algebra
polynomial
function
number theory solved
number theory

I found this problem in the problem seminar of D. J. Newman which can be found on Kalva (problem 17):

Given $n$, find the smallest degree of a monic polynomial which takes values in $n\mathbb Z$ in all integers. I think that if $n=p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_k^{a_k}$, then the required polynomial is $LCM((x^{p_1}-x)^{a_1},\ldots,(x^{p_k}-x)^{a_k})$. However, I couldn’t manage to express the degree of this polynomial in a simpler form as a function of $n$ (didn't actually try that hard :)), and I don't even know whether this is correct or not. What do you guys think?

The problem says that the polynomials have real coefficients, but I think it can be shown that if a monic polynomial takes integer values in all integers, then it has integer coefficients.

Your last conjecture is false Grobber : The polynomial $P(x)=x^3 + \frac {x(x+1)} 2$ is a monic one and only takes integer values for $x$ integer, without having all its coefficients which are integers.

Concerning the problem :
The polynomials $P_k(x) = C_x^k = \frac {x(x-1) \cdots (x-k+1)} {k!}$ form a basis of $R[x]$.
Use the fact that a polynomial only takes integers values for integer values of $x$ if and only if it is an integer linear combination of the $P_k$.

Thus, if a monic polynomial $P$ is such that $P(Z) \subset nZ$ then $P =n( \sum_{k=0}^m a_kP_k)$ where all the $a_k$'s are integers.
Considering the leading coefficient, we see that $1= \frac {n \cdot a_m} {m!}$, qo that $\frac {m!} n = a_m$ is an integer, which forces $n$ to divide $m!$.

Conversely, if $n$ divides $m!$ then just choose $P(x)= n \cdot (\frac {m!} n )P_m(x)$ to get a polynomial with the desired properties.

Thus, the minimal degree is $m$, where $m$ is the least integer such that $n$ divides $m!$.

Pierre.

I see.. Thank you. I was totally off :). Besides, that thing about integer coefficients was really stupid :?.


calculus
integration
ceiling function
floor function
inequalities
combinatorics proposed
combinatorics

[quote="Valentin Vornicu"]A positive integer given in decimal representation $\overline{ a_na_{n-1} \ldots a_1a_0 }$ is called [i]monotone[/i] if $a_n\leq a_{n-1} \leq \cdots \leq a_0$. Determine the number of monotone positive integers with at most 1993 digits.[/quote]
$\binom{2001}{1993}-1$
by recurrences

you really don't need any recurrences; just start with any integral non negative solution to $x_0+x_1+\cdots+x_9=1993$. here $x_i$ denotes the number of digits that are equal to $i$. Then arrange those digits in ascending order. since we finally seek only positive integers, we throw away the solution corresponding to $x_0=1993$ and then we are through.

[quote="jin"]$\binom{2003}{9}-1$[/quote]

I agree with this one.

[hide]
We create these magic things called "counters" that make the next digit larger than the last. We can place counters at the very beginning, at the very end, or anywhere in between that we want. We also happen to be able to place multiply counters in the same space between two successive of the 1993 digits (note that "at least 1993" can be ignored if we just allow leading zeroes).

Let $\frac{\lceil n \rceil}{\lfloor k \rfloor}$ denote the number of ways to place counters.

Then we see that $\frac{\lceil n \rceil}{\lfloor k \rfloor} = \frac{\lceil n-1 \rceil}{\lfloor k \rfloor} + \frac{\lceil n \rceil}{\lfloor k-1 \rfloor}$

You can start drawing out a pasca-like triangle to descompose it, and you can see and also easily prove that $\frac{\lceil n \rceil}{\lfloor k \rfloor} = \binom{k-1}{0} + \binom{k}{1} + \binom{k+1}{2} + ... + \binom{n+k-1}{n} = \binom{k-1}{k-1} + \binom{k}{k-1} + \binom{k+1}{k-1} + ... + \binom{n+k-1}{k-1} = \binom{n+k}{k}$.

We wish to calculate $\frac{\lceil 1994 \rceil}{\lfloor 9 \rfloor}$, and this is of course $\binom{2003}{9}$. Subtract one because zero is invalid, and you get the answer of $\binom{2003}{9}-1$.
[/hide]

My apologies for reviving an old topic, but I believe that the answer above is incorrect.  Suppose that you have at most $ d$ digits.  Then it follows that you can construct a $ (d-1)\times 9$ grid and find the number of ways to reach $ (d-1,1),\hdots, (d-1,9)$. The the answer is clearly $ \sum_{k=1}^{9}\binom{d+k-1}{k}=\binom{d-1+9+1}{9}-1=\binom{d+9}{9}-1$.  For $ d=1993$, we have $ \binom{2002}{9}-1$.

Any positive integer having at most $ 1993$ digits can be written as $ \overline{a_{1}a_{2}\cdots a_{1993}}$, where the $ a_{i}$ are digits and [b]leading zeroes are allowed[/b]. Such a positive integer is monotone if and only if the numbers $ b_{j}$, given by $ b_{j}\equal{} a_{j}\plus{}j$, satisfy the chain of inequalities $ b_{1}< b_{2}< b_{3}<\cdots < b_{1993}$. Since $ b_{1}\geq 1$ and $ b_{1993}\leq 1993\plus{}9 \equal{} 2002$, there is a bijection between the nonnegative monotone integers having at most $ 1993$ digits, and the strictly increasing sequences having exactly $ 1993$ terms, belonging to the set $ \{1,2,\,\cdots,2002\}$. Since there are (obviously) $ {2002\choose 1993}\equal{}{2002\choose 9}$ such sequences, it follows that the answer is $ {2002\choose 9}\minus{}1$. (Indeed, the sequence $ 1,2,\,\cdots,1993$ "represents" the monotone integer $ 0$, which is not allowed since it is not positive. Sorry for bumping this, but I like this approach, and I didn't see it in the other posts.

This is just balls and urns, for any set of 1993 digits we pick we can order them one way, and then we can visualize 1993 digits with 9 dividers separating each type of digit, 0 counted so it would be $ {2002 \choose 9}$, but we cant have 000000......
Why are all these solutions so long and complicated? (I know this is similar to arne's but with balls and urns it just involves the digits 0-9 rather than that strictly increasing sequence arne had).

[quote="dgreenb801"]Why are all these solutions so long and complicated?[/quote]  I see three correct solutions on this thread, which range in length from 3 to 8 lines of text.  None of them is particularly long or complicated.  They are slightly longer than yours because other people provided some explanation or proof where you just used the words "balls and urns" and expected your reader to understand.  $ \textrm{Less wordy} \not\equiv \textrm{Better}$ in general.


Imagine that you are on the surface of the moon, playing golf.

So, is it possible to hit a golf ball on the surface of the moon and have it achieve a stable orbit around the moon?

Theoreticaly yes. Required velocity is v=sqrt(GM/R).

But the problem is that the orbit would return to the point at which the ball was struck (well, the equivalent point on the surface of the moon after the moon rotates during the period of the orbit). The collision with the surface would be an obstacle to stability.

We would need to work out the exit velocity required by gravitational forces.

If we tee'd it up on the highest point, the highest mountain of the moon, there might be a chance to achieve this, right ?

But, I remember that the moon has light gravity.

It would be possible to do it with Newton's cannon ball. This is not the point.
I would like to know if a golfer can do it, without using a cannon ball.

Let $R$ be the radius of the moon. It's about 1700 km or $1.7\times 10^{6}$ meters. Let $g$ be the acceleration of gravity at the moon's surface: it's about $1.7\text{ m/s}^{2}.$

We achieve circular orbit right at the surface if the acceleration of gravity exactly matches the acceleration needed to maintain uniform circular motion: that is, if $g=\frac{v^{2}}{R}$ or $v=\sqrt{gR}.$ This would be about $1700$ meters per second, which is probably 40 times the speed any human golfer could swing a club. (And remember that 40 times the speed is 1600 times the kinetic energy.)

This is not within the right order of magnitude.


inequalities
geometry
real analysis
real analysis unsolved

Let f : [0, 1] -----> [0, 1] × [0, 1] be C1 (differentiable with a continuous
derivative). Show that:

1. f is not surjective.

2. The range of f doesn't contain any open ball .

Note that the range of $ f$ is measure zero in $ \mathbb{R}^2$, since $ f$ is Lipschits on $ [\epsilon, 1\minus{}\epsilon]$ for any sufficiently small $ \epsilon > 0$. This proves both 1 and 2.

[hide="Solution"]
Let $ \epsilon \equal{} \tfrac{1}{m}$ ($ m \geq 3$) and put $ I_m \equal{} [\epsilon, 1\minus{}\epsilon]$. Since $ \| f' \|$ is continuous on $ I_m$, there is $ M > 0$ such that $ \| f' \| \leq M/\sqrt{2}$ on $ I_m$.

Applying mean value theorem to each component of $ f' \equal{} (f_1', f_2')$ yields the inequality $ \| f(x) \minus{} f(y) \| \leq M |x \minus{} y|$.

Now let $ \eta > 0$ be given and $ \Gamma \equal{} \{ x_0 \equal{} \epsilon, x_1, \cdots, x_{n\minus{}1}, x_{n} \equal{} 1\minus{}\epsilon \}$ be the partition which divides $ I_m$ into $ n$ subintervals with equal length $ r \equal{} \frac{1\minus{}2\epsilon}{n}$ so that $ r < \frac{\eta}{2\pi M^2}$.

Put $ B_i \equal{} B_{rM}(f(x_i))$. We claim that $ \{ B_0, \cdots, B_n \}$ covers $ f(I_m)$.

If $ t \in I_m$, then $ |t \minus{} x_i| < r$ for some $ i$. Then $ \| f(t) \minus{} f(x_i) \| \leq M |t \minus{} x_i| < rM$, hence $ f(t) \in B_i$.

This shows that $ f(I_m) \in B_0 \cup \cdots \cup B_n$. Also,

$ (n\plus{}1) \cdot \pi (Mr)^2 \equal{} \frac{(n\plus{}1)}{n} \pi M^2 (1 \minus{} 2\epsilon) r  < 2\pi M^2 r < \eta$

shows that $ f(I_m)$ can be put into a set with area less than $ \eta$.

Since $ \eta$ is arbitrary, we can replace $ \eta$ with $ 2^{2\minus{}m} \eta$.

Finally, note that $ f([0, 1]) \equal{} \{ f(0) , f(1) \} \cup \bigcup_{m\equal{}3}^{\infty} f(I_m)$ and $ \sum_{m\equal{}3}^{\infty} 2^{2\minus{}m} \eta \equal{} \eta$.

Any open ball $ B_{R}(p)$ has area $ \pi R^2 > 0$, and if we choose $ \eta$ sufficiently small so that $ \eta < \pi R^2$, then $ f(I)$ cannot contain $ B_{R}(p)$. This proves both 1 and 2.[/hide]


Conside a sequence as follow
$ a_{1}\equal{}a$
$ a_{n\plus{}1}\equal{}\frac{1}{2}.log(a_{n}^{2}\plus{}1)\minus{}2004$
Prove that this sequence converge .

Consider $ f(x) \equal{}\frac{1}{2}ln(x^{2}\plus{}1)\minus{}2004$, then $ f'(x) \equal{}\frac{x}{x^{2}\plus{}1}$ and $ |f'(x)|\le frac 12 < 1$.
Therefore exist unique fixed point $ x_{0}$ (root of equation $ x \equal{} f(x)$) and $ \lim_{n\to\infty }a_{n}\equal{} x_{0}$.
Obviosly $ |a_{n}\minus{}x_{0}|<C*2^{\minus{}n},\ C\equal{}|a_{0}\minus{}x_{0}|$.


try this easy but time consuming problem :D

  The crossed ladder problem. In an alley a ladder length $3$ is propped up against one wall with its foot at the bottom of the other wall. It crosses another ladder length $2$ leaning the other way a distance $1$ from the ground. How wide is the alley? [In other words we are given a quadrilateral ABCD. $\angle{ADC}$ and $\angle{DCB}$ are right angles. $AC = 2$, $BD = 3$ and $AC$, $BD$ intersect at $X$ a distance $1$ from $CD$. Find $CD$.]

