trigonometry inequalities geometry circumcircle perimeter calculus ratio

Find the maximum of $ \cos{A} \plus{} \cos{B} \plus{} \cos{C}$, where $ A$, $ B$, and $ C$ are the angles of triangle $ ABC$.

Find the maximum of $ \sin{A} \plus{} \sin{B} \plus{} \sin{C}$, where $ A$, $ B$, and $ C$ are the angles of triangle $ ABC$.(Is this the same as last question?)

Find the maximum of $ \cos{(A \plus{} B)} \plus{} \cos{(B \plus{} C)} \plus{} \cos{(A \plus{} C)}$, where $ A$, $ B$, and $ C$ are the angles of triangle $ ABC$.

Lagrange's method makes this quite easy, although it uses calculus.

jensen's inequality can solve all even the cosines by changing cosA to sin(90-A)

I actually don't know the answers to these problems, I just made them, and I was wondering if they had nice solutions.

The thing with max/min problems is that ordinarily, they may be somewhat difficult or require some little trick but once you know calculus, particularly Lagrange multipliers, the problems become much much easier.

[b]Outline:[/b] We make this a two variable problem, since the three angles add to $ \pi$. The two problems are to maximize:

i) $ \cos x \plus{} \cos y \minus{} \cos (x \plus{} y)$

ii) $ \sin x \plus{} \sin y \plus{} \sin (x \plus{} y)$

For each, set $ x \plus{} y$ to some number $ \phi \; : \; 0 \leq \phi \leq \pi$ find the optimal relationship between $ x$ and $ y$. You will get $ x \equal{} y \equal{} \frac {\phi}{2}$

Now, you have a one-variable thing in terms of $ \phi$, and vary $ \phi$ between $ 0$ and $ \pi$.

You will find that in both cases, the maximum occurs when $ x \equal{} y \equal{} \pi \minus{} x \minus{} y \equal{} \frac {\pi}{3}$

[quote="dgreenb801"]For the second problem, we know sinA=a/2R, where R is the circumradius, so the sum of all sines is (a+b+c)/2R. So in any circle with radius R, we want to maximize the perimeter of the inscribed triangle. This intuitively seems to be an equilateral triangle, but I can't rigorously prove it. Is this the answer?[/quote]

Where does the relationship between sine and circumradius come from?

Anyway, lets consider the situation.  If we have a circle of radius R, what is the longest perimeter triangle that can be inscribed?

Choose three points on the circle that form an equilateral triangle.  Basically, I will prove that moving one of the points decreases the perimeter and therefore the max perimeter is an equilateral triangle.

Lets always keep two of these fixed, while sliding around the other one.  The distances from each of the other points are x and y.  Since AM= max possible GM, we have max(sqrt(xy)) = (x+y)/2 or 2max(sqrt(2A/sin(x - y angle))) = x+y where A is the area of the triangle.  Now, since A =bh/2 where h is the height of the triangle and b is the base (which we are holding constant), we have 2max(sqrt(bh/sin(60))) = x+y.  Note that the angle has to be 60 degrees (because the measure of the opposite arc is not changing and the angle's measure is the arc's measure)

To maximize this quantity, we must maximize h.  h is maximized at x=y (the max h occurs when h bisects the arc, because moving in either direction from there would force h to shrink with the circle).  So we have an equilateral triangle and cannot move a point without decreasing the x+y and the perimeter along with x and y; therefore, the max perimeter is achieved with an equilateral triangle.

I think this has no flaws, but tell me if you agree with my reasoning or not.

[quote="facis"]Where does the relationship between sine and circumradius come from?[/quote]

This is the [url=http://mathworld.wolfram.com/LawofSines.html]extended Law of Sines[/url] and can be proven by dropping perpendiculars from the circumcenter.

[quote="facis"]darn, I wish I knew that last weekend!  I got stuck on a problem on a math contest where I had to find the radius of the circumcircle of an odd triangle, and after I got the side lengths didn't know what to do.  That relationship is really cool - something that was never pointed out in my math classes.[/quote]

Ouch. That's what I felt that if I had known pappus's, then I would have been able to own some random math competition somewhere.

[quote="1=2"]@facis: Complain! Also, what was the problem?[/quote]

The lenth of the diagonal of square ABCD is $ 2\sqrt{2}$.  Point P is on BC such that the ratio of PC to PB is 3.  Determine the radius of the circle that passes through A, P, and the center of the square. (GPML/mathleague Target Test Round 10807)

I did it all without coordinates, but that way I had no way find the circumradius of the triangle.  Got all the side lengths for the triangle, but didn't know what to do with them.  Looking back, I could have used law of cosines, converted one of the cosines to a sine, and then used this version of the law of sines.  Messy, but it could have worked.  By the way, there is only about 5-7 minutes for this problem, so going back to set up a coordinate system in the last 2 minutes wasn't really an option.

@DonkeyKingKong - This one problem wouldn't have been quite enough on it's own.  But if I hadn't labeled ABC a 60 degree angle when told $ AB\equal{}2BC$ ($ AB \perp AC$) it might have been enough.  By the way, what's pappus's?


Given that a, b, c are positive numbers.
Find the minimum value of $\frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}$.

[hide]
Use a suitable transformation, the whole expression will be simplified.
[/hide]

[hide]Make b arbitrarily small and set a equal to c and make both arbitrarily large.  The value tends to 2+0-2=0.  Unless this expression can be negative, that should be the minimum.[/hide]

[quote="mathmanman"][quote="socrates"][quote="Soarer"]posted and solved by namdung I think.[/quote]

[url=indeed]http://www.mathlinks.ro/Forum/viewtopic.php?t=15779[/url]

how can i type it better(to get only [color=blue]indeed[/color])  :D[/quote]

This way : [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15779]indeed[/url]. ;)[/quote]

what's the code??
 :)

If you look through the quoted part of your last message (in the edit window or by quoting again), you can look and see.  The easiest way is using the button labelled "[u]URL[/u]" which is in among the Quote, Code, Img, etc. buttons above the text-entry box.


Hi, I did the 03 Nationals tests, but I don't have the answers to them.  Could someone post or PM the answers for this so I could check them?   Thanks.

Same with me, I did the tests, but don't have the answers...


[hide]first time, he made 85 - 20 or 65 dollars, and the second time, he made 135 - 110 or 25 dollars. Adding, it gives him 90 dollars of profit.[/hide]

[quote="MCrawford"]An antique dealer buys a chair for $\$20 and sells it for $\$85.  The dealer later buys back the same chair for $\$110 and sells it for $\$135.  How many dollars total profit did the dealer make on this chair?[/quote]

[hide]$-20+85-110+135=65+25=90$ dollars[/hide]

[quote="MCrawford"]An antique dealer buys a chair for $ \$20 and sells it for$\$85.  The dealer later buys back the same chair for $ \$110 and sells it for$\$135.  How many dollars total profit did the dealer make on this chair?[/quote]
This is plain easy: (85-20)+(135-110)=65+25=90 dollars.

this is a simple question that is super easy to solve. you do (80-25) + (135-110) =

the answer is

[hide] 90 dollars.[/hide]

Or y'know..
We could just like not post the same solution over 10 times?


Vote for the most beautiful girl in the IMO 2006 Slovenia  :D

Let's try and avoid obviously sexist threads, hm?

Go [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14274]here[/url] and have fun. You can discuss it in different forums, e.g. the round table.


calculus
geometry
3D geometry
sphere
integration
rotation
function

I know this may seem easy, but I'm only a fourth grader by age and I'm trying to learn calculus, so here's the question:

[hide]To prove the volume of a sphere is $\frac{4}{3} \pi r^3$, $\displaystyle\int^{x=r}_{x=0} 4 \pi x^{2}\,dx=4\pi \left(\frac{x^3}{3}\right)^r_0 =\frac{4}{3} \pi r^3$.
I get the last part, but not the first one.[/hide]

Hmm, well I'm not completely sure if this is the right interpretation, but anyway. $4 \pi x^2$ is the surface area of the spherical "shell" at any radius $x$. If you integrate that from $x = 0$ to $x = r$, the result will be the sum of the areas of all the spherical shells, or the volume.

Of course this assumes you have already derived the formula for the surface area of a sphere.

I can prove it using a similar method, although it is slightly different.

The equation of the top-half of a circle centered around the origin is $\displaystyle\sqrt{r^2-x^2}$.

When rotating the region bound by this curve and the x-axis around the x-axis, you get a sphere.  Note that the curve intersects the x-axis at $x=r$.

Using the disk method, the volume of this solid of revolution will be $\displaystyle\int^{r}_{-r}\pi(\sqrt{r^2-x^2})^2 dx$.  Since this solid is symmetrical with respect to the y-axis, this can be rewritten as $2\displaystyle\int^{r}_{0}\pi(\sqrt{r^2-x^2})^2 dx$.  (This may make it a bit easier).

This integral can be rewritten further as $2\displaystyle\int^{r}_{0} \pi (r^2-x^2) dx$,
or $2\displaystyle\int^{r}_{0} \pi r^2- \pi x^2 dx$.

This is equivalent to $2(\pi r^3 - \frac{1}{3} \pi r^3)$,
which is equal to $2(\frac{2}{3} \pi r^3)$
or $\frac{4}{3} \pi r^3$.

Does this make it clearer?

Imagine a solid ball of radius $r$ as being composed of a bunch of concentric, thin spherical shells, each having thickness $\Delta x$.  If one of those shells has inner radius $x_i$, then the volume of that shell will be [i]approximately[/i] $4\pi x_i^2 \Delta x$.  (This is assuming we already know that the surface area of a sphere of radius $x$ is $4\pi x^2$.)  To approximate the volume of the solid ball of radius $r$, you can add up the approximate volumes of each of those spherical shells that you are imagining the solid ball as being composed of.  And what do you get?  You get a Riemann sum.  Repeating this approximation process again and again, using more and more spherical shells each time in order to get better and better approximations, the Riemann sums that we get will approach a certain integral.  That integral is $\int_0^r 4\pi x^2 dx$.  (This is similar to what Paladin said, but slightly different.)

One comment I have, though, is how are we supposed to know that the surface area of a sphere of radius $x$ is $4 \pi x^2$, if we don't already know the formula for the volume of a solid ball?  (Again, Paladin made a related remark.)

In my mind, the way that I derive the formula for the surface area of a sphere is by first finding the volume of a ball (using Seesaw's method), and then saying this:

$\frac{4}{3}\pi r^3 = \int_0^r S(x) dx$, where $S(x)$ is the surface area of a sphere of radius $x$.  Then I differentiate both sides of that equation with respect to $r$, which gives me $4\pi r^2 = S(r)$.

So to me it seems backwards to use the formula for the surface area of a sphere in order to derive the formula for the volume of the ball.  In my mind, I have always used the formula for the volume of a ball to derive the formula for the surface area of a sphere.

Edit:  For some reason I was accidentally saying "cylindrical" when I meant "spherical"...I changed that.

Model the volume of a sphere as the area bounded by a circle centred at the origin in the positive $y$ quadrants rotated around the $x$ axis by $2\pi$ radians.

The equation of the circle is $x^2+y^2=r^2$, so $y^2=r^2-x^2$
This clearly intercepts the $x$ axis at $\pm(r)$ where $r$ is the radius of the circle.

So, from volume of revolution:
$V=2\pi\int_{0}^{r}(y^2)dx$  as the circle is symetrical.

$V=2\pi\int_{0}^{r}(r^2-x^2)dx$
$=2\pi[r^2x-\frac{x^3}{3}]_{0}^{r}$

$=2\pi[r^3-\frac{r^3}{3}]$
Hence $V=\frac{4}{3}\pi(r^3)$

[quote="mgao"]And how do you derive the formula "surface area= $4 x^2 \pi$?"  I don't get why you're supposed to derive it using this.  coffeym's answer is more reasonable, but hte textbook said the first one's how you're supposed to do it.[/quote]

Well you could rotate the same function to find the surface area of revolution... but that seems like a lot of unnecessary work just to find the volume.


vector
algebra
polynomial
topology
analytic geometry
Functional Analysis
real analysis

Let $ X$ be a linear topological vector space, $ A\colon X\to X$ --- injective linear continuous operator.
Suppose that range $ R(A)$ is closed. Does this imply that $ A$ is [i]closed[/i]?
([i]closed[/i] means that $ A$ maps closed sets to closed sets)

This counterexample is, probably, complicated far beyond necessary but it works.

Start with any normed space $ Y$ and a non-closed injective continuous linear operator $ S: Y\to Y$. Then $ R(S)$ is a subspace in $ Y$. Choose some algebraic basis $ \{v_\alpha\}_{\alpha\in A}$ of unit vectors $ v_\alpha\in Y$ over $ R(Y)$. Let $ Z$ be the set of sequences $ \{z_\alpha\}_{\alpha\in A}$ in which only finitely many $ z_\alpha\ne 0$ with $ \ell^1$ norm (this space is very far from complete, of course). Let $ X\equal{}Y\times Z$ and $ T(y,z)\equal{}(S(y)\plus{}\sum_\alpha z_\alpha v_\alpha,0)$. Then the full range of $ T$ is $ Y\times \{0\}$, which is closed, but the image of  $ Y\times \{0\}$ is $ R(S)\times \{0\}$, which is not.

Note that the result is true in a Banach space, due to the bounded inverse theorem.

Consider as $ X$ the space of all polynomials $ p(t) \equal{} a_0 \plus{} a_1 t \plus{} \ldots \plus{} a_n t^n$, with $ \|p\|: \equal{}|a_0| \plus{} \ldots \plus{}|a_n|$. Define $ A\colon X \to X$ as follows:
\[ A(p)\equal{} a_0 \plus{} \frac{1}{2}a_1 t \plus{} \ldots \plus{} \frac{1}{n}a_n t^n.\] This operator is linear and bounded. $ A(X) \equal{} X$ is closed. The inverse operator $ A^{\minus{}1}(p) \equal{} a_0 \plus{} 2 a_1 t \plus{} \ldots \plus{} n a_n t^n$ is unbounded, so $ A^{\minus{}1}$ is discontinuous. This means that there is a closed set $ G$, such that $ A(G) \equal{} (A^{\minus{}1})^{\minus{}1}(G)$ is not closed. For example one can take $ G\equal{}\{t^n\}_{n \in \mathbb N}$.

