: "If so, you can find an opportune definition of order and arrange the books as you wish". That's correct. I assumed that order means something like alphabetical order. And another thing... the history books are usually big: you can put other books on them and also put them horizontally. That's another question I guess, if not I just say I'm not >129.
Bye, Gil.


Hey guys..could you help me with this question?

8^(3x-3)-8^3x=448

Thanks.

Take a closer look.
[hide]$ 8^{3x\minus{}3}\minus{}8^{3x} \equal{} 8^{3x\minus{}3}(1\minus{}8^3) \equal{} 448$, so $ (8^3)^{x\minus{}1} \equal{} \minus{}\frac {448} {8^3 \minus{} 1}$, but all this is useless, since LHS is negative while RHS is positive, so never any real solution[/hide]

$ 8^{3x \minus{} 3} \minus{} 8^{3x} \equal{} 448$

let $ 8^{3x}\equal{}t$

$ t/8^{3} \minus{} t \equal{}448$

$ t\equal{}\minus{}32768/73$

$ 8^{3x}\equal{}\minus{}32768/73$

so there is no real solution to the equation


calculus
integration
geometry
algebra
binomial theorem
calculus computations

Let $ A$ be the region : $ \left\{(x,\ y)|x\geq 0,\ y\geq 0, \left(\frac{x}{a}\right)^{\frac{1}{2}}\plus{}\left(\frac{y}{1\minus{}a}\right)^{\frac{1}{2}}\leq 1\right\}$ for $ 0<a<1.$

(1) Find the area of $ A.$

(2) Find the maximum volume of the solid obtained by revolving $ A$ around the $ x$ axis.

[hide]Solve for y to obtain

$ y \equal{} (1\minus{}a)\left(1 \minus{} 2(x/a)^{1/2} \plus{} (x/a)\right), \quad 0 \le x \le a \le 1$.

With the change of variable

$ u \equal{} x/a, \quad du \equal{} 1/a \, dx$, we obtain

$ A(a) \equal{} a(1\minus{}a) \int_{u\equal{}0}^1 (1 \minus{} 2 \sqrt{u} \plus{} u) \, du \equal{} \frac{a(1\minus{}a)}{6}$.

The volume V obtained by revolving the region A about the x-axis is given by

$ V(a) \equal{} \int_{x\equal{}0}^a \pi y(x)^2 \, dx \equal{} \pi a (1\minus{}a)^2 \int_{u\equal{}0}^1 (1 \minus{} \sqrt{u})^4 \, du$.

Since the integral is merely a constant, we can immediately differentiate the above with respect to a to obtain the value that maximizes V:

$ V'(a) \sim (1\minus{}a)(1\minus{}3a) \equal{} 0$

implies V attains its maximal volume for a = 1/3.  (The only other critical points are a = 0 and a = 1, both of which give V(a) = 0.)  Now to compute the integral, we employ another substitution (to avoid expanding via the binomial theorem):

$ t \equal{} 1 \minus{} \sqrt{u}$,
$ u \equal{} (1 \minus{} t)^2$,
$ du \equal{} 2(t\minus{}1) \, dt$;

$ \int_{u\equal{}0}^1 (1\minus{}\sqrt{u})^4 \, du \equal{} \int_{t\equal{}1}^0 2(t\minus{}1)t^4 \, dt \equal{} 2 \Bigl[\frac{t^6}{6} \minus{} \frac{t^5}{5}\Bigr]_{t\equal{}1}^0 \equal{} \frac{1}{15}$,

and therefore

$ V\left(\frac{1}{3}\right) \equal{} \frac{4\pi}{405}$.[/hide]

In view of double integral, we see that \[ v\left( A \right) \equal{} \int_0^{1 \minus{} a} {\int_0^{a\left( {1 \minus{} \sqrt {\frac{y}
{{1 \minus{} a}}} } \right)^2 } {dx} dy}  \equal{} a\int_0^{1 \minus{} a} {\left( {1 \minus{} \sqrt {\frac{y}
{{1 \minus{} a}}} } \right)^2 dy}  \equal{} \cdots
\]

[quote="atomicwedgie"][hide]Solve for y to obtain

$ y \equal{} (1 \minus{} a)\left(1 \minus{} 2(x/a)^{1/2} \plus{} (x/a)\right), \quad 0 \le x \le a \le 1$.

With the change of variable

$ u \equal{} x/a, \quad du \equal{} 1/a \, dx$, we obtain

$ A(a) \equal{} a(1 \minus{} a) \int_{u \equal{} 0}^1 (1 \minus{} 2 \sqrt {u} \plus{} u) \, du \equal{} \frac {a(1 \minus{} a)}{6}$.

The volume V obtained by revolving the region A about the x-axis is given by

$ V(a) \equal{} \int_{x \equal{} 0}^a \pi y(x)^2 \, dx \equal{} \pi a (1 \minus{} a)^2 \int_{u \equal{} 0}^1 (1 \minus{} \sqrt {u})^4 \, du$.

Since the integral is merely a constant, we can immediately differentiate the above with respect to a to obtain the value that maximizes V:

$ V'(a) \sim (1 \minus{} a)(1 \minus{} 3a) \equal{} 0$

implies V attains its maximal volume for a = 1/3.  (The only other critical points are a = 0 and a = 1, both of which give V(a) = 0.)  Now to compute the integral, we employ another substitution (to avoid expanding via the binomial theorem):

$ t \equal{} 1 \minus{} \sqrt {u}$,
$ u \equal{} (1 \minus{} t)^2$,
$ du \equal{} 2(t \minus{} 1) \, dt$;

$ \int_{u \equal{} 0}^1 (1 \minus{} \sqrt {u})^4 \, du \equal{} \int_{t \equal{} 1}^0 2(t \minus{} 1)t^4 \, dt \equal{} 2 \Bigl[\frac {t^6}{6} \minus{} \frac {t^5}{5}\Bigr]_{t \equal{} 1}^0 \equal{} \frac {1}{15}$,

and therefore

$ V\left(\frac {1}{3}\right) \equal{} \frac {4\pi}{405}$.[/hide][/quote]

