vector
linear algebra
matrix
combinatorics unsolved
combinatorics

Let $k$ and $N$ be positive real numbers which satisfy $k\leq N$. For $1\leq i \leq k$, there are subsets $A_i$ of $\{1,2,3,\ldots,N\}$ that satisfy the following property.

For arbitrary subset of $\{ i_1, i_2, \ldots , i_s \} \subset \{ 1, 2, 3, \ldots, k \} $, $A_{i_1} \triangle A_{i_2}   \triangle ... \triangle A_{i_s}$ is not an empty set.

Show that a subset $\{ j_1, j_2, .. ,j_t \} \subset \{ 1, 2, ... ,k \} $ exist that satisfies $n(A_{j_1} \triangle A_{j_2} \triangle \cdots \triangle A_{j_t})  \geq  k$. ($A \triangle B=A \cup B-A \cap B$)

Associate each set $A_i$ with its characteristic vector (a row vector with $n$ components, s.t. vector $i$ has $1$ on its $j$'th component if $j\in A_i$ and $0$ otherwise). We regard these $k$ vectors as having elements from $\mathbb Z_2$.

The condition tells us that our $k$ vectors are linearly independent over $\mathbb Z_2$. The matrix having the $k$ vectors as rows has rank $k$, which means that we can find $k$ columns s.t. the corresponding $k\times k$ matrix is invertible.

We now have to solve the following problem: having $k$ vectors with $k$ components each, forming a linearly independent system over $\mathbb Z_2$, show that we can find a subset whose sum has $k$ components equal to $1$. This is clear, since the vectors are linearly independent and we have as many as the dimension of the space we're looking at (which is now $\mathbb Z_2^k$), so every vector with $k$ elements in $\mathbb Z_2$ can be obtained as a sum of vectors in our set.

[quote="lightrhee"]
By the way, how do you type "not equal to" sign with latex?(I'm not familar with Latex yet)[/quote]

I would say \neq between two $.  :)

Pierre.

For $I\subseteq\{1,\cdots,k\}$, define $A_I = \triangle_{i\in I} A_i$. ($A_\emptyset = \emptyset$)
For any $I\neq J\subseteq\{1,\cdots,k\}$, we have
$$A_I \triangle A_J = A_{I\triangle J} \neq \emptyset \Longrightarrow A_I \neq A_J$$
[b]Claim[/b] $U$ is a finite set. $\mathcal{P}(U)$ with binary operation $\triangle$ is an abelian group and $\mathcal{A}$ is a subgroup. Then there exists $A\in \mathcal{A}$ such that $2^{|A|}\geq |\mathcal{A}|$.
pf. Since $\mathcal{A}$ divide $|\mathcal{P}(U)|=2^{|U|}$, $|\mathcal{A}|=2^m$ for some integer $0\leq m\leq |U|$.
Let's prove by induction on $|U|$. If $U=\emptyset$, then it's obviously true.
Suppose the claim is true for $|U|=n-1$. Let's prove the case $|U|=n$.
If $m=n$, then $\mathcal{A}=\mathcal{P}(U)$ so $U\in\mathcal{A}$.
If $m>n$, then there exists $x\in U$ such that $\{x\}\not\in\mathcal{A}$. If $A-\{x\} = B-\{x\}$ for distinct $A,B\in\mathcal{A}$, then $A\triangle B = \{x\}$. Contradiction! Hence, $\mathcal{A}'=\{A-\{x\} : A\in\mathcal{A}\}$ is a subgroup of $\mathcal{P}(U- \{x\})$ and $|\mathcal{A}'| = |\mathcal{A}|$. Proof is done by induction hypothesis. $\qquad\blacksquare$

Set $U=\{1,\cdots,N\}$, $\mathcal{A}=\{A_I:I\subseteq\{1,\cdots,k\}\}$ for the claim. There exists $I\subseteq\{1,\cdots,k\}$ such that $2^{|A_I|} \geq |\mathcal{A}| = 2^k$.


What is $ mod$. I have seen this in many posts . Can you explain please.

AoPS volume one has a clear explanation of mods, so you might want to check that out.
a brief explanation:

for instance counting in modulo 4 would be like this 0,1,2,3,0,1,2,3,0,1,2,3...

for example, if you are asked to find 20 mod 4, you could always just count in that manner, or

find the largest number that when multiplied by 4 will give a result of 20 or less. Then you subtract what you got from 20, and that is the answer.
In this case; 4*5 = 20 and 20-20 = 0

so 20 mod 4 equals 0

Note that 20 is a multiple of 4, any number if a multiple of the mod you count in will be 0.
(we also use $ \equiv$ instead of = when refering to mods, in this case: $ 20 \equiv 0 (mod 4)$

edit: Check this article in the aopswiki for more information
http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction

so mods are much differnt than bases?

because I learned that 20 base 4 is

0110

Mod is basically
The remainder when a number is divided by another number.
So like
$ 12 \equiv 2\mbox{ (mod 5)}$
This means that
When you divide 12 by 5, you get a remainder of 2.
And that's it.

Then explain how $ 12\equiv \minus{} 3\mbox{ (mod 5)}$.

Anyways, I suppose you could say $ a\equiv b\mbox{ (mod m)}$ if $ m|(a \minus{} b)$.

Fine. It's a number that can be created by subtracting by another number x times.
When x can be anything.
Happy?
 :mad:

How about an easier explanation.  For integers $ a\text{, }b\text{, and }c$, saying $ a\equiv b\: (\text{mod }c)$ is equivalent to saying $ a\minus{}b\equal{}ck$ for an integer $ k$.

[quote="zolojetto"]I understand the concept but, but how do u use mods???[/quote]
Most common way would be to find the last x digits of a big number.

Many number theory problems are made easier using mods.  Like if you were asked to find a solution $ (a,b)$ so that when $ a^{2}\plus{}b^{2}$ is divided by 4, the remainder is 3.  If you know mods, you can very easily tell this problem has no solutions

Mods, like algebra, were created so we can more easily 'see' a problem in terms of numbers. Other than that you could do all your problems without them if you wished.

To Alanchou:

Why aren't there any solutions? For example, if $ a^2$ + $ b^2$ = 15, then wouldn't the remainder be 3 when divided by 4. Maybe you mean integeric solutions?

Yeah, seeing as it's number theory, integer solutions are what we are looking for.


Hi all,
I wondering what scores in different states were. What was needed to make nats? Thanks

4th place written in CA had a 45. (And everyone was like "OMG YOU FAIL.")

Wow..where I live I got a 44..AND I GOT FIRST!!!
I feel sorry for the ppl in CA.... :(

[quote="pythag011"][quote="modularmarc101"]1st place in Puerto Rico 26 points, yeah not really as hard to win as in the US.[/quote]

And 4th place?[/quote]

Umm a 16 i think...

[quote="USAMO13"]Ppl there are poor, and therefore cannot afford education.....
26 is a good score for them![/quote]

Where do you live?

[quote="USAMO13"]Ppl there are poor, and therefore cannot afford education.....
26 is a good score for them![/quote]

Discrimination is bad.

People were laughing at the fact that they got low scores..so i supported them...thats not called discrimination..
@AIME15...the next time you post something...make it useful l please. I would really appreciate it.

Sorry, I don't see how telling someone to stop discriminating against a territory is not as useful as someone saying "People can't afford education in Puerto Rico".

Also, looking at your posts, you seem to double-post and enjoy flaming a certain nikeballa96.

I'm not gonna say anything cuz I don't want my account to be deleted or anything.

MI: 44,42,39,37

there was a 38 but he got pwned in cd.
our skill is actually a lot higher than our scores would suggest...i'm the 44, btw.

man I was horrible

I got a 19.

I told my coach I was thinking my score was a 35.

[quote="echen"]Yea, I heard a lot of people boasting that they had like a 25 on sprint and stuff and were going to get like a 40 but ended up getting like a 20 overall.[/quote]

i know. and then the people who said that they had two wrong "as of now" but said that they rushed and were bound to make more stupid mistakes got 44s. whoopdie doo =].

[quote="Temperal"]Yuo Chen got 23 and 22nd place.[/quote
hahahaah!  :rotfl:  :rotfl:  :rotfl:
aren't typos wonderful? it is so serendipity that there actually was someone named yuo! so funny... btw, did you memorize everyone's score and place, because i don't see how you could have known that... :wink:

[quote="nikeballa96"][quote="Temperal"][quote="jimmyw8888"]Oklahoma's top scores were 43,37,31,29. Big gap... :|[/quote]

whoa, I guess you win in terms of that. nebraska still had a bigger gap between first and second, though.[/quote]

assuming you got first, what was the gap between yuo/second?[/quote]

Well Oklahoma is really bad.

West Virginia was
1st - 46   BEN TAYLOR
2nd - 38
3rd - 35 I think
4th - 30 something
5th - 30 something
6th - 29
7th - 25
12th - 19
13th-18

I think that's right and if you know what rank I got you know how bad I failed.....
I can't be sure about Ben, but my teacher said "I'm pretty sure Ben got a perfect".

alright, so im gonn take this time to show that benjamin taylor should be in the conversation for the national champion.  although many people got 46's, this 7th grader was the only one to make usamo from our state.  so, yeah, go wv @ mc...


The units digit of $3^{1001}7^{1002}13^{1003}$ is

$ \textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9 $

[hide="Answer"]$3^{1001}\equiv 3\bmod{10}$, $7^{1002}\equiv 9\bmod{10}$, and $13^{1003}\equiv 7\bmod{10}$, so then $3\cdot 9\cdot 7\equiv 9\bmod{10}\Rightarrow \boxed{E}$.[/hide]

Everyone has a different solution, and they're all cool! :D

Theres also the non-mod solution that requires looking at the patterns of the repeating powers units digit.  :lol:

That's what most of us did... we just used $\bmod{10}$ so as not to have to explain that.

yes.... i was being sarcastic  (  :lol:  - sarcasm indicator )

Notice the patterns:

Powers of $3$ go like this: $3,9,7,1$
Powers of $7$ go like this: $7,9,3,1$
Powers of $13$ go like this: $3,9,7,1$

So knowing that we can see that:
$3^{1001} \Longrightarrow 3$
$7^{1002} \Longrightarrow 9$
$13^{1003} \Longrightarrow 7$

So therefore the $3 \cdot 9 \cdot 7 \Longrightarrow \boxed{9}$

13 was easy because its units digit was the same as 3.

[quote="mathwizarddude"]How do we prove that $ n^{a} \equiv n^{a \mod 4} \mod 10$?[/quote]

$ \varphi(10)\equal{}4$, so by Euler's totient theorem, $ n^{4}\equiv 1\pmod{10}$. Therefore, if we have $ n^a$, and $ a\equal{}4q\plus{}r$, $ n^{4q}\equiv 1\pmod {10}$, so $ n^{a}\equiv n^{r}\equal{}n^{a\mod 4}\pmod{10}$.

This only applies to $ n$ coprime to $ 10$.

If $ a \ge 1$, it applies to all $ n$.  (This is easy to see by CRT.)


modular arithmetic
algorithm
algebra
polynomial
greatest common divisor
number theory
Euclidean algorithm

[hide]Because $n+10$ is a factor of $n^3+10^3$, we think to add "zero" to the given expression. $n^3+100=n^3+10^3-900=(n+10)(n^2-10n+100)-900$. $(n+10)(n^2-10n+100)$ is always divisible by $n+10$, so we need the largest value of $n$ such that $900$ is divisble by $n+10$. Thus, our answer is $n=\boxed{890}$.
[/hide]

[hide]Using synthetic division, we get that $\frac{n^3+100}{n+10} = n^2-10n-1000+\frac{900}{n+10}$, so we want the greatest positive integer such that $\frac{900}{n+10}$ is an integer. Obviously, $n=\boxed{890}$.[/hide]

I don't quite understand the first solution (Euclidean Algorithm). I understand Euclidean Algorithm, but I don't understand how to apply to polynomials. For example, why did we choose to divide by $n^2$ and not any other number in the first simplification:
$ \gcd(n^3+100,n+10)=\gcd(-10n^2+100,n+10) $, and why does $n+10$ always stay in the same spot (as the divisor). I thought Euclidean Algorithm worked only when the divisor is switched for the dividend the next equation (and the remainder as the divisor)?

First of all, forget about the names. Secondly, all this is doing is saying that if two expressions have a common divisor, then so does their difference. If you understand that, then you should be able to see how doing it with polynomial a is equivalent.

Sorry, I don't understand what you are saying. The Euclidean Algorithm deal with $GCF$s, and in the solution, they didn't just subtract the polynomials. Can you explain in more detail, please?

They did subtract the polynomials! Just a bunch of operations at a time to reduce the degree.

[quote="4everwise"]What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?[/quote]
[i]Solution.[/i] We need $n^3+100\equiv 0\pmod {n+10}$. Notice that $n\equiv -10\pmod {n+10}$, thus $n^3+100\equiv -10^3+100\equiv -900\pmod {n+10}$. So we need $900\equiv 0\pmod {n+10}$.
Consequently, $n+10\mid 900\implies n+10\le 900\implies n\le \boxed{890}$. $\blacksquare$

[quote=4everwise]What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?[/quote]

[hide="My solution"]
First, it is easy to see that, $n+10 \mid (n+10)(n^2-10n+10^2)=n^3+1000.$ Now, assume that, $n+10$ devide $n^3+100$ then
$$n+10 \mid (n^3+1000)-(n^3+100)=900.$$
So we deduce that the largest positive integer $n$ such the problem is, $n=900-10=890.$
[/hide]

Okay I guess I really did get better because I now find this fairly easy.