The advantage of Fedya's example is that the corresponding closed set is in fact a [b]linear subspace[/b].

Thank you, guys, for your examples.
It is interesting, though, that in both examples space $ X$ is incomplete.
Therefore it would be interesting to construct an example in which $ X$ is complete.
It seems to me this won't be easy. At least, according to "Topological Vector Spaces"
by A. P. Robertson, open mapping theorem is valid for $ \mathcal D'(\mathbb R)$
(and for homomorphic images of this space, too).

When you start talking about completeness, that means that $ X$ is equipped with a (complete) metric, so you probably want to talk about F-spaces. Now, if memory serves me right, a version of the bounded inverse theorem is true for F-spaces and readily implies that the operator in question is closed, but you should wait for an answer from someone who still remembers this material. :)

Yes, the corresponding theorems for Banach spaces (closed graph, etc.) can be easily generalized to Frechet spaces (by just taking the proofs and writing the distance instead of the norm everywhere).

Ok, but $ \mathcal D'(\mathbb R)$ is complete but it is not F-space, is it?
When I talk about completeness I mean that every Cauchy sequence has a limit.

One more question on the first counterexample: is the operator $ T$ injective?
For example take $ y\in Y$ for which $ S(y)\neq 0$. Since $ v_\alpha$ is algebraic basis in $ Y$ there exists a unique finite sequence $ z \equal{} \{z_n\}$ such that $ S(y) \equal{} \sum_n z_n v_n$. Therefore $ T(y, \minus{} z) \equal{} 0$.

"Basis of $ Y$ over $ R(S)$" is not the same as "basis of $ Y$"! Just read attentively. :)

When you are talking about Cauchy sequences in non-metric spaces, do you just mean $ x_n\minus{}x_m\to 0$ as $ \min(m,n)\to\infty$ in the topology of the space or something else?

[quote="fedja"]"Basis of $ Y$ over $ R(S)$" is not the same as "basis of $ Y$"! Just read attentively. :)

When you are talking about Cauchy sequences in non-metric spaces, do you just mean $ x_n \minus{} x_m\to 0$ as $ \min(m,n)\to\infty$ in the topology of the space or something else?[/quote]
Yes, I mean convergence in the topology of the space.

I do not understand what do you mean by "Basis of $ Y$ over $ R(S)$". Maybe you mean basis of $ Y/R(S)$?

It is almost the same, only "basis of $ Y/R(S)$" formally consists of classes of vectors and "basis of $ Y$ over $ R(S)$" consists of vectors. I'm not sure if my English is correct here but what I mean is the set $ V$ of vectors such that each vector in $ Y$ can be uniquely written as a finite linear combination of several vectors in $ V$ and a vector in $ R(S)$. Clearly, it is the same as to say that the classes $ v\plus{}R(S)$ ($ v\in V$) form a basis of the factor-space $ Y/R(S)$.

[quote]When I talk about completeness I mean that every Cauchy sequence has a limit.
[/quote]

This is called [i]sequential completeness[/i]. An easy example of non-metrizable sequentially complete space is the Hilbert space $ \ell_2$ (or any reflexive Banach space) equipped with weak topology. Then the identity map $ I: (\ell_2, strong) \to (\ell_2, weak)$ is continuous, but the inverse map is discontinuous.

Take $ X \equal{} (\ell_2, strong) \times (\ell_2, weak)$ in the standard product topology. One can map bijectively $ X$ to $ X$ as follows:

Write  $ (\ell_2, weak)$ as $ (\ell_2, weak)_1 \times  (\ell_2, weak)_2$ - the product of two copies of the same space (the first one corresponts to even coordinates, the second - to odd ones). Do the same with $ (\ell_2, strong) \equal{} (\ell_2, strong)_1 \times (\ell_2, strong)_2$. Now map $ (\ell_2, weak)$ isomorphically to $ (\ell_2, weak)_1$, $ (\ell_2, strong)_1$ to   $ (\ell_2, strong)$ and map $ (\ell_2, strong)_2$ continuously and bijectively but not isomorphically to $ (\ell_2, weak)$. This defines the bijective continuous map $ T: X \to X$ which inverse is discontinuous.

Outside of sequential completeness there is a concept of completeness in TVS. It uses filters instead of sequences. The above example is not complete in that sense, but one can construct a very similar example using the strongest locally convex topology on Hilbert space.

Thank you, [b]volodya[/b]! This is what I was looking for saying "Therefore it would be interesting to construct an example in which X is complete." I think I can draw your map this way:
\[ \begin{array}{ccccc}
(\ell_2, \, strong)_1 &\times& (\ell_2, \, strong)_2 &\times& (\ell_2, \, weak) \\
\downarrow &  & \downarrow & & \downarrow \\
(\ell_2, \, strong) &\times& (\ell_2, \, weak)_1 &\times& (\ell_2, \, weak)_2
\end{array}\]


I just noticed all the animation in the smilies. Thats so cool! :D
 :starwars:

hmm, i realized that a while back. My computer goes between periods of showing the animation and then not showing it. Yesterday was the second time in 4 or so months i finally saw the animation. Its gone today. :(
Another 4 months to wait now.

Never mind, it was just the firewall. :D

:rotfl: can u see the rotfl moving around? rotfl


I saw a listing in my school's math comps for Amercian Math Competition 8 Geometry. I've taken AMC 8 before, but is this anything different?

Perhaps, it is like geometry in a creative way, where you can use what you know from algebra concepts to solve the problem?

that is an interesting discovery. i will look into it!

The AMC 8 contest is November 13, 2007, with additional dates November 14-21 allowed.  All the important basic information about the AMC 8 is available in the brochure, available online at:
http://www.unl.edu/amc/b-registration/b1-archive/2007-2008/2007-AMC8Bro.pdf

There is no "AMC 8 Geometry", although of course, geometry problems appropriate for the middle-level grades do appear on the AMC 8.  This entry is probably a mistake in the school listing.

Steve Dunbar
MAA Director, American Mathematics Competitions

Oh... I guess my school was just trying to jazz up the name.


induction
arithmetic sequence
algebra proposed
algebra

Find all $ f: \mathbb{R}\to\mathbb{R}$ which follows two conditions
(i) $ f(x\plus{}f(y))\plus{}f(x)\plus{}y\equal{}2f(x\plus{}y)$ for all $ x,y\in\mathbb{R}$
(ii) $ \frac{f(x)}{x}$ is bounded for all $ x\in\mathbb{R}\minus{}\{0\}$

Ignore the second condition for now. Put in $ y \equal{} 0$ to get $ f(x\plus{}f(0)) \equal{} f(x)$, so $ f$ is periodic with period $ f(0)$ (which may mean nothing if $ f(0) \equal{} 0$). In particular this implies $ f(f(0)) \equal{} f(0)$. Now plug in $ y \equal{} f(0)$ to get $ f(x\plus{}f(f(0))) \plus{} f(x) \plus{} f(0) \equal{} 2f(x\plus{}f(0))$, which implies that $ f(0) \equal{} 0$. Next, put in $ x \equal{} 0$ to get $ f(f(y)) \minus{} f(y) \equal{} f(y) \minus{} y$ (1). Hence $ y, f(y), f(f(y)), \ldots$ is an arithmetic progression for all $ y$. (1) also shows that $ f$ is injective: if $ f(a) \equal{} f(b)$ then $ a \equal{} 2f(a) \minus{} f(f(a)) \equal{} 2f(b) \minus{} f(f(b)) \equal{} b$.

For real $ a$, suppose $ f(a) \equal{} a\plus{}b$ for some real $ b$ (this is obviously possible for any $ a$). Then $ f(a\plus{}b) \equal{} a\plus{}2b$ from the above. Putting $ x \equal{} a, y \equal{} b$ into the original equation gives $ f(a\plus{}f(b)) \plus{} f(a)\plus{}b \equal{} 2f(a\plus{}b)$ which simplifies to $ f(a\plus{}f(b)) \equal{} f(a\plus{}b)$. By $ f$ injective, $ f(b) \equal{} b$. Putting $ y \equal{} b$ into the original equation gives $ f(x)\plus{}b \equal{} f(x\plus{}b)$ (2) after some cancelling. Now we show by induction that $ f(ka) \equal{} k(a\plus{}b)$. The base case was our original assumption. $ x \equal{} a, y \equal{} ka$ gives $ f((k\plus{}1)a\plus{}kb) \plus{} (k\plus{}1)a\plus{}b \equal{} 2f((k\plus{}1)a)$. Using (2) on the first term $ k$ times and cancelling, we get $ f((k\plus{}1)a) \equal{} (k\plus{}1)(a\plus{}b)$, completing the induction. In particular we have shown $ f(kx) \equal{} kf(x)$ and $ f(x) \minus{} x$ is a fixed point of $ f$.

Now it's time to use the second condition. Suppose $ b$ is a fixed point of $ f$ as found above. Then $ f(qb) \equal{} qb$ for any rational $ q$ since $ f(kx) \equal{} kf(x)$ for all $ x$. Furthermore, we have that $ f(x\plus{}qb) \equal{} f(x)\plus{}qb$. Suppose we have a real $ c$ such that $ f(c) \equal{} c \plus{} d$ for some nonzero $ d$. Obviously $ c$ is not a rational multiple of $ b$. Then for any rational $ q$, $ f(c \minus{} q b) \equal{} f(c) \minus{} qb \equal{} (c\minus{}q b)\plus{}d$. So for $ x \equal{} c \minus{} q b$, $ \frac{f(x)}{x} \equal{} 1 \minus{} \frac{d}{c\minus{}q b}$. But we can make $ c \minus{} q b$ arbitrarily small and $ d$ is a fixed nonzero number, so $ \frac{f(x)}{x}$ is unbounded, contradiction. Thus $ d \equal{} 0$ and $ f(x) \equal{} x$ for all $ x$.

nice solution, thanks  :lol:
Indeed, we can get $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ by using only first condition. Someone told me in other website..


Maybe posting Outstanding solutions for Interesting problems Would be a good idea for saving the forum.

Below, Is a beautiful solution for a rather Hard Number theory Problem.

[color=blue]Problem[/color]
Prove that there exist infinitly many natural numbers like $k$ which :
\[\pi(k) | k \]
$\pi(k)$ is the number of primes $\leq k$

[color=Blue]Solution[/color]
Lets prove that for any $m \geq 2$ we can find $n$ such that $m\pi(n)=n.$

Take $1<m$. Since$\lim_{k \rightarrow \infty}\frac{\pi(mk)}{mk}=0$ , there is a maximal $k$ such that $\frac{1}{m}\leq \frac{\pi(mk)}{mk}, k=1$ verifies this inequality.
If we have equality, we are done. Otherwise,$k<\pi(mk)$. Since $k$ is maximal, we have $\frac{\pi(mk+m)}{ (mk+m)}< \frac{1}{m}$ ,hence
$k < \pi(km) \leq \pi(km+m) \leq k$.Contradiction.

How on the #%^#% Can someone find such a thing  :D

Wow that is such a great solution! I am very impressed. :)


absolute value
combinatorics unsolved
combinatorics

Given the expression
$P = \frac{16-15-14-13-12-11-10-9}{8-7-6-5-4-3-2-1}$
we can place as many brackets as we want anywhere on the numerator, but we must place at the same places on the denominator, and the denominator must not equal $0$.
[color=blue](a)[/color] Find the least possible value of $P$
[color=blue](b)[/color] Find the least integer value of $P$

Obviously we need to make the absolute value of the numerator as large as possible, the absolute value of the denominator as small as possible, and make sure that $P$ turns out to be negative. Checking modulo 2, we can find that both the denominator and the numerator are even no matter where the brackets are placed. It can then be trivially shown that \[\displaystyle P=\frac{16-(15-14-13-12-11-10-9)}{8-7-6-(5-4-3-2)-1}\] yields the answer to both questions to be $-35$.

My mistake. I did not see the full problem... sorry.

Now that I think of it, I don't think it's hard to prove that $P$ is negative only when \[\displaystyle P= \frac{16-(15-14-13)-12-11-10-9}{8-(7-6-5)-4-3-2-1},\] yielding an answer of $-7$.
Please tell me if I'm right or not! I make many mistakes...

Could you show me how to do that?
(To be honest, I created this problem out of nothing, so I don't know what the answer should be)


algebra
polynomial
calculus
integration
algebra unsolved

(1) show that if f(x) is a polynomial of degree n with real coefficients, which takes integral values on a certain set of n+1 consecutive integers, then f(x) is integer-valued.
(2)show that if f(x) is an integer-valued polynomial of degree n, then n!f(x) is a polynomial with integral coefficients.
(3)suppose that f(x) is an integer-valued polynomial of degree n and that g=g.c.d(f(0),f(1),,,,,,,,,f(n)). show that g divides f(k) for all integers k.

Following lemmas is well-known:

[b]lemma 1[/b].
If $ C_x^n \equal{} \frac {x(x \minus{} 1)\cdots(x \minus{} n \plus{} 1)}{n!}$ then $ C_x^n\in\mathbb{Z}\,\forall x\in\mathbb{Z}$.

[b]lemma 2[/b].
a)Any polynomial $ f$ of the degree $ n$ can write as $ f(x) \equal{} \sum_{i \equal{} 0}^nb_i\cdot C_x^i$.
b)$ f(x)\in\mathbb{Z}\forall x\in\mathbb{Z}$ if and only if $ b_i\in\mathbb{Z}\,\forall i \equal{} \overline{0,n}$.


The competition defined a lucky number to be a number such that the sum of the digits in the decimal representation is equal to 19 times the number. How many such numbers are there? Positive integers btw.

let $ s(n)=$the sum of the digits in the decimal representation of $ n$
if $ n$ contain $ k$ digits.
so $ 9k\ge s(n)=19n\ge 19.10^{k-1}$
but $ \forall k: \ 9k<19.10^{k-1}$
there's no such numbre

That can't be true because I can at least generate solutions for a certain amount of digits. But my problem was that it took a while to grind out each of the solutions, I want a quick way.