That's the correct answer. :)

In case of $ \sqrt {\frac {x}{a}} \plus{} \sqrt {\frac {y}{b}} \equal{} 1\ (a > 0,\ b > 0),$

$ S \equal{} \frac {1}{6}ab,\ V_x \equal{} \frac {\pi}{15}ab^2,\ V_y\equal{}\frac{\pi}{15}a^2b.$


trigonometry
inequalities
inequalities unsolved

prove or disprove:
$ ( \sin \frac A 2 \plus{} \sin \frac B 2 \plus{} \sin \frac C 2)(\frac 1 {\sin \frac A 2} \plus{} \frac 1 {\sin \frac B 2} \plus{} \frac 1 {\sin \frac C 2}) \ge \frac 9{\sqrt {8\sin \frac A 2\sin \frac B 2\sin \frac C 2}}$
where $ A,B,C$ are  three angles of triangle.

[quote="darij grinberg"][quote="zhaobin"]prove or disprove:
$( \sin \frac A 2+ \sin \frac B 2+ \sin \frac C 2)(\frac 1 { \sin \frac A 2}+\frac 1 { \sin \frac B 2}+\frac 1 {\sin \frac C 2}) \ge \frac 9{\sqrt{8\sin \frac A 2\sin \frac B 2\sin \frac C 2}}$
where $A,B,C$ are  three angles of triangle.[/quote]

Alas, this one isn't true. There are cases where it holds and there are cases where it doesn't, and both can be arbitrary close to A = B = C.

  Darij[/quote]
can you expain more?I can't understand :(

[quote="zhaobin"]can you expain more?I can't understand :([/quote]

I mean that
[b]1.[/b] it is wrong
and
[b]2.[/b] (apparently - I have just checked with numeric data,) for every $\epsilon>0^{\circ}$, there exists a triangle ABC whose angles are all in the interval $\left[60^{\circ}-\epsilon;\;60^{\circ}+\epsilon\right]$ for which the inequality is true, and there exists a triangle ABC whose angles are all in the interval $\left[60^{\circ}-\epsilon;\;60^{\circ}+\epsilon\right]$ for which the inequality is wrong. This means that we can't make the inequality correct by demanding that triangle ABC should be acute-angled or something like that.

  Darij

[quote="darij grinberg"][quote="zhaobin"]can you expain more?I can't understand :([/quote]

and there exists a triangle ABC whose angles are all in the interval $\left[60^{\circ}-\epsilon;\;60^{\circ}+\epsilon\right]$ for which the inequality is wrong.   Darij[/quote]
thank you, :)


integration
function
calculus
calculus computations

[quote="tariq1440"]
\[ Evaluate : \int_{0}^{1}\sqrt [3]{2x^3 \minus{} 3x^2 \minus{} x \plus{} 1 } dx
\]
[/quote]

Hi tariq1440 :) !!!

we have: $ \sqrt [3]{2x^3 \minus{} 3x^2 \minus{} x \plus{} 1} \equal{} \sqrt [3]{2}\sqrt [3]{(x \minus{} \frac {1}{2})( x \minus{} \phi) (x \minus{} 1 \plus{} \phi)}$

with $ \phi \equal{} \frac {1 \plus{} \sqrt {5}}{2}$

by changing the variable $ x \equal{} t \plus{} \frac {1}{2}$

we find that:

$ \int_{0}^{1}\sqrt [3]{2x^3 \minus{} 3x^2 \minus{} x \plus{} 1 } dx \equal{} \sqrt [3]{2}\int_{ \minus{} \frac {1}{2}}^{\frac {1}{2}}\sqrt [3]{t(t^2 \minus{} \frac {5}{4}) } dx\equal{}0$

because

$ t \rightarrow t(t^2 \minus{} \frac{5}{4})$ are odded (impair) functions

$ .......$

and thanks

yes :) Indochina you're right :) !!

if : $ f(x) \equal{} \sqrt[3]{2x^3 \minus{} 3x^2 \minus{}x \plus{}1}$

we find that $ \Omega ( \frac{1}{2} ; 0)$ are a center of symetry

the same of my integral
.....

[[/quote]

Hi tariq1440 :) !!!

we have: $ \sqrt [3]{2x^3 \minus{} 3x^2 \minus{} x \plus{} 1} \equal{} \sqrt [3]{2}\sqrt [3]{(x \minus{} \frac {1}{2})( x \minus{} \phi) (x \minus{} 1 \plus{} \phi)}$

[/quote]

how do you know this , please ?

thanks

 :lol:  :)

[quote="IndoChina"]Let $ u \equal{} 1 \minus{} x$

$ \int_{0}^{1}\root{3}\of{2x^{3} \minus{} 3x^{2} \minus{} x \plus{} 1}dx \equal{} \allowbreak \minus{} \int_{0}^{1}\root{3}\of{2u^{3} \minus{} 3u^{2} \minus{} u \plus{} 1}\,du$

$ \int_{0}^{1}\root{3}\of{2x^{3} \minus{} 3x^{2} \minus{} x \plus{} 1}dx \equal{} 0$[/quote]

very nice , thanks

[quote="tariq1440"]

[quote="mathema*"]

Hi tariq1440 :) !!!

we have: $ \sqrt [3]{2x^3 \minus{} 3x^2 \minus{} x \plus{} 1} \equal{} \sqrt [3]{2}\sqrt [3]{(x \minus{} \frac {1}{2})( x \minus{} \phi) (x \minus{} 1 \plus{} \phi)}$

[/quote]

how do you know this , please ?

thanks

 :lol:  :)[/quote]

Ok tariq !

equation $ 2x^3 \minus{} 3x^2 \minus{} x \plus{} 1 \equal{} 0$ have as oubvious solution $ x \equal{} \frac {1}{2}$ than

$ (2x^3 \minus{} 3x^2 \minus{} x \plus{} 1) \equal{} 2(x \minus{} \frac {1}{2})(x^2 \minus{} x \minus{} 1) \equal{} ....$

:)

[color=green]Moderator edit: redundant quote removed.[/color]

thanks , my friend


induction
quadratics
Gauss
modular arithmetic
number theory
prime numbers
number theory proposed

For all integers $n\geq 1$ we define $x_{n+1}=x_1^2+x_2^2+\cdots +x_n^2$, where $x_1$ is a positive integer. Find the least $x_1$ such that 2006 divides $x_{2006}$.

