I don't know if this is the right place for the following one:

The distance between Cairo and Damascus is 1000 miles.
Your mission is to move a 10,000 kilograms load of grass from Cairo to Damascus using your camel.
You encounter two problems:
1) The camel won't budge unless you let it to continuously chew grass - it consumes 1 kilogram of grass per mile.
2) The camel's maximum load is 1000 kilograms.
Q1. Can you manage to get ANY of the grass to Damascus?
Q2. What is the maximum amount of grass that you can get there?

HAVE FUN!

For Q1.
You can load 1000kg and transport it to 250km. So you will have 500kg in that position and the camel in the initial point. Doing that we will have more than 5000kg in 250km apart from the initial position. Repeating that we will have 2500kg in 500km apart. Reapeat again and you will have at least 1000kg in 750km apart.

What is your solution for Q2?

I'm not very good with solving problems with recursion, but here is an equation for this problem.
$L_i=\frac{L_{i-1}}{1000}(L_{i-1}-2d_{i-1})$ where $L_0=10000$ and $\Sigma d=1000$.
$d_i$ is the distance traveled on the $i^{th}$ trip and $L_{i+1}$ is the amount of grass left after that trip.
I am of course assuming that it is most beneficial to take 1000 kilos at a time an move it all to the same place with several trips.
If you could implement this in a solution, I'd be interested to see how you did it.

I got for $Q2$ at most $1399,766...$ kilograms as:
$2000-3(1000-1000(\frac{1}{19}+\frac{1}{17}+\frac{1}{15}+...+\frac{1}{7}+\frac{1}{5}))=$
$1000(3(\frac{1}{19}+\frac{1}{17}+\frac{1}{15}+...+\frac{1}{7}+\frac{1}{5})-1)=1399,766...$
Clearly the load of grass cannot be distributed at any time in more than $2$ points (not optimal).
In two such consequently points, if the difference between the entirely load of grass that might be in these points
is greater than $1000$, there is another intermediate point which will increase the final amount of grass.
If I remember well this was my solution for $Q2$ as:
$10000 -> 9000 -> 8000 -> ... -> 2000 -> 1399,766...$
and distances:
$\frac{1000}{19}, \frac{1000}{17}, ... , \frac{1000}{5}, 1000-1000(\frac{1}{19}+\frac{1}{17}+\frac{1}{15}+...+\frac{1}{7}+\frac{1}{5})$


Let $ a,b,c\geq0$, no two of them are $ 0$.Prove that:
$ \dfrac{1}{a^2\plus{}bc}\plus{}\dfrac{1}{b^2\plus{}ca}\plus{}\dfrac{1}{c^2\plus{}ab}\geq\dfrac{3}{ab\plus{}cb\plus{}ca}$

[quote="Inequalities Master"]Let $ a,b,c\geq0$, no two of them are $ 0$.Prove that:
$ \dfrac{1}{a^2 \plus{} bc} \plus{} \dfrac{1}{b^2 \plus{} ca} \plus{} \dfrac{1}{c^2 \plus{} ab}\geq\dfrac{3}{ab \plus{} cb \plus{} ca}$[/quote]
It's equivalent to $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}\sum_{cyc}(a^4bc\plus{}a^3b^2c\plus{}a^3c^2b\plus{}a^2b^2c^2)\geq0.$


prove that

[img]http://www.mathramz.com/xyz/latexrender/pictures/4896bc5de6077a9bc48ef3117cb649c4.png[/img]

Don't explicitly evaluate this determinant.  Use Gaussian elimination to get it into a nicer form, preferably diagonal.  Otherwise, just try to get as many 0's as possible.


function
analytic geometry
algebra solved
algebra

Find all $f: \mathbb{R}^+ \to \mathbb{R}^+$ that for $x>y>0$
\[ f(x)f(y) \leq (x-y)^2\]

Strange... it is not continuous...  :maybe:

I think that $ R^+ $ is the set of positive real numbers, right? Then its nice and easy. Take any closed interval [a,b] from $ R^+$. Take the sets $ A_n = \left\{x \in [a,b] \mid f(x)>\frac{1}{n} \right\} $. The union of these sets is [a,b], thus at least one of them, let us say $ A_N $ is infinite. Take any infinite sequence from it and a subsequence which is convergent to L in [a,b], let us call it $ x_k $. Then $ \frac{1}{N^2}\leq f(x_k)f(x_m)\leq (x_k-x_m)^2 $ for all m and k. Since this sequence converges, the above cannot hold, so such a function doesnt exist. Anyway, if we replace the set of positive real numbers with the set of nonnegative real numbers, then either I am really idiot or the problem becomes much more difficult. I managed to show that the image of any such function on any interval is at most numerable and that it is integrable. I dont know how to continue.

the problem is basicly equivalent to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=105193#105193]this[/url] one. Take $f(x)$ to be the radius of the circle tangent to the real line in $x$ and you get something very similar. Actually, you'll get $|x-y|\ge 2\sqrt{f(x)f(y)}$.

Consider all circles with center (x,f(x)) and diameter f(x) this circles doesnot intersect.Because each circle has a rational point(point with rational coordinates)
Number of circles is countable.Contradiction


abstract algebra
group theory
superior algebra
superior algebra solved

Let G be an abelian group with n elements and p a prime such that p|n but p^2 doesn't divide n. Find the number of solutions of the equation x^(n/p)=e.
   Harazi

[url]http://www.mathlinks.ro/viewtopic.php?t=1772[/url]

It's a particular case of your problem, the one for p=2. We still haven't solved that one, so... :D I abanodned that problem after some struggle, but I'm willing to try again :)

Noe, it's not a particular case. Mine is easier, since the group is abelian!

We use the abelian group theorem, which in our case is reduced to saying that G is isomorphic to Zpx\piZni, where p doesn't divide any of the ni's (i goes from 1 to k).

What was your solution? You said you proposed this for the local phase of the olympiad and they thought it was too tough, right? I suppose it's not this easy to dismiss if we're not allowed to use that theorem (which I don't know how to prove, anyway :D).

I didn't think about using this powerful result. My solution is the fllowing:
   Consider the set A={x^n/p| x in G}. A is a subgroup of G and also x^p=e for all x in A. Thus, |A| is a power of p and also a divisor of n. It follows that its cardinality is 1 or p. Suppose it is 1. Then x^n/p=1 for all x in G. But it follows that there aren't elements of order p, false. So, |A|=p. Then A is cyclic . Let A={1,eps, eps^2,...,eps^p-1}. If we denote M_k={x in G| x^n/p=eps^k} one can easily prove  (via a bijection of the form f(x)=ax) that all these sets have the same cardinality, so each has cardinality n/p. In particular, M_0. The solution ends here.
   Regarding the use of that theorem in a contest, I don't know what to say. The rule is that you can use without proof any theorem that is in a document elaborated by SNEE. I do not know if this theorem is in that document.

My teacher showed me a paper saying what theorems we can use in the olympiads, and I don't remember seeing this theorem, so..


What's the second conditon(isn't too clear)Is it p divides the whole thing(no bracket?)

Anyway...

One condition is that p=2,then everything follows...
$Nowp%Error. "other" is a bad command.
%Error. "than" is a bad command.
\2....$
 :?:

[quote="shadysaysurspammed"]What's the second conditon(isn't too clear)Is it p divides the whole thing(no bracket?)[/quote]

Well, what else could it mean?  $(p|m^3)+n^3-4$ doesn't make any sense (where you are saying that $p$ divides just the $m^3$).

I was asking if p only divides m^3 or the whole thing.Yes only it dividing m^3 doesn't make sense,that's why i was just clarifying....nvm about that,lets get back to the question. :arrow:

[quote="Grt"][quote="amirhtlusa"]find all prime numbers, $p$, such that there exist integers $m,n$, satisfy the following two conditions:

$p=m^2+n^2$
$p|m^3+n^3-4$[/quote]
only  p=2 ;)[/quote]

nope, there is more

[hide="hint"]
there are 3 values of $p$.[/hide]

I hope there's  no fermat involved,please give the solution.

Sorry to interrupt the suspense..  :P :
http://www.mathlinks.ro/Forum/viewtopic.php?t=39312
http://www.mathlinks.ro/Forum/viewtopic.php?t=66912


Points A, B, C, D, E, and F lie in that order on line AF dividing it into five segments, each of length 1. Point G is not on line AF. Point H lies on GD and point J lies on GF. The line segments HC, JE, and AG are parallel. Find HC/JE.

If I interpreted it correctly...

 [geogebra]5bd5cbbac6ce126fe3152251cd7f489e9bb2a295[/geogebra]

From the diagram, we have $ \triangle CDH\sim \triangle ADG$, so $ \frac{CH}{AG}\equal{}\frac13$, and

$ \triangle EFJ\sim \triangle AFG\implies\frac{EJ}{AG}\equal{}\frac15$, so dividing the first relation by the second, we have

$ \frac{CH}{EJ}\equal{}\boxed{\frac53}$.


compute:
 sup( min( |p-q*sqrt(3)|, p+q=n) ,n=1..infinity)
   where p,q>=0 integers.

I have a quite complicate solution but i'm sure there is a nicer one ...

This was an Oral Examination Ecole Normale Suprieure Lyon 2003

hi moubi, have you a nice solution for this one because mine use some results on continued fractions ... and as you said, this problem comes from an oral examination so it must exits a nicest soluions  :)

You must be kidding, or I understand nothing!

Solving $p-q\sqrt 3=0$ we find "ideal" $p$ and $q$, namely $q=\frac{n(\sqrt 3-1)}{2}$, $q=\frac{n(3-\sqrt 3)}{2}$. Let $a=\{q\}$, then \[\min_{p+q=n}|p-q\sqrt 3|=\min(a(\sqrt 3+1),(\sqrt 3+1)-a(\sqrt 3+1)).\] It follows that $\sup \min_{p+q=n}|p-q\sqrt 3|\leq \frac{\sqrt 3+1}{2}$ and $\sup$ approaches this value as $a$ approaches $\frac{1}{2}$. Clearly, fractional part of $q=\frac{n(3-\sqrt 3)}{2}$ approaches any number on $[0,1]$.


In the history of France between 1700 and 1900 two important victories were ntoed on the days of April 22nd. Based on the following two information, find the years of these victories:

a) The second victory happened 4382 days later than the first one.
b) The sum of the digits of the year of the first victory is 23.

how many answers are there supposed to be, because the way a see it, there can be multiple answers.Please clarify :?

because there are only 2 extra days in those 12 years, it must pass through 1800 which is not a leap year.

Thus it is 17xy where x and y are digits.

1 + 7 = 8

23 - 8 = 15

thus, x+y = 15. This narrows it to:
(6, 9)
(7, 8)
(8, 7)
(9, 6)

Because the only one that is less than 12 years before 1800 is (9, 6), the victories were on April 22, 1796 and April 22, 1808

[quote="Hamster1800"]make sure you have the leap years right. 1700, 1800, and 1900 are NOT leap years[/quote]

What!?  :o Why not? They're divisible by 4 aren't they?  :?  ;)

[quote="Hamster1800"]If a year is divisible by 100 but not by 400 it is not a leap year. That is the key in narrowing it down to one solution[/quote]

Oh. http://www.google.com/search?hl=en&q=define%3Aleap+year agrees with you so I agree with you now too  :lol:

[quote="Hamster1800"]If a year is divisible by 100 but not by 400 it is not a leap year. That is the key in narrowing it down to one solution[/quote]
oh, so there is really only one answer.


induction
real analysis
real analysis theorems

My professor mentioned in class today that for the statement:
(' = complement, all of these deal with DeMorgan's)

(union of A_n for n = 1 to infinity)' = (intersection A_n' for n = 1 to infinity) cannot be proven by induction because of the infinity.  He proved it by showing inclusion both ways.  At the time I didn't think about it much, but now that I am, I can't really understand why you can't use induction to prove that.  This question came up because a hw problem was proving DeMorgans for n sets:

(A1 U A2 U ... U An) ' = A1' intersect A2' intersect ... intersect An'

using induction, which was easy to do.

Can anyone clarify why you aren't able to do it when the n goes to infinity?

Thanks alot,

Induction proves that the statement is true for any finite $ n$.  Unfortunately, an infinite collection of sets does not necessarily behave in the same way as an arbitrarily large finite collection of sets.  In other words, induction proves statements for natural numbers and infinity is [b]not a natural number.[/b]

An example of a statement that is true for finite sets and not true for infinite sets is that any surjection from a finite set to itself is a bijection, which can be proven using induction (if you really want).


1. An ordinary deck of 52 cards and 4 aces shuffled, and then the cards are drawn one by one until the first ace appears. On average, how many cards are drawn?

