Let $ x_n$ be the unique positive solution of the equation$ x^n\minus{}nx\minus{}1\equal{}0$ wehere $ n \geq 2$.
Find the limit of the sequence $ x_n$.

Clearly, $ \forall n: \,x_n>1$. From $ x_n\plus{}nx_n>\underbrace{1\plus{}nx_n}_{\equal{}x_n^n}>nx_n$ we conclude $ (n\plus{}1)^{\frac{1}{n\minus{}1}}>x_n>n^{\frac{1}{n\minus{}1}}$, therefore the limit of the convergent series $ x_n$ is $ 1$.


greatest common divisor
inequalities proposed
inequalities

Let $ a_1,a_2,...,a_n$be positive integers such that the following two properties hold:
1) $ a_1 \le 9999 ( i\equal{}1,2,...,n)$
2) The greatest common divisor of any pair of numbers is greater than $ 9999$.
Prove that: $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}<2$
 :lol:

To mod: Please lock this topic because this problem is from Mathematics And Youth Magazine in Viet Nam. The deadline is 15-05-2008.


geometry
3D geometry
sphere
trigonometry
vector

Here's a very nice problem:

- A sphere of radius $R$ is fixed on an horizontal plane. A smaller sphere, with radius $r$, initially at rest on the top of the larger sphere, is abandoned and rolls without slip on the surface of the larger sphere. The radii of the spheres are related according to $r = 0.31R$. Compute the height $h$, measured from the horizontal plane to the center of the small sphere, at which the small sphere looses contact with the surface of the large sphere.

Indeed, this is a classic, and I think we had it on the forum some time ago. Anyway, by conservation of energy, \[\frac{mv^{2}}{2}+\frac{I\omega^{2}}{2}= mg(R+r)(1-\cos\phi),\] or \[\frac{7v^{2}}{10}= g (R+r) (1-\cos\phi),\] where $\phi$ is the angle between the radius vector of the ball and the vertical, with the center of the bigger sphere taken as the origin. The sphere leaves the surface once the normal reaction force, given by \[N = mg\cos\phi-mv^{2}/(R+r),\] becomes zero. So \[\cos\phi = \frac{10}{7}(1-\cos\phi) \Rightarrow \cos\phi = \frac{10}{17}.\] The height at which this happens is \[H = (r+R) (1-\cos\phi) = \frac{7}{17}(r+R) \approx 0.54 R.\]

Very nice, but the answer is $h = 1.77R$.

Huh? What height is that? Mine is measured with respect to the center of the fixed ball. Anyway, the $7/17$ factor is correct, so any mistake I made must've come from my wrong interpretation of what height we're looking for...

[quote="Djole"]Mine is measured with respect to the center of the fixed ball.[/quote]

Then, $H = (r+R)\cos\phi$.

Yes, you're right  :oops: Than it's ok :)


[quote="goldendomer"]Here is my top ten:

1. Ohio State
2. Michigan
3. West Virginia
4. Louisville
5. Texas
6. Notre Dame
7. USC
8. California
9. Auburn
10. Florida

What are you guys' rankings?[/quote]

you should just post this college football topic.

After all, two mid-season college football topics is too much.


Let $ A$ be a real square matrix with order $ n$ ,$ A^n$=$ 0$ and $ A^{n\minus{}1}$ not equal $ 0$.Prove that there exist a square matrix $ P$ such that $ P^{\minus{}1}AP\equal{}B$ where $ B$=$ [b_{ij}]$  $ b_{ij}\equal{}1$  if $ j\equal{}i\plus{}1$ and equal $ 0$ with other positions.

[quote="minhthutt"]Let $ A$ be a real square matrix with order $ n$ ,$ A^n$=$ 0$ and $ A^{n \minus{} 1}$ not equal $ 0$.Prove that there exist a square matrix $ P$ such that $ P^{ \minus{} 1}AP \equal{} B$ where $ B$=$ [b_{ij}]$  $ b_{ij} \equal{} 1$  if $ j \equal{} i \plus{} 1$ and equal $ 0$ with other positions.[/quote]

Let T be a linear transformation corresponding to A for some basis U.

Obviously there exists $ u \in U$ such as $ T^{n \minus{}1}u \not\equal{} 0$ while $ T^n\equal{}0$

First we will prove that the set {$ {u, Tu, T^2u,...T^{n\minus{}1}u}$} forms a basis of U.

We have:
$ c_1u\plus{}c_2Tu\plus{}...c_nT^{n\minus{}1}u\equal{}0 \rightarrow T(c_1u\plus{}c_2Tu\plus{}...c_nT^{n\minus{}1}u)\equal{}0 \\
\rightarrow c_1Tu\plus{}c_2T^2u\plus{}...c_{n\minus{}1}T^{n\minus{}1}u\equal{}0$

Repeating this we get $ c_1T^{n\minus{}1}u\equal{}0$ which yields $ c_1\equal{}0$. A simple substitution gives $ c_i\equal{}0 \forall i\equal{}1,..n$

Finally notice that the matrix of T for basis {$ {T^{n\minus{}1}u, T^{n\minus{}2}u,...Tu,u}$} is the required one since:

$ T(T^i)\equal{}T^{i\plus{}1}$ meaning the ith column of the matrix is zero except for the element in row i-1..


An equilateral triangle is inscribed in a circle. The perimeter of the triangle is 36 inches. How many square inches are in the area of the circle. (Express your answer in terms of $ \pi$.)

[asy]size(200);

pair o=(0, 2*sqrt(3)), a=(-6,0), b=(6, 0), c=(0, 6*sqrt(3));

draw (a--b--c--cycle);
draw (a..b..c..cycle);
draw (a--o--c, dashed+1);
draw (o--midpoint(a--b), dashed);

label ("$12$", midpoint(a--b), S);
label ("$12$", midpoint(b--c), NE);
label ("$12$", midpoint(c--a), NW);
label ("$6$", midpoint(a--midpoint(a--b)), N);
label ("$6$", midpoint(midpoint(a--b)--b), N);
label ("$4\sqrt{3}$", midpoint(o--c), E);
label ("$2\sqrt{3}$", midpoint(o--midpoint(a--b)), E);
label ("$4\sqrt{3}$", midpoint(o--a), N);[/asy]
Thus, $ (4\sqrt{3})^2\pi \equal{} \boxed{48\pi}$.

How did you know the fraction of the altitude was 4 sqrt(3)?

Because we bisected the angle with $30$ degrees and then use the $30-60-90$ triangle ratio.


Points $ A$ and $ B$ are midpoints of the sides of the square shown. What is the ratio of the area of the shaded region to the area of the square? Express your answer as a common fraction.
[asy]import olympiad; size(75); import geometry_dev; import graph; defaultpen(linewidth(0.8));
draw(unitsquare);
fill((0,0.5)--(0.5,1)--(0,1)--cycle);
label("$B$", (0,0.5),W); label("$A$", (0.5,1),N);[/asy]

You could let the side of the square be $ 2s$, but in my opinion, it's easier to let it be something like 10. That means the shaded triangle has legs of 5 and 5, so it's area is $ 12.5$. Since the total area of the square is 100, we have $ \frac{1}{8}$.


calculus
integration
symmetry
geometry
calculus computations

Compute the integral: $\int_{0}^{1} x\sqrt{x-x^{2}} dx.$

Here's a method:

The first thought is that $\sqrt{x-x^2}$ has to do with the equation for a circle. The center lies at $x=\frac12.$ More specifically, we can complete the square to get that $x-x^2=\frac14-\left(x-\frac12\right)^2.$

So I do a "little" substitution just to make it easier for me to see what is going in. This isn't necessary, but on a human scale, very useful.

Let $u=2x-1$, so $x=\frac{u+1}2,\,dx=\frac{du}2.$

$\int_0^1x\sqrt{x-x^2}\,dx=\int_{-1}^1\frac{u+1}2\sqrt{\frac14 -\left(\frac u2\right)^2}\,\frac{du}2$

$=\frac18\int_{-1}^1(u+1)\sqrt{1-u^2}\,du.$

We can finish this quickly. $\int_{-1}^1u\sqrt{1-u^2}\,du=0$ by symmetry, and $\int_{-1}^1\sqrt{1-u^2}\,du$ is half the area of a circle of radius 1, thus $\frac{\pi}2.$

This gives kunny's answer.


algebra
polynomial
geometry
rectangle
algorithm
linear algebra
matrix

Let a_n be the number of ways to form an integer with the digits 1,3,4 such that the sum of the digits of that integer is n. Prove that a_{2n} is always a square

[hide="what I have"]
1 1
2 1
3 2
4 4
5 6
6 9
7 15
8 25
9 40
10 64

I think the pattern is $ a_{2n}\equal{}F_n^2$. Also, I think $ a_{2n \plus{} 1}^2 \equal{} a_{2n}a_{2n \plus{} 2}$ and many other recursive patterns,[/hide] but I have no clue how to prove this. Can anyone give me a hint?

Show that $ a_{n \plus{} 4} \equal{} a_{n \plus{} 3} \plus{} a_{n \plus{} 1} \plus{} a_n$.  What are the roots of the polynomial $ x^4 \equal{} x^3 \plus{} x \plus{} 1$?

Alternately, it should be possible to do this directly.  Show that $ F_n$ (with the numbering $ F_1 \equal{} 1, F_2 \equal{} 2$, which is how you are doing it but which is not standard) is the number of ways to form an integer with the digits $ 1, 2$ such that the sum of the digits is $ n$.  How can you combine two such representations to find a number with digits $ 1, 3, 4$ whose sum of digits is $ 2n$?  (This is not obvious.)

Alternately alternately, figure out a recursion that $ F_n^2$ satisfies and show that $ a_{2n}$ satisfies the same recursion.

oh i see thanks. Casework on the leftmost digit (either 1,3,4) gives that recursion and then we can explicit formula bash.

we can also do the casework method to prove the fibonacci recursion of 1's and 2's, but making into a 1,3,4 digits number part eludes me

Can you use Characteristic Equations with Complex Roots?

not sure....I tried it with $ x^2 \plus{} x \plus{} 1$, but i think i messed up the algebra

[hide="Direct method - bijection using the tiling model"]$ F_n$ is the number of tilings of a $ 1 \times n$ rectangle with $ 1 \times 1$, $ 1 \times 2$ tiles, and $ a_n$ is the number of tilings of a $ 1 \times n$ rectangle with $ 1 \times 1$, $ 1 \times 3$, $ 1 \times 4$ tiles. We want $ a_{2n} \equal{} F_n^2$.

Consider arbitrary tilings of two $ 1 \times n$ rectangles $ A$ and $ B$ with squares and dominoes. There are $ F_n^2$ ways to do this. Begin with an empty $ 1 \times 2n$ rectangle $ C$ which we will tile with squares, trominoes, and tetrominoes (all $ 1 \times ?$ shaped). Starting with the first tile in each, perform the following algorithm.

1. If both $ A$ and $ B$ have squares for the current tiles, put two squares in $ C$. Move to the next tiles in $ A$ and $ B$.

2. If both $ A$ and $ B$ have dominoes for the current tiles, put a tetromino in $ C$. Move on.

3. If $ A$ has a domino but $ B$ has a square, put a tromino in $ C$. Look at the next tile in $ B$.
3.1. If it is a square, put a square in $ C$ and move on.
3.2. If it is a domino, look at the next tile in $ A$.
3.2.1. If it is a square, put a tromino in $ C$ and move on.
3.2.2. If it is a domino, put a tetromino in $ C$ and look at the next tile in $ B$, going back to 3.1.

4. If $ A$ has a square but $ B$ has a domino, put a square in $ C$. Look at the next tile in $ A$.
4.1. If it is a square, put a tromino in $ C$ and move on.
4.2. If it is a domino, put a tetromino in $ C$, then look at the next tile in $ B$.
4.2.1. If it is a square, put a square in $ C$ and move on.
4.2.2. If it is a domino, look at the next tile in $ A$, going back to 4.1. (don't add anything yet)

Proving this is an injection (thus a bijection) is pretty blah, so I'll leave that for you to take on faith.

Despite the complicated details, the algorithm is based on a simple concept. You basically line up your two tilings $ A$ and $ B$ and consider sections between where a tile boundary in $ A$ is matched by a tile boundary in $ B$. For instance, in this tiling

[code]A: DDD.S|S|DDD|S.DDD.DDD|S.S
B: S.DDD|S|DDD|DDD.DDD.S|DDD[/code]
the | marks the sections I'm referring to. There's four ways you can begin each section, since each board can begin the section with either a square or a domino. If the tiles are the same it's fairly obvious. If not, your section has to have two squares on each end, and then offset dominoes everywhere else. You basically just snake around the tiling from the left square to the other. In the fourth section from the above example, we would do
[code]
A: S.DDD.DDD
.. V >>V ^>>
.. V ^ V ^ V
.. >>^ V>> V
B: DDD.DDD.S[/code]

Then you just take the pairs of dominoes you hit to be tetrominoes, and merge the first domino with the square (making a tromino) based on whether or not the first square was in $ A$ or $ B$. In the algorithm above we do this merging if $ B$ had the initial square. Finally, if there's a domino left over merge it with the last square to make a tromino, otherwise the last square stays a square.

(as a final note: the example A and B tiling above gives 33114144113)[/hide]

Very nice!

[hide="Some comments"] $ F_n$ describes walks in a graph with adjacency matrix $ \left[ \begin{array}{cc} 1 & 1 \\
1 & 0 \end{array} \right]$.  A walk from the first vertex to the second vertex and back is a $ 1 \times 2$ tile and a walk from the first vertex to itself is a $ 1 \times 1$ tile.  Now I claim that the [url=http://en.wikipedia.org/wiki/Tensor_product_of_graphs]tensor product[/url] of this graph with itself, which has adjacency matrix

$ \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \end{array} \right],$

has walks described by $ F_n^2$ (and a few related sequences).  This is because walks on a tensor product correspond to 'simultaneous' walks on the two graphs that make up the product.  (This is a pretty quick proof that the pointwise product of two sequences defined by linear recurrences is also defined by linear recurrences.)  On the other hand, the natural graph to talk about $ a_n$ on is the [i]directed[/i] graph with adjacency matrix

$ \left[ \begin{array}{cccc} 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \end{array} \right]$.