Lets say for 3 digits, if the decimal expansion was xyz, then the actual value is 100x + 10y + z. Set that value to 19 times the sum of the digits so we have:

$ 100x+10y+z = 19(x+y+z)$.

We can reduce that down to y+2z=9x. We can generate a finite number of solutions, knowing that x,y and z can only be digits between 0-9. So one such solution would be say, z=1, x=1, y=7.

So one lucky number would be 171, which is 19 times the sum of its digits, 9.


abstract algebra
group theory
superior algebra
superior algebra unsolved

I would like to know whether part [b](a)[/b] (and, thus, my proof of part [b](b)[/b]) is right.
I have proved both, but I am not 100% sure whether I didn't lose some index or minus sign during computation...

[b](a)[/b] Let $ K$ be an abstract simplicial complex. Define a chain complex $ \widehat{C}_{\ast}\left(K\right)$ as follows:

For every $ n\in\mathbb{N}$, let $ \widehat{C}_n\left(K\right)$ be the free abelian group generated by the (ordered!) $ n$-tuples $ \left(i_1,i_2,...,i_n\right)$ such that $ \left\{i_1,i_2,...,i_n\right\}\in K$ (note that there may be several equal elements among $ i_1$, $ i_2$, ..., $ i_n$, and that $ \left(i_1,i_2,i_3,i_4,...,i_n\right)$ and $ \left(i_2,i_1,i_3,i_4,...,i_n\right)$ are different $ n$-tuples in general).

For every $ n\in\mathbb{N}$, define the boundary operator $ \partial_n: \widehat{C}_{n}\left(K\right)\to\widehat{C}_{n \minus{} 1}\left(K\right)$ by

$ \partial_n\left(i_1,i_2,...,i_n\right) \equal{} \sum_{k \equal{} 1}^{n}\left( \minus{} 1\right)^{k \minus{} 1}\left(i_1,i_2,...,\widehat{i_k},...,i_n\right)$ for every $ n$-tuple $ \left(i_1,i_2,...,i_n\right)$ among the generators of $ \widehat{C}_n\left(K\right)$,

and continued by linearity to the whole abelian group $ \widehat{C}_n\left(K\right)$.

Let $ R_{\ast}: \widehat{C}_{\ast}\left(K\right)\to\widehat{C}_{\ast}\left(K\right)$ be a chain homomorphism such that $ R_1 \equal{} 0$, and such that for every generator $ \left(i_1,i_2,...,i_n\right)$ of $ \widehat{C}_n\left(K\right)$, the $ n$-chain $ R_{\ast}\left(i_1,i_2,...,i_n\right)$ is a linear combination of $ n$-chains of the form $ \left(i_{a\left(1\right)},i_{a\left(2\right)},...,i_{a\left(n\right)}\right)$, with $ a_1$, $ a_2$, ..., $ a_n$ being elements of $ \left\{1,2,...,n\right\}$.

Prove that $ R_{\ast}$ is chain homotopic to the zero homomorphism.

[b](b)[/b] Assume that the set of vertices of $ K$ is totally ordered. Then, prove that there exists a canonical isomorphism between the homologies of the chain complexes $ \widehat{C}_n\left(K\right)$ and $ C_n\left(K\right)$.

Here are my answers, if someone wishes to check them:

[b](a)[/b] For every $ n\in\mathbb{N}$, define a homomorphism $ f: \widehat{C}_n\left(K\right)\to \widehat{C}_{n \plus{} 1}\left(K\right)$ by

$ f\left(x_1,x_2,...,x_n\right) \equal{} \sum_{k \equal{} 1}^{n} \left( \minus{} 1\right)^{k \minus{} 1} \left(x_1,x_2,...,x_k\right) @ R\left(x_k,x_{k \plus{} 1},...,x_n\right)$,

where the partial linear map $ @ : \widehat{C}_a\left(K\right)\times \widehat{C}_b\left(K\right)\to \widehat{C}_{a \plus{} b}\left(K\right)$ (infix notation) is defined by $ \left(u_1,u_2,...,u_a\right) @ \left(v_1,v_2,...,v_b\right) \equal{} \left(u_1,u_2,...,u_a,v_1,v_2,...,v_b\right)$ (and by linear continuation). (This map $ @$ is not always defined everywhere, but $ \left(x_1,x_2,...,x_k\right) @ R\left(x_k,x_{k \plus{} 1},...,x_n\right)$ is always defined by the condition of the problem).

Compute $ \partial f\plus{}f\partial$.

[b](b)[/b] For every $ n\in\mathbb{N}$, define a homomorphism $ \Pi_n: \widehat{C}_n\left(K\right)\to C_n\left(K\right)$ by

$ \Pi_n\left(i_1,i_2,...,i_n\right) \equal{} \mathrm{sign}\pi\cdot\left\{i_{\pi\left(1\right)},i_{\pi\left(2\right)},...,i_{\pi\left(n\right)}\right\}$, if there exists a permutation $ \pi\in S_n$ such that $ i_{\pi\left(1\right)} < i_{\pi\left(2\right)} < ... < i_{\pi\left(n\right)}$ (of course, such a permutation exists if and only if the vertices $ i_1$, $ i_2$, ..., $ i_n$ are pairwise distinct);

$ \Pi_n\left(i_1,i_2,...,i_n\right) \equal{} 0$ otherwise,

and continued by linearity to the whole abelian group $ \widehat{C}_n\left(K\right)$.

Then, $ \Pi_{\ast}$ is a chain homomorphism.

For every $ n\in\mathbb{N}$, define a homomorphism $ I_n: C_n\left(K\right)\to\widehat{C}_n\left(K\right)$ by

$ I_n\left(\left\{i_1,i_2,...,i_n\right\}\right) \equal{} \left(i_1,i_2,...,i_n\right)$, where $ \left\{i_1,i_2,...,i_n\right\}\in K$ is a set in normal form (where "normal form" means that $ i_1 < i_2 < ... < i_n$),

and continued by linearity to the whole abelian group $ \widehat{C}_n\left(K\right)$.

Then, $ I_{\ast}$ is a chain homomorphism.

We have $ \Pi_{\ast}\circ I_{\ast} \equal{} \mathrm{id}$. Besides, the chain homomorphism $ I_{\ast}\circ\Pi_{\ast}$ is chain homotopic to $ \mathrm{id}$, as we can see by applying [b](a)[/b]... if it is correct.

  darij


Prove that there exists a natural number n such that

sin(n) > 2005/2006

[quote="pavel kozlov"]It's true, since {$n\frac{1}{2\pi}$} is everywhere dense function.
P.S. I think you should not post it in Geometry section.[/quote]

Yes,it's right,but i want an "elementary" solution,for the 9-th grade, without density considerations.


hello
i am new member and i hope to find the solution of my problem

the problem is

If
P(A1) = 5P(A2) , AC S.
P(A/A1)= 0.2 ,
P(A/A2) = 0.3
what is the value of P(A1 /A) ?

i am trying to solve it but i can not find the relation between the rules...
From the rules

P(A/A2)=P(A ∩A2) / P(A2)
0.3= P(A ∩A2) / P(A2)

P(A/A1)=P(A ∩A1) / P(A1)
0.2= P(A ∩A1) / P(A1)
= P(A ∩A1) / 5 P(A2)

i know it is somthing missing but i do not know it
do you know it???
Thank you.......

what do you know about the events $ A_1, A_2$? because, for example, if you know that $ A_1, A_2$ partition your event space, then $ P(A_1|A) \equal{} \frac{P(A_1\cap A)}{P(A)} \equal{} \frac{P(A_1)P(A|A_1)}{P(A|A_1)P(A_1)\plus{}P(A|A_2)P(A_2)}$

which you could easily compute given your information

i did not if A1,A2  partition event space..
i post the question without change.
if A1,A2 not partition event space ,can i apply this rule ???
or there is another way??
Thank you pleurestique so much..

no, you can't apply this rule if $ A_1, A_2$ don't partition the event space. because what if, for example, there were a third event $ A_3$ such that $ A_1, A_2, A_3$ partitioned the space? then the formula would become

$ P(A_1|A) \equal{} \frac {P(A_1)P(A|A_1)}{P(A_1)P(A|A_1) \plus{} P(A_2)P(A|A_2) \plus{} P(A_3)P(A|A_3)}$

so the probability $ P(A_1|A)$ would be smaller.


This forum needs a moderator. BTW, don't elect me, I don't want to be mod for this.

I believe that this has already been answered by levans. He said no mods for this forum or the other FTW forum (not the questions one though)

If you think something needs cleaning up badly, private message levans or another admin, and he'll take care of it.


Hey. I am wondering if anyone knows the state cutoffs for Massachusetts from 2000-2005.

Any help is appreciated.
max.

I have no clue about 2006, but in 2005 the cutoff to make the national team was 39, or f my memory is completely failing me, it was 38.

I was on the MA team in 06. Here were the top 4 scores:
1. 41
2. 39(me)
3. 38
4. 38

There may have been other 38s, I'm not sure.

I was on the MA team for 2005. I don't remember the scores, but I got a 41 and I was third. I think the cutoff was either 39 or 38.

In 2003, the cutoff was 38 (I got 37, got 7th place).

It's usually around 38-39. 40 would be almost a guaranteed spot on the state team.

The cutoff for 2005 was 39. I was the fourth place person in that year.


function
limit
real analysis
real analysis unsolved

Let $f: [a,b]\to \mathbb{R}$ a function such that $\lim_{x\tox_{0}}f(x)$ is real for any $x_{0}$ in $[a,b]$. Prove that f is integrable.

Well...I loved this problem....Maybe someone could give me a source for similar problems :)

Nobody in the mood for such a problem? I am really curious to see your solutions...hope they will be different from the one I have :)

By this [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=countable+discontinuity&t=31694]problem[/url], $f$ is continuous, with the exception of an at most countable set.

Hence $f$ is continuous almost everywhere, so by Lebesgue's Integrability Criterion we have that $f$ is Riemann integrable.

Note that we only need the existence of only one of the lateral limits in every $x \in [a,b]$.

I think that a more direct proof can be given, but...


If a prime $ p$ satisfies the term below, we call $ p$ a pillai prime.

[term]There exists a natural number $ n$ that $ n$ does not divide $ p \minus{} 1$, and $ p$ divides $ n! \plus{} 1$.

Prove that there are infinitly many pillai primes.

P.S.Some pillai primes are 23, 29, 59, 61, 67, 71, 79, 83, 109, 137, 139, 149, 193, 227, 233, 239, 251, 257, 269, 271, 277, 293, 307, 311, 317, 359, 379, 383, 389, 397, 401, 419, 431, 449, 461, 463, 467, 479, 499.


Show that the following inequality holds for any $ x,y\ge 1$ :

$ (xy\minus{}x\minus{}y)^2\plus{}(6\sqrt{3}\minus{}10)xy\ge 6\sqrt{3}\minus{}9$

$ x \equal{} a \plus{} 1$, $ y \equal{} b \plus{} 1$, $ a$, $ b$ positive gives us our answer immediately because $ 6\sqrt3 \minus{} 9 > 1$.


geometry
circumcircle
trigonometry
circles
isosceles
IMO
IMO 1962

Consider an isosceles triangle. let $R$ be the radius of its circumscribed circle and $r$ be the radius of its inscribed circle. Prove that the distance $d$ between the centers of these two circle is \[ d=\sqrt{R(R-2r)} \]

The triangle does not have to be isosceles. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31633]Triangles with common Circumcircle, Incircle [Inversion?][/url] for a proof of Euler's formula by inversion.

$\triangle ABC,\ w=C(O,R),\ \{A,A'\}\subset AI\cap w.$

$IA=\frac{r}{\sin \frac A2},\ IA'=BA'=CA'=2R\sin \frac A2\Longrightarrow$

$R^2-OI^2=-p_w (I)=IA\cdot IA'=2Rr\Longrightarrow OI^2=R^2-2Rr.$

[u]Remark.[/u] $p_w (I)$ is the power of point $I$ w.r.t. the circumcircle $w$.

Check here for proof: [url]https://www.cut-the-knot.org/triangle/EulerIO.shtml[/url]

[quote=ZETA_in_olympiad]Euler's formula holds for all triangle, so I'll just prove Euler's formula in the classical way:

$\textbf{LEMMA: For all real } x, \textbf{ we have, } e^{ix}=\cos x+i\sin x.$
Proof. Consider the function $$f(\theta)= e^{-i\theta} (\cos \theta+ i\sin \theta).$$
Differentiating $f$ yields $$e^{-i\theta} (i\cos \theta-\sin \theta)- ie^{-i\theta} (\cos \theta +i\sin \theta)=0.$$ Therefore, $f$ is constant and our claim follows. $\blacksquare$[/quote]

Ummm this isn't the euler's formula the problem is talking about, you might want to check the problem statement again.

[quote=Quidditch][quote=ZETA_in_olympiad][quote=BVKRB-]

Ummm this isn't the euler's formula the problem is talking about, you might want to check the problem statement again.[/quote]

It's the special case of Euler's formula for isosceles triangles. Please see #2.[/quote]

The euler's formula #2 was talking about is that for any triangle $ABC$, the distance from incenter to circumcenter is $R(R-2r)$. (Or is this a troll?)[/quote]

I apologize I quoted the wrong Euler's formula. No troll though.


geometry
geometric transformation
rotation
circumcircle
vector

1. Prove that if a rotation maps $ A$ to $ C$ and $ B$ to $ D$, then it maps segment $ AB$ to segment $ CD$.

2. Let $ \triangle{ABC}$ be an isosceles triangle with $ AB\equal{}AC$, $ \angle{BAC}\equal{}\alpha$, and circumcenter $ O$.  Prove that there exist rotations about both $ A$ and $ O$ which carry segment $ AB$ to segment $ AC$.

3. Let $ \triangle{ABC}$ be as in the prior problem, and suppose points $ M$ and $ N$ on $ AB$ and $ AC$ respectively are such that $ BM\equal{}AN$.  Compute the angle of rotation (in terms of $ \alpha$) needed to map segment $ AB$ to $ AC$ with center $ O$.