2. Given a deck of 52 cards, extract 26 of the cards place them on top of the deck in the same relative order as they were before being selected. What is the expected number of cards that now occupy the same position in the deck as before?

3. Given any sequence of $n$ disdtinct integers, we compute its "swap number" in the following way: Reading from left to right, whenever we reach a number which is less than the first number in the sequence, we swap its position with the first number in the sequence. We continue in this way until we get to the end of the sequence. The swap number of the sequence is the total number of swaps. Find the average value ofthe swap numbers of all the possible permutations of 1,2,3,4,5,6,7.

I can find the answers to these problems intuitively, but it's more difficult to justify rigorously.

1. Let $X_{1}$ - be the number of cards from the beginning of the deck to the first ace. $X_{2}$ - from the first ace to the second ace etc.
$X_{1}$, $X_{2}$, ... $X_{5}$ are dependent but identically distributed.
So $E(X_{1})+1=\frac{48}{5}+1$.

I obtained a slightly different result. My problem with your solution is that you claim that $X_{1}, X_{2}, ...X_{5}$ are equally distributed, but this is actually rather asymmetrical. Arrange all the cards in a circle and you see why. Your gap between the last ace and the first ace is twice as large as the gap between any two other aces.

I'am back from holydays!

Yes, the gap is twice larger.

Let's consider the experiment A:
1. Add a Fake Ace to the deck.
2. Arrange all cards in a circle.
3. Shuffle many times.
4. Consider Fake Ace as a delimeter of the deck.
Conclusion:
It's obvious that the resulting deck is random.
Because Fake Ace is shuffled as normal ace the 5 distances between aces (including the fake one) are identically distributed. But the Fake Ace is the beginning (and the end) of the deck.

Let's consider the experiment B:
1. Make a deck consisting only of 4 aces.
2. Shuffle it.
3. Randomly add a new card in one of 5 locations.
4. Randomly add a new card in one of 6 locations.
etc.
By induction we can see that $X_{1}$, ... $X_{5}$ are identically distributed after each step.

I was also impressed by this fact. I've convinced myself with small example of 4 cards 2 of wich are aces and all permutations of this small deck.

The second one...
Card occupies the same position => The card was located after the last picked for shuffling card.
As for the first problem we see:
26 delimeters => 27 identically distributed intervals.
The mean length is $\frac{52-26}{27}=\frac{26}{27}$

For the third one, one nice method is to count the total number of switches among all permutations of $\{1, 2, \ldots, n+1\}$ in terms of the total number of switches among all permutations of $\{1, 2, \ldots, n\}$.


group theory
abstract algebra
symmetry
superior algebra
superior algebra unsolved

Is there some easy way to compute the minimal generating set of a given group? I was asked to compute the minimal generating sets of $ S_3$ and $ S_4$ so is the correct way to do the problem by taking one element $ s_1\in S_3$, check which subgroup it generates, then take another element $ s_2\in S_3\setminus \{ s_1 \}$, check what group is $ \langle s_1,s_2\rangle$ and so on. Can we say something general about the generating set of $ S_n$, like how many elements it has, the number of generating sets (probably asymptotical approximation), or if $ S_n\subset S_m$ then a generating set of $ S_n$ can be extended to a generating set of $ S_m$?

I got a hint that the cycles $ s\equal{}(1,2,\ldots ,n)$, and $ t\equal{}(1,2)$ generate $ S_n$ but how this helps to find every generating set of $ S_3$ or $ S_4$?

you only have to give one minimal generating set, and $ (12...n),(12)$ is clearly minimal (otherwise $ S_n$ would by cyclic, i.e. $ n \geq 2$).

I thought to wrote the minimal generating sets, not a minimal generating set but forgot that s. So $ S_3$ has a minimal generating set $ (1,2),(1,2,3)$ and $ S_4$ has a minimal generating set $ (1,2),(1,2,3,4)$. It looks like $ (1,2)(1,3)(1,4)$ is another minimal generating set.

If you really mean "minimal," i.e. no subset is a generating set, rather than the generating sets with the smallest cardinality, then the construction you gave in your first post is probably the best way to go--just systematically traverse the tree of subsets, stopping when you have enough elements.  Exploiting symmetry a bit, this should be possible by hand for smallish groups.

This is probably the best strategy--I suspect that this is a hard problem in general.


geometry
trapezoid
ratio
inradius
similar triangles

a trapezoid has bases of lengths 8 and 12 and an altitude of 6. find the distance from the intersection of its diagonals to the longer base.

[hide="Pretty large hint:"]
Similar triangles.  :wink: [/hide]
(I'm assuming that the bases are parallel, as there would be different desired lengths were one vertex of the smaller base kept six units from the larger, and the second vertex rotated.)

If the bases weren't parallel, it wouldn't be a trapezoid.
[img]8025[/img]
[hide="Solution"]
WLOG, let the trapezoid be isoceles. This is valid because the interesections of the diagonals will always remain a constant height in a trapezoid, by the law of cosines.

So we have a base of 8 on top of a base of 12.
The two parts on the right and left of the central bottom base section are 2 units long, and the altitude is given as 6.

So set up two similar triangles. The first one has vertices at the lowe left point of the trapezoid, the upper right point, and the point formed by the right altitude and the bottom base. It has a base of 2 + 8 and a height of 6. Now set up similar triangle. Drop an altitude from the point of the diagonal's intersection to the lower base. This forms a triangle similar to the first with a base of $2+\frac{8}{2}$ and a height of $x$

By similar triangles, we have $\frac{10}{6}= \frac{6}{x}$ and thus $x = 3.6$.[/hide]

[quote="cincodemayo5590"][hide="Pretty large hint:"]
Similar triangles.  :wink: [/hide][/quote]
every time i see a geometry probelm, the first thing i think is "similar triangles"

but i'm really terrible at deciding how to use similar triangles.  :blush:

not necessary...

let ABCD, with AB=8, CD=12, AB//CD, AC and BD intersect at Y

let L be the line that is perpendicular to AB (and CD), that passes through Y

L intersects AB at X, and CD at Y

show that XAY is similar to ZCY
XBY is similar to ZDY

then XY/YZ=AX/CX=BX/DX=(AX+BX)/(CX+DX)=AB/CD=8/12

XY+YZ=6 [the height]

now you can solve

---

note: that was a lot of unnecessary work; you can just say that in similar triangles, the corresponding lengths of anything are in ratio with another pair of corresponding lengths, in this case, the altitudes are in a ratio with the bases, you can use this with inradius, and other things like that...

[quote="Altheman"]note: that was a lot of unnecessary work; you can just say that in similar triangles, the corresponding lengths of anything are in ratio with another pair of corresponding lengths, in this case, the altitudes are in a ratio with the bases, you can use this with inradius, and other things like that...[/quote]
how exactly does inradius help?  :oops:

EDIT: nevermind i think you were referring to in general.

[quote="iamagenius"][quote="DiscreetFourierTransform"]WLOG, let the trapezoid be isoceles. This is valid because the interesections of the diagonals will always remain a constant height in a trapezoid, by the law of cosines.[/quote]
wow so you can just assume that...?[/quote]

My visual proof of that is to imagine sliding the top base along a line. As it slides over, so does each diagonal by an equal amount, which moves the intersection point horizontally by that amount, but with zero vertical shift. This basically keeps the intersection point of the diagonals fixed vertically, which is necessary for the solution to the problem.

[quote="DiscreetFourierTransform"]This basically keeps the intersection point of the diagonals fixed vertically, which is necessary for the solution to the problem.[/quote]

But it isn't necessary at all.

All you need to see is that, [hide=" as Alex has pointed out,"] the triangles formed with each base are similar by AA, so their heights are in the ratio of the bases.[/hide]


ARML
geometry
calculus
integration
trigonometry
ratio
3D geometry

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

[hide="Solution to Problem 7"]

We can factorize $n!+(n+1)!+(n+2)!$ as

$n!(1+n+1+n^2+3n+2)$

Which equals $n!(n^2+4n+4)$

Which equals $n!(n+2)^2$

Now let us take a look. We know that $7^2=49$ So we know that there has to be two sevens in the expression somewhere.

So $n!$ must contain two factors of $7$ or $n+2$ must be a multiple of $7$.

We know that $n=5$ is one solution.

We also find that $n=12$ is a solution.

And everything $14$ and up has two factors of $7$ since $n!$ has a $7$ and a $14$ in it

So $n!+(n+1)!+(n+2)!$ is an integral multiple of $49$ if...

$n=5, 12, 14, 15,$ or $16$ [/hide]

[hide="#2"]
Suppose $ a+b=x $, $ b+c=y $, and $ c+a=z $. We need to find the value of 14-x, so we can substitute y and z into the equation $ x+y+z= 28 $. We can figure out from the ratio that $ y= \frac{7x}{6} $ and $ z= \frac{8x}{6} $ (Leave 6 in denominator for easy addition later). Substituting into the equation above, $ x+\frac{7x}{6}+\frac{8x}{6}=28 $. Solving for x, $ 7x/2= 28 $, which means that x=8, and 14-x= $ \boxed{6} $ [/hide]

[hide="#1"]
Since $ \triangle{ABC} $ is right, and $ \angle{C} = 90^{\circ} $, by Pythagorean we know that $ a^2 + b^2 = c^2 $.

Thus using some basic identities, we get:

$ (a + b)^2 - 2ab = c^2   \implies   (\sqrt{17})^2 - 2ab = 3^2   \implies   2ab = 8   \implies   ab = 4 $.

What we want is $ \frac{ab}{2} $. Thus, our answer is simply $ \frac{4}{2}   \implies   \boxed{2} $.

[/hide]

[hide="#6"]Note that in order for an integer to equal the product of its proper divisors, it must have 4 divisors of which two are prime and multiply to the number, and the other two are 1 and the number.
Therefore we only need to look at the numbers which are the product of only 2 primes  ($ p_1 \cdot p_2 $) and are perfect cubes.
For $ p_1 = 2 $ we can have $ p_2 $ as 3,5,7,11,13,17,19, and 23 (8 numbers)
For $ p_1 = 3 $ we can have $ p_2 $ as 5,7,11, 13 and 17 (5 numbers)
For $ p_1 = 5 $ we can have $ p_2 $ as 7 (1 number)
For cubes, we can only have $ 2^3 $ and $ 3^3 $ (2 numbers)
$ 8+5+1+2= \boxed{16} $[/hide]

[quote="Aang"]I have replied to this thread before and it got deleted so Im guessing your not supposed to post here :P[/quote]
I wonder why it got deleted. Maybe it's because you like trashcans and hate doges and post off-topic stuff.
Relevant to topic: We can solve #2 using guess and check

[hide="#5"]

Consider either one of the circles. Note that the radius is the perpendicular bisector of the chord. This creates a a $ 3 - 4 - 5 $ right triangle. The same applies to the other circle as well. Since the two circles are perpendicular, we simply have an isosceles right triangle with legs of length of $ 3 $ and what we want is the hypotenuse of this right triangle which is $ \boxed{3\sqrt{2}} $.

[/hide]

[hide="8"]

By the Extended Law of Sines: $ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R $, we have:

$ \frac{a^3 + b^3 + c^3}{\frac{a^3}{D^3} + \frac{b^3}{D^3} + \frac{c^3}{D^3}} = 7 $ where we have made the substitution $ 2R = D$.

This simplifies to $ D^3 = 7    \implies     D = 	\sqrt[3]{7} $. Since the longest side of the triangle cannot exceed the circumdiameter, our maximum possible value of $ a $ is simply $ \boxed{\sqrt[3]{7}} $. $ \blacksquare $

[/hide]

Wow I just realized these were the problems we did at that last practice.....

@Iamteehee

lolol yeah they were, but luckily, I only did the super easy problems beforehand so the amount of problems I saw beforehand was minimal :P.

Let's finish this off.

[hide="#4"]

For easier calculations, fix the point on the circle as any one of the vertices of the $2n$-gon. We want to find the sum of the squares of the distances from any one vertex of the $2n$-gon to the other vertices.
First, since the polygon has an even number of sides, the longest diagonal is actually the diameter of the circle it is inscribed in. Thus, the longest distance we want is $ 10 $ and the square of that is $ 100 $.
Now we need to find the sum of the squares of the other distances (diagonals), call them $ d_1 ^2, d_2 ^2, d_3 ^2, ..., d_{2n-2} ^2 $ such that their sum is $ 500 $. By symmetry, we actually only need to find $ d_1 ^2 + d_2 ^2 + d_3 ^2 + ... + d_{n-1} ^2 = 250 $.