I think $x_1=1.$ But i am wrong.
Let y=x(x+1)(mod 1003). If y=0, then x=0 or x=-1 (mod 1003). If y=-1=x(x+1) we have $(2x+1)^2=3 (mod 1003)$. It give contradition, because $(\frac{3}{1003})=-(\frac{1003}{3})=-1$. Therefore if $x_1\not= 0,-1$, then $1003\not | x_n$ for all n>1.

You prove: if $x_1=0,-1(mod 1003)$, then  $2006|x_n$ for n>1. But you don't prove, that $x_1\not =0,-1$ don't work.

I agree that $x_{n+1}=x_n^2+x_n$ but only for $n\geq 2$ . Since from the question we can see that

$x_2=x_1^2\not =x_1+x_1^2$

@Rust : The question does say " Positive Integer " , so I think we doesnt have to prove the case for $x_1=0,-1$ .

[quote="shyong"]
$\implies x_{k-1}(x_{k-1}+1)\equiv -1\mod 59$
Here is the part I dont know how to prove but i think there is no solution for $x_{k-1}$ which satisfy this ?

Someone plz complete the "incomplete" part for me to make this solution valid :P[/quote]

You need to show that the equation
$x^2 +x +1\equiv 0 (\mod 59 )$
does not has solution in $Z_{59}$
This is equivalent to
$(2x +1)^2 = -3 (\mod 59)$

Now $(\frac{56}{59})= (\frac{7}{59}) (\frac{2}{59})^3= (\frac{59}{7})=(\frac 37)=-1$ , because $(\frac{2}{prime(8k+3)} )=-1.$

Hence -3 is not quadratic residue modulo 59 !

[quote="xirti"]We have that $x_{n+1}=x_n+x_n^2$ hence all $x_n$ are even. We will prove by a backward induction that $x_2$ must be divisible by $1003$.

$x_{n}\equiv 0 (\mod 1003)\Leftrightarrow x_{n-1}^2+x_{n-1}\equiv 0 (\mod 1003)\Leftrightarrow x_{n-1}\equiv -1$ or $x_{n-1}\equiv 0 (\mod 1003)$.

We will prove that $x_{n-1}\equiv -1$ cannot hold for $n-1\ge 2$:

$x_{n-1}=x_{n-2}^2+x_{n-2}\equiv -1 (\mod 1003)\Leftrightarrow (2x_{n-2}+1)^2\equiv -3 (\mod 1003)$.

But from quadratic reciprocity law we get that $3$ is quadratic residue $\mod 1003$, but since $1003=4\cdot 250+3$ we have that $-1$ is not a quadratic residue $(\mod 1003)$. Hence contradiction.

Since $1003\mid x_2$, we have that $x_1\equiv 0,-1 (\mod 1003)$. Hence the least value for $x_1$ is 1002, which works.[/quote]

Hy I can't get something 1003 is not a prime number so how can you apply Gauss lemma in your first line ?

[quote="shyong"]There is some part not complete since I cant prove it , but i think it is true .

I claim that $x_1=472$

From question , $x_2=x_1^2$ and $x_n=x_{n-1}(1+x_{n-1})$ for $n\geq 3$

Or $x_n=x_1^2(x_2+1)(x_3+1)\cdots (x_{n-1}+1)$ ,$n\geq 3$ .....(i)

Since any $(x_i+1,x_j+1)=1$ and $(x_i+1,x_1)=1$ , the terms are all pairwise coprime .

We also notice that $x_n$ is even for all $n\geq 3$ and $x_1^2(x_2+1)$ is even .

If $x_{2006}$ is divisible by $2006=2\cdot 17\cdot 59$ , then we know that there must be ONLY one $x_i+1$ or $x_1$ who has the factor of $17$ or $59$ .

If it is $x_k+1$ ($k\geq 3)$ who has the factor of $17$ or $59$  , then
(i) $x_k+1\equiv 0\mod 59$ $\implies x_{k-1}(x_{k-1}+1)\equiv -1\mod 59$
Here is the part I dont know how to prove but i think there is no solution for $x_{k-1}$ which satisfy this ?

if that is true , then we can say that it is either $x_2+1$ or $x_1$ who has the factor of $59$ . But $x_2+1\equiv 0\mod 59$ $\implies x_1^2\equiv -1\mod 59$ (anyway to prove no solution for this other that brute force all cases?) . Hence we conclude $x_1\equiv 0\mod 59$......(i)

(ii)As for the factor of $17$ , we do the same way and found that it must also be either $x_2+1$ or $x_1$ who has the factor of $17$ . But $x_2+1\equiv 0\mod 17$ $\implies x_1\equiv \pm 4\mod 17$ .....(ii) (another case is $x_1\equiv 0\mod 17$ but that will give us $x_1\equiv 0\mod 1003$ when combine with (ii) )

Hence by (i),(ii) , we use Chinese Remainder Theorem and conclude that either $x_1\equiv 472\mod 1003$ or $\equiv 531\mod 1003$ . So the smallest such $x_1$ is $\boxed{472}$