Thanks!

I'm not very good at rotations, so don't completely trust my solutions

[hide="1"]
Consider the point P, which the rotation is centered at. We can see that $ AP \equal{} CP$ and $ BP \equal{} DP$. Call the angle of the rotation $ \alpha$. We can see that $ \triangle BAP\cong\triangle DCP$ by SAS, and since $ BPD \equal{} \alpha$, the same rotation maps $ \triangle BAP$ to $ \triangle DCP$, so therefore, it maps AB to CD.[/hide]

Only a case needs to be checked... what if $ A,B,P$ are collinear?

If A, B, and P are collinear, then D, C, and P will have to collinear. We can see that AB = CD. Consider the circumcircles of APC and BPD. These are 2 concentric circles, with length of the AB and CD equal to the distance between the 2 circles. From this, we can see that the rotation also maps AB to CD.

I'm not sure if this is rigorous...Is there a better way?

You should to keep two things in mind:

$ \blacktriangleright$ Any rotation is a [url=http://en.wikipedia.org/wiki/Linear_map]linear map[/url].
$ \blacktriangleright$ Three points are collinear iff, if we write two of them as vectors $ \vec{A}$ and $ \vec{B}$, the third can be expressed in the form $ c_1 \vec{A} \plus{} c_2 \vec{B}$, where $ c_1 \plus{} c_2 \equal{} 1$.

I don't think that I'm understanding the problem correctly.  Can someone please draw a diagram?  Thanks!

I have no idea which problem you're asking about, but if it's the first, then all it's saying is that:

If you rotate a figure so that point A moves to point C, and it moves point B to point D, prove that the segment connecting A and B becomes the segment connecting C to D.

If it's the 2nd:

ABC is a triangle. The length of the segment connecting A to B is equal to the segment connecting A to C. Angle BAC (the one where point A is at the vertex) is labeled $ \alpha$. The center is O. Show that you can rotate the triangle around both points A and O so that segment AB moves to where segment AC used to be.

If it's the third:

Assume the triangle is the same as before. Assume that there is point M on segment AB, and point N on AC so that the distance from B to M equals the distance from A to N. How many degrees do you need to rotate the triangle (in terms of $ \alpha$) with center O so that segment AB moves to where AC was?

OK.  Let's start with the first one since the other two seem to be based off of it.  I'm kind of slow, so I haven't gotten any of the solutions at all.  :wink:  Could someone please help me?

Thanks a lot!

This question is from AoPS Volume 1.  I've posted it here for more than 2 weeks, and still don't get it, including the solution.  Could someone please help me?

To a moderator - Did I post this topic in the wrong forum?  Looks like no one's answering.

I guess there's just a lack of interest, or people just don't know the solutions.

As for the first one, I can't see that there's much to prove...just think about it... :|


number theory
least common multiple
number theory proposed

It is known that $m$ and $n$ are positive integers, $m > n^{n-1}$, and all the numbers $m+1$, $m+2$, \dots, $m+n$ are composite. Prove that there exist such different primes $p_1$, $p_2$, \dots, $p_n$ that $p_k$ divides $m+k$ for $k = 1$, 2, \dots, $n$.

[i]Proposed by C. A. Grimm [/i]

We use marriage theorem, choose $t$ of our numbers and suppose their product only has $t-1$ primes, then we can easily see some of them divides the product of the others, but this implies some of them divides a number smaller than $n!$, however they are all greater than $n^n$

I don't understand that. Why a number dividing the product of t-1 others should divide something smaller than n! ? Obviously we can take any number, however large, and find anothers with product divisible by it...
   By the way, I don't know any solution of this problem really using marriage lemma (though the temptation is strong; in fact, the only participant of the olympiad who solved the problem managed to insert this lemma into his solution. It was funny to outline a large part of his solution which could be deleted with no harm at all.)
   So, I'd be really glad to hear a solution using the lemma.

Here is my solution. First the numbers m+i which have at least n prime divisors are no problem because we can always take one prime divisor which is not already taken.
Now for the other numbers of the form m+i which have n-1 or smaller number of prime divisors.
Let  $m+i=p_{1}^{a_{1}}.p_{2}^{a_{2}}...p_{k}^{a_{k}}$ where $k\leq n-1$. Now for this nubmer choose $p_{i}$ such that $p_{i}^{a_{i}}$ is maximal.
The claim is that this primes are all different. Suppose p is chosen for m+i and m+j. And let $p^{a}$ and $p^{b}$ divides m+i and m+j respectively.
WLOG $a\geq b$. So $p^{b}$ divides $m+j-(m+i)\leq n-1$. But since p is chosen that way we have
$p^{b}\geq \sqrt[n-1]{m+j}\geq n$
This contradiction proves the statement.

By the way the general statement (with no $m>n^{n-1}$) is Grimm hypothesis and i think it is still open question. If you are interesting in this kind of questions look here:
http://www.math.spbu.ru/user/aih/papers/grimm.zip

My solution(I holp it is correct):
Let $q_{1},...q_{k}$ be all the primes not greater than $n$
then Let $q_{i}^{\alpha_{i}}\parallel LCM[1,2,...,n]$
Define $f: \{m+1,...m+n\}: \to P$ as the following:
if $m+l=\prod_{i=1}^{k}{q_{i}^{\beta_{i}}}*q,q_{i}\not |q$
Then 1)  $f(l)=$one of the prime fractor of $q$,when $q>1$
2) $f(l)=q_{t}$ when $q=1$ and $\beta_{t}>\alpha_{t}$(for some $t$)
and it is easy to check that if $\beta_{t}\leq \alpha_{t},\forall t$ and $q=1$,then $m+l\leq n^{k}\leq n^{n-1}$ (becuase$q_{t}^{\alpha_{t}}\leq n$)
so there are only two case 1),2)
and we easily check that $f(1),...f(n)$ are distinct( if $f(i)=f(j)=q_{u}$ then $q_{u}^{\alpha_{u}+1}|(m+i)-(m+j)$ but $q_{u}^{\alpha_{u}+1}>n$
contradition.)
Edit: I don't know why if no$m>n^{n}-1$ it is still true? if $m+1=2,m+3=4$ then there is no two distinct prime which divide $m+1$ and $m+4$?I

[quote="Baronov"]Hawk Tiger, for the general statement you must have all m+1,m+2..m+n be composite and here m+1=2 is prme, so this is not counterexample for Grim's hypothesis.[/quote]
Thanks.And I think my proof do not need all the mumber is not a prime.

[quote="Baronov"]Here is my solution. First the numbers m+i which have at least n prime divisors are no problem because we can always take one prime divisor which is not already taken.
Now for the other numbers of the form m+i which have n-1 or smaller number of prime divisors.
Let  $ m \plus{} i \equal{} p_{1}^{a_{1}}.p_{2}^{a_{2}}...p_{k}^{a_{k}}$ where $ k\leq n \minus{} 1$. Now for this nubmer choose $ p_{i}$ such that $ p_{i}^{a_{i}}$ is maximal.
The claim is that this primes are all different. Suppose p is chosen for m+i and m+j. And let $ p^{a}$ and $ p^{b}$ divides m+i and m+j respectively.
WLOG $ a\geq b$. So $ p^{b}$ divides $ m \plus{} j \minus{} (m \plus{} i)\leq n \minus{} 1$. But since p is chosen that way we have
$ p^{b}\geq \sqrt [n \minus{} 1]{m \plus{} j}\geq n$
This contradiction proves the statement.[/quote]

By the same method we can prove that the statement is true for all $ m\geq (n\minus{}1)!$

[quote=ciprian]It is known that $m$ and $n$ are positive integers, $m > n^{n-1}$, and all the numbers $m+1, m+2, \dots, m+n$ are composite. Prove that there exist such different primes $p_1,p_2, \dots, p_n$ that $p_k$ divides $m+k$ for $k = 1,2, \dots, n$.

[i]Proposed by C. A. Grimm [/i][/quote]

Recall that in the expression $p^a$, $a$ is called the exponent, $p$ the base and $p^a$ is called a power. With this terminology, we proceed as follows: Define $f(\ell)$ as the largest prime power dividing $\ell$, i.e., if $$\ell=q_1^{\alpha_1}\dots q_s^{\alpha_s}$$ then $$f(\ell)=\operatorname{max}\left(q_1^{\alpha_1},\dots, q_s^{\alpha_s}\right)$$ Set $p_i=f(m+i)$ for $1 \le i \le n$. If all the bases are different, we are done. Otherwise, there exists an index $j$ such that $p_j=f(m+a)=f(m+b)$. If at least one of $m+a$ or $m+b$ has more than $(n-1)$ distinct prime divisors, then we may replace $p_j$ for that number by some prime number other than those in $\{p_1,\dots,p_n\}$. If neither does, then $$n<\sqrt[n-1]{m}<\min \left(\sqrt[n-1]{m+a},\sqrt[n-1]{m+b}\right)<p_j^{\min(v_{p_j}(m+a),v_{p_j}(m+b))} \mid |(m+a)-(m+b)|<n$$ and we get the desired contradiction!


[quote="Yuriy Solovyov"]Let $ a,b,c,x,y,z\geq 0$. Prove the inequality:
$ \sqrt[3]{a(b+1)yz}+\sqrt[3]{b(c+1)zx}+\sqrt[3]{c(a+1)xy}\leq \sqrt[3]{(a+1)(b+1)(c+1)(x+1)(y+1)(z+1)}.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89595


Υπάρχουν άπειρες τριάδες $(x,y,z),$ με $x,y,z\in\mathbb{Z},$ ώστε $x^2+y^2+z^2=x^3+y^3+z^3;$

Θέτουμε $z=-x$ .Τότε η σχέση γίνεται : $2x^2=(y-1)y^2$
δηλαδή ο $\frac{y-1}{2}$ πρέπει να είναι τέλειο τετράγωνο .
Έστω   $\frac{y-1}{2}=k^2$ . Τότε $y=2k^2+1$  και $x=k(2k^2+1)$  άρα έχουμε άπειρες τριάδες της μορφής

$(x,y,z)=(k(2k^2+1)\ ,2k^2+1\ ,-k(2k^2+1))$

ειχα βρει μια λυση, θετωντας το ιδιο ,δηλ χ=-ζ,υπαρχει αλλη λυση ?


vector
linear algebra
linear algebra unsolved

Can someone provide an example of a set of vectors that are not linearly independent but are orthogonal?  (Use any vector space and any inner product.)

Maybe I should clarify that my book (Friedberg, Insel, Spence) has that there exists an orthogonal set of vectors that is not linearly independent.  But it also states that every orthonormal set of vectors is linearly independent.  Hence, my confusion and why I'm interested in an example.

I don't think it's possible...

Let $v_{0}$ be a nonzero vector in the set. If $v_{0}= a_{1}v_{1}+a_{2}v_{2}+\cdots+a_{n}v_{n}$ with $a_{i}\neq 0$ for some $i$, the set is linearly dependent. Then if we take the dot product of both sides with $v_{0}$, we would get

$v_{0}\cdot v_{0}= |v_{0}|^{2}= 0$,

which means $v_{0}$ is the zero vector, a contradiction.

We first have to clarify what field we are working over. I'll answer this on the assuption that these are vectors over the real numbers (or possibly over the complex numbers.) [In cases where you are working over a field of finite chacteristic, a number of items of terminology are different.]

I would prefer to define an [b]orthogonal set[/b] to be a set of [i]nonzero[/i] vectors $\{v_{k}\}$ such that $\langle v_{j},v_{k}\rangle=0$ for $j\ne k.$ In fact we can state this as follows:

$\langle v_{j},v_{k}\rangle\ \begin{cases}>0,&j=k\\=0,&j\ne k\end{cases}$

If we use that definition of an orthogonal set, then we can easily prove that any orthogonal set is independent.

Now, what could your textbook be doing differently? The only thing I can see is that they might have left out the word "nonzero." This is the same thing as omiting the "$>0$" first line of my definition-by-cases. If that's the case, a set which includes the zero vector might count as orthogonal; it's certainly not independent.

I wouldn't do that, and I don't know why your textbook would - but that's all I can think of. (That, or something involving fields of finite characteristic.)

Yeah, thank you that clears it up in my mind a lot.  In fact, I felt this had to be true intuitively.  Here is what my book has precisely:
[quote]
True/False
...
(f) Every orthogonal set is linearly independent.
(g) Every orthonormal set is linealy independent.
[/quote]
The answers are in the back for these problems and the first one is false, the second true. It may be that the first (orthogonal) can include zero, but you cannot normalize the zero vector, thus it cannot be in an orthonormal set.


\item Quadrilateral $ ABCD$ has $ AD\equal{}BC$ where $ \angle ABC < 90^{\circ}$ and $ \angle BCD < 90^{\circ}$. Given that $ \sin\angle ABC \equal{} \frac{4}{9}$ and $ \sin\angle BCD \equal{}\frac{5}{6}$, compute $ \displaystyle\frac{[ADC]}{[BDC]}$, where $ [H]$ denotes the area of $ H$.

[quote="ragnarok23"]\item Quadrilateral $ ABCD$ has $ AD \equal{} BC$ where $ \angle ABC < 90^{\circ}$ and $ \angle BCD < 90^{\circ}$.[/quote] I think this is impossible.

oops. that's supposed to be $ \angle ADC$ rather than $ \angle ABC$.


A neat game, where you start with a word, say math. Then you anagram the word and either leave it alone, add a letter, subtract a letter, or change a letter. A result would be mat. We put the current word in bold. Example:

fan
man
mean
dean
dear
dare
dart
rat
art
at
cat
[b]cart[/b]

Some rules:

1. You can't have cat, and then have cats.
2. The words must be in ENGLISH.
3. The word must be in a dictionary.
4. No abbrev.

 Let's see how far we get.

[b]Star[/b]

Let's see if anyone can find the name of a popular movie(series of movies).

Star
Rates
[b]State[/b]

This seems quite similar to (but not exactly like) "Stretch" over in GFF. Maybe this thread should go there.