By LOC, we have

\begin{align*} d_1 ^2 = 5^2 + 5^2 - (2)(5)(5)(\cos \frac{\pi}{n})     \implies     d_1 ^2 = 50 - 50\cos \frac{\pi}{n} \\ d_2 ^2 = 50 - 50\cos \frac{2\pi}{n} \\ d_3 ^2 = 50 - 50\cos \frac{3\pi}{n} \\ d_4 ^2 = 50 - 50\cos \frac{4\pi}{n} \\ ... \\ d_{n - 1} ^2 = 50 - 50\cos \frac{(n - 1)\pi}{n}\end{align*}

Thus, after factoring, we have \begin{align*} 50(n - 1) -50(n - 1)( \cos \frac{\pi}{n} + \cos \frac{2\pi}{n} + \cos \frac{3\pi}{n} + ... + \cos \frac{(n - 1)\pi}{n}) = 250 \\ \implies (n - 1) - (n - 1)( \cos \frac{\pi}{n} + \cos \frac{2\pi}{n} + \cos \frac{3\pi}{n} + ... + \cos \frac{(n - 1)\pi}{n}) = 5 \\ \implies   (n - 1) - (n - 1)(\cos \frac{\pi}{2}) = 5 \\ \implies   n - 1 = 5 \\                                  \implies \boxed{n = 6}.  \end{align*}

Remarks: My solution looks somewhat ugly and is definitely bashier than the official solution which is quite nice and easy to understand. But the key part is noticing that the $ \cos $ terms all die due to the cosines of supplementary angles summing to zero.
[/hide]


function
induction
algebra unsolved
algebra

Find all function $ f: \mathbb{Q}^{\plus{}}\rightarrow\mathbb{Q}^{\plus{}}$ such that $ f(x)\plus{}f(\frac{1}{x}\equal{}1$ and $ f(f(x))\equal{}\frac{f(x\plus{}1)}{f(x)}$ for all $ x\in\mathbb{Q}^\plus{}$

For $ x\equal{}1$, using the first condition (say (1)), we deduce that $ f(1)\equal{} \frac 1 2$.
Now, using the second conditon (say (2), we deduce that $ f(\frac 1 2) \equal{} 2f(2)$. Using (1), it follows that $ f(2) \equal{} \frac 1 3$ and $ f(\frac 1 2 )\equal{} \frac 2 3$.

By induction on $ q$, we prove that $ f(\frac p q) \equal{} \frac q {p\plus{}q}.$
Assume that is holds for each $ \frac a b$ where $ b<q$.
Now, for $ 0<p<q$, we have $ q\equal{}p\plus{}r$ so that $ \frac p q \equal{} \frac p {p\plus{}r} \equal{} f(\frac r p)$ by the induction hypothesis since $ p \leq q\minus{}1$.
It follows from (2) that $ f(\frac p q) \equal{} f(f(\frac r p)) \equal{} \frac {f(\frac {r\plus{}p}{p})}{f(\frac r p)}\equal{}\frac {p}{r\plus{}2p} \cdot \frac {r\plus{}p}{p} \equal{} \frac {q}{p\plus{}q}$, as desired.
Thus, the induction step is proved for $ p<q$. The case $ p\equal{}q$ is already done.
Now assume that $ p\equal{}q\plus{}r$ with $ r \leq q$ : then $ \frac {p}{q} \equal{} \frac r q \plus{}1$.
Using (2), we deduce that $ f(\frac p q) \equal{} f(\frac r q )f(f(\frac r q))$ so that $ f(\frac {q\plus{}r} q) \equal{} \frac q {r\plus{}q} \cdot f(\frac q {r\plus{}q})$.
Using (1), we deduce that $ f(\frac {r\plus{}q} {q} )\equal{} \frac q {r\plus{}2q}$, which is $ f(\frac p q ) \equal{} \frac q {p\plus{}q}$.
Another induction on $ n$, shows easily that $ f(\frac {nq\plus{}r}{q}) \equal{} \frac q {(n\plus{}1)q\plus{}r})$ for each positive integer $ n$ and $ 0 \leq r \leq q$. Thus we deduce the induction step in the general case, and we are done.

Conversly, it is easy to verify that $ f(\frac p q ) \equal{} \frac q {p\plus{}q}$ is a solution.

Pierre.


Let $ p$ be a prime number such that $ p\minus{}1$ is a perfect square. Prove that the equation
$ a^{2}\plus{}(p\minus{}1)b^{2}\equal{}pc^{2}$
has infinite many integer solutions $ a$, $ b$ and $ c$ with $ (a,b,c)\equal{}1$

p=n^2 +1

a = n^2 x +ny , b = x- ny , c = n z(where integer triples (x,y,z) are primitive Pythagorean triples)

[quote="trejos"]Let $ p$ be a prime number such that $ p \minus{} 1$ is a perfect square. Prove that the equation
$ a^{2} \plus{} (p \minus{} 1)b^{2} \equal{} pc^{2}$
has infinite many integer solutions $ a$, $ b$ and $ c$ with $ (a,b,c) \equal{} 1$[/quote]
I think that all solutions are given by:

$ a \equal{}\sqrt{p\minus{}1}(k^2p\minus{}2km\minus{}k^2) \plus{} 2km(p\minus{}1) \minus{} 2m^2$

$ b \equal{} (k^2 \plus{} 2km)\sqrt{p\minus{}1} \plus{} 2km \plus{} 2m^2$

$ c \equal{} k^2(p\minus{}1) \plus{} 2km\sqrt{p\minus{}1} \plus{} 2m^2$

where $ k,m \in \mathbb{Z}$

I have been wondering: Why should $ p$ be a prime?! Isn't it only interesting if you should find all solutions?!

you have to first show that  the $ sequence$ $ n^2\plus{}1$ contains infinitely many $ primes$ otherwise your solution is noit correct and its still an $ open\minus{}question$

no, but that is not required. [b]we are given[/b] a prime of the form $ n^2 \plus{}1$ and we must show that there are [b]infinitely many values of $ (a,b,c)$[/b]. nothing in the problem or any of the solutions require us to prove that their are infinitely many primes in the sequence $ n^2\plus{}1.$ since i have not yet gone through and verified their solution, i cannot say whether it is right or wrong, but on this count, all of them(including the question), are perfectly fine

all that may be required is to show that there is atleast one prime of that form. trivial examples would be $ 5,17,37,101$

yes brother,but i knew that it was just a $ prime$

the conditions are all waste!
let $b=1$,the Pell equation $x^2-py^2=-p+1$
has one solution $(1,1)$
so it has infinitely many solutions,proved!

Pretty strange condition in this task. The strange thing is that for all numbers $"p"$ this equation has a solution.

$a^2+(p-1)b^2=pc^2$

Solutions can be written:

$a=pk^2+2pkn+n^2$

$b=pk^2-n^2$

$c=pk^2+2kn+n^2$

$k,n$ - integers.

Generally for such equations is a formula, it is necessary to substitute the coefficients and obtain a formula describing their solutions. You can see in this thread.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=60&t=583648


For each even positive integer $x$, let $g(x)$ denote the greatest power of $2$ that divides $x$.  For example, $g(20)=4$ and $g(16)=16$.  For each positive integer $n$, let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$  Find the greatest integer $n$ less than $1000$ such that $S_n$ is a perfect square.

[hide]Consider the list of integers $2$, $4$, $6$, $\cdots$, $2^n$. For all integers $i$ such that $1\le i\le n-1$, we want to count how many integers in this list are divisible by $2^i$ but not by $2^{i+1}$. There are $\frac{2^n}{2^i}$ integers divisible by $2^i$, but $\frac{2^n}{2^{i+1}}$ of them are also divisible by $2^{i+1}$. Hence we must subtract $\frac{2^n}{2^{i+1}}$ from $\displaystyle\frac{2^n}{2^i}$ and multiply it by $2^i$, since each remaining integer would give a value of $2^i$. We sum it from $i=1$ to $i=n-1$ to get
$\sum_{i=1}^{n-1}{2^i(\frac{2^n}{2^i}-\frac{2^n}{2^{i+1}})}=\sum_{i=1}^{n-1}{(2^n-2^{n-1})}=\sum_{i=1}^{n-1}{2^{n-1}}=(n-1)2^{n-1}$. For $i=n$, there is only one integer that would give a value of $2^n$, namely $2^n$ itself. Thus we add it to the sum to get that $S_n=(n-1)2^{n-1}+2^n=(n+1)2^{n-1}$, so the greatest integer $n<1000$ such that $S_n$ is a perfect square is $\boxed{899}$.[/hide]

A recursive way:

[hide]
From the fact that

$g(2x) = g(2(x-2^{n-1}))$ for $2^{n-1} < x<2^n$,

we can establish a recursive relation $S_{n+1}=2S_n + 2^n$ with $S_1=2$ which gives:

$S_n=2^{n-1}(n+1)$.

[/hide]

Listing out the terms, we get $2, 4, 2, 8, 2, 4, 2, 16, 2, 4, 2 \ldots$. We can see that in this sum there will be $2^{n-2}$ twos, $2^{n-3}$ threes, and so on until one $2^{n-1}$. We also have one $2^n$ term (the final term). Therefore, $S_n = (n-1) \cdot 2^{n-1} + 1 \cdot 2^n = (n+1) \cdot 2^{n-1}$. If $n$ is even, then $n+1$ has no factors of $2$ and $2^{n-1}$ is $2$ to an odd power, so this cannot be a perfect square. Therefore, $n$ must be odd. This means $2^{n-1}$ is already a perfect square, so we just need $n+1$ to be a perfect square and $n$ to be odd. This gives an answer of $30^2 - 1 = \boxed{899}$.

We can express $$S_n=2^1a_1+2^2a_2+2^3a_3+ \cdots + 2^{n-1}a_{n-1}+2^n.$$ Note that $a_1$ is the number of odd integers less than or equal to $2^{n-2},$ $a_2$ is the number of odd integers less than $2^{n-3},$ and so on. Thus, $a_k=2^{n-k-1}$ and $2^k \cdot a_k = 2^{n-1}$ therefore $$S_n=(n-1) \cdot 2^{n-1}+2^n=(n+1) \cdot 2^{n-1}.$$ By inspection, $S_n$ is a perfect square if and only if $n$ is an odd number and therefore if $(n+1)$ is an even perfect square. The greatest even perfect square less than $1000$ is $n+1=30^2=900,$ so $n=\boxed{899}.$

Dang tricky with the $2^n$ term...

Not too bad if you just write everything out and do the computation carefully
[hide]Note that $g(2k) = 2g(k)$ so $S_n = 2\sum_{k = 1}^{2^{n-1}}g(k)$. Note that each term with $\nu_2(k) = c$ for $0\le c\le n-1$ contributes $2^c$ to the sum. Since there are exactly $\left\lfloor\frac{2^{n-1}}{2^c}\right\rfloor-\left\lfloor\frac{2^{n-1}}{2^{c+1}}\right\rfloor$ terms in the sum with $\nu_2(k) = c$, we can write
\[ S_n = 2\sum_{c = 1}^{n-1}2^c\left(\left\lfloor\frac{2^{n-1}}{2^c}\right\rfloor-\left\lfloor\frac{2^{n-1}}{2^{c+1}}\right\rfloor\right)\]
\[ \iff S_n = 2\left(2^{n-1} +  \sum_{c = 1}^{n-2}2^c(2^{n-1 - c} - 2^{n - 2 - c})\right)\]
\[ \iff S_n = 2\left(2^{n-1} + \sum_{c = 1}^{n-2}(2^{n-1} - 2^{n-2})\right) = 2(2^{n-1} + (n-2)2^{n-1} - (n-2)2^{n-2}).\]
Thus, $S_n = 2((n-1)2^{n-1} - (n-2)2^{n-2}) = 2^{n-1}(n + 1)$. If $n$ is even, then $\nu_2 (S_n) = n- 1$, which is odd, meaning $S_n$ won't be a perfect square. If $n$ is odd, then $2^{n-1}$ is a perfect square, so we just need $n + 1$ to be a perfect square. Since $n$ must be odd, $n = 30^2 - 1$ is the largest $n < 1000$ that works, so the answer is $\boxed{899}$.[/hide]

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

It is possible to just evaluate $S_n$ explicitly, by considering how many values satisfy $i$ satisfy $g(i) = 2^j$ for positive integer $j$:
$$
S_n = 2 \cdot 2^{n-2} + 2^2 \cdot 2^{n-3} + \ldots + 2 \cdot 2^{n-2} + 2^{n-1} + 2^n = 2^{n-1} (n+1).
$$

If $n$ is odd, then $n+1$ has to be an even perfect square, the largest being $900$, giving $n = \fbox{899}$. If $n$ is even, then $n+1$ has to be twice a perfect square, giving the largest value of $n$ being $881$, which is smaller.