[hide=Solution]Dividing gives $\frac{n^3+100}{n+10}=(n^2-10n+100)-\frac{900}{n+10}$. The largest factor of $900$ is $900$, for an answer of $900-10=\boxed{890}$.[/hide]

Well, no wonder - this was from 1984.

[quote=4everwise]What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?[/quote]

Rewriting numenator as $(n^3+1000)-900$
Now $\frac{(n^3+1000)-900}{n+10}$
$n^3+1000 $ is divisible by $n+10$
So being an integer, $(n+10)$ should divide $900$
Largest factor is $900$
$(n+10)=900 \implies n=\boxed{890}$

If $P(x)={{x}^{3}}+100$ then on he basis of the Bézout theoreme the rest it is $P(-10)=890$


how do you write piecewise function in latex. for example,

$ fn(x) \equal{} n^2x; 0 \leq x \leq \frac{1}{n}$
$ fn(x) \equal{} n^2(x \minus{} \frac{2}{n}); \frac{1}{n} \leq x \leq \frac{2}{n}$
$ fn(x) \equal{} 0; \frac{2}{n} \leq x \leq 1$

Use cases environment:
$ f_n(x) \equal{}
\begin{cases}
n^2x; & 0 \leq x \leq \frac {1}{n}
\\
n^2(x \minus{} \frac {2}{n}); & \frac {1}{n} \leq x \leq \frac {2}{n}
\\
0; & \frac {2}{n} \leq x \leq 1
\end{cases}$

Except I'm pretty sure you meant

$ f_n(x) \equal{} \begin{cases} n^2x; & 0 \leq x \leq \frac {1}{n} \\
\minus{}n^2(x \minus{} \frac {2}{n}); & \frac {1}{n} \leq x \leq \frac {2}{n} \\
0; & \frac {2}{n} \leq x \leq 1 \end{cases}$

Note that if you're over-defining it at $ \frac1n$ and $ \frac2n$, the redundant definitions had better agree.


function
logarithms
inequalities
inequalities unsolved

assuming a^3+b^3+c^3=1, now brute force! It's not THAT bad... :D

Write x = (b/a)^3, y = (c/b)^3, z = (a/c)^3; we have xyz = 1 and wish to prove sum 1/(1+x) >= 3/8, but at least one of x, y, z (say WLOG x) is <= 1, in which case 1/(1+x) >= 1/2, and 1/2 >= 3/8, so this is trivial. Did I make a mistake?

sorry!

with a,b,c>0
prove that
(a^3 / (a+b)^3) + (b^3 / (b+c)^3 ) + (c^3 / (c+a)^3 )  >= 3/8

sorry everyone !

Now we want to prove that if $xyz = 1$, we have $\sum \frac{1}{(1+x)^3} \ge \frac{3}{8}$. Well, if one of $x, y, z$, say $x$, satisfies $x \le \frac{1}{3}$, then $\frac{1}{(1+x)^3} \ge \frac{1}{(1+\frac{1}{3})^3} = \left(\frac{3}{4}\right)^3 = \frac{27}{64} > \frac{3}{8}$, so we are done. If not, i.e. $x, y, z \ge \frac{1}{3}$, then consider the function $f(t) = \frac{1}{(1 + e^t)^3}$. We have $\frac{df}{dt} = -\frac{3e^t}{(1+e^t)^4}$, and $\frac{d^2f}{dt^2} = \frac{12e^{2t}(1+e^t)^3 - 3e^t(1+e^t)^4}{(1+e^t)^8} = \frac{3e^t}{(1+e^t)^5} \cdot (3e^t - 1)$, which is positive for $\ln x, \ln y, \ln z$, so Jensen's on these three values yields the desired result.

another well known problem..........
It is well known that $\sum \frac{1}{(1+x)^2} \ge 1$ if $xyzt=1$. The trick is $\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2} \ge \frac{1}{1+xy}$.
Now put one of the variables to be 1 and use power mean.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=36463 . [EDIT: Or http://www.mathlinks.ro/Forum/viewtopic.php?t=36567 .]

  darij

Does anyone have the full contest and another idea?
Please post it on th forum
I need them!
Thank you

I guess that is sufficient, we don't need any more spam...

[quote="hitech"]I thinks the problem wasn't solved.

All the contest is not right!
I wish some people solved it !

Math is greatest![/quote]

I think this solution
http://www.mathlinks.ro/Forum/viewtopic.php?p=239084#p239084
the one I have been talking about, is [b]perfectly alright[/b], please read it carefully. Thanks.[/b]


geometry
conics
ellipse
number theory proposed
number theory

Find all rational solutions to \[ x^{2}\plus{}y^{2}\equal{}x\plus{}y\]

$ (2x\minus{}1)^{2}\plus{}(2y\minus{}1)^{2}\equal{}2$
we use Geometry method to find all rational point on elipse $ x^{2}\plus{}y^{2}\equal{}2$
This is easy to find all of them

But there are more elementary and purely number theoretic solutions...

A solution use complex integer but I don't no it very much.
This proof can solve for every elliptic curve degree 2.

Here is another problem of the same kind:

Find all rational solutions to the following equation:\[ x^{3}\plus{}y^{3}\equal{}x\plus{}y\]

[quote="nayel"]Here is another problem of the same kind:

Find all rational solutions to the following equation:
\[ x^{3}\plus{}y^{3}\equal{}x\plus{}y\]
[/quote]
It is simple $ (x,y)\equal{}(\minus{}t,t),\ (x,y)\equal{}(\frac{t(1\minus{}2t)}{t^{2}\minus{}t\plus{}1},\frac{1\minus{}2t}{t^{2}\minus{}t\plus{}1}),\ t\in Q.$

General
Find all rational point on curve
$ ax^{2}\plus{}bx\plus{}c\equal{}0$ ,known that it has a rational point$ (x_{0},y_{0})$

[quote="TTsphn"]$ (2x\minus{}1)^{2}\plus{}(2y\minus{}1)^{2}\equal{} 2$
we use Geometry method to find all rational point on elipse $ x^{2}\plus{}y^{2}\equal{} 2$
This is easy to find all of them[/quote]
For example $ x\equal{}\frac{t^{2}\plus{}t}{t^{2}\plus{}1}$, $ y\equal{}\frac{t^{2}\minus{}t}{t^{2}\plus{}1}$, $ t\in\mathbb Q$


vector
analytic geometry
abstract algebra
absolute value
linear algebra

Hi , i want to ask you something about vectors, that i can  not understand.
I have read in a book that if are given two points P(xp,yp) and R(xr,yr) , and  also the point Q(xp,yp). If we want to know whether the point Q is on the left side of PR ( i explain myself "that Q is  on the left of PR if Q is located over PR",) we have to check the sinus of the angle between  beam PR and the vector PQ has positive sign.And this is equivalent that  the vector multiplication (i mean ->  ->
                                                            |PRxPQ| =|PQ|.|PR|.|sin(anglebetweenthem))|
has positive sign , that is equivalent to

(yq-yp)(xr-xp) - (xq-xp)(yr-yp) >0. (*)
I cannot understand   why if Q is on the left of PR, the above is true?
Using the definition for vector multiplication i see that (*) is  exactly this ( i use that z coordinates of every opint is 0), but why the vector multiplication to determine  that angle is positive or negative?
The only place , that there is  SIN in the definition is  in module( absolute value, i mean |-3|=3, u understand...) ? Help me pls... u know that it is easy, but i cannot understand it.

This is a good question, but not on topic for the College Forum, so I will move it to a forum where I hope it will fit better.

Excuse me about  my ignorance... I am for first time here and i didin't know where to put the question.


Let $AC$ and $BD$ be two chords of a circle with center $O$ such that they intersect at right angles inside the circle at the point $M$. Suppose $K$ and $L$ are midpoints of the chords $AB$ and $CD$ respectively. Prove that $OKML$ is a parallelogram.

If  line $LM$  meets $AB$ at  point  $N$  , then $\angle NMB = \angle DML = \angle MDL$.Since $\angle NBM = \angle MCD$ and $\angle MCD + \angle MDC = 90$ , we obtain $LM \perp AB$ . In the same way we have $KM \perp DC$ . Thus $ML // OK$ and $MK // OL$.

[u] Babis[/u]

Best way is to trig bash

Let $\angle AOB=y$ and $\angle COD=x $
Since, $\Delta AMD $ is right angled, it follows that, $\frac {x+y}{2}=90^{\circ} \implies \boxed {x+y=180^{\circ}} $
$$OL=R\cos \frac {x}{2} \text{ and } AK=KM=R\sin \frac {y}{2} =R\cos \frac {x}{2} =OL $$
Similarly, we can prove $OK=ML $ which follows that the opposite side in the quadrilateral $KOML $ are equal, hence, it's a parallelogram

Dear mathlinkers：this problem is easy to calculate with complex numbers


function
inequalities unsolved
inequalities

Let $ F(x,y) \equal{} |x|^{\alpha}x \plus{} |y|^{\beta}|x|^{\gamma}x \plus{} |y|^{\eta}y$ for $ (x,y)\in\mathbb{R}^2$. Find all $ (\alpha,\beta,\gamma,\eta)\in \mathbb{R}^4_ \plus{}$ such that
\[ |F(x_1,y_1) \minus{} F(x_2,y_2)|\le |f(x_1,x_2)| .|y_1 \minus{} y_2|\]
for all $ (x_i,y_i)\in\mathbb{R}^2$, where $ f$ is a function on $ \mathbb{R}^2.$


analytic geometry
combinatorics unsolved
combinatorics

Prove that one can paint all points of the plane using 2 colours such that every three collinear points have different colours.

This is nonsensical -- what is it supposed to be?

Why do you think it's nonsensical :?:    :(

[quote="JBL"]The only interpretation I can give it that makes the question sensible is,
"Prove that one can paint every point in the plane one of two colors such that among every three collinear points, both colors are used."  And this is false, since lines have more than 4 points.[/quote]

Owned :!:

For every complex number in the plan if it has both coordinates rational color the number in red if it has at least one irational color the number black.

Then the line $\{z \mid \textrm{Re}(z) = \sqrt{2}\}$ contains infinitely many (and in particular three) black points.

[quote="derezin"][quote="andyciup"]Maybe what you mean is: Prove that one can colour the plane in two colors, such that there isn`t any mono-colored line(i.e all lines must have two points of different colours).[/quote]

Exactly  :blush:[/quote]

Then, paint any point such that $x^{2}+y^{2}=n^{2}$, $\forall n\in\mathbb{N}$ in one colour and the remainder of the plane in the other colour.

--
Patrick


Could you implement a special id enclosing each contribution in "User Contributions", as well as each edit summary? It would make writing a script to count and give information contributions ever so much easier.  :)

I have no idea what you're asking me to do, but I certainly will not do any modifications that affect the wiki database files because it's just a mess to maintain the files at each upgrade.

No, I mean for example, each contribution on an arbitrary "user contributions page" is surrounded by, for example, <span id="contrib"> </span> so that tallying them is easier.


calculus integration inequalities algebra function domain geometry

In $xyz$ plane find the volume of the solid formed by the points $(x,\ y,\ z)$ satisfying the following system of inequalities.

\[0\leq z\leq 1+x+y-3(x-y)y,\ 0\leq y\leq 1,\ y\leq x\leq y+1.\]

[quote="cckek"][quote="kunny"]In $xyz$ plane find the volume of the solid formed by the points $(x,\ y,\ z)$ satisfying the following system of inequalities.
\[0\leq z\leq 1+x+y-3(x-y)y,\ 0\leq y\leq 1,\ y\leq x\leq y+1. \]
[/quote]

    $\frac{7}{4}?$[/quote]

Maple agrees with your answer: nice.

[quote="cckek"][quote="kunny"]In $xyz$ plane find the volume of the solid formed by the points $(x,\ y,\ z)$ satisfying the following system of inequalities.
\[0\leq z\leq 1+x+y-3(x-y)y,\ 0\leq y\leq 1,\ y\leq x\leq y+1. \]
[/quote]

    $\frac{7}{4}?$[/quote]

That's right, how did you solve it?

[quote="kunny"]That's right, how did you solve it?[/quote]

Is that really necessary? You did set up the inequalities perfectly for

\[V=\int_{0}^{1}\int_{y}^{y+1}\int_{0}^{1+x+y-3(x-y)y}\ dz\ dx\ dy\]

which is not myterious or difficult.

Yes, this solution is the one of them.
Can anyone solve in other solution?

[quote="maokid7"]couldnt you this just be solve as if it were $z=f(x,y)$ and you are trying to find the volume under $z=f(x,y)$ and above $z=0$ bouned by $0\le{y}\le{1}$ and $y\le{x}\le{y+1}$
so we have

$V=\int_{0}^{1}\int_{y}^{y+1}f(x,y)dxdy$

*note that $f(x,y)=1+x+y-3(x-y)y$[/quote]

Which is, incidentally, exactly the same as the triple integral posted above.

Here is the solution for Japanese high school students.