[quote="KrazyNerd"]Star
Rates
State
[b]Skate[/b]

This seems quite similar to (but not exactly like) "Stretch" over in GFF. Maybe this thread should go there.[/quote]

Star
Rates
State
Skate
Dates
[b]Taste[/b]

Star
Rates
State
Skate
Dates
Taste
[b]Haste[/b]

[hide]STAR
TARTS
TASTER
RESTART
STAR TREK[/hide]   :maybe:

[quote="AiW"]STAR
TARTS
TASTER
RESTART
STAR TREK[/quote]This is much like the game Stretch, which can be found in GFF.

[quote="KMN"]Star
Rates
State
Skate
Dates
Taste
Haste
[b]Hate[/b]
hater[/quote]

what... is friese a word I thought it was frieze

I don't know you should ask arnav I bet he would know these kinda things

Another rule: You can only add one word at a time.

Also, I didn't know where this would go.

tar
Rates
State
Skate
Dates
Taste
Haste
Hate
hater
[b]TARE[/b]

tar
Rates
State
Skate
Dates
Taste
Haste
Hate
hater
tare
TAKE

tar
Rates
State
Skate
Dates
Taste
Haste
Hate
hater
tare
take
tack

Star
Rates
State
Skate
Dates
Taste
Haste
Hate
hater
tare
Take
Tape
[b]Pet[/b]

Star
Rates
State
Skate
Dates
Taste
Haste
Hate
hater
tare
Take
Tape
Pet

[b]Met?[/b]


rotation
trigonometry
trig identities
Law of Cosines

Triangle $ABC$ has right angle at $B$, and contains a point $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$.  Find $PC$.

[asy]
pair A=(0,5), B=origin, C=(12,0), D=rotate(-60)*C, F=rotate(60)*A, P=intersectionpoint(A--D, C--F);
draw(A--P--B--A--C--B^^C--P);
dot(A^^B^^C^^P);
pair point=P;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$P$", P, NE);[/asy]

[hide="hint"]Law of Cosines, perhaps rotating 60 degrees about B would work (not sure, but there is an interesting property associated with that)[/hide]

here is a solution since i double posted  :P
[hide="solution"]using a system for law of cosines, we have

$AB^2=10^2+6^2+10*6$
$BC^2=6^2+x^2+6x$
$AC=10^2+x^2+10x$
since it is right, $AB^2+BC^2=AC^2$, substitute,
$10^2+6^2+10*6+6^2+x^2+6x=10^2+x^2+10x$
$72+60=4x \implies x=33$[/hide]

[hide="Ugh, Same thing"]Because $\angle APB=\angle BPC = \angle CPA$ and the sum to $360^\circ$, each angle is $120^\circ$. Thus, their cosines are all $-\frac12$. Notice that we can now find expressions for $AB$, $BC$, and $AC$ involving $PC$ by using the Law of Cosines:

$AB^2=10^2+6^2+10\cdot6$, $BC^2=6^2+PC^2+6PC$, and $AC^2=10^2+PC^2+10PC$

Since $ABC$ is a right triangle, $AB^2+BC^2=AC^2$. Substituting, \[ 10^2+6^2+10*6+6^2+PC^2+6PC=10^2+PC^2+10PC. \]
Solving this system yields $x=33.$[/hide]

Is that a challenge?

EDIT: [url=https://artofproblemsolving.com/community/q1h57p172]This[/url] might be the oldest unbumped thread yet. If bumped, it would stand at 6121 days, surely a top 3.

[quote=vvluo][quote=AopsUser101]Here is my work:[/quote]

This bump of nearly 14 years has to be one of the longest in history(top 100)[/quote]

Top 5 i guess, but it was a good bump

[hide=Solution]Since $\angle APB + \angle BPC + \angle CPA = 360$ and $\angle APB = \angle BPC = \angle CPA$, we get $\angle APB = \angle BPC = \angle CPA = 120^\circ$. Applying Law of Cosines, we get $AB=14$. Let $PC=x$. Then from Law of Cosines, we get $BC=\sqrt{x^2+6x+36}$ and $AC=\sqrt{x^2+10x+100}$. From the Pythagorean Theorem, we obtain $(x^2+6x+36)+(196)=(x^2+10x+100)$, or $132=4x \implies x = \boxed{33}$.[/hide]

It took me 10 minutes to solve :sob: anyway, $\angle APB = \angle BPC = \angle CPA = 120^{\circ}.$ Now, if we do LoC, $AB=14$, let $CP=x$, now using LoC again $AC=\sqrt{x^2+10x+100}$ and $BC=\sqrt{x^2+6x+36}$, using PT, $(x^2+6x+36)+(196)=(x^2+10x+100)$, now $x=\boxed{33}.$


Arkadaşlar Aralık sınavının tarihini bilen var mı?

16-17 Aralık 2006 tarihlerinde... herkese başarılar


function
quadratics
algebra
polynomial
Rational Root Theorem
real analysis
real analysis unsolved

Hi I'd like to know how to find the zeros in the function

16x^3 - 36x^2 + 24

I used a graphing calculator and realized that none of the x intercepts were whole numbers (the method I was taught needed at least one whole number). Any suggestions on how to solve for x would be greatly appreciated.

Thx :D

Use  [url=http://en.wikipedia.org/wiki/Cubic_equation]Cardano's method[/url]

Before jumping to Cardano, let's exhaust the possibilities of the Rational Root Theorem. Take out the common factor and write this as $4x^{3}-9x^{2}+6=0.$ By the Rational Root Theorem, the possible rational roots of this are $\pm1,\pm2,\pm3,\pm6,\pm\frac12, \pm\frac32,\pm\frac14,\pm\frac34.$

Trying these out with synthetic division, we hit paydirt with $\frac32:$

$\begin{array}{ccccc}\frac32&4&-9&0&6\\ &&9&-\frac92&-\frac{27}4\\ &4&-3&-\frac92&-\frac34\end{array}$

[hide="Edited out"]So $4x^{3}-9x^{2}+6=\left(x-\frac32\right)(4x^{2}-6x-9).$

We've got one root, and we can get the other two by the quadratic formula.[/hide]

[color=green][Yeah, kunny, that was a mistake. But someone should try the other possibilities.][/color]

You seem to make a mistake, Kent Merryfield :maybe:

[quote=N.T.TUAN][quote="Kent Merryfield"]Before jumping to Cardano, let's exhaust the possibilities of the Rational Root Theorem. [/quote]
You are right!  :wink: I think+ed he must try this before![/quote]

RRT is helpful in various problems of this sort, because no bashing or cubic formula is needed - [url]https://www.youtube.com/shorts/wKpmfnyKeeM[/url]


trigonometry
geometry
number theory
relatively prime

Triangle $QRS$ has the following properties:

(1) $Q$ is a right angle
(2) $9\sin S = 8 \cot R$
(3) The length of one of the legs is 13

Let the two possible areas of the triangle be $a$ and $b$. The expression $\displaystyle \left|\frac {1}{a} - \frac {1}{b}\right|$ can be written as $\frac {t \sqrt{u}}{v}$, where $t$ and $v$ are relatively prime positive integers and $u$ is not divisible by the square of any prime. Compute the number of positive integer divisors of the product $tuv$.

I would post my solution but it follows the same sort of process and arrives at the same answer  :D


I recently got an invitation to join the math camp at the university of ottawa.  :lol: Inside the package it said that sample problems were enclosed, but I found no problems inside. I was wondering if any of  you have sample problems could you please post it here because I am wondering what level is it at the camp.

I would also appreciate some advice given on whether it is useful to attend the math camp. Is it worth paying the $ 250.00 for the five days?

Could someone please answer the post.  I need to make a decision by the end of his weekend.

[quote="siva.eas"]
I would also appreciate some advice given on whether it is useful to attend the math camp. Is it worth paying 250 for the five days? [/quote]

Well, you could eat enough food there to make the camp worth it. I don't know of any Ottawa camps, but I've been to a Regional math camp at Brock - they invited me there after the National, so it must have been a mess-up. They taught some cool stuff, though not always contest problems. If 250 is too much, you could ask them to waive a portion of your fee based on financial need - I've done that before. They want to give a chance to everybody, so they try to help people who can't come because of money. Based on what did they invite you there?

It all depends on the Math level of the camp. Last year I went to national camp also 5 days it cost over 250 dollars but it was so worth it. If I didn't go there I probably wouldn't have even known about Olymon and the Olympiads. It was great help. I think that if you think you'll learn a lot from the camp go there!

are there any non-invite ( or easy to get invited too ) camps in/around the GTA ?


algorithm
number theory
Euclidean algorithm
number theory proposed

Let n be odd positive integer and r be positive rational ,prove that there are positive integer $a_1,a_2,a_3,b_1,b_2,b_3$ such that $r=\frac{a_1^{n}+a_2^{n+1}+a_3^{n+2}}{b_1^n+b_2^{n+1}+b_3^{n+2}}$

[quote="kueh"]$gcd(n, n+2)|2$. Since $n$ is odd, $gcd(n, n+2)=1$. Thus we can find $a, b, \in \mathbb{N}$ such that $an - b(n+2) = 1$ (by the euclidean algorithm - for example $a = \frac{n+1}{2}, b = \frac{n-1}{2}$). Then it is obvious that $a_1 = r^{a}$, $a_2 = r$, $a_3 = r$, $b_1 = r$, $b_2 = r$, $b_3 = r^{b}$ works. But only when $r$ is an integer of course. I am trying to think of how to generalise for $\frac{p}{q}$[/quote]

Let $r=\frac{p}{q}$, using the above method, let $a_1 = r^{a}q^{k(n+1)(n+2)}$, $a_2 = rq^{kn(n+2)}$, $a_3 = rq^{kn(n+1)}$, $b_1 = rq^{k(n+1)(n+2)}$, $b_2 = rq^{kn(n+2)}$, $b_3 = r^{b}q^{kn(n+1)}$ for some $k$ such that $k(n+1)(n+2) \geq a$ and $kn(n+1) \geq b$. It is obvious that it all works out to integers.

[quote="kueh"]It is obvious that it all works out to integers.[/quote]
Obvious for you? :)
Well done!

Is that the same Achilleas Sinefakopoulos who attends Cornell as a grad student?


function
algebra
polynomial
algebra unsolved

Determine all functions $ f: \mathbb{N}_0 \rightarrow \mathbb{N}_0$ such that $ f(f(n))\plus{}f(n)\equal{}2n\plus{}6$ for all $ n \in \mathbb{N}_0$.

[hide]Define a sequence $ a_0 \equal{} R$ for some $ R \in \mathbb{N}_0$, and $ a_{m \plus{} 1} \equal{} f(a_m)$ for all $ m \in \mathbb{N}_0$.

Then, we have the relation:
$ a_{m \plus{} 2} \plus{} a_{m \plus{} 1} \equal{} 2a_{m} \plus{} 6$

Also:
$ a_{m \plus{} 1} \plus{} a_{m} \equal{} 2a_{m \minus{} 1} \plus{} 6$

Letting $ d_m \equal{} a_{m} \minus{} a_{m \minus{} 1}$, and taking the difference between the above recursions yields:
$ d_{m \plus{} 2} \plus{} d_{m \plus{} 1} \equal{} 2d_{m}$

The characteristic polynomial of this recursion is $ \lambda^2 \plus{} \lambda \minus{} 2$, which has the roots: $ \lambda \equal{} \minus{} 2$ and $ \lambda \equal{} 1$.

Thus, $ d_{m} \equal{} P( \minus{} 2)^m \plus{} Q(1)^m \equal{} P( \minus{} 2)^m \plus{} Q$ for some constants $ P,Q \in \mathbb{R}$.

Therefore, $ a_m \equal{} (d_m \plus{} d_{m \minus{} 1} \plus{} \cdots \plus{} d_2 \plus{} d_1) \plus{} a_0 \equal{}$ $ \dfrac{2}{3}P(( \minus{} 2)^m \minus{} 1) \plus{} Qm \plus{} R \equal{} A( \minus{} 2)^m \plus{} Bm \plus{} C$ for some constants $ A,B,C \in \mathbb{R}$

If $ A \neq 0$, then there exists some $ M \in \mathbb{N}_0$ such that $ a_M < 0$. This is a contradiction since $ f: \mathbb{N}_{0}\rightarrow\mathbb{N}_0$.

Thus, $ A \equal{} 0$, and $ a_m \equal{} Bm \plus{} C$ for all possible sequences.

So, $ a_2 \equal{} 2a_1 \minus{} a_0 \Rightarrow f(f(n)) \equal{} 2f(n) \minus{} n$. (For all possible values of $ n$ and $ f(n)$).

Plugging this into the original equation yields:
$ (2f(n) \minus{} n) \plus{} f(n) \equal{} 2n \plus{} 6 \Rightarrow 3f(n) \equal{} 3n \plus{} 6 \Rightarrow f(n) \equal{} n \plus{} 2$ for all $ n \in \mathbb{N}_0$.[/hide]

@Allnames: Thanks for catching my error, I hope that this is correct now.

[quote="Xantos C. Guin"]
So, $ a_0 \equal{} C$, and $ a_1 \equal{} B \plus{} C$. Therefore, $ a_1 \equal{} f(a_0) \equal{} a_0 \plus{} B$ for any $ a_0 \in \mathbb{N}_0$

][/quote]
Let forgive me if my question is too silly!.
But dear Xantos C. Guin, Why you can get $ f(a_0) \equal{} a_0 \plus{} B$ since only $ a_0 \equal{} C$, and $ a_1 \equal{} B \plus{} C$ (1).
As you proved above, $ A;B;C$ are constants; so that I can also show another function satisfying your (1).
Indeed ,for example, $ f(a_0)\equal{}4a_0\plus{}D$; where $ D$ is a constant (in the fact $ D\equal{}B\minus{}3C$ which is constant). And of course this result doesn't satisfying the initial property.
Sorry for my silly question!

very nice [b]Xantos C. Guin[/b]
my solution:
substitution $ f(n)\equal{}n\plus{}2\plus{}g(n)$ gets $ 0\equal{}g(n)\plus{}g(n\plus{}2\plus{}g(n))$. so $ g(n)  \ge \minus{}n\minus{}2$. next substitution: $ g(n)\equal{}\minus{}n\minus{}2\plus{}h(n)$ gets equation $ h(h(n))\equal{}n\plus{}4$. it's not very hard to prove that $ h(n)\equal{}n\plus{}2$ then $ g(n)\equal{}0$ and finally  $ f(n)\equal{}n\plus{}2$  :)


Thanx to arne this post became useles. I can't even read. :blush:

[quote="spider_boy"]$x(x+1)(x+2)(x+3)\dots(x+2n-1) + (x+2n+1)(x+2n+2)\dots(x+4n)=0$[/quote]

You misread the question, Kondr, do you see it now?