I think I have it.
[hide]For brevity let [tex]|AC|=2=a,\,  |BD|=3=b,\, \angle{ACD}=\alpha,\, \angle{BDC}=\beta[/tex] additionally let's project [tex]X[/tex] onto [tex]CD[/tex] to get [tex]E[/tex]. Then [tex]|XE|=1=d,\, |CE|=x,\, |ED|=y[/tex]. What we are to find is [tex]|CD|=x+y=h[/tex]. So we have:

[tex]\frac{h}{a}=\cos\alpha,\ \ \frac{h}{b}=\cos\beta[/tex]

[tex]\frac{x}{d}=\cot\alpha,\ \ \frac{y}{d}=\cot\beta[/tex]

[tex]h=x+y=d (\cot\alpha +\cot\beta )=[/tex]
[tex]=d \left (\frac{\cos\alpha}{\sqrt{1-\cos^2\alpha}}+\frac{\cos\beta}{\sqrt{1-\cos^2\beta}}\right )=[/tex]
[tex]=d \left (\frac{h}{\sqrt{a^2-h^2}}+\frac{h}{\sqrt{b^2-h^2}}\right )[/tex]

[tex]\Longleftrightarrow\frac{1}{\sqrt{9-h^2}}+\frac{1}{\sqrt{4-h^2}}=1[/tex]

It looks like nobody here likes quartic equations. I don't like them either, so I am satisfied with a numerical result ;) . [tex]h\approx 1.23[/tex][/hide]

I have done this sum before by an easier way, very easy using parallels and ratios....
without quartic equations.


Personally, I've never had the situation of getting one wrong on math, but I'm pretty sure that a 52.75/54 raw score scales to either a 780 or a 790, depending on the administration and difficulty. So I guess it's possible for a January 2008 taker to have gotten a 790  :( .


A dog's weight equals half its weight plus 20 pounds. How many pounds does the dog weigh?

$ \frac{x\plus{}20}{2}\equal{}x \iff x\equal{}\boxed{20 \ \text{pounds}}$
or
$ \frac{x}{2}\plus{}20\equal{}x \iff x\equal{}\boxed{40 \ \text{pounds}}$


function
limit
integration
real analysis
real analysis unsolved

Let $ f$ and $ g$ be continous functions on $ R$ such that $ f(x \plus{} 1) \equal{} f(x),g(x \plus{} 1) \equal{} g(x)$ for all $ x \in R$
Prove that :
    $ \lim_{n \to \plus{} \infty}{\int_{0}^{1}{f(x)g(nx)dx}} \equal{} \int_{0}^{1}{f(x)dx}\int_{0}^{1}{g(x)dx}$
 :chief:

sos404, by my understanding, you did not use the conditions : $ f(x \plus{} 1) \equal{} f(x),g(x \plus{} 1) \equal{} g(x)$ :?:

I think he doesn't use the fact $ g$ is periodic, but in third "=" he uses the assumption about $ f$

No. It is essential that $ g$ is periodic, but not that $ f$ is.

This is a variant of the Riemann-Lebesgue lemma, discussed many times here.


Αν $ x,y,z$ μη αρνητικοί πραγματικοί να δείξετε ότι $ x^4\plus{}y^4\plus{}z^4\minus{}x^2y^2\minus{}y^2z^2\minus{}z^2x^2\geq 2\left(x^3y\plus{}y^3z\plus{}z^3x\minus{}xy^3\minus{}yz^3\minus{}zx^3\right)$

Hint:[hide]Use Vasc's: $ (x^2 \plus{} y^2 \plus{} z^2)^2\geq3(x^3y \plus{} y^3z \plus{} z^3x)$[/hide]


inequalities
Euler
inequalities solved
wrong inequality

Here is an easier one.
Let ABC be a triangle. Prove that:

R \sqrt 2pabc \leq abc

your inequality I think is wrong.. or not correct posted.. or I am missing something.. :(
I am trying to prove that:
 Rsqrt(2pabc)<=abc but it is false!
we can write it like so:
2*R^2*p*a*b*c<=(a*b*c)^2 which gives: 2*R^2*p<=a*b*c
we have abc=4prR ( S=pr=abc/(4R)
and we get: R<=2r, but Euler says that R>=2r so the inequality should be
Rsqrt(2pabc)>=abc. :D

tschuess

Yes, you are wright. I'm sorry but I was in a hurry when I composed it and I also didn't check it before posting. My apologiases came so late beacause I was away for a while. Sorry again.


probability
real analysis
probability and stats

Somebody knows how to construct a non-measurable set without use of axiom choice?
I've seen in Willliams book "probability with martingales" the claim that this is possible. But how?

There's a result due to Solovay that states that there is a model of the real numbers consistent with the ZF axioms in which every set is Lebesgue measurable.  So it seems that you need the axiom of choice in order to get Lebesgue non-measurable sets.

Beware !!!
It depends on the sigma field you are palcing yourself

If you are in the Borel sigma-field then I think it is possible to create a set which is not measurable without he use of AoC but if you are in its completion known as the Lebesgue Sigma field then you have to use Axiom of choice

Wow! I didn't know that, thought it was the same no matter what sigma-field. Can you please give an example ?

Here is a non-constructive existence proof. All Borel sets are [url=http://en.wikipedia.org/wiki/Suslin_set]Suslin sets[/url]. Every Suslin set is defined by a countable collection of closed sets. Closed sets have cardinality continuum. Hence Borel sets have cardinality continuum. So, they cannot be all subsets of $ \mathbb R$. Actually, this proof can be turned into an explicit construction: first, create an explicit surjection $ F$ from $ \mathbb R$ to Suslin sets (it is rather tedious but feasible). Now take the set of all points $ x\in \mathbb R$ such that $ x\notin F(x)$, as usual.

[quote="fedja"] Closed sets have cardinality continuum. [/quote]
How about this set: $ X\equal{}\{0\}$???

[quote="fedja"]Here is a non-constructive existence proof. All Borel sets are [url=http://en.wikipedia.org/wiki/Suslin_set]Suslin sets[/url]. Every Suslin set is defined by a countable collection of closed sets. Closed sets have cardinality continuum. Hence Borel sets have cardinality continuum. So, they cannot be all subsets of $ \mathbb R$. [/quote]
[quote="fedja"]Sorry for the ambiguity. I meant "the set of all closed subsets of $ \mathbb R$ has cardinality continuum" but was in hurry and typed a phrase that, indeed, had a second (unintended) meaning.[/quote]
fedja, if you have time, please review these posts again  :maybe:

Now I am confused :?. The phrase I mentioned and the next one are, indeed, ambiguous but the rest is perfectly fine. What exactly is your objection?

Most people probably already know this but a similar argument cannot work for the set of lebesgue measurable sets - the cantor set is a set of continuum cardinality with measure $ 0$, so all of its subsets are (lebesgue) measurable, so ...

Hi there
Here is an example (constructive) of a non Borelian Set which analytic and then Lebesgue measurable.
Take on $ \mathbb{R}$ the set containing all contiunuing fraction of the following form
$ a(0) \plus{} \frac {1}{a(1) \plus{} \frac {1}{a(2) \plus{} \dots}}$

with $ a(n)$ dividing $ a(n \plus{} 1)$ for all $ n\in \mathbb{N}$.

This set is non borel and yet analytic as demonstrated by Lusin in an article of 1927.

Can you please say where can I find this article? ( Maybe, there a copy on-line...  :maybe: )

En français dans le texte :

N. N. Luzin [N. Lusin], "Sur les ensembles analytiques", Fund. Math.,10, 1–95 (1927).

A link with reference in english :
http://en.wikipedia.org/wiki/Non-Borel_set

enjoy


What's wrong with the following?

Problem: Solve sqrt(x) + sqrt(x - a) = 2, where a is a real number.
Solution: The equation implies that sqrt(x) - sqrt(x - a) = a / ((sqrt(x) + sqrt(x - a)) = a/2. Adding this equation with the given, we have 2*sqrt(x) = 2 + (a/2) and so x = (1 + a/4)^2. Thus, when a = 8 we have x = 9.

i think when you squared you added a few extraneous solutions where sqrt(x-a) is actually -sqrt(x-a) and sqrt(x) is actually -sqrt(x). I believe the first one is when a > 0 and the second one for a < 0.


You have a box with seven cards labelled 1 to 7. You reach in and randomly pick one card, then put it back into the box and pick randomly pick a card again. This process is continued.
Question:
a) If you pick seven cards, what is the probability that you picked each and every single card out of the box.
b) How many times do you need to pick the cards so that you have a 50% probability of picking each and every card from the box.

[hide="EDIT Wrong"]
Probability of all seven cards $\frac{1}{7^{7}}$
Number of draws for $50 \%$ probablity $\frac{7^{7}}{2}= 411771.5$ or $411772$ draws.
[/hide]

[quote="AstroPhys"][hide]
Probability of all seven cards $\frac{1}{7^{7}}$
Number of draws for $50 \%$ probablity $\frac{7^{7}}{2}= 411771.5$ or $411772$ draws.
[/hide][/quote]

Can you please explain how you got your solution?

[quote="hunter34"]A is actually:
$\frac{7}{7}\cdot \frac{6}{7}\cdot \cdots \cdot \frac{2}{7}\cdot \frac{1}{7}= \frac{7!}{7^{7}}$

Explanation:
It does not matter what card is pulled. So there are $7$ cards to chose from first, $6$ for the second and so on.[/quote]

I agree with your answer in part a, thanks.

hunter34's solution is correct
[hide="For the Second Part"]
$\frac{6! \cdot x}{7^{6}}= \frac{1}{2}$
$x = \frac{ 7^{6}}{2 \cdot 6!}$
Then $82$ draws.
Which makes more sense then my answer  :oops: .
[/hide]

You're slapping things into formulas without any indication that you're thinking about what you're doing -- the calculations you've made so far don't have any obvious meaning.

For the second question, you need to calculate the probability that after one draw, two draws, three draws, etc., you have drawn all seven cards, and keep going until you get a number larger than 50%.  There isn't any good way to avoid the calculations involved here, although you can make them more reasonable with smart bookkeeping (also known as Markov chains).  When you do it all out, the answer turns out to be [hide]17 (although it's very, very close to 50% after 16 draws).[/hide]


Find all positive integers $ n$ such that $ 1 \plus{} 2n \plus{} 3n^2$ is a perfect square.

Write $ 3n^2 \plus{} 2n \plus{}1 \equal{} m^2 \Leftrightarrow 9n^2 \plus{} 6n \plus{} 3 \equal{} (3n \plus{} 1)^2 \plus{} 2 \equal{} 3m^2 \Leftrightarrow (3n\plus{}1)^2 \minus{} 3m^2 \equal{} \minus{}2$.

This is a [url=http://en.wikipedia.org/wiki/Pell_equation]Pell's equation[/url] whose general solution is a subset of $ (3n_k\plus{}1) \plus{} \sqrt{3} m_k \equal{} (1 \plus{} \sqrt{3})(2 \plus{} \sqrt{3})^k$.  We only need the rational term to be congruent to $ 1 \bmod 3$, which I believe occurs whenever $ k$ is even.


An object of mass m is thrown vertically upward from the surface of the earth with initial velocity vo. We will calculate the value of vo, called the escape velocity, with which the object can escape the pull of the gravity and never return to earth. Since the object is moving far from the surface of the earth, we must take into account the variation of gravity with altitude. If the acceleration due to gravity at sea level is g, and R is the radius of earth, the gravitational force, F, on the object of mass m at an altitude h above the surface of earth is given by

F = mgR^2/(R=h)^2

a. suppose v is the velocity of the object at that time. Use Newtons law of motion to show that dv/dt = -gR^2/(R+h)^2

The gravitational force (magnitude) is given by
\[F=G\frac{Mm}{r^{2}}=G\frac{Mm}{(R+h)^{2}}\]
Newtons second gives $a=F/m$ hence
\[\frac{dv}{dt}=g_{h}=F/m=G\frac{M}{(R+h)^{2}}\]
but you used $g$ at sea-level in your expression ($g_{sea}=G\frac{M}{R^{2}}$):
\[\frac{dv}{dt}=G\frac{M}{(R+h)^{2}}\cdot g \cdot \frac{R^{2}}{MG}=\frac{gR^{2}}{(R+h)^{2}}\]
as desired.
Then of course one could attach a minus sign.