Someone plz complete the "incomplete" part for me to make this solution valid :P[/quote]
You're right, that is the right value. My solution is similar to xirti one, but the problem in his solution is that he forget some Ideas, this is a correct form.
[hide=" Solution"]
First, not that the proprety $x_{n+1}=x_{n}+x_{n}^2$ is true only for $n\geq 2$ and so for all $n\geq 3$, $x_n$ is even. Now let $p\in \{17,59\}$, it's then sufficient to find the smallest $x_1$ for which $p|x_{2006}$. Let $n\geq 3$ if $p|x_{n+1}$ the $p|x_n$ or $p|x_n+1$. If $p|x_{n}$ then $p|x_{n-1}^2+x_{n-1}+1$ or $p|(2x_{n-1}+1)^2+3$ which is wrong because, $-3$ is not a quadratic residue modulo $17$ or $59$ (that's a simplificated way to use reciprocity law on prime numbers), and then $p|x_{n}$, so what we have shown is that, for every integer $n\geq 3$, if $p|x_{n+1}$ consequently: $p|x_{n}$, so $p|x_3=x_1^4+x_1^2$
therefore $p|x_1^2(x_1^2+1)$. In order to minimise $x_1$, we have to find the smallest that verify one of the following cases:
[b]case 1.[/b] $17|x_1^2$ and $59|x_1^2+1$
[b]case 2.[/b] $59|x_1^2$ and $17|x_1^2+1$
The first case is obviously impossible since $-1$ isn't a quadratic residu modulo $59$.
On the second case, we put $x_1=59k$ remark that $59^2\equiv 13\pmod{17}$ so, $17|13k^2+1$ and so, $k^2\equiv 13 \pmod{17}$ which is equivalent with $k\equiv 8\pmod{17}$ or $k\equiv 9 \pmod{17}$. So $k_{min}=8$ and hence the smallest $x_1$ verifying the problem is $8\times 59$ or, $x_1=472$.
The reciproc is obvious.
[/hide]

I find that[b] x_1=118 [/b]satisfies the problem, (since x-2=118*119=7*2006) thus all x_i greater than 1 is divisible by 2006. for this reason consider the first index d such that 17 or 59 ( no matter because they have [b]reminder -1 while dividing by 6[/b])divide  x_d . thus we must have x_d =-1 (mod 17 or 59) and (x_d-1)^2 +x_d-1+1=0 (mod 17 or 59) but all of primes dividing x^2 +x+1 is of the form 6k+1. contradiction. thus we must have this cases:[b] x_1 =-1(mod 59,17) x_1=0 (mod 17) and -1 (mod 59) and x_1=-1(mod 17) and x_1=0 (mod 59)[/b]. after some calculations we can see that 118 obtained from the third cases.

I think you made a mistake  :)  $ x_{2}=x_{1}^{2} $  :wink:


Is there going to be Intermediate Num Theo or something of that nature?

We expect to offer that next year some time.


We were asked to measure the density of water. however, when the water was transfered to the graduated cylinder (to obtain the volume) we were asked to compute the density of water using the mass and weight excluding the 1mL...why is this so? thanks

Excluding which 1 mL?  And density is based on mass, not weight.

ok. let me explain to you this. We were given an unknown. We were asked to compute the density of the liquid. The instructions are as follows: obtain 1 mL of that liquid. record the mass of the liquid. Now, add 1mL to the 1mL liquid (in other words, now, there are 2 mL of the liquid). Record the new mass of the two mL liquid. Compute the density of the liquid by excluding the initial 1mL's mass and volume.

So like, if this is the data:
                  Weight of graduated cylinder + 1mL           Weight of graduated cylinder + 2mL   Final Volume
                             30.67                                                                 31.57                        2

So, the density of the liquid (Excluding the initial 1mL's volume and mass) is: (31.57-30.67)/(2-1)=0.9

in other words, why couldnt we just have weighed the initial 1mL and use the mass and volume directly to get the density? that's my questions...

I'm not sure if I completely understand it yet.  In order to measure the density, essentially all you have to do is mass a graduated cylinder, pour out 1 mL, and mass the combination.  That should be it.

oh never mind. someone told me that the base of the qraduated cylinders were not standardized. that's why


Find n0 \in N* such that for all n \in N this ineq holds:

284^n>n^24

I tried using the function 284^x - x^24 and to aproximate it's solutions using Newton's methods.. but i'm not sure it is good...

i'm sorry for deleting Kuba's messege, but this problem is in a ROmanian Gazzette and it's too soon to answer it!

I also locked the topic!


i realy want to see your proof for these:
$a,b,c$ are real numbers prove that:
1)$(-a)(-b)=ab$
2)$(-a)(b)=-ab$
3)$\dfrac{a}{a}=1$
4)$a \times \b=0 \rightarrow a=0 \bigvee b=0$
5)a(b+c)=a.b+ac
there is more but i think it is enough for now.please dont say these are obvious ,please prove these even they are obvious and send me your proof. :D

come on boys,you can do it and you will like it,so dint waist the time and prove these.i promisse you will lose nothing.

Good morning,

http://smard.cqu.edu.au/Database/Junior/Number/Integers/jn_in002.txt

sums it up nicely.

-- Owlbert

Please pay greater attention to the level of each forum.  This isn't the first time I've explained that you are posting inappropriately difficult material in the middle school forums.  If you continue, I'll have to take away your ability to post in the middle school area.

i though this problems are easy and so i possed it here if i knew they are not i won't post it here.i am realy sorry.and thanks you.
 :D


If the last three digits of $1978^{m}$ & $1978^{n}$ are the same with $1\leq m<n$ , then find the least value of $m+n$.

This was one of the contest problems on IMO '78. Thus I'm sure this was posted before :)


vector
trigonometry
linear algebra
linear algebra unsolved

Let $ A \in M_2(\mathbb{Z})$ such that there exists $ n \in \mathbb{N}^{*}$, $ (n,6)\equal{}1$, and $ A^n\equal{}I_2$. Prove that $ A\equal{}I_2$.