Is it obvious that the square of the second matrix is similar to the first?  Is the similarity given by a matrix that encodes a bijection? [/hide]


Don't you think trig is a bit hard for the Middle School forums?

uhh..that is written in a differnt language...I CANNOT do it.

I don't think trig is really needed for a simple equation like this?
Or am I overlooking somthing?

 :surf:

I think there's an easier way to do it.  The numerator and denominator of the first fraction cancel out!

We now have $ 1/x^2 \plus{} x^2 \minus{}5 \equal{}0$ or $ 1/x^2 \plus{} x^2 \equal{} 5.$.

There must be a way to easily work with an equation like that, but I've read Chapter 22 in the Intro to Algebra Textbook and haven't found it out yet.

[quote="maybach"]I think there's an easier way to do it.  The numerator and denominator of the first fraction cancel out![/quote]
Could explain that? How do they cancel?

oh gees.  Never mind.  I was thinking we were multiplying the fractions.  Oops :blush: I have no idea how to solve it now.

hello, the most importend step is to show that
$ \frac{(1\plus{}x)^2}{x^2}\plus{}\frac{1}{(1\plus{}x)^2}\plus{}x^2\minus{}5\equal{}\frac{(\minus{}1\minus{}2x\plus{}x^2\plus{}x^3)^2}{x^2(1\plus{}x)^2}$
Sonnhard.

I think I did something wrong... I got the square root of 2

The most important step is to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=234326]search.[/url]

The $ \text{cis}$ thing used to work, but then it didn't anymore, so sorry.

Getting a common denominator:

$ \frac {x^4 + 4x^3 + 6x^2 + 4x + 1 + x^2 + x^6 + 2x^5 + x^4 - 5x^4 - 10x^3 - 5x^2}{x^2(x + 1)^2} = 0$

$ \frac {x^6 + 2x^5 - 3x^4 - 6x^3 + 2x^2 + 4x + 1}{x^2(x + 1)^2} = 0$

$ \frac {(x^3 + x^2 - 2x - 1)^2}{x^2(x + 1)^2} = 0$

Thus we need to find solutions to $ x^3 + x^2 - 2x - 1 = 0$

Letting $ x = y + \frac {1}{y}$

$ \left(y + \frac {1}{y}\right)^3 + \left(y + \frac {1}{y}\right)^2 - 2\left(y + \frac {1}{y}\right) - 1 = 0$

$ y^3 + 3y + \frac {3}{y} + \frac {1}{y^3} + y^2 + 2 + \frac {1}{y^2} - 2y - \frac {2}{y} - 1 = 0$

$ y^3 + y^2 + y + 1 + \frac {1}{y} + \frac {1}{y^2} + \frac {1}{y^3} = 0$

Multiplying by $ y^4 - y^3$ ,

$ y^7 - 1 = 0$

Thus our solutions are the primitive roots of unity, i.e. $ y = \text{cis}\left(\frac {2k\pi}{7}\right)$

Substituting back into our $ x$ ,

$ x = y + \frac {1}{y}$

$ x = \text{cis}\left(\frac {2k\pi}{7}\right) + \left( \text{cis}\left(\frac {2k\pi}{7}\right)\right) ^{ - 1}$

By Demoivre’s Theorem,

$ x = \cos{\left(\frac {2k\pi}{7}\right)} + \sin{\left(\frac {2k\pi}{7}\right)} + \cos{\left( - \frac {2k\pi}{7}\right)} + \sin{\left( - \frac {2k\pi}{7}\right)}$

As $ \cos{ - a} = \cos{a}$ and $ \sin{ - a} = - \sin{a}$,

$ x = 2\cos{\frac {2k\pi}{7}}$

Also, as $ \cos{2\pi - a} = \cos{a}$ the 3 distinct values of $ x$ are $ 2\cos\frac{2\pi}{7}$ , $ 2\cos\frac{4\pi}{7}$ , and $ 2\cos\frac{6\pi}{7}$ which match up with Kouichi’s answers.

I would recommend that this thread be moved to a more suitable forum.

Hello, this is not the Olympiad forum. Please do not follow my example in the link, and hide your solution.

By the way, this should indeed be moved to Pre-Olympiad. It is too elegant to be in the Middle School forum.


x^y+y^z+z^x=1 Solve th equation for
a) x,y,z are positive real numbers
b) x,y,z are integers

1) This has nothing to do with geometry.

2) Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=256512[/url]


This might be a random topic, but hows the weather in your country/region? From where I am..its snowing!!! :D (just might cancel school..estimated 5-10 inches :))

oh yeah 3x, i live in Massachusets, so i guess we don't

splintercell, you got 2 topics on the same stuff (there is a duplicate one!!!), please delete one as it is considered SPAM. umm I'm hungry, I like SPAM's canned ham  :D  and yeah it is snowing here.

Are we living in the same area splintercell? New Jersey USA here. bye

Another Masschusetts guys here. Are you one of those Lexington kids splintercell?

[quote="splintercelll990"]actually nope, im not i live in shrewsbury...but if you live in lexingtion you have one of the best high schools (they win at like every math competition :D-WPI, WOCOMAL..everything including the annual Harvard-MIT math competition)[/quote]

Naw, I'm not from Lexington either.

Since this is a USA dominated post . . . I'll post too.  Ummmm . . . Maryland is a two part harmony . . . we get lots of snow in the winter and really hot weather in the summer.

yeah i guess you are right...:D only USA is posting (anyway, i have today off as school's closed  :D  :D  :D ). Anyway, it seems like maryland is somewhat similar to Massachusets-here we have a very cold winter, a hot summer but that does not always happen like that-most of the time it is very volatile :?

[quote="grobber"]Time to break the USA barricade around this topic :D:D

Winter is almost gone here. I still have to wear a hat, since my ears are pretty sensitive to cold, but basicly we have a nicew mild winter end.[/quote]

thats pretty good-how cold can it/does it get in Romania? Hows your summer there?

it's now spring here.
Temperature ranges from 17-18C to 24-25C
Usually sunny, while sometimes there is drizzle

nice :)...spring in the USA "officialy" starts this saturday or sunday

Please register or at least say who you are. I just don't like 'Guest' postings. :(  :(  :(

[quote="Anonymous"]Splintercell: I go to St. John's... do you go to Shrewsbury High?  Do you do ARML?[/quote]

who is this guest, and when did I say I go to Saint John's?

and by the way grobber,

[quote="grobber"]Summers can get pretty hot. In the last few years there have been some sumers with temperatures up to 40-42 degrees. There were cases of people fainting or even worse. This however, doesn't happen that often. Winters aren't that cold either.[/quote]

thats pretty cool, here the summers can get really hot (sometimes 65, 70 degrees in mid july :D)

That was me - i forgot to log on.

You can't go to Shrewsbury High, cause they had a 2 hour delay.  So it's most likely Mass Academy or St. John's, and i'm pretty sure you don't go to St. John's.


geometry
geometric transformation
rotation
complex numbers

[quote="b-flat"][hide="am I missing anything?"]Isn't this just he basic definition of rotation?[/hide][/quote]
I think you have assumed that $O$ is on $\ell$, but that is not necessarily the case.

Yes, but all that means is that you have to extend the lines with parrallel lines...

You could do a nice proof with complex numbers, but that would be beyone HSB.

Draw the perpendicular from $O$ to $A$ and let $A'$ be the rotation of $A$ about $O$ through an angle of $\alpha$. From the quadrilateral, we find that the obtuse angle formed by the intersection of the two lines is $360-90-90-\alpha=(180-\alpha)^{\circ}$. The acute angle formed by the intersection of the two lines is thus $\alpha$.

[quote="jli"][quote="rnwang2"]Show that if a line $\ell$ is rotated about any point $O$ through an angle with measure $\alpha$ to a new line $\ell '$, then lines $\ell$ and $\ell '$ intersect in an angle with measure $\alpha$.[/quote]

please site the source nextime.  i believe it was from march 1993 mandelbrot team round part 4...i think? :ninja:[/quote]

uhh...this is a pretty standard problem...the mandelbrot people did not really make this up, they just took the classic problem and put it on their contest


A problem that I found in my math book that I couldn't solve...

What is the sum of all the real numbers [i]x[/i] that satisfy

((2^x)-4)^3 +((4^x)-2)^3 =((4^x) + (2^x) -6)^3     ?

(A)3/2

(B)2

(C)5/2

(D)3

(E)7/2

Hint: the equation takes the form $ a^3\plus{}b^3\equal{}(a\plus{}b)^3$, which is equivalent to $ ab(a\plus{}b)\equal{}0$.

Putting it in latex:
$ (2^{x} \minus{} 4)^{3} \plus{} (4^{x} \minus{} 2)^{3} \equal{} (4^{x} \plus{} 2^{x} \minus{} 6)^{3}$

[hide]This can be rewritten to $ (2^x \minus{} 4)^3 \plus{} (4^x \minus{} 2)^3 \equal{} (2^x \minus{} 4 \plus{} 4^x \minus{} 2)^3$
Using [b]pluristiq[/b]'s hint, set $ 2^x \minus{} 4 \equal{} a$ and $ 4^x \minus{} 2 \equal{} b$ to obtain $ a^3 \plus{} b^3 \equal{} (a \plus{} b)^3$.
$ a^3 \plus{} b^3 \equal{} a^3 \plus{} 3a^2b \plus{} 3ab^2 \plus{} b^3$
$ 3ab(a \plus{} b) \equal{} 0$ or $ ab(a \plus{} b) \equal{} 0$. This implies that either $ a$ or $ b$ is $ 0$, or $ a \equal{} \minus{} b$.
Solving each case results in:
$ 2^x \equal{} 4$, $ x \equal{} 2$
$ 4^x \equal{} 2$, $ x \equal{} \frac {1}{2}$
$ 2^x \minus{} 4 \equal{} 2 \minus{} 4^x$, $ 2^{2x} \plus{} 2^x \minus{} 6 \equal{} 0$, $ (2^x \plus{} 3)(2^x \minus{} 2) \equal{} 0$, so disregard $ 2^x \equal{} \minus{} 3$ and we get from this that $ x \equal{} 1$. Thus the sum is $ \frac{5}{2}$ $ C$
[/hide]

the answer is 7/2, i actually tried to solve the problem, it takes a combination of the binomial theorem and other things.... You forgot to add the "1" in your answer...


modular arithmetic
number theory
greatest common divisor
relatively prime
number theory unsolved

Suppose that positive integers $x$, $y$ satisfy the equation $2x^{2}-1=y^{15}$. Prove that if $x>1$ than $x$ is divisible by $5$.

[hide]
Clearly $y$ is odd. The problem can then be factored into
\[x^{2}=\left( \frac{y+1}{2}\right) (y^{4}-y^{3}+y^{2}-y+1)(y^{10}-y^{5}+1)\]
Now if $y\equiv-1\pmod{5}$, then we're done. Otherwise, assume there is a solution, and note the following:

-The remainders on dividing $y^{4}-y^{3}+y^{2}-y+1$ and $y^{10}-y^{5}+1$ by $y+1$, respectively, are 5 and 3 (just plug in -1 into the polynomials). Therefore the first and second factors are relatively prime (as $y\not\equiv-1\pmod{5}$), and the gcd of the first and third factors divide 3.

-Observe that
\[y^{10}-y^{5}+1=(y^{5}+1)^{2}-3(y^{5}+1)+3\]
so the gcd of the second and third factor divides 3. On the other hand, the second factor is never divisible by 3, so they are relatively prime.

Now if $y\equiv-1\pmod{3}$ then $\frac{y+1}{2}$ and $y^{10}-y^{5}+1$ are both divisible by 3, so that
\[\frac{y+1}{6},\ y^{4}-y^{3}+y^{2}-y+1,\ \frac{y^{10}-y^{5}+1}{3}\]
are pairwise relatively prime and therefore all squares. Otherwise,
\[\frac{y+1}{2},\ y^{4}-y^{3}+y^{2}-y+1,\ y^{10}-y^{5}+1\]
are pairwise relatively prime and therefore all squares. Either way, $y^{4}-y^{3}+y^{2}-y+1$ is a square. But then
\[\left( y^{2}-\frac{y}{2}\right)^{2}=y^{4}-y^{3}+\frac{y^{2}}{4}<y^{4}-y^{3}+y^{2}-y+1\]
as $3y^{2}-4y+4=2y^{2}+(y-2)^{2}>0$. Therefore, we have
\[\left( y^{2}-\frac{y}{2}+\frac{1}{2}\right)^{2}\leq y^{4}-y^{3}+y^{2}-y+1\Rightarrow (y-1)(y+3)\leq 0\]
But if $x>1$, then $y>1$ so $(y-1)(y+3)>0$, a contradiction. Therefore $y\equiv-1\pmod{5}$ and 5 divides $x$, as desired.
[/hide]

I participated in this olympiad as guest from Bulgaria and my solution  is similar to the one above. THe interesting thing about this equation is that it has no solution with x>1.


Show that for any $\triangle{ABC}$, $\cot{A}+\cot{B}+\cot{C}\geq\sqrt{3}$.

umm, no, Jensen only works for the acute triangles.
[hide="this hint should work"]
look at $\cot A\cot B+\cot B\cot C+\cot C\cot A$
[/hide]

Oh oops...somehow I was not visualizing the graph of cot correctly.


What is the probability of rolling three number cubes, each with digits 1-6, and getting the same number on top of all three cubes? Express your answer as a common fraction.

What number you roll on the first cube doesn't matter, it's just that the other two cubes have to match that number, so the answer is simply $ \frac{1^2}{6^2}\equal{}\frac{1}{36}.$

[hide="my answer i think"]teh 1st # does not really matter it can be anything from 1-6

but the next 2 do

1^2/6^2=1/36

so our answer is 1/36 or $\boxed{1/36}$[/hide]

[b][i][u]please rate my post thanks![/u][/i][/b]

please be fair though [=

 :welcome:  :welcome:  :welcome:  :welcome:


inequalities
logarithms
inequalities proposed

Let $h_{a},h_{b},h_{c}$ be three altitude of triangle $ABC$ with side $a,b,c$, resp.Prove that
$(\frac{h_{a}^{t}+h_{b}^{t}+h_{c}^{t}}{3})^\frac{1}{t}\ge \frac{\sqrt{3}}{2}(\frac{a^{t}+b^{t}+c^{t}}{3})^\frac{1}{t}$
hold iff $\frac{-\ln 4}{\ln 4-\ln 3}\le t\le\frac{\ln 4}{\ln 4-\ln 3},t\not=0$

[color=purple]May be it's in Crux and was proposed by Walther Janous.If you need solution i'll post it.[/color]

ok I need a solution, post it if you can !