Replace $x$ by $x-2n$ to obtain
$P(x)=(x-2n)(x-2n+1)\dots(x-1) + (x+1)(x+2)\dots(x+2n)=0$

Observe that $P(-x)=P(x)$, hence the coefficients of the odd degree terms are $0$. On the other hand, in is easy to see that after multiplying, all coefficients of the even degree terms are positive, hence the equation $P(x)=0$ has no real roots.

Thanks  very much Arne for writing in latex and Enescu for solution.


[quote="Inequalities Master"]Let $ a,b,c\geq 0$.Prove that:
$ \frac {a}{a \plus{} b}\sqrt {4a^2 \plus{} 5ab} \plus{} \frac {b}{b \plus{} c}\sqrt {4b^2 \plus{} 5bc} \plus{} \frac {c}{c \plus{} a}\sqrt {4c^2 \plus{} 5ca}\leq \frac {25}{16}(a \plus{} b \plus{} c)$[/quote] $ a\equal{}0,b\equal{}1, c\equal{}1234444444445^{2345555555555555555}$


We will send a formal letter with the USAMO total score to each USAMO participant later this week.

This is just as it says in the S-I letter, which all USAMO participants already  received: "A copy of the USAMO with solutions and results (including your score) will be mailed to your USAMO Exam Manager on Wednesday, May 9th.  Also included will be a USAMO Certificate of Participation.... Students should not inquire about results before Friday, May 11th, 2007."

Steve Dunbar
AMC Director

Sorry to bother you, but will the scores be available online by student ID anonymously as they have been in the past on Wednesday as well? Thanks.

yeah, and while we're at it, let's not inquire about scores until [b]Friday, May 11[/b] as requested.  when will blue MOP participants be called/emailed/announced?

EDIT: At least you did so very politely, Ignite168.

And while we're at it, let's also read the other announcements by amcdirector...

[quote]Emails have been sent to all students who meet these categories.  We don't have the time right now for phone calls to these students, so the email will have to suffice.  A formal paper letter of invitation will be going out this week.[/quote]

Please don't inquire about the timing of when results will be released.  The AMC is very busy, and the information will be released as it is available.

Thanks to Steve Dunbar and the AMC office for posting the information here.

As mentioned elsewhere, the individual scores are now available on the [url=http://www.unl.edu/amc]AMC website[/url].


The positive difference of the cube of an integer and the square of
the same integer is 100. What is the integer?

We have $ n^3 \minus{} n^2 \equal{} 100 \implies n^2(n \minus{} 1) \equal{} 100$. A little trial and error yields $ \boxed{5}$ as our answer.

BTW, $ n^3\minus{}n^2\minus{}100\equal{}0$ factors as $ (n\minus{}5)(n^2\plus{}4n\plus{}20)$


geometry
perpendicular bisector
geometry unsolved

Let $A,B$ be two points on circle $O$. Let $X$ be the intersection of the tangent lines of $O$ at $A,B$. Choose two points $C,D$ on $O$ such that $C,D,X$ are collinear in that order and $CA,BD$ are perpendicular. Let $F=CA\cap BD, G=CD\cap AB, H=BD\cap$perpendicular bisector of $GX$. Show that $X,F,G,H$ are cyclic.

http://www.mathlinks.ro/Forum/viewtopic.php?t=33818

  darij


Angle bisectors of $\angle{B}$ , $\angle{C}$ of$\triangle{ABC}$ intersect sides $AC$ , $AB$ at $P$ and $Q$respectively. $I$ is intersection of $BP$ and $CQ$.  $IP=IQ$ and ${AB}\not={AC}$ and $O$ is circumcenter of$\triangle{ABC}$.Find $\angle{OIB}=?$

Circle (I) touches AB,AC at E,F
Easily triangle IQE and IPF are congruent so angles IPA=IQE
This means A,P,I,Q are cyclic
We have 180=BAC+PIQ=BAC+90+BAC/2
so BAC=60
BIC=BOC=120 so B,I,O,C are cyclic
OIB=OIC+BIC=OBC+BIC=150


function
trigonometry
calculus
calculus computations

If $ f$ is a continuous function for all real number, such that $ f$ statisfies for all $ x$ and $ y$,there exist $ k$ such that,$ |f(x)\minus{}f(y)| \leq k|x\minus{}y|$. And $ lim (\frac {f(x)}{x})\equal{}0$, for $ x$ attains infinity or $ x$ attains minus infinity, does it follow that $ f$ is a constant function ?
Thank you for the solution before.

Isn't $ f(x) \equal{} \sin x$ a counterexample?

[quote="Qruni"]If we change the statement "$ f$ statisfies for all $ x$ and $ y$,there exist $ k$ such that,$ |f(x) \minus{} f(y)| \leq k|x \minus{} y|$" with the statement : "f is a convex function" ?.[/quote]

yes, it's true.  if $ f$ is nonconstant and convex, then it grows at least linearly as $ x\to\infty$ or as $ x\to \minus{}\infty$.


Twenty-five men sit around a circular table. Every hour there is a vote, and each must respond $ yes$ or $ no$. Each man behaves as follows: if his response on the $ n$-th vote is the same as the response of at least one of the two people he sits between, then he will respond the same way on the $ (n\plus{}1)$-th vote; otherwise he will change his response. Prove that however the men responded on the first vote, there will be a time after which nobody's response will ever change.

[hide]We will say that at the $ n$-th vote a person is "fixed" iff the person always casts the same vote on all turns $ n,n\plus{}1,\dots$.  Note that a sufficient condition for a person being fixed is that one of the neighbors casts the same vote on the $ n$-th turn, that neighbor being on its turn fixed, since for each one of the two of them, at least one of their neighbors (the other fixed neighbor) has the same response at the $ n$-th vote, hence both will vote the same in the $ n\plus{}1$-th vote.  Clearly, at the first vote there are at least $ 2$ fixed people; otherwise, number the voters clockwise starting from a person that voted $ yes$, hence all odd people vote $ yes$ and all even people vote $ no$, but $ 1$ and $ 25$ sit next to one another, reaching a contradiction.

Pick at random, at turn $ n$, a fixed person, and move clockwise until finding the first non-fixed person $ A$.  This person clearly has a fixed neighbor $ B$ (otherwise it would not be the first that we find) that voted differently in the turn $ n$, and clearly the other neighbor (fixed or not) also voted differently from $ A$.  Thus $ A$ changes vote in the $ n\plus{}1$-th turn, but $ B$ doesn't, and $ A$ becomes fixed.  The number of fixed people thus increases every vote, and since it cannot increase without bounds, all voters end up being fixed.

Note that this solution readily generalizes to any odd number of voters, but an even number of voters may be distributed such that each person that votes $ yes$ sits between two people that vote $ no$ and viceversa, thus leading to everybody changing votes every turn.  Note however that, as soon as a fixed person appears in a table, regardless of the parity of the number of people, all voters will eventually be fixed.[/hide]


I live in western MA. do you know when the amc 10a results for the schools are sent out?
(i took it on tue. 2/6)

I just asked this in another thread. The AMC Director responded with 3 weeks.


IMO 2007 is definitely a highlight of my life so far despite my disappointing performance.

Here's a link to my pictures. I wanted to have all the teams in the opening ceremony, but my camera did not cooperate. Please help to identify the teams and faces that I could not, and correct errors that I might have made.

http://pictures.aol.com/ap/viewAlbum.do?user=guest&mode=guestView&shareInfo=KnVVBplenSM9s%252B%252BUgMLT%252BozLZ4S3UN9x63sNFxIMIy5WGd%252BBXiXp6w%253D%253D&albumId=10415.1813.1186381317467.3

(select Slide Show)

really cool, Teddy and you did a good job on IMO(^-^*)

Teddy, I don't think I've ever seen you without that camera.  So this is why.  Nice pictures!

And don't sweat your performance: you did great, remember that anyone's performance can vary a lot.


Hello everyone,

My question is this and is very specific:

100000 euros deposited on 1 January in a bank at 8% annually.
What is the value at the end of each quarter of the year?

Is this correct?:

$ Amount \equal{} C(1 \plus{} i*n)$.

$ n$ taking the value of $ 3$ to calculate the value of the first quarter.

But I would not know if $ i \equal{} 0.08.$

A greeting

Unfortunately, your question is incomplete. You also should specify the interest compounding method (which can range from "annually" to "continuously" with plenty of options in between...). You seem to assume monthly compounding. If so, there are two wrong things with your solution:
1) The monthly interest rate is not 0.08 but __________
2) After the first month, you'll also gain interest on interest, and so on, so the money do not increase by the same amount each month but are _______ by the same number, which is _________


Let a,b,c be poirtive real numbers  which satisfy:
 \[ a+b+c=3 \]
Find the maximum value of P and Q:
\[ P= \frac{ \sum a^{2}}{ \sum a^{3}}\]

This is my solution:
Applying Bunyakovsky-Cauchy-Schwarz Inequality we have
\[ (a+b+c)(a^{3}+b^{3}+c^{3})\geq\(a^{2}+b^{2}++c^{2})^{2}\]
equivalent
\[ a^{3}+b^{3}+c^{3}\geq\frac{(a^{2}+b^{2}+c^{2})^{2}}{3}\]
Then we have
\[ P\leq\frac{3}{a^{2}+b^{2}+c^{2}}\leq\frac{9}{(a+b+c)^{2}}=1 \]
Equation when (a,b,c)=(1,1,1)
I think find minimum P the equation will happen when (a,b,c)=(0,0,1/2) and the permutation.
Then you can easily to prove by using SOS

[size=125][color=orange] Keep it simple: As simple as possible, but no simpler
__Albert Einstein__[/color][/size]
[size=125][color=red]I'm Nguyễn Tâm Đăng , Vietnam[/color][/size]


linear algebra
matrix
linear algebra unsolved

Dear Forum

There exists a matrix A(n,p) such that A'A symmetric.

Then show that A'A is nonnegative definite.

Thank U in advance.

Walkman


geometry
incenter
circumcircle
geometry proposed

Let $ ABC$ ba a triangle with circumcircle $ W$. $ D$ is a point on $ [BC]$.
$ W_{1}$ is a circle which tangents to $ [DC]$ at $ K$ and tangents to $ W$ and $ [AD]$.
$ Q$ is a point on $ W_{1}$ such that tangent to $ W_{1}$ at $ Q$ is parallel to $ AB$.
Prove that incenter of $ ADC$, $ K$ and $ Q$ are collinear.

Dear friends!

Let $ I,I'$ be the incenters of the triangles $ ABC$ and $ ADC$.
$ \left\{E\right\} \equal{} W_{1}\cap AD$, $ \left\{F\right\} \equal{} W_{1}\cap W$ and $ \Gamma$ is the circumcircle of the triangle $ AEF$.
The points $ K,I,E$ are collinear and $ I\in\Gamma$. (see here [url]http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf[/url]).

Let's prove that $ I'\in\Gamma$.
$ \angle I'IK \equal{} 90^{\circ} \minus{} \angle DI'I \equal{} 90^{\circ} \minus{} \left(\angle I'CD \plus{} \angle I'DC\right) \equal{} 90^{\circ} \minus{} \frac {\angle ACD \plus{} \angle ADC}{2} \equal{}$
$ \equal{} 90^{\circ} \minus{} \frac {180^{\circ} \minus{} \angle DAC}{2} \equal{} \frac {\angle DAC}{2} \equal{} \angle DAI'\Rightarrow\angle I'IK \equal{} \angle DAI'\Rightarrow$
$ \Rightarrow$ the quadrilateral $ AEII'$ is cyclic $ \Rightarrow I'\in\Gamma$.

Let $ O_{1}$ be the center of the circle $ W_{1}$ and $ V$ is the intersection point of the tangent to $ W_{1}$ at $ Q$ with $ AD$.
$ \angle EKQ \equal{} \frac {\angle EO_{1}Q}{2} \equal{} \frac {\angle AVQ}{2} \equal{} \frac {\angle BAD}{2} \equal{} \frac {\angle BAC \minus{} \angle DAC}{2} \equal{}$
$ \equal{} \angle IAC \minus{} \angle I'AC \equal{} \angle IAI' \equal{} \angle IEI' \equal{} \angle EKI'\Rightarrow\angle EKQ \equal{} \angle EKI'\Rightarrow$
$ \Rightarrow$ the points $ K,I',Q$ are collinear.

P.S.
1. $ \angle EI'Q \equal{} 2\cdot\angle EKQ \equal{} \angle EO_{1}Q\Rightarrow$ the quadrilateral $ EI'O_{1}Q$ is cyclic. The quadrilateral $ EO_{1}QV$ is cyclic,
so the points $ V,E,I',O_{1},Q$ are concyclic. They all lie on the circle with diameter $ VO_{1}$.
2. $ \angle QI'V \equal{} \angle QEV \equal{} \angle EQV \equal{} \angle EI'V \equal{} \angle EKQ \equal{} \angle KEI' \equal{} \frac {\angle BAD}{2}$.
So $ I'V$ is the bisector of the angle $ \angle EI'Q$ and $ I'V\parallel{}KE$.
3. If $ \left\{A,X\right\} \equal{} AB\cap \Gamma$ then $ \angle EI'X \equal{} \angle EAX \equal{} \angle AVQ \equal{} \angle EI'Q\Rightarrow Q\in I'X$.
So the points $ K,I',Q,X$ are collinear.

Best regards, Petrisor Neagoe :)


Find all functions $f, g, h : R =\Rightarrow  R$ such that
$f(x+y^{3})+g(x^{3}+y) = h(xy)$ for all reals numbers $x, y$.