From $ A^n\equal{}I$ follows that if $ \lambda , \overline{\lambda} \in\mathbb{C}$ are eigenvalues of $ A$ then they must be roots of the equation $ z^n\equal{}1$.Suppose that $ \lambda\equal{} \cos \frac{2k\pi}{n} \plus{}i \sin \frac{2k\pi}{n}$ and since $ tr(A)\equal{}\lambda \plus{}\overline{\lambda} \equal{}2\cos \frac{2k\pi}{n} \in \mathbb{Z}$ we must have $ 2\cos \frac{2k\pi}{n} \in \{\pm 1,\pm 2, 0\}$.And since $ (n,6)\equal{}1$ it follows that $ \cos \frac{2k\pi}{n}$ can only be equal to $ 1$. So both eigenvalues of $ A$ are equal to $ 1$.Therefore, from Cayley-Hamilton theorem we get that $ A^2\equal{}2A\minus{}I$ , and so on, inductively we get that $ A^n\equal{}nA\minus{}(n\minus{}1)I$ which from the hypothesis we know that it is equal to $ I$. So $ A^n\equal{}nA\minus{}(n\minus{}1)I\equal{}I$ $ \Rightarrow$ $ A\equal{}I$.


trigonometry
geometry
rectangle
circumcircle
geometry unsolved

Let ABC be a right triangle with right angle at A has altitude AH > Let N,M be the midpoints AB,AC, respectively. The two lines BM and AH intersect at D and the two lines CN and AH intersect at E .Prove that $ \hat{ACD}\equal{}\hat{ABE}$

[size=134][color=darkred][b]Nice problem ![/b][/color][/size]

[quote="tranquoc"] Let $ ABC$ be a $ A$ - right triangle . Denote $ \left\|\begin{array}{c} H\in BC\ ,\ AH\perp BC \\
 \\
N\in [AB]\ ,\ NA \equal{} NB \\
 \\
M\in [AC]\ ,\ MA \equal{} MC\end{array}\right\|$ and $ \left\|\begin{array}{c} E\in CN\cap AH \\
 \\
D\in BM\cap AH\end{array}\right\|$ . Prove that $ \widehat{ACD}\equiv\widehat{ABE}$ . [/quote]
Prove easily that $ \boxed {\ \tan\widehat {ACD} \equal{} \tan\widehat {ABE} \equal{} \frac {bc}{a^2}\le \frac 12\ }$ .

We denote the points $ K\equiv BE\cap ON$ and $ L\equiv BC\cap AK,$ where $ O$ is the midpoint of the side-segment $ BC$ of the given triangle $ \bigtriangleup ABC.$

Because of the line segment $ ON,$ is the midperpendicular of $ AB,$ we have that $ \angle ABE \equal{} \angle BAL$ and so, it is enough to prove that $ \frac {LH}{LB} \equal{} \frac {DH}{DA},$
because of the similar right triangles $ \bigtriangleup HAB$ and $ \bigtriangleup HCA,$ with $ \angle HAB \equal{} \angle HCA.$

Applying the [b][size=100]Menelaus theorem[/size][/b], we can say that:

From the right triangle $ \bigtriangleup HBE,$ with transversal $ AKL$ $ \Longrightarrow$ $ \frac {LH}{LB}\cdot \frac {KB}{KE}\cdot \frac {AE}{AH} \equal{} 1$ $ \Longrightarrow$ $ \frac {LH}{LB} \equal{} \frac {KE}{KB}\cdot \frac {AH}{AE}$ $ ,(1)$

From the triangle $ \bigtriangleup BEC,$ with transversal $ OKN$ $ \Longrightarrow$ $ \frac {KE}{KB}\cdot \frac {OB}{OC}\cdot \frac {NC}{NE} \equal{} 1$ $ \Longrightarrow$ $ \frac {KE}{KB} \equal{} \frac {NE}{NC}$ $ ,(2)$

From the right triangle $ \bigtriangleup HEC,$ with transversal $ ANB$ $ \Longrightarrow$ $ \frac {AH}{AE}\cdot \frac {NE}{NC}\cdot \frac {BC}{BH} \equal{} 1$ $ \Longrightarrow$ $ \frac {AH}{AE} \equal{} \frac {NC}{NE}\cdot \frac {BH}{BC}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {LH}{LB} \equal{} \frac {BH}{BC}$ $ ,(4)$

From the right triangle $ \bigtriangleup HAC,$ with transversal $ BDM$ $ \Longrightarrow$ $ \frac {DH}{DA}\cdot \frac {MA}{MC}\cdot \frac {BC}{BH} \equal{} 1$ $ \Longrightarrow$ $ \frac {DH}{DA} \equal{} \frac {BH}{BC}$ $ ,(5)$

From $ (4),$ $ (5)$ $ \Longrightarrow$ $ \frac {LH}{LB} \equal{} \frac {DH}{DA}$ $ ,(6)$ and the proof is completed.

Kostas Vittas.

[quote="tranquoc"] [color=darkred]Let $ ABC$ be a $ A$ - right triangle . Denote $ \left\|\begin{array}{c} H\in BC\ ,\ AH\perp BC \\
 \\
N\in [AB]\ ,\ NA \equal{} NB \\
 \\
M\in [AC]\ ,\ MA \equal{} MC\end{array}\right\|$ and $ \left\|\begin{array}{c} E\in CN\cap AH \\
 \\
D\in BM\cap AH\end{array}\right\|$ . Prove that $ \widehat{ACD}\equiv\widehat{ABE}$ .[/color] [/quote]

[color=darkblue][b][u]Proof I (synthetic).[/u][/b] Let $ \left\|\begin{array}{ccc} P\in AB\cap CD \\
 \\
R\in AC\cap BE\end{array}\right\|$ and $ \left\|\begin{array}{c} AH \equal{} h \\
 \\
\phi \equal{} m(\widehat {ACD}) \\
 \\
\psi \equal{} m(\widehat {ABE})\end{array}\right\|$ . Since $ D$ , $ E$ belong to the medians $ BM$ , $ CN$ respectively obtain :

$ \boxed {\ \left\|\begin{array}{c} HP\parallel AC \implies\ HP\perp AB\implies PHA\sim ABC\implies\frac {AP}{CA} \equal{} \frac {AH}{CB}\implies\tan\phi \equal{} \frac {h}{a} \\
 \\
HR\parallel AB \implies HR\perp AC\implies RHA\sim ACB\implies\frac {AR}{BA} \equal{} \frac {AH}{BC}\implies\tan\psi \equal{} \frac {h}{a}\end{array}\right\|\implies\widehat{ACD}\equiv\widehat {ABE}\ }$ .