John has five children.  What is the probability that at least half of them are girls?

[quote=maseratirex][quote=math154]Notice that there is no case in which the number of boys and girls are equal.

Thus, by symmetry, the probability is $ \boxed{\frac12}$.[/quote]

Hide...[/quote]

He will not see this post 8 years after posting the solution.

[quote=NotMyRealNameII]Never have gotten this answer. It just doesn't click.[/quote]

Here is a more legit/hopefully understandable [hide=solution]We can count the number of ways of at least half of the children being girls with casework.

[b]Case 1: 3 girls[/b]: The family looks like BBGGG and there are $\binom{5}{2}=10$ ways for this to happen.

[b]Case 2: 4 girls[/b] The family looks like BGGGG and there are $\binom{5}{1}=5$ ways for this to happen.

[b]Case 3: 5 girls[/b] The family looks like GGGGG and there are $\binom{5}{0}=1$ way for this to happen.

So there are $16$ ways for at least half of the children to be girls.

The total number of ways is $2^5=32$ because each of the five children can be either a girl or boy.

And $\frac{16}{32}=\boxed{\frac{1}{2}}$[/hide]

Why it is $\frac{1}{2}:$

[hide=solution]

If we substitute boys instead of girls in the problem, we still solve it the same way. Since there are $5$ children, there must be either more girls or boys. Therefore, it is equally likely that there are more girls or more boys.

So, the answer is $\boxed{\frac{1}{2}}.$
[/hide]


geometry
incenter
circumcircle
geometry unsolved

By simple calculation you can get that $\angle DAC=20^o,\angle ACB=40^o$ and $\angle ADC=130^o$. Let $F$ be the point on side $BC$
s.t. $\angle BAF=20^o$, then $\angle FAC=40^0$ so triangle $AFC$ is
isosceles, whit $\angle AFC=100$, so $F$ is a circumcentre of
triangle $ACD$. $\angle BFA=80^o$ so $AB=AF$ and $\angle AFD=2\cdot \angle ACD=60^o$ and $AF=FD=radiusOfCicumcircle$ so triangle $AFD$
is equilateral. Now we obtain $AB=AF=DA$ so $\angle BAD=50^o$. Like
you said, just angles :)

bye

[quote="socrates"]in a quadrilateral $ABCD$ it is $\angle A=\angle B=80 ^\circ$
$\angle BAC=60^\circ$ and $\angle ACD=30 ^\circ$

find $\angle ADB$

 :)[/quote]

Don't know why I keep enjoying so much this little ones  :lol: . Even though I'm quite busy these days, just couldn't resist, maybe because they are the ones I can solve, :) By the way, nice solution m@re, just angles... here is mine

1) Take $I$ the incenter of triangle $ABC$. Then $\angle AIC = 30 = \angle ACD$, $\angle ACI = 20 = \angle DAC$. Hence $AICD$ parallelogram.
2) $AD = CI$
3) Let $E$ be the point where $BI$ meets the circumcircle of $ABC$, then $\angle CBE = 40 \Rightarrow AB = EC$
4) It is inmediate to find that triangle $CIE$ is equilateral so $EC = CI$ and using 2) and 3) one has $AB = AD$ hence $\angle ADB = 50$

Daniel


number theory
prime numbers
number theory unsolved

Let $ p, q$ be distinct positive prime numbers, and let $ g \equal{} \gcd(p\minus{}1, q\minus{}1)$.
1. If $ a \in \mathbb{Z}: \gcd(a, pq) \equal{} 1$, then $ a^{\frac{(p\minus{}1)(q\minus{}1)}{g}} \equiv 1 \mod pq$

2. Let $ e, d \in \mathbb{Z}$ with $ e, d \geq 1$ so that $ \gcd\left(e, (p\minus{}1)(q\minus{}1)\right) \equal{}1$ and $ de \equiv 1 \mod (p\minus{}1)(q\minus{}1)$. If $ c \in \mathbb{Z}$ with $ \gcd(c, pq) \equal{}1$, then $ x \equiv c^d \mod pq$ is the unique solution to the congruence equation $ x^e \equiv c \mod pq$.

1-we just need to prove that $ a^{\frac{(p\minus{}1)(q\minus{}1)}{g}} \equiv 1\mod q$ and also $ a^{\frac{(p\minus{}1)(q\minus{}1)}{g}}\equiv 1\mod p$ and it is really easy since $ g$ divides both $ p\minus{}1$ and $ q\minus{}1$.
2-we know that $ x^{e}\equiv c\mod q$ (and similarly for $ p$) since $ q$(and $ p$) has a primitive root so due to the well know theorem and the condition $ (e,q\minus{}1)\equal{}1$ (and similarly for $ p\minus{}1$)
this congruency has only one answer and since we know that $ c^d$ is one of the answer so it is the unique answer.

Nice and short solutions. shoki! Thank you.


Lightning is unfair. If you are racing online and lightning strikes, you could be the unlucky racer to get hit while crossing a gap, so you fall. Thus you lose the race due to bad luck, despite the fact that you are good.

ok.. whats the point of this topic though. was it neccessary? :mad:

Well, this topic is a bit random, but it is the Fun Factory after all.

@OP:  I assume you're talking about Mario Kart here.  That's what all items in the game are designed to do:  give racers who are behind a chance to win, even if they are somewhat less skilled than their opponents.  In my opinion, the unpredictability that items create is part of what makes Mario Kart such a fun game to play.

You see, that is why Mario Kart is not played as competitively, as, say, Halo, or SSBM. For a game to be able to be played competitively, it must be fair, and for it to be fair, there must be as little random factors as possible.

And yes, what do you hope to achieve by posting this hear? Are you asking us to discuss anything?

And I'm assuming you guys are talking about the NintendoFtlii


Express
$ \ln{(x^{2n} \minus{} y^n)} \equal{} 1$
in terms of $ n$.

Edit: whoops, it's now an actual equation.

[quote="JRav"]I'm not sure it can be done.  First, I can't see a way to get the $ n$ out of the logarithm, and second, this is just an expression, there's no equal sign or anything to express it in terms of anything.[/quote]

[hide="Hint"]
$ \ln{a\minus{}b}\equal{}\frac{\ln{a}}{\ln{b}}$
$ \ln{a^n}\equal{}n\ln{a}$[/hide]

[quote="madness"][quote="Anaxerzia"][quote="JRav"]I'm not sure it can be done.  First, I can't see a way to get the $ n$ out of the logarithm, and second, this is just an expression, there's no equal sign or anything to express it in terms of anything.[/quote]

[hide="Hint"]
$ \ln{a \minus{} b} \equal{} \frac {\ln{a}}{\ln{b}}$
$ \ln{a^n} \equal{} n\ln{a}$[/hide][/quote]

Your first hint is wrong  :huh:  its only for lna - lnb  :)[/quote]The first hint is totally wrong. The correct identity for 'log of a quotient' is: $ \ln\frac{a}{b}\equal{}\ln a\minus{}\ln b$

This problem cannot be solved algebraically, just as $ 4^n \minus{} 3^n \equal{} 3$ cannot be solved algebraically.


If $ a,b,c>0$, then
$ \sum\frac{a^2c^2}{b^4\plus{}ab^2c\plus{}a^2c^2}\geq1$

Setting $ ab\equal{}x$, $ bc\equal{}y$, $ ca\equal{}z$, we have to show
$ \sum\frac{x^4}{x^4\plus{}x^2yz\plus{}y^2z^2} \geq 1$.

By Cauchy, we get
$ \sum\frac{x^4}{x^4\plus{}x^2yz\plus{}y^2z^2} \cdot \sum(x^4\plus{}x^2yz\plus{}y^2z^2) \geq (x^2\plus{}y^2\plus{}z^2)^2$.

Therefore, it suffices to show that $ (x^2\plus{}y^2\plus{}z^2)^2 \geq \sum(x^4\plus{}xyz^2\plus{}x^2y^2) \Leftrightarrow \sum x^2y^2 \geq \sum x^2yz$, which is easy by Am-Gm  :lol: .

[quote="lipako"]If $ a,b,c > 0$, then
$ \sum\frac {a^2c^2}{b^4 \plus{} ab^2c \plus{} a^2c^2}\geq1$[/quote]
After substitution $ x \equal{} \frac {a^2}{bc},$ $ y \equal{} \frac {b^2}{ac}$ and $ z \equal{} \frac {c^2}{ab}$ we need to prove that $ \sum_{cyc}\frac {1}{x^2 \plus{} x \plus{} 1}\geq1,$ where $ xyz \equal{} 1,$
which is equivalent to $ \sum_{cyc}(x \minus{} y)^2\geq0.$

it's also equivalente with:
$ \frac{m^2}{p^2\plus{}pm\plus{}m^2}\plus{}\frac{n^2}{m^2\plus{}nm\plus{}n^2}\plus{}\frac{p^2}{p^2\plus{}pn\plus{}n^2}\geq 1$  :wink:


algebra
polynomial
superior algebra
superior algebra unsolved

Is it possible to find an infinite field of characteristic $ 7$?  Naturally I would think to start with the field $ \mathbb{Z}/p\mathbb{Z}$, but can we use this to construct a field of rational polynomials mod $ 7$?

Yes, if $ K \equal{} \mathbb{Z}/p\mathbb{Z}$ doesn't $ K(x)$ have characteristic $ p$?

$ K(x)$ is the field of quotients of $ K[x]$

Or simply the algebraic closure of $ \mathbb{Z}/7\mathbb{Z}$. ;)


This was conjectured by somebody on Physics Forums and I thought you guys might have some idea about how to prove it (or if it's false, to find a counterexample and suggest how to repair the idea into a true conjecture).

[i]Suppose that $ p_k$ is the $ k^{th}$ prime, $ k \ge 3$. Then for each prime $ q$ with $ p_{k \minus{} 1}^2 < q < p_k^2,$ there exist positive integers $ m, n$ with $ q \equal{} m \minus{} n$ and $ p_j \mid mn \iff j < k.$[/i]

For example, if $ k \equal{} 3, 4,$ we have:
27-16=11
16-3=13
81-64=17
27-8=19
27-4=23
---
45-16=29
81-50=31
45-8=37
45-4=41
45-2=43
50-3=47

Observe that if $ p_j \mid m$ or $ p_j \mid n$ for each $ j < k$ and $ m \minus{} n < p_k^2$ then $ m \minus{} n$ is prime. The question is whether we can obtain all primes that way.

[b]The statement of the conjecture was edited to account for Albanian Eagle's objection. Before that it didn't specify that $ p_j$ couldn't divide $ m$ or $ n$ if $ j \ge k.$[/b]

I don't quite get what you're asking:
couldn't n be the product of all primes up to q and m=n+q ?!

Sorry $ m$ and $ n$ should not be divisible by any primes bigger than $ p_k$.

Another way of stating the condition is:

$ p_j \mid mn \iff j < k$

I fixed the original post accordingly.


function
topology
Support
geometry
real analysis
real analysis unsolved

Here is a really beautiful, but difficult result: let $X,Y$ 2 compact metric spaces and $A,B$ the algebras of continuous functions from $X$ to $R$ and from $Y$ to $R$. Prove that if $A$ and $B$ are homeomorphic, then $X$ and $Y$ are homeomorphic.

I think you mean that the function algebras are isomorphic.
It is true for compact Haussdorff topological spaces, or locally compact if the algebras of functions are the ones with compact support ($f = 0$ outside a compact set).

This is a very popular fact, proved by Gelfand, giving a duality between topological spaces and operator algebras.  It is repeated like a mantra in papers on "noncommutative geometry" which is even more popular than Gelfand's lemma.

By "isomorphic", do you mean algebraic isomorphism, or algebra isomprphism that's also a homeomorphism?

Algebra isomorphism.  I don't know what it means for two algebras to be homeomorphic
(I mean, you can imagine placing a topology on the algebras, but it is perverse to consider maps between two topological algebras that do not respect the algebra structure, which in this case means isomorphisms as we ask for invertible maps).

Maybe there is some canonical topology on $C(X)$ in the case when $X$ is a metric space, that you have in mind?  If $X$ is compact you can define a metric (and so a topology) by $d(f,g) = max |f-g|$, and there is a theorem that I can't remember (Arzela-Ascoli) that characterizes the compact subsets in this topology.

Oh, sorry, I was thinking about the algebra of both continuous and bounded functions with the $\sup$ norm.

I see.  Our messages crossed during the editing.


