function
geometry
rectangle
parallelogram
vector
real analysis
logarithms

Given an interval, there are infinitely many rationals and infinitely many irrationals, yet irrationals are often described as being more "numerous," so to speak, or rather more "dense" in an interval. Similiarly, we can apply density to primes. While I'm certain my terminology for this is not the one typically used, out of total lack of better knowledge I'm going to call these densities "linear" as they are simply distinguishing within $ \mathbb{R}$ or $ \mathbb{N}$.

So my question is, what else can we apply this density to? Once again, it may appear that I have no idea what I'm talking about (and I probably do not) so here is the example that got me started:

Among the set quadrilaterals, is there a describable comparison of "density" between cyclic quadrilaterals and non-cyclic quadrilaterals, just as there is among the set of reals a comparison between the densities of rationals and irrationals?

For simplicities sake, I'm going to define a quadrilateral by its angles (so that density will not be affected by size as size does not affect cyclicity (?) in relation to angles, and angles are the easiest way to define a quadrilateral. I also know that one can have two different quadrilaterals with the same angle by "stretching it," so to speak. I think angles is still the best way to describe quadrilaterals as this "stretching operation seems to apply the same way to ALL quadrilaterals, regardless of cyclicity, though I have no idea how to rigorously prove this.). This makes out to be a 4 variable function (I think: we can always describe the last angle as simply 360 minus the sum of the other angles, so I think this quadrilateral "angle" function can possibly be actually a 3 variable cyclic function. However, defining the last angle this way seems to make a few complications, i.e. counting the same quadrilateral multiple times. Again, it seems to affect all quadrilaterals the same way, and again, I have onyl a vague idea on how to rigorously prove this.)

So in this 4 variable (possibly 3) function, I am looking for the density of a specific kind, i.e. those that produce cyclic quads. Does this "density" apply to more than one variable (again, I don't know the terminology)?

Actually this was a terrible example of what I was asking. The conceived convex quadrilateral function would be a 3 variable cyclic function f(a,b,c) that maps all a,b,c such that $ a\plus{}b\plus{}c\le 360 \forall a,b,c\le 180$. I think this can be described as a [b]space[/b] with a few finite flaws (finite in the sense if I am overcounting, I think I am only doing it a finite number of times, i.e. if worst comes to worst in my function one quadrilateral gets counted a finite amount of times [though it should be 1]), whilst the cyclic function is in the form f(a,b,180-a) and simply depends on two independent variables making it's mapping a [b]plane[/b]. So cyclic functions do not appear dense. Once again, I may have no idea of what I'm talking about.

I think the general question can still be answered, however.

A heuristic explanation is that within a set, a subset is a [url=http://en.wikipedia.org/wiki/Dense_set]dense subset[/url] if any element of the main set can be "well approximated" by an element of the subset. This is true of the rational numbers within the set of real numbers because for any real number X, there exists a rational number within any arbitrarily small neighborhood around X *.

This is clearly not the case with the subset of cyclic quadrilaterals within the set of quadrilaterals -- for example, with the exception of the rectangle, no parallelogram can be "well-approximated" by a cyclic quadrilateral.

You can, however, have the concept of density with vectors; for example, the set of ordered pairs of rational numbers is dense within the set of ordered pairs of real numbers. **

[i]* - For all $ X \in \mathbb{R}$ and $ \varepsilon > 0$ (no matter how small), there exists a rational number $ r \; \; (r \in \mathbb{Q} )$ such that $ |X \minus{} r| < \varepsilon$[/i]

[i]** - For all $ \mathbf{x} \in \mathbb{R}^2$ and $ \varepsilon > 0$, there exists a $ \mathbf{r} \in \mathbb{Q}^2$ such that $ \| \mathbf{x} \minus{} \mathbf{r} \| < \varepsilon$[/i]

When you mention the density of primes, I imagine you are talking about the [url=http://en.wikipedia.org/wiki/Prime_number#The_distribution_of_the_prime_numbers]distribution of the prime numbers[/url]. This is quite a different concept.

the topological notion of denseness is not relevant here.

there are a couple of ways to describe the rationals being less numerous than the irrationals.

one is [url=http://en.wikipedia.org/wiki/Measure_theory]measure[/url]; the rationals in $ [0,1]$ have (lebesgue) measure 0, the irrationals in $ [0,1]$ have (lebesgue) measure 1.

another is [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url]; the rationals in $ [0,1]$ are countable whereas the irrationals are uncountable.

there are other 'topological' notions of largeness; these have to do with [url=http://en.wikipedia.org/wiki/Baire_space]baire spaces[/url] or the [url=http://en.wikipedia.org/wiki/Baire_category_theorem]baire category theorem[/url].

these different notions of size are not pairwise compatible though;

1) every countable set has lebesgue measure 0, but not every uncountable set has positive lebesgue measure
2) there are open dense sets (topologically large) with arbitrary small lebesgue measure and closed nowhere dense sets in $ [0,1]$ (topologically small) with lebesgue measure arbitrarily close to $ 1$.

As for the primes, we know via the prime number theorem what the [url=http://en.wikipedia.org/wiki/Natural_density]natural density[/url] is of the primes in the naturals ($ 0$, of course), and of course it gives us quantitatively a lot more than that (like that $ p_n\sim n\log n$)

[quote="blahblahblah"]the topological notion of denseness is not relevant here.[/quote]

 :roll:

[quote="Anaxerzia"]So my question is, what else can we apply this [color=blue]density [/color]to? [/quote]

[quote="Anaxerzia"]Does this "[color=blue]density[/color]" apply to more than one variable (again, I don't know the terminology)?[/quote]

Anaxerzia is using the phrase 'density' in a heuristic, nonmathematical way which does not correspond to how it is used in mathematics.  This is obvious from reading his/her post, but nevertheless the concept being referred to exists in mathematics and can probably be quantified by introducing a measure on a (hopefully compact) set of quadrilaterals and looking at what measure the set of cyclic quadrilaterals have within this (almost certainly $ 0$)

I'll say it again -- the topological notion of denseness is not relevant here, and the fact that he/she used a word in a nontechnical sense that just happens to be a word with a specific technical meaning in mathematics is just a coincidence.

So you've probably figured out by now that there are a lot of ways to define "density" (whatever that might mean) depending on context.  The motivation behind each definition is that it has desirable properties; for example, if we do not want to use cardinality to talk about subsets of the natural numbers and we want to usefully distinguish between certain sets, we might instead prefer [url=http://en.wikipedia.org/wiki/Schnirelmann_density]Schnirelmann density[/url], and if we're talking about primes we might want [url=http://en.wikipedia.org/wiki/Dirichlet_density]Dirichlet density[/url].

Your reasoning about cyclic quadrilaterals is fairly intuitive.  Among the quadrilaterals, the cyclic quadrilaterals are found by imposing the additional constraint that pairs of opposite angles are supplementary, and this constraint forces the cyclic quadrilaterals to have lower "dimension" among the quadrilaterals.  In other words, a "random" quadrilateral is cyclic with probability $ 0$.  (Analogously, a "random" real is rational with probability $ 0$.)  Of course, you need to pick an appropriate notion of "random," but your reasoning about degrees can be made fully rigorous as follows:

Consider the hypercube $ (a, b, c, d), 0 \le a, b, c, d < 360$ (in other words, $ a, b, c, d$ are reasonable angles).  The locus of points within the hypercube such that $ a \plus{} b \plus{} c \plus{} d \equal{} 360$ is a hyperplane of dimension $ 3$ (one less than the hypercube).  On the other hand, the locus of points within the hypercube such that $ a \plus{} b \equal{} 180, c \plus{} d \equal{} 180$ (or permutations) is an intersection of hyperplanes of dimension $ 3$, hence in fact a plane (of dimension $ 2$).  So among the set of quadrilaterals (which we are describing as a $ 3$-dimensional [url=http://en.wikipedia.org/wiki/Polytope]polytope[/url]), which has some ($ 3$-dimensional) volume, the set of cyclic quadrilaterals has volume (that is, [url=http://en.wikipedia.org/wiki/Lebesgue_measure]measure[/url]) zero.

Yes I do not think I meant topology :). Thanks for your response anyway xD

And yes the I had figured this particular argument was almost certainly 0, line out of a plane has to be... But cyclic quads have restrains that reduce the variables: other functions or properties of things may not. What about, say, the distribution of, say, trapezoids? I don't need an answer to this particular example, but I was wondering if there was a specific field of study for distributions in higher dimensions (if thats what you call it...). Seems very interesting...  :D  :D

If you can associate real numbers like angles or side lengths with an object in a collection, you can associate a Lebesgue measure with the collection in some $ \mathbb{R}^n$.  So I suppose the proper field to call this is [url=http://en.wikipedia.org/wiki/Measure_theory]measure theory[/url].

So these "distributions" (it seems to be the correct term) would be related to this field? Thanks everyone! :)

As a side note, this is my second independent question that led to measure theory, so I think I should probably pursue it now...


vector
function
trigonometry
parameterization
calculus
calculus computations

find a vector valued function to represent the graph of the intersection between the semiellipsoid x^2/12 + y^2/24 + z^2/4 = 1 z>0 and the parabolic cylinder y=x^2

i tried solving for x^2 in the first function and setting
y = 12 - y^2/2 - 3z^2
y^2 + 2y + 6z^2 - 24 = 0
(y+1)^2 - 1 - 12 + 6z^2 = 0
(y+1)^2 + 6z^2 = 13
(y+1)^2/13 + 6z^2/13 = 1

then again as Ci + (sqrt(13)sinT-1)j + (sqrt(13/6)cosT)k

would this be the right way to go about this problem? and how would i find C?

That's not $C$: it's a function which you need to calculate.

As you wrote, we have $y(t)=x(t)^{2}=12-\frac{y(t)^{2}}{2}-3z(t)^{2}$
$(y(t)+1)^{2}+6z(t)^{2}=25$. (you made a slight mistake here)
You chose to parametrize $y$ and $z$ with $\sin$ and $\cos$:
$y(t)=5\sin t-1$ and $z(t)=\sqrt{\frac{25}{6}}\cos t$

Now, what's $x$? We have $x(t)^{2}=y(t)=5\sin(t)-1$, so $x(t)=\pm\sqrt{5\sin(t)-1}$. Both signs are meaningful, and only some values of $t$ will work. This is ugly.

If we want a smooth parametrization on an interval, we need a different method. We have $2x^{2}+6z^{2}-24=-y^{2}=-x^{4}$, so $(x^{2}+1)^{2}+6z^{2}=25$. Since $x^{2}\ge0$, we must have $z^{2}\le 4$. I'd set $z(t)=2\cos t$ (interval to be specified later); this gives $x^{2}+1=\sqrt{25-24\cos^{2}t}=\sqrt{1+24\sin^{2}t}$ and $x(t)=\pm\sqrt{\sqrt{1+24\sin^{2}t}-1}$, so $y(t)=\sqrt{1+24\sin^{2}t}-1$.
To make this into a single-valued function which traces the curve once, work on the interval $(-\pi,\pi]$ and declare $x(t)=\begin{cases}\sqrt{\sqrt{1+24\sin^{2}t}-1}&t>0\\-\sqrt{\sqrt{1+24\sin^{2}t}-1}&t\le 0\end{cases}$

This latter answer is actually smooth despite appearances; it has a series expansion near zero. I don't think there's any way to get a single formula that does this.


integration
algebra
polynomial
vector
calculus
derivative
linear algebra

Find the smallest $\lambda \in \mathbb{R}$ s.t. :
$\forall P\in{\mathbb{R}}_{2}[X]$ $\int_{-1}^{1}\left( P^{\prime }\left( t\right) \right)^{2}dt\leq \lambda \int_{-1}^{1}\left( P\left( t\right) \right)^{2}dt$
I'd be glad if someone can help me , thanks  :lol:

By $\mathbb{R}_{2}[X],$ you mean real polynomials of degree at most 2, right?

For convenience, write the norm on the vector space this way:
\[\|P\|^{2}=\frac12\int_{-1}^{1}(P(x))^{2}\,dx \]
Now use the following orthonormal basis:
\[\{1,\sqrt{3}x,\frac{\sqrt{5}}2(3x^{2}-1)\}\]
In this basis, the derivative as an operator has the matrix
\[M=\begin{bmatrix}0&\sqrt{3}&0\\ 0&0&\sqrt{15}\\ 0&0&0\end{bmatrix}\]
Then, $M^{*}M=\begin{bmatrix}0&0&0\\0&3&0\\ 0&0&15\end{bmatrix}$, which has a spectral radius of $15.$

It looks like your desired $\lambda=15,$ and is achieved when $P(x)=c(3x^{2}-1).$

I'm sorry I don't understand   :blush:
what is the relation between the matrix $M^{*}M$ and the norm you defined above ?