[b][u]Proof II (metric).[/u][/b] Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {BDM}$ and the triangle $ ACH$ :

$ \frac {BH}{BC}\cdot\frac {MC}{MA}\cdot\frac {DA}{DH} \equal{} 1\implies$ $ \boxed {\ \frac {DA}{DH} \equal{} \frac {a^2}{c^2}\ }$ .  Apply an well-known relation to the cevian $ [CD$ in the triangle $ ACH$ :

$ \frac {DA}{DH} \equal{} \frac {CA}{CH}\cdot\frac {\sin\widehat {DCA}}{\sin\widehat {DCH}} \equal{}$ $ \frac {a\sin\phi}{b\sin (C \minus{} \phi)}$ $ \implies$ $ \boxed {\ \frac {DA}{DH} \equal{} \frac {a^2\tan\phi}{b(c \minus{} b\tan\phi )}\ }$ . Thus, $ \frac {1}{c^2} \equal{} \frac {\tan\phi}{b(c \minus{} b\tan\phi )}$ $ \implies$ $ \tan\phi \equal{} \frac {bc}{b^2 \plus{} c^2}$ .

Analogously obtain $ \tan\psi \equal{} \frac {bc}{b^2 \plus{} c^2}$ . In conclusion, $ \widehat{ACD}\equiv\widehat{ABE}$ . Shortly, $ \tan\phi \equal{} \tan \psi \equal{} \frac {bc}{a^2} \equal{} \frac {h_a}{a}$ .[/color]

[color=darkred][b][u]Remark.[/u][/b] Construct, outside of $ \triangle ABC$ , the rectangle $ BCST$ , where $ BT \equal{} CS \equal{} h$ . Denote the circumcircle $ \Gamma$ of this rectangle.

Prove easily that $ S\in PD$ , $ T\in RD$ , $ \widehat {CBS}\equiv\widehat {ACD}\equiv\widehat {ABE}$ , i.e. the points $ P$ , $ R$ belong to the circle $ \Gamma$ .

If $ \{X,Y\} \equal{} AH\cap \Gamma$ , then $ XY \equal{} h\sqrt 3$ and $ AX \equal{} \frac {\sqrt 7 \minus{} \sqrt 3}{2}\cdot h$ .[/color]

Here is a solution similar to [b]Virgil Nicula's[/b], but differs at the end:
[hide="Solution"]
Extend $ CD$ and $ BE$ to meet $ AB$ and $ AC$ at $ Y$ and $ X$ respectively.  Then, Ceva's Theorem gives that $ \frac {YB}{AY}\cdot \frac {AM}{MC}\cdot \frac {CH}{BH} \equal{} 1$.  Since $ AM\equal{}MC$, we have that $ \frac {YB}{AY} \equal{} \frac {BH}{CH}$ so $ YH\parallel AC$.  Similarly, $ HX\parallel AB$.  Now, this means that since $ AC\perp AB$, we have that $ HY\perp AB$.  Similarly, $ HX\perp AC$.  Since $ \angle AHB\equal{}90$ and $ HY\perp AB$, we have that $ \angle YHA \equal{} \angle ABC$.  Yet, $ \angle HYA \plus{} \angle HXA \equal{} 90 \plus{} 90 \equal{} 180$, so $ YHXA$ is cyclic, which implies that $ \angle YHA \equal{} \angle YXA$.  This means that $ \angle YBC \equal{} \angle YXA$, so $ BYXC$ is cyclic, so $ \angle ABE \equal{} \angle YBX \equal{} \angle YCX \equal{} \angle ACD$. [/hide]


When creating user-defined functions, which should be used - \newcommand or \def?

What are the differences between these two commands (apart from \newcommand having the restriction that the name of your function can't begin with "end")? Are there specific cases where one should be used over the other?

Thanks.

\def is a TeX primitive whilst \newcommand is a \LaTeX macro which, amongst other things, checks for clashes of names with other commands and will allow commands like \par in their definition. Thus \newcommand calls \def at some point but does other things, so is preferred if possible.


geometry
3D geometry
analytic geometry
geometric transformation
rotation
reflection

[hide]

I cannot find any set formula for unfolded cubes, but I have noticed two different trends for acceptable arrangements:

The first is based off the traditional 'cross' arrangement; given an arrangement of 5 squares that form a 'T' (three high, a cross bar three wide, with a shared square at the top), the 6th square can be placed anywhere on the top or bottom of the 'T'.

For example:

0

000

-0

-0

(the hyphens are place holders)

The second comes from the idea that each square is involved in forming a corner.  If each square is part of a three-square right angle (a square can be part of more than one), provided two adjacent groups don't mirror each other to form a 'u', then a square can be formed.

For example:

-0

000      works

--00

but

-0

000      doesn't

0  0

[/hide]

I am definitely open for someone to come up with a more generalized explanation.

[quote="Ravi B"]Have you ever seen those problems about folding a group of squares into a cube?  For example, in the diagram below, the first group of squares can be folded into a cube, whereas the second group of squares cannot.  (Each O represents a square.  Sorry for the crude picture.)