[color=purple]You want to challenge me don't you skywalker ? I don't have time to post the solution but I can show the number of Crux.It's in Crux 1/2004 page 44 [/color]

[quote="evarist"]If you need solution i'll post it.[/quote]
[quote="evarist"]You want to challenge me don't you skywalker ?  I don't have time to post the solution...[/quote]
You said the stupid thing, I post it and I don't wait for your reply, but you said if I need you will post solution and now when I said I need you said I chellenge you and you don't have time.. :rotfl:  :rotfl:
I don't have Crux!
An advice for some Vietnameses, you should only reply if you can post solution or hinr or similar related thing or link to similar problem on this forum, don't spam stupid things!

[color=purple]I don't spam please ! I think it's not stupid ! In the morning I think I have time so i reply  but now i don't hev time to post enough solution so i must show the number of Crux .And I think it's enough.
       You can criticize me but can not criticize all Việt Nam are member of ML!
         I'll post the pdf file of this Crux [/color]

[quote="evarist"]You can criticize me but can not criticize all Việt Nam are member of ML! [/quote]
You are right I criticize you but I don't criticize all Việt Nam are member of ML!
[quote="skywalker"]
An advice for some Vietnameses, you should only reply if you can post solution or hinr or similar related thing or link to similar problem on this forum, don't spam stupid things!
[/quote]
I only say about some Vietnamese!
Actually I know many Vietnameses are very good at mathematic and I admire them! But some things that some Vietnameses show on mathlinks recently when I join mathlinks, I feel disappointed for them, If you want they respect yourself you should do more than say!

[color=purple]Ok ! This is the Crux have the proff.[/color]

[color=purple] Do you sastisfy skywalker ? And do you have more ? [/color]

ok nice solution in Crux, I still have a problem do it if you can
$(\frac{a^{t}+b^{t}+c^{t}}{3})^\frac{1}{t}\le \sqrt{3}R$ is valid iff $t\le \frac{\ln 9-ln 4}{\ln 4-ln 3}$  :D

[quote="skywalker"]
I only say about some Vietnamese!
Actually I know many Vietnameses are very good at mathematic and I admire them! But some things that some Vietnameses show on mathlinks recently when I join mathlinks, I feel disappointed for them, If you want they respect yourself you should do more than say![/quote]
I'm so sorry about that, skywalker. :(
I hope in next time, you will have a better look at Vietnamese members. I'm so sorry. :(


advanced fields
advanced fields theorems

[b]Alexendroff Subbasis theorem :[/b]

In order that a space $ X$ be compact ,it's necessary and sufficient that there exist  a subbasis $ S$ for $ X$ s.t. every open cover of $ X$ by members of $ S$ has a finite subcover.

Does anyone have or know an elegant proof for this theorem ?

Thank you in advanced .

I have a proof, but I don't consider it to be particularly elegant. Do you already have one and are looking for a better one? I don't want to post it if you already have it.

[quote="Kalle"]What is wrong with the standard proof with Zorn's lemma? I don't know if it's elegant, but it's totally standard, and just the usual deal with Zorn's lemma.[/quote]

No problem and that what I have ...

to be more clear , I'm just looking to another proof without using Zorn's lemma.

Of course , need it to be elegant one.

Anyway , Could I see the proof you have Hilbert may be will be nice for me.

Thank you guys so much.

Scofield


The book says the domain of $\sqrt{\frac{x-2}{x+4}}$ is $(-\infty, -4)$ $or$ $[2,\infty)$.
The graph of this only shows the portion where x is positive, because you can't square root a negative number...but:
$\sqrt{\frac{x-2}{x+4}}=\frac{\sqrt{x-2}}{\sqrt{x+4}}$ and now, you can't cancel out the negatives inside the square root, therefore it supports the graph, but is this reasonable?  (Is the book right or is the graph right?)

if x is a negative number less than -4 then two imagary square roots will be created. take the $i$ out of both square roots and cancel them out giving you a real solution.

My graphing calculator graphs out the negatives... (i.e. it shows $(-\infty,-4)$)

Oops, that was my problem.  The domain of the graph was too small.  Well.  I understand now.  Thank you furious and WindSlicer.


geometry
perimeter
Euler
similar triangles
geometry proposed

Let ${T}=AD \cap BE$ and ${S}=AB \cap DE$. Then $TF$ is the polar of $S$. But because $DE$ is the polar of  $C$ and $S$ is on $DE$ we get that $C$ lies on the polar of $S$ too. So $CF \perp AB$  :)

Hi

I don't know what is a "polar"...

Can you give another way without that?

It is easy to prove this problem, by Pascal’s theorem.

We denote as $T,$ the intersection point of $AD,$ $BE$ and so, because of $AE\perp TB$ and $BD\perp TA,$ we conclude that the point $F,$ is the orthocenter of the triangle $\bigtriangleup TAB.$ Hence, we have that $TF\perp AB.$

We consider now, the cyclic non-convex quadrilateral $ADBE,$ as the special case of the  non-convex hexagon $ADD'BEE',$ of which $D'\equiv D$ and $E'\equiv E.$

So, by Pascal’s theorem, we have that the points $T,$ as the intersection point of the opposite sides $AD,$ $BE$ of $ADD'BEE'$ and $C,$ as the intersection point of the opposite sides $DD',$ $EE'$ $($ here the tangents of semicircle $(O)$ at $D,$ $E$ $)$ and $F,$ as the intersection point of the opposite sides $AE,$ $BD,$ are collinear.

Hence, we conclude that $CF\perp AB$ and the proof is completed.

Kostas Vittas.

PS. As in the configuration of a convex hexagon, we say as opposite two sides of a non-convex one, when between the end of the one side and the start of the other $($ in the order of drawing the hexagon's perimeter $),$ two it’s adjoined sides, are inserted.

Thanks

I'm not advanced in geometry 'cause I don't know how to apply Pascal's Theorem.

I'd really appreciate if you may give another way using simply geometry, I mean similar triangles, angles, things like this.

[quote="MathTeX"]Let $AB$ be the diameter of a semicircle. Let $\overline{CD}$ and $\overline{CE}$ be the tangents to it from any point $C$. Let $F = \overline{AE}\cap \overline{BD}$, then prove that $\overline{CF}\perp \overline{AB}$[/quote]
[color=red]
 Please , see the attachment![/color]

Obviously points $D$ and $E$ are on a circle $k$ with diameter $CO$ where $O$ is the midpoint of $AB$. Let $P$ be the other intersection of $k$ and $AB$. Angle $OPC$ is right.
Let $T$ be the intersection of $AD$ and $BE$. Now we can see that $k$ is the nine point (Euler) circle of $ATB$ as it passes thru $D$ snd $E$ (feet of altitudes $AE$ and $BD$) and the midpoint $O$ of $AB$. This meens that $P$ is also a feet of the altitude so points $T$, $C$ and $F$ are colinear and additionally we proved $FC$=$CT$.


In a row written in increasing order all the irreducible positive rational numbers, such that the product of the numerator and the denominator is less than 1988. Prove that any two adjacent fractions $\frac{a}{b}$ and $\frac{c}{d},$ $\frac{a}{b} < \frac{c}{d},$ satisfy the equation $b \cdot c - a \cdot d = 1.$

Well, if they would be in $(0,1)$ then http://en.wikipedia.org/wiki/Farey_sequence . I think it's not hard to generalize it to our case.


Hmm... For some reason, when I try to edit my signature and click "Submit", a blank page appears and my signature hasn't changed. I am using the latest Firefox.

And also for the blogs, how do you get the white strip in the middle to be transparent to the background with CSS?

[quote="Anaxerzia"]Hmm... For some reason, when I try to edit my signature and click "Submit", a blank page appears and my signature hasn't changed. I am using the latest Firefox.

[/quote]
Exact same problem for me.  I use Internet Explorer 6.  It doesn't happen with other profile options.


geometry
cyclic quadrilateral
geometry unsolved

Quadrilateral $ ABCD$ is inscribed in circle with $ AC$ a diameterof the circle and $ BD \bot AC$. Diagonals $ AC$ and $ BD$ intersect at $ E$. Extend segment $ DA$ through $ A$ to $ F$. Extend segment $ BA$ through $ A$ to $ G$ such that $ DG\| BF$. Extend segment $ GF$ through $ F$ to $ H$ such that $ CH \bot GH$. Prove that points $ B;E; F;H$ lie on a circle.

[quote="livetolove212"]$ FHBE$ is cyclic quadrilateral iff $ \angle FHE \equal{} \angle FBE \equal{} 180^o \minus{} \angle EDG$
$ \Leftrightarrow GHED$ is cyclic
$ \Leftrightarrow \angle GHD \equal{} \angle GED$
But $ FHCD$ is cyclic then $ \angle FHD \equal{} \angle FCD$
Therefore we need to prove that $ \angle FCD \equal{} \angle GED$
$ \Leftrightarrow \angle AFC \equal{} 90^o \minus{} \angle FCD \equal{} 90^o \minus{} \angle GED \equal{} \angle AEG$
It's easy to see that $ \angle FAC \equal{} \angle EAG$
Since $ BF//DG$ we get $ \frac {AF}{AD} \equal{} \frac {AB}{AG}$
$ \Rightarrow AF.AG \equal{} AD.AB \equal{} AB^2 \equal{} AE.AC$
$ \Rightarrow \frac {AF}{AC} \equal{} \frac {AE}{AG}$
$ \Rightarrow \Delta FAC\sim \Delta EAG \Rightarrow \angle AFC \equal{} \angle AEG$
$ \rightarrow$ QED[/quote]
Nice solution :D
thank you very much.


Solve:
∂^2z/∂x^2	= (1/a)^2  ∂^2z/∂y^2

Given that z is never infinite for any real values of x or y and that z=0 when x=0 or y=0.

$ \frac{\partial^2 z}{\partial x^2} \equal{} \frac1{a^2} \frac{\partial^2 z}{\partial y^2}$.

$ z(x\equal{}0,y) \equal{} z(x, y\equal{}0) \equal{} 0$.

This is the one dimensional [url=http://en.wikipedia.org/wiki/Wave_equation]wave equation[/url]. In particular, look at d'Alembert's formula.

hello, see also here
http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node12.html
Sonnhard.

Thank you

But it is an ellementary problem. I got this problem in H T H Piaggio's book on Differential Equations in Art.45(see http://www.archive.org). i don't think we should approach this problem in such a complex way.


function
trigonometry
calculus
derivative
floor function
algebra

[quote="Yogenzaga"]What does cos^2 represent? I have a question here in my calc that includes this function, but, as funny as it may seem, somehow I never encountered a trigonomic function to the second power, soooo... help? ^_^[/quote]

$\cos^{2}(x) = (\cos(x))^{2}\neq \cos(x^{2})$

For example, if $x = \frac{\pi}{6}$, then you can clearly see why the above is true.

But,
$\cos^{-1}(x) = \arccos(x) \neq (\cos(x))^{-1}$

Function notation can be very confusing because in other contexts the notation $f^{2}(x)$ can actually mean $f(f(x))$.  The notation $\cos^{2}x = (\cos x)^{2}$ is particular to trigonometric functions.

Just to add to the confusion, $f^{2}(x)$ can also mean the second derivative of $f(x)$ wrt $x$

Well, I'd say that standard notation for the second derivative of a function $f(x)$ is $f''(x)$... well, once I found in a mathematical contest $f^{(2)}(x)$, that meant $f(f(x))$...

The standard notation for the $n^{th}$ derivative of a function $f(x)$ is $f^{(n)}(x)$ (pay careful attention to the parentheses).  The parenthetical notation for function composition is not standard, as far as I know.

stand composite is $f{x}\circ g(x)$

-I think.  :maybe:

You have a choice between $(f\circ g)(x)$ and $f(g(x))$. I prefer the later.

I guess that some things would just have to be defined in context.  For example, the Greatest Integer Function with brackets. $[\,]$.  Although I don't see why they can't just use $\lfloor\,\rfloor$.

But with something like $f^{3}(x)$, it would probably say whether it means $f(f(f(x)))$, $(f(x))^{3}$, or the third derivative of $f(x)$ with respect to $x$.  Still, for the third one, it'd probably say $f'''(x)$.

Personally, I would use $f^{3}(x)$ to denote $f(f(f(x)))$.  It's consistent with the inverse notation, $f^{-1}(x)$.

I guess when you think about it, it is kind of weird to use superscripts with trig functions to mean different things.  For example, $\sin^{2}\theta=(\sin\theta)^{2}\neq\sin(\sin\theta)$, while $\sin^{-1}\theta=\arcsin\theta\neq\csc\theta$.

[quote="7h3.D3m0n.117"][quote="Yogenzaga"]What does cos^2 represent? I have a question here in my calc that includes this function, but, as funny as it may seem, somehow I never encountered a trigonomic function to the second power, soooo... help? ^_^[/quote]

Well what is anything to the second power represent? For example, if it was

$\cos 30^{\circ}$.

The answer is $\frac{\sqrt{3}}{2}$.

$\cos^{2}30^{\circ}= \left(\frac{\sqrt{3}}{2}\right)^{2}= \frac{3}{4}$[/quote]
Actually, it's very uncommon for $f^{2}(x)$ to be the same as $f(x^{2})$ for $f(x)$.


calculus
integration
trigonometry
calculus computations

Evaluate $\int_0^{2\pi}(\sin^2{2x}-\sin^42x)dx$=?

I thouhgt it's $\frac{\pi}4$ while the answer in my book is $0$. Which is correct?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find  3  groups  K ,  H , G such  that : K is a subgroup of H and  H of G.  and K is normal  in H , H  is normal  in  G  but K  is not  normal  in  G.

Consider $ D_8$, the dihedral group of order $ 8$.  Use the fact that if $ H
\leq G$ and $ |G: H|\equal{}2$, then $ H\lhd G$.