48 marbles are divided into 3 groups .Then, some marbles are removed from the first group and put in the second group so that the number of marbles in the second group is doubled .Some marbles are removed from the second group and put in the third group so that the number of marbles in the third group is doubled and some marbles are removed from the third group and put in the first group so that the number in the first group is doubled.As aresult , there are equal number of marbles in each group.Find the number of marbles in each group

[hide] If you mean the marbles at the  beginning, I get First Group:22,2nd:14,and 3rd:12
At the end there were 16 marbles in each group. I made a chart of values
16    16       16
Before, there were
8      16       24

8      28       12

22    14        12

[/hide]

i agree.  22 in first group, 14 in second, 12 in last

okay...so u have to start backwards. 16 marbles each at the end, and 8 marbles were removed from the 3rd bag and added to the 1st bag, which means that:
8, 16, and 24.
Then bag 3 got 12 marbles from bag 2 (in order to make it doubled), so:
8, 28, and 12.
Then Bag 2 got 14 marbles from Bag 1 (again so it would be doubled), so:
22, 14, 12


A metal stick, with length $l$, rotates in a plane $\pi$, with the angular freq. $\omega$ around one of its ends. Perpendicular to $\pi$ there is a homogeneous magnetic field $B$.
Determine the voltage between the ends of the stick.

[hide]

$V=\frac{1}{2}\omega Bl^{2}$

It's Hall effect.

[/hide]

Is this advanced physics?

Well I am not sure were to post an easy Olympiad problem, but I dosent make it right not to
appreciate the problem. Moderators could move it.
Btw, you haven't shown a full solution. And the phenomenon
is rather called Electromagnetic induction.

[hide="Here is a solution"]
\[e=\frac{d\phi}{dt}=\frac{BdA}{dt}=\frac{B\cdot l\cdot l\cdot d\theta}{2dt}=\frac{Bl^{2}}{2}\omega \]
[/hide]

now you see, those problems should go into the introductory physics section. come on, its no thinking required just substituting into a formula. :D


algorithm
calculus
integration
algebra
function
domain
Ring Theory

Let $ a$ and $ b$ be integers. Prove that if a prime $ p$ divides $ ab$, then $ p$ divides either $ a$ or $ b$.

This is a corollary to The Fundamental Theorem of Arithmetic. It does not have a proof that is too complex in 104 NT.

You probably mean prove that if $ p|ab$, then $ p|a$ or $ p|b$.

If $ p|a$, then we're done. So assume $ p\nmid a$. Since $ \gcd{(p,a)}\equal{}1$, we can find integers $ x$ and $ y$ such that $ px\plus{}ay\equal{}1$, or $ pxb\plus{}ayb\equal{}b$. Since both terms on the left are divisible by $ p$, it follows that their sum (i.e. $ b$) is also divisible by $ p$.

[quote="LeFromage"]You probably mean prove that if $ p|ab$, then $ p|a$ or $ p|b$.[/quote]

Oops. Typed it in wrong  :blush:

[quote="LeFromage"][hide]If $ p|a$, then we're done. So assume $ p\nmid a$. Since $ \gcd{(p,a)}\equal{} 1$, we can find integers $ x$ and $ y$ such that $ px\plus{}ay \equal{} 1$, or $ pxb\plus{}ayb \equal{} b$. Since both terms on the left are divisible by $ p$, it follows that their sum (i.e. $ b$) is also divisible by $ p$.[/hide][/quote]

Thanks. Excellent Proof :)

[quote="LeFromage"]assume $ p\nmid a$. Since $ \gcd{(p,a)}\equal{} 1$[/quote]

Edit:  [hide="Some justification"] Since $ gcd(p, a) | p$ and $ gcd(p, a)\neq p$, it follows from the usual definition of primality that $ gcd(p, a) \equal{} 1$. [/hide]

[quote="Mersenne"]This is a corollary to The Fundamental Theorem of Arithmetic. [/quote]

I'd call it a necessary lemma in the proof of the Fundamental Theorem of Arithmetic.  How do you propose to prove FTA without proving this first?

In abstract algebra, this is actually the definition of a prime. An irreducible is a number such that if $ p\equal{}ab$ then either $ a$ or $ b$ is a unit ($ \pm1$ in the integers). It can be shown that every prime is irreducible, but irreducibles might not be prime.

This problem actually asks to show that in the case of the integers, all irreducibles are primes.

[quote="tjhance"]In what cases would this occur?[/quote]

The proof that all irreducible elements are prime requires only the division algorithm; hence, in an integral domain without a division algorithm this is not necessarily true.

[quote="tjhance"][quote="Hamster1800"]but irreducibles might not be prime.
[/quote]

In what cases would this occur?[/quote]

Not all irreducibles are primes. You need a [url=http://en.wikipedia.org/wiki/Principal_ideal_domain]principal ideal domai[/url]  :wink: (it was also mentioned like 2 times before).


The difference between two number is 9, and the sum of the squares of each number is 153. What is the value of the product of the two numbers?

We can factor $ a^2\plus{}b^2\equal{}(a\minus{}b)^2\plus{}2ab$. Setting this value equal to $ 153$ and the value of $ (a\minus{}b)\equal{}9$, we have $ 153\equal{}81\plus{}2ab\implies2ab\equal{}72\implies ab\equal{}\boxed{36}$. Although this method might not be as fast as finding the individual numbers (since they are relatively small, and produce integer solutions), knowing how to do it algebraically helps a lot when the problems get harder.


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

proove that if $f:[0,1]\to R$ a continous function then:
a)$\int_0^\pi xf(sin(x))dx=\frac{\pi}{2}\int_0^\pi f(sin(x))dx$
b)calculate $\int_0^\pi \frac{xsin(x)}{1+cos^2(x)}dx$

Well.. let's see now:

a) the problem is equivalent to: $\displaystyle \int_{0}^{\pi} \left(x - \frac{\pi}{2} \right) f(\sin x) dx = 0$. Now you make the substitution: $y= x- \pi / 2$.

Then: $\displaystyle \int_{0}^{\pi} \left(x - \frac{\pi}{2} \right) f(\sin x) dx = \int_{a}^{b} y \cdot f(\cos y) dy=0$.  (this is because $y \cdot f( cos y)$ is odd on [a,b])

where $a=-\pi/2$ and $b=\pi/2$.

b) you apply a) and get $\frac{\pi^{2}}{4}$.

cheers!! :D :D

P.S. are you heading for the Mathematics-Informatics University in Bucharest? Because it seems u are posting problems from their book? :D

[quote="Lagrangia"]

P.S. are you heading for the Mathematics-Informatics University in Bucharest? Because it seems u are posting problems from their book? :D[/quote]

Could you post more problem from this book ?
thx

no problem
just look at all my posts and see them
i'm posting problems from that book constantly
the ones i don't know how to solve or the one's i haven't found a similar
solution in another book.


In the regular hexagon to the right, how many degrees are in the exterior angle indicated?

[asy]size(101);
draw((0,0)--(2,0)--(3,sqrt(3))--(2,2sqrt(3))--(0,2sqrt(3))--(-1,sqrt(3))--cycle);
draw((2,0)--(4,0));
pair arrowstart = (2,0) + .75 expi(pi/7);
draw(arrowstart--arrowstart + expi(pi/7),BeginArrow);[/asy]

That's true... I've encountered this problem several times and the picture never showed up...
However, we know that all exterior angles of a hexagon is $ \frac{360}6 \equal{} \boxed{60}$...
but this might not be that obvious without the picture.


Let a and b be positive integers with no common factors other than 1. Show that there are no integers x and y such that ab=ax+by.

Check your Problem statement once as
(b,0) satisfies second equation irrespective of a,b being co-prime.

The second statement must be no positive integer solutions. Then
a(b-x)=by
As, a,b are coprime, (b-x) should be multiple of b, which is not possible
as x is a positive integer.


Could some one please help me by giving some links or books on
dynamic programming and Graph theory

Thanks
MM

"Introduction to Algorithms" by Cormen, Leiserson, Rivest, Stein (CLRS) is a good reference for a lot of stuff. In particular, it covers both dynamic programming and many different graph algorithms (shortest path, network flow, minimum spanning tree, etc).

For online stuff, I'd recommend the USACO training gateway; it has tutorials and nice practice problems for various topics. The only drawback is that you have to progress through it in order so you can't skip around to the topics you're looking for specifically.

Thank you for helping.

I got the book (Intro. to algorithms) and it seems really useful..
I have started training on the USACO gateway and I agree that getting stuck at a problem is very annoying...
I am currently struggling on the Packing rectangles question..  Guess I'll have to wait until I reach the graph algorithms section on this extremely useful training gate

Any other resources will be appreciated.


Prove that: $ x(x \plus{} y)^3 \plus{} y(y \plus{} z)^3 \plus{} z(z \plus{} x)^3\geq \frac {8}{27}(x \plus{} y \plus{} z)^4,\ \forall\ x,\ y,\ z\in\mathbb{R}$.

hello, are $ x,y,z$ positive real numbers?
Sonnhard.

if $ x,y,z$ are just positive reals, it follows after AM-GM  :D

otherwise, just express $ LHS\minus{}RHS$ as a sum of squares

Sorry, because I didn't mention. $ x,\ y,\ z\in\mathbb{R}$.

If $ x + y + z = 0$, then

$ S = x(y + z)^3 + y(z + x)^3 + z(x + y)^3 = x( - x)^3 + y( - y)^3 + z( - z)^3$
$ = - (x^4 + y^4 + z^4) \leq 0 = \frac {8}{27}(x + y + z)^4$ as required.

If not, then since both sides of the inequality are homogeneous with even degree, we can scale $ x,y$ and $ z$ to obtain $ x + y + z = 1$. The inequality can then be written as
$ S = (1 - a)a^3 + (1 - b)b^3 + (1 - c)c^3 \leq \frac {8}{27}$, where $ (a,b,c) = (y + z,z + x,x + y)$. We then have to prove this for all $ a + b + c = 2$.

If we let $ f(x) = (1 - x)x^3$, then we have to prove that $ f(a) + f(b) + f(c) \leq 3f(\frac {a + b + c}{3})$ for all $ a + b + c = 2$.
Now $ f''(x) = 6x(1 - 2x)$, which is negative on $ ( - \infty,0)$ and $ (\frac {1}{2},\infty)$ and positive on $ (0,\frac {1}{2})$ so if $ a,b$ and $ c$ are all not less then $ \frac {1}{2}$, then we can use Jensen's inequality to prove the above.

Now for all real $ x$, $ \frac {3}{4}(\frac {4x^4}{3}) + \frac {1}{4}(\frac {3^3}{4^4}) \geq (\frac {4x^4}{3})^{(\frac {3}{4})}(\frac {3^3}{4^4})^{\frac {1}{4}} = x^3$ by AM-GM, so $ f(x) \leq \frac{3^3}{4^4}$ for all $ x \in \mathbb{R}$. Also, $ f(x) \leq 0 %Error. "doubleleftrightarrow" is a bad command.
x \eq 0 or x \geq 1$, so if x, y, or z is negative, (wlog assume it is z) then
$ S \leq f(x) + f(y) \leq$


Would half be even?

if 5#s add up to an odd number

ooooo     100% odd
oooee      60%  odd
oeeee      20%  odd

so 60% odd? There might be like 4,9,9,9,9 can only be arranged in 5 ways but these cases might cancel out.

[quote="G-UNIT"]Take all those 5-digit numbers in which the sum of the digits is 37. Out of these numbers, how many are even and how many are odd?[/quote]

7x4
10x1

5, 1 even, 4 odd

7x2
8x3

10, 6 even, 4 odd

7x3
8x1
9x1

20, 4 even, 16 odd

I believe 11 even and 24 odd. Just an educated guess, though.

You can get the number of numbers from the percentages.

[quote="math92"]are you guys really twins?[/quote]

no

yes we are!

what the heck!

you are a negative twin now chess64

[quote="superman"]
you are a negative twin now chess64[/quote]

you mean a triplet? sorry  :(  :?  :oops_sign:  :stretcher:


geometry
MATHCOUNTS
number theory
AMC 10

Maybe start with a chapter that interests you first, then delve on to your challenging chapters...