[code]
OO
 OO
  OO

 O
OO
 OOO
[/code]

Anyway, I think the usual way of solving these problems is to try to visualize the folding in 3 dimensions.  Because I have trouble visualizing in 3D, and for curiousity's sake, I was wondering whether there is a method to figure out the answer by sticking to 2 dimensions.  Is there some geometric or graph-theoretic principle that would allow us to determine when a group of squares can be folded into a cube?[/quote]

We mathematicians regularly work in four and more dimensions and most of us cannot visualize in 4D. We maneuver by coordinates instead.

The following solution to your question is longer and more tedious than required. There are simpler ways to tell if a pattern of squares can be folded into a cube. But this is a chance to teach the power of coordinate systems.

If a flat pattern of squares were folded into a cube, one of those squares would be the bottom face of the cube. So imagine that the folded cube is in Cartesion space with its corners at (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), and (1,1,1). The bottom face has the corners (0,0,0), (0,1,0), (1,0,0), and (1,1,0), which I will abbreviate as 000, 010, 100, and 110. The edge between 000 and 100 is the set {([i]x[/i],0,0) | 0 :le: [i]x[/i]  :le: 1 }, which I will abbreviate as x00. The edge between 100 and 110 is the set {(1,[i]y[/i],0) | 0 :le: [i]y[/i]  :le: 1 }, which I will abbreviate as 1y0. And so on. The square itself I will call xy0.

[tex]\begin{picture}(100,80)
\put(20,10){\framebox(60,60){}}
\put(0,0){$000$}
\put(81,0){$100$}
\put(81,72){$110$}
\put(0,72){$010$}
\put(40,37){$xy0$}
\put(40,0){$x00$}
\put(81,37){$1y0$}
\put(40,72){$x10$}
\put(0,37){$0y0$}
\end{picture}[/tex]

Now suppose we add another square to the pattern.

[tex]\begin{picture}(160,80)
\put(20,10){\framebox(60,60){}}
\put(0,0){$000$}
\put(71,0){$100$}
\put(71,72){$110$}
\put(0,72){$010$}
\put(40,37){$xy0$}
\put(40,0){$x00$}
\put(81,37){$1y0$}
\put(40,72){$x10$}
\put(0,37){$0y0$}
\put(80,10){\framebox(60,60){}}
\end{picture}[/tex]

The edge in the new square that touches vertex 100 will be perpendicular to the original square xy0 when the pattern is folded. So it must go in the z direction. Thus, the third digit in its label is z. The other two digits we can take from the point 100, so the edge is 10z.

Likewise the top edge of the new square starts at 110 and points in the z direction after folding, so it is 11z.

The other end of 10z must be 101 and the other end of 11z must be 111. The edge between 101 and 111 changes in the y coordinate, so it is 1y1. This makes it parallel to edge 1y0, which makes sense because opposite sides of a square are parallel.

Finally, the whole square has edges in the y direction and the z direction, so it must be in the yz-plane. The x coordinate of all its vertices is 1, so the square is 1yz.

[tex]\begin{picture}(160,80)
\put(20,10){\framebox(60,60){}}
\put(0,0){$000$}
\put(71,0){$100$}
\put(71,72){$110$}
\put(0,72){$010$}
\put(40,37){$xy0$}
\put(40,0){$x00$}
\put(81,37){$1y0$}
\put(40,72){$x10$}
\put(0,37){$0y0$}
\put(80,10){\framebox(60,60){}}
\put(141,0){$101$}
\put(141,72){$111$}
\put(105,0){$10z$}
\put(141,37){$1y1$}
\put(105,72){$11z$}
\put(105,37){$1yz$}
\end{picture}[/tex]

Let us add another squares above the second square. I already labeled it in the diagram below. I had to leave out some old labels, because apparently the TeX simulator in this forum has a limit in how many picture elements it can handle.

[tex]\begin{picture}(160,140)
\put(20,10){\framebox(60,60){}}
\put(62,72){$110$}
\put(0,72){$010$}
\put(40,37){$xy0$}
\put(40,72){$x10$}
\put(81,37){$1y0$}
\put(80,10){\framebox(60,60){}}
\put(141,0){$101$}
\put(141,72){$111$}
\put(105,0){$10z$}
\put(141,37){$1y1$}
\put(105,72){$11z$}
\put(105,37){$1yz$}
\put(80,70){\framebox(60,60){}}
\put(141,132){$011$}
\put(141,97){$x11$}
\put(105,132){$01z$}
\put(105,97){$x1z$}
\put(62,132){$010$}
\put(61,97){$x10$}
\end{picture}[/tex]

In this diagram, two edges and two vertices have the same label. But that is fine, because when the sides are folded, those edges and vertices are going to be glued together into a single edge and vertex.

We can see that it is impossible to now add a new square above the original square xy0. That square would have two perpendicular edges both labeled x10. Since x-direction edges are parallel, they cannot be perpendicular. Also, that new square would end up labeled x1z (if we labeled via its bottom edge) or xy0 (if we labeled via its right edge), and we already have squares labeled x1z and xy0. Repeating labels on edges is fine, but repeating labels on squares means that those squares will end up on the same face when the pattern is folded into a cube.

If we added three squares to the left of the second square, for a total of six squares, we would have enough squares to make a cube. But does that pattern fold into a cube? The diagram below shows the labels on the squares. No square folds over onto face x0z, so this cannot form a cube.

[tex]\begin{picture}(125,50)
\put(0,0){\framebox(25,25){}}
\put(25,0){\framebox(25,25){}}
\put(50,0){\framebox(25,25){}}
\put(75,0){\framebox(25,25){}}
\put(100,0){\framebox(25,25){}}
\put(25,25){\framebox(25,25){}}
\put(3,8){$xy0$}
\put(28,8){$1yz$}
\put(53,8){$xy1$}
\put(78,8){$0yz$}
\put(103,8){$xy0$}
\put(28,33){$x1z$}
\end{picture}[/tex]

On the other hand, if the extra xy0 square on the left is moved to beneath the square it touches, it becomes labeled as the x0z square that we need. The pattern below will fold into a cube.