Cut this solid by the plane $y=t\ (0\leq t\leq 1),$ we have the domain of the cross-section such that $0\leq z\leq (1-3t)x+3t^{2}+t+1$ and $t\leq x\leq t+1.$
Let $S(t)$ be the area this domain we have $S(t)=\frac{1}{2}\{(1+2t)+(2-t)\}(t+1-t)=\frac{1}{2}(t+3),$ which is the area of a trapezoid, but for some case a rectangular.
Thus desired volume is $\int_{0}^{1}S(t)dt=\frac{1}{2}\int_{0}^{1}(t+3)dt=\frac{7}{4}.$

How about this one? [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=exam+entrance+university+tokyo&t=113004[/url]


modular arithmetic
algebra
polynomial
number theory
divisibility tests
number theory proposed

Find all positive integers $ k$ with the following property: There exists an integer $ a$ so that $ (a\plus{}k)^{3}\minus{}a^{3}$ is a multiple of $ 2007$.

I'm sure there must be an easier solution, but here's mine:
\[ k(k^{2}\plus{}3ka\plus{}3a^{2})\equiv 0\pmod{2007}.\]
Looking modulo 3, we observe that $ k\equiv 0\pmod 3$. Therefore let $ k \equal{} 3x$, $ x$ being a positive integer. We can prove now that all multiples of 3 are good. Check http://www.mathlinks.ro/Forum/viewtopic.php?p=934678#934678

A small addendum/correction to Valentin's nice solution:

There is some confusion with k being divisible by 3 and k being divisible by 9:
Valentin correctly shows that k=3x, and that x must be a multiple of 223.
Hence, k must be a multiple of 669. (That's the gist if his argument.)

Every occurrence of "2007" after the second line of Valentin's proof
should be replaced by "669". (a repeated typo).

$ k^{2}\plus{}3ka\plus{}3a^{2}\equiv 0(\mod 223)$
You can use algebra number
$ x\equiv (\frac{a}{k})(\mod 223)$
$ 3x^{2}\plus{}3x\plus{}1\equiv 0(\mod 223)$
Use quadric reside.

[quote="test20"]Every occurrence of "2007" after the second line of Valentin's proof
should be replaced by "669". (a repeated typo).[/quote]Oops, I was in a hurry. Obviously $ 3\times 223 \equal{} 669\neq 2007$ :D

You are right limsk1, your answer is right and your solution is very nice.

Try a more general problem, with a natural number b instead of 2007 and natural number c instead of 3.

I mean for given numbers b and c, find all numbers k, such that there exist a number a, that:
$ a^{c}\equiv (a\plus{}k)^{c}(mod b)$

The Polynomial congruence quite hard to solve
$ f(x)\equiv 0(mod p)$
$ x \equal{} (\frac{k}{a})(\mod b)$
You can find out all b such that exist $ (a,k)$ and $ \gcd(a,b) \equal{} 1$
$ (a\plus{}k)^{k}\minus{}a^{3}\equiv 0(\mod b)$

It seems that I  did a computational mistake in my problem, as the congruence is solvable  :blush: :)
Thanks for pointing that out limsk1

Another solution:

[b]Proposition: [/b] Let $ p\geq 3$ be a prime number and let $ d > 1$ be a divisor of $ p \minus{} 1$. Then there exists an integer $ a$ so that $ (a \plus{} k)^d\equiv a^d\pmod p$.

[i]Proof:[/i] if $ p\mid k$ then it's obvious so suppose that $ p\nmid k$. Let us furthermore assume that $ p\nmid a$. Then notice that the congruence is equivalent to
\[ \left(1 \plus{} \frac {k}{a}\right)^d \equiv 1 \pmod p
\]
(if anyone is not familiar with fractions in modular arithmetic - there are enough posts about it on ML).
Let $ g$ be a primitive root modulo $ p$. Then we see that
\[ a\equiv \frac {k}{g^{\frac {p \minus{} 1}{d}} \minus{} 1} \pmod p
\]
is a solution since
\[ \left(1 \plus{} \frac {k}{a}\right)^d \equiv \left(1 \plus{} \frac {k\cdot (g^{\frac {p \minus{} 1}{d}} \minus{} 1)}{k}\right)^d \\
\equiv g^{p \minus{} 1} \equiv 1 \pmod p.
\]
$ \Box$

It's obvious that $ 3\mid k$ is a neccessary and sufficient condition for divisibility by $ 9$, so from the above, we see that the solutions are the multiples of $ 3$.

it suffices to consider mod 9,223.
by$9|(a+k)^3-a^3$
we get $3|k^3$
so$3|k$ easy to check it's sufficient.
when mod 223,if $223|k$ it's trivial
if $(k,223)=1$ let g be an original root mod 223
then$a+k\equiv ag^{74} or ag^{148}$
it's easy to calculate {g^74,g^148}={39,183}
hence $k=38a or 182a(mod 223)$
since$ (38,223)=(182,223)=1$
so for any resicue class mod 223,we can choose an appropriate a satisfying these conditions.
so $3|a$ is the answer.


Evaluate dy/dt for the function at the point.
 x^3 + y^3 = 9  ; dx/dt = -3 , x = 1 , y = 2


integration
calculus
real analysis
real analysis unsolved

Prove that

$\sum_{n=0}^\infty \left(\frac{1}{3n+1}-\frac{1}{3n+2}\right) = \frac{\pi}{3\sqrt{3}}$

$\sum_{n=0}^{\infty}\left(\frac1{3n+1}-\frac1{3n+2}\right)$

$=\sum_{n=0}^{\infty}\left(\int_{0}^{1}(x^{3n}-x^{3n+1})\,dx \right)$

$=\int_{0}^{1}\sum_{n=0}^{\infty}(x^{3n}-x^{3n+1})\,dx$

$=\int_{0}^{1}\frac{1-x}{1-x^{3}}\,dx$

$=\int_{0}^{1}\frac1{1+x+x^{2}}\,dx$

Now we have a routine calculus problem.

Several of those steps do need to be justified; there's at least one use of Abel's Theoerm in there.

Maybe this one is not as easily killed by that method:

$\sum_{n=0}^\infty (\frac{1}{7n+1}+\frac{1}{7n+2}-\frac{1}{7n+3}+\frac{1}{7n+4}-\frac{1}{7n+5}-\frac{1}{7n+6}) = \frac{\pi}{\sqrt{7}}$

I think you will have to integrate the expression
$\frac{x^{4}+2x^{3}+x^{2}+2x+1}{x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1}$
which probably works by the same method but is not fun by hand! ;) :lol:


[quote="13375P34K43V312"]Circle $A$ and circle $B$ are externally tangent, and their radii are 3 and 2, respectively. Line $l$ is drawn externally tangent to both circles and line $m$ is drawn tangent to $A$ such that $l\parallel m$. Circle $C$ is drawn tangent to line $m$ and circles $A$ and $B$. Find the radius of circle $C$.[/quote]

$\sqrt{(3+r)^{2}-(3-r)^{2}}+\sqrt{(2+r)^{2}-(4-r)^{2}}=\sqrt{24}$
solving that I get $\frac{9}{8}$

Its just drawing th perpendicular from the diameter of circle A to B, which is $\sqrt{24}$.  Then you can from two right triangles using the center C and the sum of the two legs of the two right triangle should be the perpendicular distance from the diameter of A to that of B.


$A= \{ x \in \mathbb{R}| y,z$ exist and are real so as $x-\sqrt y = y-\sqrt z = z-\sqrt x = 1 \}$
my question is ... does A have any elements at all ?
and if it has, are they rational, irational .. ?

Also notice that $x,y,z$ cannot be distinct integers since $f\in\mathbb{Z}[X]$.  Now, what can you conclude about $x,y,z\in\mathbb{Q}$?  :wink:

[quote="perfect_radio"]Let $f : \mathbb R \to \mathbb R$, $f(t) = (t-1)^{2}$.

We have that $y = f(x), \, z = f(y), \, x = f(z)$, so that $x = \left( f \circ f \circ f \right) (x)$. You can easily see that $A$ is composed exactly of the (non-negative) solutions of that equation. Some solutions can be found by solving $x = f(x)$, but for the others I'm not willing to factorize that polynomial.[/quote]

Next step : it's rather easy to show that $x = \left( f \circ f \circ f \right) (x)$ has exactly two real roots.
Hence these roots are the two real roots of $f(x)=x$

And A has just two elements : $A=\{\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}\}$


Find the largest value of $ n$ such that $ 3x^2 \plus{}nx \plus{} 72$ can be factored as the product of two linear factors with integer coefficients.

Since we want n to be as large as possible, n should be the sum of the two farthest apart factors, which are 1 and 72. In order to maximize n, we wan the 3 to multiply the 72, so the largest possible of n factorization is $ (3x\plus{}1)(x\plus{}72)\equal{}0$. If we expand that out, we get n=72*3+1=217


The Binomial Expansion is valid for exponents that are not integers.  That is, for all real numbers $ x, y,$ and $ r$ with $ |x| > |y|,$
\[ (x \plus{} y)^r \equal{} x^r \plus{} rx^{r \minus{} 1} \plus{} \frac {r(r \minus{} 1)}2x^{r \minus{} 2}y^2 \plus{} \frac {r(r \minus{} 1)(r \minus{} 2)}{3!}x^{r \minus{} 3}y^3 \plus{} \cdots
\]
What are the first three digits to the right of the decimal point in the decimal representation of $ \left(10^{2002} \plus{} 1\right)^{10/7}?$


I probably should be more worried about the AIME than the USAMO, but...
What amount of formalism should someone display during the USAMO?
The question is vague, but I want a wide range of answers and maybe examples.

I'm not really sure, but judging from the "official" solutions, each USAMO is on average about 1.5 pages long typed, which would be about 3 pages written. These are the ones that would most likely get perfect scores, so they have about the desired amount of detail. Of course, the amount of detail varies from problem to problem, and on your method of solution.

Not necessarily, most problems can be written on 1-2 pages of paper if a good solution is found.

Where do you find these official solutions?  Or a list of problems for that matter...
Links are good.

You can buy them from the [url=http://www.unl.edu/amc]AMC website[/url].

The official solutions are well written but generally longer than what would be "necessary".

It's hard to describe how much detail is sufficient. Basically there shouldn't be any nonobvious jumps from one idea to another.

There are some free solutions on the AMC website...
Here are the solutions for the 2005 USAMO, as an example:
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/Solutions/05SOL.pdf

And remember to write as if the graders were mean and stupid.

[quote="probability1.01"]But don't think they're too stupid or they might get mad at you. I put a ton of detail in one of my WOOT tests proofs and the grader didn't like it very much :P. Make sure all your arguments are rigorous too.[/quote]

I think WOOT graders prefer short solutions. So far most of the comments I got include "solution could be shorter/more concise"

[quote]
I think WOOT graders prefer short solutions. So far most of the comments I got include "solution could be shorter/more concise"[/quote]

Well it's because

(a) We have to grade a ton of stuff in one afternoon

(b) Zhai generalized this problem for no reason using like 800000 variables forcing me to go step by step confirming the validity of each step, and then ended up going back and just solving the problem the way it was written the last four lines

It's hard to describe what a good solution is. I don't know why, but when the WOOT graders get together, we all seem to know independently how long/concise a solution should be, which parts should have been left out, etc., but it's just not the kind of thing that can be described easily. It's more of a feel/intuition thing that you develop.

[quote="FMako"]Thanks for the information everyone!  How should go about acquiring this intuition?  I want a reasonably rigorous study of USAMO style proofs but I have yet to find a documentation of these things.  I am willing to practice, but I should probably practice AIME style stuff harder than I am now.[/quote]

I hope you remember the 2004 TJUSAMO practices! I put a lot of effort into writing good, complete solutions for every problem set we worked.


calculus
integration
trigonometry
calculus computations

hello, using integration by parts we get
$ \int\arcsin(\sqrt{1\minus{}x^2})\,dx\equal{}x\arccos(\sqrt{1\minus{}x^2})\minus{}\int\frac{x^2}{\sqrt{1\minus{}x^2}\sqrt{x^2}}\,dx\equal{}$
$ \equal{}x\arccos(\sqrt{1\minus{}x^2})\plus{}x\frac{\sqrt{1\minus{}x^2}}{\sqrt{x^2}}$
The value of your integral is $ \frac{\pi}{2}\minus{}1$
Sonnhard.

Thanks, Sonnhard.

We could also let $ x\equal{}\sin \theta$ to get

$ I \equal{} \int \cos^{\minus{}1} (\cos \theta) \cos \theta \ d\theta$

$ \equal{}\int \theta \cos \theta \ d\theta \equal{} \theta sin \theta \minus{} \int \sin \theta\ d\theta$

$ \equal{}\theta \sin \theta \plus{} \cos \theta \equal{} x \sin^{\minus{}1} x \plus{} \sqrt {1\minus{}x^{2}}$

$ \equal{}x \cos^{\minus{}1} \sqrt{1\minus{}x^{2}} \plus{} \sqrt{1\minus{}x^{2}}$

So, putting in the limits $ 0$ and $ 1$, we get $ \frac{\pi}{2} \minus{} 1$.


What is the greatest integer n such that a square cannot be partitioned into n smaller squares, where overlapping squares are not counted twice?

Is brute force the best method here?

Yes, I did try searching for an explanation in the forum, but I didn't get any good results.

i think it's 5...