Alternatively:
Set $ G \equal{} A_{4}, H \equal{} \{ e, (12)(34), (13)(24), (14)(23) \}, K \equal{} \{e, (12)(14)\}$.

Then H, which is isomorphic to Klein four group is normal in G. K a subgroup of an abelian group H must be normal or it is just normal because it has index 2 in H. You only need to show that K is not normal in G.


inequalities
vector
linear algebra
matrix
geometry
3D geometry
geometric transformation

This goes to the geometric meaning of the deteminant. Three vectors $\vec{v}_i,\,i=1,2,3$ in $\mathbb{R}^3$ determine a parallelopiped. The points of this parallelopiped are $\left\{\alpha_1\vec{v}_1+\alpha_2\vec{v}_2+\alpha_3\vec{v}_3: 0\le \alpha_1,\alpha_2,\alpha_3\le 1\right\}.$

If we then create a $3\times 3$ matrix whose rows (or columns) are the thee vectors in question, then the absolute value of the determinant is the volume of the parallelopiped.

It is clear that for specified magnitudes of the three vectors, the volume would be maximized by making the three vectors orthogonal (hence making the parallelopiped a rectangular prism). In Cezar Lupu's problem, the three vectors are specified as being of unit length, so the maximal volume would be that of a cube of volume 1.

What about the general inequality?

[color=blue][b]Hadamard inequality.[/b] If $\displaystyle A = \left( a_{ij} \right)_{1 \leq i,j \leq n} \in \mathcal M_n \left( \mathbb C \right)$ then $\displaystyle \left| \det \, A \right|^2 \leq \prod_{i=1}^n \left( \sum_{j=1}^n \left| a_{ij} \right|^2 \right)$.[/color]

Is there a miraculous solution to this one? PlanetMath gives a long proof... is it the only one?

Another solution: try Lagrange's multipliers. ;)

[quote="Kent Merryfield"]
It is clear that for specified magnitudes of the three vectors, the volume would be maximized by making the three vectors orthogonal (hence making the parallelopiped a rectangular prism). In Cezar Lupu's problem, the three vectors are specified as being of unit length, so the maximal volume would be that of a cube of volume 1.[/quote]

How do you prove this? (A proof that is generalizable to n-dimension?)

I think it is clear in dimension $3$, but for higher dimensions it is precisely Hadamard inequality that shows that. (And there are far better proofs that Lagrange multipliers  :D )

Yes, but the one using Lagrange multipliers is also a super nice solution. :)

can anyone tell me what the Lagrange multipliers thing is about ? and eventually solve this exercise posted here ...

Just think geometrically :).

Given a complex matrix $A$, we can make it upper triangular by multiplying it with unitary matrices (which we can regard as "rotations") to the left. We can prove this by induction on $n$. Suppose we've shown it for $n-1$, and let $U$ be a unitary matrix which takes the first column of $A$ onto a complex multiple of $e_1$ ($e_i,\ i=\overline{1,n}$ is the standard basis of $\mathbb C^n$). Now $UA$ will have the form $\begin{pmatrix}a&T\\0&B\end{pmatrix}$, where $a$ is a complex number, and $B$ is an $(n-1)\times(n-1)$ matrix. By the induction hypothesis, there is a unitary $(n-1)\times(n-1)$ matrix $V$ such that $VB$ is upper triangular. Now, if $\widetilde U=\begin{pmatrix}1&0\\0&V\end{pmatrix}$, then $\widetilde UUA$ will be upper triangular, as desired.

The operation above does not change $|\det A|$, and it does not change the lengths $\sqrt{\sum_{i=1}^n|a_{ij}|^2}$ of the columns of $A$. Since the inequality $|\det A|^2\le\prod_{j=1}^n\left(\sum_{i=1}^n|a_{ij}|^2\right)\ (*)$ is obvious for an upper triangular matrix, it's true for any complex matrix.

P.S.

The stated inequality was actually $(*)$ with $i$ and $j$ interchanged, but there's no difference. We can multiply $A$ with unitary matrices to the right, or just apply the reasoning above to the transpose $A^t$ of $A$.


If, for all $x$, $f(x) = f(2a)^x$ and $f(x+2)=27f(x)$, then find $a$.

[hide]If the inverse of f(x) is a function, a=1/2.
When x=1 f(1)=f(2a) so a=1/2 since the inverse is a function.
If the inverse is not a function, let's assume f(1)=f(2a) where a is not 1/2, (since we know 1/2 must be true)
We set x=2a.  Then, f(2a)=f(2a)^(2a).  We can remove the f(2a)s and immediately get 2a=1 so a=1/2.  Therefore, a can't be anything else.  [/hide]


Hey, just wonder....k...does anyone have the arbelos journals? I just ordered them off the net, and ya..before they arrive, just wanting an opinion of their content....good? bad? useful? useless? easy? hard? does it have any training?

Thanks!

Sorry, I found myself a totally stranger to all these books or authors :oops:
just googled and it seems to be  fairly nice

I didn't like them, but that's just my opinion.

There are better reads/resources.  Most of them are free ^.^.


trigonometry
arithmetic sequence
Pythagorean Theorem
geometry
national olympiad

1 July 2006 ; Time : 4.5 hrs

1. In the triangle $ABC,\angle A=\frac{\pi}{3},D,M$ are points on the line $AC$ and $E,N$ are points on the line $AB$ such that $DN$ and $EM$ are the perpendicular bisectors of $AC$ and $AB$ respectively. Let $L$ be the midpoint of $MN$. Prove that $\angle EDL=\angle ELD$

2. Show that any representation of 1 as the sum of distinct reciprocals of numbers drawn from the arithmetic progression $\{2,5,8,11,...\}$ such as given in the following example must have at least eight terms: \[1=\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{20}+\frac{1}{41}+\frac{1}{110}+\frac{1}{1640}\] 3. Consider the sequence $p_{1},p_{2},...$ of primes such that for each $i\geq2$, either $p_{i}=2p_{i-1}-1$ or $p_{i}=2p_{i-1}+1$. Show that any such sequence has a finite number of terms.

4. Let $n$ be positive integer. Let $S_{1},S_{2},...,S_{k}$ be a collection of $2n$-element subsets of $\{1,2,3,4,...,4n-1,4n\}$ so that $S_{i}\cap S_{j}$ contains at most $n$ elements for all $1\leq i<j\leq k$. Show that \[k\leq 6^{(n+1)/2}\] 5. Let $a,b,n$ be positive integers. Prove that $n!$ divides \[b^{n-1}a(a+b)(a+2b)...(a+(n-1)b)\]

Solution for Question 1:
1. $DN$, $EM$ are perpendicular bisectors of $AC$, $AB$ hence $E$,$D$ are midpoints of $AB$, $AC$ so $ED=\frac{BC}{2}$   (*)
2. $ME$ is perpendicular to $AB$, so $\triangle{ENM}$ is rectangular so if $L$ is midpoint of $NM$, $EL=NL=ML=\frac{NM}{2}$.  (**)
3. $\angle{ADN}=\frac{\pi}{2}$, $\angle{BAC}=\frac{\pi}{3}$ hence $ND=AD \times \sqrt{3}$.
4. $\angle{AEM}=\frac{\pi}{2}$, $\angle{BAC}=\frac{\pi}{3}$ hence $AM= 2AE$ hence $DM=|2AE-AD|$.
5. $\angle NDC=\frac{\pi}{2}$ so by Pythagorean theorem,
$NM=\sqrt{ND^{2}+DM^{2}}$
$=\sqrt{3AD^{2}+4AE^{2}+AD^{2}-4AE\times AD}$
$AD=\frac{AC}{2}, AE=\frac{AB}{2}$, so:
$NM=\sqrt{4(\frac{AC}{2})^{2}+4(\frac{AB}{2})^{2}-4\frac{AB}{2}\times\frac{AC}{2}}$
$NM=\sqrt{AC^{2}+AB^{2}-AC\times AB}$.   (***)
6. By Cosine Law in $\triangle{ABC}$:
$BC=\sqrt{AB^{2}+BC^{2}-2AC\times BC cos{BAC}}$
$BC=\sqrt{AB^{2}+BC^{2}-AC\times AB}$.
Using (***), get $BC=NM$.
7. From (*) and (**):
$ED=\frac{BC}{2}$ and  $EL=\frac{NM}{2}$.
But $BC=NM$ so $ED=EL$ so $\triangle{DEL}$ is isosceles and so $\angle ELD=\angle EDL$, QED.


[quote="onlylove_math"]Solve system :

$ \{\begin{aligned} 2b^3 - a^3 - c^3 = 3b^2c - 3c^2a \\
2c^3 - b^3 - a^3 = 3c^2a - 3a^2b \\
2a^3 - c^3 - b^3 = 3a^2b - 3b^2c \end{aligned}$[/quote]

Just a comment but it's kinda extraneous when you give a system that one equation can be gained by adding the two other :D

[quote="onlylove_math"]Oh I 'm sorry  :( . I don't see my problem careful way . It 's too easy . We only need add 3 equality , we 'll have a=b=c[/quote]
It not too easy,my friend,If we dont know the ineq that you used ,your problem is quite hard, :lol: ,At all,I like yours

[quote="onlylove_math"] We only need add 3 equality , we 'll have a=b=c[/quote]
Sorry it 's wrong  :blush: . Allnames, you are true. We must base a analysis and a known inequality :wink:

It's a homogeneous system.
$ c\equal{}0$ implies $ a\equal{}b\equal{}0$.
$ c\equal{}1$ implies
 $ (a^2\plus{}a\plus{}1)(a^3\plus{}3a^2\minus{}1)(a\minus{}1)^3\equal{}0,6b\equal{}\minus{}5a^6\minus{}11a^5\plus{}10a^4\plus{}8a^3\plus{}17a^2\minus{}10a\minus{}3$.
Thus $ a\in\{1,u\approx{\minus{}2.88},v\approx{\minus{}0.65},w\approx{0.53}\}$ and if $ a\equal{}1$ then $ b\equal{}1$ else $ b\equal{}\minus{}a^2\minus{}2a\plus{}1$.
Conclusion: the solutions $ (a,b,c)$ are:
$ \{(t,t,t),(ut,(\minus{}u^2\minus{}2u\plus{}1)t,t),(vt,(\minus{}v^2\minus{}2v\plus{}1)t,t)),(wt,(\minus{}w^2\minus{}2w\plus{}1)t,t)\}$ where $ t\in\mathbb{R}$.

[quote="loup blanc"]It's a homogeneous system.
$ c \equal{} 0$ implies $ a \equal{} b \equal{} 0$.
$ c \equal{} 1$ implies
 $ (a^2 \plus{} a \plus{} 1)(a^3 \plus{} 3a^2 \minus{} 1)(a \minus{} 1)^3 \equal{} 0,6b \equal{} \minus{} 5a^6 \minus{} 11a^5 \plus{} 10a^4 \plus{} 8a^3 \plus{} 17a^2 \minus{} 10a \minus{} 3$.
Thus $ a\in\{1,u\approx{ \minus{} 2.88},v\approx{ \minus{} 0.65},w\approx{0.53}\}$ and if $ a \equal{} 1$ then $ b \equal{} 1$ else $ b \equal{} \minus{} a^2 \minus{} 2a \plus{} 1$.
Conclusion: the solutions $ (a,b,c)$ are:
$ \{(t,t,t),(ut,( \minus{} u^2 \minus{} 2u \plus{} 1)t,t),(vt,( \minus{} v^2 \minus{} 2v \plus{} 1)t,t)),(wt,( \minus{} w^2 \minus{} 2w \plus{} 1)t,t)\}$ where $ t\in\mathbb{R}$.[/quote]
I think it isn't still finish . Who can give mo another solution?

Use sum of square(like SOS)
ex:$ 2b^3 \plus{} c^3 \minus{}3b^2c\equal{}2c^3\plus{}a^3\minus{}3c^2a$

[quote="Math pro"]Use sum of square(like SOS)
ex:$ 2b^3 \plus{} c^3 \minus{} 3b^2c \equal{} 2c^3 \plus{} a^3 \minus{} 3c^2a$[/quote]
I don't think so , my friend. Can you detail and finish it :wink: ?

Hi onlylove_math,
you write too much and you don't think enough.


let a,b >0 prove that
\[ \frac {{a \plus{} b \plus{} \sqrt {ab} }}{3} \le \mathop {\sqrt {\frac {{\sqrt [3]{{\mathop a\nolimits^2 b}} \plus{} \sqrt [3]{{a\mathop b\nolimits^2 }}}}{2}} }\nolimits^3
\]

For this problem you can get $ a\equal{}kb$


Let $B$ be the set of $k$ sequences which have $n$ terms equal to 1 or -1. The product of two sequences $(a_1, a_2, \ldots , a_n)$ and $(b_1, b_2, \ldots ,b_n)$ is defined as $(a_1b_1, a_2b_2, \ldots , a_nb_n)$. Prove that there exists a sequence $(c_1, c_2, \ldots , c_n)$ such that the intersection of $B$ and the set containing all sequences from $B$ multiplied by $(c_1, c_2, \ldots , c_n)$ contains at most $k^2 \cdot 2^{-n}$ sequences.

The probabilistic method works well here.  Let $c$ be a random sequence from ${\{-1, 1\}}^n$.  Let's compute the expected size of the intersection:
[tex]\begin{eqnarray*}
E(|B \cap cB|) & =  & \sum_{x \in B} \Pr(x \in cB) \\
                      & = &  \sum_{x \in B} \Pr(c \in xB) \\
                      & = & \sum_{x \in B} \frac{k}{2^n} \\
                      & = & \frac{k^2}{2^n}.
\end{eqnarray*}
[/tex]
Hence there exists a specific sequence $c$ such that $|B \cap cB| \geq k^2 / n$.

By the way, where did this problem come from?

[quote="Ravi B"]
By the way, where did this problem come from?[/quote]

It comes from ISL**. Look at ISLs I posted in the download section and you can look the year up yourself. BTW for for 1983-2002 all ISL problems have solutions. Sometimes even ILL problems are available.  :)


Line $ l$ touches circle $ k$ at point $ A$. Let $ AB$ be diameter of circle $ k$. Take any point $ C$ on $ k$ and draw perpendicular $ CD$ to $ AB$, with $ D$ on $ AB$. On line $ CD$, beyond $ D$, take point $ E$ such that $ ED\equal{}BC$. From point $ E$ draw tangents to circle $ k$ and denote by $ N$ and $ K$ intersections points of these tangents with line $ l$. Prove that $ KN\equal{}AB$.