[tex]\begin{picture}(100,75)
\put(0,25){\framebox(25,25){}}
\put(25,25){\framebox(25,25){}}
\put(50,25){\framebox(25,25){}}
\put(75,25){\framebox(25,25){}}
\put(25,50){\framebox(25,25){}}
\put(75,0){\framebox(25,25){}}
\put(3,33){$xy0$}
\put(28,33){$1yz$}
\put(53,33){$xy1$}
\put(78,33){$0yz$}
\put(28,58){$x1z$}
\put(78,8){$x0z$}
\end{picture}[/tex]

The fastest method of determining whether a pattern of six squares will fold into a cube is to memorize the ones that do. There are only eleven of them, not counting rotations and reflections (I checked rigorously). I leave it as a challenge to the rest of you to find them all.

Erin Schram

Thanks for the explanation and the pictures!  I will think more about your coordinate method.


I got an okay score. At least I beat my goal of qualifying for AIME by a positive multiple of 10.

Hmm Yongyi I am sure your score was "okay."

I got 144. Yay for failing similar triangles on #17.

I don't deserve to talk about my score.

So I won't :(.

Well, either do I^

I feel so disappointed...now I need like a 9 on the AIME inorder to pass because I guessed away 7.5 points !!!!!!!!!!!!!!!!!!

Sad thing was, if I had done 22 in the time limit (I did it in lunch in my head), I wouldve not guessed on the 4 remaining and wouldve gotten 132.

Which woulda beaten Lyndon!!  Too bad I didnt and only got a 120...127.5 would sure look nice.

Apparently Lyndon epic phailed compared to last year?

xpmath, I coordinate bashed number 17 :D

138 for me (trivialized 22 even though it was epicly trivial, put 1/6 and failed number 25).  My brother got number 25 right because he thought 2^7=64  :|

I hope 7 on AIME will qualify for USAMO.

Coordinate bashing on #17 should be automatic disqualification from AIME. :P

Seriously, I solved it in like 10 seconds with similar triangles.

YES WELL SOME OF US AREN'T AS GOOD AS YOU ARE.

Ahem. So apparently this AMC 10 was relatively easy, according to everybody? Then again I thought last year's 10A was around the same difficulty level, so...

Darn I still haven't gotten rid of my spamming urge. I'll go back to my own forum now.

i took AMC 12A. and phailed. i got like a 130.5

114. I hope to do better.  Actually, I will do better!!


function
complex analysis
complex analysis unsolved

$ K\equal{}\{z\in\mathbb{C}: |z|\le 1\},\quad T\equal{}\{z\in\mathbb{C}: |z|\equal{}1\},\quad |a|<1$

define $ h: K\to C$ by $ h(z)\equal{}\frac{z\minus{}a}{1\minus{}\overline{a}z}$

find $ \min_{z\in T}|h(z)|$

Note that $ h$ is a Mobious Transformation, which maps circles to circles and quick calculation shows
\[ |h(e^{i\theta})| \equal{} \left|\frac {e^{i\theta} \minus{} a}{1 \minus{} \overline{a}e^{i\theta}}\right| \equal{} \left|\frac {e^{i\theta} \minus{} a}{e^{ \minus{} i\theta} \minus{} \overline{a}}\right| \equal{} 1.
\]
Hence $ h(T) \equal{} T$. It can be further shown that the maps $ \phi_a(z) : \equal{} \frac {z \minus{} a}{1 \minus{} \overline{a}z}$ for $ |a| < 1$ are all of the conformal maps of the open unit disk onto itself (up to rotation).


function
quadratics
calculus
algebra
quadratic formula
algebra unsolved

1. Maria did a study of babysitting tares in her neighborhood over the last three years and discovered that the function b(t)=6.25(1.2)^t represents the change in pricing, where t is the number of years since 2000. What was the babysitting cost per hour in Maria's neighborhood in 2000?
[color=darkred]My answer was: first, b=6.25(1.2)^3=10.8 --> pricing change. But i don't know how to find her cost per hour?[/color]

2. An archer shoots an arrow into the air such that its hieght at any time, t, is given by the function h(t)=-16t^2+kt+3. If the maximum height of the arrow occurs at time t=4, what is the value of k? [color=darkred] I totally have no idea.[/color]

Thanks a lot for answering!

This question should be posted in a different forum, but I will try to answer them anyway.

For your first part, I might not understand correctly, but I would believe that the question is just asking for $6.25\cdot (1.2)^{0}=6.25$ as the answer.  I don't think there is enough info for the question to be asking anything else....

The second part is easy: the apex of a quadratic occurs at $\frac{-b}{2a}$ (just look at the quadratic formula, or use some differential calculus), so we must have $\frac{-k}{2\cdot (-16)}=4$ which means $k=128$.


geometry
3D geometry
tetrahedron
function
Putnam

Is there any generalization of Pick's Theorem to 3+ dimensions?

Consider the tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), and (1,1,k) for any integer k. Each of these has four vertex lattice points, no edge or face or interior lattice points, but they all have different volume. So volume can't be a function of number of lattice points.

Maybe there's one that has to do with faces passing through "lattice squares" (I just made that term up; I mean squares of side 1 that have all four vertices as lattice points) as well as lattice points?

It would probably be more like lattice lines.  Lattice points are 0 dimensional, so you'd want something 1 dimensional.

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

Though I doubt that will be useful for any math contests...

Cool.  I was just curious if such a formula existed.  I wasn't worried about contests.

There is a preprint of a book on this subject at [url]http://math.sfsu.edu/beck/papers/ccd.pdf[/url].  It should be of interest to members of this forum, and even refers to a problem from a past Putnam contest