[hide]use mod 3? ...idk[/hide]

[hide]it is possible to partition into any even number of squares by having n squares in the top row, and n-1 number of squares in the left column. and then a big square to fill up the rest of the space. that makes n+n-1+1=2n squares
so that means we can have any even number of squares except 2. and to make odd number of squares, all you have to do is take the big square in the even partition, and divide that big square into 4 differnet pieces. so the big square turn into 4 little squares, and that add 3 more squares. And you originally had an even number of squares, now you have all the odd number of squares as well except 1 and 3 and 5.[/hide]


trigonometry
function
algebra
difference of squares
special factorizations

Hey could someone help me with these trig equations?
csc 2 x - cot 2 x
--------------------------                       =    cos 2 x
1 + tan 2 x

I know you do difference of squares on the top, and the denominator is equal to sec 2 x, so that means the numberator has got to be equal to 1 right?  I just don't know how to do it.

     cscx .................. cscx
-------------     -     -------------      =      2secx
secx - tan ......... secx + tanx

This one you get a common denominator right? Which is sec 2 x - tan 2 x which is just 1 but i dont know what do do in the numerator.

Thanks

Try writing everything in terms of sin and cos, and then simplfying.

Often, when faced with the "uglier" trig functions like tan, cot, sec, and csc, it's easier to write everything in terms of sin and cos.

For the first equation, the numerator can be written as 1, instead of using difference of squares. The definition of sec A is 1/cos A, so that's proved.

This one is easy when you look at it right.

The denominator is just 1 + tan^2x = sec^2x by pythagoras.

So this gives us cos^2x(csc 2 x - cot 2 x) = cos^2x, so we have to prove that that part is equal to 1.

But hey, csc^2x - cot^2x = 1, again by pythagoras.

So it's proven.


Does the bellow depicted figure fit into a square $5\times5$.

I was indeed missing something..  :blush:

You are missing something  :)  The figure above must lie within the square $5\times 5$.

[hide]No?? Unless you can cut the figure and move the pieces.  :D
The diagonal length is $4\sqrt{2}$ which is greater than 5. Is that what you're asking??[/hide]

[quote="JBL"]greengrapes, that is not sufficient: a rectangle that's $4\sqrt2 \times 1$ fits easily inside a $5\times5$ square.[/quote]

Oh, yeah.  :blush:
[hide] The answer still [u]is[/u] no, right? Well, if you try to put it in a square, there's no hope of filling in the 10 half-squares surrounded on two sides on the egdes, right? So that makes 23 squares. It's close. Of course, I don't think this is going to help in the proof. :( [/hide]

[quote="Peter VDD"]Is this a TST for IMO?[/quote]

Yes, it is, and it is the last problem on the TST (so it is problably intended to be the hardest one).

In my opinion they weren't ordered by increasing difficulty. Yes it was a TST for IMO.  :)

Just an idea of the proof:
[hide]Just imagine the figure is leaning against a wall and resting on the ground.
Let this figure slide with different inclinations.
Then calculate the length and breadth of the smallest rectangle, with sides parallel to the wall and the ground repectively,  that required to enclose this figure.
With all possible inclinations, either the length or the breadth will exceed 5.[/hide]

Just saw this problem and I'll add my solution. I don't think Lithuanian tst's belongs to the preolympiad section, anyway,

1) Notice that if the figure is to fit inside the square then it is optimum to put the center of the figure in the center of the 5x5 square. This is because the figure is symmetric with respect to its center, so if we put the center elsewhere, we're actually trying to fit it in a rectangle which in turn fits in the 5x5 square.

2) Now consider the two diameters of the figure only. They are two segments with length $\sqrt{40}$ which form an angle.  It is optimum again to have the big diagonal in the square bisecting the angle among them. This means, that the figure in the first post is actually placed in such a way that all we need to do is try to fit it into a 5x5 square with sides parallel to the sides of the figure. But this clearly can't be done since the figure is 6 squares high and long.

Daniel[/url]

Consider the two column 3 and 4, we can easily see that one of them has to be the diagonal of the square since both of them have 5 boxes each. However, the length of the diagonal is $4\sqrt{2}$ which is greater than $5$. So, it is not possible to fit the figure into $5 x 5$ square.

Consider WLOG the smallest rectangle $ABCD$ with $AB$ and $CD$ at angle $\theta\leq\frac{\pi}{4}$ clockwise from horizontal. Also let the center be $(0,0)$.

$AB$ and $CD$ go through $\pm(1,3)$. The width of the rectangle is then $2\sqrt{10}\cos\left(\arctan\frac{1}{3}-\theta\right)$.

This is increasing for $\theta<\arctan\frac{1}{3}$ and decreasing for $\theta>\arctan\frac{1}{3}$ because $\arctan\frac{1}{3}-\theta\subset(-\pi,\pi)$.

At $\theta=0$ the width is $6$. At $\theta=\frac{\pi}{4}$ the width is $4\sqrt{2}<6$. So every rectangle envelope must have one dimension of size at least $4\sqrt{2}$. But $4\sqrt{2}=\sqrt{32}>5$. So a the shape won't fit in a $5\times 5$ square.

Solved with: Kevin Wu, Isaac Zhu, Chris Qiu, Reyna Li, Ryan Yang, Elliot Liu, Alex Zhao, Eric Shen, David Dong, and Alan Bu.


A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, then find the median of the triangle.

[quote="h_s_potter2002"]Don't you mean, median of the trapezoid?[/quote]
I think he meant the median of the trapezoid...because I can change shape of triangle and with same area and different lengths for the median of triangle.

[hide]The area of a triangle is $\frac{bh}{2}$, and the area of a trapezoid is $mh$, where m is the median.

$\frac{bh}{2}=mh$
$18h=2mh$
$9h=mh$
$m=9$[/hide]

[hide="Ans. Solution"]Call the height of the triangle $x$, the area of the triangle would be $9x$
Call the bases of the trapezoid $a$ and $b$, the area of the trapezoid would be $\frac{abx}{2}$

$\frac{abx}{2} = 9x$

$ab = 18$

$\frac{ab}{2} = 9$

The Median is $9$ inches long.[/hide]

This problem seems fimiliar, I think i've done it somewhere, but im not sure where

You can regard a triangle as a trapezoid with upper base=0.

Generally the length of the median (or mid-line) is arithemetic mean of the bases, so it's $\frac{18}{2}=9$.


Wow, so my conjecture was true: the minimum must be $\frac{21}{2}$. I will try to prove that. I guess it is not difficult. As for your problems with n even, I have found them for a long time and I have very simple solutions. Indeed, using the Cauchy-Schwarz inequality we can write $(\sum {\frac{2S-na_1}{a_1}})^2\leq\(\sum{a_i^2}\cdot\sum {\frac{1}{a_i^2}})$ from where the conclusion. The problem for n odd is far more difficult in my opinion. Anyway, since noone seems to be interested in more than 3 lines inequalities ( except Darij, of course), I guess I will let you post your solution after some days. Meanwhile, please let us try a little bit to prove your inequality a). Seems to be damn tough, with the strange case of equality 2,2,1.

This is just incredible! Thanks, Vasc, for a magnificient gift. Lets prove that $ (\sum {\frac{b+c}{a}}-2)^2\leq 2\cdot\sum {a^2}\cdot\sum {\frac{1}{a^2}}-2$ (which is obviously equivalent to a)). Squaring in a smart way, we can reduce it to the inequality
   $\sum {(\frac{b-c}{a})^2}\geq 2\cdot\frac{3abc+\sum {a^3}-\sum {ab(a+b)}}{abc}$. Now, using the identity $\sum {a^3}+3abc-\sum {ab(a+b)}=\sum {a(a-b)(a-c)}$, we can reduce it to $\sum {(\frac{b-c}{a})^2}\geq 2\cdot\sum{\frac{(a-b)(a-c)}{bc}}$. You have already recognized the form $(\sum{\frac{b-c}{a}})^2\geq 0$.

Very nice solution, Harazi  :)

Thank you very much

Indeed Harazi, the case n-even is a particular case for a known inequality, which was discussed in this forum and in the book Old and New Inequalities.
In my solution for n=3, I used the substitutions $u=a/b+b/c+c/a$ and $v=b/a+c/b+a/c$.

[quote=harazi]Let $a,b,c>0$ be such that $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$. Prove that $\frac{b^2+c^2}{a^2}+\frac{c^2+a^2}{b^2}+\frac{a^2+b^2}{c^2}>\frac{31}{3}$.
    I dont have a solution for the following: what is the minimal value of the expression in the left-hand side of the inequality?[/quote]
http://artofproblemsolving.com/community/c6h1133750p5282061
http://artofproblemsolving.com/community/c6h1181691p5722099


Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}$.  Find the number of possible values of $S$.

im not that good with explaining how to get to an answer.  i hope this works.
[hide]i brute forced this problem.  since $S$ has to be a 90-element subset of $A$  it is easy to see the possible subsets.
$\{1...90\}~\{2...91\}$
and so on...until you reach $\{11...100\}$
this leaves 11 possibilities.  you can now do the same thing except skip in each set.  this gives you:
$\{1...91\}~\{2..92\}$
until you get to $\{10..100\}$
in each new subset, you can skip 89 numbers.  so do $89\cdot10=890$
add 11 to that from the first part to receive 901[/hide]

[hide]The easiest way to do is to find the max and min S of this set.  You can do this by $\sum{n}$ from 1 to 100 and then subtracting $\sum{n}$from 1-10.  This gives us the max value, to find the min $\sum{n}$ from 1-90.  The rest is simple intuition to obtain 901. Remember that $\sum{n}=\frac{n(n+1)}{2}$[/hide]

[hide]
The simplest way to solve this problem is to think of the min and max value of any 90 element subsets. The max value is $\frac{90(100+11)}{2}= 4995$. The minimum value is $\frac{90(90+1)}{2}= 4095$. Subtracting, you get 900. However, since this situation is inclusive, the answer is $900+1=\fbox{901}$.[/hide]

[quote="Arnizzle"]Hey guys, i just found this forum - my first post!
[/quote]

 :welcomeani:

If anyone couldn't prove to themselves every intermediate sum was attainable, i thought of it like this:

[hide]1-90 is clearly the least subset sum. It is possible to get one more than this by summing 1-89 and then 91. And again one more by summing 1-89 and then 92. Continue this till you reach 1-89 and then 100. However, now, we repeat by sliding the 89th number over one digit at a time, as in the subset containing 1-88, 90, and 100. Basically, its like a huge abbacus. By sliding the beads over one at a time, we see that every sum is attainable.[/hide]

[quote="Arnizzle"]Hey guys, i just found this forum - my first post!
[hide] I did this problem like the rest of you, except I considered all sets of 10 integers NOT included in the subset. This works, too, and the numbers aren't quite so big. You get that all possible sums of numbers not included are from 55 to 955, inclusive, or 901 numbers.[/hide][/quote]

I did it this way as well. But to check my work later I did it the way with the bigger numbers.

[quote="chess64"][hide="Solution"]
The smallest sum possible is $1+2+3+\cdots+90$ and the largest sum possible is $11+12+13+\cdots+100$, and all sums in between can clearly be attained as well. So our answer is \[ 10\cdot 90 + 1 = \boxed {901}.  \]
[/hide][/quote]

Just a note about this solution, you actually didn't have to calculate any of the sums, you could just subtract corresponding terms of the sequence and you get 10 for each pair. There are 90 pairs. Add on the 1 to account for the fact that you need the total INCLUSIVE.

OMG this was such an easy problem and I missed it.  I am going to kill myself.  *Goes to kill himself, then realizes he's hungry...*

[hide=Solution]
The minimum possible sum is $1+\dots+90$ and the maximum possible sum is $11+\dots+100$.

So our answer is $(11+\dots+100)-(1+\dots+90)+1=10\cdot90+1=\boxed{901}$.
[/hide]

[hide]
We just need to remove $10$ elements from the set since it's $100$ numbers. The total sum is $5050$ of all $100$ numbers and the least sum that you can remove from it is $1 + 2 + 3 \dotsc + 10$ which is $55$. The largest that you can remove is $91 + 92 + 93 + 94 \dotsc + 100 = 955$. We can then assume that you can get the numbers in between by changing some of the numbers. Which means we have $955 - 55 + 1  = 901$ possible values.
[/hide]

The minimum possible value is $1+2+3+\ldots+90 = 4095$, and the maximum possible value is $11+12+\ldots+100 = 4995$. All numbers between these two numbers are attainable, so the answer is $4995-4095+1=\boxed{901}$.

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]DottedCaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLscelesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

$S$ ranges from $X = \sum_{i=1}^{90}i$ to $Y=\sum_{i=11}^{100}i$, and all values in between are achievable by incrementing individual elements in the minimal configuration until the maximal configuration is reached. Hence, the number of possible values of $S$ is

$$Y - X = 1 + \sum_{i=1}^{10} (90+i)-i = \boxed{901}.$$


[quote="dhuang614"]I sent out an entry form around late August, but the USAMTS site says that I haven't sent in my entry form yet.  Should I send in another one?

Thanks.[/quote]

Please do so, just in case.  Also, make sure you are registered online.  If you participated last year, you only have to go to your Profile in the USAMTS page and set the grade to the correct grade you are in now, and this will complete your online registration for Year 18 (but we still need the Entry Form).