Once you've found an orthonormal basis for a finite dimensional inner product space, you might as well be working in $\mathbb{R}^{n}$ (or $\mathbb{C}^{n}$) with the Euclidean inner product and norm. Any linear operator can then be given as an $n\times n$ matrix $M.$ It is well known in linear algebra that, using the Euclidean norm, the operator norm of a matrix $M$ is the square root of the spectral radius of $M^{*}M.$

One could probably capture the essence of my proof without quoting quite so much linear algebra, but orthogonality must remain a key point. The idea would be to show that $3x^{2}-1$ is orthogonal to every lower degree polynomial, and then use that to show that it is maximal for the norm of this operator.

The same ideas can be used for $\mathbb{R}_{d}(X)$ for $d\ge 3,$ but the computations become more complex. Assuming that we continue to use the (normalized) Legendre polynomials as our orthonormal basis, then the matrix $M$ starts acquiring nonzero entries in places other than the immediate superdiagonal, which would lead to $M^{*}M$ not being diagonal, which would lead to us needing to compute the dominant eigenvalue rather than just reading it off.

We would clearly get that $\lambda\to\infty$ as $d\to\infty.$


linear algebra
matrix
IMC
college contests

6. If $ p,q$ are rationals, $r=p+\sqrt{7}q$, then prove there exists a matrix
$\left(\begin{array}{cc}a&b\\c&d\end{array}\right) \in M_{2}(Z)- ( \pm I_{2})$
 for which $\frac{ar+b}{cr+d}=r$ and $det(A)=1$

This problem becomes pretty obvious if you show that $p+q{\sqrt{7}}$ has periodic continious fractions.

I edited in the initial  post $det(A)=1$ was missing.


Compute the probability that a point chosen randomly inside a circle is closer to the center than the circle.

[hide]It has to be inside the concentric circle with radius of 1/2 r

So it's (1/4 r :^2:  :pi: )/(r :^2:  :pi: )

= 1/4???[/hide]


function
real analysis
real analysis unsolved

Let f be real valued function defined for all x in [0,1].
Suppose there is a positive number M such that
for every choice of a finite number of points x_1,...,x_n in [0,1]

 |f(x_1) + ... + f(x_n)| <= M.

Let S be the set of points in [0,1] such that f(x) is not equal to 0.
Prove that S is countable.

Hint: prove that the set $ \{x\colon f(x)>\epsilon\}$ is finite for any $ \epsilon>0$.

What is the direction of the proof.  Directly or by contrapositive or contradiction?


If we let $ n$ be some integer, then this problem boils down to solving $ n^{2}\equal{}144\minus{}\sqrt[3]{x}$.  In other words, which values of $ x$ will make this expression a perfect square?  If I did my math correctly, the only value that fits is the trivial solution $ x\equal{}0$.

Umm, no.

[hide]
Substitute $ u \equal{} \sqrt [3]{x}$. Then it boils down to $ n^2 \equal{} 144 \minus{} u$, or in other words, how many squares are there that are less than or equal to 144? ($ u$ can be 0). Obviously, there are 13 (don't forget $ 0^2$!).[/hide]


100 weights of 1, 2, ... , 100 grams are placed on the two pans of a balance so that equilibrium is achieved. Prove that two weights can be removed from each pan so that equilibrium will be preserved.

We have to find two weights from left pan and two weights from right pan s.t. they sum to same weight.
Suppose there are $k$ weights in the left pan forming set $A$ and $100-k$ weights in the right pan forming set $B$.
Suppose $A$ contains $1,\cdots m$ and $B$ contains $m+1$.
Now we divide the problem into two subcases:
1)$100 \in A$. Let $100-l \in B$ and $101-l,\cdots 100$ in $A$.
We can easily see that if $a<b$ are two consecutive elements from $A$ (consecutive means there are no elements from $A$ between them) then $a\geq b-2$ (or else choose $a+1,b-1$ from $B$ and eliminate the four numbers)
This implies that $k \geq \frac {100+m+l}2$ and by a similar argument from $B$ we get $100-k\geq \frac{100-m-l}2$.
Summing we get $100\geq 100$ hence equalities hold everywhere.
This means $A$ has form $1,2,\cdots m,m+2,\cdots, 100-l+1,\cdots 100$ and $B$ has form $m+1,m+3,\cdots 100-l$. Now we easily see that the sum of elements in $B$ is less then sum of elements in $A$ contradiction.
2)$100 \in B$. Let $100-l \in A$ and $101-l,\cdots 100 \in B$. By a similar argument like above, we see that $A$ has form
$1,\cdots m,m+2,m+4, \cdots,100-l$ and $B$ has form $m+1,m+3 \cdots 99-l,101-l,102-l,\cdots 100$.
Note that $l,m$ have same parity. Hence the sum of elements in $A$ is $\frac {m(m-1)}2+\frac{(102-l-m)(100-l+m)}{4}$. So we get $10100=2m^2-2m+(101-l)^2-(m+1)^2=(101-l)^2+(m-2)^2-3$ so $10103=(101-l)^2-(m-2)^2=(99+m-l)(103-m-l)$. Now we see that $10103$ can't be represented as product as two numbers less than $100$ contradiction

[b]Edit[/b] [Edited according to Remike's 2 observations  below]

It's beautiful solution I suppose.

[quote="iura"]Finally, if $m \leq 2$ we easily see the sum of elements from $A$ is smaller than the sum of elements from $B$ contradiction[/quote]
I just wanna observe that it is true only for $m=1$. For $m=0$ or $m=2$ the sum of elements from $A$ is bigger than the sum of elements from $B$. But I think it's understandable and rather obvious.

Exuse me, but I see more serious mistake now.

I think you should consider the case $l,\ldots,100$ are in one set (in $A$ or in $B$). Otherwise you have been mistaken with the bound $k \geq \frac{100+m}{2}$.

So, in your solution we may replace 100 to any integer number. But the statement of the problem can't be true for 119, since there is a partition $A=\{1,\ldots,84\}$ and $B=\{85,\ldots,119\}$ so that the sum of elements from $A$ is equal to the sum of elements from $B$ and we can't remove two elements from $A$ and two elements from $B$ preserving the equilibrium (since any element from $A$ is smaller than any element from $B$).

The number 119 is not a random. If we consider a sequence $\{a_n\}$ so that $a_0=0$, $a_1=3$ and $a_{n}=6a_{n-1}-a_{n-2}+2$ then it is easy to see that for every $n$ we have $a_n^2+(a_n+1)^2=m^2$ for some $m \in \mathbb{N}$. Hence we may assume $k=(1+m)/2$ and consider $A=\{1,\ldots,k-1\}$ and $B=\{k,\ldots,a_n\}$. It is obvious that in such a case the sum of elements from $A$ is equal to the sum of elements from $B$, but the statement of the problem can't be true.

So, $a_3=119$ and my previous counterexample becomes clear.


Given [tex]a \in R^+;a>1[/tex] and [tex]f:R^+\to R^+[/tex] satisfying the following properties
i)[tex]2f(x)\le x+af(\frac{x}{a}) \forall x>0[/tex]
ii)[tex]f(x)<2005 \forall x<\frac{1}{2005}[/tex]
a) prove that if [tex]1<a<2[/tex] then [tex]f(x)\le x \forall x\in R^+[/tex]
b) Prove that if [tex]a\ge 2[/tex] then exist [tex]f |f(x)>x \forall x>0[/tex]

a)Fix  $x\in\mathbb R^{+}$.We have $f(x)-x\leq{\frac a 2}\big(f(\frac x a)-{\frac x a}\big)\leq...\leq (\frac a 2)^{n}\big(f(\frac x {a^n})-{\frac x {a^n}}\big)$
For all $n$ sufficient large we have $\frac x {a^{n}}<\frac 1 {2005}$,then $f(x)-x\leq(\frac a 2)^{n}.2006\forall n$large.But $a<2$ then $f(x)\leq x$
b)$f(x)=x+\frac 1 2$.


I think mathcounts should either sell or give away all mathcounts tests from 1983 to 1998, because you cannot buy them online. In my state, the same school wins every year, because they are the only ones that have all the tests, and they are not willing to share. Other kids have a math talent, but they can't show it, because the same school is at a bigger advantage. Buying the Educaide software is useless if you are a kid and/or a individual where you have to buy it yourself. Most kids don't have $500.

You cannot buy it even if you have the money.  Educaide sells their software only to schools - not individuals.

The tests that are available are plenty for practice.   Visit http://mathcounts.saab.org/mc.cgi to get some practice problems that are adapted from past Mathcounts tests.

I have a zip file of all national problems, about 15.5 MB. AIM at metzphan001 if you're desperate.

If you guys wnat to share the documents with others, please don't communicate that over the forums. Sure, lots of people do it, but shareware is illegal, and that would be one definition. You claim one team wins your state all the time because they have the tests, but the truth is, schools and individuals can buy those tests from Mathcounts. If I were coaching a Mathcounts team, 5 tests would be enough. You have 5 Chapter tests, 5 States tests, and 5 Nationals tests. If you were to meet once a week, and not during vacations, October to March would only be about 20 or so weeks maximum. Even if you only used the states and nationals tests, that still allows for one practice tests every two weeks.

So please do NOT post old tests over the forum. That is a violation of copyright, and we don't want the forums to get into any trouble.

[quote="mathmom"][quote="GoBraves"]Have you tried- http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm[/quote]

Thanks for that resource!  I was kicking myself for not saving the 2003 materials before they disappeared![/quote]

Your welcome

my school wins every year and it's not cause we have the questions. It's becasue the students in our school are jsut smarter all around, no offense, since we have to take a pretty hard test to get in. So for all you complaining that you don't have the previous tests, stop whining, the tests hardly help.

And... you have four tests for each year and level... a countdown, sprint, target, and team, which really makes about [b][u]60 tests[/u][/b], which in reality is a lot. It would be nice if they published the older years (they are a bit more challenging) but you can make do with what you have. And I completely agree with AznXtreme418.  [color=darkblue][size=75]<snip>[/size][/color].

Hey, hockeyadi, your website doesn't work on Macs.

It doesn't work if you use C+A+D alot on XP. Now on previous versions of Windows.... your website works wonderfully/terribbly!

Man, no wonder our school never wins!  :P


A classroom of students is seated in an $n\times n$ array of seats. The teacher asks the students to move one seat over (up, down, left, right) so that each students is moved (no one remains in his place). Show that this is possible if $n$ is even and impossible if $n$ is odd.

I think I solved it, I just dont like my solution so much. Tell me how much you will grade it on the scale 1 to 6.

[hide]
The even part is really easy. Just pair adjacent students in each row which is possible since $n$ is even and ask them to interchange positions. The odd part is one that takes some work. Let $u,d,l,r$ represent the number of up, down, left and right movement done by all the students. Since each student moves exactly once and every there are $n^{2}$ students we see that $u+d+l+r = n^{2}$. Now the crucial part is that $u=d$ and $l=r$, without lose of generality let us show that $u=d$. Say there is a student that moves up 1 row. Then he has occupied the next row, that means of the students in that new row needs to leave. They cant leave by moving left nor right so one of the students needs to move up or down. If he moves down then we are done, we shown for each student who moves up there is one that moves down. But if not (since one of the students in the new row can move up) then the new student who moved up will occupy the next above row. So the next row must have a student moving up or down. If down we are done. Otherwise he moves up. Eventually this continues until the top is reach and we must have a student moving down. This shows for each student moving up there is one moving down. Similarly for each moving left there is one moving right. That means the quantity $u+d+l+r$ is even whereas $n^{2}$ is odd, a contradiction.
[/hide]

[hide="Shorter Solution"]
$n$ is odd: Color the seats alternating black and white, like a checkerboard, with the four corners black. A "move" changes the color of a student's seat. There will be one more student in a black seat than there will be white seats; thus, after a "move" the white seats cannot accommodate all originally black-seated students.

$n$ is even: Pair up adjacent columns, starting from one edge, and swap.[/hide]


limit
Euler
continued fraction
calculus
calculus computations

[b]Lemma:[/b] Let $ a_1,a_2,\ldots$ and $ b_0,b_1,\ldots$ - two sequences. Now let $ P_n$ and $ Q_n$ alsow are sequences, such that $ P_0 \equal{} 0, \; P_1 \equal{} b_0,\; Q_0 \equal{} 1,\; Q_1 \equal{} a_1,$

$ P_{n\plus{}1} \equal{} a_{n\plus{}1}P_n \plus{} b_{n}P_{n\minus{}1},$
$ Q_{n\plus{}1} \equal{} a_{n\plus{}1}Q_n \plus{} b_{n}Q_{n\minus{}1}$

for $ n\geq 1.$ Then $ \frac{b_0}{a_1\plus{}\frac{b_1}{a_2\plus{}\ldots\plus{}\frac{b_{n\minus{}1}}{a_n}}}\equal{}\frac{P_n}{Q_n}.$
This lemma could be easily proved by induction. In our problem we have $ a_n\equal{}6$ and $ b_n\equal{}(2n\plus{}1)^2.$ So:

$ P_{n\plus{}1} \equal{} 6P_n \plus{} (2n\plus{}1)^2P_{n\minus{}1},$
$ Q_{n\plus{}1} \equal{} 6Q_n \plus{} (2n\plus{}1)^2Q_{n\minus{}1}$

with $ P_0 \equal{} 0,\; P_1 \equal{} 1$ and $ Q_0 \equal{} 1,\; Q_1 \equal{} 6.$ Left to prove that $ \lim_{n\to \infty}{\frac{P_n}{Q_n}}\equal{}\pi \minus{} 3.$ I spent a lot of time to find close froms for $ P_n$ and $ Q_n$ or represent them throw each other. Maybe someone could do this.