:blush: Sorry,but in my figure,KN>BA  Could you draw us a figure?

Dear plane geometry, I'm sure that you think E is on line CD, beyond C. But,
[quote="delegat"]On line $ CD$, beyond $ D$, take point $ E$ such that $ ED \equal{} BC$[/quote]

Nice problem!
Suppose A,N,K are on the tangent of (O) in order that. Denote M,I be respectively the tangency point of EN,EK and (O), F=CE∩(O).
$ EM^2\equal{}EF.EC\equal{}ED^2\minus{}CD^2\equal{}BC^2\minus{}CD^2\equal{}BD^2\Rightarrow EM\equal{}EI\equal{}BD$
S(ENK)=S(OAKI)-S(OANM)-S(OIEM)=R.AK-R.EN=R.(NK-EM)=R(NK-BD)
$ \Rightarrow NK.AD\equal{}2R(NK\minus{}BD)$, so 2R.BD=2R.NK-NK.AD=NK.BD, hence NK=2R=AB.

Oh, i guess i misunderstand this statement.
Thank you dear mr.danh, not only for pointing out my mistake,but also for the nice solution! :lol:

[quote="Ahiles"][color=darkblue]Everything is normal.[/color]
[img]http://i074.radikal.ru/0807/3e/333aaff18df0.png[/img][/quote]Are you proving it without words? I can't see how the figure proves it. :maybe:


In the expression $ xy^2$, the values of $ x$ and $ y$ are each decreased $ 25\%$; the value of the expression is:
$ \textbf{(A)}\ \text{decreased } 50\% \qquad\textbf{(B)}\ \text{decreased }75\%\\
\textbf{(C)}\ \text{decreased }\frac{37}{64}\text{ of its value} \qquad\textbf{(D)}\ \text{decreased }\frac{27}{64}\text{ of its value}\\
\textbf{(E)}\ \text{none of these}$

It's $ \frac34x\frac9{16}y^2\equal{}\frac{27}{64}xy^2$, so the decrease is 37/64

Answer: C


Suppose $ a,b,c$ are positive real numbers ,prove that

$ \frac{a^3}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}ca\plus{}a^2}\ge \frac{a\plus{}b\plus{}c}{3}$


[quote="parinaz.sa"]you mean : if$1\le x,y,z\le1.5$prove it:$xy\sqrt{3-2z}+xz\sqrt{3-2y}+yz\sqrt{3-2x}\ge x^{3}+y^{3}+z^{3}$[/quote]
It has been flipped.

PF
By AM-GM,$x^{3}+y^{3}+z^{3}\ge 3xyz=xyz+yzx+zxy$.
Also, since $x\ge1, x^{2}+2x-3=(x-1)(x+3)\ge 0$, $x\ge\sqrt{3-2x}$.
ITS, $y\ge\sqrt{3-2y}, z\ge\sqrt{3-2z}$.
$\therefore x^{3}+y^{3}+z^{3}\ge 3xyz \ge xy\sqrt{3-2z}+xz\sqrt{3-2y}+yz\sqrt{3-2x}$
Equality occurs when $x=y=z=1$.

\[\text{In general }1\leq x,y,z\leq \frac{n}{n-1}\\ : \ \ x^{3}+y^{3}+z^{3}\geq xyz+xyz+xyz \\ (x-1)(x+n)\geq 0\implies x^{2}\geq n-(n-1)x\implies 3xyz\geq xy\sqrt{n-(n-1)z}+xz\sqrt{n-(n-1)y}+yz\sqrt{n-(n-1)x}\\ x^{3}+y^{3}+z^{3}\geq xy\sqrt{n-(n-1)z}+xz\sqrt{n-(n-1)y}+yz\sqrt{n-(n-1)x}\]


[quote="sorry iloveyourfriend"]$ ab + bc + ca = \frac {1}{3}$
prove:$ \sum {\frac {1}{a^2 - bc + 1} \le \ 3}$
with a,b,c>0[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30719


[hide]If you multiply out and cancel the sixth power terms, you get $3a^{4}b^{2}+3a^{2}b^{4}=2a^{3}b^{3}$ ie $3a^{2}+3b^{2}=2ab$ since $a^{2}b^{2}\neq 0$ Dividing by 3ab yields $\frac{a}{b}+\frac{b}{a}=\frac{2}{3}$[/hide]


4 mutually tangent spheres with radius $x$ rest on a plane. What is the distance from the top of the spheres to the plane?

3 mutually tangent spheres with radius $x$ rest on a plane. A fourth sphere of radius $y$ is placed on top. What is the distance from the top of the spheres to the plane?

PS. The subject reminds me of the movie "Snakes on a Plane."  :rotfl:

[hide]The first part is very easy.  That's just $2x$.

For the second part, you use a tetrahedron with vertices at the centers of each of the spheres.  Since the distance from a vertex of the base to the base's centroid is $\frac{x\sqrt{3}}{3}$, you can use a right triangle with a leg of length $\frac{x\sqrt{3}}{3}$ and a hypotenuse with length $x$ to find the the height of the tetrahedron, which is $\frac{x\sqrt{2}}{3}$.  Then you simply add $x$ to that, giving you $\boxed{x+\frac{x\sqrt{2}}{3}}$, or $\boxed{x \left( 1+\frac{\sqrt{2}}{3}\right)}$[/hide]

Oops.  Is that better?

[quote="DanK"][hide]...you can use a right triangle with a leg of length $\frac{x\sqrt{3}}{3}$ and a hypotenuse with length $x$ to find the the height of the tetrahedron, which is $\frac{x\sqrt{6}}{3}$...[/hide][/quote] Check that again.  :D


[quote="jaka"]Let us have a triangle $ABC$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Show that if $MC$ and $NB$ are perpendicular, then
[list]$\cos \beta + \cos \gamma \geq \frac 23$[/list][/quote]

Should we be assuming that $\beta$  is angle B, and $\gamma$ is angle C?  If not what angles do the $\beta$ and $\gamma$ represent?

Sorry...
$\angle ABC = \beta$, $\angle ACB = \gamma$ and it's a mistake - show that $\cot \beta + \cot \gamma \geq \frac 23$

WLOG, represent the vertices $A, B, C$ with complex numbers $a = x+ iy,\ b = -1,\ c = +1$. Then $BC = 2$ and

$\cos \beta = \frac{\text{Re}(a + 1)}{|a + 1|},\ \ \sin \beta = \frac{\text{Im}(a + 1)}{|a + 1|},\ \ \cot \beta = \frac{\text{Re}(a + 1)}{\text{Im}(a + 1)}$

$\cos \gamma = -\frac{\text{Re}(a - 1)}{|a - 1|},\ \ \sin \gamma = \frac{\text{Im}(a - 1)}{|a - 1|},\ \ \cot \gamma = -\frac{\text{Re}(a - 1)}{\text{Im}(a - 1)}$

$\cot \beta  + \cot \gamma = \frac{\text{Re}(a + 1)}{\text{Im}(a + 1)} - \frac{\text{Re}(a - 1)}{\text{Im}(a - 1)} = \frac{x + 1}{y} - \frac{x - 1}{y} = \frac 2 y$

Let $D$ be the foot of the altitude from the vertex $A$ to the side line $BC$, then $y = AD$. Since the medians $CM \perp BN$ are perpendicular, their intersection $G$ is on the unit circle $(O, 1)$ with center $O$ at the midpoint of the segment $BC = 2$ and radius $r = \frac{BC}{2} = 1$. The centroid $G$ divides the 3rd median $AO$ in the ratio $OG : GA = 1 : 2$. Since $r = OG = 1$ is fixed as radius of the unit circle $(O, 1)$, length of the 3rd median is $OA = 3$. The locus of vertices $A$ is then a circle $(O, 3)$ with center $O$ and radius $r' = OA = 3$. It is clear that the altitude $AD$ is maximum, when its foot $D$ coincides with the midpoint $O$ of the segment $BC$, i.e., when the triangle $\triangle ABC$ is isosceles with $AB = AC$. The altitude $AD$ then coincides with the median $AO$ and $AD = 3$. Thus $\max y = \max{AD} = 3$, $y \le 3$ and

$\cot \beta  + \cot \gamma = \frac 2 y \ge \frac 2 3$

Prove that if the medians $BN \perp CM$ are perpendicular,

$\cos \beta  + \cos \gamma < \frac 2 3$

alternatively...

(G being the centroid), let $ GN = b $, $ GM = c $; then $ BG = 2b $, $ CG = 2c $.  then

$ \cot \beta = \frac{1-\frac cb \frac {c}{2b}}{\frac cb + \frac {c}{2b}} = \frac {2b^2-c^2}{3bc} $

and similarly

$ \cot \gamma = \frac {2c^2-b^2}{3bc} $

so $ \cot \beta + \cot \gamma = \frac {b^2+c^2}{3bc} \geq \frac {2bc}{3bc} = \frac23 $

Why is $\displaystyle  \cot \beta = \frac{1-\frac cb \frac {c}{2b}}{\frac cb + \frac {c}{2b}}$?

the well-known trig formula...

$ \tan (x + y) = \frac {\tan x + \tan y}{1- \tan x \tan y} $, so $ \cot (x + y) = \frac {1- \tan x \tan y} {\tan x + \tan y} $

we let $ x = \angle GBC $ and $ y = \angle GBA $, etc...

This is well known problem, and appeared on Canadian Olympiad '93 , problem 3.

(http://cms.math.ca/Competitions/CMO/examt/english93.html)

--
soln can be found here: http://www.kalva.demon.co.uk/canada/casoln/csol933.html


function
real analysis
advanced fields
advanced fields unsolved

Gelfand-Naimark theorem states that every commutative C-* algebra is isometric to $ C(M)$, the ring of continuous functions over its spectrum. Is the theorem true for ordinary commutative Banach semisimple algebras, i.e. without *? Every proof that the Gelfand transform is an isometry uses the fact that in C-* algebra $ |xx^{*}| \equal{} |x|^{2}$, so I wonder whether it is true when we don't have the * and that identity. If it is not isometric, is it isomorphic?

no. a natural counterexample is studied in

$ L^1(I, X)$ with Order Convolution
# R. K. Dhar and H. L. Vasudeva
# Proceedings of the American Mathematical Society, Vol. 83, No. 3 (Nov., 1981), pp. 499-505

I'll take a look at that paper if I find it, but meanwhile I have some questions. If I have a semisimple algebra $ B$ and take its Gelfand transform $ f(B)$, then $ f(B)$ is dense in $ C(M)$ - so I guess that in your counterexample $ B^{^}$ isn't closed in $ C(M)$? How do you prove it?


I'm thinking of a number!  It lies between 1 and 10.  Its square is rational and lies between 1 and 10.  The number is larger than pi.  Correct to two decimal places, name the number.

[hide="Click me!"]$3.16$  :wink: [/hide]

[quote="mysmartmouth"][hide="What about"]3.15?[/hide][/quote]

Hmm... Maybe you are right. I think he meant that the square of his integer is in between $1$ and $10$, exclusive. Interval, could you clarify?

[quote="vishalarul"]I think he meant that the square of his integer is in between $1$ and $10$, exclusive.[/quote]

It doesn't matter.  $3.16^{2}< 10$, strictly.

Does this question have several answers?

Yeah, it has an infinite amount of answers, but they can be rounded to only 2 different answers, since you only asked for 2 digits...

[hide="w00t, the first solution"]Note that $\sqrt{10}\approx3.162$, and $\pi\approx3.141$, so the only two possible numbers, rounded to two decimal places are $3.15$ and $3.16$.[/hide]


limit
real analysis
real analysis unsolved

Prove that {$ x_n$} is a sequence that satisfies
$ \mid x_n \mid\le\dfrac{1\plus{}n}{1\plus{}n\plus{}2n^2}$
for all n$ \in$ N, then {$ x_n$} is Cauchy.

Note, that $ b_n\equal{}\frac{1\plus{}n}{1\plus{}n\plus{}2n^2}$ is decreasing and apply the definition !

$ \forall \epsilon >0\exists n\in\mathbb{N}$ such that $ b_n<0.5\epsilon$, i.e. $ \forall, k,m>n$ we have $ |x_k\minus{}x_m|\leq |x_k|\plus{}|x_m|\leq b_k\plus{}b_m\leq 2 b_n<\epsilon$

Another comment: $ \lim_{n\to0}x_n\equal{}0,$ so the sequence converges. A convergent sequence is necessarily a Cauchy sequence.


Is there an equation for this problem? Or is the only way to solve it by counting the possibilities?

Set A has all even integers from 1 - 100 inclusive.

If a random number is chosen from set A, what is the probability that the number is less than 40?

Set A has 50 in it (2,4,6,8...100)

There are 19 even numbers < 40 (2,4,6...38)

The answer is 19/50.

Counting the possibilities [i]is[/i] an equation.  Specifically,

$ \text{Probability of something} \equal{} \frac{ \text{Number of ways something can occur} }{ \text{Total number of events} }$.

In this case, there is one event for each different random number you can choose (this is assuming each event occurs with the same probability) and $ 19$ of those events out of $ 50$ are the "something" that we want.

[quote="t0rajir0u"]Counting the possibilities [i]is[/i] an equation.  Specifically,

$ \text{Probability of something} \equal{} \frac {\text{Number of ways something can occur} }{\text{Total number of events} }$.

In this case, there is one event for each different random number you can choose (this is assuming each event occurs with the same probability) and $ 19$ of those events out of $ 50$ are the "something" that we want.[/quote]

Thank you. I didn't think so, but I prefer to be sure.