It was posted here
http://www.mathlinks.ro/Forum/viewtopic.php?t=109448&sid=033eca92bbfbc1c927733818ef6d314c

and no solution was given .


geometry
geometric transformation
Intel STS

Does anyone know some unanswered questions in Chemistry? Also, currently, what are some exciting fields of research?

Thanks.

I don't know how many "unanswered questions" there are like there are in physics.  I think now, most of the research is in creating new things, nanotechnology, and trying to improve existing things.

What about designing proteins? How much do chemists understand the workings of the protein as in how to create functioning protein molecules of the desired nature?

[quote="JRav"]Well I'm not a chemist but I would say that's more in the realm of biology (transcription, translation, that stuff).  Look at some of the Intel STS finalists for some recent topics.[/quote]

I think the 2006 Nobel Chemistry Prize was awarded to works on the Eukaryotic transcription mechanism (or something similar).

Very nice link T0rajir0u, thank you very much.

I read a big thing nowadays is stereoselective methods in synthesis.


algebra
function
domain
calculus
complex analysis
complex analysis unsolved

Let $ \Omega$ be a domain which contains the closed unit disk $ \overline D$. Suppose that $ f$ is analytic in $ \Omega$, $ |f(z)| >1$ and $ f(0)\equal{}1$. Must $ f$ have a zero in $ \overline D$?

Can you solve it?  :D

We have [url=http://www.artofproblemsolving.com/Forum/index.php?f=347]forums[/url] for complex analysis. Stop posting these here [in Calculus- the topic has since been moved].

What does $ |f(z)|>1$ mean? For all $ z\in\Omega$? Suppose $ f(z_0)\equal{}0$ for some $ z_0\in\Omega$, then $ |f(z_0)|\equal{}0<1$ a contradiction? So it can never have a zero in $ \Omega$.


advanced fields
advanced fields unsolved

Does the space $ X\equal{}\prod_{n\equal{}1}^\infty S^1$ aquire a universal covering?

no, it is not semi-locally simply connected.

if $ x \in X$ has a neighborhood $ U$ such that $ \pi_1(U) \to \pi_1(X)$ is trivial, then we may assume that $ U$ is the product of small balls, which are almost all $ S^1$. but the inclusion $ 0 \times ... \times 0 \times \mathbb{Z} \times \mathbb{Z} \times ... \to \mathbb{Z} \times \mathbb{Z} \times ...$ is not trivial.


How do you make a hexagon and label it with LaTeX?  Is it even possible?

Thanks.

the link doesn't work. can you fix it. I want to learn it too.

The site has been just been reorganised so the page is now at http://tug.org/TeXnik/mainFAQ.cgi?file=symbols/hexagon

Unfortunately some of the links on that page haven't been adjusted so don't work! So try http://www.devel.lyx.org/~rae/www-user/help/symbols/hexagon.html instead which leads to http://www.devel.lyx.org/~rae/www-user/help/symbols/hexagon1.gif and http://www.devel.lyx.org/~rae/www-user/help/symbols/wasysym.html

It uses the wasysym package and you'll find details of this at http://texcatalogue.sarovar.org/entries/wasysym.html


pigeonhole principle
number theory proposed
number theory

All positive integers $ 1,2,...,121 $ are arranged in the squares of a $ 11*11 $ table. Dima found the product of numbers in each row and Sasha found the product of the numbers in each column. Could they get the same set of $ 11 $ numbers?

Nope.

There are at least twelve primes between 61 and 121 and these primes do not occur as primefactors in any of the other numbers. By the pigeonhole principle there are at least two such primes in one row and since they are in the same row they cannot be in the same column. Thus all the column product are different from that row product.

Done.

Very easy. A harder question is:
How many of Dima's and Sasha's products can be equal at most? :ninja:


function
vector
real analysis
real analysis solved

Find the real function f such that $ f(x+y)=f(x)+f(y), f(1)=1$ and $ f(x)f(1/x)=1$.

Since f is a linear (additive) function we have f(y*n) =y*f(n).
Choose y = 1 => f(n) = n.
The only problem is that I don't understand what to use f(x)*f(1/x) = 1 for.

[quote]Since f is a linear (additive) function we have f(y*n) =y*f(n).[/quote]
You can only say this for rational $y$.

Note that nothing was said about the continuity (or even measurability) of $f$. If we were to leave out the last condition and have only the condition $f(x+y)=f(x)+f(y)$, then $f$ would be a linear mapping of $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$. As such we can take a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ and make an assignment of $f$ on the elements of this Hamel basis that is quite arbitrary (with the one exception that $f(1)=1$ fixes the assignment on the one-dimensional subspace $\mathbb{Q}$). Hence there are a very large number of solutions, most of them discontinuous everywhere (and non-measurable).

The real question being posed here: does that extra condition $f(x)f(1/x)=1$ restrict all of those wild solutions away, or are there still many (discontinuous, non-measurable) solutions? I can't tell yet, although so far I'd guess that there are many solutions.

It was an  Oral Examination Ecole Normale Suprieure Paris 1996

Hints:

The first two conditions gives $f(r)=r$ for all $r\in Q$

With the third condition try to prove that $f$ is bounded on a neighbourg of $0$
and conclude that $f(x)=x$ for all $x \in R$

See RMS Revue Mathmatiques Spciales 1997 Mai-Juin n9-10, page 860-862

i solved it two years ago: actually, let $|x|\geq 2$. then there is an $a$ such that $x=a+\frac{1}{a}$. it follows that $|f(x)|=|f(a)+\frac{1}{f(a)}|\geq 2$, or equivalently, for $|x|leq \frac{1}{2}$, we have $|f(x)|\leq \frac{1}{2}$. that gives the desired result.

Peter

Here's something which doesn't use boundedness at all. It's longer maybe, but the fact that the problem has such different approaches is, I think, really cool :).

I know for sure, from a Newman problem, that multiplication can be expressed in terms of $-$ and $a^{-1}$ operations alone. I couldn't do this, but I managed to express $x^2$, and this is enough to show that $f(x^2)=f^2(x)$. This, in turn, solves the probelm, because $f^2(x)+2f(xy)+f^2(y)=$ $f(x^2)+f(2xy)+f(y^2)=f((x+y)^2)=f^2(x+y)=(f(x)+f(y))^2=$ $f^2(x)+2f(x)f(y)+f^2(y)$, and this shows $f(xy)=f(x)f(y)$, and from here it's easy.

In order to express $x^2$ only in terms of $+$ (and $-$) and $a^{-1}$ operations, we observe that $x^2=\frac 1{[(1-x)^{-1}-x]^{-1}+[x^{-1}-(1-x)]^{-1}}+x-1$. Now apply $f$ to both sides to get $f(x^2)=f^2(x)$.


It is given a triangle ABC with $\angle ACB=132^{\circ}$ and AC=BC. Let AL be the bisector of $\angle BAC$ and let D be on the $AC\longrightarrow$ such that CD=AL. Prove that $\angle BDC<60^{\circ}$

Actually it isn't hard, all is calculation!
Now we can let $\angle ACB=2\alpha=132^{0}$ and $CA=CB=a\Rightarrow AB=2a\sin\alpha$
$AL=\frac{2a.2a\sin\alpha\cos\frac{\pi-2\alpha}{2}}{a+2a\sin\alpha}=4a\frac{\sin^{2}\alpha}{1+2\sin\alpha}$
here $\alpha=66^{0}\Rightarrow AL\approx 1.18a$
$BD^{2}=CD^{2}+BC^{2}-2BCCD\cos\2\alpha=(1.18a)^{2}+a^{2}-2\cdot 1.18a\cdot a(-0.67)\approx 4a^{2}\Rightarrow BD\approx 2a$
Now $\cos BDC=\frac{CD^{2}+BD^{2}-BC^{2}}{2DC.DB}={(1.18a)^{2}+4a^{2}-a^{2}}{2\cdot 1.18a\cdot 2a}\approx 0.93\Rightarrow \angle BDC\approx 22^{0}$
Thus we can show it is less than $30^{0}$
 :)


inequalities
trigonometry
inequalities unsolved

For $ p_1\in (0,1)$, $ p_2\equal{}1\minus{}p_1$, $ \alpha,\beta,\gamma \in [0,2\pi]$ and $ R_1,S_1,R_2,S_2 \in [0,1]$, prove that

\[ p_1p_2\left[R_1 S_2 \cos{\left(\beta\plus{}\gamma\right)}\plus{} R_2 S_1 \cos{\left(\alpha\plus{}\gamma\right)} \minus{} R_1 S_1 \cos{\gamma} \minus{} R_2 S_2 \cos{\left(\alpha\plus{}\beta\plus{}\gamma\right)}\right] \geq p_1\sqrt{1\minus{}R_1^2}\sqrt{1\minus{}S_1^2}\plus{}p_2\sqrt{1\minus{}R_2^2}\sqrt{1\minus{}S_2^2}\minus{}\sqrt{1\minus{}p_1^2R_1^2\minus{}p_2^2R_2^2\minus{}2p_1p_2R_1 R_2 \cos{\alpha}}\sqrt{1\minus{}p_1^2S_1^2\minus{}p_2^2S_2^2\minus{}2p_1p_2 S_1 S_2 \cos{\beta}}\]

Any help would be greatly appreciated!

This inequality arises from the same context as the inequality proposed in
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=189287[/url] (proved by solyaris -- Thanks!).

I believe the same approach would be useful here, but haven't been able to make progress so far.


[quote="miyomiyo"]An isosceles triangle has integer side lengths.  A cevian separates the base into two segments of length $ 41$ and $ 49$.  Find the smallest possible integer length of the cevian.[/quote]

[hide="solution"]
 [geogebra]4ad458dfb8a460965290c63d4150b6eba2637b8f[/geogebra]
Since $ ABC$ is isosceles, point $ A$ must lie on the perpendicular bisector of $ BC$. Therefore the smallest intiger distance of the cevian occurs when $ A$ is on the line $ BC$. Therefore $ \overline{AP}\equal{}49\minus{}45 \equal{} 4$

Unfortunately, we no longer have a triangle, just a line... so

Let $ AP\equal{}k, AB\equal{}AC\equal{}a, \rightarrow a,k \in \mathbb{Z}$ and$ AM\equal{}h, h\in \mathbb{R}$
Now we set up two equations
$ h^2 \plus{}45^2\equal{}a^2 \qquad(1)$
$ h^2 \plus{} 4^2 \equal{} k^2 \qquad(2)$

$ h$ does not have to be an intiger so
$ a^2 \minus{} 45^2 \equal{} k^2 \minus{}4^2$
$ a$ and $ k$ however, must be intigers and we have to minimise $ k$
Letting $ a\equal{}k\plus{}n$ we find

$ n(2k\plus{}n) \equal{} 45^2\minus{}4^2 \equal{} 41\cdot 49$
Now $ k$ will be minimised when the two factors $ n$ and $ (2k\plus{}n)$ are close together.
$ \Rightarrow n\equal{}41, 2k\plus{}n\equal{}49 \Longrightarrow k \equal{} 4$

But this is the same answer we got before!! so it cannot be correct  :(
The next solution is
$ n\equal{}7, 2k\plus{}n\equal{}287 \Rightarrow k\equal{}140$

So the smallest intiger length of the cevian is $ 140$ unit. :)

[/hide]


For positive real numbers $ a$, $ b$, $ c$ and $ d$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$  prove that

$ a^2b^2cd \plus{} \plus{}ab^2c^2d \plus{} abc^2d^2 \plus{} a^2bcd^2 \plus{} a^2bc^2d \plus{} ab^2cd^2 \le 3/32,$

and determine the cases of equality.

[quote="bnick17"]For positive real numbers $ a$, $ b$, $ c$ and $ d$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$  prove that

$ a^2b^2cd \plus{} \plus{} ab^2c^2d \plus{} abc^2d^2 \plus{} a^2bcd^2 \plus{} a^2bc^2d \plus{} ab^2cd^2 \le 3/32,$

and determine the cases of equality.[/quote]
$ a^2b^2cd \plus{} \plus{} ab^2c^2d \plus{} abc^2d^2 \plus{} a^2bcd^2 \plus{} a^2bc^2d \plus{} ab^2cd^2\equal{}$
$ \equal{}abcd(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)\leq\frac{1}{16}\cdot\frac{3}{2}\equal{}\frac{3}{32}.$
The equality occurs when $ a\equal{}b\equal{}c\equal{}d\equal{}\frac{1}{2}.$

[quote="GioMott"][quote="arqady"]$ abcd(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)\leq\frac {1}{16}\cdot\frac {3}{2}$[/quote]

Why? Did you get it via AM-GM or something similar?[/quote]
we have with AM-GM $ a^2\plus{}b^2\plus{}c^2\plus{}d^2 \ge 4\sqrt{abcd}$,and
$ \frac{a^2\plus{}b^2\plus{}a^2\plus{}c^2\plus{}a^2\plus{}d^2\plus{}b^2\plus{}c^2\plus{}b^2\plus{}d^2\plus{}c^2\plus{}d^2}{2} \ge (ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)$ ,
so $ abcd \le \frac{1}{16}$ and $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \le \frac{3}{2}$ ,multiplying we get the result.

This is really weak... (abcd) is weaker than SUM a^2, and SUM ab  is weaker than SUM a^2...


x=2pi/11  then find $cosx+cos2x+cos3x+cos4x+cos5x=?$

$\cos\frac{2\pi}{11}+\cos\frac{4\pi}{11}+\cos\frac{6\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{10\pi}{11}$
$=-\cos\frac{\pi}{11}+\cos\frac{2\pi}{11}-\cos\frac{3\pi}{11}+\cos\frac{4\pi}{11}-\cos\frac{5\pi}{11}$
You can arrive at this by looking at the reference angles.
Now consider the attached triangle, where the red segments have length $x$. The numbers are angle measures, in $\frac{\pi}{11}$ radians (1 means $\frac{\pi}{11}$, 2 means $\frac{2\pi}{11}$,...). Note that the triangle is isosceles. Then the sides must be congruent.
$x+2x\cos\frac{2\pi}{11}+2x\cos\frac{4\pi}{11}=2x\cos\frac{\pi}{11}+2x\cos\frac{3\pi}{11}+2x\cos\frac{5\pi}{11}$
$\Rightarrow -\cos\frac{\pi}{11}+\cos\frac{2\pi}{11}-\cos\frac{3\pi}{11}+\cos\frac{4\pi}{11}-\cos\frac{5\pi}{11}=-\frac12$

Check out my Paint skills.  :oops:

or u can solve it in complex number method

consider $x^{11}=1$ then $x=\cos\frac{2k}{11}\pi+i\sin\frac{2k}{11}\pi$ where $k=-5-4,...,4,5$

From Vieta Theorem , we know that

$\sum^5_{k=-5}\cos\frac{2k}{11}\pi=0$

$\implies 2\sum^5_{k=1}\cos\frac{2k}{11}\pi=-\cos 0=-1$    (using cos (-x) = cos x )

$\implies \sum^5_{k=1}\cos\frac{2k}{11}\pi=\frac{-1}{2}$

?