We know that $ \frac{4}{x\plus{}\frac{1^2}{2x\plus{}\frac{3^2}{2x\plus{}\frac{5^2}{2x\plus{}\ldots}}}}\equal{}\frac{\Gamma ^2 \left(\frac{x\plus{}1}{4}\right)}{\Gamma ^2 \left(\frac{x\plus{}3}{4}\right)}.$
If put here $ x\equal{}3$ we get $ \frac{4}{3\plus{}\mathbb{F}}\equal{}\frac{\Gamma ^2 \left(1\right)}{\Gamma ^2 \left(\frac{3}{2}\right)}$, where $ \mathbb{F}$ is your infinite fraction. Now easy to find that $ \mathbb{F}\equal{}\pi \minus{} 3.$

[quote="Yuriy Solovyov"]We know that $ \frac {4}{x \plus{} \frac {1^2}{2x \plus{} \frac {3^2}{2x \plus{} \frac {5^2}{2x \plus{} \ldots}}}} \equal{} \frac {\Gamma ^2 \left(\frac {x \plus{} 1}{4}\right)}{\Gamma ^2 \left(\frac {x \plus{} 3}{4}\right)}.$[/quote]
I don't know  :huh: .
Can you post solution?

This is a (straightforward?) application of [url=http://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula]Euler's continued fraction formula[/url].

[quote="t0rajir0u"]This is a (straightforward?) ... .[/quote]
I don't think so, since if you try to find such continued fraction for $ \pi$ you will find  this much times (without proof ), but it was obtaind only by
L. Lange,  in 1999.
hence this one is new.
here is reference
 L. Lange, An elegant new continued fraction for  $ \pi$ mer. Math. Monthly 106 (1999) 456-
458.
here you can find same result
http://arxiv.org/abs/0807.0872v2

This is not new formula. It was proved by Euler ;) Proof is not elementar. More general formula is:
Suppose that either $ n$ is an odd integer and $ x$ is any complex number or that $ n$ is any complex number and $ Re (x) > 0.$ Then
\[ \frac {\Gamma\left(\frac {1}{4}(x \plus{} n \plus{} 1)\right)\Gamma\left(\frac {1}{4}(x \minus{} n \plus{} 1)\right)}{\Gamma\left(\frac {1}{4}(x \plus{} n \plus{} 3)\right)\Gamma\left(\frac {1}{4}(x \minus{} n \plus{} 3)\right)} \equal{} \frac {4}{x \minus{} \frac {n^2 \minus{} 1^2}{2x \minus{} \frac {n^2 \minus{} 3^2}{2x \minus{} \frac {n^2 \minus{} 5^2}{\ldots}}}}
\]
Now put here $ n \equal{} 0$ and we get above formula.


Find the sum of all positive two-digit integers that are divisible by each of their digits.

:o Do you have any idea of what you just said?!?  b=0 !!!??? If there were math gods, they would most certainly unleash their wrath upon all living things for that.

But seriously...do you see WHY b cannot possibly be 0?

Hello,

I understand nearly all of [b]t0rajir0u[/b]'s systematic solution.

However, I could not follow this part of the solution:

$ n \equal{} (10\plus{}k)a$ must be divisible by $ b \equal{} ka$, so we know $ 10\equiv 0\bmod k$.

Can you guys help explain it to me?

Thanks!  :)

[quote="lightning"]Hello,

I understand nearly all of [b]t0rajir0u[/b]'s systematic solution.

However, I could not follow this part of the solution:

$ n \equal{} (10 \plus{} k)a$ must be divisible by $ b \equal{} ka$, so we know $ 10\equiv 0\bmod k$.

Can you guys help explain it to me?

Thanks!  :)[/quote]

I got lost on most of t0rajir0u's solution, but I understand this statement.
If $ (10\plus{}k)a$ is divisible by $ ka$, then that means $ 10\plus{}k$ is divisible by $ k$ and therefore $ 10\equiv 0\bmod k$. Hope that clears things up!

Oh, ok. That makes much more sense to me.

So with the equation

$ 10\plus{}k \equal{} 0$, we take the $ \bmod k$ of each side, and get

$ 10\equiv 0\bmod k$,

and from there, realize that $ k \equal{} 1,2,5$, but not $ 10$ because $ 10$ is a double digit, and would not make $ b$ a single digit, and then complete the casework.

Thanks!!! :)

[quote=t0rajir0u][hide="Solution"]
We want a number $n = 10a + b$ such that $a, b | n$.  Taking the equation $\bmod a$, we see that $0 \equiv b \bmod a$, and taking the equation $\bmod b$, we see that $0 \equiv 10a \bmod b$.  The first equation tells us that $b = ka$ for some integer $k$, so we rewrite $n = (10+k)a$.  This number must still be divisible by $b = ka$, so we know $10 \equiv 0 \bmod k$, giving us $k = 1, 2, 5$ ($k \neq 10$ because then $b$ would not be a digit).

For $k = 1$, we have $n = 11a$ for nine possibilities, giving us a sum of $11 \frac{9(10)}{2} = 495$.

For $k = 2$, we have $n = 12a$ for four possibilities (the higher ones give $b > 9$), giving us a sum of $12 \frac{4(5)}{2} = 120$.

For $k = 5$, we have $n = 15a$ for one possibility (again, higher ones give $b > 9$), giving us a sum of $15$.

If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$.  Taking the case $b = 0$ into account makes this sum $> 1000$, so it is not a valid AIME question  :P [/hide][/quote]

You can't divide a number by $0$


geometry
incenter
search
geometry unsolved

Let $I_{XYZ}$ denote the incenter of triangle $XYZ$. Given the
quadrilateral $ABCD$ with $P$ as its diagonals intersection, prove
that $ABCD$ is cyclic iff $I_{ABP}I_{BCP}I_{CDP}I_{DAP}$ is cyclic.

I am also wondering the same question, because I heard that problem a long time ago and I am not sure did I remember it right...

Nevertheless, where did you draw that picture, I mean, in wich programe?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=21758 for the correct question and solution.

  darij

[quote="SpongeBob"]Nevertheless, where did you draw that picture, I mean, in wich programe?[/quote]
This is called the Geometer's Sketchpad. Do a search on this forum and you shall get a lot of informations.


floor function
inequalities
induction
combinatorics proposed
combinatorics

Let $ X$ be a set of distinct positive real numbers. Prove that there exist no more than $ \frac {2^n}{\sqrt {n}}$ subsets of $ X$ such that the sum of their elements is $ 1$.

if $ A_1$ and $ A_2$ are subsets of $ X$ such that sum of the elements are $ 1$,

it is obvious that $ A_1 \not\subset A_2$ and $ A_1 \not\supset A_2$

so the number of subsets of $ X$, that the sum of the elements is 1, is less than

the number of subsets of $ X$, that there are no two sets which one contains the other.

because the maximum of the number of subsets of $ X$, that there are no two sets which one contains the other

is $ \dbinom{n}{\lfloor \dfrac{n}{2} \rfloor}$, it is enough to show $ \dfrac{2^n}{\sqrt{n}} \ge \dbinom{n}{\lfloor \dfrac{n}{2} \rfloor}$.

we can easily check it until $ n\equal{}3$

for $ n\ge 4$, divide into 2 cases.

[b]Case1)[/b] when $ n$ is even

we will show more powerful inequality, $ \dfrac{(n\minus{}1)2^n}{n\sqrt{n}} \ge \dbinom{n}{\dfrac{n}{2}}$

using induction.

for $ n\equal{}4$, $ L.H.S\equal{}6$ and $ R.H.S\equal{}6$ so satisfaction.

when we assume that the inequality is right for $ n\equal{}2k\minus{}2$,

it is enough to show $ 4 \ge \dfrac{(2k\minus{}1)(2k)}{k^2} \dfrac{2k\minus{}3}{(2k\minus{}2)\sqrt{2k\minus{}2}} \dfrac{2k\sqrt{2k}}{2k\minus{}1}$

it is same with $ (2k\minus{}2)\sqrt{2k\minus{}2} \ge (2k\minus{}3)\sqrt{2k}$.

when we expand it, it is right inequality.

by induction, proof for [b]Case)1[/b] is ended.

[b]Case2)[/b] when $ n$ is odd

at [b]Case1)[/b], we knew that the inequality is right for all even positive numbers.

so when we put $ n\equal{}2k\plus{}1$,

it is enough to show that $ 2 \ge \dfrac{2k\plus{}1}{k\plus{}1} \dfrac{\sqrt{2k\plus{}1}}{\sqrt{2k}}$

it is same with $ 4k^2\plus{}2k\minus{}1 \ge 0$, which is true.

from Case1) and Case2), the proof is ended

The given result fails if we take a "set" of nonnegative reals, allowing many (distinct) copies of the same number -- take the set with one $ 1$ and $ n \minus{} 1$ $ 0$s for a total of $ 2^{n \minus{} 1}$ such subsets.

What if we require the elements to be distinct, but allow them to be negative?


Here is an use full result :
${ v_{p}(n!) = \frac {n - S_{p}(n)}{p - 1}}$ where $ S_{p}(n)$ is sum of digit of n when represent in base p
[b]Proof [/b] Let $ n = \sum_{i = 1}^k a_ip^i$ where $ 0\leq a_i\leq p - 1$
$ v_{p}(n!) = \sum_{i = 1}^{ + \infty}[\frac {n}{p^i}] = \frac {n - {\sum_{i = 1}^k a_i}}{p - 1}$
This law has a name but I don't remember.
Apply this lemma for $ p = 2$ we has :
$ v_2(n!) = n - S_{2}(n)$
To $ 2^{n - 1}|n!$  then $ n - 1\leq n - S_{2}(n)$ imply that $ S_2(n)\leq 1$ so $ S_2(n) = 1$ imply that $ n = 2^k$ where $ k\in N$

[quote=Peter]Find all $n \in \mathbb{N}$ such that $ 2^{n-1}$ divides $n!$.[/quote]

By Kummer's thereom,we need $\frac{n-S_(n)}{n-1}$ which happens only for $n=2^k$


function algebra domain limit calculus topology trigonometry

I have seen in a book or here [url]http://www.toolingu.com/definition-800310-41529-discontinuous-function.html[/url] the following case of discontinuity.
If a function is not defined at a point, then it is discontinuous at that point. Is that true? In my mind, if the definition of continuity is applied on the domain of the function, then discontinuity is also applied on the domain. Why would you make affirmations about points which are not in domain?

A function $ f$ is continuous at $ a$ if and only if $ \lim_{x \to a}f(x) \equal{} f(a)$. This implies three things:

$ 1) f(a)$ exists
$ 2) \lim_{x \to a} f(x)$ also exists
$ 3) \lim_{x \to a} f(x) \equal{} f(a)$

Since your $ f$ fails to satisfy $ 1)$ at $ a$, $ f$ is discontinuous at $ a$.

Now, to answer your question, this is my take: When you say that a function is discontinuous at $ a$, you are not talking about a point on the graph (which, in this case, does not exist) but the line $ x\equal{}a$; the function breaks as it crosses that line. Does this make things clearer?

Yes, it is clear. But I think the problems that ask to determine the discontinuities where the function is not defined are bad problems. The notion of continuity is essentially more than that. It might confuse the student who just learned about the limits and continuous functions.
 For example, consider this problem:
"Determine the point(s) at which the given function is not continuous.
f(x)= 5 csc (7x) "
Why wouldn't we ask the same for differentiable functions?

The "AP calculus" definition says that a function is discontinuous at a point if it isn't defined there.  But I agree that a student who wants to take real analysis and/or topology seriously should only consider continuity with respect to the subspace topology on the domain on which the function is defined.

This has been a source of confusion to me too, because for f to be called a function, it must be defined at every point in its domain. So what is the point in discussing discontinuity at a point not in the domain at all

Nearly every function you're interested in when studying calculus can be defined over all of $ \bf R$.  It seems to me useful to have a way of distinguishing those functions that aren't so-defined.

I think it is [color=red]ridiculous[/color] for a textbook to discuss, define, consider, etc, discontinuity at a point not in the domain of the function. The only issue that may occur (and it is an important issue) is if the function [color=blue]may be prolonged by continuity[/color] to a point not in its domain.