[quote="now a ranger"]I have volume 1 the basics as a school math team textbook. I haven't really studied it at all throughout this semester, although I have had it in my bookbag for heaven knows how long. It seems so tedious to actually go through a book like that. What should I study, and what will help me score higher or learn basic concepts in the book? Examples of chapters include triangles, quadrilaterals... the list goes on

          :?:[/quote]
Study the whole book. I read it very thoroughly over the summer and got like -2 times smarter.

I've heard you strive to improve in geometry. Well the book is more than 1/2 geometry. Study those if you feel comfortable with the little nt and combinatorics they provide, as well as the 4-5 algebra chapters.

[quote="ragnarok23"]
Study the whole book. I read it very thoroughly over the summer and got like -2 times smarter.[/quote]

well let's hope you were negative smart before last summer...

[quote="now a ranger"]good idea. and It is useful for improving in geometry math tournaments? I like studying usually.[/quote]

If you are interested in geometry, it definitely is a good choice, but I would also recommend the Introduction to Geometry book, which is obviously enough, entirely on geometry.

[quote="k8reindeer"][quote="now a ranger"]It seems so tedious to actually go through a book like that.

         [/quote]

It seems tedious but once you start it's actually pretty engaging.
It also helps to get motivation: learn what you weakness is in math team and then study up on that, or wait till there's a promlem you can't get and learn the concepts behind it.[/quote]

this person knows what they are talking about it is my experience..

now a ranger, try things other than geometry man.

Maybe some algebra, or some number theory?

what is number theory? is it like cryptarithms etc. I haven't done algebra in ages as in practicing, though I saw it on boring math leagues

[quote="mustafa"]Number theory is mods. The AoPS wiki probably has something to teach about number theory :wink:[/quote]
It's not only mods, it's basically the study of integers.

[quote="now a ranger"]oh that sounds fun. Are there practice tests of those on mu alpha theta? I've heard of mods . I think theyare like something a number than mod something, which has remainder and stuff[/quote]
Try some amc 10 and 12 questions, I prefer them much more than mao questions.

[quote="JavaMan"]Many mu-alpha-theta problems don't feel like problem solving, to me anyway.[/quote]

Yeah. They seem more like busy work.

If it feels tedious go to the back of the problem set for a chapter. If you can do those skip the chapter...

[quote="now a ranger"]amc 10 and 12? what are those? are they similar to amc 8. I am sort of new to these since I have never done any amc or mathcount stuff before[/quote]

Take a look at the [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]AMC Forum[/url] or check out their [url=http://www.unl.edu/amc/]website[/url].

well i already started the triangles chapter since it is my weakness.

I wish I had solutions, and a bit more details over a few things. I Guess i will just surf the web for stuff.  :P


symmetry conics parabola quadratics algebra

Find all real numbers x such that
$x=\sqrt(x-1/x)+ \sqrt(1-1/x)$

[hide] So and obvious first step in this this is to isolate something square and hope for the best, it works out best if i isolate $\sqrt(1-(1/x))$ , this gives $x^2-2\sqrt(x^3-x)+x=1$. Isolate that bothersome square root term again and square getting $x^4+2x^3-x^2-2x+1=4x^3-4x$ factoring a little gives $x(x-2)(x-1)(x+1)+1=0$ the solutions manual said that this equation is symmetric about x=1/2, but how in the world did they discover that? theres another method of solution involving perfect squares however im really interested in this symmetry thing.[/hide]

[quote] im really interested in this symmetry thing[/quote]
Obviously  x(x-1) is symmetric around (1/2) ... (Either look at the parabola of x^2-x or look at it as (x-1/2+1/2)(x-1/2-1/2).
Samething with the extra two factors. (x-2) and (x+1)

Also  if you put ($y=x-1/x$ in your equation $x^2+1/(x^2) - 2 (x - 1/x) -1 = 0$ (Which is same as $x^4-2x^3 \dots = 0$ you get a second degree equation.

Last but not the least... for fun ..

Also FWIW:

(Assuming $x$  is not zero etc...)
Let $a=\sqrt{(1-1/x)}$   and $b=\sqrt{(x-1/x)}$
Then $x=a+b$
and $b-a = (b^2-a^2)/(a+b) = 1-1/x$
From these we get $2b = x-1/x+1 = b^2+1 \implies b=1$

so $x -1/x = 1$ which is just a quadratic equation :)  (Only solution which will work  $x= (1+\sqrt 5)/2$

okay so the $+1$ at the end only changes the y-value so it has nothing to do with the symmetry

looking at $x(x-1)(x-2)(x+1)$ we have $x=-1,0,1,2$ as the "roots" or the values that make that whole thing 0

from there pairing 0,1 and -1,2 we can see that those "roots" are symmetric about $x=1/2$


calculus
integration
number theory unsolved
number theory

assume that x,y both odd, is a integer solution to x^2 - d*y^2 =±4.

form the rational numbers u,v to:
u +v*sqrt(d ) = ((x + y*sqrt(d))/2)^3

Show that u,v is a interger solution to u^2 - d*v^2 = ±1

We don't need them to be odd...
$u+v \sqrt d = \left(\frac{x+y \sqrt d}{2}\right)^3$ by definition, thus $u-v \sqrt d = \left(\frac{x-y \sqrt d}{2}\right)^3$ by taking conjugates (or simple calculating out).
Now
$u^2-v^2d = (u+v \sqrt d)(u-v \sqrt d)=\left(\frac{x+y \sqrt d}{2}\right)^3 \left(\frac{x-y \sqrt d}{2}\right)^3 =$
$\left(\frac{(x+y \sqrt d)(x-y \sqrt d)}{4}\right)^3 = \left(\frac{x^2-y^2d}{4}\right)^3 = (\pm 1)^3 = \pm 1$.

So left to show that $u,v$ are integral. But $8u=x^3+3dxy^2 \equiv x(x^2+3dy^2) = x(\pm 4 + 4dy^2) \equiv 0 \mod 8$, thus $u$ is integral, and so also $v$.


[img]http://img.photobucket.com/albums/v194/mukmasterxps/formula.gif[/img]

I don't know what the difficulty of this proof is, but I'm fairly certain it belongs here.

outline for a proof I am too lazy to actually write out well:

[hide]
Base cases: blah
Inductive hypothesis: for some m<n  left side sums to the right side

now consider case m+1:

notice that the the sum has 2x each of the m case numbers, plus F(m+1)-1.  so by inductive hypothesis simplify the first ones into twice the left side of the m case. Simplify down to:

2^(m+1) - 2F(m+3) + F(n+1) +1

now note that
F(m+3) = F(m+1) + F(m+2)
F(m+3) = F(m+4) - F(m+2)
add them together to get 2F(m+3), substitute it in poof! it looks right.yay[/hide]

also one may use binet's formula (although the algebra is messy, it is easy to evalutate because of the geometric series formula)


Hi guys I have solved some vector problems and I just want to know if I had got the right answers.

1) Solve for the distance of $P(-3,1,2)$ to the line with equation $l: x = 1+t$   $y = 2+t$  $z = 3+t$.

The answer I got here was $\frac{\sqrt{66}}{3}.$

2) Solve for the equation of the line passing thru $(5,0,-2)$ and parallel to the intersection of equations $x-4y+2z = 0$ and $2x+3y-z+1 = 0$.

The equation of the line i got here were $x = 5-2t$  $y = 5t$  $z =-2+11t$.

3) Solve for the general equation of the plane that is perpendicular to $x+y-2z = 1$ and passing thru the intersection of $4x-y+z = 2$ and $2x+y-2z =3$.

My answer here was $x+y-2z-\frac{13}{6}= 0$.

4) Get the distance between the skew lines $x = y = z$ and $x+1 = \frac{y}{2}= \frac{z}{3}.$

What I got here was $d = \frac{\sqrt{6}}{6}.$

Hope you could help me verify my answers. Thanks.

1) incorrect [hide="Answer"]$\sqrt{6}$, when $t=-2$.[/hide]
2) correct
3) [hide="Maybe"]$26x-8y+9z=11$[/hide]
4) correct

Um sir thanks for the reply about the questions.

With no. 1, just a minor mistake in the signs made my answer that way. Now I also get $\sqrt{6}$.

Regarding no. 3, I am not really sure with my answer. I got mine by getting the a point in the intersection of the two planes (a line). I then got the general equation by substituting the values in the formula $A(X-X_{0})+B(Y-Y_{0})+C(Z-Z_{0})+D = 0$ where the plane passes thru pt. $P(X_{0},Y_{0},Z_{0})$ and normal to the vector $N = Ai+Bj+Ck$ and $D = AX_{0}+BY_{0}+CZ_{0}$

Did I use the wrong method sir? Hope you could help.

The line of intersection in #3 is $(t,10t-7,6t-5)$. If you wanted to find a plane perpendicular to $(1,1,-2)$ containing this line, you would fail; every plane perpendicular to that vector intersects the line in exactly one point.

We want instead a plane perpendicular to $x+y-2z=1$, which means that its normal is perpendicular to $(1,1,-2)$. Its normal is also perpendicular to $(1,10,6)$, which is where kunny's equation comes from.


:(  :blush: hey, i'm ashley and i'm in grade 7, i have an easy question on my algebra sheet and i'm stumped, because we had not worked on algebra for a quite a few months, so can sombody answer this for me?

 10x+75-5x=110

thanks amillion

Hi Ashley,

Here is my take on the problem.

[hide]

The solution to this question is basically moving stuff around so that our variable (x) is on one side of the equation and the other side has NO varibles (only a numerical expression like 1/2 or 36 or something like that).  This is an important skill to gain so lets see how we would isolate x here.

10x+75-5x=110

Before we start any rearranging, we can combine both of our variable terms 10x and 5x.  10x is added, and 5x is subtracted so the result is 5x.  Now we can rewrite the statement:

5x + 75 = 110

Our aim is to get x alone, so we want to get rid of the "numbers" that are next to it.  Lets get rid of the 75 first.  Since it is being added, we can subtract it away from both sides so that it cancels out on the left side:

5x + 75 - 75 = 110 - 75

5x + 0 = 35

5x = 35.

Now lets get rid of the next number, 5.  Since it is being multiplied, we can divide it away from both sides so that it cancels out on the left side.

5x / 5 = 35 / 5

x = 7!

To make sure we made no mistakes, we can check our work by plugging x = 7 into the original equation.

10x+75-5x=110
10*(7) + 75 - 5*7 = 110
70 + 75 - 35 = 110
145 - 35 = 110
110 = 110!

Since both sides came out equal, we made no mistakes and we are sure out answer is x = 7.
[/hide][/hide]

thanks, i'm dumb for forgetting that one! lol, i remeber now. :rotfl:

[quote="236factorial"]When you first did the exclamation marks, Barnacle, I thought that the problem answer was 7 factorial!!

What you get from seeing factorials for so long, and having a screenname involving factorials.[/quote]
hahaha thats what i thought because I don't know.... I guess I'm used to seeing them too. By the way, if you don't know what 7 factorial is, it is 7x6x5x4x3x2x1. In another case,  5! = 5x4x3x2x1 and 10! = 10x9x8x7x6x5x4x3x2x1

hahah I went back to look at my own answer and I also mistook it for a factorial....  :o

lol, i got confused with the factorial sign too. And  :welcome: AshleyMD.


This is a monstruous inequality, with an ulgy solution published in Crux. Does anyone have other solution:
   For any $ a,b,c,d>0 $ which add up to 2, we have $\sum\frac{a^2}{(1+a^2)^2}\leq\frac{16}{25} $

Here is some progress for this question:

Let $f(x) = \frac{48}{125}x - \frac{4}{125}, g(x) = \frac{x^2}{(1+x^2)^2}$

Case 1: $a,b,c>0.12$.
In this case, $h(x) = f(x) - g(x) = \frac{48}{125}x - \frac{4}{125} - \frac{x^2}{(1+x^2)^2} = \frac1{125x^2}(2x-1)^2 (12x^3 + 11x^2 + 32x - 4) \geq 0$ since the cubic factor has no roots exceeding $0.12$.  It follows that $\sum \frac{a^2}{(1+a^2)^2} \leq \sum (\frac{48}{125}a - \frac{4}{125}) = \frac{96}{125} - \frac{16}{125} = \frac{16}{25}$, as required.