Fie triunghiul $ABC$ şi $I$ centrul cercului său înscris. Notăm cu $M$ şi $N$ mijloacele laturilor $AB$ şi respectiv $AC$, iar cu $P$ şi $Q$ intersecţiile dreptei $MN$ cu dreptele $BI$ şi respectiv $CI$. Să se arate că $PQ=\frac{AI}{2}$ dacă şi numai dacă $m(\angle{PAQ})=30^{o}$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=457[/img]
Cum $MN$ e linie mijlocie, avem $MN\parallel BC$, deci $\angle MPB=\angle PBC=\angle PBM$(prima egalitate din alterne-interne, a doua din faptul ca $BI$ e bisectoare). Deci $\triangle MPB$ e isoscel $\Rightarrow MP=MB\Rightarrow MA=MP=MB$, deci triunghiul $APB$ e dreptunghic, cu $\angle APB=90^o$. Analog $\angle AQC=90^o$
Deci $AQIP$ e inscriptibi, iar cercul circumscris are diagonala $[AI]$. Fie O mijlocul lui $[AI]$(adica centrul cercului)
"$\Rightarrow$" Daca $\angle PAQ=30^o$, atunci $\angle POQ=60^o$. Cum $OQ=OP$, avem ca $\triangle POQ$ este echilateral, deci $PQ=OP=AO=\frac{AI}{2}$
"$\Leftarrow$" Daca $PQ=\frac{AI}{2}$, atunci $PQ=OP=OQ$, deci $\triangle POQ$ e echilateral, $\Rightarrow\angle POQ=60^o\Rightarrow\angle PAQ=30^o$

[hide="Aici este un comentariu didactic asupra abordarii unei probleme de geometrie."][color=blue]Aceasta problema si solutia oferita de Frozen imi ofera prilejul sa mentionez un [b]pas obligatoriu de trecere[/b] in debutul demonstratiei unei probleme de geometrie (asa cum in algebra este celebrul domeniu de existenta sau sens real). Dupa parerea mea, in primul rand trebuie precizate pozitiile punctelor $P$ si $Q$ in raport cu punctele $M$ si $N$ pe dreapta $MN$ in situatia particulara $2\cdot PQ=AI$. Se observa ca (intotdeauna) $h_a>2r^{\ (*)}$, $\frac{PQ}{a}=\frac{\frac{h_a}{2}-r}{r}$, adica $\boxed {PQ=p-a}$ si $AI=\frac{bc}{p}\cos \frac A2$. Prin urmare, $2\cdot PQ=AI$ $\Longleftrightarrow$ $2p(p-a)=bc\cos \frac A2$ $\Longleftrightarrow$ $2\cos^2\frac A2=\cos \frac A2$ $\Longleftrightarrow$ $\cos \frac A2=\frac 12$ $\Longleftrightarrow$ $A=120^{\circ}$. Asadar, $a>b$, $a>c$ si punctele $P$ si $Q$ apartin segmentului deschis $(MN)$ (in figura din reply - ul lui Frozen pozitiile punctelor $P$ si $Q$ sunt gresite). In continuare, se arata usor ca $MQ=\frac{a-b}{2}$, $NP=\frac{a-c}{2}$, $QA\perp QC$, $PA\perp PB$ si patrulaterul $APIQ$ este inscris in cercul de diametru $AI$. Deci $PQ=AI\sin \widehat {PAQ}$. Astfel, $A=120^{\circ}$ $\Longleftrightarrow$ $2\cdot PQ=AI$ $\Longleftrightarrow$ $m(\widehat {PAQ})=30^{\circ}$.[/color]

[color=darkred]In incheiere, imi cer scuze lui Frozen ca am indraznit sa ma leg de desenul care insoteste demonstratia sa si tot odata apreciez ca pe o figura gresita a dovedit corect echivalenta propusa. Ceea ce am facut eu aici as numi [b]"studiu de caz".[/b][/color]

$(*)$ [b]Observatie.[/b] $h_a>2r\Longleftrightarrow S>ar\Longleftrightarrow p>a\Longleftrightarrow b+c>a\ .$[/hide]

Domnule profesor, va multumesc pentru observatie.

Cu deosebita placere ! Poate imi spui printr-un mesaj privat cine esti (cum te cheama, in ce clasa si de unde esti). Poate ne vom intalni vreo data la olimpiada sau vreo interjudeteana (in general, nu prea stau pe acasa de sarbatorile astea ale Matematicii).


calculus
derivative
inequalities proposed
inequalities

[quote="shaaam"]Given that $ a,b,c$ are real numbers.

What is the minimum value of:

a) $ \left(\frac {a \minus{} b}{b \minus{} c}\right)^4 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^4 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^4$

Notice that, the problem is variation of [url]http://www.mathlinks.ro/viewtopic.php?t=295199[/url][/quote]

The minimum is very ugly. It is about $ .753733827$
My solution is straightforward. Let
$ x \equal{} \frac {a \minus{} c}{a \minus{} b}$ , $ y \equal{} \frac {b \minus{} a}{b \minus{} c}$,  $ z \equal{} \frac {c \minus{} b}{c \minus{} a}$
then

$ xyz \equal{} \minus{} 1$ and $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} xz \plus{} 3$
Now, let $ p \equal{} x \plus{} y \plus{} z$. With some manipulations, you got
$ x^4 \plus{} y^4 \plus{} z^4 \equal{} p^4 \minus{} 4p^3 \plus{} 14p^2 \minus{} 16p \plus{} 18 \equal{} f(p)$
To find the min of $ \sum x^4$, take derivative wrt p and solve for p, you get $ .753733827$

[quote="fjwxcsl"][quote="shaaam"]Given that $ a,b,c$ are real numbers.

What is the minimum value of:

a) $ \left(\frac {a \minus{} b}{b \minus{} c}\right)^4 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^4 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^4$

b) $ \left(\frac {a \minus{} b}{b \minus{} c}\right)^6 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^6 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^6$

Notice that, the problem is variation of [url]http://www.mathlinks.ro/viewtopic.php?t=295199[/url][/quote]

for $ a),f(a,b,c)_{min} \equal{} 12.5037895...,$ it is a root of the equation$ k^3 \minus{} 7k^2 \plus{} 3k \minus{} 898 \equal{} 0.$
for $ b),f(a,b,c)_{min} \equal{} 32.5668090 ...,$it  is a root of the equation $ \minus{} 2253543 \minus{} 309028k \minus{} 19191k^2 \minus{} 17k^3 \minus{} 3k^4 \plus{} k^5 \equal{} 0.$[/quote]
for $ c),f(a,b,c)\equal{}\left(\frac {a \minus{} b}{b \minus{} c}\right)^8 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^8 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^8,$
$ f(a,b,c)_{min} \equal{} 85.1220762...,$it  is a root of the equation $ \minus{} 76004879517 \plus{} 8196929634k \minus{} 120392192k^2 \minus{} 15278707k^3 \minus{} 354594k^4 \plus{} 21k^5 \minus{} 11k^6 \plus{} k^7.$

[quote="fjwxcsl"][quote="fjwxcsl"][quote="shaaam"]Given that $ a,b,c$ are real numbers.

What is the minimum value of:

a) $ \left(\frac {a \minus{} b}{b \minus{} c}\right)^4 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^4 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^4$

b) $ \left(\frac {a \minus{} b}{b \minus{} c}\right)^6 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^6 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^6$

Notice that, the problem is variation of [url]http://www.mathlinks.ro/viewtopic.php?t=295199[/url][/quote]

for $ a),f(a,b,c)_{min} \equal{} 12.5037895...,$ it is a root of the equation$ k^3 \minus{} 7k^2 \plus{} 3k \minus{} 898 \equal{} 0.$
for $ b),f(a,b,c)_{min} \equal{} 32.5668090 ...,$it  is a root of the equation $ \minus{} 2253543 \minus{} 309028k \minus{} 19191k^2 \minus{} 17k^3 \minus{} 3k^4 \plus{} k^5 \equal{} 0.$[/quote]
for $ c),f(a,b,c) \equal{} \left(\frac {a \minus{} b}{b \minus{} c}\right)^8 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^8 \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^8,$
$ f(a,b,c)_{min} \equal{} 85.1220762...,$it  is a root of the equation $ \minus{} 76004879517 \plus{} 8196929634k \minus{} 120392192k^2 \minus{} 15278707k^3 \minus{} 354594k^4 \plus{} 21k^5 \minus{} 11k^6 \plus{} k^7.$[/quote]

for $ d),f(a,b,c) \equal{} \left(\frac {a \minus{} b}{b \minus{} c}\right)^{10} \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^{10 } \plus{} \left(\frac {c \minus{} a}{a \minus{} b}\right)^{10},$
$ f(a,b,c)_{min} \equal{} 222.644285 ...,$it  is a root of the equation
$ \minus{} 34104229028388226\plus{}972127795927737k\minus{}224739119524671k^2\plus{}4165968288737k^3\minus{}17147861646k^4\minus{}891009832k^5\minus{}6339816k^6\plus{}36k^7\minus{}13k^8\plus{}k^9\equal{}0.$

Thank you, fjwxcsl.  Can you also find minimum of the following: (with the same condition)

\[ \left(\frac {(a \minus{} b)^2}{b \minus{} c}\right)^2 \plus{} \left(\frac {(b \minus{} c)^2}{c \minus{} a}\right)^2 \plus{} \left(\frac {(c \minus{} a)^2}{a \minus{} b}\right)^2\]

:maybe:

[quote="shaaam"][quote="fjwxcsl"]This is clearly,$ [\sum{({\frac {(a - b)^2}{b - c})^{2}}]_{min} = 0^{ + },[\sum{({\frac {(a - b)^2}{b - c})^{2}}]_{max} = { + \infty}.}}$[/quote]

In what values of $ a,b,c$ , the minimum value approaches to $ 0$ ?[/quote]

 Let $ a=3x,b=2x,c=x\to{0}.$


modular arithmetic
number theory
relatively prime

Let $n$, $k$ be positive integers such that $n$ is not divisible by $3$ and $k\ge n$. Prove that there exists a positive integer m which is divisible by $n$ and the sum of its digits in the decimal representation is $k$.

Let $ n \equal{} a*b$ where $ b$ is not divisible by $ 2$ or $ 5$.
Then, $ a$ is of form $ 2^{k}*5^{l}$ so the number $ S \equal{} 10^{k\plus{}l}$ will be divisible by $ a$ as $ 10^{k\plus{}l}\equal{} 2^{k\plus{}l}5^{k\plus{}l}$

Now, because $ b$ is not divisible by $ 2$ or $ 5$, then $ \gcd(10,b) \equal{} 1$ hence if $ r$ is the number of integers from $ 1$ to $ b$ that are relatively prime to $ b$, then by Euler's Theorem, $ 10^{r}\equiv 1\pmod b$ so $ 10^{kr}\equiv1\pmod b$, and also $ 10^{kr\plus{}1}\equiv 10\pmod b$.
Then, the number $ T$ of form $ \sum_{i \equal{} 1}^{x}10^{ir}\plus{}\sum_{j \equal{} 1}^{y}10^{jr\plus{}1}\equiv x\plus{}10y\pmod b$ and the sum of its digits is $ x\plus{}y$. Then, if we can find positive integers $ x,y$ such that $ x\plus{}y \equal{} k$ and $ b|(x\plus{}10y)$ we will find the number $ T$ with sum of digits $ k$ and divisible by $ n$.
Now, $ b|(x\plus{}10y)\Leftrightarrow b|(9y\plus{}(x\plus{}y))\Leftrightarrow b|(9y\plus{}k)\Leftrightarrow 9y\equiv\minus{}k\pmod b$. But $ n$ is not divisible by $ 3$ so $ \gcd(b,9) \equal{} 1$ so there exists a positive integer $ z < b\leq k$ such that $ 9z\equiv\minus{}k\pmod b$. Then, if we take $ y \equal{} z$ and $ x \equal{} k\minus{}z$ then the integer $ T \equal{}\sum_{i \equal{} 1}^{x}10^{ir}\plus{}\sum_{j \equal{} 1}^{y}10^{jr\plus{}1}$ is divisible by $ b$ and has the sum of digits $ k$.
Then, $ TS \equal{} T*10^{k\plus{}l}$ has the sum of digits $ k$ and is divisible by $ 10^{k\plus{}l}b$ so it is divisible by $ n$, and the result follows.


topology
real analysis
real analysis unsolved

Hi, i´m having little trouble in solving this problem, please help me,
the problem goes this way:
Let К subset of R(real set) consist of 0 and the numbers 1/n, for n = 1, 2, 3, .... Prove that К is
compact directly from the definition (without using the Heine-Borel theorem).
I find difficult to get a finite subcover of an arbitrary open covering of K.
Thanks.