For example, $ f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ given by $ f(x) \equal{} x\sin\frac {1} {x}$ may be prolonged to a continuous function $ \tilde{f}$ on $ \mathbb{R}$ by taking $ \tilde{f}(0) \equal{} 0$, while $ g : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ given by $ g(x) \equal{} \sin\frac {1} {x}$ may not.

pardon me  but

The term removable discontinuity is sometimes (improperly) used for cases in which the limits in both directions exist and are equal, while the function is undefined at the point x0

is it not true??

Yes, but that's not the issue.  The issue is when you claim that, for example, $ \frac{1}{x}$ is discontinuous at $ x \equal{} 0$.

[quote]
discontinuous function: A function that has a break, hole, or jump in the graph.
[/quote]
Yeap, with such "definitions" we can discuss this issue forever :roll:. Just don't take what they write in low-level calculus books too seriously. Almost all of them are written under the premise that the reader is incapable of understanding mathematics anyway.

As to the terminology, there are 2 schools here. One maintains that the statement "$ f$ is continuous at $ a$" should be either true or false for all $ f$ and $ a$ and, since you clearly cannot claim that $ f$ is continuous at $ a$ if it is not defined there, you are forced to say that "it is false that $ f$ is continuous at $ a$" in such case, which is just equivalent to saying that $ f$ is discontinuous there. The other claims that the notion of continuity is defined only for the points in the domain, so the statement "$ f$ is (dis)continuous at $ a$" just doesn't make any sense if $ f$ is not defined at $ a$. I personally prefer the second approach but many textbook authors choose the first one.

Thus, as long as your teacher doesn't give you a quiz consisting of ten questions like "Is number 2 commutative?" with the choice of an answer being

a) Yes, because $ 2$ times any other number is the same as that other number times $ 2$.
b) Yes, because you cannot find $ x,y\in 2$ such that $ xy\ne yx$.
c) No, because $ 2$ is neither a group, nor a ring.

you can just safely ignore all such purely linguistic questions and concentrate on more meaningful ones. If he insists that $ f$ is discontinuous at the points where it is not defined, he's just telling you which of the two possible schools he belongs to so that you can easily please him. The problem can arise only if you both start insisting that your point of view is the only correct one (which would merely mean that you both misunderstand what mathematical definitions are really for).

[quote]A function $ f$ is continuous at $ a$ if and only if $ \lim_{x \to a}f(x) \equal{} f(a)$.[/quote]
If $ a$ is an isolated point of the domain of definition of $ f$, then $ f$ is continuous at $ a$, although $ \lim_{x \to a}f(x)$ is not defined.

One could say that this is a matter of convention too, but the topological definition of continuity ("preimages of open sets are open") leaves no other choice.

But if you restate it as "[i]For every sequence $ x_k$ in the domain such that $ x_k\to a$, one has $ \lim_{k\to\infty}f(x_k)\equal{}f(a)$"[/i], you'll be fine again, even at the isolated points. :)

In this definition, fedja, would you assume that $ a$ is in the domain of $ f$, or allow it to possibly not be?

If $ a$ wasn't in the domain, what would $ f(a)$ even mean?

[quote="fedja"]you'll be fine again, even at the isolated points. :)[/quote]
But not in a general topological space.  :mad:

It is nice that this discussion has taken place.I had posted this question (refer to http://www.mathlinks.ro/viewtopic.php?t=259134) and I had this very doubt in my mind,but the discussion did not materialise.

Hey All.
Jenny new to MathLinks. I was searching for some maths formula solution and I find this nice website. I am maths lover and right now I am an engineering student. There is a very good query and discussion for a topic. I would like to follow to this unit. Thank you for information.


limit
trigonometry
calculus
calculus computations

Can somebody do this one without L'Hôspital?
$E=\lim_{x\to0}\Big(\frac{\tan5x}{\tan3x}\Big)$

...Massive use of trig addition/double angle formulas?

Looks rather nasty to do that way... But you could eventually get it down to a nasty expression in terms of $\tan x$ that would hopefully turn out nicely...

[quote="jmerry"]You just need the basic limit $\lim_{x\to 0}\frac{\tan x}{x}=1$. We're looking at $\lim_{x\to 0}\frac53\cdot\frac{\tan 5x}{5x}\cdot\frac{3x}{\tan 3x}$.[/quote]

...That would be quite a bit easier, wouldn't it.  :fool:


Solve the equation $y^3 = x^3 + 8x^2 - 6x +8$, for positive integers $x$ and $y$.

$(x+2)^3 < x^3 + 8x^2 - 6x +8 < (x+3)^3$ for $x > 9$

So only solution is for $x \leq 9$.

Checking manually gives only $9,11$ as solution

[quote="tµtµ"]$(x+2)^3 < x^3 + 8x^2 - 6x +8 < (x+3)^3$ for $x > 9$
[/quote]

How do you prove that?

Me and my dumb calculations,i got x>3 :blush:

I think this is an RMO question (and not inmo) :P
My solution is similar to tµtµ.

[quote=Googlu15]Another solution: $x=0,y=2$[/quote]

It says positive solutions

[quote=duohead]This problem was in Engel's Problem Solving Strategies.[/quote]
I think this question is from Germany 1995 paper.

This is an Olympiad?!

[hide=Solution]We use the well known technique of bounding. Note that $2^3=8$, so we start with $(x+2)^3$. This expands to $x^3+6x^2+12x+8$, which is less than $x^3+8x^2-6x+8$ for $x>9$. Assume $x>9$. Also, $x^3+8x^2-6x+8<(x+3)^3$. So we must have integer $x$ but the expression is in between $(x+2)^3$ and $(x+3)^3$, which is not possible. Hence our assumption of $x>9$ was false. Now we simply check for values of $x \leq 9$, giving $\boxed{(x,y)=(9,11)}$.[/hide]

There would be a (x+2)^3 -18 x+2x^2  ... For that we can prove that every next or previous element of differ by 3(i^2)( x) + 3(i) (x^2)+ i^3

(x+2)^3 being a perfect cube
Do not allows the 2x^2-18x =3(i^2)( x) + 3(i) (x^2)+ i^3

As we dont have constant term

So (x+pst k)^3 cant be a result to this y expression
       Nor can (x-ngt k)^3 be a result to this y expression

Thus y can only be equal to (x+2)^3
So remaining expression y=(x+2)^3 - (2 x)(9-x)

Says us x can be 0 or 9
(9,11) is our soln


In Mathland, the area codes consist of three digits such that the
first is a digit from 2 to 9, inclusive, the second is either 0 or 1,
and the third is a digit other than 0. How many area codes are
possible?

The first digit has $ 8$ possibilities, the second digit has $ 2$ possibilities, and the third digit has $ 9$ possibilities, so the answer is $ 8 \times 2 \times 9$ or $ \boxed{144}$.


Please add as many formulas as possible.

Sum of an infinite geometric sequence is a/(1-r) where a is first term and r is ratio between terms (like 1st and 2nd term) though you probably already knew that

Sum of an finite geometric sequence is (a(1-r^n))/(1-r). Same as above except you multiply the first term by one minus the ratio to the nth power where n is the number of terms.

here was a discussion on formula with a lot of formulas:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28871[/url]


I'm going to "promote" this to another forum in view of its difficulty level.

[quote="pakagawa"]what's the range of solutions to:

1.

x <= :sqrt:  x <=  :^2:  :sqrt: x <=  :^3:  :sqrt: x <=  :^4:  :sqrt: x...

2.

x >= x :^2:  >= x :^3:  >= x :^4: ....[/quote]

Do you mean
$1. \quad x \leq \sqrt{x} \leq \sqrt[2]{x} \leq \sqrt[3]{x} \leq \sqrt[4]{x} \leq \ldots$

$2. \quad x \geq x^2 \geq x^3 \geq x^4 \geq \ldots$?

(In that case, the $\sqrt[2]{x}$ part isn't necessary, since $\sqrt{x} = \sqrt[2]{x}$.)

Yes I can't use LATEX, even if you let me download it; I'm not allowed

O_O

Actually, all you need to do is type the thing in LaTeX between $'s.

And my answer for both of those problems are [hide]0 to 1 inclusive[/hide]

between $'s?

I don't get it.

And don't try.

-don't reply to this anymore unless important-


PEOPLE JOIN THE PERFECT COMPETITION ITS DYING AND ITS AN AWESOME COMP WITH AWESOME QUESTIONS JOIN NOW

No need to start a new thread...
Can a mod delete or lock this...

There's no need to delete this topic. I'll just move it to the Games and Fun Factory.


Is it possible two quardilaterals have equal angles, equal diagonals , equal perimeters and not to be equal at all ? Justify your answer.

i have seen a problem:

if we have the four sides of a quadrilateral and also the sum of the two angles (not the nearby angles),

we can draw it.

and according to u problem i think if the quadrilaterals have equal sides and angles.

the quadrilaterals will be equal.

Ashegh you did wrong.
Ashegh bad jooob zadi.

[quote="saeid"]
Ashegh bad jooob zadi.[/quote]

You must not mock the others . All of us make mistakes. ;)

oh i made a mistake :blush:

i will think a bout it again ;) ...


If x = 2 tany, express sin 2y as a function of x.

$\sin 2y=\frac{2\sin y\cos y}{\sin ^ 2 y+\cos ^ 2 y}$

$\sin2y=\frac{2\sin y\cos y}{\sin ^ 2 y+\cos ^ 2 y}=\frac{2\tan y}{1+\tan ^ 2 y}$


calculus
derivative
topology
real analysis
real analysis unsolved

$f \in C[a,b]$(means f is continuous on $[a,b]$),and we have $f'(x)>0$ except for $x \in \{x_{n}\}\subseteq [a,b]$.($\{x_{n}\}$ is a infinite sequence).Prove that $f(x)$  is strictly increasing on $[a,b]$

Notice
1.the same meaning $f'(x)>0$ except for a  countable set.
2.$f'(x_{n})$ may not exist.
3.Please not use measure.

Thanks for help.

[hide="Solution"]Suppose that there are two points $c,d\in [a,b]$ such that $c<d$ but $f(c)>f(d)$. For each number $t$ between $f(c)$ and $f(d)$ consider the set $U_{t}=\{x\in[c,d]\colon f(x)\ge t\}$. Since $U_{t}$ is a closed set, it contains its least upper bound $s$. Note that $f(s)=t$ and $f(x)<t$ for all $s<x\le d$. Therefore, $f$ cannot have a positive derivative at $s$. But we have uncountably many values of $t$ between $f(c)$ and $f(d)$, and different values of $t$ yield different values of $s$. This is a contradiction.

The above proves that $f$ is nondecreasing. If $f$ is not strictly increasing, then it must be constant on some open interval. But then $f'$ is identically zero on that interval, a contradiction.[/hide]


For each n is positive integer ,let P(n) is product of numbers $\leq n$ and relatively prime of n.Caculator P(n)

$n=\prod_{i=1}^{k} p_i^{a_i}$
$T_{p_i}=$ the set of multiples of $p_i$ wich are $\leq n$
$G=\bigcup_{i=1}^{k} T_{p_i},$
then $P(n)=\frac{n!}{\prod_{x \in G} x}$


If there are 2 out comes higher or lower and 80 % of the time its higher that means there is a 20% chance of it being lower. What is the chance of it being lower 2,3 and 4 times is a row?[/b]

It depends on what you mean precisely.  If we have [i]independent[/i] repetitions of an event (so the first outcome doesn't affect the probabilities associated to the second outcome), to calculate the probability of a particular sequence of outcomes you just multiply the probabilities.  In your case, the probability of the lower outcome twice in a row is $ 20\% \cdot 20\% \equal{} 4\%$, and so on.

One note (which may not be relevant at all to you): if one performs a longer series of events (say, in this example, ten runs instead of just two) then the probability of getting two in a row [i]at some point[/i] is going to require a substantially more involved calculation (and will end up quite a bit higher).

If you have 10 spaces in roulette, 8 are black and 2 are red.

If you spin once the chance of getting red is 20% where as the chance of getting black is 80%.

What will the percentage be if red is hit once of getting red again?

What will the percentage be if red is hit twice of getting red again?

What will the percentage be if red is hit 3 times, 4 times, 5 times and 6 times of getting red again?

Also what is the formular? How does it work if its 70% and 30%?

Its anoying me so much and i'm not even blonde, haha.

If anyone can help i'd appreciate it.

Spins of a roulette wheel are in principle (if not always in practice ;)) independent, so the outcome of one spin doesn't affect the outcomes of the other spins.  Thus, given that red has been hit before, the probability that you get red next is still 20%.  (On the other hand, if you really do see six reds in a row right off the bat (again, this is different from seeing, say, six reds in a row somewhere in a string of 1000 spins), you might consider the possibility that the game is being weighted ;)

(This question is quite different from the question you asked initially.)

I have used rullette as away to help me explain the question.