Case 2: $ a \leq 0.12 \leq b, c , d $.
Since $h$ is increasing for $a>0$, it assumes a minimum value of $h(0) = -\frac{4}{125}$.  However, $a \leq 0.12$ implies $max(b,c,d) \geq \frac{2-0.12}{3} = \frac {1.88}{3}$ and therefore, $h(max(b,c,d)) + h(a) \geq h(\frac{1.88}{3}) + h(0) > 0$.  The result then follows by application of the previous case.

The cases where 2 or 3 of $a, b, c$ and $d$ do not exceed $0.12$ are handled similarly and are even weaker.

Hopefully, this solution suffices.

A very nice solution to a super-hard problem. The solution published in Crux is absolutely monstruous. Congratulations!

Beautiful solution !!! I think you should send this solution to Crux . Keep going!


hi,
if i have triangle ABC congruent to triangle DEF, and triangle DEF is congruent to triangle GHB, is it safe to say that triangle ABC is congruent to triangle GHI??

i would say yes, cause by saying triangle ABC is congruent to triangle DEF then i meant that corresponding sides and angles are congruent... if the corresponding sides and angles of triangle DEF are congruent to the corresponding sides and angles of triangle GHI, then we have by the transitive theorem of congruences that corresponding sides and angles of triangle ABC are congruent to corresponding sides and angles of triangle GHI, and therefore, by definition of congruence and correspondence, we can say that triangle ABC is congruent to triangle GHI.

wow that felt good...am i right? can someone confirm, tnx.

i have another proof assignment....prove that the midpoints of sides of an equilateral triangle also form an equilateral triangle. my proof is very long and cumbersome, i proved one triangle containing one side of the new triangle congruent to another triangle, and then i proved the second triangle congruent to the third triangle, and so on...then i concluded that because these triangles are congruent, the sides of the new triangles are congruent as well by CPCTC! But one thing bother me....can i say that the other smaller triangles may not be equilateral? I hope so....i can't prove they're equilateral, or maybe i dunno how.

also, my teacher are ganna start to introduce new questions: prove/disprove...how do i disprove an alleged theorem?

tnx and God bless

[quote="AIME15"]For your first one, I'll assume that you have $ \triangle ABC \cong \triangle DEF$ and $ \triangle DEF \cong \triangle GHI$, and you want to prove that $ \triangle ABC \cong \triangle GHI$.

Then we have $ \overline{AB} \cong \overline{DE}$, $ \overline{BC} \cong \overline{EF}$, and $ \overline{CA} \cong \overline {FD}$.  Likewise, $ \overline{DE} \cong \overline{GH}$, $ \overline{EF} \cong \overline{HI}$, and $ \overline{FD} \cong \overline{IG}$.  Then, by the transitive property of congruence, we have $ \overline{AB} \cong \overline{GH}$, $ \overline{BC} \cong \overline{HI}$, and $ \overline{CA} \cong \overline{IG}$.  By SSS Triangle Congruence Postulate, we have $ \boxed{\triangle ABC \cong \triangle GHI}$, as desired.

I'm not going to solve your second one, but use the midsegment theorem.

Also, to disprove anything, you just need to find [b]1[/b] counterexample.  Just one.[/quote]
hi,
i did not ask anyone to solve anything. i would like no one to solve my questions pls, just give me a tip that i may build on and come up with the proof myself.

you did not say if my proof is correct. so i take it by you posting your proof that mine is incorrect.

and i have not proved the midsegment theorem yet...but i want to know if my approach would be corect.

only one counterexample? whoa, pretty easy to disprove than to prove...ah

tnx and God bless

Well, I basically just restated what you said, pretty much.  Except I put it in a more "mathy" form.

And think about disproving something. If there is even 1 counterexample, then it obviously isn't true for all cases.

hi AIM15,
tnx for you help :D

hi mewto,
now i think of it makes sense....tnx

[quote="nick42"]ahh...geometry.  :lol:

you are correct on the first problem. if i am not mistaken, the other smaller triangles on the 2nd problem [i]are[/i] equilateral. but it helps to know the midsegment theorem here.

here's a tip: from my current geometry experience, you have to continually draw back on previous information. so don't forget all those old theorems and postulates...they could come in handy later.[/quote]
ah thanks...the midsegment theorem was mentioned twice so i looked for it and wow it is really handy...but i guess the assignment was to prove with what you got.i was not able to prove the other smaller triangles are equilateral with what i got, but now with midsegment theorem i can....

thx for your tip, i always keep the previous theorems in mind and i have made good use of them from previous chapters in later chapters...

geometry is really nice...

tnx and God bless


Does anyone here use programs such as mathematica to make cool graphics/art/anything? Or have you ever tried/considered trying it? For example, the avatar I put up is a graphic I made with mathematica. Its just curvy tubes weaving with each other in rhombicosadodecahedral symmetry, but it looks cool (to me, anyway :)).

I constructed all of the planes defined by 3 vertices of a cube, other than the faces of the cube itself.  It makes a nice looking star shape.

Here's the full view. It's construction was devastatingly difficult, but I'll try to explain it when I have more time. (I'm in the middle of a ginormous history research project right now )


[hide]The length of the belt on the small wheel is $5\pi.$

The length of the belt on the big wheel is $25\pi.$

The length of the belt between the wheels is $40\sqrt{5}.$

Adding that all up, we get $30\pi+40\sqrt{5},$ or $\approx183.69.$[/hide]

[quote="bfrock2000"]a belt is tightly stretched over 2 flywheels with radii 5 and 25. find the length of the belt if the centers are 40 units apart.[/quote]

[hide="Solution"]
the belt stretches over half of the first and second flywheels.
$\frac{10\pi}2+\frac{50\pi}2$
the other sections are congruent to each other
one of them is $\sqrt{20^{2}+40^{2}}$
so it is:
$\frac{10\pi}2+\frac{50\pi}2+2\sqrt{20^{2}+40^{2}}$
$30\pi+2\sqrt{2000}$
$30\pi+40\sqrt{5}$
[/hide]

i believe this is not correct.

i happen to know the answer is

[hide]110pi/3+40*root3[/hide]

but as i guesstimated this on the test bfrock got it from, im not sure what the correct way to solve this is.

[quote="daermon"]you "guesstimated" $\frac{110\pi}{3}+40\sqrt{3}$?  :roll:
pretty good guess  :D

could you please find a solution?  I still think the given solution is correct.[/quote]
Well, it was a multiple choice MAO test.
Eli's good at stuff like that.
I have seen the solution, but I can't read what the hell the person wrote.

[quote="sonny"][quote="i_like_pie"][quote="arsemesomething"]when the belt stretches over the smaller wheel it doesn't go the entire half-circle. for the larger one it goes over the half-circle.[/quote]
I thought about that, but isn't the missing part on the small wheel equal to the extra part on the big wheel?[/quote]

Why?[/quote]
I dunno. I just thought so. :D


probability
combinatorics unsolved
combinatorics

What is the probability for an n-gon polygon to be convex?

How do you select the vertexes?
Seems that the probability is $0$ if you choose the vertexes in C evenly.


There are 3 houses (H1 , H2 , H3 ) and 3 utilities (Water,Gas,Electricity)... Distribute all the uilities to every hose without intersecting the pipes. You cannot make a pipe from one house to another.

 H1 H2 H3

  W  G   E

I heard rummors that it is possible to solve it ... I say it is not possible  :ninja:

i heard people saying its impossible.
where did you hear the rumor?

There is another discussion of this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45997]here[/url], including a proof that it is impossible.

No, it is not possible. All but one can connect, but the last must necessarily overlap with one of the others.

It is known that:
$K_5$ is not planar.
$K_{3,3}$ is not planar.
And every non-planar graph contains $K_5$ or $K_{3,3}$.

The thing you are trying to do is to get planar graph of $K_{3,3}$, which is impossible. :(

[quote="JBL"][quote="lightrhee"]And every non-planar graph contains $K_5$ or $K_{3,3}$.[/quote]

Semi-natural but rather complicated definition of "contains."[/quote]
true :)  I cound have said 'subgraph'.

Actually, I think JBL was referring to the fact that you can easily take a edge and stick a vertex right in the middle of it.

Yes -- there are many graphs which do not contain a $K_{3,3}$ or $K_5$ subgraph but which are still non-planar.  A nice example is the [url=http://mathworld.wolfram.com/PetersenGraph.html]Petersen Graph[/url].  The word "contains" really means "contains as a graph minor" (or as a topological graph minor, which is different but equivalent), a more complicated relationship that just "containing a subgraph."

Unless it's in 3 dimensions... in which case all graphs are "planar" (although the term should really be '3-spacial' or something like that).

Planar is only used to mean "embeddable in a plane," not "embeddable in whatever my local space is."   The interesting extension is not to three dimensions (since, as you note, every finite graph can be embedded in 3-space with no crossings) but instead to other surfaces (sphere, torus, etc.).


probability counting derangement factorial

There is a bag of 16 marbles.  There are 8 different colors and 2 marbles of each color.  Suppose an octopus reaches into the bag and randomly grabs two marbles for each arm.  What is the probability that no arm contains two of the same color?

since you know that there are 8 different colors , and 2 marbles of each color, you can deduce that there are 8 pairs of 2 marbles of the SAME color; there are 16 marbles. since an octopus has 8 arms, and each arm grabs 2 marbles for each arm, you can say that the probability would be 1/8. if that's not right, i don't know what else would be the answer...this is my best guess! good luck!

need additional help? go to [url]http://www.mathworld.com[/url]for more details on probabillity.

[hide="Hint"] Derangements [/hide]
[hide="Answer?"] $\frac{!8}{8!}=\frac{2119}{5760}$???

Note: As the number of marbles $\to\infty$, the answer should $\to\frac{1}{e}$ [/hide]

whoaaah what have I missed in my combinatorics education??? I've never seen a factorial infront of a number  :blush:.

EDIT: looked at wolfram... you learn something new everyday.

[quote="max_tm"]whoaaah what have I missed in my combinatorics education??? I've never seen a factorial infront of a number  :blush:.[/quote]

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

[hide] Sorry, my above post doesn't seem very helpful.
We will call the octopi A, B, C, ... , H along with their respective first colors:
Octopi : [u]A B C D E F G H[/u]
Color 1: A B C D E F G H
Color 2: _ _ _ _ _ _ _ _
Now we must rearrange the sequence A, B, C, D, E, F, G, H so that no letter is in its original place. This is called a derangement. There are $!n$ ways to derange $n$ objects, where $!n=\left[\frac{n!}{e}\right]$
So, the answer is $\frac{!8}{8!}$ since $8$ objects can be arranged a total of $8!$ ways.

PS. "Octopi" should be replaced with "Arms." :D [/hide]
lol, I just realized this solution is incorrect. It doesn't count for situations like:
A B C D E F G A
B C D E F G H H  :blush:

oh wait..never mind...i already looked at mathworld/wolfram..whatever you want to call it!  :heli:

I got 2 to the 8 divided by 16C8 as the answer. Since you look at it as a permutation of 1's and 2's, each representing a different color. There's 16C8 ways to arrange them, and you must have the first 2 as 1 and 2, so 2 ways for each of the 8 sets, thus 2 to the 8th. I don't know though.


Find the sum of the first six terms in the geometric sequence $ \frac12,\frac14,\frac18,\dots$. Express your answer as a common fraction.

The general formula for the sum of the first n terms of the sequence a,ar,ar^2...

$ S\equal{}\frac{a(1\minus{}r^n)}{1\minus{}r}$

This can be easily derived.

More specifically, letting

$ S\equal{}a\plus{}ar\plus{}ar^2\plus{}\cdots\plus{}ar^{n\minus{}1}$ where there are $ n$ terms and the common ratio is $ r$, multiply both sides by $ r$ to get

$ rS\equal{}ar\plus{}ar^2\plus{}\cdots\plus{}ar^{n\minus{}1}\plus{}ar^n$, and subtracting, we get

$ (1\minus{}r)S\equal{}a\minus{}ar^n\equal{}a(1\minus{}r^n)\implies S\equal{}\boxed{\frac{a(1\minus{}r^n)}{1\minus{}r}}$.