[b]how can i create a blog for myself[/b] :blush: [/quote][/i]

Read Frequently Asked Questions:

http://www.mathlinks.ro/viewtopic.php?t=280499


let $ x$ , $ y$ , $ z$ $ \in ]0,\plus{}\infty[$ , prove that :

$ \frac {x^2 \plus{} xy \plus{} y^2}{x^2 \plus{} zx \plus{} xy \plus{} yz \plus{} y^2}$ $ \plus{}$ $ \frac {y^2 \plus{} yz \plus{} z^2}{y^2 \plus{} xy \plus{} yz \plus{} zx \plus{} z^2}$ $ \plus{}$ $ \frac {z^2 \plus{} zx \plus{} x^2}{z^2 \plus{} yz \plus{} zx \plus{} xy \plus{} x^2}$ $ \geq$ $ \frac{xyz}{x^3\plus{}y^3\plus{}z^3}$

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ , $ z$ $ \in ]0, \plus{} \infty[$ , prove that :

$ \frac {x^2 \plus{} xy \plus{} y^2}{x^2 \plus{} zx \plus{} xy \plus{} yz \plus{} y^2}$ $ \plus{}$ $ \frac {y^2 \plus{} yz \plus{} z^2}{y^2 \plus{} xy \plus{} yz \plus{} zx \plus{} z^2}$ $ \plus{}$ $ \frac {z^2 \plus{} zx \plus{} x^2}{z^2 \plus{} yz \plus{} zx \plus{} xy \plus{} x^2}$ $ \geq$ $ \frac {xyz}{x^3 \plus{} y^3 \plus{} z^3}$[/quote]

$ \sum\frac {x^2 \plus{} xy \plus{} y^2}{x^2 \plus{} zx \plus{} xy \plus{} yz \plus{} y^2}\ge\sum\frac {3xy}{x^2 \plus{} zx \plus{} xy \plus{} yz \plus{} y^2}$

$ \ge 3xyz\sum\frac {1}{x^2z \plus{} z^2x \plus{} xyz \plus{} yz^2 \plus{} y^2z}$

$ \ge 27xyz\left(\sum x^2z \plus{} z^2x \plus{} xyz \plus{} yz^2 \plus{} y^2z\right)^{\minus{}1}$

$ \equal{} 27xyz\left(4(x\plus{}y)(y\plus{}z)(z\plus{}x)\minus{}xyz\right)^{\minus{}1}$

$ \ge \frac{27xyz}{4}\left(\frac{2x\plus{}2y\plus{}2z}{3}\right)^{\minus{}3}$

$ \ge \frac{27xyz}{32}\left(\frac{x^3\plus{}y^3\plus{}z^3}{3}\right)^{\minus{}1}$

$ \equal{} \frac{81}{32}\cdot\frac{xyz}{x^3\plus{}y^3\plus{}z^3}$

$ \ge \frac{xyz}{x^3\plus{}y^3\plus{}z^3}$


Hiya everyone, newB here.  :D

While trawling through the net today, I came across the following theory for calculating the value of pi.  Sound's a little cooky to me (due to the arrived at value if nothing else), but thought I'd post it up for discussion anyway.  :roll:

The text below is what I can make out of the hand written text below the sketch in the image

ABCD= A SQUARE
AB=1
Inscribe a circle, diameter=1, radius = ½.
Divide the Circle-Square system into 32 segments.
1-16 are a segments.
7-32 are b segments.
Area of a segments = (6-[sqrt2])/128 = 0.035826456
Area of b segments = (2+[sqrt2])/128 = 0.026673543
Area of square = (16.a)+(16.b)=1
Area of circle=(16.a)+(8.b)=0.78661164
Pi=Area of circle/(radius*Radius)=1/4*(14-[sqrt2]) = 3.1464466…..

 :gathering:
Have fun,
Alagon.

Alagon,

I do not think that the method can be undestood from your brief description.

I did not get what to divide into 32 segments and how.

The link does not provide the information either.

Could you please try to give a more detailed description?

Thanks,
Andras

Apparently this guy is serious... well I don't see how he arrives at the value for the areas of the A and B segments. And in order to get something like that he must have used a straight line approximation somewhere... but anyway the guy is delusional.


Let be $ a,b,c,d > 0$ and $ e,f,g,h < 0$ . Show that the inequalities $ ae \plus{} bc > 0\ ,\ ef \plus{} cg > 0\ ,\ fd \plus{} gh > 0\ ,\ da \plus{} hb > 0$ can't be all true .

[hide]Put $ E\equal{}\minus{}e, F\equal{}\minus{}f, G\equal{}\minus{}g, H\equal{}\minus{}h$. Then we have $ bc>aE, EF>cG, GH>dF, ad>bH$. Since $ a,b,c,d,E,F,G,H$ are all positive, by multiplying the last inequalities we get $ abcdEFGH>abcdEFGH$, which is impossible.[/hide]


superior algebra
superior algebra unsolved

Suppose $G$ is a group.and $a,b\in{G}$.and
$m,n\in{\mathbb{N}},gcd(m,n)=1,a^{m}b^{m}=b^{m}a^{m},a^{n}b^{n}=b^{n}a^{n}$
prove that $ab=ba$


Each triangle is a 30-60-90 triangle, and the hypotenuse of one triangle is the longer leg of an adjacent triangle. The hypotenuse of the largest triangle is 8 centimeters. What is the number of centimeters in the length of the longer leg of the smallest triangle? Express your answer as a common fraction.

[asy]pair O; for(int i = 0; i < 5; ++i){
draw(O--((2/sqrt(3))^i)*dir(30*i));
}
for(int g = 0; g < 4; ++g){
draw( ((2/sqrt(3))^g)*dir(30*g)-- ((2/sqrt(3))^(g+1))*dir(30*g+30));
}
label("8 cm", O--(16/9)*dir(120), W);
label("$30^{\circ}$",.4*dir(0),dir(90));
label("$30^{\circ}$",.4*dir(25),dir(115));
label("$30^{\circ}$",.4*dir(50),dir(140));
label("$30^{\circ}$",.4*dir(85),dir(175));
draw((.9,0)--(.9,.1)--(1,.1));[/asy]


induction
strong induction
arithmetic sequence

Prove that for every real number $M$ there exists an infinite arithmetical progression of positive integers such that [list] [*] the common difference is not divisible by $10$, [*] the sum of digits of each term exceeds $M$. [/list]

Let $ s(n)$ denote the sum of the digits of $ n$.

[b]Lemma.[/b] For positive integers $ a_1,a_2,...,a_n$ with sum $ S$, we have $ s(a_1)\plus{}s(a_2)\plus{}\cdots \plus{}s(a_n)\ge s(S)$
[b]Proof.[/b] When summing the $ a_i$, if there is no "carrying" required, then the sum of the digits of $ S$ will be the same as the sum of the digits of all the $ a_i$. If any carrying is done, it will decrease the sum of the digits of $ S$ (by subtracting $ 10$ and adding $ 1$ to another digit), hence $ s(S)\le s(a_1)\plus{}s(a_2)\plus{}\cdots \plus{}s(a_n)$ as desired.

Now, let $ a\equal{}10^n\minus{}1$, that is, the number with $ n$ digits, all nines. I claim that $ s(am)\ge 9n$ for any positive integer $ m$. I will prove this by strong induction on $ m$.

The base case is $ m\equal{}1$. $ s(a)\equal{}9n$, as desired.

Inductive step: Assume that this is true for all positive integers less than $ m$. Let:

\[ am\equal{}d_0\plus{}d_1\cdot 10^n\plus{}d_2\cdot 10^{2n}\plus{}\cdots\plus{}d_k\cdot 10^{kn}\]

with $ d_i<10^n$ for $ 0\le i\le k$. In other words, $ d_0,d_1,d_2,...$ are the digits in the base $ 10^n$ representation of $ am$. Since $ 10^n\equiv 1\mod a$, we have

\[ d_0\plus{}d_1\plus{}\cdots\plus{}d_k\equiv d_0\plus{}d_1\cdot 10^n\plus{}d_2\cdot 10^{2n}\plus{}\cdots\plus{}d_k\cdot 10^{kn}\equiv am\equiv 0\mod a\]

Thus there is some positive integer $ m'<m$ with $ am'\equal{}d_0\plus{}d_1\plus{}\cdots\plus{}d_k$, so by the inductive hypothesis $ s(d_0\plus{}d_1\plus{}\cdots\plus{}d_k)\ge 9n$. Finally,

\[ s(am)\equal{}s(d_0)\plus{}s(d_1)\plus{}\cdots s(d_k)\ge s(d_0\plus{}d_1\plus{}\cdots\plus{}d_k)\ge 9n\]

completing the inductive step. Thus, if we choose $ n>\frac{M}{9}$, then the arithmetic sequence $ a,2a,3a,...$ will have the sum of the digits of each term exceed $ M$.


When $P(x)$ is divided by $x-1$ the remainder is $1$, and when divided by $(x-2)(x-3)$, the remainder is 5. Find the remainder when $P(x)$ is divided by $(x-1)(x-2)(x-3)$.
Please show your work.

doesn't any1 wanna solve this? :(
some1 try it please... :)

People can still try it but it is already solved in this topic:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=100259[/url]


According to EV-Theorem, it suffices to consider the case $b=c=1$.
The inequality becomes
$(a-1)^{2}(4a^{6}+5a^{5}-9a^{3}+8a+10) \geq 0$,
which is clearly true.

[quote="who"]whats EV theorem?[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=360537#360537[/url].


Hi all!

I'm having some difficulties with a problem for a course in Complex Analysis that I'm
taking. The problem is stated as follows:

Is there any function $ f(z)$ analytic on $ |z|<1$ such that
$ f\left(\frac{1}{2}\right) \equal{} \frac{1}{3}$, and $ f \left( \frac{1}{2n\plus{}1} \right) \equal{} \frac{1}{2n\plus{}1}$, for $ n \equal{} 1,2,3,4,...$?

I'm really at a loss on where to begin on this one. I've heard some rumours that it
might have to do with the function (or some variation of it) having some sort of
accumulation point or possibly non-isolated zeros, resulting in the fact that such
a function f(z) can't be analytic.

Any ideas?

Claim: The criteria $ f(\frac{1}{2n\plus{}1})\equal{}\frac{1}{2n\plus{}1}$ implies $ f(z)\equal{}z$.

We note that $ f(z)\minus{}z$ has infinitely many zeros inside the unit circle, which creates a sequence converging to $ 0$. Assume $ f(z)\minus{}z\equal{}a_0\plus{}a_1z\plus{}a_2z^2\plus{}\ldots$ and let $ a_n$ be the first nonzero coefficient. Then $ f(z)\minus{}z\equal{}a_nz^n\plus{}z^{n\plus{}1} g(z)$ with $ g$ analytic. Since $ g$ is analytic, it is bounded in the circle with radius $ 1/2$, say by $ M$. $ |f(z)\minus{}z|\ge |a_nz^n|\minus{}|z^{n\plus{}1}| M\equal{}|z^n|(|a_n|\minus{}M|z|)$. Now pick $ 1/(2n\plus{}1)$ so that $ 1/(2n\plus{}1)<|a_n|/M$. Then $ |f(1/(2n\plus{}1))|>0$ which is a contradiction. Hence all $ a_n$ must be zero and $ f(z)\minus{}z\equal{}0$. Hence $ f(z)\equal{}z$ which contradicts $ f(1/2)\equal{}1/3$.

Do you see how to generalize this?

Wow, answers come quickly on this forum!

Honestly, thanks a lot Alfred. I had a hunch that the solution would be much more easy than I initially imagined, but sometimes it's a bit hard find the starting point; obviously studying $ f(z) \minus{} z$ was the smart thing to do. Practice makes perfect I guess...

Anyway, thanks again - I really appreciate it! Not entirely sure on how to generalize it though, but I've got a (somewhat) similar problem to do as well, so maybe I'll get back to you...  :)

What I had in mind as a generalization is something along the following line:
If $ f,g$ are both analytic and for $ a_1,a_2,\ldots$ with $ \lim a_i\equal{}a$ we have $ f(a_i)\equal{}g(a_i)$, then $ f\equal{}g$.


I was working on solutions to a bunch of problems. Each problem should have a header called "problem 4.x" where (4.x is the problem #)

the \backslash thispagestyle command would only allow me to change the header for one page. I don't know how long my solution for each problem would be though...

e.g. header: problem 4.1 for a few pages

then header: problem 4.2 for a few pages...

I read the fancyhdr documentation and don't find anything that will do that I think... or maybe I miss something?

You may want to look into using the \markright (or \markboth) command.

yes \markright works fine. silly me... Thank you for your help :)


function
integration
logarithms
real analysis
real analysis unsolved

Suppose $ f$ is a complex measurable function on a measure space $ X$, and $ \mu$ is a positive measure on $ X$.
Define $ \phi (p) \equal{} \int_{X} \left| f \right| ^{p} d \mu \equal{}  \left\| f \right\| ^{p}_{p}$   for $ 0 < p < \infty$.
Let $ A \equal{} \left\{ p : \phi (p) < \infty \right\}$.

Is $ A$ necessary open? closed? can $ A$ consist of a single point? Can $ A$ be any connected subset of $ (0, \infty )$.

For the closed part, i found a counter example for it.

Is it the same thing for the others? To find a counter-example?

Thanks  :)

What $ L^p$ spaces does $ f(x)\equal{}\frac1{x(1\plus{}\log^2x)}$ belong to on $ (0,\infty)?$

It was discussed here http://www.mathlinks.ro/viewtopic.php?t=74987 .


Let (a_n) be a sequence of positive numbers with the properties:
a)a_n *a_(n+2)<(a_(n+1))^2 for any n natural;
b)there exits lim(n-->oo) a_(n+1)/(a_(n-1))=0
Show that there exists lim(n-->oo) (a_(n+1)/a_(n-1))*(a_(n-1))^(1/n) and compute it.

cheers! :D :D

a_(n+1)/a_(n-1)->0 (*)
and lim{n->oo}a_(n-1)^{1/n}=lim{n->oo}a_n/a_(n-1) = l
from a) we have that  a_(n+2)/a_(n+1)<a_(n+1)/a_n.
So the sequence x_n=a_n/a_(n-1) is strictly decreasing=> it is bounded(since it's positive), so it's limit is finit=>from(*) our limit is 0.


algebra
polynomial
linear algebra
matrix
superior algebra
superior algebra solved

Let $f(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}$ and let $x_1,x_2,\ldots x_n$ be its roots.
Prove that the Vandermonde determinand $V_n(x_1,x_2,\ldots x_n)$  has the property that $V_n^2=(-1)^{\frac{n(n+3)}2}\cdot n!^n$.