I'm finding it tough to explain my question.

I need to find out what is the chance of hitting the 20% minority 2,3,4,5 and 6 times in a row?

Do you know?

I think you should try to read and understand my previous two posts better.  If the events are independent, the probabilities multiply.  So if each event occurs independently with probability .2, the probability that $ n$ occur in a row is $ (.2)^n$.

Thanks for taking the time to answer the question.

If each spin is independed it would be halfed evertime, eg 20%,10%,5% 2.5% etc. to hit red in a row if i understand correctly.

many thanks.

Glad to be of service.  However, [quote="caroline-q"]if i understand correctly.[/quote]  You apparently do not.  The probabilities multiply, so the probability of getting a red twice in a row is (as I said in my first post on this topic) $ 20 \% \times 20\% \equal{} 4\%$.  In order to compute the probability of getting red three times in a row, you multiply by $ 20\%$ again, and so on.

Hi,

that make's perfect sence, thank you very much. Thats all i needed to know.

caroline, Xx.


Let $ABC$ be an acute-angled triangle, and let $D$, $E$ be the feet of the perpendiculars from $A$ and $B$ to $BC$ and $AC$ respectively. Let $P$ be the point where $AD$ meets the semicircle constructed outwards at $BC$, and let $Q$ be the point where the line $BE$ meets the semicircle constructed outwards at $AC$. Prove that $CP = CQ$.

Triangles $AQC$ and $CPB$ are right angled, thus $QC=\sqrt{CE\cdot{AC}}, CP=\sqrt{CD\cdot{CB}}$. But  $CE\cdot{AC}=CD\cdot{CB}$ due to the similarity of the triangles $CED$ and $ABC$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=19881 .

  darij


conics
parabola
function
trigonometry
arithmetic sequence
national olympiad

the 2nd Southeast China Mathenatical Olympiad was held on July 10-11, 2005, Fuzhou.

[color=red]You can view all the problems nicely written and download the pdf file at http://www.mathlinks.ro/Forum/resources-37-68-2005.html[/color]

$Day I$

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=44849
Let $a \in R$ be a parameter.
(1) Prove that the curves of $y = x^{2}+(a+2)x-2a+1$ pass through a fixed point; also, the vertices of these parabolas all lie on the curve of a certain parabola.
(2) If the function $x^{2}+(a+2)x-2a+1 = 0$ has two distinct real roots, find the value range of the larger root.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=44853
Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=44851
Let $n$ be positive integer, set $M = \{ 1, 2, ..., 2n \}$. Find the minimum positive integer $k$ such that:
For any subset $A$ (with $k$ elements) of set $M$, there exist four pairwise distinct elements in $A$ whose sum is $4n+1$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=44856
Find all positive integer solutions $(a, b, c)$ to the function $a^{2}+b^{2}+c^{2}= 2005$, where $a \leq b \leq c$.

$Day II$

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=44854
Line $l$ tangents unit circle $S$ in point $P$. Point $A$ and circle $S$ are on the same side of $l$, and the distance from $A$ to $l$ is $h$ ($h > 2$). Two tangents of circle $S$ are drawn from $A$, and intersect line $l$ at points $B$ and $C$ respectively.
Find the value of $PB \cdot PC$.

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=44852
Let $P(A)$ be the arithmetic-means of all elements of set $A = \{ a_{1}, a_{2}, ..., a_{n}\}$, namely $P(A) = \frac{1}{n}\sum^{n}_{i=1}a_{i}$. We denote $B$ "balanced subset" of $A$, if $B$ is a non-empty subset of $A$ and $P(B) = P(A)$.
Let set $M = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$. Find the number of all "balanced subset" of $M$.

7. http://www.mathlinks.ro/Forum/viewtopic.php?t=44850
(1) Find the possible number of roots for the equation $|x+1|+|x+2|+|x+3| = a$, where $x \in R$ and $a$ is parameter.
(2) Let $\{ a_{1}, a_{2}, ..., a_{n}\}$ be an arithmetic progression, $n \in N$, and satisfy the condition $\sum^{n}_{i=1}|a_{i}| = \sum^{n}_{i=1}|a_{i}+1| = \sum^{n}_{i=1}|a_{i}-2| = 507$.
Find the maximum value of $n$.

8. http://www.mathlinks.ro/Forum/viewtopic.php?t=44855
Let $0 < \alpha, \beta, \gamma < \frac{\pi}{2}$ and $\sin^{3}\alpha+\sin^{3}\beta+\sin^{3}\gamma = 1$.
Prove that $\tan^{2}\alpha+\tan^{2}\beta+\tan^{2}\gamma \geq \frac{3 \sqrt{3}}{2}$.

Following is the PDF file!

Thanks [b]zhaoli[/b] for the problems !

Anyway isn't it a little bit strange to organise Mathematical Olympiad just before IMO ?  :?

This mathematical olympiad was held intend to test the students at grade 9-10, who just enter senior high school.

This is a regional olympiad.

Zhaoli, im one of the participants in the contest from the philippines, so were we really battling with math juggernauts that just finished their senior year in high school? our team is composed of three senior students and me, a 3rd year high school student... junior year... so that's why the contest was quite difficult...

[quote="siuhochung"][quote="Micchi"]I just want to know, how many participants attended this contest. I'm only sure that there were 4 from the Philippines and 4 from India. Maybe 8 or 12 from Hong Kong... and im really not sure about the others...[/quote]
should be 8 from hong kong.
I don't know much about the hong kong team but i know that, several of them are imo hk team substitutes. Furthermore, one of them is grade 7 and it looks as if he has got a gold. One final thing is that, they have very good potentials and they should be able to enter hong kong imo team in several years.[/quote]
maybe you refer to another contest.
For southeast MO, hk sent 12 people who are of grade 9 and 10. They got 4 silvers and 8 bronzes in total.

What a pity,the problem 2 is just the IMO shortlist of 1994. :(

[quote="Soarer"]oh yes? I don't know then. I heard that chung bing ngai got a gold in a recent contest, but i don't know which.[/quote]
Hey Soarer please don't mix up the Kaohsiung Inivtational World Youth Mathematics Intercity Competition (through Po Leung Kuk) and Southeast China MO(through hkmos).
And for KIWYMIC, none of us is Hong Kong team reserve member for IMO.

[quote="shobber"][quote="georgew_bush"][quote="shobber"]9-10 grade!? I think they look like IMO problems...[/quote]
MY GOD,shobber chen,i think it is just the same level as china northern mo! :huh:
but the 7th problem is good in my opinion. :roll:[/quote]
Northern is much easier in my opinion.[/quote]
I mean all the problem is not so good.We should not let PingshengTao give any problem in future.The problems he gave are all bad enough,I think.


$AB$ and $CD$ are parallel lines which are cut by a transversal at $E$ and $F$ respactively. The trisectors of angles $FEB$ and $EFD$ intersect to give the points $P$ and $Q$ [the trisectors near $EF$ meet at $P$ the others at $Q$] $PG$ is drawn parallel to $FQ$ to meet $EQ$ at $G$, and $PH$ is drawn parallel to$EQ$ to meet $FQ$at $H$.
$GH$ extended in both directions meets $AB$ at $J$ and $CD$ at $K$.
Show that $G$ and $H$ trisect $KJ$

Denote $\epsilon = \angle FEB,\ \phi = \angle EFD,\ \epsilon+\phi = \pi$ Let O be the intersection (at infinity) of the parallel lines $AB \parallel CD$ and $\omega = \angle EOF = 0.$ The trisectors of the angle $\omega = 0$ are lines parallel to $AB \parallel CD,$ such that any transversal of $AB \parallel CD$ is trisected by them, in particular the transversal JK. Let the trisector of $\omega$ closer to AB meet EQ at U and let the trisector of $\omega$ closer to CD meet FQ at V. Then $\triangle PUV$ is the equilateral Morley triangle of the degenerate triangle $\triangle EFO.$

$\angle EQF = \pi-\frac{2 \epsilon}{3}-\frac{2 \phi}{3}= \frac{\pi}{3}$

$\angle EUP = \angle FVP = \frac{\omega+\pi}{3}= \frac{\pi}{3}$

(see the beginning of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25115]Simson line tangents to nine point circle[/url]). Therefore, $FQ \parallel PU,\ EQ \parallel PV$ which means that $U \equiv G,\ V \equiv H$ are identical.


[quote="northofnorth01"]this is fairly easy, from a friend...

1/x+1/y=2/35  (x and y are positive integers)

find the minimum value of x+y[/quote]

[hide="Solution."]
=====
First, notice that, for all positive rationals $ \frac {x}{y}$, we have $ \frac {x}{y} \plus{} \frac {y}{x}\geq 2$ by AM-GM, with equality at $ x\equal{}y$. Thus:

$ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {2}{35}$
$ \Leftrightarrow \left(\frac {1}{x} \plus{} \frac {1}{y}\right)(x \plus{} y) \equal{} \frac {2}{35}(x \plus{} y)$
$ \Leftrightarrow 1 \plus{} 1 \plus{} \frac {x}{y} \plus{} \frac {y}{x}\leq \frac {2}{35}(x \plus{} y)$
$ \Leftrightarrow 4\leq \frac {2}{35}(x \plus{} y)$
$ \Leftrightarrow x \plus{} y\geq \boxed{70}$,

with equality at $ x \equal{} y \equal{} 35$.
=====
[/hide]

sorry the question is

1/x+1/y=2/35 where x and y are two DIFFERENT positive integers......find the min value of x+y

would this be better in Pre Olympiad. If so mods feel free to move it.

[hide]Is it just x=30, y=42, x+y=72?
I did guess and check on a few values to find this.[/hide]

This is quite easy to do with a solution not just an answer.

[hide] think about fractions.[/hide]

all two different positive integers means is that $ x\plus{}y >70$ - the actual $ 70$ cannot be reached.  Having said that, we should still expect an answer in the low $ 70$'s.

$ 1/x\plus{}1/y \equal{} \frac{x\plus{}y}{xy} \equal{} 2/35$
$ x\plus{}y\equal{}2/35 xy$
, so $ 35|xy$ and $ 2|x\plus{}y$.  Thus $ x\plus{}y\equal{}71$ is unattainable; $ x\plus{}y\equal{}72$ appears possible.  So lets test it.

$ 72 \equal{} 2/35xy$
$ 36*35\equal{}xy$
$ 2^2*3^2*5*7 \equal{} xy$

so we have a total of 12 valid choices for $ x$ and $ y$ (each has a factor of 2).

So we find $ xy \equal{} 42*30$ and $ 42\plus{}30\equal{}72$ so 72 is our answer.

Your solution is correct. Yet your statement that there are 12 possible factors is ambiguous. There is only one solution so what are you trying to say? maybe i misread what you were trying to say?

Dude just use [[Simon's Favorite Factoring Trick]].

[hide="solution"]$ 2xy\minus{}35x\minus{}35y\equal{}0\Rightarrow 4xy\minus{}70x\minus{}70y\plus{}1225\equal{}1225\Rightarrow (2x\minus{}35)(2y\minus{}35)\equal{}1225$. Now $ x\plus{}y$ is minimized when they are very close together, the closest that the two factors can get without being the same is 49 and 25. Thus $ x\equal{}42$ and $ y\equal{}30$, and vice versa.[/hide]


I have modified the expression of the problem :

G={1，2，3…… 1989} is a set of integers.
 A is a sub-set of G.
 a,b are two elements of  A chosen by your wish and  a-b≠4 or 7.
Try to find out the maximum of ㄧAㄧ.

begin from#10.

What sense should group like in group theory make¿

[quote="mysteryfigure"][quote="mathisfun1"]By "group" do you mean "set" or the "group" in the context of group theory?[/quote]
the latter.[/quote]
¿¿¿
What group structure then¿
 :?

I am sorry I don't get what you mean.
But please focus on the problem itself.

You said that it is agroup like in group theory (seem very curious to me).
So what group¿ The one $\mod 1989$ makes it rather trivial...

I still somehow think that there is no group structure and your statement in post #4 is wrong...

I think we may have some difficulty in communicating（sorry for my poor english :blush: ）,I just can't make out what you are getting at "group structure" or "set".The word "group" in this case to me means a group of integers just like it means in the expression of "a group of people”or "a group of sheep"and so forth.

Yes, so $G$ is a set, and $A$ is a sub-set of $G$.
That's all, now, [b]mathisfun1[/b], you can try to solve the problem.


algebra
polynomial
calculus
calculus computations

I was gone yesterday during math class due to a math competition and I missed partial fraction decomposition. I was able to figure out how to do it when the degree of the numerator is less than the denominator but how do you do it when the degree of the numerator is more than the denominator.
example:
x^4/((x-1)^3)

When the degree of the numerator is at least as big as the degree of the denomnator, there will be a polynomial part:
$\frac{x^4}{(x-1)^3}=x+3+\frac6{x-1}+\frac4{(x-1)^2}+\frac1{(x-1)^3}$

The polynomial part can be found by long division.


let f be a continius function in R.
prove that if the equation f(x)=x does not have a solution then f(f(x))=x does not admit too a solution

Since $f(x)-x$ does not have a solution on $\mathbb R$, it follows that $f(x)-x$ has constant sign (since it's continuous).