What do you know about open subsets of $ \mathbb{R}$? Given an open covering of $ K$, at least one of the sets in this cover has to contain $ 0 \in K$. How many points of $ K$ can lie outside of this open set?

I think that a finite number of points lie outside of the open set of 0, i do not know why it is true, the problem is when i try to find another open subset that contains points of K i don´t know which one is good, the idea is to find another open subset that contains some points that don´t lie on the subset containig the 0.
That´s my problem, any idea?.

[quote="cashimingo"]I think that a finite number of points lie outside of the open set of 0, i do not know why it is true[/quote]
If $ 0$ is in an open set, it is an interior point, so there exists a ball of positive radius $ \epsilon > 0$ around $ 0$ contained entirely in this open set.  How many points lie outside this ball?

That would be the rest of the points that aren´t in the ball of radius $ \epsilon$, in the open set, that´s the thing, but i don´t know how many they are, i suppose they are  infinite points.

Infinitely many? Are you sure?
The answer may depend on $ \epsilon$, but we're in trouble if it's ever infinite; for each point other than zero, we can cover it with an open set that doesn't cover any other points in $ K$.


I am trying to write an algorithm which, given two or more strings, searches to find any characters from A to Z which are common to at least two of the strings. Once it finds a match, it needs to output a string with the relation in it, such as "string_1[2]==string_2[5]". For example here is a sample input/output:

Input:
"*_A436B56"
"0231A15534846B"
"A4"

Output:
"string_1[2]==string_2[4]==string_3[0]"
"string_1[6]==string_2[13]"

One last thing I forgot to mention: If a letter is contained in a string, then all letters preceding it alphabetically must also be contained in a string. i.e. if 'D' is in a string, then A,B, and C must also be inside some string.

I mainly need speed for this algorithm, since I am going to try to implement it on a 15MHz device. I was thinking of just searching each string for A and finding matches, then B, and so on. This is pretty slow if there are a lot of characters, though. Any ideas for how to do it faster?

Well why don't u do this:

for each char from A to Z keep array flag[ max number of strings]

that meens u r to declrae array: flag[26][max # of strings]

then flagging :P

I'm not sure if there is a faster way, but as you search for each letter you could cut down the string length, for example:

Create an array of LinkedLists or something similar one representing each letter
Each time a letter is found you can add it to its corresponding linked list in the array and shorten the length of the string:
"*_A436B56"
Would become
"(3)436B56"
After the first search
Then
"(7)56"
So you are not searching the same values over and over

This is basically just using the array of linked lists and then you will only have to search through the string once.  When you find a letter add it to its corresponding linked list.

Basically, you have to use the Dynamic Programming Principle.
Using this technique, you don't need to search each of the given strings for every member of the alphabet string (which takes a huge amount of energy). Search everything once only and store it for later use.

If any letter after 'A' is in the string, then 'A' is in the string, right? So why don't you just search the string for 'A's? Or do you have to output all matches?


geometry
incenter
conics
hyperbola
Asymptote
trigonometry
geometric transformation

Consider triangle $ABC$, and construct its incenter center $I$. Draw ray $AI\rightarrow$ and choose a variable point $X$ on this ray. Drop perpendiculars from $X$ to $AB$ and $AC$ and call them $X_{1}, X_{2}$. Let $Y$ be the intersection of $X_{1}I$ and $XX_{2}$, extended.

What is the locus of $Y$?

[hide="Think you got it?"]
[img]6764[/img]
Not such an easy locus...what could it be? My first guess was part of a Rouleaux Triangle...
[/hide]

It is a part of a hyperbola, the asymptotes are AC and a line L which is perpendicular to AB and cut AB be at a point such that the line joining this point and I is perpendicular to AC.

Very nice, however is that just a guess or do you have proof?

And I think when you say AC it is really AB in my diagram and when you say AB it is BC in my diagram, i.e. your statement should read:

[i]
It is a part of a hyperbola, the asymptotes are AB and a line L which is perpendicular to AC and cut AC be at a point such that the line joining this point and I is perpendicular to AB.
[/i]

It is easier as such:
[img]6808[/img]

Really interesting!

It seems that the axis of the hyperbola makes $45^{o}$ angle with $AI$, independently of the angle $A$ and independently of the position of $I$

Well, I didn't use the points $B,C$ because they are not necessary.

I used as constants:
The point $A(0,0)$, the point $I(c,0)$ and the angle $\phi=\frac{A}{2}$

and the variable point $X(r,0)$

But I got a very ugly system of equations

$x = \frac{2rc-a(r+c)}{(r+c)-2a}$ and $y = \frac{b(c-r)}{(r+c)-2a}$

at which we should replace
[color=red] EDIT:[/color]$a=r\cos^{2}\phi$ and $b = r\sin\phi\cos\phi$.

Then it becomes more ugly

Pontios, you seem to know a lot about loci from your other posts. Tell me, have you seen a hyperbola as a simple locus in triangle based geometry before?

[hide="me@home"]
I have not see some problem which asks to prove that the locus is a hyperbola.

But I remember that, some times, when I try to solve a problem (not necessarily locus problem), the locus of some point happens to be a hyperbola.

For example,
Let $(A)$ and $(B)$ two circles touched at point $D$.
Let $(C)$ be a third circle wich touches the two circles.
Then the locus of $C$ is a hyperbola with focuses $A,B$ and $D$ one vertice[/hide]

Well, this problem obstinated me. I wanted howsoever to find a solution. But the result doesn't make me happy. You see, I expected that the answer would be simple.

In the solution below I used much constants, for example $K=\cos\phi\cos2\phi$, in order to be more readable.

[color=blue][size=134]Solution[/size][/color]
I worked on the above formulas and I tried to show that, after the rotation about $45^{o}$ they satisfy the equation
$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$

Accurately

$x_{p}= \frac{2rc-a(r+c)}{(r+c)-2a}$ and $y_{p}= \frac{b(c-r)}{(r+c)-2a}$

where $a=r\cos^{2}\phi$ and $b=r\sin\phi\cos\phi$

[b]Rotation[/b] [size=134][color=red](* Error; but it's ok. See below)[/color][/size]

$(\begin{array}{c}x\\ y \end{array}) (\begin{array}{cc}\cos 45^{o}& \sin 45^{o}\\ \sin 45^{o}&-\cos 45^{o}\end{array}) (\begin{array}{c}x_{p}\\ y_{p}\end{array})$

The rotated hyperbola has $B=0$.
Also, $A(0,0)$ belongs to the locus, so $F=0$

After [b][u]a lot[/u][/b] of calculations I found the system

$\sin^{2}\theta \cdot A+\cos^{2}\theta \cdot C =0$

$vD+wE=0$

$c(v^{2}A+w^{2}C)-2\cos\phi (D\sin\theta+E\cos\theta)=0$

$c(K-\sqrt{8}\sin\theta)A+c(K-\sqrt{8}\cos\theta)C+\cos 2\phi(D\sin\theta+E\cos\theta)=0$

where:

$\theta =\phi+\frac{\pi}{4}$
$Q = \sqrt{2}\cos\phi$
$v = 2-Q\cos\theta$
$w = 2-Q\sin\theta$
$K = \cos\phi\cos 2\phi$

The above system is homogeneous and has $D=0$. The last equation can be eliminated.

Using $A$ as parameter we find:

$\begin{array}{l}C =-A\tan^{2}\theta \\ D =-AwT \\ E = AvT\\ \end{array}\}\ \ (1)$

where $T=c\cdot\frac{v\cos\theta+w\sin\theta}{2\cos\phi\cos^{2}\theta}$

So we have

$A x^{2}-A\tan^{2}\theta y^{2}-AwT x+AvT y=0 \Rightarrow$

$x^{2}-\tan^{2}\theta y^{2}-wT x+vT y=0$

Completing the squares we get

$(x-\frac{wT}{2})^{2}-\tan^{2}\theta(y-\frac{vT}{2\tan^{2}\theta})^{2}=(\frac{T}{2\tan^{2}\theta})^{2}((w\tan^{2}\theta)^{2}-v^{2})$

Clearly, if $(w\tan^{2}\theta)^{2}-v^{2}<0$ then the axis of the hyperbola is the axis $y$

We can see that the center of the hyperbola is the point $\frac{T}{2}\cdot(w,\frac{v}{\tan^{2}\theta})$

[size=134][color=blue]Rotate back[/color][/size]
[size=134][color=red](*)[/color][/size] Well, I made a mistake, the rotation above is not only a rotation, but rotation followed by a reflection on y-axis.

But there is not a problem, we can simply multiply again by the same matrix to return to the old system

For example, the  center goes to the point $\frac{T\sqrt{2}}{4\tan^{2}\theta}\cdot(w\tan^{2}\theta+v,w\tan^{2}\theta-v)$

Similarly, we can calculate the parameters $a,b,c$ and the focuses of the hyperbola. Rotating back we can find the original positions of the focuses.

[hide="tested"]The above results are tested.
I put in Microsoft Excel the points
$x_{p}= \frac{2rc-a(r+c)}{(r+c)-2a}$ and $y_{p}= \frac{b(c-r)}{(r+c)-2a}$
I used $r$ as variable. Any line below I added to $r$ some constant value.
The Excel fie:
At any line, it calculates $x_{p}$ and $y_{p}$.
Then it applies the rotation specified above.
With 5 points we can find the equation of any conic, that is, to find the values of $A,B,C,D,E,F$
The results satisfy the equations $(1)$.[/hide]
[hide="the cost"]I spent:
about 1.5 books of 50 pages
3 days of my life
some thousands of cerebral cells  :D

and it was hard because I was always making mistakes[/hide]

[color=red]EDIT[/color]
We can prove that $v\cos\theta+w\sin\theta=Q$.

So, we can simplify
$T=\frac{c_{o}}{\cos^{2}\theta}$ and $\frac{T}{\tan^{2}\theta}=\frac{c_{o}}{\sin^{2}\theta}$

where $c_{o}=c\frac{\sqrt 2}{2}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=129849]Here[/url] is a nice problem with a hyperbola as locus

Wow! I am amazed by your comprehensive treatment of this problem, I must thank you greatly for respecting my problem so much!
Also, thank you for the other hyperbola locus problems.


I can't see post ratings, neither for my nor other people's posts (though I CAN see them for some very old posts) - is that some thing on the board itself, or it has something to do with my browser (Opera 9.5)?

You should be able to see your own average post rating, and individual posts that have been rated enough times have ratings visible to everyone. Moderators can see any poster's ratings. Most posts don't get rated at all, so don't worry too much about that.

[quote="jmerry"]You should be able to see your own average post rating, and individual posts [b]that have been rated enough times[/b] have ratings visible to everyone. Moderators can see any poster's ratings. Most posts don't get rated at all, so don't worry too much about that.[/quote]

That must be it.

[quote="kunny"]I can see your rating 4.8. and posts of 2514. :)[/quote]

Average rating isn't the issue :) but the individual post ratings. But I can't see YOUR average rating. Why?

So, regular users don't see other people's ratings at all, if I understand correctly.

[quote="kunny"][quote]="Farenhajt
Average rating isn't the issue :) but the individual post ratings. But I can't see YOUR average rating. Why?[/quote]

What's the individual ratings? I don't know it.

By the way, my average rating hasn't been changed, it reamined to 3.9 for 3 years.

But I don't care at all. :lol:[/quote]

Individual rating is the row of six hexagonal cells along the top of each post, where you can rate other users' individual posts, or see how a particular post of yours has been rated.

Well you could go back and see your own posts...


rotation
geometry
geometric transformation

A ferris wheel with a radius of 32m makes 2 revolutions every minute. a) Find the exact angular velocity in radians per second. I was able to figure that out but i cannot figure out that  if the ferris wheel rotates for 3 min then what the distance the rider travels would be.

I'm probably missing something :P ... but
[hide]2 revolutions per min. for 3 min.  That's 6 revolutions.
The rider goes along the circumference of the wheel 6 full times
The circumference is 2(pi)r
That's 64(pi) per rotation or, for 6 rotations, 384(pi)[/hide]

$\boxed{\nu=\frac{\theta}{t}}$, where $\nu$ is the angular velocity, $\theta$ is the angle (one revolution is $2\pi$), nad $t$ is the time elapsed. Find the angular velocity, then you can figure out the angle through which the rider travles in 3 minutes. What do you think you should do now, to get the actual distance traveled?

(hint) How far would you travel if you traveled once around the circle ($2\pi^{rad}$)? Think about circumference!