Another thing this leads to is that when |r|<1, the sum of the infinite series becomes:

$ \frac{a(1\minus{}r^n)}{1\minus{}r}\equal{}\frac{a(1\minus{}0)}{1\minus{}r}\equal{}\frac{a}{1\minus{}r}$

...which can also be "proven" with a technique similar to the one in my last post (not rigorously).


If  x,y,z are  positive numbers, then

$ \frac{x}{\sqrt[3]{y}}\plus{}\frac{y}{\sqrt[3]{z}}\plus{}\frac{z}{\sqrt[3]{x}}\ge\sqrt[3]{xy}\plus{}\sqrt[3]{yz}\plus{}\sqrt[3]{zx}$

[quote="soruz"]If  x,y,z are  positive numbers, then
$ \frac{x}{\sqrt [3]{y}}\plus{}\frac{y}{\sqrt [3]{z}}\plus{}\frac{z}{\sqrt [3]{x}}\ge\sqrt [3]{xy}\plus{}\sqrt [3]{yz}\plus{}\sqrt [3]{zx}$[/quote]
By AM-GM inequality, we have:
$ \frac{x}{\sqrt [3]{y}}\plus{}\sqrt[3]{xy}\ge 2\sqrt[3]{x^{2}}$
Similarly, we have:
$ \frac{x}{\sqrt [3]{y}}\plus{}\frac{y}{\sqrt [3]{z}}\plus{}\frac{z}{\sqrt [3]{x}}\ge 2\sum\sqrt[3]{x^{2}}\minus{}\sum\sqrt[3]{xy}$
Also by AM-GM:
$ \sum\sqrt[3]{x^{2}}\ge\sum\sqrt[3]{xy}$.
And we are done!

[quote="soruz"]If  x,y,z are  positive numbers, then

$ \frac{x}{\sqrt [3]{y}}\plus{}\frac{y}{\sqrt [3]{z}}\plus{}\frac{z}{\sqrt [3]{x}}\ge\sqrt [3]{xy}\plus{}\sqrt [3]{yz}\plus{}\sqrt [3]{zx}$[/quote]
Why did you post the same problem three-time,it is the same problem as in the "inequality-inequality" topic :mad:


abstract algebra
superior algebra
superior algebra unsolved

Please, I need some help with this execise:

Consider $ M=\mathbb{Q}/\mathbb{Z}\oplus \mathbb{Q}/\mathbb{Z}\oplus \mathbb{Q}$ a $ \mathbb{Z}$-module.

1) Calculate  $ T(M),$ the set of all elements of torsion in $ M.$
2) Prove that $ T(M)$ is a direct sumand of $ M.$

 I really appreciate eny help

Do you see why $ T\left(M\right)=\left\{\left(x,y,0\right)\mid x,y\in\mathbb{Q}\slash\mathbb{Z}\right\}$ ?

  Darij

all you need are the trivial facts that $ \mathbb{Q}/\mathbb{Z}$ is torsion, $ \mathbb{Z}$ is torsionfree and $ Tor(A \oplus B) = Tor(A) \oplus Tor(B)$.


geometry
circumcircle
trigonometry
geometry solved

Let $BD$ and $CE$ be the bisectors of the interior angles $< B$ and $<C$,
respectively ($D \subset AC, E\subset AB$). Consider the circumcircle of $ABC$ with
center O and the excircle corresponding to the side BC with center $I_a$.
These two circles intersect at points $P$ and $Q$.
Prove that $PQ$ is parallel to $DE$.

[quote="mohamadhosein"]Let $ BD$ and $ CE$be the bisectors of the interior angles $ < B$and$ < C$,
respectively ($ D \subset AC, E\subset AB$). Consider the circumcircle of $ ABC$ with
center O and the excircle corresponding to the side BC with center $ I_a$.
These two circles intersect at points $ P$ and $ Q$.
Prove that $ PQ$ is parallel to $ DE$.[/quote]

The circumcircle and the A-excircle of triangle ABC have the centers O and $ I_a$, respectively, and intersect at the points P and Q. Since the line joining the two points of intersection of two circles is always perpendicular to the line joining their centers, we thus have $ PQ\perp OI_a$. Thus, in order to prove that PQ || DE, it remains to show that $ DE\perp OI_a$. This was done in [url=http://www.cip.ifi.lmu.de/~grinberg/Schroeder/Schroeder.html#6]the post #6 of my Schröder points database[/url] or on http://www.mathlinks.ro/Forum/viewtopic.php?t=15729 .

  Darij

I think there exist also less "clever" argument to this one - as Darij pointed - it is equivalent to proving that $OI_A \perp DE$ which is equivalent to:

$I_AD^2-I_AE^2=OD^2-OE^2$
$(DB^2+BI_A ^2)-(CE^2+CI_A ^2)= (R^2+DB^2-2R\cdot DB \cos OBD)-(R^2+CE^2-2R\cdot CE \cos OCJ)$
$CI^2-BI^2=2R(CE \cos OCI - DB \cos OBI)$

which doesn't look so messy and as everything here can be easy written in terms of sides and angles of triangle $ABC$ it can be easily checked.

But that was just a remark  :D


Jack, Jill, and John play a game in which each randomly picks and then $ replaces$ a card from a standard $ 52$ card deck, until a spades card is drawn. What is the probability that Jill draws the spade? (Jack, Jill, and John draw in that order, and the game repeats if no spade is drawn).

[hide]
Let drawing spades win the game.

Jack wins in these cases: $ S, OOOS, OOOOOOS\ldots$ (where $ S$ is spades and $ O$ is other)
Jill wins in these cases: $ OS, OOOOS, OOOOOOOS\ldots$
John wins in these cases: $ OOS, OOOOOS, OOOOOOOOS\ldots$

Let probability to draw $ S$ be $ p\equal{}\frac{1}{4}$ and to draw $ O$ be $ q\equal{}\frac{3}{4}$.
Let $ P_1$ be the probability that Jack wins, $ P_2$ probability that Jill wins and $ P_3$ probability that John wins.

$ P_1\equal{}p\plus{}pq^3\plus{}pq^6\plus{}\ldots$
$ P_2\equal{}pq\plus{}pq^4\plus{}pq^7\plus{}\ldots$
$ p_3\equal{}pq^2\plus{}pq^5\plus{}pq^8\plus{}\ldots$

from which we have:
$ P_1\equal{}\frac{P_2}{q}$
$ P_3\equal{}q\cdot P_3$

$ P_1\plus{}P_2\plus{}P_3\equal{}1 \implies \frac{P_2}{q}\plus{}P_2\plus{}q\cdot P_3\equal{}1 \implies P_2\equal{}\boxed{\frac{12}{37}}$[/hide]

[hide]
Probability that Jill draws a spade in any given round is $ \dfrac{3}{4} * \dfrac{1}{4}$
Probability that any person draws a spade in a round is $ \dfrac{1}{4} \plus{} \dfrac{3}{4} * \dfrac{1}{4} \plus{} \dfrac{3}{4} * \dfrac{3}{4} * \dfrac{1}{4}$.  If no one draws a spade, the game just repeats.

Thus, the probability that Jill draws a spade is
$ \dfrac{\dfrac{3}{4} * \dfrac{1}{4}}{\dfrac{1}{4} \plus{} \dfrac{3}{4} * \dfrac{1}{4} \plus{} \dfrac{3}{4} * \dfrac{3}{4} * \dfrac{1}{4}} \equal{} \dfrac{12}{37}$
[/hide]

[hide="Another Solution"]The probability of Jill winning in the first round is $ P_1=\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16}$.

The probability of Jill winning in any round $ n$ is $ P_n=P_1\cdot\left(\frac{3}{4}\right)^{3(n-1)}$.

Thus, the probability of Jill winning is
\begin{align*}P&=\sum_{k=1}^\infty P_k\\\\
&=\frac{3}{4}\cdot\frac{1}{4}\cdot\left(1+\left(\frac{3}{4}\right)^3+\left(\frac{3}{4}\right)^9+\ldots+\left(\frac{3}{4}\right)^{3n}\right)\\\\
&=\frac{3}{16}\cdot\frac{1}{1-\left(\frac{3}{4}\right)^3}\\\\
&=\frac{3}{16}\cdot\frac{1}{1-\frac{27}{64}}\\\\
&=\frac{3}{16}\cdot\frac{64}{37}\\\\
&=\boxed{\frac{12}{37}}\end{align*}[/hide]

Sorry for reviving this thread, but I think my solution, although it probably comes from the same idea, is much simpler to think about.

P(winning a given round) = P(winning current round) + P(winning next round)

P = (3/4)(1/4) + (3/4)(3/4)(3/4)P

P = 12/37


geometry
circumcircle
geometric transformation
reflection
rotation
homothety
geometry unsolved

let $ABC$ be a triangle and $P$ be a point on the circumcircle of $ABC$,
let $f(P)=AP+BP+CP$.
find the minimum value of $f$ and determine the place of $P$.

After some ugly computation, I found the following :
The greatest excircle (centered at J) touches the nine point circle at F; P is the is the reflection of H (orthocenter) in F and AP + BP + CP = 2 OJ
Note that: Max (AP+BP+CP) = 2 OJ = 2 root(R(R+2r)) where R = circumradius, r = radius of the greatest excircle.
I'm sure that someone will be able to find a short proof of this result; presently, I cannot.
Kind regards. Jean-Pierre

[color=red][Moderator edit: Combined the double post into a single one.][/color]

J is the center of the gratest excircle.
Note that I was solving Max(AP + BP + CP)

well i got something for when the triangle is equilateral :
without loss of generality assume that A  is the vertice farthest from P  then we have AP=BP+CP so  F(P)=2AP  hence we have our minimum when AP  is minimum that when P=B  or P=C.  i dont know how to generalize i haven´t really tried it  :)  :D

marko

[quote="Magnara"][quote="DusT"]I am not a geometry ace, but I think P is the Fermat-Toricelli point, so if all the angles in the triangles are less than $\frac{2\pi}{3}$, then this is the point that sees all sides in the same angle, $\frac{2\pi}{3}$. The proof is not too hard :?[/quote]

You may wish to read the problem again.   :)[/quote]
 :oops: Indeed, sorry, I didn't read the problem statement carefully, I skipped the first sentence  :blush:  :oops:
However, I think that the problem can still be solved by a similar idea with that of the Fermat-Toricelli point, that would be a rotation of 60 degrees around point A, and see where the points P, B, C will go and try to notice that $f(P)$ will actually be a skew line, which we want to make as short as possible.....Maybe this is an idea  :oops: Sorry, please forgive me for not reading carefully  :blush:

[quote="payman_pm"][quote="mecrazywong"]f(P) attains its min value when P=max{A,B,C}. Try to prove it, it's not hard.[/quote]
what is the meaning of  P=max{A,B,C}.[/quote]
max{A,B,C} means the max. of angles A,B,C.

That's the reason why I was trying to solve Max(AP+BP+CP) which is 2 root(R(R+2R')) whre R = circumradius and R' = radius of the greatest excircle; P is the reflection of the orthocenter in the contact point of the 9P circle with the greatest excircle

In fact, if A', B', C' are the midpoints of BC, CA, AB and M a point of the arc B'C' - not containing A' - of the 9P circle, the power of M wrt the A-excircle is (A'M+B'M+C'M)^2 (this can be proved using, for instance, a barycentric parametrisation of the arc B'C'). This proves that A'M+B'M+C'M can be extremal only at B', C' or at the reflection of Fa in the 9P center (Fa = contact point of the 9P circle and the A-excircle)
Now, using the homothecy (G,-2), it follows that AP+BP+CP (with P on the circumcircle) is maximal in the case above and minimal at a vertex of ABC.
Kind regards. Jean-Pierre

I'm not fully understand your solution, anyway, the following is mine:
WLOG, suppose P lies on arc BC. Then obviously BP+CP attains its min value when P=BorC. Also, AP attains the min. when P=BorC. The remaining is trivial.

As for DusT's misunderstanding, it is not totally "useless" because when $\max \{ A,B,C\}\ge120^\circ$ then the Fermat point is just exactly that vertex ;)


Hello everyone,

I am not sure how to solve for a (Once I have a, I can solve b) in the following system.