If $f(x) > x, \, \forall x$, then, $f(f(x)) > f(x) > x$.
Otherwise, $f(x) < x, \, \forall x$, so, $f(f(x)) < f(x) < x$.


function
advanced fields
advanced fields unsolved

Let $ n_1, \dots, n_N \in \mathbb{N}\setminus\{0\}$. Show that
$ \frac{1}{(n_1 \plus{} \cdots \plus{} n_N)(n_1\cdots n_N)} \equal{} n!\zeta(n\plus{}1)$

I don't understand your question. What is $ n$? The result seems not to be true at least for odd $ n$ since if $ n_1,\ldots,n_N\in\mathbb{N}\setminus \{0\}$ then $ \frac {1}{(n_1 \plus{} \ldots \plus{} n_N)n_1\cdot \ldots \cdot n_N}\in \mathbb{Q}$ but $ n!\zeta (n \plus{} 1)\in\mathbb{R}\setminus\mathbb{Q}$.

It's supposed to ask you to show $ \sum_{a_1 \equal{} 1}^\infty \cdots \sum_{a_n \equal{} 1}^\infty \frac {1}{(a_1 \plus{} \cdots \plus{} a_n)(a_1\cdots a_n)} \equal{} n!\zeta(n \plus{} 1)$.


There are positive integers that have these properties:
I. The sum of the squares of their digits is 50, and
II. Each digit is larger than the one on its left.
What is the product of the digits of the largest integer with both properties?

i got 1236 for the largest integer with the properties (assuming each digit has to be strictly larger, otherwise 11111...)

There are just seven numbers, $ 1,2\cdots 7$ that can be a digit of the integer that satisfying two conditions(as a matter of fact, $ 7$ is useless). I tested all the possible cases, and I found $ 1236$ which shallow mentioned above is the right largest number we want. The answer is $ 36$.


Find a closed form for the expression:

$1+\sin x + \sin 2x + \sin 3x +..... \sin nx$

I was thinking along the lines of DeMoivre's or something....?

Using your suggestion,

$1 + \Im \sum_{k=1}^{n} e^{i k x} = 1 +\Im \frac{e^{i x (n+1)} - e^{ix}}{e^{ix} - 1} = 1 + \Im \frac{\cos{(n+1)x} + i \sin{(n+1)x} - \cos x - i \sin x}{\cos x + i \sin x - 1} = 1 + \frac{(\cos x - 1)(\sin{(n+1)x} - \sin x) - \sin x (\cos {(n+1)x} - \cos x)}{(\cos x - 1)^2 + \sin^2 x} = 1 + \frac{\sin (nx) + \sin x - \sin ((n+1)x)}{2-2\cos x}$

The form looks a little messy, maybe someone else could make the result nicer looking.

Note, $\sin {(n+1)x} = \sin ((n+1)x)$

Nice problem.

You can rewrite it as $1+\frac{\sin\frac{n+1}{2}x \sin\frac{nx}{2}}{\sin\frac{x}{2}}$.

Also see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50721 (Post #5)

Complex numbers have wide applicability when finding the sines or cosines of stuff like this.


If the numbers 1, 2, 3 and 4 are each used once to replace the
variables $ a$, $ b$, $ c$ and $ d$, what is the minimum positive value of the
expression $ \frac{a\plus{}b}{c\minus{}d}$? Express your answer as a common fraction.

To minimize the value, we want the numerator to be as small as possible and the denominator as large as possible.

The largest possible numerator and denominator combination is $ \frac {2 \plus{} 3}{4 \minus{} 1}$, or $ \boxed{\frac {5}{3}}$.

EDIT: Fixed. :blush:

minimum = numerator is smallest and denominator is largest

denominator = 4-1=3
numerator=2+3=5

answer : $ \frac {5}{3}$

EDIT; blah beaten to it

[quote="GameBot"]If the numbers 1, 2, 3 and 4 are each used once to replace the
variables $ a$, $ b$, $ c$ and $ d$, what is the [b]minimum[/b] positive value of the
expression $ \frac {a \plus{} b}{c \minus{} d}$? Express your answer as a common fraction.[/quote]

I'm confused. isabella2296 was looking for the maximum value and FantasyLover was looking for the minimum value and they got the same answer.

[quote="isabella2296"]To maximize the value, we want the numerator to be as [b]small[/b] as possible and the denominator as [b]large[/b] as possible.

The largest possible numerator and denominator combination is $ \frac {2 \plus{} 3}{4 \minus{} 1}$, or $ \boxed{\frac {5}{3}}$.[/quote]

Fixed izzy's mistake.  Obviously, the largest answer would be $ 7$, but that's besides the point.


Find the sum 1*2+2*2^2+3*2^3+...+21*2^21, where each term is in the form n*2^n.

call that sum S. now think about what 2S looks like. further hint: [hide]now think about what S-2S looks like[/hide]

Here's a trick worth remembering, especially if you like calculus. Consider just the 2^n terms, and replace them by x^n. When we sum x^n from n=0 to k we get (x^(k+1)-1)/(x-1). Now we differentiate. Each x^n term turns into n*x^(n-1). Now that's not quite what we had in mind. We wanted n*x^n. But that's not a problem; we just multiply by x. So we differentiate (x^(k+1)-1)/(x-1) and then multiply by x (which I'm too lazy to do). Then plug in your x and k at the end. You probably can't solve many problems this way that you can't solve by pure ingenuity, but it's definitely nice to know you can solve a problem somehow (without actually adding them up). Also, you don't have to know a ton of different methods to solve problems this way. It's just one super-method that solves tons of problems.

...somehow i knew you'd say that, simon. ;)

this does give some cool identites tho, especially when you're using generating functions i believe

and it gives the cool identity 0*C(n,0)+1*C(n,1)+2*C(n,2)+3*C(n,3)+...+n*C(n,n)=n*2^(n-1)

try to prove it using simon's method! and then prove it some other ways too, if you want to.

It's so awesome. How could I not say it?

Wait wait wait, could you explain the solution more thoroughly please?

yeah, when I first saw the problem i was thinking of the identity you were talking about mystic.


Prove that any subset of 55 numbers chosen from the set {1,2,....100} must contain 2 numbers which differ by 10, need not contain a pair differing by 11, but must contain a pair differing by 12.

I don't know what the solution is. Help!

Here's a hint.  Partition the numbers from 1 to 100 according to their remainder mod 10.  You'll get 10 groups of 10 numbers.  In any particular group, how many numbers can you choose without having two numbers being 10 apart?


[hide]the sum is 10000. there are 100 numbers and the sum of the first and the last is 200. therefore, 200*100/2 gives us the sum, which is 10000[/hide]

Correct. Do you know the formula, the sum of the first $n$ odd natural numbers is $n^2$?

no, i didn't know. you learn something new everyday. :D

Always look on the side toolbar. They have some important stuff there and its proof.

[quote="anirudh"][quote="236factorial"]What is 1+3+5+7+9+...+199?[/quote]

20000[/quote]

No.

[quote="236factorial"]What is 1+3+5+7+9+...+199?[/quote]

[hide]
There are 200/2=100 odd numbers. So the sum is 100 x100=10000
[/hide]


algebra
polynomial
Euler
complex numbers
superior algebra
superior algebra unsolved

$ \Phi_n$ denotes the $ n$-th cyclotomic ploynomial and $ \varphi$ is the Euler function.
And for $ n\geqslant 2$ denote by $ P_n$, the polynomial $ P_n(X)\equal{}\Phi_n(X)\minus{}(X\minus{}1)^{\varphi(n)}$.
If $ z$ is a root of unity and $ P_n(z)\equal{}0$, show that $ z^6\equal{}1$.

[quote="JC_math"]$ \Phi_n$ denotes the $ n$-th cyclotomic ploynomial and $ \varphi$ is the Euler function.
And for $ n\geqslant 2$ denote by $ P_n$, the polynomial $ P_n(X) \equal{} \Phi_n(X) \minus{} (X \minus{} 1)^{\varphi(n)}$.
If $ z$ is a root of unity and $ P_n(z) \equal{} 0$, show that $ z^6 \equal{} 1$.[/quote]

Hello.
The statement is obvious if $ n \equal{} 2$, thus we may assume $ n \geq 3$. Then neither $ 1$ nor $ \minus{} 1$ is a root of $ \Phi_n$. Thus we can choose a system $ S$ of $ \varphi(n)/2$ roots of  $ \Phi_n$ such that a root and its inverse never are both in this system.
Then $ ((z \minus{} 1)^{2})^{\varphi(n)/2} \equal{} \prod_{\rho}((z \minus{} \rho)(z \minus{} \rho^{ \minus{} 1}))$,
where $ \rho$ ranges over $ S$.
This can be written
$ ((z \minus{} 1)^{2})^{\varphi(n)/2} \equal{} \prod_{\rho}(z^{2} \minus{} (\rho \plus{} \rho^{ \minus{} 1})z \plus{} 1)$, thus
$ (\vert z \minus{} 1 \vert ^{2})^{\varphi(n)/2} \equal{} \prod_{\rho} \vert z^{2} \minus{} (\rho \plus{} \rho^{ \minus{} 1})z \plus{} 1 \vert$
There is thus at least one $ \rho$ such that
$ \vert z \minus{} 1 \vert ^{2} \leq \vert z^{2} \minus{} (\rho \plus{} \rho^{ \minus{} 1})z \plus{} 1 \vert$. ..... (1)
Put $ \rho \equal{} r \plus{} si$, with $ r, s$ real, $ i^{2} \equal{} \minus{} 1$. Since $ \rho$ is a root of unity, it has modulus $ 1$, thus $ \rho$ and $ \rho^{ \minus{} 1}$ are complex conjugates, thus $ \rho \plus{} \rho^{ \minus{} 1} \equal{} 2r$.
Put $ z \equal{} a \plus{} bi$, with $ a$ and $ b$ real. Our relation (1) can be written
$ (a \minus{} 1)^{2} \plus{} b^{2} \leq \vert z^{2} \minus{} 2rz \plus{} 1 \vert$.
Replacing $ z$ with $ a \plus{} bi$ in the right member, elevating to second power and using $ a^{2} \plus{} b^{2} \equal{} 1$ gives
$ (1 \minus{} r)(r \plus{} 1 \minus{} 2a) \leq 0$
But, since $ \rho$ has modulus $ 1$ and is different from $ 1$, we have $ 1 \minus{} r > 0$, thus
$ r \plus{} 1 \minus{} 2a \leq 0$. Now $ r \plus{} 1 > 0$, thus $ a > 0$.

Thus we have proved that $ z$ has real part $ > 0$. Let $ d$ be the order of the root of unity $ z$. The hypothesis $ P_n(z) \equal{} 0$ remains true if $ z$ is replaced with any primitive $ d$-th root $ z'$ of unity (use an isomorphism which applies $ z$ on $ z'$).
Thus each primitive $ d$-th root of unity has real part $ > 0$.
Thus there is no integer $ x$ prime to $ d$ such that $ d/4 \leq x \leq 3d/4$.
For $ d > 1$, this is possible only if $ d \equal{} 6$. (If $ d$ is odd, $ (d \plus{} 1)/2$ is prime with $ d$ and lies in the interval $ d/4 \leq x \leq 3d/4$. If $ d$ is divisible by $ 4$, $ d/2 \plus{} 1$ is prime to $ d$ and lies in the interval. If $ d$ is congruent to $ 2$ modulo $ 4$ and $ > 6$, then $ d/2 \minus{} 2$ is prime to $ d$ and lies in the interval.)

Thus $ z$ is a primitive root of unity of order $ 6$.

Friendly,
P.


$ \bigtriangleup{ABC}$ with incenter $ I$ is inscribed in a circle. The segment $ AI$ is extended so it passes through the circle at point $ D$. Prove that point $ D$ is the circumcenter of $ \bigtriangleup{BIC}$.

(reworded from memory)

We must prove that $ DB=DC=DI$ , now u can easily prove that $ \angle DIB=\angle DBI=90-C/2$ and get $ DI=DB$ , similiary get $ DI=DC$ .


hey guys,

do u think summer break is too long?

ours is two and a half months, we get bored

Summer break can [i]never[/i] be too long! It always ends so quickly!

yea... especially this summer b/c i went for a camp that took up most of summer but it was worth it.

Summer break doesn't seem too long if you schedule camps to take up chunks of time. But if you have nothing planned, it can get boring sometimes.

i went to two camps

hhsc which was two weeks which was fun
town summer camp which is so freakin boring its worse than doin nothin

i should book more exciting things i guess

the rest of the time i was doing nothin, i got REALLY BORED

Take up math as a hobby. You'll never be bored again. ;)

Yeah over the summer I've read 782 pages of a calculus book...I only got bored on occaision.