The determinant of Van Der Monde is
$V_n =  \begin {array}{cc}\prod \\ i<j \end{array}(x_j - x_i)$  and
$V_n^2 = \begin{array}{cc} \prod \\i<j \end{array}(x_j - x_i) ^2$ is the discriminant of the given equation.


linear algebra
matrix
linear algebra unsolved

This is one of the problems given to my students in their exam:

Let $ A\in M_n(F)$ be a diagonal(izable) matrix with $ p(x)\equal{}(x\minus{}a_1)^{n_1}(x\minus{}a_2)^{n_2}\dots (x\minus{}a_k)^{n_k}$ as its characteristic polynomial. Find dimension of $ C(A)$, ie the set of matrices that commute with $ A$.

WLOG, $ A$ is diagonal - or to be more specific, block-diagonal with $ a_iI$ (of the appropriate size) as the diagonal blocks. What can you obtain from that?

If $ A$ is diagonalizable, then $ dim(\mathcal{C}(A))\equal{}n_1^2\plus{}\cdots\plus{}n_k^2$.

If I am not wrong this problem is from IMC


This problem is not hard. You can suppose that $x_{5}\leq x_{4}\leq x_{3}\leq x_{2}\leq x_{1}$
Now, you should consider ${\min \{x_{i}-x_{j})^{2}>0}$, it follows $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}>1$ (The rest is very easy  :wink: )

I don't know how to do then? Can you explain?
No one observe this problem? Or it's too easy for you?


Show that the series $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\minus{}1}}{x^2\plus{}n}$ converges uniformly on $ \mathbb{R}$ although the convergence is conditional at every point.

For all $ x$, that series can be compared to

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

the conditional convergence of which you can show in a variety of ways, such as the alternating series test.


I came up with this today, and with a bit of effort proved it.

Given an acute triangle ABC and a point H inside the figure, prove that H is the orthocentre iff the circumcircles of triangles ABH, BCH and CAH have equal radii.

[hide]

some angle chasing yeilds <BHC=180-A, <AHB=180-C, <AHC=180-B

the sine law gives us that the inradii are equal

the reverse is true also because we have only one unique triangle ABC.

[/hide]

Surely you meant that "circumradii are equal"? ;)

But a nice (and easy too) solution! This proof also establishes that the circumradii of the orthogonal triangles are equal to the circumradius of the original triangle.

Now, for the bored, does anyone have a geometric proof? :D

Let $AH$ meet $BC$ at $D$ and meet circumcircle at $H'$. Then $\triangle{BCH} \cong \triangle{BCH'}$. Hence the circumradius of BCH is equal to the circumradius of BCH', which is equal to the circumradius of ABC. Do that for the other 2 triangles and we are done.


Sort lowest to highest the following list

$ e^\pi,\;\pi^e,\;e^3,\;3^e,\;e^{e\plus{}1},\;(e\plus{}1)^e$

Greetings  :D

One can evaluate those numbers. One decimal approximation is enough. I got $ 3^e<e^3<\pi^e<e^\pi<(e\plus{}1)^e<e^{e\plus{}1}$.

[quote="TALENTO"]Sort lowest to highest the following list

$ e^\pi,\;\pi^e,\;e^3,\;3^e,\;e^{e \plus{} 1},\;(e \plus{} 1)^e$

Greetings  :D[/quote]

[b]trying to do this without evaluating, I got[/b]

using the fact that $ f(x) \equal{} x^{\frac {1}{x}}$ takes its maximum at $ e$, we get

$ e^{e \plus{} 1} > (e \plus{} 1)^{e}$
$ e^{\pi} > \pi^{e}$
$ e^3 > 3^e$

I don't know what to do next... :(

[quote="earldbest"]I don't know what to do next... :([/quote]
Without evaluating those exactly, one can approximate transcendent numbers by rationals. For example $ e^3 < \pi^e$ as $ e^3 < 2.72^3 < 3.14^{2.71} < 3.14^e < \pi^e$.


function
trigonometry
algebra proposed
algebra

Prove that for every $ \epsilon\in [0;1]$, there exists a continous function $ f: \mathbb R\rightarrow\mathbb R$ satisfying $ f(x)\minus{}\epsilon \sin f(x)\equal{}x$

[hide="solution"]
Let $ f(x)\equal{}c$, where $ c$ is an arbitrary constant. Then $ f(x)$ is continuous. Solving for $ \epsilon$, we have $ \epsilon \equal{} x\csc{c}$, which is onto $ [0,1]$. Q.E.D.
[/hide]


Let $ a,b,c,d\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$.Prove that:
$ (abc)^3\plus{}(bcd)^3\plus{}(cda)^3\plus{}(dab)^3\leq4$

There seems to be a solution here:

[url]http://reflections.awesomemath.org/2006_6/2006_6_mixing.pdf[/url]

However, the argument used seems a bit incomplete (?)

[quote="Inequalities Master"]But I'm sure this can be done easier.

arqady, where are you :?: [/quote]
I am here always!   :D
I didn't solve this problem. Accordingly, I don't want to post the Vasc's solution.  :wink:

I'm glad. :D

But if you finish all your ideas unsuccesfully, then please post Vasc's solution, arqady.


Gerbert stole a box of hydrogen bombs from Caddlebot. The box has three faulty bombs and five working bombs. If Gerbert straps three bombs picked at random to himself and sets them off, what's the probability that he doesn't die [b]PROVIDED HE IS WEARING IMMORTALITY RINGS[/b]?

[hide]There are 8!/5!3! ways to choose the bombs. The only one in which he will survive is the case where all three are duds. So 8*7*6/3*2      8*7       1/56[/hide]

Not something I would try but I suppose you wouldn't even feel it... :D

[quote]..... what's the probability that he doesn't die?[/quote]
Probablity is zero.

[hide][size=150]Sooner of later he will die ..That is,  assuming, [b]No one[/b] is immortal.[/size]

ok so the moderator went and edited an  additional line about "immortility ring"  so I am editing my answer too..
... But if he is..umm ..if  these "rings" are envolved ... would a hydrogen bomb make any difference one way or the other?:?

On the second thought I will still stick to my answer .. probablity that those rings will make ANY difference is zero :)  [/hide]

[quote="Gyan"][quote]..... what's the probability that he doesn't die?[/quote]
Probablity is zero.

[hide][size=150]Sooner of later he will die ..That is,  assuming, [b]No one[/b] is immortal.[/size][/hide][/quote]

I don't think you saw the thread on immortality rings.

[hide]3/8*2/7*1/6, cross cancel as much as you can and you get 1/56[/hide]


[quote="mysmartmouth"][b]Problem 1[/b]:

In triangle ABC, let $AC=7$, $AB=6$, and $BC=10$.  Let the angle bisector of CAB hit BC at point D.  What is the ratio of the area of $\triangle ADC$ to the area of $\triangle ADB$?

[b]Problem 2[/b]:

AB and CD are chords in a circle that intersect at point E.  The radius of the circle is 10, and the distance from the center to AB is 6.  What are the lengths of segments AE and BE?[/quote]

[hide="1"]$\frac{7}{6}$ by the angle bisector theorem[/hide]

[hide="2"]I don't think there's enough info to find AE and BE but I did find AB to be 16.[/hide]

Can you give a detailed solution for #1?

Oh, OK, for some reason I thought you had written 1/6; I guess I misread your answer...sorry!


geometry
incenter
perimeter
circumcircle
trigonometry
area of a triangle
Heron\u0027s formula

So what are the side lengths? And what do you mean by "length of the area?"

[quote="yiyizhang"]Ok, i have a question bout this question that shows up a lot, but i can't seem to get it.
You are given triangle ABC, and the incenter along with the incircle. You then are given the lengths of the three sides, but what you are looking for is the length of the area of the circumscribed circle. so? Any idea? I bet there a lo of ppl who understand this problem out there, how about some help...[/quote]
First of all, this should be in HSB, because most people in this forum don't have a clue what to do. The formula you are looking for is $R=\frac{ABC}{4K}$. The K represents the area of the triangle. A,B, and C are the side lengths of the triangle. Then you can figure out the area of the circle since you now have the radius.

And if you don't know how to get the area of a triangle from three sides, use Heron's Formula:
$K=\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter (half the perimeter) and $a, b,$ and $c$ are sides of the triangle.

The formula ckck gave us can also give you the area of the triangle from the three sides and the circumradius; knowing your formulas for the area of a triangle really helps.

also, if you are given the radius of the incircle, say $r$, the are of the triangle is $\frac{1}{2}(a+b+c)r=sr$ where $a$, $b$, and $c$ are the side length of the triangle and $s$ is just the semiperimeter of the triangle.

The formula $R=\frac{ABC}{4K}$ can be proven easily inscribing the circle in the circumcircle and drawing chords from each vertex of the triangle that goes through the center of the circle (which essenciall makes a diameter) and using $\sin$. This also proves law of sine and many other things about triangles.


I am a bit confused over whether this Harriet Miers is
a good thing or not.

Here's what is confusing ... Bush recommended her ..
so I expected to see all of the conservatives jumping
up for joy .. but that is not really the case.  I did
a bit of searching around and found that the heavy
duty conservatives @ Townhall.com seem to be a bit
pissed off:
http://www.townhall.com/blogs/c-log/J%20Garthwaite/story/2005/10/03/159236.html

And ... on my side .. there doesn't seem to be much
joy .. so what's the deal with this chick?

Mirza :roll:

Harriet Miers, who had NO PREVIOUS HISTORY OF BEING A JUDGE, is a texan and one of bushes friends
[url=http://cagle.msnbc.com/news/HarrietMiers/main.asp]look at this[/url]

well, if bush nominated an ultra-conservative, which i'm sure he'd like to do, the nominee would not get passed in congress.

[quote="chess64"]whre did harriet miers come from? nobody knows about her. bush took the politically easy way out. ;)[/quote]

Yep. It wasn't exactly a smart thing to do though, to nominate a close friend, because then the non-Bush-admirers can cry nepotism. We'll see if she goes through. As for me, I'm against it.

I'm kind of suprsied I haven't seen that much about Harriet Miers being Bush's personal lawyer.  Judges aren't supossed to have strong opinions on things, right?  As far as I know, Judges are supossed to be neutral, and just interpret the law.  But since Miers is Bush's lawyer, she would obviously biased towards what Bush feels, which I don't think is right.

But about the part about Miers not having any previous experience, haven't there been a lot of previous judges that no one knew about?


superior algebra
superior algebra solved

[color=olive]Show that if a^3 l b^2, then a l b.[/color]

Thanks very much guys.

If $p^{n}$ divides $a$ where $p$ is prime and $n$ is maximal, then $p^{3n}$ divides $b^{2}$. If the power of $p$ in the decomposition of $b$ is less than $n$, than the power of $p$ in the decomposition of $b^{2}$ is less than $2n$ (impossible).


The function f satisfies f(x) + f(x-1) = x :^2: for all x.  If f(19) = 94, find the remainder when f(94) is divided by 1000.


Does anyone know of a tool that allows one to convert LaTeX into an image file (GIF, BMP, JPG, etc.)?  A search on CTAN turned up [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/mimetex.html]mimeTeX[/url], but that seems to be restricted to running in an HTML document - you put the mimeTeX files in the cgi-bin, then in generating HTML files, use <img src=.../cgi-bin/mimetex.cgi?[insert LaTeX here]> and an image is automatically put in place of the img src command.  LatexRender works similarly, requiring php instead.

Is there any tool that doesn't require use of a webpage - that is, one where you make a small text file, run the program on it, and get an image?

Thanks.

Although LatexRender uses PHP, that is just a wrapper around the basic commands, so you can do them independently. You'll need LaTeX and ImageMagick to do so.
It's done in 3 stages:
1. Create a DVI file using latex
2. Convert DVI file to Postscript (PS) using dvips (tetex)
3. convert, trim and add transparency by using 'convert' from the
  imagemagick package.

The commands in LatexRender (stripped of the PHP wrapper) are:
[code]1. latex filename.tex
2. dvips -E filename.dvi -o filename.ps
3. convert -density 120 -trim -transparent "#FFFFFF" filename.ps filename.gif[/code]

Put these in a batch file and you will have your own conversion program.
There are a number of other converters out there (Google for tex to gif) but they may use other programs, such as GhostScript and Netpbm, to do the conversion.

You can use mimeTeX without using an HTML page if you use the -e switch. Quoting from the manual at http://www.forkosh.com/mimetexmanual.html
[code]-e   Like -d but writes the actual gif (or xbitmap) directly  to
     export_file, e.g.,
               ./mimetex  -e expression.gif  x^2+y^2
     also creates expression.gif containing an image of x^2+y^2[/code]
This means that mimeTeX is the only program required, with no need for LaTeX or anything else.

I hope that helps.


Putnam
pigeonhole principle
function
group theory
abstract algebra
algebra
polynomial

hey, I was thinking:  i have done two putnams already, and practiced with few past years exams, and it seems they always have pigeonhole, taylor expansion, a little group theory, a lot of analysis, and almost never euclidean geometry.  is there any good book that can prepare one for putnam?  arts of problem solving isn't very good, by jon scholes' opinion, and unfortunately when i took a putnam class here at johns hopkins, that's the book we used.  i looked over some websites by other colleges, and they seem more problem- oriented than subject- oriented.  but i think i have some serious math background deficiency because i knew next to nothing about convex functions and convex analysis, and it was on the test!

and oh, they say they don't require any deep knowledge of higher math, come on, there are problems that involve group actions and sylow p-subgroups, which isn't usually covered in undergraduate algebra courses.

i can only think of so many topics, but i have only done it for two years.  i am sure there are more experienced people on this website.  so any suggestions of good practice books, and how about a detailed list on topics that one should know: i have given a head start, everyone should add to the list:

holder ineq, taylor expansion, pigeonhole, x^2-dy^2=1, everything about polynomials, gauss lemma on convex hull in complex numbers,

[url=http://books.google.com/books?id=qFNZIUQ_MYUC&dq=problem+solving+through+problems+larson&pg=PP1&ots=YLdCg4yP99&sig=qa6zDn64iIDUWFDA35mc270uy20&hl=en&prev=http://www.google.com/search%3Fhl%3Den%26q%3DProblem%2BSolving%2Bthrough%2BProblems%2BLarson&sa=X&oi=print&ct=title&cad=one-book-with-thumbnail]Problem Solving through Problems[/url] by Loren Larson

[url=http://books.google.com/books?id=QZ1QY4CWZv4C&printsec=frontcover&dq=William+Lowell+Putnam+Math&ei=ZzYgSIaRIJXujAHcqLCHBQ&sig=iA-8fH8kpXP95TPF25QydLjupLs#PPA1,M1]The William Lowell Putnam Mathematical Competition 1985-2000[/url] by Kiran Kedlaya, Bjorn Poonen, and Ravi Vakill

[quote="sexynsmartjenny"]and oh, they say they don't require any deep knowledge of higher math, come on, there are problems that involve group actions and sylow p-subgroups, which isn't usually covered in undergraduate algebra courses.[/quote]
Any group theory class should at least mention Sylow's theorems. In my first algebra class that was like the 4th week. (OK I admit my school's math department has an unusual style, though.)