(sorry, html is not working for some reason)


At the same moment,two particles start respectively from vertices B and C of triangle ABC which has a right angle at C.The particles move at constant speeds and arive at vertex A at the same moment.If the time of travel is equal to (AB/AC)^(3/2) and the length of BC is equal to the sum of the constant speeds,determine the time of travel.

denoting by v1 the speed of point moving from B to A and v2 the second speed, i got that k:=(v1/v2) must be a solution of equation:
x^3-x^2+2x-1=0.
Then,once you know v1/v2 you know that
AB/v1=AC/v2
from where you get that AB/AC=k and so the time is
k^(3/2)


Well the rmo results are out!!! They're calling up those who got selected, but you can call up hbcse to find out... But i reallydont understand how they selected students, coz three of my friends and i solved 5 and a half questions....but one didnt get in...and i got in even though i had used calculus and she hadnt...and i know people who solved three qs who got selected!

congratulations
well it also matters how you write the solutions
one minor detail missing and you get less than or equal to half mark alloted for that sum :)

congrats esconie
are results in any other region out yet
it is strange the results are out but the official solutions are not up :rotfl:

ny website whr they have put up th results..??

Thanks...oh i didnt know that, its my first olympiad...i dont know about other regions, even here they havent finished calling up everyone and mailing the letters, we just know cos we called them up to ask...

anybody knows if results are out yet in Kolkata? :maybe:

btw welcome to aops esconie...and also congrats

[quote="rituraj007"]congrats
gaurav what abt u?? :D[/quote]
i cleared it last year! :)

[quote="rituraj007"]does that by ne way means u were in 12th last year? :roll:[/quote]He was the topper of INMO 08. And he is only in 11th now. He'll only have to keep touch with postal coaching to be called at the IMOtc I think.
[quote="Arnav...!!!"]is there anything to ask ritu from gaurav??? :rotfl:
i am sure u know the answer of the ques u asked :P[/quote]I don't think so. Then he wouldn't have said that.
BTW congrats esconie and others who've qualified.

Akshanil's right I am in 11th standard now and am in touch with professors with the postal coaching [When I say i am in touch I ASSUME that Indian postal service is delivering all I give it!!]
hope to see you all at IMOTC this year

hey are the results out in uttar pradesh

there r few bengalis lik keshav(perhaps)

but i thought the hindi meaning :rotfl:

me also  :rotfl:

what is its meaning????

wats the hindi meaning anyway :|
IN bengali it mean hen and is generally used to blame someone  as foolish :P

it means hen only in Hindi

ofc the languages arent too different :P

even though i am from bengal... i am not bengali...

n i lyk his username... bcoz i use that word very often...
just to irritate th pple arnd me!!

All hail! Murgi!

PS: I have a friend in school who is called "murgi" by his friends just for fun ( not that they insult him).....
$ e^{i (murgi)} \equal{} hen$

lol I have a friend whom we call 'chidhiya' No murgis though...
P.S. I LUUUUVVVV eating murgis!

I was never good at bio, but still i think you might catch bird flue :rotfl:

How can I catch a bird after it 'flue' in the skies?  :rotfl:

you can if you are a HAWK {i think you are} :rotfl:  :rotfl:

:o As a matter of fact I am a COW! So its the birds who catch me.
Derivation-
My real name=Gaurav
Dividing both sides by 2,
     real    me=Gau
But as Gau=Cow
Result--> I am a COW!  :rotfl:
P.S. Do I need to latex this?

that result also applies to me hence, i am of the same family (cow family)
thanks i now know what am i

That means I am a bull.  :huh:
Pulkit/2=pul
pul==bull :rotfl:

then what am i ??chitu :rotfl:

@gaurav---bio says cow family even includes buffalo and bull :P  :rotfl:


Find all $k \in R$ such that $\exists$ continuous function $f: R \rightarrow R$ satisfying $f(f(x))=kx^{25} \forall x\in R$.

This is similar than Putnam 1979-A/2 :
http://www.kalva.demon.co.uk/putnam/psoln/psol792.html

Proof is the same.

Pierre.


if$A$ and $I$ are two square matrixs such that
 $det ( A+I) = trace A+1$
  where $I$ is unit matrix and $rankA =1$
 with trade A is principal diagonal element sum

[quote="tranthanhnam"]if$A$ and $I$ are two matrixs have[color=blue] row and column equal.[/color]
prove that :
 $det ( A+I) = trace A+1$
  where $I$ is unit matrix[/quote]
I don't understand. What they are?


Hello everyone, I figured this was the best section of the forum to post this question in so enjoy!

"There are nine men to every ten women at Big State University. There are 26,239 students at the university. How many women are there?"


inequalities
function
logarithms
inequalities unsolved

Since $\frac{3}{2} \ge x+y+z \Rightarrow 1 \ge \frac{4}{9}(x+y+z)^2$, so L.H.S $\ge \frac{4}{9}\frac{(x+y+z)^2}{x^2+y^2+z^2}+\frac{4}{9}\frac{(x+y+z)^3}{xyz}$, hence it suffices to prove that
\[\frac{4}{9}\frac{(x+y+z)^2}{x^2+y^2+z^2}+\frac{4}{9}\frac{(x+y+z)^3}{xyz} \ge \frac{40}{3} \Leftrightarrow\]
\[\frac{(x+y+z)^2}{x^2+y^2+z^2}+\frac{(x+y+z)^3}{xyz} \ge 30 \Leftrightarrow\]
\[\displaystyle\sum_{cyc}x^5 + 3 \displaystyle\sum_{sym}x^4y + 4\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^2y^2z \ge 23\displaystyle\sum_{cyc}x^3yz.\]
By AM-GM inequality, $\frac{x^5+x^3y^2}{2} \ge x^4y$ and etc. so
\[\displaystyle\sum_{cyc}x^5 + \frac{1}{2}\displaystyle\sum_{sym}x^3y^2 \ge \displaystyle\sum_{sym}x^4y,\]
and hence
\[\displaystyle\sum_{cyc}x^5 + 3 \displaystyle\sum_{sym}x^4y + 4\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^2y^2z \ge 4 \displaystyle\sum_{sym}x^4y + \frac{7}{2}\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^2y^2z.\]
By AM-GM inequality, $x^4z+y^4z +2x^2y^2z \ge 2x^3yz + 2xy^3z$ and etc., so
\[4\displaystyle\sum_{sym}x^4y + 8\displaystyle\sum_{cyc}x^2y^2z \ge 8\displaystyle\sum_{cyc}x^3yz,\]
and hence
\[4 \displaystyle\sum_{sym}x^4y + \frac{7}{2}\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^2y^2z \ge \frac{7}{2}\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^3yz.\]
Finally by Muirhead's theorem, since $(3,2,0) \succ (3,1,1)$, so
\[\frac{7}{2}\displaystyle\sum_{sym}x^3y^2 \ge 15 \displaystyle\sum_{cyc}x^3yz,\]
and therefore
\[\displaystyle\sum_{cyc}x^5 + 3 \displaystyle\sum_{sym}x^4y + 4\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^2y^2z \ge \frac{7}{2}\displaystyle\sum_{sym}x^3y^2+8\displaystyle\sum_{cyc}x^3yz \ge 23 \displaystyle\sum_{cyc}x^3yz.\]

By Cauchy's inequality, we have
\[
\frac{1}{x^2+y^2+z^2}+\sum_{\text{cyc}}\left(\frac{1}{3xy}+\frac{1}{3xy}+\frac{1}{3xy}\right)\geq\frac{10^2}{x^2+y^2+z^2+9(xy+yz+zx)}.
\]
But
\[
x^2+y^2+z^2+9(xy+yz+zx)\leq(x+y+z)^2+\frac{7}{3}(x+y+z)^2\leq\frac{30}{4},
\]
we get the proof.

I think the following stronger inequality holds:
If x,y,z>0 and $x+y+z=3$, then

$\frac{12}{x^2+y^2+z^2}+\frac{1}{yz}+\frac{1}{xy}+\frac{1}{xz}\geq 7$.  ;)

[quote="sjk"]I think the same trick as above, using Cauchy's inequality, still work with Vasc's inequality.  :)[/quote]
I am always happy to see short nice solutions!  :)

[quote="sjk"]I think the same trick as above, using Cauchy's inequality, still work with Vasc's inequality.  :)[/quote]

can you show me?

So sorry, I'm totally wrong. :(  The old trick does not give a solution to Vasc's inequality. But, using linearization, I obtain the following one:
\[
\frac{12}{x^2+y^2+z^2}+\frac{8}{3}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\geq12.
\]
I'm wondered how to lower the factor 8/3 to 1.  ;)

what vasc proposed seems just terribly hard for me.
anyone managed to prove it?

[quote="cuong"]indeed , this is very hard , but vac said that we can use his n-1EV varriables to prove this kind of ineq;

i don't understand much about that method so i don't want to
 solve it :D[/quote]

I think inequalities lost its beauty if you always use methods like (n-1)ev, lagrange multiplier..

maybe this ineq can only be solved in the (n-1)EV way. i don't understand it too well either.

[b]n-1 Equal Variable Principle[/b] ( http://www.mathlinks.ro/Forum/topic-7942-140.html)
Let $a_1,...,a_n$ ($n \ge 3$) be given positive numbers,let $p\in(-\infty,1)\cup(1,\infty)$ and let $x_1,...,x_n$ be positive numbers such that
$x_1+...+x_n=a_1+...+a_n$ and $x_1^p+...+x_n^p=a_1^p+...+a_n^p$
($p=0$ means $x_1x_2...x_n=a_1a_2...a_n$).
Let $f(x)$ be a differentiable function on $(0,\infty)$ so that $g(x)=f'(x^{\frac{1}{p-1}})$ is strictly convex, and let $F_n(x_1,...,x_n)=f(x_1)+...+f(x_n)$.
(a) If $p \le 0$, then $F$ is minimal for $x_1\le x_2=x_3=...=x_n$, and is maximal for $x_1=x_2=...=x_{n-1}\le x_n$;
(b) If $p>0$, then $F$ is maximal for $x_1=x_2=...=x_{n-1}\le x_n$, and is minimal for $0=x_1=...=x_{k-1}<x_k\le x_{k+1}=...=x_n$, where $1\le k\le n$.

From (n-1)EV Principle it follows that:
(1) for $x+y+z=3$ and $x^2+y^2+z^2=constant$, the product $xyz$ is maximal for $x=y \le z$
($p=2$, $f(x)=\ln{x}$, and the function $g(x)=f'(x)=1/x$ is clearly convex);
(2) for $x+y+z=3$ and $xyz=constant$, the expression $x^2+y^2+z^2$ is maximal for $x=y \le z$
($p=0$, $f(x)=x^2$, and the function $g(x)=f'(1/x)=2/x$ is clearly convex).

[quote="Vasc"]
If x,y,z>0 and $ x \plus{} y \plus{} z \equal{} 3$, then

$ \frac {12}{x^{2} \plus{} y^{2} \plus{} z^{2}} \plus{} \frac {1}{yz} \plus{} \frac {1}{xy} \plus{} \frac {1}{xz}\geq 7$. [/quote]
More way:
Let $ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^2$ and $ xyz \equal{} w^3.$
Hence, $ \frac {12}{x^{2} \plus{} y^{2} \plus{} z^{2}} \plus{} \frac {1}{yz} \plus{} \frac {1}{xy} \plus{} \frac {1}{xz}\geq 7\Leftrightarrow$
$ f(w^3)\geq0,$ where $ f(w^3) \equal{} 9u^5 \minus{} 6u^3v^2 \minus{} (17u^2 \minus{} 14v^2)w^3.$
But $ f$ is linear function of $ w^3.$
Thus, the minimum of $ f$ ( if it exists ) holds on boundary of $ w^3,$
id est, when $ y \equal{} z$ or $ w^3\rightarrow 0^ \plus{} ,$ which easily to check.


induction
inequalities
inequalities unsolved

I'm really stumped on this.  We've done proofs by induction on things with equals signs, but then we get this:

Prove by mathematical induction that:

1 + 1/2^2 + 1/3^2 + ... + 1/n^2 less than or equal to 2 - 1/n

for all positive integers n.

I'm really confused... how are you suppose to do induction on inequalities?  What's different about it than doing induction on normal equal sign questions?