Also, I played a lot and got a lot better at minesweeper, I only got bored when my hand started cramping  :P

Excuse me for asking, but should this be moved to Fun & Games along with the other "summer" topic?

[quote="mtms5467"]What do you guys do during your summer? (besides math, that is :P )[/quote]

READ!!!! :)

and sleep alot.

[quote="mtms5467"]What do you guys do during your summer? (besides math, that is :P )[/quote]
nothing :rotfl:
[size=59](jk, i also read and sleep and play basketball)[/size]

i can read it!

ha ha rem

size 0 may get past me though

I'm with [b]rem[/b], but replace basketball by ultimate frisbee. ;) Some days I did more math than sleep. D:

[quote="mtms5467"]What do you guys do during your summer? (besides math, that is :P )[/quote]
let's see... TV, video games, basketball, i'm preparing for a debate competition saturday, write essays, write some more, and more math.

[quote="mtms5467"]What do you guys do during your summer? (besides math, that is :P )[/quote]
[size=0]tennis, basketball, read[/size]
Nothing.
Summer's so boring and too long and gets boring after the first 2 days.

I like  being bored and having nothing to do all the time.
So much better than going to school

the best part at school is lunch

half an hour to eat and chill with ur friends

besides math of course...

do u like it, or do u wish it was longer?

if i was u, that would a bit small

[quote="nutz_for2.718281828"]My summer break is
[hide="please put your jaws back in place when you read this."]one month.[/hide]

I'm telling the truth.

I'm cold.[/quote]

Unless you go to a regular school 11 months a year (excluding winter/spring holidays), I don't really think you can say you have 1 year breaks...

[quote="nutz_for2.718281828"]What are you talking about?

I'm lucky that dad didn't take away my break altogether![/quote]

o_O what do you do when you should be on break?

Does it involve a basement? Does it involve devices? Does it involve your dad? *snicker*

Heh, for the last 8 years my break's been 1 1/2 months. This year it's 2 months, and for the next 3 years, it's 2 1/2 months.

Oh, and I voted no. How can a break from school be too long?

I voted yes- I still have another month of summer break (~2.5 months total) and I can't wait for school to start.  And no, I am not sane.

I wanna do school!!! WAAAA!!!!!

 :(    :o  :D   DAD'S BUYING ICECREAM!!!! YAYY!!!!!!!!!!!!!!!!!!!!!  *head pops*

mom: "EWWWW!!! Honey!"
dad: "What?"
mom: "His head popped again. Please inflate it." :rotfl:

Summer break is very short!

I work two-three days a week at a realestate company, building databases. I work out 3-4 times a week, which sucks up about 2 hours that day. I see my girlfriends 4-5 times a week. I see my closest friends twice a week. I see some larger groups of friends maybe once every week and half. I TRY and read fiction, but there is no time, when I can spend every other waking moment MATHING.

Summer is short. I didn't want to go back to school untill I became a math olympian. Two months is too short for that though!!@  :D


calculus
integration
trigonometry
calculus computations

Using calculus, how to find derive the formula of the volume of a prolate spheroid, which is Volume = (4/3) * pi * a * b2

Thank you

Why you need calculus if we can immediately derive it from ball's volume $\frac{4}{3}\pi R^3$?

For your problems we are to calculate $2\int_{0}^a\pi \frac{b^2x^2}{a^2}dx$. It is a common way for all solids of rotation. :?

Okay. Let us consider the volume of solid with the surface of
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$.

The parametic equation is
$x=a\sin\phi\cos\theta,y=b\sin\phi\sin\theta,z=c\cos\phi,\quad 0\leq \theta\leq 2\pi, 0\leq \phi\leq \pi$
Hence
$\frac{\partial(x,y)}{\partial(\phi,\theta)}=ab\sin\phi\cos\phi$

By Gauss's Theorem,
$V=\iiint_\Omega dxdydz=\iint_{\partial\Omega} zdxdy = \iint_{0\leq \theta\leq 2\pi\atop 0\leq \phi\leq \pi} c\cos\phi\frac{\partial(x,y)}{\partial(\phi,\theta)}d\phi d\theta = abc \iint_{0\leq \theta\leq 2\pi\atop 0\leq \phi\leq \pi} \sin\phi\cos^2\phi d\phi d\theta = \int_0^{2\pi}d\theta\int_0^\pi \sin\phi\cos^2\phi d\phi=\frac{4}{3}\pi abc$.

Or let us use
\[x = ar\sin\phi\cos\theta,\ y=br\sin\phi\sin\theta,\ z=cr\cos\phi,\quad 0\leq\theta\leq 2\pi, 0\leq r\leq 1, 0\leq\phi\leq \pi\]
The Jacobi determinant
\[
\frac{\partial(x,y,z)}{\partial(r,\phi,\theta)} = abcr^2\sin\phi
\]
Hence the volume
$\iiint_\Omega dxdydz = \iiint_{\Omega_1} abcr^2\sin\phi dr d\phi d\theta = abc\int_0^1 r^2dr \int_0^{2\pi}d\theta + \int_0^\pi \sin\phi\d\phi = \frac{4\pi}{3}abc$


How many squares have all four vertices on the 5 by 5 rectangular grid of dots below? Two such squares are displayed.
[asy]size(50);
for(int i = 0; i < 5; ++i){
	for(int j = 0; j < 5; ++j){
		dot((i,j));
	}
}
draw((0,4)--(1,4)--(1,3)--(0,3)--cycle,linewidth(0.7));
draw((2,0)--(4,1)--(3,3)--(1,2)--cycle,linewidth(0.7));[/asy]

Identify all the possible sizes of squares and count the number of squares of each size separately.

\[ \begin{array}{cc}\text{Size}&\text{number of squares}\\ \hline\rule{0pt}{12pt}1\times 1 & 16\\ 2\times 2 & 9\\ 3\times 3 & 4\\ 4\times 4 & 1\\ \sqrt{2}\times\sqrt{2}& 9\\ \sqrt{5}\times\sqrt{5}& 8\\ \sqrt{8}\times\sqrt{8}& 1\\ \sqrt{10}\times\sqrt{10}& 2\end{array}\]

The sum of the numbers in the second column is .
Note: the possible side lengths of a square drawn on a square grid with  dots are the real numbers of the form $ \boxed{50}$ where x and y are nonnegative integers satisfying $ x\plus{}y\leq n\minus{}1$.

wait... how are you supposed to know what types of side lengths there are?

[quote="Python54"]Look in the Introduction to Counting & Probability textbook. This exact same problem is in it (Section 2.2 Casework) and it has a $2\frac{1}{2}$-page solution.[/quote]
Don't forget, alcumus is aligned with introduction books.

[quote=KingSmasher3]Let us first count the number horizontal-vertical squares. There are $ 16$ one by ones, $ 9$ two by twos, $ 4$ three by threes, and $ 1$ four by four.

There are $ 9$ one by one diagonals, and $ 1$ two by two diagonal.

There are $ 8$ squares with $ 2 \times 1$ diagonals.

There are $ 2$ squares with $ 3 \times 1$ diagonals.

Adding these together gives us a total of $ \boxed{60}$ possible squares.[/quote]

wait... which is the correct answer? this or the one below?

[quote=Aplusaccount]How do make a normal text hidden text?[/quote]

I think this is probably discussed somewhere else, but I will give you the answer here. You go to the concealed eye icon above your post. Then, you can click on that. Or, you can follow the [ hide =insert title here]insert text here[/hide] without the space in between the hide and the "=" sign.

or just do [ then type hide and then ]. so it will look like $[hide]$ don't put the dollar signs around it and no spaces. then the txt that you want to hide you type after the $[hide]$. [hide] ur welcome.

[quote=oiler8]You would do

[code][hide=Hi!]Hello[/hide][/code]
That is
[hide=Hi!]Hello[/hide][/quote]

that works too if ya wanna make it more fancy. :D

[quote=apolloheo][hide=@above]Maybe look at the one with the better quality.

Quality is # of up-votes-# of down-votes.[/quote]

Um... no...

Solution:
Identify all the possible sizes of squares and count the number of squares of each size separately. \[
\begin{array}{cc}
\text{Size} & \text{number of squares} \\ \hline
\rule{0pt}{12pt}1\times 1 & 16 \\
2 \times 2 & 9 \\
3 \times 3 & 4 \\
4 \times 4 & 1 \\
\sqrt{2}\times\sqrt{2} & 9 \\
\sqrt{5}\times\sqrt{5} & 8 \\
\sqrt{8}\times\sqrt{8} & 1 \\
\sqrt{10}\times\sqrt{10} & 2
\end{array}
\] The sum of the numbers in the second column is boxed{50}.

Identify all the possible sizes of squares and count the number of squares of each size separately. \[
\begin{array}{cc}
\text{Size} & \text{number of squares} \\ \hline
\rule{0pt}{12pt}1\times 1 & 16 \\
2 \times 2 & 9 \\
3 \times 3 & 4 \\
4 \times 4 & 1 \\
\sqrt{2}\times\sqrt{2} & 9 \\
\sqrt{5}\times\sqrt{5} & 8 \\
\sqrt{8}\times\sqrt{8} & 1 \\
\sqrt{10}\times\sqrt{10} & 2
\end{array}
\] The sum of the numbers in the second column is $\boxed{50}$.

Note: the possible side lengths of a square drawn on a square grid with $n^2$ dots are the real numbers of the form $\sqrt{x^2+y^2}$ where $x$ and $y$ are nonnegative integers satisfying $x+y\leq n-1$.

^ I think that the [b]exact[/b] same solution was already posted $3$ times in this thread.

I have seen this problem so many times that I have memorized it. I hate these types of problems because of how grindy they are. You can know  the different side lengths of the squares really easily but the amount of them is hard to tell and hurts my eyes. Also, why are all of the solutions here the official alcumus solutions?

EDIT: Having thought about these more I see that they are actually really interesting espisiously with other shapes!

The general formula for this: In an $n$ by $n$ grid, the answer is $\frac{n^4-n^2}{12}$.


[quote] Here. You can just click here. That's my website. [/quote]

you have your own website?!?!?!?!
do you run it yourself?!?!!?
How old are you?!?!?!?
 :D

[quote="chess64"]It's not hard to make a website! I've known HTML since I was 7. ;)[/quote]

Really? I've never known html before (well, only a slight amount).

Well, html is really easy, then you'll have to play with javascript, and, a bit more interesting : php :)

[quote="chess64"]It's not hard to make a website! I've known HTML since I was 7. ;)[/quote]

wow. That is so cool. :D

html is not that difficult. It's css that really stumps me.

[quote="chess64"][quote="hwenterprise"]html is not that difficult. It's css that really stumps me.[/quote]

CSS is even easier! :roll:[/quote]

Except it doesn't work on my computer  :o
Now, where did HTML come from Mathcounts resources?

well, i said that i keep updating mathcounts resources on [url=http://4aunlimited.co.nr]my website[/url], and then math92 just thought that makeing websites is a great deal, when it wasn't and also, you don't need to know html (even though i do), but i didn't use it, because i think it takes for ever to write in html, for the sites i make. Also, in some cases, html is better.  :)

[quote="solafidefarms"]css is a bit harder... learn javascript, and css will follow.

Incidentally, way, way off-topic, does anyone know about classes in Javascript? Not the kind that teach you Javascript, the kind that is used in OOP?[/quote]

Do you mean Java?

No, Javascript; I picked up a book on Javascript and saw the chapter on classes - I don't know how to construct them.

yeah. i menat javascript, html, i don't really care what you call them. but this stuff:

[code]<HTML>
<HEAD>
My website
</HEAD>
<HTML>[/code]

[quote="ToBe"]I'm learning PHP, that stuff is wicked.. its so cool what you can do w/ it.. (ANYTHING!). I haven't quite mastered it (at all) but.. its cool stuff. I don't know javascript though, i probably should. :blush:  Anyone interested in building a site need any PHP work?[/quote]

I started using asp, and i got a win xp home that didnt have iis :x.  so i got stuck with apache and php, and it's really fun.  Try flipping randomly in the manual [php.net].  it's easy once you master the structure.  Dont forget to always have a ; after a statement.  That caused most of my errors when i started. (Asp didnt have semicolons) Good Luck.

[quote="ToBe"]I'm learning PHP, that stuff is wicked.. its so cool what you can do w/ it.. (ANYTHING!). I haven't quite mastered it (at all) but.. its cool stuff. I don't know javascript though, i probably should. :blush:  Anyone interested in building a site need any PHP work?[/quote]

i do . can you help????  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)  :)

What kind of thing do you want? I can try to build something small for you (in about 2 weeks actually, when i get back to my reg. computer) , i need the practice. what do you have in mind?

like a password protection thingie(a strong one that doesn't require MySQL.