If I forgot my ID on my entry form when I sent it with my solutions (it was by e-mail so my ID was with them), is that a problem?  I sent in the entry form with the ID later.

I also have a question. I am going to e-mail my solutions, is there a way to e-mail the entry form with them as well? How should I e-mail it?

I scanned the signed copy...but an administrator would be the one to answer our questions in order to be certain

We prefer that the entry form be mailed or faxed.


inequalities
trigonometry
inequalities proposed

Prove that :
$Sin^{2}A+Sin^{2}B+Sin^{2}C \leq \frac{9}{4}$
With $A+B+C=\pi$ $A,B,C>0$

This inequality is equivalent to
\[(\cos{C}-\frac{1}{2}\cos (A-B))^{2}+\frac{1}{4}(1-\cos^{2}(A-B))\geq 0. \]


geometry
circumcircle
geometric transformation
reflection
power of a point
geometry solved

Let ABC be a triangle with orthocenter H. Let M be the midpoint of [BC], A1 the intersection of AM with the circumcircle of ABC. Finally, let A 2 be the reflection of A1 in M. Prove that A2H is perpendicular to AM.

It seems to be pretty hard only using "classic geom". Maybe you should settle for that inversion proof I gave you a long tima ago? :D

Finally, a classic solution (not by me btw).

Let F be the foot of the altitude of triangle ABC which passes through A.

Let P be the intersection of AF with the circumcircle of ABC.
It will suffice to show that MA2HF is cyclic.
So it suffices to show that AA2.AM = AH.AF by power of a point. Now
AA2.AM
= AM.(AM - MA1)
= AM2 - AM.MA1
= AM2 - MB.MC
= AF2 + FM2 - MB2
= AF2 + (FM - MB)(FM + MB)
= AF2 - FB.FC
= AF2 - AF.FP
= AF.AH
after some applications of power of a point and Pythagoras. So we're done.


function
trigonometry
algebra unsolved
algebra
functional equation

Find all functions $ f: \mathbb{R}\to \mathbb{R}$ such that :

$ i)$    $ f(2x) \equal{} f(\sin(\frac {\pi}{2}x \plus{} \frac {\pi}{2}y) \plus{} f(\sin(\frac {\pi}{2}x \minus{} \frac {\pi}{2}y)$

$ ii)$    $ f(x^2 \minus{} y^2) \equal{} (x \plus{} y)f(x \minus{} y) \plus{} (x \minus{} y)f(x \plus{} y)$

no body ?? , if it's so hard i post the solution  :)

$ x: \equal{} {a \plus{} b\over 2}$ and $ y: \equal{} {a \minus{} b\over 2}$ in i) gives

[b]1)[/b] $ f(a \plus{} b) \equal{} f(\sin{\pi \over 2}a) \plus{} f(\sin{\pi \over 2}b)$
for all $ a,b\in\mathbb{R}$.

Then $ a\equal{}b\equal{}0$ gives $ f(0)  \equal{} 0$.
$ b \equal{} 0$ in [b]1)[/b] gives
[b]2)[/b] $ f(a) \equal{} f(\sin{\pi \over 2}a)$ for all $ a\in\mathbb{R}$.

Then [b]1)[/b] and [b]2)[/b] gives the Cauchy equation
[b]3)[/b] $ f(a \plus{} b) \equal{} f(a) \plus{} f(b)$ for all $ a,b\in\mathbb{R}$
and therefore $ f$ is $ \mathbb{Q}$-linear.

Take $ \alpha\in [0,1]$ with $ \sin{\pi\over 2}\alpha\equal{}{3\over 5}$. Then $ \cos{\pi\over 2}\alpha\equal{}{4\over 5}$
and for all $ x\in\mathbb{R}$ we have
$ f(\alpha)\plus{}f(x)\equal{}f(\alpha\plus{}x)\\
\hspace*{0.72in}\equal{}f(\sin{\pi \over 2}(\alpha\plus{}x))\\
\hspace*{0.72in}\equal{}f({4\over 5}\sin{\pi \over 2}x\plus{}{3\over 5}\cos{\pi \over 2}x)\\
\hspace*{0.72in}\equal{}{4\over 5}f(\sin{\pi \over 2}x)\plus{}{3\over 5}f(\sin{\pi \over 2}(1\minus{}x))\\
\hspace*{0.72in}\equal{}{4\over 5}f(x)\plus{}{3\over 5}f(1\minus{}x)\\
\hspace*{0.72in}\equal{}{1\over 5}f(x)\plus{}{3\over 5}f(1)$
$ \Rightarrow f(x)\equal{}{3\over 4}f(1)\minus{}{5\over 4}f(\alpha)$ constant for all $ x\in\mathbb{R}$.
So $ f\equiv 0$ and I have not used ii) at all.

I will solve this problem not neccesary condition ii .
As above we have rerult :
[b]1[/b] $ f(2x) \equal{} f(\sin{\pi x }$ and $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$
Easy to check $ f(0) \equal{} f(1) \equal{} 0$ so $ f(x) \equal{} 0,\forall x\in Q$
From  condition we have :
$ f(x \plus{} 1) \equal{} f(\sin\frac {{\pi.(x \plus{} 1)}}{2}) \equal{} f(\cos{\frac {{\pi x}}{2}})$
But $ f(1) \equal{} 0$ so $ f(x) \equal{} f(x \plus{} 1)$
So $ f(x) \equal{} f(\cos{\frac {\pi x}{2}})$
So $ f(x) \equal{} f( \minus{} x)$
Take $ x \equal{} \minus{} x$ we have $ f(x \plus{} y) \equal{} f(y \minus{} x)$
So $ f(x)\equiv 0 ,\forall x \in R$


I've read the definitions of substance, molecule, atom, element, and compound many times, yet I am still a little confused.

For example, let's consider NO (nitric oxide).

It is made up of N and O atoms, so it is a molecule. However, I believe it is also a compound, right? And, by definition, compounds are made up of elements. So N and O must be elements. But "atom" and "element" are not synonymous... Is this because if you have a container of NO gas, it would contain many molecules of NO, and would be considered a compound?

Also, what if you had like $F_{2}$? It is made up of the element fluorine, so is it a compound? But you can also have $F_{2}$ molecules. From a previous topic, I saw that $O_{2}$ was an element, so $F_{2}$ would also be an element?

 :?:

The other question I have is about the law of definite proportion (a given compound always has the exact same proportion of elements by mass). How exactly did it lead Dalton to propose that elements were made up of atoms? Like if you had a gallon of $H_{2}O$ and 2 gallons of $H_{2}O$, both would have 2 g of H for every 16 g of O. But where does the idea of atoms come from?

[b]Element[/b] is a word we use to classify atoms based on how many protons they have.  For example, one can have an atom of the element oxygen, and an atom of the element nitrogen.  In other words,

$\text{Element : Atom : : Brand : Car}$

[b]Compound[/b] is a word we use to classify molecules based on what elements make them up.  For example, one can have a molecule of the compound $NO$.  Again,

$\text{Compound : Molecule : : Brand : Car}$

The use of the word "element" to describe diatomic molecules like $F_{2}$ is, in my opinion, not entirely correct.

[quote="aspiringchemlete"]The other question I have is about the law of definite proportion (a given compound always has the exact same proportion of elements by mass). How exactly did it lead Dalton to propose that elements were made up of atoms? Like if you had a gallon of $H_{2}O$ and 2 gallons of $H_{2}O$, both would have 2 g of H for every 16 g of O. But where does the idea of atoms come from?[/quote]

To the first question, it's most likely the fact that the proportions were very nice ratios of small integers.

To the second, the idea of atoms extends back to the musings of the Greek philosopher [url=http://en.wikipedia.org/wiki/Democritus]Democritus[/url], although in my view he just got lucky.

[quote="T0rajir0u"]Element is a word we use to classify atoms based on how many protons they have.[/quote]

The word "element" is also used to name the naturally occurring form of a given element.

[quote="T0rajir0u"]Compound is a word we use to classify molecules based on what elements make them up.[/quote]

A more correct definition would be "Compound is a word used to classify molecules and ionic substances based on what elements make them up, the number of each element in the molecule or ion, and the tridimensional array of those elements in the molecule or ion."
This because isomers have the same number and type of atoms but different chemical and physical properties, and so they are different compounds.

[quote="aspiringchemlete"]If you had a glass of pure water, would that glass of water be considered a compound? The glass of water would be made up of individual water molecules. Would those individual water molecules be considered compounds?[/quote]

The water in the glass is a substance composed of [b]the[/b] compound $H_{2}O$.  The individual water molecules are molecules of [b]the[/b] compound $H_{2}O$.

The glass of water would be better described as a [i]sample of the compound water[/i].

[quote="t0rajir0u"]The water in the glass is a substance composed of [b]the[/b] compound $H_{2}O$.  The individual water molecules are molecules of [b]the[/b] compound $H_{2}O$.[/quote]
Are you saying that "compound" is really just an abstract idea? For example, if you looked at some $H_{2}O$, you could say "I see molecules of a compound," but you could not say "I see a compound".

Correct.  You would say, perhaps, "I see a substance."


This year a French student broke a world record of mental arithmetic. By pure mental calculation, he found the 13th root of a 200-digit number, which is a 16 digit number, in 267.77 seconds.
    Does anyone know how those people perform these tasks? What method or algorithm do they use? I don't even have a clue about how to find the 13th root by paper and pencil!

well I'm assuming the 13th root is like an integer, or it would be pretty hard to find all the digits... but anyway if it's an integer you can use like all kinds of random NT stuffs... basically estimate I think is the best way.

Let me ask another question : is mental arithmetic a learned abililty or natural?

[url=http://www.13throot.com/accueilang.htm]Click for a site on 13th roots.[/url]

HINT: Newton's Method of finding roots will help, but I don't know how much time it'll require for such large numbers.

[quote="krustyteklown"]newton's method for finding roots...
that's for functions : |[/quote]
[url=http://www.karlscalculus.org/calc5_2.html]Yes it is. You write what you want as a function[/url]

what do you mean?
aren't they given a random 200-digit number that's a perfect 13th power and asked to factor?

if you can do $3*3$ in 5 seconds mentally, you it's good enough not to even debate about this topic  :D

If there perfect 13th powers, they probably are able to block off a certain range of numbers just based on seeing the number, and then they probably raised numbers to the 13th power to test how far off they are instead of trying to take the 13th root.

[quote="krustyteklown"]newton's method for finding roots...
that's for functions : |[/quote]
Example:

$\sqrt[13]{n}$
Then $x^{13}=n\implies x^{13}-n=0$.  The rest is up to Newton's Method.

Computational Example:
$\sqrt[6]{2}$
$\\f(x)=x^6-2\implies f'(x)=6x^5\\ x_{n+1}=x_n-\frac{x_n^6-2}{6x_n^5}$.
$y=x$ approximately intersects $f(x)=x^6-2$ at $x_1=1$. We then obtain \[ \\x_2=1.16666667\\x_3=1.12644368\\x_4=1.1249707\\x_5=1.12246205\\x_6=1.12246205 \]
$\therefore$ $\sqrt[6]{2}=1.12246205$ to eight decimal places. :)

Masoud Zargar

The Trachtenberg Speed System can be used for very fast arithmetic (addition, subtraction, multiplication, division, squaring and square roots) but I believe finding the 13th root is a specialised technique which people take to the extreme.


geometry
3D geometry
HMMT
conics
ellipse
perimeter
trigonometry

The light rays from a spotlight spread out to form a solid cone, with the light rays making an angle of up to 30 degrees with the centerline of the beam. If the spotlight is 10 feet above the floor, and the centerline of the beam hits the floor at a 60 degree angle of inclination, what is the area the spotlight covers on the floor?

A) $25\pi$
B) $75\pi$
C) $\frac{100\pi}{3}$
D) $\frac{400\pi}{9}$
E) $25\sqrt{6}\pi$

This was the only problem that nobody could get at the Flora practice session today.  Mr. R said that he could solve it with a formula that he hadn't memorized.  So I think that Shobhit should be able to get this.  :rotfl:

EDIT:  I'll post a better diagram later.

Mr. R said they should have included a diagram as well.  He didn't get it on his first try, but I assume he's working on it.  The problem might be flawed.  It makes perfect sense to me what their asking, I just don;t know how to find the answer. :maybe:

yeah, this problem really needs a diagram... This is one of the reasons we should go to HMMT instead of College of Charleston.. :wink:

[hide="PWNED"]Ok, well finding the major axis isn't too hard to find (compared with finding the major axis...) so I'll leave out the details.  You find it to be $10\sqrt{3}$.  Now for the beast...the minor axis.

Ok, so the minor axis goes through the center $O$ of the ellipse and hits the ellipse at two points.  Let one of these points be $Q$ (we can just focus on one side so I won't even bother labelling the other one).  Let $OQ = x$. Now, using the $P$ from Shobhit's diagram, draw in right triangle $OPQ$.  It has legs of length $x$ and $\frac 5{\sqrt{3}}$.  Using Pythagoras we find that $PQ = \sqrt{x^{2}+\frac{25}3}$.