Thanks.  :)

This is an example of the principle of [i]strengthening the induction hypothesis.[/i]  What you are actually going to prove is that

$ \sum_{k\equal{}1}^{n\minus{}1}\frac{1}{k(k\plus{}1)}\equal{} 1\minus{}\frac{1}{n}$

by induction, which is easy.  Note then that

$ \sum_{k\equal{}1}^{n\minus{}1}\frac{1}{(k\plus{}1)^{2}}<\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{k(k\plus{}1)}$

The conclusion follows.  :)

[quote="t0rajir0u"]This is an example of the principle of [i]strengthening the induction hypothesis.[/i]  What you are actually going to prove is that

$ \sum_{k \equal{} 1}^{n\minus{}1}\frac{1}{k(k\plus{}1)}\equal{} 1\minus{}\frac{1}{n}$

by induction, which is easy.  Note then that

$ \sum_{k \equal{} 1}^{n\minus{}1}\frac{1}{(k\plus{}1)^{2}}<\sum_{k \equal{} 1}^{n\minus{}1}\frac{1}{k(k\plus{}1)}$

The conclusion follows.  :)[/quote]

Nice!!, notice that we can proove $ \sum_{k \equal{} 1}^{n\minus{}1}\frac{1}{k(k\plus{}1)}\equal{} 1\minus{}\frac{1}{n}$ without induction:

$ \sum_{k \equal{} 1}^{n\minus{}1}\frac{1}{k(k\plus{}1)}\equal{}\sum_{k \equal{} 1}^{n\minus{}1}\frac{1\minus{}k\plus{}k}{k(k\plus{}1)}\equal{}\sum_{k \equal{} 1}^{n\minus{}1}\frac{1}{k}\minus{}\frac{1}{k\plus{}1}$

[quote="future_phd"]
I'm really confused... how are you suppose to do induction on inequalities?  What's different about it than doing induction on normal equal sign questions?

Thanks.  :)[/quote]

Its exactly the same :)

[quote="FOURRIER"][quote="future_phd"]
I'm really confused... how are you suppose to do induction on inequalities?  What's different about it than doing induction on normal equal sign questions?

Thanks.  :)[/quote]

Its exactly the same :)[/quote]

One warning:  at no point can you subtract inequalities.  This is not valid.  What I mean is that

$ a > b$
$ c > d$

does [b]not[/b] imply that

$ a\minus{}c > b\minus{}d$

[quote="t0rajir0u"]
One warning:  at no point can you subtract inequalities.  This is not valid.  [/quote]

Wich inequalities? please explain

[b]Any[/b] inequalities.  I merely mean in general.  Subtraction is a valid operation when trying to prove an induction on an equality, but the same is not true of inequalities.  That's all.  :)

I think that the exact solution this question requires is this:
Let us assume that $ 1\plus{}\frac{1}{2^{2}}\plus{}\frac{2}{3^{2}}\plus{}...\plus{}\frac{1}{n^{2}}\leq 2\minus{}\frac{1}{n}$
Also $ \frac{1}{(n\plus{}1)^{2}}<\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$
Adding the two inequalities we get $ 1\plus{}\frac{1}{2^{2}}\plus{}\frac{2}{3^{2}}\plus{}...\plus{}\frac{1}{n^{2}}\plus{}\frac{1}{(n\plus{}1)^{2}}\leq 2\minus{}\frac{1}{n\plus{}1}$
and that completes the induction. :wink:


function
calculus
integration
real analysis
real analysis unsolved

Let $ (X, \mathcal{B}, \mu)$ be a measure space.  Prove that if $ (A_n)_{n \geq 1}$ is a sequence in $ \mathcal{B}$ with $ \sum_{n\equal{}1}^{\infty} \mu(A_n)< \infty$ then there exists a set $ E$ with $ \mu(E)\equal{}0$ such that if $ x \not \in E$ then $ x$ belongs to at most finitely many of the $ A_n$'s.

Here, I really don't see how to answer this question.  I am really struggling with this one.  Any hints on where to start would be appreciated.

Let $ \mathbf{1}_{A_n}$ be the characteristic function of $ A_n$, $ 1$ on that set and zero elsewhere. The function $ f \equal{} \sum_n \mathbf{1}_{A_n}$ is nonnegative and measurable. What is its integral, and what does this say about the measure of the set on which $ f$ is infinite? [Edit- and then I see your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316656]other thread[/url], posted just before this.]


ratio articles geometry rectangle algebra polynomial

There is SO much that could be posted about it.  I suggest reading [url=http://www.amazon.com/exec/obidos/tg/detail/-/0767908163/qid=1116095004/sr=8-1/ref=pd_csp_1/002-3546049-8072022?v=glance&s=books&n=507846][i]The Golden Ratio[/i][/url] by Mario Livio if you get a chance.

Aside from occuring just about everywhere in nature...

I'm pretty sure our eyes see in the golden ratio...

And I think TV lengths are made to follow the ratio as well....

But yea, read that book...

That's on my list of books to read this summer

I read an article in the college journal of mathematics (from MAA) about all this "Golden Ratio" hype. I'll look it up when I get back home but basically the guy says that "phi" is just a special case of a whole class of numbers with the interesting properties of the golden ratio and that its appearance in nature, art, music, architecture, etc is highly exaggerated.

here is from Mathworld:
http://mathworld.wolfram.com/GoldenRatio.html
read this one too, it's about golden rectangle:
http://mathworld.wolfram.com/GoldenRectangle.html

Thanks guys. I may try reading the book, since summer vacation is on the horizon and I'll have a lot of time on my hands.

[quote="4everwise"]Could people post some information about the golden ratio? I am interested in the topic but know little about it.[/quote]

Basically the Golden Ratio is one of those transcendental numbers like, $\pi$ or $e$ that keep popping-up in all kinds of unexpected places.

Try this, draw a line segment on a piece of paper.

Label one of the end points P; the other, Q.

Find the point R between P and Q such that

$\frac{PR}{RQ} =\frac{PQ}{PR}$

Now measure $PR$ and $RQ$.

Did you find that their ratio, $\frac{PR}{RQ} =1.61803...?$

Measure the height of your belly button.

Now take your whole height and divide by the height of your belly button.

What number did you get? Was it 1.61803 ...?

As you might have guessed this is the Golden Ratio, viewed by the ancients as the divine proportion, kind of a signature of God sort of thing.

*Don't be too disappointed if you measure your belly button and your height doesn't turn out to be exactly 1.61803 times this number. The golden ratio is related to its own reciprocal, so I just measured mine to make sure I didn't have it upside down. I turned out to be a smidge taller than the golden ratio. Nevertheless, this was DaVinci's ideal of beauty. His famous drawing of man with the palms turned up, and arms and legs outstretched is scaled to the Golden Ratio.

[quote="Mariag"][quote="Dr. No"]Basically the Golden Ratio is one of those transcendental numbers like, $\pi$ or $e$ that keep popping-up in all kinds of unexpected places.[/quote]
the golden ratio is not transcendental... IT is the solution of a polynomial equation. it does come up often though :)[/quote]

Pardon me, my bad  :blush:

[quote="white_horse_king88"]I personally heard it was ratio of height to wingspan though I may be wrong.[/quote]

That too, though I haven't seen too many winged humans lately  ;)

[quote="NoSoupForYou"]$\phi=\frac{1+\sqrt{5}}{2}$. Thus $\phi$ is not transcendental.[/quote]

That doesn't, in itself, show that $\phi$ not transcendental.  Saying that $\phi$ is a root of $x^2-x-1=0$ would be much better.


limit
logarithms
calculus
calculus computations

Consider the limit as h approaches 0 of [ln(e+h)-1]/h is

a.  f'(e), where f(x) = lnx
b.  f'(e), where f(x) = lnx/x
c.  f'(1), where f(x) = lnx
d.  f'(1), where f(x) = ln(x+e)
e.  f'(0), where f(x) = lnx

Can someone please help me and show all work?  Thanks so much!

$\lim_{h \to \infty} \frac{ f(x+h) - f(x) }{h} = f'(x)$
Set $f(x) = \ln x$.  Then we want $f'(e)$.

can you explain it more in depth at how you arrived your answer to be choice a?

never mind, i think i get it..thanks for the help


Given are vectors:
$ a=(0,2\lambda,\lambda),b=(2,2,1),c=(-1,-2,-1)$
find vector $ \vert{d}$ if : $ a \times b = c \times d$ and $ a \times c = b \times d$.
Any idea  how to find $ d$ ? maybe  it is too easy but i couldn't solve it  :blush:

hello, computing
$ (2,2\lambda,\lambda)\times(2,2,1)=(0,2\lambda-2,4-4\lambda)$ (1)
$ (2,2\lambda,\lambda)\times(-1,-2,-1)=(0,2-\lambda,2\lambda-4)$ (2)
let $ d=(x,y,z)$ then (1) must be
$ (-2z+y,-x+z,-y+2x)$
and (2) must be
$ (2z-y,x-2z,2y-2x)$
Sonnhard.


This is from 2004-2005 handbook warmup 18
One standard quarter and one special quarter with Head on both sides. If select coin random and without look at it first, flip three times. If flip a Head three straight times. what is the probability he selected the special quarter?
(I got answer 8/27, but the actuall answer is 8/9, can anyone help out?)

There is a 1/2 chance you will select the special quarter and get 3 heads.  There is a 1/16 chance you will select the standard quarter and get 3 heads.

Out of to total 9/16 chance you will get 3 heads, 1/2 is from the special quarter.

So the answer is (1/2)/(9/16)=8/9.

Or, you can make a tree chart thing with branches with H and T.  You can count there are 9 total paths with 3 heads and 8 of them come from the special quarter (I don't know how to put images in latex).


(i) Find a positive number A which is a four digit number where the cube of the sum of its digits is equal to A itself.

(ii) Prove that for any seven digit number ( a total of 7! combinations) made up of $2,3,4,5,6,7,8$ (non-repeat) , it cannot be a complete square.

[quote="violetcraze"][quote="t0rajir0u"][hide="ii"]Such a number is $\equiv 8 \bmod 9$, but squares are $\equiv 0, 1, 4, 7 \bmod 9$.  [/hide][/quote]

well, what i meant is, for any seven-digit-number made out of $2,3,4,5,6,7,8$. you can arrange them in any way as long as the digits. don't repeat.E.g. 2678345, 6475823........   :)[/quote]

the remainder mod 9 is just the sum of the digits, so it does not matter

[hide="i"]
$s(x)$ = sum of the digits of $x$. $s(x)^3 = x \equiv s(x) \pmod{9}$ so $s(x) \equiv 1,8 \pmod{9}$ by checking. Cubing a couple of values we're like yay $4913 = (4+9+1+3)^3 = 17^3$.[/hide]


function
geometry
3D geometry
sphere
vector
topology
combinatorial geometry

Here is a very easy and nice problem: let $U$ be an open set in $R^n$ and $f$ be a convex function on $U$. Does it follow that $f$ is continuous?  ;)

It should follow because in $\Re$ the statement is true.  :roll:

Just a brief sketch:

Fix $x_0\in U$. Choose some points $a_1,\ldots,a_k$ on the unit sphere centered at $x_0$ such that every point inside the unit ball centered at $x_0$ which is sufficiently close to $x_0$ can be written as a convex combination $\lambda_0x_0+\sum_{i=1}^k\lambda_ia_i\ (*)$. If $x$ is sufficiently close to $x_0$, in the expression $(*)$ we can take $\lambda_0$ close to $1$. For such $x$ we then have, by the convexity of $f,\ f(x)\le\lambda_0f(x_0)+\sum_{i=1}^k\lambda_if(a_i)$, which will be very close to $f(x_0)$. In other words, this proves that $\limsup_{x\to x_0}f(x)\le f(x_0)\ (1)$.

Assume now we're in the same setting as in the previous paragraph (we choose those $a_i$'s and so on). Also fix a ray starting in $x_0$, of the form $\mathcal R_u=\{p_t=x_0+tu\ |\ t\in[0,1]\}$ for some vector $u$. If $x$ is close enough to $x_0$, then we will be able to find $t_x\in(0,1)$ such that $p_{t_x}$ lies in the convex hull of $x$ and the $a_i$'s. Moreover, as $x\to x_0$, this $t_x$ can be taken arbitrarily close to $0$, so $f(p_{t_x})\to f(x_0)$, because $f$ is continuous on $\mathcal R_u$. Also, when we express $p_{t_x}$ as a convex combination of $x$ and the $a_i$'s, the coefficient of $x$ can be taken very close to $1$ (this "very close" depending only on the distance from $x$ to $x_0$). This means that we get something of the form $f(p_{t_x})\le\lambda_0f(x)+\sum_{i=1}^k\lambda_if(a_i)$, with $f(p_{t_x})$ close to $f(x_0)$ and $\lambda_0$ close to $1$ for all $x$ with $\|x-x_0\|$ sufficiently small. In other words, $\liminf_{x\to x_0}f(x)\ge f(x_0)\ (2)$.

$(1)$ and $(2)$ now give the desired conclusion.

there is an extremely nice and short proof of this result:

A. W. Roberts and D. E. Varberg, Another proof that convex functions are locally Lipschitz, Amer. Math. Monthly 81 (1974), 1014-1016.