Well, to password protect something, the correct answers have to be stored somewhere, so it would require some database, MySQL would be what I'd use.l If it were only one user or so its a very small database, yet you still need it. Is there a reason you dont want MySQL used? I would also need the info on where the password protection will lead to.

i don't know how to install it properly, and it doesn't work. Another thing: I don't want passwlrd protection. I want a script that could let users upload files to my webspace. If you know how to install MySQL, can you tell me how?


this is an easy one but one of my favorites:
Let $A$ and $B$ be disjoint sets whose union is the set of naturals. SHow that for every natural $n$ there exist distinct $a,b>n$ such that $ \{a,b,a+b \} \subseteq A$ or $ \{a,b,a+b \} \subseteq B$

A rather informal proof:
Certainly, if all integers $x>n$ are in one set, this will work.

For the rest of the parts, let's try contradiction: if $A,B$ never had two consecutive elements, WLOG, we would have:
$A: n+1, n+3$
$B: n+2, n+4$
Then using the numbers that cannot be in $A$ or cannot be in $B$ and shoving them in the other set, we get:
$A: n+1, n+3, 2n+6, 3n+6, 3n+8, 3n+9$
$B: n+2, n+4, 2n+4, 3n+7, 3n+9$
Since $3n+9$ cannot be in either of the sets, yet must be in one, we reach a contradiction.

Thus, there must be three numbers (WLOG), such that $m+1, m+2\in A$ and $m+3\in B$.
$A: m+1, m+2, 3m+6, 5m+10, 5m+11, 6m+11$
$B: m+3, 2m+3, 4m+7, 4m+8, 6m+11$
Since $6m+11$ cannot be in either of the sets, yet must be in one, we reach a contradiction.
And the proof is completed?

there is a much more direct and simple proof :D

I think the proof is not quite complete...

and what's your "direct and simple" proof?

If $|A|$ is finite then it hast a largest element say $n$. Therefore we can choose $a+1, a+2$ in $B$ and their sum will also be in $B$ for all $a>n$.
So we can assume both $A$ and $B$ are infinite.
Asumme the claim is false, that is for any $a,b>n$ , $\{a,b,a+b\}$ is not in $A$ nor in $B$. We now choose $x,y,z$ in $A$ such that $x>y>z>n$ and $y-z>n$ this is posible since $A$ is infinite. Then $\{x+y, y+z, z+x\}$ are in $B$ but then $y-z$ has no place to go. contradiction!

The statement is true even if instead of $A$ and $B$ we have finitely many sets which partition $\mathbb N$. It's a classical result of Ramsey Theory that if $k$ is fixed, $n$ is large enough, and we partition $\{1,2,\ldots,n\}$ in $k$ subsets, then one of them contains three numbers $a,b,a+b$. To deduce our problem from here, all we need to do is apply it to the multiples of $n$ for some large $n$ (i.e. instead of $\mathbb N$ we work with $n\mathbb N$, so that all the involved numbers will automatically be at least $n$).

Manuel, where did you find it, exactly? I think it was used a very long time ago in a Romanian NMO (in the 80's or something like that). I remember seeing it in a book with all the Romanian NMO's, and it said there it had been proposed by Marius Cavachi (a great Romanian teacher and problem proposer).

Grobber, i found it in one of titu andreescu's book, 102 combinatorial problems. the general version that you said is really nice too
btw, that book you are talking about with all romanian olympiads... is it in english? or is there an english version?

I really doubt there's an English version, unfortunately :(. It's rather old. It doesn't contain [b]all[/b] the NMO's, as I said above; that was an exaggeration :); it has all of them from the beginning, which was sometime in the 50's, up until 1990, so I guess that's about when it was written.

P.S.

Just to make it clear: the problem I saw in the book was exactly this one, not the general version.


Dear everyone.

For an acute triangle ABC, draw a point D on the extension of C on BC such that AC=CD.
Let the circle with its diameter BC, the circumcircle of triangle ACD be O(1), O(2) respectively.
O(1) and O(2) meets each other at G, which is not C.
AB and CG meet at F. BG and CA meet at E.
Prove that F, E, D are collinear.

Let $ AG\cap BC\equal{}\{J\}$
$ D,E,F$ are collinear iff $ (BCJD)\equal{}\minus{}1 (1)$. But $ \angle BGC\equal{}90^o$ so $ (1)\Leftrightarrow GC$ is the bisector of $ \angle JGD$.
Since $ AGCD$ is cyclic we get $ \angle JGC\equal{}\angle CDA\equal{}\angle CAD\equal{}\angle CGD$. Therefore $ GC$ is the bisector of $ \angle JGD$ and we are done!


In book "bất đẳng thức suy luận và khám phá" Phạm Văn Thuận.Help me prove :
Given $ a,b,c \ge 0$Prove that:
$ \frac{a^2}{a^2\plus{}ab\plus{}b^2}\plus{} \frac{b^2}{b^2\plus{}bc\plus{}c^2}\plus{} \frac{c^2}{c^2\plus{}ca\plus{}a^2}\plus{} \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \le 2$

For this problem it's best to prove $ \displaystyle\sum_{cyc} \dfrac{a^2}{a^2 \plus{} ab \plus{} b^2} \le 1$
and $ \dfrac{ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \le 1$ separately and then add the inequalities together.
For the first inequality let $ x \equal{} \dfrac{a}{b}$, $ y \equal{} \dfrac{b}{c}$, $ z \equal{} \dfrac{c}{a}$, so then we have to prove $ \displaystyle\sum_{cyc}\dfrac{x}{x^2 \plus{} x \plus{} 1} \le 1$.
By AM-GM, $ \displaystyle\sum_{cyc}\dfrac{x}{x^2 \plus{} x \plus{} 1} \equal{} \displaystyle\sum_{cyc}\dfrac{1}{x \plus{} 1 \plus{} \frac {1}{x}}\le \displaystyle\sum_{cyc} \left(\dfrac{1}{3}\right) \equal{} 1$
The second inequality is equivalent to $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca \equal{} \dfrac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{2} \ge 0$.

[quote="Mij"]For this problem it's best to prove $ \displaystyle\sum_{cyc} \dfrac{a^2}{a^2 \plus{} ab \plus{} b^2} \le 1$
[/quote]
It's not trues, my friend. Try $ c\equal{}0$, $ a, b >0.$

[quote="babyloverain"]In book "bất đẳng thức suy luận và khám phá" Phạm Văn Thuận.Help me prove :
Given $ a,b,c \ge 0$Prove that:
$ \frac {a^2}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^2}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^2}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \le 2$[/quote]
$ 2\minus{}\frac{ab\plus{}ac\plus{}bc}{a^2\plus{}b^2\plus{}c^2}\minus{}\sum_{cyc}\frac{a^2}{a^2\plus{}ab\plus{}c^2}\equal{}\frac{(ab\plus{}ac\plus{}bc)\sum(a^5c\minus{}a^3c^2b)}{(a^2\plus{}b^2\plus{}c^2)\prod(a^2\plus{}ab\plus{}b^2)}.$


Fie $f: [a,b]\to R$ convexa ai $f(\frac{a+b}{2})=\frac{f(a)+f(b)}{2}$. Atunci f este liniara.

Iese destul de simplu, nu?

Presupunem că există $t \in \left( 0, \frac12 \right)$ astfel încât $f(ta+(1-t)b) < t f(a)+(1-t) f(b)$. Fie $m = \frac{a+b}2, \, x=ta+(1-t)b$.
Ne iese că $\frac{f(m)-f(x)}{m-x}> \frac{f(m)-f(a)}{m-a}= \alpha$.
Dar cum $f$ e convexă, trebuie să avem $\frac{f(m)-f(x)}{m-x}\leq \frac{f(b)-f(m)}{b-m}= \alpha$, contradicţie, deci $f(ta+(1-t)b) = t f(a)+(1-t) f(b), \, \forall t \in \left( 0, \frac12 \right)$.
În mod asemănător iese şi cazul $t \in \left( \frac12, 1 \right)$, deci $f$ îi liniară.


algebra
function
domain
combinatorial geometry
advanced fields
advanced fields unsolved

The plane is divided into domains by $ n$ straight lines in general position, where $ n \geq 3$. Determine the maximum and minimum possible number of angular domains among them. (We say that $ n$ lines are in general position if no two are parallel and no three are concurrent.)

Minimum is 3. It is not hard to draw an example where
there are only 3 angular regions. There must be always
at least 3 angles because the convex hull of all intersection
points is at least triangle and each vertex of this convex hull
is the vertex of some angular region.

Maximum is n if n is odd. This maximum achieves if we
take sides of the regular n-gon. It is not possible to have more
than n angles because there are 2n unbounded regions and no
2 consequtive are angles.
For even n one can construct an example with n-1 angles,
so it remains to prove that [b][i]for an even n it is impossible to
have n angles[/i][/b].
Perhaps somebody else can finish it.


The set $ \{3,6,9,10\}$ is augmented by a fifth element $ n$, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of $ n$?

$ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$

[hide="Solution"]The mean of the numbers after augmenting by n is $\frac{28+n}{5}$.

If you put $n < 3$, the median is 6.
If you put $3 < n < 6$, the median is 6.
If you put $6 < n < 9$, the median is $n$.
If you put $9 < n < 10$, the median is 9.
Finally, if you put $10 < n$, the median is 9.

Thus, we have three equations that we can solve:

$\frac{28+n}{5}= 6$
$\frac{28+n}{5}= n$
$\frac{28+n}{5}= 9$

$28+n = 30$
$28+n = 5n$
$28+n = 45$

Thus, we get $n = \boxed{2, 7, 17}$
The sum of these values is $\boxed{26}$, or choice (E)
[/hide]

[quote="King_James23"]What are you saying[/quote]

elaborate on your question please

i agree to the answer [b]26[/b]
but how did you get it

Depending on where you place N, you will have several different medians (three cases).

Because the mean is equal to the median, we have three cases of what N could possibly be.

As I said earlier, the mean will always be $\frac{28+n}{5}$, and because it can equal three different medians, there must be three different N values.
I solve all three equations, and add the resulting N values to get 26.


geometry
3D geometry
prism
calculus
pyramid
rectangle

see question 11 on the following link how do u solve it
http://web.archive.org/web/20040530211115/www.kalva.demon.co.uk/aime/aime83.html
i keep getting an answer way bigger then wut it says
i got 472 for my ans

A solution is available [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1983&p=292534]here[/url].

You could calculate the volume of a larger or smaller prism and then subtract or add the volume of "shape adjustment" pyramids.
If you know a bit of calculus, you could use calculus.
No matter how you do it, it's easy to make calculation errors. You may forget to divide by 2, or 3, or double a quantity twice.

The reference given by jvenezuela716 calculates the volume of the larger prism and then subtracts the volume of two rectangular-base pyramids. It has the right result, but may be confusing due to some typos. The volume of the two pyramids removed is 144, but the formula given would yield 288 as a result. Also, it's clear that 4everwise means "volume" when "area" is mentioned.

The most important part of solving this problem is understanding the shape and figuring out how to handle it. You can see that it is symmetrical. You can see how a triangular base prism can be made out of that solid by adding something to the end or by "chopping off" the ends. The base (or cross-section) of that prism is a triangle with a base measuring $ 6\sqrt{2}$ and a height of $ 6$. The area of that triangular cross-section is $ 18\sqrt{2}$.

LARGE PRIM: If you chose to use the larger prism (as 4everwise did), the height will be $ 12\sqrt{2}$, and the volume will be $ (12\sqrt{2})(18\sqrt{2})\equal{}432$. To "adjust" the volume to the desired solid, you have to "carve out" two pyramids with a rectangular base $ (6\sqrt{2}$ by $ 3\sqrt{2})$ and a height of $ 6$. The volume of each pyramid is $ 72$ (area of the base, times height, divided by 3). Subtracting the volume of the two pyramids from the volume of the large prism you get $ 432\minus{}144\equal{}288$.

SMALL PRISM: If you chop off the ends of the solid in the problem, you have a prism height of $ 6\sqrt{2}$. The volume of the smaller prism is $ (6\sqrt{2})(18\sqrt{2})\equal{}216$. The pyramid "chopped out" at each end has a base that was the crosss section triangle $ (area\equal{}18\sqrt{2})$, and a height of $ 3\sqrt{2}$. Its volume is $ (18\sqrt{2})(3\sqrt{2})/3\equal{}36$. If you add the volumes of the small prism and the two triangle-based pyramids, you get the volume of the solid $ 216\plus{}72\equal{}288$.

USING CALCULUS: Using calculus, you could calculate the area of a "horizontal" cross section a distance $ x$ from the "top spine" of the solid. It's the area of a rectangle of length $ (12\sqrt{2}\minus{}x\sqrt{2})$ and width $ x\sqrt{2})$. Integrating that area over the "vertical" distance from $ x\equal{}0$ to $ x\equal{}6$, you get the volume of the solid.