Then we can find that $QB^{2}= x^{2}+75$.  From right triangle $ABQ$ we get that $AQ = \sqrt{x^{2}+175}$.  Now we have triangle $APQ$ with $m\angle QAP = 30^\circ$ (the light that hits the actual perimeter of the ellipse is shining out at a full $30^\circ$ from the centerline).  Thus we have a triangle $APQ$ which we can use LoC on to get \begin{eqnarray*}{(\sqrt{x^{2}+\frac{25}3}})^{2}&=& (\sqrt{x^{2}+175})^{2}+(\frac{20}{\sqrt{3}})^{2}-2\cdot\frac{20}{\sqrt{3}}\cdot\sqrt{x^{2}+175}\cdot \cos 30^\circ\\ x^{2}+\frac{25}3 &=& x^{2}+175+\frac{400}3-20\sqrt{x^{2}+175}\\ x &=& 5\sqrt{2}\end{eqnarray*} (er, forgot to mention that you can find $AP=\frac{20}{\sqrt{3}}$...it's just a simple application of Pythagoras)

Ok, so half of major axis is half of $10\sqrt{3}$ or just $5\sqrt{3}$ and half of minor axis is $5\sqrt{2}$.  Thus our area is $25\sqrt{6}\pi$.  QED

Hm, Mr R...which of those formulas was too hard for you to memorize? [/hide]

hm. yes mr. r, pythagorus and law of cosines are quite hard to memorize... :noo:

remark, this problem is very similar to to this year's USAMTS problem (#5, round 4) Here is a link to the problem:
http://www.usamts.org/Solutions/Solution5_4_17.pdf
In my opinion--as well as a few others I talked to about this problem :wink:--I would consider this to be a #9ish AIME, so I have no idea how this would be on College of Charleston! I think that most people are just not good at working with ellipses and other conic sections, as shown by the rarity of this kind of problems on most math contests. But of course, after seeing this type of problem twice in one week, I should look into it! :D

I think Joe elaborates well on this issue:
[quote="joml88"]BTW, I am completely outraged they would put a problem like this on the level 2 test at C of C. They were totally off base when judging the difficulty level of this problem. It might have made a good problem for level 1...even then I'm not so sure it's even worth the students' time to solve such a trivial problem. Luckily the problem shouldn't take more than 35 seconds or so (maybe 36 seconds if you have to rederive cos 30).[/quote]


logarithms calculus derivative function limit algebra polynomial

Prove that the limit of $\frac{log x}{x^{n}}= 0$  for all n>0.

The context makes it clear that you're talking about the limit as $x\to\infty.$

Here's a different approach. Let $f_{n}(x)=\frac{\ln x}{x^{n}}.$ For all $x>1,\ f_{n}(x)>0.$ We calculate the derivative:

$f_{n}'(x)=\frac{1-n\ln x}{x^{n+1}}$

We have that $f_{n}'(x)<0$ for $x>e^{\frac1n}.$ So $f_{n}$ is an eventually decreasing function that is bounded below by $0.$ Hence it has a limit. Define $L_{n}=\lim_{x\to\infty}f_{n}(x).$

But now choose $m$ such that $0<m<n.$ (Side issue: I was never assuming $n$ was an integer - simply that it was a postive real number. And $m$ is another positive real number.) Then

$L_{n}=\lim_{x\to\infty}f_{n}(x)= \lim_{x\to\infty}\frac1{x^{n-m}}\cdot f_{m}(x)=0\cdot L_{m}=0.$

A very similar approach can be made to the question of showing that for $a>0,$ $\lim_{x\to0^{+}}x^{a}\ln x=0.$

[color=blue][Unnecessary quote removed by moderator.][/color]

Are there other approaches aside from derivative?

${\lim_{x\to \infty }\, \frac{\log (x)}{x^{n}}}$

${\lim_{x\to \infty }\, \log (x)\to \infty }$
${\lim_{x\to \infty }\, x^{n}\to \infty }$

L'Hospital Rule

${\lim_{x\to \infty }\, \frac{\log (x)}{x^{n}}=\lim_{x\to \infty }\, \frac{\frac{\partial \log (x)}{\partial x}}{\frac{\partial x^{n}}{\partial x}}}$
${\lim_{x\to \infty }\, \frac{\log (x)}{x^{n}}=\lim_{x\to \infty }\, \frac{1}{n x^{n}}=\frac{\lim_{x\to \infty }\, \frac{1}{x^{n}}}{n}}$

${n>0}$

${\lim_{x\to \infty }\, \frac{\log (x)}{x^{n}}=0}$

Make the substitution $x=e^{t}$; as $x\to\infty$ clearly $t\to\infty$.

So you need to show that $\frac{t}{e^{nt}}\to 0$ as $t\to\infty$.

Since $1+nt+\frac{(nt)^{2}}{2}<e^{nt}$, this is clear.

you could replace $\ln(x)$ with $\ln(1+x)$ and use taylor polynomials i guess.

One can use that for all $x>0$ we have
$\frac{x-1}{x}\le \ln x \le x-1$


integration
algebra
polynomial
complex analysis
complex analysis unsolved

let $ f$ have a power series espansion $ f(z)\equal{}\sum_{n\equal{}0}c_{n}z^{n}$ in $ D(0;R)$. Prove that, for $ r<R$,
\[ \int^{2\pi}_{0}|f(r\exp(it))\minus{}P(r\exp(it))|dt\] attains its minimumm over all polynomials $ p$ of degree $ k$ when $ p(z)\equal{}\sum^{k}_{n\equal{}0} c_{n}z^{n}$ :blush:

I'm pretty sure that you forgot the square: as written, the statement is just false ;)

Powers of $ z$ are orthogonal on each circle centered at the origin- it's the same as Fourier series in a slightly different context.


Hi,,,

um. since AMC8 is coming, I started thinking about preparing for it.. and I know AMC8 is not really discussion materials for this forum, but since it's AMC I'm posting it here

um... what should kind of materials should I study with and sample "mock" AMC8 will be highly appreciated..

and.. by they way, recommanding some mathbooks related to AMC8 will be highly appreciated as well..

best regards
from a kid

The best way to study for the AMC 8 would probably be to practice. You can use old tests, Mathcounts, Math League, etc. Also, if you haven't already, study the AoPS Vol 1 book, and take an Intro AoPS Course for a subject you're weak in, if possible.

Good luck!  :)

MATHCOUNTS is a great way to study for the AMC 8. Old tests will help greatly as well. I didn't start using the AoPS books till the AMC 10 (big mistake:P) but there's no doubt that those will help you. Also, u might want to take a look at some easier AMC 10 questions, as some of those questions will be in the middle-harder section of the AMC 8 (difficulty wise).


function
calculus
derivative
algebra
polynomial
inequalities
inequalities unsolved

for $0 \leq a,b,c \leq 1$ and $a+b+c=t$
find the minimum of $a^2b+b^2c+c^2a$
thanks

I will use some useful ideas from [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5E2y%2By%5E2z%2Bz%5E2x&t=24113]this page[/url].

Denote $f(a,b,c)=a^2b+b^2c+c^2a$. It's cyclic, so we can assume that $a$ is the biggest one. Now we show useful thing:
$f(a,b,c)-f(a,c,b)=a^2b+b^2c+c^a-a^2c-c^2b-b^2a=(a-b)(a-c)(b-c)$
It means that if $b>c$ then $f(a,b,c)$ is bigger and if $b<c$ then $f(a,c,b)$ is bigger. So we may assume $a\geq b\geq c$.

Let's have $0< z\leq c$, then:
$f(a,b,c) < f(a+z,b,c-z)$
It's because
$f(a+z,b,c-z)-f(a,b,c)=z(z^2+z(a+b-2c)+(b-c)(2a-b-c))$
and it's clearly $>0$ (every term is nonegative or positive cause of $a\geq b\geq c$).
So we have to only look for such $a,b,c$ where we can't do this mixing with $z$. It obtains cases where $a=1$ (we cannot add nothing to $a$) and if $c=0$ (we cannot make $c$ much smaller):

1. Let $c=0$. Then we have to maximize $f(a,b,0)=a^2b$, with condition $a+b=t$. If there isn't another condition $a,b\leq 1$, we can do this estimate: $a^2b=4\frac{a}{2}\frac{a}{2}b\leq 4\left(\frac{\frac{a}{2}+\frac{a}{2}+b}{3}\right)^3=\frac{4t^3}{27}$ with equality only when $a=\frac{2t}{3},b=\frac{t}{3}$. This is all ok when $t\leq \frac{3}{2}$. if $2\geq t>\frac{3}{2}$ we must look at condition $a,b\leq 1$. Let's look at function $g(x)=\left(\frac{2t}{3}-x\right)^2\cdot \left(\frac{t}{3}+x\right)=x^3-tx^2+\frac{4t^3}{27}$. It's derivative is $g'(x)=3x^2-2tx=3x(x-\frac{2t}{3})$. It means that this function is decreasing on interval $(0,\frac{2t}{3})$ and it means also that $f(a,b,0)<f(a+\epsilon,b-\epsilon,0)$ ($\epsilon > 0$) and it means that the biggest value of $f(a,b,0)$ with $\frac{3}{2}<t<2$ is attained when $a=1$ and $b=t-1$, so $f(1,t-1,0)$ is maximal.

2. Now suppose that $a=1$ => we are trying to maximize $f(1,b,c)=b+b^2c+c^2$. If $1\leq t < 2$ then we can prove this: $f(1,b,c)=b+b^2c+c^2\leq b+c$ (it's equivalent to $c(b^2+c-1)\leq 0$ and it's true, cause $b^2+c-1\leq b+c -1 \leq 0$, equality occurs when $c=0$ and $b=t-1$). So if $1\leq t \leq 2$ we have $f(1,b,c)\leq b+c=t$.

Now we just have to look at case $t>2$. We can prove $f(1,b,c)=b+b^2c+c^2\leq 1+(b+c-1)+(b+c-1)^2$ (it's equivalent to $(b-1)^2(c-1)\leq 0$ => it's true for all possible $b,c$. Also equality occurs when $b=1$ and $c=t-2$ or $b=t-2$ and $c=1$).

Adding all these informations together we get (I will describe only one solution for each case, althought there are cyclic ones...):
For $0<t$ or $t>3$ there are no such $a,b,c$.
For $0\leq t \leq \frac{3}{2}$ there is maximum for $a=\frac{2t}{3},b=\frac{t}{3},c=0$. $f(a,b,c)=\frac{4t^3}{27}$.
For $\frac{3}{2}< t < 2$ there is maximum for $a=1, b=t-1, c=0$. $f(a,b,c)=t-1$
For $2 \leq t \leq 3$ there are maximums for $a=1,b=t-2,c=1$ and $a=1,b=1,c=t-2$ (hmm, they are cyclic ;)) and $f(a,b,c)=1+(t-2)+(t-2)^2$

yeah,your solution is a nice solution for the maximum,but i want to find the minmum. ;)
for $0 \leq t \leq 1$ the answer is 0, but what about $1 \leq t \leq 3$

uuups  :blush:  :blush: , I didn't realize it :), sorry, but it would be more complex (having upper and lower bounds together ;) )

never mind ;)
and anybody try it,
i haven't solved it,either :oops:

I have got it. Can someone chceck it, please? I'll detail it better if it's unclear. It's very similar to finding maximal value (so better read also my previous post :D ). In there, we were supposing $a\geq b\geq c$. Now we can similary (WLOG) suppose that $a\leq b\leq c$. Also assume $t>1$ ($t\leq$ is easy). Denotation $f(a,b,c)$ remains. Now ($z>0$):

$f(a,b,c)-f(a-z,b,c+z)=a^2b+b^2c+c^2a-(a-z)^2b-b^2(c+z)-(c+z)^2(a-z)=z(z^2+z(2c-a-b)+(c-b)(c+b-2a))>0$
Which is certainly true. So we have $f(a,b,c)>f(a-z,b,c+z)$. It means that we just have to check those values $a,b,c$ where $a=0$ or $c=1$:

1. let $a=0$. Then we have to minimize $f(0,b,c)=b^2c$ with $b+c=t$. If $t\leq 1$ it's trivial. If not, we will prove that $f(0,b,c)=b^2c \geq (b+c-1)^2$. It's true, cause it's equivalent to $(1-c)(b(1-b)+(b+c-1))\geq 0$. Equality occurs only when $c=1$. So maximal value is attained when $b=t-1,c=1$.

2. Let $c=1$. Then we have to minimize $f(a,b,1)=a^2b+b^2+a$ with $a+b=t-1$ (and denote $t-1=m$) and $a,b<1$. There are three cases:

If $m<\frac{1}{2}$. Then $f(a,b,1)=a^2b+b^2+a\geq (a+b)^2$, cause it's equivalent to $a(ab+1-a-2b)\geq 0$. Equality occurs when $a=0, b=m$.

If $1<m\leq 2$ we can say it like this: We are trying to find minimum of polynomial $g(x)=x^2(m-x)+(m-x)^2+x$ (it's like $a=x,b=m-x$) on interval $m-1,1$. We have $g(m-1)=g(1)=1+(m-1)+(m-1)^2$. $g'(x)=-3x^2+2(m+1)x+(1-2m)$. $g'(x)=0$ have two solutions: $x=1$ and $x=\frac{2m-1}{3}$. Also we can show that $g\left(\frac{2m-1}{3}\right)\leq g(m-1)=g(1))$. So in interval $m-1,1$ we have minimum when $a=\frac{2m-1}{3},b=\frac{m+1}{3}$.