What they mean is that almost all of the topics are accessible based on material from introductory classes: typically a first course in real analysis and linear algebra are sufficient for most problems, and I've seen plenty of problems which require knowledge from a first course in algebra or complex analysis or graph theory/combinatorics or differential equations. There was also a problem (2005 B5 maybe?) which was a reasonably basic harmonic analysis problem, but maybe they thought that all you really needed for that problem was some basic analysis and cleverness with polynomials. But it was kinda lame because anyone who had taken a course which covered harmonic analysis (as some advanced undergrad analysis classes or many intro graduate analysis classes do) could easily do that problem. But notice that the advantage to more advanced students is generally slim -- the top scorers on Putnam are often weighted towards the younger end because the more advanced students spend most of their time studying things that would never appear on Putnam.

Also, it's really better to practice for Putnam by doing problems, rather than by trying to learn subjects. That's why all the book recommendations you're going to get are problem books.

[quote="Xevarion"]
Any group theory class should at least mention Sylow's theorems. In my first algebra class that was like the 4th week. (OK I admit my school's math department has an unusual style, though.) [/quote]

My friend, you may have a warped view of undergraduate math classes because you go to princeton.

But thanks for the suggestions otherwise

You could have looked just a few lines down in this forum to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=182436]this topic[/url].


Can someone discribe me the way to find the largest summand in the expansion
for instance
$ 1) (1\plus{}\sqrt{2})^{100}$ and
$ 2) (\sqrt2\plus{}\sqrt{3})^{50}$  :?:

You can write out the general term, yes?  Now consider the ratio between successive terms.


$ a\in R\plus{}$ and $ 2^a\in Z,3^a\in Z$,then is $ a$ necessarily an integer?

According to [url=http://www.mathlinks.ro/viewtopic.php?p=1593378#1593378]this post[/url], this is an open problem. Another relevant discussion is [url=http://www.mathlinks.ro/viewtopic.php?t=29085]here.[/url]

Thank you so much for your nice link :lol:
By the way,happy new year !:D


Denote by $S$ the set of all primes $p$ such that the decimal representation of $\frac 1p$ has the fundamental period divisible by $3$. For each $p\in S$,we can write $\frac 1p=0.\overline{c_1c_2...c_{3r}}$, where $3r$ is the fundamental period of $p$ ;we define $f(k,p)=c_k+c_{k+r}+c_{k+2r}$ for every $k=1,2,...,r$. Determine the maximum possible value of $f(k,p)$.

It's 1999 to be exact: [url=http://www.kalva.demon.co.uk/short/soln/sh99n4.html]www.kalva.demon.co.uk/short/soln/sh99n4.html[/url]


Find all positive integers $ n$ such that $ n^n\plus{}1,(2n)^{2n}\plus{}1$ are primes.


ABCDEF is a convex hexagon with $\angle B + \angle D + \angle F = 360^{\circ}$ and $AB\cdot CD\cdot EF = BC\cdot DE\cdot FA$. Show that $BC\cdot DF\cdot EA = EF\cdot AC\cdot BD$.

This is ISL 1998 geometry problem 6. See

http://www.mathlinks.ro/Forum/viewtopic.php?t=1117

  Darij


Hi everybody, post how you went in the UNSW junior/senior division here.  :D

Can someone post the placings for seniors?

Snr
1. Graham
2. me
3. Morris

Jnr
1. Max
2. Congrats you guys
5. Random grammar guy?


It is $ \plus{}\infty$.

What did you mean?

[quote="kihe_freety5"]The sequence $ a_{1},a_{2},\cdots,a_{2010}$ has the following properties:

(1) each sum of the 20 successive values of the sequence is nonnegative,

(2) $ |a_{i}a_{i \plus{} 1}| \leq 1$ for $ i \equal{} 1,2,\cdots,2009$.

Determine the maximal value of the expression $ \sum_{i \equal{} 1}^{2010}a_{i}$[/quote]

This is a problem from the Austrian-Polish competition 2001.
The correct formulation is about integers, with a constraint on differences, and minimization:

----------------------------------------------------------------------------------------------------
A sequence $ a_1,a_2,\ldots,a_{2010}$ of integers has the following properties:
(1) each sum of 20 successive values of the sequence is nonnegative,
(2) $ |a_i\minus{}a_{i \plus{} 1}| \le1$ for $ i \equal{} 1,2,\ldots,2009$.

Determine the smallest possible value of $ \sum_{i\equal{}1}^{2010}a_i$ over all such sequences.
----------------------------------------------------------------------------------------------------


Find all positive integers x,y satsfying
$X^{2}Y = X+4 Y+8$

[hide]$y(x-2)(x+2)=8+y$
So $(x+2)|(8+x)$ and $x\ge 3$
$8+x\ge 2(2+x)\leftrightarrow 4\ge x$
$x\in{3,4}$
We can easily get $x=4$ and $y=1$[/hide]

[hide]$X^{2}Y = X+4 Y+8 \Longrightarrow Y(X-2) = 1+\frac{6}{X+2}\Longrightarrow X \in \{0,1,4\}$. Since $0$ is not positive,  and $1$ leads to $Y$ being negative, we are left with $X=4$ and $Y=1$.[/hide]

[quote="ashrafmod"]Find $\{x,y\}\subset N^{*}$ satsfying
$x^{2}y = x+4 y+8$[/quote]
$y(x^{2}-4)=x+8>0\ ,\ y>0\Longrightarrow x\ge 3$.

See easily that $x=3\Longrightarrow y=\frac{11}{5}\not\in N^{*}$. Thus, $\boxed{x\ge 4}$.

$y=f(x)=\frac{x+8}{x^{2}-4}=\frac{1}{x-2}+\frac{6}{x^{2}-4}$.

If $x>4\Longrightarrow$ $y=f(x)<f(4)=\frac{1}{2}+\frac{6}{12}=1\Longrightarrow$

$y<1\Longrightarrow y\not\in N^{*}$, what is falsely. Thus, $\boxed{x\le 4}$.

Therefore exists only the solution $(4,1)$, i.e. $x=4$, $y=1$.


[quote="DiscreetFourierTransform"]I'm assuming those are a cyclic sum/product?[/quote]

Oh, sorry, that was vague. I'll change it.

[quote="cincodemayo5590"]Prove:
\[\alpha, \beta, \gamma \in \mathbb{R}^{+}\text{ and }\alpha+\beta+\gamma = \frac{ \pi}2 \ \Longrightarrow \cot \alpha+\cot \beta+\cot \gamma= \cot \alpha \cot \beta \cot \gamma \]
[/quote]

[hide]Hint:
Make a substitution, of $\gamma=\frac{\pi}{2}-\alpha-\beta$[/hide]


Find min $e^{x}-cosx+\frac{x^{2}}{2}$ and prove that $e^{x}-cosx+\frac{x^{2}}{2}$ has only 2 root

This belongs in the Calculus forums.  I also don't know why you think this is so interesting; in particular, that minimum surely cannot be expressed with elementary functions.


arithmetic sequence
combinatorics proposed
combinatorics

The two players, liyi and x^4t^2, take turns, liyi first, inserting their choice of one of the signs +,-, or x in any of the empty  places between the terms 1 2 3 4 ... 99 100 until an arithmetic progression S is determined at the end of 99 turns. Prove that liyi can make S turn out be odd or even as he wishes.

The sequence is equivalent to 1 0 1 0 1 0 ... 1 0. Inserting an x is equivalent to deleting the number 1 next to the sign. Inserting + is equivalent to inserting -.

First liyi inserting a sign in the leftmost space: ± if he wish to get an odd sum; x if he wishes to get an even sum. If x^4t^2 inserts an x, liyi inserts an x in the leftmost free space. If x^4t^2 inserts a ±, liyi deletes the number 1 next to the sign if it is still undeleted, otherwise liyi inserts an x in the leftmost free space. Therefore the 1's except the first are deleted at the end of the game.


It means you're using the wrong minus sign. Just hit the "-" key - no shift, no option, no alt. If you hit "–" instead, the longer minus sign, you get what happened above. (Those really were different.)

$ 3-5$ (with the ordinary minus sign)

$ 3-5$ (using the option-minus on my Mac keyboard)


A corner lot has dimensions 50 by 100 yards. The city plans to take a strip of uniform width along the two sides bordering the streets to widen these roads. How wide should the strip be if the remainder of the lot is to have an area of 4851 square yards?

Set up the problem by labeling the unknowns with variables of your choice.
width of strip =  ?????

Translate the given information into mathematical language and find an equation that will produce the solution to the problem. You need not solve this equation. Use your variable from the previous part.
  ?????? = 4851

Wooooow, I love you  :rotfl: thank you so much

Dude...is this your homework?

And this isn't even geometry...


function
vector
real analysis
real analysis unsolved

Assume that $ \varphi$ is a real function on $ (a,b)$ such that
$ \varphi\displaystyle(\frac{x\plus{}y}{2})\leq \frac{\varphi(x)}{2}\plus{}\frac{\varphi(y)}{2}$ for all $ x$ and $ y\in (a,b)$.
Prove that exist $ \varphi$ that is not convex.

$ \mathbb{R}$ is a vector space over $ \mathbb{Q}$. Let $ v_{0} \cup \{v_{i}\}_{i \in I}$ be a Hamel basis for the vector space. An application like $ \varphi(v_{0}) \equal{} v_{0}$, $ \varphi(v_{i}) \equal{} 0 \mbox{ if } i \neq 0$ might be of some help.

:wink:


calculus
function
limit
abstract algebra
calculus computations

How is the L'Hospital rule proved? It isn't in my textbook.

Nine million applications of the mean value theorem.

I know it sounds a bit flippant, but I don't know a tidy proof of it that takes in every case.  Every elementary analysis text will have it, though, so you may want to find one.

The correct proof is a little technical, and for a first calculus class, that technicality is a little distracting.

The following proof is WRONG, but this is an instance in which I find it instructive to start with the wrong proof, then fix the hole.

As $x\to a$ we have both $f(x)\to 0$ and $g(x)\to 0.$ If these functions aren't defined at $a$, then let's go ahead and define them both to be zero there.

Now look at $\dfrac{f(x)}{g(x)}=\dfrac{f(x)-f(a)}{g(x)-g(a)}.$

Now, $f$ is continuous on $[a,x]$ (or $[x,a]$) and differentiable on $(a,x)$, so by the Mean Value Theorem (Lagrange's Theorem), there exists $y$ in between $a$ and $x$ such that $f(x)-f(a)=f'(y)(x-a).$

We can make the same claim for $g$: there exists $y$ in between $a$ and $x$ such that $g(x)-g(a)=g'(y)(x-a).$

Now, we would like to say the following:

$\dfrac{f(x)}{g(x)}=\dfrac{f'(y)(x-a)}{g'(y)(x-a)}=\dfrac{f'(y)}{g'(y)}.$

And letting $x\to a$ also forces $y\to a$.

Unfortunately, we can't say that. The hole in this proof is that on two independent occasions, I said "there exists $y$". I can't just assume that those two $y's$ are the same. They should really be $y_1$ and $y_2,$ but if they're different, then the rest of the argument breaks down.

So the correct proof begins with stating and proving a souped-up generalization of the Mean Value Theorem (which uses the original of either the Mean Value Theorem or Rolle's Theorem in its proof), and then using that on the original problem. As I said, a bit technical. Look for it in books with titles like "Introduction to Real Analsis" or "The Theory of Calculus."

For $\frac 00$ case, apply Cauchy's Mean-value Theorem directly.

For $\frac \ast\infty$ case, it would be helpful to write
$\dfrac{f(x)}{g(x)} = \dfrac{f(x)-f(A)}{g(x)-g(A)}\cdot \dfrac{g(x)-g(A)}{g(x)} + \dfrac{f(A)}{g(x)}$
then apply Cauchy's Mean-value Theorem to $\dfrac{f(x)-f(A)}{g(x)-g(A)}$.

Here's a second wrong proof. Actually, it's the right proof, but of a theorem whose hypotheses are more restrictive than the usual hypotheses of L'Hôpital's rule.

Theorem: Suppose $f$ and $g$ are both differentible at $a$ and $f(a)=g(a)=0.$ Suppose also that $g'(a)\ne 0.$

Then $\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{f'(a)}{g'(a)}.$

This proof is simple. $f(x)=f(0)+f'(a)(x-a)+o(x-a)=f'(a)(x-a)+o(x-a)$

and $f(x)=g(0)+g'(a)(x-a)+o(x-a)=g'(a)(x-a)+o(x-a)$

Then $\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(a)(x-a)+o(x-a)} {g'(a)(x-a)+o(x-a)}$

$=\lim_{x\to a}\dfrac{f'(a)+o(1)}{g'(a)+o(1)}=\dfrac{f'(a)}{g'(a)}.$

[quote="Kent Merryfield"]Theorem: Suppose $f$ and $g$ are both differentible at $a$ and $f(a)=g(a)=0.$ Suppose also that $g'(a)\ne 0.$

Then $\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{f'(a)}{g'(a)}.$
[/quote]
We don't really need Taylor's expansion with Peano's remainder.
In fact, taking limits on both sides of
$\dfrac{f(x)}{g(x)}= \dfrac{f(x)-f(a)}{g(x)-f(a)}=\dfrac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}$
is enough.