Could anyone please give me a hint/tip?

Thanks!

P.S. : Could someone please tell me how to align both equations to the left, so there's no space in front of $ (1)$?

---

1. $ \left\{\begin{aligned} a + b = 170 - (1)\\
a^2 + b^2 = 130^2 -(2) \end$

[u]My Work:[/u]

Squaring first equation: $ a^2 + 2ab + b^2 = 28900$

Subtract $ (1)$ from $ (2)$ : $ 2ab = 12000$

$ b = 6000a$

[quote="catalan"]$ 2ab \equal{} 12000$

$ b \equal{} 6000a$[/quote]

The second line should be $ b \equal{} \frac{6000}{a}$.  Plug this into $ (1)$ and you have a quadratic in $ a$.

Let $ S\equal{}a\plus{}b, \ P\equal{}ab$. Then the system becomes
$ \displaystyle\left\{\begin{array}{ll}S\equal{}170\\S^2\minus{}2P\equal{}130^2\end{array}\right.$
We get $ S\equal{}170, \ P\equal{}6000$.
$ a$ and $ b$ are the roots of the quadratic equation $ t^2\minus{}St\plus{}P\equal{}0\Rightarrow t_1\equal{}50,t_2\equal{}120$.
Then the solutions of the system are $ (50,120), \ (120,50)$

Thank you very much for your help, red_dog and t0rajir0u.

I didn't get $ b \equal{} 6000/a$, so I wasn't able to find the roots of this simple system.

you could also look at it as a triangle. a and b are the legs. 130 is the hypotenuese. A+b=170. its somewhat obvious that is a triple.

[quote="abacadaea"]its somewhat obvious that is a triple.[/quote]

This assumption is only valid because homework problems tend to work out nicely, but in the general case this isn't a useful assumption.


trigonometry
AMC
AMC 12
geometry
AoPS Books
trig identities
Law of Sines

I know there is trig in the AMC 12's, but about how many questions each test are there that require it?  The reason is I currently don't know trig, i'm only taking alg II.  Also, what would be the best way to learn it?

The number of questions depends.  I would guess that there would be about 2 straight trig ones and two geometry ones that use trig (i.e. law of sines, cosines etc.).  Learning trig before AMC would be pretty hard to do now.  If you have the AoPS books try going through the trig sections.

There aren't too many questions on trig so you could always focus these last couple weeks on strengthening the other areas.


geometry
perimeter
algebra
polynomial
complex numbers
algebra proposed

In the complex plane, consider squares having the following property: the complex numbers its vertex correspond to are exactly the roots of integer coefficients equation $ x^4 \plus{} px^3 \plus{} qx^2 \plus{} rx \plus{} s \equal{} 0$. Find the minimum of  square areas.

[quote="Fang-jh"]In the complex plane, consider squares having the following property: the complex numbers its vertex correspond to are exactly the roots of integer coefficients equation $ x^4 \plus{} px^3 \plus{} qx^2 \plus{} rx \plus{} s \equal{} 0$. Find the minimum of the squares.[/quote]

minimum what? area,perimeter?? :huh:  :huh:

Not really a difference, is there? Guess it matters in how you present the answer, but you can still find which square it is.

The four roots of the equation are $ \alpha \plus{} \beta, \alpha \plus{} i \beta, \alpha \minus{} \beta, \alpha \minus{} i \beta$ for some complex $ \alpha, \beta$ with $ \beta \neq 0$. We want the minimum of $ |\beta|$. Let $ P(x)$ be the polynomial given (so it has the roots $ \alpha \plus{} \beta$ etc.), then $ P(x) \equal{} x^4 \minus{} 4\alpha x^3 \plus{} 6\alpha^2 x^2 \minus{} 4 \alpha^3 x \plus{} \alpha^4 \minus{} \beta^4$ by Vieta. Since $ \minus{}4\alpha$ is an integer $ \alpha$ is clearly rational. But $ \minus{}4\alpha^3$ is also an integer, so $ \alpha$ has to be an integer as well. WLOG $ \alpha \equal{} 0$. $ x^4 \minus{} \beta^4$ has to have integer coefficients, and $ |\beta| \equal{} 1$ is the minimum.

So the minimum size square is the one inscribed in the unit circle.


Galois Theory
superior algebra
superior algebra unsolved

Let $ L=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $ G=Aut(L/\mathbb{Q})$. Prove that $ |G|=4$ and that every element of $ G$ of order $ 1$ or $ 2$. Using the Fundamental Theorem of Galois Theory, dertemine all the subfields of $ L$. Describle those fields explicitly.

Please post your solution!  :wink:

One can show (-oo- did it here some time ago) that the galoisgroup of $ \mathbb{Q}(\sqrt{p_{1}},\dots,\sqrt{p_{n}})$ with pairwise distinct primes $ p_{i}$ is isomorphic to $ \mathbb{Z}_{2}^{n}$, that this isomorphism maps the element a to the automorphism $ \sqrt{p_{i}}\mapsto(-1)^{a_{i}}\sqrt{p_{i}}$ and that a primitive element of this extension is $ \sqrt{p_{1}}+\dots+\sqrt{p_{n}}$.

It is an easy corollary from this, that the subfields of $ \mathbb{Q}(\sqrt{p_{1}},\dots,\sqrt{p_{n}})$ are exactly $ \mathbb{Q}(\sum_{i\in E}\sqrt{p_{i}})$ where $ E\subseteq\{1,\dots,n\}$.

P.S.: Your other problem (http://www.mathlinks.ro/Forum/viewtopic.php?t=158547) can be solved too if you know the above facts about $ \mathbb{Q}(\sqrt{p_{1}},\dots,\sqrt{p_{n}})$.


For how many real values of $ x$ is $ \sqrt {120 \minus{} \sqrt {x}}$ an integer?

$ \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$

[hide]

We have one $x$ that corresponds to each square number less than 120.

We have $10^2$, $9^2$,..., $0^2$.

So, there are 11 values of x that result in an integer.

E[/hide]

[hide]
Writing all the aquare numbers less than 120:
0,1,4,9,16,25,36,49,64,81,100

So we set sqrt (x) = 120, 119, 116, ... , 20 which are 11 values.

answer E[/hide]

[hide] The value of the expression has to be greater than or equal to zero (can't have negative square roots) and has to be less than or equal to the square root of 120 because the square root of x has to be non-negative and the maximmum value of 120-x where x is greater than or equal to zero is 120. Therefore the answer is all the integers between 0 and the square root of 120, inclusive, which is 11.

E.[/hide]

Yes, they test you on that all too much.  Integers: -2, -1, 0, 1, 2 ...

[quote="Hexaditidom"]I left this blank, I was unsure whether they would consider 0 to be an integer[/quote]

... Of course 0's an integer. The only thing people worry about is if 0 is a natural or if 0 is a whole number.

Its a whole number but not a natural number right?

Yeah, this question was perfectly clear...but still a lot of people (myself included) left out 0.  At least it wasn't like that Math League problem where you had to arbitrarily know that they considered 0 to be a perfect square... (tip: always consider 0 to be a perfect square)

Just watch out for 'Gaussian Integers'  :P

[hide="Answer"]$120-\sqrt{x}=n^2$, so $x=k^2$, and $k<120$. We have $11$ perfect squares less than $120$, and therefore $11$ values of $x$. $\boxed{E}$[/hide]

how is $ \sqrt {119}$ an integer?

I thought this was asking how many numbers were integer results of $ \sqrt {120 \minus{} \sqrt {x}}$
so I got for eg. $ \sqrt {120 \minus{} \sqrt {400}}$ = 10, all the way to 1, so I still got 11, but yea, some of you I think had bad reasoning to get there

Edit: yea i know this is 4 years old, but I was confused...

If we let $ x\equal{}k^2$, like JesusFreak did, we have $ \sqrt{120\minus{}k}\equal{}n$, so $ 120\minus{}k\equal{}n^2$. Since $ k^2$ is nonnegative, $ n^2$ must be a perfect square less than $ 120$. From there, he got that there are $ 11$ of them.

[quote=RedFireTruck]$\lfloor\sqrt{120}\rfloor+1=\boxed{11}$[/quote]

Nice 11 year bump

[quote=RedFireTruck]$\lfloor\sqrt{120}\rfloor+1=\boxed{11}$[/quote]

Congratulations bumping  an 11 year thread!

Edit:bruhh snipedd

[quote=kante314][quote=RedFireTruck]$\lfloor\sqrt{120}\rfloor+1=\boxed{11}$[/quote]

Nice job. You are now a bumper car. [hide=click it noowwwwwww] You have more than 2.5k posts. WHYYYYYYY[/quote]

What's wrong with bumping these threads

[quote=v_Enhance]3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.[/quote]

It's is okay to bump a problem as long you give a solution for it. There's nothing wrong with what RFT is doing. For example, saying this: [quote]This problem took me two hours to solve![/quote]
doesn't really contribute anything, but saying:[quote]Factoring, we get $(x+3)(x+5)$ so the roots are $-3$ and $-5$.[/quote] is perfectly valid like @above said.

So $120-\sqrt{x}$ has to be a perfect square. There are $\boxed{\textbf{(E)}\ 11}$ perfect squares less than or equal to $120$.


let $f: R\to R$ and $g: R\to R$ be continiuos and bijectives such that:

        $\forall{x}\in\mathbb{R}: fog^{-1}\left(x\right)+gof^{-1}\left(x\right)=2x$
       prove that if $\exists{x_{0}}\in\mathbb{R}$ .such $f\left(x_{0}\right)=g\left(x_{0}\right)$. then $\forall{x}\in\mathbb{R}: f\left(x\right)=g\left(x\right)$ :roll:

Beautiful problem!
Let $h: =gof^{-1}$ which is as well continuous and bijective. Then $h(x)+h^{-1}(x)=2x\ \forall x\in\mathbb R$.

Suppose there are $x\ne y$ such that $h(x)=y.$ Then $h^{-1}(y)=x$, thus $h(y)=2y-x$.

By induction in both directions, we get, putting $a: =y-x\ne 0$, \[h(x+ka)=x+(k+1)a \ \forall k\in\mathbb Z.\]
Now, because of the continuity, $h$ is either monotonously increasing or decreasing.
Wlog we can suppose $h$ increasing and $a>0$ (otherwise replace $h$ by $-h$ and/or $h^{-1}$). Now just choose $k\in\mathbb Z$ such that $x+ka<x_{0}<x+(k+1)a$. Contradiction to $h(x_{0})=x_{0}$ :)  :)

So we conclude $h(x)=x$ for all $x$ and we are done.

also nice solution:but we have $h$ straicly incraising  :lol:


An odd, three-digit positive integer has three distinct digits. The integer is a perfect cube, and the sum of its digits is also a perfect cube. What is the three-digit integer?

Let's take a look at odd perfect squares:
\[ 1^3\equal{}1\\
3^3\equal{}27\\
5^3\equal{}125\]
We see that $ 125$ is a three digit perfect cube with distinct digits, and the sum of its digits, $ 1\plus{}2\plus{}5\equal{}8\equal{}2^3$, is also a perfect cube.  Thus, our answer is $ \boxed{125}$

[quote="splatyango"]Doesn't 512 also work?[/quote]

512 satisfies all the criteria, except the number we're looking for has to be odd.


geometry
geometric transformation
rotation

I need some help!

I need to know how to prove that Fermats point really does work. How do i prove that this point minimizes the sum of 3 verticies of a triangle to this point??

Please help!!!! :!:

 :lol:

Try some rotation and the triangle inequality. Or just look at [url=http://www.cut-the-knot.org/Generalization/fermat_point.shtml]here[/url].

that rotation proof is so elegant!!! i always use that technique, but i did not think of it this time for some reason


Here is a hard problem:
  Let $x_0=0, x_1=1$ and $a,b$ integers greater than 1. Suppose that $\frac{x_n+x_{n-2}}{x_{n-1}}$ is equal to $a$ if $n$ is even and with $b$ otherwise. Then $ x_{n+m}x_{n+m-1}...x_{n+1}$ is a multiple of $ x_m x_{m-1}$ for all $m,n$.

Well, we can certainly characterise the sequence.  I claim it is composed of two recurrent sequences taken alternately.