And at last if $\frac{1}{2}\leq m \leq 2$. We will use our polynomial $g(x)$, but now we are trying to minimize it's value on interval $0,m$. With the same argumentation we deduce that minimum is attainde in the same point: $a=\frac{2m-1}{3},b=\frac{m+1}{3}$

Together we have:
For $t<0,t>3$ there are no such $a,b,c$.
For $0\leq t\leq 1$ minimum is $0$.
For $1<t\leq \frac{3}{2}$ minimum is $(t-1)^2$
For $\frac{3}{2}<t\leq 3$ minimum is $\frac{(2m-1)^2}{3^2}\frac{m+1}{3}+\frac{(m+1)^2}{3^2}+\frac{2m-1}{3}=\frac{(2t-3)^2t}{27}+\frac{t^2}{9}+\frac{2m-3}{3}=\frac{4t^3-9t^2+27t-27}{27}$

[quote="zhaobin"]thanks ;)[/quote]
I was also interested in this result ;) It's quite strange, how the minimum and maximum depends on $t$, there can be done several generalizations, but I don't know how hard are they ;)

yes,inequality has so much secrets and interestings :)  :)

Hmm... I was trying to minimize $a^3b+b^3c+c^3a$ with same conditions but if I am not mistaken it's harder...

maybe this can help a little bit
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5Eny%2By%5Enz%2Bz%5Enx&t=2549


If $ 500 \minus{} 90 \plus{} 7 \equal{} 400 \minus{} 60 \plus{} 5 \plus{} N$, what is the value of $ N$?

Simplify, to get $ 417 \equal{} 345 \plus{} n$, then solve for N.

$ \boxed{n \equal{} 72}$


The  current Wiki word of the week says it's a word of the week until the 21st. It's the 22nd right now.

I could point out a number of other things.

-The past six weeks of WotW have no diffs, and for four weeks before that I myself put in the diffs.
-Azjps's suggestion about the WotW box has not been responded to, and is several months old.
-My month-old topic about unprotection of all pages hasn't been responded to, except by a moderator simply correcting a point in my argument.

I'm beginning to wonder if the admins still think the AoPSWiki is a good idea.

We still think it's a good idea and I've made some improvements to it some time ago (common login, latex and asymptote like on the forum, etc.) However we do have a lot of stuff going on right now that needs out attention so the Wiki has a low priority.

Okay, I just thought it might be too much, an extra thing on top of what already appears to be a mountain...


geometry
circumcircle
incenter
geometric transformation
homothety
ratio
perpendicular bisector

Suppose that a triangle $ABC$ has circumcenter $O$, and let $N$ be the center of the nine-point circle of the triangle $ABC$.
Choose a point $N^{\prime}$ such that $\measuredangle N^{\prime}BA = \measuredangle NBC$ and $\measuredangle N^{\prime}AB = \measuredangle NAC$.
Suppose the perpendicular bisector of the segment $OA$ meets $BC$ in $A^{\prime}$.
Define two points $B^{\prime}$ and $C^{\prime}$ similarly.
[b](a)[/b] Prove that the points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ lie on a line $l$.
[b](b)[/b] Prove that $l$ is perpendicular to $ON^{\prime}$.

Hmmm... I wouldn't say that yet. I get the feeling (b) is going to take a LONG time. :D  :D  :D

Yeah, the official solution ... it is really weird and long.

But maybe it might be a good one for you ... your favourite technique, you know ?

Ok, I do have a solution for (a), but I used a theorem which i'm not going to prove here (you can look it up on the Net; the proof isn't all that hard, it's just a basic application of Ceva's thm):

KARIYA'S THEOREM: Let I be the incenter of ABC and let D, E, F the projections of I on BC, CA, AB respectively. Take A1 on (ID, B1 on (IE and C1 on (IF s.t. IA1=IB1=IC1. then AA1, BB1, CC1 are concurrent.

Now for (a): Let A1 be the intersection between the perpendicular bisectors of OB and OC. By the same method we obtain pts B1 and C1. Then O is the incenter of A1B1C1 and by using Kariya's thm for triangle A1B1C1 we get AA1, BB1, CC1 concurrent. Then, by using Desargue's thm for triangles ABC and A1B1C1 we get the fact that the intersection pts of the pairs of lines (BC,B1C1), (CA, C1A1), (AB, A1B1) are collinear, which is exactly what we wanted.

This isn't exactly what I understand by a nice solution, but it's still a solution (I'm really not happy with it, it has no "poetry"  :D ). The bad thing is that I really don't know how to use any of this to prove (b). Maybe we need a different type of soln for (a), one which would help us approach (b)...

I've discovered a really interesting thing (by using a dynamic geometry program - Euklides):

I've defined A1, B1, C1 in my previous post. I've also proved that AA1, BB1, CC1 are concurrent. Well, guess what their concurrency point is. That's right, it's N' (unfortunately I don't have a soln for this, I've just observed it). This is the only thing I have to prove in order to have a complete soln (it's not obvious from what I've said so far; the rest of the solution involves an inversion; I'll post it when or if I can prove the thing Ive observed).

Man! I feel like I'm so close, yet so far...

I've managed to prove that thing I mentioned in my previous post, but let's assume it known for now (it's a computational proof and I don't know if I'll post it; I'll post the main idea though).

So we assume we know that N' is on AA1, BB1, CC1. Let T, U, V be the midpts of OA, OB, OC respectively, and X, Y, Z be the midpts of BC, CA, AB respectively. Now consider the inversion of pole O and power R^2/2. The circumcircles of XOT, YOU, ZOV have a second common point O' because they have a common point O and their centers are collinear (the line is obtained from l by a homothety of center O and ration 1/2). What we must prove is that N' is on OO'. By the inversion I've mentioned the circles XOT, YOU, ZOV turn into the lines AA1, BB1, CC1 respectively. These lines intersect at N', which must then be the image of O' (the other common point of the 3 circles which have the 3 lines concurrent at N' as images), so O, N', O' are collinear, which is what we want to prove.

The idea to prove that AA1, BB1, CC1 are concurrent at N':

Let's denote by (b,c)(X) the ratio between the distance from X to the side b and the distance from X to the side c. It would be enough to prove that (b,c)(N') = (b,c)(A1), which means that N' is on AA1 (it's obvious that we would use an analogous method to prove N' to be on BB1, CC1). Since N' and N are isogonal points (this means that thing abt the angles) we get    (b,c)(N')=1/(b,c)(N). (b,c)(N) isn't at all hard to calculate, it's just boring (we get it in terms of cos and sin of A, B, C). The same could be said abt (b,c)(A1). To calculate this one I took the intersection point between the perpendicular at B to OB and the one at C to OC (call it K) for which we can easily calculate the distances to the sides b and c. A1 is the midpt of OK, so we can calculate d(A1,b) (for example) as (d(O,b)+d(K,b))/2 etc.

I hope it's clear enough.

[quote="Arne"]Suppose that a triangle $ABC$ has circumcenter $O$, and let $N$ be the center of the nine-point circle of the triangle $ABC$.
Choose a point $N^{\prime}$ such that $\measuredangle N^{\prime}BA = \measuredangle NBC$ and $\measuredangle N^{\prime}AB = \measuredangle NAC$.[/quote]

This means that the point N' is simply the isogonal conjugate of the nine-point center N of triangle ABC. This point N' is called the [b]Kosnita point[/b] of triangle ABC; its main property of this point is the following: If X, Y, Z are the circumcenters of triangles BOC, COA, AOB, then the lines AX, BY, CZ concur at N'.

[quote="Arne"]Suppose the perpendicular bisector of the segment $OA$ meets $BC$ in $A^{\prime}$.
Define two points $B^{\prime}$ and $C^{\prime}$ similarly.
[b](a)[/b] Prove that the points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ lie on a line $l$.
[b](b)[/b] Prove that $l$ is perpendicular to $ON^{\prime}$.[/quote]

The solution is immediate using the [url=http://www.mathlinks.ro/viewtopic.php?t=4337]Sondat theorem[/url]. Indeed, consider the circumcenters X, Y, Z of triangles BOC, COA, AOB. Then, since the circumcenters Y and Z of triangles COA and AOB both lie on the perpendicular bisector of the segment OA, the line YZ is the perpendicular bisector of the segment OA, and hence this line meets the line BC at the point A'. Similarly, the lines ZX and XY meet CA and AB at the points B' and C'.

Now, since the line YZ is the perpendicular bisector of the segment OA, we have $YZ\perp OA$, and similarly $ZX\perp OB$ and $XY\perp OC$, so that the perpendiculars from the vertices A, B, C of triangle ABC to the sidelines YZ, ZX, XY of triangle XYZ concur at the point O.

On the other hand, the circumcenters O and X of triangles ABC and BOC both lie on the perpendicular bisector of the segment BC. Hence, $BC\perp OX$, and again we have by analogy $CA\perp OY$ and $AB\perp OZ$. Thus, the perpendiculars from the vertices X, Y, Z of triangle XYZ to the sidelines BC, CA, AB of triangle ABC concur at the point O, too.

Therefore, the triangles ABC and XYZ are orthologic, and both of their orthologic centers coincide with O. Now, since the lines AX, BY and CZ concur at the point N', it follows from the Sondat theorem that the intersections of the lines YZ, ZX, XY with the lines BC, CA, AB - i. e. the points A', B', C' - lie on one line l, and the points O, O and N' lie on one line perpendicular to l. In other words, the line ON' is perpendicular to l. This solves the problem.

But I don't know any elegant proof for the Sondat theorem. Do you know any good proof for your problem, so I could see whether it generalizes?

  Darij

I know this is probably too much to ask, but does a simpler solution exist? cuz this question is also listed in a hmwrk problem at MOP 2002, as a "medium hard" problem, and by the looks of it, it should be VERY hard.....

see [i]http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2002-ma/hwpages/02mosphw.pdf[/i]


calculus
trigonometry
complex analysis
real analysis
real analysis unsolved

I did pretty okay on this practice test, but I never took complex analysis, though I've tried to teach myself the rudiments. Maybe that's not even the problem, but this seems kind of basic, something I shouldn't be missing. So I'm looking for help with how to solve this problem, and also suggestions for preparing for the exam so that I don't miss this category of problem come test day. Here it is:

If z = e^(2*pi *i / 5), then 1 + z + z^2 + z^3 +5*z^4 + 4*z^6 + 4*z^7 +4*z^8 +5*z^9 =

a)0

b) 4*e^(3*pi*i / 5)

c) 5*e^(4*pi*i / 5)

d) -4*e^(2*pi*i / 5)

e) -5*e^(3*pi*i / 5)

Thank you everyone!

This is not a question in complex analysis, but simply in the use of complex numbers.  (Complex analysis is, roughly, what happens when you do calculus on the complex numbers.)  One needs the following

[b]Theorem[/b] (or definition, or whatever):  $ e^{ix} \equal{} \cos x \plus{} i \sin x$.

$ z \equal{} e^{ \frac{ 2 \pi i}{5} }$ is then the complex number $ \cos \frac{2 \pi}{5} \plus{} i \sin \frac{2 \pi}{5}$.  It is possible to evaluate these sines and cosines, but not useful, for there is a second

[b]Theorem[/b] (de Moivre):  $ (\cos x \plus{} i \sin x)^n \equal{} e^{inx} \equal{} \cos nx \plus{} i \sin nx$.

This allows you to compute $ z^2 \equal{} e^{ \frac{ 4 \pi i}{5} }$, etc.  The important thing to note here is that

$ z^5 \equal{} e^{2 \pi i} \equal{} \cos 2\pi \plus{} i \sin 2 \pi \equal{} 1$

and therefore that in the problem we have $ 4z^6 \equal{} 4z, 4z^8 \equal{} 4z^3$, and so forth.  It is also useful to know that because $ z^5 \equal{} 1$ and $ z \neq 1$ we know that

$ \frac{z^5 \minus{} 1}{z \minus{} 1} \equal{} 1 \plus{} z \plus{} z^2 \plus{} z^3 \plus{} z^4 \equal{} 0$.

I believe that's all you need to do this problem.  Is anything still unclear?

You don't need complex analysis for this question - just complex arithmetic, the complex exponential, and roots of unity. (I plan to put a few days on just that into the later part of my Calculus II class this semester.) And you need to know about finite geometric sums, which a good mathematics student should really have learned in high school.

The important thing is that if $ z \equal{} e^{2\pi i/5},$ then $ z^5 \equal{} 1.$

What you have there is the sum of two geometric sum:

$ \sum_{k \equal{} 0}^4z^k \plus{} 4\sum_{k \equal{} 4}^8z^k \plus{} 5z^9 \equal{} \frac {1 \minus{} z^5}{1 \minus{} z} \plus{} \frac {4(z^4 \minus{} z^9)}{1 \minus{} z} \plus{} 5z^9.$

Since $ z^5 \equal{} 1,$ we have both $ 1 \minus{} z^5 \equal{} 0$ and $ z^4 \minus{} z^9 \equal{} 0.$ That leaves the sum equaling $ 5z^9 \equal{} 5z^4.$ That's actually answer e), which takes some careful looking.

[color=green][Was composing while T0rajir0u posted. He and I saw the same things.][/color]

One thing I have noticed about such problems is that there is always a time-saving way to discover the answers to these test questions. That is a trick to keep in mind.