I don't understand what bodom is saying

[quote="maokid7"]i assume that u used sum-product to get to the last equality right?[/quote]

yep , i use $2\cos A\sin B=\sin (A+B)-\sin (A-B)$

[quote="1234567890"][quote="bodom"]or you can do the simplest thing:just multiply by 2sinx/2 and you'll get the result ;)[/quote]

Adding, not multiplying. ;)[/quote]
no,multiplying ;)

[quote="shyong"][quote="gotztahbeazn"]ive seen methods where u cleverly multiply by something such as $\sin \frac{2x}{11}$ to use the sum-product identities and then have the sum telescope nicely, but i dont think $\sin \frac{2x}{11}$ works in this case[/quote]

Letting $\frac{\pi}{11}=a$ then we want to find

$\cos 2a+\cos 4a +\cos 6a+\cos 8a+\cos 10a$

So multiply it with $\frac{2\sin a}{2\sin a}$ to get

$\frac{\sin 11a-\sin a}{2\sin a}=-\frac{1}{2}$ ( because $\sin 11a=\sin \pi=0$)[/quote]
that's just what i have said

[quote="gotztahbeazn"]ive seen methods where u cleverly multiply by something such as $\sin \frac{2x}{11}$ to use the sum-product identities and then have the sum telescope nicely, but i dont think $\sin \frac{2x}{11}$ works in this case[/quote]
it's not clever,it's well-known method and sin( pi/11) surely works in here


For any three positive reals $a, b, c$, we have
$\sum\frac{1}{2a^2+3a(b+c)+bc}\ge\frac{1}{ab+bc+ca}$

Replace $a,b,c$ with $1/a,1/b,1/c$ respectively, then use C-S Inequality.
Nice inequality, xtar! :lol:

[quote="leibniz"]:oops: I don't know if I m right but there is no role for [tex] \sum  [/tex] in this inequality!!! :?[/quote]
he means cyclic sum.


calculus function inequalities ratio induction logarithms algebra

$ n\in\mathbb{N}$, $ n\ge 3$.

can you prove this one:

$ n^n(n\minus{}2)^{n\minus{}2}>(n\minus{}1)^{2(n\minus{}1)}$

or this:

$ (n\minus{}2)^{n\minus{}2}(n\plus{}1)^{n\plus{}1}>n^{n\plus{}1}(n\minus{}1)^{n\minus{}2}$

 without taking log of both sides and differentiating obtained function $ f$ (difference of both sides) to see that $ f$ increasing and $ f(3)>0$ so inequality holds ?

The idea I mentioned works fine, but I'm interesting in a proof using some known inequalities, without differentiating

[b]Note:[/b] Write $ \frac {n^n}{(n \minus{} 1)^{n \minus{} 1}} > \frac {(n \minus{} 1)^{n \minus{} 1}}{(n \minus{} 2)^{n \minus{} 2}}$

Now, the statement is the same as saying that $ n^n$ has faster than exponential growth [i](the ratio of successive terms is increasing)[/i], which is intuitive.

Rigorizing that, take an $ n$ out of the LHS and divide out $ n\minus{}1$ from both sides so that we end up with \[ \frac{n}{n\minus{}1} \cdot \left( \frac{n}{n\minus{}1} \right)^{n\minus{}1} > \left( \frac{n\minus{}1}{n\minus{}2} \right)^{n\minus{}2}\]
\[ \frac{n}{n\minus{}1} \cdot \left( 1 \plus{} \frac1{n\minus{}1} \right)^{n\minus{}1} > \left( 1 \plus{} \frac1{n\minus{}2} \right)^{n\minus{}2}\]

You can prove using the extended binomial theorem that $ f(x) \equal{} (1\plus{}\frac1x)^x$ is strictly increasing, using the fact that if $ a > b$ then $ (a\minus{}k)/a > (b\minus{}k)/b$. This finishes the problem. In general the fact that the $ f$ I just gave is increasing will solve most problems like the ones you gave.

Thanks MellowMelon, nice proof! the only thing your proof needs is to know that the sequence $ (1\plus{}\frac{1}{n})^n$ is increasing, we don't have to use the function

We could also use that $ \frac {n^n}{(n \minus{} 1)^{n \minus{} 1}} \minus{} \frac {(n \minus{} 1)^{n \minus{} 1}}{(n \minus{} 2)^{n \minus{} 2}}$ approaches $ e$ as $ n$ goes to infinity.

[quote="TZF"][b]Note:[/b] Write $ \frac {n^n}{(n \minus{} 1)^{n \minus{} 1}} > \frac {(n \minus{} 1)^{n \minus{} 1}}{(n \minus{} 2)^{n \minus{} 2}}$[/quote]

It turns out induction is trickier to get to work than it looks. The problem comes from the fact that you have to produce a base one higher AND an exponent one higher. Also, if you look at the equivalent form TZF gave, the problem reduces to showing that $ f(n\plus{}1) > f(n)$ for positive integers $ n$ where $ f(x) \equal{} \frac {x^x}{(x \minus{} 1)^{x \minus{} 1}}$. In general it is not easy to prove that functions are increasing by induction, at least compared to other methods, unless the function is recursive in nature.

[u]Two things:[/u]

[b](1)[/b] The following doesn't suffice as a proof, but it can be made more rigorous to show that $ f(x)$ is increasing for sufficiently large (which isn't actually that large) $ x$.  I know Stirling's Approximation is a calculus result, but it is helpful to know:

$ x! \approx \sqrt {2 \pi x} \left( \frac {x}{e} \right) ^x \implies x^x \approx \frac {x! e^x}{\sqrt {2 \pi x}}$

$ \therefore f(x) \equal{} \frac {x^x}{(x \minus{} 1)^{x \minus{} 1}} \approx \frac {x! e^x / \sqrt {2 \pi x}}{(x \minus{} 1)! e^{x \minus{} 1} / \sqrt {2 \pi (x \minus{} 1)}} \equal{} ex \sqrt {\frac {x}{x \minus{} 1}}$

This result implies what [b]nayel[/b] claimed in post #5.

[b](2)[/b] The log can be written as $ (x \minus{} 1) \log \left( 1 \plus{} \frac {1}{x \minus{} 1} \right) \plus{} \log x$.

The second term is obviously increasing. The first can be shown to be increasing with calculus, but I'd think we could use some non-calculus approach. In particular, if we could show that $ \log (1 \plus{} y) \approx y$ for $ y \approx 0$, then we'd be all set.

You can use the weighted AM-GM inequality to prove $ (1\plus{}\frac{1}{n\plus{}1})^{n\plus{}1}$ > $ (1\plus{}\frac{1}{n})^n$


How can i compute this limit :
$ \lim_{ n->{\infty}} \sum_{k=1}^{n} \frac{k}^{(k+1)!} $

You can use telescoping method, nemesis :)

Let's determine the value of $a,b$ such that $\frac{k}{(k+1)!}=\frac{a}{k!}+\frac{b}{(k+1)!}$

now i understand :
$ \lim_{ n->{\infty}} \sum_{k=1}^{n} \frac{k}^{(k+1)!}$ = $ \lim_{ k->{\infty}}  1-\frac{1}{(k+1)!}  $
and the limit is '1'.

But I don't understand how did you know to use
$ \frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!} $


Suppose $ S$, a Lebesgue measurable set, has the following property: $ x, y \in S, x \neq y$ implies $ \frac{x\plus{}y}{2} \notin S$. Show that $ \lambda(S) \equal{} 0$.

[quote="PluTo"]Suppose $ S$, a Lebesgue measurable set, has the following property: $ x, y \in S, x \neq y$ implies $ \frac {x \plus{} y}{2} \notin S$. Show that $ \lambda(S) \equal{} 0$.[/quote]

If $ \lambda(S) > 0$, then $ S$ contains a point of density, which you can assume is $ 0$. Now consider $ S\cap\minus{}S$.


LaTeX
limit
logarithms
function
calculus
calculus computations

[quote="mathbeauty"]Evaluate lim of {(x(x^3 - 2y^2) + y^2 * (3 - 2x))/(x^4 + 3y^2)}^{x^3/(3y^2) + 1/x} as x, y -> 0, if
 the limit exists.[/quote]
I will not give you the answer, but consider two different linear paths focusing on the first term.

In $ \text{\LaTeX}$ just to make sure I've read it correctly:
\[ \lim _{(x,y)\to (0,0)}\left(\frac {x(x^3 - 2y^2) + y^2(3 - 2x)}{x^4 + 3y^2}\right)^{\frac {x^3}{3y^2} + \frac {1}{x}}
\]
You can find what the limit must be ([u]if[/u] it exists, which in this case is the more difficult question to resolve) with Mathstud28's method by looking, for example, at the limit along the line $ y = x$, or any other line (the limit happens to be $ e^{ - \frac {4}{3}}$).  However, this does not prove that the limit exists.

i understand that the limit must be the same along all paths in order for it to exist, but how did you get that the limit is e^-4/3 along the line y=x?

Just set $ x\equal{}y$ and use single-variable techniques. For any exponential limit, taking a logarithm first usually helps.

The first problem to deal with is that along the $ x$ and $ y$ axes, the function is undefined as a literal $ 1^{\infty}$. That already breaks the usual definition, since it isn't defined on a punctured neighborhood of the origin. We still have some faint chance that the limit exists when restricted to the domain.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Calculate
$ \int \frac{1}{(2\minus{} \sin x)(3 \minus{} \sin x)}\,dx.$

$ \int \frac{1}{\cos^8x \minus{} \sin^8x}\,dx.$

 $ \int \frac{\sin x \minus{} \sin^3x}{1 \minus{} \sin^3x}\,dx.$

Hi aktyw19.

Just spotted these. For the first one, I'd try the substituion $ u \equal{} \tan (\frac {x}{2})$ and for the others I'm thinking that factorising might be a start. Sorry if this is not much help.

hello,for the first one, setting $ z\equal{}\tan(x/2)$ and $ \sin(x)\equal{}\frac{2z}{1\plus{}z^2}$ we get the integral
$ \int\frac{1\plus{}z^2}{(1\minus{}z\plus{}z^2)(3\minus{}2z\plus{}3z^2)}\,dz$
Sonnhard.

ok thanks
and $ \int \frac{\sin x \minus{} \sin^3x}{1 \minus{} \sin^3x}\,dx.$?

hello, to the second integral show that
$ \cos^8(x)\minus{}\sin^8(x)\equal{}4\cos(x)^6\minus{}6\cos(x)^4\plus{}4\cos(x)^2\minus{}1$
and set $ dx\equal{}\frac{2dz}{1\plus{}z^2}$ and $ \cos(x)\equal{}\frac{1\minus{}z^2}{1\plus{}z^2}$
Sonnhard.

hello, for the third integral setting $ \sin(x)\equal{}\frac{2z}{1\plus{}z^2}$ and $ dx\equal{}\frac{2dz}{1\plus{}z^2}$ we get the integral
$ \int\frac{4z(1\plus{}z)^2}{(1\plus{}z^2)(1\plus{}2z\plus{}6z^2\plus{}2z^3\plus{}z^4)}\,dz$
Sonnhard.


Can anybody help me how to learn geometry from basics to higher level - i have got only a month to prepare to a future examination. Please tell me which books to study. Thank you. :!:

[hide="Some books are..."]
1)Plane Geometry by Sharygin
2)Plane Trigonometry by S.L.Loney
3)Geometry Unbound by Kiran Kedlaya
4)Geometry Revisited by Coxeter and Greitzer[/hide]
Out of these i've only used the 2nd and 3rd....but have heard quite a lot about the other two....

thank you, i had all of these books, but (2) on my HDD, but didn't notice them. Thanx, "I'll TRY".  :)


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calculus
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real analysis
calculus computations

Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on [0, 4].

lim                      4/n  (  root(4/n) + root(8/n) + ... + root(4n/n))
n->oo

how do i do this?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=369679316&t=71992]Here[/url] is a related thread. The general principles for thinking about Riemann sums are in there.

i think i understand that topic, but in this one, i dont get what the formula is

i know the xi = 4i/n  and xn = 4/n,,, then u sub xi in x for the equation, and multiply by xn, and solve the limit with integrals, but i cant get what the basic functions is, is it 4/n ?, and how do i use the statement in the brackets?

A Riemann sum of the function $ f(x)$ over $ [a,b]$ is

$ \lim_{n \rightarrow \infty} \frac{b\minus{}a}{n} \sum_{k\equal{}1}^n f\left(a \plus{} \frac{b\minus{}a}{n} k\right) \equal{} \int_{x\equal{}a}^b f(x) \, dx$.

In this case, $ a \equal{} 0, b \equal{} 4$, and we choose $ f(x) \equal{} \sqrt{x}$:

$ \lim_{n \rightarrow \infty} \frac{4}{n} \sum_{k\equal{}1}^n \sqrt{\frac{4k}{n}} \equal{} \int_{x\equal{}0}^4 \sqrt{x} \, dx$.

The left-hand side is equal to the given sum.  Therefore, the desired value is found by evaluating the integral:

$ \int_{x\equal{}0}^4 \sqrt{x} \, dx \equal{} \left[\frac{2}{3} x^{3/2} \right]_{x\equal{}0}^4 \equal{} \frac{16}{3}$.