Let's consider
\[
x_{2k} = a x_{2k-1} - x_{2k-2}
\]
\[
x_{2k-1} = b x_{2k-2} - x_{2k-3}
\]
\[
x_{2k-2} = a x_{2k-3} - x_{2k-4}
\]
So
\[
x_{2k} = a (b x_{2k-2} - x_{2k-3}) -x_{2k-2} = (ab-1)x_{2k-2} -a x_{2k-3} = (ab-1)x_{2k-2} - (x_{2k-2}+x_{2k-4})
\]
from which we deduce that the subsequences of odd and even terms each satisfy the recurrence
\[
z_k = (ab-2)z_{k-1} + z_{k-2} ,
\]
with initial values $0,a$ and $1,ab-1$ respectively.


Okay, the website is not fully edited yet, but if you have nothing to do and want to pretend to live in Candyopia (imaginary nation) then go to candyopia.webs.com and create a membership (free)!!!!!!!

that reminds me, is there a html editor in freewebs? because i don't like the lack of customization the "customizable" premade stuff is. and if you program a java game or something, is there a way to add that into the applet?

no idea. I was just getting the word out about Candyopia.

Candytopia?  Is this another one of those little kids games like club penguin on miniclips?  Or is it actually cool?

miniclip?

o yah, commando 2.............now that game is AWESOME

and how did clup penguin get to be #1 on the list????

commando 2 was #1 for a little while......

Candyopia is kinda dumb, but if you want to pretend you live in a fake country, its awesome. It's a completely post-based game. Meaning no actual pictures or real-time stuff. BUT JOIN. The discussions are sometimes fun.

Is the "warrior X" person some AoPSer? I really want to know who that is!!!!

Oh, so Candyopia is just a big discussion, not a game?

I guess you could argue that BattleFront is not really a discussion. It is very game-like. But yes, it is a forum, not an actual game program of some kind. We already tried that, but we got bought out, so...........

We are trying to get shops online too, to make it more realistic, but having trouble.


trigonometry
integration
calculus
calculus computations

cos (piex/3)

how do i antidifferentiate this problem

$\frac{d}{dx}\sin \frac{\pi x}{3}=\cdots$, thus you can get $\int \cos \frac{\pi x}{3}dx=\cdots$.


Greetings! I am hosting a mock MATHCOUNTS Masters' Round Contest on AoPS. Sign-ups are open from now to May 23, 2008 at 11:59 P.M. forum time.

The competition will be on May 24, 2008 from 10:00 A.M. forum time to 11:30 A.M. forum time.

Here's how things will go:
1. You have from now until May 23, 2008 at 11:59 P.M. forum time to post in this thread to sign-up for the contest.
2. After 11:59 P.M. forum time on May 23, 2008, sign-ups will no longer be accepted. There are no exceptions to this rule!!!
3. At 10:00 A.M. forum time on May 24, 2008, I will PM the questions out to the entrants, all at once. The entrants then have until 11:30 A.M. forum time to read, completely answer the questions (all work, including computation, shown), and PM their answers to me.
4. Any late submissions after 11:30 A.M. forum time on May 24, 2008 will not be accepted!!! If I receive your submission even 1 minute late, I will not accept it.
5. I will then read through all the submissions and decide on the top 3 entries. The names of the top 3 entrants will be posted, along with the questions and the winner's answers by May 30, 2008, on the forums.

Rules:
1. The decisions of the judges are final.
2. Rules are subject to change at any time.
3. The official time used in the contest is the forum time. Although the forum time is very close to PST, the official time used is the time shown on the forums. If you submit your answers to me at 11:31 A.M. forum time on May 24, 2008, I will NOT except them, even if 11:31 A.M. forum time is 11:29 A.M. PST. There are no exceptions to this rule. If you do not understand, ask below.
4. No discussing of the problems with anyone else, anywhere, during the competition. Do not PM someone else about them. Do not discuss them with someone else on these forums.
5. You may use a calculator and other aids (book, etc.).
6. Have fun!!!

So are you interested? If you are, then sign-up now!!! Good luck and have fun!!!

Post below to sign up or ask questions.

Oh, and I'm the judge (for those who were wondering...). I do not need anyone else to help me judge or look over the problems (I've already had my friend help me in proofreading the problems). If you really, really want to be a judge or question-writer/proofreader, you can always host your own contest.

-12markkram34

[u]Current Entrants[/u]:
1. stevenmeow
2. moogra
3. Brut3Forc3
4. isabella2296
5. lotsofmath
6. tinytim
7. Yongyi781
8. math154
9. georgiamathdude
10. jackdillon

Reserved for the announcement of the winners.

Who are the judges? I think that determines how this is going to turn out.

I should definitely be a judge. Since I'm an idiot. Not that the judges are idiots (sorry if it was interpreted that way xD)

So mathcrazed and I will be judges.

Oh, and ... hi mathcrazed  :D  :ninja:  :D

Oh, I'm the judge (guess it wasn't obvious).

i guess ill sign up...btw wut day is may 24? im too lazy 2 look at calendar right next to me

I guess I'll join even though I'm in high school...
What's forum time? I'm in PST, so... uh...

The current forum time is listed at the top of every forum page in something that looks like this:

You last visited on Yesterday, at 8:01 pm
The time now is Sun May 18, 2008 12:56 am
All times are UTC - 7 (DST in action)
AoPSWiki WoW: Squeeze Play Theorem, Math Day at the Beach

Do you understand now? Feel free to ask again if you still don't understand.

To clear up some confusion, the Part II Questions refer to the interior diagonal of the solid, not the diagonal of the face the ant is standing on.

I'm confused: I thought that the competition would be 1 1/2 hours. Now, form what I gather, we were supposed to get the problems at 1 and turn them in by 1:30, which is what I did (maybe I turned mine in at 1:31). Can someone please explain?

So sorry!!!!!!!!!! You have until 2:30!!!!!!!! I typoed -.-. :oops:

If I have an extra 50 minutes left, can I PM you a message that elaborates on some of my solutions because of the misunderstanding?

Yes, that is fine, as long as I receive it before 2:30 P.M. forum time.

15 minutes left until solutions are due.

I thought you said 1:30 forum time? At the time you said that, we would have had 18 minutes left...

I meant approx. 15 mins left, sorry.

The competition is officially over!!!!!

Thanks to everyone wh participated. The winner will be announced by Friday in a new thread.

Once more, sorry for any misunderstandings and typos.

Yeah anyway thanks a lot!

edit: Wait, are you going to just put a winner, or also top 3, like you said at the beginning of this?

Sorry that I didn't get a chance to look through 6 pages of posts, but I was wondering if the problems would be up after the competition just for fun. I didn't get a chance to do it, and if I could get them PM'd, I think that'd be fun. :lol: Thanks!

Same here. I'd like to take a look at the problems too, if possible.

Wow I was being stupid I forgot about this...

Ah, I will post the problems in a new thread along with the announcement of the winners. There are you look at and discuss the problems as much as you want.


number theory
relatively prime
number theory open

Define a number n to be good if it can be expressed as n = a1+a2+a3 ... +a(k) for some k, and any two of a1, a2 ... a(k) are mutually relatively prime. Prove almost all numbers are good (i.e. with finitely many exceptions).

(note when k=2, we should be able to do it if the goldbach conjecture works).

Well, if you don't ask for some more conditions, the problem seems to be very easy and does not need Goldbach's conjecture to be true....

First, you may consider that $k = n$ and $a_i = 1$ for each $i$.
Since I think that it is not what you are asking, just note that the following decompositions work :

- If $n = 2k+1$ then $n = k + (k+1)$
- If $n = 4k$ then $n = (2k+1) + (2k-1)$
- If $n = 4k+2$ then $n = 2 + (2k-1) + (2k+1)$

This gives decompositions for each $n > 2$ if you allow that at most one of the summands is $1$, and for $n \geq 10$ otherwise.

Pierre.

I think he ment that for any fixed k, all but finitely many numbers can be expressed as the sum of k muttualy relatively prime numbers. I think I saw a particular case ( 3 relatively prime and $\geq 17$ ) in Sierpinski and it wasn't very trivial/ So, the general case shouldn't be easy at all.

yes that was was i meant - for a fixed k almost all k can be expressed as the sum of k relatively prime numbers. sorry for the ambiguity.
for k=2 it is extremely easy; no need for goldbach's sorry. harazi, how would you the k=3 case? i would be interested. also, i think it would also be slightly easier when k is a power of 2.

This problem is closed. In fact it was closed long time ago, by (who else?) Erdos, in an article published in Acta Arithmetica. This question, raised by Sierpinski, has the following answer:
   There is a constant c such that for any number s we have $ f(s)\leq c\cdot s\cdot log s+p_1+...+p_s $, where $ p_n $ is the sequence of primes and f(s) is the least number k such that any number greater than k is the sum of k distinct primes. Anyway, do not try to prove this, since it uses deep results in number theory and also the geniality of Erdos!


Lizzie folded a piece of 8-inch by 10-inch paper in half again
and again (without ever unfolding it) until she ended up with
a folded piece that measured 1 inch by 2.5 inches. How many
times did she fold the paper in half?

[asy]size(202);
picture p,q,r,s;
real radius = .2;
draw((0,0)--(0,10)--(8,10)--(8,0)--cycle,linewidth(1));
draw((4,0)--(4,10),linetype("2 4"));
label("10",(0,5),W,fontsize(8pt)); label("8",(4,10),N,fontsize(8pt));
draw(p,(0,5)--(4,5),linewidth(1),EndArrow(4));
add(shift(8.5*right)*p);
draw(q,(0,0)--(0,10)--(4,10)--(4,0)--cycle,linewidth(1));
draw(q,(0,5)--(4,5),linetype("2 4"));
draw(r,(0,5)--(0,10)--(4,10)--(4,5)--cycle,linewidth(1));
fill(s,circle((0,5),radius)); fill(s,circle((3radius,5),radius)); fill(s,circle((6radius,5),radius));
add(shift(13.5*right)*q); add(shift(23*right)*r); add(shift(32.5*right)*s); add(shift(18*right)*p); add(shift(27.5*right)*p);
label("1 fold",(10.5,5),N,fontsize(8pt)); label("2 folds",(20,5),N,fontsize(8pt)); label("3 folds",(29.5,5),N,fontsize(8pt));[/asy]

Each time she folds it one of the sides decreases by a factor of 2. The 8 inch side has decreased by a factor of 8, so it has been folded 3 times that way, and the 10-inch has been reduced by a factor of 4, so it has been folded twice that way, for a total of 3+2=5 folds.


A rectangle ABCD has side lengths AB = 4 and BC = 4. Find the side length of the rhombus AXYZ, where X lies on AB , Y lies on BC and Z lies on BD.

Why is ABCD not a square then?

If ABCD is a square i am getting a really obvious answer.


number theory
relatively prime
number theory unsolved

For an odd positive integer $n>1$, let $S$ be the set of integer $x$, $1\leq x\leq n$, such that both $x$ and $x+1$ are relatively prime to $n$ . Show that $\prod_{x\in S}x\equiv\ 1 (mod n)$.


Sequence $\{a_n\}$ satisfies:$a_1=3,a_2=7,a_{n+1}=\frac {(a_n)^2+5}{a_{n-1}}, n \geq 2$.If $a_{n}+(-1)^n$ is a prime ,prove that there exist a nonnegative integer m such that $n=3^m$

Is it normal that $a_3$, $a_4$ etc aren't integers, or is that a typo?


ok, so i have this mathfinal coming up, and i need to know/remember the log properties...something along a list would really help me. Thanks in advance!

Here are a few

$\log_ab + \log_ac = \log_a(bc)$

$\log_ab - \log_ac = \log_a{\frac{b}{c}}$

$c\log_ab = \log_a{b^c}$

$\log_ab = \frac{\log_cb}{\log_ca} = \frac{\ln b}{\ln a}$

I can't remember if there's any others.

$(\log_a b)(\log_c d)=(\log_a d)(\log_c b)$
This is just from $\log_a b=\frac{\log b}{\log a}$ so then you can switch the denominators in the multiplication and convert back

$\log_{a^n} b^n=\log_a b$
$\log_a b=\frac{1}{\log_b a}$