I'll disagree with Kent though and say no one ever taught me how to calculate geometric sums in HS. IMO every math student is stunted by American public high schools.

Here's another kind of silly way to get the answer without any real computation.

Reducing the exponents mod $ 5$, we get that the sum is equal to $ 1\plus{}5z\plus{}5z^2\plus{}5z^3\plus{}10z^4$.  We can write the sum as $ [1\plus{}(5z\plus{}5z^4)\plus{}(5z^2\plus{}5z^3)]\plus{}5z^4$, where it's clear that the sum in square brackets is real.  (It's actually equal to $ 0$, but we don't need that).

Now equate imaginary parts to get that the answer is e).


geometry
quadratics
algebra
quadratic formula
system of equations

What should a good school offer for math classes? Disregard budget and stuff.

I think it should be half regular classes like what they have now (Algebra II, Geometry, etc.), and half problem solving. There's no point in learning a bunch of concepts and formulas if you don't know how to use them. Besides, problem solving is fulfilling and interesting.

I have little idea what the teaching of maths is like currently in the US, but I can't really agree with this idea. Sure, the most able (perhaps the top two to five per cent) would really benefit from problem solving, but the remaining 95% would then find maths even more boring, and would drop maths earlier.

Again, I don't know how maths is assessed outside the UK, but here everyone takes a GCSE paper at age 16, before having the choice whether to carry it on for another two years for an A level (typically, one would take 3 or 4 subjects for A level, compared with 9-11 for GCSE). The content of GCSE seems to be learning a load of formulas and concepts (like the quadratic formula) and using them to answer about 20 meaningless questions. I don't think this helps anyone very well.

What most people need from school mathematics education (at least when it is compulsory) is to become 'numerate'. Nearly everyone will need, during their life, to be able to add/multiply/subtract in their head easily, to work out interest rates and stuff (with a calculator) and be able to manipulate measurements for things like DIY. Most people, I think should learn numeracy in school- in fact, much the opposite of problem solving.

For the few who are able mathematicians, and those who find problem solving 'stimulating and interesting' (though I can't think why you'd think that if you weren't good at it), there should be an alternative, more 'real maths'- proofs, problem solving and that kind of thing like you said. Of course, even out of those, few will become actual mathematicians, but those who don't will benefit from a kind of mental stimulation from problem solving.

Basically, I think maths education should be much more two-tier, a kind of 'maths required for life' option for the majority, and a 'problem solving' option for the best, though you would get a choice between the two, rather than it being based entirely on your ability. (I know this idea would never be accepted in the UK, because it would be seen as elitist).

You have to realize that the normal math classes are really hard for normal people. They have enough trouble just doing the normal stuff, using it creatively is a real challange. To be able to problem solve, you have to have creativity that the normal person doesn't have.

Normal math is hard for most people because it isn't taught well. Most great mathematicians do not become teachers. Professors, yes, but not high school teachers. Thus, a lot of high school math teachers aren't able to convey the real meaning behind all the math, and then math seems like just syntactic manipulation. ("How do I solve this system of equations? I have to do a bunch of funny operations on matrices. Why? No clue.") I've noticed that the main skill necessary for success in high school math is memorization. A lot of people aren't good at memorization, so they don't do well in math. The problem is that if the curriculum is changed to focus more on problem solving, then, as mustafa said, there will still be people who don't have the necessary skills - in this case, creativity.

My proposed solution is to offer both kinds of math class as electives. The people who need math - those who want to become physicists, statisticians, economists, engineers, etc. - can take the ordinary math classes. The people who enjoy math, such as the AoPSers, can take the problem solving classes (and the ordinary math classes as well if their topics are not covered in the problem solving classes). If people won't need math for their profession and don't enjoy math, why should we force them to take it?

The common answer to this question is that math is like exercises for the brain. But ordinary math isn't. Ordinary math is, as I already stated, mainly memorization. But if we want to give students "brain exercises", it may be possible to have classes that are accessible to everyone but encourage creativity more than regular math classes do. I'm thinking about the same idea behind recreational math books, like those by Martin Gardner, which are accessible even to people without much mathematical background, and are interesting enough that they've had pretty good sales, compared to more math-intensive books. Perhaps a math class with the same focus as such books could encourage creative mathematical thinking without being too difficult. This would not be a "class" in the conventional sense, as the goal would not be to teach subject matter but rather to foster creativity and encourage exploration, although the students would learn the mathematical concepts behind the subjects. An idea for a more conventional class would be to choose a topic related to math and use this topic to motivate the related mathematical topics. For example, my university offers an introductory math course which teaches mathematical concepts in the context of cryptology.

No matter how good your school gets, a math class is a math class.  It is impossible to rely on a math class for any type of math competition.

I think the reason they don't do "problem solving" classes is the level of interest.  Yes, the AoPS community (as well as me and you, hopefully) feels that "problem solving is fulfilling and interesting," but the number of people in one school who feel that may be, rounded to the nearest 10, zero.  The point of a school is to prepare you for the real world.  The real world seems to not realize that "(Algebra II, Geometry, etc.)" is not the way to go in terms of stimulating the math part of your mind.

If you want to know about a good school, I think Andover has postings online or something.  I go to a pretty good private school, and they have "regular classes" plus college level "regular classes."

Problem solving is very important.  I mean, the odds that you'll run a problem like the one below in everyday life is very unlikely.

Solve for $x$: $1.02^{32}x-x=1,000,000$

You may want to know how much money you'd need to put into a bank that offers 2% interest per year to earn 1,000,000 dollars in 32 years.  Of course, you'd use the equation above to find the answer.

Pre-Algebra--AlgebraII, pre-calc, calc, geometry(all levels), number theory(all levels), trigonometry(all levels), counting a d probability(all levels) and problem solving(all levels)

[quote="13375P34K43V312"]Disregard budget and stuff.[/quote]
You never can. "Budget and stuff" will always be an issue, along with the question of what teachers are available to teach the courses in question. The supply of teachers who would be comfortable teaching a problem-solving course may be limited in some areas. And even if you have a teacher or two, they're likely to also be the best teachers of the conventional courses - what is the highest and best use of those teachers?

A good school needs to be practical. Disregarding things isn't going to help with that.

On the subject of being practical, make an estimate of how many students in your school would be interested in taking a "problem solving" course. Most people are struggling enough with regular scohol math and don't have time to pursue problem solving.


Solve for "x" in the following equations.

a. (4-x)3=36

b. 6x+5=53

c. -4x+9=17

d. -0.25x-9=17

[hide="a"]-8[/hide]
[hide="b"]8[/hide]
[hide="c"]-2[/hide]
[hide="d"]-104[/hide]
[hide="e"] :sqrt: 1 =  :pm: 1[/hide]

I like the extension problem..

a)[hide]-8
[/hide]
b)[hide]8[/hide]
c)[hide]-2
[/hide]
d)[hide]-104[/hide]
e)[hide]x= :pm: 1[/hide]

SoccerBrainy40,

Thank you for providing the steps.  I think that's far more helpful to other students than simply declaring an answer.

[hide]a.
x=-8
b.
x=8
c.
x=-2
d.
x=-104
e.
x=1 or -1[/hide]
[color=darkred][size=75][Edited by moderator rcv the font color thief to hide answers.][/size][/color]

[quote="Treething"]extension f

Make a generalization for the answer to this equation:

x^n - 1 = 0

n is an even number[/quote]

Not sure if this is right, but...
[hide]
The only number that you can subtract one from and get zero is one. So now we have to figure out what number, to an even power, will give us one. Obviously, the two numbers are 1 and -1.

Therefore, if n is an even number in x^n - 1 = 0, then x must be 1 or -1.
[/hide]

Just curious...is there that symbol with the addition sign on the top and the subtraction sign on the bottom in the Emoticons panel. If there is, I can't seem to locate it. :lol:

[quote="shinwoo"]actually.
[hide="watchy soccerbrainy40"]it could be any number if n was a zero....[/hide][/quote]

Ahh, true! Didn't think of that!

shinwoo has the right answer! Mine is wrong! :P

$n=0$ fails for $x=0$ because $0^0$ is undefined.

Hello. I think you guys should learn to use the $LaTeX language so that everyone can understand what you are writing more clearly. :)
This language isn't hard so pls try it!!
Thanks!!

[b]@ ^^, hey, I'll try it.

a. $(4-x)3=36$. Expanding gives, $12-3x=36$. Solving gives us, $x=-8$

b. $6x+5=53$. Moving $5$ to the other side gives $6x=48$. Thus, $x=8$

c.  $-4x+9=17$. Same process. Conclusion: $x=-2$

d.  $-0.25x-9=17$. Adding $9$ to both sides, $-0.25x=26$. Then again, $x=-104$.

e. $x^2=1$. Take the square root of both sides gives $x=\pm\sqrt{1}$. Therefore, $x=1$, or $x=-1$.
[/b]


[quote="Albanian Eagle"]find the smallest real k so that the inequality
$a^{3}+b^{3}+c^{3}<k(a+b+c)(ab+bc+ca)$
holds for all a,b,c sides of a triangle.[/quote]
Maybe not very hard.
k=1
Let $a=x+y,\dots$
the inequality is equivlent to
$5\sum (x^{2}y+y^{2}x)+18xyz\geq 0$
and if Let $a=b=1,c\to 0$ $k\geq 1$

no it's not hard at all! :)
but it's from Bulgarian olympiad.

note for k=1
it's equivalent to $3abc+\sum a^{2}(b+c-a)\geq 0$ which is obvious.


A circular ceiling fan rotates at a constant speed of 80 revolutions per minute. A point halfway between the center of the fan and the outer edge of the fan travels 97,968 inches in 15 minutes. How far does a point on the outer edge of the fan travel in 30 minutes?

Make that triple posting.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=323335
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=323337

lol  my middle school makes its own school round...
because that question want too hard


How many 1'' cubes will pack into a rectangular box with dimensions 8'' by 6'' by 6.5''?

Since you can't pack 1/2 a cube into the box, the number of cubes you can pack in is $ 8 \times 6 \times 6$ or $ 288$.

I see, i forgot about the part about only a whole #.  I never rounded 6 to 6.5

[hide=Formal Answer]The dimensions of the box are $8"$ by $6"$ by $6.5"$

The number of cube lengths that can fit in the $6"$ is $6$.

The number of cube lengths that can fit in the $6.5"$ is also $6$.

The number of cube lengths that can fit in the $8"$ is $8$.

The total number of cubes: $6*6*8=288$ cubes
[/hide]


email
LaTeX
calculus
derivative
calculus computations

im having trouble with these extra credit problems from the teacher..

[img]http://myspace-878.vo.llnwd.net/00703/87/81/703681878_l.jpg[/img]
[img]http://myspace-830.vo.llnwd.net/00703/03/83/703683830_l.jpg[/img]
[img]http://myspace-456.vo.llnwd.net/00703/65/45/703685456_l.jpg[/img]

I would really appreciate any ones help on solutions not just answers
please email me

 got_bass_69@hotmail.com

First you need to do your own homework or extra credit. its good for you. $%Error. "texbfl" is a bad command.
{2^{nd}}$ use latex!

[hide="Hint"]$v\left(t\right)=s'\left(t\right)$

What do you know about max. / min.?  What happens to the first and second derivatives?  It should be pretty straightforward after that.[/hide]


[quote="DaReaper"]Rajan owns a special magical unicorn.  To ensure that the unicorn does not iscape, Rajan ties the unicorn to a rope and the rope to the corner of a square barn. If the side of the barn is 10 meters and the rope length is 16 meters, what is the area of the space the unicorn can roam?

**i have the answer just no clue how to solve it** :|[/quote]
[hide]So you draw a square.(I would draw one, but I'm too tired.) Then from one corner draw a line that is 16. Then draw a circular area to the adjacent vertices. SO it should like 3/4 of a circle. But since that side is 10 it will go over by 6 and form a quarter circle. So you have one 3/4 circle with radius of 16, and two quarter circles with radii 6. So solving for the area of the 3/4 one is $192\pi$. one of the quarter circles is $9\pi$. So multiplying that by two and adding that to the area of the first one the area is $210\pi$.[/hide]

thats what i got but the answer is $256\pi$

[quote="Quevvy"][quote="DaReaper"]Rajan owns a special magical unicorn.  To ensure that the unicorn does not iscape, Rajan ties the unicorn to a rope and the rope to the corner of a square barn. If the side of the barn is 10 meters and the rope length is 16 meters, what is the area of the space the unicorn can roam?

**i have the answer just no clue how to solve it** :|[/quote]
[hide]The 3/4 of the circle's area is $\frac{(\pi 16^{2})3}{4}=192\pi$
Then there area of the 2 quarter circles is $\frac{\pi 6^{2}}{4}*2=18\pi$ Then you add $192+18$ to get $210\pi$

If the answer was $256\pi$, then that would mean that you could go through the barn but usually you wouldn't be able to do that...[/hide][/quote]
Well, Dareaper never stated if you could go through the barn or not. I guess for this problem you can...

Well, probably that's what the problem means when they say he has a special unicorn.  Whoever wrote it (manufacturer) should make it more clear, if the unicorn couldn't, the answer you got is completely correct.

yeah. they should make it more clear. If this was from a MAO test I would definitely dispute this question.